If $\phi: G_1 \rightarrow G_2$ is a continuous homomorphism of real Lie groups, then $\phi$ is automatically a smooth. Is this also true for complex (analytic) Lie groups?

For example, if $G$ is a complex Lie group, and $\pi: G \rightarrow \operatorname{GL}(V)$ a continuous homomorphism for $V$ a finite dimensional complex vector space, is $\pi$ necessarily analytic?

I believe that for smooth Lie groups, this is a corollary of the theorem that a closed subgroup of a Lie group is automatically a Lie subgroup. So it suffices to show that the analogue for analytic Lie groups holds.

  • $\begingroup$ "However, these requirements are a bit stringent; every continuous homomorphism between real Lie groups turns out to be (real) analytic.[9]" See en.wikipedia.org/wiki/Lie_group etc. So, given that a complex Lie group can also be viewed as a real Lie group it would seem you at least have real analytic. Surely complex analytic follows from that. That said, this is not a proof. $\endgroup$ – James S. Cook May 30 '18 at 22:03

Consider the case when $G_1=G_2=(\mathbb{C},+)$ and $\phi$ is complex conjugation. Then $\phi$ is not complex analytic.

Also, the fixed points of $\phi$ form a closed subgroup of dimension $1$, whereas any complex manifold has even dimension.


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