Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 19I (Representation Theory)

A compact topological group is a group $G$ equipped with a compact Hausdorff topology for which multiplication $G\times G\to G$ and inversion $G\to G$ are continuous.

The special unitary group isIt is closed in the finite-dimensional real vector space $M_n(\mathbb{C})$ because the equations $U^{\ast }U=I$ and $\det U=1$ are closed conditions. It is bounded because every column of a unitary matrix has norm one. The Heine-Borel theorem therefore makes it compact. Matrix multiplication is continuous, and inversion is continuous because $U^{-1}=U^{\ast }$. Hence $SU(n)$ is a compact topological group.

A continuous homomorphism $G\to O(n)$ is immediately an $n$-dimensional real group representation by composing with the inclusion $O(n)\subset G{L}_n(\mathbb{R})$.

Conversely, let $\pi :G\to GL(V)$ be a continuous real representation with ${\dim }_{\mathbb{R}}V=n$. Choose any positive-definite inner product $(,{)}_0$ on $V$ and normalized Haar measure $dg$ on $G$. DefineThis is bilinear and symmetric. It is positive definite because the continuous nonnegative integrand $(\pi (g)v,\pi (g)v{)}_0$ is positive at every $g$ whenever $v\ne 0$. Right invariance of normalized Haar measure on a compact group givesChoose an orthonormal basis for this averaged form. In that basis every $\pi (h)$ lies in $O(n)$, proving the orthogonalization of a compact-group representation and the stated equivalence.

LetFor $U\in SU(2)$ defineConjugation preserves trace, and unitarity givesso $V$ is invariant. This gives a continuous real-linear representation $\rho :SU(2)\to GL(V)$.

The condition $A+A^{\ast }=0$ forces the diagonal entry $a$ to be purely imaginary and $c=-\overline{b}$. Thus every $A\in V$ has a unique expressionThe map $A\mapsto (x,y,z)$ is a real-linear isomorphism, so

DefineIt is symmetric because $\mathrm{tr}(AC)=\mathrm{tr}(CA)$. Writing $A=iH$ and $C=iK$ with $H,K$ Hermitian shows that $B(A,C)=\mathrm{tr}(HK)$ is real and bilinear over $\mathbb{R}$. Moreover,for $A\ne 0$, so the form is positive definite. Finally, cyclicity of the trace givesHence $B$ is an invariant positive-definite symmetric bilinear form, and after identifying $V$ with Euclidean ${\mathbb{R}}^3$ we have $\rho (SU(2))\subseteq O(3)$.

Because the bilinear form from part (ii) is invariant, $\rho (SU(2))\subseteq O(3)$. The mapis continuous. Since $SU(2)$ is connected and $\det \rho (I_2)=1$, it is identically one. Hence

To find the kernel, suppose $U$ commutes with every traceless skew-Hermitian matrix. Commutation withforces $U$ to be diagonal. Commutation withthen forces its two diagonal entries to agree. Thus $U=z{I}_2$. Since $U\in SU(2)$,and therefore

It remains to prove surjectivity. The derived representation at the identity isIf this vanishes for every $A$ in the SU(2) Lie algebra, the same commutant calculation makes $X$ scalar; tracelessness then gives $X=0$. Thus $d{\rho }_{I_2}$ is injective. Its domain and codomain are both three-dimensional real Lie algebras, so it is an isomorphismThe inverse function theorem now shows that the image of $\rho$ contains a neighbourhood of the identity in $SO(3)$. Therefore the image is an open subgroup. The special orthogonal group $SO(3)$ is connected, since every rotation can be deformed continuously to the identity by reducing its rotation angle. A connected topological group has no proper open subgroup, soTogether with the kernel calculation, this is the Adjoint double cover from SU(2) to SO(3).