11
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Note that formula $(\circ)$ expresses the character of $\mathbb{S}_\lambda(U) = V\otimes_{\mathbb{C}[H]}W$ as a polynomial in certain values of the character $\chi_W$ of $W$ and the character $\chi_U$ of the fundamental $G$-module $U$, instead of the character $\chi_V = (\chi_U)^n$ of $V = U^{\otimes n}$ itself. $\endgroup$ – ivanpenev Jun 7 '14 at 16:04
  • $\begingroup$ @ivanpenev In light of your comment I've narrowed my question down. $\endgroup$ – blue Jun 25 '14 at 22:05
14
+100
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ Awesome, thanks. I'm glad I was right about inflating to the isotypic component, but I didn't think to inflate further to the full tensor power using the group algebra idempotent that way. I will arrange for you to get some more rep. $\endgroup$ – blue Jul 8 '14 at 18:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.