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The formal character identity ${\chi }_{{\wedge }^{2}U}(x)=\frac{1}{2}({\chi }_U(x{)}^2-{\chi }_U(x^2))$ counts unordered pairs of distinct basis vectors. Substituting ${\chi }_{2,1}=h_2(x)h_1(x^{-1})-h_1(x)$ and the Weyl character formula gives the displayed multiplicity-free decomposition. Its dimensions sum to $21+42+24+15+3=105=\left(\genfrac{}{}{0ex}{}{15}{2}\right)$. The sl3 interlacing character formula provides an exact weight-enumeration check.

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.