Simply connected manifold with nonpositive curvature has no more than one geodesic between points
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For the sake of clarity, completeness of the manifold is not an assumption - this is not Cartan-Hadamard, and the proof of that theorem usually deals with the number of geodesics by concluding that exp is a diffeomorphism, hence injective. That does not apply, here.
In particular, I believe I have a proof of the statement, but there are some claims I made along the way that I would like confirmation of. The source of the problem is a recent UMD Topology/Geometry qualifying exam.
Let $M$ be a simply connected Riemannian manifold with nonpositive sectional curvature. Let $p,q$ be distinct points in $M$. Then there is, at most, one geodesic through $p$ and $q$.
Proof: By way of contradiction, assume not. Then we have distinct points $p,qin M$ such that we have two distinct geodesics through $p$ and $q$, say $gamma_1$ and $gamma_2$. Parametrize the geodesics so $p=gamma_1(0)=gamma_2(0)$, $q=gamma_1(L)=gamma_2(L)$.
Claim 1 (and the one that I am least certain of): because $M$ is simply connected, we can take a smooth, proper variation $f$ from $gamma_1$ to $gamma_2$ (essentially, a parametrized surface extending slightly beyond $gamma_1$ and $gamma_2$). Say $gamma_1(t)=f(0,t)$ and $gamma_2(t)=f(1,t)$, possibly by reparametrizing $f$. Let $V=frac{df}{ds}$.
Claim 2 (straight from do Carmo, chapter 9): Since $gamma_1$ is a geodesic and $f$ is a proper variation, $E'(0)=0$.
Claim 3 (also from do Carmo): $frac{1}{2}E''(s)=I_L(V,V)+<frac{D}{ds}V,frac{df}{dt}>|^L_{t=0}$, which, since $f$ is proper, is just the first term, $int^L_0<V',V'>-<R(frac{df}{dt},V)frac{df}{dt},V>dt=int^L_0<V',V'>-K(frac{df}{dt},V)*||frac{df}{dt}wedge V||^2dtgeq0$, since $|V'|^2geq0$ and $Kleq0$, by assumption.
Claim 4 (from calculus): since $gamma_1$ and $gamma_2$ are distinct, $E''(s)$ is strictly positive for some portion of $sin(0,1)$, hence $E'$ is strictly positive after that point (since $E'(0)=0$ and $E''(s)geq0$). Therefore, we have $E'(1)>0$, contradicting the fact that $f$ is a proper variation of $gamma_2(t)=f(1,t)$.
differential-geometry curvature geodesic
asked Jan 16 at 15:18
PepperPepper
508
$endgroup$
add a comment |
$begingroup$
For the sake of clarity, completeness of the manifold is not an assumption - this is not Cartan-Hadamard, and the proof of that theorem usually deals with the number of geodesics by concluding that exp is a diffeomorphism, hence injective. That does not apply, here.
In particular, I believe I have a proof of the statement, but there are some claims I made along the way that I would like confirmation of. The source of the problem is a recent UMD Topology/Geometry qualifying exam.
Let $M$ be a simply connected Riemannian manifold with nonpositive sectional curvature. Let $p,q$ be distinct points in $M$. Then there is, at most, one geodesic through $p$ and $q$.
Proof: By way of contradiction, assume not. Then we have distinct points $p,qin M$ such that we have two distinct geodesics through $p$ and $q$, say $gamma_1$ and $gamma_2$. Parametrize the geodesics so $p=gamma_1(0)=gamma_2(0)$, $q=gamma_1(L)=gamma_2(L)$.
Claim 1 (and the one that I am least certain of): because $M$ is simply connected, we can take a smooth, proper variation $f$ from $gamma_1$ to $gamma_2$ (essentially, a parametrized surface extending slightly beyond $gamma_1$ and $gamma_2$). Say $gamma_1(t)=f(0,t)$ and $gamma_2(t)=f(1,t)$, possibly by reparametrizing $f$. Let $V=frac{df}{ds}$.
Claim 2 (straight from do Carmo, chapter 9): Since $gamma_1$ is a geodesic and $f$ is a proper variation, $E'(0)=0$.
Claim 3 (also from do Carmo): $frac{1}{2}E''(s)=I_L(V,V)+<frac{D}{ds}V,frac{df}{dt}>|^L_{t=0}$, which, since $f$ is proper, is just the first term, $int^L_0<V',V'>-<R(frac{df}{dt},V)frac{df}{dt},V>dt=int^L_0<V',V'>-K(frac{df}{dt},V)*||frac{df}{dt}wedge V||^2dtgeq0$, since $|V'|^2geq0$ and $Kleq0$, by assumption.
Claim 4 (from calculus): since $gamma_1$ and $gamma_2$ are distinct, $E''(s)$ is strictly positive for some portion of $sin(0,1)$, hence $E'$ is strictly positive after that point (since $E'(0)=0$ and $E''(s)geq0$). Therefore, we have $E'(1)>0$, contradicting the fact that $f$ is a proper variation of $gamma_2(t)=f(1,t)$.
differential-geometry curvature geodesic
asked Jan 16 at 15:18
PepperPepper
508
$endgroup$
add a comment |
$begingroup$
For the sake of clarity, completeness of the manifold is not an assumption - this is not Cartan-Hadamard, and the proof of that theorem usually deals with the number of geodesics by concluding that exp is a diffeomorphism, hence injective. That does not apply, here.
In particular, I believe I have a proof of the statement, but there are some claims I made along the way that I would like confirmation of. The source of the problem is a recent UMD Topology/Geometry qualifying exam.
Let $M$ be a simply connected Riemannian manifold with nonpositive sectional curvature. Let $p,q$ be distinct points in $M$. Then there is, at most, one geodesic through $p$ and $q$.
Proof: By way of contradiction, assume not. Then we have distinct points $p,qin M$ such that we have two distinct geodesics through $p$ and $q$, say $gamma_1$ and $gamma_2$. Parametrize the geodesics so $p=gamma_1(0)=gamma_2(0)$, $q=gamma_1(L)=gamma_2(L)$.
Claim 1 (and the one that I am least certain of): because $M$ is simply connected, we can take a smooth, proper variation $f$ from $gamma_1$ to $gamma_2$ (essentially, a parametrized surface extending slightly beyond $gamma_1$ and $gamma_2$). Say $gamma_1(t)=f(0,t)$ and $gamma_2(t)=f(1,t)$, possibly by reparametrizing $f$. Let $V=frac{df}{ds}$.
Claim 2 (straight from do Carmo, chapter 9): Since $gamma_1$ is a geodesic and $f$ is a proper variation, $E'(0)=0$.
Claim 3 (also from do Carmo): $frac{1}{2}E''(s)=I_L(V,V)+<frac{D}{ds}V,frac{df}{dt}>|^L_{t=0}$, which, since $f$ is proper, is just the first term, $int^L_0<V',V'>-<R(frac{df}{dt},V)frac{df}{dt},V>dt=int^L_0<V',V'>-K(frac{df}{dt},V)*||frac{df}{dt}wedge V||^2dtgeq0$, since $|V'|^2geq0$ and $Kleq0$, by assumption.
Claim 4 (from calculus): since $gamma_1$ and $gamma_2$ are distinct, $E''(s)$ is strictly positive for some portion of $sin(0,1)$, hence $E'$ is strictly positive after that point (since $E'(0)=0$ and $E''(s)geq0$). Therefore, we have $E'(1)>0$, contradicting the fact that $f$ is a proper variation of $gamma_2(t)=f(1,t)$.
differential-geometry curvature geodesic
asked Jan 16 at 15:18
PepperPepper
508
$endgroup$
For the sake of clarity, completeness of the manifold is not an assumption - this is not Cartan-Hadamard, and the proof of that theorem usually deals with the number of geodesics by concluding that exp is a diffeomorphism, hence injective. That does not apply, here.
In particular, I believe I have a proof of the statement, but there are some claims I made along the way that I would like confirmation of. The source of the problem is a recent UMD Topology/Geometry qualifying exam.
Let $M$ be a simply connected Riemannian manifold with nonpositive sectional curvature. Let $p,q$ be distinct points in $M$. Then there is, at most, one geodesic through $p$ and $q$.
Proof: By way of contradiction, assume not. Then we have distinct points $p,qin M$ such that we have two distinct geodesics through $p$ and $q$, say $gamma_1$ and $gamma_2$. Parametrize the geodesics so $p=gamma_1(0)=gamma_2(0)$, $q=gamma_1(L)=gamma_2(L)$.
Claim 1 (and the one that I am least certain of): because $M$ is simply connected, we can take a smooth, proper variation $f$ from $gamma_1$ to $gamma_2$ (essentially, a parametrized surface extending slightly beyond $gamma_1$ and $gamma_2$). Say $gamma_1(t)=f(0,t)$ and $gamma_2(t)=f(1,t)$, possibly by reparametrizing $f$. Let $V=frac{df}{ds}$.
Claim 2 (straight from do Carmo, chapter 9): Since $gamma_1$ is a geodesic and $f$ is a proper variation, $E'(0)=0$.
Claim 3 (also from do Carmo): $frac{1}{2}E''(s)=I_L(V,V)+<frac{D}{ds}V,frac{df}{dt}>|^L_{t=0}$, which, since $f$ is proper, is just the first term, $int^L_0<V',V'>-<R(frac{df}{dt},V)frac{df}{dt},V>dt=int^L_0<V',V'>-K(frac{df}{dt},V)*||frac{df}{dt}wedge V||^2dtgeq0$, since $|V'|^2geq0$ and $Kleq0$, by assumption.
Claim 4 (from calculus): since $gamma_1$ and $gamma_2$ are distinct, $E''(s)$ is strictly positive for some portion of $sin(0,1)$, hence $E'$ is strictly positive after that point (since $E'(0)=0$ and $E''(s)geq0$). Therefore, we have $E'(1)>0$, contradicting the fact that $f$ is a proper variation of $gamma_2(t)=f(1,t)$.
differential-geometry curvature geodesic
differential-geometry curvature geodesic
asked Jan 16 at 15:18
PepperPepper
508
asked Jan 16 at 15:18
PepperPepper
508
edited Jan 16 at 15:24
Pepper
asked Jan 16 at 15:18
PepperPepper
508
asked Jan 16 at 15:18
PepperPepper
508
asked Jan 16 at 15:18
PepperPepper
508
508
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