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Is there anything known about isolated conics in a del Pezzo surface: their number, arrangement, and the corresponding elements of the class group of surface's minimal desingularization? (Isolated means not belonging to a continuous family of conics in the surface.)

A description similar to the one for isolated lines would be of most interest: "A del Pezzo surface has only finitely many lines. They correspond to curves E such that E^2 = E·K = −1 (so-called -1-curves) on the desingularization."

More specifically, the question is about del Pezzo surfaces of degrees 5 and 6. References not requiring much background in algebraic geometry are greatly appreciated.

[Edit] And if we have a surface in C^3, whose linear normalization is a degree 5 or 6 Del Pezzo surface, can we say anything about isolated conics in this situation?

[Edit2] I have found the following related result in the literature:

"Any surface is a projection from its linear normalization. The projection is birational, and it preserves the degree of the surface and the degree of any curve not contained in the singular locus."

Notice that the conics contained in the singular locus are also interesting for me.

Additional question about surfaces in C^3 still unanswered.

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While the number of lines on Del Pezzo surfaces are finite, the number of conics is infinite. More precisely, there are finitely many families $X\to P^1$ whose fibers are plane conics. Let me explain this in more detail.

As you probably know, a degree $d$ Del Pezzo surface $X$ can be realized as the blow-up of $P^2$ in $r=9-d$ points in general position. The Picard group of $X$ has rank $r+1$ and is generated by the classes of the exceptional divisors $E_1,\ldots, E_r$ and $L$ which is the pullback of a general line in $P^2$ via the blow-up morphism $\pi:X\to P^2$. The intersection form on $N^1(X)=\mbox{Pic }X$ is given by $$ E_i\cdot E_j=-\delta_{ij}, \qquad E_i\cdot L=1, \qquad L^2=1. $$Also, the anticanonical class equals $-K=3L-E_1-\ldots-E_r$ in this basis.

If $X$ has degree $\ge 4$, then $-K$ is very ample, and the conics on $X$ correspond precisely to the effective divisor classes such that $$ -K.D=2 \mbox{ and } D^2=0 $$Examples are $L-E_i$ (pullback of a line through the point $p_i$) and $2L-E_1-E_2-E_3-E_4$ (pullback of a conic avoiding $p_5$). Using the AM-GM inequality, one can show that the number of such classes is finite.

In fact it is easy to see that any conic can be written as the sum of two exceptional curves (which form the generators for the effective cone $\overline{NE}(X)$). So $D=E+F$ for some $E,F$ with $E.F=1$. Moreover, using this description, it is not hard to verify that the conic divisors $D$ are even base-point free and so by Riemann-Roch, define morphisms $X\to \mathbb{P}^1$. These morphisms are conic bundles, i.e., every fiber is isomorphic to a plane conic in $X$.

On the other hand, the lines on $X$ correspond to classes satisfying $-K.E=1 \mbox{ and } E^2=-1$ so they don't 'move' in linear systems like the conics do, which explains why their number is finite.

EDIT: There can not be any isolated conics on $X$, since if $D$ is any isolated rational curve, then $D^2<0$ and the adjunction formula implies that $D^2=-1$, so $D$ is an exceptional curve, i.e., a line.

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Thank you very much for your detailed answer. Can there be any ISOLATED conics on the surface in addition to families of conics you mention? Does equation −K.D=2 and D^2=0 hold for effective divisor classes of ALL conics, or only of those belonging to continuous families? Sorry if asking something obvious. –  mikhail skopenkov May 4 '11 at 10:45
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There can be no isolated conics on the Del Pezzo surface. I added an explanation in the answer above. –  J.C. Ottem May 4 '11 at 10:50
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Well, if you embed X by -2K instead of -K, then isolated lines will embed as isolated conics. Measuring the degree of your curves is relative to the embedding. –  mdeland May 4 '11 at 12:58
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If you embed X by -2K, and then project down to P^3 - you can see isolated conics on the (singular) image, right? Do you have a bound on the degree of your surface in 3-space? –  mdeland May 8 '11 at 22:42
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again, if your surface has N "-1 curves", and then you embed by -2K, project to P^3, then the result will have N isolated conics. Since they came from -1 curves, you would have already computed their divisor classes, etc. What is it that you want to know? –  mdeland May 9 '11 at 18:56
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If by a del-Pezzo surface you mean what is written here : http://en.wikipedia.org/wiki/Del_Pezzo_surface, i.e. surface such that $-K$ is ample, then there are no isolated conics on such surfaces at all. Indeed a smooth rational curve $C$ is isolated on a surface iff $C^2<0$. On the other hand by adjunction formula we have $(K+C)C=-2$, i.e., $-KC= 2+C^2$. Hence $-KC$ is positive only on a rational curve $C$ only if $C^2\ge -1$. But if $C^2=-1$, then $C$ is a line, if $C^2\ge 0$ it is not isolated.

Sometimes by del-Pezzo surface people mean rational surface with $-K$ semi-ample and with $K^2>0$ (thanks to Artie for making this precise), as it is in the following article http://www.staff.science.uu.nl/~looij101/coble6.pdf . More standard terminology for such surfaces are weak del-Pezzo surfaces. They indeed have exceptional curves $C$ with $C^2=-2$. The number of such curves is finite too. This is described, for example in the book of Dolgachev topics in classical algebraic geometry, beginning of chapter 8, http://www.math.lsa.umich.edu/~idolga/topics.pdf

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Just one comment, by a conic , in algebraic geometry people often mean a rational curve, also, often one speaks of "conic bundles". –  Dmitri May 4 '11 at 11:04
    
Aah... And your answer covers this as well! –  mikhail skopenkov May 4 '11 at 11:40
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Hi Dmitri, a very minor nitpick: the first sentence of your second paragraph "Sometimes by del-Pezzo surface people mean rational surface with −K semi-ample, as..." is a bit misleading. If you look at the link, they require the extra condition that (-K)^2 is between 1 and 9, in particular not 0. That rules out things like rational elliptic surfaces, which certainly should not count as del Pezzo. (Of course you know all this stuff, but maybe a casual reader could get the wrong idea by skimming your answer without looking at the link.) –  Artie Prendergast-Smith May 11 '11 at 8:17
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