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

For a divisor $D$ on a smooth complex projective surface $X$, the stable fixed part is the maximal effective divisor $E$ which, for every $n \in \mathbb{N}$, is contained in every memeber of the complete linear series $|nD|$.

Question. Is there a surface $X$ possessing nef and big divisors $D$ whose stable fixed part:

(a) has arbitrarily high degree?
(b) has an arbitrarily high number of components?

Here, the surface $X$ is to be fixed, and $D$ is allowed to run through all the nef and big divisors on $X$.

ADDED following Mark's answer: What if we drop the restriction that $D$ be nef (considering all big divisors at once)? A negative answer to this would imply, in particular, the following:

Question 2. Let $X$ be a surface. As $D$ runs through all big divisors on $X$, is the number of curves $C \subset X$ with $D.C < 0$ bounded?

Note: I have changed the title to reflect this latter question, which appears to be slightly more interesting. I hope this is OK.

share|improve this question
    
Dear @Vesselin Dimitrov: Is the new tag 'linear-series' supposed to be distinct from the existing tag 'divisors'? Even if so, it is probably a good idea to add the well-established tag 'divisors' anyway. –  Ricardo Andrade Jul 16 at 18:56
    
@RicardoAndrade: I just replaced the tag. –  Vesselin Dimitrov Jul 16 at 19:45

1 Answer 1

up vote 1 down vote accepted

I think the answer to (b) should be "no". The stable base locus you're considering is contained in the augmented base locus $\mathbf B_+(D) = \bigcap_{\text{$A$ ample}} \mathbf B(D-A)$. By Nakamaye's theorem, since $D$ is big and nef, we have $\mathbf B_+(D) = \text{Null}(D) = \bigcup \{ C : D \cdot C = 0 \}$. (This is just saying that $D$ must be $0$ on any curve in your locus -- probably there's an exact sequence that gives this more easily.)

But $D$ is big and nef, so $D^2 > 0$. That means that the intersection form on $D^\perp$ is negative-definite, and its dimension is buonded by $\rho(X)-1$. So I think the number of components of $\mathbf B_+(D)$ should be bounded by $\rho(X)-1$.

I bet that there are examples of (a), but I'll have to think about it.


For the new question:

If $D$ is a pseudoeffective (including big) divisor on a surface, it has a Zariski decomposition $D=P+N$ with $P$ nef and $N$ effective. The only things it can possibly be negative on are curves in the support of $N$. But the intersection matrix on these curves is negative-definite (this is a property of Zariski decomposition), and so their number is bounded above by $\rho(X)-1$. (Note: on threefolds Zariski decomposition doesn't work, and all bets are off)

share|improve this answer
    
I see. Thanks! I wonder what would the answer be if we remove the restriction that $D$ be nef. Might the number of curves $C$ with $D.C < 0$ be arbitrarily large as $D$ runs through the big divisors? –  Vesselin Dimitrov Jul 16 at 19:49
1  
No: if $D$ is a pseudoeffective divisor, it has a Zariski decomposition $D =P+N$ with $P$ nef and $N$ effective. The only things it can possibly be negative on are curves in the support of $N$. But the intersection matrix on these curves is negative-definite, and so their number is bounded above by $\rho(X)$. (But on threefolds this is possible!) –  Mark Jul 16 at 20:16

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.