Lectures in Representation Theory

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 28 August 2013

Lecture 22

Recall our setting from last time. Suppose $A$ and $B$ are algebras and $V$ is a completely decomposable $A\text{-module,}$ with decomposition $V={\sum }_{\lambda \in \hat{V}}{m}_{\lambda }{V}^{\lambda }\text{.}$ Suppose that $V$ is also a module for $B$ and that $V\left(B\right)=\overline{V\left(A\right)}\text{.}$

We showed that $B$ has certain irreducible modules $W^{\lambda }$ (called $C^{\lambda }$ last time), each of dimension $m_{\lambda }$ for $\lambda \in \hat{V}\text{.}$ The module $V$ is a completely decomposable $B\otimes A$ module, with decomposition into irreducible $B\otimes A$ modules $$V\cong \sum _{\lambda \in \hat{V}}{W}^{\lambda }\otimes V^{\lambda }\text{.}$$

Corollary 3.22 Let ${\chi }_B^{\lambda }$ be the character of $W^{\lambda },$ and let ${\chi }_A^{\lambda }$ be the character of $V^{\lambda }\text{.}$ Then for $b\in B$ and $a\in A$ $${\text{tr}}_V(b\otimes a)=\sum _{\lambda \in \hat{V}}{\chi }_B^{\lambda }\left(b\right){\chi }_A^{\lambda }\left(a\right)$$

		Proof.
	Let ${\left\{w_i^{\lambda }\right\}}_{1\le i\le m_{\lambda }}$ be a basis for each $W^{\lambda }$ and similarly choose bases ${\left\{v_j^{\lambda }\right\}}_{1\le j\le d_{\lambda }}$ for the $V^{\lambda }\text{.}$ Then by the above decomposition, the set $$\{w_i^{\lambda }\otimes v_j^{\lambda }\hspace{0.17em}|\hspace{0.17em}1\le i\le m_{\lambda },1\le j\le d_{\lambda },\lambda \in \hat{V}\}$$ forms a basis for $V\text{.}$ Thus $$\begin{array}{ccc}{\text{tr}}_V(b\otimes a)& =& \sum _{\lambda \in \hat{V}}\sum _{i=1}^{m_{\lambda }}\sum _{j=1}^{d_{\lambda }}\left((b\otimes a)(w_i^{\lambda }\otimes v_j^{\lambda })\right){|}_{w_i^{\lambda }\otimes v_j^{\lambda }}\\ & =& \sum _{\lambda \in \hat{V}}\sum _{i=1}^{m_{\lambda }}\sum _{j=1}^{d_{\lambda }}(\left(b{w}_i^{\lambda }\right)\otimes \left(a{v}_j^{\lambda }\right)){|}_{w_i^{\lambda }\otimes v_j^{\lambda }}\\ & =& \sum _{\lambda \in \hat{V}}\sum _{i=1}^{m_{\lambda }}\sum _{j=1}^{d_{\lambda }}\left(b{w}_i^{\lambda }\right){|}_{w_i^{\lambda }}\left(a{v}_j^{\lambda }\right){|}_{v_j^{\lambda }}\\ & =& \sum _{\lambda \in \hat{V}}\left(\sum _{i=1}^{m_{\lambda }}\left(b{w}_i^{\lambda }\right){|}_{w_i^{\lambda }}\right)\left(\sum _{j=1}^{d_{\lambda }}\left(a{v}_j^{\lambda }\right){|}_{v_j^{\lambda }}\right)\\ & =& \sum _{\lambda \in \hat{V}}{\chi }_B^{\lambda }\left(b\right){\chi }_A^{\lambda }\left(a\right)\end{array}$$ since ${\chi }_B^{\lambda }={\text{tr}}_{W^{\lambda }}$ and ${\chi }_A^{\lambda }={\text{tr}}_{V^{\lambda }}\text{.}$ $\square$

Consider the regular representation $L_A:A\to M_d\left(ℂ\right)$ $(d=\text{dim}\hspace{0.17em}A)$ of an algebra $A$ defined by $a\mapsto L_a,$ the transformation defined by left multiplication by $a\text{.}$ Recall that we often denoted this action by $a_1·\overrightarrow{a_2}=\overline{a_1}\overrightarrow{a_2},$ using the vector notation to distinguish elements of the vector space $\overrightarrow{A}$ from transformations in $A\text{.}$

We showed that $\bar{A}=R_A,$ that is, those transformations which commute with all left multiplications consist precisely of the right multiplications by elements of $A\text{.}$ Let $B=A^{\text{op}},$ the opposite algebra of $A\text{.}$ The algebra $A^{\text{op}}$ is defined to have $\overrightarrow{A}$ as the underlying vector space, but with multiplication defined by $$a_1\circ a_2=a_2{a}_1$$ where the multiplication of the right hand side is the usual multiplication in $A\text{.}$

The algebra $B=A^{\text{op}}$ acts on $\overrightarrow{A}$ by right multiplication: for $b\in B$ and $a\in A$ $$b\overrightarrow{a}=\overrightarrow{ab}$$

Problem 3.23 Verify that this defines a module action of $A^{\text{op}}$ on $\overrightarrow{A}\text{.}$

Then $B$ acts as $R_A=\overline{\overrightarrow{A}\left(A\right)},$ as in the statement of the theorem. If $\overrightarrow{A}$ is completely decomposable as an $A\text{-module}$ (i.e. if $A$ is a semisimple algebra), then as a $B\otimes A$ module, $$\overrightarrow{A}\cong \underset{\lambda \in \hat{A}}{⨁}{B}^{\lambda }\otimes A^{\lambda }$$ where $B^{\lambda }$ and $A^{\lambda }$ are the irreducible $B$ and $A$ modules respectively. Note that in this case, each irreducible $A\text{-module}$ appears in the decomposition of $V=\overrightarrow{A}\text{.}$

We knew this already, however. If $A$ is semisimple, then $A\cong {\oplus }_{\lambda }{M}_{d_{\lambda }}\left(ℂ\right),$ and the irreducible (left) $A\text{-modules}$ are the column spaces of dimension $d_{\lambda }$ for each simple component. Then, by taking transposes, we see that the irreducible $A^{\text{op}}$ modules are the row vectors for each simple component (Note that they have the same dimension as the corresponding $A\text{-module).}$ Then $M_{d_{\lambda }}\left(ℂ\right)\cong B^{\lambda }\otimes A^{\lambda }$ and we have the above decomposition.

We do get one useful piece of information from this situation, however. Continuing with $A$ semisimple, suppose $A^{\lambda }$ has character ${\chi }^{\lambda },$ and denote the representation $A^{\lambda }:A\to M_{d_{\lambda }}\left(ℂ\right)$ by $a\mapsto A^{\lambda }\left(a\right)\text{.}$ Then the irreducible modules for the opposite algebra $B^{\lambda }:B\to M_{d_{\lambda }}\left(ℂ\right)$ are defined by $B^{\lambda }\left(b\right)={\left(A^{\lambda }\left(b\right)\right)}^t\text{.}$ Thus, $$\begin{array}{ccc}{\chi }_B^{\lambda }\left(b\right)& =& \text{tr}\left(B^{\lambda }\left(b\right)\right)\\ & =& \text{tr}\left(A^{\lambda }{\left(b\right)}^t\right)\\ & =& \text{tr}\left(A^{\lambda }\left(b\right)\right)\\ & =& {\chi }_A^{\lambda }\left(b\right)\end{array}$$

Applying the corollary, $${\text{tr}}_{\overrightarrow{A}}(b\otimes a)=\sum _{\lambda \in \hat{A}}{\chi }_B^{\lambda }\left(b\right){\chi }_A^{\lambda }\left(a\right)=\sum _{\lambda \in \hat{A}}{\chi }_A^{\lambda }\left(b\right){\chi }_A^{\lambda }\left(a\right)\text{.}$$

We next evaluate this trace in another way. Choose a basis $ℬ={\left\{g_i\right\}}_{1\le i\le \text{dim}\hspace{0.17em}A}$ and a nondegenerate trace $\overrightarrow{t}=\left(t_{\lambda }\right)\text{.}$ Recall that $⟨b,a⟩=\overrightarrow{t}\left(ba\right)$ is then a non-degenerate symmetric bilinear form on $A,$ so we may choose a dual basis ${ℬ}^{*}=\left\{g_j^{*}\right\}$ with respect to this form.

We defined $\left[a\right]={\sum }_{i=1}^{\text{dim}\hspace{0.17em}A}{g}_{i}a{g}_i^{*}\text{.}$ Evaluating the trace above, $$\begin{array}{ccc}{\text{tr}}_{\overrightarrow{A}}(b\otimes a)& =& \sum _i(b\otimes a)g_i{|}_{g_i}\\ & =& \sum _i\left(a{g}_{i}b\right){|}_{g_i}\\ & =& \sum _i⟨a{g}_{i}b,g_i^{*}⟩\\ & =& \sum _i\text{tr}\left(a{g}_{i}b{g}_i^{*}\right)\\ & =& \text{tr}\left(a\sum _i{g}_{i}b{g}_i^{*}\right)\\ & =& ⟨a,\left[b\right]⟩\text{.}\end{array}$$ We have shown previously that the $[·]$ mapping is an adjoint map with respect to this form, hence ${\text{tr}}_{\overrightarrow{A}}(b\otimes a)=⟨\left[a\right],b⟩\text{.}$ Combining these two calculations,

Proposition 3.24 Let $A$ be a semisimple algebra with irreducible representations $A^{\lambda }$ indexed by $\lambda \in \hat{A},$ and denote the character of $A^{\lambda }$ by ${\chi }^{\lambda }\text{.}$ Then $$\sum _{\lambda \in \hat{A}}{\chi }^{\lambda }\left(b\right){\chi }^{\lambda }\left(a\right)=⟨\left[a\right],b⟩=⟨a,\left[b\right]⟩\text{.}$$

Example 3.25 Let $G$ be a finite group, and let $A=ℂG$ be the group algebra over $ℂ\text{.}$ By Maschke’s theorem, $A$ is semisimple. By definition, the elements of the group $G$ form a basis for $A\text{.}$ We have shown that the linear function $\overrightarrow{t}:A\to ℂ$ defined by $$\overrightarrow{t}\left(a\right)=a{|}_1$$ is a nondegenerate trace on $A,$ and that the dual basis with respect to the associated bilinear form is $\left\{g^{-1}\hspace{0.17em}\right|\hspace{0.17em}g\in G\}\text{.}$ Recall also that if $h\in G$ then $$\left[h\right]=\sum _{g\in G}gh{g}^{-1}=s_h\sum _{g\in {𝒞}_h}g$$ where $s_k$ is the order of the stabilizer of $h$ under conjugation, and ${𝒞}_h$ is the conjugacy class of $h$ (so $s_k={\displaystyle \raisebox{1ex}{$\left|G\right|$}\!\left/ \!\raisebox{-1ex}{$\left|{𝒞}_h\right|$}\right.}\text{).}$ Applying the proposition in this context yields:

Proposition 3.26 (Second Orthogonality Condition) Let $G$ be a finite group. If $g,h\in G,$ then $$\sum _{\lambda \in \hat{A}}{\chi }^{\lambda }\left(g\right){\chi }^{\lambda }\left(h^{-1}\right)=⟨\left[g\right],h^{-1}⟩=\{\begin{array}{cc}{s}_g& \text{if}\hspace{0.17em}h\in {𝒞}_g\\ 0& \text{otherwise}\end{array}$$

3.1$$Schur-Weyl Duality

Let $v_1,v_2,\dots ,v_n$ be non-commuting variables and let $V$ be the $ℂ$ vector space with basis $\left\{v_i\right\}\text{.}$ Let $V^{\otimes m}$ be the vector space with basis $\left\{v_{i_1}{v}_{i_2}\cdots v_{i_m}\hspace{0.17em}\right|\hspace{0.17em}1\le i_j\le m\}\text{.}$ Then ${𝒮}_m$ acts on $V^{\otimes m}$ by permuting the subscripts of the basis vectors $$\sigma ·v_{i_1}{v}_{i_2}\cdots v_{i_m}=v_{i_{\sigma \left(1\right)}}{v}_{i_{\sigma \left(2\right)}}\cdots v_{i_{\sigma \left(m\right)}}\text{.}$$

What is $\overline{V^{\otimes m}\left(ℂ{𝒮}_m\right)}\text{?}$ We know that $ℂ{𝒮}_m$ is semisimple by Maschke’s theorem, so $V^{\otimes m}$ is completely decomposable. To use the theorems of the last few classes, we’d like to find an algebra $B$ acting on $V^{\otimes m}$ such that $V^{\otimes m}\left(B\right)=\overline{V^{\otimes m}\left(ℂ{𝒮}_m\right)}\text{.}$ Then, $$V^{\otimes m}=\underset{\underset{\ell \left(\lambda \right)\le n}{\lambda ⊢m}}{⨁}{S}_{\lambda }\otimes V^{\lambda }$$ where $S_{\lambda }$ are the irreducible $ℂ{𝒮}_m$ modules and $V^{\lambda }$ are the irreducible modules for $B\text{.}$

Of course, there is a natural action of the general linear group $\text{GL}\left(n\right),$ the set of all $n\times n$ invertible matrices, on $V$ given by $$g·v_i=\sum _{j=1}^n{v}_j{g}_{j,i}$$ where $g=\left(g_{i,j}\right)\text{.}$ We may extend this action to $V^{\otimes m}$ as the linear extension of $$g·v_{i_1}{v}_{i_2}\cdots v_{i_m}=\left(g{v}_{i_1}\right)\left(g{v}_{i_2}\right)\cdots \left(g{v}_{i_m}\right)\text{.}$$ This is a group representation, so we may consider the representation of the group algebra $ℂ\text{GL}\left(n\right)\text{.}$ A word of caution is necessary: since this is a group algebra, the invertible matrices form a basis and the addition operation is formal addition which differs from the usual addition of matrices.

It is easy to see that $V^{\otimes m}\left(ℂ\text{GL}\left(n\right)\right)\subseteq \overline{V\left(ℂ{𝒮}_m\right)},$ that is, the action of $\text{GL}\left(n\right)$ commutes with the action of ${𝒮}_m$ on $V^{\otimes m}$ (we shall prove it). That the reverse inclusion holds is as surprising as it is beautiful.

Theorem 3.27 (Schur-Weyl Duality) With the above notation $$V^{\otimes m}\left(ℂ\text{GL}\left(n\right)\right)=\overline{V^{\otimes m}\left(ℂ{𝒮}_m\right)}\text{.}$$ This is equivalent to the classical Fundamental Theorem of Invariant Theory.

Lemma 3.28 $V^{\otimes m}\left(ℂ\text{GL}\left(n\right)\right)\subseteq \overline{V^{\otimes m}\left(ℂ{𝒮}_m\right)}\text{.}$

		Proof.
	Let $g=\left(g_{i,j}\right)\in \text{GL}\left(n\right)$ and suppose $w=v_{i_1}{v}_{i_2}\cdots v_{i_m}$ is a basis word in $V^{\otimes m}\text{.}$ Then $$\begin{array}{ccc}g·w& =& \left(g{v}_{i_1}\right)\cdots \left(g{v}_{i_m}\right)\\ & =& \sum _{1\le j_1,j_2,\dots ,j_m\le n}{v}_{j_1}{v}_{j_2}\cdots v_{j_m}{g}_{j_1,i_1}{g}_{j_2,i_2}\cdots g_{j_m,i_m}\end{array}$$ Then $$\begin{array}{ccc}g\sigma ·w& =& g·v_{i_{\sigma \left(1\right)}}\cdots v_{i_{\sigma \left(m\right)}}\\ & =& \sum _{1\le j_1,j_2,\dots ,j_m\le n}{v}_{j_1}{v}_{j_2}\cdots v_{j_m}{g}_{j_1,i_{\sigma \left(1\right)}}{g}_{j_2,i_{\sigma \left(2\right)}}\cdots g_{j_m,i_{\sigma \left(m\right)}}\end{array}$$ Acting by $\sigma \in {𝒮}_m,$ we have $$\begin{array}{ccc}\sigma g·w& =& \sigma ·\sum _{1\le j_1,j_2,\dots ,j_m\le n}{v}_{j_1}{v}_{j_2}\cdots v_{j_m}{g}_{j_1,i_1}{g}_{j_2,i_2}\cdots g_{j_m,i_m}\\ & =& \sigma ·\sum _{1\le j_1,j_2,\dots ,j_m\le n}{v}_{j_{{\sigma }^{-1}\left(1\right)}}{v}_{j_{{\sigma }^{-1}\left(2\right)}}\cdots v_{j_{{\sigma }^{-1}\left(m\right)}}{g}_{j_{{\sigma }^{-1}\left(1\right)},i_1}{g}_{j_{{\sigma }^{-1}\left(2\right)},i_2}\cdots g_{j_{{\sigma }^{-1}\left(m\right)},i_m}\end{array}$$ after replacing $j_k$ with $j_{{\sigma }^{-1}\left(k\right)}\text{.}$ Acting by sigma, we obtain $$\begin{array}{ccc}\sigma g·w& =& \sum _{1\le j_1,j_2,\dots ,j_m\le n}{v}_{j_1}{v}_{j_2}\cdots v_{j_m}{g}_{j_{{\sigma }^{-1}\left(1\right)},i_1}{g}_{j_{{\sigma }^{-1}\left(2\right)},i_2}\cdots g_{j_{{\sigma }^{-1}\left(m\right)},i_m}\\ & =& \sum _{1\le j_1,j_2,\dots ,j_m\le n}{v}_{j_1}{v}_{j_2}\cdots v_{j_m}{g}_{j_1,i_{\sigma \left(1\right)}}{g}_{j_2,i_{\sigma \left(2\right)}}\cdots g_{j_m,i_{\sigma \left(m\right)}}\\ & =& g\sigma ·w\end{array}$$ Hence $V^{\otimes m}\left(ℂ\text{GL}\left(n\right)\right)\subseteq \overline{V^{\otimes m}\left(ℂ{𝒮}_m\right)}\text{.}$ $\square$

The remaining direction of the theorem will take some work. Although we have not yet proven that $\text{GL}\left(n\right)$ generates the full centralizer of $ℂ{𝒮}_m$ on $V^{\otimes m},$ we do know that the actions of $\text{GL}\left(n\right)$ and ${𝒮}_m$ commute, hence, we may view $V^{\otimes m}$ as a $ℂ\text{GL}\left(n\right)\otimes ℂ{𝒮}_m$ bimodule. Here the action is given by $$g\otimes \sigma ·v_{i_1}\cdots v_{i_m}=g\sigma \left(v_{i_1}\cdots v_{i_m}\right)\text{.}$$

Let us compute the trace of $g\otimes \sigma$ on $V^{\otimes m},$ $$\text{tr}(g\otimes \sigma )=\sum _{1\le i_1,i_2,\cdots ,i_m\le m}\left(g\sigma \left(v_{i_1}{v}_{i_2}\cdots v_{i_m}\right)\right){|}_{v_{i_1}\cdots v_{i_m}}\text{.}$$ For $\pi \in {𝒮}_m$ and $A\in \text{GL}\left(n\right),$ we have $$\begin{array}{ccc}\text{tr}(Ag{A}^{-1}\otimes \pi \sigma {\pi }^{-1})& =& \text{tr}\left(Ag{A}^{-1}\pi \sigma {\pi }^{-1}\right)\\ & =& \text{tr}\left(Ag\pi \sigma {\pi }^{-1}{A}^{-1}\right)\text{Actions Commute}\\ & =& \text{tr}\left(A^{-1}Ag\pi \sigma {\pi }^{-1}\right)\text{Trace Prop.}\\ & =& \text{tr}\left(\pi g\sigma {\pi }^{-1}\right)\text{Actions Commute}\\ & =& \text{tr}\left(g\sigma {\pi }^{-1}\pi \right)\text{Trace Prop.}\\ & =& \text{tr}(g\otimes \sigma )\text{.}\end{array}$$ Thus we may choose to calculate using convenient conjugates of $g$ and $\sigma \text{.}$ For $\sigma \in {𝒮}_m,$ there exists $\pi \in {𝒮}_m$ such that $$\pi \sigma {\pi }^{-1}={\gamma }_{\mu }$$ where $\mu$ is the cycle type of $\sigma \text{.}$ Moreover, there exists $S\in \text{GL}\left(n\right)$ such that $A=Sg{S}^{-1}$ is in Jordan Canonical Form. In particular, the eigenvalues $x_1,\dots ,x_n$ lie along the diagonal. These are nonzero, since $g$ is invertible.

We have $\text{tr}(\sigma \otimes g)=\text{tr}({\gamma }_{\mu }\otimes A)\text{.}$

Homework Problem 3.29 Let $\mu =({\mu }_1,{\mu }_2,\dots ,{\mu }_k)⊢m\text{.}$ Then $$\text{tr}({\gamma }_{\mu }\otimes A)=\prod _{i=1}^k\text{tr}\left({\gamma }_{{\mu }_i}A\right)$$

So we need only compute $\text{tr}({\gamma }_r\otimes A)$ on $V^{\otimes m},$ with $A$ in Jordan Canonical Form. Write $A=\left(a_{i,j}\right)$ so $a_{i,i}=x_i$ and $a_{i,j}=0$ for $j>i\text{.}$ Then $$\begin{array}{ccc}\text{tr}({\gamma }_r\otimes A)& =& \sum _{1\le i_1,i_2,\cdots ,i_r\le n}\left({\gamma }_{r}A\left(v_{i_1}{v}_{i_2}\cdots v_{i_r}\right)\right){|}_{v_{i_1}{v}_{i_2}\cdots v_{i_r}}\\ & =& \sum _{1\le i_1,i_2,\cdots ,i_r\le n}{\gamma }_r\sum _{1\le j_1,j_2,\cdots ,j_r\le n}{v}_{j_1}\cdots v_{j_r}{a}_{j_1,i_1}\cdots a_{j_r,i_r}{|}_{v_{i_1}{v}_{i_2}\cdots v_{i_r}}\\ & =& \sum _{\overrightarrow{i},\overrightarrow{j}}{v}_{j_r}{v}_{j_1}{v}_{j_2}\cdots v_{j_{r-1}}{a}_{j_1,i_1}{a}_{j_2,i_2}\cdots a_{j_r,i_r}{|}_{v_{i_1}{v}_{i_2}\cdots v_{i_r}}\\ & =& \sum _{\overrightarrow{i}}\sum _{j_k}{v}_{j_r}{v}_{j_1}\cdots v_{j_{r-1}}{a}_{j_1,i_1}{a}_{j_2,i_2}\cdots a_{j_r,i_r}{|}_{v_{i_1}\cdots v_{i_r}}\\ & =& \sum _{\overrightarrow{i}}{a}_{i_2,i_1}{a}_{i_3,i_2}\cdots a_{i_r,i_{r-1}}{a}_{i_1,i_r}\end{array}$$ Note that the terms in the last summation are zero unless $$i_1\le i_r\le i_{r-1}\le \cdots \le i_3\le i_2\le i_1\text{.}$$ Hence $\text{tr}({\gamma }_r\otimes A)={\sum }_{i=1}^n{a}_{i,i}^r={\sum }_{i=1}^n{x}_i^r$ is a symmetric function in the eigenvalues $x_i$ (called the $r\text{th}$ power symmetric function $p_r(x_1,\dots ,x_n)\text{).}$ To summarize,

Proposition 3.30 If $\sigma \in {𝒮}_m$ has cycle type $\mu$ and $g\in \text{GL}\left(n\right)$ has eigenvalues $x_1,x_2,\dots ,x_n,$ then

• $\text{tr}(\sigma \otimes g)$ depends only on $\mu$ and the $x_i\text{.}$
• $\text{tr}({\gamma }_{\mu }\otimes g)=p_{\mu }(x_1,x_2,\dots ,x_n),$ the $\mu \text{-power}$ symmetric function.

Recall that can write $$p_{\mu }(x_1,\dots ,x_n)=\sum _{\underset{\ell \left(\lambda \right)\le n}{\lambda ⊢m}}{\chi }^{\lambda }\left(\mu \right)s_{\lambda }(x_1,\dots ,x_n)$$ where $s_{\lambda }$ is the Schur function and ${\chi }^{\lambda }$ is the character of $ℂ{𝒮}_m$ labeled by $\lambda \text{.}$

Aside. Note that not all characters of $\text{GL}\left(n\right)$ are rational functions of the eigenvalues. Let $V:\text{GL}\left(n\right)\to M_2\left(ℂ\right)$ be the representation defined by $$V\left(g\right)=\left(\begin{array}{cc}1& \text{log}\hspace{0.17em}\left|\text{det}\left(g\right)\right|\\ 0& 1\end{array}\right)$$ whose character has constant value 2.

Representations of $\text{GL}\left(n\right)$ in which the matrix $V\left(g\right)$ is a rational function of the entries of $g$ are called rational representations. If the entries of $V\left(G\right)$ are, in fact, polynomial functions in the entries of $g,$ the representation $V$ is said to be a polynomial representation. Since $\text{tr}(·)$ is a polynomial in the entries of $V\left(g\right),$ one can equally refer to rational or polynomial characters. Of course, the polynomial representations are a subset of the rational representations.

The proof of the fundamental theorem via symmetric functions, as suggested last time, can be found in paper 59 of Schur’s collected works. The proof we offer here is due to Curtis and Reiner [CR62].

Theorem 3.31 Fundamental Theorem of Invariant Theory. $$V^{\otimes m}\left(ℂ\left[{\text{GL}}_n\right]\right)={\text{End}}_{ℂ\left[S_m\right]}\left(V^{\otimes m}\right)\text{.}$$

		Proof.
	We know that $V^{\otimes m}\left(ℂ\left[{\text{GL}}_n\right]\right)\subseteq {\text{End}}_{ℂ\left[S_m\right]}\left(V^{\otimes m}\right),$ so we show the reverse inclusion. Let $c\in {\text{End}}_{ℂ}\left(V^{\otimes m}\right),$ Then the action of $c$ on a word $v_{i_1}{v}_{i_2}\cdots v_{i_m}\in V^{\otimes m}$ is $$c·v_{i_1}{v}_{i_2}\cdots v_{i_m}=\sum _{1\le j_1,\dots ,j_m\le n}{v}_{j_1}{v}_{j_2}\cdots v_{j_m}{c}_{\overrightarrow{j},\overrightarrow{i}},$$ where $\overrightarrow{j}$ denotes the sequence $j_1,\dots ,j_m,$ $\overrightarrow{i}$ denotes the sequence $i_1,\dots ,i_m,$ and $c_{\overrightarrow{j},\overrightarrow{i}}\in ℂ\text{.}$ If $c\in {\text{End}}_{ℂ\left[S_m\right]}\left(V^{\otimes m}\right),$ then for any $\sigma \in S_m,$ we have $$c·v_{i_1}{v}_{i_2}\cdots v_{i_m}={\sigma }^{-1}c\sigma ·v_{i_1}{v}_{i_2}\cdots v_{i_m}\text{.}$$ Therefore, $$\begin{array}{ccc}\sum _{\overrightarrow{j}}{v}_{j_1}\cdots v_{j_m}{c}_{\overrightarrow{j},\overrightarrow{i}}\hspace{0.17em}& =& c·v_{i_1}{v}_{i_2}\cdots v_{i_m}\\ & =& {\sigma }^{-1}c\sigma ·v_{i_1}{v}_{i_2}\cdots v_{i_m}\\ & =& {\sigma }^{-1}c·v_{i_{\sigma \left(1\right)}}{v}_{i_{\sigma \left(2\right)}}\cdots v_{i_{\sigma \left(m\right)}}\\ & =& {\sigma }^{-1}\sum _{\overrightarrow{j}}{v}_{j_1}\cdots v_{j_m}{c}_{\overrightarrow{j},\sigma \left(\overrightarrow{i}\right)}\\ & =& \sum _{\overrightarrow{j}}{v}_{j_{{\sigma }^{-1}\left(1\right)}}\cdots v_{j_{{\sigma }^{-1}\left(m\right)}}{c}_{\overrightarrow{j},\sigma \left(\overrightarrow{i}\right)}\\ & =& \sum _{\overrightarrow{j}}{v}_{j_1}\cdots v_{j_m}{c}_{\sigma \left(\overrightarrow{j}\right),\sigma \left(\overrightarrow{i}\right)}\end{array}$$ Thus, we conclude that $$c\in {\text{End}}_{ℂ\left[S_m\right]}\left(V^{\otimes m}\right)\iff c_{\overrightarrow{j},\overrightarrow{i}}=c_{\sigma \left(\overrightarrow{j}\right),\sigma \left(\overrightarrow{i}\right)}$$ for all sequences $\overrightarrow{i}$ and $\overrightarrow{j}$ and all $\sigma \in S_m\text{.}$ Let $\Omega$ be the set of pairs of sequences $(\overrightarrow{j},\overrightarrow{i})m$ which we view as a two line array $$(\overrightarrow{i},\overrightarrow{j})=\left(\begin{array}{ccc}{j}_1& \cdots & j_m\\ i_1& \cdots & i_m\end{array}\right),$$ satisfying (1) $1\le i_k\le n,$ (2) $1\le j_1\le j_2\le \cdots \le j_n\le n,$ (3) if $j_k=j_{k+1}$ then $i_k\le i_{k+1}\text{.}$ Then if $c\in {\text{End}}_{ℂ\left[S_m\right]}\left(V^{\otimes m}\right),$ then $c$ depends only on the values $c_{\overrightarrow{j},\overrightarrow{i}}$ for pairs $(\overrightarrow{j},\overrightarrow{i})\in \Omega \text{.}$ We define a basis of ${\text{End}}_{ℂ\left[S_m\right]}\left(V^{\otimes m}\right)$ as follows: for each pair $(\overrightarrow{j},\overrightarrow{i})\in \Omega ,$ let $c^{\overrightarrow{j},\overrightarrow{i}}\in {\text{End}}_{C\left[S_m\right]}\left(V^{\otimes m}\right)$ be defined by $$c^{\overrightarrow{j},\overrightarrow{i}}·v_{k_1}\cdots v_{k_m}=v_{j_{\sigma \left(1\right)}},\cdots ,v_{j_{\sigma \left(m\right)}}$$ if $(k_1,\dots ,k_m)=(v_{i_{\sigma \left(1\right)}},\dots ,v_{i_{\sigma \left(m\right)}})$ for some $\sigma \in S_m,$ and by $$c^{\overrightarrow{j},\overrightarrow{i}}·v_{k_1}\cdots v_{k_m}=0$$ if $(k_1,\dots ,k_m)\ne (v_{i_{\sigma \left(1\right)}},\dots ,v_{i_{\sigma \left(m\right)}})$ for any $\sigma \in S_m\text{.}$ Then $c^{\overrightarrow{j}\overrightarrow{i}}\in {\text{End}}_{ℂ\left[S_m\right]}\left(V^{\otimes m}\right)$ and every element $p\in {\text{End}}_{ℂ\left[S_m\right]}\left(V^{\otimes m}\right)$ can be written as $$p=\sum _{(\overrightarrow{j},\overrightarrow{i})}{p}_{\overrightarrow{j}\overrightarrow{i}}{c}^{\overrightarrow{j}\overrightarrow{i}}$$ for $p_{\overrightarrow{j}\overrightarrow{i}}\in ℂ\text{.}$ Homework Problem 3.32 Show that $\text{dim}\hspace{0.17em}{\text{End}}_{ℂ\left[S_m\right]}\left(V^{\otimes m}\right)=\left(\begin{array}{c}{n}^2+m-1\\ m\end{array}\right)\text{.}$ In particular, if $g\in {\text{GL}}_n,$ and $g$ acts on $V^{\otimes m}$ as the transformation $g^{\otimes m},$ then $$g^{\otimes m}\left(v_{i_1}\cdots v_{i_m}\right)=\sum _{j_1,\dots ,j_m}{v}_{j_1}\cdots v_{j_m}\left(g_{j_1,i_1}\cdots g_{j_m,i_m}\right),$$ and $g^{\otimes m}\in {\text{End}}_{ℂ\left[S_m\right]}\left(V^{\otimes m}\right)$ so $$g^{\otimes m}=\sum _{(\overrightarrow{j},\overrightarrow{i})\in \Omega }{g}_{\overrightarrow{j},\overrightarrow{i}}{c}^{\overrightarrow{j},\overrightarrow{i}}=\sum _{(\overrightarrow{j},\overrightarrow{i})\in \Omega }{g}_{j_1,i_1}\cdots g_{j_m,i_m}{c}^{\overrightarrow{j},\overrightarrow{i}}\text{.}$$ Therefore, $V^{\otimes m}\left(ℂ\left[{\text{GL}}_n\right]\right)=\text{span}\left\{g^{\otimes m}\hspace{0.17em}\right|\hspace{0.17em}g\in {\text{GL}}_n\}\subseteq {\text{End}}_{ℂ\left[S_m\right]}\left(V^{\otimes m}\right)\text{.}$ Now define an inner product on ${\text{End}}_{ℂ\left[S_m\right]}\left(V^{\otimes m}\right)$ by $$⟨c^{\overrightarrow{j},\overrightarrow{i}}{c}^{\overrightarrow{j}\prime ,\overrightarrow{i}\prime }⟩=\{\begin{array}{cc}0& \text{unless}\hspace{0.17em}\overrightarrow{j}\prime =\overrightarrow{j}\hspace{0.17em}\text{and}\hspace{0.17em}\overrightarrow{i}\prime =\overrightarrow{i},\\ 1& \text{if}\hspace{0.17em}\overrightarrow{j}\prime =\overrightarrow{j}\hspace{0.17em}\text{and}\hspace{0.17em}\overrightarrow{i}\prime =\overrightarrow{i},\end{array}$$ for all $(\overrightarrow{j},\overrightarrow{i}),(\overrightarrow{j},\overrightarrow{i})\in \Omega \text{.}$ This makes the basis of $c^{\overrightarrow{j},\overrightarrow{i}}$ orthonormal with respect to $⟨\text{.},\text{.}⟩\text{.}$ Now consider $p\in V^{\otimes m}{\left({\text{GL}}_n\right)}^{\perp }$ where $$V^{\otimes m}{\left({\text{GL}}_n\right)}^{\perp }=\{p\in {\text{End}}_{ℂ\left[S_m\right]}\left(V^{\otimes m}\right)\hspace{0.17em}|\hspace{0.17em}⟨p,g^{\otimes m}⟩=0\hspace{0.17em}\text{for all}\hspace{0.17em}g\in {\text{GL}}_n\}\text{.}$$ Then we write $p$ and $g\in {\text{GL}}_n$ in terms of the basis of $c^{\overrightarrow{j},\overrightarrow{i}}$ as $$\begin{array}{ccc}p& =& \sum _{(\overrightarrow{j},\overrightarrow{i})\in \Omega }{p}_{\overrightarrow{j},\overrightarrow{i}}{c}^{\overrightarrow{j},\overrightarrow{i}}\\ g& =& \sum _{(\overrightarrow{j}\prime ,\overrightarrow{i}\prime )\in \Omega }{g}_{\overrightarrow{j}\prime ,\overrightarrow{i}\prime }{c}^{\overrightarrow{j}\prime ,\overrightarrow{i}\prime }\end{array}$$ and we have $$0=⟨p,g^{\otimes m}⟩=⟨\sum _{(\overrightarrow{j},\overrightarrow{i})\in \Omega }{p}_{\overrightarrow{j},\overrightarrow{i}}{c}^{\overrightarrow{j},\overrightarrow{i}},\sum _{(\overrightarrow{j}\prime ,\overrightarrow{i}\prime )\in \Omega }{g}_{\overrightarrow{j}\prime ,\overrightarrow{i}\prime }{c}^{\overrightarrow{j}\prime ,\overrightarrow{i}\prime }⟩=\sum _{(\overrightarrow{j},\overrightarrow{i})\in \Omega }{p}_{\overrightarrow{j},\overrightarrow{i}}{g}_{\overrightarrow{j}\prime ,\overrightarrow{i}\prime }\text{.}$$ Thus $p\in V^{\otimes m}{\left({\text{GL}}_n\right)}^{\perp }$ if $$\sum _{(\overrightarrow{j},\overrightarrow{i})\in \Omega }{p}_{\overrightarrow{j},\overrightarrow{i}}{g}_{j_1,i_1}\cdots g_{j_m,i_m}=0$$ for all $g=\left(g_{ij}\right)\in {\text{GL}}_n\text{.}$ In other words, we must have $$\sum _{(\overrightarrow{j},\overrightarrow{i})\in \Omega }{p}_{\overrightarrow{j},\overrightarrow{i}}{g}_{j_1,i_1}\cdots g_{j_m,i_m}=0$$ for all choices of $g_{i,j}\in ℂ$ for which $$\text{det}\left(g\right)=\sum _{\tau \in S_n}\epsilon \left(\tau \right)g_{1,\tau \left(1\right)}\cdots g_{n,\tau \left(n\right)}\ne 0\text{.}$$ Let $x_{1,1},x_{1,2},\dots ,x_{n,n}$ be commuting variables, and define polynomials $$P\left(x_{ij}\right)=\sum _{(\overrightarrow{j},\overrightarrow{i})\in \Omega }{P}_{\overrightarrow{j},\overrightarrow{i}}{x}_{j_1,i_1}\cdots x_{j_m,i_m}\in ℂ[x_{1,1},\dots ,x_{n,n}]$$ and $$\text{det}\left(x_{ij}\right)=\sum _{\tau \in S_n}\epsilon \left(\tau \right)x_{1,\tau \left(1\right)}\cdots x_{n,\tau \left(n\right)}\in ℂ[x_{1,1},\dots ,x_{n,n}]\text{.}$$ Then the product $P\left(x_{ij}\right)\text{det}\left(x_{ij}\right)$ is zero when evaluated at all ${\alpha }_{ij}\in ℂ,$ so, by the fundamental theorem of algebra, $P\left(x_{ij}\right)\text{det}\left(x_{ij}\right)=0\text{.}$ But $\text{det}\left(x_{ij}\right)\ne 0$ in $ℂ[x_{1,1},\dots ,x_{n,n}],$ and $ℂ[x_{1,1},\dots ,x_{n,n}]$ is an integral domain, so $P\left(x_{ij}\right)=0\text{.}$ Moreover, for $(\overrightarrow{j},\overrightarrow{i})\in \Omega ,$ the monomials $x_{j_1,i_1}\cdots x_{j_m,i_m}$ are distinct, since we ordered them. Thus $p_{\overrightarrow{j},\overrightarrow{i}}=0$ for all $(\overrightarrow{j},\overrightarrow{i})\in \Omega$ implying that $p=0,$ and therefore, $V^{\otimes m}\left({\text{GL}}_n\right)={\text{End}}_{ℂ\left[S_m\right]}\left(V^{\otimes m}\right)\text{.}$ $\square$

Now let $A=ℂ\left[S_m\right]$ and $B=ℂ\left[{\text{GL}}_r\right]\text{.}$ Then $V^{\otimes m}$ is a module for both $A$ and $B$ with the property that $$V^{\otimes m}\left(B\right)=\overline{V^{\otimes m}\left(A\right)}\text{.}$$ Moreover, $V^{\otimes m}$ is completely decomposable as an $A\text{-module}$ (by Maschke’s theorem), so $B\otimes A$ acts on $V^{\otimes m},$ and we have the decomposition $$V^{\otimes m}\cong \underset{\lambda \in {\hat{V}}^{\otimes m}}{⨁}{S}_{\lambda }\otimes m{V}^{\lambda },$$ where $S_{\lambda }$ is an irreducible $ℂ\left[S_m\right]\text{-module}$ and $V^{\lambda }$ is an irreducible $ℂ\left[{\text{GL}}_n\right]\text{-module.}$ We showed that if $\sigma \in S_m$ with cycle type $\mu$ and $g\in {\text{GL}}_n$ with eigenvalues $x_1,\dots ,x_n,$ then $$\text{Tr}\left(\sigma g\right)=p_{\mu }(x_1,\dots ,x_r)=\sum _{\underset{\ell \left(\lambda \right)\le n}{\lambda ⊢m}}{\chi }^{\lambda }\left(\mu \right)s_{\lambda }(x_1,\dots ,x_n),$$ where ${\chi }^{\lambda }\left(\mu \right)$ is the irreducible character of $S_m$ evaluated on the conjugacy class labeled by $\mu$ and $s_{\lambda }$ is the Schur function corresponding to $\lambda \text{.}$ It follows that:

Corollary 3.33

(1)	${\text{GL}}_n$ has some irreducible representations that can be indexed by the partitions $\lambda$ having $\ell \left(\lambda \right)\le n\text{.}$
(2)	The character ${\eta }^{\lambda }\left(g\right)$ of the ${\text{GL}}_r\text{-representation}$ indexed by $\lambda$ can be given by ${\eta }^{\lambda }\left(g\right)=s_{\lambda }(x_1,\dots ,x_n)\text{.}$

Notes and References

This is a copy of lectures in Representation Theory given by Arun Ram, compiled by Tom Halverson, Rob Leduc and Mark McKinzie.

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