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Let $B$ be a curve (integral but not necessarily smooth) and let $\pi: C --> B$ be a family of curves such that each fiber is a rational curve with $g$ many elliptic tails attached.

Let $\omega$ be the relative dualizing sheaf.

Question: Why is the pushforward $\pi_* \omega$ trivial (as a vector bundle)?

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1 Answer 1

up vote 6 down vote accepted

I don't think this is quite right. Here is the right statement: let $E_1, ..., E_g$ be the tails, with maps $q_i: E_i --> B$. Then

$\pi_* \omega_{C/B} = \bigoplus (q_i)_* \omega_{E_i/B}$.

So, if your tails don't vary with B, this bundle is trivial.

Explanation: $\omega_{C/B}$ can be described explicitly: a section of $\omega_{C/B}$ is a one-form on each component of $C$, with simple poles at the nodes of $C$, so that at every node the residues of the form on the two components match.

Now, on a curve of genus $1$, a one-form with only a single simple pole, must in fact have no poles. So the sections of $\omega_{C/B}$, restricted to the $E_i$, are sections of $\omega_{E_i/B}$. Moreover, the sections of $\omega_{C/B}$ restricted to the rational components are one-forms with no poles, and are hence $0$. So to give a section of $\omega_{C/B}$ is simply to give a section of $\omega_{E_i/B}$ on each $E_i$. QED.

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