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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.

Normalize the cubic anomaly coefficient of the $SU(N)$ fundamental to $+1$. The two-index antisymmetric representation has anomaly coefficient $N-4$, while an antifundamental contributes $-1$. Cancellation of the $SU(N{)}^3$ gauge anomaly requires

The electric quarks give $2N_c$ copies of the $SU(2N_f)$ fundamental, soIn the magnetic theory the $2{\tilde{N}}_c$ color components of $q$ transform as flavor antifundamentals and contribute $-2{\tilde{N}}_c$. The singlet $M^{ij}$ is the two-index antisymmetric representation of $SU(2N_f)$, whose cubic anomaly coefficient is $2N_f-4$. HenceThis is a direct anomaly check of Seiberg duality.

Quantum consistency requires cancellation of the $SU(N{)}^3$ gauge anomaly. The two-index symmetric square has cubic anomaly coefficient $A(S)=N+4$, whereas each antifundamental has coefficient $-1$. Therefore

Choose the flavor orientation in which each electric $\tilde{\Phi }$ is a fundamental of $SU(p)$. Its $N$ color components then giveIn the magnetic theory, the eight Spin(8) components of $q$ transform in the antifundamental and contribute $-8$. The singlet $M$ transforms in the symmetric representation of $SU(p)$, whose cubic anomaly coefficient is $p+4$, while $t$ and $U$ are flavor singlets. Thereforewhere $p=N+4$ was used. The two $SU(p{)}^3$ 't Hooft anomalies match, as required by Seiberg duality.

The two-index antisymmetric representation of $SU(N)$ is the exterior square of its fundamental representation. For $N\ge 3$, it has dimension $N(N-1)/2$, Dynkin index $N-2$, and cubic anomaly coefficient $N-4$ in the same normalization. For $N=2$, the Dynkin index formula still applies in the stated normalization, but every symmetric cubic trace vanishes, so the stated fundamental-normalized cubic anomaly coefficient is not defined.

The two-index symmetric representation of $SU(N)$ is the symmetric square of its fundamental representation. For $N\ge 3$, it has dimension $N(N+1)/2$, Dynkin index $N+2$, and cubic anomaly coefficient $N+4$ when the fundamental index and anomaly are normalized to one. For $N=2$, the Dynkin index formula still applies in the stated normalization, but every symmetric cubic trace vanishes, so the stated fundamental-normalized cubic anomaly coefficient is not defined.