Why do manifolds with negative sectional curvature not have conjugate points?












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$begingroup$


I'm trying to understand why manifolds with negative sectional curvature not have conjugate points. In fact for me it is sufficient to understand it for surfaces, but of course i'd be interested in the general case.



many regards,
and thanks in advance,
Leo










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$endgroup$

















    2












    $begingroup$


    I'm trying to understand why manifolds with negative sectional curvature not have conjugate points. In fact for me it is sufficient to understand it for surfaces, but of course i'd be interested in the general case.



    many regards,
    and thanks in advance,
    Leo










    share|cite|improve this question









    $endgroup$















      2












      2








      2


      3



      $begingroup$


      I'm trying to understand why manifolds with negative sectional curvature not have conjugate points. In fact for me it is sufficient to understand it for surfaces, but of course i'd be interested in the general case.



      many regards,
      and thanks in advance,
      Leo










      share|cite|improve this question









      $endgroup$




      I'm trying to understand why manifolds with negative sectional curvature not have conjugate points. In fact for me it is sufficient to understand it for surfaces, but of course i'd be interested in the general case.



      many regards,
      and thanks in advance,
      Leo







      differential-geometry dynamical-systems






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Apr 24 '13 at 18:28









      MarloMarlo

      455414




      455414






















          2 Answers
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          5












          $begingroup$

          Let $gamma:[0,epsilon]rightarrow M$ be a geodesic curve, $epsilon>0$ and $J$ be a Jacobi vector field along $gamma$ with $J(0)=J(epsilon)=0$.



          I remind you that :




          1. $J$ being a Jacobi vector field means that $dfrac{D^2}{dt^2}J+R(gamma',J)gamma'=0$ with the usual notation for the covariant derivative $D$ and $R$ the Riemannian curvature tensor.

          2. $left[|X|^2|Y|^2-langle X,Yrangle^2right] kappa(X,Y)=langle R(X,Y)X,Yrangle$ for any $X,Yin TM$ with $kappa$ the sectional curvature.


          Now look at the map $phi:tmapsto |J(t)|^2$. Its second derivative is easy to compute and using the two remarks you get :
          $$begin{align} phi''(t) & = langle dfrac{D^2}{dt^2}J,Jrangle + langle dfrac{D}{dt}J,dfrac{D}{dt}Jrangle\ & = -langle R(gamma',J)gamma',Jrangle + |dfrac{D}{dt}J|^2 \ & = -left[|X|^2|Y|^2-langle X,Yrangle^2right]kappa(gamma',J) + |dfrac{D}{dt}J|^2.end{align}$$



          In particular, if all sectional curvatures are non-positive then $phi''geq 0$. So $phi$ is a convex map and since $phi(0)=phi(epsilon)=0$, $phi(t)=0$ for any $tin [0,epsilon]$. It follows that $J$ has to be trivial.



          So we can't have conjugate points in $M$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            thanks a lot Bebop for the very nice answer. So the argument goes: 1. the Jacobi field has $frac{d^2}{dt^2}||J(t)||^2>0$, i.e. the length squared is a convex function on $[0,epsilon]$ 2. $||J(0)||^2=||J(epsilon)||^2$ implies that J is trivial. But why do we require $J(0)=J(epsilon)$? Is that the condition for conjugate points?
            $endgroup$
            – Marlo
            Apr 24 '13 at 23:04












          • $begingroup$
            I forgot to write that $J(0)=J(epsilon)=0$, otherwise the conclusion would be that $phi$ is constant. But don't you know the definition of conjugate points ? This is all about your question ? Go on wikipedia to get the full definition.
            $endgroup$
            – Bebop
            Apr 24 '13 at 23:19



















          1












          $begingroup$

          This is very late, but to provide an alternative proof:



          The setup for the Rauch Comparison Theorem is very wordy, but the claim follows immediately from comparison with $mathbb{R}^m$.



          Assume that $M$ is a Riemannian manifold with nonpositive sectional curvature, that $pin M$, and that $gamma:left[0,Tright]to M$ is any unit speed geodesic originating at $p$. Let $m=dim(M)$. Take any Jacobian vector field $J$ along $gamma$ with $J(0)=0$ and $||J'(0)||>0$.



          Let $tildegamma(t)=(t,0,0,...,0)$ be a unit-speed geodesic in $mathbb{R}^m$. Let $tilde{J}$ be the Jacobian vector field along $tildegamma$ satisfying $tilde{J}(0)=0$, $tilde{J}'(0)=(<gamma'(0),J'(0)>,(||J'(0)||^2-<gamma'(0),J'(0)>^2)^frac{1}{2},0,0,...,0)$. Then $J(0)=0=tilde{J}(0)$, $||tilde{J}'(0)||=||J'(0)||$, and $<tilde{gamma}'(0),tilde{J}'(0)>=<gamma(0),J'(0)>$.



          For all $tinleft[0,Tright]$, $max(K(gamma'(t),X))leq0=min(tilde{K}(tilde{gamma}'(t),tilde{X}))$ and $dim(mathbb{R}^m)leq dim(M)$ so, by the Rauch Comparison Theorem, $||J(t)||geq||tilde{J}(t)||$ for all $tinleft[0,Tright]$. In particular, since $mathbb{R}^m$ has no conjugate points, $tilde{J}'(0)not=0$ guarantees that $||tilde{J}||>0$ for all positive $t$, hence $||J(t)||>0$ for all positive $t$, thus $p$ has no conjugate points along $gamma$.



          Since $p$ and $gamma$ were arbitrary, $M$ has no conjugate points.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks @Pepper! That's also a nice way of showing it :)
            $endgroup$
            – Marlo
            Dec 25 '18 at 17:59











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          2 Answers
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          2 Answers
          2






          active

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          active

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          active

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          5












          $begingroup$

          Let $gamma:[0,epsilon]rightarrow M$ be a geodesic curve, $epsilon>0$ and $J$ be a Jacobi vector field along $gamma$ with $J(0)=J(epsilon)=0$.



          I remind you that :




          1. $J$ being a Jacobi vector field means that $dfrac{D^2}{dt^2}J+R(gamma',J)gamma'=0$ with the usual notation for the covariant derivative $D$ and $R$ the Riemannian curvature tensor.

          2. $left[|X|^2|Y|^2-langle X,Yrangle^2right] kappa(X,Y)=langle R(X,Y)X,Yrangle$ for any $X,Yin TM$ with $kappa$ the sectional curvature.


          Now look at the map $phi:tmapsto |J(t)|^2$. Its second derivative is easy to compute and using the two remarks you get :
          $$begin{align} phi''(t) & = langle dfrac{D^2}{dt^2}J,Jrangle + langle dfrac{D}{dt}J,dfrac{D}{dt}Jrangle\ & = -langle R(gamma',J)gamma',Jrangle + |dfrac{D}{dt}J|^2 \ & = -left[|X|^2|Y|^2-langle X,Yrangle^2right]kappa(gamma',J) + |dfrac{D}{dt}J|^2.end{align}$$



          In particular, if all sectional curvatures are non-positive then $phi''geq 0$. So $phi$ is a convex map and since $phi(0)=phi(epsilon)=0$, $phi(t)=0$ for any $tin [0,epsilon]$. It follows that $J$ has to be trivial.



          So we can't have conjugate points in $M$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            thanks a lot Bebop for the very nice answer. So the argument goes: 1. the Jacobi field has $frac{d^2}{dt^2}||J(t)||^2>0$, i.e. the length squared is a convex function on $[0,epsilon]$ 2. $||J(0)||^2=||J(epsilon)||^2$ implies that J is trivial. But why do we require $J(0)=J(epsilon)$? Is that the condition for conjugate points?
            $endgroup$
            – Marlo
            Apr 24 '13 at 23:04












          • $begingroup$
            I forgot to write that $J(0)=J(epsilon)=0$, otherwise the conclusion would be that $phi$ is constant. But don't you know the definition of conjugate points ? This is all about your question ? Go on wikipedia to get the full definition.
            $endgroup$
            – Bebop
            Apr 24 '13 at 23:19
















          5












          $begingroup$

          Let $gamma:[0,epsilon]rightarrow M$ be a geodesic curve, $epsilon>0$ and $J$ be a Jacobi vector field along $gamma$ with $J(0)=J(epsilon)=0$.



          I remind you that :




          1. $J$ being a Jacobi vector field means that $dfrac{D^2}{dt^2}J+R(gamma',J)gamma'=0$ with the usual notation for the covariant derivative $D$ and $R$ the Riemannian curvature tensor.

          2. $left[|X|^2|Y|^2-langle X,Yrangle^2right] kappa(X,Y)=langle R(X,Y)X,Yrangle$ for any $X,Yin TM$ with $kappa$ the sectional curvature.


          Now look at the map $phi:tmapsto |J(t)|^2$. Its second derivative is easy to compute and using the two remarks you get :
          $$begin{align} phi''(t) & = langle dfrac{D^2}{dt^2}J,Jrangle + langle dfrac{D}{dt}J,dfrac{D}{dt}Jrangle\ & = -langle R(gamma',J)gamma',Jrangle + |dfrac{D}{dt}J|^2 \ & = -left[|X|^2|Y|^2-langle X,Yrangle^2right]kappa(gamma',J) + |dfrac{D}{dt}J|^2.end{align}$$



          In particular, if all sectional curvatures are non-positive then $phi''geq 0$. So $phi$ is a convex map and since $phi(0)=phi(epsilon)=0$, $phi(t)=0$ for any $tin [0,epsilon]$. It follows that $J$ has to be trivial.



          So we can't have conjugate points in $M$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            thanks a lot Bebop for the very nice answer. So the argument goes: 1. the Jacobi field has $frac{d^2}{dt^2}||J(t)||^2>0$, i.e. the length squared is a convex function on $[0,epsilon]$ 2. $||J(0)||^2=||J(epsilon)||^2$ implies that J is trivial. But why do we require $J(0)=J(epsilon)$? Is that the condition for conjugate points?
            $endgroup$
            – Marlo
            Apr 24 '13 at 23:04












          • $begingroup$
            I forgot to write that $J(0)=J(epsilon)=0$, otherwise the conclusion would be that $phi$ is constant. But don't you know the definition of conjugate points ? This is all about your question ? Go on wikipedia to get the full definition.
            $endgroup$
            – Bebop
            Apr 24 '13 at 23:19














          5












          5








          5





          $begingroup$

          Let $gamma:[0,epsilon]rightarrow M$ be a geodesic curve, $epsilon>0$ and $J$ be a Jacobi vector field along $gamma$ with $J(0)=J(epsilon)=0$.



          I remind you that :




          1. $J$ being a Jacobi vector field means that $dfrac{D^2}{dt^2}J+R(gamma',J)gamma'=0$ with the usual notation for the covariant derivative $D$ and $R$ the Riemannian curvature tensor.

          2. $left[|X|^2|Y|^2-langle X,Yrangle^2right] kappa(X,Y)=langle R(X,Y)X,Yrangle$ for any $X,Yin TM$ with $kappa$ the sectional curvature.


          Now look at the map $phi:tmapsto |J(t)|^2$. Its second derivative is easy to compute and using the two remarks you get :
          $$begin{align} phi''(t) & = langle dfrac{D^2}{dt^2}J,Jrangle + langle dfrac{D}{dt}J,dfrac{D}{dt}Jrangle\ & = -langle R(gamma',J)gamma',Jrangle + |dfrac{D}{dt}J|^2 \ & = -left[|X|^2|Y|^2-langle X,Yrangle^2right]kappa(gamma',J) + |dfrac{D}{dt}J|^2.end{align}$$



          In particular, if all sectional curvatures are non-positive then $phi''geq 0$. So $phi$ is a convex map and since $phi(0)=phi(epsilon)=0$, $phi(t)=0$ for any $tin [0,epsilon]$. It follows that $J$ has to be trivial.



          So we can't have conjugate points in $M$.






          share|cite|improve this answer











          $endgroup$



          Let $gamma:[0,epsilon]rightarrow M$ be a geodesic curve, $epsilon>0$ and $J$ be a Jacobi vector field along $gamma$ with $J(0)=J(epsilon)=0$.



          I remind you that :




          1. $J$ being a Jacobi vector field means that $dfrac{D^2}{dt^2}J+R(gamma',J)gamma'=0$ with the usual notation for the covariant derivative $D$ and $R$ the Riemannian curvature tensor.

          2. $left[|X|^2|Y|^2-langle X,Yrangle^2right] kappa(X,Y)=langle R(X,Y)X,Yrangle$ for any $X,Yin TM$ with $kappa$ the sectional curvature.


          Now look at the map $phi:tmapsto |J(t)|^2$. Its second derivative is easy to compute and using the two remarks you get :
          $$begin{align} phi''(t) & = langle dfrac{D^2}{dt^2}J,Jrangle + langle dfrac{D}{dt}J,dfrac{D}{dt}Jrangle\ & = -langle R(gamma',J)gamma',Jrangle + |dfrac{D}{dt}J|^2 \ & = -left[|X|^2|Y|^2-langle X,Yrangle^2right]kappa(gamma',J) + |dfrac{D}{dt}J|^2.end{align}$$



          In particular, if all sectional curvatures are non-positive then $phi''geq 0$. So $phi$ is a convex map and since $phi(0)=phi(epsilon)=0$, $phi(t)=0$ for any $tin [0,epsilon]$. It follows that $J$ has to be trivial.



          So we can't have conjugate points in $M$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Apr 24 '13 at 23:20

























          answered Apr 24 '13 at 22:45









          BebopBebop

          2,805714




          2,805714












          • $begingroup$
            thanks a lot Bebop for the very nice answer. So the argument goes: 1. the Jacobi field has $frac{d^2}{dt^2}||J(t)||^2>0$, i.e. the length squared is a convex function on $[0,epsilon]$ 2. $||J(0)||^2=||J(epsilon)||^2$ implies that J is trivial. But why do we require $J(0)=J(epsilon)$? Is that the condition for conjugate points?
            $endgroup$
            – Marlo
            Apr 24 '13 at 23:04












          • $begingroup$
            I forgot to write that $J(0)=J(epsilon)=0$, otherwise the conclusion would be that $phi$ is constant. But don't you know the definition of conjugate points ? This is all about your question ? Go on wikipedia to get the full definition.
            $endgroup$
            – Bebop
            Apr 24 '13 at 23:19


















          • $begingroup$
            thanks a lot Bebop for the very nice answer. So the argument goes: 1. the Jacobi field has $frac{d^2}{dt^2}||J(t)||^2>0$, i.e. the length squared is a convex function on $[0,epsilon]$ 2. $||J(0)||^2=||J(epsilon)||^2$ implies that J is trivial. But why do we require $J(0)=J(epsilon)$? Is that the condition for conjugate points?
            $endgroup$
            – Marlo
            Apr 24 '13 at 23:04












          • $begingroup$
            I forgot to write that $J(0)=J(epsilon)=0$, otherwise the conclusion would be that $phi$ is constant. But don't you know the definition of conjugate points ? This is all about your question ? Go on wikipedia to get the full definition.
            $endgroup$
            – Bebop
            Apr 24 '13 at 23:19
















          $begingroup$
          thanks a lot Bebop for the very nice answer. So the argument goes: 1. the Jacobi field has $frac{d^2}{dt^2}||J(t)||^2>0$, i.e. the length squared is a convex function on $[0,epsilon]$ 2. $||J(0)||^2=||J(epsilon)||^2$ implies that J is trivial. But why do we require $J(0)=J(epsilon)$? Is that the condition for conjugate points?
          $endgroup$
          – Marlo
          Apr 24 '13 at 23:04






          $begingroup$
          thanks a lot Bebop for the very nice answer. So the argument goes: 1. the Jacobi field has $frac{d^2}{dt^2}||J(t)||^2>0$, i.e. the length squared is a convex function on $[0,epsilon]$ 2. $||J(0)||^2=||J(epsilon)||^2$ implies that J is trivial. But why do we require $J(0)=J(epsilon)$? Is that the condition for conjugate points?
          $endgroup$
          – Marlo
          Apr 24 '13 at 23:04














          $begingroup$
          I forgot to write that $J(0)=J(epsilon)=0$, otherwise the conclusion would be that $phi$ is constant. But don't you know the definition of conjugate points ? This is all about your question ? Go on wikipedia to get the full definition.
          $endgroup$
          – Bebop
          Apr 24 '13 at 23:19




          $begingroup$
          I forgot to write that $J(0)=J(epsilon)=0$, otherwise the conclusion would be that $phi$ is constant. But don't you know the definition of conjugate points ? This is all about your question ? Go on wikipedia to get the full definition.
          $endgroup$
          – Bebop
          Apr 24 '13 at 23:19











          1












          $begingroup$

          This is very late, but to provide an alternative proof:



          The setup for the Rauch Comparison Theorem is very wordy, but the claim follows immediately from comparison with $mathbb{R}^m$.



          Assume that $M$ is a Riemannian manifold with nonpositive sectional curvature, that $pin M$, and that $gamma:left[0,Tright]to M$ is any unit speed geodesic originating at $p$. Let $m=dim(M)$. Take any Jacobian vector field $J$ along $gamma$ with $J(0)=0$ and $||J'(0)||>0$.



          Let $tildegamma(t)=(t,0,0,...,0)$ be a unit-speed geodesic in $mathbb{R}^m$. Let $tilde{J}$ be the Jacobian vector field along $tildegamma$ satisfying $tilde{J}(0)=0$, $tilde{J}'(0)=(<gamma'(0),J'(0)>,(||J'(0)||^2-<gamma'(0),J'(0)>^2)^frac{1}{2},0,0,...,0)$. Then $J(0)=0=tilde{J}(0)$, $||tilde{J}'(0)||=||J'(0)||$, and $<tilde{gamma}'(0),tilde{J}'(0)>=<gamma(0),J'(0)>$.



          For all $tinleft[0,Tright]$, $max(K(gamma'(t),X))leq0=min(tilde{K}(tilde{gamma}'(t),tilde{X}))$ and $dim(mathbb{R}^m)leq dim(M)$ so, by the Rauch Comparison Theorem, $||J(t)||geq||tilde{J}(t)||$ for all $tinleft[0,Tright]$. In particular, since $mathbb{R}^m$ has no conjugate points, $tilde{J}'(0)not=0$ guarantees that $||tilde{J}||>0$ for all positive $t$, hence $||J(t)||>0$ for all positive $t$, thus $p$ has no conjugate points along $gamma$.



          Since $p$ and $gamma$ were arbitrary, $M$ has no conjugate points.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks @Pepper! That's also a nice way of showing it :)
            $endgroup$
            – Marlo
            Dec 25 '18 at 17:59
















          1












          $begingroup$

          This is very late, but to provide an alternative proof:



          The setup for the Rauch Comparison Theorem is very wordy, but the claim follows immediately from comparison with $mathbb{R}^m$.



          Assume that $M$ is a Riemannian manifold with nonpositive sectional curvature, that $pin M$, and that $gamma:left[0,Tright]to M$ is any unit speed geodesic originating at $p$. Let $m=dim(M)$. Take any Jacobian vector field $J$ along $gamma$ with $J(0)=0$ and $||J'(0)||>0$.



          Let $tildegamma(t)=(t,0,0,...,0)$ be a unit-speed geodesic in $mathbb{R}^m$. Let $tilde{J}$ be the Jacobian vector field along $tildegamma$ satisfying $tilde{J}(0)=0$, $tilde{J}'(0)=(<gamma'(0),J'(0)>,(||J'(0)||^2-<gamma'(0),J'(0)>^2)^frac{1}{2},0,0,...,0)$. Then $J(0)=0=tilde{J}(0)$, $||tilde{J}'(0)||=||J'(0)||$, and $<tilde{gamma}'(0),tilde{J}'(0)>=<gamma(0),J'(0)>$.



          For all $tinleft[0,Tright]$, $max(K(gamma'(t),X))leq0=min(tilde{K}(tilde{gamma}'(t),tilde{X}))$ and $dim(mathbb{R}^m)leq dim(M)$ so, by the Rauch Comparison Theorem, $||J(t)||geq||tilde{J}(t)||$ for all $tinleft[0,Tright]$. In particular, since $mathbb{R}^m$ has no conjugate points, $tilde{J}'(0)not=0$ guarantees that $||tilde{J}||>0$ for all positive $t$, hence $||J(t)||>0$ for all positive $t$, thus $p$ has no conjugate points along $gamma$.



          Since $p$ and $gamma$ were arbitrary, $M$ has no conjugate points.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks @Pepper! That's also a nice way of showing it :)
            $endgroup$
            – Marlo
            Dec 25 '18 at 17:59














          1












          1








          1





          $begingroup$

          This is very late, but to provide an alternative proof:



          The setup for the Rauch Comparison Theorem is very wordy, but the claim follows immediately from comparison with $mathbb{R}^m$.



          Assume that $M$ is a Riemannian manifold with nonpositive sectional curvature, that $pin M$, and that $gamma:left[0,Tright]to M$ is any unit speed geodesic originating at $p$. Let $m=dim(M)$. Take any Jacobian vector field $J$ along $gamma$ with $J(0)=0$ and $||J'(0)||>0$.



          Let $tildegamma(t)=(t,0,0,...,0)$ be a unit-speed geodesic in $mathbb{R}^m$. Let $tilde{J}$ be the Jacobian vector field along $tildegamma$ satisfying $tilde{J}(0)=0$, $tilde{J}'(0)=(<gamma'(0),J'(0)>,(||J'(0)||^2-<gamma'(0),J'(0)>^2)^frac{1}{2},0,0,...,0)$. Then $J(0)=0=tilde{J}(0)$, $||tilde{J}'(0)||=||J'(0)||$, and $<tilde{gamma}'(0),tilde{J}'(0)>=<gamma(0),J'(0)>$.



          For all $tinleft[0,Tright]$, $max(K(gamma'(t),X))leq0=min(tilde{K}(tilde{gamma}'(t),tilde{X}))$ and $dim(mathbb{R}^m)leq dim(M)$ so, by the Rauch Comparison Theorem, $||J(t)||geq||tilde{J}(t)||$ for all $tinleft[0,Tright]$. In particular, since $mathbb{R}^m$ has no conjugate points, $tilde{J}'(0)not=0$ guarantees that $||tilde{J}||>0$ for all positive $t$, hence $||J(t)||>0$ for all positive $t$, thus $p$ has no conjugate points along $gamma$.



          Since $p$ and $gamma$ were arbitrary, $M$ has no conjugate points.






          share|cite|improve this answer









          $endgroup$



          This is very late, but to provide an alternative proof:



          The setup for the Rauch Comparison Theorem is very wordy, but the claim follows immediately from comparison with $mathbb{R}^m$.



          Assume that $M$ is a Riemannian manifold with nonpositive sectional curvature, that $pin M$, and that $gamma:left[0,Tright]to M$ is any unit speed geodesic originating at $p$. Let $m=dim(M)$. Take any Jacobian vector field $J$ along $gamma$ with $J(0)=0$ and $||J'(0)||>0$.



          Let $tildegamma(t)=(t,0,0,...,0)$ be a unit-speed geodesic in $mathbb{R}^m$. Let $tilde{J}$ be the Jacobian vector field along $tildegamma$ satisfying $tilde{J}(0)=0$, $tilde{J}'(0)=(<gamma'(0),J'(0)>,(||J'(0)||^2-<gamma'(0),J'(0)>^2)^frac{1}{2},0,0,...,0)$. Then $J(0)=0=tilde{J}(0)$, $||tilde{J}'(0)||=||J'(0)||$, and $<tilde{gamma}'(0),tilde{J}'(0)>=<gamma(0),J'(0)>$.



          For all $tinleft[0,Tright]$, $max(K(gamma'(t),X))leq0=min(tilde{K}(tilde{gamma}'(t),tilde{X}))$ and $dim(mathbb{R}^m)leq dim(M)$ so, by the Rauch Comparison Theorem, $||J(t)||geq||tilde{J}(t)||$ for all $tinleft[0,Tright]$. In particular, since $mathbb{R}^m$ has no conjugate points, $tilde{J}'(0)not=0$ guarantees that $||tilde{J}||>0$ for all positive $t$, hence $||J(t)||>0$ for all positive $t$, thus $p$ has no conjugate points along $gamma$.



          Since $p$ and $gamma$ were arbitrary, $M$ has no conjugate points.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 24 '18 at 16:44









          PepperPepper

          508




          508












          • $begingroup$
            Thanks @Pepper! That's also a nice way of showing it :)
            $endgroup$
            – Marlo
            Dec 25 '18 at 17:59


















          • $begingroup$
            Thanks @Pepper! That's also a nice way of showing it :)
            $endgroup$
            – Marlo
            Dec 25 '18 at 17:59
















          $begingroup$
          Thanks @Pepper! That's also a nice way of showing it :)
          $endgroup$
          – Marlo
          Dec 25 '18 at 17:59




          $begingroup$
          Thanks @Pepper! That's also a nice way of showing it :)
          $endgroup$
          – Marlo
          Dec 25 '18 at 17:59


















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