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The two extreme Young permutation modules

Example

For $n \geq 1$ the two extreme partitions of $n$ give the two extreme Young permutation modules:

for $λ = (n)$ there is exactly one tabloid, so $M^{(n)}$ is one-dimensional, and every $σ \in S_{n}$ fixes that tabloid; hence $M^{(n)}$ is the trivial representation $C$ of $S_{n}$ ;
for $λ = (1^{n})$ the row-equivalence classes are singletons and the tabloids are the $n!$ tableaux of shape $(1^{n})$ , so $M^{(1^{n})}$ has a basis indexed by $S_{n}$ on which $S_{n}$ acts by left multiplication; hence $M^{(1^{n})}$ is the regular representation $C [S_{n}]$ .

The row module is generated by its single tabloid and the column module has dimension $n! = \dim C [S_{n}]$ . For $n = 1$ the two partitions coincide, $(1) = (1^{1})$ , and both descriptions give the one-dimensional module; for $n = 0$ the only partition is $\emptyset$ , there is one empty tabloid, and $M^{\emptyset} = C$ is again the one-dimensional trivial module, which is also the regular module of the trivial group $S_{0} = {1}$ .

Facts & Assumptions

Given: For $n \geq 1$ , the partitions $(n)$ and $(1^{n})$ of $n$ and their standard row-filled tableaux $t_{0}$ ; separately, for $n = 0$ , the empty partition and empty tableau.

[F1]

For $λ ⊢ n$ the standard Young subgroup $S_{λ}$ is the subgroup preserving each consecutive block of sizes $λ_{1}, \dots, λ_{k}$ ; the Young permutation module $M^{λ}$ is the complex vector space with the tabloids ${t}$ of shape $λ$ as basis, $S_{n}$ acting by $σ \cdot {t} = {σ \cdot t}$ ; the stabilizer of ${t}$ is $R_{t}$ and the stabilizer of the tabloid ${t_{0}}$ of the standard row-filled tableau is $S_{λ}$ , so $Ω_{λ} = {{σ \cdot t_{0}} : σ \in S_{n}}$ (Young subgroups, tabloids, and permutation modules).

[F2]

For every $λ ⊢ n$ there is an isomorphism of complex $S_{n}$ -modules $M^{λ} ≅ C [S_{n} / S_{λ}]$ , the permutation representation of $S_{n}$ on the left coset set $S_{n} / S_{λ}$ , and hence $M^{λ} ≅ {Ind}_{S_{λ}}^{S_{n}} 1$ ; this holds also for $n = 0$ (Young permutation modules are induced trivial modules).

[F3]

The trivial representation of a finite group $G$ over $C$ is the one-dimensional representation on $C$ in which every $g$ acts as the identity, and the regular representation is the representation on $C [G]$ given by left multiplication by the basis units, $g \cdot [h] = [g h]$ (The trivial representation, the regular representation, and permutation representations from finite $G$ -sets).

Verification

technique · direct

1.1

For $λ = (n)$ the blocks of [F1] are the single block ${1, \dots, n}$ , so $S_{(n)} = S_{n}$ , and every $λ$ -tableau has the single row set ${1, \dots, n}$ ; hence any two $λ$ -tableaux are row equivalent, there is exactly one tabloid, and $M^{(n)}$ is one-dimensional with that tabloid as basis.

givenF1

1.2

For $λ = (1^{n})$ the blocks are the $n$ singletons ${1}, \dots, {n}$ , so $S_{(1^{n})} = {1}$ , and the row set of row $i$ of a tableau is the singleton carrying its single entry, so two tableaux are row equivalent exactly when they agree entrywise and each tabloid is a singleton class; a $λ$ -tableau is a bijection from the $n$ boxes onto ${1, \dots, n}$ , so there are $n!$ of them, and the map $σ \mapsto σ \cdot t_{0}$ from $S_{n}$ to the tableaux is a bijection, being injective because $t_{0}$ is surjective and surjective because $σ (t_{0} (i, j)) : = t (i, j)$ defines a permutation, whence $σ \mapsto {σ \cdot t_{0}}$ is a bijection onto the tabloids with $τ \cdot {σ \cdot t_{0}} = {τ σ \cdot t_{0}}$ for all $τ \in S_{n}$ .

givenF1

2.1

For $λ = (n)$ and any $σ \in S_{n}$ the tableau $σ \cdot t_{0}$ again has the single row set ${1, \dots, n}$ , so $σ \cdot {t_{0}} = {t_{0}}$ : every $σ$ acts as the identity on the one-dimensional $M^{(n)}$ , which by [F3] is the trivial representation, in agreement with [F2] because $S_{n} / S_{n}$ is a single coset.

step 1.1F2F3

2.2

Transporting the basis along the bijection of step 1.2 turns the action of [F1] into left multiplication on $C [S_{n}]$ , which by [F3] is the regular representation, so $M^{(1^{n})} ≅ C [S_{n}]$ ; the coset description of [F2] reduces to the same statement because $S_{(1^{n})} = {1}$ and the cosets $σ {1}$ are the singletons of $S_{n}$ .

step 1.2F1F2F3

3.1

For $n = 1$ the two partitions coincide, $(1) = (1^{1})$ , and both conclusions above give the same one-dimensional module; for $n = 0$ the only partition is $\emptyset$ , whose tabloid set consists of the single empty tabloid, so $M^{\emptyset} = C$ is one-dimensional with trivial action and coincides with the regular module $C [S_{0}]$ of the trivial group $S_{0} = {1}$ . Thus $M^{(n)}$ is the trivial module and $M^{(1^{n})}$ the regular module for every $n \geq 1$ , and for $n = 0$ both descriptions give the same one-dimensional module. ∎

step 2.1step 2.2F1F3

Depends on

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Sources

Charlotte Chan, Representation Theory of Symmetric Groups - Example 3.7(a)(b), printed p. 12 (PDF p. 13) (standard reference, not scraped)
David Craven, Groups, Geometries and Representation Theory - Section 1.6, printed pp. 13-14 (PDF pp. 15-16) (standard reference, not scraped)