## Virtual coherent cohomology

I gave a talk yesterday where I attempted to draw parallels between the cohomology of (arithmetic) 3-manifolds and weight one modular forms. It was natural then to think about whether there was an analogue of the virtual Betti number conjecture. Recall the following:

Theorem [Agol] Let $M$ be a compact hyperbolic 3-manifold. Then $\mathrm{dim} H^1(N,\mathbf{Q})$ is unbounded as $N$ ranges over all finite covers $N \rightarrow M.$

(There’s an analogous version for finite volume hyperbolic manifolds with cusps.) What is the corresponding conjecture in coherent cohomology? Here is a first attempt at such a question.

Question: Let $X$ be a proper smooth curve of genus $g \ge 2$ defined over $\mathbf{Q}.$ Let $\mathscr{L}$ denote a line bundle such that $\mathscr{L}^{\otimes 2} = \Omega^1_X.$ As one ranges over all (finite etale) covers $\pi: Y \rightarrow X,$ are the groups

$H^0(Y,\pi^* \mathscr{L})$

of unbounded dimension?

One might ask the weaker question as to whether there is a cover where this space has dimension at least one (and in fact this is the first question which occurred to me). However, there are some parity issues. Namely, Mumford showed the dimension of $H^0(X, \mathscr{L})$ is locally constant in $(X,\mathscr{L})$, and this dimension is odd for precisely $2^{g-1}(2^g + 1)$ choices of $\mathscr{L}$ (there are $2^{2g}$ such choices and the choices are a torsor for 2-torsion in the corresponding Jacobian). But I think this means that one can always make $\pi^* \mathscr{L}$ effective for some degree 2 cover, and thus produce at least one dimensions worth of sections. For example, when $g = 1,$ then $\Omega^1_X = \mathcal{O}_X$, and $\mathscr{L} = \mathcal{O}_X$ has global sections whereas the other square-roots correspond literally to 2-torsion points. But those sections become trivial after making the appropriate 2-isogeny.

Another subtlety about this question which is worth mentioning is that I think the result will have to be false over the complex numbers, hence the deliberate assumption that X was defined over $\mathbf{Q},$ or at least over a number field. Specifically, I think it should be a consequence of Brill-Noether theory that the set of X in $\mathscr{M}_g$ such that

$\mathrm{dim} H^0(Y, \pi^* \mathscr{L}) > 1$

for any choice of $\mathscr{L}$ and any cover $\pi: Y \rightarrow X$ of degree bounded by a fixed constant D will be a finite union of proper varieties of positive dimension. And now the usual argument shows that, as D increases, any countable union of varieties cannot exhaust $\mathscr{M}_g.$ But it can, of course, exhuast all the rational points, and even all the algebraic points.

There’s not really much evidence in favor of this question, beyond the following three very minor remarks.

1. The only slightly non-trivial case one can say anything about is when $X$ is a Shimura curve over $\mathbf{Q},$ and then the answer is positive because there exist lots of weight one forms (which one can massage to have the right local structures after passing to a finite cover).
2. The analogy between $H^0(X,\mathscr{L})$ and $H^1(M,\mathbf{Q})$ is fairly compelling in the arithmetic case, so why not?
3. There doesn’t seem to be any a priori reason why the virtual Betti number conjeture itself was true, and it is certainly false in for related classes of groups (groups with the same number of generators and relations, word hyperbolic groups), so, by some meta-mathematical jiu-jitsu, one can view the lack of a good heuristic in the hyperbolic case as excusing any real heuristic in the coherent case.
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### 7 Responses to Virtual coherent cohomology

1. voloch says:

If X is defined over a field of characteristic two, then there is a natural line bundle like in your question that you can take. Namely, in terms of divisors, the divisor of an exact differential dx is of the form 2D. Use the line bundle corresponding to D and consider first etale covers of X with degree a power of 2. If X is ordinary your H^0’s will be trivial and if X is not ordinary they will grow linearly with the degree. The proof is not difficult but is too long for a blog comment. Now, if X is ordinary but has a non-ordinary etale cover, then you get unbounded H^0 by taking covers of that cover. Does that work for all X? Maybe, I don’t know. Unfortunately, this does not help in characteristic zero even if X has good reduction at 2, as the inequalities for H^0 go the wrong way.

Are you in Melbourne? Would you like to visit Christchurch? Get in touch if you do.

• Interesting comment! I’m in Melbourne yes, but on a fairly tight schedule. My next stop is the annual general meeting at MSRI on March 3 (the subject of my next post).

• voloch says:

Replying to myself to record that I ran across a result of Pop and Saidi that says that an ordinary curve with absolutely simple Jacobian has non-ordinary etale covers.

2. JSE says:
• I remember that post! What I didn’t remember was the similar rhetorical device of using some analogy to virtual bigness to justify a possibly idle (but completely blog-worthy) question.

3. JSE says:

And of course the comment about what goes wrong over C reveals that this is yet another example of a charming kind of question, “when can a “natural” countable union of proper subvarieties cover all the Qbar-points of a variety?” I scratched my chin about this one too https://quomodocumque.wordpress.com/2008/12/15/some-visitors-and-countable-unions/
but I think your example goes even beyond the class of examples I discussed there!

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