Finite dimensional algebras

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

Last update: 6 November 2012

Finite dimensional algebras

The trace, $\text{tr}\left(a\right),$ of a $d\times d$ matrix $a=\mid a_{ij}\mid$ is the sum of its diagonal elements, $\text{tr}\left(a\right)=\sum _i{a}_{ii}$ A trace $\overrightarrow{t}$ on an algebra $A$ is a $ℂ\text{-linear}$ map $\overrightarrow{t}:A\to ℂ$ such that for all $a,b\in A,$

$$\begin{array}{cc}\overrightarrow{t}\left(ab\right)=\overrightarrow{t}\left(ba\right)\text{.}& \text{(2.1)}\end{array}$$

Every representation $V$ of $A$ determines a trace ${\overrightarrow{t}}_V$ on $A$ given by ${\overrightarrow{t}}_V\left(a\right)=\text{tr}\left(V\left(a\right)\right)$ where $a\in A\text{.}$ A trace is nondegenerate if for each $a\in A,$ $a\ne 0,$ there exists $b\in A$ such that $\overrightarrow{t}\left(ba\right)\ne 0\text{.}$ A trace $\overrightarrow{t}$ on $A$ determines a symmetric bilinear form $⟨,⟩$ on $A$ given by

$$\begin{array}{cc}⟨a,b⟩=\overrightarrow{t}\left(ab\right)\text{.}& \text{(2.)}\end{array}$$

Suppose $A$ is finite dimensional and let $B=\{b_1,b_2,\dots ,b_s\}$ be a basis of $A\text{.}$ A basis $B^{*}=\{b_1^{*},b_2^{*},\dots ,b_s^{*}\}$ of $A$ is dual to $B$ with respect to the form $⟨,⟩$ if

$$⟨b_i^{*},b_j⟩={\delta }_{ij}\text{.}$$

The Gram matrix of $A$ is the matrix

$$\begin{array}{cc}G=\mid ⟨b_i,b_j⟩\mid \text{.}& \text{(2.3)}\end{array}$$

Suppose that $B^{*}$ exists and that $C=\mid c_{ij}\mid$ an $s\times s$ matrix such that

$$\begin{array}{cc}{b}_i^{*}=\sum _k{c}_{ik}{b}_k\text{.}& \text{(2.4)}\end{array}$$

Then

$$⟨b_i^{*},n_j⟩=\sum _k{c}_{ik}⟨b_k,b_j⟩={\delta }_{ij}\text{.}$$

In matrix notation this says that $CG=I_s\text{.}$ So $C$ must be $G^{-1}\text{.}$ Conversely, if $C=G^{-1}$ then defining $b_i^{*}$ by (2.4) determines a dual basis $B^{*}\text{.}$ This shows that $B^{}$ exists if and only if $G$ is invertible and that if it exists it is unique.

(2.5) Proposition. If $\overrightarrow{t}$ is a trace on a finite dimensional algebra $A$ with basis $B=\{b_1,b_2,\dots ,b_s\}$ and $⟨,⟩$ is given by (2.2) then the Gram matrix $G$ is invertible if and only if $\to$ is nondegenerate.

		Proof.
	The trace $\overrightarrow{t}$ is degenerate if and only if there exists $a,b\in A$ such that $\overrightarrow{t}\left(ab\right)=0$ for all $a\in A\text{.}$ This is the same as saying that $\overrightarrow{t}\left(b_{1}b\right)=0$ for each basis element $b_i\text{.}$ If $b=\sum _j{c}_j{b}_j,$ $c_j\in ℂ,$ we have that the $c_j$ satisfy the system of equations $$\overrightarrow{t}\left(b_{i}b\right)=\sum \overrightarrow{t}\left(b_i{b}_j\right)c_j=0\text{.}$$ This system has a nontrivial solution if and only if the matrix $G=\mid \overrightarrow{t}\left(b_i{b}_j\right)\mid$ is singular. $\square$

Symmetrization

Let $A$ be a finite dimensional algebra with a nondegenerate trace $\overrightarrow{t}$ and let $B$ be a basis of $A\text{.}$ Let $B^{*}$ be the dual basis to $B$ with respect to the form $⟨,⟩$ given by (2.2). For $g\in B$ let $g^{*}$ denote the element of $B^{*}$ such that $\overrightarrow{t}\left(g{g}^{*}\right)=1\text{.}$ Let $V_1$ and $V_2$ be representations of $A$ of dimensions $d_1$ and $d_2$ respectively.

(2.6) Proposition. Let $C$ be any $d_1\times d_2$ matrix with entries in $ℂ\text{.}$ If

$$\left[C\right]=\sum _{g\in B}{V}_1\left(g\right)C{V}_2\left(g^{*}\right),$$

then, for any $a\in A,$

$$V_1\left(a\right)\left[C\right]=\left[C\right]V_2\left(a\right)\text{.}$$

		Proof.
	Let $a\in A\text{.}$ Then $$\begin{array}{ccc}{V}_1\left(a\right)\left[C\right]& =& \sum _g{V}_1\left(ag\right)C{V}_2\left(g^{*}\right)\\ & =& \sum _{g\in B}{V}_1\left(\sum _{h\in B}⟨ag,h^{*}⟩h\right)C{V}_2\left(g^{*}\right)\\ & =& \sum _{g,h\in B}⟨ag,h^{*}⟩V_1\left(h\right)C{V}_2\left(g^{*}\right)\\ & =& \sum _{g,h\in B}{V}_1\left(h\right)C\overrightarrow{t}\left(ag{h}^{*}\right)V_2\left(g^{*}\right)\\ & =& \sum _{h\in B}{V}_1\left(h\right)C\sum _{g\in B}\overrightarrow{t}\left(h^{*}ag\right)V_2\left(g^{*}\right)\\ & =& \sum _{h\in B}{V}_1\left(h\right)C{V}_2\left(\sum _{g\in B}⟨h^{*}a,g⟩g^{*}\right)\\ & =& \sum _{h\in B}{V}_1\left(h\right)C{V}_2\left(h^{}a\right)\\ & =& \left[C\right]V_2\left(a\right)\text{.}\end{array}$$ $\square$

If $V_1$ and $V_2$ are irreducible then Schur's lemma gives that $\left[C\right]=0$ if $V_1$ and $V_2$ are inequivalent and that if $V_1=V_2$ then $\left[C\right]=c{I}_{d_1}$ for some $c\in ℂ\text{.}$

Let $A$ be a finite dimensional algebra. The action of $A$ on itself by multiplication on the left turns $A$ into an $A\text{-module.}$ The resulting representation is the regular representation of $A$ and we denote it by $\overrightarrow{A}\text{.}$ The set is the same as the set $A,$ but we distinguish elements of $\overrightarrow{A}$ by writing $\overrightarrow{a}\in \overrightarrow{A}\text{.}$ As usual we denote the algebra of this representation by $\overrightarrow{A}\left(A\right)\text{.}$ We denote the trace of this representation by tr. Notice that the trace tr of the regular representation can be given by

$$\begin{array}{cc}\text{tr}\left(a\right)=\sum _{g\in B}ag{\mid }_g,& \text{(2.7)}\end{array}$$

(2.8) Theorem. If $A$ is a finite dimensional algebra such that the regular representation $\overrightarrow{A}$ has nondegenerate trace then every representation $V$ of $A$ is completely decomposable.

		Proof.
	Let tr denote the trace of the regular representation. Let $B$ be a basis of $A$ and for each $g\in B$ let $g^{*}$ denote the element of the dual basis to $B$ with respect to the trace tr such that $\text{tr}\left(g{g}^{*}\right)=1\text{.}$ Let $V$ be a representation of $A$ of dimension $d$ and let $V_1$ be an irreducible invariant subspace of $V\text{.}$ Let $P:V\to V$ be an arbitrary projection of $V$ onto $V_1\text{.}$ Define $$P_1=\sum _{g\in B}V\left(g\right)PV\left(g^{*}\right)\text{.}$$ Then, by (2.6), we know that $$V\left(a\right)P_1=P_{1}V\left(a\right)\text{.}$$ Since $V_1$ is an $A\text{-invariant}$ subspace, $P_{1}V\subseteq V_1\text{.}$ Since $V_1$ is irreducible $P_{1}V$ is either 0 or $V_1\text{.}$ Let $e=\sum _{g\in B}g{g}^{*}\text{.}$ If $a\in A$ then $$\begin{array}{ccc}\text{tr}\left(ae\right)& =& \text{tr}\left(\sum _{g\in B}ag{g}^{*}\right)\\ & =& \sum _{g\in B}⟨ag,g^{*}⟩\\ & =& \sum _{g\in B}ag{\mid }_g\\ & =& \text{tr}\left(a\right)\text{.}\end{array}$$ This shows that $\text{tr}\left(a(e-1)\right)=0$ for all $a\in A\text{.}$ Since tr is nondegenerate we have that $$\begin{array}{cc}e=\sum _{g\in B}g{g}^{*}=1\text{.}& \text{(2.9)}\end{array}$$ Now let $v\in V_1\text{.}$ Then since $V\left(g^{*}\right)v\in V_1$ we have $$\begin{array}{ccc}{P}_{1}v=& =& \sum _{g\in B}V\left(g\right)P\left(V\left(g^{*}\right)v\right)\\ & =& \sum _{g\in B}V\left(g\right)V\left(g^{*}\right)v\\ & =& V\left(\sum _{g\in B}g{g}^{*}\right)v\\ & =& V\left(1\right)v\\ & =& v\text{.}\end{array}$$ So $P_{1}V=V_1$ and $P_1{P}_{1}V=P_{1}V\text{.}$ Let $P_1^{\prime }=I_d-P_1$ and let $V_2=P_1^{\prime }V\text{.}$ Notice that $V\left(a\right)P_1^{\prime }=P_1^{\prime }V\left(a\right)$ for all $a\in A\text{.}$ So $V_2$ is an $A\text{-invariant}$ subspace of $V\text{.}$ Since, for every $v\in V,$ $v=P_{1}v+(I_d-P_1^{\prime })v=P_{1}v+P_1^{\prime }v,$ we have $V=P_{1}V+P_1^{\prime }V\text{.}$ If $v\in P_{1}V\cap P_1^{\prime }V$ then $v=P_{1}v=P_1{P}_1^{\prime }v=P_{\prime }(I_d-P_1^{\prime })v=0\text{.}So$ P_{1}V\cap P_1^{\prime }V=0\text{.}$Thus\; we\; see\; that$ V=P_{1}V\oplus P_1^{\prime }V\text{.}$$ If $P_1^{\prime }V$ is irreducible then we are done. If not apply the same process again with $P_1^{\prime }V$ in place of $V\text{.}$ Since $V$ is finite dimensional continuing this process will eventually produce a decomposition of $V$ into irreducible representations. $\square$

Now let $A$ be a finite dimensional algebra such that the trace tr of the regular representation $\overrightarrow{A}$ of $A$ is nondegenerate. Let $B$ be a basis of $A$ and for each $g\in B$ let $g^{*}$ denote the element of the dual basis to $B$ with respect to the trace tr such that $\text{tr}\left(g{g}^{}\right)=1\text{.}$ Let $V$ be a faithful representation of $A\text{.}$ By (2.8) we know that $V$ can be completely decomposed into irreducible representations. Choose a maximal set $\left\{W_{\lambda }\right\}$ of nonisomorphic irreducible representations appearing in the decomposition of $V\text{.}$ Let $d_{\lambda }=\text{dim}{W}_{\lambda }$ and define $M_{\bar{d}}\left(ℂ\right){\oplus }_{\lambda }{M}_{d_{\lambda }}\left(ℂ\right)\text{.}$ We view $M_{\bar{d}}\left(ℂ\right)$ as an algebra of block diagonal matrices with one $d_{\lambda }\times d_{\lambda }$ block for each $\lambda \text{.}$ $V\left(A\right)\cong {\oplus }_{\lambda }{W}_{\lambda }\left(A\right)$ is a subalgebra of $M_{\bar{d}}\left(ℂ\right)$ in a natural way. Let $E_{ij}^{\lambda }$ denote the $d\times d$ matrix with 1 in the $(i,j)$ entry of the $\lambda \text{th}$ block and 0 everywhere else and let $I_{\lambda }$ be the matrix which is the identity on the $\lambda \text{th}$ block and 0 everywhere else.

For each $g\in B$ let $W_{ij}^{\lambda }\left(g^{*}\right)$ denote the $(i,j)$ entry of the matrix $W_{\lambda }\left(g^{*}\right)\text{.}$ Then

$$k\text{th row of}\left(W_{ji}^{\lambda }\left(g^{*}\right)W_{\lambda }\left(g\right)\right)=j\text{th row of}\left(W_{\lambda }\left(g^{*}\right)E_{ik}^{\lambda }{W}_{\lambda }\left(g\right)\right)\text{.}$$

So

$$\begin{array}{cc}\begin{array}{ccc}k\text{th row of}\left(\sum _{g\in B}{W}_{ji}^{\lambda }\left(g^{*}\right)W_{\lambda }\left(g\right)\right)& =& j\text{th row of}\left(\sum _g{W}_{\lambda }\left(g^{*}\right)E_{ik}^{\lambda }{W}_{\lambda }\left(g\right)\right)\\ & =& j\text{th row of}\left(c{I}_{\lambda }{\delta }_{ik}\right)\text{.}\end{array}& \text{(2.10)}\end{array}$$

So the $i\text{th}$ row of $\sum _g{W}_{ji}^{\lambda }\left(g^{}\right)W_{\lambda }\left(g\right)$ is all zeros except for $c$ in the $j\text{th}$ spot and all other rows of $\sum _{g\in B}{W}_{ji}^{\lambda }\left(g^{*}\right)W_{\lambda }\left(g\right)$ are zero. So

$$\begin{array}{cc}\sum _g{W}_{ji}^{\lambda }\left(g^{*}\right)W_{\lambda }\left(g\right)=c{E}_{ij}^{\lambda }& \text{(2.11)}\end{array}$$

for some $c\in ℂ\text{.}$ We can determine $c$ by setting $i=k$ to get

$$\begin{array}{ccc}c{d}_{\lambda }& =& \text{tr}\left(c{I}_{\lambda }{\delta }_{ii}\right)\\ & =& \text{tr}\left(\sum _g{W}_{\lambda }\left(g^{*}\right)E_{ii}^{\lambda }{W}_{\lambda }\left(g\right)\right)\\ & =& \text{tr}\left(\sum _g{W}_{\lambda }\left(g\right)W_{\lambda }\left(g^{*}\right)E_{ii}\right)\\ & =& \text{tr}\left(W_{\lambda }\left(\sum _{g}g{g}^{*}\right)E_{ii}^{\lambda }\right)\text{.}\end{array}$$

Since the trace of the regular representation was used to construct the $g^{*}$ we have, (2.9), that $\sum _{g}g{g}^{*}=1,$ giving

$$\begin{array}{ccc}\text{tr}\left(W_{\lambda }\left(\sum _{g}g{g}^{*}\right)E_{ii}^{\lambda }\right)& =& \text{tr}\left(W_{\lambda }\left(1\right)E_{ii}^{\lambda }\right)\\ & =& \text{tr}\left(I_{\lambda }{E}_{ii}^{\lambda }\right)\\ & =& 1\text{.}\end{array}$$

So $c{d}_{\lambda }=1$ and we can write (2.11) as

$$d_{\lambda }\sum _g{W}_{ji}^{\lambda }\left(g^{*}\right)W_{\lambda }\left(g\right)=E_{ij}^{\lambda }\text{.}$$

Since we have expressed each $E_{ij}^{\lambda }$ as a linear combination of basis elements of $V\left(A\right)$ we have that $E_{ij}^{\lambda }\in V\left(A\right)$ for every $i$ and $j\text{.}$ But the $E_{ij}^{\lambda }$ form a basis of $M_{\bar{d}}\left(ℂ\right)\text{.}$ So $M_{\bar{d}}\left(ℂ\right)\subseteq V\left(A\right)\text{.}$ Then $A\cong V\left(A\right)=M_{\bar{d}}\left(ℂ\right)\text{.}$ We have proved the following theorem.

(2.12) Theorem. (Artin-Wedderburn) If $A$ is a finite dimensional algebra such that the trace of the regular representation of $A$ is nondegenerate, then, for some set of positive integers $d_{\lambda },$

$$A\cong {\oplus }_{\lambda }{M}_{d_{\lambda }}\left(ℂ\right)\text{.}$$

Examples

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,$$\mathrm{tr}$ is nondegenerate. If we set $\overrightarrow{t}\left(a\right)=a{|}_1$ then $\overrightarrow{t}$ is a trace on $A$ and ${\left\{g^{-1}\right\}}_{g\in G}$ is the dual basis to the basis ${\left\{g\right\}}_{g\in G}$with respect to the trace.
4. Let $\overrightarrow{t}$ be the trace of a faithful realisation $\phi$ of an algebra $A$ (ie for each $a\in A,\overrightarrow{t}\left(a\right)$ is given by the styandard trace of $\phi \left(a\right)$ where $\phi$ is an injective homomorphism $\phi :A\to M_d\left(ℂ\right)$). Let $\sqrt{A}=\left\{a\in A|\overrightarrow{t}\left(ab\right)=0\text{for all}b\in A\right\}.$ $\sqrt{A}$ is an ideal of $A.$ Let $a\in \sqrt{A}.$ Then $\mathrm{tr}\left(a^{k-1}a\right)=\mathrm{tr}\left(a^k\right)=0$ for all $k.$ If ${\lambda }_1,\dots ,{\lambda }_d$ are the eigenvalues of $\phi \left(a\right)$ then $\overrightarrow{t}\left(a^k\right)={\lambda }_1^k+{\lambda }_2^k+\dots +{\lambda }_d^k=p_k\left(\lambda \right)=0$ for all $k>0$, where $p_k$ represents the $k$-th power symmetric functions [Mac1979]. Since the power symmetric functions generate the ring of symmetric functions this means that the elementary symmetric functions $e_k\left(\lambda \right)=0$ for $k>0$, [Mac1979] p17, 2.14. Since the characteristic polynomial of $\phi \left(a\right)$ can be written in the form $${\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 $A.$

Notes and References

The approach to the theory of semisimple algebras that is presented in this section and the following section follows closely a classical approach to the representation theory of finite groups, see for example [Ser1977] or [Ham1962]. Once one has the analogue of the symmetrization process for finite groups, the only nontrivial step in the theory that is not exactly analogous to the theory for finite groups is formula (2.9).

I discovered this method after reading the sections of [CRl] concerning Frobenius and symmetric algebras. Frobenius and symmetric algebras were introduced by R. Brauer and C. Nesbitt, [BNe1937] and [Nes1938]. T. Nakayama [Nak1941] has a version of Theorem (2.6) and R. Brauer [Bra1945] proves analogues of the Schur orthogonality relations that are analogous to formula (2.10). Ikeda [Ike1953], and Higman [Hig1955], following work of Gaschütz [Gas1952], construct "Casimir" type elements similar to those in (2.9) and §3 Ex. 7. In [CRe1981] §9 Curtis and Reiner use a similar approach but with different proofs, communicated to them by R. Kilmoyer, to obtain theorems (3.8) and (3.9) for split semisimple algebras (over fields of characteristic 0). N. Wallach has told me that essentially the same approach works for finite dimensional Lie algebras.

This approach is useful for studying semisimple algebras that have distiguished bases. The recent interest in quantum deformations is producing a host of examples of semisimple algebras that are not group algebras but that do have distinguished bases. Some examples are Hecke algebras associated to root systems, the Brauer algebra, and the Birman-Wenzl algebra [BWe1989]. For an approach to the Hecke algebras that is essentially an application of the general theory given here see [GUn1989] and [Cr3] §68C.

I would like to thank Prof. A. Garsia for suggesting that I try to find an analogue of the symmetrization process for finite groups for the Brauer algebra. It was this problem that resulted in my discovery of this approach. I would like to thank Prof. C.W. Curtis for his helpful suggestions in locating literature with a similar approach. I would also like to thank Prof. Garsia for showing me the proofs of Exs. 4 and 5.

Notes and References

This is an excerpt from the unpublished first chapter of Arun Ram's dissertation entitled Representation Theory, written July 4, 1990.

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