$M$ 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
- $tildeJ(0)=J(0)=0$
$|tildeJ'(0)|=|J'(0)|$
$langle tildeJ'(0),gamma'(0) rangle = langle J'(0), gamma'(0) rangle$
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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$M$ 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
- $tildeJ(0)=J(0)=0$
$|tildeJ'(0)|=|J'(0)|$
$langle tildeJ'(0),gamma'(0) rangle = langle J'(0), gamma'(0) rangle$
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
Compare $M$ with $tildeM=T_pM.$
â 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 $t=0$.
â 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 $tildeJâÂÂ(0)=w.$
â Frieder Jäckel
Sep 4 at 19:28
add a comment |Â
up vote
1
down vote
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up vote
1
down vote
favorite
$M$ 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
- $tildeJ(0)=J(0)=0$
$|tildeJ'(0)|=|J'(0)|$
$langle tildeJ'(0),gamma'(0) rangle = langle J'(0), gamma'(0) rangle$
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
$M$ 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
- $tildeJ(0)=J(0)=0$
$|tildeJ'(0)|=|J'(0)|$
$langle tildeJ'(0),gamma'(0) rangle = langle J'(0), gamma'(0) rangle$
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
riemannian-geometry
edited Sep 1 at 21:56
asked Aug 31 at 3:16
TuoTuo
1,558513
1,558513
Compare $M$ with $tildeM=T_pM.$
â 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 $t=0$.
â 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 $tildeJâÂÂ(0)=w.$
â Frieder Jäckel
Sep 4 at 19:28
add a comment |Â
Compare $M$ with $tildeM=T_pM.$
â 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 $t=0$.
â 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 $tildeJâÂÂ(0)=w.$
â Frieder Jäckel
Sep 4 at 19:28
Compare $M$ with $tildeM=T_pM.$
â Frieder Jäckel
Aug 31 at 14:50
Compare $M$ with $tildeM=T_pM.$
â 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 $t=0$.
â 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 $t=0$.
â 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 $tildeJâÂÂ(0)=w.$
â 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 $tildeJâÂÂ(0)=w.$
â Frieder Jäckel
Sep 4 at 19:28
add a comment |Â
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Compare $M$ with $tildeM=T_pM.$
â 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 $t=0$.
â 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 $tildeJâÂÂ(0)=w.$
â Frieder Jäckel
Sep 4 at 19:28