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}.


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:


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


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)?


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.


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.



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:

PropositionV_\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.

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