Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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. – Georges Elencwajg Jun 26 '12 at 9:08

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).

share|cite|improve this answer
I have heard (and not verified) that in the real/complex case, the decomposition of projective space into schubert varieties is actually a cell decomposition of a CW-space. While a cell decomposition itself does not give the homology/cohomology without the degrees of the attaching maps, in the complex case you only have cells of even degree, and hence each cell becomes a basis element in the homology/cohomology (using cellular homology/cohomology). Unfortunately, I don't know how things work algebro-geometrically for the computation. – Aaron Jun 26 '12 at 14:57
@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 $V_{i} \subset V_{i+1}$). – Autolatry Jun 28 '12 at 16:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.