Question

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

Answer 1

Here is a an example, albeit with $Y$ non reduced:

Let $C$ be a smooth projective curve of genus $g > 0$ over a field $k$ and let $C_{\epsilon} = C \times_{k} Spec(k[\epsilon])$ where $k[\epsilon] = k[x]/(x^2)$ is the ring of dual numbers. Let $\mathcal{L}$ be a non-trivial line bundle on $C_{\epsilon}$ such that its restriction to $C$ is trivial. Such line bundles exist since $g > 0$.

Multiplication by $\epsilon$ gives us an exact sequence of sheaves on $C$ $$ 0 \to \epsilon \mathcal{L} \to \mathcal{L} \to \epsilon\mathcal{L} \to 0 \ .$$

Since $\epsilon \mathcal{L} = \mathcal{L}/\epsilon \mathcal{L}$, it follows from the construction that this sheaf is just $\mathcal{O}_C$. Since $\mathcal{L}$ is non-trivial the boundary map in the long exact sequence of cohomology from $H^0(C, \mathcal{O}_C)$ to $H^1(C, \mathcal{O}_C)$ is non-trivial. Thus $H^0(C, \mathcal{L}) = H^0(C, \mathcal{O}_C)= k$.

If we let $X = C_{\epsilon}$, $Y = Spec(k[\epsilon])$, $f$ the natural map $C_{\epsilon} \to Spec(k[\epsilon])$, and $E = \mathcal{L}$, it follows from the above that $f_*E$ is not locally free.

Answer 2

Here is an example with $Y$ smooth!

Let $C$ be a smooth projective curve of genus $g \geq 3$. Denote by $Y$ the 'translated Jacobian' $J = {\rm Pic}^{2g-2}C$. Set $X = Y \times C$. Let $f \colon X \to Y$ and $g \colon X \to C$ be the two projections. Let $L$ be a Poincare bundle on $X$, namely a universal line bundle of degree $2g-2$. Denote by $O \in Y$ the point corresponding to the canonical bundle $K_C$. I claim that $f_*L$ is not locally free at $O$.

Let $y$ denote a point in $y$ and $L|_y$ the restriction of $L$ to $y \times C$, the fiber of $f$ over $y$. Note that \begin{equation} h^0(L|_y) = \begin{cases} g-1 & \text{ if $y \neq O$} \\ g & \text{ if $y = O$} \end{cases}. \end{equation} Similarly, \begin{equation} h^1(L|_y) = \begin{cases} 0 & \text{ if $y \neq O$} \\ 1 & \text{ if $y = O$} \end{cases}. \end{equation}

To see that $f_*L$ is in fact not locally free at $O$, choose a point $p \in C$ and let $P = Y \times p$. We have an exact sequence

\begin{equation} 0 \to L \to L \otimes O_Y(P) \to L \otimes O_Y(P)|_P \to 0. \end{equation}

Applying $f_*$, we get

\begin{equation} 0 \to f_* L \to f_*(L \otimes O_Y(P)) \to f_*(L \otimes O_Y(P)|_P) \to R^1f(L) \to 0 \end{equation}

The zero on the far right is because $R^1f_*(L \otimes O_Y(P)) = 0$, by Grauert's theorem. Indeed, $H^1(L|_y \otimes O_C(P)) = 0$ for any $y \in Y$, since $L|_y \times O_C(p)$ is a line bundle on $C$ of degree $2g-1$. Also, note that the middle two terms $f_*(L \otimes O_Y(P))$ and $f_*(L \otimes O_Y(P)|_P)$ are locally free on $Y$. For the first, we again apply Grauert's theorem and for the second we see that $f \colon P \to Y$ is an isomorphism. Finally, the jumping of $h^0$ and $h^1$ mentioned above and a theorem on cohomology and base change (i.e. Hartshorne Ch 3 Thm 12.11) implies that $R^1f(L)$ is supported (set-theoretically) at $O$.

I claim that if $f_*L$ were locally free, we would get a contradiction. Namely, we will have a resolution of a sheaf supported at a point that is too small in length. To make this precise, we use the Auslander-Buchsbaum formula. If $f_*L$ is locally free, then the projective dimension of $R^1f(L)$ is at most two. Hence its depth is at least $g-2 \geq 1$. However, the dimension of its support is zero! This is a contradiction. We conclude that $f_*L$ is not locally free.

Answer 3

EDIT: the example below does not answer the question because the map is not smooth. (I had not read the question carefully, sorry!). I don't remove the answer since it might still be useful to somebody.

Let $Y\subset \mathbb A^3$ be the quadric cone, defined by $xy-z^2=0$ and take the map $X={\mathbb A}^2\to Y$ given by $(u,v)\mapsto (u^2,v^2, uv)$. This is a (non flat) double cover and the direct image of ${\mathcal O}_X$ is of the form ${\mathcal O}_Y\oplus F$, where $F$ is a rank 1 reflexive sheaf that is not locally free.

Answer 4

When $E$ is trivial and $X$ is smooth, the following result of De Bois gives a positive answer:

Let $f:X\to Y$ be a proper, flat morphism of algebraic varieties over a field of characteristic zero, such that the fibers $X_y$ have du Bois singularities for all $y\in Y$ (this holds in particular if $f$ is smooth). Then $R^if_*\mathcal{O}_X$ is locally free for all $i\ge 0$.

See Theorem 4.6 of Du Bois, P. Complexe de de Rham filtré d'une variété singulière. Bulletin de la Société Mathématique de France, 109 (1981), p. 41-81

Answer 5

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.