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

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.

• mathoverflow.net/questions/76565/… provides a counter example. For assumptions that would make it true, look up cohomology and base change theorems. Mar 17, 2020 at 17:29
• @SamirCanning that looks like an answer - would you care to record it below? Mar 17, 2020 at 22:39

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.