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I'm having trouble proving that every complex Lie algebra whose realification is a compact Lie algebra must be abelian. Can someone help?

My idea is the following. Let $\mathfrak{g}$ be a complex Lie algebra with $\mathfrak{g}^\mathbb{R}$ compact. Then $\mathfrak{g}^{\mathbb{R}} = \mathfrak{z} \oplus \mathfrak{s}$ where $\mathfrak{z}$ is the center of $\mathfrak{g}$ and $\mathfrak{s}$ is semisimple with negative definite Cartan-Killing form. I think $\mathfrak{s}$ must be zero, but I can't prove it.

Thanks!

Marcos Lages
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  • By "realification" you mean $\mathfrak g$ considered as real Lie algebra, i.e. scalar restriction, i.e. what you get is a real Lie algebra whose real dimension is $2$ times the complex dimension of the original $\mathfrak g$? – Torsten Schoeneberg Feb 19 '25 at 20:40
  • And your definition of compact real Lie algebra is what you wrote, reductive with a semisimple part that has a negative definite Killing form? – Torsten Schoeneberg Feb 19 '25 at 20:43

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As per comment, I assume for you a (real) Lie algebra is called compact by definition iff it is reductive with the semisimple part having negative definite Killing form. Cf. this related question.

So assume that $\mathfrak g^\mathbb R \simeq \mathfrak{s} \oplus \mathfrak{z}$ (as real Lie algebras) with $\mathfrak{s}$ semisimple, and $\mathfrak{z}$ the centre of $\mathfrak{g}$ = the centre of $\mathfrak{g}^{\mathbb R}$. Note that $\mathfrak{s} = [\mathfrak{g}^{\mathbb R}, \mathfrak{g}^{\mathbb R}] = [\mathfrak{g}, \mathfrak{g}]$ (not as trivial as it seems, but taking the real or complex span here gives the same result). So $\mathfrak{s}$ is actually a complex semisimple Lie algebra, so in fact this decomposition is one of complex Lie algebras as well.

If $B(x,y)$ is the Killing form of any complex Lie algebra, then the Killing form of its realification is given by twice the real part of that,

$B^{\mathbb R}(x,y) = 2\cdot Re(B(x,y))$ (cf. discussion in comments here).

Now as soon as $B(x,x) <0$, we get $B(ix,ix) >0$, so the Killing form on $\mathfrak s$ (be it $B$ or $B^{\mathbb R}$) has no chance to be negative definite unless $\mathfrak s=0$.

Torsten Schoeneberg
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