Let $e={\sum }_{g_i\in G}{g}_i{g}_i^{*}\text{;}$ we wish to show that $e=1\text{.}$

Let $a\in A$ be arbitrary, and denote the coefficient of $g_i$ in any element $a$ by $a{|}_{g_i}\text{.}$ Then $$\text{tr}\left(ae\right)=\sum _{g_i\in G}\text{tr}\left(a{g}_i{g}_i^{*}\right)=\sum _{g_i\in G}⟨ag,g^{*}⟩=\sum _{g_i\in G}\left(a{g}_i\right){|}_{g_i}=\text{tr}\left(a\right)$$ using the dual basis property. Hence $\text{tr}\left(a(1-e)\right)=0$ for all $a\in A,$ and the nondegeneracy of tr implies $1-e=0,$ that is $e=1\text{.}$

Hence $P_1:V⟶V_1$ is a projection.

Furthermore, $V_2=(I-P_1)V$ is an $A\text{-module.}$ Indeed, by the lemma from last time, $A$ commutes with $P_1,$ hence with $I-P_1\text{.}$ Therefore, for any $a\in A$ and $v\in V,$ $a·(I-P_1)v=(I-P_1)a·v\in V_2,$ hence $I-P_1$ is an $A\text{-module}$ homomorphism mapping $V$ onto $V_2\text{.}$

Finally, we assert that $V=V_1\oplus V_2\text{.}$ Any $v\in V$ may be written $v=P_{1}v+(I-P_1)v,$ hence $V=V_1+V_2\text{.}$ To show the sum is direct, note first that $P_1^2=P_1$ since $P_1$ fixes $V_1=\text{im}\hspace{0.17em}{P}_1$ pointwise. Suppose $x\in V_1\cap V_2,$ so we may write $x=(I-P_1)v$ for some $v\in V\text{.}$ Then $P_{1}x=x$ as $x\in V_1,$ and hence $$x=P_{1}x=P_1(I-P_1)v=P_{1}v-P_1^{2}v=0\text{.}$$ which completes the proof of $\left(1\right)\Rightarrow \left(2\right)\text{.}$

Next, prove $\left(1\right)\Rightarrow \left(3\right)\text{.}$ By the previous case, we may also assume that the all finite dimensional modules are completely decomposable.

Let $V=\overrightarrow{A}$ be the hideously denoted left regular representation of $A\text{.}$ Since ${\text{dim}}_{ℂ}\hspace{0.17em}A<\infty ,$ we may write $V\cong {\oplus }_{\lambda \in \hat{V}}{\left(W^{\lambda }\right)}^{\otimes m_{\lambda }},$ where $\hat{V}$ is some (finite) index set.

Note that the representation $V:A⟶M_d\left(ℂ\right)$ is injective (sometimes called faithful) as $V\left(a\right)·\overrightarrow{1}=\overrightarrow{a}$ for all $a\in A\text{.}$ Thus $A\cong V\left(A\right)\cong {\oplus }_{\lambda \in \hat{V}}{\left(W^{\lambda }\left(A\right)\right)}^{\otimes m_{\lambda }}$ as algebras. The number $m_{\lambda }$ are called the multiplicity of the irreducible representation $W^{\lambda }$ in $V\left(A\right)\cong A\text{.}$

For convenience, write $W={\oplus }_{\lambda \in \hat{V}}{\left(W^{\lambda }\left(A\right)\right)}^{\otimes m_{\lambda }}\text{.}$ Note that an arbitrary element of $W$ is of the form $$W\left(a\right)=\left(\begin{array}{c}{W}^{\lambda }\left(a\right)\\ & \ddots & \multicolumn{2}{c}{}& 0\\ \multicolumn{2}{c}{}& W^{\lambda }\left(a\right)\\ \multicolumn{3}{c}{}& W^{\mu }\left(a\right)\\ & 0& \multicolumn{2}{c}{}& \ddots \\ \multicolumn{5}{c}{}& W^{\nu }\left(a\right)\end{array}\right)$$ for some $a\in A,$ where there are $m_{\gamma }$ occurrences of each $W^{\gamma }\left(a\right)$ in the block diagonal decomposition of $V\left(a\right)\text{.}$ The product in the algebra $W\left(A\right)$ is componentwise, hence the map $W⟶{\oplus }_{\lambda \in \hat{V}}{W}^{\lambda }$ given by deleting duplicate copies of irreducibles, i.e. $W\left(a\right)\mapsto {\oplus }_{\lambda \in \hat{V}}{W}^{\lambda }\left(a\right)$ is an (onto) algebra homomorphism.

It is not hard to see that these algebras are isomorphic. Let $\{g_1^{\lambda },g_2^{\lambda },\dots ,g_{d_{\lambda }}^{\lambda }\}$ be a basis for $W^{\lambda },$ and let $g_{i,j}^{\lambda }\in W$ be the corresponding element to $g_i^{\lambda }$ in the $j\text{th}$ copy of $W^{\lambda }$ in $W\text{.}$ The set $$\{h_i^{\lambda }=\sum _{j=1}^{d_{\lambda }}{g}_{i,j}^{\lambda }\hspace{0.17em}|\hspace{0.17em}1\le i\le d_{\lambda },\hspace{0.17em}\lambda \in \hat{V}\}$$ is a basis for $W,$ since an element of $W$ must have identical matrices in all blocks indexed by a given $\lambda \text{.}$ Hence $W\left(A\right)$ has the same dimension as ${\oplus }_{\lambda \in \hat{V}}{W}^{\lambda }\left(A\right)\text{;}$ indeed, the homomorphism above sends $h_i^{\lambda }$ to $g_i^{\lambda }$ (following the same indexing scheme) and is clearly invertible.

Setting $\hat{A}=\hat{V},$ we have $A\cong V\left(A\right)\cong {\oplus }_{\lambda \in \hat{A}}{W}^{\lambda }\left(A\right)\text{.}$ It remains to show that $W^{\lambda }\left(A\right)\cong M_{d_{\lambda }}\left(ℂ\right)\text{.}$ This follows from the next extremely useful lemma:

Lemma 1.49 (Schur) Suppose $V$ and $W$ are irreducible representations of $A$ of dimensions $d_1$ and $d_2,$ respectively. If $B$ is a $d_1\times d_2$ matrix such that $V\left(a\right)B=BW\left(a\right)$ for all $a\in A,$ then either $B=0$ or $V$ is equivalent to $W$ and $B=cI\text{.}$

		Proof.
	Let $V={ℂ}^{d_1}$ and $W={ℂ}^{d_2}$ represent the corresponding $A\text{-modules}$ of these representations, and assume $B\ne 0\text{.}$ Let $\{w_1,\dots ,w_{d_2}\}$ be the standard basis of $W,$ so that $B$ represents a linear transformation $W⟶V$ given by $v_i=B{w}_i\text{.}$ The condition $V\left(a\right)B=BW\left(a\right)$ becomes $B(a·w)=a·B\left(w\right),$ i.e. $B$ is an $A\text{-module}$ homomorphism. Since $B\ne 0,$ the simplicity of $W$ forces $\text{ker}\hspace{0.17em}B=0\text{.}$ Therefore $B$ is an isomorphism of modules, hence $d_1=d_2$ and the original representations are equivalent. Observe that for all $c\in ℂ,$ $V\left(a\right)(B-cI)=(B-cI)W\left(a\right),$ so $B-cI$ is either zero or an isomorphism. Let $c$ to be an eigenvalue of $B$ (this requires that our field be algebraically closed), hence $B-cI$ is not invertible. Then $B-cI=0$ or $B=cI\text{.}$ $\square$

We have $W^{\lambda }\left(A\right)\subseteq M_{d_{\lambda }}\left(ℂ\right)\text{;}$ to show that $W^{\lambda }\left(A\right)=M_{d_{\lambda }}\left(ℂ\right),$ we will show that $W^{\lambda }\left(A\right)$ contains the basis of matrix units. Note that ${\sum }_{g_i\in G}{W}^{\lambda }\left(g^{*}\right)E_{i,m}{W}^{\lambda }\left(g\right)$ commutes with $W^{\lambda }\left(A\right)$ by the lemma of last time. By Schur’s lemma (with $V=W=W^{\lambda }\text{)},$ this element must be of the form $c{I}_{d_{\lambda }}$ for some $c\in ℂ$ (possibly zero). We calculate traces; using the trace property:

On the other hand, $\text{tr}\left(c{I}_{d_{\lambda }}\right)=c{d}_{\lambda },$ so $c=0$ for $i\ne m$ and $c={(1/d_{\lambda })}^2$ if $i=m\text{.}$

Continued in next lecture.