Timeline for The First Homology Group of Configuration Space and Knot Theory
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 26, 2013 at 19:56 | vote | accept | Samuel Reid | ||
| Dec 3, 2012 at 8:07 | answer | added | Tarje Bargheer | timeline score: 1 | |
| Dec 1, 2012 at 23:20 | comment | added | Autumn Kent | @Samuel: Yes, that's essentially my comment. Mariano's objection is that you are really interested in $C_n(\mathbb{R}^2)$ and $C_{\hat n}(\mathbb{R}^2)$, not $C_n(\mathbb{R})$ and $C_{\hat n}(\mathbb{R})$. | |
| Dec 1, 2012 at 20:40 | comment | added | Samuel Reid | @Richard Kent, $C_{n}([0,1]) = \Delta^{n}$ and $C_{\hat{n}}([0,1]) = \coprod_{i=1}^{n!} \Delta^{n}$. | |
| Dec 1, 2012 at 14:48 | comment | added | Autumn Kent | @Dan, isn't the configuration space of $n$ unordered points in the line a cell? (Each component of the configuration space of $n$ ordered points in the line is an intersection of half-spaces in $\mathbb{R}^n$, which means it's convex. The symmetric group just permutes these.) | |
| Dec 1, 2012 at 6:41 | answer | added | David C | timeline score: 1 | |
| Dec 1, 2012 at 6:14 | comment | added | Dan Petersen | Mariano: the configuration space of $n$ points in $\mathbb R^m$ is not contractible for any $n > 1$. I think you meant to write "simply connected". | |
| Dec 1, 2012 at 3:44 | comment | added | Mariano Suárez-Álvarez | You probably want to do configurations in $\mathbb R^2$; otherwise your configuration spaces are contractible. | |
| Dec 1, 2012 at 3:41 | comment | added | Autumn Kent | $H_{1}(C_{n}(\mathbb{R})) = H_1(B_n) = \mathbb{Z}$ and $H_{1}(C_{\hat{n}}(\mathbb{R})) = H_1(PB_n) = \mathbb{Z}^{n(n-1)/2}$, as can be seen by abelianizing. This can be seen using the usual presentations (planetmath.org/BraidGroup.html). | |
| Dec 1, 2012 at 3:09 | history | asked | Samuel Reid | CC BY-SA 3.0 |