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Specht restriction has a removable-corner filtration over every field

Statement

Let $n \geq 1$ , let $λ ⊢ n$ , let $F$ be any field, and let $r_{1} < \dots < r_{m}$ be the rows of the removable corners of $λ$ , so that $m \geq 1$ , the corners $x_{i} = (r_{i}, λ_{r_{i}})$ are listed from top to bottom, and the partitions $λ^{(i)}$ of $n - 1$ are obtained by deleting $x_{i}$ (Ordered removable corners and tabloid deletion maps, Removable and addable nodes). Let $V_{i} : = {span}_{F} {e_{t} : t$ a standard $λ$ -tableau whose entry $n$ lies in one of the rows $r_{1}, \dots, r_{i}}$ for $1 \leq i \leq m$ and $V_{0} : = 0$ , inside the Specht module $S_{F}^{λ}$ (Integral and field-valued Specht modules). Then the restriction ${Res}_{S_{n - 1}}^{S_{n}} S_{F}^{λ}$ of $S_{F}^{λ}$ to the subgroup $S_{n - 1} \leq S_{n}$ of permutations fixing $n$ (The sign representation of $S_{n}$ and the restriction ${Res}_{H}^{G} (V)$ of a representation to a subgroup) has a filtration by $S_{n - 1}$ -submodules $0 = V_{0} ⊊ V_{1} ⊊ \dots ⊊ V_{m} = S_{F}^{λ}$ whose successive quotients are $V_{i} / V_{i - 1} ≅ S_{F}^{λ^{(i)}} (1 \leq i \leq m)$ as $S_{n - 1}$ -modules. Equivalently, the restriction of $S_{F}^{λ}$ has a filtration whose successive quotients are $S_{F}^{λ^{(1)}}, \dots, S_{F}^{λ^{(m)}}$ , one for each removable corner of $λ$ , taken from top to bottom. No splitting of this filtration is asserted.

Facts & Assumptions

Given: an integer $n \geq 1$ , a partition $λ ⊢ n$ , a field $F$ , the removable rows $r_{1} < \dots < r_{m}$ with corners $x_{i}$ and partitions $λ^{(i)}$ , the Specht module $S_{F}^{λ}$ , and its subspaces $V_{i}$ defined in the Statement.

[F1]

Each $V_{i}$ is stable under the action of $S_{n - 1}$ on $S_{F}^{λ}$ by restriction, and $V_{0} \subseteq V_{1} \subseteq \dots \subseteq V_{m} = S_{F}^{λ}$ (The corner-filtration subspaces of a Specht module are S_(n-1)-invariant).

[F2]

For every $1 \leq i \leq m$ the deletion map $θ_{i}$ restricts to a surjection $V_{i} \to S_{F}^{λ^{(i)}}$ with kernel $V_{i - 1}$ , hence induces an isomorphism of $S_{n - 1}$ -modules $V_{i} / V_{i - 1} ≅ S_{F}^{λ^{(i)}}$ (Deletion identifies each Specht branching quotient).

[F3]

For every partition $ν$ of $n - 1$ the Specht module $S_{F}^{ν}$ is nonzero: for $ν = \emptyset$ one has $S_{F}^{\emptyset} = F$ , and for $ν \neq \emptyset$ the polytabloid $e_{t}$ of any $ν$ -tableau $t$ has coefficient $1$ at the tabloid ${t}$ , whence $e_{t} \neq 0$ (Integral and field-valued Specht modules).

[F4]

A subspace of a representation stable under the action of a subgroup is a subrepresentation of the restricted representation (Subrepresentations, direct sums of representations, and irreducibility, The sign representation of $S_{n}$ and the restriction ${Res}_{H}^{G} (V)$ of a representation to a subgroup).

[F5]

For $n \geq 1$ a partition of $n$ has at least one removable corner, so $m \geq 1$ and the list $r_{1} < \dots < r_{m}$ is finite and nonempty (Ordered removable corners and tabloid deletion maps, Removable and addable nodes).

Proof

technique · constructive

1.1F1F4construct

[construct] By [F1] the subspaces $V_{i}$ are stable under the action of $S_{n - 1}$ and satisfy $0 = V_{0} \subseteq V_{1} \subseteq \dots \subseteq V_{m} = S_{F}^{λ}$ ; by [F4] each $V_{i}$ is therefore an $S_{n - 1}$ -submodule of ${Res}_{S_{n - 1}}^{S_{n}} S_{F}^{λ}$ .

1.2F2algebra

By [F2], for every $1 \leq i \leq m$ the quotient $V_{i} / V_{i - 1}$ is isomorphic to $S_{F}^{λ^{(i)}}$ as an $S_{n - 1}$ -module; in particular the successive quotients of the chain are the Specht modules of the deletion shapes, in the order $i = 1, \dots, m$ of the corners from top to bottom.

2.1F2F3step 1.1step 1.2algebra

The inclusions are strict: by [F3] the quotient $S_{F}^{λ^{(i)}}$ is nonzero, so $V_{i} / V_{i - 1} \neq 0$ and $V_{i - 1} \neq V_{i}$ for every $i$ . Hence $0 = V_{0} ⊊ V_{1} ⊊ \dots ⊊ V_{m} = S_{F}^{λ}$ is a filtration of the $S_{n - 1}$ -module $S_{F}^{λ}$ with successive quotients $S_{F}^{λ^{(1)}}, \dots, S_{F}^{λ^{(m)}}$ . This proves the displayed filtration and the identification of its quotients.

3.1F3F5givenstep 2.1discharge-construct∎

Boundary, field and choice audit. For $n = 1$ one has $λ = (1)$ , $m = 1$ , $r_{1} = 1$ , $λ^{(1)} = \emptyset$ and $S_{F}^{(1)} = F e_{t}$ with a single standard tableau $t$ , so the filtration reads $0 ⊊ V_{1} = S_{F}^{(1)}$ with quotient $S_{F}^{\emptyset} = F$ , in agreement with the statement. More generally the list of corners is finite and nonempty by [F5], and every removable corner of $λ$ occurs exactly once, as the corner $x_{i}$ for the unique $i$ with $r_{i}$ its row. The argument is uniform in $F$ : the two quoted lemmas use the standard-polytabloid bases and the field-uniform deletion maps of [F2], with no division and no characteristic hypothesis, and [F3] gives nonzero quotients over every field, including $char F = 2$ . Nothing here asserts that the filtration splits or that the quotients are simple or irreducible, and in positive characteristic the restriction need not be semisimple; only the existence of the filtration with the stated quotients is claimed. The subspaces $V_{i}$ are explicitly defined spans of finite standard-polytabloid sets determined by $λ$ , and the corner list is determined by $λ$ , so no choice principle is invoked. This proves the theorem.

Remarks

Why the order matters. The standard-polytabloid filtration uses the top-to-bottom corner order of Ordered removable corners and tabloid deletion maps; arbitrary reordering can destroy invariance. For $λ = (2, 1)$ , putting the bottom corner first gives the line spanned by $v = {12 ∣ 3} - {23 ∣ 1}$ . But $(12) v = {12 ∣ 3} - {13 ∣ 2}$ lies outside that line, by independence of the three tabloids. The specified order is the one used in Deletion identifies each Specht branching quotient.
What is not claimed. No direct sum decomposition ${Res}_{S_{n - 1}}^{S_{n}} S_{F}^{λ} ≅ ⨁_{i} S_{F}^{λ^{(i)}}$ is asserted; over $C$ such a splitting does follow from complete reducibility, which is the content of the complex branching rule proved later on this page, but in positive characteristic the filtration genuinely need not split.
Provenance caveat. The statement of the filtration is classical (Chan Theorem 4.16, printed pp. 18-19; Wildon Section 6, printed pp. 26-33); the proof above is assembled from the two preceding lemmas, whose arguments are field-uniform and avoid the Robinson-Schensted-Knuth correspondence.

Depends on

Used by

The complex Specht restriction branching rule Corollary
The branching filtration need not split in modular characteristic Counterexample

Dependency tree · two levels

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Sources

Charlotte Chan, Representation Theory of Symmetric Groups, Theorem 4.16, printed pp. 18-19, and Theorem 6.8, printed p. 26 (standard reference, not scraped)
David A. Craven, Groups, Geometries and Representation Theory, Sections 2.2 and 2.4, printed pp. 22-23 and 28-31 (standard reference, not scraped)
Mark Wildon, Representation Theory of the Symmetric Group, Section 6, printed pp. 26-33 (standard reference, not scraped)