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For $V=L({\omega }_1)$ with the G2 root system numbered short-first,In the chain basis of the crystal of the seven-dimensional G2 representation, normalize $v_1=f_1{v}_0$, $v_2=f_2{v}_1$, $v_3=f_1{v}_2$. Then $v_0\wedge v_1$ and $v_0\wedge v_3-2v_1\wedge v_2$ are nonzero highest-weight vectors of weights ${\omega }_2$ and ${\omega }_1$; the identities $e_1{v}_1=v_0$, $e_1{v}_3=2v_2$, $e_2{v}_2=v_1$ verify both raising conditions. The Weyl complete reducibility theorem and G2 dimension polynomial exhaust the twenty-one dimensions of the exterior square. In the symmetric square, $v_0^2$ generates the twenty-seven-dimensional summand. Self-duality supplies a Lie-invariant bilinear form, which is symmetric because a nondegenerate alternating bilinear form cannot have odd dimension; its inverse gives the remaining invariant line.

In the irreducible sl2 Lie algebra module $V(m)$, use $v_j=f^j{v}_0$, with $0\le j\le m$. The Lie-invariant bilinear form $B(v_j,v_k)=(-1{)}^j{1}_{\{j+k=m\}}$ is nondegenerate: its anti-diagonal is nonzero. Invariance follows from $h{v}_j=(m-2j)v_j$, $f{v}_j=v_{j+1}$ and $e{v}_j=j(m-j+1)v_{j-1}$. Its transpose equals $(-1{)}^{m}B$, so it is a symmetric bilinear form when $m$ is an even number, and an alternating bilinear form when $m$ is an odd integer. The Weyl complete reducibility theorem gives a nondegenerate invariant form on any finite-dimensional module by taking the orthogonal direct sum of these forms.

Use the classification of finite-dimensional sl2 representations. In the irreducible representation $V(m)$, choose $v_j=f^j{v}_0$, $0\le j\le m$, sowith vectors beyond the endpoints interpreted as zero. The invariant form on an irreducible sl2 module isIts anti-diagonal entries are nonzero, so it is nondegenerate. Lie-invariant bilinear form invariance under $h$ follows from the sum of the two weights. For $f$, the two potentially nonzero terms are $(-1{)}^{j+1}$ and $(-1{)}^j$. For $e$, their coefficients coincide when $j+k=m+1$, and their signs are opposite. These checks prove invariance under the generators of the sl2 Lie algebra.

Interchanging $j,k$ multiplies the form by $(-1{)}^m$, givingEquivalently it is a symmetric bilinear form in odd integer dimension and an alternating bilinear form in even number dimension. The Weyl complete reducibility theorem expresses any finite-dimensional representation as a direct sum of these irreducibles. Give each summand the displayed form and make different summands orthogonal. The resulting form is invariant and nondegenerate. On a reducible representation with both parities, this orthogonal sum need not itself be symmetric or alternating; the dichotomy in the preceding part required irreducibility.

For the finite-dimensional Lie algebra representation specified in the PDF, a Lie-invariant bilinear form satisfiesThe dual Lie algebra representation has action $(x\varphi )(w)=-\varphi (xw)$, so the map $T_B:V\to V^{\ast }$ defined by $T_B(v)=B(v,-)$ is an intertwining operator. If $V$ is irreducible and $B\ne 0$, its kernel is zero. Since $V,V^{\ast }$ have equal finite dimension, $T_B$ is an isomorphism, and $B$ is a nondegenerate bilinear form.

For $B_1\ne 0$, the composition $T_{B_1}^{-1}{T}_{B_2}$ is a representation endomorphism. Over an algebraically closed field, the Schur lemma makes it scalar, provingTransposing $B$ gives another Lie-invariant bilinear form. For $B\ne 0$, write $B^T=\mu B$; transposing twice gives ${\mu }^2=1$. In field characteristic different from two, this means $\mu =1$ or $\mu =-1$. Thus the form is a symmetric bilinear form or an alternating bilinear form, respectively. The zero form has both properties. In the second case $B(v,v)=-B(v,v)$ forces $B(v,v)=0$, using the same characteristic assumption.

Choose the chain basis $v_0,\dots ,v_6$ from the preceding part, normalized by $v_1=f_1{v}_0$, $v_2=f_2{v}_1$, $v_3=f_1{v}_2$. The associated sl2 Lie algebra strings give $e_1{v}_1=v_0$, $e_1{v}_3=2v_2$ and $e_2{v}_2=v_1$; other raising actions used below vanish because their proposed weights do not occur.

In the exterior square, $v_0\wedge v_1$ is a highest-weight vector of weight $2{\omega }_1-{\alpha }_1={\omega }_2$: $e_1$ produces $v_0\wedge v_0=0$, and $e_2$ kills both factors. Alsois a nonzero highest-weight vector of weight ${\omega }_1$. Its two $e_1$ terms cancel, and its $e_2$ image is zero. The Weyl complete reducibility theorem supplies irreducible summands with those weights; their dimensions $14+7=21$ exhaust $\dim {\Lambda }^{2}V$. ThusFor the symmetric square, $v_0\otimes v_0$ is a highest-weight vector of weight $2{\omega }_1$, giving a 27-dimensional summand. The longest Weyl-group element of $G_2$ acts as minus the identity, so the highest weight of a dual representation is $-w_0{\omega }_1={\omega }_1$: $V$ is self-dual. A nonzero intertwiner $V\to V^{\ast }$ gives a nondegenerate Lie-invariant bilinear form. It is symmetric, since an alternating bilinear form cannot be nondegenerate in odd dimension. Its inverse is a nonzero invariant vector in $S^{2}V$, supplying the scalar summand. Since $27+1=28=\dim S^{2}V$,These are the exterior and symmetric squares of the seven-dimensional G2 representation; all highest-weight and dimension claims are specified rather than relying on an unstated cross-product identity.