Elementary divisors of Gram matrices of certain Specht modules (jt. w. G. Nebe and A. Mathas)

Notation

S_{n}  symmetric group
λ ⊢ n  partition of n
λ′  transpose of λ
S^{λ}  integral Specht module over ZS_{n}

n_{λ} := rk_{Z} S^{λ}

S^{λ,∗}  Z-linear dual
S_{A}^{λ} := A ⊗_{Z} S^{λ} for a commutative ring A
(=, −)  the S_{n}-invariant nondegenerate

Z-bilinear form on S^{λ}

Problem

ZS_{n}-linear map:

[figure: S^{λ}; η; S^{λ,∗}; ξ; (ξ, −)]

Determine quotient S^{λ,∗}/S^{λ} as a finite abelian group. Equivalently, determine el. div. of Gram matrix of (=, −).

Motivation

I) Simple modules Ordinary:

{S_{C}^{λ} | λ ⊢ n} = {simple CS_{n}-modules}

Modular: p prime.

D_{F_{p}}^{λ} := Im ((S^{λ} [figure: η] S^{λ,∗}) ⊗_{Z} F_{p})

James proves:

{D_{F_{p}}^{λ} | λ ⊢ n p-regular} = {simple F_{p}S_{n}-modules} So

dim_{F_{p}} D_{F_{p}}^{λ} = #{el. div. of S^{λ} not divisible by p}

However, we will make extensive use of known such dimensions rather than derive them.

II) Quasiblocks How to understand the integral group ring ZS_{n}? Wedderburn isomorphism:

QS_{n} ^{∼ ^{\overline{}}}∏_{λ⊢n} \underbrace{End_{Q} S_{Q}^{λ}}_{≃ Q^{n_{λ}×n_{λ}}}

Restrict to Wedderburn embedding.

ZS_{n} [figure] ∏_{λ⊢n} \underbrace{End_{Z} S^{λ}}_{≃ Z^{n_{λ}×n_{λ}}}

Describe ZS_{n} as image of the Wedderburn embedding.

Example:

[figure: ZS_{3} ≃; Z; ^{( \begin{matrix}Z & Z \\ 3Z & Z \end{matrix} )}; Z; 3; 3; 2]

Call Q^{(2,1)} = (\begin{matrix}Z & Z \\ 3Z & Z \end{matrix}) a quasiblock of ZS_{3}.

Orthogonal decomposition into rational primitive central idempotents:

1_{QS_{n}} = ∑_{λ⊢n} ε_{λ}

General definition of a quasiblock :

Q^{λ} := ZS_{n}ε_{λ} ( [figure] End_{Z} S^{λ} ≃ Z^{n_{λ}×n_{λ}} )

Unknown: index of this inclusion (as abelian groups).

Necessary condition on φ ∈ End_{Z} S^{λ} to lie in image:

η^{−1}φ η has to be integral

(η^{−1} = rational inverse of η). Not sufficient in general.

The Smith form of S^{(2,1)} is (\begin{matrix}1 & 0 \\ 0 & 3 \end{matrix}), so for Q^{(2,1)} = (\begin{matrix}Z & Z \\ 3Z & Z \end{matrix}) this necessary condition is in fact sufficient.

In general, elementary divisors of S^{λ} partially describe Q^{λ}.

Method

I) Jantzen filtration Let p be a prime, let i ⩾ 0. Define S_{F_{p}}^{λ}(i) by

[figure: Kern; S^{λ}; η; S^{λ,∗}/p^{i}S^{λ,∗}; S_{F_{p}}^{λ}(i); S_{F_{p}}^{λ}]

The Jantzen filtration is the finite filtration

S_{F_{p}}^{λ} = S_{F_{p}}^{λ}(0) ⊇ S_{F_{p}}^{λ}(1) ⊇ S_{F_{p}}^{λ}(2) ⊇ ⋯, We have S_{F_{p}}^{λ}(0)/S_{F_{p}}^{λ}(1) ≃ D_{F_{p}}^{λ}. The Jantzen filtration gives Z_{(p)}-linear elementary divisors:

S_{Z_{(p)}}^{λ,∗}/S_{Z_{(p)}}^{λ} ≃ ⨁_{i⩾0}(Z/p^{i}Z)^{dim_{Fp} S_{Fp}^{λ}(i)/S_{Fp}^{λ}(i+1)} .

So to get elementary divisors:

• Determine composition factors of Jantzen subquotients (using Jantzen’s Lemma and Schaper’s Theorem).

• Use results of James et al. on the dimensions of simple modules.

II) Jantzen’s Lemma p prime, λ ⊢ n arbitrary, μ ⊢ n p-regular.

ϑ_{i} := [S_{F_{p}}^{λ}(i)/S_{F_{p}}^{λ}(i + 1) : D_{F_{p}}^{μ}] .

Jantzen’s Lemma:

\begin{aligned}[S_{F_{p}}^{λ} : D_{F_{p}}^{μ}] & = & ϑ_{0} & + & ϑ_{1} & + & ϑ_{2} & + & ⋯ \\ [S_{Z_{(p)}}^{λ,∗}/S_{Z_{(p)}}^{λ} : D_{F_{p}}^{μ}] & = & 0ϑ_{0} & + & 1ϑ_{1} & + & 2ϑ_{2} & + & ⋯ \end{aligned}

If λ = μ, then ϑ_{0} = 1, and ϑ_{i} = 0 for i > 0. If λ ≠ μ, we have ϑ_{0} = 0. Then :

(i) If [S_{F_{p}}^{λ} : D_{F_{p}}^{μ}] = 1, then

ϑ_{i} = \{ \begin{matrix}1 & for i = [S_{Z_{(p)}}^{λ,∗}/S_{Z_{(p)}}^{λ} : D_{F_{p}}^{μ}] \\ 0 & else. \end{matrix}

(ii) If [S_{F_{p}}^{λ} : D_{F_{p}}^{μ}] = 2, then : If [S_{Z_{(p)}}^{λ,∗}/S_{Z_{(p)}}^{λ} : D_{F_{p}}^{μ}] = 2, then ϑ_{1} = 2, others = 0. If [S_{Z_{(p)}}^{λ,∗}/S_{Z_{(p)}}^{λ} : D_{F_{p}}^{μ}] = 3, then ϑ_{1} = 1, ϑ_{2} = 1, others = 0. If [S_{Z_{(p)}}^{λ,∗}/S_{Z_{(p)}}^{λ} : D_{F_{p}}^{μ}] = 4, then (ϑ_{1} = 1, ϑ_{3} = 1, others = 0) or (ϑ_{2} = 2, others = 0). Etc. So there is still ambiguity.

III) Schaper’s Theorem Even if all occurring decomposition numbers [S_{F_{p}}^{λ} : D_{F_{p}}^{μ}] are known, we still need [S_{Z_{(p)}}^{λ,∗}/S_{Z_{(p)}}^{λ} : D_{F_{p}}^{μ}]. Schaper’s Theorem is a formula in the Grothendieck group:

[S_{Z_{(p)}}^{λ,∗}/S_{Z_{(p)}}^{λ}] = ∑_{λ⊴ ν} α_{ν}[S_{F_{p}}^{ν}] ,

with combinatorially determined coefficients α_{ν} ∈ Z. So for our purposes, knowledge of further decomposition numbers [S_{F_{p}}^{ν} : D_{F_{p}}^{μ}] is required.

IV) Transposition Jantzen filtration of S^{λ} [figure] Jantzen filtration of S^{λ′}

Some results

Table of calculated Jantzen filtrations. Partially depending on conjectures on dec. numbers. Partially with ambiguity.

λ Remarks

(n − m, 1^{n−m}) hooks, two methods: Jantzen-Schaper and directly; independently done by James-Mathas

(n − m, m) two-row; independently done by Fayers

and by Murphy

(n − 3, 2, 1)

(n − 4, 2^{2}) if p = 2: depending on a conj. on [S_{F_{2}}^{(n−4,2^{2})} : D_{F_{2}}^{(n)}];

if p = 3: with ambiguity

(n − 4, 3, 1) if p = 2: with ambiguity

(n − 4, 2, 1^{2}) if p = 2: depending heavily on conj. on

dec. nos., and with ambiguity;

if p = 3: depending on conj. on dec. nos.

Case n − λ_{1} < p using Kleshchev’s Modular Branching and Carter-Payne

How to proceed further?

Let e^{μ} be a primitive idempotent of Z_{(p)}S_{n} belonging to D_{F_{p}}^{μ} and let ε^{λ} be the primitive central idempotent of QS_{n} belonging to S_{Q}^{λ}.

The embedding η decomposes into a direct sum of embeddings

S_{Z_{(p)}}^{λ}e^{μ} [figure: e^{μ}ηe^{μ}] S_{Z_{(p)}}^{λ,∗}e^{μ} ,

so we get a block diagonalisation of the Gram matrix, with blocks of edge length

rk_{Z_{(p)}} S^{λ}e^{μ} = [S_{F_{p}}^{λ} : D_{F_{p}}^{μ}] .

It remains to

calculate the Smith form of e^{μ}ηe^{μ}.

Schaper’s Theorem ess. allows to calculate the determinant:

v_{p}(det(e^{μ}ηe^{μ})) = [S_{Z_{(p)}}^{λ,∗}/S_{Z_{(p)}}^{λ} : D_{F_{p}}^{μ}]

Assume [S_{F_{p}}^{λ} : D_{F_{p}}^{μ}] ⩾ 1. Let

Q_{(p)}^{λ:μ} := e^{μ}Q_{Z_{(p)}}^{λ}e^{μ} ( [figure] Z_{(p)}^{[S_{Fp}^{λ}:D_{Fp}^{μ}] × [S_{Fp}^{λ}:D_{Fp}^{μ}]} ) .

• Q ⊗_{Z_{(p)}} Q_{(p)}^{λ:μ} is simple.

• Q_{(p)}^{λ:μ} is local (for e^{μ}Q_{Z_{(p)}}^{λ} is indecomposable).

• e^{μ}ηe^{μ} is an inclusion of simple Q_{(p)}^{λ:μ}-lattices.

If we could classify all inclusions of simple lattices over Q_{(p)}^{λ:μ}, we would know the Smith form of e^{μ}ηe^{μ} to be one of the Smith forms of these inclusions (moreover, one of a given determinant).

Example: Q_{(2)}^{(3,1^{2}):(5)} ≃ \{(\begin{matrix}a & b \\ c & d \end{matrix}) ∈ Z_{(2)}^{2×2} | b ≡_{2} 0 , a ≡_{2} d\},

with simple lattices X := (\begin{matrix}Z_{(2)} \\ Z_{(2)} \end{matrix}) and Y := (\begin{matrix}2Z_{(2)} \\ Z_{(2)} \end{matrix}).

Inclusions of determinant 4: only the scalar inclusions X [figure: 2] X and Y [figure: 2] Y, both with Smith form (\begin{matrix}2 & 0 \\ 0 & 2 \end{matrix}).

Need: lattices over Q_{(p)}^{λ:μ}. Note: rk_{Z_{(p)}} Q_{(p)}^{λ:μ} = [S_{F_{p}}^{λ} : D_{F_{p}}^{μ}]^{2}. So to ask for Q_{(p)}^{λ:μ} can be viewed a refined version of the question for the decomposition numbers.

Elementary divisors over Z[q, q^{−1}]?

For the Specht module over the Hecke algebra H with ground ring Z[q, q^{−1}], we may also consider the Gram matrix of the invariant bilinear form. Distinguish (also over Z_{(p)}[q, q^{−1}]):

• divisibly diagonalizable matrices (each diagonal entry divides successor)

• diagonalizable, but not divisibly diagonalizable matrices

• non-diagonalizable matrices Andersen remarked that not every Gram matrix of a Specht module is diagonalizable. For λ = (n − l, 1^{l}): bases divisibly diagonalizing the Gram matrix (the q causing difficulties). For λ = (3, 3, 2) : not diagonalizable over Z_{(2)}[q, q^{−1}]. For λ = (4, 2, 1, 1): neither diagonalizable over Z_{(2)}[q, q^{−1}] nor over Z_{(3)}[q, q^{−1}]. Method: if diagonalizable, then there is a connection between the elementary divisors over Q[q, q^{−1}] and F_{p}[q, q^{−1}].

We do not know (not even conjecturally) a criterion for λ to decide on diagonalizability.