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In the finite-dimensional setting, let $R(a,b)$ denote the irreducible ${\mathfrak{sl}}_3$ highest-weight representation of highest weight $a{L}_1-b{L}_3$, where $a,b$ are nonnegative integers. Thus $(a,b)$ are its Dynkin labels. The target $\lambda =L_1-2L_3$ has labels $(1,2)$. Write the highest-weight difference in simple roots asThe dominant weight multiplicity formula for sl3 gives zero unless $r,s$ are nonnegative integers, and otherwise givesFor completeness, this multiplicity is an actual count. In the sl3 interlacing character formula, use top row $(a+b,b,0)$ and bottom entry $s+3$; the target diagonal exponents are $(s+3,s+2,s)$. The middle row $(P,Q)$ obeys $P+Q=2s+5$ andEach allowable integral $P$ gives one Gelfand–Tsetlin basis vector. Subtracting the lower bound from the upper bound gives the minimum of $a,s,r,r+3,b,s+2$, which reduces to $\min (a,b,r,s)$. This proves the displayed multiplicity formula in the case at hand.

Since both summands have a nonzero weight space and their dimensions at $\lambda$ sum to three, their multiplicities must be one and two. For multiplicity one, the minimum is zero. The cases $a=0$ and $b=0$ give $(0,4+3j)$ and $(5+3j,0)$; the remaining cases $r=0$ or $s=0$ give $(1,2)$ and $(3,1)$. Hence the full family isFor multiplicity two, the minimum is one. The cases $a=1$ and $b=1$ give $(1,5+3j)$ and $(6+3j,1)$; the remaining cases $r=1$ or $s=1$ give $(2,3)$ and $(4,2)$. ThusThe minimum conditions exhaust all cases and each listed module has the stated multiplicity. All candidate unordered pairs are $\boxed{\{U,V\}\text{ with }U\in {\mathcal{C}}_1,V\in {\mathcal{C}}_2}$. There are infinitely many at this stage; the question imposes no upper bound on their highest weights.

Use the ${\mathfrak{sl}}_3$ weight of a representation convention from Question 1: $L_1+L_2+L_3=0$, with simple roots $L_1-L_2$ and $L_2-L_3$. This identification is necessary: the general preceding setup does not by itself specify which root system the symbols $L_i$ refer to.

The Weyl group is $S_3$, acting by permutations of the $L_i$. Let $s_1=(12)$ and $s_2=(23)$, with composition read from right to left. ThenHenceFor direct verification, the diagonal coefficients of $\mu$ are $(1,0,-2)$, so its two Dynkin labels are $1-0=1$ and $0-(-2)=2$, both strictly positive. Thus $\mu$ is in the open Weyl chamber. The original three coefficients $(-2,1,0)$ are distinct, so its stabilizer is trivial and this permutation is unique.

The commuting Hermitian operators $H_1,H_2$ admit a weight-space decomposition. The six root vectors shift their simultaneous eigenvalues byor their negatives. Put $h_i=2{\alpha }_i\cdot H$, $e_i=\sqrt{2}{E}_{+}^i$ and $f_i=\sqrt{2}{E}_{-}^i$. The supplied relations giveThese are three sl2 triples, coming from the complexifications of the three SU(2) Lie algebra root-pair subalgebras. They act through homomorphisms into $\mathrm{End}(V)$. For a nontrivial representation, simplicity of the SU(3) Lie algebra makes its kernel zero, so these images are actual ${\mathfrak{sl}}_2(\mathbb{C})$ subalgebras; a trivial representation gives zero actions. The real compact generators can be taken as $i{h}_i$, $i(E_{+}^i+E_{-}^i)$ and $E_{+}^i-E_{-}^i$.

Decomposing under any one of these subalgebras gives finite weight strings in its root direction. The two simple-root highest weights must be nonnegative integers. In the SU(3) highest-weight coordinates used here, this condition isHere $(a,b)$ are the Dynkin labels; $(p,q)$ are the eigenvalues of $H_1,H_2$. An irreducible representation is determined by these labels and has dimension $(a+1)(b+1)(a+b+2)/2$. An arbitrary finite-dimensional anti-Hermitian representation is a direct sum of such representations: the orthogonal complement of an invariant subspace remains invariant.

The weight diagram has Weyl group symmetry, namely reflections and rotations by $120$ degrees. Its convex boundary is a hexagon with alternating side lengths $a,b$, degenerating to a triangle when one label vanishes. Generic representations need not have $60$-degree symmetry. Boundary weight multiplicities are one. Moving inward through hexagonal layers increases multiplicity by one, up to $\min (a,b)+1$; triangular inner layers retain that maximum. An exact version of this rule is the dominant weight multiplicity formula for sl3: move a weight by the Weyl group to dominant Dynkin coordinates $(c,d)$ and setIts multiplicity is zero unless $r,s$ are integers, and otherwise is $\max \{0,1+\min (a,b,r,s)\}$. These rules construct the diagram while counting coincident weights correctly.

For the Casimir calculation, set $t_3=H_1$, $t_8=H_2$, and for each root pair use $(E_{+}+E_{-})/\sqrt{2}$ and $(E_{+}-E_{-})/(i\sqrt{2})$. In the defining representation these eight generators satisfy $\mathrm{tr}(t_a{t}_b)={\delta }_{ab}/2$. Their summed squares are precisely the question's $X$. The invariant trace form makes their structure constants of a Lie algebra antisymmetric, sothe antisymmetric coefficient contracts a symmetric anticommutator. Thus $X$ is a quadratic Casimir operator and is scalar on an irreducible representation by the Schur lemma.

Let $v$ be a highest-weight vector, annihilated by all $E_{+}^i$. On $v$, each $E_{-}^i{E}_{+}^i$ vanishes and each $E_{+}^i{E}_{-}^i$ reduces to the root commutator. Summing those three commutators gives $H_1+\sqrt{3}{H}_2$. Henceand therefore the SU(3) quadratic Casimir eigenvalue isFor a reducible representation the same calculation applies separately to its irreducible blocks.

For light-hadron flavour symmetry, the up quark, down quark and strange quark form $3$, and antiquarks form $\overline{3}$. Conventional mesons have $3\otimes \overline{3}=1\oplus 8$. Conventional three-quark baryons haveobtained from $3\otimes 3=6\oplus \overline{3}$, $6\otimes 3=10\oplus 8$ and $\overline{3}\otimes 3=8\oplus 1$. Their Dynkin labels are $(0,0)$, $(1,1)$ and $(3,0)$, respectively. Thus the conventional meson octet, baryon octet, baryon decuplet and singlet multiplets give $\boxed{X=0,3,6}$, with the same values for conjugate antibaryon multiplets. More generally, the zero flavour triality of a light-quark hadron imposes $a+2b\equiv 0(\mathrm{mod}3)$ on multiplets constructed entirely from light quarks, antiquarks and gluons. Their Casimir values are given by the same formula for these labels. Higher multiquark representations can therefore have other eigenvalues; the illustrated $27$, of labels $(2,2)$, has $X=8$.

Start with a nonzero highest-weight vector $v_0=|n⟩$ and set $v_r=E_{-}^r{v}_0$. The sl2 highest-weight lowering formula follows inductively from the commutator relations:Indeed $[H,E_{-}]=-2E_{-}$ gives the weight formula, and $E_{+}{E}_{-}=E_{-}{E}_{+}+H$ turns the coefficient $r(n-r+1)$ at step $r$ into $(r+1)(n-r)$ at step $r+1$. The nonzero $v_r$ have distinct $H$ eigenvalues, so are linearly independent. Finite dimensionality therefore forces a last nonzero vector $v_N$. Applying $E_{+}$ to $v_{N+1}=0$ gives $(N+1)(n-N)v_N=0$, hence $N=n$. This also proves that no earlier vector can vanish. Thus the cyclic highest-weight representation generated by $|n⟩$ isIt is irreducible: spectral projections in $H$ select a weight vector from any nonzero invariant subspace, and its nonzero raising and lowering coefficients then generate the whole string. This gives the classification of finite-dimensional sl2 representations used for each root direction below.

For the SU(3) Lie algebra, the repeated-index relation is the trace constraint ${\sum }_i{T}^i{}_i=0$. The displayed generators are a basis of its complexification, not nine independent generators or three individually vanishing diagonal elements. Realize them by matrix units as $T^i{}_j=E_{ij}-{\delta }^i{}_{j}I/3$; the compact real algebra is obtained from anti-Hermitian combinations. Thenand $E_{1+}=E_{12}$, $E_{1-}=E_{21}$, $E_{2+}=E_{23}$, $E_{2-}=E_{32}$. Substituting into the given commutators givesThus both root pairs obey the required sl2 Lie algebra relations. More generally, their Cartan matrix isA weight vector with eigenvalues $(h_1,h_2)$ moves by $(-2,1)$ under $E_{1-}$ and by $(1,-2)$ under $E_{2-}$. The third negative-root operator is $[E_{2-},E_{1-}]=E_{31}$ and moves it by $(-1,-1)$. In the required weight diagram coordinatesthese moves are $(-2,0)$, $(1,-\sqrt{3})$ and $(-1,-\sqrt{3})$ respectively. Successive lowerings from a highest-weight vector, with linear dependencies removed, construct the following bases.

For Dynkin labels $(1,0)$, use the defining action of SU(3) on ${\mathbb{C}}^3$. Its highest vector is $e_1$; lowering gives $e_2=E_{21}{e}_1$ and $e_3=E_{32}{e}_2$. There are no further independent vectors. The weight diagram of the defining SU(3) representation, in the present rescaled coordinates, isand $\boxed{\dim R(1,0)=3}$. Any invariant weight subspace raises to $e_1$ and lowers to all three vectors, proving irreducibility.

For $(1,1)$, use the adjoint representation of SU(3) on traceless matrices, with $X$ acting by $Y\mapsto [X,Y]$. The highest vector is $E_{13}$: both $[E_{12},E_{13}]$ and $[E_{23},E_{13}]$ vanish, while $[H_i,E_{13}]=E_{13}$. Lowering gives $[E_{21},E_{13}]=E_{23}$ and $[E_{32},E_{13}]=-E_{12}$. Further lowering produces $[E_{32},E_{23}]=-H_2$ and $[E_{21},E_{12}]=-H_1$, then $[E_{21},H_1]=2E_{21}$ and $[E_{32},H_2]=2E_{32}$, and finally $[E_{32},E_{21}]=E_{31}$. Consequently the basis is the six off-diagonal matrix units together with $H_1,H_2$:The six nonzero weights have weight multiplicity one, and zero has multiplicity two. Thus $\boxed{\dim R(1,1)=6+2=8}$. Its irreducibility follows also because an adjoint-invariant subspace is an ideal and ${\mathfrak{sl}}_3(\mathbb{C})$ is a simple Lie algebra.

For $(3,0)$, take the symmetric cubic representation of SU(3), ${\mathrm{Sym}}^3{\mathbb{C}}^3$, realized by homogeneous degree-three polynomials. Hereso $z_1^3$ is a highest-weight vector with eigenvalues $(3,0)$. For $a+b+c=3$,which is nonzero and constructs every monomial. These monomials form a basis and have distinct weights $(a-b,b-c)$. If an invariant subspace is nonzero, joint spectral projections in $H_1,H_2$ select one monomial; repeated $E_{2+}$ transfers its $z_3$ powers to $z_2$, and repeated $E_{1+}$ transfers all its $z_2$ powers to $z_1$, reaching $z_1^3$. Lowering then generates the whole basis, proving irreducibility without merely assuming the dimension formula. The plotted weights arewith multiplicity one at every point. Hence $\boxed{\dim R(3,0)=\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10}$.

For approximate flavor symmetry, identify $e_1,e_2,e_3$ with the up quark, down quark and strange quark. The defining representation is their flavour triplet. The baryon octet realizes $R(1,1)$ and contains the spin-$1/2$ nucleons and the $\Lambda ,\Sigma ,\Xi$ multiplets; its central double weight distinguishes $\Lambda$ and ${\Sigma }^0$. The baryon decuplet realizes the symmetric flavour representation $R(3,0)$, containing the spin-$3/2$ $\Delta ,{\Sigma }^{\ast },{\Xi }^{\ast },\Omega$ multiplets. In standard isospin and flavor hypercharge normalization, $I_3=h_1/2$ and $Y=(h_1+2h_2)/3$, so these plots use $x=2I_3$ and $y=\sqrt{3}Y$. Three-quark spin and flavour combine subject to the Pauli constraint on three-quark flavour multiplets and the antisymmetric three-quark colour singlet. The adjoint flavour representation also describes a meson octet. A distinct, exact colour SU(3) gauge symmetry places quarks in its defining representation and gluons in its adjoint octet; colour and flavour here are different physical group actions.

Use Dynkin labels $(a,b)$ for the highest weight $a{\omega }_1+b{\omega }_2$ of the complex special linear Lie algebra ${\mathfrak{sl}}_3$. The A2 root system has ${\alpha }_1=(2,-1)$ and ${\alpha }_2=(-1,2)$ in these coordinates. In the drawings, ${\omega }_1$ and ${\omega }_2$ have equal lengths and angle $60^{\circ }$; a label at a point records its weight multiplicity, not a further copy at a different position.

The defining fundamental representation has the three weightsFor ${\Gamma }_{2,1}$, lower from its highest weight by the simple roots, retaining multiplicities. One convenient way to calculate them is the sl3 interlacing character formula: for shape $(3,1,0)$ the integer patterns satisfy $1\le p\le 3$, $0\le q\le 1$, $q\le r\le p$, and contribute the weightEnumerating these patterns gives the weight diagramIts dimension is $3\cdot 2+9=15$. The diagram below draws all twelve distinct positions, with the three inner multiplicities equal to two. The extra panel gives the symmetric square used in the calculation.

The six symmetric monomials in the defining basis give $S^2{\Gamma }_{1,0}={\Gamma }_{2,0}$, with weightseach occurring once. Thus the tensor product has dimensionIn a tensor product of Lie algebra representations, weights add and their multiplicities multiply. In terms of formal characters, $\mathrm{ch}V=\mathrm{ch}{\Gamma }_{2,1}\mathrm{ch}{\Gamma }_{2,0}$. Consequently $m_V(\mu )={\sum }_{\nu }{m}_{2,1}(\mu -\nu )$, summing over the six weights just listed. To show the indicated dominant multiplicities explicitly, the contributions in that order areThe tensor-product weight diagram below includes every position, and highlights these dominant weights. It also records the zero-weight multiplicity nine; that multiplicity is not a count of trivial summands.

Apply the Weyl complete reducibility theorem and subtract irreducible formal characters in decreasing dominance order. The multiplicities at these five dominant positions in the potential summands areThese entries can be obtained by the same interlacing enumeration or by weight strings. Starting with $(1,2,4,3,7)$, subtracting ${\Gamma }_{4,1}$ leaves $(0,1,2,2,5)$; subtracting ${\Gamma }_{2,2}$ leaves $(0,0,1,1,3)$; then the two ten-dimensional modules leave a single copy of the dominant weight $(1,1)$. This is highest-weight character subtraction. ThereforeEvery summand occurs once. The Weyl dimension formula gives $35+27+10+10+8=90$, exhausting the dimension of $V$ and ruling out further irreducible summands. Computing the complete formal character also leaves no residual weight multiplicities.

Define the exterior square over the arbitrary field $k$ by ${\bigwedge }^{2}V=(V\otimes V)/\mathrm{span}\{v\otimes v:v\in V\}$. Write the image of $u\otimes v$ as $u\wedge v$. Then $v\wedge v=0$ and $u\wedge v=-v\wedge u$; the basis is $v_i\wedge v_j$ with $i<j$, in characteristic two as well. The exterior-power Lie algebra representation isThe tensor product of Lie algebra representations descends to this quotient: $x\cdot (v\otimes v)=xv\otimes v+v\otimes xv$ is a linear combination of square tensors, namely $(v+xv)\otimes (v+xv)-v\otimes v-xv\otimes xv$. On the wedge basis its coefficients follow directly from the original representing matrices. Expanding both actions shows $[{\rho }_{{\wedge }^2}(x),{\rho }_{{\wedge }^2}(y)]={\rho }_{{\wedge }^2}([x,y])$.

The inclusions of $V$ and $W$ into their direct sum define the mapA combined basis shows that this sends a basis to the pure-$V$, pure-$W$ and mixed wedge basis vectors, with no overlap and no omission. The defining action on each wedge proves equivariance. Hence the exterior square of a direct sum givesNo division by $2$ is involved, so the proof remains valid in characteristic two.

For the complex special linear Lie algebra ${\mathfrak{sl}}_3$, write ${\Gamma }_{a,b}$ for the irreducible highest-weight representation with Dynkin labels $(a,b)$, and $V={\Gamma }_{1,0}$. First the sl3 decomposition of the symmetric-square dual tensor product isIndeed contraction $c((uv)\otimes f)=f(u)v+f(v)u$ is a surjective Lie algebra representation homomorphism onto $V$. Its 15-dimensional kernel contains the highest-weight vector $e_1^2\otimes e_3^{\ast }$ of weight $(2,1)$. The Weyl dimension formula gives $\dim {\Gamma }_{2,1}=15$, and the Weyl complete reducibility theorem identifies the kernel and splits the map. With $B={\Gamma }_{2,1}$, the direct-sum identity reduces the requested calculation to ${\bigwedge }^{2}B$, $B\otimes V$ and ${\bigwedge }^{2}V$.

For completeness, these decompositions can be checked entirely by formal characters. Let $x_1{x}_2{x}_3=1$, $h_1=x_1+x_2+x_3$, $h_1^{\ast }=x_1^{-1}+x_2^{-1}+x_3^{-1}$ and $h_2={\sum }_i{x}_i^2+{\sum }_{i<j}{x}_i{x}_j$. Then ${\chi }_B=h_2{h}_1^{\ast }-h_1$. The Weyl character formula takes the determinant formSubstitute ${\chi }_B$ into ${\chi }_{{\wedge }^{2}B}(x)=\frac{1}{2}({\chi }_B(x{)}^2-{\chi }_B(x_1^2,x_2^2,x_3^2))$ and collect the determinant characters. This gives the exterior square of the sl3 representation of highest weight (2,1) and the sl3 highest-weight tensor rule:The first line has dimensions $21+42+24+15+3=105=\left(\genfrac{}{}{0ex}{}{15}{2}\right)$; the tensor product has dimension $24+15+6=45$. Equivalently, enumerate weights with the sl3 interlacing character formula and subtract characters from the highest weight downwards. FinallyIts dimension is $21+42+48+30+6+6=153=\left(\genfrac{}{}{0ex}{}{18}{2}\right)$. The exterior square here is taken of the entire 18-dimensional tensor product; taking it only of $S^{2}V$ would be a different representation.

Start with the tensor product of Lie algebra representations, whose action isWriting $T_x=\rho (x)\otimes I+I\otimes \rho (x)$, the two tensor factors commute, so $[T_x,T_y]=T_{[x,y]}$. This verifies the Lie algebra representation identity over every field.

The exterior square and symmetric square are the quotient vector spacesIn the exterior square, expanding $(v+w)\otimes (v+w)$ shows that $v\wedge w=-w\wedge v$, including in characteristic two. Both defining relation spaces are invariant under the tensor product action: $x(v\otimes v)=xv\otimes v+v\otimes xv$ is an exterior relation, and the image of a symmetric relation is a sum of symmetric relations. Thus the quotient actions are well-defined and satisfyFor the printed basis $v_1,\dots ,v_n$, bases are $v_i\wedge v_j$ with $i<j$ and $v_i{v}_j$ with $i\le j$. Their dimensions are $n(n-1)/2$ and $n(n+1)/2$ respectively.

If $2$ is invertible in $k$, $V\otimes V\cong {\bigwedge }^{2}V\oplus S^{2}V$ as representations. Define the flip $\tau (v\otimes w)=w\otimes v$. It commutes with the Lie algebra action and satisfies ${\tau }^2=I$. Thereforeare complementary invariant linear projections. The mapsidentify $S^{2}V$ with $\mathrm{im}{P}_{+}$ and ${\bigwedge }^{2}V$ with $\mathrm{im}{P}_{-}$. Their inverses are the corresponding quotient maps restricted to these subspaces. This proves the assertion for every field of odd characteristic, and also for characteristic zero.

Over every field, $S^2(V\oplus W)\cong S^{2}V\oplus S^{2}W\oplus (V\otimes W)$. The symmetric square of a direct sum isomorphism sends the first two summands into products within $V$ and within $W$, and sends $v\otimes w$ to the mixed product $vw$. If $v_i$ and $w_j$ are bases, the monomial basis of $S^2(V\oplus W)$ is the disjoint unionThus the map is bijective, with no division by $2$ needed. The Leibniz rule for the action preserves each of these three summands and agrees with its usual Lie algebra representation action, proving equivariance.

For $\mathfrak{g}=\mathfrak{sl}(3)$, $V={\Gamma }_{0,0}$ is trivial and $W={\Gamma }_{2,1}$ has dimension $15$. HenceIt remains to find the irreducible representations in the symmetric square of the sl3 representation of highest weight (2,1). We give the formal character calculation explicitly.

Let $E={\mathbb{C}}^3$ be the defining special linear Lie algebra representation. In Dynkin labels, its weights are $(1,0)$, $(-1,1)$, and $(0,-1)$; the dual representation has their negatives. The equivariant contractionis surjective. Its kernel has dimension $18-3=15$. The tensor $e_1^2\otimes e_3^{\ast }$ is a highest-weight vector of highest weight $(2,1)$ in that kernel. By the Weyl complete reducibility theorem, the kernel contains the irreducible representation ${\Gamma }_{2,1}$, whose Weyl dimension formula gives dimension $15$; therefore the kernel equals ${\Gamma }_{2,1}$. This yieldsMultiplying the six weights of $S^{2}E$ by the three weights of $E^{\ast }$ and subtracting those of $E$ gives the following full weight multiplicity list:The multiplicities sum to $9+2\cdot 3=15$.

For any finite-dimensional weight-space decomposition, a weight $\mu$ of multiplicity $m_{\mu }$ contributes $m_{\mu }(m_{\mu }+1)/2$ to weight $2\mu$ in its symmetric square. Distinct weights $\mu ,\nu$ contribute $m_{\mu }{m}_{\nu }$ to $\mu +\nu$. Equivalently,Applying this to the displayed list gives all dominant weight multiplicities in the second column below. The remaining columns are the weight multiplicities of the candidate irreducible representations:For an explicit way to compute each irreducible column, set $z_1{z}_2{z}_3=1$ and use the Weyl character formula in the formA monomial $z_1^{n_1}{z}_2^{n_2}{z}_3^{n_3}$ has Dynkin labels $(n_1-n_2,n_2-n_3)$. Equivalently, the quotient is enumerated by Semistandard Young tableaux of shape $(a+b,b)$ with entries $1,2,3$, weakly increasing across rows and strictly increasing down columns; the exponents count the three entries.

The five irreducible columns sum to the $S^{2}W$ column. These are all its dominant weights, and all five candidate characters have no other dominant weights. Every Weyl group orbit meets the dominant chamber, and weight multiplicities are constant on Weyl group orbits. Thus the table proves equality of the full formal characters, and the Weyl complete reducibility theorem givesThe Weyl dimension formula checks the result:Consequently the requested decomposition isIts total dimension is $1+15+120=136=16\cdot 17/2$.

Let ${\alpha }_i^{\vee }=2{\alpha }_i/({\alpha }_i,{\alpha }_i)$ be the coroots. With a chosen set of simple roots, the root lattice and weight lattice areThe fundamental weights ${\omega }_j$ are defined by $⟨{\omega }_j,{\alpha }_i^{\vee }⟩={\delta }_{ij}$, and form an integral basis of $P$. We use the Cartan matrix convention $A_{ij}=⟨{\alpha }_j,{\alpha }_i^{\vee }⟩$.

For the A2 root system, inversion of its Cartan matrix yields the A2 fundamental weights and weight lattice:Equivalently ${\alpha }_1=2{\omega }_1-{\omega }_2$ and ${\alpha }_2=-{\omega }_1+2{\omega }_2$. The root lattice has index three in the weight lattice, since the change-of-basis matrix has determinant three. For a planar realization takeThe weight lattice is triangular. In fundamental weight coordinates $(a,b)$, a point is in the root lattice exactly when $a-b$ is divisible by three. The following sketch marks both bases and the sublattice: