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Among the representations in c, the $10$ of $SU(5)$ occurs in the $15$ of $SU(6)$, while the simplest source of $\bar{5}$ is $\bar{6}$. Thus the economical direct sum containing the prescribed family isThe extra $SU(5)$ multiplets are one $5$ and one trivial representation. Calling $R$ a representation does not require it to be an irreducible representation; none of the five individual choices in c contains both required $SU(5)$ summands.

There is a further physical qualification if this is to be the entire chiral matter content of a consistent grand unified theory. Merely containing one Standard Model family does not guarantee cancellation of the cubic $SU(6)$ gauge anomaly. In the convention $A(6)=1$, one has $A(\bar{6})=-1$ and $A(15)=2$, so the displayed sum has $A(R)=1$. With these types of representations, the minimal anomaly-free completion is insteadBeyond one family this adds a vectorlike gauge spectrum $5\oplus \bar{5}$ and two trivial representations. The conjugate pair can receive a mass after the appropriate symmetry breaking, without removing the remaining family.^[ref] The first boxed answer addresses representation containment; the second includes the extra consistency condition for a chiral gauge theory.

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.