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.
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.