Examples in representation theory

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

and

Department of Mathematics
University of Wisconsin, Madison
Madison, WI 53706 USA
ram@math.wisc.edu

Last updates: 8 March 2010

Basic theory

1. Let $A$ be an algebra of $d\times d$ matrices. Since all matrices in $A$ commute with all elements of $\stackrel{-}{A},$ $$A\subseteq \stackrel{-}{\stackrel{-}{A}}.$$ Also, $$$$ I n A = M n A and $$$$ M n A = I n A . Hence $$$$ I n A = I n A = .
2. Schur's Lemma. Let $W_1$ and $W_2$ be irreducible representations of $A$ of dimensions $d_1$ and $d_2.$ If $B$ is a $d_1\times d_2$ matrix such that $$W_1\left(a\right)B=B{W}_2\left(a\right),\text{for all}a\in A,$$ then either
1. $W_1\ncong W_2$ and $B=0$, or
2. $W_1\cong W_2$ and if $W_1=W_2$ then $B=c{I}_{d_1}$ for some $c\in ℂ.$

3. 		Proof.
   	$B$ determines a inear transformation $B:W_1\to W_2$. Since $Ba=aB$ for all $a\in A$ we have that $$B\left(a{w}_1\right)=Ba{w}_1=aB{w}_1=aB\left(w_1\right),$$ for all $a\in A$ and $w_1\in W_1.$ Thus $B$ is an $A$-module homomorphism. $\ker B$ and $\mathrm{im}B$ are submodules of $W_1$ and $W_2$ respectively and are therefore equal to either 0 or equal to $W_1$ or $W_2$ respectively. If $\ker B=W_1$ or $\mathrm{im}B=0$ then $B=0$. In the remaining case $B$ is a bijection, and thus an isomorphism between $W_1$ and $W_2.$ In this case we have that $d_1=d_2.$ Thus the matrix $B$ is square and invertible. Now suppose that $W_1=W_2$ and let $c$ be an eigenvalue of $B.$ Then the matrtix $c{I}_d-B$ is such that $$W_1\left(a\right)\left(c{I}_{d_1}-B\right)=\left(c{I}_{d_1}-B\right)W_1\left(a\right)\text{for all}a\in A.$$ The arument preceding shows that $\left(c{I}_{d_1}-B\right)$ is either invertible or 0. But if $c$ is an eigenvalue of $B$ then $\det \left(c{I}_{d_1}-B\right)=0.$ Thus $\left(c{I}_{d_1}-B\right)=0.$ □

3. Suppose that $V$ is a completely decomposable representation of an algebra $A$ and that $V\cong {\oplus }_{\lambda }{W}_{\lambda }^{\oplus m_{\lambda }}$ where the $W_{\lambda }$ are nonisomorphic irreducible representations of $A.$ Schur's lemma shows that the $A$-homomorphisms from $W_{\lambda }$ to $V$ form a vector space $${\mathrm{Hom}}_A\left(W_{\lambda },V\right)\cong {ℂ}^{\oplus m_{\lambda }}.$$ The multiplicity of the irreducible representation $W_{\lambda }$ un $V$ is $$m_{\lambda }=\dim {\mathrm{Hom}}_A\left(W_{\lambda },V\right).$$
4. Suppose that $V$ is a completely decomposable representation of an algebra $A$ and that $V\cong {\oplus }_{\lambda }{W}_{\lambda }^{\oplus m_{\lambda }}$ where the $W_{\lambda }$ are nonisomorphic irreducible representations of $A$ and let $\dim W_{\lambda }=d_{\lambda }.$ Then $$V\left(A\right)\cong {\oplus }_i{W}_{\lambda }^{\oplus m_{\lambda }}\left(A\right)\cong {\oplus }_{\lambda }{I}_{m_{\lambda }}\left(W_{\lambda }\left(A\right)\right)\cong {\oplus }_{\lambda }{W}_{\lambda }\left(A\right).$$ If we view elements of ${\oplus }_{\lambda }{I}_{m_{\lambda }}{W}_{\lambda }\left(A\right)$ as block diagonal matrices with $m_{\lambda }$ blocks of size $d_{\lambda }\times d_{\lambda }$ for each $\lambda$, then by using Ex 1 and Schur's lemma we get that $$\begin{array}{r}\end{array}$$ V A ≅ ⊕ λ I m λ W λ A = ⊕ λ M m λ W λ A = ⊕ λ M m λ I d λ ℂ .
5. Let $V$ be an $A$-module and let $p$ be an idempotent of $A.$ Then $pV$ is a subspace of $V$ and the action of $p$ on $V$ is a projection from $V$ to $pV.$ If $p_1,p_2\in A$ are orthogonal idempotents of $A$ then $p_{1}V$ and $p_{2}V$ are mutually orthogonal subspaces of $V,$ since if $p_{1}v=p_{2}v\text{'}$ for some $v,v\text{'}\in V$ then $p_{1}v=p_1{p}_{1}v=p_1{p}_{2}v\text{'}=0.$ So $V=p_{1}V\oplus p_{2}V.$
6. Let $p$ be an idempotent in $A$ and suppose that for every $a\in A,pap=kp$ for some constant $k\in ℂ.$ If $p$ is not minimal then $p=p_1+p_2,$ where $p_1,p_2\in A$ are idempotents such that $p_1{p}_2=p_2{p}_1=0.$ Then $p_1=p{p}_{1}p=kp$ for some constant $k\in ℂ.$ This implies that $p_1=p_1{p}_1=k{p}_1{p}_1=k{p}_1,$ giving that either $k=1$ or $p_1=0.$ So $p$ is minimal.
7. Let $A$ be a finite dimensional algebra and suppose that $z\in A$ is an idempotent of $A.$ If $z$ is not minimal then $z=p_1+p_2$ where $p_1$ and $p_2$ are orthogonal idempotents of $A.$ If any idempotent in this sum is not minimal we can decompose it into a sum of orthogonal idempotents. We continue this process until we have decomposed $z$ as a sum of minimal orthogonal idempotents. At any particular stage in this process $z$ is expresed as a sum of orthogonal idempotents, $z={\sum }_i{p}_i.$ So $zA={\sum }_i{p}_{i}A.$ None of the spaces $p_{i}A$ is 0 since $p_i=p_i.1\in p_{i}A$ and the spacers $p_{i}A$ are all mutually orthogonal. Thus, since $zA$ is finite dimensional it will only take a finite number of steps to decompose $z$ into minimal idempotents. A partition of unity is a decomposition of 1 into minimal orthogonal idempotents.

Finite dimensional algebras

1. Let $𝒜=\left\{a_i\right\}$ and $ℬ=\left\{b_i\right\}$ be two bases of $A$ and let $𝒜*=\left\{a_i*\right\}$ and $ℬ*=\left\{b_i*\right\}$ be the associated dual bases with respect to a nondegenerate trace $\overrightarrow{t}$ on $A.$ Then $$b_i=\sum _j{s}_{ij}{a}_j,\text{and}$$ $$b_i*=\sum _j{t}_{ij}{a}_j*,\text{and}$$ for some constants $s_{ij}$ and $t_{ij}.$ Then $$\begin{array}{rcl}{\delta }_{ij}=⟨b_i,b_j*⟩& =& ⟨\sum _k{s}_{ik}{a}_k,\sum _l{t}_{jl}{a}_l*⟩\\ & =& {\sum }_{k,l}{s}_{ik}{t}_{jl}⟨a_k,a_l*⟩\\ & =& {\sum }_{k,l}{s}_{ik}{t}_{jl}{\delta }_{kl}\\ & =& {\sum }_k{s}_{ik}{t}_{jk}.\end{array}$$ In matrix notation this says that the matrices $S=\parallel s_{ij}\parallel$ and $T=\parallel t_{ij}\parallel$ are such that $$S{T}^t=I.$$ Then, in the setting of Proposition 2.6, $$\begin{array}{rcl}\sum _i{V}_1\left(b_i\right)C{V}_2\left(b_i*\right)& =& \sum _i\left(\sum _j{s}_{ij}{V}_1\left(a_j\right)\right)C\left(\sum _k{t}_{ik}{V}_2\left(a_k*\right)\right)\\ & =& \sum _{j,k}\left(\sum _i{s}_{ij}{t}_{ik}\right)V_1\left(a_j\right)C{V}_2\left(a_k*\right)\\ & =& {\sum }_{j,k}{\delta }_{jk}{V}_1\left(a_j\right)C{V}_2\left(a_k*\right)\\ & =& \sum _j{V}_1\left(a_j\right)C{V}_2\left(a_j*.\right)\end{array}$$ This shows that the matrix $\left[C\right]$ of proposition 2.6 is independent of the choice of basis.
2. Let $A$ be the algebra of elements of the form $c_1+c_{2}e,c_1,c_2\in ℂ,$ where $e^2=0.$ $A$ is commutative and $\overrightarrow{t}$ defined by $\overrightarrow{t}\left(c_1+c_{2}e\right)=c_1+c_2$ is a nondegenerate trace on $A.$ The regular representation $\overrightarrow{A}$ of $A$ is not completely decomposable. The subspace $ℂ\overrightarrow{e}\subseteq \overrightarrow{A}$ is invariant and its complemenetary subspace is not. The trace of the regular representation is given explicitly by $\mathrm{tr}\left(1\right)=2$ and $\mathrm{tr}\left(e\right)=0.$ $\mathrm{tr}$ is degenerate. There is no matrix representation of $A$ that has trace given by $\overrightarrow{t}.$
3. Suppose $G$ is a finite group and that $A=ℂG$ is its group algebra. The the group elements $g\in G$ form a basis of $A.$ So, using 2.7, the trace of the regular representation can be expressed in the form $$\begin{array}{rcl}\mathrm{tr}\left(a\right)& =& \sum _{g\in G}ag\underset{g}{|}\\ & =& \sum _{g\in G}a{|}_1\\ & =& \left|G\right|a{|}_1,\end{array}$$ where 1 denotes the identity in $G$ and $a{|}_g$ denotes the coefficient of $g$ in $a.$ Since $\mathrm{tr}\left(g^{-1}g\right)=\left|G\right|\ne 0$ for each $g\in G,$$\mathrm{tr}$ is nondegenerate. If we set $\overrightarrow{t}\left(a\right)=a{|}_1$ then $\overrightarrow{t}$ is a trace on $A$ and ${\left\{g^{-1}\right\}}_{g\in G}$ is the dual basis to the basis ${\left\{g\right\}}_{g\in G}$with respect to the trace.
4. Let $\overrightarrow{t}$ be the trace of a faithful realisation $\phi$ of an algebra $A$ (ie for each $a\in A,\overrightarrow{t}\left(a\right)$ is given by the styandard trace of $\phi \left(a\right)$ where $\phi$ is an injective homomorphism $\phi :A\to M_d\left(ℂ\right)$). Let $\sqrt{A}=\left\{a\in A|\overrightarrow{t}\left(ab\right)=0\text{for all}b\in A\right\}.$ $\sqrt{A}$ is an ideal of $A.$ Let $a\in \sqrt{A}.$ Then $\mathrm{tr}\left(a^{k-1}a\right)=\mathrm{tr}\left(a^k\right)=0$ for all $k.$ If ${\lambda }_1,\dots ,{\lambda }_d$ are the eigenvalues of $\phi \left(a\right)$ then $\overrightarrow{t}\left(a^k\right)={\lambda }_1^k+{\lambda }_2^k+\dots +{\lambda }_d^k=p_k\left(\lambda \right)=0$ for all $k>0$, where $p_k$ represents the $k$-th power symmetric functions [Mac]. Since the power symmetric functions generate the ring of symmetric functions this means that the elementary symmetric functions $e_k\left(\lambda \right)=0$ for $k>0$, [Mac] p17, 2.14. Since the characteristic polynomial of $\phi \left(a\right)$ can be written in the form $${\mathrm{char}}_{\phi \left(a\right)}\left(t\right)=t^d-e_1\left(\lambda \right)t^{d-1}+e_2\left(\lambda \right)t^{d-2}+\dots \pm e_d\left(\lambda \right),$$ we get that ${\mathrm{char}}_{\phi \left(a\right)}\left(t\right)=t^d.$ But then the Cayley-Hamilton theorem implies that ${\phi \left(a\right)}^d=0.$ Since $\phi$ is injective we have that $a^d=0.$ So $a$ is nilpotent. Let $J$ be an ideal of nilpotent elements and suppose that $a\in J.$ For every element $b\in A,ba\in J$ and $ba$ is nilpotent. This implies that $\phi \left(ba\right)$ is nilpotent. By noting that a matrix is nilpotent only if in Jordan block form the diagonal contains all zeros we see that $\overrightarrow{t}\left(ba\right)=0.$ Thus $a\in \sqrt{A}.$ So $\sqrt{A}$ can be defined as the largest ideal of nilpotent elements. Furthermore, since the regular representation of $A$ is always faitful, $\sqrt{A}$ is equal to the set $\left\{a\in A|\mathrm{tr}\left(ab\right)=0\text{for all}b\in A\right\}$ where $\mathrm{tr}$ is the trace of the regular representation of $A.$
5. Let $𝒜$ be the basis and $$\overrightarrow{t}$$ the trace of a faithful realisation of an algebra $A$ as in Ex3 and let $G\left(𝒜\right)$ be the Gram matrix with respect to the basis $𝒜$ and the trace $\overrightarrow{t}$ as given by 2.2 and 2.3. If $ℬ$ is another basis of $A$ then $$G\left(ℬ\right)=P^{t}G\left(𝒜\right)P,$$ where $P$ is the change of basis matrix from $𝒜$ to $ℬ.$ So the rank of the Gram matrix is independent of the choice of the basis $𝒜.$

   Choose a basis $\left\{a_1,,,a_2,,,\dots ,,,a_k\right\}$ of $\sqrt{A}$ ($\sqrt{A}$ defined in Ex 3) and extend this basis to a basis $\left\{a_1,a_2,\dots ,a_k,b_1,\dots ,b_s\right\}$ of $A.$ The Gram matrix with respect to this basis is of the form $$\left(\begin{array}{rc}0& 0\\ 0& G\left(B\right)\end{array}\right)$$ where $G\left(B\right)$ denotes the Gram matrix on $\left\{b_1,,,b_2,\dots ,b_s\right\}.$ So the rank of the Gram matrix is certainly less than or equal to $s$.

   Suppose that the rows of $G\left(B\right)$ are linearly dependent. Then for some contants $c_1,c_2,\dots ,c_s,$ not all zero, $$c_1\overrightarrow{t}\left(b_1{b}_i\right)+c_2\overrightarrow{t}\left(b_2{b}_i\right)+\dots +c_s\overrightarrow{t}\left(b_s{b}_i\right)=0$$ for all $1\le i\le s.$ So $$\overrightarrow{t}\left(\left(\sum _j{c}_j{b}_j\right),b_i\right)=0,\text{for all}i.$$ This implies that ${\sum }_j{c}_j{b}_j\in \sqrt{A}.$ This is a contradiction to the construction of the $b_j.$ So the rows of $B\left(B\right)$ are linearly independent.

   Thus the rank of the Gram matrix is $s$ or equivalently the corank of the Gram matrix of $A$ is equal to the dimension of the radical $\sqrt{A}.$ Thus the trace $\mathrm{tr}$ of the regular representation of $A$ is nondegenerate iff $\sqrt{A}=\left(0\right).$

6. Let $W$ be an irreducible representation of an arbitrary algebra $A$ and let $d=\dim W.$ Denote $W\left(A\right)$ by $A_W.$ Note that representation $W$ is also an irreducible representation of $A_W$( $W\left(a\right)=a$ for all $a\in A_W$ ).

   We show that $\mathrm{tr}$ is nondegenerate on $A_W,$ ie that if $a\in A_W,a\ne 0$, then there exists $b\in A_W$ such that $\mathrm{tr}\left(ba\right)\ne 0.$ Since $a$ is a nonzero matrix there exists some $w\in W$ such that $aw\ne 0.$ Thus $Aaw=W.$ So there exists some $w\in W$ such that $aw\ne 0.$ Now $Aaw\subseteq W$ is an $A$-invariant subspace of $W$ and not 0 since $aw\ne 0.$ Thus $Aaw=W$. So there exists some $b\in A_W$ such that $baw=w.$ This shows that $ba$ is not nilpotent. So $\mathrm{tr}\left(ba\right)\ne 0.$ So $\mathrm{tr}$ is nondegenerate on $A_W.$ This means that $A_W={\oplus }_{\lambda }{M}_{d_{\lambda }}\left(ℂ\right)$ for some $d_{\lambda }.$ But since by Schur's lemma $$ A W = I d ℂ , where $d=\dim W,$ we see that $W\left(A\right)=A_W=M_d\left(ℂ\right).$

7. Let $A$ be a finite dimensional algebra and let $\overrightarrow{A}$ denote the regular representation of $A.$ The set $\overrightarrow{A}$ is the same as the set $A$, but we distinguish elements of $\overrightarrow{A}$ by writing $\overrightarrow{a}\in A.$

   A linear transformation $B$ of $\overrightarrow{A}$ is in the centraliser of $\overrightarrow{A}$ if for every element $a\in A$ and $\overrightarrow{x}\in \overrightarrow{A},$ $$Ba\overrightarrow{x}=aB\overrightarrow{x}.$$ Let $B\overrightarrow{1}=\overrightarrow{b}.$ Then $$\begin{array}{rcl}B\overrightarrow{a}& =& Ba\overrightarrow{1}\\ & =& aB\overrightarrow{1}\\ & =& a\overrightarrow{b}\\ & =& \overrightarrow{\left(ab\right)}.\end{array}$$ So $B$ acts on $\overrightarrow{a}\in \overrightarrow{A}$ by right multiplication on $b.$ Conversely it is easy to see that the action of right multiplication commutes with the action of left mutliplication since $$\left(a\overrightarrow{x}\right)b=a\left(\overrightarrow{x}b\right),$$ for all $a,b\in A$ and $\overrightarrow{x}\in \overrightarrow{A}.$ So the centraliser algebra of the regular represnetation is the algebra of matrices determined by the action of right multiplication of elements of $A.$

Matrix units and characters

1. If $A$ is commutative and semisimple then all irreducible representations of $A$ are one dimensional. This is not necessarily true for algebras over fields which are not algebraically closed (since Schur's lemma takes a different form).
2. If $R$ is a ring with identity and $M_n\left(R\right)$ denotes $n\times n$ matrices with entries in $R$. the ideals of $M_n\left(R\right)$ are of the form $M_n\left(I\right)$ where $I$ is an ideal of $R.$
3. If $V$ is a vector space over $ℂ$ and $V*$ is the space of $ℂ$-valued functions on $V$ then $\dim V*=\dim V.$ If $B$ is a basis of $V$ then the functions ${\delta }_b,b\in B,$ determined by $${\delta }_b\left(b_i\right)=\left\{\begin{array}{rc}1,& \text{if }b=b_i,\\ 0,& \text{otherwise,}\end{array}\right\}$$ for $b_i\in B,$ form a basis of $V*.$ If $A$ is a semisimple algebra isomorphic to $M_d\left(ℂ\right)={\oplus }_{\lambda \in \tilde{A}}{M}_{d_{\lambda }}\left(ℂ\right),$ $\tilde{A}$ an index set for the irreducible representations $W_{\lambda }$ of $A,$ then $$\dim A=\sum _{\lambda \in \tilde{A}}{d}_{\lambda }^2,$$ and the functions $W_{ij}^{\lambda }(W_{ij}^{\lambda }\left(a\right)$ the $\left(i,j\right)$-th entry of the matrix $W_{\lambda }\left(a\right),a\in A)$ on $A$ form a basis of $A*.$ The $W_{ij}^{\lambda }$ are simply the functions ${\delta }_{e_{ij}^{\lambda }}$ for an appropriate set of matrix units $\left\{e_{ij}^{\lambda }\right\}$ of $A.$ Thi shows that the coordinate functions of the irreducible representations are linearly independent. Since ${\chi }^{\lambda }=\sum _i{W}_{ii}^{\lambda },$ the irreducible characters are also linearly independent.
4. Let $A$ be a semisimple algebra. Virtual characters are elements of the vector space $R\left(A\right)$ consisting of the $ℂ$-linear span of the irreducible characters of $A.$ We know that there is a one-to-one correspondence between the minimal central idempotents of $A$ and the irreducible characters of $A.$ Since the minimal central idempotents of $A$ form a basis of the center $Z\left(A\right)$ of $A,$ we ca define a vector space isomorphism $\phi :Z\left(A\right)\to R\left(A\right)$ by setting $\phi \left(z_{\lambda }\right)={\chi }^{\lambda }$ for each $\lambda \in \tilde{A}$ and extending linearly to all of $Z\left(A\right).$

   Given a nondegenerate trace $\overrightarrow{t}$ on $A$ with trace vector $\left(t_{\lambda }\right)$ it is more natural to define $\phi$ by setting $\phi \left(z_{\lambda }/t_{\lambda }\right)={\chi }^{\lambda }.$ Then, for $z\in Z\left(A\right),$ $$\phi \left(z\right)\left(a\right)=\overrightarrow{t}\left(za\right),$$ since $$\begin{array}{rcl}\phi \left(z_{\mu }/t_{\mu }\right)\left(a\right)& =& \overrightarrow{t}\left(z_{\mu }/t_{\mu }a\right)\\ & =& \overrightarrow{t}\left(\left(\frac{1}{t_{\mu }}\right)z_{\mu }a\right)\\ & =& \left(\frac{1}{t^{\mu }}\right)\left(t_{\mu }{\chi }^{\mu }\left(a\right)\right)\\ & =& {\chi }^{\mu }\left(a\right).\end{array}$$

5. If $A$ is a semisimple algebra isomorphic to $M_b\left(ℂ\right)={\oplus }_{\lambda \in \tilde{A}}{M}_{d_{\lambda }}\left(ℂ\right),\tilde{A}$ an index set for the irreducible representations $W_{\lambda }$ of $A,$ then the right regular representation decomposes as $$\overrightarrow{A}\cong {\oplus }_{\lambda \in A}{W}_{\lambda }^{\oplus d_{\lambda }}.$$ If matrix units $e_{ij}^{\lambda }$ are given by (3.7) then $$\mathrm{tr}\left(e_{ii}^{\lambda }\right)=\mathrm{tr}\left(d_{\lambda }{E}_{ii}^{\lambda }\right)=d_{\lambda }.$$ So the trace of the regular representation of $A,$$\mathrm{tr},$ is given by the trace vector $\overrightarrow{t}=\left(t_{\lambda }\right),$ where $t_{\lambda }=d_{\lambda }$ for each $\lambda \in \tilde{A}.$
6. Let $A$ be a semisimple algebra and let $B*=\left\{g*\right\}$ be a dual basis to $B=\left\{g\right\}$ of $A$ with respect to the tracde of the regular representation of $A.$ We can define an inner product on the space $R\left(A\right)$ of virtual characters, Ex 4, of $A$ by $$⟨\chi ,\chi \text{'}⟩=\sum _{g\in B}\chi \left(g\right)\chi \text{'}\left(g*\right).$$ The irreducible characters of $A$ are orthonormak with respect to this inner product. Nate that $\chi ,\chi \text{'}$ are characters of representations $V$ and $V\text{'}$ respectively, then, by Ex4 and Theorem 3.9, $$⟨\chi ,\chi \text{'}⟩=\dim {\mathrm{Hom}}_A\left(V,V\text{'}\right).$$ If ${\chi }^{\lambda }$ is the character of the irreducible representation $W_{\lambda }$ of $A$ then $⟨\chi ,\chi \text{'}⟩$ gives the multiplicity of $W_{\lambda }$ in the representation $V$ as in Section 1, Ex 3.
7. Let $A$ be a semisimple algebra and $\overrightarrow{t}=\left(t_{\lambda }\right)$ be a non-degnerate trace on $A.$ Let $B$ be a basis of $A$ and for each $g\in B$ let $g*$ denote the element of the dual basis to $B$ ith respect to the trace $\overrightarrow{t}$ such that $\overrightarrow{t}\left(gg*\right)=1.$ For each $a\in A$ define $$\left[a\right]=\sum _{g\in B}gag*.$$ By Section 2, Ex 1, the element $\left[a\right]$ is independent of the choice of the basis $B.$ By using a set of matrix units $e_{ij}^{\lambda }$ of $A$ we get $$\begin{array}{rcl}\left[a\right]& =& \sum _{i,j,\lambda }\left(\frac{1}{t_{\lambda }}\right)e_{ij}^{\lambda }a{e}_{ji}^{\lambda }\\ & =& \sum _{i,j,\lambda }\left(\frac{1}{t_{\lambda }}\right)a_{jj}^{\lambda }{e}_{ii}^{\lambda }\\ & =& \sum _{\lambda }\left(\frac{1}{t_{\lambda }}\right)\left(\sum _j{a}_{jj}^{\lambda }\left(\sum _i{e}_{ii}^{\lambda }\right)\right)\\ & =& \sum _{\lambda }\left(\frac{1}{t_{\lambda }}\right){\chi }^{\lambda }\left(a\right)z_{\lambda }.\end{array}$$ So ${\chi }^{\lambda }\left(\left[a\right]\right)=\left(\frac{d_{\lambda }}{t_{\lambda }}\right){\chi }^{\lambda }\left(a\right).$ By 3.9 $$\sum _{g\in B}\left(\frac{t_{\lambda }^2}{d_{\lambda }}\right){\chi }^{\mu }\left(g*\right)\left[g\right]=\sum _{\lambda }\sum _{g\in B}\left(\frac{t_{\lambda }^2}{d_{\lambda }}\right)\left(\frac{1}{t_{\lambda }}\right){\chi }^{\lambda }\left(g\right){\chi }^{\mu }\left(g*\right)z_{\lambda }=\sum _{\lambda }{\delta }_{\lambda \mu }{z}_{\lambda }=z_{\mu }.$$ Thus the $\left[g\right],g\in B,$ span the center of $A.$
8. Let $G$ be a finite group and let $A=ℂG$. Let $\overrightarrow{t}$ be the trace on $A$ given by $$\overrightarrow{t}\left(a\right)=a{|}_1,$$ where 1 is the identity in $G.$ By Ex 5 and Section 2 Ex 3 the trace vector of $\overrightarrow{t}$ is given by $t_{\lambda }=\left(\frac{d_{\lambda }}{\left|G\right|}\right)$ where $d_{\lambda }$ is the dimnesion of the irreducible representation of $G$ corresponding to $\lambda .$

   If $h\in G,$ then the element $$\left[h\right]=\sum _{g\in B}ghg*=\sum _{g\in B}gh{g}^{-1}$$ is a multiple of the sum of the elements of $G$ that are conjugate to $h.$ Let $\Lambda$ be an index set of the conjugacy classes of $G$ and for each $\lambda \in \Lambda$, let $C_{\lambda }$ denote the sum of the elements in the conjugacy class indexed by $\lambda$. The $C_{\lambda }$ are linearly independent elements of $ℂG$. Furthermore by Ex 7 they span the center of $ℂG.$ Thus $\Lambda$ must also be an index set for the irreducible representations of $G.$ So we see that the irreducible representations of the group algebra of a finite group are indexed by the conjugacy classes.

9. Let $G$ be a finite group and let $C_{\lambda }$ denote the conjugacy classes of $G.$ Note that since $$\mathrm{tr}\left(V\left(hg{h}^{-1}\right)\right)=\mathrm{tr}\left(V\left(h\right)V\left(g\right)V{\left(h\right)}^{-1}\right)=\mathrm{tr}\left(V\left(g\right)\right)$$ for any representation $V$ of $G$ and all $g,h\in G,$ characters of $G$ are constant on conjugacy classes. Using theorem 3.8, $$\begin{array}{rcl}\left|G\right|{\delta }_{\lambda \mu }& =& \sum _g{\chi }^{\lambda }\left(g\right){\chi }^{\mu }\left(g^{-1}\right)\\ & =& \sum _{\rho }\sum _{g\in C_{\rho }}{\chi }^{\lambda }\left(g\right){\chi }^{\mu }\left(g^{-1}\right)\\ & =& \sum _{\rho }\left|C_{\rho }\right|{\chi }^{\lambda }\left(\rho \right){\chi }^{\mu }\left(\rho \text{'}\right),\end{array}$$ where $\rho \text{'}$ is such that $C_{\rho \text{'}}$ is the conjugacy class which contains the inverses of the elements in $C_{\rho }.$ Define matrices $\Xi =\parallel {\Xi }_{\lambda \rho }\parallel$ and $\Xi \text{'}=\parallel {\Xi \text{'}}_{\lambda \rho }\parallel$ by ${\Xi }_{\lambda \rho }={\chi }^{\lambda }\left(\rho \right)$ and ${\Xi }_{\lambda \rho }\text{'}=\left|C_{\rho }\right|{\chi }^{\lambda }\left({\rho }^{\text{'}}\right).$ By Ex 8 these matrices are square. In matrix notation the above is $$\Xi {\Xi }^{\text{'}t}=\left|G\right|I,$$ but then we also have that ${\Xi }^{\text{'}t}\Xi =\left|G\right|I,$ or equivalently that $$\sum _{\lambda }{\chi }^{\lambda }\left({\rho }^{\text{'}}\right){\chi }^{\lambda }\left(\tau \right)=\left(\frac{\left|G\right|}{\left|C_{\rho }\right|}\right){\delta }_{\rho \tau }.$$
10. This example gives a generalisation of the preceding example. Let $A$ be a semisimple algebra and suppose that $B$ is a basis of $A$ and that there is a partition of $B$ into classes such that if $b$ and $b\text{'}\in B$ are in the same classes then for every $\lambda \in \tilde{A}$, $${\chi }^{\lambda }\left(b\right)={\chi }^{\lambda }\left(b^{\text{'}}\right).$$ The fact that the characters are linearly independent implies that the number of classes must be the same as the number of irreducible characters ${\chi }^{\lambda }.$ Thus we can inbox the classes of $B$ by the elements of $\tilde{A}.$ Assume that we have fixed such a correspondence and denote the classses of $B$ by $C_{\lambda },\lambda \in \tilde{A}.$

    Let $\overrightarrow{t}$ be a nondegenerate trace on $A$ and let $G$ be the Gram matrix with respect to the basis $B$ and the trace $\overrightarrow{t}.$ If $g\in B,$ let $g*$ denote the element of the dual basis to $B$, with respect to the trace $\overrightarrow{t}$, such that $\overrightarrow{t}\left(gg*\right)=1.$ Let $G^{-1}=C=\parallel c_{gg\text{'}}\parallel$ and recall that $g*={\sum }_{g\text{'}\in B}{c}_{gg\text{'}}g\text{'}.$ Then Collecting $g,g\text{'}\in B$ by class size gives where ${\chi }^{\lambda }\left(\rho \right)$ denotes the value of the charactwr ${\chi }^{\lambda }\left(\rho \right)$ at elements of the class $C_{\rho }.$ Now define a matrix $$ C =∥ c ρτ ∥ with entries $${}_{}$$ c ρτ = ∑ g∈ C ρ ,g'∈ C τ c gg' , and let $\Xi =\parallel {\Xi }_{\lambda \rho }\parallel$ and $\Xi \text{'}=\parallel {\Xi \text{'}}_{\lambda \rho }\parallel$ be matrices given by ${\Xi }_{\rho \lambda }={\chi }^{\lambda }\left(\rho \right)$ and ${\Xi \text{'}}_{\lambda \rho }=\left(\frac{t_{\lambda }}{d_{\lambda }}\right){\chi }^{\lambda }\left(\rho \right).$ Note that all of these matrices are square. Then the above gives that $$I=\Xi$$ C Ξ' t . So $$I=$$ C Ξ' t Ξ, or equivalently that $$\begin{array}{rcl}{\delta }_{\rho \tau }& =& \sum _{\sigma ,\lambda }{}_{}\end{array}$$ c ρσ t λ d λ χ λ σ χ λ τ = ∑ σ,λ ∑ g∈ C ρ ,g'∈ C σ c gg' t λ d λ χ λ σ χ λ τ = ∑ λ ∑ g'∈B ∑ g∈ C ρ c gg' χ λ σ χ λ τ = ∑ g∈ C ρ ∑ λ χ λ g* χ λ τ .

Double centraliser nonsense

1. Let $G$ be a group and let $V$ and $W$ be two representations of $G$. Define an action of $G$ on the vector space $V\otimes W$ by $$g\left(vw\right)=\left(gv\right)\left(gw\right),$$ for all $g\in G,v\in V$ and $w\in W$ (see also Section 5 Ex 4). In matrix form, the representation $V\otimes W$ is given by setting $$\left(V{\otimes }_{d}W\right)\left(g\right)=V\left(g\right)\otimes W\left(g\right),$$ for each $g\in G.$ Note, however, that if we extend this action to an action of $A=ℂG$ on $V\otimes W,$ then for a general $a\in A,$ $a\left(vw\right)$ is not equal to $\left(av\right)\left(aw\right)$ and $\left(V{\otimes }_{d}W\right)\left(a\right)$ is not equal to $V\left(a\right)\otimes W\left(a\right).$
2. Theorem 4.6 gives that there is a one-to-one correspondence between minimal central idempotents $z_{\lambda }^C$ of $C$ and characters ${\chi }_A^{\lambda }$ of irreducible representations of $A$ of $A$ appearing in the decomposition of $V$. Let ${\chi }_C^{\lambda }$ be the irreducible characters of $C$ and for each $\lambda$ set $d_{\lambda }^C={\chi }_C^{\lambda }\left(1\right),$ so that the $d_{\lambda }$ are the dimensions of the irreducible representations of $C.$ The Frobenius map is the map $$\begin{array}{rcll}F:& Z\left(C\right)& \to & R\left(A\right)\\ & \left(\frac{1}{d_{\lambda }^C}\right)z_{\lambda }^X& \mapsto & {\chi }_A^{\lambda }.\end{array}$$ Let $t:C\otimes A\to ℂ$ be the trace of the action of $C\otimes A$ on the representation $V.$ By taking traces on each side of the isomorphism in Theorem 4.11 we have that $$t\left(q,a\right)=\sum _{\lambda }{\chi }_C^{\lambda }\left(q\right){\chi }_A^{\lambda }\left(a\right).$$ Let ${\overrightarrow{t}}_C=\left(t_{\lambda }^C\right)$ be a nondegenerate trace on $C$, let $B$ be a basis of $C$ and for each $g\in B$ let $g*$ be the element of the dual bsis to $B$ with respect to the trace ${\overrightarrow{t}}_C$ such that ${\overrightarrow{t}}_C\left(gg*\right)=1.$ Then, for any $z\in Z\left(C\right),$ the center of $C,$ $$F\left(z\right)=\sum _{g\in B}{\overrightarrow{t}}_C\left(zg*\right)t\left(g,.\right),$$ since, using 3.8 and 3.9, $$\begin{array}{rcl}F\left(\frac{z_{\mu }^C}{d_{\mu }^C}\right)& =& \sum _g\left(\frac{1}{d_{\mu }^C}\right){\overrightarrow{t}}_C\left(z_{\mu }^{C}g*\right)t\left(g,.\right)\\ & =& \sum _g\left(\frac{t_{\mu }^C}{d_{\mu }^C}\right){\chi }_C^{\mu }\left(g*\right)t\left(g,.\right)\\ & =& \sum _g\left(\frac{t_{\mu }^C}{d_{\mu }^C}\right){\chi }_C^{\mu }\left(g*\right)\sum _{\lambda }{\chi }_C^{\lambda }\left(q\right){\chi }_A^{\lambda }\left(.\right)\\ & =& \sum _g\left(\frac{t_{\mu }^C}{d_{\mu }^C}\right){\delta }_{\mu \lambda }\left(\frac{d_{\lambda }^C}{t_{\lambda }^C}\right){\chi }_A^{\lambda }\left(.\right)\\ & =& {\chi }_{\mu }^A\left(.\right).\end{array}$$

   If we apply the inverse $F^{-1}$ of the Frobenius map to (4.13) we get $$F^{-1}\left(t\left(q,.\right)\right)=\sum _{\lambda }{\chi }_C^{\lambda }\left(q\right)\left(\frac{z_{\lambda }^C}{d_{\lambda }^C}\right).$$ Formula 3.13 shows that $$F^{-1}\left(t\left(q,.\right)\right)=\left(\sum _{\lambda }\left(\frac{t_{\lambda }^C}{d_{\lambda }^C}\right)z_{\lambda }^C\right)\left[q\right].$$ In the case that ${\overrightarrow{t}}_C$ is the trace of the regular representation $\sum _{\lambda }\left(\frac{t_{\lambda }^C}{d_{\lambda }^C}\right)z_{\lambda }^C=1$ and $F^{-1}\left(t\left(q,.\right)\right)=\left[q\right].$

Centralisers

1. Let $A,B$ and $C$ be vector spaces. A map $f:A\times B\to C$ is bilinear if $$f\left(a_1+a_2,b\right)=f\left(a_1,b\right)+f\left(a_2,b\right),$$$$f\left(a,b_1+b_2\right)=f\left(a,b_1\right)+f\left(a,b_2\right),$$$$f\left(\alpha a,b\right)=f\left(a,\alpha b\right)=\alpha f\left(a,b\right),$$ for all $a,a_1,a_2\in A,b,b_1,b_2\in B,\alpha \in ℂ.$

2. The tensor product is given by a vector space $A\otimes B$ and a map $i:A\times B\to A\otimes B$ such that for every bilinear map $f:A\times B\to C$ there exists a linear map $\stackrel{-}{f}:A\otimes B\to C$ such that the following diagram commutes:

   
   

   One constructs the tensor product $A\otimes B$ as the vector space of elements $a\otimes b,a\in A,b\in B,$ with relations $$\left(a_1+a_2\right)\otimes b=a_1\otimes b+a_2\otimes b,$$$$a\otimes \left(b_1+b_2\right)=a\otimes b_1+a\otimes b_2,$$$$\left(\alpha a\right)b=a\otimes \left(\alpha b\right)=\alpha \left(a\otimes b\right),$$ for all $a,a_1,a_2\in A,b,b_1,b_2\in B$ and $\alpha \in ℂ.$ The map $i:A\times B\to A\otimes B$ is given by $i\left(a,b\right)=a\otimes b.$ Using the above universal mapping property one gets easily that the tensor product is unique in the sense that any two tensor products of $A$ and $B$ are isomorphic.

   If $R$ is an algebra and $A$ is a right $R$-module (a vector space that affords an antirepresentation of $R$) and $B$ is a left $R$-module them one forms the vector space $A{\otimes }_{R}B$ as above except that we require a bilinear map $f:A\times B\to C$ to satisfy the additional condition $$f\left(ar,b\right)=f\left(a,rb\right)$$ for all $r\in R.$ Then the tensor product $A{\otimes }_{R}B$ once again is constructed by using the vector space of elements $a\otimes b,a\in A,b\in B,$ with the relations above and the additional relation $$ar\otimes b=a\otimes rb,$$ for all $r\in R.$

2. Let $A\subseteq B$ be semisimple algebras such that $A$ is a subalgebra of $B$ Let $\tilde{A}$ and $\tilde{B}$ be index sets of the irreducible representations of $A$ and $B$ respectively, and suppose that $\left\{f_{ij}^{\mu }\right\},\mu \in \tilde{A},$ is a complete set of matrix units of $A$

   [Bt] There exists a complete set of matrix units $\left\{e_{rs}^{\lambda }\right\},\lambda \in \tilde{B},$ of $B$ that is a refinement of the $f_{ij}^{\mu }$ in the sense that for each $\mu \in \tilde{A}$ and each $i$, $$f_{ii}^{\mu }=\sum e_{rr}^{\lambda },$$ for some set of $e_{rr}^{\lambda }$.

   		Proof.
   	Suppose that $B\cong {\oplus }_{\lambda \in \tilde{B}}{M}_{d_{\lambda }}\left(ℂ\right)$. Let $z_{\lambda }^B$ be the minimal central idempotent of $B$ such that $I_{\lambda }=B_{z_{\lambda }}$ is the minimal ideal corresponding to the $\lambda$ block of matrices in ${\oplus }_{\lambda }{M}_{d_{\lambda }}\left(ℂ\right).$ For each $\mu \in \tilde{A}$ and each $i$ decompose $f_{ii}^{\mu }$ into minimal orthogonal idempotents of $B$ (Section 1, Ex 7), $f_{ii}^{\mu }=\sum p_j.$ Label each $p_j$ appearing in this sum by the element $\lambda \in \tilde{B}$ which indexes the minimal ideal $I_{\lambda }=B{p}_{j}B$ of $B$. Then $$1=\sum _{\mu ,i}{f}_{ii}^{\mu }=\sum _{\lambda \in \tilde{B}}\sum _{j=1}^{d_{\lambda }}{p}_j^{\lambda }.$$ Now $$B=1.B.1=\sum _{\lambda ,\mu \in \tilde{B}}\sum _{1\le i\le d_{\lambda },1\le j\le d_{\mu }}{p}_i^{\lambda }B{p}_j^{\mu }.$$ If $\lambda \ne \mu$ then the space $p_i^{\lambda }B{p}_j^{\mu }=p_i^{\lambda }B\left(z_{\mu }^B{p}_j^{\mu }\right)=p_i^{\lambda }{z}_{\mu }^{B}B{p}_j^{\mu }=0$ for all $i,j.$ Since $p_i^{\lambda }=p_i^{\lambda }.1.p_i^{\lambda }\in p_i^{\lambda }I{p}_i^{\lambda }$ and $p_i^{\lambda }B{p}_j^{\lambda }{p}_j^{\lambda }B{p}_i^{\lambda }=p_i^{\lambda }{I}_{\lambda }{p}_i^{\lambda }\ne 0$, we know that $p_i^{\lambda }B{p}_j^{\lambda }$ is not zero for any $1\le i,j\le d_{\lambda }.$ Furthermore, since the dimension of $B$ is ${\sum }_{\lambda }{d}_{\lambda }^2$ each of the spaces $p_i^{\lambda }B{p}_j^{\lambda }$ is one dimensional. For each $p_i^{\lambda }$ define $e_{ii}^{\lambda }=p_i^{\lambda }.$ For each $\lambda$ and each $1\le i<j\le d_{\lambda }$ let $e_{ii}^{\lambda }$ be some element of $p_i^{\lambda }B{p}_j^{\lambda }.$ Then choose $e_{ii}^{\lambda }\in p_j^{\lambda }B{p}_i^{\lambda }$ such that $e_{ij}^{\lambda }{e}_{ji}^{\lambda }=e_{ii}^{\lambda }.$ This defines a complete set of matrix units of $B.\square$

3. Let $G$ be a finite group and let $H$ be a subgroup of $G.$ Let $R=\left\{g_i\right\}$ be a set of representatives for the left cosets $gH$ of $H$ in $G.$ The action of $G$ on the cosets of $H$ in $G$ by left multiplication defines a representation ${\pi }_H$ of $H$ in $G.$ This representation is a permutation representation of $G.$ Let $g\in G.$ The entries ${\pi }_H{\left(g\right)}_{i\text{'}i}$ of the matrix ${\pi }_H\left(g\right)$ are given by ${\pi }_H{\left(g\right)}_{i\text{'}i}={\delta }_{i\text{'}k}$ where $k$ is such that $g{g}_i\in g_{k}H.$

   Let $V$ be a representation of $H.$ Let $B=\left\{v_j\right\}$ be a basis of $V.$ Then the elements $g\otimes v_j$ where $g\in G,v_j\in B$ span $ℂG{\otimes }_{ℂH}V.$ The fourth relation in 5.1 gives that the set $\left\{g_i\otimes v_j\right\},g_i\in R,v_j\in B$ forms a basis of $ℂG{\otimes }_{ℂH}V.$

   Let $g\in G$ and suppose that $g{g}_i=g_{k}h,$ where $h\in H$ and $g_k\in R.$ Then $$\begin{array}{rcl}g{g}_i\otimes v_j& =& g_{k}h\otimes v_j& & =& g_k\otimes h{v}_j\\ & =& \sum _j{g}_k\otimes v_{j\text{'}}V{\left(h\right)}_{j\text{'}j}\\ & =& \sum _{i\text{'},j\text{'}}{g}_{i\text{'}}\otimes v_{j\text{'}}V{\left(h\right)}_{j\text{'}j}{\delta }_{i\text{'}k}\\ & =& \sum _{i\text{'},j\text{'}}{g}_{i\text{'}}\otimes v_{j\text{'}}V{\left(h\right)}_{j\text{'}j}{\pi }_H{\left(g\right)}_{i\text{'}i}.\end{array}$$ Then $$\begin{array}{rcl}{\chi }_{V{\uparrow }_H^G}\left(g\right)& =& \sum _{g_i\in R,v_j\in B}g{g}_i\otimes v_j{|}_{g_i\otimes v_j}\\ & =& \sum _{g_i,v_j,g{g}_i\in g_{i}H}V{\left({g_i}^{-1}g{g}_i\right)}_{jj}.\end{array}$$

   Since characters are constant on conjugacy classes we have that $$\begin{array}{rcl}{\chi }_{V{\uparrow }_H^G}\left(g\right)& =& \left(\frac{1}{\left|H\right|}\right)\sum _{h\in H}\sum _{g_i;h^{-1}{g_i}^{-1}g{g}_{i}h\in H}{\chi }_V\left(h^{-1}{g_i}^{-1}g{g}_{i}h\right)\\ & =& \left(\frac{1}{\left|H\right|}\right)\sum _{a\in H,a\in C_g}{\chi }_V\left(a\right),\end{array}$$ where $C_g$ denotes the conjugacy class of $g.$ This is an alternate proof of Theorem 5.8 for the special case of inducing from a subgroup $H$ of a group $G$ to the group $G.$

4. Define $ℂG{\otimes }_dℂG$ to be the subalgebra of the algebra $ℂG\otimes ℂG$ consisting of the span of the elements $g\otimes g$, $g\in G.$ Then $ℂG\cong ℂG{\otimes }_dℂG$ as algebras.

   Let $V_1$ and $V_2$ be representations of $G.$ Then the restriction of the $ℂG\otimes ℂG$ representation $V=V_1\otimes V_2$ to the algebra $ℂG{\otimes }_dℂG$ is the Kronecker product (Section 4, Ex 1) $$V_1{\otimes }_d{V}_2=V_1\otimes V_2{\downarrow }_{ℂG\otimes ℂG}^{ℂG{\otimes }_dℂG}$$ of $V_1$ and $V_2.$ Since $ℂG\cong ℂG{\otimes }_dℂG$ we can view $V_1{\otimes }_d{V}_2$ as a representation of $G.$

   Let $V_{\lambda }$ and $V_{\mu }$ be irreducible representations of $G$ such that $V_{\lambda }\otimes v_{\mu }$ appears as an irreducible component of the $ℂG\otimes ℂG$ representation $V_1\otimes V_2$. The decomposition of the Kronecker product $$V_{\lambda }{\otimes }_d{V}_{\mu }=V_1\otimes V_2{\downarrow }_{ℂG\otimes ℂG}^{ℂG{\otimes }_dℂG}\cong {\oplus }_{\nu }{g}_{\lambda \mu }^{\nu }{V}_{\nu }$$ into irreducible representations $V_{\nu }$ of $G$ is given by the branching rule for $ℂG\otimes ℂG\supset ℂG{\otimes }_dℂG.$ Let $C_1$ and $C_2$ be the centralisers of the representations $V_1$ and $V_2$ respectively. Let $C$ be the centraliser of the $ℂG\otimes ℂG$ representation $V=v_1\otimes V_2.$ Applying Theorem 5.9 to $V$ where $A=ℂG\otimes ℂG$ and $ℂG{\otimes }_dℂG=B\cong G$ shows that the $g_{\lambda \mu }^{\nu }$ are also given by the branching rule for $C_1\otimes C_2\subset C.$

References

[BG] A. Braverman and D. Gaitsgory, Crystals via the affine Grassmanian, Duke Math. J. 107 no. 3, (2001), 561-575; arXiv:math/9909077v2, MR1828302 (2002e:20083)

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