Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
Show us what you've tried and progress that you've made so we can nudge you along... –  JohnD Feb 8 '13 at 20:45
Coul you give me some ideas? –  ArthurStuart Feb 8 '13 at 20:53
What does $(\cdot,\cdot)$ mean here? The Killing form? –  Eric O. Korman Feb 8 '13 at 21:25
@EricO.Korman That is a generic bilinear form. –  ArthurStuart Feb 8 '13 at 21:38
Lies and killing are bad.... –  goblin Feb 8 '13 at 23:00

1 Answer 1

up vote 5 down vote accepted

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 non-degenerated 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 non-degenerated.

The bilinear forms $(\cdot,\cdot)$ and $\langle\cdot,\cdot\rangle$ are non-degenerated (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-degenerated, we have that $P\circ\mathrm{ad}(X)\circ P^{-1} =\mathrm{ad}(X)$, for all $X\in L$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.