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Embed SU(2) into SU(3) by $U\mapsto \mathrm{diag}(U,1)$. Its defining representation decomposes as an irreducible doublet and a trivial representation. The Schur lemma forces every commuting matrix to be $\mathrm{diag}(z{I}_2,w)$. Unitarity and unit determinant imply $|z|=1$ and $w=z^{-2}$, proving the formula. With Hermitian generator $X=\mathrm{diag}(1,1,-2)$, the defining representation restricts as $2_1\oplus 1_{-2}$ and the Adjoint representation as $3_0\oplus 2_3\oplus 2_{-3}\oplus 1_0$. The U(1) charges in the Adjoint representation are differences of defining charges. The generated connected subgroup is $(SU(2)\times U(1))/{\mathbb{Z}}_2$, since the two factors meet in $\mathrm{diag}(-I_2,1)$.

In four spacetime dimensions, the pure local gauge anomaly for a circle group is proportional to the sum of cubed U(1) charges over left-handed Weyl fermions, counting a right-handed Weyl fermion through its left-handed charge conjugate. A consistent anomaly-free gauge theory requires this sum to vanish, absent an additional anomaly-cancelling mechanism. Cancellation of the mixed gauge-gravitational anomaly additionally requires ${\sum }_a{Q}_a=0$. If the same spectrum is restricted to a finite cyclic group, $q_a=Q_a\mathrm{mod}n$ implies ${\sum }_a{q}_a^3=0\mathrm{mod}n$ and, when imposed, ${\sum }_a{q}_a=0\mathrm{mod}n$. These elementary consequences alone do not classify intrinsic discrete gauge anomalies or guarantee an anomaly-free integer lift.

The map $(z,U)\mapsto zU$ identifies the unitary group with $(U(1)\times SU(n))/K$, where $K=\{({\zeta }^k,{\zeta }^{-k}{I}_n):\zeta =e^{2\pi i/n}\}$. If an irreducible SU(n) tensor transformation has $b$ boxes in its Young diagram, the central scalar $\zeta I_n$ acts as ${\zeta }^b$. Therefore, for integer $q$, $(z,U)\mapsto z^{q}R(U)$ is trivial on $K$ precisely when ${\zeta }^{q-b}=1$, proving the criterion. Adding a full $n$-box column changes $b$ by $n$ and leaves the condition unchanged. In particular a lift with trivial first-factor U(1) charge descends only if its central action is trivial; this does not prevent other lifts of the same special unitary group representation from descending.

Suppose fields with integer U(1) charges $Q_i$ acquire nonzero vacuum expectation values, with at least one $Q_i\ne 0$. The stabilizer in the circle group is exactly ${\mathbb{Z}}_d$, where $d=\gcd (|Q_1|,\dots ,|Q_r|)$. Indeed, $e^{i\theta }$ preserves every condensate if and only if $e^{i{Q}_i\theta }=1$ for every $i$. The Bezout identity implies $e^{id\theta }=1$, and conversely $d\mid Q_i$ proves every $d$th root of unity preserves every condensate. Neutral fields impose no constraint; if all condensing fields are neutral, the whole circle group survives.

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.

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.

Write $X=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$. Commuting with all $\mathrm{diag}(U_1,U_2)$ requiresFor $n_1,n_2\ge 2$, the defining complex representations of each factor are irreducible. Schur lemma gives $A=z{I}_{n_1}$ and $D=w{I}_{n_2}$. The two block vector spaces are inequivalent representations of the product group, even when $n_1=n_2$: the first factor acts nontrivially on just the first block. Thus their off-diagonal intertwiners vanish, $B=C=0$. Unitarity and the unit determinant give

The assertion that this full group is always a circle needs correction. Put $d=\gcd (n_1,n_2)$, $a=n_1/d$, $b=n_2/d$. The continuous group homomorphismmaps this centralizer onto the $d$th roots of unity. Its kernel is the identity component of a Lie group, parametrized injectively by $e^{i\theta }\mapsto (e^{ib\theta },e^{-ia\theta })$. Coprimality of $a,b$ proves injectivity; a Bezout identity proves that this parametrization exhausts the kernel. Extending the primitive integer row $(a,b)$ to a determinant-one integer matrix gives coordinates in which the constraint is simply $t^d=1$. ConsequentlyFor a concrete counterexample, take $n_1=n_2=2$. The constraint is $(zw{)}^2=1$, giving two separate circles $zw=1$ and $zw=-1$. The element $\mathrm{diag}(I_2,-I_2)$ lies in the second component, so this centralizer cannot be isomorphic as a Lie group to the connected circle group.

For the intended specialization of this centralizer of two special unitary blocks, $d=\gcd (2,3)=1$ and therefore the full centralizer really is a circle group:We use this primitive integer U(1) charge normalization below. If exactly one $n_i$ equals $1$ and the other is at least $2$, the same block-scalar formula still holds. If $n_1=n_2=1$, both factors are trivial and the centralizer is all of SU(2), another exception to the printed claim.

Write the defining space as $V=A\oplus B$, where $A\cong {\mathbb{C}}^2$ and $B\cong {\mathbb{C}}^3$. Use ordered pairs for the SU(2) and SU(3) factors, in that order. The generator $X_c$ in solution gives U(1) charges $3$ on $A$ and $-2$ on $B$, soTaking the conjugate representation reverses the U(1) charges. The SU(2) doublet is equivalent to its dual representation, while the SU(3) triplet becomes the antitriplet:

For the two-index antisymmetric representation, split the antisymmetric tensor $T^{IJ}=-T^{JI}$ into $T^{ab}$, $T^{a\mu }$ and $T^{\mu \nu }$, with $a,b=1,2$ and $\mu ,\nu =3,4,5$. The exterior square of a direct sum givesThe first term is an SU(2) singlet, the middle is $(2,3)$, and ${\Lambda }^{2}3\cong \overline{3}$ by contraction with the invariant alternating tensor. Adding the U(1) charges of the two indices givesThe dimensions $1+6+3=10$ check the branching of a representation. In the convention $Q_{\mathrm{em}}=T_3+Y$, weak hypercharge is $Y=X_c/6$.

The complexification of a Lie algebra in the Adjoint representation of SU(5) consists of traceless endomorphisms of $V=A\oplus B$. Splitting an endomorphism into its four blocks gives traceless endomorphisms of $A$ and $B$, one relative block-scalar direction, and the two off-diagonal spaces $\mathrm{Hom}(B,A)$ and $\mathrm{Hom}(A,B)$.

The first three pieces transform as $(3,1{)}_0$, $(1,8{)}_0$ and $(1,1{)}_0$. The off-diagonal pieces have U(1) charges equal to output minus input charge, namely $3-(-2)=5$ and $-2-3=-5$. ThusThe dimensions are $3+8+1+6+6=24$. The first three summands are precisely the gauge fields of the local gauge group $H\times U_c(1)$. The extra gauge fields of the SU(5) gauge theory transform asHermiticity relates the two off-diagonal blocks, so these describe twelve real gauge fields, or six complex fields and their conjugates. Globally, the subgroup generated by $H$ and $U_c(1)$ inside SU(5) is $(SU(2)\times SU(3)\times U(1))/{\mathbb{Z}}_6$, since the parametrization $(U_2,U_3,e^{i\theta })\mapsto \mathrm{diag}(e^{3i\theta }{U}_2,e^{-2i\theta }{U}_3)$ has a six-element kernel. This finite quotient does not alter the stated branching of a representation.

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.

Number the simple roots of the $A_4$ root system consecutively along its Dynkin diagram. For a Young diagram with row lengths $r_i$, its highest weight has Dynkin labels $a_i=r_i-r_{i+1}$. The three irreducible representations have the following Young diagrams, which can be filled increasingly down the column to give Young tableaux:Here $i=1,\dots ,5$, $v^i\mapsto U^i{}_j{v}^j$, and $T^{ij}\mapsto U^i{}_k{U}^j{}_l{T}^{kl}$. Contraction with the invariant alternating tensor identifies ${\tilde{v}}_i=(1/4!){ϵ}_{i{i}_1{i}_2{i}_3{i}_4}{w}^{i_1{i}_2{i}_3{i}_4}$, giving the dual representation of $5$. The dimensions follow from $\left(\genfrac{}{}{0ex}{}{5}{1}\right)=5$, $\left(\genfrac{}{}{0ex}{}{5}{4}\right)=5$, $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$.

Use the centralizer and circle group normalization of solution. The first three components have U(1) charge $-2$ and the last two have U(1) charge $3$, so the branching of a representation isFor the dual representation, both U(1) charges reverse and the $SU(3)$ representation is conjugated; the $SU(2)$ doublet is equivalent to its conjugate. Taking the exterior square of the first decomposition gives three types of antisymmetric tensor components: $T^{ab}$, $T^{a\alpha }$, and $T^{\alpha \beta }$, where $a,b=1,2,3$ and $\alpha ,\beta =4,5$. Their U(1) charges add, and ${\Lambda }^{2}3=\overline{3}$, ${\Lambda }^{2}2=1$. HenceThe dimensions are $3+6+1=10$. These product labels describe genuine representations of the quotient group in solution: every element of its ${\mathbb{Z}}_6$ kernel acts trivially on each summand.

Choose the first $SU(2)$ as $S{U}_w(2)$ and embed $S{U}_c(3)$ in the upper-left three-by-three block of $SU(4)$. The fundamental representations then restrict as $4\to 3\oplus 1$ and $\overline{4}\to \overline{3}\oplus 1$. Thus the representation in solution, forgetting hypercharge, restricts toThis contains every required $S{U}_w(2)\times S{U}_c(3)$ representation, with one extra singlet.

For the hypercharge embedding, let $X=\mathrm{diag}(1,1,1,-3)$ in the $SU(4)$ fundamental representation, and $T_{3R}=\mathrm{diag}(1/2,-1/2)$ in the second $SU(2)$. These commute with $S{U}_w(2)\times S{U}_c(3)$. The integer normalization in the paper isExplicitly, this circle group embeds in the covering group byBoth nontrivial blocks have determinant one, and the map is injective since the first three entries of the $SU(4)$ block recover $z$.

On $(2,1,4)$, $T_{3R}=0$, so the branching of a representation gives $(2,3{)}_1\oplus (2,1{)}_{-3}$. On $(1,2,\overline{4})$, the U(1) charges under $X$ are $-1$ on $\overline{3}$ and $3$ on $1$, while $6T_{3R}$ takes values $\pm 3$. HenceCombining the two summands givesThis is the Pati-Salam charge embedding, with $X=3(B-L)$. In terms of the common Standard Model convention $Q_{\mathrm{electric}}=T_{3L}+Y_{\mathrm{SM}}$, the paper's integer hypercharge is $Y=6Y_{\mathrm{SM}}$.

The extra neutral singlet can be interpreted as the left-handed charge conjugate ${\nu }_R^c$ of a right-handed neutrino. It is a sterile neutrino under the Standard Model gauge group, and completes the sixteen Weyl fermions of one family in this half-spin representation. Its neutrality permits a Majorana mass term without violating the remaining Standard Model gauge group.

For the circle group parameterized by $e^{i\theta }$ with period $2\pi$, a one-dimensional representation acts as $e^{i\theta }\mapsto e^{iQ\theta }$, and its integer label $Q$ is its charge. Periodicity forces $Q\in \mathbb{Z}$. A tensor product representation adds U(1) charges, a dual representation negates them, and restriction to the finite cyclic group generated by $e^{2\pi i/n}$ retains only $Q\mathrm{mod}n$. Integer normalization must be fixed before discussing a largest finite subgroup left by a condensate.