Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 4 / 15F (Representation Theory)

A one-dimensional continuous group representation is a continuous group homomorphism $\rho :S^1\to {\mathbb{C}}^{\times }$. Its image is compact. If some $|\rho (z)|$ differed from one, its positive or negative powers would be unbounded, so the image lies in $S^1$. Lift $t\mapsto \rho (e^{it})$ to a continuous real argument $a(t)$ with $a(0)=0$. Then $a(s+t)-a(s)-a(t)$ is a continuous integer multiple of $2\pi$ and vanishes at $(0,0)$, hence is always zero. The continuous Cauchy functional equation gives $a(t)=ct$. Periodicity forces $e^{2\pi ic}=1$, so $c=n\in \mathbb{Z}$. Consequently $\boxed{\rho (z)=z^n}$; conversely every such map is a continuous one-dimensional group representation.

The character of a representation is ${\chi }_V(g)=\mathrm{tr}{\rho }_V(g)$. Average any positive inner product over the compact group using Haar measure to obtain a unitary representation. On the maximal torus $\mathrm{diag}(z,z^{-1})$, the commuting unitary operators admit a simultaneous eigenbasis. Each eigenline carries a circle character of a representation $z^n$ by part (a). Hencewith finitely many nonzero coefficients. Every element of special unitary group is conjugate to an element of this torus, so this describes the full character of a representation via its torus restriction. Here the coefficient notation includes zero multiplicities.

The matrix $w=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\in \mathrm{SU}(2)$ conjugates $\mathrm{diag}(z,z^{-1})$ to $\mathrm{diag}(z^{-1},z)$. A character of a representation is invariant under conjugation, since the trace is invariant under similarity. Therefore $\boxed{{\chi }_V(z)={\chi }_V(z^{-1})}$, and opposite torus weights have equal multiplicities.

Write $V_r={\mathrm{Sym}}^r{\mathbb{C}}^2$ for the irreducible SU(2) representation of dimension $r+1$, with character of a representation $z^r+z^{r-2}+\cdots +z^{-r}$. Here $V=V_3$, with weights $3,1,-1,-3$. Squaring its character of a representation and successively removing the highest-weight irreducible characters givesThe dimensions $7+5+3+1=16$ check the decomposition; complete reducibility follows from the invariant inner product.

For the exterior square, add distinct pairs of weights. Its character of a representation is $z^4+z^2+2+z^{-2}+z^{-4}={\chi }_{V_4}+{\chi }_{V_0}$. Equivalently it is $({\chi }_V(z{)}^2-{\chi }_V(z^2))/2$. Hence $\boxed{{\Lambda }^{2}V\cong V_4\oplus V_0}$, of dimensions five and one. The complementary symmetric square is $V_6\oplus V_2$.