Pushforward of tangent bundle. Can it be $0$

algebraic-geometryschemessheaf-theoryvector-bundles

Let $f:X\to Y$ a smooth flat map between, integral projective varieties such that $\dim(Y)<\dim(X)$. Let's assume by semplicity that everything is over $\mathbb C$. So we are in the situation of "a nice" family of smooth varieties, i.e. the fibres of $f$, varying smoothly on the "parameter space" $Y$.

Now, it is well known that the pushforward of a vector bundle $f_\ast\mathscr E$ can be $0$ on $Y$. Let $\mathcal T_X$ the tangent bundle of $X$. Can it happen that $f_\ast T_X$ is zero on an open subset $U$ of $Y$? What does it tell about $f$? I have the feeling that in this case we have isomorphic fibres over $U$

Best Answer

Let $f \colon X \to Y$ be a smooth family of proper curves of genus $g \ge 2$. Then $f_* \mathcal{T}_X = 0$, but the family does not need to have isomorphic fibers.

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