For the definition of moduli space of Riemann surface, it seems that it is not based on the definiton in terms of functor(representable...) but rather put a topology on the set of isormophism classes. Is there any difference between these? Or if I miss something? Is there any reference ? Thank you!
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$\begingroup$ Possibly this might help? en.wikipedia.org/wiki/Equidistribution_theorem $\endgroup$– Talvi WatiaSep 30, 2010 at 3:10
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3 Answers
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In the complex analytic setting one can also work in terms of representing functors and so on; unfortunately, just as in the algebraic setting, it will turn out that the natural functor is not representable without either imposing level structure or passing to a larger category of stack-like objects.
However, the complex analytic study of Riemann surfaces and their moduli has its own traditions, coming from Teichmuller theory, the theory of the mapping class group, the analytic theory of Ahlfors and Bers, and so on, and these traditions come with their own view-points, in which emphasizing the representability of the functor by a certain stack-like object may not be of paramount importance.
Just to see why this might be, consider that if $f: X \to S$ is a smooth proper map of complex analytic manifolds (e.g. a family of Riemann surfaces over a smooth base), then by Ehresmann's theorem, $f$ is a fibre-bundle in the $\mathcal C^{\infty}$-category. So one sees that the fibres of $f$ don't vary topologically, and one can try to consider directly how the complex structure varies. (This is the Kodaira--Spencer view-point on deformation theory.) The fact that one has these vivid geometric view-points available means that there is less pressure to use categorical language to describe the situation.
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1$\begingroup$ Historically, the analytic theory also had a much earlier start, so in addition to sharper technical (local) tools, no incentive to recast the theory in complicated algebraic formalism. I think that recently some results in enumerative geometry appeared first in the algebraic setting, e.g., physicists conjecture formula on G-W invariants in analytic setting, mathematical proof of more general formula is carried out in algebraic context. So dichotomy of methods may be less clear today. $\endgroup$– T..Sep 20, 2010 at 6:11
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$\begingroup$ I just want to see the proof of equivalence of two definitions,one is in terms of representable functor and one is "direct" analytic construction. I even don't know the precise definitions of those two. $\endgroup$– abcSep 22, 2010 at 23:57
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$\begingroup$ @abc: Dear abc, I don't know if there is such a proof written in an accessible way in the literature. You could try looking at Deligne--Mumford: in one of their proofs of irreducibility of $\mathcal M_g$, they compare with the complex case and appeal to the known structure there, which (if I'm remembering correctly) they obtain by considering the analytic picture. If you can't find a reference this way, you might consider asking your question on MO, where it is more likely to be read by someone who knows the relevant literature well. $\endgroup$– Matt ESep 23, 2010 at 0:48
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From the mention of representability I assume you mean Riemann surfaces as algebraic curves, not the complex-analytic or conformal moduli problems.
Representing the moduli functor would mean you have object in a "reasonable" category (e.g., a variety, scheme, or algebraic space) that is a universal family of curves of whatever kind you are considering. But this doesn't exist for algebraic curves without further restrictions (eg., stability) and rigidification data (such as marked points) to get a "fine moduli space". However, if by topology you mean a Grothendieck topology then you can define such a thing, i.e., a suitable category of coverings associated to the problem, as a virtual moduli object that you can consider as a space in the sense that cohomology and other geometrical invariants can be associated to it.
The canonical and much admired exposition that explained this to the world is "Picard Groups of Moduli Problems" by David Mumford. Among other achievements it shows how to work with moduli stacks, without using the word "stack" (which may have been coined later, I don't know).
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$\begingroup$ T.., if I consider complex-analytic or conformal moduli problems, we can still consider representability of functor, is it right? for example, we can still define "family of Riemann surface"...why in the literature it seems that it just defines a topology on the isomorphism set then call it "moduli space"?Can we fit the latter into the "representability of functor "picture? Thank you! $\endgroup$– abcSep 20, 2010 at 3:55
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1$\begingroup$ Dear T.., The word "stack" was coined by Deligne--Mumford in their paper on irreducibility of $M_g$, I believe. (It was offered as a translation of the French word "champ", since the literal translation ("field") was already overloaded in mathematical English.) This paper is from the late 60s, and comes after Mumford's "Picard groups" paper, I think. $\endgroup$– Matt ESep 20, 2010 at 4:19
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1$\begingroup$ @abc: Yes, and there is also in that case the lack of a universal family, with a similar remedy: consider orbifolds instead of manifolds. However, the difference is that the situation is more concrete and does not need Grothendieck topologies to allow definition and computation of related geometrical objects (and natural "coverings" might be weaker, topological only and not carry enough information). I think the answer to why place coordinates, and thus a topology, on the isomorphism classes is "because we can" -- analysis makes this possible, unlike situation for varieties or schemes. $\endgroup$– T..Sep 20, 2010 at 4:41
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1$\begingroup$ @Matt: thanks, I had a look at Deligne-Mumford paper and in addition to coining "stack" it is they who retroactively designate Mumford's Picard groups paper as being about "moduli stacks" of elliptic curves. According to Dan Edidin's What-is-a-Stack article, stack and sheaf were both under consideration as translations of "gerbe", leaving stack as the only possibility for new terminology. $\endgroup$– T..Sep 20, 2010 at 6:26
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$\begingroup$ @T.., Edidin claims that "stack" is a possible translation of "gerbe," but I see no evidence of that in a couple of French dictionaries I checked. $\endgroup$– ColinApr 27 at 13:18
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The two viewpoints (analytic and algebro-geometric) lead to the same coarse moduli space, regarded as a complex analytic variety. The same applies even to their Deligne-Mumford compactification (which was originally constructed algebro-geometrically but also has an analytic description). See
John Hubbard and Sarah Koch, An analytic construction of the Deligne-Mumford compactification of the moduli space of curves, J. Differential Geometry, 98 (2014) 261-313.
answered Jan 27, 2018 at 20:05