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.