Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Is I consider an ind scheme such as $G(k((t)))$ for a reductive connected group over $k=\bar{k}$

I have the conjugacy action of $G(k[[t]])$.

In what category can I make the quotient $[G(k((t))/ad(G(k[[t]])]$?

In the category of presheaves? The category of fppf sheaves?

And if I want to make the fiber product of two such object over $k$, same question.

Moreover, if $f:X\rightarrow Y$ is a morphism of presheaves, i.e. contravariant functors from the category of schemes to sets, is the notion of formall smoothness (by using the infinitesimal lifting property as the definition of formal smoothness) stable under base change?

share|improve this question
    
It would be better if you give a real definition of the "ind scheme $G(k((t)))$", such as via a specific functor on a specific category of $k$-algebras (and thereby make it clearer if the fppf topology is adequate or perhaps you need the fpqc topology, if you consider $A \otimes_k k[[t]]$ or $A[[t]]$, etc.). As it stands, you are writing down a set of $k$-valued points and nothing more, which is inadequate to do anything rigorously. And to answer your questions (apart from the final one) in a useful way, it would help to know for what purpose you would apply answers. –  user28172 Nov 29 '12 at 2:49
    
To have a ind-scheme structure on G(k((t))) ,for a k-algebra R , the R-points will be $R$ G(R((t)). –  prochet Nov 29 '12 at 3:38

1 Answer 1

up vote 1 down vote accepted

Not sure what you are really asking. You can always do quotients in fppf sheaves, as explained by ancient greeks. Just quotient presheaves and sheafify around.

The point about the affine grassmanian is that it is turning out to be an ind-scheme, which is not true for quotients of general ind-schemes.

I am not sure about the last question. You can say that your thingy is formally smooth if you have the lifting property along the usual schemes. But I do not know what it means, what the examples and utility are, and whether anyone looked into them. In general, you should probably avoid even thinking about it: consider spending your life in fppf sheaves!

share|improve this answer
2  
Hey Doc, the last one's essentially a tautology. Recall that for representable (contravariant) functors base change along $S' \rightarrow S$ (when functor is equipped with a map to a given scheme $S$) corresponds to restriction of the functor to the category of $S'$-schemes within the category of $S$-schemes, so we take the latter as the definition of base change for preheaves (i.e., contravariant functors). For ${\rm{Spec}}(A)$ over $S'$ (so over $S$), $I \subset A$ with $I^2=0$, and $F' = F|_{S'}$, $F'(A) = F(A)$ and same for $A/I$. Thus, trivially base change preserves formal smoothness. –  user29283 Nov 28 '12 at 7:51

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.