For which $n$ does there exist a (topological, smooth, PL, complex) manifold $M^n$ such that $\partial M = \mathbb{R}\mathbb{P}^m$. Obvously, $m = n -1 $ (at least an in the real case). There are a couple of questions:
- Does there always exist a bounding projective space. And if not, what are the demands on your $M^n$ to have a bounding projective space (that is, apart from the obvious ones) ?
- When does non-orientability of the bounding projective space implies non-orientability of the $M^n$ (Obviously, when $\partial M^{2n+1}=\mathbb{R} \mathbb{P}^{2n}$, then the boundary is non-orientable) ?
- When does a complex projective space $\mathbb{C}\mathbb{P}^n$ bounds ? And is there any "aftereffect" (i.e. are there some specific properties that such a (probably smooth) $2n-1$-manifold has because of the complex structure of the bounding $\mathbb{C}\mathbb{P}^n$) visible in the manifold that it bounds because of the complex structure of $\mathbb{C}\mathbb{P}^n$ ?
- Does the fake complex projective plane bounds anything ?
I know this is a multitude of questions spanning probably a multitude of disciplines.
As was alluded to in the mathoverflow post Grigory posted, $\mathbb R P^n$ is a boundary if and only if $n$ is odd. You can see that $\mathbb R P^{2n}$ cannot be a boundary by using Stiefel-Whitney numbers; for a manifold to be a boundary you need all its Stiefel-Whitney numbers to vanish, which is the case for only the odd projective spaces.
A good reference for this is p. 51-53 of Milnor and Stasheff's ``Characteristic Classes."