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The A2 root system representation $R(2,0)={\mathrm{Sym}}^2{\mathbb{C}}^3$ has dimension six and weights $(2,0),(0,1),(1,-1),(-2,2),(-1,0),(0,-2)$ in Dynkin label coordinates, each with multiplicity one.

The complex Adjoint representation of SU(3) is the action by conjugation on traceless $3\times 3$ matrices. With $H_1=\mathrm{diag}(1,-1,0)$ and $H_2=\mathrm{diag}(0,1,-1)$, its highest-weight vector is $E_{13}$ of Dynkin labels $(1,1)$. The six off-diagonal matrix units have weights $(2,-1),(-2,1),(-1,2),(1,-2),(1,1),(-1,-1)$, and the two-dimensional Cartan subalgebra has weight zero. In coordinates $(h_1,(h_1+2h_2)/\sqrt{3})$, the weight diagram is a regular hexagon together with a central point of weight multiplicity two. This representation describes both the gluon colour multiplet and the approximate flavour baryon octet, under different physical SU(3) symmetries.

For simple roots $(1,0),(-1,1)$, Dynkin labels of a Euclidean weight $(x,y)$ are $(2x,-x+y)$. The highest root $(1,1)$ gives the Adjoint representation, of dimension ten. Its eight nonzero weights are the eight roots of $B_2$; the zero weight has multiplicity two. The nine distinct label pairs are $(2,0),(0,1),(-2,2),(2,-1),(0,0),(-2,1),(2,-2),(0,-1),(-2,0)$.

For B3 root system simple roots $e_1-e_2,e_2-e_3,e_3$, choose the D3 root system simple roots in chain order $e_2-e_3,e_1-e_2,e_2+e_3$. The Dynkin labels project by $(a,b,c)\mapsto (b,a,b+c)$. This separates the vector weights $\{0,\pm e_i\}$ into the six-dimensional exterior square of the defining SU(4) representation and a singlet representation. The spinor cube vertices $\frac{1}{2}(\pm e_1\pm e_2\pm e_3)$ split into its defining representation and its dual representation.

A weight is dominant when all its Dynkin labels are nonnegative.

Dominance is a nonnegativity condition on a Weight; integrality additionally requires lattice membership.

For the irreducible highest-weight representation of the special linear Lie algebra ${\mathfrak{sl}}_3$ with Dynkin labels $(a,b)$, let $(p,q)$ be a dominant integral weight. Put $r=(2a+b-2p-q)/3$ and $s=(a+2b-p-2q)/3$. If these are not integers, the weight multiplicity is zero; otherwise it is $\max \{0,1+\min (a,b,r,s)\}$.

To derive the formula, use the sl3 interlacing character formula with top row $(a+b,b,0)$. The target diagonal exponents are $(s+p+q,s+q,s)$, and an interlacing middle row $(P,Q)$ has $P+Q=2s+p+2q$, with bottom entry $s+p+q$. ThusEvery integral $P$ in this interval gives exactly one Gelfand–Tsetlin basis vector. The upper bound minus the lower bound is the minimum of six differences: $a,s,r,r+p+q,b,s+q$. Since $p,q\ge 0$, this minimum reduces to $\min (a,b,r,s)$. Counting the interval proves the formula, including absent weights of a representation. This converts a two-dimensional weight diagram calculation into four elementary boundary distances.

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.

The coordinates of a highest weight record its eigenvalues on a chosen basis of a Cartan subalgebra. They depend on that basis. Dynkin labels instead use the simple coroots; orthonormal Cartesian coordinates and Dynkin labels must be converted before applying a dimension or Casimir eigenvalue formula.

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 ${\Lambda }^{3}U\cong U^{\ast }$.

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

Use the convention $A_{ij}=⟨{\alpha }_i,{\alpha }_j^{\vee }⟩$, so the rows of the supplied Cartan matrix are the simple roots in Dynkin label coordinates. Realize the B2 root system in an orthonormal plane byThus the first simple root is long and the second is short. The Dynkin diagram has two nodes joined by a double bond, with its arrow pointing to the short node:A Euclidean weight $w=x{e}_1+y{e}_2$ has Dynkin labelsIn particular the fundamental weights are ${\omega }_1=e_1$ and ${\omega }_2=(e_1+e_2)/2$. This long-first ordering is essential: sources that put the short simple root first interchange the two Dynkin labels.

The highest weight $(0,1)$ is the spinor representation. Its Euclidean weights are $(\pm \frac{1}{2},\pm \frac{1}{2})$, whose Dynkin labels areA lowering chain subtracts ${\alpha }_2=(-1,2)$, then ${\alpha }_1=(2,-2)$, then ${\alpha }_2$ again, passing through these four weights in the displayed order. All weight multiplicities are one.

The highest weight $(1,0)$ is the defining vector representation, with Euclidean weights $e_1,e_2,0,-e_2,-e_1$. ThusHere a lowering chain subtracts ${\alpha }_1,{\alpha }_2,{\alpha }_2,{\alpha }_1$. Each weight multiplicity is again one.

Finally $(0,2)$ is the highest root $e_1+e_2={\alpha }_1+2{\alpha }_2$, giving the Adjoint representation. Its nonzero weights are the eight roots $\pm e_1,\pm e_2,\pm (e_1-e_2),\pm (e_1+e_2)$. The Cartan subalgebra supplies zero with weight multiplicity two, soThe dimensions are $4,5,10$. Independently the long-first Weyl dimension formula iswhich gives those same values. These are Lie algebra representations; the four-dimensional spinor representation integrates to the Spin group $\mathrm{Spin}(5)$, and does not descend to the special orthogonal group $SO(5)$. A concrete obstruction is a $2\pi$ rotation: it is the identity in $SO(5)$ but acts as $-I_4$ on the spinor representation, as verified in solution.

Use upper indices for the defining representation of SU(3) and lower indices for its dual representation. A Young diagram with row lengths $({\ell }_1,{\ell }_2)$ has highest weight Dynkin labels $(p,q)=({\ell }_1-{\ell }_2,{\ell }_2)$; a full column of height three is a determinant representation and can be removed for SU(3). The Weyl dimension formula reduces toThe six irreducible representations of smallest dimensions are $1,3,\overline{3},6,\overline{6},8$. The following box arrays draw their Young diagrams; filling boxes by tensor-slot labels gives representative Young tableaux:Their associated tensors, in the same order, can be taken as follows:

These low-dimensional SU3 tensor representations have dimensions $1,3,3,6,6,8$ by direct component counting. To check that none were missed, if $p+q\ge 3$ then $(p+1)(q+1)\ge p+q+1$, so the displayed dimension is at least $(p+q+1)(p+q+2)/2\ge 10$. The pairs with $p+q\le 2$ are precisely the six listed pairs.

The complexification of a Lie algebra gives $\mathfrak{su}(4{)}_{\mathbb{C}}={\mathfrak{sl}}_4(\mathbb{C})$, of type A3 root system. Choose simple roots ${\alpha }_1=e_1-e_2$, ${\alpha }_2=e_2-e_3$, ${\alpha }_3=e_3-e_4$, with squared lengths two. The Cartan matrix and Dynkin diagram areEach line represents a single bond. In the Dynkin label convention $w_a=⟨w,{\alpha }_a^{\vee }⟩$, the simple roots have coordinates equal to the rows of $A$.

In the defining representation, apply the simple coroots $E_{aa}-E_{a+1,a+1}$ to the four coordinate basis vectors. Their weights areEquivalently these are obtained by subtracting ${\alpha }_1,{\alpha }_2,{\alpha }_3$ successively from the highest weight. The dual representation has the negative weights, henceAll listed weight multiplicities are one, so both dimensions are four.

The highest weight $(1,0,1)$ is ${\alpha }_1+{\alpha }_2+{\alpha }_3$, the highest root. Its irreducible representation is therefore the Adjoint representation. Under the action on traceless matrices by conjugation, the off-diagonal matrix unit $E_{ij}$ has weight $v_i-v_j$, while the diagonal Cartan subalgebra has zero weight. ThusThe superscript $[3]$ indicates weight multiplicity three. The twelve nonzero weights each have weight multiplicity one, giving $12+3=15$ in total. This also follows from the tensor product representation $4\otimes \overline{4}=15\oplus 1$: the sixteen differences $v_i-v_j$ contain four zero weights, and removing the identity matrix removes one of them.

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.

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

For $SU(4)$, Dynkin labels $(a,b,c)$ correspond to a Young diagram with row lengths $(a+b+c,b+c,c)$. Numbered boxes below give representative Young tableaux; they specify the shapes, not additional component values:Here the dimensions are $4$, $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$, $\left(\genfrac{}{}{0ex}{}{4}{3}\right)=4$, and $4^2-1=15$. The third is the dual representation $\overline{4}$: the invariant volume tensor identifies $b_l={ϵ}_{lijk}{B}^{ijk}/3!$. The last is the Adjoint representation, equivalent to the shape $(2,1,1)$ Young symmetrizer construction after converting its three-box column to one lower index and removing the trace.

For $U\in SU(4)$, the tensor transformation laws areThese preserve antisymmetry and tracelessness. The Adjoint representation can be treated as complex traceless matrices; a real adjoint scalar field is conventionally represented by Hermitian traceless matrices with the same conjugation action.

Adding one box to the one-box Young diagram gives row shapes $(2)$ and $(1,1)$. Thus the first tensor product splits into the symmetric square and exterior square:Their tensors are respectively $S^{ij}=S^{ji}$ and $A^{ij}=-A^{ji}$. Adding one box to the three-box column gives shapes $(2,1,1)$ and $(1,1,1,1)$. The full four-box column is the determinant representation, trivial because $\det U=1$. ConsequentlyFor a mixed tensor $T^i{}_j$, the corresponding pieces are $T^i{}_j-\frac{1}{4}{\delta }^i{}_j{T}^k{}_k$ and $\frac{1}{4}{\delta }^i{}_j{T}^k{}_k$.

Normalize the simple roots to $({\alpha }_i,{\alpha }_i)=2$. The A2 root system has $({\alpha }_1,{\alpha }_2)=-1$, so its Dynkin diagram is two nodes joined by one bond:The fundamental weights satisfy $({\omega }_i,{\alpha }_j)={\delta }_{ij}$. Hence the Gram matrix in the fundamental weight basis, the metric for Dynkin label coordinates, isFor weights of a representation with Dynkin labels $a$ and $b$, their inner product is $a^{T}Gb$. In the simple root basis the metric is instead $A$ itself; stating the basis removes this apparent convention ambiguity.