ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck pass

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Normal and good nodes for the partition (6,4,4,3,2)

Example

Let $F$ be a field and $q \in F^{\times}$ , and use the Hecke algebra $H (r)$ with $(T_{j} - q) (T_{j} + 1) = 0$ (The generic type-A Hecke algebra). Here $ℓ = e (F, q)$ is its quantum characteristic (The quantum characteristic of an invertible Hecke parameter), taken in turn to be $3$ or $4$ . The case $ℓ = 4$ means a parameter of order $4$ and is not a field of characteristic $4$ ; both cases occur over $F = C$ with $q$ a primitive root of the specified order. Use Brundan's restricted DJ labels: $S^{λ}$ here denotes the ordinary right Specht module labelled by $λ^{'}$ and $D^{λ}$ its simple head, as in his §2.1. Right modules may be carried to left modules through the Hecke anti-involution fixing the generators; this preserves socles and direct sum decompositions. These source labels must not be silently identified with the same-shape Murphy labels of the group-theoretic suppliers.

Work Brundan's Example 2.8 (complete 34-page author manuscript, p. 5) in his conventions: $λ = (6, 4, 4, 3, 2)$ , padded by $λ_{6} = 0$ with $n = 6$ , is $ℓ$ -restricted for $ℓ = 3$ and $ℓ = 4$ , residues are $r - c m o d ℓ$ in the source's orientation (the negative of the $c - r$ convention of this page, Residue signatures, normal, good, conormal and cogood nodes), $R (λ) = {1, 3, 4, 5}$ is the set of removable rows, and for $i \in R (λ)$ one compares $B_{i, 6} (λ) \cap R (λ)$ with $C_{i, 6} (λ) \cap R (λ)$ in the lattice order $↓$ , where $B_{i, j} (λ) = {k \in [i . . j) : {res}_{ℓ} (i, λ_{i}) = {res}_{ℓ} (k + 1, λ_{k + 1} + 1)}, C_{i, j} (λ) = {k \in (i . . j) : {res}_{ℓ} (i, λ_{i}) = {res}_{ℓ} (k, λ_{k})},$ and $A ↓ B$ means that there is an injection $θ : A \to B$ with $θ (a) \leq a$ for all $a$ .

For $ℓ = 3$ the removable nodes have residues $1, 2, 1, 0$ for $i = 1, 3, 4, 5$ and $(B \cap R, C \cap R) = (\emptyset, {4}), ({4, 5}, \emptyset), (\emptyset, \emptyset), (\emptyset, \emptyset);$ hence $R^{n o r m a l} (λ) = {1, 4, 5}$ and, since rows $4$ and $1$ share residue $1$ , $R^{g o o d} (λ) = {1, 5}$ . For $ℓ = 4$ the residues are $3, 3, 1, 3$ and $(B \cap R, C \cap R) = (\emptyset, {3, 5}), (\emptyset, {5}), ({5}, \emptyset), (\emptyset, \emptyset);$ hence row $4$ is normal-refuted by ${5} ↓̸ \emptyset$ and $R^{n o r m a l} (λ) = {1, 3, 5}$ , and since rows $1, 3, 5$ share residue $3$ , $R^{g o o d} (λ) = {1}$ . By the branching theorems the restriction of $D^{λ}$ to $H (r - 1)$ is therefore decomposable for $ℓ = 3$ (socle $D^{(5, 4, 4, 3, 2)} \oplus D^{(6, 4, 4, 3, 1)}$ ) and indecomposable for $ℓ = 4$ (socle $D^{(5, 4, 4, 3, 2)}$ ).

Facts & Assumptions

Given: The partition $λ = (6, 4, 4, 3, 2)$ of $r = 19$ , the field and invertible Hecke parameters with quantum characteristics $ℓ = 3$ and $ℓ = 4$ , and Brundan's index sets and lattice order as displayed.

[F1]

The source's definitions: $R (λ) = {i : λ_{i} \neq λ_{i + 1}}$ is the set of removable rows, the node $(i, λ_{i})$ has residue $i - λ_{i} m o d ℓ$ , $R^{n o r m a l} (λ) = {i \in R (λ) : B_{i, n} (λ) ↓ C_{i, n} (λ)}$ with $n = 6$ , and $R^{g o o d} (λ) = {i \in R^{n o r m a l} (λ) :$ no $j \in R^{n o r m a l} (λ)$ with $j < i$ and the same residue $}$ ; there is at most one good row per residue (Brundan §2.2-2.4 and Remark 2.7, printed pp. 3-5). These are this page's normal and good nodes on the same partition, with residue names negated; Remark 2.4(II) instead compares Brundan's convention with Kleshchev's older convention on the transpose (Residue signatures, normal, good, conormal and cogood nodes).

[F2]

The actual general-Hecke endpoint General Hecke residue branching endpoints from normalized Schur cells gives the multiplicity-free good-deletion socle for Murphy restricted labels. Brundan's source-labelled simple here is $D_{s o u r c e}^{λ} ≅ D_{M} (λ)^{θ}$ , where $θ (T_{j}) = q - 1 - T_{j}$ : this is the actual unit-scaled ordinary module/form dictionary applied at ordinary shape $λ^{'}$ (Ordinary Specht modules are the signed conjugate Murphy cells). The automorphism respects the tower, so twisting both sides transfers the socle and all direct-sum conclusions without renaming a Murphy simple as an untwisted source simple. Distinct good-deletion source simples are separated by the central operator $θ (\sum_{a = 1}^{r - 1} L_{a})$ : on $D_{M} (μ)^{θ}$ its scalar is the actual Murphy scalar $\sum_{A \in [μ]} [c (A)]_{q}$ of Central Hecke Jucys–Murphy scalars separate q-Schur residue contents. For two deletions of different residues their scalar difference is the difference of their distinct q-contents, hence nonzero. Polynomial primary projectors in this central operator separate the full restricted module, not merely its socle. This proves the required separation without a sufficient same-content block-classification premise.

[F3]

A nonzero finite-dimensional module has nonzero socle: choose a nonzero submodule of least positive dimension. Thus a module with simple socle is indecomposable, because the socle of a direct sum is the direct sum of the summand socles (The radical, socle, head, and Loewy series of a finite-dimensional module). No self-duality of cell modules is assumed.

Verification

technique · direct

1.1givenF1

Removable rows and residues. $λ = (6, 4, 4, 3, 2)$ has consecutive differences $2, 0, 1, 1, 2$ , so $R (λ) = {1, 3, 4, 5}$ and $λ$ is $ℓ$ -restricted for $ℓ = 3$ and $ℓ = 4$ ; the residues $i - λ_{i}$ are, modulo $3$ , $1, 2, 1, 0$ for $i = 1, 3, 4, 5$ , and modulo $4$ , $3, 3, 1, 3$ for the same rows.

2.1step 1.1F1

Pad by $λ_{6} = 0$ . For $k = 1, \dots, 5$ , the integers $k - λ_{k + 1}$ are $(- 3, - 2, 0, 2, 5)$ and $k - λ_{k}$ are $(- 5, - 2, - 1, 1, 3)$ . To compute $B_{i, 6}$ use indices $i \leq k < 6$ with the first integer congruent to $i - λ_{i}$ ; for $C_{i, 6}$ use $i < k < 6$ with the second integer congruent to $i - λ_{i}$ , then intersect with $R = {1, 3, 4, 5}$ . For $i = 1, 3, 4, 5$ , respectively, this gives $(B \cap R, C \cap R) = (\emptyset, {4}), ({4, 5}, \emptyset), (\emptyset, \emptyset), (\emptyset, \emptyset)$ modulo $3$ , and $(\emptyset, {3, 5}), (\emptyset, {5}), ({5}, \emptyset), (\emptyset, \emptyset)$ modulo $4$ . These are exactly the corrected displayed pairs.

3.1step 1.1step 2.1F1

For $ℓ = 3$ , rows $1, 4, 5$ pass the lattice test: their demand sets are empty. Row $3$ fails because ${4, 5} ↓̸ \emptyset$ . Hence $R^{n o r m a l} = {1, 4, 5}$ , with residues $1, 1, 0$ ; taking the least normal row in each residue gives $R^{g o o d} = {1, 5}$ . For $ℓ = 4$ , rows $1, 3, 5$ pass, while row $4$ fails because ${5} ↓̸ \emptyset$ . The normal rows all have residue $3$ , so $R^{g o o d} = {1}$ .

4.1step 3.1F2

$ℓ = 3$ : decomposability. By [F2] the actual source-labelled restriction $M = D^{λ} ↓_{H (r - 1)}$ has the two simple socle factors $D^{λ (1)} = D^{(5, 4, 4, 3, 2)}$ and $D^{λ (5)} = D^{(6, 4, 4, 3, 1)}$ . Their removed Murphy node contents are $5$ and $- 3$ , of residues $2$ and $0$ modulo $3$ , so [F2]'s central operator has distinct scalars on them. Its minimal polynomial on $M$ has distinct linear factors at those two scalars; the polynomial CRT produces a central primary projector $b$ selecting the first and annihilating the second (other factors are included in $1 - b$ ). Thus $M = b M \oplus (1 - b) M$ is a sum of submodules and both are nonzero because they contain the respective simple socle factors. This proves decomposability by actual full-module projections.

4.2step 3.1F2F3

$ℓ = 4$ : indecomposability. By the proved branching input [F2], the socle of $D^{λ} ↓_{H (r - 1)}$ is the single simple module $D^{λ (1)} = D^{(5, 4, 4, 3, 2)}$ . By [F3] this makes the restriction indecomposable: a decomposition into two nonzero modules would give two nonzero summand socles. No cell-module self-duality is used.

5.1step 1.1step 2.1step 3.1step 4.1step 4.2F2∎

At $ℓ = 4$ , rows $1, 3, 5$ have residue $3$ and are all normal, whereas row $4$ has residue $1$ and fails ${5} ↓ \emptyset$ . At $ℓ = 3$ , both removable rows of residue $1$ are normal. Thus these tables do not exhibit same-residue removable nodes of different normality; they do show that good rows are the least normal rows of each residue, rather than rows whose residue occurs once. The branching conclusions in steps 4.1–4.2 use the actual general-Hecke supplier and source-label transport [F2], with $ℓ = 4$ socle label $D^{(5, 4, 4, 3, 2)}$ .

The general-Hecke input is the actual constructed endpoint supplier in [F2], including quantum characteristic four. The source's negative residue orientation affects color names but leaves residue equality and the selected row unchanged; the actual theta-twist dictionary transfers the displayed module labels and decomposability conclusions. The finite tables verify the concrete signature calculation; the independent endpoint proof supplies its module-theoretic meaning.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

42 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.