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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1$\begingroup$ Didn't you ask a really similar question the other day? And didn't you get a really similar answer? $\endgroup$– Todd TrimbleCommented Nov 13, 2012 at 15:25
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$\begingroup$ 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! $\endgroup$– FluxCommented Nov 13, 2012 at 15:40
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$\begingroup$ @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 $n$. $\endgroup$– MishaCommented Nov 13, 2012 at 16:35
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1$\begingroup$ Misha, I don't think that's not quite right. See this MO answer: mathoverflow.net/questions/84950/metrizable-space/84974#84974 $\endgroup$– Todd TrimbleCommented Nov 13, 2012 at 17:48
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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.
answered Nov 13, 2012 at 12:16
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$\begingroup$ 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? $\endgroup$– FluxCommented Nov 14, 2012 at 14:15
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$\begingroup$ Clearly, I refer to the punctured Poincare' homology sphere. $\endgroup$– FluxCommented Nov 14, 2012 at 14:19
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$\begingroup$ Every manifold embeds in some $\mathbf R^n$ (Whitney). $\endgroup$ Commented Nov 14, 2012 at 15:38