Polynomials and Representations XIX

Representations of the Symmetric Group

Let [d] be the set {1,…,d}, and Sbe the group of bijections g: [d] \to [d]. From here on, we shall look at the representations of S_d. Note that this requires a good understanding of representation theory (character theory) of finite groups.

To start, let us first find some representations.

Definition. Given a partition \lambda \vdash d, let X_\lambda be the set of partitions of [d] into disjoint subsets A_1 \coprod \ldots \coprod A_l such that |A_i| =\lambda_i for each i. The group G = S_d then acts on X_\lambda via:

(A_1, A_2, \ldots, A_l) \overset {g \in S_d} \mapsto  (g(A_1), g(A_2), \ldots, g(A_l)).

Note

If \lambda_i = \lambda_{i+1}, the partitioning is considered different when we swap A_i and A_{i+1}. For example if d = 4 and \lambda = (2, 2), then X_\lambda has 6 elements:

\begin{aligned}( \{1,2\}, \{3,4\}), \quad (\{1,3\}, \{2,4\}), \quad (\{1,4\}, \{2,3\}), \\ (\{2,3\}, \{1,4\}), \quad (\{2,4\}, \{1,3\}), \quad (\{3,4\}, \{1,2\}).\end{aligned}

In general,

\displaystyle |X_\lambda| = \frac {d!} {\lambda_1 ! \lambda_2 ! \ldots \lambda_l !}.

As X_\lambda is a G-set, we see that U_\lambda := K[X_\lambda] is a K[G]-module for any field K. Two special cases are of interest.

  • \lambda = (d): X has a single element so U_\lambda is the trivial representation;
  • \lambda = (1,1,\ldots)X is just the set of permutations of [d] so U_\lambda is the regular representation.

Exercise

Show that for \lambda = (d-1, 1)U_\lambda is the standard representation of S_d on [d].

Let us study such representations in a more general setting.

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Representations From G-Sets

Let G be a group and XG-set; then K[X] is naturally a K[G]-module. We then have the following isomorphisms of K[G]-modules:

Lemma.

  • K[X]^\vee \cong K[X] (i.e. K[X] is isomorphic to its dual);
  • K[X] \otimes_K K[Y] \cong K[X\times Y];
  • \text{Hom}_K(K[X], K[Y]) \cong K[X\times Y];
  • K[X]^G = \{v\in K[X] : \forall g\in G, gv = v\} has dimension |G\backslash X|, the number of orbits of X.

Note

In general, a representation is not isomorphic to its dual.

Proof

Take the bilinear map K[X] \times K[X] \to K induced linearly from the map:

X \times X\to K, \quad (x, y) \mapsto \delta_{xy} = \begin{cases} 1, \quad &\text{if } x=y,\\ 0,\quad &\text{if } x\ne y.\end{cases}

This induces an isomorphism K[X] \to K[X]^\vee of vector spaces. It is also G-equivariant since the bilinear map is G-equivariant, which proves the first statement.

For the second, linear algebra gives K[X]\otimes_K K[Y] \cong K[X \times Y] as vector spaces, taking x\otimes y \mapsto (x, y). One easily checks that it is G-equivariant.

The third statement follows from the first two since generically, we have \text{Hom}(V, W) \cong V^\vee \otimes_K W for any K[G]-modules V and W. Thus:

\text{Hom}_K(K[X], K[Y]) \cong K[X]^\vee \otimes_K K[Y] \cong K[X] \otimes_K K[Y] \cong K[X\times Y].

Finally, if \sum_{x\in X} c_x x\in K[X] is invariant under every g\in G, then c_x = c_{g x} for all g\in G and x\in X. Thus c_x is constant when x\in X runs through an orbit. ♦

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Now assume that char(K) = 0.

Question. Given \lambda, \mu \vdash d, what is the following value:

\displaystyle \left<U_\mu, U_\lambda\right>_G = \dim_L\text{Hom}_{L[G]}(L[X_\mu], L[X_\lambda])

where L is the algebraic closure of K?

Note

The reader may assume K =\mathbb{C} to simplify matters, in which case LK and when we mention “absolutely irreducible”, just replace it with “irreducible”.

Answer

By the above lemma, we have:

\text{Hom}_{K[G]} (K[X_\mu], K[X_\lambda]) =\text{Hom}_K (K[X_\mu], K[X_\lambda])^G \cong K[X_\lambda \times X_\mu]^G

which has dimension |G \backslash (X_\lambda \times X_\mu)|. Writing \lambda = (\lambda_1, \ldots, \lambda_l) and \mu = (\mu_1, \ldots, \mu_{m}), let us pick (A_1, \ldots, A_l) \in X_\lambda and (B_1, \ldots, B_m)\in X_\mu. Then the orbit of the pair (g(X_\lambda), g(X_\mu)) satisfies:

|g(A_i) \cap g(B_j)|_{i, j} = |g(A_i \cap B_j)|_{i, j} \quad \text{ for } 1\le i \le l, 1 \le j \le m,

so the collection of these numbers is invariant over all g\in G.

Conversely, suppose (A_i), (B_j) and (A_i'), (B_j') are partitions of [d] satisfying:

|A_i| = |A_i'| = \lambda_i, \quad |B_j| = |B_j'| = \mu_j,\quad |A_i \cap B_j| = |A_i' \cap B_j'|

for all i, j. Then we can find g\in S_d mapping the pairwise intersection sets A_i \cap B_j to A_i' \cap B_j' so the pairs (A_i), (B_j) and (A_i'), (B_j') are in the same orbit. Hence the number of orbits is the number of solutions to

\sum_{i=1}^l a_{ij} = \mu_j, \qquad \sum_{j=1}^m a_{ij} = \lambda_i

in non-negative integers a_{ij}.

In particular, the above holds when we replace K by L so:

Conclusion. We have

\left< U_\mu, U_\lambda\right>_G = M_{\lambda\mu}.

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