Timeline for Does the holonomy map define a homomorphism $\pi_k(X)\to\pi_{k-1}(Hol(\nabla))$?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Mar 26, 2014 at 10:38 | vote | accept | geodude | ||
| Mar 26, 2014 at 10:37 | history | edited | geodude | CC BY-SA 3.0 |
Corrected a mistake: the map in the end is not an isomorphism, it *induces* an isomorphism!
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| Mar 20, 2014 at 17:49 | history | edited | geodude | CC BY-SA 3.0 |
edited title
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| Mar 20, 2014 at 17:48 | answer | added | geodude | timeline score: 5 | |
| Mar 20, 2014 at 17:43 | history | edited | geodude | CC BY-SA 3.0 |
fixed typos
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| Mar 20, 2014 at 11:46 | comment | added | Konrad Waldorf | @abx: nice! Another remark: the map from $\Omega X$ to $Hol(\nabla)$ is a "group homomorphism"; usually, this imposes conditions on the possible maps. | |
| Mar 20, 2014 at 10:06 | comment | added | geodude | @Ulrich Right. Except in some interesting cases, like the sphere (see the edit). Is the rest of the reasoning true? Where has it been done? | |
| Mar 20, 2014 at 9:52 | history | edited | geodude | CC BY-SA 3.0 |
added 211 characters in body; added 48 characters in body
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| Mar 20, 2014 at 9:47 | comment | added | abx | I think what you are saying is that the holonomy defines a continuous map of the loop space $\Omega X$ onto $Hol(\nabla)$, so there are induced homomorphisms $\pi _k(\Omega X)=\pi _{k+1}(X)\rightarrow \pi _k(Hol(\nabla))$. As Ulrich points out, I am afraid that doesn't give you any restriction on the possible holonomy groups. | |
| Mar 20, 2014 at 9:34 | comment | added | Ulrich Pennig | I think the implication provides no real restriction -- at least in the way it is stated, since the quotient of $\pi_{k+1}(M)$ contained in $\pi_k(Hol(\nabla))$ might be the trivial group. | |
| Mar 20, 2014 at 9:22 | history | asked | geodude | CC BY-SA 3.0 |