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Hello, I'am reading the paper "knots, links, and 4 manifolds" In this paper, they computed the second homology of E(1)-F, E(2)-F where F is a generic fiber.

They claim that these are given as following, H_2 (E(2)-F) = 2E8 + 2H + 3(0)

H_2 (E(1)-F) = E8 + 3(0)

Is there any referernce for this except 4 manifols and kirby calculus? We want explicit calculation or clear topological description of it.

Thanks

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    $\begingroup$ Are you talking about this paper? You really ought to give an actual reference when asking specific questions about papers. arxiv.org/abs/dg-ga/9612014 $\endgroup$ Feb 2, 2012 at 6:40
  • $\begingroup$ Yes, That's right. Thanks for mentioning that. I did not know rules. $\endgroup$ Feb 2, 2012 at 6:44
  • $\begingroup$ Actually, this statement is in page 20. $\endgroup$ Feb 2, 2012 at 6:45
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    $\begingroup$ Hi lowdimension: I think this question would be vastly improved if you included a brief discussion defining, at the minimum, your notation. By asking people to search to page 20 in some paper, you are limiting the number of responses you'll get. Moreover, practicing writing clean definitions is valuable. Mostover, I (and many others) probably cannot help answer your question, but I would be interested in listening in to (and maybe even participating in) the discussion, but I don't know what the notation means (lots of things are called E(n)). Please look over mathoverflow.net/howtoask $\endgroup$ Feb 2, 2012 at 8:12
  • $\begingroup$ lowdimension, Theo is right. But I do know what the notation means, and it seems to me that what's involved will be first a Mayer-Vietoris calculation, then a calculation of the orthogonal complement to the fibre-class in the known intersection lattices of $E(1)$ (i.e., the projective plane blown up 9 times) or $E(2)$ (the K3 surface). That's a really pretty topology exercise, followed by an equally nice algebra exercise - do you really want to leave the fun to someone else? ;) $\endgroup$
    – Tim Perutz
    Feb 2, 2012 at 17:26

1 Answer 1

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For the audience, $E(1)$ is the underlying smooth 4-manifold for a rational elliptic surface. i.e. $\mathbb{C}P^2$ blown up 9 times, while $E(2)$ is the underlying smooth manifold for a $K3$ surface and is diffeomorphic to the symplectic fiber sum of two copies of $E(1)$ along an elliptic fiber. ($E(n)$ is the fiber sum of $n$-copies of $E(1)$)

I assume that you don't find the change of basis in $H_2(E(n))$ method in Gompf and Stipsciz sufficient for your purposes? An alternative way of finding the intersection forms is to look at Kodira's classification of singular fibers in an elliptic fibration. (See for example Barth-Hulek-Peters-Van de ven.) From this we can obtain an elliptic fibration with two or three singular fibers one of which has intersection form $-E8$, the others being either a cusp or two nodes. The regular neighborhood $W$ of the $-E8$ singular fiber is diffeomorphic to a plumbing of $-2$-spheres in a configuration given by the Dynkin diagram for $E8$ and has boundary equal to the Poincare homology sphere $\Sigma(2,3,5)$. The intersection for the other side, called the nucleus $N(1)$, is $\left(\begin{matrix} 0 & 1 \\ 1 & -1 \end{matrix}\right)$.

The homology of $E(1)$ is generated by $h,e_1,\ldots, e_9$ (the generators of $\mathbb{C}P^2$ and the $\overline{\mathbb{C}P}^2$s. Then $H_2(N(1))$ is generated by $[F]=3h-\sum_{i=1}^8 e_i$ and $e_9$ where $F$ is the elliptic fiber and $H_2(W)$ is generated by $e_1-e_2,e_2-e_3,\ldots,e_7-e_8,-h+e_6+e_7+e_8$.

You can now reconstruct your statement using the Meyer-Vietoris and relative homology sequences. The three square $0$ classes end up being the 3 "rim tori" in the $T^3$ boundary of $E(n)\setminus N(F)$.

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