Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2018 / Paper 2 / 10

A vector subspace $W\subseteq {\mathbb{C}}^n$ is an invariant subspace for the group representation $D$ ifBecause $g^{-1}$ is also in the group and $D(g)$ is invertible, the inclusion is equivalently $D(g)W=W$ for every $g$.

For $n>0$, the representation is an irreducible representation when its only invariant vector subspaces areThus no nonzero proper subspace is preserved by all representing matrices.

Every $D_2(g)$ is invertible because both inversion and the conjugate transpose preserve invertibility. For $g,h\in G$, the group homomorphism property of $D_1$ and reversal of multiplication under the adjoint giveAlso $D_2(e)=I$. Hence $D_2$ is a group representation. It is the inverse-adjoint representation of $D_1$; the inverse and adjoint each reverse product order, so together they preserve it.

Use the standard Hermitian inner product $⟨w,v⟩=w^{†}v$. Since $W$ is invariant under every $D_2(g)$, it is in particular invariant underLet $v\in W^{\perp }$ and $w\in W$. Thenbecause $D_1(g{)}^{†}w\in W$ and $v$ is orthogonal to $W$. Thus $D_1(g)v$ is orthogonal to every $w\in W$, provingTherefore $W^{\perp }$ is an invariant subspace for $D_1$. If $D_1$ is irreducible, $W^{\perp }$ is either $\{0\}$ or ${\mathbb{C}}^n$. In finite-dimensional inner product space, $(W^{\perp }{)}^{\perp }=W$, so these alternatives force $W={\mathbb{C}}^n$ or $W=\{0\}$. HenceNo unitarity of $D_1$ was assumed; the inverse-adjoint representation is exactly what makes this orthogonal complement argument work.