0
$\begingroup$

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 $Y$?

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ 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. $\endgroup$
    – Bort
    Commented Mar 6, 2020 at 13:42
  • 2
    $\begingroup$ In any case, a map of the kind you describe between smooth (or normal) surfaces is necessarily an isomorphism. $\endgroup$
    – Bort
    Commented Mar 6, 2020 at 15:02

1 Answer 1

Reset to default
2
$\begingroup$

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.

Improve this answer
$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$
    – Armando j18eos
    Commented Mar 6, 2020 at 12:57
  • 1
    $\begingroup$ 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... $\endgroup$
    – Enrico
    Commented Mar 6, 2020 at 13:42
  • $\begingroup$ I don't understand the edit: what is $\mathbf{CP}^2$ an example of? $\endgroup$
    – Bort
    Commented Mar 6, 2020 at 13:55
  • $\begingroup$ 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... $\endgroup$
    – Nick L
    Commented Mar 6, 2020 at 14:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .