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Let $G(r, n)$ be the Grassmannian of the set of all $r$-planes in a $n$-dim vector space. How to show that the cohomology of a Grassmannian has a basis consisting of the equivalent classes represented by Schubert cycles? I am confused since I don't know how to compute the cohomology of a variety (how to construct the co-chain complex). I am really appreciate if you can compute the cohomology of, for example, $G(2,4)$ and show that it has a basis consisting of the equivalent classes represented by Schubert cycles.

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If you mean singular cohomology over $\mathbb{Z}$ of the complex Grassmannian, I think you can use cellular cohomology, using the fact that the complex Grassmannian has a decomposition into even-dimension cells so all of the differentials in the cellular cochain complex are trivial. I think the details are worked out in this blog post:… – Qiaochu Yuan Feb 28 '12 at 4:03
This is a great comment. – Kerry Feb 28 '12 at 4:24
The cellular cohomology of the complex Grassmannian is also talked about here. See in particular Wikilinks given by WCKronholm. – Jyrki Lahtonen Feb 28 '12 at 4:27
up vote 1 down vote accepted

I am not an expert and this is probably not an answer, but your question is essentially a classical problem. You may refer to Hatcher's proof in his book or notes by Michael Hopkins in here. I am sorry that I could not address your question on $Gr(2,4)$ as I have not done the computation for years. The essential tool would be celluar decomposition, calculation of the degree, and Poincare duality. This used to be my final problem when I was learning algebraic topology years ago.

There may be slicker ways to carry out the computation of Schubert calculus but as far as I know it is usually a troublesome process and has only been solved by Ravi Vakil by his paper in Annals. This is also related to some extent with Schubert polynomials, which I personally know very little. Generally Schubert calculus offer a way of getting 'universal' objects that might represent specific objects.

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