$M$ a complete Riemannian manifold with nonpositive curvature. Show that $|d(exp_p)_v(w)| geq |w|$.

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$M$ a complete Riemannian manifold with nonpositive sectional curvature. Show that $|d(exp_p)_v(w)| geq |w|$ for all $v in T_p M$, and all $w in T_v(T_p M)$.



Update:



Based on the hints I've gotten, here is my attempted solution.



We will compare $M$ with $tildeM=T_p M approx T_v T_p M$ via the Rauch comparison theorem. Let $gamma:[0,1] rightarrow M$ be a geodesic and let $tildegamma: [0,1] rightarrow tildeM$ be a comparison geodesic (which means that $gamma$ and $tildegamma$ have the same speed). Note that $tildegamma$ also has no conjugate points because curvature zero. Write $gamma(0)=p$ and $gamma'(0)=v$.



Let $J(t)$ be the geodesic along $gamma$ with $J(0)=0$ and $J'(0)=w$; then $J(t)$ can be written



$J(t)=d(exp_p)_tgamma'(0)(tJ'(0))$



so we see that $J(1)=d(exp_p)tv(tw)$.



The Rauch comparison theorem gives us that, for a Jacobi field $tildeJ$ along $tildegamma$ with




  1. $tildeJ(0)=J(0)=0$

  2. $|tildeJ'(0)|=|J'(0)|$


  3. $langle tildeJ'(0),gamma'(0) rangle = langle J'(0), gamma'(0) rangle$




we have $|tildeJ| leq |J|$, so if we had a Jacobi field $tildeJ$ along $tildegamma$ with $tildeJ(1)=w$ we'd be done.




There is a theorem that says if there is no conjugate points along $tildegamma$ then there is a unique Jacobi field $tildeJ$ along $tildegamma$ with $tildeJ(0)=0$ and $tildeJ(1)=w$ but I am unsure how how to get that the highlighted conditions (2) and (3) above hold.











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  • Compare $M$ with $tildeM=T_pM.$
    – Frieder Jäckel
    Aug 31 at 14:50






  • 1




    Hint: $d(exp_p)_v(w)$ is equal to $J(1)$, where $J$ is a Jacobi field along the geodesic $tmapsto exp_p(tv)$ that vanishes at $t=0$.
    – Jack Lee
    Aug 31 at 18:48










  • $tildeM=T_pM$ has constant metric, so the curvature tensor vanishes and the covariant derivative along curves is the usual derivative. Therefore $tildeJ(t)=tw$ is the Jacobifield (Uniqueness!) along $tto tv$ with $tildeJ(0)=0$ and $tildeJ‘(0)=w.$
    – Frieder Jäckel
    Sep 4 at 19:28















up vote
1
down vote

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$M$ a complete Riemannian manifold with nonpositive sectional curvature. Show that $|d(exp_p)_v(w)| geq |w|$ for all $v in T_p M$, and all $w in T_v(T_p M)$.



Update:



Based on the hints I've gotten, here is my attempted solution.



We will compare $M$ with $tildeM=T_p M approx T_v T_p M$ via the Rauch comparison theorem. Let $gamma:[0,1] rightarrow M$ be a geodesic and let $tildegamma: [0,1] rightarrow tildeM$ be a comparison geodesic (which means that $gamma$ and $tildegamma$ have the same speed). Note that $tildegamma$ also has no conjugate points because curvature zero. Write $gamma(0)=p$ and $gamma'(0)=v$.



Let $J(t)$ be the geodesic along $gamma$ with $J(0)=0$ and $J'(0)=w$; then $J(t)$ can be written



$J(t)=d(exp_p)_tgamma'(0)(tJ'(0))$



so we see that $J(1)=d(exp_p)tv(tw)$.



The Rauch comparison theorem gives us that, for a Jacobi field $tildeJ$ along $tildegamma$ with




  1. $tildeJ(0)=J(0)=0$

  2. $|tildeJ'(0)|=|J'(0)|$


  3. $langle tildeJ'(0),gamma'(0) rangle = langle J'(0), gamma'(0) rangle$




we have $|tildeJ| leq |J|$, so if we had a Jacobi field $tildeJ$ along $tildegamma$ with $tildeJ(1)=w$ we'd be done.




There is a theorem that says if there is no conjugate points along $tildegamma$ then there is a unique Jacobi field $tildeJ$ along $tildegamma$ with $tildeJ(0)=0$ and $tildeJ(1)=w$ but I am unsure how how to get that the highlighted conditions (2) and (3) above hold.











share|cite|improve this question























  • Compare $M$ with $tildeM=T_pM.$
    – Frieder Jäckel
    Aug 31 at 14:50






  • 1




    Hint: $d(exp_p)_v(w)$ is equal to $J(1)$, where $J$ is a Jacobi field along the geodesic $tmapsto exp_p(tv)$ that vanishes at $t=0$.
    – Jack Lee
    Aug 31 at 18:48










  • $tildeM=T_pM$ has constant metric, so the curvature tensor vanishes and the covariant derivative along curves is the usual derivative. Therefore $tildeJ(t)=tw$ is the Jacobifield (Uniqueness!) along $tto tv$ with $tildeJ(0)=0$ and $tildeJ‘(0)=w.$
    – Frieder Jäckel
    Sep 4 at 19:28













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$M$ a complete Riemannian manifold with nonpositive sectional curvature. Show that $|d(exp_p)_v(w)| geq |w|$ for all $v in T_p M$, and all $w in T_v(T_p M)$.



Update:



Based on the hints I've gotten, here is my attempted solution.



We will compare $M$ with $tildeM=T_p M approx T_v T_p M$ via the Rauch comparison theorem. Let $gamma:[0,1] rightarrow M$ be a geodesic and let $tildegamma: [0,1] rightarrow tildeM$ be a comparison geodesic (which means that $gamma$ and $tildegamma$ have the same speed). Note that $tildegamma$ also has no conjugate points because curvature zero. Write $gamma(0)=p$ and $gamma'(0)=v$.



Let $J(t)$ be the geodesic along $gamma$ with $J(0)=0$ and $J'(0)=w$; then $J(t)$ can be written



$J(t)=d(exp_p)_tgamma'(0)(tJ'(0))$



so we see that $J(1)=d(exp_p)tv(tw)$.



The Rauch comparison theorem gives us that, for a Jacobi field $tildeJ$ along $tildegamma$ with




  1. $tildeJ(0)=J(0)=0$

  2. $|tildeJ'(0)|=|J'(0)|$


  3. $langle tildeJ'(0),gamma'(0) rangle = langle J'(0), gamma'(0) rangle$




we have $|tildeJ| leq |J|$, so if we had a Jacobi field $tildeJ$ along $tildegamma$ with $tildeJ(1)=w$ we'd be done.




There is a theorem that says if there is no conjugate points along $tildegamma$ then there is a unique Jacobi field $tildeJ$ along $tildegamma$ with $tildeJ(0)=0$ and $tildeJ(1)=w$ but I am unsure how how to get that the highlighted conditions (2) and (3) above hold.











share|cite|improve this question















$M$ a complete Riemannian manifold with nonpositive sectional curvature. Show that $|d(exp_p)_v(w)| geq |w|$ for all $v in T_p M$, and all $w in T_v(T_p M)$.



Update:



Based on the hints I've gotten, here is my attempted solution.



We will compare $M$ with $tildeM=T_p M approx T_v T_p M$ via the Rauch comparison theorem. Let $gamma:[0,1] rightarrow M$ be a geodesic and let $tildegamma: [0,1] rightarrow tildeM$ be a comparison geodesic (which means that $gamma$ and $tildegamma$ have the same speed). Note that $tildegamma$ also has no conjugate points because curvature zero. Write $gamma(0)=p$ and $gamma'(0)=v$.



Let $J(t)$ be the geodesic along $gamma$ with $J(0)=0$ and $J'(0)=w$; then $J(t)$ can be written



$J(t)=d(exp_p)_tgamma'(0)(tJ'(0))$



so we see that $J(1)=d(exp_p)tv(tw)$.



The Rauch comparison theorem gives us that, for a Jacobi field $tildeJ$ along $tildegamma$ with




  1. $tildeJ(0)=J(0)=0$

  2. $|tildeJ'(0)|=|J'(0)|$


  3. $langle tildeJ'(0),gamma'(0) rangle = langle J'(0), gamma'(0) rangle$




we have $|tildeJ| leq |J|$, so if we had a Jacobi field $tildeJ$ along $tildegamma$ with $tildeJ(1)=w$ we'd be done.




There is a theorem that says if there is no conjugate points along $tildegamma$ then there is a unique Jacobi field $tildeJ$ along $tildegamma$ with $tildeJ(0)=0$ and $tildeJ(1)=w$ but I am unsure how how to get that the highlighted conditions (2) and (3) above hold.








riemannian-geometry






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share|cite|improve this question













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edited Sep 1 at 21:56

























asked Aug 31 at 3:16









TuoTuo

1,558513




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  • Compare $M$ with $tildeM=T_pM.$
    – Frieder Jäckel
    Aug 31 at 14:50






  • 1




    Hint: $d(exp_p)_v(w)$ is equal to $J(1)$, where $J$ is a Jacobi field along the geodesic $tmapsto exp_p(tv)$ that vanishes at $t=0$.
    – Jack Lee
    Aug 31 at 18:48










  • $tildeM=T_pM$ has constant metric, so the curvature tensor vanishes and the covariant derivative along curves is the usual derivative. Therefore $tildeJ(t)=tw$ is the Jacobifield (Uniqueness!) along $tto tv$ with $tildeJ(0)=0$ and $tildeJ‘(0)=w.$
    – Frieder Jäckel
    Sep 4 at 19:28

















  • Compare $M$ with $tildeM=T_pM.$
    – Frieder Jäckel
    Aug 31 at 14:50






  • 1




    Hint: $d(exp_p)_v(w)$ is equal to $J(1)$, where $J$ is a Jacobi field along the geodesic $tmapsto exp_p(tv)$ that vanishes at $t=0$.
    – Jack Lee
    Aug 31 at 18:48










  • $tildeM=T_pM$ has constant metric, so the curvature tensor vanishes and the covariant derivative along curves is the usual derivative. Therefore $tildeJ(t)=tw$ is the Jacobifield (Uniqueness!) along $tto tv$ with $tildeJ(0)=0$ and $tildeJ‘(0)=w.$
    – Frieder Jäckel
    Sep 4 at 19:28
















Compare $M$ with $tildeM=T_pM.$
– Frieder Jäckel
Aug 31 at 14:50




Compare $M$ with $tildeM=T_pM.$
– Frieder Jäckel
Aug 31 at 14:50




1




1




Hint: $d(exp_p)_v(w)$ is equal to $J(1)$, where $J$ is a Jacobi field along the geodesic $tmapsto exp_p(tv)$ that vanishes at $t=0$.
– Jack Lee
Aug 31 at 18:48




Hint: $d(exp_p)_v(w)$ is equal to $J(1)$, where $J$ is a Jacobi field along the geodesic $tmapsto exp_p(tv)$ that vanishes at $t=0$.
– Jack Lee
Aug 31 at 18:48












$tildeM=T_pM$ has constant metric, so the curvature tensor vanishes and the covariant derivative along curves is the usual derivative. Therefore $tildeJ(t)=tw$ is the Jacobifield (Uniqueness!) along $tto tv$ with $tildeJ(0)=0$ and $tildeJ‘(0)=w.$
– Frieder Jäckel
Sep 4 at 19:28





$tildeM=T_pM$ has constant metric, so the curvature tensor vanishes and the covariant derivative along curves is the usual derivative. Therefore $tildeJ(t)=tw$ is the Jacobifield (Uniqueness!) along $tto tv$ with $tildeJ(0)=0$ and $tildeJ‘(0)=w.$
– Frieder Jäckel
Sep 4 at 19:28
















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