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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:

The integers called Dynkin indices here are the Dynkin labels of the highest weight:For finite-dimensional Irreducible Lie algebra representations, they are nonnegative integers. All weight coordinates below are these Dynkin label coordinates, not simple-root coordinates.

A weight-string enumeration algorithm gives the set of weights without their weight multiplicities. Begin with $\Lambda$ and process known weights by increasing height below $\Lambda$. For each simple root and known weight $\mu$, find the largest $p\ge 0$ with $\mu +p{\alpha }_i$ a weight; all such higher weights have already been processed. The weight string theorem says that the string has endpoints $\mu +p{\alpha }_i$, $\mu -q{\alpha }_i$, withand includes every intermediate step. Append $\mu -{\alpha }_i,\dots ,\mu -q{\alpha }_i$ and repeat until no new weights appear. This terminates in finite dimension and supplies all weights; every nonhighest weight can be reached by simple-root lowering. A string can contain contributions from several sl2 summands, so the resulting set does not by itself determine weight multiplicities.

For the explicit calculation we use the following general facts: finite-dimensional representations of a complex semisimple Lie algebra are completely reducible by the Weyl complete reducibility theorem; the symmetric powers of the defining sln representation are irreducible of highest weight $m{\omega }_1$; and weights in a tensor product of Lie algebra representations add with multiplicities multiplied. We also use the highest-weight representation classification: each finite-dimensional irreducible has a unique dominant integral highest weight, and a nonzero vector killed by all simple-root raising operators supplies an irreducible summand with that highest weight in a completely reducible module. For $A_2$, the Weyl dimension formula specializes toFor $A_2\cong {\mathfrak{sl}}_3(\mathbb{C})$, the defining fundamental representation has weights $(1,0),(-1,1),(0,-1)$. Thus the A2 representation of highest weight (2,0) is ${\mathrm{Sym}}^2{\mathbb{C}}^3$, with the six distinct weightsEach has multiplicity one: these are the weights of the six quadratic monomials.

For the tensor square of the A2 representation of highest weight (2,0), add every ordered pair of elements of $W_6$. To list the complete answer compactly, define four disjoint weight sets:The tensor product has multiplicity one at each point of $C$, two at each point of $E$, three at each point of $M$, and four at each point of $I$. These fifteen distinct weights account for $3+12+9+12=36$ states.

We next identify the irreducible summands rather than just their dimensions. Split the tensor square of the six-dimensional space into its symmetric power and exterior power, of dimensions $21$ and $15$. The square of a highest-weight vector in the symmetric part has highest weight $(4,0)$, giving $R(4,0)$ of dimension $15$.

Let $v_a,v_b$ be vectors of weights $a=(2,0)$ and $b=(0,1)=a-{\alpha }_1$. In the exterior part $v_a\wedge v_b$ has weight $(2,1)$ and is killed by both simple-root raising operators: raising $v_b$ along ${\alpha }_1$ gives a multiple of $v_a$, whose wedge with itself is zero, and the other raising actions vanish. Thus it is a highest-weight vector. The Weyl dimension formula gives $\dim R(2,1)=15$, so the entire exterior part is this irreducible summand.

The remaining symmetric part has dimension six. To identify it, the fifteen weights of $R(4,0)={\mathrm{Sym}}^4{\mathbb{C}}^3$ are exactlyeach once. Subtract them from the unordered-pair weights of ${\mathrm{Sym}}^{2}R(2,0)$. The residual weights are $M\cup I$, each once, with highest weight $(0,2)$. They are the negatives of $W_6$, so this six-dimensional summand is $R(0,2)$. ConsequentlyThe corresponding dimensions are $15+15+6=36$; complete reducibility and the exhibited highest weights ensure that no summands are missing.

The full weight multiplicities in the A2 tensor square of highest weight (2,0) are summarized below. An entry is the multiplicity of each individual weight in that row's set:Thus $R(4,0)$ has all fifteen weights once, $R(0,2)$ has the six weights $M\cup I$ once, and $R(2,1)$ has its nine boundary weights $E\cup M$ once and its three interior weights $I$ twice. Only the three interior weights of $R(2,1)$ are degenerate among the irreducible components.