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The branching of a group representation $V$ to a subgroup $H$ is its decomposition into irreducible representations of $H$, when that restriction of a representation is a semisimple representation: $V{|}_H={⨁}_{\lambda }{m}_{\lambda }{W}_{\lambda }$. The nonnegative integers $m_{\lambda }$ are the branching multiplicities. For finite groups over the complex numbers, they are the character inner products of the restricted character with the irreducible characters of $H$. Restriction to an arbitrary subgroup need not be completely reducible.

An unbroken subgroup $H$ must fix the vacuum expectation value itself. Thus a nonzero $H$-invariant vacuum expectation value can exist only if the restriction of a representation to $H$ contains a trivial representation. Applying the singlet criterion for an unbroken subgroup to solution givesThe two “no” entries are forbidden for every nonzero vacuum expectation value. Each “yes” means that the indicated subgroup can remain unbroken, subject to the scalar potential choosing such a vacuum. It does not by itself prove that the indicated subgroup is the entire stabilizer.

For exact spontaneous symmetry breaking to the stated subgroup, the distinction matters. A complex $4$ admits an exact $H_3$ stabilizer; a complex $6$ admits an exact $H_2$ stabilizer. A single $15$ preserving $H_2$ or $H_3$ necessarily preserves an additional continuous $U(1)$, as computed in solution, so neither exact endpoint is possible with that single field. Also, if the $6$ scalar field is constrained to the real structure of the SU4 exterior square, its nonzero stabilizer is larger than $H_2$. The paper does not specify whether this reality constraint is imposed, so both cases are distinguished below.

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.

For the standard A4 root system inside the A5 root system, let $p$ delete the fifth Dynkin label. Put $v_i=p(w_i)$ for $i\le 5$. These are precisely the five weights of the defining $SU(5)$ fundamental representation:Since $p(w_6)=0$, the six projected weights of $\overline{6}$ are $-v_1,\dots ,-v_5,0$. The first five form the dual representation $\overline{5}$, with highest weight $(0,0,0,1)$; the last is a singlet. ThusFor $15$, split the fifteen weights in b into the ten pairs with $1\le i<j\le 5$ and the five pairs $i6$. The former project to $v_i+v_j$, the complete weight decomposition of ${\Lambda }^{2}5=10$, with highest weight $(0,1,0,0)$. The latter project to $v_i$, the complete weight decomposition of $5$. ThereforeThis restriction of a representation can also be checked without a table: $V{\downarrow }_{SU(5)}=5\oplus 1$, so its dual representation is $\overline{5}\oplus 1$, while its exterior square is ${\Lambda }^{2}5\oplus (5\otimes 1)$. The dimensions match $6=5+1$ and $15=10+5$.

One generation of left-handed Standard Model fermions fits into $\overline{5}\oplus 10$ of an SU(5) grand unified theory. With hypercharge defined by $Q=T_3+Y$, their restriction of a representation to $SU(3{)}_c\times SU(2{)}_L\times U(1{)}_Y$ isHere $Q$ in the multiplet list denotes the left-handed quark doublet, and the superscript $c$ labels left-handed charge-conjugate fields of the corresponding right-handed fermions.

The results of c therefore giveThus the required matter representations are present. The extra singlet has Standard Model representation $(1,1{)}_0$ and can describe the left-handed charge-conjugate field of a right-handed neutrino. Its gauge singlet character permits a Majorana mass term; a Yukawa interaction with a lepton doublet and a suitable Higgs field can then supply neutrino masses. The branching alone does not fix those interactions or their scales.

The extra $5$ hasIf it describes Weyl fermions, these are additional coloured and electroweak fermions in the conjugate Standard Model representations to the ordinary $\overline{5}$. With an additional $\overline{5}$, they can form a vectorlike gauge spectrum and acquire invariant masses. Pairing them with the only ordinary $\overline{5}$ would also remove that generation's $d^c,L$ fields, so a realistic heavy pairing needs extra matter. If the $5$ instead describes a scalar field, it has the quantum numbers of the usual $SU(5)$ Higgs field: a weak doublet and a coloured triplet. A single fermionic $15$ does not turn its extra fermions into a scalar Higgs field merely by branching.

There is a further consistency condition if these are the entire chiral matter content of an $SU(6)$ gauge theory. Normalize the cubic anomaly coefficient by $A(6)=1$. The dual representation has $A(\overline{6})=-1$, and the two-index antisymmetric representation has $A(15)=6-4=2$. Thus one copy of each has nonzero gauge anomaly, $A_{\mathrm{total}}=1$. Their inclusion of the desired $SU(5)$ multiplets is therefore not, by itself, a complete consistent chiral unified theory.

For example, anomaly cancellation holds for $15\oplus 2\overline{6}$, whose restriction of a representation isThis leaves one ordinary generation, a potentially massive vectorlike gauge spectrum, and two possible singlet neutrinos. This anomaly-free matter choice is also used in explicit $SU(6)$ model constructions.^[ref] These interpretations require additional choices of matter statistics, spontaneous symmetry breaking, and interactions; the weight systems determine the multiplet content.

The restriction of a representation $\rho :G\to \mathrm{GL}(V)$ to $H$ is the same vector space and operators with the domain limited to $H$; its character is ${\chi }_H(h)=\chi (h)$. Averaging a positive Hermitian inner product over $H$ makes it invariant, so invariant subspaces have invariant orthogonal complements. Repeating gives a direct sum of irreducible $H$-spaces and hence ${\chi }_H={\sum }_i{d}_i{\psi }_i$ with nonnegative integer multiplicities.

By character inner product orthogonality,since $\chi$ is irreducible and has normalized squared norm one on $G$. Equality holds precisely when the omitted nonnegative terms vanish, namely $\boxed{\chi (g)=0\text{ for every }g\notin H}$. This is the restriction multiplicity bound.

Write $A=kG$ and $J=J(A)$, its Jacobson radical. First suppose the characteristic is $p>0$. We prove the counting statement through the cocenter of a finite-dimensional algebrawhere $[A,A]$ is the linear span of commutators, not an ideal. In a group algebra, this span is exactly the span of differences of conjugate group elements: $gh-hg$ is such a difference, and $xg{x}^{-1}-g=[x,g{x}^{-1}]$. Consequently $C(A)$ has one basis vector for each conjugacy class.

There is a well-defined Frobenius map on an algebra cocenterHere $F$ is additive and $F(c[a])=c^{p}F([a])$. To prove additivity, expand $(a+b{)}^p$ into words of length $p$. Cyclic rotations of a word are equal modulo commutators. Every nonconstant word has a rotation orbit of length $p$, so its orbit contributes zero; only $a^p,b^p$ remain. Alsobecause the two products are cyclic rotations. Additivity then shows that the whole commutator subspace maps to zero, proving well-definedness on the quotient.

Let $s$ be the number of simple modules up to isomorphism. The semisimple algebra $A/J$ is ${\prod }_{\nu =1}^s{M}_{d_{\nu }}(k)$: over an algebraically closed field, a finite-dimensional division algebra is the field itself, since the minimal polynomial of any element splits into linear factors and a division algebra has no zero divisors. In a matrix algebra, the commutator subspace is precisely the traceless matrices. Off-diagonal matrix units are commutators, and $E_{ii}-E_{jj}=[E_{ij},E_{ji}]$; these span that subspace in every characteristic. ThusThe map induced by $F$ on each coordinate is $z\mapsto z^p$, since $\mathrm{tr}(B^p)=(\mathrm{tr}B{)}^p$. This is bijective because $k$ is a perfect field.

The natural surjection $C(A)\twoheadrightarrow C(A/J)$ has kernel $(J+[A,A])/[A,A]$. Every kernel class has a representative in $J$. Since $J$ is nilpotent, $F^N$ kills that kernel for sufficiently large $N$. On the quotient it remains bijective, so its image has dimension exactly $s$. On the conjugacy-class basis, however,For large $N$, every $g^{p^N}$ is a p-regular element; moreover the $p^N$-power map permutes the p-regular conjugacy classes, because an inverse exponent can be chosen modulo the prime-to-$p$ part of the exponent of $G$. The stable image therefore has precisely those conjugacy classes as its basis. Comparing dimensions provesThis argument uses dimensions of semilinear images: a bijective scalar Frobenius map preserves dimension, so the usual rank count applies.

In characteristic zero every group element is relevant. Averaging a linear projection onto a submodule over $G$ makes that projection $G$-equivariant, proving Maschke's theorem and semisimplicity of $kG$. The same matrix-algebra cocenter calculation now gives $s=\dim C(kG)$, the total number of conjugacy classes.

For the second request, $M$ is a relative projective module for $H$ when every $G$-equivariant epimorphism onto $M$ which splits after restriction to $H$ already splits over $G$. Equivalently $M$ is a direct summand of $kG{\otimes }_{kH}M$, the induced representation of the restriction of a representation to $H$. We can verify the defining splitting property directly: if $q:N\twoheadrightarrow M$ has an $H$-linear section $u$, choose representatives $T$ of the left cosets $G/H$ and setThe sum is independent of representatives by $H$-linearity. Reindexing $gt=t^{\prime }h$ proves $G$-linearity, while $qv(m)=[G:H{]}^{-1}{\sum }_{t}m=m$. Hence every $M$ is relatively $H$-projective when the index is invertible in $k$.

For completeness the canonical induced-module epimorphism $\epsilon (g\otimes m)=gm$ has the explicit $G$-linear sectionIt exhibits the promised direct summand and is the projectivity detected on a subgroup of invertible index construction.

If a group representation of $G$ has character $\chi$, its restriction of a representation to $H$ has character $\chi {|}_H$. For an $H$-representation $V$ with character $\psi$, the induced representation is $\mathbb{C}[G]{\otimes }_{\mathbb{C}[H]}V$, with $G$ acting on the first factor. Decomposing into coset copies of $V$ shows that its character isIndeed only cosets fixed by $g$ contribute to the trace, and each coset has $|H|$ representatives.

The Frobenius reciprocity says $⟨{\mathrm{Ind}}_H^G\psi ,\chi {⟩}_G=⟨\psi ,{\mathrm{Res}}_H^G\chi {⟩}_H$. Directly, with the usual character inner product,The first equality substitutes $g=xh{x}^{-1}$ in the finite sum, and the second uses conjugacy invariance. This proves the theorem.

For a finite group $G$, subgroups $H,K$, and a complex group representation $V$ of $H$, the Mackey restriction formula isChoose one representative $g$ of each double coset $KgH$. The conjugate subgroup representation ${}^{g}V$ is a representation of $gH{g}^{-1}$ on the same vector space, with $(gh{g}^{-1})v=hv$. Restriction of a representation keeps the vector space and restricts the acting subgroup. induced representation is $\mathbb{C}G{\otimes }_{\mathbb{C}H}V$, and analogously for each subgroup of $K$. The displayed isomorphism is an isomorphism of $K$-representations. These definitions specify every summand and its subgroup.

An unbroken subgroup $H$ must fix the vacuum expectation value $v$ of a scalar field in a group representation $V$. Thus a nonzero such $v$ exists exactly when $V^H\ne \{0\}$. For a compact group, complete reducibility of compact-group representations identifies this fixed vector subspace with the sum of the trivial representations in the restriction of a representation $V{|}_H$. Every $v\in V^H$ has $H\subseteq \mathrm{Stab}(v)$; equality requires a separate stabilizer calculation. The criterion neither determines the full unbroken group nor guarantees that a given scalar potential has its minimum at that direction.