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. 



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> **Question**: Why is the pushforward $\pi\_\* \omega$ trivial (as a vector bundle)?
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Question about a family of semistable curves
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.


