Integral curves over a closed interval.











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Let $[0,b]$ be an interval and let $t$ denote the standard coordinate on $[0,b].$Suppose that $pi:E rightarrow [0,b]$is a vector bundle over $[0,b]$ with connection (the definition of connection I use is as in this question). Let $tilde{partial}$ denote the horizontal lift of $partial/partial{t}.$



Suppose we knew the following two facts:




  1. If $c:[0,a] rightarrow E$ is an integral curve of $tilde{partial},$ then $c(a) in E_a.$ (this is not hard to show).

  2. Let $0 leq t_0 < b.$Then there is a fixed $epsilon >0$ depending only on $t_0$ such that all maximal integral curves of $tilde{partial}$ originating in the fiber $E_{t_0}$ are defined at least on $[t_0,epsilon).$


Given these two facts, how do I prove that all integral curves of $tilde{partial}$ have domain equal to $[0,b]?$



This is an exercise from the book "Manifolds and Differential Geometry" by J.M. Lee






differential-geometry







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    Let $[0,b]$ be an interval and let $t$ denote the standard coordinate on $[0,b].$Suppose that $pi:E rightarrow [0,b]$is a vector bundle over $[0,b]$ with connection (the definition of connection I use is as in this question). Let $tilde{partial}$ denote the horizontal lift of $partial/partial{t}.$



    Suppose we knew the following two facts:




    1. If $c:[0,a] rightarrow E$ is an integral curve of $tilde{partial},$ then $c(a) in E_a.$ (this is not hard to show).

    2. Let $0 leq t_0 < b.$Then there is a fixed $epsilon >0$ depending only on $t_0$ such that all maximal integral curves of $tilde{partial}$ originating in the fiber $E_{t_0}$ are defined at least on $[t_0,epsilon).$


    Given these two facts, how do I prove that all integral curves of $tilde{partial}$ have domain equal to $[0,b]?$



    This is an exercise from the book "Manifolds and Differential Geometry" by J.M. Lee






    differential-geometry







    share|cite|improve this question


























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      Let $[0,b]$ be an interval and let $t$ denote the standard coordinate on $[0,b].$Suppose that $pi:E rightarrow [0,b]$is a vector bundle over $[0,b]$ with connection (the definition of connection I use is as in this question). Let $tilde{partial}$ denote the horizontal lift of $partial/partial{t}.$



      Suppose we knew the following two facts:




      1. If $c:[0,a] rightarrow E$ is an integral curve of $tilde{partial},$ then $c(a) in E_a.$ (this is not hard to show).

      2. Let $0 leq t_0 < b.$Then there is a fixed $epsilon >0$ depending only on $t_0$ such that all maximal integral curves of $tilde{partial}$ originating in the fiber $E_{t_0}$ are defined at least on $[t_0,epsilon).$


      Given these two facts, how do I prove that all integral curves of $tilde{partial}$ have domain equal to $[0,b]?$



      This is an exercise from the book "Manifolds and Differential Geometry" by J.M. Lee






      differential-geometry







      share|cite|improve this question















      Let $[0,b]$ be an interval and let $t$ denote the standard coordinate on $[0,b].$Suppose that $pi:E rightarrow [0,b]$is a vector bundle over $[0,b]$ with connection (the definition of connection I use is as in this question). Let $tilde{partial}$ denote the horizontal lift of $partial/partial{t}.$



      Suppose we knew the following two facts:




      1. If $c:[0,a] rightarrow E$ is an integral curve of $tilde{partial},$ then $c(a) in E_a.$ (this is not hard to show).

      2. Let $0 leq t_0 < b.$Then there is a fixed $epsilon >0$ depending only on $t_0$ such that all maximal integral curves of $tilde{partial}$ originating in the fiber $E_{t_0}$ are defined at least on $[t_0,epsilon).$


      Given these two facts, how do I prove that all integral curves of $tilde{partial}$ have domain equal to $[0,b]?$



      This is an exercise from the book "Manifolds and Differential Geometry" by J.M. Lee






      differential-geometry




      differential-geometry






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      edited Nov 23 at 16:43

























      asked Nov 22 at 18:12









      Dedalus

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