Let $\mathfrak{g}{}$ be a Lie algebra, $V$ be an irreducible representation of $\mathfrak{g}{}$. Suppose $V$ is isomorphic to $V^*$. I'm then supposed to show that the induced bilinear form on $V$ is either symmetric or antisymmetric.

I have found the following argument which shows that this form is always antisymmetric, which doesn't seem quite right.

For any $x \in \mathfrak{g}{}$, $v,w \in V$, we have $x \cdot (v,w) =-(v,x\cdot w)$, by definition of the action of $\mathfrak{g}{}$ on $V^*$. On the other hand, $x \cdot (v,w)=(x\cdot v,w)$, since the form induces an isomorphism $V \rightarrow V^*$. Set $v=w$ to obtain $(x\cdot v,v)+(v,x\cdot v)=0$ for all $x$. Since $V$ is irreducible, $x \cdot v$ span $V$, and therefore $(w,v)+(v,w)=0$ for all $v,w \in V$, proving the form is antisymmetric.

I'd like to know if this argument is correct, which seems strange given the wording of the question, and if it isn't I'd appreciate help finding the correct one. Thanks!


The fault in your argument is that for $v\in V$, $\{x\cdot v \;|\;x\in \mathfrak{g}\}$ does not necessarily equal $V$ - you might need to apply elements of $\mathfrak{g}$ multiple times to reach some vectors in $V$.

The question can be solved as follows. Let the $\mathfrak{g}$-module isomorphism $V\rightarrow V^*$ be given by a matrix $A$ wrt some basis in $V$ and the dual basis in $V^*$. Then the induced bilinear form is given by $(v,w)=v^T Aw$.

The fact that $A$ preserves the $\mathfrak{g}$-action is equivalent to the following: for any $x\in\mathfrak{g}$ acting on $V$ by matrix $X$, we have $$\forall v,w\in V\;,v^T AXw=-(Xv)^T Aw=-v^T X^T Aw\\ \text{which is the same as }\; AX=-X^T A$$

The trick is to notice that by taking transposes we have that $A^T$ also satisfies this property, and so $A+A^T$ and $A-A^T$ do as well. These induce symmetric and antisymmetric bilinear forms respectively which sum to $2A$, so we are done if we can show that one of these is zero. But they map between irreducible $\mathfrak{g}$-modules, so by Schur's lemma, if neither is zero then they are both isomorphisms. But then $(A+A^T)^{-1}(A-A^T)$ is a $\mathfrak{g}$-automorphism of $V$, so its eigenspaces are submodules, and irreducibility of $V$ implies that it is a multiple of the identity matrix. Thus $A+A^T =\lambda (A-A^T)$ for some $\lambda\neq0$, which is a contradiction because a nonzero matrix cannot be both symmetric and antisymmetric.


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