So I have seen that every semi-simple complex lie algebra has a split and compact real form, where the compact real forms correspnding to semi-simple compact real lie algebras hence we can classify all possible complex semi simple algebras (and label them by their coreesponding compact semi simple real form).

Now Im wondering if it's possible to classify all real semi-simple lie algebras by real forms of the complex semi simple algebras. That is for any real semi simple $\mathfrak{g}$, does it correspond to a real form of one of the classified complex semi simple algebras $\mathfrak{g}^{\mathbb{C}}$?

  • $\begingroup$ Your first paragraph strikes me as being reversed from the usual direction: In all sources I know, one classifies the compact real LAs via the complex semisimple LAs (whose classification boils down to root systems, which boils down to Dynkin diagrams), not the other way around. I would not know how to first classify those compact real ones, without (explicitly or implicitly) already using complexification. $\endgroup$ Commented Sep 29, 2022 at 16:15
  • $\begingroup$ That being said, that every real semisimple LA is a real form of a complex semisimple LA is easy to see, cf. answer and comment. In my view, the fact you casually quote at the beginning, that conversely each complex semisimple LA has at least two real forms, is more profound. In general, of course, there is much more to say, and if you're interested, check out the references in math.stackexchange.com/q/3121110/96384 and math.stackexchange.com/q/1024861/96384. $\endgroup$ Commented Sep 29, 2022 at 16:19

1 Answer 1


Yes. See Theorem 6.94 of Knapp, Lie groups, beyond an introduction.

This states: if $\mathfrak{g}$ is a real simple Lie algebra then either its complexification $\mathbf{C} \otimes_{\mathbf{R}} \mathfrak{g}$ is simple or its complexification is not simple, $\mathfrak{g}$ is a complex simple Lie algebra, and its complexification is isomorphic to $\mathfrak{g} \oplus \mathfrak{g}$. In either case, $\mathfrak{g}$ is a real form of a semi-simple Lie algebra.

  • $\begingroup$ Then for real semi simple Lie algebras, $\mathfrak{g}$, it follows that its complexification is semi-simple (since the orginal semi simple real algebra can be written as the decomposition of simple real algebras)? $\endgroup$ Commented Sep 29, 2022 at 15:10
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    $\begingroup$ @SheldonCooper Yes, precisely. $\endgroup$
    – Stephen
    Commented Sep 29, 2022 at 15:21
  • $\begingroup$ It is a basic lemma that if $\mathfrak g$ is any Lie algebra over any characteristic $0$ field $K$, and $E \vert K$ any field extension, then $\mathfrak g$ is semsimple if and only if the scalar extension $\mathfrak g \otimes E$ (a Lie algebra over $E$) is semisimple. One can e.g. prove this via the Cartan criterion with Killing form. And all we need here is one direction (arguably the easier one) of this lemma. $\endgroup$ Commented Sep 29, 2022 at 16:07
  • $\begingroup$ Of course this Theorem gives much more precise information about what can happen in case of a simple real Lie algebra, but that lies somewhat deeper and might be confusing for beginners. (For this matter, cf. e.g. math.stackexchange.com/q/3974731/96384, and the links in comments to math.stackexchange.com/q/4517453/96384.) $\endgroup$ Commented Sep 29, 2022 at 16:09
  • $\begingroup$ @TorstenSchoeneberg Hmmm. I am not sure that I agree that the "basic lemma" you mention is simpler than Theorem 6.94 from Knapp (have a look at the proof: it requires almost no machinery, and in particular nothing about the Cartan criterion and the Killing form). $\endgroup$
    – Stephen
    Commented Sep 29, 2022 at 16:42

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