5
$\begingroup$

Recently I had to compute the rational singular cohomology ring of a hypersurface in a product of projective spaces. I managed to do it, but only by interpreting this variety as a blowup of projective space (so I wasn't using the fact that it was a hypersurface in a toric variety).

My question is:

Given a nonsingular projective toric variety $X$ and a line bundle $L \to X$, is there a simple way of writing down the singular cohomology ring of the hypersurface cut out by $L$?

By "simple" I am thinking of something not much more involved than the presentation of the rational cohomology ring of a symplectic toric manifold in terms of the Stanley-Reisner ideal. I would be happy to know the answer in the special case that $X$ and the hypersurface are both Fano.

The original question that came to my mind was whether there's an easy way of computing cohomology for a nonsingular complete intersection in a nonsingular projective toric variety, but after doing some browsing around I would guess that this is too much to hope for.

Improve this question
$\endgroup$
6
  • $\begingroup$ Your question specializes to «is there a simple way of writing down the singular cohomology of a hypersurface in $P^n$?» One can compute the dimensions of the rational cohomology groups if the surface is smooth, I think, but I don't know if the ring structure comes out as easily. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented May 12, 2013 at 3:12
  • $\begingroup$ OK, thanks Mariano. I find it really surprising that the ring structure of a projective hypersurface isn't easy to compute! $\endgroup$
    – Nathaniel Bottman
    Commented May 12, 2013 at 4:02
  • $\begingroup$ Since you're happy with Fano, this also specializes to "write down the cohomology of a smooth anticanonical hypersurface", which is about the first sort of Calabi-Yau manifold physicists really grappled with. (After that they moved on to complete intersections, as you are, and beyond. But it was never easy.) $\endgroup$
    – Allen Knutson
    Commented May 12, 2013 at 4:22
  • $\begingroup$ By the Lefchetz hyperplane theorem, in every degree except for middle cohomology, the cohomology of the hypersurface is the same as that of the toric variety which, as you say, is easy to present. $\endgroup$
    – David E Speyer
    Commented May 12, 2013 at 23:58
  • 1
    $\begingroup$ @Nate Bottman: Yes. Let $T$ be the toric variety, let $X$ be the toric hypersurface and let $\omega \in H^2(T)$ be the class of $X$. Let $K \subset H^{\ast}(T)$ be the kernel of multiplication by $\omega$. Then the image of $H^{\ast}(T)$ in $H^{\ast}(X)$, as a ring, is $H^{\ast}(T)/K$. $\endgroup$
    – David E Speyer
    Commented May 14, 2013 at 15:13

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .