TITLE: Cohomology of $G_3(\mathbb{R}^5)$
QUESTION [8 upvotes]: This is in some sense a specialization of the question integral or rational cohomology of real grassmannians. Let $G_3(\mathbb{R}^5)$ denote the real Grassmannian of (unoriented) $3$-planes in $\mathbb{R}^5$, which is a non-orientable closed manifold of dimension $6$. I would like to know the integral cohomology ring
$$
H^*(G_3(\mathbb{R}^5);\mathbb{Z}).
$$
Does anyone know of a reference where this is worked out, or how to go about doing so?

REPLY [12 votes]: The cohomology groups of the Grassmann manifold are worked out in combinatorial terms in Luis Casian and Yuji Kodama's paper, http://arxiv.org/pdf/1309.5520v1.pdf; they make a conjecture at the end about the multiplicative structure.  The authors actually do the example you ask about; they compute 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).