# pushforward of locally free sheaf is locally free?

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

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Take $Y$ to be an elliptic curve and let $X=Y\times Y$, with $f$ a projection. Let $Z\in X$ denote the zero section, $\Delta$ the diagonal and let $E=O(\Delta-Z)$. I'll let you compute what happens. –  Donu Arapura Sep 27 '11 at 21:59
Actually, I realize the above ex. doesn't work, the derived dir, image $Rf_*E$ isn't loc. free, but $0$th is. I'll let someone else take over -- too much on my plate these days... –  Donu Arapura Sep 27 '11 at 23:00
According to Corollary II.13.1 of Mumford's Abelian varieties, $E$ has no global sections (or any cohomology at all outside of degree 1), which wouldn't be the case if $pr_{2*}E$ were a skyscraper sheaf. –  Keerthi Madapusi Pera Sep 28 '11 at 2:47
$f_*E$ is torsion free so there cannot be a counterexample with $Y$ a smooth curve. –  ulrich Sep 28 '11 at 4:33
It's local on the target, not the source. –  Keerthi Madapusi Pera Sep 28 '11 at 12:08

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.

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Thanks for the answer. Why is the boundary map nonzero? –  Nicolás Sep 28 '11 at 19:37
Line bundles on $C_{\epsilon}$ with a trivialisation over $C$ correspond bijectively to $Ext^1_C(\mathcal{O}_C,\mathcal{O}_C) = H^1(C, \mathcal{O}_C)$. This identification is given precisely by the boundary map, so a line bundle is trivial iff the corresponding boundary map is zero. –  ulrich Sep 29 '11 at 5:16

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 $$h^0(L|_y) = \begin{cases} g-1 & \text{ if y \neq O} \\\\ g & \text{ if y = O} \end{cases}.$$ Similarly, $$h^1(L|_y) = \begin{cases} 0 & \text{ if y \neq O} \\\\ 1 & \text{ if y = O} \end{cases}.$$

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

$$0 \to L \to L \otimes O_Y(P) \to L \otimes O_Y(P)|_P \to 0.$$

Applying $f_\*$, we get

$$0 \to f_\* L \to f_\*(L \otimes O_Y(P)) \to f_\*(L \otimes O_Y(P)|_P) \to R^1f(L) \to 0$$

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.

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I read rita's comment after writing all this! It's basically the same idea, but I need a curve of genus $g \geq 3$. –  anand Sep 29 '11 at 2:32
Very nice example! –  Rex Sep 29 '11 at 6:40
In fact in the meantime I had also realized that one needs $g>2$. I had made a very down to earth computation that amounted to your argument. It's nice to have it all sorted out. –  rita Sep 29 '11 at 7:40
I concur, it's very nice. –  Donu Arapura Sep 29 '11 at 12:19

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.

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This is also very nice. –  Donu Arapura Sep 29 '11 at 18:32

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.

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In the question $f$ is supposed to be a smooth proper morphism. –  ulrich Sep 28 '11 at 10:38
I had missed the smooth''. Thank you! –  rita Sep 28 '11 at 11:01

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

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I haven't looked at that paper in a while, but don't need to assume that $f$ is semistable? Should be OK in general when $\dim Y=1$, by semistable reduction. –  Donu Arapura Sep 28 '11 at 17:41
For $i=0$ only, a similar result is given in EGA III, 7.8.7. –  user2035 Sep 28 '11 at 17:45
–  user2035 Sep 28 '11 at 17:54
JC: I'm sorry if this sounds critical. In fact, I would be happier if it was true unconditionally since I use these sorts of results, but note the sentence: "Si pour tout point geometrique s de la fleche $O_{X_s}\to \Omega_{X_s}^0$ est isomorphisme" Nowadays, we would say the fibres are Du Bois. It is quite a strong condition. –  Donu Arapura Sep 28 '11 at 18:30
Sure, with smooth maps there is no problem. It is also OK for semistable maps, which is why I remembered it that way. –  Donu Arapura Sep 28 '11 at 18:47