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Why is the realification of a simple complex Lie algebra a semisimple real Lie algebra?

The realification here means to consider the complex Lie algebra as a real Lie algebra of twice the dimension.

The statement was used in the proof of Proposition 12.46 in https://doi.org/10.1016/S0079-8169(08)61672-4.

glS
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richardfatman
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2 Answers2

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Actually, a stronger statement holds true: Denoting the "realification" of a complex Lie algebra $L$ by $Res_{\mathbb C\vert \mathbb R}L$, then

$L$ is simple iff $Res_{\mathbb C\vert \mathbb R}L$ is simple.

Namely, for the non-trivial direction, say we've already shown that $Res_{\mathbb C\vert \mathbb R}L$ is semisimple like in José Carlos Santos' answer. Let $S$ be a simple component of $Res_{\mathbb C\vert \mathbb R}L$. Now for any $0 \neq \lambda \in \mathbb C$, we have that $\lambda S$ is also an ideal of $Res_{\mathbb C\vert \mathbb R}L$, and $[S, \lambda S] = \lambda S \neq 0$ hence by simplicity $S \subseteq \lambda S$. But then we actually have equality $S=\lambda S$, because they have the same real dimension. But that means the complex span of $S$ in $L$ is $S$ itself, hence it is a non-zero ideal of $L$, hence all of $L$ by $L$ being simple, and hence as a set it is also all of $Res_{\mathbb C\vert \mathbb R}L$.

This statement and proof, including the step provided by José Carlos Santos' answer, works far more generally for any finite extension of characteristic $0$ fields $K\vert k$ instead of $\mathbb C \vert \mathbb R$, and I took the proof from Bourbaki's Lie Groups and Algebras, chapter I §6 no. 10.

If one digs deeper into the theory, actually the following holds true: If the Dynkin diagram to $L$ is of type $R$ (i.e. $A_n, B_n, C_n, D_n, E_6, E_7, E_8, F_4$ or $G_2$), then $Res_{\mathbb C\vert \mathbb R}L$ is a quasi-split real Lie algebra whose Satake diagram consists of two copies of the Dynkin diagram of $L$, with complex conjugation flipping those two copies. That real Lie algebra is simple, but not "absolutely simple", since if one complexifies it again, one gets a direct sum of two copies of the original $L$. Again, this holds true in greater generality, compare section 4.1 (especially p. 67) of my thesis.

Torsten Schoeneberg
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By the Cartan criterion, a Lie algebra $\mathfrak g$ is semisimple if and only of its Killing form is non-degenerate. So, if $\mathfrak g$ is a semisimple complex Lie algebra, its Kiling form is non-degenerate, but that Killing form is also the Killing form of the realification of $\mathfrak g$.

José Carlos Santos
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  • Thank you. I wonder if there would be some explicit decomposition of the realification into simple components. – richardfatman Jul 02 '20 at 16:56
  • I would like to help there, but I don't know. – José Carlos Santos Jul 02 '20 at 17:01
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    A small issue is that the Killing form on a complex Lie algebra is complex-bilinear and complex-valued, while on a real Lie algebra it should be real-valued... – paul garrett Jul 02 '20 at 17:06
  • @paulgarrett I am not very familiar with the Killing form as I just started to learn about Lie algebra. I wonder if the issue mentioned can be fixed in some easy way. – richardfatman Jul 02 '20 at 17:14
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    Letting $B(x,y)$ be the Killing form on the complex Lie algebra, the Killing form on the realification is $B'(x,y)=\Re(B(x,y))$. The non-degeneracy of $B'$ is not hard to prove from that of $B$: given $x\not=0$, there must be $y$ such that $B'(x,y)=1$. (Other, equivalent definitions just say $B'(x,y)\not=0$.) Well, the same $y$ works as did for $B$ itself. – paul garrett Jul 02 '20 at 17:18
  • @richardfatman, José: It's actually simple (pun included), cf. my answer. – Torsten Schoeneberg Jul 02 '20 at 19:08
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    @paulgarrett: Actually $B'(x,y) = \color{red}{2} \cdot \mathfrak{R}(B(x,y))$, which of course does not interfere with our main goal here ... (Bourbaki's Lie volume, ch. I §3 exercise 7.) – Torsten Schoeneberg Mar 30 '22 at 17:05
  • Doesn't Cartan's criterion apply only for finite-dimensional Lie algebras? – Jianing Song Mar 31 '22 at 01:53
  • @JianingSong Yes, but semisimple Lie algebras are finite-dimensional by definition. – José Carlos Santos Mar 31 '22 at 06:57
  • @JoséCarlosSantos What's your definition of semisimplicity here? What I learned is that $\mathfrak{g}$ is called semisimple if it does not have a nonzero solvable ideal. No finite dimensionality assumed here. – Jianing Song Mar 31 '22 at 10:04
  • @JianingSong I use the Wikipidia definition: it's a finite-dimensional Lie algebra over a field of characteristic $0$ with no nonzero solvable ideals. – José Carlos Santos Mar 31 '22 at 10:38
  • @JoséCarlosSantos OK, but what happens in the general case is a question. :) – Jianing Song Mar 31 '22 at 12:29