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Let $G, H$ be complex Lie groups. Then $\mathfrak{g}=\operatorname{Lie}(G)$, $\mathfrak{h}=\operatorname{Lie}(G)$ are complex vector spaces and complex Lie-algebras also. $f: G \rightarrow H$ -- Lie group homomorphism, and $\phi$ is corresponding homomorphism of Lie algebras (its differential at point $e$).

Does it follow somehow evidently that $\phi$ is complex Lie algebras homomorphism (as it is in real case)? And if so, how to deal with the Lie group homomorphism $(\mathbb{C},+)$ to itself by complex conjugation, from which it follows that the corresponding Lie-algebras homomorphism is only $\mathbb{R}$- linear?

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  • $\begingroup$ It depends on your definition of a homomorphism between complex Lie groups, there are at least three inequivalent defintions. $\endgroup$ – Moishe Kohan Dec 8 '17 at 19:55
  • $\begingroup$ @Moishe Cohen, i thought of it as a homomorphism of vector spaces over $\mathbb{C}$ and equation determines compatibility with lie-bracket. $\endgroup$ – Sasha Mayer Dec 8 '17 at 20:00
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Yes, if $f\colon G\longrightarrow H$ is a Lie group homomorphism, then $\phi$ is a Lie algebra homomorphism. In particular, it is a $\mathbb C$-linear map.

What is there to deal with when our Lie group homomorphism is the conjugation? The conjugation is a group homomorphism from $(\mathbb{C},+)$ into itself, yes, but it is not an analytic map. Therefore, the Lie theory doesn't apply here.

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  • $\begingroup$ Thank you. I was a little bit confused because it is smooth if regarded as the map between real manifolds. Everything fell into place now. $\endgroup$ – Sasha Mayer Dec 8 '17 at 20:11

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