In a complete Riemannian manifold $M$, do any two points $p,qin M$ admit a length minimizing geodesic?
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
Clearly, completeness gives us the existence of at least one geodesic $gamma:[0,1]to M$ such that $gamma(0)=p$ and $gamma(1)=q$, however, this geodesic need not be length minimizing. I strongly suspect length minimizing geodesics must exist, but I can't quite construct the argument for their existence in full detail. This is what I've come up with.
If we let $r=d(p,q)$, and $U= B(p,2r)cup B(q,2r)$, then by completeness, $U$ is a precompact open in $M$. If we define $Gamma$ to be the set of smooth (or sufficiently regular) paths $gamma:[0,1]to M$ with b.c. $gamma(0)=p$ and $gamma(1)=q$, and define $Gamma_0subset Gamma$ to contain those paths $gamma $ with image in $overline U$, then we may apply the direct method of calculus of variations to minimize the energy functional
$$
E:Gammatomathbb R,quad E[gamma]=int_0^1! |dotgamma(t)|,mathrm dt.
$$
To do so, we define a weak topology on $Gamma$ under which $Gamma_0$ is weakly compact and $E$ is lower-semicontinuous, so that we may obtain a geodesic of minimal length.
The tricky thing here, of course, is to find a suitable weak topology, on a suitable candidate space $Gamma$, such as $H^1(0,1;M)$ with boundary conditions. Naïvely, I would $H^1(0,1;M)$ to suffice, with a weak topology defined via coordinate charts, but before going through the calculations, I'm wondering if there's already a reference for this result, with a clean proof.
differential-geometry pde metric-spaces riemannian-geometry calculus-of-variations
asked Sep 10 at 9:04
Monstrous Moonshine
2,6571530
add a comment |Â
up vote
2
down vote
favorite
Clearly, completeness gives us the existence of at least one geodesic $gamma:[0,1]to M$ such that $gamma(0)=p$ and $gamma(1)=q$, however, this geodesic need not be length minimizing. I strongly suspect length minimizing geodesics must exist, but I can't quite construct the argument for their existence in full detail. This is what I've come up with.
If we let $r=d(p,q)$, and $U= B(p,2r)cup B(q,2r)$, then by completeness, $U$ is a precompact open in $M$. If we define $Gamma$ to be the set of smooth (or sufficiently regular) paths $gamma:[0,1]to M$ with b.c. $gamma(0)=p$ and $gamma(1)=q$, and define $Gamma_0subset Gamma$ to contain those paths $gamma $ with image in $overline U$, then we may apply the direct method of calculus of variations to minimize the energy functional
$$
E:Gammatomathbb R,quad E[gamma]=int_0^1! |dotgamma(t)|,mathrm dt.
$$
To do so, we define a weak topology on $Gamma$ under which $Gamma_0$ is weakly compact and $E$ is lower-semicontinuous, so that we may obtain a geodesic of minimal length.
The tricky thing here, of course, is to find a suitable weak topology, on a suitable candidate space $Gamma$, such as $H^1(0,1;M)$ with boundary conditions. Naïvely, I would $H^1(0,1;M)$ to suffice, with a weak topology defined via coordinate charts, but before going through the calculations, I'm wondering if there's already a reference for this result, with a clean proof.
differential-geometry pde metric-spaces riemannian-geometry calculus-of-variations
asked Sep 10 at 9:04
Monstrous Moonshine
2,6571530
4
This fact is usually included as part of the Hopf-Rinow theorem, so you can find a (relatively) elementary proof in many introductory texts in Riemannian geometry. For example, it is the very first part of Hopf-Rinow addressed by Do Carmo in theorem 2.8.
– Anthony Carapetis
Sep 10 at 10:27
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Clearly, completeness gives us the existence of at least one geodesic $gamma:[0,1]to M$ such that $gamma(0)=p$ and $gamma(1)=q$, however, this geodesic need not be length minimizing. I strongly suspect length minimizing geodesics must exist, but I can't quite construct the argument for their existence in full detail. This is what I've come up with.
If we let $r=d(p,q)$, and $U= B(p,2r)cup B(q,2r)$, then by completeness, $U$ is a precompact open in $M$. If we define $Gamma$ to be the set of smooth (or sufficiently regular) paths $gamma:[0,1]to M$ with b.c. $gamma(0)=p$ and $gamma(1)=q$, and define $Gamma_0subset Gamma$ to contain those paths $gamma $ with image in $overline U$, then we may apply the direct method of calculus of variations to minimize the energy functional
$$
E:Gammatomathbb R,quad E[gamma]=int_0^1! |dotgamma(t)|,mathrm dt.
$$
To do so, we define a weak topology on $Gamma$ under which $Gamma_0$ is weakly compact and $E$ is lower-semicontinuous, so that we may obtain a geodesic of minimal length.
The tricky thing here, of course, is to find a suitable weak topology, on a suitable candidate space $Gamma$, such as $H^1(0,1;M)$ with boundary conditions. Naïvely, I would $H^1(0,1;M)$ to suffice, with a weak topology defined via coordinate charts, but before going through the calculations, I'm wondering if there's already a reference for this result, with a clean proof.
differential-geometry pde metric-spaces riemannian-geometry calculus-of-variations
asked Sep 10 at 9:04
Monstrous Moonshine
2,6571530
Clearly, completeness gives us the existence of at least one geodesic $gamma:[0,1]to M$ such that $gamma(0)=p$ and $gamma(1)=q$, however, this geodesic need not be length minimizing. I strongly suspect length minimizing geodesics must exist, but I can't quite construct the argument for their existence in full detail. This is what I've come up with.
If we let $r=d(p,q)$, and $U= B(p,2r)cup B(q,2r)$, then by completeness, $U$ is a precompact open in $M$. If we define $Gamma$ to be the set of smooth (or sufficiently regular) paths $gamma:[0,1]to M$ with b.c. $gamma(0)=p$ and $gamma(1)=q$, and define $Gamma_0subset Gamma$ to contain those paths $gamma $ with image in $overline U$, then we may apply the direct method of calculus of variations to minimize the energy functional
$$
E:Gammatomathbb R,quad E[gamma]=int_0^1! |dotgamma(t)|,mathrm dt.
$$
To do so, we define a weak topology on $Gamma$ under which $Gamma_0$ is weakly compact and $E$ is lower-semicontinuous, so that we may obtain a geodesic of minimal length.
The tricky thing here, of course, is to find a suitable weak topology, on a suitable candidate space $Gamma$, such as $H^1(0,1;M)$ with boundary conditions. Naïvely, I would $H^1(0,1;M)$ to suffice, with a weak topology defined via coordinate charts, but before going through the calculations, I'm wondering if there's already a reference for this result, with a clean proof.
differential-geometry pde metric-spaces riemannian-geometry calculus-of-variations
differential-geometry pde metric-spaces riemannian-geometry calculus-of-variations
asked Sep 10 at 9:04
Monstrous Moonshine
2,6571530
asked Sep 10 at 9:04
Monstrous Moonshine
2,6571530
asked Sep 10 at 9:04
Monstrous Moonshine
2,6571530
asked Sep 10 at 9:04
Monstrous Moonshine
2,6571530
asked Sep 10 at 9:04
Monstrous Moonshine
2,6571530
2,6571530
4
This fact is usually included as part of the Hopf-Rinow theorem, so you can find a (relatively) elementary proof in many introductory texts in Riemannian geometry. For example, it is the very first part of Hopf-Rinow addressed by Do Carmo in theorem 2.8.
– Anthony Carapetis
Sep 10 at 10:27
add a comment |Â
4
This fact is usually included as part of the Hopf-Rinow theorem, so you can find a (relatively) elementary proof in many introductory texts in Riemannian geometry. For example, it is the very first part of Hopf-Rinow addressed by Do Carmo in theorem 2.8.
– Anthony Carapetis
Sep 10 at 10:27
4
4
This fact is usually included as part of the Hopf-Rinow theorem, so you can find a (relatively) elementary proof in many introductory texts in Riemannian geometry. For example, it is the very first part of Hopf-Rinow addressed by Do Carmo in theorem 2.8.
– Anthony Carapetis
Sep 10 at 10:27
This fact is usually included as part of the Hopf-Rinow theorem, so you can find a (relatively) elementary proof in many introductory texts in Riemannian geometry. For example, it is the very first part of Hopf-Rinow addressed by Do Carmo in theorem 2.8.
– Anthony Carapetis
Sep 10 at 10:27
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
Jurgen Jost also gives a proof in Riemannian Geometry and Geometric Analysis, chapter 1.5. It's only for compact manifolds (complete manifolds are in chapter 1.7), but you already figured out how to deal with this problem.
In short, it's enough to take $Gamma$ as the class of Lipschitz curves $gamma colon [0,1] to U$ with constant speed. One can easily modify such curves to be piecewise geodesic, and then the compactness argument is very elementary, since every curve is represented just by a finite collection of points.
answered Sep 12 at 8:45
Michał Miśkiewicz
2,663616
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Jurgen Jost also gives a proof in Riemannian Geometry and Geometric Analysis, chapter 1.5. It's only for compact manifolds (complete manifolds are in chapter 1.7), but you already figured out how to deal with this problem.
In short, it's enough to take $Gamma$ as the class of Lipschitz curves $gamma colon [0,1] to U$ with constant speed. One can easily modify such curves to be piecewise geodesic, and then the compactness argument is very elementary, since every curve is represented just by a finite collection of points.
answered Sep 12 at 8:45
Michał Miśkiewicz
2,663616
add a comment |Â
up vote
0
down vote
Jurgen Jost also gives a proof in Riemannian Geometry and Geometric Analysis, chapter 1.5. It's only for compact manifolds (complete manifolds are in chapter 1.7), but you already figured out how to deal with this problem.
In short, it's enough to take $Gamma$ as the class of Lipschitz curves $gamma colon [0,1] to U$ with constant speed. One can easily modify such curves to be piecewise geodesic, and then the compactness argument is very elementary, since every curve is represented just by a finite collection of points.
answered Sep 12 at 8:45
Michał Miśkiewicz
2,663616
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Jurgen Jost also gives a proof in Riemannian Geometry and Geometric Analysis, chapter 1.5. It's only for compact manifolds (complete manifolds are in chapter 1.7), but you already figured out how to deal with this problem.
In short, it's enough to take $Gamma$ as the class of Lipschitz curves $gamma colon [0,1] to U$ with constant speed. One can easily modify such curves to be piecewise geodesic, and then the compactness argument is very elementary, since every curve is represented just by a finite collection of points.
answered Sep 12 at 8:45
Michał Miśkiewicz
2,663616
Jurgen Jost also gives a proof in Riemannian Geometry and Geometric Analysis, chapter 1.5. It's only for compact manifolds (complete manifolds are in chapter 1.7), but you already figured out how to deal with this problem.
In short, it's enough to take $Gamma$ as the class of Lipschitz curves $gamma colon [0,1] to U$ with constant speed. One can easily modify such curves to be piecewise geodesic, and then the compactness argument is very elementary, since every curve is represented just by a finite collection of points.
answered Sep 12 at 8:45
Michał Miśkiewicz
2,663616
edited Sep 13 at 10:20
answered Sep 12 at 8:45
Michał Miśkiewicz
2,663616
answered Sep 12 at 8:45
Michał Miśkiewicz
2,663616
answered Sep 12 at 8:45
Michał Miśkiewicz
2,663616
2,663616
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2911712%2fin-a-complete-riemannian-manifold-m-do-any-two-points-p-q-in-m-admit-a-leng%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
4
This fact is usually included as part of the Hopf-Rinow theorem, so you can find a (relatively) elementary proof in many introductory texts in Riemannian geometry. For example, it is the very first part of Hopf-Rinow addressed by Do Carmo in theorem 2.8.
– Anthony Carapetis
Sep 10 at 10:27