Past exam of the groups and representations course of the University of Oxford / 2020 / 1 / a / Solution

A group representation on a vector space $V$ over a field $K$ is a group homomorphism $R:G\to \mathrm{GL}(V)$. In particular, $R(gh)=R(g)R(h)$, $R(e)=I$ and $R(g^{-1})=R(g{)}^{-1}$. Choosing a basis identifies it with a matrix representation.

Two group representations $R$ on $V$ and $S$ on $W$ are equivalent representations when an invertible linear map $T:V\to W$ satisfies $TR(g)=S(g)T$ for every $g$. Thus $S(g)=TR(g)T^{-1}$; they differ only by a choice of basis.

A nonzero group representation is a reducible representation if there is a vector subspace $U$ with $0\ne U\ne V$ and $R(g)U\subseteq U$ for every $g$. It is an irreducible representation if it has no such invariant subspace. Reducibility alone does not assert the existence of an invariant complementary vector subspace; that stronger conclusion says that the representation is a semisimple representation.

The Schur lemma states that an intertwiner $T$ between irreducible representations is either zero or an isomorphism. Consequently an intertwiner between inequivalent irreducible representations is zero. For a finite-dimensional irreducible representation over the complex numbers, every commuting endomorphism is scalar:For the first assertion, the kernel and image of $T$ are invariant subspaces, so a nonzero $T$ is injective and surjective. For the scalar conclusion, choose an eigenvalue $\lambda$ of $T$. The intertwiner $T-\lambda I$ has a nonzero kernel, so it must vanish. The finite-dimensional complex hypothesis matters: the scalar conclusion need not hold over the real numbers, as a real planar rotation through $\pi /2$ is an irreducible representation of the cyclic group of order four, while the rotation itself is a commuting nonscalar endomorphism.