4
$\begingroup$

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)?

$\endgroup$

1 Answer 1

8
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.