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The Lagrangian Grassmannian is an important example in symplectic geometry, see here or here for details. It shares many similarities with the ordinary Grassmannians (as one would expect from the name). As is well-known, the cohomology ring of the Grassmannians has a very nice combinatorial description in terms of partitions, see this very nice M.O. answer for example. Does there exist an analogous description for the cohomology ring of the Lagrangian Grassmannians?

More precisely:

(i) How many generators does the ring have? (ii) What are its dimensions? (iii) What is its multiplicative structure?

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  • $\begingroup$ 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 $\endgroup$
    – Sam Hopkins
    Commented Feb 10, 2017 at 23:15
  • $\begingroup$ Can one at least say that whether the ring has one generator or more then one generator? $\endgroup$
    – Lars Pettersen
    Commented Feb 10, 2017 at 23:54
  • $\begingroup$ 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. $\endgroup$
    – Jason Starr
    Commented Feb 11, 2017 at 11:30

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I'm assuming you mean in $\mathbb{C}^{2n}$ (the answer for real Lagrangian Grassmannians is trickier). In this case, the answer is easy:

The cohomology is isomorphic to the symmetric polynomials in $n$ variables modulo the relation killing all positive degree symmetric polynomials in the squares of the variables.

I'm reasonably certain that you need $n$ generators, since the ideal killed lies in the square of the unique graded maximal ideal.

Like the usual Grassmannian, the cohomology of the Lagrangian Grassmannian is generated by the Chern classes of the tautological bundle which correspond to the elementary symmetric functions, but the relations are a bit different. They come from the observation that there is a trivial vector bundle filtered by the tautological bundle $T$ and its dual, so we get the vanishing relations from Whitney sum formula. For isotropic Grassmannians, you don't kill all the symmetric polynomials in the squares, just the complete symmetric polynomials of high enough degree.

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  • $\begingroup$ I am sure you know this, but for $n=2$ the number of generators is $2$. $\endgroup$
    – Jason Starr
    Commented Feb 11, 2017 at 11:32

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