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If an endomorphism of $\mathbb{C}^*$ is assumed to be holomorphic, then it is not difficult to show that it must be of the form $z^n$ for some $n\in \mathbb Z$. I was wondering if this is true without any topological or analytic assumptions on the endomorphism. My intuition says that there should be lots of them, but I'm usually working with endomorphisms which are at least holomorphic if not a true morphism of varieties, so I'm not sure.

edited Feb 5 at 11:15

Andrej Bauer
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asked Feb 4 at 21:09

HNuer
915●2●7

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$C^\times$ is isomorphic as a group to $S^1\times\mathbb R$, and $\mathbb R$, as any infinite dimensional vector space over $\mathbb Q$, has lots of endomorphisms as an abelian group! – Mariano Suárez-Alvarez Feb 4 at 21:16

Although I think you mean $R_{>0}$, I see what you're saying, so thanks for the help! – HNuer Feb 4 at 21:27

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No, I mean $\mathbb R$, just because the multiplicative group $\mathbb R_{>0}$ and the additive group $\mathbb R$ are isomorphic. – Mariano Suárez-Alvarez Feb 4 at 21:27

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@HNuer: Well, you did not specify yourselves the kind of structure on $\mathbf C^*$ for which you wanted to determine the endomorphisms. – ACL Feb 4 at 22:45

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All continuous endomorphisms are of the form $z\mapsto z^n\bar z^m$, for some $n,m \in \mathbb Z$. – Andreas Thom Feb 9 at 10:44

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Chris Heunen · Answer 1 · 2013-02-05 10:17:27Z UTC

There are many "wild" automorphisms of the complex numbers, that preserve addition and multiplication, but not much else. These are hard to write down, however, as their existence seems to rely on the axiom of choice. See e.g. the great expository paper "Automorphisms of the complex numbers" by P. Yale, Mathematics Magazine 39(3), 1966.

Are all endomorphisms of C^* just power maps?

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