# For what manifold is boundary given odd-dimensional projective space?

Take projective real space $\mathbb P_n (\mathbb R)$ of ODD dimension. It is easy to proof that all his Stiefel-Whitney numbers are zero . So according Thom theorem there must exists manifold $M$ with boundary such that boundary is $\partial M= \mathbb P_n (\mathbb R)$. I should like to see directly such $M$, without using Thom Theorem . For example if $n=1$ evident choice is $M=$ closed disk.
I have no idea in general case. Can some one help please ?

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mathoverflow.net/questions/8829/… gives the answer. –  Jason DeVito Nov 12 '10 at 20:44
@Jason De Vito: Wonderfull! Thanks very much for very quick answer. –  evgeniamerkulova Nov 12 '10 at 21:21

## 1 Answer

See the equivalent question on mathoverflow: What manifolds are bounded by RP^odd?

(Since it seems there is a good reason to have the answer recorded as such (and, borrowing from the suggestion here), I'm moving my comment here. Since all I'm doing it linking to another place, I don't want to gain reputation for this, so I'm making it community wiki.)

However, in an effort to personally gain something from this, I'll provide a link to a similar question I asked on MO which still hasn't been answered. The question is: What manifold has $\mathbb{H}P^{odd}$ as a boundary? Incidentally, the case of $\mathbb{C}P^{odd}$ is covered in my question.

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If I took the wrong line of action here, please let me know! –  Jason DeVito Nov 14 '10 at 20:49
I think that's the right thing to do. I hate when I'm searching unanswered questions and see that some were already answered in comments. –  Nuno Nov 14 '10 at 21:24