Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 19J (Representation Theory)

WriteThe homogeneous polynomial representation of SU2 is the symmetric-power action determined byand extended multiplicatively to $V_n$.

In the ordered basis $(x^3,x^{2}y,x{y}^2,y^3)$, the matrix for $g$ is

The character of a representation $(V,\rho )$ is ${\chi }_V(g)=\mathrm{tr}\rho (g)$. Every $g\in S{U}_2$ is conjugate to $\mathrm{diag}(z,z^{-1})$ with $|z|=1$, and hence the character of the homogeneous polynomial representation of SU2 isFor $z=e^{i\theta }$ this is $\sin ((n+1)\theta )/\sin \theta$, interpreted by continuity when $\sin \theta =0$.

The classification of finite-dimensional representations of SU2 says that every irreducible is one of the $V_n$, and compactness gives complete reducibility. The standard alternating formis $S{U}_2$-invariant and identifies $V_1$ with $V_1^{\ast }$. Taking symmetric powers gives $V_n\cong V_n^{\ast }$ for every $n$. Therefore, ifthenThis proves self-duality of finite-dimensional SU2 representations.

For an irreducible $V$, Schur lemma makes the central element $-I$ act by a scalar $\epsilon$. Since $(-I{)}^2=I$, $\epsilon =\pm 1$, and on $V\otimes V$ it acts by ${\epsilon }^2=1$. Thus $\{\pm I\}$ lies in the kernel and the action factors through $S{U}_2/\{\pm I\}$, as in central parity on SU2 tensor products.

This fails for reducible representations. Take $V=V_0\oplus V_1$. On the summand $V_0\otimes V_1\subset V\otimes V$, the element $-I$ acts as $-1$, so the tensor-square action does not factor through the quotient.

For $g$ conjugate to $\mathrm{diag}(z,z^{-1})$, multiplication of the weight sums givesIndeed, the coefficient of $z^{2n-2m}$ on the left counts pairs $(r,s)$ with $0\le r,s\le n$ and $r+s=m$; the nested weight strings on the right give the same multiplicity. Characters determine finite-dimensional representations of the compact group $S{U}_2$, soThis weight-counting argument proves the required case of the Clebsch-Gordan decomposition for SU2 rather than merely quoting it.

If the eigenvalues of $\rho (g)$ are ${\lambda }_1,\dots ,{\lambda }_d$, then those of $g$ on ${\bigwedge }^{2}V$ are ${\lambda }_i{\lambda }_j$ for $i<j$. Therefore the Character of an exterior square is

Let $v_k=x^{n-k}{y}^k$. In the summand $V_{2n-2j}$ from part (d), a highest-weight vector isThe raising operator $x{\partial }_y$ annihilates this sum because consecutive terms cancel. Interchanging the tensor factors sends $w_j$ to $(-1{)}^j{w}_j$. Since each summand in part (d) has multiplicity one, the flip acts by that scalar on the whole summand. The alternating square therefore consists exactly of the odd-$j$ summands. This flip parity in the SU2 tensor square gives