Simply connected manifold with nonpositive curvature has at most one geodesic between points
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Once it's clear that any two points with multiple connecting geodesics are conjugate points, it's a straightforward application of the Jacobi equation to get that there are no conjugate points, as seen in Why do manifolds with negative sectional curvature not have conjugate points?. However, I wasn't sure I had a correct inclusion of simply connected. My argument is as follows:
Take two geodesics connecting $p$ and $q$, say $exp_p(tv)$ and $exp_p(tw)$, with $vnot=cw$ for any positive $c$ and $q=exp_p(v)=exp_p(w)$. Then $gamma_s(t)=exp_p(t*(v+s*(w-v))$ is a single-parameter family of geodesics, and $w-vinker(d(exp_p)_v)$, hence $s(w-v)inker(d(exp_p)_v)$, so $gamma_s(1)=$q for all $s$.
Simply connected should come into play in ensuring that changing $s$ is a smooth transition from one geodesic to the other, but I wasn't sure if it was enough to just say that or if there was more to be proven.
differential-geometry curvature geodesic
asked Dec 23 '18 at 21:34
PepperPepper
508
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add a comment |
$begingroup$
Once it's clear that any two points with multiple connecting geodesics are conjugate points, it's a straightforward application of the Jacobi equation to get that there are no conjugate points, as seen in Why do manifolds with negative sectional curvature not have conjugate points?. However, I wasn't sure I had a correct inclusion of simply connected. My argument is as follows:
Take two geodesics connecting $p$ and $q$, say $exp_p(tv)$ and $exp_p(tw)$, with $vnot=cw$ for any positive $c$ and $q=exp_p(v)=exp_p(w)$. Then $gamma_s(t)=exp_p(t*(v+s*(w-v))$ is a single-parameter family of geodesics, and $w-vinker(d(exp_p)_v)$, hence $s(w-v)inker(d(exp_p)_v)$, so $gamma_s(1)=$q for all $s$.
Simply connected should come into play in ensuring that changing $s$ is a smooth transition from one geodesic to the other, but I wasn't sure if it was enough to just say that or if there was more to be proven.
differential-geometry curvature geodesic
asked Dec 23 '18 at 21:34
PepperPepper
508
$endgroup$
$begingroup$
Two comments: (1) Unfortunately, two different definitions of the Riemann curvature tensor are in common use in Riemannian Geometry, which differ by a sign. I suspect the poster in the linked M:SE has the opposite sign convention from the one you are using. (I do not think it is accurate to call this answer "incorrect".) (2) From your argument, it's not obvious to me why $w - v$ is in the kernel of $d(exp_p)_v)$. (This happens to be true, by the thing you are trying to prove; but it won't be true in general if $M$ is not simply connected.)
$endgroup$
– mollyerin
Dec 24 '18 at 15:25
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1) Now that I know to look for it, you're correct that the other poster is using the other sign convention, and I'll amend my post. 2) I believe you are also correct about where simply connected is being used, although I have yet to fix it - certainly, as soon as $w-v$ is in the kernel, the remainder follows.
$endgroup$
– Pepper
Dec 24 '18 at 16:03
$begingroup$
In particular, $w-v$ is in the kernel as soon as there is a 1-parameter family of geodesics through the two endpoints, because then it's equivalent to a Jacobi vector field. Simply connected must be used to prove that such a family can be produced from the original two geodesics.
$endgroup$
– Pepper
Dec 24 '18 at 17:07
$begingroup$
Your proof is incorrect, all what you know from the simply-connected assumption is that there is a family of smooth curves interpolating between two given geodesics between $p$ and $q$. The actual proof is more complicated, take a look at this question. Lastly, you forgot the assumption that your manifold is complete.
$endgroup$
– Moishe Cohen
Dec 25 '18 at 15:40
$begingroup$
No, I did not forget to include completeness. I am familiar with the Cartan-Hadamard theorem. I don't doubt that my proof is incorrect, though. For reference, the problem is #4 on the January 2018 UMD Geometry/Topology qualifying exam, seen here www-math.umd.edu/images/pdfs/quals/Topology/…
$endgroup$
– Pepper
Dec 26 '18 at 15:11
add a comment |
$begingroup$
Once it's clear that any two points with multiple connecting geodesics are conjugate points, it's a straightforward application of the Jacobi equation to get that there are no conjugate points, as seen in Why do manifolds with negative sectional curvature not have conjugate points?. However, I wasn't sure I had a correct inclusion of simply connected. My argument is as follows:
Take two geodesics connecting $p$ and $q$, say $exp_p(tv)$ and $exp_p(tw)$, with $vnot=cw$ for any positive $c$ and $q=exp_p(v)=exp_p(w)$. Then $gamma_s(t)=exp_p(t*(v+s*(w-v))$ is a single-parameter family of geodesics, and $w-vinker(d(exp_p)_v)$, hence $s(w-v)inker(d(exp_p)_v)$, so $gamma_s(1)=$q for all $s$.
Simply connected should come into play in ensuring that changing $s$ is a smooth transition from one geodesic to the other, but I wasn't sure if it was enough to just say that or if there was more to be proven.
differential-geometry curvature geodesic
asked Dec 23 '18 at 21:34
PepperPepper
508
$endgroup$
Once it's clear that any two points with multiple connecting geodesics are conjugate points, it's a straightforward application of the Jacobi equation to get that there are no conjugate points, as seen in Why do manifolds with negative sectional curvature not have conjugate points?. However, I wasn't sure I had a correct inclusion of simply connected. My argument is as follows:
Take two geodesics connecting $p$ and $q$, say $exp_p(tv)$ and $exp_p(tw)$, with $vnot=cw$ for any positive $c$ and $q=exp_p(v)=exp_p(w)$. Then $gamma_s(t)=exp_p(t*(v+s*(w-v))$ is a single-parameter family of geodesics, and $w-vinker(d(exp_p)_v)$, hence $s(w-v)inker(d(exp_p)_v)$, so $gamma_s(1)=$q for all $s$.
Simply connected should come into play in ensuring that changing $s$ is a smooth transition from one geodesic to the other, but I wasn't sure if it was enough to just say that or if there was more to be proven.
differential-geometry curvature geodesic
differential-geometry curvature geodesic
asked Dec 23 '18 at 21:34
PepperPepper
508
asked Dec 23 '18 at 21:34
PepperPepper
508
edited Dec 24 '18 at 16:04
Pepper
asked Dec 23 '18 at 21:34
PepperPepper
508
asked Dec 23 '18 at 21:34
PepperPepper
508
asked Dec 23 '18 at 21:34
PepperPepper
508
508
$begingroup$
Two comments: (1) Unfortunately, two different definitions of the Riemann curvature tensor are in common use in Riemannian Geometry, which differ by a sign. I suspect the poster in the linked M:SE has the opposite sign convention from the one you are using. (I do not think it is accurate to call this answer "incorrect".) (2) From your argument, it's not obvious to me why $w - v$ is in the kernel of $d(exp_p)_v)$. (This happens to be true, by the thing you are trying to prove; but it won't be true in general if $M$ is not simply connected.)
$endgroup$
– mollyerin
Dec 24 '18 at 15:25
$begingroup$
1) Now that I know to look for it, you're correct that the other poster is using the other sign convention, and I'll amend my post. 2) I believe you are also correct about where simply connected is being used, although I have yet to fix it - certainly, as soon as $w-v$ is in the kernel, the remainder follows.
$endgroup$
– Pepper
Dec 24 '18 at 16:03
$begingroup$
In particular, $w-v$ is in the kernel as soon as there is a 1-parameter family of geodesics through the two endpoints, because then it's equivalent to a Jacobi vector field. Simply connected must be used to prove that such a family can be produced from the original two geodesics.
$endgroup$
– Pepper
Dec 24 '18 at 17:07
$begingroup$
Your proof is incorrect, all what you know from the simply-connected assumption is that there is a family of smooth curves interpolating between two given geodesics between $p$ and $q$. The actual proof is more complicated, take a look at this question. Lastly, you forgot the assumption that your manifold is complete.
$endgroup$
– Moishe Cohen
Dec 25 '18 at 15:40
$begingroup$
No, I did not forget to include completeness. I am familiar with the Cartan-Hadamard theorem. I don't doubt that my proof is incorrect, though. For reference, the problem is #4 on the January 2018 UMD Geometry/Topology qualifying exam, seen here www-math.umd.edu/images/pdfs/quals/Topology/…
$endgroup$
– Pepper
Dec 26 '18 at 15:11
add a comment |
$begingroup$
Two comments: (1) Unfortunately, two different definitions of the Riemann curvature tensor are in common use in Riemannian Geometry, which differ by a sign. I suspect the poster in the linked M:SE has the opposite sign convention from the one you are using. (I do not think it is accurate to call this answer "incorrect".) (2) From your argument, it's not obvious to me why $w - v$ is in the kernel of $d(exp_p)_v)$. (This happens to be true, by the thing you are trying to prove; but it won't be true in general if $M$ is not simply connected.)
$endgroup$
– mollyerin
Dec 24 '18 at 15:25
$begingroup$
1) Now that I know to look for it, you're correct that the other poster is using the other sign convention, and I'll amend my post. 2) I believe you are also correct about where simply connected is being used, although I have yet to fix it - certainly, as soon as $w-v$ is in the kernel, the remainder follows.
$endgroup$
– Pepper
Dec 24 '18 at 16:03
$begingroup$
In particular, $w-v$ is in the kernel as soon as there is a 1-parameter family of geodesics through the two endpoints, because then it's equivalent to a Jacobi vector field. Simply connected must be used to prove that such a family can be produced from the original two geodesics.
$endgroup$
– Pepper
Dec 24 '18 at 17:07
$begingroup$
Your proof is incorrect, all what you know from the simply-connected assumption is that there is a family of smooth curves interpolating between two given geodesics between $p$ and $q$. The actual proof is more complicated, take a look at this question. Lastly, you forgot the assumption that your manifold is complete.
$endgroup$
– Moishe Cohen
Dec 25 '18 at 15:40
$begingroup$
No, I did not forget to include completeness. I am familiar with the Cartan-Hadamard theorem. I don't doubt that my proof is incorrect, though. For reference, the problem is #4 on the January 2018 UMD Geometry/Topology qualifying exam, seen here www-math.umd.edu/images/pdfs/quals/Topology/…
$endgroup$
– Pepper
Dec 26 '18 at 15:11
$begingroup$
Two comments: (1) Unfortunately, two different definitions of the Riemann curvature tensor are in common use in Riemannian Geometry, which differ by a sign. I suspect the poster in the linked M:SE has the opposite sign convention from the one you are using. (I do not think it is accurate to call this answer "incorrect".) (2) From your argument, it's not obvious to me why $w - v$ is in the kernel of $d(exp_p)_v)$. (This happens to be true, by the thing you are trying to prove; but it won't be true in general if $M$ is not simply connected.)
$endgroup$
– mollyerin
Dec 24 '18 at 15:25
$begingroup$
Two comments: (1) Unfortunately, two different definitions of the Riemann curvature tensor are in common use in Riemannian Geometry, which differ by a sign. I suspect the poster in the linked M:SE has the opposite sign convention from the one you are using. (I do not think it is accurate to call this answer "incorrect".) (2) From your argument, it's not obvious to me why $w - v$ is in the kernel of $d(exp_p)_v)$. (This happens to be true, by the thing you are trying to prove; but it won't be true in general if $M$ is not simply connected.)
$endgroup$
– mollyerin
Dec 24 '18 at 15:25
$begingroup$
1) Now that I know to look for it, you're correct that the other poster is using the other sign convention, and I'll amend my post. 2) I believe you are also correct about where simply connected is being used, although I have yet to fix it - certainly, as soon as $w-v$ is in the kernel, the remainder follows.
$endgroup$
– Pepper
Dec 24 '18 at 16:03
$begingroup$
1) Now that I know to look for it, you're correct that the other poster is using the other sign convention, and I'll amend my post. 2) I believe you are also correct about where simply connected is being used, although I have yet to fix it - certainly, as soon as $w-v$ is in the kernel, the remainder follows.
$endgroup$
– Pepper
Dec 24 '18 at 16:03
$begingroup$
In particular, $w-v$ is in the kernel as soon as there is a 1-parameter family of geodesics through the two endpoints, because then it's equivalent to a Jacobi vector field. Simply connected must be used to prove that such a family can be produced from the original two geodesics.
$endgroup$
– Pepper
Dec 24 '18 at 17:07
$begingroup$
In particular, $w-v$ is in the kernel as soon as there is a 1-parameter family of geodesics through the two endpoints, because then it's equivalent to a Jacobi vector field. Simply connected must be used to prove that such a family can be produced from the original two geodesics.
$endgroup$
– Pepper
Dec 24 '18 at 17:07
$begingroup$
Your proof is incorrect, all what you know from the simply-connected assumption is that there is a family of smooth curves interpolating between two given geodesics between $p$ and $q$. The actual proof is more complicated, take a look at this question. Lastly, you forgot the assumption that your manifold is complete.
$endgroup$
– Moishe Cohen
Dec 25 '18 at 15:40
$begingroup$
Your proof is incorrect, all what you know from the simply-connected assumption is that there is a family of smooth curves interpolating between two given geodesics between $p$ and $q$. The actual proof is more complicated, take a look at this question. Lastly, you forgot the assumption that your manifold is complete.
$endgroup$
– Moishe Cohen
Dec 25 '18 at 15:40
$begingroup$
No, I did not forget to include completeness. I am familiar with the Cartan-Hadamard theorem. I don't doubt that my proof is incorrect, though. For reference, the problem is #4 on the January 2018 UMD Geometry/Topology qualifying exam, seen here www-math.umd.edu/images/pdfs/quals/Topology/…
$endgroup$
– Pepper
Dec 26 '18 at 15:11
$begingroup$
No, I did not forget to include completeness. I am familiar with the Cartan-Hadamard theorem. I don't doubt that my proof is incorrect, though. For reference, the problem is #4 on the January 2018 UMD Geometry/Topology qualifying exam, seen here www-math.umd.edu/images/pdfs/quals/Topology/…
$endgroup$
– Pepper
Dec 26 '18 at 15:11
add a comment |
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$begingroup$
Two comments: (1) Unfortunately, two different definitions of the Riemann curvature tensor are in common use in Riemannian Geometry, which differ by a sign. I suspect the poster in the linked M:SE has the opposite sign convention from the one you are using. (I do not think it is accurate to call this answer "incorrect".) (2) From your argument, it's not obvious to me why $w - v$ is in the kernel of $d(exp_p)_v)$. (This happens to be true, by the thing you are trying to prove; but it won't be true in general if $M$ is not simply connected.)
$endgroup$
– mollyerin
Dec 24 '18 at 15:25
$begingroup$
1) Now that I know to look for it, you're correct that the other poster is using the other sign convention, and I'll amend my post. 2) I believe you are also correct about where simply connected is being used, although I have yet to fix it - certainly, as soon as $w-v$ is in the kernel, the remainder follows.
$endgroup$
– Pepper
Dec 24 '18 at 16:03
$begingroup$
In particular, $w-v$ is in the kernel as soon as there is a 1-parameter family of geodesics through the two endpoints, because then it's equivalent to a Jacobi vector field. Simply connected must be used to prove that such a family can be produced from the original two geodesics.
$endgroup$
– Pepper
Dec 24 '18 at 17:07
$begingroup$
Your proof is incorrect, all what you know from the simply-connected assumption is that there is a family of smooth curves interpolating between two given geodesics between $p$ and $q$. The actual proof is more complicated, take a look at this question. Lastly, you forgot the assumption that your manifold is complete.
$endgroup$
– Moishe Cohen
Dec 25 '18 at 15:40
$begingroup$
No, I did not forget to include completeness. I am familiar with the Cartan-Hadamard theorem. I don't doubt that my proof is incorrect, though. For reference, the problem is #4 on the January 2018 UMD Geometry/Topology qualifying exam, seen here www-math.umd.edu/images/pdfs/quals/Topology/…
$endgroup$
– Pepper
Dec 26 '18 at 15:11