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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 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 squares of $I,-I,\pm A,\pm B$ are $I$, whereas $(\pm C{)}^2=-I$. Therefore ${\chi }_{\rho }(g^2)=(2,2,2,2,-2)$. The symmetric square and exterior square formulas from solution giveComparison with the character table gives the Clebsch-Gordan decomposition of group representationsFor explicit bases, put $e_1=(1,0{)}^T,e_2=(0,1{)}^T$. The symmetric tensors $e_1\otimes e_1+e_2\otimes e_2$, $e_1\otimes e_2+e_2\otimes e_1$, and $e_1\otimes e_1-e_2\otimes e_2$ transform as $L_{++},L_{+-},L_{-+}$ respectively. The antisymmetric tensor $e_1\otimes e_2-e_2\otimes e_1$ transforms as $L_{--}$, also the determinant representation of $\rho$. Adding these direct sums recovers $\rho \otimes \rho ={⨁}_{s,t}{L}_{st}$.

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

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 $[r_1,r_2,\dots ]$ for a Young diagram with row lengths $r_i$, and depict an empty box by $\square$. Appending the box of the fundamental representation to another single box gives a row or a column:The corresponding symmetric tensors and antisymmetric tensors have dimensions $4\cdot 5/2=10$ and $4\cdot 3/2=6$, soFor the conjugate representation, the invariant volume tensor identifies $\bar{4}\cong {\Lambda }^{3}4$, represented by a column of three boxes. Adding one box givesA column of four boxes is the determinant representation, which is trivial on $SU(4)$. The remaining Young diagram has Dynkin labels $(1,0,1)$ and dimension fifteen. For example the hook-content formula givesEquivalently, a mixed tensor $M^i{}_j$ transforms by $M\mapsto UM{U}^{†}$ and splits into its trace and traceless matrix parts. Thereforewhere $15$ is the Adjoint representation.

Use the standard chain ordering of the five simple roots of $SU(6)$. A Young diagram of row lengths $r_1\ge \cdots \ge r_6=0$ has highest weight Dynkin labels $a_i=r_i-r_{i+1}$. A full column of six boxes can be removed because it is the trivial determinant representation. The required data areIn the same order, the Young tableaux areEach upper index transforms by $U$, and each lower index by $U^{-1}$. In particular $M\mapsto UM{U}^{-1}$ is the complex Adjoint representation. The column of five boxes can be represented either by $w_i$ or by the totally antisymmetric tensor $B^{i_1\dots i_5}$, using the invariant six-index alternating tensor:Thus it really is the conjugate representation, rather than a second independent six-dimensional choice. Similarly converting the lower index of $M^i{}_j$ into five alternating upper indices gives the displayed hook Young diagram; the removed trace would be its full six-box column component.

The dimension counts areThe first two count independent entries of symmetric tensors and antisymmetric tensors; the last removes one trace from a mixed tensor. The hook-content formula independently gives these dimensions from the Young diagrams.