Let $\overline{\mathcal M}_{g,n}$ be the compactified Deligne-Mumford moduli stack (although I don't think taking the coarse moduli space will make much of a difference here). If we decompose $g = 1 + \frac{ \sum_{i=1}^k n_i}{2}$ for positive natural numbers $n_1,\dots, n_k$, then there are several natural maps

$$ \overline{\mathcal M}_{1,n_1} \times \dots \times \overline{\mathcal M}_{1, n_k} \to \overline{\mathcal M}_{g}$$

one for each perfect matching of the $\sum_{i=1}^k n_i$ marked points that connects the $k$ elliptic curves.

Is the functoriality map $H^i\left(\overline{\mathcal M}_{g}, \mathcal O_{\overline{\mathcal M}_{g}}\right) \to H^i\left( \overline{\mathcal M}_{1,n_1} \times \dots \times \overline{\mathcal M}_{1, n_k}, \mathcal O_{ \overline{\mathcal M}_{1,n_1} \times \dots \times \overline{\mathcal M}_{1, n_k}}\right)$ ever nontrivial for $i>0$?

The motivation is that this would imply that some piece of the motive of $\overline{\mathcal M}_g$ is easy to understand, because it will be a tensor product of the motives of modular forms. You may of course substitute "Hodge structure" or "Galois representation" in for "motive" if you prefer.

I think not because it seems like that would make it "too easy" to understand part of the cohomology, but maybe it's difficult just because the map is sometimes zero and sometimes nonzero and there's no good way of telling when.

  • $\begingroup$ In the "stable range", doesn't Mumford's conjecture / the Madsen-Weiss theorem tell us that the cohomology is all $(p,p)$? $\endgroup$ Commented Jul 26, 2015 at 23:28
  • $\begingroup$ @JasonStarr Yes, I think so. So we need to take $i \geq (g-1)/2$ for the left side to be nontrivial. The right side can be nontrivial for $i$ all the way up to $2g$. $\endgroup$
    – Will Sawin
    Commented Jul 27, 2015 at 0:06
  • $\begingroup$ Is the left hand side actually known to be nonzero for some $g$ and $i$? Presumably $g$ would have to be large enough so that the moduli space is not rationally connected, but even assuming that it is of general type it is not obvious (to me) that such $g$, $i$ exist. $\endgroup$
    – naf
    Commented Jul 27, 2015 at 5:36
  • $\begingroup$ One more consideration: any non-$(p,p)$ cohomology of the Satake compactification of $A_g$ will pullback to zero on $\overline{\mathcal{M}}_{1,n_1}\times \dots \times \overline{\mathcal{M}}_{1,n_k}$. This is because the Torelli map restricted to this locus factors through any of the many projections to the product $\overline{\mathcal{M}}_{1,1}\times \dots \overline{\mathcal{M}}_{1,1}$. As far as $\mathbb{Q}$-cohomology is concerned, this stack is essentially a product of copies of $\mathbb{P}^1$, thus has only $(p,p)$-cohomology. $\endgroup$ Commented Jul 27, 2015 at 11:22
  • 1
    $\begingroup$ @JasonStarr Does this give you the right answer for stacks? If you try it for $\overline{\mathcal M}_{1,1}$, won't you get $-1/12$ or $11/12$ or something? $\endgroup$
    – Will Sawin
    Commented Jul 29, 2015 at 5:28

1 Answer 1


Not an answer but a comment. There's a paper by Pikaart, "An orbifold partition of $\overline M_g^n$", in which he proves (among other things) the following result. Consider the boundary stratum in $\overline M_g$ parametrizing a genus one curve and a curve of genus $g-11$ meeting in $11$ nodes. The corresponding pushforward map $$ H^0(\overline M_{g-11,11}) \times H^{11}(\overline M_{1,11}) \to H^{33}(\overline M_g)$$ is injective for $g$ large enough. This is Corollary 4.7 in his paper. This doesn't answer your question since this gives cohomology of type $(22,11)$ and $(11,22)$, an $11$-fold Tate twist of the motive attached to the cusp form $\Delta$, but morally it seems very similar.

It might be possible to answer your precise question by reading Pikaart's paper more carefully. Also, I believe (but I never compared the two carefully) that Pikaart's construction was essentially reinvented by Teleman and is described in Section 5 of his paper on the classification of 2d semisimple field theories. In fact, Teleman writes that this section is the key part of the whole argument. You might find Teleman's paper easier reading.

  • $\begingroup$ Interesting! I'll look into this... $\endgroup$
    – Will Sawin
    Commented Jul 30, 2015 at 13:55

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