Generalizing Newton's identities: Trace formula for Schur functors

Question

We work over $\mathbb C$. A general linear group ${\rm GL}(V)$ acts diagonally on the tensor power $V^{\otimes n}$ as

$$(A^{\otimes n})(v_1\otimes\cdots\otimes v_n):=(Av_1)\otimes\cdots\otimes (Av_n).$$

And the symmetric group $S_n$ acts on $V^{\otimes n}$ from the right as

$$(v_1\otimes \cdots \otimes v_n)\sigma=v_{\sigma(1)}\otimes\cdots\otimes v_{\sigma(n)}.$$

The actions of ${\rm GL}(V)$ and $S_n$ commute with each other. More strongly, Schur-Weyl duality states that the complex subalgebras of ${\rm End}(V^{\otimes n})$ generated by these two actions are each other's centralizers, in addition to providing the decomposition of $V^{\otimes n}$ into irreducible representations of ${\rm GL}(V)$, which are the Schur functors applied to $V$, given explicitly by ${\Bbb S}_\lambda(V)=V^{\otimes n}\otimes_{\Bbb C[S_n]}M_\lambda$ where $M_\lambda$ are the irreducible reps of $S_n$ corresponding to integer partitions $\lambda\vdash n$. The assignment $V\mapsto\Bbb S_\lambda(V)$ is indeed functorial, which mean ${\rm GL}(V)$ also acts on $\Bbb S_\lambda(V)$.

GroupProps provides the following trace formula, but does not provide any reference or discussion:

$${\rm tr}_{\Bbb S_\lambda(V)}A=\frac{1}{n!}\sum_{\sigma\in S_n} \chi_\lambda(\sigma) \prod_{i=1}^n {\rm tr}_V(A^i)^{c_{\large i}(\sigma)}, \tag{$\circ$}$$

where $\chi_\lambda$ is the character of $M_\lambda$ and $c_i(\sigma)$ the number of length $i$ cycles in $\sigma$. According to the relation to representation theory section on Wikipedia's article on schur polynomials, it follows from something called the Weyl character formula, but I have no idea how they're related and I'm not really familiar with Lie algebras anyway. I can prove the formula when $\lambda=(1,\cdots,1)$ and $M_\lambda$ is the $1$-dimensional sign representation and $\Bbb S_\lambda(V)$ is the $n$th alternating/exterior power $\Lambda^n(V)$. Here, the trace of $g\in{\rm GL}(V)$ acting on $\Bbb S_\lambda(V)$ is the $n$th elementary symmetric polynomial in the original eigenvalues of $g$. My argument uses generating functions and Vieta's formulas, the latter which is particular to the $e_n$s, so I doubt it generalizes to arbitrary $\lambda$.

I'd like to see a self-contained proof of $(\circ)$, which doesn't assume anything about Schur polynomials, and ideally uses as little representation theory specific to $S_n$ as possible. (For instance, perhaps we could compute the trace of $g\in{\rm GL}(V)$ on $V^{\otimes n}\otimes_{\Bbb C[S_n]}(\Bbb C[S_n]\sum_{\sigma\in S_n}\chi_\lambda(\sigma)\sigma)$ and then divide by $\chi_\lambda(1)$, without ever having to actually know the values $\chi_\lambda(\sigma)$, but that might be optimistic). Your thoughts?

Answer 1

Put $m = \dim V$, and let $\{e_1,\ldots,e_m\}$ be a basis of $V$ over $\mathbb{C}$. Let $\pi : \operatorname{GL}(V) \to \operatorname{GL}(V^{\otimes n})$ and $\rho : S_n \to \operatorname{GL}(V^{\otimes n})$ denote the commuting representations of $\operatorname{GL}(V)$ and $S_n$ on $V^{\otimes n}$, respectively. In particular, for $\sigma \in S_n$ and $1 \leq i_1,\ldots,i_n \leq m$ we have $$ \rho(\sigma)(e_{i_1} \otimes \cdots \otimes e_{i_n}) = e_{i_{\sigma^{-1}(1)}} \otimes \cdots \otimes e_{i_{\sigma^{-1}(n)}}.$$ Fix a partition $\lambda \vdash n$. The operator $P_\lambda \in \operatorname{End}(V^{\otimes n})$, given by $$ P_\lambda = \frac{\dim M_\lambda}{n!}\sum_{\sigma\, \in\, S_n}\overline{\chi_\lambda(\sigma)}\;\rho(\sigma), \tag{1}$$ is the projection onto the $\lambda$-isotypic component of $V^{\otimes n}$ as a module over $S_n$ (cf. e.g. Cor. 4.3.11 on pp. 213-214 in Symmetry, representations, and invariants (2009) by Goodman and Wallach). By Schur-Weyl duality, the $\lambda$-isotypic component of $V^{\otimes n}$ is isomorphic to $\mathbb{S}_\lambda(V)\otimes_{\mathbb{C}} M_\lambda$, as a module over $\operatorname{GL}(V)\times S_n$. Therefore, one can relate the character of $\mathbb{S}_\lambda(V)$ to the character of $V^{\otimes n}$ as follows: $$ \operatorname{tr}_{\mathbb{S}_\lambda(V)}\pi(A) = \frac{1}{\dim M_\lambda}\operatorname{tr}_{V^{\otimes n}}\big(\pi(A)\,P_\lambda\big). \tag{2}$$ For our purposes, it will be more convenient to rewrite $(1)$ with $\sigma$ replaced by $\sigma^{-1}$. Thus, $$P_\lambda = \frac{\dim M_\lambda}{n!}\sum_{\sigma\, \in\, S_n}\chi_\lambda(\sigma)\;\rho(\sigma^{-1}). \tag{3}$$ In view of $(2)$ and $(3)$, in order to prove formula $(\circ)$ in the question above, we have to show that $$\operatorname{tr}_{V^{\otimes n}}\big(\pi(A)\,\rho(\sigma^{-1})\big) = \prod_{k=1}^n \operatorname{tr}_V\big(A^k\big)^{c_k(\sigma)} \tag{4}$$ for $A \in \operatorname{GL}(V)$ and $\sigma \in S_n$. Since the semi-simple elements are everywhere dense in $\operatorname{GL}(V)$ and the characters are polynomial class functions, it suffices to consider only the case when $A$ is diagonal with respect to the chosen basis $\{e_1,\ldots,e_m\}$ of $V$. Thus, let $A(e_i) = x_ie_i$ with $x_i \in \mathbb{C}^\times$. If $\{e_1^*,\ldots,e_m^*\}$ is the dual basis of $V^*$, then we have $$ \begin{multline}\operatorname{tr}_{V^{\otimes n}}\big(\pi(A)\,\rho(\sigma^{-1})\big) = \sum_{1\leq i_1,\ldots,i_n \leq m} \langle e_{i_1}^*\otimes \cdots \otimes e_{i_n}^*,\, A(e_{i_{\sigma(1)}})\otimes \cdots \otimes A(e_{i_{\sigma(n)}})\rangle =\\ \sum_{1\leq i_1,\ldots,i_n\leq m} \delta_{i_1,i_{\sigma(1)}}\cdots \delta_{i_n,i_{\sigma(n)}}\cdot x_{i_1}\cdots x_{i_n},\end{multline} \tag{5}$$ where $\delta_{ij}$ denotes Kronecker's symbol. Fix $\sigma \in S_n$, and let $\sigma = \zeta_1 \cdots \zeta_s$ be its decomposition into a product of disjoint cycles (including those of length $1$). For $1\leq j\leq n$, let $\ell_j$ denote the length of the cycle $\zeta_j$, and fix $p_j \in \{1,\ldots,n\}$ so that $\zeta_j = \left(p_j,\sigma(p_j),\ldots,\sigma^{\ell_j-1}(p_j)\right)$. The coefficient in front of $x_{i_1}\cdots x_{i_n}$ in $(5)$ is non-zero if and only if $i_q = i_{\sigma(q)}$ for all $1 \leq q \leq n$, i.e. if and only if $i_{\sigma^k(p_j)} = i_{p_j}$ for all $1\leq j\leq s$, $0< k <\ell_j$. Thus, the non-zero terms of $(5)$ are of the form $$ x_{i_1} \cdots x_{i_n} = \prod_{j=1}^s \left(\prod_{k=0}^{\ell_j-1}x_{i_{\sigma^k(p_j)}}\right) = x_{i_{p_1}}^{\ell_1}\cdots x_{i_{p_s}}^{\ell_s}.$$ This means that in $(5)$, the values of the indices $i_{p_1},\ldots,i_{p_s}$ can be chosen arbitrarily and independently of each other in $\{1,\ldots,m\}$, while the values of the remaining indices are determined by those of the former ones, according to the cycle structure of $\sigma$. Hence, putting $r_j = i_{p_j}$ for $1\leq j \leq s$, we obtain $$ \operatorname{tr}_{V^{\otimes n}}\big(\pi(A)\,\rho(\sigma^{-1})\big) = \sum_{1\leq r_1,\ldots,r_s\leq m} x_{r_1}^{\ell_1} \cdots x_{r_s}^{\ell_s}.$$ The right hand side of the above formula is nothing but $$ \prod_{j=1}^s\left(\sum_{r=1}^m x_r^{\ell_j}\right) = \prod_{j=1}^s \operatorname{tr}_V\left(A^{\ell_j}\right).$$ Grouping together the contributions from cycles of the same length in this product, we obtain the right hand side of $(4)$.

The idea for this calculation was taken from http://mathnt.mat.jhu.edu/zelditch/LargeNseminar/archive/SchurWeyl.pdf, slides 53-54