Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $i>0$?

share|improve this question
    
What are the assumptions on $X$? Is it a scheme or what? –  IMeasy Nov 17 '13 at 20:49
2  
You stipulate that $i > 0$. Do you know a counter-example when $i=0$? –  Daniel Loughran Nov 17 '13 at 20:50
    
You might find it useful to read about the geometry/topology of "Lefschetz fibrations". –  Dan Petersen Nov 18 '13 at 7:50
    
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. –  Karl Schwede Nov 18 '13 at 15:02
    
@Schwede: Could you give some example or reference. In my case $\mathcal{F}$ is the normal sheaf of $X$. –  Jana Nov 19 '13 at 0:03
show 4 more comments

1 Answer

up vote 12 down vote accepted

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.

share|improve this answer
    
Thanks for the answer. –  Jana Nov 18 '13 at 0:27
    
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? –  Jana Dec 5 '13 at 6:07
    
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. –  abx Dec 5 '13 at 6:21
    
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? –  Jana Dec 5 '13 at 6:41
    
It is of course, but not a separable algebra. –  abx Dec 5 '13 at 6:59
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.