Monday, 14 January 2008

ag.algebraic geometry - Stacks and sheaves

Let me see if I understand your example correctly: you are fixing X and Y, families
of curves over S, and now you are considering the functor which maps an S-scheme T
to the set of T-isomorphisms fXtofY (where f is the map from T to S).



If I have things straight, then this functor shouldn't be so bad to think about, because it is actually representable, by an Isom scheme. In other words, there is an S-scheme
IsomS(X,Y) whose T-valued points, for any f:TtoS, are precisely the T-isomorphisms
from fX to fY. (One can construct the Isom scheme by looking inside a
certain well-chosen Hilbert scheme.)



One way to think about this geometrically is as follows: one can imagine that two
curves over k (a field) are isomorphic precisely when certain invariants coincide
(e.g. for elliptic curves, the j-invariant). (Of course this is a simplification,
and the whole point of the theory of moduli spaces/schemes/stacks is to make it precise,
but it is a helpful intuition.) Now if we have a family X over S, these invariants
vary over S to give a collections of functions on S (e.g. a function j in the
genus 1 case), and similarly with Y. Now X and Y will have isomorphic
fibres precisely at those points where the invariants coincide, so if we look
at the subscheme Z of S defined by the coincidence of the invariants,
we expect that fX and fY will be isomorphic precisely if the map f
factors through Z. Thus Z is a rough approximation to the Isom scheme.



It is not precisely the Isom scheme, because curves sometimes have non-trivial
automorphisms, and so even if we know that Xs and Ys are isomorphic for
some sinS, they may be isomorphic in more than one way. So actually the
Isom scheme will be some kind of (possibly ramified) finite cover of Z.



Of course, if one pursues this line of intuition much more seriously, one will
recover the notions of moduli stack, coarse moduli space, and so on.



Added: The following additional remark might help:



The families X and Y over S correspond to a map phi:StomathcalMgtimesmathcalMg. The stack which maps a T-scheme to IsomT(fX,fY) can then seen to be the fibre product of the map phi and the diagonal
Delta:mathcalMgtomathcalMgtimesmathcalMg.



In the particular case of mathcalMg the fact that this fibre product is representable is part of the condition that mathcalMg be an algebraic stack.



But in general, the construction you describe is the construction of a fibre product
with the diagonal. This might help with the geometric picture, and make the relationship to Mike's answer clearer. (For the latter:note that the path space into X has a natural
projection to XtimesX (take the two endpoints), and the loop space is the fibre product
of the path space with the diagonal XtoXtimesX.)

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