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Oct 15, 2012 at 6:19 vote accept Mark Bell
Oct 13, 2012 at 23:33 comment added Fernando Muro Ah, I see, ok, ok.
Oct 13, 2012 at 23:31 comment added Johannes Ebert @Fernando: I meant that I could embed a skeleton of $BG$, not all of it.
Oct 13, 2012 at 22:38 comment added Fernando Muro No way if $BG$ has infinite cohomological dimension.
Oct 13, 2012 at 17:45 comment added Johannes Ebert The (now deleted) answer answered a different question, namely if one could have a noncompact manifold without boundary. That is certainly true and not so difficult: if $G$ is countable, then the standard $BG$ has countable skeleta that can be deformed and embedded into some $R^k$ as in Mishas comment. Take an open neighborhood.
Oct 13, 2012 at 13:32 comment added Donu Arapura It's always awkward when answers reach different conclusions, but I am convinced by this. I guess one could (almost) rephrase the first part by saying that if the group has infinitely generated homology in some degree $< n$, then no such compact manifold can exist.
Oct 13, 2012 at 13:14 history answered Johannes Ebert CC BY-SA 3.0