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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 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 $Y=X_c/6$.

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.