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I would like to consider/prove the following problems: let $k$ be a field, $g$ a finite-dimensional simple Lie algebra over $k$ with Killing form $B$.

If $\sigma:g\times g\rightarrow k$ is a symmetric bilinear form such that $\sigma([x,y],z)=\sigma(x,[y,z])$, then there is $c\in k$ with $B=c\sigma$. I know that for $k$ algebraically closed, this follow from a version of Schur's lemma.

Problem 1: what about the case $k=\mathbb{R}$? Is this result still true, and how can it be proven if it is? By some kind of base change argument perhaps?

Apparently this can be generalized to semisimple $g$; in this case the obvious base change argument might actually work (as $g$ is semisimple iff its complexification is, which is, as far as I know, false for simple $g$). However, it is still not obvious why $c\in\mathbb{R}$ instead of merely $c\in\mathbb{C}$?

Actually I found the following exercise in Bourbaki's book: it suggests that the result might be false.

Problem 2 if $g$ is a simple subalgebra of $gl_n(\mathbb{R})$ and $\sigma(x,y)=\mathrm{Tr}(xy)$, then $B=c\sigma$ for some $c\in\mathbb{R}$.

Is Problem 2 not even true?

Henri
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1 Answers1

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The Killing form of simple real Lie algebras is again a multiple of the trace form, see here for explicit results for real classical simple Lie algebras. This can be computed directly, e.g., $$ B(x,y)=2n \cdot tr(xy) $$ for the simple Lie algebra $\mathfrak{sl}_n(\mathbb{R})$. In general, I think, without using the classification, one could relate the Killing form of the real forms of simple complex Lie algebras to the Killing form of the complex Lie algebras.

Dietrich Burde
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  • Thanks for the answer. Something which is not clear to me, however, is how you do it for general $g$ as in the problem that I am supposed to solve. Are you using some kind of classification result for simple Lie algebras? Also what is not clear to me is how a negative answer to my first problem (note the edit) does not imply a negative answer to the second one. For by Ado's theorem we may embedd $g$ into $gl_n(\mathbb{R})$. Perhaps that property (space of invariant bilinear forms has dimension 1) is not invariant under embeddings? – Henri Jan 13 '16 at 10:37
  • or has it to do with the special form of $\sigma$? perhaps it is true for that $\sigma$, but not for others? – Henri Jan 13 '16 at 10:40
  • how can it be done in general? how can I use the special form of the bilinear form $tr(xy)$? – Henri Jan 15 '16 at 20:47
  • The formulas for simple complex Lie algebras are clear. We need to see that the formulas only change by a nonzero constant, for all real forms, i.e., for the compact form and the non-compact forms. I will look for a reference. – Dietrich Burde Jan 16 '16 at 09:55