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This is a followup to Spaces of Finite Subspaces. Just for convenience, $\exp_nX$ is the space whose underlying set is the set of nonempty subsets $S\subseteq X$ with $|S|\le n$.
As Alex Suciu pointed out in his answer to the previous post (referencing Chris Tuffley), $\exp_{2k-1}S^1$ and $\exp_{2k}S^1$ are homotopy equivalent to $S^{2k-1}$. And as for embeddings, $\exp_nS^1-\exp_{n-2}S^1$ has the homotopy type of an $(n-1,n)$-torus knot complement!

Then can anything be said about homeomorphic properties? Cliff Wagner showed in his thesis that $\exp_nS^1$ is a closed manifold iff $n=1$ or $n=3$, so generically they are not spheres. For instance, $\exp_2S^1$ is the Mobius band, yet this information is lost under homotopy; so it would be interesting to know what is lost in higher dimensions.

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up vote 4 down vote accepted

Yes, the manifold question is completely answered here. Namely $\exp_n X^k$ (where $X^k$ is a $k$-manifold) is a manifold if and only if $k=1, n=3$ or $k=n=2.$ This is Theorem 1.3 in the referenced paper. EDIT Also, of course if $n=1,$ though the authors overlook this...

EDIT By the way, some very nice papers on the subject have been written by Chris Tuffley (a couple seem to be in AGT).

ANOTHER EDIT In particular, Tuffley gives the simplicial complex structure of $\exp_n S^1$ explicitly, and also describes the "complement" of $\exp_{n-2}S^1$ in $\exp_n S^1,$ which is already interesting in the case $n=3$ (it's the trefoil knot complement). From Tuffley's thing you can, at least in principle, answer all homeomorpism-related questions (I am referring to the paper with the illogical title: "Finite Subset Spaces of $S^1.$"

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If you are ONLY interested in $S^1$ your comment is correct. Their result is for ANY closed manifold. – Igor Rivin Dec 29 '11 at 14:12
But this is part of the 'starting point' of my question, because for $S^1$ the result is that $\exp_nS^1$ is a closed manifold iff $n=1,3$. For other $n$ it could still be some other nice-enough space (like $\exp_2S^1$). – Chris Gerig Dec 29 '11 at 14:12
(sorry edited my comment right after you responded to it)... Yes for this followup question I am only interested in $S^1$. – Chris Gerig Dec 29 '11 at 14:13
See the second edit... – Igor Rivin Dec 29 '11 at 14:16

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