Polynomials and Representations XXI

We have established that all irreps of $S_d$ are defined over $\mathbb{Q}$ and hence any field of characteristic 0. For convenience we will fix $K = \mathbb{C}$ .

Twists

For any group G and representation $\rho : G \to \text{GL}(V)$ over $\mathbb{C},$ if $\chi : G \to \mathbb{C}^*$ is a group homomorphism, we can twist $\rho$ as follows:

$\rho \otimes \chi : G\to \text{GL}(V), \qquad g \mapsto \rho(g)\chi(g).$

Sometimes, we also write $V\otimes\chi$ for the twist; this gives an action on the set of all irreducible representations of G. Note that $\dim(V\otimes\chi) = \dim(V)$ so the dimension of V is invariant among the equivalence classes of twists.

For example in $S_4$ , the character table is:

chartable_s4_2

so we have 3 equivalence classes of irreps if we twist by $\chi_{\text{alt}}$ .

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Symmetric Group

When $G = S_d,$ we let $\chi$ be the alternating character, i.e. $\chi(g) := \text{sgn}(g) \in \{+1, -1\}.$ Now we consider the modules $U_\lambda \otimes \chi = \mathbb{C}[X_\lambda] \otimes \chi.$

Question. What is the following:

$\left< U_\mu, U_\lambda \otimes \chi\right> = \dim_{\mathbb{C}} \text{Hom}_{\mathbb{C}[G]}(U_\mu, U_\lambda \otimes\chi)?$

Answer

Note that the underlying vector space for $U_\lambda \otimes\chi$ is the same as $U_\lambda$ though the action has a twist. Recall that, as $G$ -modules,

$\text{Hom}_\mathbb{C}(U_\mu, U_\lambda \otimes \chi) \cong \mathbb{C}[X_\mu] \otimes \mathbb{C}[X_\lambda] \otimes \chi\cong \mathbb{C}[X_\mu \times X_\lambda] \otimes \chi$

and we wish to find all G-invariant elements in this space.

Each element $\alpha$ on the RHS is a linear combination of elements $((A_i), (B_j)) \in X_\mu \times X_\lambda.$ Let us unwind the definition for $\alpha$ to be invariant under G. Note that:

$[d] = A_1 \coprod \ldots \coprod A_l = B_1 \coprod \ldots \coprod B_m, \qquad |A_i| = \mu_i, |B_j| = \lambda_j$

and $g\in S_n$ takes it to $\chi(g) \cdot (g(A_i), g(B_j))$ . Now if any $A_i \cap B_j$ has more than one element, then the transposition $g\in S_d$ which swaps them would flip the sign while leaving $g(A_i) = A_i$ , $g(B_j) = B_j$ . Thus we have shown:

If $|A_i \cap B_j| \ge 2$ for some $i, j$ , then the coefficient of $((A_i), (B_j))$ in $\alpha$ is zero.

Conversely, if $|A_i \cap B_j| \le 1$ for all $i, j$ , then no $g\in S_d$ takes $((A_i), (B_j))$ back to itself, so as it ranges over an orbit, the coefficient of $((A_i), (B_j))$ in $\alpha$ is uniquely determined.

Example

Suppose d=3 and $\lambda = \mu = (2,1)$ so we get

$[3] = \overbrace{\{2,3\}}^{A_1} \cup \overbrace{\{1\}}^{B_1} = \overbrace{ \{1,3\} }^{A_2} \cup \overbrace{\{2\}}^{B_2} = \overbrace{ \{1,2\} }^{A_3} \cup \overbrace{ \{3\}}^{B_3}.$

Then $X_\lambda \times X_\mu$ has 9 elements $(A_i, B_i, A_j, B_j)$ for $1\le i,j\le 3$ . Any invariant element in $\mathbb{C}[X_\mu \times X_\lambda]\otimes \chi$ must be a multiple of:

$\begin{aligned} &(A_1, B_1,A_2, B_2) - (A_1, B_1, A_3, B_3) - (A_2, B_2, A_1, B_1) \\- &(A_3, B_3, A_2, B_2) + (A_2, B_2, A_3, B_3) + (A_3, B_3, A_1, B_1).\end{aligned}$

Conclusion. We have:

$\left<U_\mu, U_\lambda\otimes \chi\right> = \text{Hom}_{\mathbb{C}[G]}(U_\mu, U_\lambda \otimes \chi) = N_{\lambda\mu},$

the number of binary matrices $a_{ij}$ such that $\sum_i a_{ij} = \lambda_j$ and $\sum_j a_{ij} = \mu_i.$

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Consequences

We saw that $\mathbf N = (N_{\lambda\mu})$ satisfies:

$\mathbf N = \mathbf K^t \mathbf J \mathbf K,$

where J is the permutation matrix which swaps $\lambda$ with its transpose $\overline \lambda.$ Writing $U_\mu = \oplus_{\nu} V_\nu^{\oplus K_{\nu\mu}}$ we get:

$\displaystyle N_{\mu\lambda} = N_{\lambda\mu} = \sum_\nu K_{\nu\mu}\left<V_\nu, U_\lambda\otimes\chi\right>.$

If $A_{\nu\lambda} = \left<V_\nu, U_\lambda\otimes\chi\right>$ then writing in matrix form gives $\mathbf K^t \mathbf A = \mathbf N.$ Hence $\mathbf A = \mathbf J\mathbf K$ and we have $\left<V_\nu, U_\lambda\otimes\chi\right> = K_{\overline\nu\lambda}.$ Thus:

$\displaystyle \bigoplus_\nu (V_\nu \otimes\chi)^{\oplus K_{\nu\lambda}}= U_\lambda\otimes\chi = \bigoplus_\nu V_\nu^{\oplus K_{\overline\nu \lambda}} = \bigoplus_\nu V_{\overline\nu} ^{\oplus K_{\nu\lambda}}.$

Since the matrix $K_{\nu\lambda}$ is invertible we have proven:

Proposition. $V_\lambda\otimes \chi \cong V_{\overline\lambda}.$

In particular, $V_\lambda\otimes\chi \cong V_\lambda$ if and only if $\lambda$ is its own transpose, e.g. (5, 3, 2, 1, 1).

Hence under the Frobenius map, the corresponding symmetric polynomial for $U_\lambda\otimes\chi$ is $\sum_\nu K_{\overline\nu \lambda}s_\nu = e_\lambda.$

Also, the involution $\omega : \Lambda^{(d)} \to \Lambda^{(d)}$ which maps $e_\lambda \leftrightarrow h_\lambda$ corresponds to $V \leftrightarrow V\otimes\chi.$ This gives an algebraic interpretation for $\omega.$