Past exam of the groups and representations course of the University of Oxford / 2022 / 4

Use the A5 root system with simple roots ${\alpha }_i=e_i-e_{i+1}$ for $1\le i\le 5$, in the hyperplane ${\sum }_{i=1}^6{x}_i=0$. The Dynkin diagram is the five-node chainIts Cartan matrix isDeleting the final node leaves the A4 root system, with simple roots ${\alpha }_1,\dots ,{\alpha }_4$. Concretely, the Lie algebra homomorphism ${\mathfrak{sl}}_5(\mathbb{C})\hookrightarrow {\mathfrak{sl}}_6(\mathbb{C})$ sends $X$ to $\mathrm{diag}(X,0)$ and preserves the Lie bracket. This realizes the required Lie subalgebra and fixes the embedding used below.

For the special unitary group $SU(N)$, denote a one-column Young tableau with $k$ empty boxes by ${\mathcal{C}}_k$. Explicitly the columns needed here areA column ${\mathcal{C}}_k$ is the exterior power ${\Lambda }^{k}V$, with highest weight ${\omega }_k$ and dimension $\left(\genfrac{}{}{0ex}{}{N}{k}\right)$. Contraction with the invariant volume tensor identifies $V^{\ast }\cong {\Lambda }^{N-1}V$ and ${\Lambda }^2{V}^{\ast }\cong {\Lambda }^{N-2}V$. This gives all the requested Dynkin labels, Young tableaux, dimensions and tensors:Here $1\le I,J\le 6$ and $1\le i,j\le 5$. Lower indices denote the dual representation, equivalent to the conjugate representation for these compact groups. The SU(6) $15$ in this question is an exterior square; the SU(5) $15$ in solution is a symmetric square. The ambient group is essential to distinguish them.

For the special unitary group $SU(N)$ define ${\epsilon }_i=e_i-\frac{1}{N}{\sum }_{j=1}^N{e}_j$. These are the weights of a representation in the defining space, with ${\sum }_i{\epsilon }_i=0$. Their Dynkin labels areThe dual representation negates each weight of a representation, and the tensor product of group representations adds them. Consequently the complete weight systems areEvery listed weight of a representation has multiplicity one: the corresponding weight basis vectors are $v_i$, $v^i$, $v_i\wedge v_j$ and $v^i\wedge v^j$. Distinct pairs give distinct projected sums for $N=5,6$. For an explicit enumeration in Dynkin labels, the two defining weight systems areThe weight systems for $\overline{6}$ and $\overline{5}$ are the negatives of the entries in their respective columns, without reversing coordinate order.

For the exterior squares, the full enumerations areThe negatives of the $A_5$ column are exactly the fifteen weights of a representation of $\overline{15}$; the negatives of the $A_4$ column are exactly the ten of $\overline{10}$. Thus the tables specify all eight weight systems, including multiplicities, rather than only their highest weights.

For the embedding in solution, restriction retains the first four Dynkin label coordinates, since the simple coroots of $A_4$ are the first four of $A_5$. In a coordinate-free version of the same projection, let ${\eta }_i$ be the five defining weights of a representation of $A_4$. Restriction gives ${\epsilon }_i\mapsto {\eta }_i$ for $1\le i\le 5$, while ${\epsilon }_6\mapsto 0$.

For the fundamental representation, its first five weights of a representation therefore form a $5$ and the sixth is the zero weight of a representation of a trivial representation. Negating the weights of a representation gives the conjugate representation:For the two-index antisymmetric representation, split the fifteen weights of a representation into the ten with $i<j\le 5$ and the five with $j=6$. Their restricted values are ${\eta }_i+{\eta }_j$ and ${\eta }_i$, respectively. These are precisely the weight systems of $10$ and $5$. Negating again givesFor instance the highest $A_5$ weight of a representation $(0,1,0,0,0)$ restricts to $(0,1,0,0)$ for the $10$, while $(1,0,0,0,-1)$ restricts to $(1,0,0,0)$ for the $5$. The restricted weight systems are disjoint unions of multiplicity-one sets in these cases, and their dimensions check as $6=5+1$ and $15=10+5$.

One left-handed Standard Model family requires the SU(5) representation content $10\oplus \overline{5}$. No single irreducible representation in solution contains both. At the level of containing the required fields, the smallest choice isThe additional SU(5) multiplets are a $5$ and a trivial representation. Another minimal choice using only two of the listed irreducible representations isIts additional multiplets form the conjugate representation of a family. More generally, for multiplicities $(m_6,m_{\bar{6}},m_{15},m_{\overline{15}})$, branching of a representation givesThus containing a family requires $m_{15}\ge 1$ and $m_{\bar{6}}+m_{\overline{15}}\ge 1$.

For a consistent chiral grand unified theory, containment alone is insufficient. Normalize the cubic gauge anomaly coefficient by $A(6)=1$. For the special unitary group $SU(N)$, $A({\Lambda }^2\mathbf{N})=N-4$, so for SU(6) it is $2$. To verify this directly, take a traceless diagonal generator with entries $x_i$: its exterior square has entries $x_i+x_j$, andThe conjugate representation reverses this cubic gauge anomaly. The combination $15\oplus \overline{6}$ has coefficient $2-1=1$, so cannot alone be a quantum-consistent chiral SU(6) gauge theory. The minimal anomaly-free choice with one net chiral gauge spectrum of a family is insteadThe additional nonsinglet fields form a vectorlike gauge spectrum that can acquire an SU(5)-invariant mass, and the two gauge singlets may serve as sterile neutrinos. More generally, precisely one net chiral gauge spectrum of a family with vectorlike extra nonsinglets requires $m_{15}-m_{\overline{15}}=1$ and $m_{\bar{6}}-m_6=2$; these conditions also cancel the cubic SU(6) gauge anomaly. The $15\oplus \overline{15}$ choice above is anomaly-free but has zero net chirality.