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Let $X$ be an algebraic surface (over the complex) with $p_g=q=0$. Is it possible to have disjoint curves $C_1,\ldots, C_b$, of positive genus, spanning $H_2(X,{\mathbb Q})$, $b=b_2(X)$?

(When $X$ is the blow-up of the projective plane at some points, the exceptional divisors and a curve coming from the projective plane not passing through the points, give an answer if we do not assume the condition on the genus.)

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  • $\begingroup$ I only have a couple of topological considerations: Donaldson's theorem forces the intersection form to be diagonal, and the assumption $p_g = q = 0$ pins it down to $\langle1\rangle+(b-1)\langle-1\rangle$ (if I'm not mistaken). The genus assumption says something about the canonical divisor (like pinning down the sign of most coefficients). $\endgroup$ – Marco Golla Mar 21 '15 at 16:44
  • $\begingroup$ Yes, that's right. All of them should have negative self-intersection except one. However, I do not know even how to prove (or if it is true) that $X$ is minimal $\endgroup$ – Vicente Munoz Mar 21 '15 at 17:19
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    $\begingroup$ This is easily seen to be impossible for $X = (\mathbb{P}^1)^2$; the genus condition is not even needed here. $\endgroup$ – ulrich Mar 22 '15 at 6:34
  • $\begingroup$ Donaldson's theorem concerns simply connected 4-manifolds. However, there are (non rational) complex surfaces with $p_g=q=0$ that are not simply connected (for instance, the classical Godeaux surface), so in this case I do not see how to conclude that the intersection form must be necessarily diagonalisable. I'm missing something? $\endgroup$ – Francesco Polizzi Mar 23 '15 at 8:51
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    $\begingroup$ This discussion is somewhat out of focus: the OP considers rational (co)homology. A quadratic form over $\Bbb{Q}$ is always diagonalizable. $\endgroup$ – abx Mar 23 '15 at 14:22
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Thank you.

The following gives an answer in the case that all the curves are assume to have genus $g=1$, and $b\geq 2$.

One of the curves, say $C=C_1$ has $C^2>0$. As $p_g=q=0$ we have $\chi(O_X)=1-q+p_g=1$. By Riemann-Roch, we have $\chi(C)=1+ \frac12 (C^2-K\cdot C)$. As $C^2+K\cdot C=2g(C)-2=0$ (by adjunction), we have $\chi(C)=1+C^2$. Also as $h^0(K)=p_g=0$ and $C$ is effective, we have $h^2(C)=h^0(K-C)=0$. From the exact sequence $0\to O_X\to O_X(C) \to O_C(C)\to 0$, taking the long exact sequence in cohomology, we have $H^1(X,O(C))=H^1(C, O_C(C))$. If $C^2>0$ then the last group vanishes, hence $h^1(C)=0$. So $\chi(C)=h^0(C)=1+C^2>1$, and there is a pencil of curves of genus $1$. If necessary, blow up at points of $C$ to arrange $C^2=1$. The bundle $O_C(C)$ is of the form $O_C(p_0)$, for some point $p_0\in C$. The pencil defined by the linear series $|C|$ is a pencil of genus $1$ curves passing through $p_0$. Blowing up at this point we get a surface $X'$ and a map $\pi:X'\to P^1$. Let $\sigma$ be the exceptional divisor, which is a rational curve of self-intersection $-1$. This gives the structure of an elliptic surface, possibly non-minimal, and containing all our genus $1$ curves. Blow down any $(-1)$-curve in a fiber of $\pi$ to get a minimal elliptic surface $X''\to P^1$. This has $q=0, p_g=0$ and it has a section (the image of $\sigma$). By the Enriques-Kodaira classification, this has to be a rational surface. Any section of a minimal elliptic surface has to have self-intersection $-1$ (let $s$ be a section; the canonical bundle is $K=-f$; then $s^2+K\cdot s=-2$, so $s^2=-1$). So $\sigma$ cannot intersect any rational curve being blown-down when going from $X'$ to $X''$, since otherwise its self-intersection increases. All the genus one surfaces $C_i\subset X$ give genus one surfaces $C_i'\subset X'$ which map down to $C_i''\subset X''$. These must be fibers (any curve not intersecting $C$ is contained in a fiber, so it is a rational curve or a fiber). This means that $C_i''$ intersect $\sigma$, and hence $C_i'$ intersects $\sigma$. When going to $X$, $C_i$ intersects $C$. Contradiction!

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The answer is clearly affirmative for $b=1$, which unfortunately corresponds only to 101 surfaces, including ${\mathbb P}^2$. I doubt that there is a single other case in which it works.

Indeed I do not know a single example with $b>1$ for which your statement holds. It clearly fails for $({\mathbb P}^1)^2$, as mentioned by ulrich, and a similar argument shows that it fails for all surfaces constructed in BCG. I can easily prove that it fails for a bunch of the surfaces constructed by me and some of my collaborators (listed in my survey published here). As noticed by Marco, when the answer is affirmative, you can contract all the negative curves, obtaining what is called ${\mathbb Q}-$homology plane. The smooth ones (corresponding to the case $b=1$) have been recently classified by Prasad-Yeung/Cartwright-Steger and give the 101 surfaces mentioned above. The classification of the singular ones is a big open problem, but the only known examples have rational singularities, so do not fill your assumption on the genus. The most recent paper on this subject is this one.

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  • $\begingroup$ Is it automatic that for all these surfaces there is a holomorphic curve representing some nontrivial homology class? Also, why can you always contract negative curves? (This could well be a standard argument I am not aware of...) $\endgroup$ – Marco Golla Mar 23 '15 at 13:04
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    $\begingroup$ These surfaces are projective, hence $H^2(S,\Bbb{Q})$ is spanned by the class of an ample divisor. And, yes, a curve of negative square on a smooth surface can always be contracted. $\endgroup$ – abx Mar 23 '15 at 14:19
  • $\begingroup$ The assumption $p_g=0$ gives $h^2(X,{\mathbb C})=h^{1,1}$ and therefore $H^2(X,{\mathbb Q}) \subset H^{1,1}$: these surfaces have automatically maximal Picard number. $\endgroup$ – Roberto Pignatelli Mar 23 '15 at 14:50

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