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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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related: mathoverflow.net/questions/8829/… –  Grigory M Aug 3 '11 at 13:08
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(relevant concepts are cobordism and characteristic numbers) –  Grigory M Aug 3 '11 at 13:12
    
Intuitively, boundaries have to be oriented... but I'm not seeing an obvious proof, so perhaps this is not true. What is obvious, though, is that there is a manifold with boundary $\mathbb{RP}^1$, since $\mathbb{RP}^1$ is a circle. I suspect $\mathbb{RP}^3$ might also be a boundary, since it is homeomorphic to $\mathrm{SO}(3)$. –  Zhen Lin Aug 3 '11 at 13:17
    
@Zhen: apparently the Klein bottle is a boundary. –  Qiaochu Yuan Aug 3 '11 at 13:29
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@Zhen: A product of manifolds $M_1\times M_2$ is orientable iff both factors are. On the other hand, the product of a manifold which bounds with anything bounds. Thus, if say $M_1$ bounds and either of $M_1$ or $M_2$ is nonorientable, then $M_1\times M_2$ is nonorientable and bounds. For a concrete example, $S^1\times\mathbb{R}P^2$ bounds $D^2\times\mathbb{R}P^2$. –  Jason DeVito Aug 3 '11 at 16:26

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up vote 3 down vote accepted

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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I know this works in the smooth category, but what about in the other categories the OP mentions? –  Jason DeVito Aug 3 '11 at 16:01
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@Jason AFAIR, $w_n$ of a (PL) manifold can be defined just as the sum of all simplices of (co)dimension n — so S-W numbers seem to give an obstruction to cobordism in PL category as well. –  Grigory M Aug 3 '11 at 16:05
    
...so MPL=MO. And MTop=MO in dim≠4 (by some surgery theory, AFAIK)... –  Grigory M Aug 3 '11 at 16:27

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