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Sep 1, 2019 at 7:06 comment added Ryan Budney @DanRamras, apologies, I somehow conflated the two spaces. Delete my 2nd sentence -- those two aren't quite the same.
Sep 1, 2019 at 1:51 comment added Dan Ramras @RyanBudney I'm confused by the claim that $SS_3 (S^1)$ is the 3-sphere. If $SS_3 (S^1)$ is really the same as the symmetric product $(S^1)^3/\Sigma_3$, then there are maps $S^1\to (S^1)^3/\Sigma_3 \to S^1$ defined by $z \mapsto [z, 1, 1]$ and $[x, y, z]\mapsto xyz$ which compose to the identity. So $(S^1)^3/\Sigma_3$ isn't simply connected.
Mar 10, 2017 at 9:42 history edited CommunityBot
replaced http://upload.wikimedia.org/ with https://upload.wikimedia.org/
Mar 13, 2011 at 22:12 comment added Sam Nead Ah - here is the question I saw - mathoverflow.net/questions/22138/…
Mar 13, 2011 at 22:08 comment added Sam Nead Your claim about taking the link of the singular point of the hypersurface $x^2 + y^2 + z^2 + w^3 = 0$, and getting an exotic five-sphere, sounds wrong to me. Is there a reference for this?
Mar 13, 2011 at 19:40 history edited Sputnik CC BY-SA 2.5
Typo.
Mar 13, 2011 at 19:01 history edited darij grinberg CC BY-SA 2.5
fixed typo
Aug 16, 2010 at 0:00 comment added Ryan Budney There's an interesting related result -- let $SS_n(X)$ denote the space of subsets of a topological space $X$ where the subsets have cardinality between $1$ and $n$. As a space, you can consider it to be $X^n / \Sigma_n$. Then $SS_3(S^1)$ is the 3-sphere. By design, $SS_n(S^1)$ has a fixed-point free $SO_2$-action, so it's a Seifert-fibred space. A non-singular orbit of this $SO_2$-action on $SS_3(S^1)$ is the trefoil knot.
Jul 15, 2010 at 19:55 comment added The Mathemagician Wow,that IS pretty whack,Csar.This example alone is a testament to the power of modern topology and geometry and the incredible connections it has uncovered.
Feb 17, 2010 at 1:16 history edited Csar Lozano Huerta CC BY-SA 2.5
added 25 characters in body
Feb 16, 2010 at 22:35 history answered Csar Lozano Huerta CC BY-SA 2.5