On a representation of $S_n$.

Question

Fix a natural number $n$, and a complex vector space $V$ of dimension $d$. Consider the representation of $S_n$ on $V^{\otimes n}$ given by $\rho(\sigma)(v_1\otimes\cdots\otimes\cdots v_n)=v_{\sigma(1)}\otimes\cdots v_{\sigma(n)}$. This is not an irreducible representation (the invariants are the symmetric powers). Is there a formula for its decomposition into irreducible representations, and so its character, for any $d$?

Answer 1

As a representation of $S_n$, $V^{\otimes n}$ is just a permutation representation: namely, if $X = \{ e_1, \dots e_d \}$ is any basis of $V$, it's the permutation representation of the permutation action of $S_n$ on the set of functions from $[n] = \{ 1, 2, \dots n \}$ to $X$.

Accordingly its character is very easy to compute: its value on a permutation $\sigma \in S_n$ is the number of fixed points of $\sigma$ acting on the set of functions $[n] \to X$, which I invite you to verify is equal to $d^{c(\sigma)}$ where $c(\sigma)$ is the total number of cycles of $\sigma$.

The decomposition into irreducibles is harder but more interesting. Note that $V^{\otimes n}$ also has an action of $GL(V)$ commuting with the natural action of $S_n$, and as it turns out, also has a decomposition into irreducible representations of $GL(V)$ as well as $S_n$. Schur-Weyl duality relates these decompositions: it says that we have a decomposition

$$V^{\otimes n} \cong \bigoplus_{\lambda} V_{\lambda} \otimes S_{\lambda}$$

where the sum runs over partitions of $n$, $S_{\lambda}$ is the corresponding Specht module, and

$$V_{\lambda} \cong \text{Hom}_{S_n}(S_{\lambda}, V^{\otimes n})$$

is by definition the corresponding Schur functor, which is either an irreducible representation of $GL(V)$ or zero. The two simplest special cases are the symmetric and exterior powers, which correspond to the trivial and sign representations of $S_n$ respectively.

The characters of algebraic representations of $GL(V)$ can be identified with symmetric functions on $d$ variables (by looking at the restriction to diagonal matrices), and accordingly the character of $V_{\lambda}$ is the corresponding Schur polynomial $s_{\lambda}$. In particular, plugging in all $1$s to the Schur polynomial gives $\dim V_{\lambda}$ and hence the multiplicity of $S_{\lambda}$ in $V^{\otimes n}$; this formula can be found here.