Smooth curves and velocity












2












$begingroup$


Let $M$ be a smooth manifold.



Let $X:M to TM$ be a global smooth vector field on $M$, and let $K$ be the support of $X$, i.e. $K=overline{{pin M: X_pnot=0}}$. Suppose $K$ a is compact subset of $M$.



Let $pnotin K$, and let $theta^p$ be the maximal integral curve of $X$ starting at $p$.




Why $theta^p$ is costant?




I know that $theta^p:U^pto M$ is a smooth map, with $U^p$ open interval in $mathbb{R}$, and for each $t in U^p$ I have $(theta^p)'(t)=X_{theta^p(t)}$.



The definition of $(theta^p)'(t)$ is $(theta^p)'(t):C^infty(M)to mathbb{R}$ defined by $(theta^p)'(t)(f)=frac{d}{dt}(fcirctheta^p)(t)$.



I know I have to use that $X_q=0$ for each $qnotin K$ but I can't see how.



For example, I don't know if $theta^p(t)$ is in $K$ or not, so when I have $(theta^p)'(t)=X_{theta^p(t)}$ I don't know if $X_{theta^p(t)}$ is $0$ or not. And, even if I succeed in showing that $(theta^p)'(t)$ is $0$ for each $tin U^p$, then I don't know how to deduce from this that $theta^p $ is in fact costant.






differential-geometry smooth-manifolds smooth-functions







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    Let $M$ be a smooth manifold.



    Let $X:M to TM$ be a global smooth vector field on $M$, and let $K$ be the support of $X$, i.e. $K=overline{{pin M: X_pnot=0}}$. Suppose $K$ a is compact subset of $M$.



    Let $pnotin K$, and let $theta^p$ be the maximal integral curve of $X$ starting at $p$.




    Why $theta^p$ is costant?




    I know that $theta^p:U^pto M$ is a smooth map, with $U^p$ open interval in $mathbb{R}$, and for each $t in U^p$ I have $(theta^p)'(t)=X_{theta^p(t)}$.



    The definition of $(theta^p)'(t)$ is $(theta^p)'(t):C^infty(M)to mathbb{R}$ defined by $(theta^p)'(t)(f)=frac{d}{dt}(fcirctheta^p)(t)$.



    I know I have to use that $X_q=0$ for each $qnotin K$ but I can't see how.



    For example, I don't know if $theta^p(t)$ is in $K$ or not, so when I have $(theta^p)'(t)=X_{theta^p(t)}$ I don't know if $X_{theta^p(t)}$ is $0$ or not. And, even if I succeed in showing that $(theta^p)'(t)$ is $0$ for each $tin U^p$, then I don't know how to deduce from this that $theta^p $ is in fact costant.






    differential-geometry smooth-manifolds smooth-functions







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$


      Let $M$ be a smooth manifold.



      Let $X:M to TM$ be a global smooth vector field on $M$, and let $K$ be the support of $X$, i.e. $K=overline{{pin M: X_pnot=0}}$. Suppose $K$ a is compact subset of $M$.



      Let $pnotin K$, and let $theta^p$ be the maximal integral curve of $X$ starting at $p$.




      Why $theta^p$ is costant?




      I know that $theta^p:U^pto M$ is a smooth map, with $U^p$ open interval in $mathbb{R}$, and for each $t in U^p$ I have $(theta^p)'(t)=X_{theta^p(t)}$.



      The definition of $(theta^p)'(t)$ is $(theta^p)'(t):C^infty(M)to mathbb{R}$ defined by $(theta^p)'(t)(f)=frac{d}{dt}(fcirctheta^p)(t)$.



      I know I have to use that $X_q=0$ for each $qnotin K$ but I can't see how.



      For example, I don't know if $theta^p(t)$ is in $K$ or not, so when I have $(theta^p)'(t)=X_{theta^p(t)}$ I don't know if $X_{theta^p(t)}$ is $0$ or not. And, even if I succeed in showing that $(theta^p)'(t)$ is $0$ for each $tin U^p$, then I don't know how to deduce from this that $theta^p $ is in fact costant.






      differential-geometry smooth-manifolds smooth-functions







      share|cite|improve this question











      $endgroup$




      Let $M$ be a smooth manifold.



      Let $X:M to TM$ be a global smooth vector field on $M$, and let $K$ be the support of $X$, i.e. $K=overline{{pin M: X_pnot=0}}$. Suppose $K$ a is compact subset of $M$.



      Let $pnotin K$, and let $theta^p$ be the maximal integral curve of $X$ starting at $p$.




      Why $theta^p$ is costant?




      I know that $theta^p:U^pto M$ is a smooth map, with $U^p$ open interval in $mathbb{R}$, and for each $t in U^p$ I have $(theta^p)'(t)=X_{theta^p(t)}$.



      The definition of $(theta^p)'(t)$ is $(theta^p)'(t):C^infty(M)to mathbb{R}$ defined by $(theta^p)'(t)(f)=frac{d}{dt}(fcirctheta^p)(t)$.



      I know I have to use that $X_q=0$ for each $qnotin K$ but I can't see how.



      For example, I don't know if $theta^p(t)$ is in $K$ or not, so when I have $(theta^p)'(t)=X_{theta^p(t)}$ I don't know if $X_{theta^p(t)}$ is $0$ or not. And, even if I succeed in showing that $(theta^p)'(t)$ is $0$ for each $tin U^p$, then I don't know how to deduce from this that $theta^p $ is in fact costant.






      differential-geometry smooth-manifolds smooth-functions




      differential-geometry smooth-manifolds smooth-functions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 6 at 17:36







      Minato

















      asked Jan 6 at 17:30









      MinatoMinato

      600314




      600314






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          Since



          $p notin K, tag 1$



          $X(p) = 0; tag 2$



          let



          $theta^p(t) tag 3$



          be the unique solution of



          ${theta^p}'(t) = X(theta^p(t)) tag 4$



          such that



          $theta^p(0) = p; tag 5$



          we note that the constant curve



          $phi^p(t) = p, ; forall t in Bbb R, tag 6$



          satisfies



          ${phi^p}´(t) = 0 = X(p) = X(phi^p(t)), ; forall tin Bbb R; tag 7$



          we note that $theta^p(t)$ and $phi^p(t)$ both satisfy the same differential equation with the same initial conditions; therefore by uniqueness of solutions (which applies since $X$ is smooth, hence Lipschitz), we must have



          $theta^p(t) = phi^p(t) = p, ; forall t in Bbb R. tag 8$






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            $theta^p$ is constant because the constant function is a solution of the differential equation.






            share|cite|improve this answer









            $endgroup$














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              2 Answers
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              2 Answers
              2






              active

              oldest

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              active

              oldest

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              active

              oldest

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              1












              $begingroup$

              Since



              $p notin K, tag 1$



              $X(p) = 0; tag 2$



              let



              $theta^p(t) tag 3$



              be the unique solution of



              ${theta^p}'(t) = X(theta^p(t)) tag 4$



              such that



              $theta^p(0) = p; tag 5$



              we note that the constant curve



              $phi^p(t) = p, ; forall t in Bbb R, tag 6$



              satisfies



              ${phi^p}´(t) = 0 = X(p) = X(phi^p(t)), ; forall tin Bbb R; tag 7$



              we note that $theta^p(t)$ and $phi^p(t)$ both satisfy the same differential equation with the same initial conditions; therefore by uniqueness of solutions (which applies since $X$ is smooth, hence Lipschitz), we must have



              $theta^p(t) = phi^p(t) = p, ; forall t in Bbb R. tag 8$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Since



                $p notin K, tag 1$



                $X(p) = 0; tag 2$



                let



                $theta^p(t) tag 3$



                be the unique solution of



                ${theta^p}'(t) = X(theta^p(t)) tag 4$



                such that



                $theta^p(0) = p; tag 5$



                we note that the constant curve



                $phi^p(t) = p, ; forall t in Bbb R, tag 6$



                satisfies



                ${phi^p}´(t) = 0 = X(p) = X(phi^p(t)), ; forall tin Bbb R; tag 7$



                we note that $theta^p(t)$ and $phi^p(t)$ both satisfy the same differential equation with the same initial conditions; therefore by uniqueness of solutions (which applies since $X$ is smooth, hence Lipschitz), we must have



                $theta^p(t) = phi^p(t) = p, ; forall t in Bbb R. tag 8$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Since



                  $p notin K, tag 1$



                  $X(p) = 0; tag 2$



                  let



                  $theta^p(t) tag 3$



                  be the unique solution of



                  ${theta^p}'(t) = X(theta^p(t)) tag 4$



                  such that



                  $theta^p(0) = p; tag 5$



                  we note that the constant curve



                  $phi^p(t) = p, ; forall t in Bbb R, tag 6$



                  satisfies



                  ${phi^p}´(t) = 0 = X(p) = X(phi^p(t)), ; forall tin Bbb R; tag 7$



                  we note that $theta^p(t)$ and $phi^p(t)$ both satisfy the same differential equation with the same initial conditions; therefore by uniqueness of solutions (which applies since $X$ is smooth, hence Lipschitz), we must have



                  $theta^p(t) = phi^p(t) = p, ; forall t in Bbb R. tag 8$






                  share|cite|improve this answer









                  $endgroup$



                  Since



                  $p notin K, tag 1$



                  $X(p) = 0; tag 2$



                  let



                  $theta^p(t) tag 3$



                  be the unique solution of



                  ${theta^p}'(t) = X(theta^p(t)) tag 4$



                  such that



                  $theta^p(0) = p; tag 5$



                  we note that the constant curve



                  $phi^p(t) = p, ; forall t in Bbb R, tag 6$



                  satisfies



                  ${phi^p}´(t) = 0 = X(p) = X(phi^p(t)), ; forall tin Bbb R; tag 7$



                  we note that $theta^p(t)$ and $phi^p(t)$ both satisfy the same differential equation with the same initial conditions; therefore by uniqueness of solutions (which applies since $X$ is smooth, hence Lipschitz), we must have



                  $theta^p(t) = phi^p(t) = p, ; forall t in Bbb R. tag 8$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 6 at 19:09









                  Robert LewisRobert Lewis

                  48.9k23168




                  48.9k23168























                      2












                      $begingroup$

                      $theta^p$ is constant because the constant function is a solution of the differential equation.






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        $theta^p$ is constant because the constant function is a solution of the differential equation.






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          $theta^p$ is constant because the constant function is a solution of the differential equation.






                          share|cite|improve this answer









                          $endgroup$



                          $theta^p$ is constant because the constant function is a solution of the differential equation.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 6 at 17:44









                          MindlackMindlack

                          4,900211




                          4,900211






























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