Timeline for Holonomy map on a connected manifold determines the connection and the bundle
Current License: CC BY-SA 4.0
12 events
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| Sep 22, 2019 at 6:43 | history | edited | Praphulla Koushik | CC BY-SA 4.0 |
added 236 characters in body
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| Sep 16, 2019 at 3:10 | comment | added | Praphulla Koushik | @DavidRoberts no problem. Please respond only when you are free | |
| Sep 16, 2019 at 1:41 | comment | added | David Roberts♦ | @PraphullaKoushik I can answer that question, but not right now. | |
| Sep 15, 2019 at 16:23 | comment | added | Praphulla Koushik | For another element $y\in X$ there exists unique $h\in G$ such that $\phi(y)=y.h$. There exists unique $r\in G$ such that $y=x.r$. Then, $\phi(y)=\phi(x).r=x.g.r=yr^{-1}gr$. This conclude $r^{-1}gr=h$; that is $gr=rh$. Similarly, $\phi(x)=\phi(y)r^{-1}=yhr^{-1}=xrhr^{-1}$. So, we have $g=rhr^{-1}$; that is $gr=rh$. In case $G$ is abelian, I can then say $gr=hr$ which would then imply $g=h$. Thus, for each $\phi:X\rightarrow X$, there is unique $g\in G$ representing $\phi$... | |
| Sep 15, 2019 at 16:22 | comment | added | Praphulla Koushik | Let $G$ be a Lie group, $X$ be a manifold. Suppose further that $G$ acts freely and transitively on $X$. Let $\phi:X\rightarrow X$ be a $G$-equivariant diffeomorphism. Fix an element $x\in X$. Choose an element $y\in X$. As $G$ acts transitively on $X$, there exists $g\in G$ such that $y=x.g$. As $\phi$ is $G$-equivariant, $\phi(y)=\phi(x.g)=\phi(x).g$. Thus, $\phi(x)$ determines the map $\phi:X\rightarrow X$. So, any element of $X$ determines the map $\phi$. Fix $x\in X$. We have $\phi(x)\in X$. As $G$ acts freely and transitively, there exists unique $g\in G$ such that $\phi(x)=x.g$. | |
| Sep 15, 2019 at 15:55 | comment | added | Praphulla Koushik | @DavidRoberts Is it not necessary for $G$ to be an abelian group? I see that the map $\phi:X\rightarrow X$ given by $\phi(x)=xg$ is not a $G$-equivariant map.. $\phi(x.h)=xgh\neq (xg).h=\phi(x).h$ unless $G$ is abelian.. Am I misunderstanding something here? | |
| Aug 19, 2019 at 3:39 | comment | added | Praphulla Koushik | Oh, ok.. I don’t know much about this $B(\Omega(M)=M$.. I can not even convince myself that this is possible.. can you give so:e reference.. Assuming this, I got your point... Holonomy map is a map $\Omega(M,x)\rightarrow G$... See it as $\Omega(M)\rightarrow G$.. Then “take” the classifying space to get the map $M\rightarrow BG$.. then pullback the principal bundle $EG\rightarrow BG$ along this $M\rightarrow BG$ to get a principal bundle $P\rightarrow M$... what you said is mostly clear except that result..can you give some reference... | |
| Aug 18, 2019 at 23:13 | comment | added | Kevin Casto | If you're happy with uniqueness up to homotopy: By taking classifying spaces, the map $\Omega M \to G$ gives us a map $B(\Omega M) = M \to BG$, which is just the classifying map of the bundle, and so determines the bundle up to homotopy. Note that $B(\Omega M) = M$ requires $M$ connected. | |
| Aug 18, 2019 at 22:48 | answer | added | Vít Tuček | timeline score: 1 | |
| Aug 18, 2019 at 9:36 | comment | added | Praphulla Koushik | @DavidRoberts That is absolutely correct... I did not realise about the restriction of automorphism group to equivariant automorphism group... So, there was some confusion.. As you said, if a Lie group $G$ acts freely, transitivley on a manifold $M$, the equivariant automorphism group is identified with the Lie group $G$... Here, $G$ acts on $\pi^{-1}(x)$ freely and transitively and the maps $\pi^{-1}(x)\rightarrow \pi^{-1}(x)$ are actually $G$-equivariant.. So, this is clear,, | |
| Aug 18, 2019 at 9:14 | comment | added | David Roberts♦ | The group of G-equivariant automorphism of a G-space with a free and transitive action is G itself, because any such automorphism is determined on any given point, and then given the original point and the point it is sent to, there is a unique element of G taking one to the other. Given any other starting point, it gives rise to the same group element. Moreover, composition of these automorphisms corresponds to group multiplication. | |
| Aug 18, 2019 at 7:40 | history | asked | Praphulla Koushik | CC BY-SA 4.0 |