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The SU(6) representations $15={\Lambda }^{2}6$ and $\overline{6}$ branch under SU(5) as $10\oplus 5$ and $\overline{5}\oplus 1$. Their cubic gauge anomaly coefficients are $2$ and $-1$ in fundamental normalization. Thus $15\oplus 2\overline{6}$ is anomaly-free and gives one Standard Model family $10\oplus \overline{5}$, a vectorlike gauge spectrum $5\oplus \overline{5}$, and two singlets. With these representations and their conjugates, precisely one net chiral family requires $m_{15}-m_{\overline{15}}=1$ and $m_{\bar{6}}-m_6=2$, so this is the smallest such choice.

Use upper indices for the defining representation of SU(6) and lower indices for its dual representation. The Young tableaux below display tensor-slot labels; their shapes determine the Young diagrams. Full columns of height six are determinant representations and can be removed for SU(6).Here all indices run from one to six. The highest weight Dynkin labels follow from the differences of successive row lengths. In the same order the row lengths are $(1)$, $(1,1,1,1,1)$, $(2)$, $(1,1)$ and $(2,1,1,1,1)$.

The defining vector transforms as $v^a\mapsto U^a{}_b{v}^b$. For a unitary representation, the conjugate representation is equivalent to the dual representation, so $\overline{6}$ can be written as a covector with $w_a\mapsto w_b(U^{-1}{)}^b{}_a$. Its five-box column is equivalently a totally antisymmetric tensor $A^{a_1\cdots a_5}$, related by the invariant alternating tensor $w_a={ϵ}_{a{a}_1\cdots a_5}{A}^{a_1\cdots a_5}/5!$.

The symmetric tensor transforms as $S^{ab}\mapsto U^a{}_c{U}^b{}_d{S}^{cd}$, with $6(6+1)/2=21$ independent components. The antisymmetric tensor has the same transformation law but obeys $A^{ab}=-A^{ba}$, giving $6(6-1)/2=15$ independent components. For the Adjoint representation, $T\mapsto UT{U}^{-1}$; the trace constraint removes one component from the $36$ entries of a mixed tensor, leaving $35$. Its six-box Young diagram is obtained by converting the five-index column to a lower index and selecting the traceless mixed tensor.

Thus the requested dimensions, in the order above, are

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.

The first coordinate is fixed by the embedded SU(5), while the last five transform in its defining representation. Therefore $6{\downarrow }_{SU(5)}=1\oplus 5$, and taking the dual representation gives $\overline{6}{\downarrow }_{SU(5)}=1\oplus \overline{5}$.

For the symmetric square, the fixed-fixed component is one trivial representation, the fixed-five mixed components give $5$, and the symmetric five-five block is $15$. For the exterior square, the fixed-fixed component vanishes, the mixed components give $5$, and the antisymmetric five-five block is $10$. Finally restrict $35=6\otimes \overline{6}-1$, using $5\otimes \overline{5}=24\oplus 1$.

The five SU6 tensor branchings to SU5 are thusThe symbols $15$ on the left and right refer to different groups: the left-hand one is the SU(6) exterior square, whereas the right-hand one is the SU(5) symmetric square. The dimension checks are $6=5+1$, $21=15+5+1$, $15=10+5$ and $35=24+5+5+1$.

Explicitly, in the Adjoint representation the lower-right traceless five-by-five block gives $24$; the lower-left column and upper-right row give $5$ and $\overline{5}$; and the diagonal direction $\mathrm{diag}(-5,1,1,1,1,1)$ gives the remaining trivial representation.

For the special unitary group $SU(N)$, denote a one-column Young tableau with $k$ empty boxes by ${\mathcal{C}}_k$. Explicitly the columns needed here areA column ${\mathcal{C}}_k$ is the exterior power ${\Lambda }^{k}V$, with highest weight ${\omega }_k$ and dimension $\left(\genfrac{}{}{0ex}{}{N}{k}\right)$. Contraction with the invariant volume tensor identifies $V^{\ast }\cong {\Lambda }^{N-1}V$ and ${\Lambda }^2{V}^{\ast }\cong {\Lambda }^{N-2}V$. This gives all the requested Dynkin labels, Young tableaux, dimensions and tensors:Here $1\le I,J\le 6$ and $1\le i,j\le 5$. Lower indices denote the dual representation, equivalent to the conjugate representation for these compact groups. The SU(6) $15$ in this question is an exterior square; the SU(5) $15$ in solution is a symmetric square. The ambient group is essential to distinguish them.

One left-handed Standard Model family requires the SU(5) representation content $10\oplus \overline{5}$. No single irreducible representation in solution contains both. At the level of containing the required fields, the smallest choice isThe additional SU(5) multiplets are a $5$ and a trivial representation. Another minimal choice using only two of the listed irreducible representations isIts additional multiplets form the conjugate representation of a family. More generally, for multiplicities $(m_6,m_{\bar{6}},m_{15},m_{\overline{15}})$, branching of a representation givesThus containing a family requires $m_{15}\ge 1$ and $m_{\bar{6}}+m_{\overline{15}}\ge 1$.

For a consistent chiral grand unified theory, containment alone is insufficient. Normalize the cubic gauge anomaly coefficient by $A(6)=1$. For the special unitary group $SU(N)$, $A({\Lambda }^2\mathbf{N})=N-4$, so for SU(6) it is $2$. To verify this directly, take a traceless diagonal generator with entries $x_i$: its exterior square has entries $x_i+x_j$, andThe conjugate representation reverses this cubic gauge anomaly. The combination $15\oplus \overline{6}$ has coefficient $2-1=1$, so cannot alone be a quantum-consistent chiral SU(6) gauge theory. The minimal anomaly-free choice with one net chiral gauge spectrum of a family is insteadThe additional nonsinglet fields form a vectorlike gauge spectrum that can acquire an SU(5)-invariant mass, and the two gauge singlets may serve as sterile neutrinos. More generally, precisely one net chiral gauge spectrum of a family with vectorlike extra nonsinglets requires $m_{15}-m_{\overline{15}}=1$ and $m_{\bar{6}}-m_6=2$; these conditions also cancel the cubic SU(6) gauge anomaly. The $15\oplus \overline{15}$ choice above is anomaly-free but has zero net chirality.

For the embedding $U\mapsto \mathrm{diag}(1,U)$, the defining SU(6) representation restricts to the defining SU(5) representation plus a trivial representation. Taking dual representations, symmetric squares, exterior squares and traceless endomorphisms gives $\overline{6}\downarrow =\overline{5}\oplus 1$, $21\downarrow =15\oplus 5\oplus 1$, $15\downarrow =10\oplus 5$, and $35\downarrow =24\oplus 5\oplus \overline{5}\oplus 1$. The fixed diagonal direction in the Adjoint representation is $\mathrm{diag}(-5,1,1,1,1,1)$. Dimensions label representations of different groups on the two sides: the fifteen-dimensional SU(6) representation is alternating, while the fifteen-dimensional SU(5) representation is symmetric.