Let $f:X \to \mathbb{P}^1$ be a projective flat morphism, $X$ is a projective scheme. Let $\mathcal{F}$ be a locally free sheaf on $X$. Are the higher direct image sheaves $R^if_*\mathcal{F}$ locally free for all $i>0$?
$\begingroup$
$\endgroup$
9
-
$\begingroup$ What are the assumptions on $X$? Is it a scheme or what? $\endgroup$– IMeasyCommented Nov 17, 2013 at 20:49
-
2$\begingroup$ You stipulate that $i > 0$. Do you know a counter-example when $i=0$? $\endgroup$– Daniel LoughranCommented Nov 17, 2013 at 20:50
-
$\begingroup$ You might find it useful to read about the geometry/topology of "Lefschetz fibrations". $\endgroup$– Dan PetersenCommented Nov 18, 2013 at 7:50
-
$\begingroup$ Can you tell us more about your sheaf $\mathcal{F}$ and your map $f$? For many sheaves related to differentials (twisted in various "nice" ways) assuming $f$ is nice enough, the statement is true. $\endgroup$– Karl SchwedeCommented Nov 18, 2013 at 15:02
-
$\begingroup$ @Schwede: Could you give some example or reference. In my case $\mathcal{F}$ is the normal sheaf of $X$. $\endgroup$– JanaCommented Nov 19, 2013 at 0:03
|
Show 4 more comments
1 Answer
$\begingroup$
$\endgroup$
4
No. For instance there is a flat, projective morphism $f:X\rightarrow \Bbb{P}^1$ such that $X_t:=f^{-1}(t)$ is a smooth rational curve for $t\neq 0$, but $X_0$ is a nodal plane cubic curve with an embedded point (see Hartshorne, III.9.8.4). Then $H^1(X_t,\mathcal{O}_{X_t})$ is zero for $t\neq 0$, but $\ \dim H^1(X_0,\mathcal{O}_{X_0})=1$ . By base change it follows that $R^1f_*\mathcal{O}_X$ is the skyscraper sheaf with 1-dimensional fiber at 0.
-
$\begingroup$ I am a bit confused right now. I noticed in the example that you state $X_0$ is not a closed subscheme of $\mathbb{P}^2$. So I do not totally understand why $f$ is projective. Adding to my confusion is the corollary $7.8.7$ in EGA-III which if I understand correctly means that the dimension of the global sections of the structure sheaf remains contant in this example if it were projective. Could you tell me what is it that I am getting wrong? $\endgroup$– JanaCommented Dec 5, 2013 at 6:07
-
$\begingroup$ Projectivity is not a problem, look at Hartshorne, example 9.8.4. And for your EGA corollary, the key hypothesis, namely that $H^0(X_0,\mathcal{O}_{X_0})$ is a separable $\mathbb{C}$-algebra (i.e. $\mathbb{C}\times \ldots \times \mathbb{C}$) is not satisfied. $\endgroup$– abxCommented Dec 5, 2013 at 6:21
-
$\begingroup$ The last line of the example says it is not a closed subscheme (see last line on pp. 259). Isnt $H^0(\mathcal{O}_{X_0})$ a finite dimensional $\mathbb{C}$-vector space? $\endgroup$– JanaCommented Dec 5, 2013 at 6:41
-
$\begingroup$ It is of course, but not a separable algebra. $\endgroup$– abxCommented Dec 5, 2013 at 6:59