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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.

In the convention of a, the fundamental weights areFor a weight $w=(w_1,w_2,w_3)$ in this orthonormal coordinate system, its Dynkin labels areThese labels, rather than the orthonormal coordinates, are used below and in the projection in c.

The highest weight $(1,0,0)$ is $e_1$. Successive subtraction of the simple roots gives the complete stringThe two consecutive short-root steps are allowed because $e_3$ has coroot pairing two with ${\alpha }_3$. The weight system of this seven-dimensional vector representation is thereforeAll seven weight multiplicities are one, including the zero weight.

The highest weight $(0,0,1)$ is ${\omega }_3$. The Weyl group of the B3 root system acts by signed coordinate permutations. Its orbit is the eight cube vertices $w=(s_1{e}_1+s_2{e}_2+s_3{e}_3)/2$, with independent $s_i=\pm 1$. Starting from $(+,+,+)$, lowering by ${\alpha }_3$, then ${\alpha }_2$, and then either ${\alpha }_1$ or ${\alpha }_3$ generates the middle branches; further lowering reaches all remaining sign choices. Thus the eight-dimensional spin representation hasThe stipulated unit weight multiplicities give dimensions seven and eight by counting these lists.

For the branching from B3 to A3, apply the supplied weight projection matrix to column vectors of Dynkin labels. It sendsIt is consistent with choosing the $A_3$ simple roots in the order ${\beta }_1=e_2-e_3$, ${\beta }_2=e_1-e_2$, ${\beta }_3=e_2+e_3$. Their Dynkin diagram is the chain ${\beta }_1-{\beta }_2-{\beta }_3$. Their coroot pairings are respectively $a_2,a_1,a_2+a_3$, explaining the projection without confusing labels and Euclidean coordinates.

The seven weights in b project asThe six nonzero weights form the weight system of the SU(4) two-index antisymmetric representation of highest weight $(0,1,0)$. The zero weight is the remaining singlet representation. ConsequentlyThe eight spinor weights project asThe four weights $(1,0,0),(-1,1,0),(0,-1,1),(0,0,-1)$ constitute the defining fundamental representation $4$. The other four are their negatives, forming the dual representation $\overline{4}$ of highest weight $(0,0,1)$. HenceFor the Young tableaux and tensors, let $U={\mathbb{C}}^4$ be the defining SU(4) representation. The singlet representation is ${\Lambda }^{0}U$, with the empty Young diagram and a scalar. The other three Young diagrams are columns of one, two and three boxes, representing $U$, ${\Lambda }^{2}U$ and ${\Lambda }^{3}U$:The square brackets indicate tensor index antisymmetrization, with indices from 1 to 4. The invariant Levi-Civita symbol identifies the three-index antisymmetric tensor with a covector, $w_{\ell }={ϵ}_{\ell ijk}{T}^{ijk}/3!$, explaining ${\Lambda }^{3}U\cong U^{\ast }$.

The projection deletes the third Dynkin label, so the four weights $v_i$ of solution restrict to $(1,0),(-1,1),(0,-1),(0,0)$. The first three are the defining SU(3) weight system, and the fourth is a singlet. Negating them gives the dual representation. HenceFor the Adjoint representation, the projected nonzero weights consist of the six A2 root system roots, the three defining weights and their negatives. There are also three zero weights. Two zero weights belong to the SU(3) Adjoint representation, leaving one singlet. Therefore the SU4 adjoint branching to SU3 isOne can see all four invariant subspaces directly by writing a traceless matrix asThe traceless part of $A$ is $8$, the column $b$ is $3$, the row $c$ is $\overline{3}$, and the remaining diagonal matrix is proportional to $\mathrm{diag}(1,1,1,-3)$ and is fixed by SU(3). These have dimensions $8+3+3+1=15$. Equivalently restrict $4\otimes \overline{4}-1$ and use $3\otimes \overline{3}=8\oplus 1$.

Use Dynkin labels $w_i=⟨w,{\alpha }_i^{\vee }⟩$, so the rows of the Cartan matrix are the simple roots expressed in fundamental weight coordinates. For the A2 root system,Starting at the highest weight $(1,0)$ and subtracting successive allowed simple roots gives $(1,0)\to (-1,1)\to (0,-1)$. Negating these weights gives the dual representation. Hence the two fundamental weight systems areEach listed weight has weight multiplicity one. The highest weight $(1,1)={\alpha }_1+{\alpha }_2$ is the highest root, so its representation is the Adjoint representation. Its nonzero weights are the six roots, while the Cartan subalgebra supplies the zero weight twice:The superscript $[2]$ indicates weight multiplicity. This also follows from $3\otimes \overline{3}=8\oplus 1$: the nine tensor product weights comprise six differences of distinct fundamental weights and three zeros, one zero belonging to the trivial representation.

Order the simple roots of the G2 root system with ${\alpha }_1$ long and ${\alpha }_2$ short. This ordering is consistent with the supplied projection in solution; reversing the nodes would interchange the two fundamental Dynkin labels. Choose $({\alpha }_1,{\alpha }_1)=2$, $({\alpha }_2,{\alpha }_2)=2/3$, and $({\alpha }_1,{\alpha }_2)=-1$. With the same row convention for the Cartan matrix as in solution,The positive roots, with their Dynkin labels, areFor the highest weight $(0,1)$, a lowering chain isThe middle three terms form the length-three weight string for the short simple root. The Weyl dimension formula, with this long-first convention, isIt gives seven for $(0,1)$, so the chain exhausts the weight system, with every weight multiplicity equal to one:For $(1,0)$, observe that $2{\alpha }_1+3{\alpha }_2=(1,0)$ is the highest root. Thus this is the Adjoint representation; the twelve root spaces contribute the twelve nonzero weights once, and the rank-two Cartan subalgebra contributes two zero weights. ThereforeThe same Weyl dimension formula gives fourteen, confirming the count.

For the special unitary group $SU(N)$ define ${\epsilon }_i=e_i-\frac{1}{N}{\sum }_{j=1}^N{e}_j$. These are the weights of a representation in the defining space, with ${\sum }_i{\epsilon }_i=0$. Their Dynkin labels areThe dual representation negates each weight of a representation, and the tensor product of group representations adds them. Consequently the complete weight systems areEvery listed weight of a representation has multiplicity one: the corresponding weight basis vectors are $v_i$, $v^i$, $v_i\wedge v_j$ and $v^i\wedge v^j$. Distinct pairs give distinct projected sums for $N=5,6$. For an explicit enumeration in Dynkin labels, the two defining weight systems areThe weight systems for $\overline{6}$ and $\overline{5}$ are the negatives of the entries in their respective columns, without reversing coordinate order.

For the exterior squares, the full enumerations areThe negatives of the $A_5$ column are exactly the fifteen weights of a representation of $\overline{15}$; the negatives of the $A_4$ column are exactly the ten of $\overline{10}$. Thus the tables specify all eight weight systems, including multiplicities, rather than only their highest weights.

For the embedding in solution, restriction retains the first four Dynkin label coordinates, since the simple coroots of $A_4$ are the first four of $A_5$. In a coordinate-free version of the same projection, let ${\eta }_i$ be the five defining weights of a representation of $A_4$. Restriction gives ${\epsilon }_i\mapsto {\eta }_i$ for $1\le i\le 5$, while ${\epsilon }_6\mapsto 0$.

For the fundamental representation, its first five weights of a representation therefore form a $5$ and the sixth is the zero weight of a representation of a trivial representation. Negating the weights of a representation gives the conjugate representation:For the two-index antisymmetric representation, split the fifteen weights of a representation into the ten with $i<j\le 5$ and the five with $j=6$. Their restricted values are ${\eta }_i+{\eta }_j$ and ${\eta }_i$, respectively. These are precisely the weight systems of $10$ and $5$. Negating again givesFor instance the highest $A_5$ weight of a representation $(0,1,0,0,0)$ restricts to $(0,1,0,0)$ for the $10$, while $(1,0,0,0,-1)$ restricts to $(1,0,0,0)$ for the $5$. The restricted weight systems are disjoint unions of multiplicity-one sets in these cases, and their dimensions check as $6=5+1$ and $15=10+5$.

For the A3 root system, take simple roots ${\alpha }_i=e_i-e_{i+1}$ for $i=1,2,3$ in the hyperplane ${\sum }_{i=1}^4{x}_i=0$ of an orthonormal ${\mathbb{R}}^4$. Its Dynkin diagram is the single-bond chain $1-2-3$. The Cartan matrix has entries $C_{ij}=2({\alpha }_i,{\alpha }_j)/({\alpha }_j,{\alpha }_j)$, givingFor a weight $\mu$, use Dynkin labels $w_i=2(\mu ,{\alpha }_i)/({\alpha }_i,{\alpha }_i)$. The four weights ${\mu }_j=e_j-\frac{1}{4}{\sum }_k{e}_k$ of the defining representation have labelsThey form the weight system with highest weight $(1,0,0)$. Equivalently, subtract successive rows of $C_{A_3}$ from that highest weight. The dual representation negates every weight, so the weight system with highest weight $(0,0,1)$ isAll these weights have multiplicity one. The numbered Dynkin diagrams used in this solution and in solution are shown below.

For the D5 root system, take an orthonormal basis $e_1,\dots ,e_5$ and simple roots ${\alpha }_i=e_i-e_{i+1}$ for $i=1,2,3,4$, together with ${\alpha }_5=e_4+e_5$. The Dynkin diagram has single bonds $1-2$, $2-3$, $3-4$, $3-5$, with the fork at node $3$. Thus its Cartan matrix isThe fundamental weight ${\omega }_5=\frac{1}{2}(e_1+e_2+e_3+e_4+e_5)$ has Dynkin labels $(0,0,0,0,1)$. Its pairing with every positive coroot $e_i\pm e_j$, $i<j$, is zero or one. Hence its half-spin representation is a minuscule representation, so its weight system is a single Weyl group orbit with multiplicity one. The Weyl group of the D5 root system permutes coordinates and changes an even number of signs. Consequently its complete weight system isThere are $1+\left(\genfrac{}{}{0ex}{}{5}{2}\right)+\left(\genfrac{}{}{0ex}{}{5}{4}\right)=16$ such weights. Their Dynkin labels are $((s_1-s_2)/2,(s_2-s_3)/2,(s_3-s_4)/2,(s_4-s_5)/2,(s_4+s_5)/2)$, explicitlyHere the Lie algebra is $\mathfrak{so}(10,\mathbb{C})$, or its compact real form when discussing compact groups. Globally the half-spin representation is a representation of $\mathrm{Spin}(10)$, and does not descend to $SO(10)$; the question only specifies the Lie algebra.

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 $(-I_2,-I_2,-I_4)$.

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.

A coordinate matrix expressing restriction of weights to the Cartan subalgebra of a Lie subalgebra. With Dynkin labels as column vectors, its rows encode the subalgebra's coroots in terms of the original coroots. It need not be an idempotent projection operator: an equal-rank embedding can give an invertible matrix. Applying it to the full weight system, including weight multiplicities, permits reconstruction of the restricted irreducible representations.

The weight system of a finite-dimensional Lie algebra representation is its collection of weights with their weight multiplicities. Restriction to a Lie subalgebra projects each weight to the smaller Cartan subalgebra, retaining and combining weight multiplicities. Grouping the resulting weights into irreducible weight systems computes the branching rule.