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.
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 $v(a) = \max \{r: 2^r|a\}$.