9
$\begingroup$

I was reading a paper from 1994 which claimed that the following statement was a conjecture of Hirzerbruch:

If a complex surface X is homeomorphic to either $S^2 \times S^2$ or $\mathbb{C}P^2 \# \overline{\mathbb{C}P^2}$ then it is biholomorphic to one of the Hirzerbruch surfaces.

I would like to know if this conjecture has been resolved or what the current status about it.

Also, I would like to know the state of the art about the variant of the conjecture in which "homeomorphic" is replaced by "diffeomorphic":

If a complex surface X is diffeomorphic to either $S^2 \times S^2$ or $\mathbb{C}P^2 \# \overline{\mathbb{C}P^2}$ then is it biholomorphic to one of the Hirzerbruch surfaces? If this has been proved, what is a reference?

$\endgroup$

1 Answer 1

10
$\begingroup$

Suppose $X$ is diffeomorphic to $S^2\times S^2$ or $\mathbb{CP}^2\#\overline{\mathbb{CP}^2}$. Then $X$ is biholomorphic to a Hirzebruch surface.

Note that $b_1(X) = 0$, so $X$ admits a Kähler metric. It follows from Seiberg-Witten theory that any compact Kähler surface with non-negative Kodaira dimension does not admit a metric of positive scalar curvature, but $X$ does, so $X$ has Kodaira dimension $-\infty$. By the classification of surfaces, $X$ is either rational (birational to $\mathbb{CP}^2$) or a ruled surface over a curve of genus $g \geq 1$. The latter possibility cannot occur as $X$ is simply connected, so $X$ is rational and hence biholomorphic to $\mathbb{CP}^2$ or a Hirzebruch surface; see Proposition VI.3.3 of Compact Complex Surfaces (second edition) by Barth, Hulek, Peters, and Van de Ven. As $b_2(X) = 2 \neq 1 = b_2(\mathbb{CP}^2)$, we conclude that $X$ is biholomorphic to a Hirzebruch surface.

If $X$ is only homeomorphic to $S^2\times S^2$ or $\mathbb{CP}^2\#\overline{\mathbb{CP}^2}$, then I believe it is still not known whether $X$ must be biholomorphic to a Hirzebruch surface.

As before, $X$ must admit a Kähler metric, but $X$ need not admit a metric of positive scalar curvature a priori, so we cannot immediately determine the Kodaira dimension of $X$ as we did above. Note, if we could show that $X$ had Kodaira dimension $-\infty$, we could conclude that $X$ is biholomorphic to a Hirzebruch surface by the exact same argument. We can however use classification to show that the Kodaira dimension of $X$ is not zero or one.

We can rule out Kodaira dimension zero because any Kähler surface of Kodaira dimension zero is finitely covered by a blownup torus or blownup $K3$ surface, which is not the case for $X$. The impossibility of Kodaira dimension one follows from the fact that $c_1^2 = 0$ for minimal such surfaces, so $c_1^2(X) \in \{0, -1\}$, and $c_2(X) = \chi(X) = 4$, so $\operatorname{Td}(X) = \frac{1}{12}(c_1^2(X) + c_2(X)) \not\in \mathbb{Z}$, which is impossible. Therefore, Hirzebruch's conjecture reduces to providing a negative answer to the following question:

Is there a general type surface homeomorphic to $S^2\times S^2$ or $\mathbb{CP}^2\#\overline{\mathbb{CP}^2}$?

It is natural to ask whether anything changes if $X$ is a compact complex surface which is only homotopy equivalent to $S^2\times S^2$ or $\mathbb{CP}^2\#\overline{\mathbb{CP}^2}$. As $X$ is simply connected, it follows from Freedman's Theorem that $X$ is actually homeomorphic to $S^2\times S^2$ or $\mathbb{CP}^2\#\overline{\mathbb{CP}^2}$, so we're again in the realm of Hirzebruch's conjecture.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.