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Let $\pi:\overline{\mathcal{M}}_{g,n+1}\to \overline{\mathcal{M}}_{g,n}$ be the map that forgets the last marked point and $\omega_\pi$ the relative cotangent line bundle (Here we are identifying $\overline{\mathcal{M}}_{g,n+1}$ with the "universal" curve). On $\overline{\mathcal{M}}_{g,n+1}$ there are boundary divisors $\Delta_{i:S}$ for $i\in\{0,\dots, \lfloor g/2\rfloor\}$ and $S\subseteq \{1,\dots ,n+1\}$ (with $|S|\geq 2$ when $i=0$), whose general element is a genus $i$ curve glued to a genus $g-i$ curve with the marked points indexed by $S$ lie on the genus $i$ component.

One can consider divisors on $\overline{\mathcal{M}}_{g,n}$ of the form $\pi_*(c_1(\omega_\pi).\Delta_{i:S})$. Are the classes of these divisors known? (Meaning that, are their representation by the well known generators of $\mbox{Pic}(\overline{\mathcal{M}}_{g,n})$ known?

Any help would be appreciated.

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  • $\begingroup$ You should also allow $i$ to equal 0. It should be possible to write a linear relation involving your class via GRR applied to the invertible sheaf of that boundary divisor. $\endgroup$ Dec 26, 2015 at 13:29
  • $\begingroup$ Thanks for the correction. As for your suggestion, it was also my initial idea to try GRR, but when I do so, on the left hand side of GRR I get the first Chern class of the pushforward of the invertible sheaf of $\Delta_{i;S}$, which I could not compute... $\endgroup$ Dec 26, 2015 at 14:16
  • $\begingroup$ There is a special case when $i$ equals $0$ and $S$ has cardinality $2$. Except for that case, the pushforward $\mathcal{F}$ of $\mathcal{O}(\Delta_{i;S})$ is a torsion-free sheaf of rank $1$. Denote by $U$ the maximal open subset of $\overline{\mathcal{M}}_{g,n}$ such that $\pi^{-1}(U)$ is disjoint from $\Delta_{i;S}$. The restriction of $\mathcal{F}$ to $U$ is isomorphic to $\mathcal{O}_U$. This isomorphism gives a rational section $s$ of $\mathcal{F}$ with $\text{Div}(s)$ disjoint from $U$. The computation of $\text{Div}(s)$ is "local" at the generic points of the complement of $U$. $\endgroup$ Dec 26, 2015 at 15:49

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