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For a traceless Hermitian adjoint scalar with potential $-m^2\mathrm{tr}{\Phi }^2+a(\mathrm{tr}{\Phi }^2{)}^2+b\mathrm{tr}{\Phi }^4$, the stationary eigenvalues satisfy $4b{x}^3+(4a{R}^2-2m^2)x-\zeta =0$. The usual bound of three distinct eigenvalues requires this polynomial not to vanish identically. With $b=0$, $a,m>0$, every traceless matrix of squared norm $R^2=m^2/(2a)$ is a global minimum, including matrices with all eigenvalues distinct. In SU(5), this permits an unbroken maximal torus $U(1{)}^4$, outside the generic three-block classification.

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.

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.

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.

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.

Impose the discrete Z2 symmetry $\boxed{\Phi \mapsto -\Phi }$. The quadratic and quartic trace invariants are unchanged, while $\mathrm{tr}{\Phi }^3$ changes sign. Invariance therefore forbids a nonzero coefficient of that cubic operator. This is an additional symmetry of the scalar potential, not a consequence of SU(5) conjugation alone; cubic traces are permitted by the latter.

The spectral theorem for a Hermitian matrix gives a unitary diagonalizer $V$ with $V\Phi V^{†}=\mathrm{diag}(x_1,\dots ,x_5)$ and real $x_i$. Similarity preserves the trace, so ${\sum }_i{x}_i=0$.

To ensure that the diagonalizer belongs to SU(5) rather than only the unitary group $U(5)$, write $\det V=e^{i\alpha }$ and put $U=e^{-i\alpha /5}V$. Its determinant is one, and the scalar phase cancels from its conjugation action. ThusThe same argument shows why no additional reality condition beyond Hermiticity is needed. Permuting the real eigenvalues merely changes their order in the diagonal form.

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.