Timeline for Interpretation of the monomorphism $H^2(\pi_1(X),\mathbb{Z}) \rightarrow H^2(X,\mathbb{Z})$
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 26, 2015 at 9:54 | comment | added | Chris Gerig | This is obtained for $X=BG$ in mathoverflow.net/questions/48489/… | |
| May 19, 2015 at 11:27 | comment | added | Will Sawin | BTW, it's clear that you obtain a central extension. Clearly it's a normal subgroup, and the conjugation action of $\pi_1(X)$ on $\pi_1(S^1)$ is just the monodromy of the circle bundle, which is trivial because the cohomology class goes in the direction of the circle action, which is well-defined everywhere hence has no monodromy. | |
| May 19, 2015 at 11:25 | comment | added | Will Sawin | ACL's method does secretly give you the correct map. Because the correct map is injective but not surjective, it has an inverse that's a partial function, and I'm pretty sure that's exactly ACL's map. ACL's map is only defined when the long exact sequence map $\pi_2(X) \to \pi_1(S^1)=\mathbb Z$ vanishes. This map is determined using the map $\pi_2(X) \to H_2(X)$ and the pairing $H_2(X) \times H^2(X,\mathbb Z) \to \mathbb Z$.I'm pretty sure it vanishes exactly when the line bundle comes from a central extension, and he computes the correct central extension. | |
| May 19, 2015 at 8:04 | comment | added | Pyramid | can I ask a question: what is the cokernel of the map $$H^2\big(\pi_1(X), \mathbb Z\big) \longrightarrow H^2(X,\mathbb Z).$$ | |
| Apr 17, 2015 at 9:14 | vote | accept | Jeremy Daniel | ||
| Apr 16, 2015 at 8:50 | answer | added | Konrad Waldorf | timeline score: 8 | |
| Apr 16, 2015 at 7:56 | answer | added | Simon Henry | timeline score: 9 | |
| Apr 15, 2015 at 13:17 | comment | added | Matthias Wendt | This isn't particularly geometric, but the map on $H^2$ is induced from the map $X\to B\pi_1$ (which can be obtained by killing all higher homotopy of $X$). So the map pulls back complex line bundles from $B\pi_1$ to $X$. If you start with a central extension $\mathbb{Z}\to E\to\pi_1(X)$, then there is a fibration of classifying spaces $S^1\to BE\to B\pi_1(X)$. If the extension is central, this is a principal $S^1$-bundle, and using $BS^1\cong \mathbb{CP}^\infty$ is equivalent to a complex line bundle. Pull this back to get a complex line bundle on $X$. | |
| Apr 15, 2015 at 12:05 | comment | added | Jeremy Daniel | This would give a map in the wrong direction, right? Moreover, is it clear that the extension obtained is central? | |
| Apr 15, 2015 at 11:56 | comment | added | ACL | What about removing the zero-section and considering its fundamental group? | |
| Apr 15, 2015 at 9:54 | history | asked | Jeremy Daniel | CC BY-SA 3.0 |