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I'm looking for some fairly explicit varieties to use as (counter?-)examples for my thesis and I'd appreciate any suggestions. I need a smooth projective variety $X$ of general type that satisfies:

  • The Hodge number $h^{1,1}$ is at least 2.
  • The cohomology ring of $X$ (or at least the subring generated by $(1,1)$-classes) is explicit.
  • The Kahler cone of $X$ is known(-ish).

I want to calculate the sectional curvatures of the Riemannian metric on the Kahler cone of $X$ which is defined by the intersection product. These curvatures may be expressed by the intersection product on the subring $A$ of $H^*(X)$ genereated by degree $(1,1)$ classes, which explains the conditions I impose.

The condition on the Hodge number is necessary, since when $h^{1,1} = 1$ one ends up with the metric $g(x,y)(t) = xy/t$ on the half-line $\mathbb R_+$ and not many interesting things remain unsaid about this case. This excludes most hypersurfaces in $\mathbb P^n$, except perhaps for those in $\mathbb P^3$.

I must also exclude the example of a blowup of several points of a variety $X$ with $h^{1,1} = 1$, since one can calculate explicitly what happens in this case. Are there other relatively easy examples?

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What does "knownish" mean? –  Fernando Muro Apr 14 '12 at 11:53
    
Like pornography, I'll know it when I see it. What I have in mind is that maybe I don't have to describe exactly the Kahler cone in $H^{1,1}(X)$, but only know some open set that is definitely in the cone. –  Gunnar Magnusson Apr 14 '12 at 12:58
    
non-general K3 surfaces ? I'm not familiar with the literature, but I think this must be known in many cases. If all of $h^{1,1} is algebraic, that should work- no ? –  aginensky Apr 14 '12 at 13:38
    
and i missed 'general type in my first reading' sorry. –  aginensky Apr 14 '12 at 13:56
3  
Why don't you use a general type hypersurface (or complete intersection) in $\mathbb{P}^a \times \mathbb{P}^b$, or in any toric variety of whatever Picard number you like? If the dimension is at least 3, then you can use the Lefschetz hyperplane theorem to describe the Kaehler cone. Of course you can use adjunction to insure the variety is of general type. –  Jason Starr Apr 14 '12 at 16:11

1 Answer 1

You could look at surfaces with maximal Picard rank. A surface is said to have maximal Picard rank if $H^{1,1}(X) \cap H^1(X, \mathbb{R})$ is spanned by curve classes. So the Kahler cone is the same as the ample cone, and you can compute the intersection pairing on $H^{1,1}$ by just intersecting curves with each other.

Now, this raises the question of whether there is a surface of general type with maximal Picard rank. Some googling found me the following:

Section 3 of Quintic surfaces with maximum and other Picard numbers shows that the Fermat sextic $w^6+x^6+y^6+z^6=0$ has maximal Picard rank. Apparently, this was originally computed by Beauville.

Roulleau shows that there are infinitely many Fano surfaces with maximal Picard rank, and gives a basis for the Neron-Severi group. (Confusingly, the Fano surface is not a Fano variety, but rather is of general type.)

A warning: If you look at this paper or this one, they compute a lot of surfaces with maximal Picard. However, their method is to make a surface with an $A_n$ singularity and prove that its resolution has maximal Picard. Since this resolution has a $-2$ curve, it is automatically not of general type. Sorry. As Donu says, I was confused. $K$ will not be ample, but it will be big, and the varieties are general type.

I was coming back to edit in one more idea -- if $C$ is the Klein quartic, then I think that $C \times C$ has maximal Picard rank. But Jason Starr's comment is simpler than any of my suggestions.

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How does having a $-2$ curve make a surface not of general type? –  Dustin Cartwright Apr 14 '12 at 15:15
1  
David, I think you mean that $K$ won't be ample thanks to the $-2$ curves; Persson's examples are certainly of general type. –  Donu Arapura Apr 14 '12 at 17:28

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