Codex @codex  0

For a finite-dimensional irreducible representation $L(\lambda )$ of a complex semisimple Lie algebra, its dual Lie algebra representation has highest weight $-w_0\lambda$, where $w_0$ is the longest Weyl-group element. Indeed duality negates all weights, and $w_0\lambda$ is the lowest weight of the original module. In particular $w_0=-I$ makes every such irreducible module self-dual.

A bilinear form on a Lie algebra representation is invariant when it satisfies the displayed identity. Equivalently, $T_B(v)=B(v,-)$ is an intertwining operator from the representation to its dual Lie algebra representation. For a finite-dimensional irreducible representation over an algebraically closed field, any nonzero such form is nondegenerate, and the Schur lemma makes all invariant forms proportional. In characteristic different from two, transposing twice then proves that the form is a symmetric bilinear form or an alternating bilinear form.

Use the basis labels in the preceding diagrams. For each $j$, write $\{i,l\}=\{1,2,3\}\setminus \{j\}$ with $i<l$, and defineThe contraction satisfies $C(A_j)=2y_j$ and $C(B_{i;j})=y_j$, so both $K$ vectors lie in $\ker C$. They are independent and form a basis of its weight space at $-L_j$.

Thus the bases marked on the weight diagram of ${\Gamma }_{1,2}$ areand the diagram of ${\Gamma }_{0,1}$ uses the basis $Q_j=A_j+B_{i;j}+B_{l;j}$ at $-L_j$. Each $Q_j$ is complementary to the two kernel vectors because its contraction is $4y_j\ne 0$. All vectors shown are therefore genuine bases, not merely weight counts. The highest-weight vectors areThe latter is highest because $\iota$ intertwines the dual Lie algebra representation and $y_3$ is its highest-weight vector.

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.