Brian Bi
$\DeclareMathOperator{\End}{End} \DeclareMathOperator{\char}{char} \DeclareMathOperator{\tr}{tr} \DeclareMathOperator{\ker}{ker} \DeclareMathOperator{\im}{im} \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\span}{span} \DeclareMathOperator{\diag}{diag} \DeclareMathOperator{\Id}{Id} \DeclareMathOperator{\ad}{ad} \newcommand\d{\mathrm{d}} \newcommand\pref[1]{(\ref{#1})}$

Remark: In Theorem 4.5.1, it is asserted that the element $$P \in \mathbb{C}[G]$$ defined by $$P = \frac{1}{|G|} \sum_{g\in G} g$$ acts as the identity in the trivial representation $$G$$ and as the zero operator in any other irreducible representation of $$G$$. I had not noticed that $$P$$ is central until I saw it pointed out on Math StackExchange, although the proof does explicitly say that $$P \in Z(\mathbb{C}[G])$$. This fact is needed in order to understand the assertion. Indeed, since $$P$$ is central, it acts as a scalar in every irreducible representation, and since $$P^2 = P$$, this scalar is either 1 or 0. Travis points out that $$gP = P$$ for all $$g \in G$$, so in the case where $$P = \Id$$, we have $$\rho(g)x = \rho(g)\rho(P)x = \rho(P)x = x$$ for all $$x \in X$$, therefore $$\rho(g)$$ is also the identity for each $$g \in G$$, and $$X$$ is the trivial representation.

Additionally,I found the equalities at the end of the proof to be difficult to understand. Since $$P$$ is the projector onto the $$G$$-invariant subspace, and the trace of a projection operator is the dimension of the subspace that it projects onto, it is clear that $$\tr|_{V\otimes W^*}(P) = \dim(V \otimes W^*)^G \cong \dim \Hom(W, V)^G$$. Darij Grinberg points out that $$\Hom(W, V)^G = \Hom_G(W, V)$$, so this completes the proof. I'm not sure why $$\dim \Hom_G(\mathbb{C}, V \otimes W^*)$$ is mentioned, but oh well, there are other theorems to think about...

Problem 4.5.2

1. We follow the hint given in the text. Let $$h \in G$$. Then \begin{gather*} h\psi_i = \frac{\dim V_i}{|G|} \sum_{g \in G} \chi_{V_i}(g) h g^{-1} = \frac{\dim V_i}{|G|} \sum_{k \in G} \chi_{V_i}(k^{-1}h) k \\ \psi_i h = \frac{\dim V_i}{|G|} \sum_{g \in G} \chi_{V_i}(g) g^{-1} h = \frac{\dim V_i}{|G|} \sum_{k \in G} \chi_{V_i}(hk^{-1}) k \end{gather*} But for all $$k$$, $$\chi_{V_i}(k^{-1}h) = \chi_{V_i}(hk^{-1})$$ since the elements $$k^{-1}h$$ and $$hk^{-1}$$ are conjugates. Therefore $$h\psi_i = \psi_i h$$, and $$\psi_i$$ is central in $$\mathbb{C}[G]$$. By Schur's lemma, $$\psi_i$$ acts as a scalar operator $$\lambda_{ij} \Id$$ on each $$V_j$$.

Taking the trace in $$V_j$$ we obtain \begin{align*} \lambda_{ij} \dim V_j &= \chi_{V_j}(\psi_i) \\ &= \frac{\dim V_i}{|G|} \sum_{g \in G} \chi_{V_i}(g) \chi_{V_j}(g^{-1}) \\ &= \frac{\dim V_i}{|G|} \sum_{g \in G} \chi_{V_i}(g) \overline{\chi_{V_j}(g)} \\ &= (\dim V_i) \langle \chi_{V_i}, \chi_{V_j} \rangle \\ &= \delta_{ij} \dim V_i \end{align*} so $$\lambda_{ij} = \delta_{ij} \frac{\dim V_i}{\dim V_j}$$, which is zero when $$i \ne j$$ and unity when $$i = j$$.

2. Suppose $$i \ne j$$. Let $$V_k$$ be an irreducible representation of $$G$$. At least one of $$i, j$$ is unequal to $$k$$, so by part (a), at least one of $$\psi_i, \psi_j$$ acts as the zero operator in $$V_k$$. Therefore $$\psi_i \psi_j$$ acts as the zero operator in $$V_k$$. Therefore $$\psi_i \psi_j \in \mathrm{Rad}(\mathbb{C}[G])$$. By Maschke's theorem, $$\psi_i \psi_j = 0$$.

On the other hand, if $$i = j$$, then $$\psi_i \psi_j = \psi_i^2$$ and by part (a), in every irreducible representation of $$G$$, $$\psi_i$$ acts as the scalar 1 or 0 depending on whether $$i = k$$ or $$i \ne k$$. Thus, in all irreducible representations, $$\psi_i^2 - \psi_i$$ acts as zero. Again, by Maschke's theorem, $$\psi_i^2 - \psi_i = 0$$.

Remark: In the proof of Theorem 4.5.4, it appears that the object $$g \otimes (h^*)^{-1}$$ is an abbreviated way of writing $$\rho(g) \otimes (\rho(h)^T)^{-1}$$ where $$\rho$$ is the action for the representation $$\oplus_V V$$. This gave me a lot of headaches trying to figure it out, and perhaps there is some deeper way in which Etingof intended for $$h^*$$ makes sense by itself, but in any case, with this interpretation, the first equality follows from the identity $$(\tr A)(\tr B) = \tr(A \otimes B)$$, and the second equality follows from (I think) the canonical isomorphism between $$V \otimes V^*$$ and $$\End V$$, regarded as $$(\End(V) \otimes \End(V))$$-modules, or alternatively $$(\End V, \End V)$$-bimodules.