Does parallel transport change the subspace?












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Let $M$ be a Riemannian manifold and $N$ be an immersed submanifold. Take $gamma$ a geodesic starting at and perpendicularly to $N$. Let $X(t)$ be a vector field along $gamma$ such that $X(0) in T_{gamma(0)}N$. Assume that $X(t)$ is not necessarily parallel along $gamma$. Take an instant $t_0$ and consider $X(t_0)$. Denote by $E_X(t)$ the parallel transport of $X(t_0)$ along $gamma$. Does $E_X(0) in T_{gamma(0)}N$?






riemannian-geometry geodesic







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    $begingroup$
    I assume by the sentence, "Let $X(t)$ be a tangent vector field to $N$ at $t = 0$, you mean $X(0) in T_{gamma(0)}N$? Is that right? Cheers!
    $endgroup$
    – Robert Lewis
    Feb 2 at 20:12








  • 1




    $begingroup$
    @RobertLewis, yes, I will rephrase.
    $endgroup$
    – L.F. Cavenaghi
    Feb 2 at 20:12
















6












$begingroup$


Let $M$ be a Riemannian manifold and $N$ be an immersed submanifold. Take $gamma$ a geodesic starting at and perpendicularly to $N$. Let $X(t)$ be a vector field along $gamma$ such that $X(0) in T_{gamma(0)}N$. Assume that $X(t)$ is not necessarily parallel along $gamma$. Take an instant $t_0$ and consider $X(t_0)$. Denote by $E_X(t)$ the parallel transport of $X(t_0)$ along $gamma$. Does $E_X(0) in T_{gamma(0)}N$?






riemannian-geometry geodesic







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I assume by the sentence, "Let $X(t)$ be a tangent vector field to $N$ at $t = 0$, you mean $X(0) in T_{gamma(0)}N$? Is that right? Cheers!
    $endgroup$
    – Robert Lewis
    Feb 2 at 20:12








  • 1




    $begingroup$
    @RobertLewis, yes, I will rephrase.
    $endgroup$
    – L.F. Cavenaghi
    Feb 2 at 20:12














6












6








6


2



$begingroup$


Let $M$ be a Riemannian manifold and $N$ be an immersed submanifold. Take $gamma$ a geodesic starting at and perpendicularly to $N$. Let $X(t)$ be a vector field along $gamma$ such that $X(0) in T_{gamma(0)}N$. Assume that $X(t)$ is not necessarily parallel along $gamma$. Take an instant $t_0$ and consider $X(t_0)$. Denote by $E_X(t)$ the parallel transport of $X(t_0)$ along $gamma$. Does $E_X(0) in T_{gamma(0)}N$?






riemannian-geometry geodesic







share|cite|improve this question











$endgroup$




Let $M$ be a Riemannian manifold and $N$ be an immersed submanifold. Take $gamma$ a geodesic starting at and perpendicularly to $N$. Let $X(t)$ be a vector field along $gamma$ such that $X(0) in T_{gamma(0)}N$. Assume that $X(t)$ is not necessarily parallel along $gamma$. Take an instant $t_0$ and consider $X(t_0)$. Denote by $E_X(t)$ the parallel transport of $X(t_0)$ along $gamma$. Does $E_X(0) in T_{gamma(0)}N$?






riemannian-geometry geodesic




riemannian-geometry geodesic






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edited Feb 2 at 20:13







L.F. Cavenaghi

















asked Feb 2 at 19:50









L.F. CavenaghiL.F. Cavenaghi

2,326619




2,326619








  • 1




    $begingroup$
    I assume by the sentence, "Let $X(t)$ be a tangent vector field to $N$ at $t = 0$, you mean $X(0) in T_{gamma(0)}N$? Is that right? Cheers!
    $endgroup$
    – Robert Lewis
    Feb 2 at 20:12








  • 1




    $begingroup$
    @RobertLewis, yes, I will rephrase.
    $endgroup$
    – L.F. Cavenaghi
    Feb 2 at 20:12














  • 1




    $begingroup$
    I assume by the sentence, "Let $X(t)$ be a tangent vector field to $N$ at $t = 0$, you mean $X(0) in T_{gamma(0)}N$? Is that right? Cheers!
    $endgroup$
    – Robert Lewis
    Feb 2 at 20:12








  • 1




    $begingroup$
    @RobertLewis, yes, I will rephrase.
    $endgroup$
    – L.F. Cavenaghi
    Feb 2 at 20:12








1




1




$begingroup$
I assume by the sentence, "Let $X(t)$ be a tangent vector field to $N$ at $t = 0$, you mean $X(0) in T_{gamma(0)}N$? Is that right? Cheers!
$endgroup$
– Robert Lewis
Feb 2 at 20:12






$begingroup$
I assume by the sentence, "Let $X(t)$ be a tangent vector field to $N$ at $t = 0$, you mean $X(0) in T_{gamma(0)}N$? Is that right? Cheers!
$endgroup$
– Robert Lewis
Feb 2 at 20:12






1




1




$begingroup$
@RobertLewis, yes, I will rephrase.
$endgroup$
– L.F. Cavenaghi
Feb 2 at 20:12




$begingroup$
@RobertLewis, yes, I will rephrase.
$endgroup$
– L.F. Cavenaghi
Feb 2 at 20:12










3 Answers
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Consider the plane ${(x, y, 0)mid x,y in Bbb R}subsetmathbb{R}^3$, the line $tmapsto (0, 0, t)$ and the vector field $tmapsto(1, 1, t)$ along it. Then the parallel transport of no vector in the field along the curve will be tangent to the plane.






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$endgroup$





















    3












    $begingroup$

    Negative answer. Counter-example: $(M,g) = (Bbb R^3, langlecdot, cdot rangle)$, $N = Bbb R^2times {0}$, $gamma(t) = (0,0,t)$ and $X(t) = (0,0,sin t)$. Then $X(0) = 0 in T_{gamma(0)}N$, but the parallel transport of $X(pi/2) = (0,0,1)$ from $gamma(pi/2) = (0,0,pi/2)$ to $gamma(0) =0$ via $gamma$ is $(0,0,1) notin T_{gamma(0)}N$. The values $X(t)$ for $tneq t_0$ should not influence the result of transporting $X(t_0)$, since a priori one transports vectors and not fields.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      I really appreciate your comment about transporting vectors and not fields, this was very clarifying for some stuff I am working on, never had thought about it.
      $endgroup$
      – L.F. Cavenaghi
      Feb 3 at 3:36



















    2












    $begingroup$

    We of course understand that $dot gamma(0) ne 0$, lest $gamma(t)$ be a rather trivial geodesic.



    We observe that, since $dot gamma(0)$ is "perpendicular" to $N$, the condition



    $langle dot gamma(0), X(0) rangle = 0 tag 0$



    is necessary, but not sufficient, for



    $X(0) in T_{gamma(0)}N, tag 1$



    for (0) says that $X(0)$ has no component along $dot gamma(0)$; in the event that $N$ is a hypersurface in $M$, that is,



    $dim N = dim M - 1, tag 2$



    (0) is in fact also sufficient, for then $dot gamma(0)$ and $T_{gamma(0)}N$ span $T_{gamma}(0)N$; in fact, (2) implies



    $T_{gamma(0)}M = langle dot gamma(0) rangle oplus T_{gamma(0)}N, tag 3$



    where $langle dot gamma(0) rangle$ is the one-dimensional subspace of $T_{gamma(0)}M$ generated by $dot gamma(0)$.



    Now consider any vector field $X(t)$ along $gamma(t)$; we have



    $nabla_{dot gamma} langle dot gamma, X rangle = langle nabla_{dot gamma} dot gamma, X rangle + langle dot gamma, nabla_{dot gamma} X rangle; tag 4$



    if $X(t)$ is parallel along $gamma(t)$, then



    $nabla_{dot gamma}X = 0; tag 5$



    thus (4) becomes



    $nabla_{dot gamma} langle dot gamma, X rangle = langle nabla_{dot gamma} dot gamma, X rangle; tag 6$



    now the general formulation of the geodesic equation as I understand it is



    $nabla_{dot gamma} dot gamma = f(t) dot gamma; tag 7$



    here $f(t)$ is determined by the choice of the running parameter $t$ of $gamma(t)$; if we substitute this into (6), we obtain



    $nabla_{dot gamma} langle dot gamma, X rangle = f(t) langle dot gamma, X rangle; tag 8$



    the solution to this linear, first order differential equation for $langle dot gamma, X rangle$ with $langle dot gamma(t_0), X(t_0) rangle = langle dot gamma, X rangle(t_0)$ is



    $langle dot gamma, X rangle(t) = exp left (displaystyle int_{t_0}^t f(s) ; ds right ) langle dot gamma, X rangle (t_0); tag 9$



    since



    $forall t in Bbb R, ;exp left (displaystyle int_{t_0}^t f(s) ; ds right ) ne 0, tag{10}$



    (9) shows that



    $langle dot gamma(t), X(t) rangle = langle dot gamma, X rangle(t) = 0 Longleftrightarrow langle dot gamma, X rangle (t_0) = langle dot gamma(t_0), X(t_0) rangle = 0; tag{11}$



    of course, if $gamma(t)$ is affinely parametrized then



    $f(t) = 0, tag{12}$



    and we in fact have



    $langle dot gamma, X rangle(t) = langle dot gamma, X rangle (t_0), ; forall t in Bbb R; tag{13}$



    now taking $t_0 = 0$ yields



    $langle dot gamma(t), X(t) rangle = 0 Longleftrightarrow langle dot gamma(0), X(0) rangle = 0; tag{14}$



    if we now choose $X(t)$ such that



    $langle dot gamma(t), X(t) rangle ne 0, tag{15}$



    (14) together with (0) shows the parallel transport of $X(t)$ along $gamma(t)$ cannot satisfy



    $X(0) in T_{gamma(0)} N, tag{15}$



    since it has a component in the direction $dot gamma(0)$.






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      3 Answers
      3






      active

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      3 Answers
      3






      active

      oldest

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      active

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      active

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      4












      $begingroup$

      Consider the plane ${(x, y, 0)mid x,y in Bbb R}subsetmathbb{R}^3$, the line $tmapsto (0, 0, t)$ and the vector field $tmapsto(1, 1, t)$ along it. Then the parallel transport of no vector in the field along the curve will be tangent to the plane.






      share|cite|improve this answer











      $endgroup$


















        4












        $begingroup$

        Consider the plane ${(x, y, 0)mid x,y in Bbb R}subsetmathbb{R}^3$, the line $tmapsto (0, 0, t)$ and the vector field $tmapsto(1, 1, t)$ along it. Then the parallel transport of no vector in the field along the curve will be tangent to the plane.






        share|cite|improve this answer











        $endgroup$
















          4












          4








          4





          $begingroup$

          Consider the plane ${(x, y, 0)mid x,y in Bbb R}subsetmathbb{R}^3$, the line $tmapsto (0, 0, t)$ and the vector field $tmapsto(1, 1, t)$ along it. Then the parallel transport of no vector in the field along the curve will be tangent to the plane.






          share|cite|improve this answer











          $endgroup$



          Consider the plane ${(x, y, 0)mid x,y in Bbb R}subsetmathbb{R}^3$, the line $tmapsto (0, 0, t)$ and the vector field $tmapsto(1, 1, t)$ along it. Then the parallel transport of no vector in the field along the curve will be tangent to the plane.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Feb 2 at 20:40









          Ivo Terek

          46.7k954146




          46.7k954146










          answered Feb 2 at 20:37









          IanIan

          914




          914























              3












              $begingroup$

              Negative answer. Counter-example: $(M,g) = (Bbb R^3, langlecdot, cdot rangle)$, $N = Bbb R^2times {0}$, $gamma(t) = (0,0,t)$ and $X(t) = (0,0,sin t)$. Then $X(0) = 0 in T_{gamma(0)}N$, but the parallel transport of $X(pi/2) = (0,0,1)$ from $gamma(pi/2) = (0,0,pi/2)$ to $gamma(0) =0$ via $gamma$ is $(0,0,1) notin T_{gamma(0)}N$. The values $X(t)$ for $tneq t_0$ should not influence the result of transporting $X(t_0)$, since a priori one transports vectors and not fields.






              share|cite|improve this answer









              $endgroup$









              • 1




                $begingroup$
                I really appreciate your comment about transporting vectors and not fields, this was very clarifying for some stuff I am working on, never had thought about it.
                $endgroup$
                – L.F. Cavenaghi
                Feb 3 at 3:36
















              3












              $begingroup$

              Negative answer. Counter-example: $(M,g) = (Bbb R^3, langlecdot, cdot rangle)$, $N = Bbb R^2times {0}$, $gamma(t) = (0,0,t)$ and $X(t) = (0,0,sin t)$. Then $X(0) = 0 in T_{gamma(0)}N$, but the parallel transport of $X(pi/2) = (0,0,1)$ from $gamma(pi/2) = (0,0,pi/2)$ to $gamma(0) =0$ via $gamma$ is $(0,0,1) notin T_{gamma(0)}N$. The values $X(t)$ for $tneq t_0$ should not influence the result of transporting $X(t_0)$, since a priori one transports vectors and not fields.






              share|cite|improve this answer









              $endgroup$









              • 1




                $begingroup$
                I really appreciate your comment about transporting vectors and not fields, this was very clarifying for some stuff I am working on, never had thought about it.
                $endgroup$
                – L.F. Cavenaghi
                Feb 3 at 3:36














              3












              3








              3





              $begingroup$

              Negative answer. Counter-example: $(M,g) = (Bbb R^3, langlecdot, cdot rangle)$, $N = Bbb R^2times {0}$, $gamma(t) = (0,0,t)$ and $X(t) = (0,0,sin t)$. Then $X(0) = 0 in T_{gamma(0)}N$, but the parallel transport of $X(pi/2) = (0,0,1)$ from $gamma(pi/2) = (0,0,pi/2)$ to $gamma(0) =0$ via $gamma$ is $(0,0,1) notin T_{gamma(0)}N$. The values $X(t)$ for $tneq t_0$ should not influence the result of transporting $X(t_0)$, since a priori one transports vectors and not fields.






              share|cite|improve this answer









              $endgroup$



              Negative answer. Counter-example: $(M,g) = (Bbb R^3, langlecdot, cdot rangle)$, $N = Bbb R^2times {0}$, $gamma(t) = (0,0,t)$ and $X(t) = (0,0,sin t)$. Then $X(0) = 0 in T_{gamma(0)}N$, but the parallel transport of $X(pi/2) = (0,0,1)$ from $gamma(pi/2) = (0,0,pi/2)$ to $gamma(0) =0$ via $gamma$ is $(0,0,1) notin T_{gamma(0)}N$. The values $X(t)$ for $tneq t_0$ should not influence the result of transporting $X(t_0)$, since a priori one transports vectors and not fields.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Feb 2 at 20:39









              Ivo TerekIvo Terek

              46.7k954146




              46.7k954146








              • 1




                $begingroup$
                I really appreciate your comment about transporting vectors and not fields, this was very clarifying for some stuff I am working on, never had thought about it.
                $endgroup$
                – L.F. Cavenaghi
                Feb 3 at 3:36














              • 1




                $begingroup$
                I really appreciate your comment about transporting vectors and not fields, this was very clarifying for some stuff I am working on, never had thought about it.
                $endgroup$
                – L.F. Cavenaghi
                Feb 3 at 3:36








              1




              1




              $begingroup$
              I really appreciate your comment about transporting vectors and not fields, this was very clarifying for some stuff I am working on, never had thought about it.
              $endgroup$
              – L.F. Cavenaghi
              Feb 3 at 3:36




              $begingroup$
              I really appreciate your comment about transporting vectors and not fields, this was very clarifying for some stuff I am working on, never had thought about it.
              $endgroup$
              – L.F. Cavenaghi
              Feb 3 at 3:36











              2












              $begingroup$

              We of course understand that $dot gamma(0) ne 0$, lest $gamma(t)$ be a rather trivial geodesic.



              We observe that, since $dot gamma(0)$ is "perpendicular" to $N$, the condition



              $langle dot gamma(0), X(0) rangle = 0 tag 0$



              is necessary, but not sufficient, for



              $X(0) in T_{gamma(0)}N, tag 1$



              for (0) says that $X(0)$ has no component along $dot gamma(0)$; in the event that $N$ is a hypersurface in $M$, that is,



              $dim N = dim M - 1, tag 2$



              (0) is in fact also sufficient, for then $dot gamma(0)$ and $T_{gamma(0)}N$ span $T_{gamma}(0)N$; in fact, (2) implies



              $T_{gamma(0)}M = langle dot gamma(0) rangle oplus T_{gamma(0)}N, tag 3$



              where $langle dot gamma(0) rangle$ is the one-dimensional subspace of $T_{gamma(0)}M$ generated by $dot gamma(0)$.



              Now consider any vector field $X(t)$ along $gamma(t)$; we have



              $nabla_{dot gamma} langle dot gamma, X rangle = langle nabla_{dot gamma} dot gamma, X rangle + langle dot gamma, nabla_{dot gamma} X rangle; tag 4$



              if $X(t)$ is parallel along $gamma(t)$, then



              $nabla_{dot gamma}X = 0; tag 5$



              thus (4) becomes



              $nabla_{dot gamma} langle dot gamma, X rangle = langle nabla_{dot gamma} dot gamma, X rangle; tag 6$



              now the general formulation of the geodesic equation as I understand it is



              $nabla_{dot gamma} dot gamma = f(t) dot gamma; tag 7$



              here $f(t)$ is determined by the choice of the running parameter $t$ of $gamma(t)$; if we substitute this into (6), we obtain



              $nabla_{dot gamma} langle dot gamma, X rangle = f(t) langle dot gamma, X rangle; tag 8$



              the solution to this linear, first order differential equation for $langle dot gamma, X rangle$ with $langle dot gamma(t_0), X(t_0) rangle = langle dot gamma, X rangle(t_0)$ is



              $langle dot gamma, X rangle(t) = exp left (displaystyle int_{t_0}^t f(s) ; ds right ) langle dot gamma, X rangle (t_0); tag 9$



              since



              $forall t in Bbb R, ;exp left (displaystyle int_{t_0}^t f(s) ; ds right ) ne 0, tag{10}$



              (9) shows that



              $langle dot gamma(t), X(t) rangle = langle dot gamma, X rangle(t) = 0 Longleftrightarrow langle dot gamma, X rangle (t_0) = langle dot gamma(t_0), X(t_0) rangle = 0; tag{11}$



              of course, if $gamma(t)$ is affinely parametrized then



              $f(t) = 0, tag{12}$



              and we in fact have



              $langle dot gamma, X rangle(t) = langle dot gamma, X rangle (t_0), ; forall t in Bbb R; tag{13}$



              now taking $t_0 = 0$ yields



              $langle dot gamma(t), X(t) rangle = 0 Longleftrightarrow langle dot gamma(0), X(0) rangle = 0; tag{14}$



              if we now choose $X(t)$ such that



              $langle dot gamma(t), X(t) rangle ne 0, tag{15}$



              (14) together with (0) shows the parallel transport of $X(t)$ along $gamma(t)$ cannot satisfy



              $X(0) in T_{gamma(0)} N, tag{15}$



              since it has a component in the direction $dot gamma(0)$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                We of course understand that $dot gamma(0) ne 0$, lest $gamma(t)$ be a rather trivial geodesic.



                We observe that, since $dot gamma(0)$ is "perpendicular" to $N$, the condition



                $langle dot gamma(0), X(0) rangle = 0 tag 0$



                is necessary, but not sufficient, for



                $X(0) in T_{gamma(0)}N, tag 1$



                for (0) says that $X(0)$ has no component along $dot gamma(0)$; in the event that $N$ is a hypersurface in $M$, that is,



                $dim N = dim M - 1, tag 2$



                (0) is in fact also sufficient, for then $dot gamma(0)$ and $T_{gamma(0)}N$ span $T_{gamma}(0)N$; in fact, (2) implies



                $T_{gamma(0)}M = langle dot gamma(0) rangle oplus T_{gamma(0)}N, tag 3$



                where $langle dot gamma(0) rangle$ is the one-dimensional subspace of $T_{gamma(0)}M$ generated by $dot gamma(0)$.



                Now consider any vector field $X(t)$ along $gamma(t)$; we have



                $nabla_{dot gamma} langle dot gamma, X rangle = langle nabla_{dot gamma} dot gamma, X rangle + langle dot gamma, nabla_{dot gamma} X rangle; tag 4$



                if $X(t)$ is parallel along $gamma(t)$, then



                $nabla_{dot gamma}X = 0; tag 5$



                thus (4) becomes



                $nabla_{dot gamma} langle dot gamma, X rangle = langle nabla_{dot gamma} dot gamma, X rangle; tag 6$



                now the general formulation of the geodesic equation as I understand it is



                $nabla_{dot gamma} dot gamma = f(t) dot gamma; tag 7$



                here $f(t)$ is determined by the choice of the running parameter $t$ of $gamma(t)$; if we substitute this into (6), we obtain



                $nabla_{dot gamma} langle dot gamma, X rangle = f(t) langle dot gamma, X rangle; tag 8$



                the solution to this linear, first order differential equation for $langle dot gamma, X rangle$ with $langle dot gamma(t_0), X(t_0) rangle = langle dot gamma, X rangle(t_0)$ is



                $langle dot gamma, X rangle(t) = exp left (displaystyle int_{t_0}^t f(s) ; ds right ) langle dot gamma, X rangle (t_0); tag 9$



                since



                $forall t in Bbb R, ;exp left (displaystyle int_{t_0}^t f(s) ; ds right ) ne 0, tag{10}$



                (9) shows that



                $langle dot gamma(t), X(t) rangle = langle dot gamma, X rangle(t) = 0 Longleftrightarrow langle dot gamma, X rangle (t_0) = langle dot gamma(t_0), X(t_0) rangle = 0; tag{11}$



                of course, if $gamma(t)$ is affinely parametrized then



                $f(t) = 0, tag{12}$



                and we in fact have



                $langle dot gamma, X rangle(t) = langle dot gamma, X rangle (t_0), ; forall t in Bbb R; tag{13}$



                now taking $t_0 = 0$ yields



                $langle dot gamma(t), X(t) rangle = 0 Longleftrightarrow langle dot gamma(0), X(0) rangle = 0; tag{14}$



                if we now choose $X(t)$ such that



                $langle dot gamma(t), X(t) rangle ne 0, tag{15}$



                (14) together with (0) shows the parallel transport of $X(t)$ along $gamma(t)$ cannot satisfy



                $X(0) in T_{gamma(0)} N, tag{15}$



                since it has a component in the direction $dot gamma(0)$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  We of course understand that $dot gamma(0) ne 0$, lest $gamma(t)$ be a rather trivial geodesic.



                  We observe that, since $dot gamma(0)$ is "perpendicular" to $N$, the condition



                  $langle dot gamma(0), X(0) rangle = 0 tag 0$



                  is necessary, but not sufficient, for



                  $X(0) in T_{gamma(0)}N, tag 1$



                  for (0) says that $X(0)$ has no component along $dot gamma(0)$; in the event that $N$ is a hypersurface in $M$, that is,



                  $dim N = dim M - 1, tag 2$



                  (0) is in fact also sufficient, for then $dot gamma(0)$ and $T_{gamma(0)}N$ span $T_{gamma}(0)N$; in fact, (2) implies



                  $T_{gamma(0)}M = langle dot gamma(0) rangle oplus T_{gamma(0)}N, tag 3$



                  where $langle dot gamma(0) rangle$ is the one-dimensional subspace of $T_{gamma(0)}M$ generated by $dot gamma(0)$.



                  Now consider any vector field $X(t)$ along $gamma(t)$; we have



                  $nabla_{dot gamma} langle dot gamma, X rangle = langle nabla_{dot gamma} dot gamma, X rangle + langle dot gamma, nabla_{dot gamma} X rangle; tag 4$



                  if $X(t)$ is parallel along $gamma(t)$, then



                  $nabla_{dot gamma}X = 0; tag 5$



                  thus (4) becomes



                  $nabla_{dot gamma} langle dot gamma, X rangle = langle nabla_{dot gamma} dot gamma, X rangle; tag 6$



                  now the general formulation of the geodesic equation as I understand it is



                  $nabla_{dot gamma} dot gamma = f(t) dot gamma; tag 7$



                  here $f(t)$ is determined by the choice of the running parameter $t$ of $gamma(t)$; if we substitute this into (6), we obtain



                  $nabla_{dot gamma} langle dot gamma, X rangle = f(t) langle dot gamma, X rangle; tag 8$



                  the solution to this linear, first order differential equation for $langle dot gamma, X rangle$ with $langle dot gamma(t_0), X(t_0) rangle = langle dot gamma, X rangle(t_0)$ is



                  $langle dot gamma, X rangle(t) = exp left (displaystyle int_{t_0}^t f(s) ; ds right ) langle dot gamma, X rangle (t_0); tag 9$



                  since



                  $forall t in Bbb R, ;exp left (displaystyle int_{t_0}^t f(s) ; ds right ) ne 0, tag{10}$



                  (9) shows that



                  $langle dot gamma(t), X(t) rangle = langle dot gamma, X rangle(t) = 0 Longleftrightarrow langle dot gamma, X rangle (t_0) = langle dot gamma(t_0), X(t_0) rangle = 0; tag{11}$



                  of course, if $gamma(t)$ is affinely parametrized then



                  $f(t) = 0, tag{12}$



                  and we in fact have



                  $langle dot gamma, X rangle(t) = langle dot gamma, X rangle (t_0), ; forall t in Bbb R; tag{13}$



                  now taking $t_0 = 0$ yields



                  $langle dot gamma(t), X(t) rangle = 0 Longleftrightarrow langle dot gamma(0), X(0) rangle = 0; tag{14}$



                  if we now choose $X(t)$ such that



                  $langle dot gamma(t), X(t) rangle ne 0, tag{15}$



                  (14) together with (0) shows the parallel transport of $X(t)$ along $gamma(t)$ cannot satisfy



                  $X(0) in T_{gamma(0)} N, tag{15}$



                  since it has a component in the direction $dot gamma(0)$.






                  share|cite|improve this answer









                  $endgroup$



                  We of course understand that $dot gamma(0) ne 0$, lest $gamma(t)$ be a rather trivial geodesic.



                  We observe that, since $dot gamma(0)$ is "perpendicular" to $N$, the condition



                  $langle dot gamma(0), X(0) rangle = 0 tag 0$



                  is necessary, but not sufficient, for



                  $X(0) in T_{gamma(0)}N, tag 1$



                  for (0) says that $X(0)$ has no component along $dot gamma(0)$; in the event that $N$ is a hypersurface in $M$, that is,



                  $dim N = dim M - 1, tag 2$



                  (0) is in fact also sufficient, for then $dot gamma(0)$ and $T_{gamma(0)}N$ span $T_{gamma}(0)N$; in fact, (2) implies



                  $T_{gamma(0)}M = langle dot gamma(0) rangle oplus T_{gamma(0)}N, tag 3$



                  where $langle dot gamma(0) rangle$ is the one-dimensional subspace of $T_{gamma(0)}M$ generated by $dot gamma(0)$.



                  Now consider any vector field $X(t)$ along $gamma(t)$; we have



                  $nabla_{dot gamma} langle dot gamma, X rangle = langle nabla_{dot gamma} dot gamma, X rangle + langle dot gamma, nabla_{dot gamma} X rangle; tag 4$



                  if $X(t)$ is parallel along $gamma(t)$, then



                  $nabla_{dot gamma}X = 0; tag 5$



                  thus (4) becomes



                  $nabla_{dot gamma} langle dot gamma, X rangle = langle nabla_{dot gamma} dot gamma, X rangle; tag 6$



                  now the general formulation of the geodesic equation as I understand it is



                  $nabla_{dot gamma} dot gamma = f(t) dot gamma; tag 7$



                  here $f(t)$ is determined by the choice of the running parameter $t$ of $gamma(t)$; if we substitute this into (6), we obtain



                  $nabla_{dot gamma} langle dot gamma, X rangle = f(t) langle dot gamma, X rangle; tag 8$



                  the solution to this linear, first order differential equation for $langle dot gamma, X rangle$ with $langle dot gamma(t_0), X(t_0) rangle = langle dot gamma, X rangle(t_0)$ is



                  $langle dot gamma, X rangle(t) = exp left (displaystyle int_{t_0}^t f(s) ; ds right ) langle dot gamma, X rangle (t_0); tag 9$



                  since



                  $forall t in Bbb R, ;exp left (displaystyle int_{t_0}^t f(s) ; ds right ) ne 0, tag{10}$



                  (9) shows that



                  $langle dot gamma(t), X(t) rangle = langle dot gamma, X rangle(t) = 0 Longleftrightarrow langle dot gamma, X rangle (t_0) = langle dot gamma(t_0), X(t_0) rangle = 0; tag{11}$



                  of course, if $gamma(t)$ is affinely parametrized then



                  $f(t) = 0, tag{12}$



                  and we in fact have



                  $langle dot gamma, X rangle(t) = langle dot gamma, X rangle (t_0), ; forall t in Bbb R; tag{13}$



                  now taking $t_0 = 0$ yields



                  $langle dot gamma(t), X(t) rangle = 0 Longleftrightarrow langle dot gamma(0), X(0) rangle = 0; tag{14}$



                  if we now choose $X(t)$ such that



                  $langle dot gamma(t), X(t) rangle ne 0, tag{15}$



                  (14) together with (0) shows the parallel transport of $X(t)$ along $gamma(t)$ cannot satisfy



                  $X(0) in T_{gamma(0)} N, tag{15}$



                  since it has a component in the direction $dot gamma(0)$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Feb 3 at 6:26









                  Robert LewisRobert Lewis

                  49k23168




                  49k23168






























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