2-cycle of K3 surface

Hi there,

As we know, its betti number $b_2$=22, so there will be 22 2-cycle generators.

Is there any topological way to figure out such cycles direct?

For example, in the best case, if the K3 surface is elliptic and has a global section, can we use combinations of fibre and section to represent all the 22 2-cycles?

Thanks!

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Jay, the fibers and a section of a fibration are algebraic curves, so anything you can generate will be inside the Picard group. So, in any case, the maximum rank you can get this way is 20, but in most cases much smaller. – Sándor Kovács Jun 29 '12 at 5:07
@S'andor -- What about supersingular K3 surfaces? – Jason Starr Jun 30 '12 at 16:13
I suspect that the questioner was interested in the complex case, but I could be wrong. – Donu Arapura Jun 30 '12 at 18:59
In the third paragraph of www.math.ens.fr/~wittenberg/transcendental.pdf, you'll find an explicit example of a K3 surface over $\mathbb{Q}$, with an elliptic fibration, such that the rank generated by a section and components of fibers is $20$, that is the maximum in characteristic $0$. – Olivier Benoist Feb 7 '13 at 8:54

I think the easiest place to see the $22$ is in a Kummer surface. Let $A$ be an abelian surface, so topologically $(S^1)^4$. This clearly has $h_2 = \binom{4}{2} = 6$, and there are obvious topological repreentatives for the $2$-cycles, given by $(S^1)^2$ in $6$ different ways.

Let $X$ be the quotient of $A$ by negation. This has $16$ singular points; the images of the $16$ $2$-torsion points of $A$. Let $Y$ be $X$ blown up at these $16$ points. Then $H_2(Y)$ is (ADDED rationally, see below) generated by the pushforwards of the $6$ $2$-cycles from $A$, and the $16$ $\mathbb{P}^1$'s introduced by resolving the singularities. $6+16=22$.

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There is only one problem with this: The (integer) lattice generated by all of these is not all of $H_2(Y)$. The resulting integer lattice of is index $2^{22}$; see the book Compact Complex Surfaces by Barth, Peters and Van de Ven. – Simon Rose Jun 29 '12 at 21:30
we lose some data there? – Jay Jun 30 '12 at 14:39
Essentially, yes. This is a great way to find rational generators, but it doesn't give integral ones. One quick way to see that is to look at the intersection form: For a K3 surface, the intersection form is $2E_8 \oplus 3H$, whereas the intersection form for the lattice generated by these classes is much simpler (for example, the 16 singular points are all (-2)-curves which intersect no other classes given). – Simon Rose Jun 30 '12 at 17:17

You could try Aspinwall's paper:

"K3 Surfaces and String Duality" http://arxiv.org/pdf/hep-th/9611137v5.pdf

Sections 2.3 and 2.5 are relevant to your question.

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i find a paper by Michael B. Schulz and Elliott F. Tammaro: http://arxiv.org/abs/1206.1070

from page. 11, it gives an explicit description these cycles from the point of view of the resolution of $T^4/Z_2$, seems great!

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