# 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

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.