Timeline for Can a homotopy inverse of the map from a Lie group to loops on its classifying space be given by holonomy?
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10 events
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| Jan 24, 2011 at 0:22 | comment | added | Jim Stasheff | @Klein - if we use Moore loops and a smooth bundle with connection determining the `homotopy', then multiplicative is that what Whitehead said? | |
| Jan 23, 2011 at 2:26 | comment | added | John Klein | @Todd, regarding my last comment: the issue of why the map $\Omega B \to G(F)$ is multiplicative is discussed in Whitehead's book. | |
| Jan 21, 2011 at 23:31 | comment | added | Todd Trimble | Yes, sure. I was simply saying that "the transport operation described above factors through G" does not hold for every choice of section (as "described above"). Otherwise I'm fine with what you had written (and thanks again; it was a good answer). | |
| Jan 21, 2011 at 22:27 | comment | added | John Klein | Actually, I'm not sure the map I constructed is apriori an $A_\infty$ homomorphism. I will think about how to rectify that. | |
| Jan 21, 2011 at 20:19 | comment | added | John Klein | If the fibration is actually a fiber bundle with structure group G, then the section of $E^I \to \Lambda_p$ can be chosen so that the parallel transport along the path is a homeomorphism and is given by the action of $G$. How does one chose it? By the classical holonomy of course. In any case, the space of sections of the lifting problem is contractible, so whatever section you chose is homotopic to mine and so classical holonomy $\Omega B \to G$ followed by the map in $G(F)$ is homotopic to mine. Lastly, the map I constructed is a homomorphism in the $A_\infty$ sense. | |
| Jan 21, 2011 at 16:36 | comment | added | Todd Trimble | Thank you, John. I follow you fine except for the point where you say that the transport operation can be factored through G, which I don't see since the preceding only used the fact that the bundle is a Hurewicz fibration. (Perhaps that particular point doesn't matter if all one is setting out to do is get the homotopy inverse I inquired about, but classical holonomy of a connection would give the extra information that the map $\Omega BG \to G$ takes path composition to group multiplication.) | |
| Jan 20, 2011 at 22:38 | history | edited | John Klein | CC BY-SA 2.5 |
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| Jan 20, 2011 at 21:51 | history | edited | John Klein | CC BY-SA 2.5 |
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| Jan 20, 2011 at 21:44 | history | edited | John Klein | CC BY-SA 2.5 |
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| Jan 20, 2011 at 18:18 | history | answered | John Klein | CC BY-SA 2.5 |