Timeline for Representations of \pi_1, G-bundles, Classifying Spaces
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 9, 2010 at 12:58 | vote | accept | Justin Curry | ||
| Apr 9, 2010 at 1:57 | answer | added | Paul | timeline score: 6 | |
| Apr 9, 2010 at 0:57 | comment | added | Justin Curry | sorry, to be accurate both $G$ and $\pi_1(X)$ need to be non-abelian to prevent UCT from getting the desired interpretation | |
| Apr 8, 2010 at 21:54 | comment | added | Justin Curry | Yeah, I also approached this using universal coefficient theorem, but the non-abelian-ness of either $pi_1(X)$ or $G$ means UCT provides no suitable interpretation in terms of maps from $\pi_1$. | |
| Apr 8, 2010 at 21:36 | answer | added | Ben Webster♦ | timeline score: 5 | |
| Apr 8, 2010 at 21:29 | comment | added | Tim Perutz | To amplify part of Tyler's comment (something that Angelo has also noted): in algebraic topology courses one learns that, for path-connected spaces $X$, $H_1(X)$ is the abelianisation of $\pi_1(X)$. The "dual" statement seems to be less widely appreciated: that for any (discrete) abelian group $A$, one has $H^1(X;A)=Hom(H_1(X),A)=Hom(\pi_1(X),A)$. You can prove this using universal coefficients; the Ext term vanishes because $H_0(X)$ is free abelian. | |
| Apr 8, 2010 at 20:46 | answer | added | Angelo | timeline score: 5 | |
| Apr 8, 2010 at 19:53 | comment | added | Tyler Lawson | One point to make is that for G a discrete group, homotopy classes of maps from a connected space X to K(G,1) are equivalent to homomorphisms $\pi_1(X) \to G$ if you take based homotopy classes, and conjugacy classes of homomorphisms if you take unbased homotopy classes. | |
| Apr 8, 2010 at 19:00 | history | edited | Justin Curry | CC BY-SA 2.5 |
title change, further elaborations
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| Apr 8, 2010 at 18:00 | history | edited | Justin Curry | CC BY-SA 2.5 |
added 210 characters in body
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| Apr 8, 2010 at 17:31 | history | asked | Justin Curry | CC BY-SA 2.5 |