Timeline for Homotopy groups of Fredholm operators
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 8, 2015 at 18:24 | history | edited | Paul Siegel | CC BY-SA 3.0 |
added 223 characters in body
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| Oct 14, 2014 at 8:33 | vote | accept | Chandler | ||
| Sep 30, 2014 at 12:13 | vote | accept | Chandler | ||
| Oct 14, 2014 at 8:33 | |||||
| Sep 30, 2014 at 12:13 | vote | accept | Chandler | ||
| Sep 30, 2014 at 12:13 | |||||
| Aug 20, 2014 at 5:06 | comment | added | Dan Ramras | Oh, and a reference for the bijection in the above answer is the Appendix to Atiyah's book K-theory. | |
| Aug 20, 2014 at 5:02 | comment | added | Dan Ramras | (cont'd) So $\pi_1 (\mathcal{F}_0) = 1$ (for this, it's enough to check that each loop is (unbased) nullhomotopic, which follows from $[S^1, \mathcal{F}_0] = \widetilde{K} (S^1) = 0$) and now $\pi_n (\mathcal{F}) = \pi_n (\mathcal{F}_0) = \langle S^n, \mathcal{F}_0 \rangle = [S^n, \mathcal{F}_0] = \widetilde{K}(S^n)$. Here $\langle , \rangle$ means based homotopy classes of maps, which is the same as unbased homotopy classes when the range is simply connected. | |
| Aug 20, 2014 at 4:59 | comment | added | Dan Ramras | Hi Paul, Maybe I'm missing something, but I think that $\pi_n (\mathcal{F})$ is isomorphic to the reduced groups $\widetilde{K} (S^n)$, not $K(S^n)$. Unreduced K-theory classifies (unbased) homotopy classes of (unbased) maps out of X, but for homotopy groups we're interested in based maps and based homotopy. Letting $\mathcal{F}_0$ denote the index 0 Fredholm operators (which form one connected component of $\mathcal{F}$), the bijection above restricts to $[X, \mathcal{F}_0] \cong \widetilde{K} (X)$. | |
| Aug 19, 2014 at 17:38 | comment | added | Paul Siegel | That's correct. | |
| Aug 19, 2014 at 17:20 | comment | added | Chandler | Thanks for the answer. So I can use the $K_0$-groups of $S^n$? | |
| Aug 19, 2014 at 17:16 | history | answered | Paul Siegel | CC BY-SA 3.0 |