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Let pi: \bar{Mg,1} \to \bar{M_g} be natural projection of compactified moduli stacks of curves and let omega be the relative dualizing sheaf. Then the Hodge class \lambda of \bar{M_g} is the first chern class of the pushforward \pi_*(ω). Among other things the hodge class, together with the boundary divisors, freely generates the Picard group of \bar{M_g}.

Question: Why is lambda big and nef?

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  • $\begingroup$ Eh, I think it's an interesting question that would be noticed by more people if it mentioned the main object, namely M_g, in the title. $\endgroup$
    – Ilya Nikokoshev
    Commented Oct 22, 2009 at 22:18

3 Answers 3

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Some multiple of lambda is defined on the coarse moduli space and this is the pullback of an ample bundle on \bar{A_g}, the Satake-Baily-Borel compactification of A_g. Since \bar{M_g} maps birationally onto its image in \bar{A_g}, it follows that lambda is nef and big, in fact also semi-ample (some multiple is base point free) on the coarse moduli space.

(The map to \bar{A_g } contracts the boundary divisor corresponding to irreducible nodal curves so lambda is not ample.)

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Let H be the Hodge bundle on \bar{M_g}, and let D be the boundary. Cornalba and Harris (MR0974412) show that aH - bD is ample iff a > 11b > 0. This means, for example, that 12H - D is ample, which implies that 12H is big, and so the same is true for H.

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For nef you can do the following: Take a map from some Hurwitz scheme to \bar{Mg}. Then any test curve in \bar{Mg} pulls to one in the Hurwitz scheme. You now compute the integral of the hodge class on the pullback of this test curve using Grothendieck-Riemann-Roch on the surface lying over the test curve.

I'm not sure that it's big - is it ?

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