Hi all. I'm looking for an example of a smooth projective surface $X$ and a pseudo-effective divisor $D$ on $X$ such that when I consider the Zariski decomposition $D=P+N$ there is some component $E$ of the negative part $N$ such that $(D\cdot E)\geq 0$. Can you help me? Thank you Gianni
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EDIT: New example, hopefully this one works.
Blow up $\mathbb{P}^2$ at a point $p$, then blow up the resulting surface at a point $q$ on the exceptional divisor. The resulting surface has Picard group generated by the class $H$ of a line, the proper transform $E_1$ of the first exceptional divisor, and the second exceptional divisor $E_2$. We have $E_1^2 = -2$ and $E_2^2 = -1$, while $E_1\cdot E_2 = 1$. Now consider $D = E_1 +2E_2$. The Zariski decomposition of $D$ is $P = 0$, $N = E_1+2E_2$, as no effective divisor supported on $E_1$ and $E_2$ is nef. Then $D\cdot E_1 = 0$. We can get strict inequality instead by taking $D = E_1+3E_2$, for instance.
answered Nov 16, 2011 at 14:37
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$\begingroup$ In a Zariski decomposition I think you need $P\cdot C=0$ for each component of $N$. This does not happen in your example. $\endgroup$ Nov 16, 2011 at 14:58
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$\begingroup$ It looks like, with the above notation, $H$ is the pullback of a line, so that it should be nef by itself and its Zariski decomposition is trivial. Am I wrong? $\endgroup$ Nov 16, 2011 at 15:13
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$\begingroup$ Yes, my original example was wrong. I hope this new one is OK. $\endgroup$ Nov 16, 2011 at 16:05