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I am trying to learn about linear systems of divisors. Let $X$ be a smooth projective complex surface and $L$ a line bundle. Let's write the complete linear system $|L|=F+|M|$ decomposed in its fixed and mobile part.

I don't see why is $M$ big and nef, i.e. $M^2\geq0$ and $M.C\geq0$ for all irreducible curves. Can anybody explain me this apparently obvious fact?

Heitor Fontana
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1 Answers1

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$M$ need not be big and nef: for nef divisors ($M.C \geq 0$ for all curves $C$) the condition to be big is that $M^2 >0$, a strict inequality. To see this more concretely, just take $L$ to be a basepoint-free but not big line bundle.

On the other hand, the inequalities you wrote are true for the mobile part of a linear system on a surface. They are trivial if $|M|$ is empty, so assume it isn't: since $M$ is mobile then for any curve $C$ there is a representative of $M$ that meets $C$ properly, so $M$ must have non-negative intersection with all curves (in particular with any curve representing $M$ itself).

Postscript: In fact we can say a bit more in this situation. Zariski showed that a line bundle $L$ whose base locus is zero-dimensional must actually be semi-ample (i.e. some positive multiple $L^m$ is basepoint-free). Fujita generalised this as follows: if the restriction of $L$ to the base locus is ample, then again $L$ is semi-ample. (Since any line bundle restricted to a point is ample, this implies Zariski's original statement.) As usual, Lazarsfeld's book is a good place to read about these things.

  • thanks for your answer. what do you mean with Zariski's theorem? – Heitor Fontana Oct 03 '13 at 07:25
  • There is a theorem due to Zariski that says that a complete linear system cannot have isolated basepoints, so a mobile linear system on a surface must be basepoint-free. (You can google "Zariski--Fujita" to find a more general version.) –  Oct 03 '13 at 08:44
  • I simplified the answer a little bit to remove the unnecessary reference to Zariski's theorem. –  Oct 03 '13 at 11:32
  • Dear @AsalBeagDubh, I suggest that it's better to add your edit as an addendum, and leave the mention of Zariski--Fujita (or vice-versa), for the simple (selfish) reason that I haven't heard of this theorem before reading your answer, and subsequently looked it up! Cheers – Andrew Oct 03 '13 at 15:01
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    @Andrew: ok, done! –  Oct 03 '13 at 15:26
  • Dear @AsalBeagDubh: isn't $M^2\ge 0$ just a consequence of $M$ being nef ? – Cantlog Oct 03 '13 at 16:19
  • But if $|M|$ is non-empty, you can replace $M$ be an effective divisor in $|M|$, am I wrong ? – Cantlog Oct 03 '13 at 16:24
  • @Cantlog: (Sorry, I am mangling the comment thread.) Yes, you're right; I typed too fast. Let me fix it up. –  Oct 03 '13 at 16:31
  • (this is because I edited my comment). Thanks! Nice answer! I didn't know Zariski's result, does it hold only in dimension $2$ ? – Cantlog Oct 03 '13 at 16:33
  • @Cantlog: Thanks! Originally, I was trying to explain why any nef line bundle must have $L^2 \geq 0$, but that is overkill here. Thanks for the comments! –  Oct 03 '13 at 16:35
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    @Cantlog: no, Zariski--Fujita is true in all dimensions. Unfortunately, in higher dimensions there is a gap between 0 and $n-1$, so movable divisors need not be semi-ample! But that is what makes birational geometry interesting... –  Oct 03 '13 at 16:38