Past exam of the groups and representations course of the University of Oxford / 2023 / 3

Schur's lemma states that an intertwiner $T:V\to W$ between irreducible representations is either zero or an isomorphism. In particular, inequivalent irreducible representations have no nonzero intertwiners. For a finite-dimensional irreducible representation over the complex numbers, every commuting endomorphism is scalar:Indeed, the kernel and image of an intertwiner are invariant subspaces, proving the first statement. For the second, choose an eigenvalue $\lambda$ of $T$. The intertwiner $T-\lambda I$ has nonzero kernel, so it must vanish. The scalar conclusion uses an algebraically closed coefficient field: over the real numbers, planar rotation through $\pi /2$ gives an irreducible real representation of ${\mathbb{Z}}_4$ and commutes with itself without being a real scalar matrix.

Partition $M$ into the same five-by-five blocks as $R$. The equation for block $(i,j)$ is $M_{ij}{R}_j(g)=R_i(g)M_{ij}$, so each block is an intertwiner. By Schur's lemma, blocks between the two inequivalent irreducible representations vanish, whereas blocks between identical copies are arbitrary scalar multiples of the relevant identity matrix. Thus the commutant of a semisimple representation isEquivalently, $M=(A\otimes I_{n_1})\oplus (B\otimes I_{n_2})$, where $A\in M_3(\mathbb{C})$ and $B\in M_2(\mathbb{C})$. Substitution proves every such matrix commutes with every $R(g)$. There is no invertibility requirement on $M$, and its solution space has complex dimension $3^2+2^2=13$, even if $n_1=n_2$.

The identity belongs to the centralizer $C_G(H)$. If $x,y\in C_G(H)$, then for every $h\in H$,Also $xh=hx$ implies $h{x}^{-1}=x^{-1}h$ by multiplying on both sides by $x^{-1}$. Hence $xy$ and $x^{-1}$ belong to $C_G(H)$, proving by the subgroup test that $\boxed{C_G(H)\le G}$. This argument works even when $H$ is any subset, giving the centralizer of a subset.

Write $D(z)=\mathrm{diag}(z^{-2}{I}_3,z^3{I}_2)$ and partition a matrix $X$ into $3+2$ blocks. The equation $XD(z)=D(z)X$ makes its off-diagonal blocks obey $(z^3-z^{-2})X_{12}=0$ and $(z^{-2}-z^3)X_{21}=0$ for every $z\in U(1)$. Choosing any $z$ with $z^5\ne 1$ forces both blocks to vanish. Conversely, every block diagonal matrix commutes with every $D(z)$. Intersecting with the special unitary group gives the exact centralizerIts Lie algebra is $\mathfrak{su}(3)\oplus \mathfrak{su}(2)\oplus \mathfrak{u}(1)$. At the group level there is a finite central identification. The surjective group homomorphismhas kernel $\{(z^2{I}_3,z^{-3}{I}_2,z):z^6=1\}\cong {\mathbb{Z}}_6$. Surjectivity follows by choosing $z$ with $z^{-6}=\det A^{\prime }$ for any target pair $(A^{\prime },B^{\prime })$, and setting $A=z^2{A}^{\prime }$, $B=z^{-3}{B}^{\prime }$. ThusThe six distinct kernel elements give a concrete reason the displayed map does not identify the centralizer with an unquotiented product.

Number the simple roots of the $A_4$ root system consecutively along its Dynkin diagram. For a Young diagram with row lengths $r_i$, its highest weight has Dynkin labels $a_i=r_i-r_{i+1}$. The three irreducible representations have the following Young diagrams, which can be filled increasingly down the column to give Young tableaux:Here $i=1,\dots ,5$, $v^i\mapsto U^i{}_j{v}^j$, and $T^{ij}\mapsto U^i{}_k{U}^j{}_l{T}^{kl}$. Contraction with the invariant alternating tensor identifies ${\tilde{v}}_i=(1/4!){ϵ}_{i{i}_1{i}_2{i}_3{i}_4}{w}^{i_1{i}_2{i}_3{i}_4}$, giving the dual representation of $5$. The dimensions follow from $\left(\genfrac{}{}{0ex}{}{5}{1}\right)=5$, $\left(\genfrac{}{}{0ex}{}{5}{4}\right)=5$, $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$.

Use the centralizer and circle group normalization of solution. The first three components have U(1) charge $-2$ and the last two have U(1) charge $3$, so the branching of a representation isFor the dual representation, both U(1) charges reverse and the $SU(3)$ representation is conjugated; the $SU(2)$ doublet is equivalent to its conjugate. Taking the exterior square of the first decomposition gives three types of antisymmetric tensor components: $T^{ab}$, $T^{a\alpha }$, and $T^{\alpha \beta }$, where $a,b=1,2,3$ and $\alpha ,\beta =4,5$. Their U(1) charges add, and ${\Lambda }^{2}3=\overline{3}$, ${\Lambda }^{2}2=1$. HenceThe dimensions are $3+6+1=10$. These product labels describe genuine representations of the quotient group in solution: every element of its ${\mathbb{Z}}_6$ kernel acts trivially on each summand.