An example when the direct image of a locally free sheaf is not locally free

Question

Let $X$ be a scheme over $\mathbb{C}$, and $\Sigma$ a smooth projective curve over $\mathbb{C}$. Let $E$ be a locally free sheaf on $X\times \Sigma$. Denote $\pi : X\times \Sigma \to X$ the natural projection to the first coordinate.

Is it true that the direct image $\pi_*E$ is a locally free sheaf on $X$?

In case it isn't, I would also want to know is there any simple assumption on $X$ or $E$ that would make it true. Thank you in advance.

Answer 1

This is a community wiki answer recording the discussion from the comments, in order that this question might be marked as answered (once this post is upvoted or accepted).

https://mathoverflow.net/questions/76565/pushforward-of-locally-free-sheaf-is-locally-free provides a counter example. For assumptions that would make it true, look up cohomology and base change theorems. – Samir Canning

There are several linked counterexamples. Here is one by Sasha:

There is even a more simple example (although essentially the same) than in the Anand's answer. Take $C = P^1 = P(V)$, where $\dim V = 2$, let $Y = S^2V$ and $X = C\times Y$ with the map $f:X \to Y$ being just the projection. Take $E$ to be the universal extension of $O(2)$ by $O(-2)$. This means that $E$ fits into exact sequence $$ 0 \to p^*O(-2) \to E \to p^*O(2) \to 0, $$ where $p$ is the projection $X \to C$, and the extension is given by the canonical element in $S^2V\otimes S^2V^* \subset S^2V\otimes k[S^2V] = Ext^1(p^*O(2),p^*O(-2))$. Applying the pushforward via $f$ to the above sequence one gets $$ 0 \to f_*E \to S^2V^*\otimes O_Y \to O_Y \to R^1f_*E \to 0, $$ and it is clear that the middle map is given by the canonical embedding $S^2V^* \to k[S^2V]$. Thus $R^1f_*E$ is the structure sheaf of the point $0 \in Y = S^2V$ and $f_*E$ is the second syzygy sheaf of a point on a 3-dimensional variety, which is the simplest example of a reflexive non-locally-free sheaf.

By the way, the pushforward of a vector bundle under a smooth morphism is always reflexive. This is why the above example is the simplest possible.