Past exam of the groups and representations course of the University of Oxford / 2019 / 3

For $z\in U(1)$ and $U\in SU(n)$, the matrix $zU$ is unitary, since $(zU{)}^{†}(zU)=|z{|}^2{U}^{†}U=I_n$. Thus the map has its image in the unitary group $U(n)$. Scalar matrices commute with all matrices, so the direct product of groups multiplication givesAlso $f(1,I_n)=I_n$. HenceThe special unitary group condition $\det U=1$ imposes no requirement that $zU$ itself have unit determinant; its determinant is $z^n$, as expected for a general element of the unitary group.

First the group homomorphism is surjective. For any $W\in U(n)$, choose $z\in U(1)$ with $z^n=\det W$ and set $U=z^{-1}W$. Then $U$ is unitary and $\det U=z^{-n}\det W=1$, so $W=f(z,U)$.

Next $f(z,U)=I_n$ is equivalent to $U=z^{-1}{I}_n$, and the special unitary group condition becomes $z^n=1$. With $\zeta =e^{2\pi i/n}$, the kernel of a group homomorphism is thereforeIt is the cyclic central subgroup generated by $(\zeta ,{\zeta }^{-1}{I}_n)$; the scalar and special unitary group factors are paired with inverse phases. The first isomorphism theorem now givesExplicitly the isomorphism sends the coset $[(z,U)]$ to $zU$. The $n$ possible choices of the root $z$ differ precisely by multiplication by elements of $K$, which explains why the product before quotienting is not a unique factorization.

Use upper indices for the defining representation of SU(6) and lower indices for its dual representation. The Young tableaux below display tensor-slot labels; their shapes determine the Young diagrams. Full columns of height six are determinant representations and can be removed for SU(6).Here all indices run from one to six. The highest weight Dynkin labels follow from the differences of successive row lengths. In the same order the row lengths are $(1)$, $(1,1,1,1,1)$, $(2)$, $(1,1)$ and $(2,1,1,1,1)$.

The defining vector transforms as $v^a\mapsto U^a{}_b{v}^b$. For a unitary representation, the conjugate representation is equivalent to the dual representation, so $\overline{6}$ can be written as a covector with $w_a\mapsto w_b(U^{-1}{)}^b{}_a$. Its five-box column is equivalently a totally antisymmetric tensor $A^{a_1\cdots a_5}$, related by the invariant alternating tensor $w_a={ϵ}_{a{a}_1\cdots a_5}{A}^{a_1\cdots a_5}/5!$.

The symmetric tensor transforms as $S^{ab}\mapsto U^a{}_c{U}^b{}_d{S}^{cd}$, with $6(6+1)/2=21$ independent components. The antisymmetric tensor has the same transformation law but obeys $A^{ab}=-A^{ba}$, giving $6(6-1)/2=15$ independent components. For the Adjoint representation, $T\mapsto UT{U}^{-1}$; the trace constraint removes one component from the $36$ entries of a mixed tensor, leaving $35$. Its six-box Young diagram is obtained by converting the five-index column to a lower index and selecting the traceless mixed tensor.

Thus the requested dimensions, in the order above, are

In this part the earlier variable $n$ specializes to six. A group representation of $U(1)\times SU(6)$ factors through the quotient group $U(6)$ exactly when the central subgroup $K$ in solution acts trivially. For an irreducible SU(6) tensor representation represented by a Young diagram with $b$ boxes, the scalar matrix ${\zeta }^{-1}{I}_6$ acts as ${\zeta }^{-b}$, where $\zeta =e^{2\pi i/6}$.

The prescribed representation $\tilde{R}(z,U)=R(U)$ gives no compensating phase from the first factor. Its value on the generator of $K$ is therefore $R({\zeta }^{-1}{I}_6)={\zeta }^{-b}I$. The box counts in solution giveConsequentlyConcretely, conjugation $T\mapsto WT{W}^{-1}$ defines it for every $W\in U(6)$, and the scalar phase cancels.

The qualification “prescribed lifts” matters. The defining representation of $U(6)$ certainly exists, but its pullback is $(z,U)\mapsto zU$, not $(z,U)\mapsto U$. For instance $(\zeta ,{\zeta }^{-1}{I}_6)$ and $(1,I_6)$ represent the same quotient element, while the latter prescription assigns them different matrices in the defining representation. More generally, for an integer $q$, a lift $z^{q}R(U)$ descends precisely whenThis descent criterion for representations of U(n) explains how all the listed SU(6) representations can be extended after choosing a suitable first-factor U(1) charge, even though only one descends with that charge fixed to zero.

The first coordinate is fixed by the embedded SU(5), while the last five transform in its defining representation. Therefore $6{\downarrow }_{SU(5)}=1\oplus 5$, and taking the dual representation gives $\overline{6}{\downarrow }_{SU(5)}=1\oplus \overline{5}$.

For the symmetric square, the fixed-fixed component is one trivial representation, the fixed-five mixed components give $5$, and the symmetric five-five block is $15$. For the exterior square, the fixed-fixed component vanishes, the mixed components give $5$, and the antisymmetric five-five block is $10$. Finally restrict $35=6\otimes \overline{6}-1$, using $5\otimes \overline{5}=24\oplus 1$.

The five SU6 tensor branchings to SU5 are thusThe symbols $15$ on the left and right refer to different groups: the left-hand one is the SU(6) exterior square, whereas the right-hand one is the SU(5) symmetric square. The dimension checks are $6=5+1$, $21=15+5+1$, $15=10+5$ and $35=24+5+5+1$.

Explicitly, in the Adjoint representation the lower-right traceless five-by-five block gives $24$; the lower-left column and upper-right row give $5$ and $\overline{5}$; and the diagonal direction $\mathrm{diag}(-5,1,1,1,1,1)$ gives the remaining trivial representation.