hi thank you , yes i am aware of the patching result you mentioned above and hence suspected the inclusion $B \subset \mathcal{G}(X)$. When all the parahorics under consideration are inside $G(\mathcal{O}_x)$,we have a map $\mathcal{G} \rightarrow C \times G$,so is it the case that $\mathcal{G}(X) \subset G$?

hi thank you i believe you meant f non isotrivial.

Oh yes $\theta$ could be $0$ as in the isotrivial case.

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.

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 sky-scrapper 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 $T$.

@Damian yes that is what I think too, infact I believe it will torsion free only when the fibers are reduced. Otherwise as I commented above won't it be of the form $\omega_f \otimes I_Z$ ?.

@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.

hi yes I was trying that .. :)