Timeline for Understanding the analytic index map of the Atiyah-Singer index theorem
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 8, 2011 at 11:30 | vote | accept | AlexE | ||
| Sep 8, 2011 at 11:30 | comment | added | AlexE | Ok, now I see it, you are right. First I was confused, because you put the "homogeneous of order 0" directly into the definition of the K-groups. But Lemma 13.3 tells us that we get the same groups if we put this "homogeneous-thing" into the definition or not. Thanks. | |
| Sep 6, 2011 at 8:44 | comment | added | Johannes Ebert | But this is shown in Lawson-Michelssohn, Lemma 13.3, page 245. | |
| Sep 5, 2011 at 20:38 | comment | added | AlexE | You have shown that the analytic index map $K^0(TX, TX-0) \to Z$, defined by mapping the class $u = (E_0, E_1; f)$ to the Fredholm index of F (a classical $\Psi DO$ with asymptotic symbol f), is well-defined. But I don't see how it follows that given an arbitrary elliptic operator D of order 0, the operator F we get from $[\sigma(D)] = [(E_0, E_1; f)]$ has the same Fredholm index as D. For this, we would need to show, e.g., that the symbol $\sigma(D)$ is regularly homotopic to one which is homogeneous of order 0 outside the zero section - and I don't see this yet. | |
| Aug 31, 2011 at 19:15 | history | answered | Johannes Ebert | CC BY-SA 3.0 |