Past exam of the groups and representations course of the University of Oxford / 2020 / 3

The componentwise product is closed in $G_1\times G_2$. Associativity follows componentwise from associativity in each group:The identity is $(e_1,e_2)$, and $(g_1,g_2{)}^{-1}=(g_1^{-1},g_2^{-1})$. These verify all the group axioms, so this is the direct product of groups.

Write the paper's Kronecker product $\times$ as $\otimes$. Its mixed-product identity and the representation laws of $R_1,R_2$ giveAlso $R(e_1,e_2)=I_{n_1}\otimes I_{n_2}=I_{n_1{n}_2}$, andThus the images are invertible and this is a group representation of dimension $n_1{n}_2$, called the external tensor product of group representations. No irreducibility assumption is needed for this construction.

Use upper indices for the defining representation of SU(3) and lower indices for its dual representation. A Young diagram with row lengths $({\ell }_1,{\ell }_2)$ has highest weight Dynkin labels $(p,q)=({\ell }_1-{\ell }_2,{\ell }_2)$; a full column of height three is a determinant representation and can be removed for SU(3). The Weyl dimension formula reduces toThe six irreducible representations of smallest dimensions are $1,3,\overline{3},6,\overline{6},8$. The following box arrays draw their Young diagrams; filling boxes by tensor-slot labels gives representative Young tableaux:Their associated tensors, in the same order, can be taken as follows:

These low-dimensional SU3 tensor representations have dimensions $1,3,3,6,6,8$ by direct component counting. To check that none were missed, if $p+q\ge 3$ then $(p+1)(q+1)\ge p+q+1$, so the displayed dimension is at least $(p+q+1)(p+q+2)/2\ge 10$. The pairs with $p+q\le 2$ are precisely the six listed pairs.

Under the stated SU(2) embedding, the first two components of the defining SU(3) representation form a doublet, while the third is fixed. Hence $3{\downarrow }_{SU(2)}=2\oplus 1$. Its dual representation restricts in the same way because the SU(2) doublet is self-dual.

For the Adjoint representation, restrict $8=3\otimes \overline{3}-1$. The SU(2) tensor product $2\otimes 2=3\oplus 1$ givesHere subtraction denotes removal of the trivial representation generated by the identity matrix; it is not a negative-dimensional summand. The first SU(3) factor is unchanged, so the requested branching of a representation rules areThe dimensions on the two sides are respectively $9=6+3$, $3=3$, $3=2+1$ and $8=3+2+2+1$.

An element commuting with the full first SU(3) factor has its first component in that factor's center of a group, which is discrete. In the second factor, commutation with every $\mathrm{diag}(U,1)$ forces the doublet block to be scalar and the mixed blocks to vanish, by the Schur lemma. Unitarity and unit determinant then give the connected commuting subgroupThis is the centralizer of the standard SU2 subgroup in SU3 in the second factor; the full centralizer in $G$ also includes the first factor's discrete center of a group ${\mathbb{Z}}_3$. The Lie algebra generator is $iX$ in the anti-Hermitian convention. The opposite sign or a common nonzero rescaling changes the charge convention, not the commuting direction.

In the defining representation, the U(1) charges are $1,1,-2$. In the dual representation they have the opposite signs. In the Adjoint representation a component $T^i{}_j$ has U(1) charge $X_i-X_j$. Thus its upper-right doublet has charge $3$, its lower-left doublet has charge $-3$, and its diagonal blocks have charge zero. Adding the charges to solution givesEach subscript denotes the eigenvalue of $X$, so a state of charge $q$ acquires the phase $e^{iq\theta }$. The connected subgroup generated by this circle and the second SU(2) is $(SU(2)\times U(1{)}_X)/{\mathbb{Z}}_2$: $\theta =\pi$ gives $\mathrm{diag}(-I_2,1)$, also the central element of the embedded SU(2). This finite intersection does not change the displayed Lie algebra branching of a representation rules.

The commuting direction in solution provides the correct candidate Lie algebra for a gauge group, $\mathfrak{su}(3{)}_c\oplus \mathfrak{su}(2{)}_L\oplus \mathfrak{u}(1{)}_Y$. This is a necessary structural condition for the Standard Model, but it does not by itself establish a viable grand unified theory.

The representations in solution do not give a Standard Model fermion generation with the correct weak hypercharges. Use $Q_{\mathrm{electric}}=T_3+Y$. Identifying $(3,2{)}_1$ with the left-handed quark doublet, whose weak hypercharge is $1/6$, fixes $Y=X/6$. But then the doublet in $(1,3)$ has $Y=1/6$, whereas the left-handed lepton doublet requires $Y=-1/2$. Its singlet has $Y=-1/3$, whereas a right-handed electron requires $Y=-1$. The $(\overline{3},1{)}_0$ is a further concrete mismatch: right-handed quarks written as left-handed antiquarks require $Y=-2/3$ or $1/3$, and a rescaling cannot turn zero into either value. These comparisons use the electroweak representation and hypercharge table.

The charge-$-3$ doublet in $(1,8)$ does have the required lepton weak hypercharge under this normalization, but the same Adjoint representation also brings a doublet of the opposite charge and an extra triplet. Therefore merely adding that representation does not supply the missing chiral generation without additional states and a mechanism for their masses.

Furthermore $SU(3)\times SU(3)$ is a product of two Simple Lie groups, permitting two independent gauge couplings; it does not enforce a common strong and electroweak gauge coupling. Thus the embedding exists, but the displayed matter content does not furnish a viable minimal unification of the Standard Model. A more elaborate model would require additional representations, consistent chiral matter, anomaly cancellation, and a specified spontaneous symmetry breaking mechanism. The calculations here do not prove that every possible extension using this product group is impossible.