Let $X$ and $Y$ be varieties. Let $E$ be a locally free sheaf over $X$. Let $f: X \to Y$. Is there some nice criteria which ensures that $f\_\ast E$ is still locally free? Sorry, if this is a very standard question.



When is the pushforward of a vector bundle still a vector bundle?
Under reasonable hypotheses on $X$, $Y$ and $f$, the answer is that $\dim H^0(X\_y, \, E\_y)$ is a *constant* function implies $f\_\*E$ is locally free, where


$$X\_y:=f^{-1}(y), \quad E\_y:=E|\_{X\_y}.$$


More precisely, there is the following result, whose proof can be found in [Mumford, *Abelian Varieties*, Chapter II]:



> 
> **Theorem (Base Change).** Let $f \colon X \to Y$ be a proper morphism of Noetherian schemes, with $Y$ reduced and connected, and $E$ a coherent sheaf on $X$, flat over $Y$. Then for all integers $p \geq 0$ the following conditions are equivalent:
> 
> 
> $\boldsymbol{(i)}$ $y \to \dim H^p(X\_y, E\_y)$ is a constant function;
> 
> 
> $\boldsymbol{(ii)}$ $F:=R^pf\_\*E$ is a locally free sheaf on $Y$ and, for all $y \in Y$, the natural map
> $$F \otimes\_{\mathcal{O}\_Y} k(y) \to H^p(X\_y, E\_y)$$
> is an isomorphism.
> 
> 
> 


For instance, if $f$ is a finite map and $E$ is a line bundle we obtain that $f\_\*E$ is a vector bundle with $\operatorname{rank} f\_\*E=\deg f$, whereas $R^p f\_\*E=0$ for $p >0$.


An example showing it's not enough to assume $f\_\*(E)$ is a bundle is via taking $F$ some elliptic curve, consider on $F \times F$ the line bundle $E = \mathcal{O}(\Delta - p \times F)$ for some $p \in F$, and projecting $\pi\_{1,\*}(E)$ (i.e $F \times F \to F$). Then the fibers usually have $H^0 =0$, jump at $p$, but the pushforward is $0$).


