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Suppose $\pi:X\to Y$ is a flat projective morphism with connected fibers. Suppose $Y$ is a smooth projective variety ($X$ and $Y$ are over $\mathbb C$). Let $E$ be a locally free sheaf on $X$ such that its restriction to fibers of $\pi$ is trivial. Suppose finally that for each fiber $F$ of $\pi$ the higher cohomology of its structure sheaf vanish $H^i(O_F)=0$ for $i>0$. Is it true that

$$\pi^* \pi_*(E)\cong E?$$

If yes, then what would be a reference for this statement? If no, I would like to know a general enough restriction on the morphism so that the statement holds. (In the case that is of interest to me all fibers are irreducible, and a generic fiber is a smooth Fano manifold).

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    $\begingroup$ You should require that the fibres are connected, otherwise it is definitely false (take $X\to \PP^1$ to be finite, and $E\in Pic^0(X)$ nontrivial). $\endgroup$ – Donu Arapura Nov 12 '12 at 20:34
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One possible approach to this is to divide the problem into two and figure out what conditions are needed for the two parts.

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First assume that $\pi_*E$ is locally free of the same rank as $E$ (say $r$). By adjointness of $\pi^*$ and $\pi_*$ we have a natural map $$ \nu:\pi^*\pi_*E\to E $$ If $E|_F$ is trivial, then $\nu$ is surjective (this can be checked locally, to do that at $x\in X$, take a set of local generators of $\pi_*E$ near $\pi(x)$. Their pre-images in $\pi^*\pi_*E$ will map to a set of generators in $E_x$). But then $\nu$ actually has to be an isomorphism, because $\pi^*\pi_*E$ and $E$ has the same rank, so it is an isomorphism at the generic points of $X$ and hence the kernel of $\nu$ is a torsion sheaf. Since $\pi^*\pi_*E$ is torsion-free this shows that $\nu$ is injective and hence an isomorphism.


OK, so how can we guarantee that $\pi_*E$ is locally free of the same rank as $E$?

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We certainly need that $E|_F$ is trivial, but probably a little more.

Let me say, without exploring whether weaker conditions suffice, that the following works:

If $Y$ is integral and all fibers of $\pi$ are reduced, then this is OK. This is essentially Grauert's theorem (see Hartshorne, III.12.9).

(The condition on the fibers is to ensure that $h^0(F, E|_F)$ is constant. See Mohan's comment to Qing Liu's answer).


So we have the following:

Let $\pi:X\to Y$ be a flat projective morphism and assume that $Y$ is integral and all fibers of $\pi$ are reduced and connected. Further let $E$ be a locally free sheaf on $X$ such that $E$ restricted to any fiber of $\pi$ is trivial. Then $$\pi^*\pi_*E\simeq E.$$


Note some of the assumptions you made or was willing to make are not necessary. You don't need $Y$ to be smooth or the higher cohomology of $\mathscr O_F$ to vanish.

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  • $\begingroup$ Sandor, thank you very much for this detailed answer, it is very helpful! $\endgroup$ – aglearner Nov 13 '12 at 11:45
  • $\begingroup$ Dear Sandor, I realised that I can not quite understand the following bit of your answer: "But then $\nu$ actually has to be an isomorphism, because $\pi^*\pi_*E$ and $E$ has the same rank, so it is an isomorphism at the generic points of $X$". Probably this is because you use the notion of generic point. Is this the scheme theoretic generic point? In particular, do I understand correctly that in the case $X$ is irreducible, the proof is shorter. Namely you can just say that a subjective morphism between locally free sheaves of same rank is always an isomorphism? $\endgroup$ – Dmitri Panov Nov 19 '12 at 15:03
  • $\begingroup$ Dear Dmitri, yes, essentially I am saying that a surjective morphism between locally free sheaves of same rank is always an isomorphism. $\endgroup$ – Sándor Kovács Nov 19 '12 at 16:47
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Let me first show that the canonical morphism $O_Y\to \pi_{\star} O_X$ is an isomorphism under OP's hypothesis plus the assumption that the generic fiber of $X\to Y$ is reduced (normality as in my previous edit is not needed). For any affine open subset $V$ of $Y$, the canonical map $O_Y(V)\to K:=K(Y)$ (field of rational functions) is injective. By flatness of $\pi$, the middle canonical morphism $$(\pi_\star O_X)(V)=\pi_\star O_X\otimes O_Y(V)\to \pi_\star O_X\otimes K=H^0(X_K, O_{X_K})=K$$
is injective. So $\pi_\star O_X(V)$ is a finite sub-algebra of $K$ containning $O_Y(V)$. As $Y$ is normal because smooth, $O_Y(V)\to (\pi_\star O_X)(V)$ is an isomorphism. This proves $O_Y\simeq \pi_\star O_X$.

As $H^1(X_y, O_{X_y})=0$ for all $y\in Y$, the canonical morphism $\pi_*O_X\otimes k(y) \to H^0(X_y, O_{X_y})$ is an isomorphism for all $y\in Y$ (Mumford, Abelian varieties, p. 53, Cor. 3). Hence $H^0(X_y, O_{X_y})=\mathbb C$ for all $y$.

Now let $r$ be the rank of $E$. Then for all $y\in Y$, $H^0(X_y, E_y)\simeq H^0(X_y, O_{X_y})^r\simeq \mathbb C^r$. Hence $\pi_{\star}E$ is locally free of rank $r$ (op. cit., p. 50-51, Cor. 2). This implies that $\pi^*\pi_\star E\simeq E$ as in Sándor Kovács's answer.

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  • $\begingroup$ Connectedness is not enough for $\pi__O_X=O_y$. For example, take $X=Y\times \text{Spec} k[\epsilon]$ with $\epsilon^2=0$. $\endgroup$ – Mohan Nov 12 '12 at 23:24
  • $\begingroup$ Dear Liu thank you for the answer! You write "Therefore if $\pi^* \pi_* E\cong E$, then $E$ it the pull-back by $\pi$ of some invertible sheaf". It seems to me that this is the logic opposite to the one that I need? I need to prove that $\pi^* \pi_* E\cong E$. Also, in my case $E$ is not necessarily invertible, it can have rank $>1$ (i.e. it is a vector bundle). $\endgroup$ – aglearner Nov 13 '12 at 0:04
  • $\begingroup$ (I am sorry if I misunderstood your answer) $\endgroup$ – aglearner Nov 13 '12 at 0:08
  • $\begingroup$ @Mohan, I will add explanations. @aglearner: I read trivial as isomorphic to the structural sheaf, sorry. $\endgroup$ – Qing Liu Nov 13 '12 at 6:07
  • $\begingroup$ @Mohan, I don't know why $X$ was normal in my mind. Thanks ! $\endgroup$ – Qing Liu Nov 13 '12 at 6:22
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You need to assume that $H^0(O_F)=\mathbb{C}$ for the fibers and then it is just semicontinuity theorem (Hartshorne's book for example).

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  • $\begingroup$ Dear Mohan, thank you for the answer. I will check this semicontinuity theorem and try to see why it immediately implies what I want. $\endgroup$ – aglearner Nov 13 '12 at 0:14

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