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I know that there's counterexample which a simple lie algebra whose complexification is not simple, see here.

However, I'm not sure if a simple complex lie algebra's real form is still simple or not.

If not, what's the classical way to show $\mathfrak{su}(n)$ is simple?

Kenny S
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    Complexification takes ideals to ideals, so if a real Lie algebra is not simple then its complexification is not simple. Taking the contrapositive, if the complexification is simple, the original real Lie algebra is simple. – Qiaochu Yuan Jul 06 '24 at 06:59
  • If I don’t understand wrongly, if a real Lie algebra is not simple, it can be decomposed into direct sum of simple ideals, so is it true that the complexification takes simple ideals to simple ideals? – Kenny S Jul 06 '24 at 07:30
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    @KennyS: You don't even need a decomposition into simple ideals (for which you would need to invoke the corresponding statement with "semisimple"). You just use that if the real form has any ideal besides $0$ and itself, then so does its complexificstion. – Torsten Schoeneberg Jul 06 '24 at 15:37
  • Thank you for your comment! I understand. – Kenny S Jul 06 '24 at 16:03

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