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One interesting property of the projective plane is that any two plane curves intersect. (More generally, if $V$ and $W$ are subvarieties of any projective space, and codim $V$ + codim $W \geq 0$, then $V$ and $W$ intersect.) However, the same does not seem to hold for most other easy examples of surfaces. For instance, any ruled surface $S \to C$ has non-intersecting curves (take the fibers over any two distinct points of $C$). Furthermore, any surface $S$ obtained from $\mathbb{P}^2_k$ by blowing up points $p_1, \ldots, p_n$ has two non-intersecting curves: take two lines that intersect transversely at $p_1$ and avoid $p_2, \ldots, p_n$, and lift them to curves on $S$.

Thus, my question:

Is there any nonsingular algebraic surface other than the projective plane such that any two curves on the surface have nontrivial intersection?

(Note: assume the base field is algebraically closed.)

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Yep. To avoid blow-up examples, perhaps you intend to assume minimality? What makes the projective plane work is the Neron-Severi group having rank 1 (and trivial torsion), so this leads us to look for other minimal surfaces with NS-group $\mathbf{Z}$. Consider smooth quartics in $\mathbf{P}^3$. These are K3, and have NS-rank 1 precisely when every (integral closed) curve is a complete intersection with a hypersurface in $\mathbf{P}^3$, so all are ample and hence we win. A "generic" smooth quartic has NS-rank 1 (Deligne), and such examples even exist defined over $\mathbf{Q}$ (van Luijk). –  BCnrd Jun 16 '10 at 1:03
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Some general type examples (over $\mathbb{C}$): en.wikipedia.org/wiki/Fake_projective_plane (for similar reasons to BCnrd's: any curve in a projective surface is non-trivial in rational homology, so when that has rank 1, all curves have non-zero self-intersection). –  Tim Perutz Jun 16 '10 at 1:27
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How does allowing blow-up examples make the problem any easier? –  Charles Staats Jun 16 '10 at 2:12
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This is from a slightly different direction. If the surface has positive holomorphic curvature then any two nonsingular curves will intersect. The reason is if C and D are disjoint then there are points p in C and q in D that minimize distance. Let gamma be a geodesic path realizing the minimum distance. Apply Synges variational inequality to a suitably chosen pair of vector fields along the geodesic path to see that it is not distance minimizing. –  Charlie Frohman Jun 16 '10 at 3:19
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Related question: mathoverflow.net/questions/47895 –  Artie Prendergast-Smith Dec 15 '10 at 10:52
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3 Answers 3

Generalizing BCnrd's example, almost all surfaces S of degree d ≥ 4 in P^3 have Picard group = Z, hence every curve on a general S is a complete intersection with another surface. Thus any two curves on S intersect. This is the theorem of Noether-Lefschetz.

http://www.springerlink.com/content/t754510m417u0712/

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Here is another way to construct such examples. This way one can get examples with Picard number at least up to $4$.

Notation: Let $S$ be a smooth projective surface, $\overline{NE}(S)$ the closed cone of effective curves, $Q_{\text{tot}}(S)$ be the set of numerical classes with positive self-intersection, and $Q(S)$ the closure of the connected component of $Q_{\text{tot}}(S)$ containing an ample class.

Fact (a simple consequence of Riemann-Roch): $Q(S)\subseteq \overline{NE}(S)$.

Claim Let $S$ be a smooth projective surface such that every proper curve $C$ on $S$ has postive self-intersection: $C^2>0$. Then for any two proper effective curves $C_1,C_2\subset S$ we have $C_1\cdot C_2>0$ and in particular $C_1\cap C_2\neq\emptyset$.

Remark: If the Picard number is not $1$, then the condition that "every proper curve $C$ on $S$ has postive self-intersection: $C^2>0$" is equivalent to assuming that $Q(S)=\overline{NE}(S)$ and that the boundary of $Q(S)$ is empty.

Proof By the nature of the statement we may assume that the Picard number of $S$ is at least $2$ and that $C_1$ and $C_2$ are irreducible. By the assumption $C_1^2>0$ and if it is irreducible, then this implies that it is nef. Hence the linear functional induced by $C_1\cdot(\quad)$ is non-negative on $Q(S)$, but since the boundary of $Q(S)$ is empty, it follows that $C_1\cdot(\quad)$ is actually positive on every effective class and hence the statement follows.

To see that there exists surfaces satisfying the condition in the Claim simply consider K3 surfaces that do not contain smooth rational or elliptic curves. These exist with Picard number 1-4. I suspect that one can find examples satisfying the condition in the Claim with higher Picard numbers, too.

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Nice answer. It's interesting that nobody is able to come up with an example with Picard number greater than 4 (including me, in my answer to the question linked in the comments). It would be nice to know if this is just because surfaces of general type are not well understood, or if there is a deeper reason. –  Artie Prendergast-Smith Oct 27 '11 at 19:37
    
@Artei: nice example in the linked answer. I hadn't seen your comment, because it was hidden. I guess these K3s also give counter examples to JC's question. –  Sándor Kovács Oct 27 '11 at 20:14
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I beleive an example with Picard number two is the symmetric square of a general curve. I would guess there are also examples with arbitary large Picard number, but I cannot think of any.

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