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?
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asked Aug 16, 2015 at 17:50
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1$\begingroup$ One suggestion: Use the fact that the oriented double cover of $G_3(\mathbb{R}^5)$ is the complex $3$-quadric $Q_3$. The integral cohomology of this is well-known, and you have $G_3(\mathbb{R}^5)$ as the base of a $\mathbb{Z}_2$-bundle whose total space is $Q_3$. (The $\mathbb{Z}_2$-action on $Q_3$ is just complex conjugation.) I think that you can then figure out the integral cohomology of $G_3(\mathbb{R}^5)$ using a standard spectral sequence argument. $\endgroup$– Robert BryantCommented Aug 16, 2015 at 22:31
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1 Answer
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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).
answered Aug 16, 2015 at 19:53
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1$\begingroup$ Thank you for the reference, it told me everything I needed to know. By the way, there is a second author, Yuji Kodama. $\endgroup$ Commented Aug 20, 2015 at 15:48
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$\begingroup$ I updated the answer to reflect this; must have been reading quickly! $\endgroup$ Commented Aug 20, 2015 at 20:35