# Finding cohomology group of open dense subset of Schubert variety

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!

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.
• Is it only true for smooth $X$? – Ben Jul 12 '16 at 14:27
• @Ben i doubt there would be an issue when $X$ is not smooth, but I don't have a proof handy. – Matt Samuel Jul 12 '16 at 14:31