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The root system of the complex special linear Lie algebra ${\mathfrak{sl}}_6(\mathbb{C})$ is $\{e_i-e_j:1\le i\ne j\le 6\}$ in ${\sum }_i{x}_i=0$. Its simple roots $e_i-e_{i+1}$ give a five-node An Dynkin diagram. The defining fundamental representation has six weights $w_i=e_i-\frac{1}{6}{\sum }_j{e}_j$. The dual representation has weights $-w_i$, and the two-index antisymmetric representation has fifteen weights $w_i+w_j$, $i<j$, all with weight multiplicity one.

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 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.

For a finite-dimensional complex group representation, combining inverse and conjugate transpose gives another representation: both operations reverse multiplication order. This is the complex conjugate of the algebraic dual representation in matrices. If $W$ is invariant under $D^{\#}$, its orthogonal complement is invariant under $D$, because $⟨w,D(g)v⟩=⟨D^{\#}(g^{-1})w,v⟩=0$ for $w\in W$, $v\in W^{\perp }$. Thus irreducibility is preserved. Applying the construction twice returns $D$, so the equivalence of irreducibility works in both directions for arbitrary groups in finite dimension.

For a complex group representation with matrices $R(g)$, its conjugate representation has matrices $\overline{R(g)}$ on the conjugate vector space. In a unitary representation it is equivalent to the dual representation, since $(R(g{)}^{-1}{)}^T=\overline{R(g)}$. On a diagonal maximal torus, its weights are the negatives of the original weights.

The dual of a Lie algebra representation on $V$ acts on its dual space by $(x\varphi )(v)=-\varphi (xv)$. The minus sign makes this a Lie algebra homomorphism: taking transposes reverses the order in a commutator. It is the infinitesimal version of the dual representation of a group.

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.

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

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.

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.

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 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 $3\otimes \overline{3}=8\oplus 1$.

Write a homogeneous quadratic polynomial as $p(x,y)=a{x}^2+bxy+c{y}^2$. Invariance under $B:(x,y)\mapsto (x,-y)$ forces $b=0$. Invariance under $A:(x,y)\mapsto (y,x)$ then forces $a=c$. Since $A,B$ generate the group, this proves that all homogeneous quadratic polynomial invariants areThis agrees with the single trivial representation in the symmetric square in solution; more precisely, coordinate polynomials lie in the symmetric square of the dual representation, which here is equivalent to $\rho$. If “quadratic” permits terms of smaller degree, arbitrary constants can be added. There are no invariant linear terms because $-I\in G$ negates every linear polynomial.

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.

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 $\frac{1}{4}{\delta }^i{}_j{T}^k{}_k$.

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 $Y=X_c/6$.

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.