Skip to main content

Timeline for contractible configuration spaces

Current License: CC BY-SA 3.0

4 events
when toggle format what by license comment
Mar 3, 2015 at 21:40 comment added John Klein You're correct of course. But I think you can replace the euclidean space with $D^n_+$ (the upper hemisphere). Then $F(D^n_+,k)$ becomes weakly contractible as $n$-tends to infinity by a general position argument. Argument 1 will at then at least imply that $F(S^\infty,k)$ is weakly contractible.
Mar 3, 2015 at 7:41 comment added Eric Wofsey In Argument 1, the inclusions $\mathbb{R}^n\to\mathbb{R}^{n+1}$ you obtain are not the standard inclusions, and so it is not obvious that their colimit is $\mathbb{R}^\infty$. I believe it still is, though I haven't checked all the details (I think you can identify the $\mathbb{R}^n$ appearing in your direct system with the ball of radius $n$ around the origin inside the standard $\mathbb{R}^n\subset\mathbb{R}^\infty$).
Mar 3, 2015 at 2:39 history edited John Klein CC BY-SA 3.0
added 451 characters in body
Mar 2, 2015 at 21:16 history answered John Klein CC BY-SA 3.0