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Sorry The title might not be suggestive enough.

The question is about things like the following: A reductive group scheme is defined to be a (really nice) group scheme whose geometric fibers are reductive groups. So in some sense, "reductiveness" is some kind of "geometric" notion.

So whatelse properties of schemes can be checked only on geometric fibers? What I know, for example, given a scheme over a field $k$, it is projective iff it is projective over $\bar{k}$. But can this be extended to any base scheme?

In particular, is there any reference that collects such results? And "WHY" should this work? for example, WHY geometrically-reductive group schemes turn out to be the right generalization of reductive algebraic groups?

Sorry this question might be a little to vague, and thank you in advance.

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If you have a flat, or even better a smooth morphism, then the intuition is that there is some sort of continuity along the fibres. It is then not unreasonable to make definitions by looking at fibres. (You should think about Erhesmann's theorem, which says that a smooth proper map of smooth manifolds is a fibre bundle. So if you want to have a bundle of, say, genus g surfaces over a manifold $X$, it is enough to ask for a proper smooth map $Y \to X$ each of whose fibres is a genus $g$ surface. Similarly, switching back to the language of algebraic geometry, if we want a family of genus $g$ curves, it makes sense to ask for a smooth proper map whose fibres are curves of genus $g$.)

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I just want to point out that by "smooth map of smooth manifolds" here, Emerton means a submersion (the more typical terminology in differential geometry). – Kevin H. Lin Mar 24 '10 at 21:55
Thanks, I was being lazy and using the terminology that is most comfortable for me! – Emerton Mar 24 '10 at 22:19

I'm not sure I'm answering your question, but let me say some (vague and general) things and hopefully they'll be helpful.

Algebraic geometry over algebraically closed fields is really nice, essentially because of the Nullstellensatz: points are actually points, and functions are actually functions on these points. So it's often easy (or classical) to define a property of a variety over an algebraically closed field.

But then Grothendieck comes along and sez, hey, we should say things relatively, over an arbitrary base scheme. And the procedure for that is this: if you have a class of varieties C over algebraically closed fields, you extend it to the relative situation by saying that a map of schemes X --> S is in C if it is flat (which, experience and several nice theorems tell us, amounts to saying that the fibers are continuously parametrized by S) and each geometric fiber is in C.

There are two things that this buys you right off the bat: first, C is closed under base change, and second, C satisfies fpqc descent. Both are great indications that you have a good in-families notion; for instance they are certainly necessary if you want to make a good moduli functor out of C. If you didn't use geometric points, you couldn't guarantee fpqc descent, only Zariski descent.

But it seems like you were more interested in the converse: why should, if we have a good in-families notion, it be sufficient to check it on geometric fibers? Well actually, here "geometric" has nothing to do with it: you should always be able to check on fibers, because you want to be talking about a family of elements of C parametrized by the base. It's just that over non-algebraically closed fields it's probably harder to say what it means to lie in C -- if you don't do it geometrically, you potentially lose stability under base change and descent.

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Lest someone else get confused, let me add that there are some relative properties P which are most definitely not the same as flat + fiberwise P. For instance, properness. – JBorger Mar 25 '10 at 1:25
thank you! i was misspeaking. there by "relative" i really meant "in-families", a special case of relative. – Dustin Clausen Mar 25 '10 at 1:28
@James: maybe I'm just missing an easy point here, but, you indeed made me get a bit confused...Could you please tell me an example of a (let's say, surjective) morphism which is flat and has proper fibers but it's not a proper morphism? Thank you in advance. – Qfwfq Mar 25 '10 at 10:24
@unknown (google): let X be A^1. Take X union (X minus point) -> X in the obvious way. – Allen Knutson Mar 25 '10 at 17:24

This isn't really an answer to your question, but I believe I should point out the following -- you ask:

WHY geometrically-reductive group schemes turn out to be the right generalization of reductive algebraic groups?

Careful, the notion of a "geometrically reductive" group doesn't seem to be an example of the sort of use of the term "geometric" that you describe.

In fact, the notion of a "geometrically reductive" group is due to Mumford -- see his book Geometric Invariant Theory -- and is defined using a property of linear representations of G (it is a weaker condition than "complete reducibility of representations"). Haboush proved for algebraic groups over fields of positive char. p>0 that reductive ==> geometrically reductive. [BTW: the proof of Haboush can be found in II.10 of Jantzen's book Representations of Algebraic Groups]

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ah sorry, when I say geometrically-reductive, I meant that all geometric fiber are reductive groups... – natura Mar 25 '10 at 19:25
I thought that was probably the case. Note that if S is a scheme and G --> S a group scheme, one says simply that G is reductive if it is affine and smooth over S, and if all its geometric fibers are connected reductive [SGA3 Tome 3] – George McNinch Mar 31 '10 at 16:17

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