Prove that every bilinear associative form in a simple Lie algebra is a multiple of a Killing form

Question

Let $L$ be a Lie algebra. I have to prove that if $L$ is a simple Lie algebra every bilinear associative form (e.g. $([x,y],z)= (x,[y,z])$ for all $x,y,z \in L$) is a multiple of Killing form.

Answer 1

The proof below is based in the following statement which is valid (at least) for linear spaces $V$ over fields of characteristic $0$:

If $(\cdot,\cdot)_1$ and $(\cdot,\cdot)_2$ are nondegenerate bilinear forms on $V$ then there is a linear autormorphism $P\colon V\to V$ such that $$(v,w)_1=(Pv,w)_2,$$ for all $v,w\in V$.

Also, we will also use (a consequence of) the Schur's Lemma:

If $\rho\colon L\to\mathfrak{gl}(V)$ is a irreducible representation of the Lie Algebra $L$ (over a algebraically closed field of characteristic $0$) and $P\in\mathrm{GL}(V)$ is such that $$P\circ\rho(X) = \rho(X)\circ P,$$ for every $X\in L$, then $P=\lambda I$ (where $I$ is the identity function) for some scalar $\lambda$.

I will assume (as we usually do when we talk about Killing form) that $L$ is a simple Lie algebra over a algebraically closed field of characteristic $0$. Now we begin the proof of the statement:

Every bilinear and associative form $(\cdot,\cdot)$ on $L$ is a multiple of the Killing form $\langle\cdot,\cdot\rangle$ on $L$.

Firstly, we must note that $$L^\perp:=\{X\in L\colon (X,Y)=0\text{ for all }Y\in L\}$$ is a ideal of $L$. In fact, given $X\in L^\perp$ and $Y\in L$, we have that $$([X,Y],Z)=(X,[Y,Z])=0,$$ for every $Z\in L$, and, hence, $[X,Y]\in L^\perp$.

So, since $L$ is simple, $L^\perp=L$ or $0$. In the first case we already get the result because $L^\perp=L$ implies that $(\cdot,\cdot)=0$. So, in what follows, let us suppose that $L^\perp=0$. It means, that $(\cdot,\cdot)$ is nondegenerate.

The bilinear forms $(\cdot,\cdot)$ and $\langle\cdot,\cdot\rangle$ are nondegenerate (by the Cartan's Criterion of semisimplicity) on $L$. So, let $P\in\mathrm{GL}(L)$ be such that $$(X,Y)=\langle P X,Y\rangle,$$ for every $X$ and $Y\in L$.

Next, we will show that $$P\circ\mathrm{ad}(X)\circ P^{-1} =\mathrm{ad}(X),$$ for all $X\in L$. Then, we may conclude, from Schur's Lemma, that $P=\lambda I$, for some scalar $\lambda$ and, whence, $$(X,Y)=\langle P X,Y\rangle = \lambda\langle X,Y\rangle,$$ for every $X$ and $Y\in L$. So, given $X\in L$, we have, for every $Y$ and $Z\in L$, that $$\begin{array}{rcl} \langle P\circ\mathrm{ad}(X)\circ P^{-1}Y,Z\rangle & = & ([X,P^{-1}Y],Z) \\ & = & -([P^{-1}Y,X],Z) \\ & = & -(P^{-1}Y,[X,Z]) \\ & = & -\langle Y,[X,Z]\rangle \\ & = & -\langle [Y,X],Z\rangle \\ & = & \langle\mathrm{ad}(X)Y,Z\rangle. \end{array}$$ Thus, since the Killing form on $L$ is non-degenerate, we have that $P\circ\mathrm{ad}(X)\circ P^{-1} =\mathrm{ad}(X)$, for all $X\in L$.

Answer 2

Notice that if $L$ is simple, then its adjoint representation $$\text{ad}: L\longrightarrow \mathfrak{gl}(L)$$ is irreducible (As if not, there exists a non-trivial $\text{ad}$-invariant subspace $L_1$ hence also an ideal).

Also notice that there is a natural co-adjoint representation $$\text{ad}^*: L\longrightarrow \mathfrak{gl}(L^*)$$ which is also irreducible.

Now, a non-degenerate $\text{ad}-$invariant bilinear form $B$ will induce an $L-$equivariant isomorphism $L\longrightarrow L^*$.

By Schur's Lemma: morphisms between irreducible representations are quite few——they are at most 1-dimensional. And the Killing form $\kappa$ happens to be such a bilinear form, therefore: $$B=\lambda \kappa$$