Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 1 / 19G / Solution

For $\ell \in V^{\ast }$ define $(g\ell )(v)=\ell (g^{-1}v)$. The representing matrix is the transpose of the inverse matrix on $V$, so the dual character isThe last equality follows by averaging a Hermitian inner product to make the finite-group group representation unitary. Complex finite-group representations are isomorphic exactly when their characters agree. Hence $V$ is self-dual exactly when its character is real-valued.

The invariant tensor identification $(V\otimes V{)}^G\simeq {\mathrm{Hom}}_G(V^{\ast },V)$ and Schur lemma show that, for irreducible $V$, the trivial multiplicity is zero unless $V^{\ast }\simeq V$, and one in that case. Dually, a self-duality isomorphism gives a nonzero invariant bilinear form $B$ on $V$. Its radical is an invariant subspace, so irreducibility makes $B$ nondegenerate. The space of invariant bilinear forms is one dimensional. Since the transpose $B^T$ is another such form, $B^T=cB$; transposing again gives $c^2=1$. Therefore $B$ is symmetric or alternating: in the negative-transpose case, $B(v,v)=-B(v,v)$ forces $B(v,v)=0$. Both types cannot coexist, since proportional nonzero forms cannot have both transpose signs.

In odd dimension an alternating matrix has determinant zero: $\det B=\det (-B^T)=(-1{)}^n\det B=-\det B$. Thus only the symmetric case is possible. Over $\mathbb{C}$ a nondegenerate symmetric bilinear form has a basis with matrix $I$: diagonalize it by congruence and rescale each nonzero diagonal entry. In that basis invariance says $\rho (g{)}^T\rho (g)=I$. Consequently