TITLE: pushforward of locally free sheaf is locally free?
QUESTION [9 upvotes]: Hi,
Is there an example of a proper smooth map of schemes $f:X\to Y$ and a vector bundle $E$ on $X$
such that $f_*E$ is not locally free on $Y$?
Thanks

REPLY [6 votes]: 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.