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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.

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    So a better question whould be: when you call a manifold irreucible (I am certainly this is not the right term here) if from $M_1 \times M_2$ it follows that either $M_1$ or $M_2$ is empty, look for irreducible bounding $M^n$.2011-08-05

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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."

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    ...so MPL=MO. And MTop=MO in dim≠4 (by some surgery theory, AFAIK)...2011-08-03