Let $\overline{\mathcal{M}}_{g,n}$ be the moduli stack of stable curves of genus $g$ with $n$ points.


$0 \neq \gamma \in Pic(\overline{\mathcal{M}}_{g,n})$

the first Chern class of a line bundle (note that the Picard coincides with $H^2(\overline{\mathcal{M}}_{g,n}, \mathbb{Z})$ and with the Neron-Severi group in this case)

It is well-known from Arbarello-Cornalba-Harer that the group of first Chern classes is generated by $\kappa_1$ and by $\delta$-classes and $\psi$-classes.

My question: is the multiplication by $\gamma$ an isomorphism:

$ H^{3g-3+n-1} \to H^{3g-3+n+1} \ ? $

(as it happens in the Hard Lefschetz theorem when $\gamma$ is an hyperplane section). Are there $\gamma$s for which it is not an isomorphism?

This is of course true when $\gamma$ is in the ample cone or in the antiample cone.

  • $\begingroup$ Pleasechange your username $\endgroup$ – Felipe Voloch May 11 '12 at 15:34
  • $\begingroup$ Do you like this one better? $\endgroup$ – OldMacdonaldHadaForm May 11 '12 at 16:33
  • $\begingroup$ Much better. The software running this site randomly bumps old unanswered questions from time to time. When it does so, it does under the name "MathOverflow", hence my request. $\endgroup$ – Felipe Voloch May 11 '12 at 17:43
  • $\begingroup$ Dear OMHF: Yes, there are choices of $\gamma$ such that $\gamma^k$ is zero. For instance, at least for many choices of $(g,n)$, there are nonconstant morphisms $u:\overline{M}_{g,n}\to Y$ with positive fiber dimension. If $\gamma$ is the pullback under $u$ of any divisor class from $Y$, then $\gamma^k$ will be zero for $k> \text{dim}(Y)$. $\endgroup$ – Jason Starr May 11 '12 at 18:00
  • $\begingroup$ Dear Jason Starr, thank you very much for your comment. I will rephrase the question to include only the case $k=1$, which is the one I was originarily interested in. $\endgroup$ – OldMacdonaldHadaForm May 11 '12 at 18:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.