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Let $M$ be a smooth, compact manifold without boundary. I will say that $M$ is a boundary when there is a smooth, compact manifold with boundary $W$ such that $\partial W=M$. After some lectures I know that projective spaces $\mathbb{P}^{n}(\mathbb{R})$ are NOT boudaries exactly when $n$ is even. Let us consider Grasmanians $\mathbb{G}_k(\mathbb{R}^n)$ (the space of all $k$ dimensional subspaces of $\mathbb{R}^n$). I would like to know when such a Grasmanian is a boundary of some manifold?

EDIT: I corrected my post according to the comment below, thank you.

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  • $\begingroup$ Since $RP^1$ and $RP^3$ are obviously boundaries, I believe you accidently swapped $n$/$n+1$ or even/odd at some point. $\endgroup$ – Daniel Valenzuela May 9 '15 at 1:04
  • $\begingroup$ A closed manifold $M$ is a boundary iff its Stiefel-Whitney numbers all vanish, so one way to approach the problem is to compute the Stiefel-Whitney numbers of the Grassmannians. This might get tedious though. $\endgroup$ – Qiaochu Yuan May 9 '15 at 1:25
  • $\begingroup$ @QiaochuYuan I thought the same. Turns out it is indeed tidious. And not sufficient. $\endgroup$ – Daniel Valenzuela May 9 '15 at 1:29
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So first I thought, the tangent bundle of a Grassmannian is easy to write down and as soon as you have the Stiefel Whitney classes you are done by the famous theorem of Thom, since the cohomology ring of a Grassmannian is very well behaved (hence it is easy to compute the SW numbers).

Turns out it is not that easy after all. Still not hard, but tidious. I found the following a good reference to get an overview (note that the relevant paper mentioned in there proved the relevant stuff) "Cobordism independence of Grassmann manifolds", A. Das.

To answer your question: $G_kR^n$ is boundary iff $v(k)<v(n)$ where $v: \mathbb N \to \mathbb N$ is given by $v(a) = \max \{r: 2^r|a\}$.

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