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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Feb 11, 2017 at 20:49 vote accept Lars Pettersen
Feb 11, 2017 at 11:30 comment added Jason Starr Already the Grassmannian of isotropic $2$-dimensional subspaces of a symplectic $4$-dimensional vector space is a smooth quadric hypersurface in $\mathbb{P}^4$. So the integral cohomology ring has $2$ generators in that case.
Feb 11, 2017 at 2:23 answer added Ben Webster timeline score: 4
Feb 10, 2017 at 23:54 comment added Lars Pettersen Can one at least say that whether the ring has one generator or more then one generator?
Feb 10, 2017 at 23:15 comment added Sam Hopkins I don't know the details but I think a lot of this was worked out by Pragacz in this paper: link.springer.com/chapter/10.1007/BFb0083503
Feb 10, 2017 at 23:11 history edited Lars Pettersen
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Feb 10, 2017 at 22:35 history asked Lars Pettersen CC BY-SA 3.0