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

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.

Write $P=(P_{ij}{)}_{i,j=1}^3$, where $P_{ij}:{\mathbb{C}}^{m_j}\to {\mathbb{C}}^{m_i}$. Block multiplication makes the commuting condition equivalent toThus every block is an intertwiner. For $i\ne j$, the irreducible representations are inequivalent, so the Schur lemma gives $P_{ij}=0$. On each diagonal block, the same Schur lemma gives $P_{ii}={\lambda }_i{I}_{m_i}$. ThereforeConversely, every displayed matrix commutes with $R(g)$, proving both necessity and sufficiency. This commutant of a semisimple representation has complex dimension three. The parameters may vanish: $P$ was not required to be invertible.

Now $m_1=m_2=n$ and the first two matrix representations agree in the chosen bases. Each of $P_{11},P_{12},P_{21},P_{22}$ is therefore a commuting endomorphism of that common irreducible representation, so each is scalar by the Schur lemma. All blocks connecting either of these copies with the third vanish, since the third irreducible representation is inequivalent. HenceEvery such matrix commutes with $R(g)$. In particular the off-diagonal scalars $b,c$ can mix the two identical copies. The commutant of a semisimple representation is $M_2(\mathbb{C})\oplus \mathbb{C}$, of complex dimension five. If the first two representations were merely equivalent, rather than literally equal, the mixed blocks would instead be scalar multiples of a fixed invertible intertwiner and its inverse.

All three vector spaces now have the same dimension $n$. Each of the nine blocks of $P$ commutes with the common irreducible representation, so the Schur lemma givesConversely $R(g)=I_3\otimes R_1(g)$, so the Kronecker product identity immediately verifies commutation for every $A$. The commutant of a semisimple representation acts arbitrarily on the three-dimensional multiplicity vector space while acting as a scalar on each irreducible factor. Its complex dimension is nine.