Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 2 / 19J (Representation Theory)

Start with any positive-definite Hermitian inner product $⟨-,-⟩$ on ${\mathbb{C}}^n$ and average it over the finite group:This remains positive definite. For $h\in G$, reindexing $g\mapsto gh$ givesso the form is $G$-invariant.

Choose an orthonormal basis for $⟨-,-{⟩}_G$. In this basis every $\rho (g)$ preserves the standard Hermitian form and is therefore unitary. If $P$ is the change-of-basis matrix, thendefines an isomorphic representation with ${\rho }^{\prime }(G)\le U_n$.

An invariant bilinear form $B$ defines a linear mapInvariance givesso ${\Phi }_B$ is an $G$-homomorphism. Its kernel is a $G$-subrepresentation of irreducible $V$. Hence either ${\Phi }_B=0$, so $B=0$, or ${\Phi }_B$ is injective and therefore a linear isomorphism $V\to V^{\ast }$. In the latter case $B$ is a nondegenerate.

If $B$ and $C$ are two nonzero invariant forms, thenBy Schur lemma, this endomorphism is scalar, so $B=\lambda C$. Cases involving the zero form are immediate. Thus any two invariant bilinear forms are proportional.

Given a nonzero invariant form $B$, its transposeis again nonzero and $G$-invariant. By part (i), $B^T=\lambda B$ for some nonzero scalar $\lambda$. Transposing once more givesSince $B\ne 0$, ${\lambda }^2=1$, and hence $\lambda \in \{1,-1\}$. ThereforeThus every nonzero invariant bilinear form on an irreducible complex representation is either symmetric or alternating.

The induced representation iswith $G$ acting by left multiplication on the first tensor factor. If $R$ is a set of representatives for the left cosets $G/H$, then as a vector space it is ${⨁}_{r\in R}r\otimes V$.

The operator $g$ permutes these summands. A summand indexed by $xH$ contributes to the trace exactly when $gxH=xH$, equivalently when $x^{-1}gx\in H$, and its contribution is ${\chi }_V(x^{-1}gx)$. Accounting for the $|H|$ representatives of each coset gives

For a $G$-representation $W$, restrict $W$ to $H$ in the left-hand induced representation. DefinebyIt is well-defined because both representatives of the balancing relation giveIt is $G$-equivariant, and its inverse isConsequently