*Remark:* In the proof of Proposition 4.7.1 it was not obvious to me
at first what was going on—what does \(\langle xw_i^*, w_j^*\rangle\)
mean? After some thought, I realize the induced inner product had to be
\(\langle w_i^*, w_j^*\rangle = \overline{\langle w_i, w_j\rangle}\); the
complex conjugation is needed in order to ensure conjugate linearity in the
*second* argument. Also, as pointed out in the previous section, since
we are taking the dual here with respect to the invariant Hermitian form, it
provides the isomorphism between the dual representation and the complex
conjugate representation; thus \(xw_i^* = (xw_i)^*\). Combining these facts, we
seee that \(\langle xw_i^*, w_j^*\rangle = \langle (xw_i)^*, w_j^* \rangle =
\overline{\langle xw_i, x_j\rangle}\), as required.

On \(V \otimes W^*\), the inner product is the obvious one induced by the inner products on \(V\) and \(W^*\).

The proof asserts that if \(V \ne W\), then the trivial representation doesn't occur in \(V \otimes W^*\). This follows from the fact that \(\dim \Hom_G(\mathbb{C}, V \otimes W^*) = \dim \Hom_G(W, V)\), which was mentioned in the proof of Theorem 4.5.1; the right-hand side vanishes by Schur's lemma. On the other hand, if \(V = W\), then Schur's lemma for algebraically closed fields implies that \(\dim \Hom_G(W, V) = 1\), so there is exactly one copy of the trivial representation inside \(V \otimes V^*\); this is how we know that \(P\) projects onto the first summand.

The formula for the projection of \(v_i \otimes v_{i'}^*\) to the subspace of \(V \otimes V^*\) isomorphic to \(\mathbb{C}\) is just the usual decomposition of endomorphisms into a scalar component (corresponding to \(\mathbb{C}\)) and a traceless component (corresponding to \(L\)).