## Semisimple algebras

Last update: 6 November 2012

## Semisimple algebras

An algebra $A$ is simple if $A\cong {M}_{d}\left(ℂ\right)\text{.}$

Suppose $\stackrel{\to }{t}$ is a trace on ${M}_{d}\left(ℂ\right)\text{.}$ Then

$t→(Eij) = t→ ( Ei1 E1j ) = t→ ( E1j Ei1 ) = t→(E11) δij.$

If $a=\mid {a}_{ij}\mid \in {M}_{d}\left(ℂ\right)$ then

$t→(a) = t→ ( ∑i,j aij Eij ) = ∑i,j aij t→(Eij) = ∑i,j aijt→ (E11) δij = t→(E11) (∑iaii).$

So, up to a constant factor there is a unique trace function on ${M}_{d}\left(ℂ\right),$ that given by the standard trace on matrices.

Suppose $J$ is an ideal of ${M}_{d}\left(ℂ\right)$ and that $a=\mid {a}_{ij}\mid \in J,$ with $a\ne 0\text{.}$ So ${a}_{ij}\ne 0$ for some $\left(i,j\right)\text{.}$ Since $a\in J$ and $J$ is an ideal,

$(1/aij) ∑k=1d Ekia Ejk= (1/aij) ∑kaij Ekk=Id$

is an element of $J\text{.}$ Thus $J={M}_{d}\left(ℂ\right)\text{.}$ This shows that the only ideals of ${M}_{d}\left(ℂ\right)$ are the trivial ones, 0 and ${M}_{d}\left(ℂ\right)\text{.}$ It is an immediate consequence of §1 Ex. 1 that the center of ${M}_{d}\left(ℂ\right)$ is ${I}_{d}\left(ℂ\right)=ℂ{I}_{d}\text{.}$ Furthermore, ${I}_{d}$ is the unique central idempotent in ${M}_{d}\left(ℂ\right)\text{.}$

(3.1) Proposition. There is a unique irreducible representation of ${M}_{d}\left(ℂ\right)$ given by the usual multiplication of $d×d$ matrices on all column vectors of size $d\text{.}$

 Proof. Let $V$ be the $d$ dimensional vector space of column vectors of size $d\text{.}$ The standard basis of $V$ consists of the vectors ${e}_{i}={\left(0,\dots ,0,1,0,\dots ,0\right)}^{1},$ $1\le i\le d$ where the 1 appears in the $i\text{th}$ spot. Suppose that ${V}^{\prime }$ is a nonzero invariant subspace of $V\text{.}$ Let $v=\sum _{i}{v}_{i}{e}_{i},$ ${v}_{i}\in ℂ,$ be a nonzero element of ${V}^{\prime }\text{.}$ So ${v}_{i}\ne 0$ for some $1\le i\le d\text{.}$ Then $\left(1/{v}_{i}\right){E}_{ji}v={e}_{j}\text{.}$ Since ${V}^{\prime }$ is invariant we have that ${e}_{j}\in {V}^{\prime }$ for each $1\le j\le d\text{.}$ But since the ${e}_{j}$ are a basis of $V$ this implies that $V={V}^{\prime }\text{.}$ So $V$ is an irreducible representation of ${M}_{d}\left(ℂ\right)\text{.}$ Now let W be an arbitrary irreducible representation of $A={M}_{d}\left(ℂ\right)\text{.}$ There is some vector $w\in W$ and some $a\in A$ such that $aw\ne 0,$ otherwise $W$ would be the zero representation. If $a=\mid {a}_{ij}\mid$ then $aw=\sum _{i,j}{a}_{ij}{E}_{ij}w\ne 0$ implies that ${E}_{ij}w\ne 0$ for some pair $\left(i,j\right)\text{.}$ The space ${M}_{d}\left(ℂ\right){E}_{ij}$ consists of all matrices that are 0 except in the $j\text{th}$ column and is isomorphic to $V\text{.}$ The map $ϕ: Md(ℂ)Eij ⟶ W aEij ⟶ aEij (3.2)$ is an isomorphism since both $V$ and $W$ are irreducible. $\square$

So the regular representation of ${M}_{d}\left(ℂ\right)$ decomposes as a direct sum of $d$ copies of the unique irreducible representation $V$ of ${M}_{d}\left(ℂ\right),$ one copy for each column in ${M}_{d}\left(ℂ\right)\text{.}$

An algebra $A$ is semisimple if

$A≅⊕λ∈Λ Mdλ(ℂ), (3.3)$

where $\Lambda$ is a finite index set. The vector $\stackrel{\to }{d}=\left({d}_{\lambda }\right),$ $\lambda \in \Lambda$ of positive integers is called the dimension vector of the algebra $A\text{.}$ We will use ${M}_{\stackrel{‾}{d}}\left(ℂ\right)$ as a shorthand notation for the algebra given by the right hand side of (3.3). We can view ${M}_{\stackrel{‾}{d}}\left(ℂ\right)$ as the full algebra of block diagonal matrices where the $\lambda \text{th}$ block is dimension ${d}_{\lambda }\text{.}$ We denote the matrix having 1 in the $\left(i,j\right)\text{th}$ position of the $\lambda \text{th}$ block and zeros everywhere else by ${E}_{ij}^{\lambda }\text{.}$ Denote the matrix which is the identity on the $\lambda \text{th}$ block and 0 everywhere else by ${I}_{\lambda }\text{.}$

Any trace on ${\oplus }_{\lambda \in \Lambda }{M}_{{d}_{\lambda }}\left(ℂ\right)$ is completely determined by a vector $\stackrel{\to }{t}=\left({t}_{\lambda }\right)$ of complex numbers such that $\stackrel{\to }{t}\left({E}_{11}^{\lambda }\right)={t}_{\lambda }$ for each $\lambda$ in the finite index set $\Lambda \text{.}$ The vector $\stackrel{\to }{t}=\left({t}_{\lambda }\right)$ is the trace vector of the trace $\stackrel{\to }{t}\text{.}$ A trace $\stackrel{\to }{t}$ on ${M}_{\stackrel{‾}{d}}\left(ℂ\right)$ is nondegenerate if and only if ${t}_{\lambda }\ne 0$ for all $\lambda \in \Lambda \text{.}$ The only ideals of ${M}_{\stackrel{‾}{d}}\left(ℂ\right)={\oplus }_{\lambda \in \Lambda }{M}_{{d}_{\lambda }}\left(ℂ\right)$ are of the form ${\oplus }_{\lambda \in {\Lambda }^{\prime }}{M}_{{d}_{\lambda }}\left(ℂ\right)$ where ${\Lambda }^{\prime }\subset \Lambda \text{.}$ The ${I}_{\lambda },$ $\lambda \in \Lambda$ form a basis of the center of ${M}_{\stackrel{‾}{d}}\left(ℂ\right)\text{.}$ Every central idempotent is a sum of some subset of the ${I}_{\lambda }\text{.}$ There is, up to isomorphism, one irreducible representation of ${\oplus }_{\lambda \in \Lambda }{M}_{{d}_{\lambda }}\left(ℂ\right)$ for each $\lambda \in \Lambda \text{.}$ It can be given by left multiplication on the space ${M}_{\stackrel{‾}{d}}\left(ℂ\right){E}_{ij}^{\lambda },$ for any $i,j,$ $1\le i,j\le {d}_{\lambda }\text{.}$ The decomposition of the regular representation of ${\oplus }_{\lambda \in \Lambda }{M}_{{d}_{\lambda }}\left(ℂ\right)$ into irreducibles is given by

$Md‾(ℂ) ⟶ =⊕λ∈Λ Wλ⊕dλ (3.4)$

where ${W}_{\lambda }$ denotes the irreducible representation corresponding to $\lambda \text{.}$

### Matrix units and characters

Let $A$ be an algebra and $\stackrel{^}{A}$ a finite index set such that $A\cong {\oplus }_{\lambda \in \stackrel{^}{A}}{M}_{{d}_{\lambda }}\left(ℂ\right)$ under an isomorphism $\varphi :\phantom{\rule{0.2em}{0ex}}A\to {\oplus }_{\lambda \in \stackrel{^}{A}}{M}_{{d}_{\lambda }}\left(ℂ\right)\text{.}$ (Let Md() denote the algebra ${\oplus }_{\lambda }{M}_{{d}_{\lambda }}\left(ℂ\right)\text{.)}$ Warning: The isomorphism $\varphi$ is not unique; nontrivial automorphisms of Md() do exist, just conjugate by an invertible matrix. ${z}_{\lambda }={\varphi }^{-1}\left({I}_{\lambda }\right)$ is an idempotent and an element of the center of $A\text{.}$ The zλ are the minimal central idempotents of $A\text{.}$ They are minimal in the sense that every central idempotent of $A$ is a sum of ${z}_{\lambda }\text{'s.}$ These elements are independent of the isomorphism $\varphi \text{.}$

A set of elements ${e}_{ij}^{\lambda }\in A,$ $\lambda \in \stackrel{^}{A},$ $1\le i,j\le {d}_{\lambda }$ is the set of matrix units of $A$ if

$eijλ ersμ { 0, ifλ≠μ; 0, ifλ=μ, j≠r; eisλ, ifλ=μ, j=r. (3.5)$

A complete set of matrix units of $A$ is a set of matrix units which forms a basis of $A\text{.}$ Let ${E}_{ij}^{\lambda }\in {M}_{\stackrel{‾}{d}}\left(ℂ\right)$ denote the matrix having 1 in the $\left(i,j\right)\text{th}$ position of the $A\text{th}$ block and zeros everywhere else. If $\left\{{e}_{ij}^{\lambda }\right\}$ is a set of matrix units of $A,$ the mapping ${e}_{ij}^{\lambda }⟼{E}_{ij}^{\lambda }$ determines explicitly an isomorphism $A\to {M}_{\stackrel{‾}{d}}\left(ℂ\right)\text{.}$ Conversely, an isomorphism $\varphi :\phantom{\rule{0.2em}{0ex}}A\to {M}_{\stackrel{‾}{d}}\left(ℂ\right)$ determines a set of matrix units ${e}_{ij}^{\lambda }={\varphi }^{-1}\left({E}_{ij}^{\lambda }\right)\text{.}$ Note that the ${e}_{ii}^{\lambda }$ are minimal orthogonal idempotents in $A\text{.}$

Let ${W}_{\lambda },$ $\lambda \in \stackrel{^}{A}$ denote the irreducible representations $A\text{.}$ By (3.1) and (3.2), for each $\lambda \in \stackrel{^}{A},$

$Wλ≅A eijλ, (3.6)$

for any $i,j,$ $1\le i,j\le {d}_{\lambda },$ where the action of $A$ on $A{e}_{ij}^{\lambda }$ is given by left multiplication. For each $\lambda \in \stackrel{^}{A}$ denote the character of the irreducible representation ${W}_{\lambda }$ by ${\chi }^{\lambda }$ and for each $\lambda \in \stackrel{^}{A}$ and $a\in A$ let ${W}_{ij}^{\lambda }\left(a\right)$ denote the $\left(i,j\right)\text{th}$ entry of the matrix ${W}_{\lambda }\left(a\right)\text{.}$ Note that we can view each ${W}_{\lambda }\left(a\right)$ as a matrix in ${M}_{\stackrel{‾}{d}}\left(ℂ\right)$ with all but the $\lambda \text{th}$ block 0.

Let $B$ be an arbitrary basis of $A\text{.}$ Let $\stackrel{\to }{t}=\left({t}_{\lambda }\right)$ $\lambda \in \stackrel{^}{A}$ be a nondegenerate trace on $A\text{.}$ For each $g\in B$ let ${g}^{*}$ denote the element of the dual basis to $B$ with respect to the trace $\stackrel{\to }{t}$ such that $\stackrel{\to }{t}\left(g{g}^{*}\right)=1\text{.}$

(3.7) Theorem. (Fourier inversion formula) The elements

$eijλ= ∑g∈B tλ Wjiλ (g*)g$

form a complete set of matrix units of $A\text{.}$

 Proof. Let $\varphi :\phantom{\rule{0.2em}{0ex}}A\to {M}_{\stackrel{‾}{d}}\left(ℂ\right)$ be given by $\varphi \left(a\right)={\oplus }_{\lambda }{W}_{\lambda }\left(a\right)\text{.}$ This is an isomorphism. For each $\lambda \in \stackrel{^}{A}$ and $1\le i,j\le {d}_{\lambda }$ let $eijλ= ϕ-1 (Eijλ).$ The set $B=\left\{{e}_{ij}^{\lambda }\right\}$ forms a basis of $A\text{.}$ The dual basis with respect to the trace $\stackrel{\to }{t}=\left({t}_{\lambda }\right)$ is the basis $\left\{\left(1/{t}_{\lambda }\right){e}_{ji}^{\lambda }\right\}\text{.}$ $∑g∈B tλWjiλ (g*)g = ∑k,l,μ tλ Wjiλ ( (1/tμ) elkμ ) eklμ = ∑k,l,μ tλ(1/tμ) δjl δik δλμ eklμ = eijλ.$ Notice that $kth row of ( ∑g∈B Wjiλ (g*) Wλ(g) ) =jth row of ( ∑g∈B Wλ(g*) Eikλ Wλ(g) ) .$ By § Ex. 1 we know that $\sum _{g\in B}{W}_{\lambda }\left({g}^{*}\right){E}_{ik}^{\lambda }{W}_{\lambda }\left(g\right)$ is independent of the basis $B\text{.}$ $\square$

(3.8) Theorem.

$∑g∈Btλ χλ(g*) g=zλ.$

 Proof. $zλ = ∑i=1dλ eiiλ = ∑i ∑g∈B tλWiiλ (g*)g = ∑g∈B ( ∑itλ Wiiλ (g*) ) g = ∑g∈B tλχλ (g*)g.s$ $\square$

(3.9) Theorem.

$∑g∈Bχλ (g)χμ (g*) (dλ/tλ) δλμ.$

 Proof. $dλδλμ = χλ(zμ) = χλ ( ∑g∈B tλχμ (g*)g ) = ∑g∈B tλχλ (g)χμ (g*).$ $\square$

### Examples

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 $n×n$ 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 $\mathrm{dim}V*=\mathrm{dim}V.$ If $B$ is a basis of $V$ then the functions ${\delta }_{b},b\in B,$ determined by for ${b}_{i}\in B,$ form a basis of $V*.$ If $A$ is a semisimple algebra isomorphic to ${M}_{d}\left(ℂ\right)={\oplus }_{\lambda \in \stackrel{~}{A}}{M}_{{d}_{\lambda }}\left(ℂ\right),$ $\stackrel{~}{A}$ an index set for the irreducible representations ${W}_{\lambda }$ of $A,$ then $dimA= ∑ λ∈ A ~ d λ 2 ,$ and the functions ${W}_{ij}^{\lambda }\left({W}_{ij}^{\lambda }\left(a\right)$ the $\left(i,j\right)$-th entry of the matrix ${W}_{\lambda }\left(a\right),a\in A\right)$ 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 $\lambda \in \stackrel{~}{A}$ and extending linearly to all of $Z\left(A\right).$

Given a nondegenerate trace $\stackrel{\to }{t}$ 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 $z\in Z\left(A\right),$ $φ z a = t → za ,$ since $φ z μ / t μ a = t → z μ / t μ a = t → 1 t μ z μ a = 1 t μ t μ χ μ a = χ μ a .$

5. If $A$ is a semisimple algebra isomorphic to ${M}_{b}\left(ℂ\right)={\oplus }_{\lambda \in \stackrel{~}{A}}{M}_{{d}_{\lambda }}\left(ℂ\right),\phantom{\rule{2em}{0ex}}\stackrel{~}{A}$ an index set for the irreducible representations ${W}_{\lambda }$ of $A,$ then the right regular representation decomposes as $A → ≅ ⊕ λ∈A W λ ⊕ d λ .$ If matrix units ${e}_{ij}^{\lambda }$ are given by (3.7) then $tr e ii λ =tr d λ E ii λ = d λ .$ So the trace of the regular representation of $A,$$\mathrm{tr},$ is given by the trace vector $\stackrel{\to }{t}=\left({t}_{\lambda }\right),$ where ${t}_{\lambda }={d}_{\lambda }$ for each $\lambda \in \stackrel{~}{A}.$
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 $χ χ' = ∑ g∈B χ g χ' g* .$ 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, $χ χ' =dim Hom A V V' .$ 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 $\stackrel{\to }{t}=\left({t}_{\lambda }\right)$ 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 $\stackrel{\to }{t}$ such that $\stackrel{\to }{t}\left(gg*\right)=1.$ For each $a\in A$ define $a = ∑ g∈B gag*.$ 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 $a = ∑ i,j,λ 1 t λ e ij λ a e ji λ = ∑ i,j,λ 1 t λ a jj λ e ii λ = ∑ λ 1 t λ ∑ j a jj λ ∑ i e ii λ = ∑ λ 1 t λ χ λ a z λ .$ So ${\chi }^{\lambda }\left(\left[a\right]\right)=\left(\frac{{d}_{\lambda }}{{t}_{\lambda }}\right){\chi }^{\lambda }\left(a\right).$ By 3.9 $∑ g∈B t λ 2 d λ χ μ g* g = ∑ λ ∑ g∈B t λ 2 d λ 1 t λ χ λ g χ μ g* z λ = ∑ λ δ λμ z λ = z μ .$ 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 $\stackrel{\to }{t}$ be the trace on $A$ given by $t → a =a | 1 ,$ where 1 is the identity in $G.$ By Ex 5 and Section 2 Ex 3 the trace vector of $\stackrel{\to }{t}$ 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 $h = ∑ g∈B ghg*= ∑ g∈B gh g -1$ 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 $tr V hg h -1 =tr V h V g V h -1 =tr V g$ for any representation $V$ of $G$ and all $g,h\in G,$ characters of $G$ are constant on conjugacy classes. Using theorem 3.8, $G δ λμ = ∑ g χ λ g χ μ g -1 = ∑ ρ ∑ g∈ C ρ χ λ g χ μ g -1 = ∑ ρ C ρ χ λ ρ χ μ ρ' ,$ 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 $Ξ Ξ 't = G I,$ but then we also have that ${\Xi }^{\text{'}t}\Xi =\left|G\right|I,$ or equivalently that $∑ λ χ λ ρ ' χ λ τ = G C ρ δ ρτ .$
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 $\lambda \in \stackrel{~}{A}$, $χ λ b = χ λ b ' .$ 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 $\stackrel{~}{A}.$ Assume that we have fixed such a correspondence and denote the classses of $B$ by ${C}_{\lambda },\lambda \in \stackrel{~}{A}.$

Let $\stackrel{\to }{t}$ be a nondegenerate trace on $A$ and let $G$ be the Gram matrix with respect to the basis $B$ and the trace $\stackrel{\to }{t}.$ If $g\in B,$ let $g*$ denote the element of the dual basis to $B$, with respect to the trace $\stackrel{\to }{t}$, such that $\stackrel{\to }{t}\left(gg*\right)=1.$ Let ${G}^{-1}=C=\parallel {c}_{gg\text{'}}\parallel$ and recall that $g*={\sum }_{g\text{'}\in B}{c}_{gg\text{'}}g\text{'}.$ Then $d λ t λ δ λμ = ∑ g∈B χ λ g χ μ g* = ∑ g∈B χ λ g χ μ ∑ g'∈B c gg' g' = ∑ g,g'∈B χ λ g c gg' χ μ g' .$ Collecting $g,g\text{'}\in B$ by class size gives $d λ t λ δ λμ = ∑ ρ,τ ∑ g∈ C ρ ,g'∈ C τ χ λ g c gg' χ μ g'$ where ${\chi }^{\lambda }\left(\rho \right)$ denotes the value of the character ${\chi }^{\lambda }\left(\rho \right)$ at elements of the class ${C}_{\rho }.$ Now define a matrix $C=\parallel {c}_{\rho \tau }\parallel$ 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 $I=Ξ C Ξ' t .$ So $I= C Ξ' t Ξ,$ or equivalently that $δ ρτ = ∑ σ,λ c ρσ t λ d λ χ λ σ χ λ τ = ∑ σ,λ ∑ g∈ C ρ ,g'∈ C σ c gg' t λ d λ χ λ σ χ λ τ = ∑ λ ∑ g'∈B ∑ g∈ C ρ c gg' χ λ σ χ λ τ = ∑ g∈ C ρ ∑ λ χ λ g* χ λ τ .$

### Notes and References

The Fourier Inversion formula for representations of finite groups appears in [Ser1977] p. 49. I must thank Prof. A. Garsia for suggesting the problem of finding a generalization. I know of no references giving a similar generalization. Theorems (3.7) and (3.8) are due to R. Kilrnoyer and appear in [CRe1981] (9.17) and (9.19). Ex. 3 is the Frobenius-Schur theorem. Ex. 9 is known as the second orthogonality relation for characters of finite groups (the first orthogonality relation being (3.8)), see [CRe1981] (9.26) or [Ser1977] Chap. 2, Prop. 7. The generalization given in Ex. 10 is new as far as I know. [Rl] shows that the Brauer algebra is an example of semsimple algebra that is not a group algebra with a natural basis that can be partitioned into classes such that (3.13) holds.

## Notes and References

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