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The Adjoint representation of G2 has one weight at each of its twelve roots and two at zero. Its six long roots form the A2 root system; together with the two zero weights they give the Adjoint representation of SU(3). The remaining six short roots form the weight systems of its defining representation and dual representation. Hence restriction to the long-root SU(3) gives $14=8\oplus 3\oplus \overline{3}$. This is a branching of a representation; the complement is invariant under the subgroup, not under the full G2 action.

For the special unitary group $SU(N)$, put ${\epsilon }_i=e_i-N^{-1}{\sum }_j{e}_j$. Its defining representation has weight system $\{{\epsilon }_i\}$, and its exterior square has weight system $\{{\epsilon }_i+{\epsilon }_j:i<j\}$; each has multiplicity one. Their dual representations negate the weights of a representation. The Dynkin label coordinates are $({\epsilon }_i{)}_r={\delta }_{ir}-{\delta }_{i,r+1}$. These follow by evaluating diagonal traceless generators on coordinate vectors and their exterior products.

Under the standard embedding $SU(N-1)\subset SU(N)$, the last defining weight of a representation restricts to zero and the others become the defining weights of a representation of $SU(N-1)$. Splitting pairs according to whether they contain the last index proves the branching of a representation $V_N\downarrow =V_{N-1}\oplus 1$ and ${\Lambda }^2{V}_N\downarrow ={\Lambda }^2{V}_{N-1}\oplus V_{N-1}$, including multiplicities.

Under the stated SU(2) embedding, the first two components of the defining SU(3) representation form a doublet, while the third is fixed. Hence $3{\downarrow }_{SU(2)}=2\oplus 1$. Its dual representation restricts in the same way because the SU(2) doublet is self-dual.

For the Adjoint representation, restrict $8=3\otimes \overline{3}-1$. The SU(2) tensor product $2\otimes 2=3\oplus 1$ givesHere subtraction denotes removal of the trivial representation generated by the identity matrix; it is not a negative-dimensional summand. The first SU(3) factor is unchanged, so the requested branching of a representation rules areThe dimensions on the two sides are respectively $9=6+3$, $3=3$, $3=2+1$ and $8=3+2+2+1$.

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.

The highest root is $\theta =2{\alpha }_1+3{\alpha }_2$. In the Extended Dynkin diagram, the extra node $-\theta$ is long and is joined by a single bond to the long node ${\alpha }_1$; it is orthogonal to the short node ${\alpha }_2$. Deleting the short node therefore leaves the A2 root system Dynkin diagram. Equivalently, the six long roots are a closed root subsystem, so their root spaces together with the Cartan subalgebra form an A2 root system Lie subalgebra. The corresponding compact subgroup is $SU(3)\subset G_2$.

For the particular projection in the paper, choose the A2 root system simple roots as ${\beta }_1={\alpha }_1+3{\alpha }_2$ and ${\beta }_2={\alpha }_1$. Since these are long, ${\beta }_1^{\vee }={\alpha }_1^{\vee }+{\alpha }_2^{\vee }$ and ${\beta }_2^{\vee }={\alpha }_1^{\vee }$. Consequently the Dynkin labels restrict asProjecting the six short roots gives $\pm (1,0),\pm (0,1),\pm (1,-1)$. Group these into the three weights of $3$ and the three of $\overline{3}$ listed in solution. The zero weight of $7$ is left over, soThe three positive long roots project as $(2,-3)\mapsto (-1,2)$, $(-1,3)\mapsto (2,-1)$ and $(1,0)\mapsto (1,1)$. Together with their negatives and the two zero weights, these are exactly the Adjoint representation $8$. The remaining six short roots again give $3\oplus \overline{3}$. Hence the branching of a representation is

Under the two-block subgroup $H_2$, the defining vector space splits into its first and last two coordinates. ThereforeFor the exterior square, the exterior square of a direct sum identity gives two within-block determinant representations, each trivial for $SU(2)$, and the cross-block tensor product $(2,2)$. For the Adjoint representation, use $\mathrm{End}({\mathbb{C}}^4)-1$: each diagonal two-by-two block contributes a scalar and a traceless triplet, while each off-diagonal block gives $(2,2)$ because the SU(2) doublet is self-dual. Removing the overall trace leaves one scalar. The dimension counts are $2+2=4$, $1+1+4=6$, and $3+3+4+4+1=15$.

Under $H_3$, the first three coordinates carry the defining SU(3) representation and the fourth is fixed. HenceIndeed ${\wedge }^2(3\oplus 1)={\wedge }^{2}3\oplus 3=\overline{3}\oplus 3$. The upper three-by-three block of a traceless matrix supplies the Adjoint representation $8$ and one scalar, while the upper-right and lower-left blocks give $3$ and $\overline{3}$. This establishes the branching of a representation in each case without a dimension-only guess.

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.

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.

Branching of a representation describes the irreducible representation content of a restriction of a representation:This direct decomposition applies when the restriction of a representation is a semisimple representation, as for finite groups or continuous finite-dimensional representations of compact groups. For an arbitrary group and subgroup, complete reducibility need not hold. A vector subspace preserved by every $R(h)$ for $h\in H$ need not be preserved by the additional $R(g)$ for $g\notin H$, so an irreducible representation can become reducible.

For a concrete example, take the defining representation of $G=SU(2)$ on ${\mathbb{C}}^2$ and $H=\{\mathrm{diag}(z,z^{-1}):|z|=1\}$, a subgroup isomorphic to the circle group. The defining representation is irreducible: $SU(2)$ acts transitively on unit vectors, so no complex line is preserved by all its elements. Under $H$, the two coordinate lines are invariant subspaces, andHere the two one-dimensional representations act by $z$ and $z^{-1}$.

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.

The highest root of the D5 root system in our numbering is $\theta =e_1+e_2={\alpha }_1+2{\alpha }_2+2{\alpha }_3+{\alpha }_4+{\alpha }_5$. Adjoin ${\alpha }_0=-\theta$. The Extended Dynkin diagram has an extra bond $0-2$. Deleting node $2$ leaves the two isolated nodes $0,1$ and the chain $4-3-5$, hence the root subsystem $A_1\oplus A_1\oplus A_3$.

This is a maximal Lie subalgebra, not merely a full-rank Lie subalgebra. In the orthogonal realization it is $\mathfrak{so}(4)\oplus \mathfrak{so}(6)\subset \mathfrak{so}(10)$, acting on a $4+6$ decomposition. The remaining off-diagonal part transforms as the irreducible tensor product representation $(4,6)$ of dimension $24$. Any intermediate Lie subalgebra containing one nonzero vector in that complement must contain the entire complement, because this representation is irreducible, and hence all $6+15+24=45$ dimensions. This proves maximality.

To apply the supplied projection, write a D5 root system weight as the column $w=(a,b,c,d,e{)}^T$ of Dynkin labels. ThenFor example, the highest weight $w=(0,0,0,0,1{)}^T$ maps to $(1,0;1,0,0{)}^T$, and $w=(-1,0,1,-1,0{)}^T$ maps to $(0,1;0,0,1{)}^T$. Applying the same expression to all sixteen weights in solution gives exactlywhere $F$ and $\overline{F}$ are the four-element weight systems in solution. Each set has eight distinct weights, each of multiplicity one. Since an $A_1$ doublet has weights $\pm 1$ and an $A_1$ singlet has weight zero, the branching of a representation isThe highest weights of these two summands are $(1,0;1,0,0)$ and $(0,1;0,0,1)$, and $2\cdot 1\cdot 4+1\cdot 2\cdot 4=16$ checks the dimensions. Here $SU(2)\times SU(2)\times SU(4)$ is the convenient covering group; its image inside $\mathrm{Spin}(10)$ has a common ${\mathbb{Z}}_2$ central identification. Both summands are trivial on $(-I_2,-I_2,-I_4)$.

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.