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May 14, 2013 at 15:13 comment added David E Speyer @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 $H^{\ast}(T)/K$.
May 13, 2013 at 2:32 comment added Nathaniel Bottman @David I agree you can get all the Betti numbers except in the middle dimension (and that too, if you can get the Euler characteristic). But does this say much about the ring structure?
May 12, 2013 at 23:58 comment added David E Speyer 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.
May 12, 2013 at 4:22 comment added Allen Knutson 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.)
May 12, 2013 at 4:02 comment added Nathaniel Bottman OK, thanks Mariano. I find it really surprising that the ring structure of a projective hypersurface isn't easy to compute!
May 12, 2013 at 3:12 comment added Mariano Suárez-Álvarez 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.
May 11, 2013 at 21:01 history edited Nathaniel Bottman CC BY-SA 3.0
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May 11, 2013 at 20:56 history asked Nathaniel Bottman CC BY-SA 3.0