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?
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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). $\endgroup$– abxCommented Aug 23, 2016 at 10:36
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$\begingroup$ @abx: sorry, your comment appeared to me only after posting my answer (which says essentially the same things). $\endgroup$– Francesco PolizziCommented Aug 23, 2016 at 10:43
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8$\begingroup$ @Francesco Polizzi: no problem, these are not the olympic games! $\endgroup$– abxCommented Aug 23, 2016 at 12:54
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1 Answer
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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.
answered Aug 23, 2016 at 10:42