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By the usual definition, Lie group is a manifold $G$ with a group structure on it such that the multiplication $m\colon G\times G\to G$ and taking inverse $i\colon G\to G$ are both smooth maps. But it is not difficult to proove that actually we need only the smoothness of the multiplication map $m$. The proof uses the fact that the preimage $m^{-1}(1)$ of $1\in G$ is a smooth submanifold of $G\times G$, and the inverse function theorem.

Now I am wondering, is that true for algebraic groups? It might be obvious, but it's not obvious to me, since in this case we can't use methods from the smooth case. More precisely, suppose $G$ is a variety with a group structure such that $m\colon G\times G\to G$ is a regular map. Is it true that it automatically guarantees regularity of the inversion map $i\colon G\to G$ ?

Thank you very much for your help!

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A group structure in the category of smooth manifolds, by definition, has its three structure maps smooth. I imagine your first observation is talking about finding a group structure not in the usual category of smooth manifolds, but in the larger category whose objects are smooth manifolds and whose morphisms are arbitrary functions (rather than smooth maps), and looking at the conditions that make a group structure in the larger category actually be a group in the smaller category. So for your question on algebraic groups, what is the larger category you're considering? –  Hurkyl Aug 1 '13 at 21:54
I am sorry for being not very precise. Here by group structure I mean set-theoretic group structure. –  Sasha Patotski Aug 1 '13 at 21:57

1 Answer 1

up vote 4 down vote accepted

I think that the answer is yes.

In my argument I will use the following fact: a morphism of varieties which is a bijection on points and whose target is normal (e.g. smooth) is necessarily an isomorphism of varieties. (This follows from Zariski's main theorem.)

I will first prove that $G$ is necessarily smooth. Since $G$ is a variety it is generically smooth, i.e. it contains a non-empty Zariski open subset $U$ consisting of smooth points. Let $g\in G$ be an arbitrary point, and choose some $h \in G$ such that $gh \in U$ (possible because $G$ is a group under multiplication and $U$ is non-empty).

Then $m^{-1}(U) \cap (G \times h)$ is a Zariski open subset of $G \times h$, say $V \times h$, such that $g \in V$. Then right multiplication by $h$ gives a regular map $V \to U$ which is a bijection on points, with $U$ being smooth. Thus this map must be an isomorphism of varieties, and so $V$ is also smooth. Thus every point of $G$ is smooth.

Now consider the preimage $\Gamma$ of $1$ under $m$; this is a subvariety of $G \times G$, which set-theoretically consists of the graph of $i$. In particular, each of the projection maps is a set-theoretic bijection $\Gamma \to G$.

Since $G$ is smooth (by what we proved above), the morphism $\Gamma \to G$ is a morphism of varieties which is a set-theoretic bijection with smooth target: it is thus necessarily an isomorphism.

Added: the fact about bijections requires a char. zero assumption as stated, since, in the application of ZMT that underlies it, it assumes that the quasi-finite morphism in question if generically etale (so that we can go from set-theoretic bijection to birational morphism). This is automatic in char. zero, but can fail in positive char.

My guess is that with a little more care the above argument should actually work in char. $p$ too; on the one hand mult. by $g$ induces a morphism $G \to G$ which is set-theoretically a bijection, so on function fields it gives a purely inseparable embedding $k(G) \hookrightarrow k(G)$ of some degree $p^{n_g}$. Now $n_g$ should be generically a constant, but on the other hand, composing mult. by $g$ with mult. by $h$ shows that $n_{gh} = n_g n_h$. Taking $g$ and $h$ to be generic gives that $n_g = 1$ for generic $g$, hence mult. by $g$ is generically etale of degree $1$, i.e. generically an isomorphism. This should give the generic etaleness needed for the ZMT argument to go through. I didn't think this through carefully though.

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That's nice! Thanks! I didn't know the main fact you're using. –  Sasha Patotski Aug 1 '13 at 22:11
@SashaPatotski: Dear Sasha, A basic example to remember is $\mathbb A^1$ mapping to $y^2 = x^3$ via $t \mapsto (t^2,t^3)$; this is a bijection on points, but of course not an isomorphism. But the fact I am using says this can't happen if the target is normal. The point is that the the morphism is a bijection, therefore has finite fibres (indeed, singleton fibres), therefore is quasi-finite, hence by (Grothendieck's form of) ZMT is an open immersion into a finite morphism. But since this finite morphism is a bijection on an open subset, it is birational, and a finite birational morphism ... –  Matt E Aug 1 '13 at 22:17
... with normal target must be an isomorphism. (This is more or less the definition of normal.) Thus our original map is an open immersion into an isomorphism, i.e. simply an open immersion, but since it is also a bijection on points it is in fact already an isomorphism. Regards, –  Matt E Aug 1 '13 at 22:18
Thank you very much for your wonderful comment! It's really very instructive! –  Sasha Patotski Aug 1 '13 at 22:31
Set-theoretic bijections with smooth target can fail to be isomorphisms in positive characteristic. –  Julian Rosen Aug 2 '13 at 0:16

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