If Lie algebra is equal to its commutant, is it semisimple?

Question

I am searching for a prove or a counterexample for this statement:

If finite-dimensional complex Lie algebra is equal to its commutant, then it is semisimple.

I suppose it is not true, because otherwise I would be able to find this beautiful result somewhere.

No, it is strictly weaker property. If Lie algebra is equal to its commutant, it is called perfect. Here is a counterexample: http://mathoverflow.net/questions/60498/lie-algebra-semisimple — Rafael Mrden, Nov 28 '13 at 17:25
@rafaelm Converted to an answer as it's been over 10 minutes. Let me know if you want me to delete. — Ahaan S. Rungta, Nov 28 '13 at 17:38
Well, I can confirm that the $10$ minutes are over now, after $9$ years. — Dietrich Burde, Jul 02 '22 at 19:46

score 3 · Accepted Answer · edited Apr 13 '17 at 12:58

3

No, it is a strictly weaker property. If Lie algebra is equal to its commutant, it is called perfect. See here for a counterexample.

edited Apr 13 '17 at 12:58

Community

1

answered Nov 28 '13 at 17:37

Ahaan S. Rungta

7,706

Alas, I can not ask it there, on the page of answer. I do not understand why the mentioned operation satisfies Jacobi identity: $[[(X,v),(Y,u)],(Z,w)] = [([X,Y],Xu-Yv),(Z,w)]=([[X,Y],Z], [X,Y]w-Z(Xu-Yv))=([[X,Y],Z], [Y,Z]v+[Z,X]u+[X,Y]w)$, but the sum of three summand like this one will not be zero. – evgeny Dec 04 '13 at 07:25

Dietrich Burde · Answer 2 · 2013-12-06T21:29:39.870

2

Nonsemisimple perfect Lie algebras arise when contracting, or degenerating semisimple Lie algebras. A well-known example in physics is that the semisimple Lie algebra $\mathfrak{so}(4,1)$ (the de Sitter Lie algebra) contracts to the Poincaré Lie algebra, which is perfect, but not semisimple. Of course, this Lie algebra is just $\mathbb{R}\rtimes \mathfrak{so}(3,1)$.

edited Dec 06 '13 at 21:29

answered Nov 28 '13 at 21:30

Dietrich Burde

130,978

If Lie algebra is equal to its commutant, is it semisimple?

2 Answers2