Past exam of the groups and representations course of the University of Oxford / 2022 / 3

The identity commutes with every $h\in H$, so the centralizer $C_G(H)$ is nonempty. If $x,y\in C_G(H)$, then for every $h\in H$,Thus it is closed under multiplication. From $xh=hx$, multiplication by $x^{-1}$ on both sides gives $h{x}^{-1}=x^{-1}h$, so $x^{-1}\in C_G(H)$. These are exactly the subgroup conditions; henceThe proof works for the centralizer of a subset as well; it does not require that $H$ itself be a subgroup.

Write $X=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$. Commuting with all $\mathrm{diag}(U_1,U_2)$ requiresFor $n_1,n_2\ge 2$, the defining complex representations of each factor are irreducible. Schur lemma gives $A=z{I}_{n_1}$ and $D=w{I}_{n_2}$. The two block vector spaces are inequivalent representations of the product group, even when $n_1=n_2$: the first factor acts nontrivially on just the first block. Thus their off-diagonal intertwiners vanish, $B=C=0$. Unitarity and the unit determinant give

The assertion that this full group is always a circle needs correction. Put $d=\gcd (n_1,n_2)$, $a=n_1/d$, $b=n_2/d$. The continuous group homomorphismmaps this centralizer onto the $d$th roots of unity. Its kernel is the identity component of a Lie group, parametrized injectively by $e^{i\theta }\mapsto (e^{ib\theta },e^{-ia\theta })$. Coprimality of $a,b$ proves injectivity; a Bezout identity proves that this parametrization exhausts the kernel. Extending the primitive integer row $(a,b)$ to a determinant-one integer matrix gives coordinates in which the constraint is simply $t^d=1$. ConsequentlyFor a concrete counterexample, take $n_1=n_2=2$. The constraint is $(zw{)}^2=1$, giving two separate circles $zw=1$ and $zw=-1$. The element $\mathrm{diag}(I_2,-I_2)$ lies in the second component, so this centralizer cannot be isomorphic as a Lie group to the connected circle group.

For the intended specialization of this centralizer of two special unitary blocks, $d=\gcd (2,3)=1$ and therefore the full centralizer really is a circle group:We use this primitive integer U(1) charge normalization below. If exactly one $n_i$ equals $1$ and the other is at least $2$, the same block-scalar formula still holds. If $n_1=n_2=1$, both factors are trivial and the centralizer is all of SU(2), another exception to the printed claim.

Write the defining space as $V=A\oplus B$, where $A\cong {\mathbb{C}}^2$ and $B\cong {\mathbb{C}}^3$. Use ordered pairs for the SU(2) and SU(3) factors, in that order. The generator $X_c$ in solution gives U(1) charges $3$ on $A$ and $-2$ on $B$, soTaking the conjugate representation reverses the U(1) charges. The SU(2) doublet is equivalent to its dual representation, while the SU(3) triplet becomes the antitriplet:

For the two-index antisymmetric representation, split the antisymmetric tensor $T^{IJ}=-T^{JI}$ into $T^{ab}$, $T^{a\mu }$ and $T^{\mu \nu }$, with $a,b=1,2$ and $\mu ,\nu =3,4,5$. The exterior square of a direct sum givesThe first term is an SU(2) singlet, the middle is $(2,3)$, and ${\Lambda }^{2}3\cong \overline{3}$ by contraction with the invariant alternating tensor. Adding the U(1) charges of the two indices givesThe dimensions $1+6+3=10$ check the branching of a representation. In the convention $Q_{\mathrm{em}}=T_3+Y$, weak hypercharge is $Y=X_c/6$.

The complexification of a Lie algebra in the Adjoint representation of SU(5) consists of traceless endomorphisms of $V=A\oplus B$. Splitting an endomorphism into its four blocks gives traceless endomorphisms of $A$ and $B$, one relative block-scalar direction, and the two off-diagonal spaces $\mathrm{Hom}(B,A)$ and $\mathrm{Hom}(A,B)$.

The first three pieces transform as $(3,1{)}_0$, $(1,8{)}_0$ and $(1,1{)}_0$. The off-diagonal pieces have U(1) charges equal to output minus input charge, namely $3-(-2)=5$ and $-2-3=-5$. ThusThe dimensions are $3+8+1+6+6=24$. The first three summands are precisely the gauge fields of the local gauge group $H\times U_c(1)$. The extra gauge fields of the SU(5) gauge theory transform asHermiticity relates the two off-diagonal blocks, so these describe twelve real gauge fields, or six complex fields and their conjugates. Globally, the subgroup generated by $H$ and $U_c(1)$ inside SU(5) is $(SU(2)\times SU(3)\times U(1))/{\mathbb{Z}}_6$, since the parametrization $(U_2,U_3,e^{i\theta })\mapsto \mathrm{diag}(e^{3i\theta }{U}_2,e^{-2i\theta }{U}_3)$ has a six-element kernel. This finite quotient does not alter the stated branching of a representation.

The symmetric square of a direct sum givesNow ${\mathrm{Sym}}^{2}2=3$ and ${\mathrm{Sym}}^{2}3=6$, rather than the singlet and antitriplet arising from exterior squares. ThusThe dimensions are $3+6+6=15$.

By contrast, solution gives the following left-handed Weyl fermion content of $\overline{5}\oplus 10$:These are one Standard Model family, with weak hypercharge $Y=q_c/6$. The $15$ contains $Q$ but lacks $L$, $d^c$, $u^c$ and $e^c$; instead it contains an SU(2) triplet and an SU(3) sextet. Equality of total dimensions therefore does not give the same representation content, and $15$ cannot supply the family.