Ordinary Hecke Specht Polytabloids and Signed Murphy Comparison: Examples

1 · Prerequisites

2 · Summary

Objective. This companion works the comparison of ordinary-hecke-specht-polytabloids-and-signed-murphy-comparison on the smallest explicit instances. Every item is a dependency leaf, every computation is finite and exact over $Z [q^{\pm 1}]$ before specialization, and no choice principle is used.

Examples and counterexample

The row and column Specht conventions in rank three computes the four rank-one modules of the rank-three pair: $P ((3))$ and $C_{M} ((3))$ act by $q$ , $P ((1^{3}))$ and $C_{M} ((1^{3}))$ act by $- 1$ , the self-form values are $1, q^{- 3} p_{3} (q), p_{3} (q), 1$ with $p_{3} (q) = (1 + q) (1 + q + q^{2})$ , and the comparison units are $u_{(3)} = 1$ and $u_{(1^{3})} = q^{3}$ ; at $q = 1$ the values are $1, 6, 6, 1$ and $θ$ exchanges trivial and sign.
The two-tableau pairing at a singular parameter computes the full two-tableau comparison for $λ = (2, 1)$ : the row action matrices $R_{1}, R_{2}$ , the polytabloid basis with unit embedding minor, the DJ Gram matrix $B = (\begin{matrix} q + 1 & q^{2} \\ q^{2} & q^{3} + q \end{matrix})$ , the Murphy quotient with $G = (\begin{matrix} q + 1 & - 1 \\ - 1 & q^{2} + 1 \end{matrix})$ , the isomorphism matrix $A$ with $A G A^{T} = B$ , and the perfect cross-pairing of determinant $- q^{2}$ . At the singular parameter $q = - 1$ the forms remain nonsingular, and at $q = 1$ in characteristic two they reduce to $(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})$ with zero radical, separating a singular algebra parameter from degeneracy of this module form.
Untwisted conjugation changes the module refutes the untwisted conjugation: over $Q$ at $q = 1$ and $n = 3$ the Murphy cell $C_{M} ((1^{3}))$ is the sign module while $P ((3))$ and its dual are trivial, so $C_{M} (λ^{'}) ≅ P (λ)^{\circ}$ without $θ$ fails; the correct formula is $C_{M} (λ^{'}) ≅ P (λ)^{θ}$ , and the conjugate chain reads $P (λ^{'}) ≅ P (λ)^{\circ, θ}$ .

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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Untwisted conjugation changes the module

Statement refuted

Refuted claim: the same-shape duality $C_{M} (λ) ≅ P (λ)^{\circ}$ of Murphy cells are the same-shape contragredient Specht duals may be applied without a sign twist, i.e. one may conclude $C_{M} (λ^{'}) ≅ P (λ)^{\circ}$ directly.

This is false. Over $Q$ at $q = 1$ and $n = 3$ the Murphy cell module $C_{M} ((1^{3}))$ is the sign module, in which $T_{i}$ acts by $- 1$ , while the ordinary polytabloid module $P ((3))$ and its contragredient dual are trivial, in which $T_{i}$ acts by $+ 1$ . Hence $C_{M} (λ^{'}) ≅ P (λ)^{\circ}$ without $θ$ fails in general. The correct bridge is $C_{M} (λ^{'}) ≅ P (λ)^{θ}$ of Ordinary Specht modules are the signed conjugate Murphy cells; combining it with same-shape duality gives $P (λ^{'}) ≅ P (λ)^{\circ, θ}$ , which is a different formula. No choice principle is used.

Facts & Assumptions

Given: The field $Q$ , the specialization $q = 1$ of the type- $A$ Hecke algebra $H$ , the polytabloid module $P ((3)) = z_{(3)} H$ with its DJ form, the Murphy cell module $C_{M} ((1^{3}))$ with its form, and the comparison theorem Ordinary Specht modules are the signed conjugate Murphy cells together with its corollary Polytabloid specialization and the exact signed multiplicity dictionary.

[F1]

At $q = 1$ the specialization of $H$ is the group algebra $Q [S_{3}]$ and the sign automorphism $θ$ is the pullback along $w \mapsto sgn (w) w$ ; the partitions $(3)$ and $(1^{3})$ are conjugate (Polytabloid specialization and the exact signed multiplicity dictionary).

[F2]

The comparison theorem gives $P ((3)) ≅ C_{M} ((1^{3}))^{θ}$ with the exact unit $u_{(3)} = q^{ℓ (w_{0} (S_{(1^{3})})) - ℓ (w_{(3)})}$ (Ordinary Specht modules are the signed conjugate Murphy cells).

[F3]

$P ((3)) = x_{(3)} H$ is spanned by $x_{(3)} = \sum_{w \in S_{3}} T_{w}$ , and $x_{(3)} T_{i} = q x_{(3)}$ for $i = 1, 2$ ; the contragredient dual of a one-dimensional module with trivial action is trivial (Ordinary Specht modules are the signed conjugate Murphy cells).

[F4]

In $C_{M} ((1^{3}))$ one has $f^{(1^{3})} = 1$ , the higher ideal $I^{⊳ (1^{3})}$ contains the elements $1 + T_{i}$ , and the cell form has value $G_{(1^{3})} (m_{0}, m_{0}) = 1$ (Polytabloid specialization and the exact signed multiplicity dictionary, Ordinary Specht modules are the signed conjugate Murphy cells).

Counterexample

technique · direct

1.1F3algebra

The ordinary module is trivial. At $q = 1$ one has $x_{(3)} = \sum_{w \in S_{3}} T_{w}$ and $x_{(3)} T_{i} = \sum_{w} T_{w} T_{i} = \sum_{w} T_{w s_{i}} = x_{(3)}$ for each generator, because right multiplication by $s_{i}$ permutes the group and every $T_{w s_{i}}$ is again one of the summed basis elements; with the scalar carried along this reads $x_{(3)} T_{i} = q x_{(3)} = x_{(3)}$ at $q = 1$ , and $P ((3)) = x_{(3)} H = Q x_{(3)}$ is one-dimensional. Its dual is therefore also one-dimensional with trivial action.

1.2F4algebra

The cell is sign. At $q = 1$ the element $1 + T_{i}$ lies in the higher ideal $I^{⊳ (1^{3})}$ , because $1 + T_{1} = x_{(2, 1)}$ belongs to that two-sided ideal and conjugation by the invertible braid word $T_{1} T_{2}$ sends $T_{1}$ to $T_{2}$ , so $1 + T_{2}$ also belongs to it; hence in the one-dimensional cell $m_{0} (1 + T_{i}) = 0$ , that is, $m_{0} T_{i} = - m_{0}$ . The cell form value is $1$ , and the module is the sign module.

2.1step 1.1step 1.2F1F2algebra∎

No untwisted isomorphism. A right $H$ -module isomorphism $f ⁣ : C_{M} ((1^{3})) \to P ((3))^{\circ}$ would satisfy $f (m h) = f (m) h$ for every $h$ ; taking $h = T_{i}$ and a nonzero $m$ gives $f (- m) = f (m)$ because $C_{M} ((1^{3}))$ has $T_{i}$ acting by $- 1$ and $P ((3))^{\circ}$ has $T_{i}$ acting by $+ 1$ , while $f (- m) = - f (m)$ in characteristic zero, so $- f (m) = f (m)$ forces $f (m) = 0$ , contradicting injectivity over $Q$ . Therefore no such isomorphism exists and the untwisted claim is false. The twisted statement is consistent: on $C_{M} ((1^{3}))^{θ}$ the generator $T_{i}$ acts by the ordinary action of $θ (T_{i}) = - T_{i}$ , that is by $+ 1$ , matching $P ((3))$ , and the unit is $u_{(3)} = 1$ by [F2].

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The row and column Specht conventions in rank three

Example

Let $H = H_{A, q} (S_{3})$ , $A = Z [q^{\pm 1}]$ , and put $p_{3} (q) = (1 + q) (1 + q + q^{2})$ . All four modules $P ((3)), P ((1^{3})), C_{M} ((3)), C_{M} ((1^{3}))$ have rank $1$ . On $P ((3))$ and $C_{M} ((3))$ both generators act by $q$ ; on $P ((1^{3}))$ and $C_{M} ((1^{3}))$ both act by $- 1$ . In the canonical generators $x_{(3)}$ , $y_{(3)}$ , $c_{(3)}$ , $c_{(1^{3})}$ the respective self-form values are $1$ , $q^{- 3} p_{3} (q)$ , $p_{3} (q)$ , $1$ . Thus $Φ_{(3)} (x_{(3)}) = c_{(1^{3})}$ has exact form multiplier $u_{(3)} = 1$ , whereas $Φ_{(1^{3})} (y_{(3)}) = c_{(3)}$ has $u_{(1^{3})} = q^{3}$ . In both targets the action is twisted by $θ$ , so the two module identities match. At $q = 1$ over $Q$ the respective form values are $1, 6, 6, 1$ , and $θ$ exchanges trivial and sign. These identities specialize over every commutative ring with a unit parameter. No choice principle is used.

Facts & Assumptions

Given: The type- $A$ Hecke algebra $H$ of $S_{3}$ over $A = Z [q^{\pm 1}]$ with generators $T_{1}, T_{2}$ , the shapes $(3)$ and $(1^{3}) = (3)^{'}$ , their row and polytabloid modules and Murphy cells, the sign automorphism $θ$ with $θ (T_{i}) = (q - 1) - T_{i}$ , and the comparison theorem and its corollary Ordinary Specht modules are the signed conjugate Murphy cells, Polytabloid specialization and the exact signed multiplicity dictionary.

[F1]

$x_{(3)} = \sum_{w \in S_{3}} T_{w}$ spans $P ((3)) = x_{(3)} H$ and $x_{(3)} T_{i} = q x_{(3)}$ for $i = 1, 2$ ; the distinguished basis of $M ((3))$ is ${x_{(3)}}$ with $β (x_{(3)}, x_{(3)}) = 1$ ; the Murphy cell $C_{M} ((3))$ has one basis class $c_{(3)} = x_{(3)} + I^{⊳ (3)}$ and $G_{(3)} (c_{(3)}, c_{(3)}) = P_{(3)} (q) = p_{3} (q)$ (Ordinary Specht modules are the signed conjugate Murphy cells).

[F2]

$x_{(1^{3})} = 1$ , $y_{(3)} = \sum_{w \in S_{3}} (- q)^{- ℓ (w)} T_{w}$ spans $P ((1^{3})) = y_{(3)} H$ and $y_{(3)} T_{i} = - y_{(3)}$ for $i = 1, 2$ ; the distinguished basis of $M ((1^{3}))$ is the standard basis ${T_{w}}$ and $β (y_{(3)}, y_{(3)}) = q^{- 3} p_{3} (q)$ ; the Murphy cell $C_{M} ((1^{3}))$ has one basis class $c_{(1^{3})} = 1 + I^{⊳ (1^{3})}$ with $G_{(1^{3})} (c_{(1^{3})}, c_{(1^{3})}) = 1$ and $T_{i}$ acting by $- 1$ (Ordinary Specht modules are the signed conjugate Murphy cells, Polytabloid specialization and the exact signed multiplicity dictionary).

[F3]

The comparison theorem gives base-change compatible isomorphisms $Φ_{λ} ⁣ : P (λ) \to C_{M} (λ^{'})^{θ}$ with $G_{λ^{'}}^{θ} (Φ_{λ} x, Φ_{λ} y) = u_{λ} (q) β_{P (λ)} (x, y)$ and $u_{λ} (q) = q^{ℓ (w_{0} (S_{λ^{'}})) - ℓ (w_{λ})}$ , where the twisted action is $m \cdot h = m θ (h)$ (Ordinary Specht modules are the signed conjugate Murphy cells).

[F4]

At $q = 1$ over $Q$ the specialization is the group algebra $Q [S_{3}]$ and $θ (T_{i}) = - T_{i}$ is the sign automorphism (Polytabloid specialization and the exact signed multiplicity dictionary). The $H$ -linearity of the comparison maps is supplied by [F3]; their self-form comparison is a separate assertion.

Verification

technique · direct

1.1F1algebra

Shape (3). For the one-row shape, $x_{(3)} = \sum_{w \in S_{3}} T_{w}$ and pairing $u$ and $u s_{i}$ with $ℓ (u s_{i}) = ℓ (u) + 1$ gives $(T_{u} + T_{u s_{i}}) T_{i} = q (T_{u} + T_{u s_{i}})$ , hence $x_{(3)} T_{i} = q x_{(3)}$ ; since $z_{(3)} = x_{(3)} T_{w_{(3)}} y_{(1^{3})} = x_{(3)}$ (here $w_{(3)} = 1$ and $y_{(1^{3})} = 1$ ) the module $P ((3))$ is the free rank-one module on $x_{(3)}$ with action $q$ . Squaring gives $x_{(3)}^{2} = \sum_{v, w} T_{v} T_{w} = \sum_{v} T_{v} x_{(3)} = (\sum_{v} q^{ℓ (v)}) x_{(3)} = p_{3} (q) x_{(3)}$ , because the length distribution in $S_{3}$ is $(1, 2, 2, 1)$ and $1 + 2 q + 2 q^{2} + q^{3} = (1 + q) (1 + q + q^{2})$ ; hence the Murphy cell has form value $G_{(3)} (c_{(3)}, c_{(3)}) = p_{3} (q)$ and action $q$ , and $β (x_{(3)}, x_{(3)}) = 1$ by [F1].

1.2F2algebra

Shape (1^3). For the one-column shape, $x_{(1^{3})} = 1$ and $P ((1^{3})) = y_{(3)} H$ with $y_{(3)} = \sum_{w \in S_{3}} (- q)^{- ℓ (w)} T_{w}$ . For each pair $u, u s_{i}$ with $ℓ (u s_{i}) = ℓ (u) + 1$ , its contribution is $a (T_{u} - q^{- 1} T_{u s_{i}})$ , where $a = (- q)^{- ℓ (u)}$ . The quadratic relation gives $(T_{u} - q^{- 1} T_{u s_{i}}) T_{i} = T_{u s_{i}} - q^{- 1} ((q - 1) T_{u s_{i}} + q T_{u}) = - (T_{u} - q^{- 1} T_{u s_{i}})$ . Summing the pairs gives $y_{(3)} T_{i} = - y_{(3)}$ for $i = 1, 2$ , the sign representation of the rank-one module. The distinguished basis of $M ((1^{3}))$ is the standard basis and $β$ is diagonal with entries $q^{ℓ (w)}$ , so $β (y_{(3)}, y_{(3)}) = \sum_{w} q^{- 2 ℓ (w)} q^{ℓ (w)} = \sum_{w} q^{- ℓ (w)} = q^{- 3} p_{3} (q)$ , the last equality by $w \mapsto w_{0} w$ . Modulo the higher ideal $I^{⊳ (1^{3})}$ , which contains $1 + T_{i}$ because $1 + T_{1} = x_{(2, 1)}$ is a higher Murphy element and conjugation by the invertible braid word $T_{1} T_{2}$ sends $T_{1}$ to $T_{2}$ , placing $1 + T_{2}$ in the same two-sided ideal, the class $c_{(1^{3})}$ satisfies $T_{i} = - 1$ and $G_{(1^{3})} (c_{(1^{3})}, c_{(1^{3})}) = P_{(1^{3})} (q) = 1$ by [F2].

2.1step 1.1step 1.2F3algebra

The comparison maps. By [F3] applied to $λ = (3)$ , where $λ^{'} = (1^{3})$ and $ℓ (w_{0} (S_{(1^{3})})) = ℓ (w_{(3)}) = 0$ , the isomorphism $Φ_{(3)}$ has unit $u_{(3)} = 1$ and the identity $G_{(1^{3})}^{θ} (Φ_{(3)} x_{(3)}, Φ_{(3)} x_{(3)}) = β (x_{(3)}, x_{(3)}) = 1$ is consistent because both sides are the value of the respective rank-one forms; explicitly $Φ_{(3)} (x_{(3)}) = c_{(1^{3})}$ by [F3]. Applying it to $λ = (1^{3})$ , where $λ^{'} = (3)$ and $ℓ (w_{0} (S_{(3)})) = 3$ , $ℓ (w_{(1^{3})}) = 0$ , gives $u_{(1^{3})} = q^{3}$ , and the identity $G_{(3)} (c_{(3)}, c_{(3)}) = q^{3} q^{- 3} p_{3} (q) = p_{3} (q)$ is consistent; explicitly $Φ_{(1^{3})} (y_{(3)}) = c_{(3)}$ . In the target $C_{M} ((1^{3}))^{θ}$ the generator acts by the ordinary action of $θ (T_{i}) = (q - 1) - T_{i}$ , hence by $(q - 1) - (- 1) = q$ for every unit parameter; at $q = 1$ this is $θ (T_{i}) = - T_{i}$ and scalar $+ 1$ ; in the target $C_{M} ((3))^{θ}$ it acts by the ordinary action of $(q - 1) - T_{i}$ on the class acting by $q$ , hence by $(q - 1) - q = - 1$ ; these match the source actions $q$ on $P ((3))$ and $- 1$ on $P ((1^{3}))$ computed in steps 1.1 and 1.2.

3.1step 1.1step 1.2step 2.1F1F2F3F4∎

Specialization and conclusion. At $q = 1$ over $Q$ the four form values become $1$ , $1^{- 3} p_{3} (1) = 6$ , $6$ and $1$ , and $θ$ is the sign automorphism exchanging the trivial and sign modules, in agreement with [F4]; the module actions match as in step 2.1. Every displayed identity is an identity in $A = Z [q^{\pm 1}]$ between Laurent polynomials, so all of them specialize along any ring map $A \to R$ with $q \mapsto q \in R^{\times}$ . Only the rank-one computations, the trace-free form evaluations and the comparison theorem were used; no choice principle is used.

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The two-tableau pairing at a singular parameter

Example

For $λ = (2, 1)$ put $x = 1 + T_{1}$ and $b_{0} = x$ , $b_{1} = x T_{2}$ , $b_{2} = x T_{2} T_{1}$ in the row permutation module. Its form is $diag (1, q, q^{2})$ and its generator action matrices, with coordinate row vectors, are $R_{1} = (\begin{matrix} q & 0 & 0 \\ 0 & 0 & 1 \\ 0 & q & q - 1 \end{matrix})$ , $R_{2} = (\begin{matrix} 0 & 1 & 0 \\ q & q - 1 & 0 \\ 0 & 0 & q \end{matrix})$ . The ordinary polytabloid basis is $p_{0} = z = x T_{2} (1 - q^{- 1} T_{1}) = b_{1} - q^{- 1} b_{2}$ and $p_{1} = z T_{2} = q b_{0} + (q - 1) b_{1} - b_{2}$ . Their embedding has unit determinant on its $b_{0}, b_{2}$ coordinates, and their form is $B = (\begin{array}{cc} q + 1 & q^{2} \\ q^{2} & q^{3} + q \end{array}) .$ In the Murphy quotient let $c_{0} = x$ , $c_{1} = x T_{2}$ modulo the higher shape $(3)$ . Its form and generator action matrices, with coordinate row vectors, are $G = (\begin{array}{cc} q + 1 & - 1 \\ - 1 & q^{2} + 1 \end{array}), M_{1} = (\begin{array}{cc} q & 0 \\ - 1 & - 1 \end{array}), M_{2} = (\begin{array}{cc} 0 & 1 \\ q & q - 1 \end{array}) .$ On the ordinary basis the matrices are $P_{1} = (\begin{matrix} - 1 & 0 \\ - q^{2} & q \end{matrix})$ and $P_{2} = M_{2}$ . The map $Φ (p_{0}) = c_{0}$ , $Φ (p_{1}) = (q - 1) c_{0} - c_{1}$ has matrix $A = (\begin{matrix} 1 & 0 \\ q - 1 & - 1 \end{matrix})$ ; it satisfies $P_{i} A = A ((q - 1) I - M_{i})$ and $A G A^{T} = B$ . The same-shape perfect cross-pairing on the bases $(c_{0}, c_{1})$ and $(p_{0}, p_{1})$ has matrix $Q = (\begin{matrix} 0 & q \\ q & q (q - 1) \end{matrix})$ with determinant $- q^{2}$ , a unit; its entries are the row-form coefficients before quotienting. Thus the exact self-form multiplier is $1$ . Both Gram determinants equal $q (1 + q + q^{2})$ . At $q = - 1$ over characteristic zero these forms are nonsingular with determinant $- 1$ , even though $[2]_{q} = 0$ ; at $q = 1$ in characteristic $2$ both reduce to $(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})$ and have radical $0$ . This distinguishes a singular algebra parameter from degeneracy of this particular module form. No choice principle is used.

Facts & Assumptions

Given: The Hecke algebra $H = H_{R, q} (S_{3})$ over a commutative ring $R$ with $q \in R^{\times}$ , the shape $λ = (2, 1) = λ^{'}$ , the row module $M (λ)$ with $x = x_{λ} = 1 + T_{1}$ and the distinguished basis ${b_{0}, b_{1}, b_{2}}$ , the polytabloid module $P (λ) = z_{λ} H$ with $z_{λ} = x T_{2} (1 - q^{- 1} T_{1})$ , the permutation form $β$ and the Murphy cell $C_{M} (λ)$ with classes $c_{0}, c_{1}$ and form $G$ , together with the comparison theorem and the perfect cross-pairing of Ordinary Specht modules are the signed conjugate Murphy cells and Murphy cells are the same-shape contragredient Specht duals.

[F1]

$H$ has generators $T_{1}, T_{2}$ with $(T_{i} - q) (T_{i} + 1) = 0$ , braid relation $T_{1} T_{2} T_{1} = T_{2} T_{1} T_{2}$ and standard basis ${T_{w}}$ ; $x T_{1} = q x$ (The generic type-A Hecke algebra, The row permutation module: distinguished basis and Hecke action).

[F2]

The distinguished basis ${b_{0}, b_{1}, b_{2}} = {x, x T_{s_{2}}, x T_{s_{2} s_{1}}}$ of $M (λ)$ satisfies $β (b_{i}, b_{j}) = δ_{i j} q^{i}$ and is the row basis of $D_{λ} = {1, s_{2}, s_{2} s_{1}}$ (Hecke row permutation modules and ordinary Dipper-James polytabloids, The row permutation module: distinguished basis and Hecke action).

[F3]

$C_{M} (λ) = M (λ) / (M (λ) \cap I^{⊳ λ})$ has basis $c_{t} = x T_{d (t)}$ for the standard tableaux $t$ , and its form satisfies $m_{u s} m_{t v} \equiv G (c_{s}, c_{t}) m_{u v} (m o d I^{⊳ λ})$ ; the ideal $I^{⊳ λ}$ contains $x_{(3)} = b_{0} + b_{1} + b_{2}$ (The Murphy cellular basis, the cell law and the Murphy form).

[F4]

$Φ ⁣ : P (λ) \to C_{M} (λ)^{θ}$ is an isomorphism with $Φ (z) = c_{0}$ and unit $u_{λ} (q) = q^{ℓ (w_{0} (S_{λ^{'}})) - ℓ (w_{λ})} = q^{1 - 1} = 1$ for $λ = (2, 1)$ ; the pairing $B_{λ} (m_{s}, p) = β (x T_{d (s)}, p)$ is perfect with the explicit entries $B_{λ} (c_{s}, e_{t}) = q^{ℓ (d (s))} a_{d (s), t}$ (Ordinary Specht modules are the signed conjugate Murphy cells, Murphy cells are the same-shape contragredient Specht duals).

Verification

technique · direct

1.1F1F2algebra

Row module. From [F1], $b_{0} T_{1} = x T_{1} = q x = q b_{0}$ , $b_{1} T_{1} = x T_{2} T_{1} = b_{2}$ and $b_{2} T_{1} = x T_{2} T_{1}^{2} = x T_{2} ((q - 1) T_{1} + q) = (q - 1) b_{2} + q b_{1}$ ; also $b_{0} T_{2} = b_{1}$ , $b_{1} T_{2} = x T_{2}^{2} = x ((q - 1) T_{2} + q) = (q - 1) b_{1} + q b_{0}$ and $b_{2} T_{2} = x T_{2} T_{1} T_{2} = x T_{1} T_{2} T_{1} = q x T_{2} T_{1} = q b_{2}$ , using the braid relation. In the basis $(b_{0}, b_{1}, b_{2})$ with coordinate row vectors these are exactly the matrices $R_{1}$ and $R_{2}$ of the statement, and $β (b_{i}, b_{i}) = q^{i}$ by [F2] because $ℓ (1) = 0$ , $ℓ (s_{2}) = 1$ , $ℓ (s_{2} s_{1}) = 2$ .

2.1step 1.1F2F4algebra

Polytabloids and their form. Since $z_{λ} = x T_{2} (1 - q^{- 1} T_{1})$ by [F4] and $y_{λ^{'}} = 1 - q^{- 1} T_{1}$ , the displayed expressions $p_{0} = b_{1} - q^{- 1} b_{2}$ and $p_{1} = z T_{2} = b_{1} T_{2} - q^{- 1} b_{2} T_{2} = q b_{0} + (q - 1) b_{1} - b_{2}$ follow from step 1.1. Their coordinates on $(b_{0}, b_{2})$ give the matrix $(\begin{matrix} 0 & - q^{- 1} \\ q & - 1 \end{matrix})$ with determinant $q \cdot q^{- 1} = 1$ , a unit over every $R$ ; hence the two polytabloids span a free rank-two split submodule. The row action gives $p_{0} T_{1} = - p_{0}$ , $p_{1} T_{1} = - q^{2} p_{0} + q p_{1}$ , $p_{0} T_{2} = p_{1}$ and $p_{1} T_{2} = q p_{0} + (q - 1) p_{1}$ . Thus their span is stable, has precisely the displayed matrices $P_{1}, P_{2}$ , and equals $z H$ because it contains $z = p_{0}$ and $z T_{2} = p_{1}$ . Weighted inner products against the diagonal form of [F2] give $B = (\begin{matrix} q + 1 & q^{2} \\ q^{2} & q^{3} + q \end{matrix})$ : explicitly $B_{00} = q + q^{- 2} \cdot q^{2} = q + 1$ , $B_{01} = q (q - 1) + (- q^{- 1}) (- 1) q^{2} = q (q - 1) + q = q^{2}$ , and $B_{11} = q^{2} + (q - 1)^{2} q + q^{2} = q^{3} + q$ . Its determinant is $q (q^{2} + q + 1)$ .

2.2step 1.1F3algebra

The Murphy quotient. In $C_{M} (λ)$ the higher vector $x_{(3)} = b_{0} + b_{1} + b_{2}$ vanishes by [F3], so $b_{2} = - c_{0} - c_{1}$ and $c_{1} = x T_{2} = b_{1}$ . Then $c_{0} T_{1} = q x = q c_{0}$ and $c_{1} T_{1} = b_{2} = - c_{0} - c_{1}$ , giving $M_{1}$ ; also $c_{0} T_{2} = c_{1}$ and $c_{1} T_{2} = (q - 1) b_{1} + q b_{0} = (q - 1) c_{1} + q c_{0}$ , giving $M_{2}$ . The defining products of [F3] give $x^{2} = (q + 1) x$ and $x T_{2} x = b_{1} + b_{2} = x_{(3)} - x \equiv - x$ . Also $x T_{2}^{2} x = (q - 1) x T_{2} x + q x^{2} \equiv (q^{2} + 1) x$ . Reading the coefficient of $m_{t_{0} t_{0}} = x$ therefore gives $G (c_{0}, c_{0}) = q + 1$ , $G (c_{0}, c_{1}) = - 1$ and $G (c_{1}, c_{1}) = q^{2} + 1$ ; equivalently $G = (\begin{matrix} q + 1 & - 1 \\ - 1 & q^{2} + 1 \end{matrix})$ with determinant $q^{3} + q^{2} + q$ .

3.1step 2.1step 2.2F4algebra∎

Comparison, cross-pairing and specializations. The comparison theorem of [F4] gives $Φ (z) = c_{0}$ , hence $Φ (p_{1}) = Φ (z T_{2}) = c_{0} θ (T_{2}) = (q - 1) c_{0} - c_{1}$ and $A = (\begin{matrix} 1 & 0 \\ q - 1 & - 1 \end{matrix})$ ; direct matrix multiplication verifies $P_{i} A = A ((q - 1) I - M_{i})$ for $i = 1, 2$ and $A G A^{T} = B$ , the latter being the form identity with the unit $u_{λ} = 1$ . The cross-pairing entries $B_{λ} (c_{s}, p_{t}) = β (b_{d (s)}, p_{t})$ on the standard bases give $Q = (\begin{matrix} 0 & q \\ q & q (q - 1) \end{matrix})$ with $\det Q = - q^{2} \in A^{\times}$ , so the pairing is perfect as in [F4]. The two Gram determinants agree with $q (1 + q + q^{2})$ . At $q = - 1$ in characteristic zero, $B = (\begin{matrix} 0 & 1 \\ 1 & - 2 \end{matrix})$ and $G = (\begin{matrix} 0 & - 1 \\ - 1 & 2 \end{matrix})$ have determinant $- 1$ , although $[2]_{q} = 1 + q = 0$ ; at $q = 1$ in characteristic $2$ , both reduce to $(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})$ with determinant $1$ , hence with zero radical. All identities are Laurent-polynomial identities, so they specialize to every $R$ ; no choice principle is used.

Ordinary Hecke Specht Polytabloids and Signed Murphy Comparison — Examples

1 · Prerequisites

2 · Summary

Examples and counterexample

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Untwisted conjugation changes the module

Statement refuted

Facts & Assumptions

Counterexample

The row and column Specht conventions in rank three

Example

Facts & Assumptions

Verification

The two-tableau pairing at a singular parameter

Example

Facts & Assumptions

Verification

Sources