Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 4 / 19I (Representation Theory)

The special unitary group is $\mathrm{SU}(2)=\{g\in M_2(\mathbb{C}):g^{\ast }g=I,\det g=1\}$. The special orthogonal group is $\mathrm{SO}(3)=\{R\in M_3(\mathbb{R}):R^{T}R=I,\det R=1\}$. Conjugation $A\mapsto gA{g}^{-1}$ preserves trace and skew-Hermiticity: for unitary $g$, $(gA{g}^{-1}{)}^{\ast }=g{A}^{\ast }{g}^{-1}=-gA{g}^{-1}$. It therefore defines the representation on the real space $V$ described in the following parts. The bar over the transpose in the PDF is essential; the condition is $A+A^{\ast }=0$, not ordinary antisymmetry over the complex numbers.

An orthonormal basis for the pairing proved below isPut $J_j=\sqrt{2}{E}_j$, so $J_j^2=-I$. For any unit real vector $v$, $J(v)={\sum }_j{v}_j{J}_j$ also squares to $-I$. Thus $g=\cos (\theta /2)I+\sin (\theta /2)J(v)$ belongs to $\mathrm{SU}(2)$. Conjugation fixes $J(v)$ and rotates its perpendicular plane by angle $\theta$ up to the orientation convention of this basis, as follows either by multiplying the displayed matrices or from their anticommutator relation $J(v)J(w)+J(w)J(v)=-2(v\cdot w)I$. Since every three-dimensional proper rotation has an axis and an angle, this proves surjectivity onto $\mathrm{SO}(3)$.

If an element is in the kernel, it commutes with each $E_j$. Commutation with $E_3$ forces it to be diagonal; commutation with $E_1$ forces its two diagonal entries to agree. Thus it is a scalar matrix, and determinant one and unitarity give exactly $\{I,-I\}$. Therefore $\boxed{\mathrm{SO}(3)\cong \mathrm{SU}(2)/\{\pm I\}}$.

The finite-dimensional complex irreducible representations of $\mathrm{SU}(2)$ are ${\mathrm{Sym}}^m({\mathbb{C}}^2)$, $m\ge 0$, of dimension $m+1$. The central element $-I$ acts by $(-1{)}^m$, so precisely the even ones descend. The irreducibles of $\mathrm{SO}(3)$ are $\boxed{{\mathrm{Sym}}^{2\ell }({\mathbb{C}}^2),\ell =0,1,2,\dots }$, viewed as representations of the quotient, with dimensions $2\ell +1$.

A traceless skew-Hermitian matrix has the unique formThis real-linear parametrization is bijective, proving $\boxed{V\cong {\mathbb{R}}^3}$ as real vector spaces. It is not a complex vector space: multiplication by $i$ generally turns a skew-Hermitian matrix into a Hermitian one.

The pairing is real and symmetric: $\overline{\mathrm{tr}(AB)}=\mathrm{tr}(B^{\ast }{A}^{\ast })=\mathrm{tr}(BA)=\mathrm{tr}(AB)$, and cyclicity of the trace exchanges $A,B$. For the coordinates in part (i), $A^2=-(x^2+y^2+z^2)I$, soThus it is a positive-definite real bilinear form, hence nondegenerate. Finallyproving invariance under conjugation.

The invariant positive-definite bilinear form makes every $\rho (g)$ orthogonal in an orthonormal real basis. Every element of $\mathrm{SU}(2)$ has the form $\left(\begin{array}{cc}a& b\\ -\overline{b}& \overline{a}\end{array}\right)$ with $|a{|}^2+|b{|}^2=1$, so this group is homeomorphic to $S^3$ and connected. The continuous function $\det \rho (g)$ takes values in $\{-1,1\}$ and equals one at the identity. It is therefore identically one. Hence $\boxed{\rho (\mathrm{SU}(2))\subseteq \mathrm{SO}(3)}$.