Timeline for pull-back connection
Current License: CC BY-SA 2.5
23 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Sep 7, 2019 at 22:31 | history | bounty ended | Ali Taghavi | ||
| S Sep 7, 2019 at 22:31 | history | notice removed | Ali Taghavi | ||
| Sep 3, 2019 at 18:52 | answer | added | Lev Soukhanov | timeline score: 1 | |
| S Sep 3, 2019 at 18:26 | history | bounty started | Ali Taghavi | ||
| S Sep 3, 2019 at 18:26 | history | notice added | Ali Taghavi | Improve details | |
| Sep 3, 2019 at 16:59 | review | Suggested edits | |||
| Sep 3, 2019 at 17:57 | |||||
| Jul 13, 2017 at 9:02 | comment | added | Emilio Ferrucci | For anyone stumbling upon this old question like me, see math.stackexchange.com/questions/2228111/pullback-connection and math.stackexchange.com/questions/155173/… , where it is explained that affine connections can't be pulled back (but as stated below, e.g. in Milnor-Stasheff p.292 connections can be pulled back to the pullback bundle). | |
| Nov 11, 2014 at 7:54 | answer | added | Peter Michor | timeline score: 5 | |
| Nov 11, 2014 at 7:11 | answer | added | Peter Hochs | timeline score: 9 | |
| Dec 13, 2010 at 18:45 | answer | added | Deane Yang | timeline score: 18 | |
| Dec 13, 2010 at 17:33 | comment | added | user11538 | Please read the second answer below! | |
| Dec 13, 2010 at 17:27 | comment | added | BCnrd | As an aside, it feels slightly circular (or ironic) to appeal to parallel transport intuition when creating this definition, since the very definition of parallel transport along general parametric curves $\gamma:J\rightarrow B$ (with an interval $J$ in $\mathbf{R}$ of positive length) -- whose image could be horribly "self-crossing" -- is to disentangle everything using the pullback connection $\gamma^{\ast}(\nabla)$ on the pullback bundle $\gamma^{\ast}(E)$ (with pullback metric). Over $J$ it unravels since bundles are globally trivial (via linear ODE's and a separate connection argument) | |
| Dec 13, 2010 at 17:15 | comment | added | BCnrd | Over a pt, connections vanish (since vector fields & 1-forms are 0). In general "$dF(X)$" makes no sense as a vector field on $B$. View connections as additive maps from sections of $E$ to sections of $E \otimes \Omega^1_B$ over varying opens in $B$. Local sections of $F^{\ast}(E)$ are function-linear combinations of $F$-pullbacks of local sections of $E$, so the pullback rule (using pullback of 1-forms and of local sections) and Leibniz yield uniqueness. Construction with local coords gives local existence (& yields d when $B$ is pt and $E$ trivial), so by uniqueness get global existence. | |
| Dec 13, 2010 at 17:14 | comment | added | user11538 | Emerton, you are right. | |
| Dec 13, 2010 at 17:10 | answer | added | user11538 | timeline score: 2 | |
| Dec 13, 2010 at 17:10 | comment | added | Willie Wong | I second Deane's suggestion to do it in local trivialization. Your equation does not actually directly define the pullback connection for all sections of the bundle $F^*E$. The expression only makes sense for pullbacks of sections $s$ of $E$ with $F$. So it should come as no surprise that if you plug-in $X$ such that $dF(X) = 0$, both sides evaluate to zero. What you then need to do is to use linearity of the connection plus the Leibniz rule, since starting with any local frame on $E$ you now already know how to parallel transport the frame. | |
| Dec 13, 2010 at 17:06 | answer | added | Daniel Pape | timeline score: 6 | |
| Dec 13, 2010 at 17:04 | comment | added | Emerton | I don't understand the statement "a connection on $E \to pt$ is an endomorphism of $E$''. I would have thought it is a map $E \to E\otimes \Omega^1_{\pt},$ and since $\Omega^1_{\pt} = 0$, there is a unique connection, namely the zero map. This makes sense in terms of parallel transport: over a point there is nowhere to transport anything! | |
| Dec 13, 2010 at 16:44 | comment | added | Matt Noonan | From the right-hand side of your equation, if $dF(X) = 0$ then $(F^*\nabla)_X = 0$ as well (note it should be $dF$, not $df$). | |
| Dec 13, 2010 at 16:26 | comment | added | Deane Yang | I take it back. Your equation is right. But I still recommend playing around with it using local co-ordinates and trivialization. It's a rather confusing equation (at least for me). | |
| Dec 13, 2010 at 16:17 | comment | added | Deane Yang | I don't know how to define $F^*\nabla$ functorially (i.e., without using local co-ordinates and/or trivializations). I suggest first working this out using local co-ordinates and trivializations. | |
| Dec 13, 2010 at 16:06 | comment | added | Deane Yang | What is $F^*s$? | |
| Dec 13, 2010 at 15:54 | history | asked | user11538 | CC BY-SA 2.5 |