0

I am trying to understand why for a semisimple Lie algebra $L$, we have $L = [L,L]$.

here are my thoughts.

Let $I:=[L,L]$ be a proper ideal in $L$, i.e., $I \ne L$, we then consider its orthogonal complement $I^\perp:=\{x \in L: \varkappa(x,y) = 0 \mbox{ for all $y\in I$}\}$ with respect to the Killing form $\varkappa$. Since $\varkappa$ is associative then $I^\perp$ is also an ideal.

On the other hand, since $y \in [L,L]$ then $y = \sum \alpha_{i,j}[y_i,y_j]$, $y_i, y_j \in L$, and $\alpha_{i,j}$ are scalars. We then get $\varkappa(x,y) = \sum\alpha_{i,j}\varkappa(x,[y_i,y_j]) = \sum \alpha_{i,j}\varkappa([x,y_i],y_j)$. And I then want to say that $I^\perp$ can be defiend as a kernal of $\varkappa$, because of $[x,y_i]$ can be also considered as an arbitrary element of $L$, and we thus get a conradiction because $L$ is assumed to be semisimple, but I am not sure that it is ok.

Butters Stotch
  • 51
  • 1
  • 3
  • Yes, this argument shows that $L=A\oplus A^{\perp}$, so that a f.d. complex semisimple Lie algebra is the direct sum of simple ideals, so that we immediately have $[L,L]=L$. See also here. – Dietrich Burde Jun 22 '22 at 13:28
  • Thank you very much! – Butters Stotch Jun 22 '22 at 14:17
  • We are assuming characteristic $0$ here to use the Killing form criterion for semisimplicity. For a counterexample in positive characteristic, see https://math.jhu.edu/~sakellar/automorphic-files/liestructure.pdf, remark 2.2. – Torsten Schoeneberg Oct 17 '22 at 17:52

0 Answers0