Consider an elliptic surface $f :X \rightarrow C$ with $\chi(X) > 0$ or equivalently the fibration has a reduced singular fiber apart from possibly multiple fibers (the field under consideration is $\mathbb{C}$,$X$,$C$ are smooth and projective ). Denote by $\Omega$ the cotangent bundle of $X$ and $K$ the canonical bundle of $C$. We have an inclusion \begin{equation} 0 \rightarrow f^*K \rightarrow \Omega \end{equation} Let $q(X)$ denote the dimension of $H^0(X,\Omega)$ over $\mathbb{C}$. We have $q(X) = g$ where $g$ is the genus of the curve $C$.On the other hand dimension of $H^{0}(C,K)$ is $g$ as well and hence we get (using the above inclusion of sheaves) \begin{equation} H^{0}(C,K) = H^{0}(X,\Omega). \end{equation} Now what I am trying to understand is the above equality in a better way (rather than just by using the dimension argument as above). Is there a way to see why all the 1forms are pull backs of those from the base curve?.

$\begingroup$ Please fix the markup. $\endgroup$– Serge LvovskiFeb 15, 2013 at 10:53

1$\begingroup$ You could write down what the entire exact sequence should be. If I'm not mistaken, it should be $0\to f^\ast K\to \Omega\to \omega_f\to 0$, where $\omega_f$ is the relative dualizing sheaf. This is a line bundle on $X$. It is zero because $\omega_f$ reduces to the canonical sheaf on the generic fibre of $X\to C$ , and the generic fibre is an elliptic curve, right? If you don't like "relative dualizing sheaves", then just note that the quotient of $\Omega$ by $f^\ast K$ reduces to the canonical sheaf on the generic fibre. So your "surjective map" should be an isomorphism. $\endgroup$– Ariyan JavanpeykarFeb 15, 2013 at 18:17

1$\begingroup$ @ariyan I think if I assume further there are no multiple fibers (which I should have mentioned above) only then the quotient sheaf $\Omega/ f^*K$ is torsion free. In that case I thought this quotient should be of the form $\omega_f \otimes I_Z$ where $I_Z$ is the ideal sheaf defining the singular points in the fibers. $\endgroup$– rvarmaFeb 16, 2013 at 2:38

1$\begingroup$ This proof has to take advantage of the fact that the family is not isotrivial, because the theorem is false there. One way to do this is using Hodge theory to compare sections of the tangent bundle to a cohomology group, and the Leray spectral sequence to compute the cohomology group. But presumably that's just a version of the dimension argument you don't like. So you need some more clever way to take advantage of that. As Ariyan points out, the key fact is that any section of $\Sigma$ must map to $0$ on $\omega_f$ and thus arise from $K$. $\endgroup$– Will SawinFeb 16, 2013 at 5:09

2$\begingroup$ @Ariyan: I dońt understand your exact sequence: $\omega_f$ should be the sheaf of differentials of $f$. This will not be a line bundle unless $f$ is smooth everywhere. $\endgroup$– Damian RösslerFeb 16, 2013 at 11:20
2 Answers
This is a comment starting from a slightly more general context. Most of the following material can be found in a paper of T. Saito and me (but most dealing with the positive characterisitc case).
Let $f : X\to C$ be a flat morphism of smooth (geometrically connected) projective varieties over a field $k$ of characteristic $0$. Consider the canonical exact sequence
$$ 0 \to f^*\Omega_{C/k}\to \Omega_{X/k}\to \Omega_{X/C} \to 0. $$
We have $f_{*}\mathcal O_X=\mathcal O_C$. Taking $f_*$ we get an exact sequence of sheaves on $C$:
$$ 0 \to \Omega_{C/k}\to f_{*}\Omega_{X/k} \to f_{*}\Omega_{X/C}\stackrel{\theta}{\to} R^1f_{*}(\mathcal O_X)\otimes\Omega_{C/k}. $$
We have $R^1f_{*}(\mathcal O_X)\simeq \omega_{X/C}^{\vee}$ (in characteristic $0$).
Let $T=\Omega_{X/C, \rm{tors}}$ (torsion as $\mathcal O_X$module).
A local analysis shows easily that $T$ is an invertible sheaf over the verticla divisor $D:=\sum_{s\in C} D_s$, where $D_s=X_s(X_s)_{\mathrm{red}}$ (here again we need $k$ of characteristic $0$) and we have an exact sequence
$$ 0 \to T\to \Omega_{X/C}\to \omega_{X/C}(D) \to S \to 0$$
with $S$ of finite length. Thus the $\mathcal O_C$torsion part of $f_*\Omega_{X/C}$ is exactly $f_*T$.
At the generic fiber $\theta$ is the KodairaSpencer map. It is nontrivial when $f$ is nonisotrivial, and it is injective if moreover the generic fiber has genus $1$. So under these hypothesis, we have
$$ 0 \to \Omega_{C/k}\to f_{*}\Omega_{X/k} \to f_{*}T\to 0. $$
Therefore the canonical map $H^0(C, \Omega_{C/k})\to H^0(X, \Omega_{X/k})$ is an isomorphism if
$$H^0(X, \Omega_{X/C, \mathrm{tors}})=H^0(C, f_*T)=0.$$
In a small neighborhood of a nonmultiple fiber $X_s$, one can show that $H^0(f_*T)=0$. Otherwise (especially when $X_s$ is irreducible but not reduced) I don't know.

$\begingroup$ hi prof Liu, thanks for the answer. The thought of using $f_*$ did not cross my mind :) . Anyway but in my case since $C$ is a smooth curve and $\Omega_{X/C}$ generically restricts to the canonical sheaf of the elliptic curve (say F), in particular $h^0(\Omega_{X/C}\mid_F) = 1$, the sheaf $f_*(\Omega_{X/C}$ splits $L_0 \oplus T$ where $L_0$ is a line bundle and $T$ is a skyscrapper sheaf. Now on the other hand $f_{*1}(O_x) \otimes \Omega_{C/k}$ is locally free as well. So the kernel of $\theta$ has to be $T$. $\endgroup$– rvarmaFeb 17, 2013 at 19:27

$\begingroup$ So as long as $\Omega_{X/C}$ is torsion free, we get $K_C \cong f_*(\Omega_{X/k})$ which I never knew. Tell me If I was wrong somewhere. Once again thank you. $\endgroup$– rvarmaFeb 17, 2013 at 19:28

$\begingroup$ Oh yes $\theta$ could be $0$ as in the isotrivial case. $\endgroup$– rvarmaFeb 17, 2013 at 19:32

$\begingroup$ @rvarma: yes, if $\Omega_{X/C}$ is torsion free (equivalently, all fibers of $f$ are reduced), your isomorphism is correct. But this isomorphism holds under a slightly more general situation, it is enough that there is no multiple fibers (this is proved in the paper with T. Saito in any characteristic, but probably there is a simpler proof in characteristic $0$). $\endgroup$– Qing LiuFeb 17, 2013 at 23:04
Consider the short exact sequence $$ 0\to f^*\Omega_C\to \Omega_X\to \Omega_{X/C}\to 0, $$ and the long exact sequence induced by $f_*$: $$ 0\to \Omega_C\to f_*\Omega_X\to f_*\Omega_{X/C}\to \Omega_C\otimes R^1 f_* \mathscr O_X\to \ldots $$ If $f$ is not isotrivial, then $\beta: f_*\Omega_{X/C}\to \Omega_C\otimes R^1 f_* \mathscr O_X$ is generically injective (at least where $f$ is smooth $f_*\Omega_{X/C}$ is a line bundle and $\beta$ is nonzero). If $\Omega_{X/C}$ is torsionfree, then so is $f_*\Omega_{X/C}$ and hence $\beta$ is injective everywhere. That implies, that then $\alpha:\Omega_C\to f_*\Omega_{X}$ is an isomorphism.
Therefore, $H^0(U,\omega_C)=H^0(f^{1}U,\Omega_X)$ for every $U\subseteq C$ open.

$\begingroup$ hi thank you i believe you meant f non isotrivial. $\endgroup$– rvarmaFeb 17, 2013 at 20:13