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

The Mumford Conjecture (now a theorem) says basically what is the (tautological subring)* of the rational cohomology ring of the stable moduli space of curves. Meaning that we know the ring structure (corresponding to the tautological ring)* of the cohomology of the moduli space of Riemann surfaces when the genus is very very big.

Though the problem could have been stated as an algebraic geometry one, It was worked out (as far as I know), roughly speaking by arguments coming from algebraic topology.

However, what are the implications (in general) of such a conjecture? In particular in Algebraic geometry and Algebraic topology.

*corrected, Thks.

share|improve this question
    
Maybe you will be interested in this question that I asked previously on MO: mathoverflow.net/questions/11301/… –  Kevin H. Lin Mar 11 '10 at 21:00
4  
"Mere" topological arguments...? –  Oscar Randal-Williams Mar 11 '10 at 23:13
2  
Re "we know the ring structure of the cohomology of the moduli space of Riemann surfaces when the genus is very very big.": we don't. Mumford's conjecture is on the tautological subring of the cohomology of the moduli spaces. As the genus increases, the tautological ring becomes a vanishingly small part of the whole (unstable) cohomology, see e.g. Harer, Zagier, The Euler characteristic of the moduli spaces of curves, Inv Math 85, 457-485. Moreover, Mumford's conjecture does not describe the unstable part of the tautological subrings, which, to my knowledge, is unknown in general. –  algori Mar 11 '10 at 23:26
    
@Oscar: "Mere topological arguments". I wrote it up carelessly, because I can't say I know the proof. Hope it's less rude now. –  Csar Lozano Huerta Mar 12 '10 at 4:58

1 Answer 1

One application that I know of the Mumford conjecture is Teleman's proof of Givental's conjecture in this paper. Givental's conjecture states that when the quantum cohomology of a smooth projective variety (or compact symplectic manifold) is (generically) semisimple (meaning that the algebra is semisimple for generic values of the deformation parameter), then the higher genus Gromov-Witten invariants are uniquely and explicitly determined by the quantum cohomology. Since quantum cohomology consists of genus 0 information, another way to say this is that the higher genus Gromov-Witten invariants are uniquely and explicitly determined by the genus 0 Gromov-Witten invariants, when the quantum cohomology is semisimple.

The proof, extremely roughly, uses the Mumford conjecture in the following way: semisimplicity allows you to go back and forth between low genus and high genus without loss of information. The Mumford conjecture tells you what things look like in very high genus, so if you want to know what something in some arbitrary genus looks like, you kick it up to very high genus, identify it by the Mumford conjecture, and then kick it back down to its original genus.

---Begin large parenthetical remark---

You might then ask, which smooth projective varieties (or compact symplectic manifolds) have semisimple quantum cohomology? Smooth projective toric varieties form a class of examples. Arend Bayer showed that semisimple quantum cohomology is preserved under blowing up points. There are other examples...

There is an interesting conjecture of Dubrovin which states that a variety has semisimple quantum cohomology if and only if its derived category has a full exceptional collection. (This conjecture is based heavily on mirror symmetry philosophy...) I don't know the status of this conjecture. But I think it is true, for example, that having a full exceptional collection is preserved under blowing up points. (Proved by Orlov? Bondal?)

---End large parenthetical remark---

It is also apparently possible to view the Mumford conjecture as a special case of the cobordism hypothesis. Take a look at Jacob Lurie's paper on TFTs, particularly section 2.5.

share|improve this answer
5  
I think I have to call you up on your very last statement: the Mumford conjecture is not a "special case" of the cobordism hypothesis as in Lurie's paper. What is a special case of it is the "Main Theorem" of the paper of Galatius-Madsen-Tillmann-Weiss, that identifies the homotopy type of the full cobordism category. To get from this to the Mumford conjecture one needs i) A theorem relating the full cobordism category to the "positive boundary subcategory", ii) Tillmann's generalised group completion theorem, iii) Harer stability. –  Oscar Randal-Williams Mar 11 '10 at 23:10
    
Such a nice answer, thanks a lot. That's amazing though that the fact of having semi simple QH*, means that the GW invariants in all genus are determined somehow by the degree d-rational curves. Quite interesting. That's what you are saying, if I didn't get it wrong. –  Csar Lozano Huerta Mar 12 '10 at 16:31
    
Yep, that's exactly what I'm saying. It is indeed quite amazing. –  Kevin H. Lin Mar 12 '10 at 18:50
    
Although, well, we should maybe not say "degree d rational curves" but "genus 0 GW invariants". I'm not exactly sure when GW invariants match up with actual enumerative invariants. Sometimes they do, sometimes they don't. –  Kevin H. Lin Mar 12 '10 at 18:53
    
Oscar: Thanks for the clarification! –  Kevin H. Lin Mar 12 '10 at 20:21

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.