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Can you exhibit an example of a noncontractible domain in R^n with the d-th cohomology groups trivial for all d greater or equal to 1? Thank you

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closed as off topic by Anthony Quas, David White, Oscar Randal-Williams, Todd Trimble, Steven Landsburg Nov 13 '12 at 17:11

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Didn't you ask a really similar question the other day? And didn't you get a really similar answer? – Todd Trimble Nov 13 '12 at 15:25
In the other question I need a noncontractible domain with Euler characteristic equal to 1. But only now I can understand that the older question and the new are related in some sense. Sorry, I'm not strong on this subject! – Flux Nov 13 '12 at 15:40
@Flux: Here is a general principle to remember: If $X$ is a finite-dimensional countable cell-complex, then $X$ is homotopy-equivalent to a domain in ${\mathbb R}^n$ for some $n$. – Misha Nov 13 '12 at 16:35
Misha, I don't think that's not quite right. See this MO answer: – Todd Trimble Nov 13 '12 at 17:48
up vote 4 down vote accepted

There are a million examples. You should google "acyclic space". Here is one: if you remove a point from a homology sphere you get a manifold whose cohomology is trivial in positive degrees. Take a tubular neighbourhood of it in some $\mathbf R^n$ and you get an open domain.

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If, for instance, I consider the Poincare' homology sphere, this is a particular 3-homology sphere (in particular not homeomorphic to the sphere) acyclic and noncontractible, thus a good example for my question. But I think I have read somewhere that this object cannot be smoothly embeded in R^4, right? How can I do in this case? Can I modify the object to have the embeding property but not changing the acyclic and the noncontractibility propreties? – Flux Nov 14 '12 at 14:15
Clearly, I refer to the punctured Poincare' homology sphere. – Flux Nov 14 '12 at 14:19
Every manifold embeds in some $\mathbf R^n$ (Whitney). – Dan Petersen Nov 14 '12 at 15:38

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