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I have problem to solve Exercise 5.6 in a book named "Introduction to Lie algebra and Representation Theory" from Humphreys.

I want to show that special linear algebra $ \mathfrak sl (3, F)$ modulo its center is semisimple, but its killing form is nondegenerate, where field $F$ has characteristic 3.

My attempt is to consider basis of $ \mathfrak sl (3, F)$ and calculate arbitrary solvable ideal of given lie algebra. However, this does not go well. May I ask for a help?

HooMun
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    "killing form" is a dangerous form, "Killing form" is harmless :) – Jean Marie Sep 22 '20 at 07:42
  • What exactly are you trying to prove, the part about semisimplicity or the part about the Killing form? And should it be that the Killing form is degenerate? – Torsten Schoeneberg Sep 22 '20 at 15:11
  • Hmm I guess you could show that if $\mathfrak g ^ \perp = {x :K(x,y) =0 \forall y \in \mathfrak g}$ is equal to the set of scalar matrices then you could possibly show that the Killing form after taking quotient is non-degenerate and by Cartan's criterion, the quotient Lie algebra is semisimple. Just an idea. – monikernemo Sep 22 '20 at 15:19
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    @monikernemo: Cartan's criterion does not necessarily work in positive characteristic, that's why I asked what I asked. -- HooMun: Maybe the calculations in https://math.stackexchange.com/q/467539/96384 are still helpful, just see what exactly one has to adapt in characteristic $3$. – Torsten Schoeneberg Sep 22 '20 at 15:46
  • Ah right, I forgot that Cartan's criterion for semisimplicty relied on characteristic zero. Thanks! – monikernemo Sep 22 '20 at 16:08

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The algebra $\mathfrak{psl}_3(F)$ is even simple in characteristic $3$. This can be easily seen by assuming that we have a nonzero ideal $I$ with $x\in I%$ and $x\neq 0$ and applying brackets repeatedly to obtain $I=\mathfrak{psl}_3(F)$. It is a well-known example for a simple modular Lie algebra having zero Killing form.

Remark: $\mathfrak{psl}_3(F)$ does not contain any $2$-dimensional subalgebra, and hence is simple, see Lemma $2.3$ here.

Dietrich Burde
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