Let $X, Y$ be smooth projective varieties and $f:X \times Y \to Y$ be the natural projetion map. Let $\mathcal{F}$ be a locally free sheaf on $X \times Y$. Is it true that $f_*\mathcal{F}$ is locally free on $Y$? If not true in general is there any additional condition on $X, Y$ under which this will hold true?
[Math] Push-forward of locally free sheaves
ag.algebraic-geometry
Related Solutions
The answer is "yes" (though I can't imagine a situation where one would really need this fact). More generally, if $A$ and $B$ are arbitrary commutative algebras over a field $k$ with $A$ noetherian and if $M$ is an $A \otimes_k B$-module which is locally free as such (perhaps not finitely generated) then $M$ is locally free as an $A$-module. Here, by "locally free" I meant relative to the Zariski topology.
Without loss of generality (since $A$ is noetherian), we may and do assume that ${\rm{Spec}}(A)$ is connected.
The first thing to observe is that $M$ is projective as an $A \otimes_k B$-module. Indeed, projectivity is a Zariski-local (even fpqc-local) property for modules over commutative rings, by 3.1.3 part II of Raynaud-Gruson (and the fact that faithfully flat ring maps satisfy their condition (C), using 3.1.4 part I of Raynaud-Gruson), so any locally free module over a commutative ring is projective. Thus, $M$ is a direct summand of a free $A \otimes_k B$-module, which in turn is also free as an $A$-module. Hence, $M$ is projective as a $A$-module. If $M$ is module-finite as such then it is certainly locally free (since $A$ is noetherian). But if it is not module-finite then we're again done since $A$ is noetherian with connected spectrum, as then it follows that any projective $A$-module that is not finitely generated is free! This is Bass' theorem "big projective modules are free"; see Corollary 4.5 in his paper with that title.
It is of course not true that for any finite morphism $f:X\to Y$ we have $f_*\mathcal{O}_X$ locally free : think about a closed immersion.
In fact, your question is about the important topic of "base change and cohomology of sheaves" for proper morphisms, which is treated by Grothendieck in EGA3. The simplest answer one can give, I think, is that if $f$ is proper [EDIT : and flat, as t3suji points out] and for all $y\in Y$ we have $H^1(X_y,\mathcal{O}_{X_y})=0$ then $f_*\mathcal{O}_X$ is locally free.
You may want to avoid going to find the exact reference in EGA3, since as Mumford says, that result is "unfortunately buried there in a mass of generalizations". In that case, go to chapter 0, section 5 of Geometric Invariant Theory (3rd ed.) by Mumford, Fogarty and Kirwan. This is where Mumford's comment is taken from.
Best Answer
Let $X = P^1$, $Y = P^3$. We will take $F$ to be an extension $$ 0 \to O(-2,1) \to F \to O \oplus O \oplus O \to 0. $$ Of course $F$ is locally free. Note that $$ Ext^1(O,O(-2,1)) = H^1(P^1\times P^3,O(-2,1)) = H^1(P^1,O(-2))\otimes H^0(P^3,O(1)) $$ is a 4-dimensional vector space, so we can take $F$ to be the extension corresponding to a 3-dimensional subspace in it. Pushing forward to $P^3$ then gives an exact sequence $$ 0 \to f_*F \to O \oplus O \oplus O \to O(1) \to R^1f_*F \to 0 $$ and the middle map corresponds to out choice of 3 linear functions on $P^3$, so $R^1f_*F$ is the structure sheaf of a point and $f_*F$ is the simplest example of a reflexive sheaf which is not locally free.