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How can one interprete Grassmanian in a projective space? Why it is true that $Gr(V,r)=Gr(P(V),r-1)$? How to compute the cohomology of Grassmanian?

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    Dear Jacob, your last question is too general: computing the cohomology of grassmannians is a whole sophisticated chapter of algebraic topology or algebraic geometry, for which you should consult a book like that of Bott-Tu. Users should ask specific questions on this site, to which we all are happy to welcome you.2012-06-26

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Let us start with the classical Grassmann variety $G(d, n)$, which is the set of all $d$-dimensional subspaces of a vector space $V$ of dimension $n$. The same set can be considered as the set of all $(d−1)$-dimensional linear subspaces of the projective space $P^{n−1}(V)$. In that case one could denote it by $G^{P}(d − 1, n − 1)$.

In terms of the second part of your question, do you mean the cohomology of the complex Grasmanian?

I am aware that can show that the cohomology of a Grassmannian has a basis consisting of the equivalent classes represented by Schubert cycles. this depends on the application of Schubert Calculus (of which I am not an expert)

The Grassmannian $G_{m,n}$ of m-dimensional subspaces (m-planes) in $P^{n}$ over a field $k$ has distinguished Schubert varieties;

$ \Omega_{a_{0},\dots,a_{m}}V_{∗} := \{W∈Gm,n:W∩Vaj≥j\}$

where $V_{∗}:V_{0} ⊂⋯⊂V_{n}=P^{n}$ is a flag of linear subspaces with dim$V_{j}=j$. The Schubert cycle $ \sigma_{a_{0},\dots,a_{n}}$ is the cohomology class Poincaré dual to the fundamental homology cycle of $ \Omega_{a_{0},\dots,a_{m}}V_{∗}$. The basis theorem asserts that the Schubert cycles form a basis of the Chow ring $A_{∗}G_{m,n}$ (when $k$ is the complex number field, these are the integral cohomology groups $H_{∗}G_{m,n}$) of the Grassmannian with

$ \sigma_{a_{0},\dots,a_{m}} ∈ A^{(m+1)(n+1)−\binom{m+1}{n+1}−a_{0}−⋯−a_{m}}G_{m,n}.$ (see also Grassmann manifold).

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    @Aaron - Yes this is true, the Schubert cells for $Gr(r, n)$ are defined in terms of an auxiliary flag such that when one takes the subspaces $V_{i}$ (for $V_{i} \subset V_{i+1}$).2012-06-28