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Let $(\mathbb{C}^{2p},Q)$ be a $2p$-dimensional complex vector space equipped with a nondegenerate symmetric bilinear form $Q$ where $p\geq 3$. Let $l\leq p-2$. You may assume that $l$ is odd if this helpful. Consider the Grassmannian $\bar{Y}=\mathbb{G}(2(l-1),2p)$ parametrizing $2(l-1)$-dimensional linear subspaces of $\mathbb{C}^{2p}$. Let $W$ be a nondegenerated $2(l-1)$-dimensional subspace. Let $G=\mathrm{SO}_{2p}$. The group $G$ acts on $\bar{Y}$ and we can consider the orbit of $W$ under the action of $G$, this is $$ Y:=GW=\{V\in\bar{Y}\mid V\text{ nondegenerated}\}=G/Q $$ where $Q$ is the stabilizer of $W$ in $G$ ($Q=\mathrm{Stab}_G(W)=\{g\in G\mid gW=W\}$). The homogeneous space $Y$ is a dense open subset of $\bar{Y}$ with respect to the Zariski topology.

My question is now the following: do we have an isomorphism in cohomology $H^*(Y,\mathbb{Z})\cong H^*(\bar{Y},\mathbb{Z})$ where the isomorphism is as rings, i.e. is supposed to preserve the cupproduct?

Is there maybe a general satement which implies this isomorphism? Or a specific argument to prove the isomorphism in this situation? There might be an obvious reason for this which I don't know.

I see that we cannot always hope to identify the cohomology of a space with a the cohomology of a dense open subspace, eg. $H^*(\mathbb{R},\mathbb{Z})\not\cong H^*(\mathbb{R}\setminus\{\text{pt}\},\mathbb{Z})$, but I hope that it is true at least in the situation described above.

Any help is appreciated! Thanks!

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  • $\begingroup$ You might want to work with Borel-Moore Homology for this to be true. $\endgroup$
    – Hanno
    Jan 21, 2016 at 13:25

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Let $M$ be a closed connected manifold, and $Y\subset M$ a dense open subset, missing at least one point. Suppose Y is connected (otherwise $H_0$ distinguishes $Y$ and $M$). Then $H^n(M,\mathbb{Z}/2)=\mathbb{Z}/2$ but $H^n(Y,\mathbb{Z}/2)=0$ as it is a non-compact n-dimensional manifold. If one adds orientable to the assumptions of $M$ then we get this for $\mathbb{Z}$ coefficients.

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  • $\begingroup$ It seems that you mean $Y$ instead of $M$ in the second occurence of $H^n$. So you say the isomorphism is just not true in my situation since the Grassmannian is orientable. Thanks! $\endgroup$ Jan 21, 2016 at 12:43
  • $\begingroup$ Note: I'm not really familiar with the zariski topology, this is just differential topology $\endgroup$
    – Thomas Rot
    Jan 21, 2016 at 12:52

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