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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.

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  • $\begingroup$ 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. $\endgroup$ Commented Jul 16, 2014 at 18:56
  • $\begingroup$ @RicardoAndrade: I just replaced the tag. $\endgroup$ Commented Jul 16, 2014 at 19:45

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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)

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  • $\begingroup$ 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? $\endgroup$ Commented Jul 16, 2014 at 19:49
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    $\begingroup$ 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!) $\endgroup$
    – user47305
    Commented Jul 16, 2014 at 20:16

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