Return to Revisions

The cohomology groups of the Grassmann manifold are worked out in combinatorial terms in L. Casian's paper, http://arxiv.org/pdf/1309.5520v1.pdf; he makes a conjecture at the end about the multiplicative structure. The author actually does the example you ask about; he computes the groups for $G_2(R^5)$ but that's homeomorphic to $G_3(R^5)$ by taking the perpendicular complement. The answer is simple enough that you could probably work out the ring structure by the requirements of Poincaré duality (plus some considerations about the universal coefficient theorem and Bocksteins).