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Let $RVar$ be the category which has complex algebraic verieties as objects and rational maps as morphisms [Edit: for $RVar$ to be a category, the rational maps have to be dominant]. Let's consider a group object $G$ in this category, i.e. a tuple $(G,\mu,e,\iota)$ where $\mu:G\times G \rightarrow G$ , $e:* \rightarrow G$, and $\iota: G \rightarrow G$ are rational maps satisfying the usual commutative diagrams defining a group structure. [Edit: in the light of the comments, e.g. the observation that rational maps have better be dominant, the identity $e:*\rightarrow G$ doesn't seem to make much sense; same for the diagram for the inverse, then; what people, among whom Weil, actually considered were "rational group chunks" in which there's only an associative rational $\mu$, and you can ask the same question(s)].

Just out of curiosity, two natural questions:

  • Is such a $G$ necessarily an algebraic group? That is: is it the case that for any $(G,\mu,e,\iota) \in Grp(RVar)$, there exists an algebraic group $(G',\mu',e',\iota')$ and a birational map $\varphi: G\rightarrow G'$ such that "$\varphi$ intertwines the operations of $G$ and $G'$"?

  • Analogous question in the holomorphic/meromorphic setting.

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Just a side question that shows my ignorance. Is the usual product of varieties the categorical product in RVar? It will be if there are not too many more rational maps than polynomials ones. –  Theo Johnson-Freyd Apr 21 '10 at 23:43
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Is RVar even a category? You can't always compose rational maps. –  Bjorn Poonen Apr 22 '10 at 3:37
    
Bjorn, maps are required to be dominant (e.g., $\mu$!), and in particular the definition of a birational group law involves no identity axiom and so no inverses! (The empty scheme is a birational group law.) It's all about translations and assoc. There is a finer notion of "strict birational group law", and thm is that (i) any non-empty bir. gp law on a smooth septd scheme of finite type contains dense open that is strict (for "induced" bir. gp law structure), and (ii) any non-empty strict bir. gp. law is dense open in unique actual smooth group. No control on affineness in the result. –  BCnrd Apr 22 '10 at 4:02
    
@BCnrd: Thank you for the clarifications. –  Qfwfq Apr 22 '10 at 4:09
    
@BCnrd: Does the definition of strict birational group law involve the "inverse" in some way? The problem in defining the identity instead, as far as I've understood from the comments above, is that $*\rightarrow G$ is almost never dominant. Right? –  Qfwfq Apr 22 '10 at 4:17

1 Answer 1

Yes. This result usually goes under the name of Weil's theorem on group chunks.

The original reference is:

A. Weil, On algebraic groups of transformations, Am. J. Math, 77, 1955, 355 - 381.

In model theory this theorem has taken on a life of its own and is the basis of Hrushovski's theorem on generically presented groups. A model theoretic treatment of Weil's theorem, including a weakening where one is allowed to use negative powers of the Frobenius, may be found in

E. Bouscaren, Model theoretic versions of Weil's theorem on pregroups, in The Model Theory of Groups, Notre Dame Mathematical Lectures, Number 11

The analogous results are known at least for compactifiable complex analytic groups and real algebraic groups. The following paper gives a good general treatment of these problems.

L. van den Dries, Weil's group chunk theorem: a topological setting, Ill. J. Math, 34 no. 1, 1990, 127 - 139.

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There is also a treatment in Cornell and Silverman (in Artin's article, I think), which is an updated (i.e. scheme-theoretic) treatment of Weil's original argument. –  Emerton Apr 22 '10 at 0:07
    
Artin also has an entire expose on this topic, incorporating generalization to a Dedekind base, near the end of SGA3 (which is used there to cleverly construct Chevalley schemes over Z with minimal computation). A more detailed exposition in that spirit is the very elegantly-written Chapter 5 of "Neron Models". The latter is my favorite reference on these matters. –  BCnrd Apr 22 '10 at 4:04

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