Past exam of the groups and representations course of the University of Oxford / 2023 / 1

A subgroup $H$ of a group $G$ is a nonempty subset closed under the inherited multiplication and inverses; equivalently, $x{y}^{-1}\in H$ whenever $x,y\in H$. A normal subgroup satisfies $gH{g}^{-1}=H$ for every $g\in G$.

A group representation on a vector space $V$ is a group homomorphism $R:G\to \mathrm{GL}(V)$, so $R(gh)=R(g)R(h)$ and $R(e)=I_V$. An irreducible representation has $V\ne 0$ and no invariant subspaces other than $0$ and $V$: there is no proper nonzero $W$ with $R(g)W\subseteq W$ for every $g\in G$. We use finite-dimensional complex representations, with continuity for Lie groups, throughout.

Branching of a representation describes the irreducible representation content of a restriction of a representation:This direct decomposition applies when the restriction of a representation is a semisimple representation, as for finite groups or continuous finite-dimensional representations of compact groups. For an arbitrary group and subgroup, complete reducibility need not hold. A vector subspace preserved by every $R(h)$ for $h\in H$ need not be preserved by the additional $R(g)$ for $g\notin H$, so an irreducible representation can become reducible.

For a concrete example, take the defining representation of $G=SU(2)$ on ${\mathbb{C}}^2$ and $H=\{\mathrm{diag}(z,z^{-1}):|z|=1\}$, a subgroup isomorphic to the circle group. The defining representation is irreducible: $SU(2)$ acts transitively on unit vectors, so no complex line is preserved by all its elements. Under $H$, the two coordinate lines are invariant subspaces, andHere the two one-dimensional representations act by $z$ and $z^{-1}$.

Let $\zeta =e^{2\pi i/n}$. The group homomorphism ${\mathbb{Z}}_n\to U(1)$ given by $[k]\mapsto {\zeta }^k$ is injective, identifying the finite cyclic group with the subgroup of $n$th roots of unity in the circle group.

The continuous irreducible representations of the circle group are the one-dimensional representations $R_Q(e^{i\theta })=e^{iQ\theta }$ with $Q\in \mathbb{Z}$. One way to see this is to make the representation unitary by averaging, simultaneously diagonalize its commuting unitary matrices, and impose $R_Q(e^{i(\theta +2\pi )})=R_Q(e^{i\theta })$. Its restriction of a representation isThus each irreducible representation remains one-dimensional, and representations whose integer labels differ by a multiple of $n$ become equivalent. Every irreducible representation of the finite cyclic group occurs this way.

Choose a normalization in which the circle group has period $2\pi$ and the U(1) charges ${\hat{Q}}_i$ of the complex scalar fields are integers. By restriction of a representation,Write $v_i=⟨{\varphi }_i⟩$. A vacuum expectation value $v_i\ne 0$ is preserved by $e^{i\theta }$ precisely when $e^{i{\hat{Q}}_i\theta }=1$. Therefore the specified finite cyclic group survives spontaneous symmetry breaking exactly when every condensing complex scalar field obeysOnly U(1) charges of fields with nonzero vacuum expectation values enter this test; in the paper all $N$ fields condense.

Suppose at least one condensing U(1) charge is nonzero and put $d=\gcd (|{\hat{Q}}_i|:v_i\ne 0)$. Since each such U(1) charge is a multiple of $d$, every $d$th root of unity preserves the vacuum expectation values. Conversely, the Bezout identity expresses $d$ as an integer combination of these U(1) charges, so a common stabilizer obeys $e^{id\theta }=1$. The discrete remnant of a circle symmetry is consequentlyA nontrivial retained finite cyclic group exists exactly when $d>1$, and its possible orders are the divisors $n>1$ of $d$. For example, condensing U(1) charges $6$ and $10$ leave ${\mathbb{Z}}_2$. If $d=1$, the stabilizer is trivial. If all condensing U(1) charges vanish, the entire circle group is unbroken and there is no largest finite $n$; this case does not break the symmetry assumed in the question.

Normalize the circle group as in solution and count all Weyl fermions as left-handed; a right-handed Weyl fermion is replaced by its left-handed charge conjugate. The restriction of a representation givesThe triangle coefficient of the cubic U(1) gauge anomaly is proportional to ${\sum }_a{Q}_a^3$. In a theory with this single gauge group, cancellation of its local pure gauge anomaly requiresIf absence of the mixed gauge-gravitational anomaly is also required, as when coupling to a general gravitational background, one additionally needs ${\sum }_a{Q}_a=0$. Additional gauge group factors would require their mixed gauge anomalies to cancel as well.

Choose integer representatives and write $Q_a=q_a+n{k}_a$. Expanding the cubic givesHence the inherited necessary congruence, and its gravitational counterpart when imposed, areThese statements concern the same complete fermion spectrum used in the parent theory. They are consequences of an anomaly-free integer lift, rather than a full classification of intrinsic discrete gauge anomalies or a sufficient test for such a lift. They also do not imply an ordinary equality for the chosen residues: for $n=3$, the anomaly-free U(1) charges $(1,-1)$ have residues $(1,2)$, whose cubes sum to $9$, rather than zero. Integrating out heavy fermions can require extra contributions to the effective discrete gauge anomaly accounting.