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Apr 13, 2017 at 12:19 history edited CommunityBot
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Oct 14, 2015 at 0:14 comment added Neil Hoffman I think you could be overlooking one of the conditions of Dunbar when you say 'most' quotients of $T^3$ have underlying space $S^3$. In fact, there are several quotients of $T^3$ with underlying space $S^1\times S^2$, for example the product of the pillowcase $S^2(2,2,2,2)$ and $S^1$. These all turn out to be fibered orbifolds, so Dunbar doesn't focus on them. However, it should be pointed out that $S^2(2,2,2,2) \times S^1$ can cover itself, so 'most' perhaps might be 'most (up to homeomorphism)'.
Aug 14, 2015 at 22:32 history bounty ended arivero
Aug 13, 2015 at 16:52 history edited Igor Rivin CC BY-SA 3.0
added five dimensions
Aug 12, 2015 at 17:21 comment added arivero well the underlying manifold must be a manifold, yep, but for the orbifold of course I am happy to allow for conical singularities and usual stuff (I just mention it for benefit of the casual passer-by)
Aug 12, 2015 at 16:34 history edited Igor Rivin CC BY-SA 3.0
added $n=4$ info.
Aug 11, 2015 at 22:38 history edited Igor Rivin CC BY-SA 3.0
more examples
Aug 8, 2015 at 23:46 comment added arivero Hmm having it for n=1,2,3 and possibly 4 starts to signal that dimension is not a main worry. It could have still some argument about having only conical singularities or more extended ones.
Aug 8, 2015 at 23:26 history edited Igor Rivin CC BY-SA 3.0
added 3-d thing.
Aug 8, 2015 at 11:02 history edited Igor Rivin CC BY-SA 3.0
added projective space examples.
Aug 8, 2015 at 10:17 history edited Igor Rivin CC BY-SA 3.0
made small correction, added example
Aug 7, 2015 at 22:42 history answered Igor Rivin CC BY-SA 3.0