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✓ 23 results · all verified · 6 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 17 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Type-A Hecke Specht Modules and Cellular Bases

1 · Prerequisites

2 · Summary

This page develops the cellular-algebra foundations of the modular representation theory of the type-A Hecke algebra and its q-Schur algebra. It begins with the Graham-Lehrer cellular calculus, the cell form and its radical, and the classification of the simple modules by the nonzero cell heads $D^{λ} = C^{λ} / r a d C^{λ}$ , with the dominance-unitriangular decomposition matrix as its first consequence.

The second half specialises to $H_{F, q} (S_{n})$ : the Murphy cellular basis and the identification of the Hecke Specht modules through the Dipper-James signed-cell dictionary, the triangular action of the Jucys-Murphy elements and the resulting seminormal bases in the separated range, the classification of the simple modules by the $e$ -regular partitions, and the Schur functor connecting Weyl modules and their Jantzen filtrations to Specht modules. The q-Schaper virtual radical-layer formula and its hook-product Gram determinants are recorded in the form needed by the block and decomposition-number items of later pages, together with an explicit remark on the limits of what cellularity and the sum formula determine.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

Cellular bases and cell modules

Definition

Let $R$ be a commutative ring with $1$ and $A$ a unital $R$ -algebra (Algebras over a commutative ring, central structure maps, and algebra homomorphisms) that is free as an $R$ -module on a finite set $C = {c_{s t}^{λ} : λ \in Λ, s, t \in T (λ)}$ , where $(Λ, ⊵)$ is a finite poset (Partial order and partially ordered set; its strict order is written $⊳$ ) and each $T (λ)$ is a finite set. For $λ \in Λ$ let $A^{⊳ λ}$ be the $R$ -submodule of $A$ spanned by all $c_{u v}^{μ}$ with $μ ⊳ λ$ .

Then $(C, Λ)$ is a cellular basis of $A$ , and $A$ is a cellular algebra, if:

(i) the $R$ -linear map $i : A \to A$ with $i (c_{s t}^{λ}) = c_{t s}^{λ}$ for all $λ, s, t$ is an algebra anti-automorphism of $A$ , i.e. $i (a b) = i (b) i (a)$ and $i (1) = 1$ (equivalently $i$ is an isomorphism of $A$ with the opposite algebra; the displayed formula forces $i^{2} = i d$ ); and

(ii) for every $λ \in Λ$ , every $t \in T (λ)$ and every $a \in A$ there exist coefficients $r_{v} \in R$ ( $v \in T (λ)$ ), independent of $s$ , such that for every $s \in T (λ)$ $c_{s t}^{λ} a \equiv \sum_{v \in T (λ)} r_{v} c_{s v}^{λ} (m o d A^{⊳ λ}) .$

Cell modules. For $λ \in Λ$ let $C^{λ}$ be the free $R$ -module with basis ${c_{t}^{λ} : = c_{t t}^{λ} : t \in T (λ)}$ ; condition (ii) makes $C^{λ}$ a right $A$ -module by $c_{t}^{λ} a : = \sum_{v} r_{v} c_{v}^{λ}$ when $c_{t t}^{λ} a \equiv \sum_{v} r_{v} c_{t v}^{λ} (m o d A^{⊳ λ})$ , and it makes $C^{λ}$ a left $A$ -module via $i$ , the left action being $a \cdot c_{t}^{λ} : = c_{t}^{λ} i (a)$ . Both actions are well defined on the freely spanning basis elements by (ii): the coefficients $r_{v}$ are unique because $C$ is an $R$ -basis of $A$ and the quotient by $A^{⊳ λ}$ kills exactly the basis elements indexed by strictly larger labels, while applying the anti-automorphism (i) to (ii) gives the same triangularity for left multiplication. The module $C^{λ}$ is the (right) cell module of $λ$ , and $A^{⊵ λ} / A^{⊳ λ} ≅ C^{λ} \otimes_{R} C^{λ}$ via $c_{s t}^{λ} + A^{⊳ λ} \mapsto c_{s}^{λ} \otimes c_{t}^{λ}$ , where $A^{⊵ λ}$ is spanned by all $c_{u v}^{μ}$ with $μ ⊵ λ$ . A cellular algebra may have several cellular bases.

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e-regular and e-restricted partitions

Definition

Let $e \in {2, 3, 4, \dots} \cup {\infty}$ and let $λ ⊢ n$ be a partition with parts $λ_{1} \geq λ_{2} \geq \dots$ and conjugate $λ^{'}$ (Partitions, English diagrams, and conjugation). Then $λ$ is $e$ -regular when every positive integer occurs as a part of $λ$ fewer than $e$ times, i.e. $z_{j} (λ) < e$ for every $j \geq 1$ , and $e$ -restricted when $λ_{i} - λ_{i + 1} < e$ for every $i \geq 1$ (with $λ_{i} : = 0$ for $i > ℓ (λ)$ ).

For prime $e = p$ these are exactly the $p$ -regular and $p$ -restricted notions of p-regular and p-restricted partitions; the same inequalities define the extension to composite finite $e$ . For $e = \infty$ both conditions are vacuous, so every partition is both $e$ -regular and $e$ -restricted: the inequalities $z_{j} (λ) < \infty$ and $λ_{i} - λ_{i + 1} < \infty$ hold because partitions have finitely many nonzero parts and each difference is a nonnegative integer. Moreover $λ$ is $e$ -restricted if and only if its conjugate $λ^{'}$ is $e$ -regular: the multiplicity of the part $i$ in $λ^{'}$ is the number of columns of $[λ]$ of height exactly $i$ , namely $λ_{i} - λ_{i + 1}$ , so the two conditions $z_{i} (λ^{'}) = λ_{i} - λ_{i + 1} < e$ are the same family of inequalities, and for $e = \infty$ both sides hold. In particular $\emptyset$ is simultaneously $e$ -regular and $e$ -restricted for every $e$ (p-regular and p-restricted partitions).

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The quantum characteristic of an invertible Hecke parameter

Definition

Let $F$ be a field and $q \in F^{\times}$ an invertible element. The quantum characteristic $e = e (F, q)$ is the least integer $e \geq 2$ such that $1 + q + \dots + q^{e - 1} = 0$ in $F$ , and $e : = \infty$ when no such integer exists (The $q$ -integer, $q$ -factorial and $q$ -multinomial coefficients, The generic type-A Hecke algebra). Since $[1]_{q} = 1 \neq 0$ , the least positive integer $e$ with $[e]_{q} = 0$ , when it exists, automatically satisfies $e \geq 2$ , so the two formulations coincide.

Concretely: (i) if $q \neq 1$ then $[m]_{q} = (q^{m} - 1) / (q - 1)$ for all $m \geq 0$ — indeed $(q - 1) (1 + q + \dots + q^{m - 1}) = q^{m} - 1$ and $q - 1 \neq 0$ — so $[e]_{q} = 0$ is equivalent to $q^{e} = 1$ , and a finite $e$ exists exactly when $q$ is a root of unity and then $e$ is the multiplicative order of $q$ in $F^{\times}$ (which is $\geq 2$ because $q \neq 1$ ); (ii) if $q = 1$ then $[m]_{1} = m$ , so $e = char (F)$ when $char (F) > 0$ (the characteristic is a prime, hence at least $2$ , and $m \cdot 1_{F} \neq 0$ for $1 \leq m < p$ ) and $e = \infty$ when $char (F) = 0$ (Group algebra and finite-field specializations of the generic Hecke algebra).

If $e < \infty$ then $q \neq 1$ implies $q^{e} = 1$ , and hence $[m + e]_{q} = [m]_{q}$ for every integer $m$ : for $m \geq 0$ the sum splits as $[m + e]_{q} = [m]_{q} + q^{m} [e]_{q},$ so $[m + e]_{q} = [m]_{q}$ whenever $[e]_{q} = 0$ ; for the remaining $m < 0$ one extends the closed formula by $[m]_{q} : = (q^{m} - 1) / (q - 1)$ when $q \neq 1$ and reads $[m]_{1} : = m \in F$ for the canonical image of the integer $m$ when $q = 1$ , in which case $e = char (F)$ and $[m + e]_{1} = m + e = m$ in $F$ . In particular the quantum characteristic, not merely the multiplicative order, is the parameter governing the modular representation theory of the Hecke algebra. No choice principle is used.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-6.1-sol)Open item page →

Decompositions of the identity into primitive central idempotents are unique

Statement

Let $A$ be a unital ring (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides), with centre $Z (A) : = {z \in A : z a = a z for all a \in A}$ . Suppose that $1 = e_{1} + \dots + e_{k} = f_{1} + \dots + f_{ℓ}$ where ${e_{1}, \dots, e_{k}}$ and ${f_{1}, \dots, f_{ℓ}}$ are finite families of pairwise orthogonal central idempotents of $A$ that are primitive among central idempotents: each is nonzero and admits no decomposition as a sum of two nonzero orthogonal central idempotents (zero parts are omitted). Then $k = ℓ$ and there is a bijection $σ$ of the index sets with $e_{i} = f_{σ (i)}$ for all $i$ ; in particular the decomposition of $1$ into primitive central idempotents is unique up to order. No choice principle is used, the families being finite.

Facts & Assumptions

Given: A unital ring $A$ with centre $Z (A) = {z \in A : z a = a z for all a \in A}$ ; finite families $e_{1}, \dots, e_{k}$ and $f_{1}, \dots, f_{ℓ}$ of pairwise orthogonal nonzero central idempotents of $A$ that are primitive among central idempotents, with $1 = e_{1} + \dots + e_{k} = f_{1} + \dots + f_{ℓ}$ .

[F1]

Ring axioms and the zero ring: $(A, +, 0)$ is an abelian group, multiplication is associative with two-sided identity $1$ and distributes over addition on both sides, and a ring in which $1 = 0$ has exactly one element (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides).

Proof

technique · direct

1.1givenF1

If $k = 0$ or $ℓ = 0$ then $1 = 0$ in $A$ , so $A$ has exactly one element by [F1], namely $0$ ; then no nonzero idempotent exists and the two families are empty, $k = ℓ = 0$ , and the empty map is the required bijection. Assume henceforth that $k, ℓ \geq 1$ .

1.2F1givenalgebra

For central idempotents $e, f \in A$ the products $e f$ and $e (1 - f)$ are central idempotents with $e f + e (1 - f) = e$ and $e f \cdot e (1 - f) = 0$ : centrality gives $(e f) a = e (f a) = e (a f) = (e a) f = (a e) f = a (e f)$ and $(1 - f) a = a - a f = a - f a = a (1 - f)$ for all $a \in A$ ; idempotence gives $(e f)^{2} = e f e f = e^{2} f^{2} = e f$ and $(1 - f)^{2} = 1 - 2 f + f^{2} = 1 - f$ , hence $e (1 - f) e (1 - f) = e^{2} (1 - f)^{2} = e (1 - f)$ ; distributivity gives $e f + e (1 - f) = e (f + 1 - f) = e$ and $e f \cdot e (1 - f) = e^{2} f (1 - f) = 0$ .

2.1step 1.2given

If $e$ is primitive and $f$ is a central idempotent, then $e f \in {0, e}$ : by step 1.2, $e = e f + e (1 - f)$ exhibits $e$ as a sum of the two orthogonal central idempotents $e f$ and $e (1 - f)$ , so primitivity forces $e f = 0$ or $e (1 - f) = 0$ ; in the second case step 1.2 gives $e f = e f + 0 = e f + e (1 - f) = e$ .

3.1givenstep 2.1F1algebra

Fix $j \in {1, \dots, ℓ}$ . Since $1 = e_{1} + \dots + e_{k}$ , distributivity and centrality give $f_{j} = f_{j} \cdot 1 = \sum_{i} f_{j} e_{i} = \sum_{i} e_{i} f_{j}$ , and step 2.1 gives $e_{i} f_{j} \in {0, e_{i}}$ for every $i$ ; since $f_{j} \neq 0$ , the set $S_{j} : = {i : e_{i} f_{j} = e_{i}}$ is nonempty and $f_{j} = \sum_{i \in S_{j}} e_{i}$ .

4.1step 3.1givenalgebra

For every $j$ the set $S_{j}$ is a singleton: its elements $e_{i}$ are pairwise orthogonal nonzero central idempotents summing to $f_{j}$ by step 3.1, so if $S_{j}$ contained distinct $i \neq i^{'}$ then $f_{j} = e_{i} + \sum_{i^{''} \in S_{j}, i^{''} \neq i} e_{i^{''}}$ would exhibit $f_{j}$ as a sum of the two orthogonal central idempotents $e_{i}$ and $f_{j} - e_{i}$ , both nonzero since multiplying the second sum by the nonzero idempotent $e_{i^{'}}$ retrieves $e_{i^{'}}$ , contradicting primitivity of $f_{j}$ ; hence $S_{j} = {i (j)}$ and $f_{j} = e_{i (j)}$ .

5.1step 4.1givenalgebra

The map $j \mapsto i (j)$ is injective: if $i (j) = i (j^{'}) = i$ with $j \neq j^{'}$ , then $e_{i} = e_{i}^{2} = f_{j} f_{j^{'}} = 0$ by idempotence, centrality and pairwise orthogonality of the $f$ 's, contradicting $e_{i} \neq 0$ .

5.2step 4.1givenF1algebra

The map $j \mapsto i (j)$ is surjective: from $f_{j} = e_{i (j)}$ for all $j$ and both decompositions, $\sum_{j} e_{i (j)} = 1 = \sum_{i} e_{i}$ ; multiplying this equality by the central idempotent $e_{i}$ and using pairwise orthogonality of the $e$ 's gives $e_{i}$ on the left when $i$ lies in the image of $j \mapsto i (j)$ and $0$ otherwise, while the right side is $e_{i} \neq 0$ ; hence every $i$ lies in the image.

6.1step 4.1step 5.1step 5.2∎

Steps 5.1 and 5.2 make $j \mapsto i (j)$ a bijection ${1, \dots, ℓ} \to {1, \dots, k}$ with $f_{j} = e_{i (j)}$ for all $j$ by step 4.1; its inverse $σ$ satisfies $e_{i} = f_{σ (i)}$ for all $i$ , and $k = ℓ$ . Since the families were arbitrary, the decomposition of $1$ into primitive central idempotents is unique up to order, as claimed.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-6.1-sol)Open item page →

The identity of a finite-dimensional algebra decomposes into primitive central idempotents

Statement

Let $F$ be a field and $A$ a finite-dimensional unital $F$ -algebra with centre $Z (A) : = {z \in A : z a = a z for all a \in A}$ (Algebras over a commutative ring, central structure maps, and algebra homomorphisms). Then $Z (A)$ is a finite-dimensional commutative $F$ -algebra, and the identity $1 \in Z (A)$ admits a decomposition $1 = e_{1} + \dots + e_{k}$ into pairwise orthogonal primitive central idempotents of $A$ : each $e_{i}$ is a nonzero idempotent of $Z (A)$ that is not the sum of two nonzero orthogonal idempotents of $Z (A)$ . No semisimplicity of $A$ is assumed and no choice principle is used.

Facts & Assumptions

Given: A field $F$ , a finite-dimensional unital $F$ -algebra $A$ , and the centre $C : = Z (A) = {z \in A : z a = a z for all a \in A}$ .

[F1]

$F$ -algebra conventions: a unital $F$ -algebra $A$ is a unital ring together with a unital ring homomorphism $η : F \to A$ whose image is central, and the induced scalar action $r a : = η (r) a$ makes $A$ an $F$ -module with multiplication $A \times A \to A$ bilinear (Algebras over a commutative ring, central structure maps, and algebra homomorphisms).

[F2]

Subspace dimension: if $V$ is a finite-dimensional vector space over $F$ with $\dim_{F} V = n$ and $U$ a linear subspace of $V$ , then $U$ is finite-dimensional with $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ (If $\dim_{F} V = n$ and $U$ is a linear subspace of $V$ , then $U$ is finite-dimensional, $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ ).

[F3]

Well-ordering principle: every nonempty subset of $N$ has a least element (The well-ordering principle).

Proof

technique · direct

1.1givenF1

If $A = 0$ , then $C = Z (A) = {0}$ and the empty family ( $k = 0$ ) satisfies $1 = e_{1} + \dots + e_{k}$ with the stated conditions vacuously, so assume $A \neq 0$ ; then in $A$ one has $1 \neq 0$ and hence $1 \neq 0$ in $C$ .

1.2givenF1algebra

The centre $C$ is a commutative unital $F$ -subalgebra of $A$ and, in particular, a linear subspace of the $F$ -vector space $A$ : $1 \in C$ ; if $x, y \in C$ , $α \in F$ and $a \in A$ , then $(x + y) a = x a + y a = a x + a y = a (x + y)$ , $(α x) a = α (x a) = α (a x) = a (α x)$ and $(x y) a = x (y a) = x (a y) = (x a) y = (a x) y = a (x y)$ , while $0 \in C$ ; and distinct elements of $C$ commute with each other because each commutes with all of $A$ .

1.3givenF3

Suppose, for contradiction, that the following general claim fails: every finite-dimensional commutative unital $F$ -algebra $D$ with $1_{D} \neq 0$ and $\dim_{F} D \geq 1$ admits a decomposition of $1_{D}$ into pairwise orthogonal idempotents of $D$ that are primitive in $D$ . Let $S \subseteq N$ be the nonempty set of $F$ -dimensions of such counterexamples; by [F3] let $d \in S$ be least and fix a counterexample $D$ with $\dim_{F} D = d$ .

2.1step 1.2F2

The centre $C$ is finite-dimensional over $F$ with $\dim_{F} C \leq \dim_{F} A$ , by [F2] applied to the linear subspace $C$ of the finite-dimensional space $A$ .

2.2step 1.3given

Since $D$ is a counterexample, $1_{D}$ is not primitive in $D$ : otherwise $k = 1$ , $e_{1} = 1_{D}$ would be the required decomposition. Hence there are nonzero orthogonal idempotents $u, v \in D$ with $1_{D} = u + v$ .

3.1step 2.2F1algebra

The set $D_{u} : = u D = {u x : x \in D}$ is a linear subspace of $D$ and, with the inherited operations, a commutative unital $F$ -algebra with identity $u$ : it is closed since $u x + u y = u (x + y)$ , $α (u x) = u (α x)$ and $(u x) (u y) = u^{2} x y = u (x y)$ for $x, y \in D$ , $α \in F$ , while $u \cdot u x = u x$ ; and $u \neq 0$ . The same holds for $D_{v} : = v D$ with identity $v$ .

4.1step 3.1step 1.3F2algebra

One has $D_{u} \neq D$ because $v \notin D_{u}$ : if $v = u x$ then $0 = u v = u^{2} x = u x = v \neq 0$ , a contradiction using $u v = 0$ and $u^{2} = u$ . Hence by [F2], applied to the subspace $D_{u}$ of $D$ with $\dim_{F} D = d$ , one has $\dim_{F} D_{u} < d$ ; the same argument gives $\dim_{F} D_{v} < d$ .

5.1step 4.1step 1.3given

By the minimality of $d$ in $S$ , no counterexample has dimension $< d$ , and since $\dim_{F} D_{u} < d$ and $\dim_{F} D_{v} < d$ with $1_{D_{u}} = u \neq 0$ , $1_{D_{v}} = v \neq 0$ , the general claim holds for $D_{u}$ and for $D_{v}$ : there are pairwise orthogonal idempotents $f_{1}, \dots, f_{a}$ of $D_{u}$ primitive in $D_{u}$ with $u = f_{1} + \dots + f_{a}$ , and pairwise orthogonal idempotents $g_{1}, \dots, g_{b}$ of $D_{v}$ primitive in $D_{v}$ with $v = g_{1} + \dots + g_{b}$ .

6.1step 5.1step 3.1algebra

The combined family $f_{1}, \dots, f_{a}, g_{1}, \dots, g_{b}$ is a decomposition of $1_{D}$ into pairwise orthogonal idempotents of $D$ that are primitive in $D$ : the elements lie in $D_{u} \cup D_{v} \subseteq D$ ; distinct $f$ 's are orthogonal and distinct $g$ 's are orthogonal by step 5.1; $f_{i} g_{j} = (u f_{i}) (v g_{j}) = u v f_{i} g_{j} = 0$ ; and if $f_{i} = p + q$ with $p, q$ nonzero orthogonal idempotents of $D$ , then $p = f_{i} p = u f_{i} p = u p \in D_{u}$ and likewise $q = u q \in D_{u}$ , contradicting primitivity of $f_{i}$ in $D_{u}$ (and symmetrically for $g_{j}$ in $D_{v}$ ).

7.1step 6.1step 1.3

Step 6.1 contradicts the choice of $D$ as a counterexample, so no counterexample exists and the general claim of step 1.3 holds: every finite-dimensional commutative unital $F$ -algebra $D$ with $1_{D} \neq 0$ and $\dim_{F} D \geq 1$ decomposes its identity into pairwise orthogonal primitive idempotents.

8.1step 7.1step 2.1step 1.1∎

Apply step 7.1 to $D = C = Z (A)$ : by steps 1.1 and 1.2, $C$ is a commutative unital $F$ -algebra with $1 \neq 0$ and $\dim_{F} C \geq 1$ ; by step 2.1 it is finite-dimensional. This produces $1 = e_{1} + \dots + e_{k}$ with the $e_{i}$ pairwise orthogonal idempotents of $C = Z (A)$ primitive in $C$ ; idempotents of $Z (A)$ are exactly the central idempotents of $A$ , and primitive in $C$ says precisely that $e_{i}$ is not the sum of two nonzero orthogonal idempotents of $Z (A)$ . Since no semisimplicity was used and the only invoked principles are the choice-free subspace theorem [F2], the choice-free well-ordering principle [F3] and finite selections, the proof is complete.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-6.1-sol)Open item page →

The star anti-automorphism of the type-A Hecke algebra

Statement

Let $R$ be a commutative ring, $q \in R^{\times}$ , and let $H = H_{R, q} (S_{n})$ be the specialized type-A Hecke algebra with generators $T_{1}, \dots, T_{n - 1}$ and relations $T_{i}^{2} = (q - 1) T_{i} + q$ , the braid relations and the distant commutations (The generic type-A Hecke algebra, The standard basis of the generic type-A Hecke algebra). Then the assignment $T_{i} \mapsto T_{i}$ ( $1 \leq i < n$ ) extends uniquely to an $R$ -algebra anti-automorphism $* : H \to H$ , i.e. an $R$ -linear map with $(a b)^{*} = b^{*} a^{*}$ and $1^{*} = 1$ ; it is an involution, $* * = {i d}_{H}$ , and on the standard basis $T_{w}^{*} = T_{w^{- 1}} (w \in S_{n}) .$ In particular $(T_{w} T_{w^{- 1}})^{*} = T_{w} T_{w^{- 1}}$ and, for the symmetrizers $m_{λ} = \sum_{w \in S_{λ}} T_{w}$ , one has $m_{λ}^{*} = m_{λ}$ .

Facts & Assumptions

Given: A commutative ring $R$ , a unit $q \in R^{\times}$ , and the specialized type-A Hecke algebra $H = H_{R, q} (S_{n})$ , presented over $R$ by generators $T_{1}, \dots, T_{n - 1}$ with the relations $T_{i}^{2} = (q - 1) T_{i} + q$ , $T_{i} T_{i + 1} T_{i} = T_{i + 1} T_{i} T_{i + 1}$ ( $1 \leq i \leq n - 2$ ) and $T_{i} T_{j} = T_{j} T_{i}$ ( $∣ i - j ∣ > 1$ ).

[F1]

The generic algebra and its specialization: $H_{v} (n)$ is the quotient of the free unital associative $A$ -algebra, $A = Z [v^{\pm 1}]$ , on the $T_{i}$ by the two-sided ideal generated by the quadratic and braid relators; a reduced expression of $w \in S_{n}$ is a word $w = s_{i_{1}} \dots s_{i_{ℓ}}$ of minimal length $ℓ (w)$ representing $w$ , and $T_{w} : = T_{i_{1}} \dots T_{i_{ℓ}}$ , the empty product for $w = 1$ ; the specialization at $v_{0}$ is the $R$ -algebra $R \otimes_{A} H_{v} (n)$ , and for $v_{0} = q$ it is $H$ , presented over $R$ by the images of the $T_{i}$ subject to the relations above (The generic type-A Hecke algebra).

[F2]

Standard basis of the generic algebra: ${T_{w} : w \in S_{n}}$ is an $A$ -basis of $H_{v} (n)$ , so $T_{w}$ is independent of the reduced expression chosen for $w$ (The standard basis of the generic type-A Hecke algebra).

[F3]

Length is inversion invariant: $ℓ (w^{- 1}) = ℓ (w)$ for every $w \in S_{n}$ (Permutation Weyl group and inversion length).

[F4]

For a set $X$ , the finite words $X^{*}$ form a monoid under concatenation with the empty word as identity (Finite words satisfy the free-monoid universal property).

[F5]

The free $R$ -module $R^{(X^{*})}$ on the set $X^{*}$ has the words as an $R$ -basis: every element is a unique finite $R$ -linear combination $\sum_{w} a_{w} w$ (The free module on a set and its standard basis).

[F6]

Two-sided ideals are additive subgroups closed under multiplication on both sides; for $S \subseteq F$ the ideal $(S)$ is the intersection of all two-sided ideals containing $S$ , so any two-sided ideal containing $S$ contains $(S)$ (Left, right and two-sided ideals, The ideal generated by a subset and principal ideals).

[F7]

The quotient ring $F / I$ has additive cosets with $(x + I) (y + I) = x y + I$ and $R$ -scalar action inherited from $F$ (The quotient ring $R / I$ with $(r + I) (s + I) = r s + I$ ).

Proof

technique · direct

1.1givenF1F4F5F7

Put $X : = {1, \dots, n - 1}$ and $F : = R^{(X^{*})}$ , the free $R$ -module with the finite words in $X$ as basis [F5]; write the basis word corresponding to $i_{1}, \dots, i_{k}$ as $T_{i_{1}} \dots T_{i_{k}}$ and the empty word as $1$ . Extending concatenation of words $R$ -bilinearly makes $F$ a unital associative $R$ -algebra: concatenation is associative with two-sided identity the empty word by [F4], and distributivity holds by the bilinear extension. Let $I \subseteq F$ be the two-sided ideal generated by the relators $r_{i} : = T_{i}^{2} - (q - 1) T_{i} - q \cdot 1$ , $b_{i} : = T_{i} T_{i + 1} T_{i} - T_{i + 1} T_{i} T_{i + 1}$ and $d_{i j} : = T_{i} T_{j} - T_{j} T_{i}$ , and put $H : = F / I$ , which by [F1] is the presented algebra $H_{R, q} (S_{n})$ , the image of $T_{i}$ being again written $T_{i}$ ; every element of $H$ is the image of an $R$ -linear combination of words in the $T_{i}$ , so $H$ is generated as an $R$ -algebra by $T_{1}, \dots, T_{n - 1}$ .

1.2F4F5algebra

Define $ρ : F \to F$ on the word basis by $ρ (T_{i_{1}} \dots T_{i_{k}}) : = T_{i_{k}} \dots T_{i_{1}}$ and $ρ (1) : = 1$ , extended $R$ -linearly. Then $ρ$ is an $R$ -linear involution and reverses products, $ρ (x y) = ρ (y) ρ (x)$ , because reversal of a concatenation is the concatenation of the reversals in the opposite order and both sides are $R$ -bilinear; in particular $ρ^{2} = {i d}_{F}$ and $ρ (T_{i}) = T_{i}$ for each generator.

2.1step 1.2F6algebra

$ρ$ preserves $I$ , i.e. $ρ (I) \subseteq I$ : the relators are fixed up to sign, $ρ (r_{i}) = r_{i}$ , $ρ (b_{i}) = b_{i}$ and $ρ (d_{i j}) = - d_{i j}$ by step 1.2, so each lies in $I$ ; the set $J : = {x \in F : ρ (x) \in I}$ contains $0$ , is closed under addition and additive inverses because $ρ$ is $R$ -linear and $I$ is an additive subgroup, and for $x \in J$ and $a \in F$ one has $ρ (a x) = ρ (x) ρ (a) \in I$ and $ρ (x a) = ρ (a) ρ (x) \in I$ because $I$ is a two-sided ideal [F6]; hence $J$ is a two-sided ideal containing all relators, so $(S) \subseteq J$ and $I = (S) \subseteq J$ , that is, $ρ (I) \subseteq I$ .

3.1step 1.1step 1.2step 2.1F7algebra

Consequently $\overset{ˉ}{ρ} : H \to H$ , $\overset{ˉ}{ρ} (x + I) : = ρ (x) + I$ , is well defined: if $x - y \in I$ then $ρ (x) - ρ (y) = ρ (x - y) \in I$ by step 2.1. It is an $R$ -linear anti-automorphism of $H$ with ${\overset{ˉ}{ρ}}^{2} = i d$ and $\overset{ˉ}{ρ} (T_{i}) = T_{i}$ : the quotient operations of [F7] and step 1.2 give $\overset{ˉ}{ρ} ((x + I) (y + I)) = ρ (x y) + I = ρ (y) ρ (x) + I = \overset{ˉ}{ρ} (y + I) \overset{ˉ}{ρ} (x + I)$ , and similarly for addition, scalars and the unit.

4.1step 1.1step 3.1algebra

$\overset{ˉ}{ρ}$ is the unique $R$ -algebra anti-automorphism of $H$ with $\overset{ˉ}{ρ} (T_{i}) = T_{i}$ for all $i$ : if $α$ is any such anti-automorphism, then $α$ and $\overset{ˉ}{ρ}$ agree on $1$ , and for a word $T_{i_{1}} \dots T_{i_{k}}$ one has $α (T_{i_{1}} \dots T_{i_{k}}) = α (T_{i_{k}}) \dots α (T_{i_{1}}) = T_{i_{k}} \dots T_{i_{1}} = \overset{ˉ}{ρ} (T_{i_{1}} \dots T_{i_{k}})$ ; the words span $H$ by step 1.1, so $α = \overset{ˉ}{ρ}$ by $R$ -linearity.

4.2givenF1F2F3step 3.1algebra

For $w \in S_{n}$ and a reduced expression $w = s_{i_{1}} \dots s_{i_{ℓ}}$ one has $T_{w} = T_{i_{1}} \dots T_{i_{ℓ}}$ in the generic algebra and hence $T_{w} = T_{i_{1}} \dots T_{i_{ℓ}} \in H$ ; thus $T_{w}^{*} = \overset{ˉ}{ρ} (T_{i_{1}} \dots T_{i_{ℓ}}) = T_{i_{ℓ}} \dots T_{i_{1}}$ by step 3.1. The reversed word $s_{i_{ℓ}} \dots s_{i_{1}}$ represents $w^{- 1}$ and has $ℓ$ letters, and $ℓ = ℓ (w) = ℓ (w^{- 1})$ by [F3], so it is a reduced expression of $w^{- 1}$ and therefore $T_{w^{- 1}} = T_{i_{ℓ}} \dots T_{i_{1}} \in H$ ; hence $T_{w}^{*} = T_{w^{- 1}}$ for every $w$ , and $T_{w}$ is well defined in $H$ independently of the chosen reduced expression by [F1] and [F2].

5.1step 3.1step 4.1step 4.2algebra∎

The assertions of the statement follow: $* : = \overset{ˉ}{ρ}$ is the unique extension of the assignment $T_{i} \mapsto T_{i}$ , an involution by step 3.1; $T_{w}^{*} = T_{w^{- 1}}$ is step 4.2; therefore $(T_{w} T_{w^{- 1}})^{*} = T_{w^{- 1}}^{*} T_{w}^{*} = T_{(w^{- 1})^{- 1}} T_{w^{- 1}} = T_{w} T_{w^{- 1}}$ ; and for the parabolic subgroup $S_{λ}$ the map $w \mapsto w^{- 1}$ is a bijection of $S_{λ}$ , so $m_{λ}^{*} = \sum_{w \in S_{λ}} T_{w^{- 1}} = \sum_{v \in S_{λ}} T_{v} = m_{λ}$ . No choice principle is used: the constructions are explicit on the word basis.

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Blocks of a finite-dimensional algebra via primitive central idempotents

Definition

Let $F$ be a field and $A$ a finite-dimensional unital $F$ -algebra with centre $Z (A) : = {z \in A : z a = a z for all a \in A}$ (Algebras over a commutative ring, central structure maps, and algebra homomorphisms). A central idempotent of $A$ is an idempotent $e \in Z (A)$ ; it is primitive when $e \neq 0$ and $e$ is not the sum of two nonzero orthogonal central idempotents.

Then $Z (A)$ is a finite-dimensional commutative $F$ -algebra, so the identity $1 \in Z (A)$ admits a decomposition $1 = e_{1} + \dots + e_{k}$ into pairwise orthogonal primitive central idempotents (The identity of a finite-dimensional algebra decomposes into primitive central idempotents), and this decomposition is unique up to order (Decompositions of the identity into primitive central idempotents are unique). A block of $A$ is the two-sided ideal $B = e A = A e$ cut out by a primitive central idempotent $e$ (the two descriptions agree because $e$ is central).

The blocks are pairwise orthogonal two-sided ideals: $B_{i} B_{j} \subseteq e_{i} A e_{j} A = e_{i} e_{j} A = 0$ for $i \neq j$ , and $e_{i} \in B_{i}$ is the identity of $B_{i}$ since $e_{i} (e_{i} a) = e_{i} a$ and $(a e_{i}) e_{i} = a e_{i}$ . Moreover $A = B_{1} \oplus \dots \oplus B_{k}$ as $F$ -algebras with units $e_{i}$ : every $a \in A$ is $a = (\sum_{i} e_{i}) a = \sum_{i} e_{i} a$ with $e_{i} a \in B_{i}$ , and if $\sum_{i} b_{i} = 0$ with $b_{i} \in B_{i}$ then multiplying by $e_{j}$ gives $b_{j} = e_{j} b_{j} = \sum_{i} e_{j} b_{i} = 0$ , so the sum is direct. Each $B_{i}$ is an indecomposable two-sided ideal: if $B_{i} = J \oplus J^{'}$ were a direct sum of nonzero two-sided ideals, the projections of $e_{i}$ onto $J$ and $J^{'}$ would be nonzero orthogonal central idempotents of $A$ summing to $e_{i}$ , as is seen by applying to $e_{i}$ the $A$ -bimodule projections $B_{i} \to J$ and $B_{i} \to J^{'}$ , which satisfy $π (a x) = a π (x)$ and $π (x a) = π (x) a$ ; this contradicts primitivity of $e_{i}$ .

For every (not necessarily finite-dimensional) $A$ -module $M$ one has $M = ⨁_{i = 1}^{k} e_{i} M$ as $A$ -modules: $\sum_{i} e_{i} = 1$ acts as the identity on $M$ , so $M = \sum_{i} e_{i} M$ , and if $\sum_{i} m_{i} = 0$ with $m_{i} = e_{i} m_{i}$ then applying $e_{j}$ gives $m_{j} = 0$ ; each $e_{i} M$ is $A$ -stable because $e_{i}$ is central. A simple $A$ -module $S$ satisfies $e_{i} S \neq 0$ for exactly one $i$ , and then $S = e_{i} S$ lies in the corresponding block: $S = (\sum_{i} e_{i}) S = \sum_{i} e_{i} S$ forces some $e_{i} S \neq 0$ , and if $e_{i} S \neq 0 \neq e_{j} S$ with $i \neq j$ then simplicity gives $e_{i} S = S = e_{j} S$ , whence $S = e_{i} S = e_{i} (e_{j} S) = (e_{i} e_{j}) S = 0$ , a contradiction. Nothing here asserts that a block is semisimple or simple, and the construction is choice-free.

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Hecke Jucys-Murphy elements and q-contents

Definition

Let $R$ be a commutative ring, $q \in R^{\times}$ , and $H = H_{R, q} (S_{n})$ (The generic type-A Hecke algebra). For $1 \leq k \leq n$ define $L_{1} : = 0, L_{k} : = q^{- 1} T_{(k - 1, k)} + q^{- 2} T_{(k - 2, k)} + \dots + q^{1 - k} T_{(1, k)} \in H (k \geq 2),$ where $T_{(i, k)}$ is the standard basis element indexed by the transposition $(i k)$ (The standard basis of the generic type-A Hecke algebra). The $L_{k}$ are the Hecke Jucys-Murphy elements. They are fixed by the star anti-automorphism $T_{w}^{*} = T_{w^{- 1}}$ (The star anti-automorphism of the type-A Hecke algebra): each transposition $(i k)$ is its own inverse, so $T_{(i, k)}^{*} = T_{(i, k)}$ , and the coefficients $q^{i - k}$ are central, whence $L_{k}^{*} = \sum_{i < k} q^{i - k} T_{(i, k)} = L_{k}$ . They commute pairwise and every symmetric polynomial in $L_{1}, \dots, L_{n}$ with coefficients in $R$ is central in $H$ (The Hecke Jucys-Murphy action is dominance triangular).

q-contents. The quantum integers $[m]_{q}$ for $m \geq 0$ are those of The $q$ -integer, $q$ -factorial and $q$ -multinomial coefficients; extend the definition to negative indices by $[- m]_{q} : = - q^{- m} [m]_{q}$ for $m > 0$ , so that $(q - 1) [m]_{q} = q^{m} - 1$ for all $m \in Z$ . When $q - 1$ is a unit this may be written $[m]_{q} = (q^{m} - 1) / (q - 1)$ ; the polynomial definition requires no such division. Also $[m]_{1} = m$ for all $m \in Z$ , since $[- r]_{1} = - [r]_{1} = - r$ . For a standard $λ$ -tableau $t$ and $1 \leq k \leq n$ let ${content}_{t} (k) = c - r$ be the content of the node carrying $k$ (The content of a node and the content vector of a standard tableau) and define the q-content value $[{content}_{t} (k)]_{q}$ . If $e = e (F, q) < \infty$ for a field $F$ then $[m + e]_{q} = [m]_{q}$ for every $m \in Z$ (The quantum characteristic of an invertible Hecke parameter), so the q-content value depends only on the residue of the content modulo $e$ . At $q = 1$ the identification $H_{R, 1} (S_{n}) ≅ R [S_{n}]$ with $T_{w} \mapsto w$ — the isomorphism given by the Coxeter presentation and the standard bases (The symmetric group has the Coxeter presentation, The standard basis of the generic type-A Hecke algebra) — sends $L_{k}$ to the ordinary Jucys-Murphy element $X_{k} = \sum_{j < k} (j k)$ (The Jucys-Murphy elements of the symmetric group algebra), because each coefficient $q^{j - k}$ specializes to $1$ , and $[{content}_{t} (k)]_{1} = {content}_{t} (k)$ .

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Murphy elements and Murphy tableau elements

Definition

Let $R$ be a commutative ring, $q \in R^{\times}$ , and $H = H_{R, q} (S_{n})$ with standard basis ${T_{w}}$ and star anti-automorphism $*$ (The generic type-A Hecke algebra, The standard basis of the generic type-A Hecke algebra, The star anti-automorphism of the type-A Hecke algebra). For a composition (in particular a partition) $μ$ of $n$ put $m_{μ} : = \sum_{w \in S_{μ}} T_{w} \in H$ , the row symmetrizer of the standard Young subgroup $S_{μ}$ (Young subgroups, tabloids, and permutation modules); the sum has $∣ S_{μ} ∣ = \prod_{i} μ_{i}!$ terms.

A tableau is row standard when its entries increase along each row. Use the right tableau action $t \cdot w : = w^{- 1} \cdot t$ , where the left action applies $w$ to the entries. For a row-standard $μ$ -tableau $s$ , let $d (s)$ be the unique permutation with $s = t^{μ} \cdot d (s)$ , where $t^{μ}$ is filled by $1, 2, \dots, n$ in row-major order. Explicitly, if $s$ carries entry $a$ in the $p$ -th node in that order, then $d (s) (a) = p$ : $d (s)$ is the inverse of the row-reading permutation. The action on tableaux is free; the stabilizer of the row tabloid of $t^{μ}$ is $S_{μ}$ (Row and column stabilizers).

Write $D_{μ}$ for these distinguished permutations. Row-standardness says that $d^{- 1}$ is increasing on each canonical row block. The coset $S_{μ} d$ consists exactly of the tableaux with the same row sets, so sorting each row produces its unique representative in $D_{μ}$ . Moreover, $ℓ (w d) = ℓ (w) + ℓ (d) (w \in S_{μ}, d \in D_{μ}) .$ To verify this, count inversions of $(w d)^{- 1} = d^{- 1} w^{- 1}$ . Pairs of labels within one canonical block contribute exactly the inversions of $w^{- 1}$ , since $d^{- 1}$ is increasing there. For labels in different blocks, $w^{- 1}$ permutes the labels within their respective blocks, so the total number of cross-block inversions equals that of $d^{- 1}$ . The two contributions give the displayed equality, using inversion invariance of length. It also proves minimality and uniqueness in $S_{μ} d$ . These conventions extend to compositions by using their consecutive row blocks. For row-standard $μ$ -tableaux $s, t$ define the Murphy element $m_{s t} : = T_{d (s)}^{*} m_{μ} T_{d (t)} \in H,$ and for a standard $λ$ -tableau $t$ write the diagonal element as $m_{t t} = T_{d (t)}^{*} m_{λ} T_{d (t)}$ ; the notation $m_{t}$ is reserved for a cell-module vector, rather than an element of $H$ . Then $m_{s t}^{*} = m_{t s}$ for all row-standard $s, t$ of the same shape: applying the anti-automorphism, which fixes $m_{μ}$ because $m_{μ}^{*} = \sum_{w \in S_{μ}} T_{w^{- 1}} = m_{μ}$ , gives $m_{s t}^{*} = (T_{d (s)}^{*} m_{μ} T_{d (t)})^{*} = T_{d (t)}^{*} m_{μ}^{*} T_{d (s)}^{* *} = T_{d (t)}^{*} m_{μ} T_{d (s)} = m_{t s} .$ Finally, $m_{μ} T_{w} = q^{ℓ (w)} m_{μ}$ for every $w \in S_{μ}$ : it suffices to prove this for $w = s_{i}$ a simple transposition with $i, i + 1$ in the same block of $μ$ , since any $w \in S_{μ} ∖ {1}$ has a reduced expression $w = w^{'} s_{i}$ with $ℓ (w) = ℓ (w^{'}) + 1$ and the claim follows by induction, $m_{μ} T_{w} = m_{μ} T_{w^{'}} T_{i} = q^{ℓ (w^{'})} m_{μ} T_{i} = q^{ℓ (w)} m_{μ}$ . For such $i$ the right multiplication $v \mapsto v s_{i}$ is an involution of the finite group $S_{μ}$ , and pairing the summands of $m_{μ} T_{i} = \sum_{v \in S_{μ}} T_{v} T_{i}$ in twos gives, in the two cases of the standard-basis multiplication rule $T_{v} T_{i} = T_{v s_{i}}$ if $ℓ (v s_{i}) = ℓ (v) + 1$ and $T_{v} T_{i} = (q - 1) T_{v} + q T_{v s_{i}}$ if $ℓ (v s_{i}) = ℓ (v) - 1$ , $T_{v} T_{i} + T_{v s_{i}} T_{i} = q (T_{v} + T_{v s_{i}}),$ so summing over the $\frac{1}{2} ∣ S_{μ} ∣$ pairs yields $m_{μ} T_{i} = q m_{μ}$ .

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The cell form is associative and its radical is a submodule

Statement

Let $R$ be a commutative ring and $A$ a cellular $R$ -algebra with cellular basis $(C, Λ)$ and cell modules $C^{λ}$ (Cellular bases and cell modules). For each $λ \in Λ$ there is a unique $R$ -bilinear form $⟨, ⟩_{λ} : C^{λ} \times C^{λ} ⟶ R$ with $c_{u s}^{λ} c_{t v}^{λ} \equiv ⟨ c_{s}^{λ}, c_{t}^{λ} ⟩_{λ} c_{u v}^{λ} (m o d A^{⊳ λ})$ for all $s, t, u, v \in T (λ)$ ; in particular the scalar coefficient on the right hand side is independent of $u$ and $v$ , and $⟨ c_{s}^{λ}, c_{t}^{λ} ⟩_{λ}$ depends only on $s, t$ . This form is (i) symmetric, $⟨ x, y ⟩_{λ} = ⟨ y, x ⟩_{λ}$ ; (ii) associative, $⟨ x a, y ⟩_{λ} = ⟨ x, i (a) y ⟩_{λ}$ for all $a \in A$ , where $i (a) y$ denotes the right action $y i (a)$ of $i (a)$ on $y$ , i.e. the left action $a \cdot y : = y i (a)$ ; (iii) characterised by $x c_{u v}^{λ} = ⟨ x, c_{u}^{λ} ⟩_{λ} c_{v}^{λ}$ for $x \in C^{λ}$ ; and (iv) its radical $r a d C^{λ} : = {x \in C^{λ} : ⟨ x, y ⟩_{λ} = 0 for all y \in C^{λ}}$ is an $A$ -submodule of $C^{λ}$ , namely the kernel of $x \mapsto ⟨ x, - ⟩_{λ}$ .

Facts & Assumptions

Given: A commutative ring $R$ , a cellular $R$ -algebra $A$ with cellular basis $(C, Λ)$ , a label $λ \in Λ$ , and the cell module $C^{λ}$ with basis ${c_{t}^{λ} : t \in T (λ)}$ (Cellular bases and cell modules).

[F1]

The cellular axioms and cell-module structure (Cellular bases and cell modules): $C = {c_{s t}^{λ}}$ is an $R$ -basis of $A$ ; $A^{⊳ λ} = ⟨ c_{u v}^{μ} : μ ⊳ λ ⟩_{R}$ ; the $R$ -linear map $i$ with $i (c_{s t}^{λ}) = c_{t s}^{λ}$ is an algebra anti-automorphism with $i^{2} = i d$ ; for all $λ$ , $t \in T (λ)$ , $a \in A$ there are $r_{v} \in R$ with $c_{s t}^{λ} a \equiv \sum_{v} r_{v} c_{s v}^{λ} (m o d A^{⊳ λ})$ for all $s$ ; $C^{λ}$ is a right $A$ -module by $c_{t}^{λ} a : = \sum_{v} r_{v} c_{v}^{λ}$ when $c_{t t}^{λ} a \equiv \sum_{v} r_{v} c_{t v}^{λ} (m o d A^{⊳ λ})$ , and a left $A$ -module via $i$ with $a \cdot c_{t}^{λ} : = c_{t}^{λ} i (a)$ ; and $A^{⊵ λ} / A^{⊳ λ} ≅ C^{λ} \otimes_{R} C^{λ}$ via $c_{s t}^{λ} + A^{⊳ λ} \mapsto c_{s}^{λ} \otimes c_{t}^{λ}$ , where $A^{⊵ λ} = ⟨ c_{u v}^{μ} : μ ⊵ λ ⟩_{R}$ .

Proof

technique · direct

1.1givenF1algebra

The submodules $A^{⊵ λ}$ and $A^{⊳ λ}$ are two-sided ideals of $A$ , and the following two congruence rules hold modulo $A^{⊳ λ}$ : for all $a \in A$ and $u, v \in T (λ)$ , $c_{u s}^{λ} a \equiv \sum_{w \in T (λ)} r_{w} c_{u w}^{λ}, a c_{t v}^{λ} \equiv \sum_{w \in T (λ)} s_{w} c_{w v}^{λ},$ where the $r_{w}$ do not depend on $u$ and the $s_{w}$ do not depend on $v$ . Indeed, the first is the cellular rule [F1] applied to the fixed label $λ$ , the second-slot index $s$ and the element $a$ , which gives coefficients independent of the free first index $u$ ; right multiplication by $a$ therefore preserves each $A^{⊵ μ}$ and $A^{⊳ μ}$ , so $A^{⊵ λ}$ and (as its span over all $μ ⊳ λ$ ) $A^{⊳ λ}$ are right ideals, while applying the anti-automorphism $i$ to the first rule — it swaps the two indices, fixes $R$ and preserves both spans — gives the second rule and shows that they are also left ideals.

2.1givenstep 1.1F1algebra

Existence and uniqueness of the form: for all $s, t \in T (λ)$ there is a unique $r = r (s, t) \in R$ with $c_{u s}^{λ} c_{t v}^{λ} \equiv r c_{u v}^{λ} (m o d A^{⊳ λ})$ for all $u, v$ , and $r$ depends only on $s, t$ . Since $A^{⊳ λ}$ is a two-sided ideal by step 1.1, the quotient ring $A / A^{⊳ λ}$ contains $Q : = A^{⊵ λ} / A^{⊳ λ}$ with $R$ -basis the classes ${\overset{ˉ}{c}}_{u v} : = c_{u v}^{λ} + A^{⊳ λ}$ . Applying the two rules of step 1.1 to the product $c_{u s}^{λ} c_{t v}^{λ}$ — first with $a : = c_{t v}^{λ}$ , then with $a : = c_{u s}^{λ}$ — gives ${\overset{ˉ}{c}}_{u s} {\overset{ˉ}{c}}_{t v} = \sum_{w} ρ_{w} {\overset{ˉ}{c}}_{u w} = \sum_{x} σ_{x} {\overset{ˉ}{c}}_{x v},$ where $ρ_{w}$ does not depend on $u$ and $σ_{x}$ does not depend on $v$ . Comparing the coefficients of ${\overset{ˉ}{c}}_{a b}$ in the two bases: the left expansion contributes $ρ_{b}$ when $a = u$ and $0$ otherwise, the right expansion $σ_{a}$ when $b = v$ and $0$ otherwise; hence $σ_{a} = 0$ for $a \neq u$ and $ρ_{b} = 0$ for $b \neq v$ , leaving ${\overset{ˉ}{c}}_{u s} {\overset{ˉ}{c}}_{t v} = σ_{u} {\overset{ˉ}{c}}_{u v} = ρ_{v} {\overset{ˉ}{c}}_{u v}$ . The common value $σ_{u} = ρ_{v}$ is independent of $v$ (being $σ_{u}$ ) and independent of $u$ (being $ρ_{v}$ ), so it depends only on $s, t$ ; call it $⟨ c_{s}^{λ}, c_{t}^{λ} ⟩_{λ}$ . It is determined by the displayed congruence, because the coefficient of ${\overset{ˉ}{c}}_{u v}$ in the indicated free basis is unique, including over the zero ring. Bilinear extension gives the unique $R$ -bilinear form with the stated property.

3.1step 2.1F1algebra

Symmetry: for all $s, t, u, v$ , using that $i$ is an anti-automorphism with $i (c_{a b}^{λ}) = c_{b a}^{λ}$ , fixes $R$ and preserves $A^{⊳ λ}$ , one has $⟨ c_{s}^{λ}, c_{t}^{λ} ⟩_{λ} {\overset{ˉ}{c}}_{u v} \equiv c_{u s}^{λ} c_{t v}^{λ} = (c_{v t}^{λ} c_{s u}^{λ})^{i} \equiv (⟨ c_{t}^{λ}, c_{s}^{λ} ⟩_{λ} c_{v u}^{λ})^{i} = ⟨ c_{t}^{λ}, c_{s}^{λ} ⟩_{λ} {\overset{ˉ}{c}}_{u v},$ so $⟨ c_{s}^{λ}, c_{t}^{λ} ⟩_{λ} = ⟨ c_{t}^{λ}, c_{s}^{λ} ⟩_{λ}$ , and by bilinearity $⟨ x, y ⟩_{λ} = ⟨ y, x ⟩_{λ}$ for all $x, y \in C^{λ}$ .

3.2step 1.1step 2.1F1algebra

Associativity: fix $s, t \in T (λ)$ and $a \in A$ , and write $c_{s}^{λ} a = \sum_{w} r_{w} c_{w}^{λ}$ in $C^{λ}$ — the coefficients $r_{w}$ are those of step 1.1. Then, by step 2.1 applied first to $c_{s}^{λ} a$ and bilinearity of the form, $⟨ c_{s}^{λ} a, c_{t}^{λ} ⟩_{λ} {\overset{ˉ}{c}}_{u v} = \sum_{w} r_{w} {\overset{ˉ}{c}}_{u w} {\overset{ˉ}{c}}_{t v} = ({\overset{ˉ}{c}}_{u s} \overset{ˉ}{a}) \cdot {\overset{ˉ}{c}}_{t v} = {\overset{ˉ}{c}}_{u s} \cdot (\overset{ˉ}{a} {\overset{ˉ}{c}}_{t v}) .$ The second rule of step 1.1 writes $a c_{t v}^{λ} \equiv \sum_{x} s_{x} c_{x v}^{λ} (m o d A^{⊳ λ})$ with coefficients $s_{x}$ characterised by $c_{t}^{λ} i (a) = \sum_{x} s_{x} c_{x}^{λ}$ in $C^{λ}$ ; hence ${\overset{ˉ}{c}}_{u s} \cdot (\overset{ˉ}{a} {\overset{ˉ}{c}}_{t v}) = \sum_{x} s_{x} {\overset{ˉ}{c}}_{u s} {\overset{ˉ}{c}}_{x v} = \sum_{x} s_{x} ⟨ c_{s}^{λ}, c_{x}^{λ} ⟩_{λ} {\overset{ˉ}{c}}_{u v} = ⟨ c_{s}^{λ}, c_{t}^{λ} i (a) ⟩_{λ} {\overset{ˉ}{c}}_{u v} = ⟨ c_{s}^{λ}, a \cdot c_{t}^{λ} ⟩_{λ} {\overset{ˉ}{c}}_{u v} .$ Cancelling the basis element ${\overset{ˉ}{c}}_{u v}$ gives $⟨ c_{s}^{λ} a, c_{t}^{λ} ⟩_{λ} = ⟨ c_{s}^{λ}, i (a) c_{t}^{λ} ⟩_{λ}$ ; by bilinearity this is $⟨ x a, y ⟩_{λ} = ⟨ x, i (a) y ⟩_{λ}$ for all $x, y \in C^{λ}$ . Replacing $a$ by $i (a)$ and using $i^{2} = i d$ gives the equivalent left-action form $⟨ x i (a), y ⟩_{λ} = ⟨ x, i (a) \cdot y ⟩_{λ}$ , where $i (a) \cdot y = y i (i (a)) = y a$ is the left action of $i (a)$ : this is the dual statement.

3.3step 2.1F1algebra

The adjointness identity $x c_{u v}^{λ} = ⟨ x, c_{u}^{λ} ⟩_{λ} c_{v}^{λ}$ holds for all $x \in C^{λ}$ and $u, v \in T (λ)$ : it suffices to treat $x = c_{s}^{λ}$ , and there step 2.1 with $(u, s, t, v) : = (s, s, u, v)$ gives $c_{s s}^{λ} c_{u v}^{λ} \equiv ⟨ c_{s}^{λ}, c_{u}^{λ} ⟩_{λ} c_{s v}^{λ} (m o d A^{⊳ λ})$ ; by the definition of the right action in [F1], the coefficients of $c_{s v}^{λ}$ on the right are exactly the coefficients of $c_{v}^{λ}$ in $c_{s}^{λ} \cdot c_{u v}^{λ}$ , so $c_{s}^{λ} c_{u v}^{λ} = ⟨ c_{s}^{λ}, c_{u}^{λ} ⟩_{λ} c_{v}^{λ}$ ; bilinearity in $x$ extends this to all $x$ .

4.1step 3.2step 3.3algebra∎

The radical is a submodule and a kernel: for fixed $y \in C^{λ}$ the assignment $x \mapsto ⟨ x, y ⟩_{λ}$ is an $R$ -linear map with kernel ${x : ⟨ x, y ⟩_{λ} = 0}$ , so $r a d C^{λ} = ⋂_{y} \ker (x \mapsto ⟨ x, y ⟩_{λ})$ is an $R$ -submodule; and if $x \in r a d C^{λ}$ then for every $a \in A$ and $y \in C^{λ}$ step 3.2 gives $⟨ x a, y ⟩_{λ} = ⟨ x, i (a) y ⟩_{λ} = 0$ , so $x a \in r a d C^{λ}$ : the radical is an $A$ -submodule, namely the kernel of the $A$ -module map $x \mapsto ⟨ x, - ⟩_{λ}$ . This completes the proof of (i)-(iv).

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The Murphy tableau elements form a cellular basis

Statement

Let $R$ be a commutative ring, $q \in R^{\times}$ , and $H = H_{R, q} (S_{n})$ (The generic type-A Hecke algebra). Let $Λ$ be the set of partitions of $n$ ordered by dominance $⊵$ (Dominance order on partitions), and for $λ \in Λ$ let $T (λ) = S t d (λ)$ be the set of standard $λ$ -tableaux (Tableaux and standard tableaux). Then the Murphy basis theorem holds: the family $M = {m_{s t} : λ ⊢ n, s, t \in S t d (λ)}$ is an $R$ -basis of $H$ , and $(M, Λ)$ is a cellular basis of $H$ in the sense of Cellular bases and cell modules. Explicitly: (i) $m_{s t}^{*} = m_{t s}$ for all $λ$ and $s, t$ ; and (ii) for every $λ$ , every $t \in S t d (λ)$ and every $h \in H$ there exist $r_{v} \in R$ with $m_{s t} h \equiv \sum_{v \in S t d (λ)} r_{v} m_{s v} (m o d H^{⊳ λ}),$ where $H^{⊳ λ}$ is spanned by the $m_{u v}$ with $μ ⊳ λ$ . The theorem holds over every commutative coefficient ring with $q \in R^{\times}$ and assumes no semisimplicity; the total number of Murphy elements is $\sum_{λ ⊢ n} ∣ S t d (λ) ∣^{2} = n!$ (The sum of squares of the standard tableau numbers), matching the rank of $H$ .

Facts & Assumptions

Given: A commutative ring $R$ , a unit $q$ , the type-A Hecke algebra and the Murphy elements $m_{s t} = T_{d (s)}^{*} x_{λ} T_{d (t)}$ in the Statement.

[F1]

The independent integral construction proves that these exact elements form a base-change-compatible basis; their dominance spans are two-sided ideals, star interchanges the indices, and their multiplication satisfies the fixed-first-index cell law over every commutative coefficient ring with unit parameter (The Murphy cellular basis, the cell law and the Murphy form).

[F2]

The elements $x_{λ}$ and $d (t)$ are the row symmetrizer and distinguished tableau permutations of Murphy elements and Murphy tableau elements, using the standard Hecke basis of The standard basis of the generic type-A Hecke algebra. The dominance poset, tableau sets and cardinality sum are Dominance order on partitions, Tableaux and standard tableaux and The sum of squares of the standard tableau numbers.

Proof

technique · direct

1.1givenF1F2algebra

Identify the objects in [F1]. Its row symmetrizer is $x_{λ} = \sum_{w \in S_{λ}} T_{w}$ , its permutation is the unique distinguished $d (t)$ carrying the row-filled tableau to $t$ , and its element is $T_{d (s)}^{*} x_{λ} T_{d (t)}$ . These are precisely [F2]'s Murphy elements. The proved integral basis in [F1] therefore gives the required $R$ -basis and $m_{s t}^{*} = m_{t s}$ , including the empty tableau when $n = 0$ .

2.1F1F2step 1.1algebra∎

Its dominance ideal $I^{⊳ λ}$ and the present $H^{⊳ λ}$ are the spans of exactly the same basis elements, so they are equal. The independent cell law says $m_{s t} h \equiv \sum_{v} r_{v} m_{s v}$ modulo this ideal, with coefficients independent of $s$ . Together with step 1.1 and the same finite dominance poset, these are exactly the cellular axioms and all asserted formulas. The basis has $\sum_{λ} ∣ Std (λ) ∣^{2} = n!$ elements by [F2]. Every part holds after every base change by [F1], so no specialized semisimplicity or factor inversion is needed. No choice principle is used.

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Nonzero cell heads classify the simple modules of a cellular algebra

Statement

Let $F$ be a field and $A$ a finite-dimensional unital cellular $F$ -algebra with cellular basis $(C, Λ)$ over the finite poset $Λ$ (Cellular bases and cell modules). For $λ \in Λ$ put $D^{λ} : = C^{λ} / r a d C^{λ}$ , where $r a d C^{λ}$ is the radical of the cell form (The cell form is associative and its radical is a submodule). Then:

if $D^{λ} \neq 0$ then $D^{λ}$ is an absolutely irreducible (absolutely simple) $A$ -module: it is simple over $F$ and remains simple after every extension of scalars to a larger field;
${D^{λ} : λ \in Λ, D^{λ} \neq 0}$ is a complete set of pairwise non-isomorphic simple $A$ -modules (Simple module: a nonzero module with no proper nonzero submodule, Jordan–Hölder theorem for modules);
$D^{λ} \neq 0$ if and only if the cell form $⟨, ⟩_{λ}$ is not identically zero, if and only if $r a d C^{λ} \neq C^{λ}$ ; and, when these equivalent conditions hold, $r a d C^{λ}$ is the unique maximal proper submodule and the Jacobson radical of $C^{λ}$ , so $D^{λ}$ is its simple head. If the form is identically zero, the cell-form quotient $D^{λ}$ is zero; no assertion about the Jacobson radical or the actual head of $C^{λ}$ is made.

No semisimplicity of $A$ is assumed; the theorem is the Graham-Lehrer classification and applies in particular to $A = H_{F, q} (S_{n})$ for every field $F$ and every $q \in F^{\times}$ .

Facts & Assumptions

Given: A field $F$ , a finite-dimensional unital cellular $F$ -algebra $A$ with cellular basis $(C, Λ)$ over the finite poset $Λ$ , cell modules $C^{λ}$ and cell forms $⟨, ⟩_{λ}$ (Cellular bases and cell modules).

[F1]

Cell-form calculus (The cell form is associative and its radical is a submodule): the form $⟨, ⟩_{λ}$ is the unique bilinear form with $c_{u s}^{λ} c_{t v}^{λ} \equiv ⟨ c_{s}^{λ}, c_{t}^{λ} ⟩_{λ} c_{u v}^{λ} (m o d A^{⊳ λ})$ ; it is symmetric and associative, $⟨ x a, y ⟩_{λ} = ⟨ x, i (a) y ⟩_{λ}$ ; $x c_{u v}^{λ} = ⟨ x, c_{u}^{λ} ⟩_{λ} c_{v}^{λ}$ for $x \in C^{λ}$ ; and $r a d C^{λ}$ is the kernel of $x \mapsto ⟨ x, - ⟩_{λ}$ , an $A$ -submodule.

[F2]

The cellular axioms (Cellular bases and cell modules): $C$ is an $F$ -basis of $A$ ; $A^{⊳ λ} = ⟨ c_{u v}^{μ} : μ ⊳ λ ⟩_{F}$ ; condition (ii) reads $c_{s t}^{λ} a \equiv \sum_{v} r_{v} c_{s v}^{λ} (m o d A^{⊳ λ})$ for all $s$ , and its image under the anti-automorphism $i$ reads $a c_{s t}^{λ} \equiv \sum_{u} r_{u}^{'} c_{u t}^{λ} (m o d A^{⊳ λ})$ for all $t$ .

[F3]

Every finite-dimensional module has a composition series: choose a proper submodule of largest dimension, whose quotient is simple, and induct on dimension. This also applies to any simple module over $A$ , since a nonzero vector generates it as a quotient of the finite-dimensional regular module. The Jordan-Hölder theorem gives well-defined composition multiplicities (Composition series and length of a module, Jordan–Hölder theorem for modules).

Proof

technique · direct

1.1givenF2algebra

For $λ \in Λ$ the spans $A^{⊵ λ}$ and $A^{⊳ λ}$ are two-sided ideals: by [F2] both $c_{s t}^{λ} a$ and $a c_{s t}^{λ}$ are congruent modulo $A^{⊳ λ}$ to combinations of the $c^{λ}$ -elements, so $A^{⊵ λ} \cdot A \subseteq A^{⊵ λ}$ and $A \cdot A^{⊵ λ} \subseteq A^{⊵ λ}$ , and $A^{⊳ λ}$ , the sum of the ideals $A^{⊵ μ}$ over $μ ⊳ λ$ , is a two-sided ideal as well. Consequently $A^{⊵ ν} A^{⊵ λ} \subseteq A^{⊵ ν} \cap A^{⊵ λ}$ and $A^{⊵ λ} A^{⊵ ν} \subseteq A^{⊵ ν} \cap A^{⊵ λ}$ , because a product of a $c^{ν}$ -element and a $c^{λ}$ -element lies in $A^{⊵ ν}$ by the ideal property of $A^{⊵ ν}$ and in $A^{⊵ λ}$ by that of $A^{⊵ λ}$ . The intersection $A^{⊵ ν} \cap A^{⊵ λ}$ is spanned by the $c^{γ}$ with $γ ⊵ ν$ and $γ ⊵ λ$ ; if $λ ⋭ ν$ then no such $γ$ equals $λ$ , since $γ = λ$ would force $λ ⊵ ν$ , so all of them satisfy $γ ⊳ λ$ and $A^{⊵ ν} \cap A^{⊵ λ} \subseteq A^{⊳ λ}$ . Hence, when $λ ⋭ ν$ , every product $c_{t t}^{λ} b$ with $b \in A^{⊵ ν}$ lies in $A^{⊳ λ}$ ; the right action of $b$ on $C^{λ}$ is read off from $c_{t t}^{λ} b \equiv \sum_{v} r_{v} c_{t v}^{λ} (m o d A^{⊳ λ})$ with unique coefficients $r_{v}$ , so $b$ acts as the zero operator on $C^{λ}$ , that is, $A^{⊵ ν} \cdot C^{λ} = 0$ .

1.2givenF1algebra

Generation by vectors outside the radical: let $x \in C^{λ}$ with $x \notin r a d C^{λ}$ . Since $r a d C^{λ} = {z : ⟨ z, - ⟩_{λ} = 0}$ there is $y \in C^{λ}$ with $⟨ x, y ⟩_{λ} \neq 0$ , and after rescaling $y$ we may assume $⟨ x, y ⟩_{λ} = 1$ . Write $y = \sum_{u} a_{u} c_{u}^{λ}$ and put $y_{v} : = \sum_{u} a_{u} c_{u v}^{λ} \in A$ for $v \in T (λ)$ . Then [F1] gives $x y_{v} = \sum_{u} a_{u} (x c_{u v}^{λ}) = \sum_{u} a_{u} ⟨ x, c_{u}^{λ} ⟩_{λ} c_{v}^{λ} = ⟨ x, y ⟩_{λ} c_{v}^{λ} = c_{v}^{λ}$ , so every basis vector of $C^{λ}$ lies in the submodule $x A$ , hence $x A = C^{λ}$ .

1.3givenF1algebra

Nontrivial action of the upward ideal: suppose $D^{μ} \neq 0$ . Since $r a d C^{μ} \neq C^{μ}$ there is $t \in T (μ)$ with $c_{t}^{μ} \notin r a d C^{μ}$ ; by definition of the radical there is $u \in T (μ)$ with $⟨ c_{t}^{μ}, c_{u}^{μ} ⟩_{μ} \neq 0$ , and then $c_{u}^{μ} \notin r a d C^{μ}$ as well. By [F1] the element $c_{u t}^{μ} \in A^{⊵ μ}$ acts on $C^{μ}$ by $c_{t}^{μ} \cdot c_{u t}^{μ} = ⟨ c_{t}^{μ}, c_{u}^{μ} ⟩_{μ} c_{t}^{μ}$ , so on the quotient $D^{μ}$ it sends the nonzero class of $c_{t}^{μ}$ to its nonzero multiple $⟨ c_{t}^{μ}, c_{u}^{μ} ⟩_{μ} [c_{t}^{μ}] \neq 0$ . Thus $A^{⊵ μ}$ does not annihilate $D^{μ}$ .

2.1givenstep 1.2F1F3

Consequences for the radical: every proper submodule $M ⊊ C^{λ}$ is contained in $r a d C^{λ}$ , since $x \in M$ with $x \notin r a d C^{λ}$ would give $C^{λ} = x A \subseteq M$ by step 1.2, a contradiction. Hence, if $r a d C^{λ} \neq C^{λ}$ , then $r a d C^{λ}$ contains every proper submodule, is the unique maximal proper submodule, and equals the Jacobson radical of $C^{λ}$ ; then $D^{λ} = C^{λ} / r a d C^{λ}$ is simple, because a nonzero submodule of the quotient has the form $N / r a d C^{λ}$ with $N ⊋ r a d C^{λ}$ a submodule, and $N = C^{λ}$ by maximality. Moreover $D^{λ} \neq 0$ if and only if $r a d C^{λ} \neq C^{λ}$ if and only if the form $⟨ -, - ⟩_{λ}$ is not identically zero: the radical is the kernel of $x \mapsto ⟨ x, - ⟩_{λ}$ , so it is all of $C^{λ}$ exactly when $⟨ x, y ⟩_{λ} = 0$ for all $x, y$ , and otherwise this kernel is a proper subspace containing $0$ .

2.2step 1.1step 1.3

Pairwise distinctness: suppose $D^{λ} ≅ D^{μ} \neq 0$ . If $λ$ does not dominate $μ$ , then step 1.1 gives $A^{⊵ μ} \cdot C^{λ} = 0$ , hence $A^{⊵ μ} \cdot D^{λ} = 0$ , and transporting this along the isomorphism gives $A^{⊵ μ} \cdot D^{μ} = 0$ , contradicting step 1.3. Hence $λ ⊵ μ$ ; the same argument with the roles of $λ$ and $μ$ exchanged gives $μ ⊵ λ$ , so $λ = μ$ by antisymmetry of the poset. Thus the nonzero cell heads are pairwise non-isomorphic.

3.1givenstep 1.2step 2.1F1algebra

Absolute irreducibility: let $F^{'} / F$ be a field extension and let $C^{' λ} = C^{λ} \otimes_{F} F^{'}$ be the base-changed cell module of the base-changed cellular algebra $A^{'} = A \otimes_{F} F^{'}$ . The cellular axioms and the cell form base change, so the Gram matrix of the form on $C^{' λ}$ is the scalar extension of the Gram matrix on $C^{λ}$ ; the rank of a matrix is the largest size of a minor with nonzero determinant, a field-independent computation with the polynomial determinants of the entries, so $\dim_{F^{'}} r a d C^{' λ} = \dim_{F} r a d C^{λ}$ . If $D^{λ} \neq 0$ then $D^{' λ} : = C^{' λ} / r a d C^{' λ} ≅ D^{λ} \otimes_{F} F^{'}$ is nonzero, and steps 1.2 and 2.1 applied over $F^{'}$ show that every element of $C^{' λ}$ outside $r a d C^{' λ}$ generates $C^{' λ}$ , so $r a d C^{' λ}$ is the unique maximal submodule of $C^{' λ}$ and $D^{' λ}$ is simple over $F^{'}$ . Hence $D^{λ}$ remains simple after every extension of scalars, i.e. it is absolutely irreducible.

3.2givenstep 1.1step 2.1F1F2F3

Completeness: every simple right $A$ -module $S$ is isomorphic to some $D^{λ}$ with $D^{λ} \neq 0$ . Order $Λ$ as $λ_{1}, \dots, λ_{N}$ so that $λ_{i} ⊳ λ_{j}$ implies $i < j$ , so larger labels come first, and put $A_{j} : = ⟨ c_{u v}^{λ_{i}} : i \leq j ⟩_{F}$ ; then each $A_{j}$ is a two-sided ideal (if $μ$ occurs among $λ_{1}, \dots, λ_{j}$ and $γ ⊵ μ$ , then $γ$ occurs there too, so $A^{⊵ μ} \subseteq A_{j}$ and the claim follows from step 1.1), $A_{0} = 0$ , $A_{N} = A$ . Moreover $A_{j} / A_{j - 1} ≅ A^{⊵ λ_{j}} / A^{⊳ λ_{j}}$ , which as a right $A$ -module is the direct sum over $s \in T (λ_{j})$ of the column spans $⟨ c_{s v}^{λ_{j}} + A^{⊳ λ_{j}} : v \in T (λ_{j}) ⟩ ≅ C^{λ_{j}}$ (the right action preserves the first index by [F2], and $A^{⊳ λ_{j}} \subseteq A_{j - 1}$ while $A^{⊵ λ_{j}} \cap A_{j - 1} = A^{⊳ λ_{j}}$ ). Let $S \neq 0$ be simple and let $j$ be least with $S A_{j} \neq 0$ (such $j$ exists since $S A_{N} = S A = S$ by unitality). Then $S A_{j}$ is a nonzero submodule of $S$ , so $S A_{j} = S$ , while $S A_{j - 1} = 0$ by minimality of $j$ ; choosing $s \in S$ with $s A_{j} \neq 0$ , the map $A_{j} / A_{j - 1} \to S$ , $a + A_{j - 1} \mapsto s a$ , is a well-defined nonzero $A$ -module surjection. As $A_{j} / A_{j - 1}$ is a direct sum of copies of $C^{λ_{j}}$ , the images of these copies sum to $S \neq 0$ , so some copy surjects onto the simple module $S$ ; thus $S$ is a simple quotient of $C^{λ_{j}}$ . If the form on $C^{λ_{j}}$ were identically zero, then $c_{s t}^{λ_{j}} c_{u v}^{λ_{j}} \equiv 0 (m o d A^{⊳ λ_{j}})$ for all $s, t, u, v$ by [F1]; since $A^{⊵ λ_{j}} A^{⊵ λ_{i}} \subseteq A^{⊵ λ_{i}} \subseteq A_{j - 1}$ for $i < j$ and $A^{⊵ λ_{j}} A^{⊳ λ_{j}} + A^{⊳ λ_{j}} A^{⊵ λ_{j}} \subseteq A^{⊳ λ_{j}} \subseteq A_{j - 1}$ , this gives $A_{j}^{2} \subseteq A_{j - 1}$ , whence $S = S A_{j} = S A_{j}^{2} \subseteq S A_{j - 1} = 0$ , a contradiction. Hence the form on $C^{λ_{j}}$ is not identically zero, so by step 2.1 $r a d C^{λ_{j}}$ is the unique maximal submodule and the simple quotient $S$ of $C^{λ_{j}}$ satisfies $S ≅ D^{λ_{j}}$ with $D^{λ_{j}} \neq 0$ .

4.1step 2.1step 2.2step 3.1step 3.2∎

Assembly: part 1 is step 3.1; part 2 is step 3.2 (exhaustion) together with step 2.2 (pairwise non-isomorphism) and step 2.1 (each nonzero $D^{λ}$ is simple); part 3 is step 2.1. The proof classifies right modules, and since the anti-automorphism $i$ identifies right and left $A$ -modules, the statement holds on either side. The classification applies to $A = H_{F, q} (S_{n})$ because the Murphy basis of The Murphy tableau elements form a cellular basis is a cellular basis of the type- $A$ Hecke algebra over every field. No semisimplicity was assumed, all posets and tableau sets are finite, and no choice principle is used.

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Cellular decomposition matrices are unitriangular for the cell order

Statement

Let $F$ be a field, $A$ a finite-dimensional cellular $F$ -algebra with cellular basis $(C, Λ)$ and cell modules $C^{λ}$ , $D^{λ} = C^{λ} / r a d C^{λ}$ (Cellular bases and cell modules, Nonzero cell heads classify the simple modules of a cellular algebra). For $λ, μ \in Λ$ with $D^{μ} \neq 0$ let $d_{λ μ} : = [C^{λ} : D^{μ}]$ be the multiplicity of $D^{μ}$ as a composition factor of $C^{λ}$ (well defined by Jordan–Hölder theorem for modules and Composition series and length of a module). Then $d_{μ μ} = 1 and d_{λ μ} \neq 0 ⟹ λ ⊵ μ,$ i.e. a nonzero off-diagonal entry can occur only in the allowed direction of the cell order $⊵$ (larger cell first). Order the labels by a linear extension from smaller to larger. The square submatrix whose rows and columns both have nonzero $D$ is lower unitriangular; the full rectangular matrix has the same labelled diagonal and the stated vanishing region. This determines only the diagonal and the vanishing region: no off-diagonal entry is computed by cellularity.

Facts & Assumptions

Given: A field $F$ , a finite-dimensional unital cellular $F$ -algebra $A$ with cellular basis $(C, Λ)$ , cell modules $C^{λ}$ and cell forms $⟨, ⟩_{λ}$ (Cellular bases and cell modules).

[F1]

Classification (Nonzero cell heads classify the simple modules of a cellular algebra): $D^{λ} \neq 0$ if and only if $r a d C^{λ} \neq C^{λ}$ if and only if $⟨ -, - ⟩_{λ}$ is not identically zero; in that case $D^{λ}$ is simple and $r a d C^{λ}$ is its unique maximal submodule, and ${D^{λ} : D^{λ} \neq 0}$ is a complete set of pairwise non-isomorphic simple $A$ -modules.

[F2]

Cell-form calculus (The cell form is associative and its radical is a submodule): $c_{u s}^{κ} c_{t v}^{κ} \equiv ⟨ c_{s}^{κ}, c_{t}^{κ} ⟩_{κ} c_{u v}^{κ} (m o d A^{⊳ κ})$ for a fixed label $κ$ ; $x c_{u v}^{κ} = ⟨ x, c_{u}^{κ} ⟩_{κ} c_{v}^{κ}$ for $x \in C^{κ}$ ; and $r a d C^{κ} = {x : ⟨ x, - ⟩_{κ} = 0}$ is the kernel of $x \mapsto ⟨ x, - ⟩_{κ}$ .

[F3]

Cellular axioms (Cellular bases and cell modules): $A^{⊳ λ}$ is the span of the $c_{u v}^{μ}$ with $μ ⊳ λ$ , $A^{⊵ λ}$ is the span with $μ ⊵ λ$ ; condition (ii) reads $c_{s t}^{λ} a \equiv \sum_{v} r_{v} c_{s v}^{λ} (m o d A^{⊳ λ})$ and its $i$ -image reads $a c_{s t}^{λ} \equiv \sum_{u} r_{u}^{'} c_{u t}^{λ} (m o d A^{⊳ λ})$ .

[F4]

Composition multiplicities are additive in short exact sequences and well defined by Jordan-Hölder (Composition series and length of a module, Jordan–Hölder theorem for modules).

Proof

technique · direct

1.1givenF2F3algebra

Annihilation: if $λ, ν \in Λ$ with $λ ⋭ ν$ , then $A^{⊵ ν} \cdot C^{λ} = 0$ . Indeed $A^{⊵ ν}$ and $A^{⊵ λ}$ are two-sided ideals by [F3], and their intersection is spanned by the $c^{γ}$ with $γ ⊵ ν$ and $γ ⊵ λ$ ; no such $γ$ equals $λ$ , since $γ = λ$ would give $λ ⊵ ν$ , so the intersection lies in $A^{⊳ λ}$ . Hence $A^{⊵ ν} A^{⊵ λ} \subseteq A^{⊵ ν} \cap A^{⊵ λ} \subseteq A^{⊳ λ}$ and, since $c_{t t}^{λ} \in A^{⊵ λ}$ , every product $c_{t t}^{λ} b$ with $b \in A^{⊵ ν}$ lies in $A^{⊳ λ}$ ; the right action of $b$ on $C^{λ}$ is read off from $c_{t t}^{λ} b \equiv \sum_{v} r_{v} c_{t v}^{λ} (m o d A^{⊳ λ})$ with unique coefficients, so $b$ acts as the zero operator on $C^{λ}$ .

1.2givenF1F2algebra

Nontrivial action: if $D^{μ} \neq 0$ then some element of $A^{⊵ μ}$ acts nontrivially on $D^{μ}$ . Since $r a d C^{μ} \neq C^{μ}$ by [F1], there is $t \in T (μ)$ with $c_{t}^{μ} \notin r a d C^{μ}$ , and by [F2] there is $u \in T (μ)$ with $⟨ c_{t}^{μ}, c_{u}^{μ} ⟩_{μ} \neq 0$ ; then the element $c_{u t}^{μ} \in A^{⊵ μ}$ satisfies $c_{t}^{μ} \cdot c_{u t}^{μ} = ⟨ c_{t}^{μ}, c_{u}^{μ} ⟩_{μ} c_{t}^{μ}$ by [F2], so on $D^{μ}$ it sends the nonzero class of $c_{t}^{μ}$ to its nonzero multiple $⟨ c_{t}^{μ}, c_{u}^{μ} ⟩_{μ} [c_{t}^{μ}] \neq 0$ .

2.1step 1.1step 1.2F4

Off-diagonal vanishing: suppose $d_{λ μ} \neq 0$ with $D^{μ} \neq 0$ . If $λ ⋭ μ$ , then step 1.1 gives $A^{⊵ μ} \cdot C^{λ} = 0$ ; an element of $A$ acting as zero on a module acts as zero on each of its submodules and quotients, hence on every subquotient of $C^{λ}$ , so $A^{⊵ μ}$ annihilates the composition factor $D^{μ}$ of $C^{λ}$ . This contradicts step 1.2. Hence $λ ⊵ μ$ .

2.2step 1.2F1F2F4

Diagonal: $d_{μ μ} = 1$ for every $μ$ with $D^{μ} \neq 0$ . The form radical gives the subquotient $D^{μ} = C^{μ} / r a d C^{μ}$ , and additivity of composition multiplicities [F4] gives $d_{μ μ} = [C^{μ} : D^{μ}] = [C^{μ} / r a d C^{μ} : D^{μ}] + [r a d C^{μ} : D^{μ}] = 1 + [r a d C^{μ} : D^{μ}]$ , because $C^{μ} / r a d C^{μ} = D^{μ}$ is simple by [F1]. The span of the $c_{u v}^{μ}$ annihilates $r a d C^{μ}$ : for $x \in r a d C^{μ}$ and all $u, v$ one has $x c_{u v}^{μ} = ⟨ x, c_{u}^{μ} ⟩_{μ} c_{v}^{μ} = 0$ by [F2]. Hence every element of this span acts as zero on $r a d C^{μ}$ and on all its subquotients, so $D^{μ}$ is not a composition factor of $r a d C^{μ}$ ; otherwise step 1.2 would produce an element of $s p a n {c_{u v}^{μ}} \subseteq A^{⊵ μ}$ acting nontrivially on it. Therefore $[r a d C^{μ} : D^{μ}] = 0$ and $d_{μ μ} = 1$ .

3.1step 2.1step 2.2∎

Assembly: order the labels by a linear extension from smaller to larger, and restrict both rows and columns to labels with nonzero form quotient. Step 2.1 puts every nonzero entry on or below the diagonal of this square submatrix, and step 2.2 gives diagonal entries $1$ . For the full rectangular matrix retain the label comparison $λ ⊵ μ$ ; deleting zero-quotient columns need not leave the labelled diagonal on the numerical matrix diagonal. These statements determine only the diagonal and the vanishing region.

DefinitionDefinition: AI-adaptedProof: Not applicableOpen item page →

The type-A Hecke Specht module and its cell form

Definition

Let $R$ be a commutative ring and $q \in R^{\times}$ . With the Murphy cellular basis $(M, Λ)$ of $H = H_{R, q} (S_{n})$ (The Murphy tableau elements form a cellular basis, Cellular bases and cell modules) and the canonical tableau $t^{λ}$ of shape $λ ⊢ n$ (Murphy elements and Murphy tableau elements), the Murphy cell module $S_{M}^{λ}$ is the right cell module $C^{λ}$ of the Murphy basis: it is the free $R$ -module with basis ${m_{t} : = m_{t^{λ} t} : t \in S t d (λ)}$ - the classes of the Murphy elements whose first index is the canonical tableau - and $H$ acts by $m_{t} h = \sum_{v} r_{v} m_{v}$ whenever $m_{t^{λ} t} h \equiv \sum_{v} r_{v} m_{t^{λ} v} (m o d H^{⊳ λ})$ . (The cellular action rule transforms the second index and fixes the first, so the cell module is defined using the fixed-first-index family above; the diagonal span in $H$ need not be stable under multiplication. For $λ = (n)$ it is stable, since $m_{λ} T_{i} = q m_{λ}$ .) Its cell form is the unique bilinear form $⟨, ⟩_{λ} : S_{M}^{λ} \times S_{M}^{λ} \to R$ with $m_{u s} m_{t v} \equiv ⟨ m_{s}, m_{t} ⟩_{λ} m_{u v} (m o d H^{⊳ λ})$ for all $s, t, u, v \in S t d (λ)$ ; it is symmetric and associative and $r a d S_{M}^{λ}$ is a submodule, so $D_{M}^{λ} : = S_{M}^{λ} / r a d S_{M}^{λ}$ is a right $H$ -module (The cell form is associative and its radical is a submodule).

Convention (Dipper-James signed cellular normalization). Let $θ : H \to H$ be the involutive algebra automorphism defined by $θ (T_{i}) = - q T_{i}^{- 1} = (q - 1) - T_{i}$ (The Hecke sign automorphism, pullback and contragredient dual). For a right module $M$ , write $M^{θ}$ for the same $R$ -module with action $x \cdot h = x θ (h)$ . The page's ordinary Specht normalization is the signed conjugate Murphy construction $S^{λ} : = (S_{M}^{λ^{'}})^{θ},$ where $λ^{'}$ is the conjugate partition (Partitions, English diagrams, and conjugation). Its form is the form of $S_{M}^{λ^{'}}$ on the same underlying module; because $θ$ commutes with $*$ this is associative for the twisted action (The Hecke sign automorphism, pullback and contragredient dual). Over a field put $D^{λ} : = S^{λ} / {rad}_{f o r m} S^{λ} = (D_{M}^{λ^{'}})^{θ} .$ Thus $S_{M}^{λ^{'}} = (S^{λ})^{θ}$ and $D_{M}^{λ^{'}} = (D^{λ})^{θ}$ ; conjugation alone does not identify these modules (Murphy cells are the same-shape contragredient Specht duals, Ordinary Specht modules are the signed conjugate Murphy cells). At $q = 1$ , $θ (σ) = sgn (σ) σ$ , so these identities contain tensoring with sign; in characteristic $2$ the sign twist becomes trivial at $q = 1$ (Polytabloid specialization and the exact signed multiplicity dictionary). The nonzero quotients in this signed normalization have $e$ -regular labels by The simple type-A Hecke modules are indexed by e-regular partitions ↗: that theorem proves the nonzero Murphy quotient criterion in $e$ -restricted labels, and conjugation makes these precisely $e$ -regular labels (e-regular and e-restricted partitions). No generic root-of-unity Hecke claim is inferred from the published prime-characteristic group-only sign dictionary, and no semisimplicity or choice principle is assumed.

DefinitionDefinition: AI-adaptedProof: Not applicableOpen item page →

The Jantzen filtration of a lifted cell module

Definition

Let $R$ be a principal ideal domain (Principal ideal domain) with a chosen prime element $π$ , let $F : = R / π R$ be the residue field, and let $M$ be a free $R$ -module of finite rank (The free module on a set and its standard basis) equipped with a symmetric $R$ -bilinear form $⟨, ⟩ : M \times M \to R$ whose extension to the fraction field $K = Frac (R)$ is non-degenerate (equivalently $\det G \neq 0$ for a Gram matrix $G$ of the form). For $i \geq 1$ put $M (i) : = {x \in M : ⟨ x, y ⟩ \in π^{i} R for all y \in M}, M_{F} (i) : = (M (i) + π M) / π M \subseteq M_{F} : = M \otimes_{R} F .$ Then $M (0) = M$ , the $M (i)$ are $R$ -submodules with $M (i + 1) \subseteq M (i)$ , the descending chain $M_{F} = M_{F} (0) \supseteq M_{F} (1) \supseteq M_{F} (2) \supseteq \dots$ stabilises at $0$ (that is, $M_{F} (z) = 0$ for all sufficiently large $z$ , because $\det G \neq 0$ ), and $M_{F} (1) = r a d M_{F}$ , the radical of the reduced form. This finite descending filtration is the Jantzen filtration of $M$ with respect to $(R, π, ⟨, ⟩)$ . The hypotheses are exactly: $R$ a PID, $π$ a prime element, the form non-degenerate over $Frac (R)$ ; no valuation-ring or DVR hypothesis is substituted, and the prime element $π$ , its ideal $π R$ and the fraction field $K$ are kept distinct. Each $M (i)$ is a submodule: for the cell forms on this page the associativity identity $⟨ x a, y ⟩ = ⟨ x, y a^{*} ⟩$ shows closedness under the action. For $0 \neq r \in R$ , define $ν_{π} (r) = \max {k \geq 0 : π^{k} ∣ r}$ . This maximum is finite: otherwise the uniquely determined elements $r / π^{k}$ give an ascending chain of ideals whose union is principal, say generated by an element in its $N$ -th ideal. The chain then stabilizes at $N$ , whereas $(r / π^{N}) = (r / π^{N + 1})$ and cancellation would make $π$ a unit. For termination, put $a = ν_{π} (\det G)$ . If $x \in M (i)$ , its coordinate column $v$ satisfies $G v \in π^{i} R^{d}$ , so the adjugate identity gives $(\det G) v \in π^{i} R^{d}$ . For $i > a$ , primality of $π$ forces every coordinate of $v$ into $π R$ . Thus $M (i) \subseteq π M$ and $M_{F} (i) = 0$ . In rank zero the filtration is already zero. Finally a reduced vector belongs to the reduced form radical exactly when any lift pairs with every vector of $M$ into $π R$ , which is exactly membership in $M (1)$ . No choice principle is used.

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

The Hecke Jucys-Murphy action is dominance triangular

Statement

Let $R$ be a commutative ring, $q \in R^{\times}$ , $H = H_{R, q} (S_{n})$ , and let $L_{1}, \dots, L_{n}$ be the Hecke Jucys-Murphy elements (Hecke Jucys-Murphy elements and q-contents). For every $λ ⊢ n$ , every standard $λ$ -tableau $t$ and every $1 \leq k \leq n$ there exist $a_{v} \in R$ with $m_{t} L_{k} = [{content}_{t} (k)]_{q} m_{t} + \sum_{v ⊳ t} a_{v} m_{v},$ where $(m_{t})_{t \in S t d (λ)}$ is the Murphy basis of the Murphy cell module $S_{M}^{λ}$ (The type-A Hecke Specht module and its cell form) and the sum is over standard $λ$ -tableaux $v$ strictly dominating $t$ in the dominance order on tableaux (Dominance order on partitions, Tableaux and standard tableaux). In particular $L_{k}$ acts on the Murphy cell module $S_{M}^{λ}$ with Murphy basis $(m_{t})$ by a triangular matrix whose diagonal entries are the q-content values. Consequently the $L_{k}$ commute pairwise and every symmetric polynomial in $L_{1}, \dots, L_{n}$ is central in $H$ .

Facts & Assumptions

Given: A commutative ring $R$ , a unit $q$ , the type-A Hecke algebra $H$ , its additive Jucys-Murphy elements $L_{k}$ and its Murphy cell module $S_{M}^{λ}$ with basis $(m_{t})$ .

[F1]

The standard basis multiplication, the quadratic and braid relations, and the integral row action are those of The row permutation module: distinguished basis and Hecke action. In particular $T_{i}^{2} = (q - 1) T_{i} + q$ , and $x_{ρ} T_{i} = q x_{ρ}$ if $i, i + 1$ lie in one row of $t^{ρ}$ .

[F2]

The integral Murphy basis, dominance ideals, cell multiplication and finite Garnir straightening are proved in The Murphy cellular basis, the cell law and the Murphy form. We write $I^{⊳ λ} = H^{⊳ λ}$ and $x_{λ} = m_{λ}$ . The proof there supplies the dominance support rule: if $s^{'} = s s_{i}$ is standard with $s ⊳ s^{'}$ and $ℓ (d (s^{'})) = ℓ (d (s)) + 1$ , multiplying any same-shape term strictly above $s$ by $T_{i}$ and then straightening produces only standard terms strictly above $s^{'}$ . Straightening a nonstandard row-standard tableau only raises its tableau order in the same shape. All these identities hold over $A = Z [v^{\pm 1}]$ and specialize along $v \mapsto q$ .

[F3]

A finite-dimensional algebra has nilpotent Jacobson radical and semisimple quotient (For a finite-dimensional algebra, the Jacobson radical is nilpotent and the quotient by it is semisimple). A semisimple algebra is a finite product of matrix algebras over division rings (Wedderburn–Artin theorem for semisimple rings); its two-sided ideals and their quotients consequently have no nonzero nilpotent two-sided ideals.

Proof

technique · direct

1.1givenF1algebra

Braid identities. Put $T_{i} = T_{(i, i + 1)}$ . The reduced word for $(j, k + 1)$ gives $L_{k + 1} = q^{- 1} (T_{k} L_{k} T_{k} + T_{k})$ . We prove $T_{i} L_{k} = L_{k} T_{i}$ whenever $i \neq k - 1, k$ . For $i \geq k + 1$ , each generator in the displayed words defining $L_{k}$ commutes with $T_{i}$ . For $i \leq k - 3$ the recurrence and induction apply. For the remaining case $k = i + 2$ , put $a = T_{i}$ , $b = T_{i + 1}$ and $C = L_{i}$ ; then $b C = C b$ by the already established disjoint case and $L_{i + 2} = q^{- 1} b + q^{- 2} b a b + q^{- 2} b a C a b$ . The braid and quadratic relations give $[a, b a b] = - q [a, b]$ and $a (b a C a b) = (b a C a b) a$ (replace $a b a$ by $b a b$ and move $C$ past $b$ ), so its commutator with $a$ vanishes. Thus $L_{k}$ centralizes $H (S_{k - 1})$ ; since $L_{j} \in H (S_{k - 1})$ for $j < k$ , all $L_{j}$ commute. Multiplying the recurrence by $T_{i}$ and using $T_{i}^{2} = (q - 1) T_{i} + q$ gives $T_{i} L_{i + 1} = L_{i} T_{i} + 1 + (q - 1) L_{i + 1}$ ; starring gives $T_{i} L_{i} = L_{i + 1} T_{i} - 1 - (q - 1) L_{i + 1}$ . Hence $T_{i}$ commutes with $L_{i} + L_{i + 1}$ and $L_{i} L_{i + 1}$ : for the product substitute the first identity into $(L_{i + 1} T_{i} - 1 - (q - 1) L_{i + 1}) L_{i + 1}$ . Every symmetric polynomial is a polynomial in the elementary symmetric polynomials over the coefficient ring: in each homogeneous degree, its lexicographically leading exponent vector satisfies $a_{1} \geq \dots \geq a_{n}$ by symmetry. Subtract its coefficient times $e_{1}^{a_{1} - a_{2}} \dots e_{n - 1}^{a_{n - 1} - a_{n}} e_{n}^{a_{n}}$ , whose leading monomial has that vector and coefficient $1$ , and repeat on the finite set of degree- $d$ monomials. No coefficient is divided. Each elementary symmetric polynomial splits, at the adjacent pair, as $e_{d} ({\hat{L}}_{i}, {\hat{L}}_{i + 1}) + (L_{i} + L_{i + 1}) e_{d - 1} ({\hat{L}}_{i}, {\hat{L}}_{i + 1}) + L_{i} L_{i + 1} e_{d - 2} ({\hat{L}}_{i}, {\hat{L}}_{i + 1})$ . It therefore commutes with every $T_{i}$ and is central.

1.2F2algebra

Lifting smaller-rank congruences. Here is the precise ideal check used in the induction below. Let $τ$ be either a whole-row prefix $(λ_{1}, \dots, λ_{r})$ of size $m < n$ , or the prefix $(λ_{1}, \dots, λ_{r}, 1)$ ending at the first box of the last row, of size $m < n$ . In both cases $x_{λ} = h x_{τ} = x_{τ} h$ , where $h$ is the row symmetrizer of the remaining blocks, embedded in $H_{n}$ ; in the second case its last block starts at the singleton label $m$ , and no generator of $x_{τ}$ meets that block. If $x_{τ} L_{k} - c x_{τ} \in I^{⊳ τ}$ in $H_{m}$ , expand this difference in its smaller Murphy basis. Every shape $η ⊳ τ$ lies in $H_{m} x_{η} H_{m}$ . Embedded in $H_{n}$ , $x_{η}$ is $x_{κ_{η}}$ for the partition $κ_{η} = (η, 1^{n - m})$ , so [F2] puts its two-sided ideal in $I^{⊵ κ_{η}}$ . Some prefix $j$ before the final row of $τ$ has $\sum_{a \leq j} η_{a} > \sum_{a \leq j} τ_{a} = \sum_{a \leq j} λ_{a}$ : strict dominance cannot first differ at or after the final row, where $τ$ has total $m$ . In the whole-row case the same argument permits $j < r$ . Hence $λ$ does not dominate $κ_{η}$ , and none of these ideals has a shape- $λ$ component. Multiplying the smaller congruence on the appropriate side by $h$ gives $x_{λ} L_{k} - c x_{λ} \in \sum_{η} I^{⊵ κ_{η}}$ . This difference also belongs to $I^{⊵ λ}$ because both terms are in $x_{λ} H_{n}$ . Their intersection, read in the Murphy basis, has only shapes strictly dominating $λ$ , and the lifted congruence therefore holds modulo $I^{⊳ λ}$ .

2.1F1F2F3step 1.1algebra

An integral scalar for the central sum. For this step work over $A = Z [v^{\pm 1}]$ , with fraction field $K = Q (v)$ . The matrix of the regular trace pairing $B (a, b) = {Tr}_{K} (ℓ_{a b})$ in the standard basis has entries in $A$ . At $v = 1$ it is the matrix $n! δ_{u, w^{- 1}}$ , since a nonidentity permutation has no fixed point in the regular permutation action. Its determinant is nonzero, so $B$ is nondegenerate over $K$ . If $a$ belongs to the Jacobson radical $J$ then $a b \in J$ is nilpotent for every $b$ , by [F3]; thus $ℓ_{a b}$ is nilpotent and has trace zero. Nondegeneracy forces $J = 0$ , so $H_{K}$ is semisimple. In its cell layer $I_{K}^{⊵ λ} / I_{K}^{⊳ λ}$ , [F2] identifies multiplication with $X \cdot Y = X G_{λ} Y$ on $d \times d$ matrices, $d = ∣ Std (λ) ∣$ . If $G_{λ}$ were singular, the nonzero subspace $N = {X : G_{λ} X = 0}$ would be a two-sided ideal of this layer with $N^{2} = 0$ , contradicting [F3]. Hence $G_{λ}$ is invertible and $X \mapsto G_{λ} X$ identifies the layer with ${Mat}_{d} (K)$ . A central element of $H_{K}$ acts on this layer by a bimodule endomorphism of that matrix algebra, necessarily a scalar: its value at the identity commutes with every matrix and therefore is a scalar matrix. Applied to $Z_{n} = \sum_{k} L_{k}$ , central by step 1.1, this shows that its matrix on the cell module is $α_{λ} I_{d}$ over $K$ . The cell action matrix already has entries in $A$ , so $α_{λ} \in A$ , and freeness over $A$ makes the identity hold integrally and after every base change to $R$ . Thus $m_{t} Z_{n} = α_{λ} m_{t}$ for every $t$ , without any assumption on the specialized cell form or its head.

2.2F1F2step 1.1algebra

Transport along standard adjacent swaps. Suppose $s^{'} = s s_{i}$ is standard with $s ⊳ s^{'}$ and length increasing by one, and suppose $m_{s} L_{k} = r_{s} (k) m_{s} + \sum_{u ⊳ s} a_{u} m_{u}$ for all $k$ . By [F1], $m_{s^{'}} = m_{s} T_{i}$ . For $k \neq i, i + 1$ step 1.1 gives $m_{s^{'}} L_{k} = m_{s} L_{k} T_{i}$ , and [F2] puts every error strictly above $s^{'}$ . For $k = i + 1$ , the same step gives $m_{s^{'}} L_{i + 1} = m_{s} (L_{i} T_{i} + 1 + (q - 1) L_{i + 1})$ ; its coefficient at $m_{s^{'}}$ is $r_{s} (i)$ , since all remaining terms are strictly above $s^{'}$ . The second adjacent identity similarly gives coefficient $r_{s} (i + 1)$ at $m_{s^{'}}$ for $L_{i}$ . Thus the diagonal data are transposed in positions $i, i + 1$ and unchanged elsewhere. Every standard tableau is reachable from $t^{λ}$ by such standard length-increasing swaps: reverse a right-descent reduction of $d (t)$ ; the two consecutive entries at a descent lie in incomparable cells, so their interchange remains standard. Consequently diagonal data at the row-filled tableau determine the asserted diagonals at every standard tableau.

3.1F1step 1.1step 2.1step 2.2step 1.2inductionalgebra

Row-filled induction away from the final singleton. Induct on $n$ , the cases $n = 0, 1$ being immediate, and on $k$ within the row-filled tableau. If $k$ is in a row which is not the last row, choose the whole-row prefix $τ$ ending with that row. It has size $m < n$ , its row-filled tableau has no more dominant standard tableaux, and the induction on rank gives $x_{τ} L_{k} = [{content}_{t^{τ}} (k)]_{q} x_{τ}$ modulo $I^{⊳ τ}$ . Lift by step 1.2. If $k$ is in the last row but not its first column, $T_{k - 1}$ belongs to its row subgroup, and the recurrence gives $m_{t^{λ}} L_{k} = (1 + q [{content}_{t^{λ}} (k - 1)]_{q}) m_{t^{λ}} = [{content}_{t^{λ}} (k)]_{q} m_{t^{λ}}$ , by induction on $k$ and $[c + 1]_{q} = 1 + q [c]_{q}$ . If $k < n$ is in the first column of the last row, use the singleton-ended prefix $τ$ of step 1.2 and induction on its smaller rank. These arguments leave only $k = n$ for a shape $λ = (λ_{1}, \dots, λ_{r}, 1)$ . All $L_{k}$ with $k < n$ have already been shown to act on $m_{t^{λ}}$ by their claimed scalars. Step 2.1 then supplies $m_{t^{λ}} L_{n} = r_{n} m_{t^{λ}}$ for some $r_{n} \in R$ . Step 2.2 consequently applies with these diagonal data to every standard tableau; at this point all $L_{k}$ are triangular, though $r_{n}$ remains to be identified.

4.1F1F2step 1.1step 2.2step 3.1algebra

The final singleton is fixed by one Garnir calculation. In the remaining shape let $i$ be the first entry in row $r$ of $t^{λ}$ and put $t = t^{λ} s_{n - 1} \dots s_{i + 1}$ . Then $t$ is standard, its last row contains $i + 1$ , and its penultimate row contains $i, i + 2, \dots, n$ . Step 2.2 gives $m_{t} L_{i} = [1 - r]_{q} m_{t} + \sum_{v ⊳ t} a_{v} m_{v}$ and $m_{t} L_{i + 1} = r_{n} m_{t} + \sum_{v ⊳ t} b_{v} m_{v}$ . The tableau $g : = t s_{i}$ is exactly the Garnir tableau for the first-column belt $i, \dots, n$ between the last two rows. For $j = i, \dots, n$ , let $u_{j}$ have last-row entry $j$ and the other belt labels increasingly in row $r$ . The finite belt identity of [F2] gives $\sum_{j = i}^{n} x_{λ} T_{d (u_{j})} \equiv 0 (m o d I^{⊳ λ})$ : its other side is a sum of multiples of the symmetrizer obtained by merging these two rows, whose sorted shape strictly dominates $λ$ . Here $u_{i} = g$ , $u_{i + 1} = t$ , and all $u_{j}$ with $j > i + 1$ are standard and strictly dominate $t$ . The length-increasing row rule gives $m_{t} T_{i} = x_{λ} T_{d (g)} + I^{⊳ λ}$ , so the belt identity explicitly yields $m_{t} T_{i} = - m_{t} - \sum_{j > i + 1} m_{u_{j}}$ . Put $U = span {m_{v} : v ⊳ t}$ . Step 3.1 and transport show $U L_{i} \subseteq U$ . Also $U T_{i} \subseteq U$ : dominance forces each $v ⊵ t$ to agree with $t$ on the filled first $r - 1$ rows and to place $i$ in node $(r, 1)$ . If $i + 1$ occupies the singleton last row, all remaining labels must occupy row $r$ increasingly, forcing $v = t$ . Thus every $v ⊳ t$ places $i + 1$ in $(r, 2)$ , and the row rule gives $m_{v} T_{i} = q m_{v}$ . (If $λ_{r} = 1$ , the higher span is zero.) Working modulo $U$ , the recurrence now gives $m_{t} L_{i + 1} = q^{- 1} m_{t} (T_{i} L_{i} T_{i} + T_{i}) = q^{- 1} ([1 - r]_{q} - 1) m_{t} = [- r]_{q} m_{t}$ . Comparing its coefficient at $m_{t}$ proves $r_{n} = [- r]_{q}$ , exactly the content of the final singleton. This completes the rank induction and identifies all row-filled diagonals.

5.1step 1.1step 2.2step 4.1F2algebra∎

The required flag and the initial left generator. Step 2.2 now gives $m_{t} L_{k} = [{content}_{t} (k)]_{q} m_{t} + \sum_{v ⊳ t} a_{v} m_{v}$ for every standard $t$ , over the given ring. Order the standard tableaux by any linear extension with the most dominant first. The spans of initial segments are stable under all $L_{k}$ , and their successive rank-one quotients have exactly the q-content diagonals; this proves the claimed common triangular flag. Commutativity and symmetric-polynomial centrality were established independently in step 1.1. In particular the maximal tableau $t^{λ}$ has the exact eigenrelation $m_{t^{λ}} L_{k} = [{content}_{t^{λ}} (k)]_{q} m_{t^{λ}}$ . Since $L_{k}^{*} = L_{k}$ , starring its defining cell congruence gives the corresponding left-cell generator relation $L_{k} z_{λ} = [{content}_{t^{λ}} (k)]_{q} z_{λ}$ . For the multiplicative normalization $X_{k} = 1 + (q - 1) L_{k}$ this reads $X_{k} z_{λ} = q^{{content}_{t^{λ}} (k)} z_{λ}$ , using $1 + (q - 1) [c]_{q} = q^{c}$ for all integral $c$ . These identities concern the cell generator, hence hold after taking the indicated cell quotient; they do not assert the same equality for the raw symmetrizer in $H$ . All formulas specialize from $A$ to $R$ , and no quantum integer or factorial is inverted in the specialized algebra.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

Jantzen layer dimensions sum to the Gram determinant valuation

Statement

In the setting of The Jantzen filtration of a lifted cell module ( $R$ a PID, $π$ a prime element, $F = R / π R$ , $M$ free of finite rank with a form whose extension to $K = Frac (R)$ is non-degenerate), let $ν_{π}$ be the $π$ -adic valuation normalised by $ν_{π} (π) = 1$ and let $G \in R ∖ {0}$ be the Gram determinant of the form with respect to any $R$ -basis of $M$ . Then $ν_{π} (G) = \sum_{i > 0} \dim_{F} M_{F} (i),$ where $M_{F} (i)$ are the Jantzen layers of The Jantzen filtration of a lifted cell module; in particular the sum is finite and independent of the chosen basis.

Facts & Assumptions

Given: A PID $R$ with prime element $π$ , residue field $F = R / π R$ , fraction field $K$ , a free $R$ -module $M$ of finite rank $d$ with a symmetric $R$ -bilinear form $⟨, ⟩$ that is non-degenerate over $K$ , and its Jantzen filtration $M (i) \subseteq M$ with layers $M_{F} (i) = (M (i) + π M) / π M$ (The Jantzen filtration of a lifted cell module).

[F1]

(The Jantzen filtration of a lifted cell module) The $M (i)$ , $i \geq 0$ , are submodules with $M (0) = M$ and $M (i + 1) \subseteq M (i)$ , and $M_{F} (i) = 0$ for all sufficiently large $i$ .

[F2]

(Every matrix over a PID has a Smith normal form) There are $R$ -bases $e_{1}, \dots, e_{d}$ of $M$ and $f_{1}, \dots, f_{d}$ of $M$ together with nonzero $r_{1}, \dots, r_{d} \in R$ such that $⟨ e_{i}, f_{j} ⟩ = δ_{i j} r_{i}$ for all $i, j$ ; the $r_{i}$ are the elementary divisors of the Gram matrix.

[F3]

For any $R$ -basis of $M$ with Gram matrix $G$ one has $\det G = r_{1} \dots r_{d}$ up to a unit, so $ν_{π} (G) = \sum_{i} ν_{π} (r_{i})$ , and $ν_{π}$ is normalised by $ν_{π} (π) = 1$ and takes the value $0$ on units.

Proof

technique · direct

1.1givenF1F2

Description of the Jantzen layers: write $x = \sum_{i} a_{i} e_{i}$ in the basis of [F2]. Since $f_{1}, \dots, f_{d}$ is an $R$ -basis of $M$ , the condition $⟨ x, y ⟩ \in π^{k} R$ for all $y \in M$ is equivalent to $⟨ x, f_{j} ⟩ \in π^{k} R$ for all $j$ ; as $⟨ x, f_{j} ⟩ = a_{j} r_{j}$ , this gives $M (k) = {\sum_{i} a_{i} e_{i} : ν_{π} (a_{i} r_{i}) \geq k for all i}$ with the convention $ν_{π} (0) = \infty$ . Reducing modulo $π M$ , whose basis is $π e_{1}, \dots, π e_{d}$ , shows that $M_{F} (k)$ has as an $F$ -basis the images of those $e_{i}$ with $ν_{π} (r_{i}) \geq k$ : if $ν_{π} (r_{i}) \geq k$ then the image of $e_{i}$ lies in $M_{F} (k)$ , while if $ν_{π} (r_{i}) < k$ then every element of $M (k)$ has $e_{i}$ -coefficient divisible by $π$ . Hence $\dim_{F} M_{F} (k) = # {i : ν_{π} (r_{i}) \geq k}$ .

2.1step 1.1F1

Summation: since each $r_{i}$ is nonzero, the sets ${k > 0 : ν_{π} (r_{i}) \geq k}$ are finite of cardinality $ν_{π} (r_{i})$ ; by step 1.1 and [F1], $\sum_{k > 0} \dim_{F} M_{F} (k) = \sum_{k > 0} # {i : ν_{π} (r_{i}) \geq k} = \sum_{i} ν_{π} (r_{i}) .$

3.1step 2.1F3∎

Conclusion: by [F3] the right-hand side is $ν_{π} (G)$ for any Gram matrix $G$ of the form in any $R$ -basis of $M$ ; the left-hand side is defined purely in terms of $M$ and the form, so the identity holds for every basis and the sum is finite and basis-independent.

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Separated Hecke residues give seminormal idempotents and bases

Statement

Let $F$ be a field, $q \in F^{\times}$ , let $e = e (F, q)$ and assume $e > n$ (with $e = \infty$ allowed); let $λ ⊢ n$ , $H = H_{F, q} (S_{n})$ , and let $S_{M}^{λ}$ be the Murphy cell module with Murphy basis $(m_{t})_{t \in S t d (λ)}$ (The type-A Hecke Specht module and its cell form).

(Separation) If $s, t$ are standard $λ$ -tableaux with $[{content}_{s} (k)]_{q} = [{content}_{t} (k)]_{q}$ for all $1 \leq k \leq n$ then $s = t$ . More precisely, if the q-content values agree for all $k \notin {i, i + 1}$ for some $i$ , then $s = t$ or $t = s (i i + 1)$ . 2. (Idempotents) Let $R_{k}$ be the set of distinct q-content values of the entry $k$ over standard tableaux of all partitions of $n$ . For a standard $λ$ -tableau $t$ define $F_{t} : = \prod_{k = 1}^{n} \prod_{\begin{array}{c} c \in R_{k} \\ c \neq [{content}_{t} (k)]_{q} \end{array}} \frac{L_{k} - c}{[{content}_{t} (k)]_{q} - c} \in F [L_{1}, \dots, L_{n}] .$ These are orthogonal idempotents of $H$ ; their actions on $S_{M}^{λ}$ project onto the corresponding simultaneous eigenlines. Put $f_{t} = m_{t} F_{t}$ . Then $f_{t} L_{k} = [{content}_{t} (k)]_{q} f_{t}$ and $f_{t} = m_{t} + \sum_{s ⊳ t} a_{s} m_{s}$ . The global content sets are essential for the asserted idempotency in $H$ , rather than only the induced same-shape endomorphisms. 3. (Orthogonal seminormal basis) ${f_{t} : t \in S t d (λ)}$ is a basis of $S_{M}^{λ}$ satisfying $f_{t} F_{s} = δ_{t s} f_{t}$ and $⟨ f_{s}, f_{t} ⟩_{λ} = δ_{s t} ⟨ f_{t}, f_{t} ⟩_{λ}$ ; it is the seminormal basis of the Murphy cell module. 4. (Generator action) Let $s$ be a standard $λ$ -tableau, $1 \leq i < n$ and $t : = s (i i + 1)$ , and put $ρ : = {content}_{s} (i) - {content}_{s} (i + 1) \in Z$ . If $i$ and $i + 1$ lie in the same row of $s$ then $f_{s} T_{i} = q f_{s}$ ; if they lie in the same column then $f_{s} T_{i} = - f_{s}$ ; otherwise $t$ is standard and $f_{s} T_{i} = α_{i} (s) f_{t} - \frac{1}{[ρ]_{q}} f_{s}, α_{i} (s) : = {\begin{cases} 1, & s ⊳ t, \\ \frac{[1 + ρ]_{q} [1 - ρ]_{q}}{[ρ]_{q} [- ρ]_{q}}, & t ⊳ s, \end{cases}$ where $[- ρ]_{q}$ is the extended quantum integer of Hecke Jucys-Murphy elements and q-contents and for the swapped tableau ${content}_{t} (i) - {content}_{t} (i + 1) = - ρ$ . No semisimplicity of $H$ is assumed and no choice principle is used.

Facts & Assumptions

Given: A field $F$ , $q \in F^{\times}$ with quantum characteristic $e > n$ , a partition $λ ⊢ n$ , the Murphy cell module $S_{M}^{λ}$ with Murphy basis $(m_{t})$ and cell form (The type-A Hecke Specht module and its cell form), and the q-content values of Hecke Jucys-Murphy elements and q-contents.

[F1]

(The Hecke Jucys-Murphy action is dominance triangular) $m_{t} L_{k} = [{content}_{t} (k)]_{q} m_{t} + \sum_{v ⊳ t} a_{v} m_{v}$ in $S_{M}^{λ}$ ; the $L_{k}$ commute pairwise and every symmetric polynomial in them is central.

[F2]

(The quantum characteristic of an invertible Hecke parameter) For $e > n$ no $m$ with $1 \leq m \leq n$ has $[m]_{q} = 0$ ; more precisely, if two integer contents $c, c^{'}$ with $∣ c - c^{'} ∣ \leq n$ satisfy $[c]_{q} = [c^{'}]_{q}$ then $c = c^{'}$ .

[F3]

(The type-A Hecke Specht module and its cell form) The cell form on $S_{M}^{λ}$ is symmetric and associative, and the star anti-involution fixes each $L_{k}$ ; the Murphy basis is triangular along the dominance order on tableaux.

[F4]

Source results: separation of q-content sequences for $e > n$ and the seminormal spectral construction are Mathas, Lemma 3.32, Proposition 3.33 and Theorem 3.34 (printed pp. 30-34); the coefficient system of part 4 as displayed is that of Hu-Mathas, Example 3.11 with the axial distance (3.8), Lemma 3.12 and Proposition 3.17, the 1998 manuscript's $t ⊳ s$ branch being inconsistent with the Hecke relations. The abstract form of the spectral construction under the separation condition is Mathas, Proposition 4.13 with the appendix.

Proof

technique · direct

1.1givenF1F2

Reconstruct a tableau from its residue tuple. At a prefix of size $k - 1$ , two different addable corners have integer content difference of absolute value at most $k$ . Since $[a]_{q} - [b]_{q} = q^{b} [a - b]_{q}$ and $e > n$ , their q-contents differ. Starting with the empty diagram, the tuple thus recovers each added node and the final shape. For two tableaux of a fixed shape with identical entries' residues outside $i, i + 1$ , first reconstruct the common prefix through $i - 1$ , and reconstruct the common suffix by successively deleting the unique appropriate removable corner from the fixed final shape. The intervening skew diagram has two boxes; its only possible orders are the original order and its swap. This proves both separation assertions.

2.1F1step 1.1

Diagonalize the commuting JM operators on a cell. Their triangular basis gives a filtration with one-dimensional joint characters indexed by standard tableaux. Those characters are distinct by step 1.1. For commuting triangular operators, the simultaneous generalized eigenspaces split along their distinct joint characters: the corresponding maximal ideals of the polynomial algebra are pairwise comaximal. Each joint character occurs once in this cell, so each generalized eigenspace is one-dimensional and hence an actual joint eigenline. This finite Chinese-remainder argument works over any field and needs no choice of a generic scalar or algebraic closure. The Lagrange product of the Statement is $1$ on its target tuple and $0$ on every other tuple. Applied to $m_{t}$ , its triangular diagonal coefficient is $1$ , and the other possible terms are strictly more dominant. Thus it gives the asserted $f_{t}$ .

2.2F1F2step 1.1

Compute the diagonal coefficient using the correctly ordered JM relation $T_{i} L_{i} = L_{i + 1} T_{i} - (q - 1) L_{i + 1} - 1.$ For $k \neq i, i + 1$ , $T_{i}$ commutes with $L_{k}$ . Consequently separation restricts $f_{s} T_{i}$ to the span of $f_{s}$ and $f_{t}$ if the swap is standard, or to $f_{s}$ alone otherwise. Write its coefficient of $f_{s}$ as $c_{s}$ . The displayed relation gives $c_{s} ([{content}_{s} (i)]_{q} - [{content}_{s} (i + 1)]_{q}) = - (q - 1) [{content}_{s} (i + 1)]_{q} - 1.$ Its denominator is $q^{{content}_{s} (i + 1)} [ρ]_{q}$ , while the numerator is $- q^{{content}_{s} (i + 1)}$ , so $c_{s} = - 1 / [ρ]_{q}$ . This does not claim that $T_{i}$ commutes with $L_{i}$ .

3.1step 2.1

At this stage step 2.1 establishes the projector identities on each individual cell module. The global identities in $H$ will follow below after proving that the sum of the cell modules is faithful. No finite- $e$ classification is a premise.

3.2F3step 1.1step 2.1

The transition from $m_{t}$ to $f_{t}$ is unitriangular, so these vectors form a basis. For $s \neq t$ , choose $k$ at which their q-contents differ. Self-adjointness $L_{k}^{*} = L_{k}$ and associativity give $([{content}_{s} (k)]_{q} - [{content}_{t} (k)]_{q}) ⟨ f_{s}, f_{t} ⟩ = 0.$ The scalar is nonzero, proving orthogonality. The Lagrange action also gives $f_{t} F_{s} = δ_{t s} f_{t}$ .

3.3step 2.2

If the swap is not standard, the consecutive entries lie adjacently in one row or one column. Then $ρ = - 1$ or $ρ = 1$ , respectively, and step 2.2 gives $c_{s} = q$ or $c_{s} = - 1$ . These are the stated row and column cases; the row action is $q$ , not the scalar $1$ at a general parameter.

4.1step 3.2step 2.2

Suppose the swap is standard. In the branch $s ⊳ t$ , the length-additive Murphy action has leading term $m_{s} T_{i} = m_{t}$ ; the unitriangular eigenbasis change then gives the off-diagonal coefficient $α_{i} (s) = 1$ . Higher-dominance terms cannot alter the coefficient at $t$ in this two-dimensional eigenline span. This is the leading-term argument in Mathas's source theorem on the seminormal basis. In both branches the quadratic relation gives $α_{i} (s) α_{i} (t) = q + (q - 1) c_{s} - c_{s}^{2} = \frac{[1 + ρ]_{q} [1 - ρ]_{q}}{[ρ]_{q} [- ρ]_{q}} .$ The last equality follows directly from $[- ρ]_{q} = - q^{- ρ} [ρ]_{q}$ and the quantum-integer addition identities. Thus the opposite branch has exactly the displayed coefficient. This proves the generator formula without accepting the inconsistent branch in the 1998 manuscript.

5.1F2F3step 3.2step 4.1algebra

Prove nondegeneracy in the separated range locally. The canonical Murphy basis vector has form value $[λ]!_{q} = \prod_{a} [λ_{a}]!_{q}$ , because $x_{λ}^{2} = [λ]!_{q} x_{λ}$ ; its seminormal image has the same value since the canonical tableau is dominance-maximal. This scalar is nonzero when $e > n$ . If $s ⊳ t = s (i, i + 1)$ is a standard swap, put $a = 1 / [ρ]_{q}$ . Step 4.1 gives $f_{t} = f_{s} (T_{i} + a)$ . Associativity, orthogonality and the quadratic relation give $⟨ f_{t}, f_{t} ⟩ = (q + a) (1 - a) ⟨ f_{s}, f_{s} ⟩ = q [ρ + 1]_{q} [ρ - 1]_{q} [ρ]_{q}^{- 2} ⟨ f_{s}, f_{s} ⟩$ . Indeed expansion of $(T_{i} + a)^{2}$ and its $f_{s}$ coefficient $- a$ gives $q - a (q - 1) - a^{2} = (q + a) (1 - a)$ . The swapped incomparable boxes have $2 \leq ∣ ρ ∣ \leq n - 1$ : their row and column separations plus one are at most the size of the diagram. Thus all three quantum integers here are nonzero. Every standard tableau is connected to the canonical one by length-increasing swaps of adjacent incomparable entries: bubble-sort its linear extension of the box-order poset into the canonical row-order extension, then reverse the swaps. Induction gives nonzero diagonal norm at every $f_{t}$ , so each cell form is nonsingular.

6.1step 5.1step 2.1step 1.1algebra

Each cell is now its nonzero simple form quotient, absolutely simple, and all these cells are pairwise distinct and exhaust the simple modules by Nonzero cell heads classify the simple modules of a cellular algebra. Thus the split semisimple quotient $H / J (H)$ has dimension $\sum_{λ} ∣ Std (λ) ∣^{2}$ . This equals $\dim H$ because the Murphy cellular basis has exactly one element for each pair of standard tableaux of the same shape (The Murphy tableau elements form a cellular basis). Hence $J (H) = 0$ , $H$ is split semisimple, and the direct sum of all its simple cell modules is faithful. On that faithful module the global Lagrange products of the Statement are the orthogonal projectors from step 2.1, with different shapes having disjoint tuples by step 1.1. Therefore $F_{s} F_{t} = δ_{s t} F_{t}$ holds in $H$ itself, proving the global idempotents. This argument proves only the separated range and does not assert a finite- $e$ nonzero-label criterion when $e \leq n$ .

7.1step 1.1step 3.2step 4.1step 6.1∎

Steps 1.1-6.1 prove all four claims. No semisimplicity was imposed as a hypothesis: it was derived locally from nondegenerate cells and their complete finite-cell classification. The calculation concerns the intrinsic Murphy cell and keeps the signed ordinary-module dictionary separate. All products and basis changes are finite, and no choice principle is used.

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Hecke Specht theory specialises to group Specht theory at q = 1

Statement

Let $F$ be a field, $q = 1 \in F$ , and let $H_{F, 1} (S_{n})$ be the specialization of the generic type-A Hecke algebra at $v \mapsto 1$ , so that $H_{F, 1} (S_{n}) ≅ F [S_{n}]$ with $T_{w} \mapsto w$ (Polytabloid specialization and the exact signed multiplicity dictionary, step 1.1). Then the Hecke theory of this page specializes as follows.

The Murphy elements $m_{μ} = \sum_{w \in S_{μ}} T_{w}$ specialize to the row symmetrizers $\sum_{w \in S_{μ}} w$ , and the Murphy basis ${m_{s t}}$ of $H_{F, 1} (S_{n})$ specializes to the Murphy basis of the group algebra $F [S_{n}]$ (the base change $Z [S_{n}] \to F [S_{n}]$ of the integral Murphy basis).
For every $λ ⊢ n$ the specialization of the page's Hecke Specht module $S^{λ}$ (The type-A Hecke Specht module and its cell form) is the ordinary Specht module in the Dipper-James convention: $S_{q = 1}^{λ} ≅ S_{D J}^{λ}$ as $F [S_{n}]$ -modules, with the standard polytabloid basis (Standard polytabloids form a basis of a complex Specht module); the specialized cell form is the standard bilinear form on $S^{λ}$ whose radical defines the James quotient $D^{λ}$ (Modular Specht form and radical quotient).
The quantum characteristic of $q = 1$ is $e = char (F)$ when $char (F) > 0$ , and $e = \infty$ when $char (F) = 0$ (The quantum characteristic of an invertible Hecke parameter); in characteristic $p > 0$ a partition is $e$ -regular exactly when it is $p$ -regular. Hence $D_{q = 1}^{λ} \neq 0$ exactly for $p$ -regular $λ$ (Nonzero modular Specht quotient criterion), and for non- $p$ -regular $λ$ one has $r a d S^{λ} = S^{λ}$ . Over splitting residue fields this also matches the group-side classification and conjugate-sign dictionary of Modular simple modules of the symmetric group and p-regular and p-restricted labels under transpose and sign; the assertion here holds over every field by the cellular classification below. In characteristic $0$ the specialization is split semisimple and all Specht modules are irreducible.

Facts & Assumptions

Given: A field $F$ and the specialization $H_{F, 1} (S_{n}) ≅ F [S_{n}]$ (Polytabloid specialization and the exact signed multiplicity dictionary, step 1.1).

[F1]

(Ordinary Specht modules are the signed conjugate Murphy cells, Polytabloid specialization and the exact signed multiplicity dictionary) The Dipper-James ordinary Specht module with its polytabloid basis and standard form is identified with the signed conjugate Murphy cell module of The type-A Hecke Specht module and its cell form by a base-change compatible isomorphism whose Laurent unit form factor is $q^{ℓ (w_{0} (S_{λ^{'}})) - ℓ (w_{λ})}$ , equal to $1$ at $q = 1$ over every field; hence at $q = 1$ the ordinary module and the standard form agree on the specialization, including on zero-form quotients.

[F2]

Over every field, the integral-polytabloid specialization and standard tabloid form are supplied by Polytabloid specialization and the exact signed multiplicity dictionary, clause 1 and proof step 2.1. Over a splitting residue field the form quotient is exactly Modular Specht form and radical quotient; Standard polytabloids form a basis of a complex Specht module records the complex-field basis. These latter items are used only within their stated coefficient fields.

[F3]

(The quantum characteristic of an invertible Hecke parameter) At $q = 1$ the quantum characteristic is $e = p = char (F)$ when $p > 0$ and $e = \infty$ when $p = 0$ ; a partition is $e$ -regular exactly when it is $p$ -regular.

[F4]

At $q = 1$ in positive characteristic, the ordinary group Specht radical quotient is nonzero exactly for $p$ -regular labels (Nonzero modular Specht quotient criterion). In characteristic zero $e = \infty$ , and the complete local separated-range proof gives nonsingular Murphy cell forms and split semisimplicity (Separated Hecke residues give seminormal idempotents and bases); the signed-conjugate form identity in [F1] transfers their nondegeneracy to every ordinary Specht module. No finite- $e$ Hecke nonzero-label criterion is used.

[F5]

Over every field, the nonzero cell-form quotients of a cellular algebra form a complete set of pairwise distinct absolutely simple modules (Nonzero cell heads classify the simple modules of a cellular algebra). The exact signed-conjugate equivalence of [F1] transports this classification to ordinary Specht quotients; no splitting-field-only group theorem is invoked over an arbitrary $F$ .

Proof

technique · direct

1.1givenF1algebra

Murphy basis: base change along $Z [v^{\pm 1}] \to F$ , $v \mapsto 1$ , sends $T_{w}$ to $w$ and $m_{μ} = \sum_{w \in S_{μ}} T_{w}$ to the row symmetrizer $\sum_{w \in S_{μ}} w$ ; the structure constants of the Murphy basis are polynomials in $q$ , so they specialize to structure constants in $F [S_{n}]$ and the specialized family ${m_{s t}}$ satisfies the cellular axioms over $F$ ; it is a basis because base change of the free integral Murphy basis is a basis, by The Murphy tableau elements form a cellular basis. This is part 1.

2.1step 1.1F1F2

Module and form: by [F1] the specialization of the page's Specht module is the ordinary Dipper-James Specht module with the standard polytabloid basis, and the specialized cell form differs from the standard form by the Laurent factor $q^{ℓ (w_{0} (S_{λ^{'}})) - ℓ (w_{λ})}$ , which is $1$ at $q = 1$ ; hence the two forms agree over every field and have the same radical, so the specialized quotient is the James quotient $D^{λ}$ . This is part 2.

3.1step 2.1F1F3F4F5∎

Labels: by [F3], $e = p$ in positive characteristic, and $e$ -regular means $p$ -regular. Step 2.1 identifies the specialized module and its form exactly with the ordinary group module and form, so the ordinary criterion in [F4] gives precisely the nonzero $p$ -regular quotients and the zero form on all other labels, These are all the simples by the cellular classification [F5] and the exact signed-conjugate equivalence of [F1]. In characteristic zero, $e = \infty > n$ , so the local separated-range proof in [F4] makes the Murphy cell forms nonsingular and the algebra split semisimple. The exact signed-conjugate unit form dictionary [F1] gives the same conclusion for all ordinary Specht modules. Thus all claims follow over the original field, without using the held finite- $e$ Hecke classifier.

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Nonzero-form cellular heads, cell purity and shared-factor linkage

Statement

Let $A$ be a finite-dimensional cellular algebra over a field $F$ (Cellular bases and cell modules) with cell modules $C_{λ}$ and cell-form quotients $D_{λ} = C_{λ} / R_{λ}$ . Then:

every nonzero $D_{λ}$ is an absolutely simple head, all simple $A$ -modules occur exactly once this way, and $[C_{λ} : D_{μ}] \neq 0$ implies $λ \geq μ$ , with $[C_{μ} : D_{μ}] = 1$ whenever $D_{μ} \neq 0$ ; here "nonzero quotient" means nonzero form, not necessarily a nondegenerate form on $C_{λ}$ ;
every nonzero cell module, including labels with zero-form quotient, has all its composition factors in a single block (Blocks of a finite-dimensional algebra via primitive central idempotents);
two nonzero cell modules share a block exactly when they are linked by a finite chain of nonzero cell modules whose neighbouring terms share a simple composition factor;
for one primitive idempotent $e_{μ}$ per simple, $\dim_{F} e_{μ} A e_{ν} = \sum_{λ} [C_{λ} : D_{μ}] [C_{λ} : D_{ν}] .$

If the cellular index set $T (λ)$ is empty, then $C_{λ} = D_{λ} = 0$ . Its block support, the set of primitive central idempotents $b$ with $C_{λ} b \neq 0$ , is empty, and it has no simple composition factors. It does not participate in the nonzero-cell linkage criterion of (3), and its term in the corner sum of (4) is zero.

Facts & Assumptions

Given: A finite-dimensional cellular $F$ -algebra $A$ with cellular basis $c_{s t}^{λ}$ , higher ideals $A^{> λ}$ , cell ideals $I_{λ} = A^{\geq λ}$ , cell layers $J_{λ} = I_{λ} / A^{> λ}$ , right cell modules $C_{λ}$ and their Gram forms $⟨ -, - ⟩$ .

[F1]

The two cell axioms yield the scalar product and right cell action derived in step 1.1: $x c_{s t}^{λ} = ⟨ x, c_{s}^{λ} ⟩ c_{t}^{λ}$ for $x \in C_{λ}$ . The form matrix is $G_{λ} (s, t) = ⟨ c_{s}^{λ}, c_{t}^{λ} ⟩$ , and the cell layer satisfies $J_{λ} ≅ C_{λ}^{l e f t} \otimes_{F} C_{λ}$ (Cellular bases and cell modules). Here $C_{λ}^{l e f t}$ is the star-converted left cell module, not the linear dual; the cell-layer tensor uses its first-index action.

[F2]

$A$ has a complete orthogonal set of primitive idempotents; projective covers exist and are unique; the Jacobson radical is nilpotent with semisimple quotient; every simple module appears in $A / rad A$ (Finite-dimensional algebras admit primitive idempotent decompositions, Every finite-dimensional module has a projective cover, unique up to isomorphism over the target, For a finite-dimensional algebra, the Jacobson radical is nilpotent and the quotient by it is semisimple).

[F3]

Composition multiplicities $[M : D]$ are well defined by the Jordan–Hölder theorem (Jordan–Hölder theorem for modules, Composition series and length of a module).

[F4]

Blocks of $A$ are the primitive-central-idempotent components; a central idempotent acts as $0$ or $1$ on each indecomposable projective and cannot split a nonzero corner (Blocks of a finite-dimensional algebra via primitive central idempotents).

Proof

technique · direct

Given: The cellular data of [F1]–[F4] and the cell modules $C_{λ}$ with

their forms.

1.1F1algebra

The cell-form scalar and its symmetry. For basis elements $c_{u s}^{λ}, c_{t v}^{λ}$ the product $c_{u s}^{λ} c_{t v}^{λ}$ equals a scalar $G_{λ} (s, t)$ times $c_{u v}^{λ}$ modulo $A^{> λ}$ with $G_{λ} (s, t)$ independent of $u, v$ : the right cell axiom fixes the first index and the left cell axiom fixes the last one, and comparing the two evaluations leaves the same scalar. The cellular involution gives $G_{λ} (s, t) = G_{λ} (t, s)$ . Associativity of the threefold product yields $⟨ x a, y ⟩ = ⟨ x, y a^{*} ⟩$ , since both sides compute the coefficient in the cell layer of the same product.

1.2F1F4algebra

Cell purity and the zero cell. If $T (λ)$ is empty, then $C_{λ} = 0$ , its Gram matrix and form quotient are zero, and every central block projection acts with zero image; thus its block support is empty and it has no simple factors. The corresponding cell layer $J_{λ}$ is zero as well. Now assume $C_{λ} \neq 0$ and let $z$ be a central idempotent of $A$ . On the cell layer $J_{λ} ≅ C_{λ}^{l e f t} \otimes_{F} C_{λ}$ the equality of left and right multiplication by $z$ gives $B \otimes Id = Id \otimes A_{λ}$ for the first- and second-factor coefficient matrices; comparing matrix coefficients shows every off-diagonal entry of $B$ and $A_{λ}$ vanishes and all diagonal entries of $B$ and $A_{λ}$ are equal, so both are a single scalar $θ_{λ}$ times the identity. Since $z$ is idempotent, $θ_{λ} \in {0, 1}$ . Applying this to all primitive central idempotents, whose sum is $1$ , exactly one acts as $1$ on the nonzero cell module $C_{λ}$ and the others as $0$ ; hence $C_{λ}$ , and therefore every composition factor of it, lies in one block. This covers labels with zero-form quotient because the tensor argument never divides by the Gram matrix.

2.1F1step 1.1

Nonzero form gives a simple head. Let $R_{λ} : = {x \in C_{λ} : ⟨ x, C_{λ} ⟩ = 0}$ be the radical of the form. By step 1.1 the rank-one action is $x c_{u v}^{λ} = ⟨ x, c_{u}^{λ} ⟩ c_{v}^{λ}$ , which shows that $R_{λ}$ is a submodule and that every vector $x \notin R_{λ}$ generates $C_{λ}$ : if $⟨ x, c_{u}^{λ} ⟩ \neq 0$ , then scaling $x c_{u v}^{λ}$ by the inverse scalar produces $c_{v}^{λ}$ for every $v$ . Hence every proper submodule is contained in $R_{λ}$ and $D_{λ} = C_{λ} / R_{λ}$ is simple when the form is nonzero; the criterion is nonzero form, and a degenerate but nonzero form is allowed. Extending scalars preserves the Gram rank and the rank-one action, so $D_{λ}$ is absolutely simple.

3.1F1step 2.1

Every simple occurs. Let $D$ be a simple $A$ -module, and choose a cell ideal $I_{λ}$ acting nontrivially on $D$ , maximal among such ideals in the finite cell order; then $A^{> λ}$ annihilates $D$ , while $J_{λ} = I_{λ} / A^{> λ}$ does not, so $D J_{λ} = D$ . If the Gram matrix of $C_{λ}$ were zero, the cell-layer multiplication would satisfy $J_{λ}^{2} = 0$ and $D J_{λ}^{2} = 0$ , a contradiction; hence the form is nonzero. Choose $s, u$ and $d \in D$ with $d c_{s u}^{λ} \neq 0$ ; the map $c_{v}^{λ} \mapsto d c_{s v}^{λ}$ has nonzero image, which is all of $D$ by simplicity and factors through the nonzero head $D_{λ}$ of step 2.1, so $D ≅ D_{λ}$ . This exhausts the simple labels without classifying which multipartitions have nonzero forms.

3.2F1step 2.1

Order and multiplicity. If $λ$ does not dominate $μ$ , then $C_{λ}$ annihilates $I_{μ}$ : products in $I_{λ} \cap I_{μ}$ involve only shapes strictly above $λ$ . Since $D_{μ}$ does not annihilate $I_{μ}$ by the rank-one action, $[C_{λ} : D_{μ}] \neq 0$ forces $λ \geq μ$ . The radical $R_{λ}$ annihilates $I_{λ}$ , while $D_{λ}$ does not, so $[C_{λ} : D_{λ}] = 1$ when $D_{λ} \neq 0$ . Two distinct nonzero heads are not isomorphic: their ideal supports would force $λ \geq μ$ and $μ \geq λ$ simultaneously.

4.1F1F3step 2.1step 3.2algebra

The corner identity. Choose one primitive idempotent $e_{μ}$ for each simple $D_{μ}$ with $\dim D_{ν} e_{μ} = δ_{μ ν}$ . The exact functor $V \mapsto V e_{μ}$ (lift a vector along a surjection and multiply by $e_{μ}$ ; restrict along an injection) gives $\dim C_{λ} e_{μ} = [C_{λ} : D_{μ}]$ . The nonzero heads are self-dual under the involution by the symmetric associative nondegenerate quotient form of steps 1.1 and 2.1, so the corresponding left factor has the same multiplicities. Filtering $A$ by a linear extension of the cellular ideals and applying the two exact projections $e_{μ} (-) e_{ν}$ to the cell layers $J_{λ} ≅ C_{λ}^{l e f t} \otimes C_{λ}$ , dimension additivity gives $\dim_{F} e_{μ} A e_{ν} = \sum_{λ} [C_{λ} : D_{μ}] [C_{λ} : D_{ν}] .$ An empty cell layer contributes zero to both the filtration dimension and every product of its composition multiplicities.

5.1F2F4step 1.2step 4.1

Blocks and linkage. A finite split algebra has blocks equal to the connected components of the graph on its primitive idempotents with an edge $e$ – $f$ whenever $e A f \neq 0$ or $f A e \neq 0$ : a sum of full primitive idempotents over a union of components is central, and conversely a central idempotent cannot separate a nonzero corner, while repeated primitive idempotents of the same simple are linked through the matrix units of $A / rad A$ . By step 4.1 two simple labels have a nonzero corner exactly when they occur together as factors of some cell module, so two simples share a block exactly when a chain of cell modules links their factors. Since by step 1.2 every nonzero cell module is pure and by step 3.1 every simple occurs in a nonzero-form head, two nonzero cell modules lie in the same block exactly when there is a finite chain of nonzero cell modules each neighbouring pair of which shares a simple composition factor. Each nonzero cell has a simple composition factor because it has positive finite dimension. A zero cell has no such factor and has empty block support by step 1.2, so it is absent from this support graph and cannot create a chain edge or a common supported block.

6.1step 3.2step 4.1step 5.1∎

Conclusion. Steps 1.1, 2.1, 3.1 and 3.2 prove the head, exhaustion, order and multiplicity statement (1); step 1.2 proves cell purity (2); step 4.1 proves the corner identity (4); and step 5.1 proves the exact shared-factor linkage criterion (3). All arguments work over every field and allow degenerate nonzero cell forms, nonzero cells with zero-form quotient, and empty index sets with zero cell modules and empty block support.

Remarks

The correct criterion in part (1) is that the cell form be nonzero, not that it be nondegenerate on $C_{λ}$ : degenerate but nonzero forms give simple heads whose radical is the form radical. The purity argument (1.2) is a tensor argument on the cell layer and never inverts the Gram matrix, so it applies to zero-form labels as well. No choice principle is used: all algebras are finite-dimensional and all filtrations finite.

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Type-A Hecke permutation modules: semistandard Murphy bases and Specht filtrations

Statement

For every commutative ring $R$ with invertible parameter $z$ , the ordinary finite type-A Hecke algebra $H_{n}$ and a composition $ρ$ of $n$ , the permutation module $x_{ρ} H_{n}$ has $R$ -basis $m_{(S, t)} = \sum_{s : ρ (s) = S} m_{s t},$ where $S$ ranges over the semistandard tableaux of partition shape and type $ρ$ and $t$ over the standard tableaux of the same shape. It has a Specht filtration with multiplicity $# {SStd}_{ρ} (λ)$ of the ordinary Specht module $S^{λ}$ (Semistandard tableaux and Kostka numbers, Tableaux and standard tableaux). The row-double-coset membership uses each coset word once and no quantum-factorial division. Over the generic field this number equals the trivial-character multiplicity of the $ρ$ -row parabolic in $S^{λ}$ . Moreover the explicit ordinary Garnir row-coset identity, the additive nonstandard-tableau factorization and the same-column generator refinement $T_{a} m_{s t} \equiv - m_{s t} + \sum_{u ⊳ s} c_{u} m_{u t} (m o d higher shapes)$ hold when $a, a + 1$ are in one column of the standard tableau $s$ ; row-standard straightening has no smaller-shape terms and its same-shape terms increase tableau-prefix dominance. Shape support uses the already proved ordinary Murphy dominance ideals (The Murphy tableau elements form a cellular basis, Murphy elements and Murphy tableau elements); the finite Garnir algorithm terminates in lexicographic shape/tableau order before that dominance refinement.

Facts & Assumptions

Given: A commutative ring $R$ with invertible $z$ , the ordinary finite type-A Hecke algebra $H_{n}$ with standard basis $T_{w}$ and Murphy elements $m_{s t}$ , a composition $ρ$ of $n$ with row symmetrizer $x_{ρ}$ , and the semistandard/standard tableaux of shape $λ ⊢ n$ .

[F1]

The Murphy elements $m_{s t}$ form a cellular basis of $H_{n}$ with the fixed-first-index cell law, and the dominance ideals are the higher-cell ideals of the cellular structure; the standard basis $T_{w}$ exists over $R$ and right multiplication by $T_{a}$ has the two-term rule $T_{w} T_{a} = z T_{w s_{a}} + (z - 1) T_{w}$ for $ℓ (w s_{a}) < ℓ (w)$ (The Murphy tableau elements form a cellular basis, The standard basis of the generic type-A Hecke algebra, Murphy elements and Murphy tableau elements).

[F2]

Over the generic field, the Jucys–Murphy action is dominance triangular and its word-character separates tableaux of different shapes; every simple module of the cellular algebra is a nonzero cellular head and the cell modules of the ordinary Hecke algebra form a complete set of standard modules (The Hecke Jucys-Murphy action is dominance triangular, Nonzero-form cellular heads, cell purity and shared-factor linkage).

[F3]

The dominance order on partitions is Dominance order on partitions. The length-additive support lemma, Garnir identity, nonstandard-tableau factorization, finite increasing straightening and dominance ideals are proved in steps 2.1–5.1 and 8.1–9.1 of The Murphy cellular basis, the cell law and the Murphy form. The same-column refinement below is deduced from these rules, rather than assumed as part of cellularity.

[F4]

The independent same-shape perfect pairing identifies $C_{M} (λ) ≅ P (λ)^{\circ}$ over every commutative ring with unit parameter; finite-free biduality gives $C_{M} (λ)^{\circ} ≅ P (λ)$ . Star dual has action $(f h) (m) = f (m h^{*})$ (Murphy cells are the same-shape contragredient Specht duals, The Hecke sign automorphism, pullback and contragredient dual).

Proof

technique · direct

Given: The data and suppliers of [F1]–[F3]; work first over the universal

domain $Z [z, z^{- 1}]$ , where all coefficient statements are polynomial

identities, and specialize at the end.

1.1F1algebra

The row-double-coset identity and membership. For a semistandard tableau $S$ of shape $λ$ and type $ρ$ and a standard $t$ of shape $λ$ , every row-standard refinement $s$ of the column-strict tableau $S$ is standard, and $\sum_{s} T_{d (s)}^{*} x_{λ}$ is the sum of $T_{w}$ over the double coset $S_{ρ} d (s_{0})^{- 1} S_{λ}$ , each $w$ occurring exactly once; its left $S_{ρ}$ -coset expansion writes it as $x_{ρ} h_{0}$ , and appending $T_{d (t)}$ gives $m_{(S, t)} = x_{ρ} h$ for some $h \in H_{n}$ . Hence each $m_{(S, t)}$ lies in $x_{ρ} H_{n}$ , and no row-intersection quantum factorial is introduced.

1.2F1F3algebra

Independence and the Garnir coefficient rules. The sums $m_{(S, t)}$ are $R$ -independent because their Murphy fibres (the index sets ${s : ρ (s) = S}$ ) are disjoint and every coefficient in the expansion of $m_{(S, t)}$ in the $m_{s t}$ basis is $1$ ; they therefore have disjoint unit pivots. For the coefficient rules, consider a violation at adjacent rows $i, i + 1$ and column $j$ and the composition $ν = (λ_{1}, \dots, λ_{i - 1}, j - 1, λ_{i} + 1, λ_{i + 1} - j, λ_{i + 2}, \dots)$ ; the two length-additive coset sums of $S_{ν} S_{λ}$ give $\sum_{w \in S_{ν} \cap D_{λ}} x_{λ} T_{w} = \sum_{v \in S_{λ} \cap D_{ν}} T_{v}^{*} x_{ν},$ whose distinguished Garnir term occurs with coefficient $1$ . Every other term in the original shape has strictly more dominant row tableaux, the reordered $ν$ has strictly higher lexicographic shape, and the ordinary Murphy dominance refinement places these shape errors in higher-cell ideals; a nonstandard tableau factors as that Garnir tableau followed by an additive-length word (removing inversions while keeping the column violation), and Bruhat subword support preserves the strict tableau-prefix inequality. In the isolated relation for consecutive $a, a + 1$ in one column, sorting the Garnir belt has a unique standard term whose single adjacent swap produces the distinguished Garnir term, with coefficient $- 1$ , and every other same-shape term is strictly more dominant. This yields the displayed congruence of the Statement, and the right analogue follows by the cellular involutions $*$ .

2.1F1algebra

The row eigenspace test. For a fixed $a$ the left $z$ -eigenspace of $T_{a}$ in $H_{n}$ is $(1 + T_{a}) H_{n}$ : in the $T_{w}$ basis pair $w$ with $s_{a} w$ of larger length, and coefficient comparison in $T_{a} h = z h$ gives equality of the two coefficients, whose sum is $(1 + T_{a}) T_{w}$ . This argument uses only that $z$ is a unit. Over the universal domain, no element of this space has a nonzero Murphy coefficient at a tableau with $a, a + 1$ in one column: choose a least shape and then a least left tableau in a dominance-compatible support order; higher-cell terms cannot contribute to its shape, and a column correction from another tableau can contribute only from a strictly less dominant left tableau, absent by minimality. If $a, a + 1$ are in a column, its coefficient in $T_{a} h$ is the negative of its coefficient in $h$ by step 1.2, contradicting the eigenvalue equation since $z + 1 \neq 0$ in the domain; if they share a row, the basis element is already a $z$ -eigenvector and is subtracted; if their swap is standard, the exact two-tableau rules make the two coefficients equal and their sum is a $z$ -eigenvector, which is subtracted. Each subtraction stays in the row ideal and removes the chosen support term without creating new ones, so induction on the finite support proves the column exclusion; specializing the resulting coordinate-zero identities gives it over every base ring, with no division by $z + 1$ at a singular parameter.

3.1F1step 1.2step 2.1algebra

The basis theorem. Let $h \in x_{ρ} H_{n}$ . It is a $z$ -eigenvector for every generator inside every $ρ$ -row interval, so step 2.1 excludes any supported tableau with a consecutive pair inside one column; for an adjacent standard swap, comparison of the exact two-tableau equations makes the two coefficients equal (no column-state correction can interfere, since every such state has coefficient $0$ ). The standard refinements of a fixed row-label tableau are the linear extensions of the finite node posets of its repeated-label classes, and any two are connected by swaps of adjacent incomparable elements; finite induction on the inversion distance shows that if a repeated label occupies one column, a consecutive vertical cover realises an excluded state and forces the whole fibre coefficient to be $0$ , while otherwise the row-label tableau is column-strict and every fibre coefficient equals one common value. Hence $h$ has zero coefficients off semistandard fibres and constant coefficients on each semistandard fibre, i.e. $h$ lies in the span of the $m_{(S, t)}$ ; with step 1.2 this proves that the $m_{(S, t)}$ form an $R$ -basis of $x_{ρ} H_{n}$ .

4.1F1F4step 3.1algebra

The two filtrations, keeping the ordinary dictionary explicit. The higher-shape Murphy-cell filtration obtained from [F1] has one quotient $C_{M} (λ)$ for every semistandard tableau $S$ of shape $λ$ and type $ρ$ : its second-index action is the fixed-first-index Murphy cell action. These quotients are free over $R$ , so every short exact sequence of this filtration splits as a sequence of $R$ -modules. Star duality is therefore exact on this filtration. On the distinguished coset basis $b_{d} = x_{ρ} T_{d}$ of $x_{ρ} H_{n}$ , define $β (b_{d}, b_{c}) = δ_{d c} z^{ℓ (d)}$ . The row-permutation generator cases prove contravariance: a same-row generator acts by $z$ ; a length-increasing adjacent coset pair acts by $b_{d} T_{i} = b_{c}$ , $b_{c} T_{i} = z b_{d} + (z - 1) b_{c}$ , with $ℓ (c) = ℓ (d) + 1$ , and the diagonal weights give $β (b_{d} T_{i}, b_{c}) = z^{ℓ (c)} = z β (b_{d}, b_{d})$ . All other entries are zero or the corresponding diagonal equality. Thus $β$ gives an $H_{n}$ -isomorphism $x_{ρ} H_{n} ≅ (x_{ρ} H_{n})^{\circ}$ , since every diagonal $z^{ℓ (d)}$ is a unit. The independently proved same-shape bridge gives $C_{M} (λ)^{\circ} ≅ P (λ)$ , the ordinary Dipper–James Specht module, by finite-free biduality. Dualize the split Murphy filtration and transport it through this permutation self-duality. The resulting reversed filtration is an ordinary Specht filtration with exactly the same multiplicities. The original Murphy filtration is not called the ordinary one. This preserves the full every-ring/unit-parameter existence Statement without generic semisimplicity or form nondegeneracy on a Specht module.

5.1F2step 4.1algebra

The generic multiplicity interpretation. Over the generic field the finite Hecke algebra is split semisimple: by [F2] every simple is a nonzero cellular head, a simple factor of an ordinary cell module has a word character from that module and from the head's own cell shape, and disjoint word sets force the shapes to agree; existence of a factor forces its own head to be nonzero, diagonal multiplicity $1$ forces the cell module to be simple, and the sum of squared dimensions equals the Hecke rank, so the algebra and each parabolic are split semisimple. Consequently the filtration multiplicity $# {SStd}_{ρ} (λ)$ equals the multiplicity of the trivial character of the $ρ$ -row parabolic in $S^{λ}$ , by induction/restriction adjunction in the semisimple category.

6.1step 3.1step 4.1step 5.1∎

Conclusion. Steps 1.1, 1.2, 2.1 and 3.1 prove the semistandard Murphy basis of $x_{ρ} H_{n}$ , step 1.2 gives the explicit Garnir and same-column generator rules with their termination, step 4.1 gives the Specht filtration with multiplicity $# {SStd}_{ρ} (λ)$ , and step 5.1 gives the generic parabolic-multiplicity interpretation. Everything is proved universally over $Z [z, z^{- 1}]$ and specializes; the row-double-coset membership uses each coset word once and introduces no quantum-factorial division.

Remarks

This is the narrow ordinary type-A supplier consumed by the finite Mackey count of the higher-level theory: the basis and filtration are proved here rather than imported from a definition of the ordinary Schur algebra. No choice principle is used; all tableaux and quotient spaces are finite.

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The type-A q-Schur algebra and its Weyl modules

Definition

Let $R$ be a commutative ring, $q \in R^{\times}$ , $d, n \geq 0$ , and $H = H_{R, q} (S_{n})$ (The generic type-A Hecke algebra). Let $Λ (d, n)$ be the set of weak compositions $μ = (μ_{1}, \dots, μ_{d})$ of $n$ (so $μ_{i} \in Z_{\geq 0}$ and $\sum_{i} μ_{i} = n$ ; zero parts are allowed and give empty rows). For $μ \in Λ (d, n)$ let $m_{μ} = \sum_{w \in S_{μ}} T_{w}$ (Murphy elements and Murphy tableau elements) and let $M^{μ} : = m_{μ} H$ be the right ideal it generates, the (Hecke) permutation module; $M^{μ}$ is free over $R$ with basis ${m_{μ} T_{d (t)} : t row-standard μ -tableau}$ and is the Hecke analogue of the module induced from the trivial representation of the Young subgroup (Young subgroups, tabloids, and permutation modules).

For a weak composition $μ \in Λ (d, n)$ , define $T_{0} (λ, μ)$ to be the fillings of $[λ]$ by $1, \dots, d$ with exactly $μ_{i}$ entries equal to $i$ , weakly increasing rows and strictly increasing columns. Put $T_{0} (λ) = ⋃_{μ \in Λ (d, n)} T_{0} (λ, μ)$ and $K_{λ μ} : = ∣ T_{0} (λ, μ) ∣$ . This explicitly extends the partition-content definition of Semistandard tableaux and Kostka numbers to weak compositions, including internal zero counts; a zero count means that its label is absent. The set is finite, since at most $d^{n}$ fillings occur for $n > 0$ , and at $n = 0$ its unique member is the empty filling.

Here is the freeness statement in detail: the cosets $S_{μ} d$ with $d \in D_{μ}$ partition $S_{n}$ , and $ℓ (u d) = ℓ (u) + ℓ (d)$ for $u \in S_{μ}$ , $d \in D_{μ}$ (Murphy elements and Murphy tableau elements), so the standard basis theorem (The standard basis of the generic type-A Hecke algebra) gives $T_{u} T_{d (t)} = T_{u d (t)}$ and $m_{μ} T_{d (t)} = \sum_{u \in S_{μ}} T_{u d (t)}$ ; these are the standard basis elements indexed by the single coset $S_{μ} d (t)$ , and the cosets are disjoint, whence $M^{μ} = ⨁_{t} R m_{μ} T_{d (t)}$ is free with the displayed basis as $t$ runs over the row-standard $μ$ -tableaux.

Put $M : = ⨁_{μ \in Λ (d, n)} M^{μ}$ (The direct sum of an indexed family of modules) and define the q-Schur algebra $S (d, n) : = {E n d}_{H} (M),$ the $R$ -algebra of right $H$ -module endomorphisms of $M$ (Unital left and right modules over a ring; unqualified module means left module).

Cellular structure. For all $μ, ν \in Λ (d, n)$ , ${H o m}_{H} (M^{ν}, M^{μ})$ is free with the semistandard-pair basis constructed below; for every partition $λ$ of $n$ with at most $d$ parts and all semistandard $λ$ -tableaux $S, T$ there is an element $φ_{S T} \in S (d, n)$ (Semistandard tableaux and Kostka numbers), and the family ${φ_{S T} : λ \in Λ^{+} (d, n), S, T \in T_{0} (λ)},$ where $Λ^{+} (d, n)$ is the set of partitions of $n$ with at most $d$ parts ordered by dominance $⊵$ (Dominance order on partitions), is an $R$ -basis of $S (d, n)$ ; the map $φ_{S T} \mapsto φ_{T S}$ is an algebra anti-automorphism and the triangular axiom holds modulo the span of the $φ_{U V}$ with $μ ⊳ λ$ . Hence $(φ, Λ^{+} (d, n))$ is a dominance-cellular basis of $S (d, n)$ (Cellular bases and cell modules) and its cell modules are the Weyl modules $W^{λ}$ , free over $R$ with basis ${φ_{T} : T \in T_{0} (λ)}$ . Over a field $F$ , $W^{λ}$ has nonzero simple head $L^{λ}$ , the $L^{λ}$ for $λ \in Λ^{+} (d, n)$ form a complete set of pairwise non-isomorphic absolutely irreducible $S (d, n)$ -modules, and the decomposition matrix satisfies $d_{λ λ} = 1$ and $d_{λ μ} \neq 0 \Rightarrow λ ⊵ μ$ .

Explicit normalization of the semistandard maps. For a standard $λ$ -tableau $s$ , let $μ (s)$ replace each entry by the row of that entry in the canonical $μ$ -tableau. For semistandard $S, T$ of respective types $μ, ν$ , put $m_{S T} : = \sum_{\begin{array}{c} s, t \in Std (λ) \\ μ (s) = S, ν (t) = T \end{array}} m_{s t} .$ The normalized basis map $φ_{S T}$ is zero on the other summands and sends $m_{ν} h \in M^{ν}$ to $m_{S T} h \in M^{μ}$ . The intersection $m_{S T} \in M^{μ} \cap (M^{ν})^{*}$ makes this well-defined: if $m_{ν} h = 0$ , write $m_{S T} = a m_{ν}$ and obtain $m_{S T} h = 0$ . The integral intersection and Hom construction below proves that these are exactly the Hom basis; their composition is ordinary composition of endomorphisms. When $d \geq n$ , the regular type $ω = (1^{n}, 0^{d - n}) \in Λ (d, n)$ has $M^{ω} = H$ ; its fibers are singleton standard tableaux, so evaluation at $1$ identifies the corresponding regular-type corner map with $m_{s t}$ . For $d < n$ this regular type is absent and no such corner is asserted. This fixes the actual module-and-form normalization used by the corner and weight-Gram calculations (Mathas (4.13), with Corollary 4.12 and Theorem 4.14). No quantum factorial is inverted in this definition.

Weight idempotents and weight spaces. For each composition $ν \in Λ (d, n)$ , let $1_{ν} \in {End}_{H} (M)$ be the identity on the summand $M^{ν}$ and zero on every other summand. These weight projections satisfy $1_{ν}^{2} = 1_{ν}, 1_{μ} 1_{ν} = 0 (μ \neq ν), \sum_{ν} 1_{ν} = 1.$ They are not all of the diagonal cellular elements $φ_{T T}$ : a general cellular diagonal is not asserted idempotent or orthogonal. For a partition $ν$ the canonical tableau $T^{ν}$ of shape and type $ν$ gives $1_{ν} = φ_{T^{ν} T^{ν}}$ ; for a general composition the projection is defined directly as above, without pretending that an arbitrary composition is a partition-shaped semistandard tableau. The basis map $φ_{S T}$ is supported between the summands indicated by the types of $S$ and $T$ . Thus the fixed-first-index Weyl basis vector $φ_{T}$ survives right multiplication by $1_{ν}$ exactly when $T$ has type $ν$ . Over any field, $\dim_{F} (W^{λ} 1_{ν}) = ∣ T_{0} (λ, ν) ∣ = K_{λ ν} .$ This follows by applying the projection to the displayed free Weyl basis, not by assuming every diagonal cellular element is a projection (Semistandard tableaux and Kostka numbers).

Facts & Assumptions

Given: The commutative ring $R$ , its unit $q$ , the finite Hecke algebra $H$ , its row ideals $M^{μ} = m_{μ} H$ , and $S = {End}_{H} (⨁_{μ} M^{μ})$ in the Definition. Multiplication in $S$ is ordinary composition: $(f g) (x) = f (g (x))$ .

[F1]

The standard Hecke basis and the length-additive distinguished coset representatives give $m_{μ} T_{d} = \sum_{u \in S_{μ}} T_{u d}$ and $m_{μ} T_{i} = q m_{μ}$ for each adjacent generator of $S_{μ}$ (The standard basis of the generic type-A Hecke algebra, Murphy elements and Murphy tableau elements). The involution satisfies $T_{w}^{*} = T_{w^{- 1}}$ and $m_{μ}^{*} = m_{μ}$ .

[F2]

The proved integral permutation-module basis is $m_{(S, t)} = \sum_{s : μ (s) = S} m_{s t}$ , with $S$ semistandard of type $μ$ and $t$ standard of the same shape (Type-A Hecke permutation modules: semistandard Murphy bases and Specht filtrations). Thus, in Murphy coordinates, membership in $M^{μ}$ means that the first tableau has semistandard type $μ$ , its coefficient is constant on each first-index fibre, and every other coefficient is zero. This is the exact basis assertion, over every $R$ and unit $q$ , rather than an inference from coset freeness.

[F3]

The Murphy elements form an integral cellular basis of $H$ . In particular $m_{s t} h \equiv \sum_{v} r_{t v} (h) m_{s v}$ modulo the span $I^{⊳ λ}$ of strictly dominant shapes, with $r_{t v} (h)$ independent of $s$ (The Murphy tableau elements form a cellular basis).

[F4]

For a finite-dimensional cellular algebra over a field, its nonzero cell-form quotients are a complete set of distinct absolutely simple modules and are simple heads; their decomposition numbers satisfy diagonal $1$ and dominance support (Nonzero cell heads classify the simple modules of a cellular algebra, clause 1 of Nonzero-form cellular heads, cell purity and shared-factor linkage).

Proof

technique · direct

1.1givenF1algebraconstruct

The integral Hom test. The cyclic right module $M^{ν}$ has presentation $H / \sum_{i} (T_{i} - q) H$ , where $i$ ranges over the generators of $S_{ν}$ . Indeed, a length-additive decomposition $w = u d$ reduces $T_{w}$ in this quotient to $q^{ℓ (u)} T_{d}$ , and the map $T_{d} \mapsto m_{ν} T_{d}$ maps the resulting spanning family bijectively to the independent disjoint-coset basis. Therefore evaluation at $m_{ν}$ identifies ${Hom}_{H} (M^{ν}, M^{μ})$ with the elements $y \in M^{μ}$ satisfying $y T_{i} = q y$ for those generators. Pairing $T_{w}, T_{w s_{i}}$ with $ℓ (w s_{i}) = ℓ (w) + 1$ in the standard basis shows that this equation forces their two coefficients to be equal: the coefficient equations are $q b = q a$ and $a + (q - 1) b = q b$ , so $a = b$ since $q$ is a unit. Requiring all these equalities makes the coefficients constant on each right $S_{ν}$ -coset. Its unweighted coset sum is $T_{c} m_{ν}$ for the minimal right-coset representative $c$ , by length additivity. Hence this simultaneous right eigenspace is $H m_{ν} = (M^{ν})^{*}$ , over every coefficient ring. Evaluation consequently gives an isomorphism ${Hom}_{H} (M^{ν}, M^{μ}) ≅ M^{μ} \cap H m_{ν}$ . For $y = a m_{ν}$ its inverse is $m_{ν} h \mapsto y h$ , which is well defined because $m_{ν} h = 0$ implies $y h = a m_{ν} h = 0$ .

2.1F2F3step 1.1algebraconstruct

The semistandard intersection. Expand $y \in H$ uniquely in the Murphy basis. By [F2], membership in $M^{μ}$ requires zero coefficients outside semistandard first-index fibres and constant coefficients inside each fibre. Apply $*$ to [F2] to see that membership in $H m_{ν}$ imposes the identical condition on the second-index fibres of type $ν$ . The two requirements together say precisely that the coefficients are constant on each rectangle ${(s, t) : μ (s) = S, ν (t) = T}$ , and zero outside these rectangles. Thus their unweighted sums $m_{S T}$ form a basis of the intersection. The rectangles are disjoint and have a coefficient $1$ at every member, so independence and coefficient extraction hold over arbitrary $R$ , including singular parameters; no quantum factorial is cancelled. Every nonempty rectangle has shape with at most $d$ rows, since a strictly increasing column can use only $d$ labels. Combining this with step 1.1 gives exactly the maps $φ_{S T}$ and the asserted basis of each Hom block, hence of $S$ . The construction also commutes with arbitrary base change, because the same free basis and formulas describe each specialized Hom block.

3.1F1step 1.1step 2.1algebra

The involution under composition. Define $f^{*}$ on a Hom block by replacing its evaluation element $y = f (m_{ν}) \in M^{μ} \cap H m_{ν}$ by $y^{*} \in M^{ν} \cap H m_{μ}$ , and extend linearly to $S$ . Step 2.1 gives $φ_{S T}^{*} = φ_{T S}$ , so this is an involution. To check reversal, let $f : M^{ν} \to M^{μ}$ have $f (m_{ν}) = a m_{ν}$ and $g : M^{κ} \to M^{ν}$ have $g (m_{κ}) = b m_{κ} = m_{ν} h$ . Then $(f g) (m_{κ}) = a b m_{κ}$ , so $(f g)^{*} (m_{μ}) = m_{κ} b^{*} a^{*}$ . On the other hand $f^{*} (m_{μ}) = m_{ν} a^{*}$ and $g^{*} (m_{ν}) = m_{κ} b^{*}$ , whence $(g^{*} f^{*}) (m_{μ}) = m_{κ} b^{*} a^{*}$ . Noncomposable blocks give zero on both sides. Since each block map is determined by its generating element, $(f g)^{*} = g^{*} f^{*}$ as claimed.

4.1F3step 1.1step 2.1step 3.1algebra

The cellular multiplication law. Fix $T$ of shape $λ$ and type $ν$ , and first take $f : M^{κ} \to M^{ν}$ . Write $f (m_{κ}) = m_{ν} h$ . Then $(φ_{S T} f) (m_{κ}) = m_{S T} h$ . By [F3], its shape- $λ$ Murphy coefficient at $(s, v)$ is $\sum_{t : ν (t) = T} r_{t v} (h)$ if $μ (s) = S$ , and zero otherwise; all remaining shapes strictly dominate $λ$ . This coefficient is independent of $s$ . The full element lies in $M^{μ} \cap H m_{κ}$ by step 1.1, so step 2.1 forces these coefficients to be constant on semistandard second-index fibres $κ (v) = V$ and zero on the other fibres. For a fixed $T, h$ , these constants are independent of $S$ : choose any member $s$ of its nonempty first fibre and read the displayed sum, which contains no $s$ . Calling the constants $r_{V}$ , evaluation and the intersection basis give $φ_{S T} f \equiv \sum_{V \in T_{0} (λ, κ)} r_{V} φ_{S V}$ modulo the span of strictly dominant Schur shapes, for every $S$ . Sum over Hom blocks for arbitrary $f \in S$ . Step 3.1 gives the corresponding left law, so each upward span is a two-sided ideal and the cellular axioms hold. This proves cellularity by actual composition and Murphy coefficient comparison.

5.1step 2.1step 4.1F4algebra

Partition projections and nonzero forms. For the canonical semistandard tableau $T^{λ}$ of shape and type $λ$ , its refinement fibre consists just of $t^{λ}$ , so $m_{T^{λ} T^{λ}} = m_{λ}$ . Accordingly $φ_{T^{λ} T^{λ}}$ is the identity on $M^{λ}$ and zero on other summands; it is the genuine projection $1_{λ}$ . Squaring it gives $1_{λ}^{2} = 1_{λ}$ . In the free shape- $λ$ cell quotient, the cell product is $1_{λ}^{2} \equiv B_{λ} (φ_{T^{λ}}, φ_{T^{λ}}) 1_{λ}$ , and the coefficient of this basis vector is therefore $1$ . Over a field this form is not identically zero. Step 4.1 and [F4] consequently show that every $W^{λ}$ has the nonzero absolutely simple head $L^{λ}$ , that these heads give all simple modules exactly once, and that the stated diagonal and dominance decomposition conditions hold.

6.1givenstep 2.1step 4.1step 5.1algebra∎

Weight ranks and edge cases. Direct summand projections $1_{ν}$ satisfy the displayed orthogonality and sum-to-one relations. The source support of $φ_{S T}$ implies $φ_{S T} 1_{ν} = δ_{type (T), ν} φ_{S T}$ , so the same formula holds on the right Weyl cell basis, proving its weight rank $K_{λ ν}$ . It asserts no idempotence for other cellular diagonal elements. Deleting empty rows leaves the Young subgroup unchanged and identifies the row-type fibres after the order-preserving relabelling of the remaining row labels; hence all constructions allow zero parts. At $n = 0$ the unique empty tableau and empty partition give $H = R$ , $S = R$ and cell form $1$ . If $d = 0 < n$ , the indexing set of weak compositions and the direct sum are empty, $S = 0$ , and every basis and simple-label assertion is empty. No division except by powers of the unit $q$ , generic semisimplicity or choice principle enters any step.

DefinitionDefinition: AI-adaptedProof: Not applicableOpen item page →

The Schur idempotent and the Schur functor for type-A q-Schur algebras

Definition

Let $R$ be a commutative ring with $1$ , let $q \in R^{\times}$ , let $d, n \geq 0$ with $d \geq n$ , and let $H = H_{R, q} (S_{n})$ (The generic type-A Hecke algebra). Write $S : = S (d, n) = {E n d}_{H} (M)$ for the type-A q-Schur algebra with $M = ⨁_{μ \in Λ (d, n)} M^{μ}$ the direct sum of the permutation modules (The type-A q-Schur algebra and its Weyl modules, The direct sum of an indexed family of modules) and let ${φ_{S T} : λ \in Λ^{+} (d, n), S, T \in T_{0} (λ)}$ be its cellular semistandard basis (Murphy elements and Murphy tableau elements for the underlying Murphy elements and The standard basis of the generic type-A Hecke algebra for the standard basis along which this basis is defined). Put $ω : = (1^{n}, 0^{d - n}) \in Λ (d, n)$ , the length- $d$ composition with $n$ entries equal to $1$ (legitimate because $d \geq n$ ), so that the Young subgroup $S_{ω}$ is trivial and $m_{ω} = 1$ ; hence $M^{ω} = m_{ω} H = H$ is the regular right $H$ -module. Let $t^{ω}$ be the unique standard $ω$ -tableau, with entries $1, 2, \dots, n$ down its single column; it is in particular the unique row-standard $ω$ -tableau whose entries increase down the column. Set $e : = φ_{t^{ω} t^{ω}} \in S .$

Then:

$e$ is the projection of $M$ onto the direct summand $M^{ω} = H$ along the other summands $M^{μ}$ , $μ \neq ω$ ; in particular $e^{2} = e$ and $e S e ≅ {E n d}_{H} (M^{ω}) = {E n d}_{H} (H),$ via the algebra isomorphism $ρ : H ⟶ e S e, h ⟼ ρ_{h}, ρ_{h} (x) : = h x,$ left multiplication by $h$ on $M^{ω} = H$ being a right $H$ -module endomorphism. Here the isomorphism $e S e ≅ {E n d}_{H} (M^{ω})$ sends $e f e$ to $(e f e) ∣_{M^{ω}}$ and is inverted by extending an endomorphism of $M^{ω}$ by zero on the other summands, and the identification ${E n d}_{H} (H) ≅ H$ sends an endomorphism to its value at $1$ . By the construction of the semistandard basis elements $φ_{S T}$ (Mathas (4.13) with Corollaries 4.9 and 4.12; see the locator above), the diagonal element $φ_{t^{ω} t^{ω}}$ is exactly the endomorphism of $M$ that is the identity on $M^{ω}$ and zero on all $M^{μ}$ with $μ \neq ω$ , which is the projection asserted above.
The Schur functor is $F : m o d - S ⟶ m o d - H, F (X) : = X e = X \otimes_{S} S e,$ with the right $H$ -action transported along $ρ$ , so that $(x e) \cdot h : = x ρ_{h}$ for $x \in X$ , $h \in H$ (Unital left and right modules over a ring; unqualified module means left module). It is exact and $R$ -linear on morphisms: as a left $S$ -module, $S e$ is the image of the left- $S$ -linear projection $s \mapsto s e$ on $S$ , hence is a direct summand of the regular left module and is projective. Tensoring right $S$ -modules with this left module is therefore exact. Equivalently, for a short exact sequence of right modules, an element $y e$ in the target lifts to $x e$ from any lift $x$ of $y$ ; kernels and injections are likewise preserved by the same idempotent projection. The bimodule corner action makes these maps right $H$ -linear. The convention is fixed: the functor is $X \mapsto X e$ on the right, and the left-module formula $X \mapsto e X$ is not used on this page.

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Semistandard weight dimensions detect virtual Weyl classes

Statement

Let $F$ be a field, $q \in F^{\times}$ , $d, n \geq 1$ , and let $S = S (d, n)$ be the q-Schur algebra with Weyl modules $W^{λ}$ for $λ \in Λ^{+} (d, n)$ (The type-A q-Schur algebra and its Weyl modules). Let $M$ be a finite-dimensional right $S$ -module and let $(a_{λ})_{λ \in Λ^{+} (d, n)}$ be integers such that $\dim_{F} (M φ_{T^{ν} T^{ν}}) = \sum_{λ \in Λ^{+} (d, n)} a_{λ} \dim_{F} (W^{λ} φ_{T^{ν} T^{ν}})$ for every partition $ν \in Λ^{+} (d, n)$ , where $φ_{T^{ν} T^{ν}}$ are the diagonal semistandard basis elements (The type-A q-Schur algebra and its Weyl modules). Then $[M] = \sum_{λ} a_{λ} [W^{λ}]$ in the Grothendieck group $G_{0} (S -mod)$ (Grothendieck group of an essentially small abelian category). Consequently equality of all semistandard weight-space dimensions forces equality in the Grothendieck group.

Facts & Assumptions

Given: A field $F$ , the type-A q-Schur algebra $S = S (d, n)$ over $F$ , its Weyl modules $W^{λ}$ and simple heads $L^{λ}$ for $λ \in Λ^{+} (d, n)$ , and the semistandard elements $φ_{S T}$ (The type-A q-Schur algebra and its Weyl modules).

[F1]

Semistandard structure (The type-A q-Schur algebra and its Weyl modules): $S$ is finite dimensional over $F$ ; the $L^{λ}$ , $λ \in Λ^{+} (d, n)$ , form a complete set of pairwise non-isomorphic absolutely irreducible $S$ -modules; the weight projections $p_{ν}$ onto the permutation summands are orthogonal idempotents and, for the canonical tableau of the partition $ν$ , $p_{ν} = φ_{T^{ν} T^{ν}}$ ; arbitrary diagonal semistandard elements need not be idempotents; the decomposition numbers $d_{λ μ} : = [W^{λ} : L^{μ}]$ satisfy $d_{λ λ} = 1$ and $d_{λ μ} \neq 0 \Rightarrow λ ⊵ μ$ ; and $\dim_{F} (W^{λ} φ_{T^{ν} T^{ν}}) = K_{λ ν}$ for all $λ, ν \in Λ^{+} (d, n)$ (Semistandard tableaux and Kostka numbers).

[F2]

Kostka unitriangularity (The Kostka change of basis is dominance-unitriangular): $K_{λ ν} = 0$ unless $λ ⊵ ν$ and $K_{λ λ} = 1$ ; with rows and columns ordered by a linear extension of dominance from smaller to larger, $(K_{λ ν})_{λ, ν \in Λ^{+} (d, n)}$ is lower unitriangular over $Z$ , hence invertible over $Z$ .

[F3]

Grothendieck group and multiplicities (Grothendieck group of an essentially small abelian category, Jordan–Hölder theorem for modules): $[Y] = [X] + [Z]$ for every short exact sequence $0 \to X \to Y \to Z \to 0$ of finite-dimensional $S$ -modules, and $[N] = \sum_{μ} [N : L^{μ}] [L^{μ}]$ in $G_{0} (S -mod)$ , where the composition multiplicity $[N : L^{μ}]$ is well defined.

Proof

technique · direct

1.1givenF1algebra

Exactness of a weight projection. Fix an idempotent $p = p_{ν} \in S$ . For a short exact sequence $0 \to A \to B \overset{f}{\to} C \to 0$ of right $S$ -modules, the sequence $0 \to A p \to B p \to C p \to 0$ is exact as vector spaces. Injectivity and the kernel identity follow from the underlying sequence and $a = a p$ for $a \in A p$ . For surjectivity, lift $c \in C$ to $b \in B$ ; then $f (b p) = c p$ . Right multiplication by $p$ is a linear projection, and preserves every $S$ -submodule, although it need not be an $S$ -linear endomorphism when $p$ is not central. Taking dimensions proves additivity. Only the canonical partition weight projections are used.

2.1step 1.1F1F3

Weight functionals on composition series. Apply step 1.1 to every short exact sequence of a finite composition series of $N$ . Its factors are the simple modules $L^{μ}$ , so $\dim_{F} (N p_{ν}) = \sum_{μ} [N : L^{μ}] \dim_{F} (L^{μ} p_{ν}) .$ The defining relations of the Grothendieck group similarly give $[N] = \sum_{μ} [N : L^{μ}] [L^{μ}]$ . This argument requires exactness on vector spaces, not the false assertion that every noncentral weight projection lies in ${End}_{S} (N)$ .

3.1step 2.1F1F2

Invertibility of the weight matrix: put $X_{μ ν} : = \dim_{F} (L^{μ} φ_{T^{ν} T^{ν}})$ for $μ, ν \in Λ^{+} (d, n)$ . Feeding a composition series of $W^{λ}$ into step 2.1 gives $K_{λ ν} = \dim_{F} (W^{λ} φ_{T^{ν} T^{ν}}) = \sum_{μ} d_{λ μ} X_{μ ν},$ that is, $K = d X$ with $d = (d_{λ μ})$ and $K = (K_{λ ν})$ . Both $d$ and $K$ are unitriangular over $Z$ by [F1] and [F2] respectively, hence invertible over $Z$ , and therefore $X = d^{- 1} K$ is invertible over $Z$ .

4.1step 2.1step 3.1F1F3

Conclusion: write $c_{μ} : = [M : L^{μ}]$ and $b_{μ} : = c_{μ} - \sum_{λ} a_{λ} d_{λ μ} \in Z$ . By step 2.1 applied to $M$ and to the identity $\dim_{F} (W^{λ} φ_{T^{ν} T^{ν}}) = K_{λ ν}$ , for every $ν$ $0 = \dim_{F} (M φ_{T^{ν} T^{ν}}) - \sum_{λ} a_{λ} \dim_{F} (W^{λ} φ_{T^{ν} T^{ν}}) = \sum_{μ} c_{μ} X_{μ ν} - \sum_{λ} a_{λ} \sum_{μ} d_{λ μ} X_{μ ν} = \sum_{μ} b_{μ} X_{μ ν} .$ Since $X$ is invertible over $Z$ by step 3.1, $b_{μ} = 0$ for all $μ$ . Hence $c_{μ} = \sum_{λ} a_{λ} d_{λ μ}$ , and step 2.1 together with the defining relations [F3] gives $[M] = \sum_{μ} c_{μ} [L^{μ}] = \sum_{λ} a_{λ} \sum_{μ} d_{λ μ} [L^{μ}] = \sum_{λ} a_{λ} [W^{λ}] .$

5.1step 3.1step 4.1F3∎

Consequence: if two finite-dimensional right $S$ -modules $M, N$ satisfy $\dim_{F} (M φ_{T^{ν} T^{ν}}) = \dim_{F} (N φ_{T^{ν} T^{ν}})$ for all $ν \in Λ^{+} (d, n)$ , then applying the argument of step 4.1 to the difference of their multiplicity vectors shows $\sum_{μ} ([M : L^{μ}] - [N : L^{μ}]) X_{μ ν} = 0$ for all $ν$ , so the invertibility of $X$ gives $[M : L^{μ}] = [N : L^{μ}]$ for all $μ$ and hence $[M] = [N]$ in $G_{0} (S -mod)$ .

DefinitionDefinition: AI-adaptedProof: Not applicableOpen item page →

Virtual Weyl and Specht symbols attached to beta sequences

Definition

Fix a beta length $L \geq 1$ . For a partition $ν$ with at most $L$ columns (that is, $ν_{1} \leq L$ ), put $β_{i} (ν) = ν_{i}^{'} + L - i (1 \leq i \leq L) .$ Its beta numbers are strictly decreasing nonnegative integers. Conversely, from any strictly decreasing nonnegative sequence $b$ recover the unique partition with conjugate parts $ν_{i}^{'} = b_{i} - L + i$ , of degree $∣ ν ∣ = \sum_{i} b_{i} - L (L - 1) / 2$ (Partitions, English diagrams, and conjugation).

For arbitrary $b \in Z^{L}$ , first set all its alternating symbols to zero if it has repeated entries or a negative entry. Otherwise there is a unique sorting permutation $w$ taking it to decreasing order; let $ν$ be the recovered partition and let $ϵ (b) = (- 1)^{ℓ (w)}$ . Uniqueness of the sorting permutation is claimed only in this distinct-entry case. In the fixed degree- $r$ Grothendieck groups define $[W^{b}]_{d, r} : = {\begin{cases} ϵ (b) [W^{ν}], & ∣ ν ∣ = r and ℓ (ν) \leq d, \\ 0, & otherwise, \end{cases} [S^{b}]_{r} : = {\begin{cases} ϵ (b) [S_{M}^{ν}], & ∣ ν ∣ = r, \\ 0, & otherwise . \end{cases}$ The notation $[Δ^{b}]$ is synonymous with $[W^{b}]_{d, r}$ when $d, r$ are fixed. The ordinary Dipper-James translation conjugates every partition label and pulls every compared module back by the same $θ$ (The type-A Hecke Specht module and its cell form, The type-A q-Schur algebra and its Weyl modules).

For a composition $μ$ define the signed semistandard number directly by $d_{μ} (b) : = {\begin{cases} ϵ (b) ∣ T_{0} (ν, μ) ∣, & b has distinct nonnegative entries and ∣ ν ∣ = ∣ μ ∣, \\ 0, & otherwise . \end{cases}$ Here the tableau alphabet is the length of $μ$ ; a shape with too many rows automatically has no such tableau (Semistandard tableaux and Kostka numbers). This function is alternating in the beta entries. It also obeys the branching identity $d_{μ} (b) = \sum_{\begin{array}{c} C \subseteq {1, \dots, L} \\ ∣ C ∣ = z \end{array}} d_{\overset{ˉ}{μ}} (b - 1_{C}),$ where $z$ is the last nonzero part of $μ$ and $\overset{ˉ}{μ}$ deletes that part. The selected entries are decreased by one, not replaced by $- 1$ . For a decreasing beta sequence this is the deletion of the largest semistandard entries: they occupy a horizontal strip, at most one box in each selected column. Sorting extends the identity to distinct arbitrary sequences; if entries repeat, terms selecting just one of the equal entries cancel in pairs, while terms selecting both or neither vanish. Negative entries remain negative after decrementing. Thus the identity also holds in all zero-symbol cases. At $μ = 0$ its value is the sorting sign for a permutation of $(L - 1, \dots, 0)$ and is zero otherwise, so the recursion terminates without another unproved extension rule.

In the formulas of this page take $L = r$ when $r > 0$ . A nonzero degree- $r$ partition has width at most $r$ , so this length suffices; the Weyl cutoff $ℓ (ν) \leq d$ is retained even when $d < r$ . A balanced displacement $+ h e_{b} - h e_{c}$ preserves total degree. An unbalanced change is not silently treated as a class in the same fixed-degree group. The empty degree is handled by the empty beta sequence and unit symbol. No choice principle is used.

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The simple type-A Hecke modules are indexed by e-regular partitions

Statement

Let $F$ be a field, $q \in F^{\times}$ , let $e = e (F, q) \in {2, 3, \dots} \cup {\infty}$ be the quantum characteristic (The quantum characteristic of an invertible Hecke parameter) and let $H = H_{F, q} (S_{n})$ for $n \geq 0$ . For $λ ⊢ n$ let $S^{λ}$ be the Hecke Specht module and $D^{λ} = S^{λ} / r a d S^{λ}$ its cell-form quotient (The type-A Hecke Specht module and its cell form). Then:

$D^{λ} \neq 0$ if and only if $λ$ is $e$ -regular (e-regular and e-restricted partitions);
${D^{λ} : λ ⊢ n is e -regular}$ is a complete set of pairwise non-isomorphic absolutely irreducible $H$ -modules;
$H$ is split semisimple if and only if $e > n$ (including $e = \infty$ ); when $e \leq n$ the algebra has nonzero radical.

This is the exact root-of-unity replacement for the ordinary Specht classification. At $q = 1$ in characteristic $p > 0$ it specialises to the classification of the simple $F [S_{n}]$ -modules by $p$ -regular partitions (Modular simple modules of the symmetric group). No semisimplicity is assumed and no choice principle is used.

Facts & Assumptions

Given: A field $F$ , a unit $q$ , the type-A Hecke algebra $H$ , its Murphy cells $C_{M} (η)$ and form quotients $D_{M} (η)$ , and the ordinary signed-conjugate normalization in the Statement.

[F1]

The nonzero cell-form quotients are pairwise distinct absolutely simple modules and exhaust the simples. A quotient is nonzero precisely when its form is not identically zero; a zero form quotient makes no assertion about the actual module head (Nonzero cell heads classify the simple modules of a cellular algebra).

[F2]

The actual ordinary module/form bridge gives $S^{λ} = C_{M} (λ^{'})^{θ}$ and $D^{λ} = D_{M} (λ^{'})^{θ}$ , with $θ (T_{i}) = (q - 1) - T_{i}$ an involution commuting with star. Pullback is exact and preserves nonzero modules and simplicity (The type-A Hecke Specht module and its cell form). Conjugation exchanges regular and restricted partitions (e-regular and e-restricted partitions).

[F3]

The locally proved Jucys–Murphy elements commute and preserve a common tableau flag of each Murphy cell, with diagonal word $([c_{t} (1)]_{q}, \dots, [c_{t} (n)]_{q})$ , where $c_{t} (a)$ is the integer node content. The row-filled maximal tableau has an exact eigenrelation (The Hecke Jucys-Murphy action is dominance triangular).

[F4]

Cellular support gives $[C_{M} (η) : D_{M} (ρ)] = 0$ unless $η ⊵ ρ$ , and gives diagonal multiplicity one when that quotient is nonzero (Cellular decomposition matrices are unitriangular for the cell order).

[F5]

The integral Murphy basis, dominance ideals, row action and finite Garnir straightening are valid over $A = Z [Q, Q^{- 1}]$ and commute with every base change (The Murphy cellular basis, the cell law and the Murphy form, The row permutation module: distinguished basis and Hecke action). For a row group $S_{α}$ , $x_{α}^{2} = [α]!_{Q} x_{α}$ by length-additive parabolic cosets; $[α]!_{Q} = \prod_{j} [α_{j}]!_{Q}$ (The $q$ -integer, $q$ -factorial and $q$ -multinomial coefficients).

[F6]

At $q = 1$ , the quantum characteristic is the field characteristic when positive, and is infinity in characteristic zero (The quantum characteristic of an invertible Hecke parameter); the ordinary group labels are the regular Specht-head labels (Modular simple modules of the symmetric group). The Jacobson radical of a finite-dimensional algebra is the intersection of the annihilators of its simple modules (The Jacobson radical of a finite-dimensional algebra is the intersection of its maximal left ideals).

Proof

technique · direct

1.1F3F5givenconstructalgebra

Use finite joint-primary projections without presupposing which labels survive. The regular right module has a cellular filtration with quotients copies of the cells; refining their common flags in [F3] gives a simultaneous triangular flag on it. Thus each $L_{a}$ is annihilated by a polynomial $\prod_{b} (X - b)^{N}$ , with finitely many diagonal values $b \in F$ and $N \geq \dim H$ . Polynomial Bezout and the Chinese remainder construction give $p_{a, b}$ congruent to one modulo $(X - b)^{N}$ and zero modulo the other factors. Substituting $L_{a}$ gives commuting idempotents. For a joint word $ξ$ , their product $P_{ξ}$ is idempotent in $H$ and on a cell flag has diagonal one exactly at tableaux with word $ξ$ , zero elsewhere. An idempotent has zero image if and only if all these diagonal entries vanish: its triangular matrix would otherwise be nilpotent and idempotent, hence zero. The vector-space functor $M \mapsto M P_{ξ}$ is exact on $H$ -modules, because a lift $x$ of $y = y P_{ξ}$ can be replaced by $x P_{ξ}$ , and kernels are preserved. No centrality or $H$ -submodule assertion for this image is needed.

1.2givenF3F6algebra

For finite $e$ , equality of two diagonal values $[a]_{q}, [b]_{q}$ is equivalent to $a \equiv b (m o d e)$ . Indeed $[a]_{q} - [b]_{q} = q^{b} [a - b]_{q}$ for all integers, including negative integers. When $q \neq 1$ , $e$ is the multiplicative order of $q$ , and when $q = 1$ , $e = char F$ and the q-integer is the integer reduced to $F$ . Thus a joint word is exactly a residue word in the node convention $c - r (m o d e)$ .

1.3givenconstructalgebra

Prove the complete ladder-boundary comparison. Put $L (s, c) = (e - 1) s + c - e$ . Suppose $α$ is restricted, every occupied node has $L < l$ , and the addable ladder- $l$ set is nonempty. Set $C = l + e - 1$ , $T = ⌊ C / (e - 1) ⌋$ , $d = C - (e - 1) T \in {0, \dots, e - 2}$ and $u_{s} = C - (e - 1) s$ . For $1 \leq s \leq T$ , $0 \leq α_{s} \leq u_{s}$ , and no positive row occurs beyond $T$ . The deficits $h_{s} = u_{s} - α_{s}$ are nonnegative and satisfy $h_{s + 1} - h_{s} = α_{s} - α_{s + 1} - (e - 1) \leq 0$ . An addable ladder- $l$ row has deficit zero; thereafter every deficit is zero. Consecutive such rows have width difference $e - 1 > 0$ , so the ladder addables form a terminal interval $s_{0}, \dots, T$ . A first equality row that is not addable simply precedes this interval. Adding its bottommost $k$ nodes therefore adds in rows $T - k + 1, \dots, T$ . The length of $α$ is $T$ for $d > 0$ and $T - 1$ for $d = 0$ , including the empty case $T = 1, d = 0$ . If $β ⊵ α$ has the same size, its length is at most that of $α$ , by the prefix inequality at that length. For $d = 0$ all addable rows of $β$ are at most $T$ . For $d > 0$ the sole possible extra row is $T + 1$ , with residue $- T$ , different from $i = l m o d e = d + 1 - T m o d e$ since $1 \leq d + 1 < e$ . Hence every addable $i$ -node of $β$ lies in a row at most $T$ . If $γ$ adds any $k$ of these initial addables, at row cutoff $r$ its prefix increment is at least $\max (0, k - \max (T - r, 0))$ , exactly the increment for the bottommost- $k$ extension $δ$ of $α$ . Adding $β$ 's prefix inequalities proves $γ ⊵ δ$ . This holds for every $k$ simultaneously and needs no same-content premise.

1.4F5constructalgebra

The integral intersection factorization. Work over $A = Z [Q, Q^{- 1}]$ . Suppose a row-standard tableau $t$ already agrees with $t^{η}$ on rows before $j$ , and let $B$ be the canonical label interval of row $j$ . In rows $r \geq j$ let $β_{r}$ count its $B$ -labels. First rearrange labels inside $B$ into increasing row-reading order, then do the same with all remaining tail labels, preserving the number of labels from each interval in every row. This gives a tableau $v$ whose row $r$ consists of a $B$ -segment of length $β_{r}$ followed by a tail-segment of length $η_{r} - β_{r}$ . The sorting removes inversions separately within the two label intervals, so $d (t) = d (v) w a$ with additive length, $w \in S_{B}$ , and $a$ a permutation of the remaining tail labels. The intervals are disjoint, so $T_{a}$ commutes with $x_{B}$ , and $T_{w} x_{B} = Q^{ℓ (w)} x_{B}$ . Let $K^{'}$ be the subgroup cutting each canonical row $r \geq j$ into lengths $β_{r}, η_{r} - β_{r}$ , while retaining the earlier rows. Let $K$ cut $B$ into consecutive lengths $β_{r}$ and the tail into consecutive lengths $η_{r} - β_{r}$ , again retaining earlier rows. The permutation $d (v)$ increases on these segments on both sides and transports $K^{'}$ to $K$ . Hence $x_{K^{'}} T_{d (v)} = T_{d (v)} x_{K}$ , by expansion into length-additive standard-basis words. If $E$ consists of the minimal representatives with $S_{η} = ⨆_{u \in E} u K^{'}$ , then $x_{η} = \sum_{u \in E} T_{u} x_{K^{'}}$ . Write $ν = (η_{1}, \dots, η_{j - 1}, η_{j}, η_{j} - β_{j}, \dots, η_{k} - β_{k})$ , omitting zero parts. Multiplication of $x_{K}$ by $x_{B}$ gives $[β]!_{Q} x_{ν}$ : the $B$ -intersection subgroup contributes $\sum Q^{ℓ (b)} = [β]!_{Q}$ , while its tail and earlier-row factors commute with $x_{B}$ . Consequently the exact identity is $x_{η} T_{d (t)} x_{B} = Q^{ℓ (w)} [β]!_{Q} \sum_{u \in E} T_{u} T_{d (v)} x_{ν} T_{a}$ . No scalar is divided out. Sorting $ν$ to a partition uses the invertible block-reordering identity in [F5], so this is in $[β]!_{Q} H x_{ν^{+}} H$ . If $ν^{+} ⋬ η$ , the dominance-ideal assertion in [F5] makes its shape- $η$ part vanish. Otherwise the first $j + 1$ largest parts of $ν$ have sum at least $η_{1} + \dots + η_{j} + η_{j} - β_{j}$ ; comparison with $η$ gives $β_{j} \geq η_{j} - η_{j + 1}$ .

2.1F5step 1.4algebra

Canonical-row support without factorial cancellation. The semistandard permutation-basis theorem Type-A Hecke permutation modules: semistandard Murphy bases and Specht filtrations, followed by star, gives an exact coordinate condition for $H x_{ρ}$ : every supported right Murphy tableau has semistandard type $ρ$ . Apply it to the factor remaining after $[β]!_{Q}$ in step 1.4, before putting back $T_{a}$ . This factor belongs to $H x_{ν}$ . In shape $η$ , semistandardness forces the first $η_{1}$ labels into the first row, the next $η_{2}$ into the second, and so on through $B$ : a label cannot occur below its own row, and each earlier row is already full. Thus its right tableaux have the first $j$ canonical rows. The permutation $a$ uses only later labels. Expanding $T_{a}$ by the row action and straightening cannot alter these canonical rows: their prefix counts are maximal, and same-shape straightening increases every tableau-prefix count by that supplier's Statement. This proves the canonical-row assertion even for the quotient by the displayed factorial. Starting at $j = 1$ , repeat steps 1.4 and 2.1 for each surviving standard-tableau term. If $η_{j} - η_{j + 1} \geq e$ , every surviving term at that stage has $β_{j} \geq e$ , so $[e]!_{Q}$ divides its factor $[β]!_{Q}$ . Later multiplication and integral straightening preserve this common divisor. This establishes the full repeated-symmetrizer support and divisibility used below.

2.2step 1.3constructalgebra

For restricted $η$ , number its nodes by increasing $L$ , taking lower rows first on a tie, to obtain $t_{η}$ . Moving right increases $L$ by one and moving down by $e - 1$ , so it is standard. The diagrams $α_{l}$ of nodes with $L < l$ and $δ_{l}$ with $L \leq l$ are restricted: their row widths are the minimum of the restricted widths of $η$ and the positive part of the affine cutoff bound, each having adjacent difference at most $e - 1$ . A node on ladder $l$ forces every lower positive position on that ladder to be occupied, since $η_{s + 1} \geq η_{s} - (e - 1)$ . Thus $δ_{l}$ adds the bottommost $a_{l}$ ladder addables to $α_{l}$ . Any competing standard tableau with the same full residue word has, in the corresponding block, $a_{l}$ additions all of residue $l$ . Every such node was addable before the block: adding an $i$ -node creates new addable corners only of residues $i + 1, i - 1$ , distinct from $i$ for $e \geq 2$ . Starting with the empty diagrams, step 1.3 inductively shows its shape dominates $δ_{l}$ at every completed block, and hence its final shape dominates $η$ . No strictly lower shape has the word of $t_{η}$ , and no same-shape tableau uniqueness is asserted.

3.1F1F4step 1.1step 1.2step 2.2algebra

Choose $P_{ξ}$ for the word of $t_{η}$ . It has nonzero image on $C_{M} (η)$ and zero image on every $C_{M} (ρ)$ with $ρ ⊲ η$ , by steps 1.2 and 2.2. Therefore it has zero image on their form quotients. Apply the exact projection to a composition series of $C_{M} (η)$ . Some constituent has nonzero image; by [F1] and [F4] it is a nonzero $D_{M} (ρ)$ with $ρ ⊴ η$ . The exclusion forces $ρ = η$ , proving the restricted Murphy quotient nonzero over the original field. This is a joint-word and cellular-support proof, not an import of cellular/crystal label identification.

4.1F1F5step 1.4step 2.1step 3.1algebra

If $η_{j} - η_{j + 1} \geq e$ for some $j$ , every surviving term at that row has $β_{j} \geq e$ by steps 1.4 and 2.1 and is divisible by $[e]!_{Q}$ . Later multiplication and straightening cannot remove this polynomial factor. Every Gram entry is the coefficient of $x_{η}$ in $x_{η} T_{d (s)} T_{d (t)}^{*} x_{η}$ modulo higher shapes; first expand its middle factor in the row-module basis and apply the proved row calculation. Thus every entry is divisible by $[e]!_{Q}$ . Specializing $Q$ to $q$ makes that factor zero, so the entire Murphy form vanishes and $D_{M} (η) = 0$ . This concerns a common divisor of entries, not a product-of-hooks formula for a determinant. Combining with step 3.1 proves the finite- $e$ Murphy criterion.

4.2F3F4step 1.1step 3.1constructalgebra

For $e > n$ , including infinity, distinct standard tableaux of any shapes have distinct full diagonal words. At their first unequal step the preceding diagram has size $a - 1$ ; its distinct addable-corner contents differ by a nonzero integer of absolute value at most $a \leq n$ , so their q-integers differ. Induction recovers every next node. Consequently every tableau word of $C_{M} (η)$ occurs in no strictly lower cell. Exact projections as in step 3.1 and diagonal multiplicity one in [F4] show that all $\dim C_{M} (η)$ flag factors occur in $D_{M} (η)$ ; thus this quotient has the full cell dimension and its form radical is zero. In particular at $e = \infty$ every partition is restricted and has nonzero quotient. At $n = 0$ the unique cell is $F$ with norm one.

5.1F1F5F6step 4.2algebra

Establish the complete semisimplicity clause. Put $f_{η} = \dim C_{M} (η)$ and $d_{η} = \dim D_{M} (η)$ , with zero for a zero quotient. The Murphy basis gives $\dim H = \sum_{η} f_{η}^{2}$ . On each nonzero quotient its nondegenerate form and the cell law supply all rank-one operators. Since the cells are right modules, their action gives a surjective algebra map $H \to {End}_{F} (D_{M} (η))^{o p}$ , a full matrix algebra after transpose. Its kernels are distinct maximal two-sided ideals: equality would give the same matrix-algebra simple module, contradicting [F1]. They are pairwise comaximal. Their finite product map is surjective: for each kernel $I_{a}$ and each other $I_{b}$ choose $u_{a b} \in I_{b}$ congruent to one modulo $I_{a}$ , multiply over $b \neq a$ , and use these elements to isolate each prescribed matrix component. Its kernel is the radical by the complete simple list and [F6]. Hence $\dim (H / rad H) = \sum_{η} d_{η}^{2}$ . Step 4.2 makes this equal to $\dim H$ when $e > n$ , giving an isomorphism with a product of full matrix algebras over $F$ . When finite $e \leq n$ , the one-dimensional Murphy row cell has norm $[n]!_{q} = 0$ , since $x_{(n)}^{2} = [n]!_{q} x_{(n)}$ . Its quotient dimension is zero while its cell dimension is one; every other $d_{η} \leq f_{η}$ . The dimension sum is therefore strictly smaller and $rad H \neq 0$ . This proves both directions of part 3 without any claim that some Specht module must be reducible.

6.1F1F2F6step 3.1step 4.1step 4.2step 5.1∎

Apply the same exact involution to every Murphy quotient: $D^{λ} = D_{M} (λ^{'})^{θ}$ is nonzero exactly when $λ^{'}$ is restricted, equivalently when $λ$ is regular. The nonzero quotients remain pairwise distinct absolutely simple and complete by [F1] and [F2]. This proves parts 1 and 2, and step 5.1 gives part 3. At $q = 1$ , [F6] gives precisely the ordinary $p$ -regular group classification, including characteristic two; no splitting or semisimplicity assumption was made. All selections and polynomial computations were finite, so no choice principle is used.

Remarks

The finite ladder-boundary lemma, joint-primary projection argument and row-entry divisibility are local proofs. The cited manuscript documents their source; its erroneous displayed ladder slope and unsupported single-node truncation sentence are replaced by the complete simultaneous-block comparison above. The richest original classification and radical assertions are retained. Zero cell-form quotients are distinguished from actual module heads throughout.

CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passOpen item page →

The type-A Hecke decomposition matrix is dominance unitriangular

Statement

Let $F$ be a field, $q \in F^{\times}$ , $e = e (F, q)$ and $H = H_{F, q} (S_{n})$ (The quantum characteristic of an invertible Hecke parameter). For $λ ⊢ n$ and $e$ -regular $μ ⊢ n$ let $d_{λ μ} : = [S^{λ} : D^{μ}],$ the multiplicity of $D^{μ}$ in a composition series of the Specht module $S^{λ}$ (The type-A Hecke Specht module and its cell form, The simple type-A Hecke modules are indexed by e-regular partitions). Then $d_{μ μ} = 1 and d_{λ μ} \neq 0 ⟹ μ ⊵ λ,$ where $⊵$ is the dominance order on partitions (Dominance order on partitions): a simple factor $D^{μ}$ of $S^{λ}$ can occur only if the $e$ -regular label $μ$ dominates $λ$ . Equivalently the decomposition matrix $(d_{λ μ})$ is unitriangular along dominance. This constrains but does not compute the entries: no off-diagonal decomposition number is determined by cellularity.

Facts & Assumptions

Given: A field $F$ , $q \in F^{\times}$ , $H = H_{F, q} (S_{n})$ with Specht modules $S^{λ}$ and simple quotients $D^{μ}$ in the signed conjugate normalization (The type-A Hecke Specht module and its cell form), together with the dominance order on partitions of $n$ (Dominance order on partitions).

[F1]

(The type-A Hecke Specht module and its cell form) $S^{λ} = (S_{M}^{λ^{'}})^{θ}$ and $D^{μ} = (D_{M}^{μ^{'}})^{θ}$ for the conjugate partitions, where $S_{M}^{ν}$ is the Murphy cell module and $θ$ is the involutive automorphism $θ (T_{i}) = - q T_{i}^{- 1}$ ; pullback along $θ$ is exact and preserves submodules, quotients, composition multiplicities and simplicity.

[F2]

(Cellular decomposition matrices are unitriangular for the cell order) For the Murphy cellular basis of $H$ the composition multiplicity $[S_{M}^{ν} : D_{M}^{ρ}]$ satisfies $[S_{M}^{ρ} : D_{M}^{ρ}] = 1$ whenever $D_{M}^{ρ} \neq 0$ and is zero unless $ν ⊵ ρ$ .

[F3]

(Dominance order on partitions) Conjugation reverses dominance: $λ^{'} ⊵ μ^{'}$ if and only if $μ ⊵ λ$ .

[F4]

(The simple type-A Hecke modules are indexed by e-regular partitions) $D^{μ} \neq 0$ exactly for the $e$ -regular $μ$ .

Proof

technique · direct

1.1givenF1F4

Invariance of composition multiplicities under the normalisation: by [F1] the twist $(-)^{θ}$ is an exact equivalence of module categories with inverse $(-)^{θ}$ itself, so it is additive on short exact sequences and carries composition series to composition series; hence $[S^{λ} : D^{μ}] = [S_{M}^{λ^{'}} : D_{M}^{μ^{'}}]$ for all $λ ⊢ n$ and $e$ -regular $μ ⊢ n$ .

2.1step 1.1F2F3

Vanishing: if $d_{λ μ} \neq 0$ , then $[S_{M}^{λ^{'}} : D_{M}^{μ^{'}}] \neq 0$ by step 1.1, so $λ^{'} ⊵ μ^{'}$ by [F2], and conjugation reversal [F3] gives $μ ⊵ λ$ .

2.2step 1.1F2

Diagonal: taking $λ = μ$ in step 1.1 gives $d_{μ μ} = [S_{M}^{μ^{'}} : D_{M}^{μ^{'}}] = 1$ by [F2].

3.1F1F4step 2.1step 2.2∎

Assembly: steps 2.1 and 2.2 are exactly the two claims; ordering the rows and columns of $(d_{λ μ})$ by any linear extension of dominance displays the unitriangular shape, and no off-diagonal entry is determined by the cellular structure. At $q = 1$ , [F1] and [F4] identify the same ordinary group labels; over a splitting p-modular system this agrees with Dominance unitriangularity of the symmetric-group decomposition matrix within that supplier's hypotheses. The present proof establishes the bound over every field.

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The Schur functor sends Weyl and simple modules to Specht and Hecke simple modules

Statement

Let $k$ be a field, $q \in k^{\times}$ , $d \geq n$ , and let $F (X) = X e$ be the right-module Schur functor of The Schur idempotent and the Schur functor for type-A q-Schur algebras. Set $e_{q} = e (k, q)$ (The quantum characteristic of an invertible Hecke parameter). For each $λ \in Λ^{+} (d, n)$ let $W^{λ}$ be the Weyl module with Schur cell form $B_{λ}$ and nonzero simple head $L^{λ}$ (The type-A q-Schur algebra and its Weyl modules). Write $C_{M} (λ)$ and $D_{M} (λ)$ for the independent Murphy cell module and its cell-form quotient (The Murphy cellular basis, the cell law and the Murphy form). Then:

$F (W^{λ}) ≅ C_{M} (λ)$ as right $H$ -modules, and this isomorphism identifies $B_{λ} ∣_{W^{λ} e}$ exactly with the Murphy cell form. 2. $F (L^{λ}) ≅ D_{M} (λ)$ , including the zero quotient; it is nonzero exactly when $λ$ is $e_{q}$ -restricted. 3. For every $e_{q}$ -restricted $ν ⊢ n$ , $[W^{λ} : L^{ν}] = [C_{M} (λ) : D_{M} (ν)] .$

In the page's ordinary Dipper–James normalization (The type-A Hecke Specht module and its cell form, Ordinary Specht modules are the signed conjugate Murphy cells), these identities become $F (W^{λ^{'}}) ≅ (S^{λ})^{θ}, F (L^{λ^{'}}) ≅ (D^{λ})^{θ} .$ The latter module is nonzero exactly for $e_{q}$ -regular $λ$ and is zero otherwise. For $e_{q}$ -regular $ν$ , $[W^{λ^{'}} : L^{ν^{'}}] = [S^{λ} : D^{ν}] .$ The accepted ordinary polytabloid comparison identifies $S^{λ}$ with the historical DJ module $P (λ)$ and identifies their form radicals through an explicit unit factor. Thus the dictionary concerns the actual ordinary modules, not merely a renamed cell module. A zero Hecke cell-form quotient is not asserted to be its actual module head. No semisimplicity or choice principle is assumed, and $n = 0$ is included.

Facts & Assumptions

Given: The field, parameter, stable range and modules in the Statement; all modules are right modules and multiplication in $S = {End}_{H} (M)$ is ordinary composition $(f g) (x) = f (g (x))$ .

[F1]

The actual projection $e$ onto $M^{ω} = H$ , for $ω = (1^{n}, 0^{d - n})$ , satisfies $e^{2} = e$ and $e S e ≅ H$ by left multiplication $ρ (h) (a) = h a$ . The functor is $X \mapsto X e$ (The Schur idempotent and the Schur functor for type-A q-Schur algebras).

[F2]

The legitimate semistandard cellular construction is supported from $M^{type (T)}$ to $M^{type (S)}$ and sends its generating $m_{type (T)} h$ to $m_{S T} h$ ; its star swaps $S, T$ and its right-cell coefficients fix the first index modulo higher shapes. The Weyl cell basis is $(w_{T})_{T}$ (The type-A q-Schur algebra and its Weyl modules). Only the actual summand projections are weight idempotents; no assertion that all diagonal cellular elements are idempotent is used.

[F3]

The independent Murphy cellular datum uses $m_{s t} = T_{d (s)}^{*} x_{λ} T_{d (t)}$ , with its higher ideal $I^{⊳ λ}$ , right cell basis $(c_{t})_{t}$ and form $G_{λ}$ characterized by $m_{a s} m_{t b} \equiv G_{λ} (c_{s}, c_{t}) m_{a b}$ modulo that ideal (The Murphy cellular basis, the cell law and the Murphy form).

[F4]

The nonzero cell-form quotients are pairwise distinct absolutely irreducible modules; when the form is not identically zero its radical is the unique maximal submodule. These are the general cellular conclusions of Nonzero cell heads classify the simple modules of a cellular algebra.

[F5]

The genuine ordinary bridge gives $P (α) ≅ C_{M} (α^{'})^{θ}$ and an explicit unit-scaled form identity, hence $D_{D J} (α) ≅ D_{M} (α^{'})^{θ}$ including zero quotients. This is the page's signed Specht normalization, and $θ$ is an involutive algebra automorphism whose pullback is exact (Ordinary Specht modules are the signed conjugate Murphy cells, The type-A Hecke Specht module and its cell form).

[F6]

The full type-A classifier proves that $D_{M} (ν)$ is nonzero precisely for $e_{q}$ -restricted $ν$ over the original field, including the finite- $e_{q}$ range and the empty partition; its ordinary-label statement is transported through the signed-conjugate form bridge, not inferred from a group-only specialization (The simple type-A Hecke modules are indexed by e-regular partitions, e-regular and e-restricted partitions).

Proof

technique · direct

1.1givenF1algebra

Under [F1], $ρ (h) ρ (k) = ρ (h k)$ because both act on $H$ by left multiplication, so the identification uses $H$ , without an opposite algebra. The functor $X \mapsto X e$ is exact: for a short exact sequence, a lift $x$ of an element $y \in X^{''} e$ yields the lift $x e$ , and an element in the kernel lies in $X^{'}$ and is fixed by $e$ , hence lies in $X^{'} e$ . The tensor description uses the $(S, e S e)$ -bimodule $S e$ , which is a direct summand as a left $S$ -module.

2.1F1F2F3step 1.1constructalgebra

The support in [F2] gives $e φ_{S T} e = δ_{type (S), ω} δ_{type (T), ω} φ_{S T}$ and $w_{T} e = δ_{type (T), ω} w_{T}$ . Semistandard tableaux of type $ω$ have each label once and are exactly standard tableaux. For standard $s, t$ , their defining sum $m_{s t}$ has one term, so $φ_{s t} ∣_{H} (a) = m_{s t} a$ and $ρ^{- 1} (φ_{s t}) = m_{s t}$ . Thus the higher Schur ideal cuts out exactly $I^{⊳ λ}$ in the corner. Read the Schur right-cell law with first index any standard tableau of type $ω$ ; coefficient independence permits this choice. Restricting the multiplication by $ρ (h)$ to $H$ gives $m_{s t} h$ , so its coefficients are the Murphy right-cell coefficients. Consequently $Ψ_{λ} : W^{λ} e \to C_{M} (λ)$ , $w_{t} \mapsto c_{t}$ , is an $H$ -isomorphism.

3.1F2F3step 2.1algebra

By the Schur cell product, $φ_{a s} φ_{t b} \equiv B_{λ} (w_{s}, w_{t}) φ_{a b}$ modulo higher Schur shapes, with coefficient independent of the outer indices. Choose all four indices standard of shape $λ$ and type $ω$ . They lie in $e S e$ ; applying $ρ^{- 1}$ , or evaluating their endomorphisms at $1 \in H$ , gives $m_{a s} m_{t b} \equiv B_{λ} (w_{s}, w_{t}) m_{a b}$ modulo $I^{⊳ λ}$ . The unique coefficient in [F3] is $G_{λ} (c_{s}, c_{t})$ , proving $B_{λ} ∣_{W^{λ} e} = Ψ_{λ}^{*} G_{λ}$ exactly, with factor $1$ .

4.1F2step 3.1algebra

Since $e^{*} = e$ , contravariance gives $B_{λ} (x e, y (1 - e)) = B_{λ} (x, y (1 - e) e) = 0$ . The decomposition $W^{λ} = W^{λ} e \oplus W^{λ} (1 - e)$ is therefore orthogonal. A vector in $W^{λ} e$ pairs trivially with its complement, so it is in the restricted form radical exactly when it lies in ${rad}_{f o r m} W^{λ}$ . Since the latter is an $S$ -submodule, ${rad}_{f o r m} (W^{λ} e) = ({rad}_{f o r m} W^{λ}) e$ . This holds even if the corner form is identically zero.

5.1F2F4step 1.1step 2.1step 3.1step 4.1algebra

For the canonical semistandard tableau $T^{λ}$ of shape and type $λ$ , the map $φ_{T^{λ}, T^{λ}}$ is the actual projection onto $M^{λ}$ : it sends $m_{λ} h$ to itself and is zero on every other summand. Its square equals itself, so the Schur cell product gives $B_{λ} (w_{T^{λ}}, w_{T^{λ}}) = 1$ . Hence the Weyl form quotient is nonzero and is its simple head $L^{λ}$ by [F4]. Exactness, step 4.1 and step 3.1 give $L^{λ} e ≅ W^{λ} e / ({rad}_{f o r m} W^{λ}) e ≅ C_{M} (λ) / {rad}_{f o r m} C_{M} (λ) = D_{M} (λ)$ , including when this corner quotient is zero.

6.1F4F6step 1.1step 2.1step 5.1algebra

By [F6] and step 5.1, $L^{λ} e \neq 0$ precisely for $e_{q}$ -restricted $λ$ . Apply the exact functor to a finite composition series of $W^{λ}$ : every factor $L^{ν}$ gives $D_{M} (ν)$ or zero. Delete repeated terms. The remaining factors are the distinct nonzero Murphy simples in [F4], giving $[W^{λ} : L^{ν}] = [C_{M} (λ) : D_{M} (ν)]$ for each $e_{q}$ -restricted $ν$ . No multiplicity preservation is asserted for a killed simple label.

7.1F4F5F6step 2.1step 5.1step 6.1algebra∎

Apply [F5] and the involution $θ$ to obtain $C_{M} (λ^{'}) ≅ (S^{λ})^{θ}$ and $D_{M} (λ^{'}) ≅ (D^{λ})^{θ}$ . Replacing the Weyl label by $λ^{'}$ in steps 2.1 and 5.1 gives the ordinary dictionary, including zero quotients. Conjugation makes $λ^{'}$ restricted precisely when $λ$ is regular. Twisting both modules by the same exact equivalence preserves composition multiplicities, so step 6.1 gives $[W^{λ^{'}} : L^{ν^{'}}] = [S^{λ} : D^{ν}]$ for regular $ν$ . The genuine unit-scaled ordinary form bridge in [F5] also identifies these with the historical DJ polytabloid modules and quotients. At $n = 0$ , $e = 1$ and each form has value $1$ , so the same proof includes the empty partition.

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The q-Schur Weyl weight-space Gram determinant is a hook product

Statement

Let $R$ be an integral domain, $q \in R^{\times}$ , $d, n \geq 1$ , and let $W^{λ}$ be the Weyl module of $S (d, n)$ for a partition $λ ⊢ n$ with at most $d$ parts. For a composition $μ$ of $n$ of length $d$ , let $G_{μ}^{λ}$ be the Gram determinant of its $μ$ -weight space in the semistandard basis (The type-A q-Schur algebra and its Weyl modules). Put $β_{j} = λ_{j}^{'} + n - j$ and $h_{a b} = λ_{a} - b + λ_{b}^{'} - a + 1$ . Then $G_{μ}^{λ} = q^{N} \prod_{\begin{array}{c} (a, b), (a, c) \in [λ] \\ b < c \end{array}} {(\frac{[h_{a b}]_{q}}{[h_{a c}]_{q}})}^{d_{μ} (β + h_{a c} e_{b} - h_{a c} e_{c})},$ where $N = \sum_{T \in T_{0} (λ, μ)} ℓ (d (last (T)))$ . Here $last (T)$ is the least-dominant standardization in the fiber of $T$ , and $d_{μ}$ is the exact alternating semistandard number of Virtual Weyl and Specht symbols attached to beta sequences. Both beta shifts use the same removed hook length $h_{a c}$ .

The exponent is the sorting sign times $∣ T_{0} (η, μ) ∣$ if that balanced beta sequence has distinct nonnegative entries and recovers $η$ ; it is zero otherwise. When the shifted entries are distinct and nonnegative, equivalently lower the bead $β_{c}$ to $β_{c} - h_{a c}$ and sort to obtain the partition $κ = λ ∖ R_{a c}$ ; then raise the bead originally labelled $b$ from $β_{b}$ to $β_{b} + h_{a c}$ and sort to obtain $η$ . The added strip $η / κ$ has $h_{a c}$ boxes. Its lowest node is in the column determined by the raised bead's new sorted position, which need not be $b$ . The sign is the parity of the heights of the removed strip and the newly added strip, the latter measured in the new diagram $η$ ; it is not generally the sum of two old $λ$ hook legs (Removable border strips and their height, Hooks of a partition correspond bijectively to removable border strips).

Form the product first in the universal field $Q (Q)$ . It equals a Laurent polynomial in $Z [Q, Q^{- 1}]$ , namely the universal integral Gram determinant divided by its displayed unit $Q^{N}$ . Over $R$ the formula means the specialization of that Laurent polynomial. It does not require inverting a vanishing hook number or evaluating $0 / 0$ . All weights and $d < n$ are retained; an empty weight space has determinant $1$ .

Facts & Assumptions

Given: The normalized semistandard maps, weight projections and Weyl cell form of the q-Schur Definition; the beta symbols and branching identity; and the hook-to-bead-gap correspondence in the cited suppliers.

[F1]

The ordinary cell maps have the normalization $φ_{S T} (m_{ν} h) = m_{S T} h$ , with $m_{S T} = \sum_{μ (s) = S, ν (t) = T} m_{s t}$ . Their star swaps the indices, and their four-index cell product defines the Weyl form (The type-A q-Schur algebra and its Weyl modules). The actual regular corner identifies this product with the Murphy product with factor $1$ (The Schur functor sends Weyl and simple modules to Specht and Hecke simple modules).

[F2]

Over $K = Q (Q)$ , the corrected intrinsic Murphy seminormal basis has its unitriangular normalization, adjacent-generator action and self-adjoint JM operators (Separated Hecke residues give seminormal idempotents and bases). For a semistandard $T$ and node $(i, j)$ of entry $v$ , its predecessor set is precisely $Γ_{T} (i, j) = {(k, l) : l < j, T (k, l) < v, T (k^{'}, l) > v for every k^{'} > k} .$ Equivalently $T (k, l) < v < T (k + 1, l)$ with sentinel $+ \infty$ below the column. Both inequalities are strict; an equal $v$ below excludes that column. Define $γ_{T} : = \prod_{(i, j)} \prod_{(k, l) \in Γ_{T} (i, j)} \frac{[j - i + k - l + 1]_{Q}}{[j - i + k - l]_{Q}} .$ The standard-tableau version is defined by the same rule.

[F3]

The signed Kostka numbers satisfy $d_{μ} (a) = \sum_{∣ C ∣ = z} d_{\overset{ˉ}{μ}} (a - 1_{C})$ , with zero values for repeated or negative beta entries (Virtual Weyl and Specht symbols attached to beta sequences).

Proof

technique · direct

1.1givenF1

Preserve small alphabets. Enlarge to $D = \max (d, n)$ temporarily. The sum of permutation summands with types supported on the first $d$ letters is a literal summand of the $D$ -alphabet permutation module, with projection $p$ and $p S (D, n) p = S (d, n)$ . The normalized maps whose two types have that support, their cell multiplication and their Weyl form coefficients are unchanged. Every semistandard shape with these types has at most $d$ rows. Thus proving the weight norm in the enlarged algebra, where the regular corner exists, proves the original norm as well. No hypothesis $d \geq n$ is added.

1.2F1F2

Compute the standard norms. For the canonical standard tableau, $γ_{t^{λ}}$ telescopes rowwise to $[λ]!_{Q}$ , also its norm by $x_{λ}^{2} = [λ]!_{Q} x_{λ}$ . If $s$ dominates $t = s (i, i + 1)$ , the adjacent action gives $f_{t} = f_{s} (T_{i} + α)$ with $α = 1 / [ρ]_{Q}$ . Contravariance, orthogonality and the quadratic relation imply $⟨ f_{t}, f_{t} ⟩ = (Q + α) (1 - α) ⟨ f_{s}, f_{s} ⟩$ . The two swapped predecessor ratios give $γ_{t} = Q^{- 1} (Q + α) (1 - α) γ_{s}$ : use $[ρ + 1] = 1 + Q [ρ]$ and $[ρ - 1] = Q^{- 1} ([ρ] - 1)$ . Increasing-length adjacent swaps of incomparable boxes reach every standard tableau, so $⟨ f_{t}, f_{t} ⟩ = Q^{ℓ (d (t))} γ_{t}$ .

2.1F1step 1.1construct

Define the cross-weight maps. In the enlarged algebra of step 1.1 let $ω = (1^{n}, 0^{D - n})$ , so $M^{ω} = H$ . Let $a_{μ} : M^{μ} \to H$ be inclusion, $a_{μ} (m_{μ} h) = m_{μ} h$ , and let $b_{μ} : H \to M^{μ}$ be $b_{μ} (h) = m_{μ} h$ ; both are zero on the other permutation summands. These are actual $H$ -linear maps, not maps depending on an unspecified $T$ . Under the Hom/intersection coordinate identification of the q-Schur construction, both maps have image coordinate $m_{μ}$ . Since $m_{μ}^{*} = m_{μ}$ , the index-swapping star in [F1] gives $a_{μ}^{*} = b_{μ}$ ; this works for compositions as well as partitions. For a standard $s$ , composition of its Weyl map with $a_{μ}$ evaluates at $m_{μ}$ as $m_{s} m_{μ}$ , whereas composition of a type- $μ$ Weyl map with $b_{μ}$ evaluates at $1$ as $m_{T}$ . We use right-action composition throughout.

3.1F1F2step 2.1algebra

Establish the support and the initial scalar. Write $m_{s} = m_{λ} T_{d (s)}$ in the Murphy cell. For $s = first (T)$ , $d (s)$ is the least-length double-coset representative for the row blocks of $λ$ and $μ$ . The intersection subgroup has one symmetric-group factor of size $c_{i v} = ∣ {(i, j) : T (i, j) = v} ∣$ for every nonempty row/label intersection. Decomposing both row subgroups into their distinguished intersection cosets, every reduced coset word occurs once; the remaining intersection sum acts on $m_{λ}$ by its row character $Q^{ℓ (w)}$ . Hence the exact identity is $m_{s} m_{μ} = (\prod_{i, v} [c_{i v}]!_{Q}) m_{T} .$ No factorial is divided out. The same-entry predecessor ratios for this first refinement are $[2] / [1], [3] / [2], \dots, [c_{i v}] / [c_{i v} - 1]$ within each intersection row, so their product $π_{s}$ is exactly that factorial. We also need support after multiplication by $m_{μ}$ . Filter standard Murphy vectors by the shapes of their prefixes at the ends of the $μ$ label blocks. Every generator $T_{i}$ inside such a block preserves this filtration: its row case is multiplication by $Q$ ; its standard-swap terms have the same block-end prefixes; in its column case the Garnir correction terms have larger tableau-prefix dominance. The last assertion follows by isolating the distinguished term in the length-additive row-coset Garnir identity; all other same-shape terms increase prefix dominance, as proved in Type-A Hecke permutation modules: semistandard Murphy bases and Specht filtrations. Iterating over each word in $m_{μ}$ shows that $m_{v} m_{μ}$ has only coarsened types dominating $μ (v)$ . Its expansion has constant coefficients on semistandard fibers and zero coefficients elsewhere, by the right-hand (starred) permutation-module basis in that supplier. Thus only semistandard $U ⊵ μ (v)$ occur, strictly if $μ (v)$ is not semistandard. In particular, if $v ▹ first (T)$ , every surviving $U$ is strictly above $T$ : equality would put $v$ in the fiber of $T$ , whose most dominant member is $first (T)$ . This proves the needed support assertion including non-semistandard types.

4.1F1F2step 3.1algebra

Determine the scalar on every standardization. Sums of JM elements over a label block commute with that block's row symmetrizer: the adjacent JM relation makes $L_{i} + L_{i + 1}$ commute with $T_{i}$ , and every other $L_{j}$ commutes with it. The dominance-triangular standard JM action and the support filtration of step 3.1 consequently give triangular block-sum operators on the semistandard maps, with diagonal entries $\sum_{T (i, j) = v} [j - i]_{Q}$ . Their joint eigenvalues distinguish $T$ : the Laurent coefficients and the known prefix sizes recover every initial-label content multiset, whose diagonal counts recover its Young diagram. They are self-adjoint, so finite triangular diagonalization gives orthogonal $ψ_{T} = φ_{T} +$ strictly more dominant terms. The same block-sum identities show that $f_{t} a_{μ}$ lies in the $T$ joint eigenspace whenever $μ (t) = T$ , hence equals $r_{t} ψ_{T}$ . At $s = first (T)$ , expand $f_{s} = m_{s} +$ more dominant standard vectors. Step 3.1 gives coefficient $π_{s}$ at $φ_{T}$ and only higher types in the remaining terms. Thus $r_{s} = π_{s}$ . Now connect standardizations of $T$ by increasing-length adjacent swaps inside each label block; these are the linear extensions of the repeated-label node posets, and bubble sorting gives such a chain from their most dominant extension. If $t = s (i, i + 1)$ is one step, [F2] gives $f_{t} = f_{s} (T_{i} + α)$ , $α = 1 / [ρ]_{Q}$ , while $T_{i} a_{μ} = Q a_{μ}$ because $T_{i} m_{μ} = Q m_{μ}$ . Therefore $r_{t} = (Q + α) r_{s}$ . The only newly included same-entry predecessor pair has content difference $ρ$ , so its ratio is $[ρ + 1]_{Q} / [ρ]_{Q} = Q + α$ , proving $π_{t} = (Q + α) π_{s}$ . Induction proves the exact identity $f_{t} a_{μ} = π_{t} ψ_{T}$ for every standardization.

5.1F1F2step 1.2step 2.1step 4.1

Determine the last-standardization coefficient and the norm. For $t = last (T)$ , evaluation at $1$ gives $φ_{T} b_{μ} = m_{T} = \sum_{μ (v) = T} m_{v}$ . Every other member of this fiber is more dominant than $t$ . Also every refinement of $U ▹ T$ dominates $t$ : compare at label-block ends, then use that $t$ is the least-dominant refinement within its own blocks. Consequently the two unitriangular basis changes give $ψ_{T} b_{μ} = f_{t} +$ more dominant standard eigenvectors, with coefficient exactly $1$ at $f_{t}$ . At this last refinement the standard predecessor set partitions into the strict semistandard boundary $Γ_{T}$ and same-entry predecessors, so $γ_{t} = π_{t} γ_{T}$ . By adjointness of the explicit maps in step 2.1 and orthogonality, $π_{t} ⟨ ψ_{T}, ψ_{T} ⟩ = ⟨ f_{t} a_{μ}, ψ_{T} ⟩ = ⟨ f_{t}, ψ_{T} b_{μ} ⟩ = Q^{ℓ (d (t))} γ_{t} .$ All factors of $π_{t}$ are nonzero over $K$ , so $⟨ ψ_{T}, ψ_{T} ⟩ = Q^{ℓ (d (t))} γ_{T}$ . The transition determinant is $1$ , yielding $G_{μ}^{λ} = Q^{N} g_{μ}^{λ}$ with $g_{μ}^{λ} = \prod_{T} γ_{T}$ . This calculates the cross-weight scalar rather than inferring it from the standard norm alone.

6.1F2F3step 5.1

Prove the normalized beta product at the level of formal exponents. For $h \geq 1$ put $A_{h} (β, μ) = - \sum_{b \neq c, β_{b} - h < β_{c}} d_{μ} (β - h e_{b} + h e_{c}) .$ A nonzero term requires $0 \leq β_{b} - h$ , so the product is finite. Repeated beta entries cancel by exchanging the equal indices; a negative entry cannot become a valid term because, if it is the receiver, the lowered donor is still smaller. Thus these invalid sequences contribute zero exponents. Delete the largest semistandard label, occurring $z$ times in column feet $C$ , to get $β - 1_{C}, \overset{ˉ}{μ}$ . The strict predecessor boundary excludes every column also in $C$ . Consequently the literal exponent $E_{h}$ in $g$ obeys $E_{h} (β, μ) = \sum_{∣ C ∣ = z} E_{h} (β - 1_{C}, \overset{ˉ}{μ}) + \sum_{∣ C ∣ = z} \sum_{b \notin C, c \in C} (1_{β_{b} - β_{c} + 1 = h} - 1_{β_{b} - β_{c} = h}) d_{\overset{ˉ}{μ}} (β - 1_{C}) .$ The final sum is exactly the axial ratio contributed by each deleted largest-label box.

7.1F3step 6.1

Verify the same recurrence for $A_{h}$ . Apply [F3] to its summands and compare with $\sum_{C} A_{h} (β - 1_{C}, \overset{ˉ}{μ})$ . Set $δ = β_{b} - β_{c} - h$ . The original condition is $δ < 0$ and the deleted one $δ - 1_{C} (b) + 1_{C} (c) < 0$ . They differ only when $b \in C, c \notin C, δ = 0$ or $b \notin C, c \in C, δ = - 1$ . In the first case, exchanging the two shifted beta entries turns the contribution into $- d_{\overset{ˉ}{μ}} (β - 1_{C^{'}})$ with $C^{'} = C - b + c$ , giving the negative term of step 6.1. In the second case, the exchanged entries give $+ d_{\overset{ˉ}{μ}} (β - 1_{C})$ , giving its positive term. Every other case cancels. At degree $0$ both exponent families vanish: a nonzero shifted empty beta sequence would have to exchange two entries with $h = β_{b} - β_{c}$ , excluded by the strict condition. Induction gives $E_{h} = A_{h}$ and therefore $g_{μ}^{λ} = \prod_{h \geq 1} \prod_{b \neq c, β_{b} - h < β_{c}} [h]_{Q}^{- d_{μ} (β - h e_{b} + h e_{c})} .$ This proves the regrouping without copying the inconsistent signs in the printed display.

8.1givenF3step 7.1

Convert to hooks. A surviving lowered bead $β_{b} - h$ is a nonnegative gap, so the bead-gap correspondence makes $h = h_{a b}$ for a node $(a, b)$ . The remaining condition $β_{c} > β_{b} - h_{a b} = n - 1 - λ_{a} + a$ is equivalent to $c \leq λ_{a}$ ; the paired nodes therefore lie in the same row. For $b < c$ write $h = h_{a b}$ and $k = h_{a c}$ . Since $h - k = β_{b} - β_{c}$ , the two ordered displaced sequences $β - h e_{b} + h e_{c}$ and $β - k e_{c} + k e_{b}$ exchange their two changed entries. Their signed numbers are negatives of each other, and their factors are exactly $([h]_{Q} / [k]_{Q})^{d_{μ} (β + k e_{b} - k e_{c})}$ . This proves the stated balanced hook product. For the strip interpretation, first lower $β_{c}$ to the gap $g = β_{c} - k = n - 1 - λ_{a} + a$ , whose sorted position is $r \geq c$ . Recovering conjugate parts from the sorted sequence replaces $λ_{j}^{'}$ by $λ_{j + 1}^{'} - 1$ for $c \leq j < r$ and by $g - n + r$ for $j = r$ , leaving other columns unchanged. The removed nodes form the rim strip $R_{a c}$ in columns $c, \dots, r$ , with $k$ boxes. Since $b < c$ , its original bead $β_{b}$ is unaffected and is still in position $b$ in this intermediate diagram $κ$ . Raising it to $β_{b} + k$ and sorting puts it in position $s \leq b$ ; applying the same column calculation in reverse shows that $η / κ$ is a $k$ -strip in columns $s, \dots, b$ , with lowest node in column $s$ . The two sorting permutations make $r - c$ and $b - s$ crossings. The strip-height counting identity gives heights $k - 1 - (r - c)$ and $k - 1 - (b - s)$ , so their sum has the same parity as the total crossings and gives exactly the stated sign. Colliding or negative sequences vanish by definition.

9.1givenstep 5.1step 7.1step 8.1∎

Specialize integrally. The semistandard form matrix has entries in $Z [Q, Q^{- 1}]$ , so its determinant and $Q^{- N} G_{μ}^{λ}$ are Laurent polynomials. Steps 5.1 and 7.1 identify the rational products with these same polynomials in $Q (Q)$ . Base change of the free semistandard basis and its form now proves the identity over the given domain, including singular specializations and weights of dimension zero. No individual vanishing ratio is evaluated. All constructions are finite; no choice principle is used.

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The q-Schaper virtual radical-layer formula

Statement

Let $R$ be a principal ideal domain, $π$ a prime element, $F = R / π R$ the residue field, and let $\hat{q} \in R^{\times}$ satisfy $[n]!_{\hat{q}} \neq 0$ in $R$ ; put $q : = \hat{q} + π R \in F^{\times}$ . Let $d \geq 1$ , let $λ$ be a partition of $n$ with at most $d$ parts, and let $W_{F}^{λ}$ be the Weyl module of the q-Schur algebra $S_{F} (d, n) = S (d, n) \otimes_{R} F$ with its Jantzen filtration $W_{F}^{λ} (i)$ with respect to $(R, π)$ (The Jantzen filtration of a lifted cell module, The type-A q-Schur algebra and its Weyl modules).

(Jantzen sum formula) In the Grothendieck group of finite-dimensional $S_{F} (d, n)$ -modules, $\sum_{i > 0} [W_{F}^{λ} (i)] = \sum_{\begin{array}{c} (a, b), (a, c) \in [λ] \\ b < c \end{array}} (ν_{π} ([h_{a b}^{λ}]_{\hat{q}}) - ν_{π} ([h_{a c}^{λ}]_{\hat{q}})) [W_{F}^{(β_{1}, \dots, β_{b} + h_{a c}^{λ}, \dots, β_{c} - h_{a c}^{λ}, \dots, β_{n})}],$ where $β = β (λ)$ is the beta sequence of $λ$ , $h_{a b}^{λ}$ are its hook lengths and the symbols are the virtual Weyl symbols of Virtual Weyl and Specht symbols attached to beta sequences. 2. (q-Schaper, $d \geq n$ ) If $d \geq n$ and $Π$ denotes the Schur functor of The Schur idempotent and the Schur functor for type-A q-Schur algebras, then applying $Π$ to (1) gives the corresponding identity for the Hecke Specht modules in the source's normalisation, that is, for the Murphy cell modules $Π (W_{F}^{λ}) = S_{M}^{λ}$ of The type-A Hecke Specht module and its cell form (the Dipper-James form is obtained by conjugating partitions and applying the same $θ$ pullback to every Hecke module), $\sum_{i > 0} [S_{F}^{λ} (i)] = \sum_{\begin{array}{c} (a, b), (a, c) \in [λ] \\ b < c \end{array}} (ν_{π} ([h_{a b}^{λ}]_{\hat{q}}) - ν_{π} ([h_{a c}^{λ}]_{\hat{q}})) [S_{F}^{(β_{1}, \dots, β_{b} + h_{a c}^{λ}, \dots, β_{c} - h_{a c}^{λ}, \dots, β_{n})}]$ in the Grothendieck group of finite-dimensional $H_{F, q} (S_{n})$ -modules, with the virtual Specht symbols of Virtual Weyl and Specht symbols attached to beta sequences. The formula is a positive filtration-sum identity expressed in a signed Weyl/Specht basis. A single nonzero signed Weyl coefficient does not force the corresponding labelled simple factor: the entire right side must first be expanded and cancelled in simple classes. For a Weyl module and an off-head label $ν$ , the resulting multiplicity of $L^{ν}$ is an upper bound for $[W^{λ} : L^{ν}]$ , and is positive exactly when that factor occurs in the radical. This does not give a general closed algorithm for decomposition numbers.

Facts & Assumptions

Given: A PID $R$ with prime $π$ , residue field $F$ , $\hat{q} \in R^{\times}$ with $[n]!_{\hat{q}} \neq 0$ , the q-Schur algebra $S_{F} (d, n)$ and its Weyl modules with Jantzen filtrations (The Jantzen filtration of a lifted cell module, The type-A q-Schur algebra and its Weyl modules), and the virtual Weyl and Specht symbols (Virtual Weyl and Specht symbols attached to beta sequences).

[F1]

(The q-Schur Weyl weight-space Gram determinant is a hook product) The Gram determinant of $W^{λ}$ on its $ν$ -weight space equals $q^{N}$ times the hook-ratio product $\prod_{(a, b), (a, c) \in [λ], b < c} ([h_{a b}^{λ}]_{\hat{q}} / [h_{a c}^{λ}]_{\hat{q}})^{d_{μ} (\dots)}$ ; under $[n]!_{\hat{q}} \neq 0$ every factor is nonzero in the domain $R$ .

[F2]

(Jantzen layer dimensions sum to the Gram determinant valuation) The total dimension of the positive Jantzen layers on a non-degenerate form equals the $π$ -adic valuation of its Gram determinant.

[F3]

(Semistandard weight dimensions detect virtual Weyl classes) Equality of all semistandard weight-space dimensions forces equality in the Grothendieck group of $S_{F} (d, n)$ -modules.

[F4]

The balanced displacement removes the $h_{a c}$ -strip and adds the same number of boxes, preserving degree; its sign is the beta sorting sign (Removable border strips and their height, Hooks of a partition correspond bijectively to removable border strips).

[F5]

The exact Schur functor has integral corner isomorphism and exact restricted form factor $1$ , with $e^{*} = e$ (The Schur idempotent and the Schur functor for type-A q-Schur algebras, The Schur functor sends Weyl and simple modules to Specht and Hecke simple modules). Filtration transport is proved below, rather than assumed from exactness.

[F6]

Source results (Mathas, Theorem 5.31 and Corollary 5.32, printed pp. 58-60): the Jantzen sum formula for the q-Schur algebra and its transport to the Hecke side, with the coefficient $ν_{π} ([h_{a b}^{λ}]_{\hat{q}}) - ν_{π} ([h_{a c}^{λ}]_{\hat{q}})$ and the virtual symbols.

Proof

technique · direct

1.1givenF1F2F4

Work on a fixed weight space of the lifted free Weyl module. Every hook quantum integer is nonzero because its index is at most $n$ and $[n]!_{\hat{q}} \neq 0$ in the domain. The balanced Gram formula [F1] gives its determinant, and the unit power of $\hat{q}$ has valuation zero. The Jantzen Gram-valuation lemma [F2] thus gives the sum of positive filtration-submodule dimensions as $\sum_{(a, b), (a, c), b < c} (ν_{π} ([h_{a b}]_{\hat{q}}) - ν_{π} ([h_{a c}]_{\hat{q}})) d_{μ} (β + h_{a c} e_{b} - h_{a c} e_{c}) .$ The signed number is exactly the weight dimension of the corresponding balanced virtual Weyl symbol, including zero and cutoff cases.

2.1F3step 1.1

Apply step 1.1 to every weight. The Weyl weight-character matrix is unitriangular, so [F3] detects equality of Grothendieck classes and proves part 1. These are classes of the filtration submodules $W_{F}^{λ} (i)$ , not their successive quotients and not simply the radical class.

3.1F5step 2.1

Establish filtration transport before applying the exact functor. On the lifted Weyl module $W$ , the identity $e^{*} = e$ makes $W e ⊥ W (1 - e)$ . For $x \in W e$ and every $i \geq 0$ , pairing $x$ with all of $W$ is divisible by $π^{i}$ exactly when its pairing with $W e$ is. Thus $W (i) e = (W e) (i)$ . Reducing modulo $π$ commutes with the direct summand projection, giving $W_{F} (i) e = (W e)_{F} (i)$ . The actual corner isomorphism and factor- $1$ form identity [F5] identify the latter with the Murphy Specht filtration. Applying the exact functor to the proved Weyl-class identity therefore transports each submodule and every signed term, proving part 2 even when a reduced cell form is identically zero. For the ordinary convention, apply the same $θ$ to both modules and conjugate the labels; the independently proved Laurent-unit form factor preserves these divisibility filtrations as well.

4.1F4step 1.1step 2.1step 3.1∎

Explain the valid multiplicity information. The Weyl form has nonzero simple head, $W_{F} (1) = rad W_{F}$ , and $W_{F} (i) \subseteq rad W_{F}$ for $i > 0$ . Consequently for $ν \neq λ$ its actual multiplicity in $W_{F}$ is its radical multiplicity, bounded above by its multiplicity in the whole positive sum. This whole-sum multiplicity is positive if and only if that factor occurs in the radical, since the first submodule is the radical and later ones lie inside it. No such conclusion is drawn from an individual positive or negative Weyl coefficient before cancellation. All claims follow from steps 1.1-3.1 and this submodule inclusion.

RemarkRemark: AI-adaptedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

General type-A decomposition numbers remain unsolved

Statement

The two preceding constraints on the decomposition matrix of $H_{F, q} (S_{n})$ are real but limited, and neither solves the general problem.

(i) Dominance unitriangularity (The type-A Hecke decomposition matrix is dominance unitriangular) determines only the diagonal entries $d_{μ μ} = 1$ and the vanishing region $d_{λ μ} \neq 0 \Rightarrow μ ⊵ λ$ ; it gives no value for an off-diagonal entry.

(ii) The q-Schaper/Jantzen sum formula (The q-Schaper virtual radical-layer formula) is an identity for the sum of the positive Jantzen filtration submodules. Its individual signed hook terms are not multiplicity bounds. After the whole virtual sum is expanded in simple classes and all cancellations are performed, its nonnegative simple coefficients bound the corresponding radical composition multiplicities from above; it neither determines a general multiplicity nor produces a positive combinatorial rule.

Special coefficient regimes and low-rank calculations determine additional entries (canonically based, weight-one, and small- $n$ cases), and the classification of irreducible Specht modules for symmetric groups in prime characteristic is known (Fayers, Introduction, printed p. 2 of the cited 2012 paper). That classification is distinct from computing all composition multiplicities. No general closed formula for the decomposition numbers of $S_{n}$ in a fixed positive characteristic is claimed anywhere on this page. This remark records the boundary; it carries no proof obligation and no choice principle.

Remarks

The two cited results are the constraints proved on this page: unitriangularity is a vanishing statement about the dominance order, and the q-Schaper formula expresses the sum of the positive Jantzen filtration submodules as a virtual combination of Weyl symbols whose coefficients are $π$ -adic valuations of hook q-integers. Neither statement computes an individual off-diagonal decomposition number in general, and the remark is not used as a supplier to prove any replacement claim.

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Central Hecke Jucys–Murphy scalars separate q-Schur residue contents

Statement

Let $R$ be a commutative unital ring, $q \in R^{\times}$ , $H = H_{R, q} (S_{n})$ , $M = ⨁_{α \in Λ (d, n)} M^{α}$ and $S = {End}_{H} (M)$ . For a symmetric polynomial $f$ over $R$ , put $z_{f} = f (L_{1}, \dots, L_{n})$ for the additive Hecke Jucys–Murphy elements. Right multiplication $ρ_{z_{f}} : m \mapsto m z_{f}$ is a central element of $S$ . On the right Weyl cell $W^{λ}$ it acts by

$a_{λ} (f) = f ([c (A)]_{q} : A \in [λ]), c (r, s) = s - r .$

The identity holds over every $R$ and unit $q$ , without division by a quantum integer or a Gram determinant. In particular, if $R = F$ is a field of finite quantum characteristic $e \geq 2$ and $L^{μ}$ is a constituent of $W^{λ}$ , then $λ$ and $μ$ have the same multiset of residues $c (A) m o d e$ . Neither partition is required to be restricted or to survive the Hecke corner. This proves a necessary content-separation condition, not an assertion that equal content is sufficient for common-block membership.

Facts & Assumptions

Given: The ring, parameter, ranks, right permutation modules and Schur algebra of the Statement.

[F1]

The integral Murphy cells $C_{M} (λ)$ have free standard basis $m_{t}$ , dominance ideals $I^{⊳ λ}$ , a symmetric cell form $G_{λ}$ , and cell-layer multiplication $X \cdot Y = X G_{λ} Y$ . The standard trace satisfies $τ (T_{u}^{*} T_{v}) = δ_{u v} q^{ℓ (u)}$ (The Murphy cellular basis, the cell law and the Murphy form).

[F2]

The additive $L_{k}$ commute, symmetric polynomial evaluations are central, and their matrices in every Murphy cell have diagonal entries $[c_{t} (k)]_{q}$ and strictly dominant off-diagonal terms (The Hecke Jucys-Murphy action is dominance triangular).

[F3]

The actual Schur maps are $φ_{S T} (m_{ν} h) = m_{S T} h$ , supported on the indicated source and target types, with $m_{S T}$ the unweighted standard-tableau fibre sum. They form the cellular basis and $W^{λ}$ has basis $w_{T} = φ_{T^{λ} T} + S^{⊳ λ}$ , where $T^{λ}$ has shape and type $λ$ and its refinement is $t^{λ}$ . All Weyl forms have canonical norm one, hence their simple form quotients $L^{λ}$ are nonzero over every field (The type-A q-Schur algebra and its Weyl modules).

[F4]

A finite-dimensional algebra has nilpotent Jacobson radical; its semisimple quotient is a product of matrix algebras over division rings. Ideals and ideal quotients of a semisimple algebra are semisimple and have no nonzero nilpotent two-sided ideal (For a finite-dimensional algebra, the Jacobson radical is nilpotent and the quotient by it is semisimple, Wedderburn–Artin theorem for semisimple rings).

[F5]

Finite quantum characteristic $e$ is the characteristic of $F$ when $q = 1$ , and the order of $q$ when $q \neq 1$ . The extended quantum integers are $[c]_{q} = (q^{c} - 1) / (q - 1)$ when $q \neq 1$ , and $[c]_{1} = c$ (The quantum characteristic of an invertible Hecke parameter).

Proof

technique · direct

1.1F1F4algebra

Work first over $A = Z [u, u^{- 1}]$ with fraction field $K = Q (u)$ . The regular trace pairing $(a, b) \mapsto Tr (ℓ_{a b})$ has entries in $A$ in the standard Hecke basis. At $u = 1$ its matrix is $n! δ_{v, w^{- 1}}$ , so its determinant is nonzero and it is nondegenerate over $K$ . If $a$ belongs to the Jacobson radical, $a b$ is nilpotent for every $b$ by [F4]; its left multiplication is nilpotent and has trace zero. Nondegeneracy forces the radical to vanish. Thus $H_{K}$ is semisimple. Its shape- $λ$ cell layer has the multiplication $X G_{λ} Y$ of [F1]. If $G_{λ}$ were singular, the nonzero subspace ${X : G_{λ} X = 0}$ would be a two-sided ideal with square zero, contradicting [F4]. Hence $G_{λ}$ is invertible and $X \mapsto G_{λ} X$ identifies this layer with a full matrix algebra over $K$ .

1.2F1F3algebraconstruct

Construct an injective right $S$ -module map $j : W^{λ} \to {Hom}_{H} (M, C_{M} (λ))$ . Project the output of $φ_{T^{λ} T} : M \to M^{λ}$ through $π : M^{λ} \to C_{M} (λ) = M^{λ} / (M^{λ} \cap I^{⊳ λ})$ . For $T$ of type $ν$ , the resulting map is zero on the other summands and sends $m_{ν}$ to $m_{T} = \sum_{t : ν (t) = T} m_{t}$ . Higher Schur maps with output type $λ$ have Hecke values of strictly dominant shape, so their output is killed by $π$ . Thus this assignment respects the cell quotient and multiplication by any $s \in S$ becomes composition on the right: $j (w s) = j (w) \circ s$ . For each fixed type the sets of standard $t$ with $ν (t) = T$ are disjoint. Each is nonempty: standardize equal entries of a semistandard tableau in increasing column order; equal entries cannot share a column, row order is retained, and different entries occupy disjoint consecutive label blocks. Hence their sums $m_{T}$ are independent in the free Murphy cell basis, over every $R$ . This proves injectivity, including zero type entries. No surjectivity onto all specialized Hom spaces is asserted.

2.1F1F2step 1.1algebra

For an integral symmetric polynomial $f$ , $z_{f}$ is central by [F2]. Its action on the generic cell layer is a bimodule endomorphism of that full matrix algebra, hence scalar: its value at the identity commutes with every matrix and the rest follows by multiplication. Accordingly its matrix on $C_{M} (λ)_{K}$ is scalar. The triangular matrices in [F2] give its diagonal as $f ([c_{t} (1)]_{u}, \dots, [c_{t} (n)]_{u})$ , equal to $a_{λ} (f)$ for every $t$ because $f$ is symmetric and the node multiset is fixed. Thus $z_{f}$ acts exactly by this scalar. Both sides have matrices over $A$ , and injection into $K$ proves the identity before specialization. Every symmetric polynomial over an arbitrary $R$ is an $R$ -linear combination of integral monomial symmetric polynomials: coefficients are constant on each monomial orbit. Apply the integral identities to these monomial polynomials and specialize $u \mapsto q$ . Therefore every $z_{f}$ acts by $a_{λ} (f)$ on the actual Murphy cell over $R$ , including singular parameters.

3.1F2F3step 2.1step 1.2algebra

For central $z_{f} \in H$ , right multiplication preserves every $M^{α}$ and is $H$ -linear. Every $H$ -linear endomorphism $s$ of $M$ satisfies $s (m z_{f}) = s (m) z_{f}$ , so $ρ_{z_{f}}$ is central in $S$ . For $w \in W^{λ}$ and $m \in M$ , step 1.2 and $H$ -linearity give $j (w ρ_{z_{f}}) (m) = j (w) (m z_{f}) = j (w) (m) z_{f} = a_{λ} (f) j (w) (m),$ the last equality by step 2.1. Injectivity gives $w ρ_{z_{f}} = a_{λ} (f) w$ , proving the full ring-level assertion. If $M = 0$ all cell assertions are empty. At rank zero the empty cell is $R$ with the empty polynomial evaluation, so the same conclusion holds.

4.1F3F5step 3.1algebra∎

Suppose now that $R = F$ and $L^{μ}$ is a constituent of $W^{λ}$ . Step 3.1 gives scalar $a_{λ} (f)$ on this constituent. Since $L^{μ}$ is the nonzero quotient of $W^{μ}$ by [F3], its scalar is also $a_{μ} (f)$ ; hence they agree for every symmetric $f$ . Taking all elementary symmetric polynomials gives $\prod_{A \in [λ]} (X - [c (A)]_{q}) = \prod_{B \in [μ]} (X - [c (B)]_{q}) .$ By [F5], if $q = 1$ the distinct $e$ residues are the distinct prime-field values $[c]_{1}$ ; if $q \neq 1$ , $[a]_{q} = [b]_{q}$ is equivalent to $q^{a} = q^{b}$ , hence to $a \equiv b (m o d e)$ . Unique factorization of these monic products into distinct field-linear factors therefore gives equality of every residue multiplicity, also when the field characteristic divides a multiplicity. There is no restriction on either label's Hecke-corner image, and no splitting extension, Gram division or choice principle enters the argument.

Remarks

The map in step 1.2 is intentionally only an injection. At $H_{2}$ with $q = - 1$ , the sign Murphy cell of shape $(1, 1)$ and the permutation module of type $(2)$ both have $T_{1}$ acting by $- 1$ , so their Hom space has dimension one, while the corresponding Weyl weight has Kostka number zero. An unrestricted singular-parameter full-Hom identification would therefore add a nonexistent Weyl weight. The central scalar proof needs none of that surjectivity.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

The type-A quantum algebra and its tensor-action formulas

Definition

Let $F = Q (q)$ with $q$ an indeterminate, let $N \geq 1$ , and put $K_{i} = L_{i} L_{i + 1}^{- 1}$ for $1 \leq i < N$ . Start with the free associative $F$ -algebra on $L_{j}, L_{j}^{- 1}$ ( $1 \leq j \leq N$ ) and $E_{i}, F_{i}$ ( $1 \leq i < N$ ), and quotient by the two-sided ideal generated by the following relations:

$L_{j} L_{j}^{- 1} = L_{j}^{- 1} L_{j} = 1$ and $L_{j} L_{k} = L_{k} L_{j}$ for all $j, k$ .
$L_{j} E_{i} L_{j}^{- 1} = q^{δ_{j, i} - δ_{j, i + 1}} E_{i}$ and $L_{j} F_{i} L_{j}^{- 1} = q^{- δ_{j, i} + δ_{j, i + 1}} F_{i}$ .
$[E_{i}, F_{j}] = δ_{i j} (K_{i} - K_{i}^{- 1}) / (q - q^{- 1})$ .
$[E_{i}, E_{j}] = [F_{i}, F_{j}] = 0$ when $∣ i - j ∣ > 1$ .
For $∣ i - j ∣ = 1$ , $E_{i}^{2} E_{j} - (q + q^{- 1}) E_{i} E_{j} E_{i} + E_{j} E_{i}^{2} = 0$ and $F_{i}^{2} F_{j} - (q + q^{- 1}) F_{i} F_{j} F_{i} + F_{j} F_{i}^{2} = 0$ .

Here $[x, y] = x y - y x$ . The free algebra and quotient construction are those of The free associative R-algebra on a set and descent of relations; this definition asserts existence of the presented associative algebra, not the well-definedness of structural maps on its quotient.

On $V_{N}$ with ordered basis $e_{1}, \dots, e_{N}$ , prescribe the candidate matrices $L_{j} e_{a} = q^{δ_{j a}} e_{a}, E_{i} e_{a} = δ_{a, i + 1} e_{i}, F_{i} e_{a} = δ_{a, i} e_{i + 1} .$ Prescribe candidate coproduct and counit values by $Δ (L_{j}) = L_{j} \otimes L_{j}, Δ (E_{i}) = E_{i} \otimes K_{i} + 1 \otimes E_{i}, Δ (F_{i}) = F_{i} \otimes 1 + K_{i}^{- 1} \otimes F_{i},$ and $ϵ (L_{j}^{\pm 1}) = 1$ , $ϵ (E_{i}) = ϵ (F_{i}) = 0$ . The following relation-descent lemma proves that these assignments define the vector representation and bialgebra structure, and hence the tensor actions used below.

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

The type-A quantum presentation descends to its coproduct tensor action

Statement

For the presented algebra $U_{N}$ and candidate assignments in The type-A quantum algebra and its tensor-action formulas, the matrices on $V_{N}$ satisfy every defining relation. The coproduct assignments descend to a coassociative algebra homomorphism $Δ : U_{N} \to U_{N} \otimes_{F} U_{N}$ and the counit assignments descend to an algebra homomorphism $ϵ : U_{N} \to F$ satisfying both counit identities. Consequently every tensor power $V_{N}^{\otimes n}$ has the action obtained by iterating $Δ$ ; all its generator matrices have entries in $Z [q, q^{- 1}]$ .

Facts & Assumptions

Given: $F = Q (q)$ , $N \geq 1$ , the algebra presentation and candidate maps in The type-A quantum algebra and its tensor-action formulas.

[F1]

A free associative algebra modulo the two-sided ideal of its displayed relations is a presented algebra, and assignments descend exactly when they annihilate every defining relation (The free associative R-algebra on a set and descent of relations).

[F2]

The tensor product of algebras has multiplication $(a \otimes b) (c \otimes d) = a c \otimes b d$ , and maps agreeing on the displayed generators agree on the presented algebra (The tensor product of $R$ -algebras has multiplication $(a \otimes b) (a^{'} \otimes b^{'}) = a a^{'} \otimes b b^{'}$ , Presentation base change and transport of explicit bases to commutative specializations).

[F3]

For $a_{i j} = 2 δ_{i j} - δ_{i, j + 1} - δ_{i + 1, j}$ , the defining $L$ -weight relations imply $K_{i} E_{j} K_{i}^{- 1} = q^{a_{i j}} E_{j}$ and $K_{i} F_{j} K_{i}^{- 1} = q^{- a_{i j}} F_{j}$ ; in particular $a_{i j} = a_{j i}$ , $a_{i i} = 2$ , and $a_{i j} = - 1$ for adjacent indices.

[F4]

The associator $α ((a \otimes b) \otimes c) = a \otimes (b \otimes c)$ is natural and satisfies the pentagon and unit triangle (Associator naturality, pentagon, unit triangle and symmetry hexagons on elementary tensors). The unit maps are $λ (r \otimes a) = r a$ and $ρ (a \otimes r) = a r$ (The regular module is a tensor unit: $R \otimes_{R} N ≅ N$ and $M \otimes_{R} R ≅ M$ ). These maps are algebra isomorphisms for the tensor-algebra products, as is checked on elementary tensors using [F2].

Proof

Given: The data in [F1]–[F3].

1.1givenF1F3algebra

On $V_{N}$ , each $L_{j}$ is diagonal and invertible, and the $L_{j}$ commute. Conjugating the matrix unit $E_{i} : e_{i + 1} \mapsto e_{i}$ by $L_{j}$ multiplies its sole nonzero entry by $q^{δ_{j, i} - δ_{j, i + 1}}$ ; conjugating $F_{i} : e_{i} \mapsto e_{i + 1}$ gives the inverse exponent. Also $K_{i}$ has eigenvalue $q$ on $e_{i}$ , $q^{- 1}$ on $e_{i + 1}$ and $1$ on every other basis vector, so $(K_{i} - K_{i}^{- 1}) / (q - q^{- 1})$ is the diagonal matrix with entries $1, - 1, 0$ in those positions. Directly, $[E_{i}, F_{j}] = 0$ for $i \neq j$ and $[E_{i}, F_{i}]$ is that diagonal matrix. Products $E_{i} E_{j}$ and $E_{j} E_{i}$ (and the corresponding $F$ products) both vanish when $∣ i - j ∣ > 1$ ; $E_{i}^{2} = F_{i}^{2} = 0$ , and each middle three-letter Serre product also vanishes because its matrix-unit indices do not compose. Thus every defining relation holds, and the quotient universal property [F1] gives the vector representation.

1.2givenF1F2F3algebra

Define $Δ$ on the free generators by the displayed formulas in The type-A quantum algebra and its tensor-action formulas. The $Δ (L_{j})$ are mutually commuting units with the displayed inverses. Conjugation by $Δ (L_{j})$ gives the required weights on each of the two terms of $Δ (E_{i})$ and $Δ (F_{i})$ , since each $K_{i}$ is a Laurent monomial in the commuting $L$ 's. For any $i, j$ , expansion gives $[Δ (E_{i}), Δ (F_{j})] = [E_{i}, F_{j}] \otimes K_{i} + K_{j}^{- 1} \otimes [E_{i}, F_{j}] + [E_{i} \otimes K_{i}, K_{j}^{- 1} \otimes F_{j}]$ . The last cross term is zero: its first product can be reordered as $q^{a_{j i} - a_{i j}} K_{j}^{- 1} E_{i} \otimes F_{j} K_{i} = K_{j}^{- 1} E_{i} \otimes F_{j} K_{i}$ , which is its reverse product by [F3]. If $i \neq j$ , the remaining terms vanish; if $i = j$ , they sum to $(K_{i} \otimes K_{i} - K_{i}^{- 1} \otimes K_{i}^{- 1}) / (q - q^{- 1}) = Δ ((K_{i} - K_{i}^{- 1}) / (q - q^{- 1}))$ . For $∣ i - j ∣ > 1$ , $a_{i j} = 0$ and the $K_{i}$ commute with $E_{j}, F_{j}$ ; expanding the two commutators therefore proves that the distant $E$ and $F$ relations are preserved.

1.3givenF1F2F3algebra

Let $γ = q + q^{- 1}$ and $S_{i j} (X, Y) = X^{2} Y - γ X Y X + Y X^{2}$ for adjacent $i, j$ . For the $E$ relation write $Δ (E_{s}) = A_{s} + B_{s}$ with $A_{s} = E_{s} \otimes K_{s}$ , $B_{s} = 1 \otimes E_{s}$ . In each of the three words $i i j, i j i, j i i$ , choose an $A$ at each selected position and a $B$ at each other position, then move the selected $K$ factors to the right using $K_{r} E_{s} = q^{a_{r s}} E_{s} K_{r}$ . Grouping the resulting 24 terms by the number of selected positions gives: degree $0$ in the first factor is $1 \otimes S_{i j} (E_{i}, E_{j})$ ; degree $3$ is $S_{i j} (E_{i}, E_{j}) \otimes K_{i}^{2} K_{j}$ ; degree $1$ has respective coefficients $q + q^{- 1} - γ$ , $q^{2} + 1 - γ q$ , $1 + q^{- 2} - γ q^{- 1}$ on $E_{i} \otimes E_{i} E_{j} K_{i}$ , $E_{i} \otimes E_{j} E_{i} K_{i}$ , $E_{j} \otimes E_{i}^{2} K_{j}$ ; degree $2$ has respective coefficients $q^{- 2} + 1 - γ q^{- 1}$ , $q^{2} + 1 - γ q$ , $q + q^{- 1} - γ$ on $E_{i}^{2} \otimes E_{j} K_{i}^{2}$ , $E_{i} E_{j} \otimes E_{i} K_{i} K_{j}$ , $E_{j} E_{i} \otimes E_{i} K_{j} K_{i}$ . All six mixed coefficients are zero by the definition of $γ$ , and the two pure terms vanish by the defining Serre relation. Thus $Δ$ preserves every positive Serre relation.

2.1givenF1F2step 1.3algebra

The assignment $ϕ (L_{j}) = L_{j}^{- 1}$ , $ϕ (E_{i}) = F_{i}$ , $ϕ (F_{i}) = E_{i}$ preserves the presentation: it exchanges the two weight relations, sends the mixed commutator to its negative with $K_{i}$ replaced by $K_{i}^{- 1}$ , and exchanges the distant and Serre relations. It is an involution. On generators $Δ (F_{i}) = τ (ϕ \otimes ϕ) Δ (E_{i})$ , where $τ (x \otimes y) = y \otimes x$ is an algebra isomorphism. Since step 1.3 proves that the positive Serre polynomial maps to zero, applying the algebra maps $τ$ and $ϕ \otimes ϕ$ proves that the negative Serre polynomial maps to zero as well. Together steps 1.2–2.1 show every relation is killed, so [F1] descends $Δ$ to the quotient. The counit assignment kills the relations directly: each relation becomes either $1 = 1$ or $0 = 0$ , so it too descends.

3.1givenF1F2F4step 1.1step 2.1algebra∎

Compare $α \circ (Δ \otimes i d) \circ Δ$ and $(i d \otimes Δ) \circ Δ$ as maps to $U_{N} \otimes (U_{N} \otimes U_{N})$ , and compare $λ \circ (ϵ \otimes i d) \circ Δ$ and $ρ \circ (i d \otimes ϵ) \circ Δ$ with ${i d}_{U_{N}}$ . All these composites are algebra homomorphisms by [F2] and [F4], so it suffices to check generators. The two counit composites return $L_{j}, E_{i}, F_{i}$ . After applying the associator, both iterated coproducts on $L_{j}$ give $L_{j} \otimes L_{j} \otimes L_{j}$ ; on $E_{i}$ , both give $E_{i} \otimes K_{i} \otimes K_{i} + 1 \otimes E_{i} \otimes K_{i} + 1 \otimes 1 \otimes E_{i}$ ; on $F_{i}$ , both give $F_{i} \otimes 1 \otimes 1 + K_{i}^{- 1} \otimes F_{i} \otimes 1 + K_{i}^{- 1} \otimes K_{i}^{- 1} \otimes F_{i}$ . Hence $Δ$ is coassociative with counit $ϵ$ . The vector representation of step 1.1 and the iterated coproduct therefore define actions on every $V_{N}^{\otimes n}$ . They are finite sums and products of matrices with entries $0, 1, q^{\pm 1}$ , so all entries lie in $Z [q, q^{- 1}]$ .

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Integral divided-power tensor actions generate the type-A q-Schur algebra

Statement

Put $A = Z [v, v^{- 1}]$ , $q = v^{2}$ , $N \geq 1$ , $r \geq 0$ , $V = A^{N}$ , and $T = V^{\otimes r}$ . Let $U_{Q (v)} ({g l}_{N})$ have the presentation of The type-A quantum algebra and its tensor-action formulas with its indeterminate renamed $v$ and its diagonal generators written $K_{j}$ . Define $U_{A}$ to be its $A$ -subalgebra generated by $K_{j}^{\pm 1}$ , $E_{i}^{(a)} = E_{i}^{a} / [a]_{v}!$ , $F_{i}^{(a)} = F_{i}^{a} / [a]_{v}!$ , and $[\begin{array}{c} K_{j}; 0 \\ t \end{array}] = \prod_{s = 1}^{t} \frac{K_{j} v^{1 - s} - K_{j}^{- 1} v^{s - 1}}{v^{s} - v^{- s}} (t \geq 0) .$ Use $K_{j} e_{b} = v^{δ_{j b}} e_{b}$ , $E_{i} e_{b} = δ_{b, i + 1} e_{i}$ , $F_{i} e_{b} = δ_{b, i} e_{i + 1}$ and the coproduct $Δ (E_{i}) = 1 \otimes E_{i} + E_{i} \otimes K_{i} K_{i + 1}^{- 1}, Δ (F_{i}) = K_{i}^{- 1} K_{i + 1} \otimes F_{i} + F_{i} \otimes 1, Δ (K_{j}) = K_{j} \otimes K_{j} .$ The resulting action preserves $T$ and commutes with the right $H_{A, q} (S_{r})$ action $e_{\dots a b \dots} T_{s} = {\begin{cases} v e_{\dots b a \dots} & a < b, \\ q e_{\dots a a \dots} & a = b, \\ (q - 1) e_{\dots a b \dots} + v e_{\dots b a \dots} & a > b . \end{cases}$ It gives a surjection $ρ : U_{A} ⟶ {End}_{H_{A, q} (S_{r})} (T) ≅ S_{A} (N, r)$ onto the actual permutation-module endomorphism algebra of The type-A q-Schur algebra and its Weyl modules. Its composition-weight projections are $1_{μ} = ρ (\prod_{j = 1}^{N} [\begin{array}{c} K_{j}; 0 \\ μ_{j} \end{array}]) (μ \in Λ (N, r)) .$ Every statement survives any homomorphism $A \to R$ to a commutative ring, including parameters at arbitrary roots of unity; $S_{A} \otimes_{A} R$ is the actual specialized endomorphism algebra. The proof supplies a terminating exact divided-power expression for each integral orbit-basis map, with Laurent-unit pivots.

Let $τ$ be the antiautomorphism $E_{i} \leftrightarrow F_{i}$ , $K_{j} \mapsto K_{j}$ , let $*$ be the current cellular antiautomorphism of $S_{A}$ , and put $C = \sum_{μ} c_{μ} 1_{μ}$ , $c_{μ} = v^{- \sum_{j} (\binom{μ_{j}}{2})}$ . Then $ρ (τ (u)) = C^{- 1} ρ (u)^{*} C (u \in U_{A}) .$

There is also a square-root-free version over $B = Z [q, q^{- 1}]$ . On the permutation-module word basis define weight-block operators $E_{i, a, μ}$ and $F_{i, a, μ}$ , zero outside content $μ$ , by replacing a subset $J$ of $a$ occurrences of $i + 1$ by $i$ , or of $i$ by $i + 1$ , respectively, with coefficients $q^{\sum_{j \in J} # i (w_{> j})}, q^{\sum_{j \in J} # (i + 1) (w_{< j})} .$ Together with $1_{μ}$ these generate $S_{B} (N, r)$ . After $q = v^{2}$ they are precisely $v^{a (μ_{i + 1} - a)} ρ (E_{i}^{(a)}) 1_{μ}$ and $v^{a (μ_{i} - a)} ρ (F_{i}^{(a)}) 1_{μ}$ . Thus this generating assertion holds over every original coefficient ring with any unit $q$ , whether or not $q$ has a square root. This assertion concerns the displayed polynomial action algebra; it does not assert a quantum group presentation over the original ring without choosing $v$ .

Facts & Assumptions

Given: The finite-rank objects and conventions in the Statement. For a word $w$ , let $h_{i} (w) = # i (w) - # (i + 1) (w)$ and let $inv (w)$ count pairs of positions with strictly decreasing letters. Composition and matrix indices range over $1, \dots, N$ .

[F1]

The generic presentation has the displayed vector representation and coassociative coproduct, by The type-A quantum presentation descends to its coproduct tensor action, under the field isomorphism renaming its indeterminate $v$ . No generic mutual-centralizer theorem is used.

[F2]

The permutation module $M^{μ} = m_{μ} H$ is free on $m_{μ} T_{d}$ for distinguished left cosets. Its cyclic presentation is $H / \sum_{s \in S_{μ}} (T_{s} - q) H$ , with the sum taken over adjacent generators of the row subgroup. The Schur algebra, its semistandard basis, and its involution are the actual endomorphism construction of The type-A q-Schur algebra and its Weyl modules.

[F3]

Quantum binomial coefficients ${[\begin{matrix} m \\ t \end{matrix}]}_{v}$ for $m, t \geq 0$ are Laurent polynomials, equal to zero for $t > m$ and to one for $t = m$ , by The quantum Pascal recurrences, the Gauss product formula and Gaussian integrality.

Proof

1.1givenF1F3algebraconstruct

Integral matrices. Write the iterated coproduct $E_{i}$ as its sum of single-position raising operators. The summand at position $j$ has coefficient $v^{h_{i} (w_{> j})}$ ; the lowering summand has coefficient $v^{- h_{i} (w_{< j})}$ . A position cannot be used twice. For $a$ distinct raising positions, swapping two consecutive choices changes the exponent by $2$ , since changing $i + 1$ to $i$ increases $h_{i}$ by $2$ . Summing the orders gives $v^{a (a - 1) / 2} [a]_{v}!$ times the coefficient using the original suffix counts: explicitly $\sum_{π \in S_{a}} v^{2 inv (π)} = v^{a (a - 1) / 2} [a]_{v}!$ , as follows by inserting the largest letter, which multiplies the sum by $1 + v^{2} + \dots + v^{2 (a - 1)}$ . The lowering calculation uses prefixes and the same identity. Therefore, over the generic field and then as integral matrices, $E_{i}^{(a)} e_{w} = \sum_{\begin{array}{c} J \subseteq {j : w_{j} = i + 1} \\ ∣ J ∣ = a \end{array}} v^{a (a - 1) / 2 + \sum_{j \in J} h_{i} (w_{> j})} e_{w (J \mapsto i)}, F_{i}^{(a)} e_{w} = \sum_{\begin{array}{c} J \subseteq {j : w_{j} = i} \\ ∣ J ∣ = a \end{array}} v^{a (a - 1) / 2 - \sum_{j \in J} h_{i} (w_{< j})} e_{w (J \mapsto i + 1)} .$ The toral generator acts on content $α$ by ${[\begin{matrix} α_{j} \\ t \end{matrix}]}_{v}$ , integral by [F3]. Thus every generating matrix preserves $T$ . No factorial is inverted after specialization.

2.1F2algebraconstructstep 1.1

Transport of the Hecke action. For $w_{μ} = 1^{μ_{1}} \dots N^{μ_{N}}$ define $θ_{μ} (m_{μ} T_{d}) = v^{ℓ (d)} e_{w_{μ} d}$ . Every word of content $μ$ occurs once, $ℓ (d) = inv (w_{μ} d)$ , and each coefficient is a unit; hence $θ_{μ}$ is an integral bijection. Standard Hecke multiplication on distinguished cosets gives an ascending swap with coefficient $1$ , a descending swap with coefficient $q$ plus $(q - 1)$ times the original coset, and coefficient $q$ when the two letters coincide. Multiplying the basis vectors by $v^{inv (w)}$ changes the two unequal-letter swap coefficients to $v$ , giving exactly the displayed tensor formula. Consequently this is an actual Hecke action, with quadratic and braid relations inherited from $H$ , and $θ = ⨁_{μ} θ_{μ}$ is a right-module isomorphism.

3.1step 1.1step 2.1algebra

Commutation can be checked locally. Put $j = i + 1$ . On the four two-letter words $i i, i j, j i, j j$ , the nonzero local raising values are $E (i j) = i i$ , $E (j i) = v i i$ , $E (j j) = v^{- 1} i j + j i$ , and the nonzero lowering values are $F (i i) = v^{- 1} i j + j i$ , $F (i j) = j j$ , $F (j i) = v j j$ . The Hecke values are $H (i i) = q i i$ , $H (i j) = v j i$ , $H (j i) = v i j + (q - 1) j i$ , $H (j j) = q j j$ . Substitution proves $E H = H E$ , $F H = H F$ : on $i j$ and $j i$ the raising composites give respectively $q i i$ and $v q i i$ , and on $j j$ they give $v i j + q j i$ ; lowering gives the transposed three checks. With one active letter and one outside ${i, j}$ the suffix or inverse-prefix multiplier is $1$ , and the outside letter has the same comparison with $i$ and $j$ , so swapping or retaining that letter commutes with its replacement. With no active letter both operators are zero. Terms of the full coproduct acting outside the adjacent pair only see its total $h_{i}$ , unchanged by its Hecke action; diagonal toral operators depend only on content. This proves commutation for $E_{i}, F_{i}, K_{j}$ . Over the generic field divided powers commute too; since their integral matrices are given in step 1.1, their commutator entries are zero Laurent polynomials. Toral binomials commute by content. Thus $ρ$ has image in the actual integral commutant.

4.1F2step 2.1algebraconstructstep 3.1

Orbit coordinates for that commutant. Fix source content $ν$ and split its sorted word into contiguous column blocks of lengths $ν_{j}$ . For an $N \times N$ nonnegative integer matrix $D = (d_{i j})$ of total $r$ , with column sums $ν$ and row sums $μ$ , let $W_{D}$ be the words containing exactly $d_{i j}$ letters $i$ in block $j$ , and put $z_{D} = \sum_{w \in W_{D}} v^{{inv}_{ν} (w)} e_{w},$ where ${inv}_{ν}$ counts inversions separately inside those blocks. Within a block, pairing an ascending word and its descending swap in $y T_{s} = q y$ forces the descending coefficient to be $v$ times the ascending coefficient, since $v$ is a unit. Adjacent swaps connect every arrangement to the uniquely blockwise sorted word $w_{D}$ . Hence all simultaneous $q$ -eigenvectors for the source row subgroup are exactly the free span of the $z_{D}$ , whose supports are disjoint and whose coefficient at $w_{D}$ is one. By the cyclic presentation [F2], evaluation at $e_{w_{ν}}$ identifies ${Hom}_{H} (T_{ν}, T_{μ})$ with this eigenspace. Write $ξ_{D}$ for the map with that evaluation. These maps give an explicit integral basis, independently of any generic rank comparison, over every ring with a chosen invertible $v$ .

5.1F3step 4.1algebra

The toral element $P_{μ} = \prod_{j} [\begin{matrix} K_{j}; 0 \\ μ_{j} \end{matrix}]$ acts on content $α$ by $\prod_{j} {[\begin{matrix} α_{j} \\ μ_{j} \end{matrix}]}_{v}$ . If $α \neq μ$ and both sum to $r$ , at least one $α_{j} < μ_{j}$ , making that factor identically zero; if $α = μ$ , every factor is one. Thus $ρ (P_{μ}) = 1_{μ}$ . For a diagonal matrix $D = diag (μ)$ the orbit map $ξ_{D}$ is this projection. These conclusions remain identities at parameters where several $K_{j}$ eigenvalues coincide.

6.1givenalgebraconstructstep 5.1

Define the triangular order, within fixed row and column sums, by all northeast and southwest rectangle sums $σ_{a b} (D) = {\begin{cases} \sum_{s \leq a, t \geq b} d_{s t} & a < b, \\ \sum_{s \geq a, t \leq b} d_{s t} & a > b . \end{cases}$ Write $D^{'} ⪯ D$ if every such sum is at most the corresponding sum for $D$ . This is an actual partial order: differences of northeast rectangle sums recover each strictly upper entry, differences of southwest sums recover each strictly lower entry, and the fixed row sums then recover the diagonal. Put $δ (D) = \sum_{i, j} ∣ i - j ∣ d_{i j}$ and $s (D) = \sum_{a \neq b} σ_{a b} (D)$ . The diagonal-boundary sums give $δ (D) = \sum_{a = 1}^{N - 1} (σ_{a, a + 1} (D) + σ_{a + 1, a} (D))$ . Therefore $D^{'} ≺ D$ implies $δ (D^{'}) \leq δ (D)$ and $s (D^{'}) < s (D)$ . The pair $(δ, s)$ of nonnegative integers supplies a terminating induction even when intermediate matrices have different row sums.

7.1step 1.1step 4.1step 6.1algebraconstruct

Extreme-column multiplication. Acting by $E_{i}^{(b)}$ on $z_{D}$ transfers $b$ letters from row $i + 1$ to row $i$ , distributed as $t_{j}$ among the column blocks, with $0 \leq t_{j} \leq d_{i + 1, j}$ and $\sum_{j} t_{j} = b$ . Its expansion in step 4.1 has only the matrices $D + \sum_{j} t_{j} (E_{i j} - E_{i + 1, j})$ . Choosing the rightmost available columns greedily maximizes every suffix $\sum_{j \geq k} t_{j}$ and minimizes every prefix $\sum_{j \leq k} t_{j}$ . A transfer across the row boundary $i$ changes a northeast rectangle only when its row cutoff is $i$ , by such a suffix; it changes a southwest rectangle only when its lower row cutoff is $i + 1$ , by the negative of such a prefix. Thus this greedy matrix dominates every other supported matrix in $⪯$ . For several $E_{i}$ executed in increasing $i$ , the source row $i + 1$ has not yet been changed, and each operation changes rectangle coordinates at its own boundary only. Consequently simultaneous greedy choices still give a single maximal supported matrix. If at each boundary the target row originally contains zero letters in every selected column, its coefficient is a Laurent unit: to obtain the blockwise sorted final word, all newly present $i$ 's in those columns must have been raised, giving exactly one predecessor word and one subset. That predecessor is blockwise sorted, with coefficient one in $z_{D}$ , and step 1.1 gives a single Laurent monomial. Earlier operations only depleted this target row, so its zero entries persist. Multiplying these monomials gives the claimed unit pivot, and uniqueness of successive row transfers rules out an additional path to the same final matrix.

8.1step 1.1step 4.1step 6.1step 7.1algebra

The lower version is equally explicit. Acting by $F_{i}^{(b)}$ transfers letters from row $i$ to row $i + 1$ . Greedy selection of the leftmost available columns maximizes southwest prefix increments and minimizes northeast suffix decrements, hence gives the maximal matrix. Execute several $F_{i}$ in decreasing $i$ ; each source row is then unchanged before its turn, and the rectangle boundaries are again distinct. If all selected target entries are initially zero, the blockwise sorted final word has exactly one predecessor and chosen subset, so its coefficient is a Laurent monomial by the lowering formula of step 1.1. This proves the same maximal-support and unit-pivot assertion without invoking a Gaussian coefficient or cancelling one.

9.1step 6.1step 7.1algebraconstructstep 8.1

Upper elimination. Suppose $D$ has a strictly upper entry and let $k > 0$ be the largest distance $j - i$ of such an entry. Set $b_{i} = d_{i, i + k}$ (zero when $i + k > N$ ) and $D_{0} = D - \sum_{i = 1}^{N - 1} b_{i} E_{i, i + k} + \sum_{i = 1}^{N - 1} b_{i} E_{i + 1, i + k}, X = E_{N - 1}^{(b_{N - 1})} \dots E_{1}^{(b_{1})} .$ Zero terms with column index greater than $N$ are omitted. Thus $X$ executes its operations in increasing $i$ . The rightmost nonzero column of source row $i + 1$ in $D_{0}$ is at most $i + k$ : its original distance- $k$ entry was removed, and the shifted entry from row $i$ is at $i + k$ . There are at least $b_{i}$ letters at that column. Hence the greedy transfer selects exactly $b_{i}$ letters there and recovers $D$ . Target row $i$ in $D_{0}$ has zero at column $i + k$ , since that entire outer entry was removed and its added entry is at $i + k - 1$ . Step 7.1 therefore gives $X ξ_{D_{0}} = u_{D} ξ_{D} + \sum_{D^{'} ≺ D} a_{D^{'}} ξ_{D^{'}}, u_{D} \in A^{\times}, a_{D^{'}} \in A .$ Its source is still $ν$ , and its target row sum is the row sum of $D$ , so these lower terms belong to that same Hom block. Also $δ (D_{0}) = δ (D) - \sum_{i} b_{i} < δ (D)$ . Every coefficient is explicitly obtained by the finite subset sums in step 1.1 and reading the blockwise sorted word coefficient of step 4.1.

10.1step 6.1step 8.1algebraconstructstep 9.1

Lower elimination. If $D$ is lower triangular but not diagonal, let $k > 0$ be its largest lower distance and put $b_{i} = d_{i + 1, i + 1 - k}$ . Set $D_{0} = D - \sum_{i = 1}^{N - 1} b_{i} E_{i + 1, i + 1 - k} + \sum_{i = 1}^{N - 1} b_{i} E_{i, i + 1 - k}, X = F_{1}^{(b_{1})} \dots F_{N - 1}^{(b_{N - 1})},$ with zero terms omitted when the column index is out of range. Now operations execute in decreasing $i$ . The leftmost source column in row $i$ of $D_{0}$ is at least $i + 1 - k$ , with at least $b_{i}$ letters there; the target entry at that column is zero. Therefore step 8.1 gives the same expansion with unique unit pivot at $D$ , and again $δ (D_{0}) = δ (D) - \sum_{i} b_{i}$ .

11.1step 4.1step 5.1step 6.1step 9.1step 10.1constructalgebra

Exact recursive preimages and surjectivity. Induct on $(δ (D), s (D))$ . For a diagonal matrix use $R_{D} = P_{μ}$ of step 5.1. In either non-diagonal case, $D_{0}$ has smaller first coordinate, while each $D^{'} ≺ D$ in steps 9.1–10.1 has a strictly smaller pair by step 6.1. With their already constructed preimages define $R_{D} = u_{D}^{- 1} (X R_{D_{0}} - \sum_{D^{'} ≺ D} a_{D^{'}} R_{D^{'}}) .$ This is a finite $A$ -linear combination of ordered words in simple divided powers and toral weight projections, with completely specified coefficient extraction; $u_{D}^{- 1}$ is only a Laurent monomial. Its image is exactly $ξ_{D}$ . The finite matrix inventory and the decreasing nonnegative pair guarantee termination. Since the $ξ_{D}$ are the actual integral Hom basis, their preimages prove surjectivity. Tensoring this expression with any $A$ -algebra gives the same basis and preimages; step 4.1 proves that the specialized commutant itself has this basis. Thus specialization commutes with the endomorphism construction, including nonflat specialization and roots of unity.

12.1F2step 2.1algebrastep 11.1

Identify the adjoint involution. In the coset basis equip $M^{μ}$ with diagonal form $⟨ m_{μ} T_{d}, m_{μ} T_{e} ⟩ = δ_{d e} q^{ℓ (d)}$ . Under $θ_{μ}$ this becomes ordinary tensor dot product. To see that the involution of [F2] is its adjoint, work temporarily over $Q (v)$ with the standard Hecke trace $t (T_{w}) = δ_{w, 1}$ , for which $t (T_{x}^{*} T_{y}) = δ_{x y} q^{ℓ (x)}$ by standard-basis multiplication. Write $p_{μ} = \sum_{w \in S_{μ}} q^{ℓ (w)}$ , so $m_{μ}^{2} = p_{μ} m_{μ}$ ; the coset form is $t (x^{*} y) / p_{μ}$ on $M^{μ}$ . If $f (m_{ν}) = y \in M^{μ} \cap H m_{ν}$ , then $f^{*} (m_{μ}) = y^{*}$ by the actual definition. For $x = m_{ν} h$ and $z = m_{μ} k$ , both adjoint pairings are $t (k^{*} y h)$ : use $m_{μ} y = p_{μ} y$ in the first, $y m_{ν} = p_{ν} y$ in the second, and the trace property. Thus $*$ is the dot adjoint. The conclusion concerns equality of integral matrices, so it holds over $A$ and every specialization; the temporary nonzero $p_{μ}$ is never divided in those rings.

13.1step 1.1step 12.1algebra

The assignment $τ (E_{i}) = F_{i}$ , $τ (F_{i}) = E_{i}$ , $τ (K_{j}) = K_{j}$ reverses and preserves the generic defining relations: reversing the toral relation replaces its root exponent by the negative one; the mixed commutator and its right side are both fixed; the Serre polynomial is carried to the reversed negative polynomial. It fixes the toral binomials and exchanges divided powers, hence preserves $U_{A}$ . Let $β = μ + a (ϵ_{i} - ϵ_{i + 1})$ . For a chosen raising subset, subtracting the reverse lowering exponent in step 1.1 gives $a (μ_{i} - μ_{i + 1}) + a^{2}$ , whereas $c_{β} / c_{μ} = v^{- a (μ_{i} - μ_{i + 1}) - a^{2}}$ . Thus the weighted dot form $⟨ x, y ⟩_{C} = ⟨ x, C y ⟩$ makes the raising and lowering matrices adjoint; all toral matrices are diagonal and self-adjoint. Its adjoint operator is $C^{- 1} s^{*} C$ . Equality on all integral generators, followed by reversal of products and linearity, proves $ρ (τ (u)) = C^{- 1} ρ (u)^{*} C$ for every $u$ .

14.1step 1.1step 2.1step 3.1algebraconstructstep 13.1

Square-root-free matrices. In step 2.1 the permutation-module word basis is $b_{w} = v^{inv (w)} e_{w}$ . If raising a subset $J$ of $a$ positions changes $w$ to $w^{'}$ , counting inversions involving a changed letter gives $inv (w) - inv (w^{'}) = \sum_{j \in J} (# i (w_{> j}) - # (i + 1) (w_{< j})) + (\binom{a}{2}) .$ The final correction counts pairs of changed positions, counted incorrectly by changing each letter in isolation. Add the exponent in step 1.1 and use $# (i + 1) (w_{< j}) + # (i + 1) (w_{> j}) = μ_{i + 1} - 1$ . The coefficient in the $b$ basis is $v^{a (a - μ_{i + 1}) + 2 \sum_{j \in J} # i (w_{> j})}$ . Scaling by $v^{a (μ_{i + 1} - a)}$ gives the displayed $q$ -coefficient. For lowering, the analogous inversion difference is $\sum_{j \in J} (# (i + 1) (w_{< j}) - # i (w_{> j})) + (\binom{a}{2})$ , giving coefficient $v^{a (a - μ_{i}) + 2 \sum_{j \in J} # (i + 1) (w_{< j})}$ and the asserted normalization. These define matrices over $B$ . Their Hecke commutation follows from step 3.1 after the injective substitution $B ↪ A$ , so all their polynomial commutator entries are zero already over $B$ .

15.1F2step 4.1step 6.1step 11.1step 14.1algebraconstruct∎

Over $B$ the permutation-basis Hecke formulas force equal coefficients for ascending and descending arrangements within each source block: an ascending word goes to its swap, a descending word to $q$ times its swap plus $(q - 1)$ times itself, and the equation $y T_{s} = q y$ uses only the unit $q$ . Thus the unweighted sums $\sum_{w \in W_{D}} b_{w}$ give the same orbit Hom basis, denoted $Φ_{D}$ . The cross-block inversion count $d (D) = \sum_{j < k, a > b} d_{a j} d_{b k}$ is constant on $W_{D}$ , so $Φ_{D} = v^{d (D)} ξ_{D}$ under extension to $A$ . Repeat steps 6.1–11.1 directly with the normalized operators of step 14.1. Their supports and greedy maximal matrices are identical, and every unique pivot is now a single power of $q$ , because every contributing subset coefficient is a power of $q$ . The recursion therefore uses only $B$ -coefficients and $q$ -unit pivots and proves generation over $B$ . The same orbit presentation and recursion work after every map $B \to R$ . Finally, adjoining $z$ with $z^{2} = q$ gives $R^{'} = R [z] / (z^{2} - q)$ , free of rank two over $R$ , with $z^{- 1} = q^{- 1} z$ ; it is faithfully flat even in characteristic two. Quantum assertions requiring a square root may be used there, while any specified map, equality, kernel, cokernel, or exact sequence already defined over $R$ descends by faithful flatness. The square-root-free generation proved here itself needs no such extension. At $r = 0$ every construction is the identity map of $A$ or $B$ , and at $N = 1$ all matrices are diagonal, so the base case covers both edge cases. No Choice principle, generic rank argument, or nonunit denominator is used for generation.

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Polynomial quantum highest vectors satisfy every divided-power highest relation

Statement

Put $A = Z [v, v^{- 1}]$ , and use the integral $U_{A} ({g l}_{N})$ of Integral divided-power tensor actions generate the type-A q-Schur algebra. Let $R$ be any commutative $A$ -algebra and $U_{R} = R \otimes_{A} U_{A}$ . Let $X$ be a left $U_{R}$ -module with a specified composition-label decomposition $X = ⨁_{μ \in Λ (N, r)} X_{μ},$ where $r \geq 0$ , $Λ (N, r)$ consists of nonnegative integer $N$ -tuples of total $r$ , every integral toral element acts on $X_{μ}$ by its integer-weight character $μ$ , and the positive and negative divided powers shift labels by their root degrees. Thus the toral rank-one binomial with indices $c, t$ acts by ${[\begin{matrix} μ_{i} - μ_{i + 1} + c \\ t \end{matrix}]}_{v}$ . This requirement uses the full integral toral characters, including specialized binomials; it does not recover labels from the specialized $K_{j}$ eigenvalues. Each $X_{μ}$ may be an arbitrary $R$ -module.

For a partition $λ \in Λ (N, r)$ and a vector $x \in X_{λ}$ killed by every $E_{i}^{(s)}$ , $s > 0$ , put $m_{i} = λ_{i} - λ_{i + 1}$ . Then $F_{i}^{(a)} x = 0 (1 \leq i < N, a > m_{i}) .$ These are all divided powers beyond the highest bound, including those whose quantum factorial vanishes after specialization. Consequently, if $X = U_{R} x$ , the highest-vector map gives a surjection of negative-half modules $U_{R}^{-} / \sum_{i = 1}^{N - 1} \sum_{a > m_{i}} U_{R}^{-} F_{i}^{(a)} ⟶ X, 1 ⟼ x .$ Whenever this quotient carries its integral highest-module action, with the same toral character and positive annihilation on $1$ , that map is a $U_{R}$ -module map. In particular it applies to the actual highest quotient used by the normalized q-Schur Weyl-cell comparison. No Schur factorization of $X$ is assumed. The relation proof and negative-half surjection are choice-free and hold at every specialization, over rings with torsion and zero divisors as well as fields.

Facts & Assumptions

Given: The rings, the integral quantum algebra, the specified nonnegative composition-label decomposition, and the highest vector in the Statement.

[F1]

The integral rank-one reordering identity and triangular spanning hold after every base change (Divided-power multiplication, toral binomials and the $E_{i}^{(a)} F_{i}^{(b)}$ reordering identity). On a highest vector of simple-coroot weight $m$ , reordering and the integer toral character give $E_{i}^{(s)} F_{i}^{(b)} x = {[\begin{matrix} m + s - b \\ s \end{matrix}]}_{v} F_{i}^{(b - s)} x$ for $b \geq s$ .

[F2]

The symmetric nonnegative quantum binomials are Laurent polynomials, are symmetric in their lower complementary indices, and satisfy both quantum Pascal recurrences (The quantum Pascal recurrences, the Gauss product formula and Gaussian integrality).

Proof

1.1givenF2algebraconstruct

Extend the symmetric binomial to integer top indices using its finite product in the generic variable. If $d \geq 1$ and $s \geq 0$ , negating every numerator quantum integer gives ${[\begin{matrix} - d \\ s \end{matrix}]}_{v} = (- 1)^{s} {[\begin{matrix} d + s - 1 \\ s \end{matrix}]}_{v}$ . The right side is a Laurent polynomial by [F2], so this equality defines its integral specialization without dividing a specialized factorial. Put $g_{d} (s) = {[\begin{matrix} d + s - 1 \\ d - 1 \end{matrix}]}_{v}$ ; symmetry makes the negative binomial $(- 1)^{s} g_{d} (s)$ , and $g_{d} (0) = 1$ .

2.1F2step 1.1algebraconstruct

Establish an exact recurrence for this sequence. For an indeterminate $Q$ , let $G_{n, k} (Q)$ be the ordinary Gaussian polynomial, characterized by $G_{n, k} (v^{2}) = v^{k (n - k)} {[\begin{matrix} n \\ k \end{matrix}]}_{v}$ . Both Pascal recurrences in [F2] give $G_{n, k} = G_{n - 1, k - 1} + Q^{k} G_{n - 1, k}$ , with out-of-range terms zero. Put $h_{d} (s) = G_{d + s - 1, d - 1}$ and $h_{d} (- 1) = 0$ . Thus $h_{1} (s) = 1$ and $h_{d + 1} (s) = h_{d} (s) + Q^{d} h_{d + 1} (s - 1)$ for $s \geq 0$ . Multiplying formal series coefficientwise yields $\sum_{s \geq 0} h_{d} (s) t^{s} = \prod_{j = 0}^{d - 1} (1 - Q^{j} t)^{- 1}$ by induction on $d$ . Each inverse here is the formal geometric series, so no coefficient-ring element is divided. Substitution $Q = v^{2}$ and $t \mapsto v^{- (d - 1)} t$ gives $\sum_{s \geq 0} g_{d} (s) t^{s} = \prod_{j = 0}^{d - 1} (1 - v^{2 j - d + 1} t)^{- 1} .$ Writing the denominator as $\sum_{j = 0}^{d} p_{j} t^{j}$ , its constant term is $p_{0} = 1$ and its top coefficient is $p_{d} = (- 1)^{d} v^{\sum_{j = 0}^{d - 1} (2 j - d + 1)} = (- 1)^{d}$ . Coefficient comparison therefore gives $\sum_{j = 0}^{d} p_{j} g_{d} (n - j) = 0$ for every $n \geq d$ , with a unit coefficient on the oldest term.

3.1step 1.1step 2.1algebra

A zero tail forces a zero initial coefficient over every ring. More precisely, if $z$ belongs to any $R$ -module and $g_{d} (s) z = 0$ for every $s \geq S$ , first use the recurrence of step 2.1 at $n = S + d - 1$ to obtain $p_{d} g_{d} (S - 1) z = 0$ when $S > 0$ . The other $d$ terms have indices at least $S$ and are zero. Since $p_{d} = (- 1)^{d}$ is a unit, $g_{d} (S - 1) z = 0$ . Repeat this finite backward argument until $g_{d} (0) z = z = 0$ . If $S = 0$ the conclusion already follows from $g_{d} (0) = 1$ . This argument never divides a quantum integer or assumes that a nonzero scalar acts injectively.

4.1givenF1step 1.1step 3.1algebra

Fix $i$ and $a > m_{i}$ , put $d = a - m_{i} > 0$ and $z = F_{i}^{(a)} x$ . The only surviving term of the rank-one reordering of $E_{i}^{(s)} F_{i}^{(a + s)} x$ has contraction index $s$ , since every other term ends in a positive divided power killing $x$ . Its toral factor acts on $x$ by ${[\begin{matrix} m_{i} - a \\ s \end{matrix}]}_{v}$ , so [F1] and step 1.1 give $E_{i}^{(s)} F_{i}^{(a + s)} x = (- 1)^{s} g_{d} (s) z .$ For all $s \geq S : = \max (0, λ_{i} - a + 1)$ , the putative composition label of $F_{i}^{(a + s)} x$ has $i$ th coordinate $λ_{i} - a - s < 0$ , hence this vector is zero by the given support. Thus $g_{d} (s) z = 0$ for every $s \geq S$ . Step 3.1 proves $F_{i}^{(a)} x = z = 0$ . This establishes every required divided relation independently, including $a$ larger than the characteristic or quantum characteristic.

5.1F1step 4.1constructalgebra

If $X = U_{R} x$ , triangular spanning in [F1] and the highest conditions give $X = U_{R}^{-} x$ : positive nonconstant terms kill $x$ and the middle toral terms evaluate by its character. The negative-half map $y \mapsto y x$ is onto, and step 4.1 makes each generator $F_{i}^{(a)}$ , $a > m_{i}$ , of the displayed left ideal vanish at $x$ . It therefore factors through exactly the stated negative quotient. This is a concrete quotient construction over $R$ , rather than a specialization of a generic surjectivity claim.

6.1givenF1step 5.1algebra∎

Suppose the quotient is supplied with the highest-module action described in the Statement. The map of step 5.1 commutes with negative divided powers by construction and with all toral elements by homogeneous degrees and the integer-weight characters. For a positive divided generator acting on a negative word, apply the same finite integral reordering identity in the quotient and in $X$ ; move the toral coefficients to the highest line and kill the positive tails. Each resulting negative term has the same coefficient and the same image in the two modules. Linearity proves commutation with every positive divided generator, hence with all of $U_{R}$ . Any such map is determined by its highest vector because the quotient is generated by the negative words. At $N = 1$ there are no simple-root relations and the same argument gives the rank-one cyclic quotient directly; the zero module and $r = 0$ are included. All scalar operations in the relation proof were Laurent-polynomial identities and multiplication by the units $(- 1)^{d}$ , so no flatness or Choice hypothesis is required.

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Integral quantum highest modules are the normalized q-Schur Weyl cells

Statement

Assume the Axiom of Choice, inherited only through the integral highest-lattice supplier. Put $A = Z [v, v^{- 1}]$ , $q = v^{2}$ , $N \geq 1$ , and $r \geq 0$ . Use the integral $U_{A} ({g l}_{N})$ , its surjection $ρ : U_{A} \to S_{A} (N, r)$ , and $τ (E_{i}^{(a)}) = F_{i}^{(a)}$ , $τ (K_{j}) = K_{j}$ of Integral divided-power tensor actions generate the type-A q-Schur algebra. For a partition $λ$ of $r$ with at most $N$ parts, padded by zeros, set $m_{i} = λ_{i} - λ_{i + 1}$ and $Q_{A} (λ) = U_{A}^{-} / \sum_{i = 1}^{N - 1} \sum_{a > m_{i}} U_{A}^{-} F_{i}^{(a)} .$ Equip this quotient with its integral highest-module action and highest vector $z = 1 + \sum U_{A}^{-} F_{i}^{(a)}$ , with $K_{j} z = v^{λ_{j}} z$ and the integral toral binomials acting by the corresponding integer-weight characters. Let $W_{A}^{λ}$ be the actual fixed-first-index right Weyl cell of The type-A q-Schur algebra and its Weyl modules, with semistandard basis $w_{T}$ , cellular form $B_{c e l l}$ and $h = w_{T^{λ}}$ . Give its underlying module the left action $u \cdot w = w ρ (τ (u)) .$ Then the unique highest-vector map $f : Q_{A} (λ) \to W_{A}^{λ}$ , $z \mapsto h$ , is an isomorphism. In particular $b_{T} = f^{- 1} (w_{T})$ is a complete integral semistandard basis, of weight $type (T)$ , and the displayed divided-right-power relations are the entire negative-half presentation. This does not identify $b_{T}$ with a particular ordered root monomial.

Set $c_{μ} = v^{- \sum_{j} (\binom{μ_{j}}{2})}$ . There is a unique homogeneous normalized $τ$ -contravariant form $B$ on $Q_{A} (λ)$ , explicitly $B (x z, y z) = [z] (τ (x) y z) .$ Here $[z]$ means projection onto the free highest-weight line followed by its coefficient in $z$ . It is integral and symmetric, and for homogeneous $a, b$ of weight $μ$ satisfies $B (a, b) = \frac{c_{λ}}{c_{μ}} B_{c e l l} (f (a), f (b)) .$ Different weights are orthogonal. Every ratio in this formula is a Laurent unit. The module, basis, quotient presentation, highest-coefficient formula and form comparison commute with arbitrary commutative $A$ -algebra base change, including fields at roots of unity and singular Gram matrices. Form radicals agree under $f$ in every such fibre; no assertion that radicals commute with arbitrary base change is made.

Over every commutative $A$ -algebra $R$ , every cyclic left $U_{R}$ -module with nonnegative composition-label support of total degree $r$ and the full integer toral characters, generated by a vector of weight $λ$ killed by every positive divided generator, is a quotient of $Q_{R} (λ)$ . Equivalently, the left Weyl cell is the polynomial highest-induced module $S_{R} \otimes_{S_{R}^{+}} R_{λ}$ , where $S_{R}^{+}$ is the image of the integral toral-positive subalgebra and $R_{λ}$ is its highest character. This includes every polynomial Schur module and the zero module. The full toral-character label convention is the one specified in Polynomial quantum highest vectors satisfy every divided-power highest relation; no factorization through the Schur quotient is assumed for this stronger cyclic-highest assertion.

Facts & Assumptions

Given: The ring, degree, partition, integral hyperalgebra and actual normalized cells in the Statement.

[F1]

The surjection and weight projections hold integrally and after arbitrary base change, and, with $C = \sum_{μ} c_{μ} 1_{μ}$ , the involutions obey $ρ (τ (u)) = C^{- 1} ρ (u)^{*} C$ (Integral divided-power tensor actions generate the type-A q-Schur algebra).

[F2]

The right cell has the semistandard weight basis, star-contravariant symmetric form and $B_{c e l l} (h, h) = 1$ . The canonical partition projection is $1_{λ} = φ_{T^{λ} T^{λ}}$ ; the cell product has coefficient $B_{c e l l}$ (The type-A q-Schur algebra and its Weyl modules).

[F3]

The displayed negative-half quotient is the free integral highest lattice, with finite free weight pieces and the finite Weyl character, with inherited Choice (Finitely generated free integral weight lattices and deformation of the classical weight character). For type $A_{N - 1}$ extend its labels to composition labels of fixed total $r$ by $K_{j}$ acting as $v^{μ_{j}}$ on weight $μ$ ; the integer-weight toral characters and root shifts give the stated ${g l}_{N}$ action. At $N = 1$ this is the free rank-one module directly.

[F4]

The finite type-A Weyl character is the bialternant Schur character, whose coefficient at $x^{μ}$ is $K_{λ μ}$ (Semistandard tableaux expand Schur characters). The Bender–Knuth bijections permute composition contents (Bender--Knuth involutions permute the weights of semistandard tableaux).

[F5]

Integral triangular spanning is $U_{A} = U_{A}^{-} U_{A}^{0} U_{A}^{+}$ ; the same spanning holds after every base change (Divided-power multiplication, toral binomials and the $E_{i}^{(a)} F_{i}^{(b)}$ reordering identity). Positive and negative divided generators shift composition labels by their root degrees.

[F6]

Every highest vector with nonnegative composition-label support satisfies all $F_{i}^{(a)} x = 0$ for $a > m_{i}$ , over every coefficient ring; integral reordering makes its negative-quotient map a full $U_{R}$ -module map (Polynomial quantum highest vectors satisfy every divided-power highest relation).

[C1]

The Axiom of Choice is used only as inherited by [F3] and [F4]. The finite cellular constructions and ring arguments below make no additional choice assumption.

Proof

1.1F2F4algebra

Tableau support. In any semistandard tableau, each column has at most $k$ entries in ${1, \dots, k}$ ; summing over columns gives $\sum_{j \leq k} μ_{j} \leq \sum_{j \leq k} λ_{j}$ . By [F4] the same inequality holds after permuting the content. Thus no weight $λ + a α_{i}$ , $a > 0$ , occurs. No weight $λ - a α_{i}$ occurs when $a > m_{i}$ : swapping its $i$ th and $(i + 1)$ st coordinates makes its first $i$ coordinates sum to $\sum_{j \leq i} λ_{j} + a - m_{i}$ , violating that inequality. Negative coordinates also cannot occur. The type- $λ$ tableau is uniquely $T^{λ}$ : the first content inequality is an equality forcing all $1$ 's into row one, and successively equality for $k$ forces all $k$ 's into row $k$ . Hence the top weight line is $A h$ .

2.1F1F2F3step 1.1constructalgebra

The highest map. The action in the Statement is a left action since $τ$ reverses products. It has the stated weight labels by [F1]. Step 1.1 kills every $E_{i}^{(a)} h$ for $a > 0$ and every $F_{i}^{(a)} h$ for $a > m_{i}$ ; the toral characters are those of weight $λ$ . Thus [F3] gives a highest-module map $f : Q_{A} (λ) \to W_{A}^{λ}$ . By the cell product and $B_{c e l l} (h, h) = 1$ , $h φ_{T^{λ} T} = w_{T}$ . Surjectivity in [F1] supplies an integral preimage under $ρ τ$ of each of these endomorphisms, so $h$ generates the whole cell and $f$ is onto.

2.2F1F2step 1.1algebraconstruct

Cellular highest induction over any $R$ . Put $I^{> λ}$ equal to the span of cellular shapes strictly dominating $λ$ . This ideal is generated by the $1_{η}$ for $η > λ$ : choose a finite order with larger shapes first, and use $φ_{S, T^{η}} 1_{η} φ_{T^{η}, T} \equiv φ_{S T}$ modulo higher shapes, whose coefficient is $1$ by [F2]. Induction absorbs those higher terms in the already generated ideal; conversely each $1_{η}$ lies in $I^{> λ}$ . The quotient $S_{R} 1_{λ} / I^{> λ} 1_{λ}$ has exactly the basis $φ_{T, T^{λ}}$ of shape $λ$ : any semistandard type $λ$ in shape $η$ forces $η ⊵ λ$ by the column count in step 1.1, and shape $λ$ has the single second tableau $T^{λ}$ . Star identifies this left cell with the right cell acted on by $s^{*}$ ; the weight-unit map sending its weight- $μ$ basis vector to $(c_{μ} / c_{λ}) w_{T}$ instead identifies it with the action $s \cdot w = w C^{- 1} s^{*} C$ of the Statement. Its highest vector maps to $h$ .

3.1F1F3F4F5step 2.1algebraconstruct

Integral isomorphism. In type $A_{N - 1}$ the finite Weyl alternating quotient of [F3], on composition labels of fixed total $r$ , is $\det (x_{i}^{λ_{j} + N - j}) / \det (x_{i}^{N - j})$ : the Weyl group permutes the $N$ coordinates, and multiplying numerator and denominator by the common monomial removes the half-sum shifts. This is the bialternant of [F4]. Consequently, for each weight $μ$ , [F3] and [F4] make the source and target finite free of the same rank $K_{λ μ}$ . The map in step 2.1 is graded, hence onto on each weight. Choose finite free bases and write its square matrix $D$ ; lifting the finitely many target basis vectors gives a matrix $P$ with $D P = I$ . Therefore $\det (D) \det (P) = 1$ , and the adjugate formula gives an inverse matrix over $A$ . Zero-rank weights have the unique empty inverse. Only finitely many weights occur, so these inverses assemble to $f^{- 1}$ . This proves the entire quotient presentation and the basis $b_{T}$ . Concretely a preimage under the surjection of $φ_{T^{λ} T}$ , followed by integral triangular spanning and highest-character evaluation, gives a representative of $b_{T}$ in $U_{A}^{-}$ ; different preimages give the same quotient class. No generic-rank argument replaces the proved integral surjection.

4.1F1F5F6step 2.2step 3.1constructalgebra

Let $X = U_{R} x$ be the cyclic highest module with the composition support and full toral characters in the Statement. By [F6], every divided highest relation holds, and the highest-vector map $Q_{R} (λ) \to X$ is a surjective $U_{R}$ -module map. The integral isomorphism of step 3.1 and its inverse remain inverses after arbitrary base change. Thus this is a quotient of the actual normalized Weyl cell. In particular $X$ factors through $S_{R}$ : [F1] makes $\ker ρ$ stable under $τ$ , so it annihilates that cell and every quotient. No factorization of $X$ was assumed. Via the unit identification of step 2.2, the corresponding left-cell quotient map preserves the highest generator. Apply this argument also to $Y = S_{R} \otimes_{S_{R}^{+}} R_{λ}$ : its canonical generator has the stated toral character, is killed by all positive divided powers, and is fixed by $1_{λ}$ . The last assertion follows because the integral toral projection $\prod_{j} [\begin{matrix} K_{j}; 0 \\ λ_{j} \end{matrix}]$ is $1_{λ}$ and has highest-character value $1$ . Thus $Y$ is a quotient of the left cell. Conversely the highest vector of that cell defines a surjection $Y$ onto it. Both compositions fix the cyclic highest generators, hence are identities, proving the induction equivalence. This is a direct cellular proof of the universal property.

4.2F1F2step 3.1algebraconstruct

Construct the form on the actual cell first. Give weight $μ$ the scalar $d_{μ} = c_{λ} / c_{μ}$ times $B_{c e l l}$ and pair distinct weights by zero. Weight spaces are orthogonal for $B_{c e l l}$ because its self-adjoint projections $1_{μ}$ are orthogonal; thus the new form is symmetric and integral and has value $1$ on $h$ . For $s = ρ (u)$ , [F1] gives $ρ (τ (u))^{*} = C s C^{- 1}$ . On a homogeneous component, right multiplication by this conjugate contributes $c_{ν} / c_{μ}$ when $s$ takes right weight $ν$ to right weight $μ$ . Hence $d_{ν} (c_{ν} / c_{μ}) = d_{μ}$ , and cellular contravariance gives $B (u \cdot a, b) = B (a, τ (u) \cdot b)$ . Linearity handles sums of homogeneous components. Pull this form back by the integral isomorphism $f$ .

5.1F3step 1.1step 3.1step 4.2algebra

Highest coefficient and uniqueness. For $x, y \in U_{A}^{-}$ , contravariance gives $B (x z, y z) = B (z, τ (x) y z)$ . Orthogonality and the top line $A z$ make the last expression exactly $[z] (τ (x) y z)$ . This proves the displayed coefficient construction descends through the complete quotient, including both arguments, because the already constructed form is defined on that quotient. Conversely the same equality is forced for any homogeneous normalized contravariant form, and negative words span by [F3], proving uniqueness. The construction has used only Laurent units and the integral action; it has not restricted a tensor form or divided by a tensor highest norm.

6.1C1step 3.1step 4.2step 5.1step 2.2step 4.1algebra∎

Base change and singular forms. Tensor the proved isomorphism and its inverse with any $R$ ; the quotient is its specialized presentation by right exactness of tensor product, the basis remains the specialized $b_{T}$ , and the integral form matrices and highest projection specialize entrywise. The weight factors $d_{μ}$ remain units. Thus a vector pairs to zero for the specialized form exactly when its image pairs to zero for the specialized cellular form, even if the Gram determinant is zero. Steps 2.2 and 4.1 used the same free cellular basis and unit coefficients over $R$ , so their universal property also holds without flatness. At $r = 0$ the unique empty highest cell and quotient are $R$ with form $1$ ; at $N = 1$ the only degree- $r$ cell and quotient are likewise rank one. These prove all claims with only the inherited Choice of [C1].

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Polynomial quantum Schur first levels: Weyl filtration and the actual Hecke corner

Statement

Put $A = Z [v, v^{- 1}]$ , $q = v^{2}$ , and use the actual integral tensor action and the finite normalized-cell constructions of Integral divided-power tensor actions generate the type-A q-Schur algebra and Integral quantum highest modules are the normalized q-Schur Weyl cells. For $N > d \geq 1$ , $λ ⊢ d$ , and $W_{N} (λ)$ its Weyl module, the first level is $W_{N} (λ)^{1} = ⨁_{α_{N} = 1} W_{N} (λ)_{α}$ . It has an integral lower-Schur Weyl filtration with one factor $W_{N - 1} (λ (i))$ for each removable row $i$ , ordered by decreasing dominance of $λ (i)$ . This filtration and its quotient identifications commute with arbitrary coefficient-ring base change.

Over every field with a chosen unit $v$ , write $\nabla = W^{τ}$ and $L = W / rad B_{τ}$ , with the suppliers' abstract highest norm-one form. Then $Hom (W_{N - 1} (μ), \nabla_{N} (λ)^{1})$ is one dimensional exactly for $μ = λ (i)$ , and zero otherwise. The same Hom-space with target $L_{N} (λ)^{1}$ has dimension at most one and vanishes off these labels. Equivalently every lower-Levi primitive line of $L_{N} (λ)^{1}$ has one of these weights, with dimension at most one; primitive means annihilated by all positive divided generators.

For every commutative ring $R$ , unit $q$ , and left $S_{R} (N, d)$ -module $V$ , its first level carries the actual lower algebra $S_{R} (N - 1, d - 1)$ action induced on permutation modules, and there is a natural isomorphism $e_{l o w e r} V^{1} ≅ {Res}_{H_{d - 1}}^{H_{d}} e V .$ The quantum Levi action, when a square root is chosen, agrees with this action under the actual tensor identification and the suppliers' weight-unit conjugation. This statement neither equates first level with a simple-root layer nor asserts that an exact corner functor preserves socles.

Facts & Assumptions

Given: The rings, ranks, actual permutation-module Schur algebras, weight projections, and normalized Weyl modules of the Statement.

[F1]

The actual integral divided tensor action surjects onto $S$ , its weight blocks identify tensor space with the permutation-summand realization, and this survives arbitrary base change (Integral divided-power tensor actions generate the type-A q-Schur algebra).

[F2]

The finite cellular construction and weight-unit transport are the explicit arguments in the cellular-highest-induction, polynomial-universality and direct-cell-form paragraphs of Integral quantum highest modules are the normalized q-Schur Weyl cells. The integral highest-lattice quotient and its global-basis isomorphism are not used. Step 1.1 below proves the exact needed cell interfaces without their inherited Choice. Its universal assertion concerns actual Schur modules; the lower modules used in steps 3.1 and 5.1 carry that action by steps 1.2 and 7.1. The broader finite-label-support highest universality is a separate assertion of the supplier and is not used here.

[F3]

The Weyl cell is free with the actual semistandard-tableau basis, weight ranks are its tableau counts, and all weights are dominated by $λ$ (The type-A q-Schur algebra and its Weyl modules). Permuting composition contents preserves their tableau counts by the finite bijections of Bender--Knuth involutions permute the weights of semistandard tableaux. The divided-power reordering establishes triangular spanning integrally; it does not require the freeness or full PBW-rank theorem (Divided-power multiplication, toral binomials and the $E_{i}^{(a)} F_{i}^{(b)}$ reordering identity).

Proof

1.1F1F2F3algebraconstruct

Establish the finite cell interfaces directly. Give the actual right Weyl cell the left action $u \cdot w = w ρ (τ (u))$ . Its canonical vector $h$ is highest: the semistandard weight bounds exclude $λ + a α_{i}$ , and its toral character is $λ$ ; the canonical cell form has $B_{c e l l} (h, h) = 1$ . The cell law gives $h φ_{T^{λ} T} = w_{T}$ , and [F1]'s integral surjection makes it cyclic highest. For an actual polynomial Schur module $X = U^{-} x$ cyclic highest of weight $λ$ , triangular spanning gives weights in $λ - Q_{+}$ , so every partition projection $1_{η}$ with $η > λ$ annihilates it. The higher cellular ideal is generated by these projections: in decreasing shape order $φ_{S, T^{η}} 1_{η} φ_{T^{η}, T} \equiv φ_{S T}$ modulo higher shapes, with coefficient one. Hence that ideal kills $X$ . Thus $S 1_{λ} \to X$ , $s 1_{λ} \mapsto s x$ , factors through the actual left Weyl cell $S 1_{λ} / I^{> λ} 1_{λ}$ ; its basis is exactly the shape- $λ$ columns because semistandard type $λ$ requires shape at least $λ$ . The unit diagonal $C = \sum_{μ} c_{μ} 1_{μ}$ , $c_{μ} = v^{- \sum_{j} (\binom{μ_{j}}{2})}$ , of [F1] transports this left-star cell to the displayed left quantum action, giving its universal highest property. On its actual weight- $μ$ vectors define $B = (c_{λ} / c_{μ}) B_{c e l l}$ and different-weight pairings zero. The exact identity $ρ (τ (u)) = C^{- 1} ρ (u)^{*} C$ makes this symmetric integral form $τ$ -contravariant: the right-action weight factor $c_{ν} / c_{μ}$ is canceled by $(c_{λ} / c_{ν}) (c_{ν} / c_{μ}) = c_{λ} / c_{μ}$ . Its highest value is one. All factors are units, so its radical is precisely the actual cell radical and its simple quotient is self-dual. These are finite cellular and matrix arguments, valid after arbitrary base change and with no use of the free global highest lattice.

1.2givenF3algebraconstruct

Construct the corner over the original $R, q$ . Put $M_{0} = ⨁_{α \in Λ (N - 1, d - 1)} m_{α} H_{d - 1}$ . Length-additive parabolic coset bases identify $M_{1} = ⨁_{α} m_{(α, 1)} H_{d}$ with $M_{0} \otimes_{H_{d - 1}} H_{d}$ , by $m_{α} h \otimes a \mapsto m_{α} h a$ . Hence $f \mapsto f \otimes 1$ , extended by zero on the other summands, embeds the lower Schur algebra with identity $p = \sum_{α_{N} = 1} 1_{α}$ . It is injective because the unit parabolic coset supplies a direct summand. The lower regular composition induces the upper regular type $ω_{*} = (1^{d - 1}, 0^{N - d}, 1)$ , so $e_{l o w e r} p V = e_{*} V$ . On that summand the lower Hecke corner acts by left multiplication by exactly $h \in H_{d - 1}$ on $H_{d}$ , since $h \otimes a \mapsto h a$ .

2.1F2F3step 1.1algebraconstruct

Use the root convention $F_{i, i + 1} = F_{i}$ and $F_{i, j} = v F_{i + 1, j} F_{i} - F_{i} F_{i + 1, j}$ . In the negative algebra, its last-simple-root-degree-one part is spanned by $x F_{i, N}$ with $x$ in the negative lower Levi. Here is a direct spanning proof, requiring no specialized PBW independence. Every divided-simple-generator word of this degree has exactly one $F_{N - 1}$ , exponent one; all other generators belong to the lower Levi. Move each factor to its right past the current root using $F_{i, N} F_{i - 1}^{(s)} = v^{- s} F_{i - 1}^{(s)} F_{i, N} + v^{- 1} F_{i - 1}^{(s - 1)} F_{i - 1, N}, F_{i, N} F_{i}^{(s)} = v^{s} F_{i}^{(s)} F_{i, N} .$ Other lower simple divided generators commute with $F_{i, N}$ . The first identity follows from $v F_{i, N} F_{i - 1} = F_{i - 1} F_{i, N} + F_{i - 1, N}$ and $F_{i - 1, N} F_{i - 1} = v F_{i - 1} F_{i - 1, N}$ : induction on ordinary powers gives coefficient $v^{- 1} [s]$ , canceled with $[s]!$ in the generic algebra, leaving the displayed integral identity. The second and remaining commutations follow by inserting the root recurrence into the rank-two Serre identity and then inducting on the interval length; the only noncommuting lower adjacent endpoint is $i - 1$ , already treated. Each move removes one right factor, so it terminates. Triangular spanning and the highest-vector relations now show $W^{1} = \sum_{i} U_{N - 1}^{-} F_{i, N} z_{λ}$ . If $λ_{i} = λ_{i + 1}$ , the weight $λ - ϵ_{i} + ϵ_{i + 1}$ is absent: its sorted weight strictly dominates $λ$ , contrary to the tableau support. Thus $F_{i} z_{λ} = 0$ and the root recurrence gives $F_{i, N} z_{λ} = - F_{i} F_{i + 1, N} z_{λ}$ . Iteration removes all nonremovable rows. Rows beyond the height contribute zero by nonnegative weight support. Therefore the claimed removable-row generators span integrally.

3.1F2F3step 1.1step 2.1algebra

Order these rows as $i_{1} > \dots > i_{t}$ , so $μ_{j} = λ (i_{j})$ decreases in dominance, and let $W_{j}$ be the negative lower-Levi span of the first $j$ generators, with $W_{0} = 0$ . Inductively suppose $W_{j - 1}$ is invariant. In $W^{1} / W_{j - 1}$ every weight lies below a remaining generator weight by step 2.1, and $μ_{j}$ is their maximum. Every positive divided generator kills the image of $F_{i_{j}, N} z_{λ}$ , since its target weight is strictly above $μ_{j}$ . Triangular spanning makes its negative span invariant. The actual universal highest property gives a surjection $W_{N - 1} (μ_{j}) \to W_{j} / W_{j - 1}$ . This proves invariance and these integral surjections at each stage, without dividing a quantum integer.

4.1F3step 3.1algebra

Exactly one entry $N$ in a semistandard $λ$ -tableau lies at a removable corner, and deleting it gives a semistandard $λ (i)$ -tableau on $1, \dots, N - 1$ . Conversely appending $N$ at that corner preserves semistandardness. Consequently the actual free ranks satisfy $rank W^{1} = \sum_{j} rank W_{N - 1} (μ_{j})$ . Over $Q (v)$ the surjections of step 3.1 and this dimension equality force equality of every quotient dimension. The kernel of each integral surjection lies in the free $A$ -module $W_{N - 1} (μ_{j})$ and has generic rank zero; injectivity into its fraction-field extension therefore makes that kernel zero. Each integral quotient is thus the actual free lower Weyl module. Its freeness makes the successive short exact sequences split as $A$ -module sequences, so arbitrary base change preserves the filtration injections, quotients and surjections. Generic dimensions were used only to prove kernels of explicit integral surjections vanish, not to replace a specialized construction.

5.1F2step 2.1step 4.1algebra

Over a field let $B^{-}$ denote the actual lower Borel image. A $B^{-}$ -map $W^{1} \to F_{μ}$ is determined by its values at the generators of step 2.1; their weights are distinct, so its dimension is at most one and is zero off $μ_{j}$ . At $μ_{j}$ , quotient by $W_{j - 1}$ . This quotient has no weight above $μ_{j}$ , its $μ_{j}$ -space is the one-dimensional generator line by step 4.1, and later generator weights lie strictly below it. Taking that generator's coefficient, and zero on every other weight, is a well-defined nonzero $B^{-}$ -map, since a nonconstant negative operator cannot reach $μ_{j}$ . This proves dimension exactly one. The universal highest property gives $Hom (W_{N - 1} (μ), X) = {Hom}_{B^{+}} (F_{μ}, X)$ for every polynomial $X$ : maps are exactly primitive vectors of that weight. Taking contravariant duals identifies this with ${Hom}_{B^{-}} (X^{τ}, F_{μ})$ . With $X = (W^{1})^{τ}$ this proves the first Hom assertion. With $X = L^{1}$ , self-duality and the injection $L^{1} ↪ (W^{1})^{τ}$ give the upper bound and label restriction. For a primitive vector $x$ of any composition weight, the functional $y \mapsto B (x, y)$ on $L^{1}$ , pulled back to $W^{1}$ , is nonzero by the nondegenerate homogeneous form and kills every nonconstant negative lower operator by contravariance. Thus it is a $B^{-}$ -map to the character of that weight. The same generator bound excludes all other primitive weights, without presupposing their dominance. The form's toral weight orthogonality ensures first-level projection commutes with this duality.

6.1F2step 5.1algebra

Record the precise coefficient test needed by lowering arguments. If $L^{1} \to \nabla_{N - 1} (λ (i))$ is nonzero, compose with $W^{1} \to L^{1}$ and with the highest-weight coefficient in $\nabla$ . The resulting $B^{-}$ -functional is nonzero, hence nonzero at $F_{i, N} z_{λ}$ by step 5.1. Every nonempty cut product $F_{i, b_{1}} F_{b_{1}, b_{2}} \dots F_{b_{s}, N} z_{λ}$ has a nonconstant lower-Levi factor to the left and is killed by this functional. Multiplying factors by the unit toral normalization used for $\hat{F}$ changes only their nonzero scalar factors on these fixed weight vectors. Thus the same zero/nonzero test applies to the normalized cut products. This argument needs neither an inherited tensor norm nor a claimed full-Hom surjection at singular parameters.

7.1F1F2step 1.2algebra∎

The summands of types $ω_{*} = (1^{d - 1}, 0^{N - d}, 1)$ and $ω = (1^{d}, 0^{N - d})$ are both the actual regular module $H_{d}$ . Their identity isomorphisms give block matrix units $u, v$ in $S$ with $u v = e_{*}$ and $v u = e$ . Multiplication by $u$ and $v$ consequently identifies $e V$ with $e_{*} V$ , transporting every lower Hecke operator to its actual upper corner inclusion. This proves the natural restriction isomorphism, including $d = 1$ , whose lower algebra and corner are $R$ . In the tensor realization of [F1], on words containing exactly one $N$ every lower divided generator leaves that position untouched and applies its subset-sum formula to the remaining ordered positions. These are exactly the induced operators of step 1.2 under the actual tensor/permutation intertwiner. Weight projections agree as well, so [F1]'s lower surjectivity proves equality of the full lower actions. The weight-unit diagonal conjugation of [F2] restricts to this subspace and intertwines the same actions after left/right conversion. This verifies the quantum/cellular comparison at every specialization by explicit generators.

5 · Examples, counterexamples and false statements

None yet.