5
$\begingroup$

Let $X$ and $Y$ be smooth del Pezzo surfaces of the same degree $K_X^2=K_Y^2$.

Are the sets $X(\mathbb{C})$ and $Y(\mathbb{C})$ homeomorphic, or at least homotopy equivalent?

$\endgroup$
3
  • 6
    $\begingroup$ Del Pezzo surfaces of degree $d\neq 8$ form a unique family, and therefore they are even diffeomorphic. For $d=8$ one finds $\mathbb{P}^1\times \mathbb{P}^1$ and $\mathbb{P}^2$ blown up at one point, which are not homotopy equivalent (the intersection form on $H^2$ is even for the former and odd for the latter). $\endgroup$
    – abx
    Aug 23, 2016 at 10:36
  • $\begingroup$ @abx: sorry, your comment appeared to me only after posting my answer (which says essentially the same things). $\endgroup$ Aug 23, 2016 at 10:43
  • 8
    $\begingroup$ @Francesco Polizzi: no problem, these are not the olympic games! $\endgroup$
    – abx
    Aug 23, 2016 at 12:54

1 Answer 1

14
$\begingroup$

Yes, with precisely one exception.

If $K^2 \neq 8$, then the del Pezzo surface is the blow-up of the plane at $9-K^2$ points, so it is homeomorphic to the connected sum of $\mathbb{CP}^2$ with $9-K^2$ copies of $\overline{\mathbb{CP}^2}$.

If $K^2=8$, then we have either the quadric $\mathbb{P}^1 \times \mathbb{P}^1$, which is clearly homeomorphic to $S^2 \times S^2$, or the blow-up of $\mathbb{P}^2$ at one point, which is homeomorphic to $\mathbb{CP}^2 \# \, \overline{\mathbb{CP}^2}$. These surfaces are not homeomorphic, since in the former case the class of $K$ is $2$-divisible in (co)homology whether in the latter case is not, and the divisibility of the canonical class is known to be a topological invariant.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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