Proving that $G=\mathrm{SL}_d(\mathbb{R})$ is semisimple

Asked Apr 04 '19 at 06:51

Active Apr 21 '19 at 12:38

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I want to find a reference to a proof of semisimplicity of the special linear group $G=\mathrm{SL}_d(\mathbb{R})$ by showing that the Lie algebra $\mathfrak{g}$ of $G$ of matrices with trace $0$ is semisimple.

One approach would be to compute the Killing form $B$ of the Lie algebra $\mathfrak{g}$ of $G$. I have seen the statement $B(X,Y)=2d\cdot\mathrm{tr}(XY)$ which would prove semisimplicity of $\mathfrak{g}$ by Cartan's criterion. Any points to a reference to a proof of $B(X,Y)=2d\cdot\mathrm{tr}(XY)$ would be appreciated.

Edit: It was not clear from the original post, but what I really want to is to prove that $G$ is semisimple by showing that $\mathfrak{g}$ is semisimple. The proofs of $B(X,Y)=2d\cdot\mathrm{tr}(XY)$ that I have found use the fact that $\mathfrak{g}$ is simple, in particular semisimple.

edited Apr 21 '19 at 12:38

Xander Henderson

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asked Apr 04 '19 at 06:51

ccornix

1

For $B(X,Y)=2d\cdot tr(XY)$ see this this duplicate. There are other posts, too. One can directly compute this on the basis of the Lie algebra. However, we actually only need that $B$ is non-degenerate to conclude that $\mathfrak{sl}_n(K)$ is semisimple. This should be in any book on Lie algebras. – Dietrich Burde Apr 04 '19 at 08:46
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Even better, $\mathfrak{sl}_n(\Bbb R)$ is simple, see this duplicate. – Dietrich Burde Apr 04 '19 at 08:52
@DietrichBurde Thanks for your help! – ccornix Apr 04 '19 at 11:48

Proving that $G=\mathrm{SL}_d(\mathbb{R})$ is semisimple

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