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Sorry for a maybe stupid long question but I'm reading the paper "Classical and minimal models of the moduli space of curves of genus two" by Brendan Hassett and I'm not able to unravel a construction he does at a certain point with families of genus 2 curves. I'm going through a self study of moduli spaces and so I'm not able to ask anyone I know for these types of clarifications.

Let me be more precise: given a family of genus 2 curves over a base scheme $S$ $$\pi:\mathcal{C} \rightarrow S$$ we can consider the relative dualizing sheaf $\omega_{\pi}$ (which in this case is globally generated), and the map $j:\mathcal{C} \rightarrow \mathbb{P}(\pi_{*}\omega_{\pi})$ such that $\psi:\mathbb{P}:=\mathbb{P}(\pi_{*}\omega_{\pi})\rightarrow S$ is a $\mathbb{P}^1-$bundle commuting with $j$ and $\pi$. It is clear to me that $j_{*}\mathcal{O}_{\mathcal{C}}$ is a rank 2 vector bundle on $\mathbb{P}$ but in the paper it is stated that using the trace (??) we get the decomposition $j_{*}\mathcal{O}_{\mathcal{C}}=\mathcal{O}_{\mathbb{P}}\oplus \mathcal{L}$ where $\mathcal{L}$ has degree -3 on the fibers of $\psi$.

Question 1: Why can we say so easily that $j_{*}\mathcal{O}_{\mathcal{C}}$ splits in that way? I'm not able to figure it out properly.

Next he says that the $\mathcal{O}_{\mathbb{P}}$ algebra structure on $\mathcal{O}_{\mathcal{C}}$ is determined by an isomorphism $\mathcal{L} \otimes \mathcal{L} \rightarrow \mathcal{O}_{\mathbb{P}}$, i.e. by a non vanishing section of $\mathcal{L}^{-2}$ with zeros along the branch locus of $j$.

Question 2: What is the meaning of the previous assertion? How can a non vanishing section of a line bundle have zeros along a subscheme?

Question 3: I'm not able to see why $\omega_{\psi}=(\psi^{*}\det\pi_{*}\omega_{\pi})(-2)$

Finally, taking the above arguments for granted, I understand that the class of a branch divisor is $$c_1(\mathcal{O}_{\mathbb{P}}(6))-2\psi^{*}c_1(\det \pi_{*}\omega_{\pi})$$

After this he uses the computation above to show that an automorphism of a genus 2 curve $C$ induces a linear isomorphism on $\Gamma(C,\omega_{C})$ that preserves the sextic on $\mathbb{P}^1$ associated to the cover $j$ restricted to $C$ up to a scalar multiple. Finally he checks that this scalar is $(\det(M)^{-2}$ where $M \in GL(2)$ is acting naturally on the space of sextics of $\mathbb{P}^1$. More in particular $(\det(M)^{-2}M\cdot F=F$ if and only if $M$ is induced by an automorphism of $C$.

Question 4: Is easy to see that the computation of the Chern classes above allows me to deduce that $\det(M)^{-2}$ is the right scalar that acts on $\Gamma(C,\omega_{C})$?

Thanks in advance for all the possible answers, suggestions and/or references.

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Q1: To get the splitting, you need maps $\mathcal O_{\mathbb P} \to j_* \mathcal O_C$ and $j_* \mathcal O_C \to \mathcal O_{\mathbb P}$ whose composition $\mathcal O_{\mathbb P} \to j_* \mathcal O_C \to \mathcal O_{\mathbb P}$ is the identity.

For a map $\mathcal O_{\mathbb P} \to j_* \mathcal O_C$, we do something purely formal: A section of $\mathcal O_{\mathbb P}$ gives a section of $\mathcal O_C$ on the inverse image, which is the same as a section of $j_* \mathcal O_C$ by definition. This can also be viewed as arising from an adjunction.

For a map $j_* \mathcal O_C \to \mathcal O_{\mathbb P}$, we take the trace: $j_* \mathcal O_C $ is a rank two vector bundle on ${\mathbb P}$, so every endomorphism of it has a trace in $\mathcal O_{\mathbb P}$. But since it has an algebra structure, every section gives an endomorphism.

Examining the composition of these two maps, we see that it is multiplication by two, since each section of $\mathcal O_{\mathbb P}$ acts as a diagonal matrix on $j_* \mathcal O_C$, whose trace is the sum of the two identical diagonal entries. So to get a splitting we must divide one of them by two.

Q2: I think it just means nonvanishing away from the branch divisor, with zeroes on the branch divisor.

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  • $\begingroup$ Thank you very much! Now it's very clear. Do you know why the type of splitting is exactly the trivial line bundle sum with a line bundle of degree -3 on the fibers? $\endgroup$
    – gigi
    Commented Mar 10, 2022 at 15:37
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    $\begingroup$ @gigi There's a number of ways to see it. One is Riemann-Roch, working fiber by fiber (so assume the base $S$ is a point). We know $H^1 ( \mathbb P, j_* \mathcal O_C) = H^1 ( C, \mathcal O_C) $ is a two-dimensional vector space and $H^1(\mathbb P, \mathcal O_{\mathbb P})=0$ so it all has to come from $\mathcal L$, which forces $\deg \mathcal L=-3$. $\endgroup$
    – Will Sawin
    Commented Mar 10, 2022 at 15:53

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