Let $\overline{\mathcal{M}}^r_{g,n}$ the space of $n$ pointed $r$-stable curves of genus $g$, endowed with a root of the canonical sheaf, and let $\mathcal{C} \to \overline{\mathcal{M}}^r_{g,n}$ be the universal curve. The family $\mathcal{C}$ is endowed with a universal root of the canonical, that I denote $\mathcal{S}$. What is the relation between $\pi_!\mathcal{S}$ and the Hodge bundle $\mathbb{E}:=\pi_*\omega_{\mathcal{C}}$ on $\overline{\mathcal{M}}^r_{g,n}$? For instance what can one say about the chern classes?
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2$\begingroup$ $\pi _*\mathcal{S}$ is zero (at least when $r>2$), because a $r$-th root of $\omega _C$ on a generic curve $C$ has no section. Maybe you mean $\pi _!\mathcal{S}$ ? $\endgroup$– abxCommented Nov 18, 2013 at 17:32
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$\begingroup$ Good point. Yes $\pi_{!}\mathcal{S}$ would be ok, I edit the question. I would be curious to see even in the derived category. $\endgroup$– IMeasyCommented Nov 18, 2013 at 20:12
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2 Answers
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A detailed computation of the Chern character of $R^\bullet \pi_\ast \mathcal S$ can be found in Alessandro Chiodo's "Towards an enumerative geometry of the moduli space of twisted curves and $r$th roots". As abx says it is a generalization of Mumford's computation of the Chern character of the Hodge bundle.
Let me also say that (at least today) this is a problem that just screams for an application of Grothendieck-Riemann-Roch (which is indeed what Mumford and Chiodo are using).
answered Nov 18, 2013 at 21:38
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$\begingroup$ Yeah, both your answer are very nice. It is clear that GRR is the way to get it right. I seem to understand anyway that it would be far too optimistic to expect a quick formula relating the two sheaves. $\endgroup$– IMeasyCommented Nov 18, 2013 at 21:43
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I assume that your $\mathcal{S}$ satisfies $\mathcal{S}^{\otimes r}\cong \omega _{\mathcal{C}/\mathcal{M}}$. Then the computation of $\pi _!\mathcal{S}$ can be done exactly as the computation of $\pi _*\omega _{\mathcal{C}/\mathcal{M}}^{\otimes n}$ in Mumford, Stability of projective varieties, L'Enseignement Mathématique, Vol. 23 (1977), §5 $-$ just put $n=\frac{1}{r} $. The result is more complicated that what you suggest, it involves the boundary divisors.
