I've gone through many texts in algebraic geometry, specifically, Schubert calculus. They all claim that the Schubert classes $[\Omega_{\lambda}]$ form a basis for the cohomology ring of the complex Grassmannian of $k$-dimensional subspaces of $n$-dimensional complex space, i.e. $H^{*}(Gr(k, n))$, but without any proof. I would think that there could be a way to somehow show this result using an argument involving linear independence and span, but I'm not sure where to begin. In the back of Fulton's $\textit{Young tableaux}$, it is suggested to construct a filtration $$Gr(k, n)=Y_0 \supset Y_1 \supset \cdots,$$ where $Y_p$ is the union of all Schubert varieties $\Omega_{\lambda}$ with $|\lambda| \geq p$. Then I would have to take the cohomology of each member of the filtration, but how does one conclude that the Schubert classes form a basis of the cohomology ring in question?
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7$\begingroup$ Since you're asking a topological question (not, say, about Chow groups), it's unlikely to be given an answer with the details you want in an algebraic geometry or combinatorics book. Read en.wikipedia.org/wiki/Cellular_homology and use the fact that each $Y_p \setminus Y_{p-1}$ is even-real-dimensional to see that the boundary maps vanish. $\endgroup$– Allen KnutsonCommented Feb 28, 2016 at 18:01
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2$\begingroup$ A different way to start getting a handle on this is to read the example given in Griffiths and Harris for $\CC$. It is done with matrices which I found more approachable to start with. $\endgroup$– B. BischofCommented Feb 28, 2016 at 18:08
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$\begingroup$ @AllenKnutson: Even-real-dimensional comes from the fact that complex space has twice the dimension of real space, correct? $\endgroup$– LibertronCommented Feb 28, 2016 at 18:11
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1$\begingroup$ Yup$ \!\!\! \! $ $\endgroup$– Allen KnutsonCommented Feb 28, 2016 at 18:14
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1$\begingroup$ Some references are given beginning on page 1078 of Kleiman and Laksov's survey math.ucr.edu/~jdolan/schubert1.pdf. $\endgroup$– Richard StanleyCommented Feb 28, 2016 at 23:21
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1 Answer
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What Fulton is suggesting is (an easy special case of) the fact that cellular and usual cohomology are equal. Consider the long exact sequence attached to the pair $(Y_p,Y_{p-1})$. We have maps $$ \cdots \to H^*(Y_{p}, Y_{p-1}; \mathbb Z)\to H^*(Y_p; \mathbb Z) \to H^*(Y_{p-1}; \mathbb Z) \to \cdots.$$
The key observation is that $Y_{p}$ with $Y_{p-1}$ smashed to a point is a wedge of $2p$ dimensional spheres, one for each Schubert cell of dimension $p$. Thus $H^i(Y_{p}, Y_{p-1}; \mathbb Z)$ is only non-zero for $i=2p$, and that degree is the free abelian group on the Schubert classes in dimension $p$. Now, induct on dimension, and the result follows.
answered Feb 28, 2016 at 23:12
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$\begingroup$ By usual do you mean simplicial? $\endgroup$ Commented Feb 28, 2016 at 23:25
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1$\begingroup$ He probably means singular (co)homology: en.wikipedia.org/wiki/Singular_homology $\endgroup$ Commented Feb 28, 2016 at 23:34
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1$\begingroup$ @Libertron All I used was the Eilenrod-Steenrod axioms (en.wikipedia.org/wiki/Eilenberg%E2%80%93Steenrod_axioms), so any cohomology theory satisfying those (such as singular cohomology) would work. $\endgroup$– Ben Webster ♦Commented Feb 29, 2016 at 0:55