# Mumford conjecture: Heuristic reasons? Generalizations? … Algebraic geometry approaches?

The Mumford conjecture states that for each integer $n$, we have: the map $\mathbb{Q}[x_1,x_2,\dots] \to H^\ast(M_g ; \mathbb{Q})$ sending $x_i$ to the kappa class $\kappa_i$, is an isomorphism in degrees less than $n$, for sufficiently large $g$. Here $M_g$ denotes the moduli of genus $g$ curves, and the degree of $x_i$ is the degree of the kappa class $\kappa_i$. This conjecture was proved by Madsen-Weiss a few years ago.

1. What are the heuristic or moral reasons for the conjecture? (EDIT: I am particularly interested in algebraic geometric reasons, if there are any. Though algebraic topologial reasons are very welcome too.) What lead Mumford to formulating the conjecture in the first place?

2. I know very little about the Madsen-Weiss proof, but I know that it mainly uses algebraic topology methods. Are there any approaches to the conjecture which are more algebraic-geometric?

3. Is there any analogous theorem or conjecture regarding the (topological) $K$-theory of $M_g$? Or the Chow ring of $M_g$? etc.

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I've added a bounty to attract some more attention, in hopes that someone can say something about the algebraic geometry aspects of my questions. Perhaps there just aren't any good algebraic geometry approaches to the Mumford conjecture --- if this is the case, then why is it so? What makes the Mumford conjecture difficult from the perspective of algebraic geometry? –  Kevin H. Lin Jan 17 '10 at 3:28
Thanks to all who answered! I wish I could accept all of them :( –  Kevin H. Lin Jan 22 '10 at 20:24
What is "wcatss"? –  Kevin H. Lin Jun 11 '10 at 21:57
about the "wcatss" tag, see the comments to mathoverflow.net/questions/7375/… –  j.c. Jun 11 '10 at 23:19

All current proofs of Mumford's conjecture in fact prove a far stronger result, the "Strong Mumford conjecture", first formulated by Ib Madsen. This says the following (where by "moduli space" in the following we mean a homotopy type classifying concordance classes of surface bundles with perhaps some extra structure): there is a stable moduli space $$\mathcal{M}_\infty := \mathrm{colim} \,\, \mathcal{M}_{g, 1}$$ where $\mathcal{M}_{g, 1}$ denotes the moduli space of genus $g$ surfaces with a single boundary component, and the colimit is formed by gluing on a torus with a single boundary component using the "pair of pants" product.

There is also a space, usually called $\Omega^\infty MTSO(2)$, which classifies cobordism classes of "formal surface bundles": that is, codimension -2 submersions with an orientation of the (stable) vertical tangent bundle. As a surface bundle is a formal surface bundle, there is a map $$\alpha: \mathcal{M}_\infty \to \Omega^\infty MTSO(2).$$ The strong Mumford conjecture says that this is an integral homology equivalence. For the record, there are currently four distinct known proofs, due to:

1. The stable moduli space of Riemann surfaces: Mumford's conjecture, Madsen and Weiss,
2. The homotopy type of the cobordism category , Galatius, Madsen, Tillmann and Weiss,
3. Monoids of moduli spaces of manifolds, Galatius and myself,
4. Madsen-Weiss for geometrically minded topologists, Eliashberg, Galatius and Mishachev.

For part 1) of your question, from this point of view (I do not know what Mumford had in mind): the map $\alpha$ can be thought of as comparable to the map which compares holonomic sections to formal sections in the statement of Gromov's $h$-principle for a sheaf. The idea is then that given a "formal surface bundle" one may begin improving it to be more and more like a bundle, but in this process one cannot really control the genus of the fibres one ends up with. This is why it gives the infinite genus moduli space. This is more or less the approach to proving the conjecture that Eliashberg, Galatius and Mishachev take. The other approaches are less direct and use more algebraic topological machinery.

So, for part 3) of your question: the map $\alpha$ is also an equivalence in any other (co)homology theory (by the Atiyah-Hirzebruch spectral sequence, for example). Thus the topological K-theory of $\mathcal{M}_\infty$ is "known" in the sense that it is the K-theory of the infinite loop space of a well-understood spectrum, which fits into various simple cofibration sequences and so on. On the other hand, it is "not known" in the sense that I don't think anyone knows what $K^0(\mathcal{M}_\infty)$ is as a group, though I once tried to compute it without success. On the other hand, even knowing this group, it does not necessarily tell you anything about what you are really interested in, $K^0(\mathcal{M}_g)$, because stability for ordinary homology does not imply stability in non-connective (co)homology theories.

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Thanks for your remarks! But sorry, I am not familiar with Gromov's h-principle? –  Kevin H. Lin Jan 10 '10 at 17:26
@Kevin Lin, the book "An Introduction to the h-principle" by Eliashberg and Mishachev is excellent for learning about the h-principle in general, although it doesn't discuss the Mumford conjecture. –  Joel Fine Jan 10 '10 at 17:40
Kevin, I would not worry too much about the h-principle stuff. I think Oscar's comment about the h-principle may be a way to think of $\alpha$ if you are already familiar with the h-principle. I could be completely wrong though. –  Sean Tilson Aug 19 '10 at 1:40
@Oscar: I think the k-theory computation isn't too hard. It is easy enough to do the computation for connective K-theory (real or complex) for the particular example of $MTSO(2)$. You can use an Adams SS for example. I think Johannes Ebert has another method to do the computation though. –  Sean Tilson May 3 '12 at 4:20

Others, more expert than I, will no doubt weigh in soon, but I'll start on 2: I think at present one definitely does not have an approach to something like the Mumford conjecture, or even Harer stability, "within" algebraic geometry. For example, I don't think one knows how to prove anything about the stable cohomology of M_g in characteristic p without appeal to characteristic 0 and using topology.

Though in this connection see the (in some respects controversial) work of Boggi for a purely algebro-geometric setup meant to analogize Harer.

Update: OA asked for more about why it was difficult to prove Mumford's theorem within algebraic geometry. Here's at least part of the answer. There are really two pieces to Mumford's theorem. On the one hand, you need to understand what the "moduli space of curvces of infinite genus" is -- i.e. you have to understand the limit given in Oscar Randall-Williams's answer. On the other hand, knowledge of this limit alone doesn't tell you that the cohomology of the individual M_g's behaves itself. You can certainly e.g. have a sequence of abelian groups whose direct limit is 0 even though none of the groups vanish. Harer's theorem guarantees that, for each M_g, the cohomology of M_g agrees with that of M_infty in a range of degrees linear in g. And for Harer's theorem you really need that the complex of curves on the genus-g surface is contractible in a linear-in-g range of degrees. But now we're talking about an topological object that, as far as I know, doesn't really have any manifestation in algebraic geometry.

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Could you elaborate on what you mean by controversial? –  Pete L. Clark Jan 10 '10 at 13:26
My understanding is that there are some mistakes (having to do with the passage between Hodge structures with Z-coefficients and Q-coefficients) which leave the proof of the main theorem incomplete, but that the "profinite teichmuller theory" setup is OK. –  JSE Jan 10 '10 at 14:04
As I understood it once, Boggi's work was meant to go further than simply understanding Harer stability in the algebro-geometric setting. It, together with some work of Looijenga, was meant to lead to a proof of the stable Mumford conjecture. I can't remember if this is related directly or not, but I recall someone once explaining to me that one (Looijenga and Boggi) could prove the Mumford conjecture in an algebraic way if one could first show that the mapping class group is 'good' in the sense that the profinite completion K(MCG,1)^ is equivalent to K(MCG^,1). Anyone know more of this? –  Jeffrey Giansiracusa Apr 1 '10 at 16:26
I'm not sure, but I know that Boggi's paper asserts that identity of profinite groups, that most people believe the identity is correct, and that people are definitely trying to write down a complete proof. –  JSE Apr 1 '10 at 18:26
I think you all are conflating two unrelated false proofs of Boggi. In one sequence of two papers (on the arXiv and entitled "Relative pro-$\ell$ completions of Teichmüller groups" and "The Teichmüller group is good") he purports to prove the Mumford conjecture. I believe that this is what Jeffrey is talking about. In another paper (entitled "Profinite Teichmuller Theory"), he purports to prove the congruence subgroup property for the mapping class group. This is what JSE is talking about. As far as I can tell, the errors have nothing to do with one another. –  Andy Putman Apr 27 '10 at 5:20

Here is a belated and perhaps too naive answer for your first question, i.e. a heuristic reason for why to believe in the Mumford conjecture.

One can think of cohomology classes on $M_g$ as characteristic classes for families of curves. That is, a class is the same thing as a rule such that whenever you are given a smooth family of curves $X \to B$ of genus $g$, this rule associates a cohomology class in $H^\bullet(B)$ in a functorial manner. (Topologically, you can also think of it as a characteristic class of oriented surface bundles of genus $g$, since $\mathrm{MCG}(\Sigma_g) \simeq \mathrm{Diff}^+(\Sigma_g)$.)

In a similar way, one might then think of a cohomology class on $M_\infty$ as a rule that assigns a cohomology class to a family of curves of arbitrary genus, or as a characteristic class of arbitrary oriented surface bundles. Of course this should be made precise since "functoriality" of such a characteristic class seems meaningless without any way of comparing surface bundles of different genus, and so one needs to define comparison maps between different moduli spaces by working with boundary components, gluing on tori/pants, etc. Nevertheless the intuitive picture is clear: a class on $M_\infty$ should be a rule that assigns in a uniform manner, to any family of curves $X \to B$ whatsoever, a cohomology class in $H^\bullet(B)$.

Now the $\kappa$ classes seem to fit the bill for being classes on $M_\infty$: given any $\pi \colon X \to B$ whatsoever, we may form the vertical tangent bundle, take its Euler class, multiply, push forward. This seems as canonical as one could hope for. Are there any others? Well, there are the $\lambda$-classes, i.e. the Chern classes of $\pi_\ast \Omega_{X/B}^1$, but Mumford showed via Grothendieck-Riemann-Roch that these are polynomials in the $\kappa$'s. It's hard to think of anything else.

Now surjectivity of $\mathbf Q[\kappa_1,\kappa_2,\ldots] \to H^\bullet(M_\infty)$ asserts that these obvious classes are really the only ones that you can write down in a uniform way, and injectivity says that there are no uniform relations between the $\kappa$'s (i.e. all relations are "low-genus accidents"). Now one might believe in the Mumford conjecture simply because people thought hard about these things and could not find any other genus-invariant characteristic classes, nor any genus-invariant relations between the $\kappa$'s. Mumford's conjecture is the simplest possible explanation for this failure.

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