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Let $Y=Gr_{m}(\mathbb{C}^n)$ be the Grassmannian of $m$-plane inside $\mathbb{C}^n$. Let $X$ and $X'$ be two Schubert varieties inside $Y$ such that $X'\subset X$ and $dim(X')<dim(X)$.

Let $Z=X\setminus X'$. It is easy to find the Euler characteristic of $Z$ since $X,X'$ are algebraic and we can just count the number of Schubert cells but finding the cohomology group of $Z$ is different. I know in general how to find the cohomology group of Schubert variety $H^*(X)=H^*(Y)/I$ where $I$ is spanned by Schubert varieties which are greater than $X$ in Bruhat order.

We don't have a good way to describe the CW structure of $Z$ because the lower dimension skeleton inside $X'$ is removed.

Can someone provide a hint on how to find the cohomology group of $Z$? Thanks a lot!

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You can use the long exact sequence of the pair $(X,X-X')$ and interpret $H^k(X,X-X')$ as the Borel-Moore homology of $X'$. The vanishing of cohomology in odd degrees makes this particularly easy.

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  • $\begingroup$ Is it only true for smooth $X$? $\endgroup$ – Ben Jul 12 '16 at 14:27
  • $\begingroup$ @Ben i doubt there would be an issue when $X$ is not smooth, but I don't have a proof handy. $\endgroup$ – Matt Samuel Jul 12 '16 at 14:31
  • $\begingroup$ Thank you for your comment! I will try to figure it out. $\endgroup$ – Ben Jul 15 '16 at 16:38

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