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If $G$ and $H$ are algebraic groups over $\mathbb{C}$, and $f : G \rightarrow H$ is an isomorphism of complex Lie groups (i.e. a biholomorphic group isomorphism), then must $f$ be algebraic? If not, are there additional hypotheses that make this true?

If $G$ and $H$ are projective then general GAGA machinery answers the question in the affirmative, but this is obviously rather restrictive (in particular, it forces $G$ and $H$ to be abelian). But if $f$ is not an isomorphism, and instead is just a holomorphic group homomorphism, then it need not be algebraic, for example the map $\mathbb{C} \rightarrow \mathbb{C}^*$ given by $\lambda \mapsto e^\lambda$.

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    $\begingroup$ Interesting question. The answer is yes if $G$ is reductive. I don't know if you are interested in non-reductive groups. The important thing being that you can prove it for semisimple groups by looking at their Lie algebras, and then extend to general reductive by working over extensions by tori. In fact, you get that every homomorphism is algebraic (note that this isn't a contradiction with your example since $\mathbb{C}$ is unipotent). $\endgroup$ – Alex Youcis Nov 5 '15 at 18:57
  • $\begingroup$ @AlexYoucis thanks for your comment. Of course it would be nice to have as general a result as possible, but certainly the reductive case is what I'm most interested in and is a great first step. Would you be willing to spell out the reductive/semisimple argument in more detail please (or give a reference)? In particular, how can you tell just from the Lie algebra morphism whether the global group morphism is algebraic? $\endgroup$ – Jez Nov 5 '15 at 20:40
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    $\begingroup$ I can of course write something up, but perhaps a reference would be more useful. I learned this fact originally from Brian Conrad's book Reductive Group Schemes. In particular, it's proposition D.2.1 here. $\endgroup$ – Alex Youcis Nov 6 '15 at 1:52
  • $\begingroup$ @AlexYoucis thanks very much for the reference! $\endgroup$ – Jez Nov 6 '15 at 7:49
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    $\begingroup$ At the other end of the spectrum, the answer is also yes for unipotent groups: a homomorphism of connected Lie groups is determined by the map on Lie algebras, and the functor $G\mapsto\mathrm{Lie}(G)$ is an equivalence of categories {unipotent algebraic groups} $\to$ {nilpotent finite dimensional Lie algebras}. $\endgroup$ – Julian Rosen Nov 8 '15 at 20:08
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An isomorphism of complex Lie groups between algebraic groups need not be algebraic in general. A counterexample is $$ \begin{align*} f:\mathbb{C}\times\mathbb{C}^\times&\xrightarrow{\sim} \mathbb{C}\times\mathbb{C}^\times,\\ (x,y)&\mapsto (x,y\cdot e^x). \end{align*} $$

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  • $\begingroup$ Thanks for this nice counter-example :) $\endgroup$ – Jez Nov 6 '15 at 20:22

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