# On the birational equivalent class of algebraic surfaces with Picard number $1$

An open subset $$U$$ of a projective surface $$Z$$ is big if $$\mathrm{codim}_Z(Z\setminus U)\geq2$$.

Let $$X$$ and $$Y$$ be smooth complex projective surface. If there exists a birational map $$f:X\dashrightarrow Y$$ which is an isomorphism between big open subsets of $$X$$ and $$Y$$, I say that $$X$$ and $$Y$$ are birational equivalent.

Does exist a smooth complex projective surface $$Y$$ which is not birational equivalent to another one $$X$$ with Picard number $$\rho(X)=1$$?

Does exist a classification of such surfaces $$Y$$?

• Let $X$ and $Y$ be smooth complex projective surface. If there exists a birational map $f: X \dashrightarrow Y$ which is an isomorphism between big open subsets of $X$ and $Y$, I say that $X$ and $X$ are birational equivalent. It is generally not such a good idea to give new definitions to terms that are already in common use.
– Bort
Commented Mar 6, 2020 at 13:42
• In any case, a map of the kind you describe between smooth (or normal) surfaces is necessarily an isomorphism.
– Bort
Commented Mar 6, 2020 at 15:02

Edit: After rereading your question, it seems that you just want one surface $$Y$$ which is not birationally equivelant to a surface with picard rank $$1$$. This is a bit easier. It is not clear if you want $$Y$$ to have Picard rank $$1$$, but in both cases there are examples. For an example with Picard rank $$1$$ take $$\mathbb{CP}^{2}$$, for an example without picard rank $$1$$ take $$\mathbb{P}^{1} \times E$$ where $$E$$ is an elliptic curve. Below I show that there can be surfaces with picard rank $$1$$ which are not birational.

Yes, for example fake projective planes https://en.wikipedia.org/wiki/Fake_projective_plane. A fake projective is a smooth projective surface with the same Betti numbers as $$\mathbb{CP}^{2}$$.

They necessarily have Kodaira dimension $$2$$, which is different from the Kodaira dimension $$\mathbb{CP}^{2}$$ i.e. $$-\infty$$. This is not so hard to see, from some classical classification results on algebraic surfaces. A projective variety with $$b_{2}=1$$ is either Fano, Calabi-Yau or general type (just consider whether $$K_{X} \in H^{2}(X,\mathbb{R})$$ is positive, 0, or negative). Fano surfaces (also called del Pezzo surfaces) are classified into 10 topological types and have no fake projective planes. Calabi-Yau surfaces with $$b_{1}=0$$ have just two topological types, K3 and Enqriques surfaces, having $$b_{2}=22$$ and $$b_{2}=10$$ respectively.

A fake projective plane $$X$$ has Picard rank $$1$$, since Picard rank is bounded above by $$b_{2}(X)=1=b_{2}(\mathbb{CP}^{2})$$ by definition and is positive because there is an ample line bundle.

• Thank you for your answer. In poor words: I'm studying a property of Higgs bundles over smooth complex projective surfaces $X$, which is invariant under previous "birational equivalence". I'm proving a theorem on such Higgs bundles which works (I guess) when $\rho(X)=1$; so I asked my self: "is $\rho(X)=1$ useless?" By your very clear examples, the answer is "It is not!" Thank you. Commented Mar 6, 2020 at 12:57
• This question is not to clear to me: a very general K3 surface and $\mathbb{P}^2$ are amongst an infinite amount of counterexamples that you can find... Commented Mar 6, 2020 at 13:42
• I don't understand the edit: what is $\mathbf{CP}^2$ an example of?
– Bort
Commented Mar 6, 2020 at 13:55
• Something with picard rank 1 which is not birational to something with picard rank one (other than itself)… Yeah, I am also confused what the OP is asking for. In any case, what is clear is that Picard rank=1 is not a strong enough assumption to put any restriction on the birational type... Commented Mar 6, 2020 at 14:18