If we have the family of complex curves $f:X\rightarrow Y$, over a complex smooth curve $Y$ , we consider a fiber $C=f^{-1}(y)$ and its tangent bundle $T_{C}$. We know that $df: f^{*}{T_{Y}}_{|C}\cong N_{C}.$ In fact both of these budnles are trivial and isomorphic to $C\times T_{y}$(Voisin pg.223). Can we give explicitely this isomorphism, i.e if we take some vector from $N_{C}$ how do we see it as an element of $f^{*}{T_{Y}}_{|C}?$
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Take a vector in $TX$ which represents the vector in $N_C$; it is unique up to adding something tangent to $C$. Now apply $f'(x)$ to it, a linear map to $T_y Y$. At a generic point of $X$, the map $f$ is a linear projection in local holomorphic coordinates, $f(x,y)=y$ and the map $f'$ is just $f$ in those coordinates. The fibers are $y=y_0$ constant, the tangent to the fibers is $\partial_x$, the normal bundle is spanned by $\partial_{y}$ mod $\partial_x$. Each element is linearly mapped by dropping the words "mod $\partial_x$", i.e. $f_* \partial_{y} \operatorname{mod} \partial_x=\partial_{y}$.
answered May 31, 2016 at 18:06
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$\begingroup$ can you give me more explicit your answer, how all this looks in coordonates? Thank you for previous answer . $\endgroup$– And RubCommented May 31, 2016 at 18:15
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$\begingroup$ I did not understand your notations $y_{i}$, these are coordonates on $Y$? Here we are in the case where $Y$ is a 1-dimesnional complex manifold,i.e. a curve and $X$ is a complex fibrated surface. $\endgroup$– And RubCommented May 31, 2016 at 23:24
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$\begingroup$ ok, fixed. This is only going to help you at a generic point. For points where $f$ does not have full rank, the story is quite different. $\endgroup$ Commented Jun 1, 2016 at 5:53