0

Let $L$ be a Lie algebra (with field $F$). Consider the killing form $$\begin{align*} k:&L \times L \to F \\ &(x,y) \to Tr(ad(x)ad(y)) \end{align*}$$ We know that $L$ is semisimple if and only if its killing form is non degenerate.

Now If $L \subseteq \mathfrak{gl}(V)$, I can define $$\begin{align*} k’:&L \times L \to F \\ &(x,y) \to Tr(xy) \end{align*} $$ I want to prove: $L$ is semisimple if and only if $k’$ is non degenerate. One implication is trivial:

Suppose $L$ semisimple. If $x\in L^{\bot}$(respect $k’$) and $y\in L$ then $Tr(xy)=0$, so in particular this is true for $y \in [L^\bot,L^\bot]$. So, by Cartan’s criterion, $L^\bot$ is solvable and so it is zero since $RadL=0$.

I don’t know how to complete the other implication. My idea is to replicate the proof of the general case (Humphreys pg 22).

Some ideas?

Tian Vlašić
  • 1,318
Mario
  • 717
  • Your previous example of $L$ was a simple Lie algebra, namely $\mathfrak{sp}(2n)$. Then, by Schur's Lemma, the Killing form is a nonzero scalar multiple of the Killing form, hence the trace form is non-degenerate. See for example here. – Dietrich Burde Jun 21 '23 at 08:06
  • For the first part, the simple Lie algebra $\mathfrak{psl}_n(\Bbb F)$, where $\Bbb F$ has characteristic $p$ dividing $n>2$, has a degenerate trace form. – Dietrich Burde Jun 21 '23 at 08:45
  • Note that you could have noticed the implication you asked about is false, by just looking at the basic case $L=F =\mathfrak{gl}(F)$. Checking examples is generally the first thing one should do if one is stuck. – Torsten Schoeneberg Jun 21 '23 at 16:25

1 Answers1

3

The trace form of the Lie algebra $\mathfrak{gl}_n(F)$ is nondegenerate over a field $F$ of characteristic zero. However, $\mathfrak{gl}_n(F)$ is not semisimple. So the converse is not true.

However, it is known, see Bourbaki's book on Lie algebras, that a Lie algebra $L$ having a nondegenerate trace form for some representation is reductive. For references see this post:

Killing form of a reductive symmetric Lie algebra

Also, if you require that all trace forms $B_{\rho}$ of $L$ are nondegenerate, then $L$ is semisimple, See this post:

Semisimplicity of Lie algebras and non-degeneracy of associated bilinear forms of representations

Dietrich Burde
  • 130,978