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On uniqueness of geodesic between two points on not necessarily complete manifold with nonpositive curvature
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I know the uniqueness of a geodesic connecting two points on a simply connected manifold $M$ with non-positive sectional curvature can be proven easily with Cartan-Hadamard theorem under the assumption that $M$ is complete. How can I go about proving this statement if the completeness of $M$ is NOT assumed?
To clarify, I'm trying to prove when $M$ is a manifold as above (simply connected, nonpositive sectional curvatures, but NOT necessarily complete), then there can be at most one geodesic between any two distinct points.
geometry riemannian-geometry
asked Jul 26 at 19:28
NaotoK
503
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up vote
2
down vote
favorite
I know the uniqueness of a geodesic connecting two points on a simply connected manifold $M$ with non-positive sectional curvature can be proven easily with Cartan-Hadamard theorem under the assumption that $M$ is complete. How can I go about proving this statement if the completeness of $M$ is NOT assumed?
To clarify, I'm trying to prove when $M$ is a manifold as above (simply connected, nonpositive sectional curvatures, but NOT necessarily complete), then there can be at most one geodesic between any two distinct points.
geometry riemannian-geometry
asked Jul 26 at 19:28
NaotoK
503
I didn't get your question. What do you want to prove?
– Anubhav Mukherjee
Jul 26 at 20:39
What happened if you remove a point from $mathbb H^3$?
– Anubhav Mukherjee
Jul 26 at 20:41
I'm trying to show that there can be at most one geodesic connecting two points on $M$ if $M$ is a manifold that is simply connected, has nonpositive curvatures, but is not necessarily complete.
– NaotoK
Jul 26 at 21:23
Also the statement still holds for $mathbbH^3 setminus textorigin$ if you were thinking about two semicircles connecting, say $(1,0,0)$ and $(-1,0,0)$. The geodesic in that case simply doesn't exist.
– NaotoK
Jul 26 at 21:43
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I know the uniqueness of a geodesic connecting two points on a simply connected manifold $M$ with non-positive sectional curvature can be proven easily with Cartan-Hadamard theorem under the assumption that $M$ is complete. How can I go about proving this statement if the completeness of $M$ is NOT assumed?
To clarify, I'm trying to prove when $M$ is a manifold as above (simply connected, nonpositive sectional curvatures, but NOT necessarily complete), then there can be at most one geodesic between any two distinct points.
geometry riemannian-geometry
asked Jul 26 at 19:28
NaotoK
503
I know the uniqueness of a geodesic connecting two points on a simply connected manifold $M$ with non-positive sectional curvature can be proven easily with Cartan-Hadamard theorem under the assumption that $M$ is complete. How can I go about proving this statement if the completeness of $M$ is NOT assumed?
To clarify, I'm trying to prove when $M$ is a manifold as above (simply connected, nonpositive sectional curvatures, but NOT necessarily complete), then there can be at most one geodesic between any two distinct points.
geometry riemannian-geometry
asked Jul 26 at 19:28
NaotoK
503
edited Jul 26 at 21:47
asked Jul 26 at 19:28
NaotoK
503
asked Jul 26 at 19:28
NaotoK
503
asked Jul 26 at 19:28
NaotoK
503
503
I didn't get your question. What do you want to prove?
– Anubhav Mukherjee
Jul 26 at 20:39
What happened if you remove a point from $mathbb H^3$?
– Anubhav Mukherjee
Jul 26 at 20:41
I'm trying to show that there can be at most one geodesic connecting two points on $M$ if $M$ is a manifold that is simply connected, has nonpositive curvatures, but is not necessarily complete.
– NaotoK
Jul 26 at 21:23
Also the statement still holds for $mathbbH^3 setminus textorigin$ if you were thinking about two semicircles connecting, say $(1,0,0)$ and $(-1,0,0)$. The geodesic in that case simply doesn't exist.
– NaotoK
Jul 26 at 21:43
add a comment |Â
I didn't get your question. What do you want to prove?
– Anubhav Mukherjee
Jul 26 at 20:39
What happened if you remove a point from $mathbb H^3$?
– Anubhav Mukherjee
Jul 26 at 20:41
I'm trying to show that there can be at most one geodesic connecting two points on $M$ if $M$ is a manifold that is simply connected, has nonpositive curvatures, but is not necessarily complete.
– NaotoK
Jul 26 at 21:23
Also the statement still holds for $mathbbH^3 setminus textorigin$ if you were thinking about two semicircles connecting, say $(1,0,0)$ and $(-1,0,0)$. The geodesic in that case simply doesn't exist.
– NaotoK
Jul 26 at 21:43
I didn't get your question. What do you want to prove?
– Anubhav Mukherjee
Jul 26 at 20:39
I didn't get your question. What do you want to prove?
– Anubhav Mukherjee
Jul 26 at 20:39
What happened if you remove a point from $mathbb H^3$?
– Anubhav Mukherjee
Jul 26 at 20:41
What happened if you remove a point from $mathbb H^3$?
– Anubhav Mukherjee
Jul 26 at 20:41
I'm trying to show that there can be at most one geodesic connecting two points on $M$ if $M$ is a manifold that is simply connected, has nonpositive curvatures, but is not necessarily complete.
– NaotoK
Jul 26 at 21:23
I'm trying to show that there can be at most one geodesic connecting two points on $M$ if $M$ is a manifold that is simply connected, has nonpositive curvatures, but is not necessarily complete.
– NaotoK
Jul 26 at 21:23
Also the statement still holds for $mathbbH^3 setminus textorigin$ if you were thinking about two semicircles connecting, say $(1,0,0)$ and $(-1,0,0)$. The geodesic in that case simply doesn't exist.
– NaotoK
Jul 26 at 21:43
Also the statement still holds for $mathbbH^3 setminus textorigin$ if you were thinking about two semicircles connecting, say $(1,0,0)$ and $(-1,0,0)$. The geodesic in that case simply doesn't exist.
– NaotoK
Jul 26 at 21:43
add a comment |Â
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I didn't get your question. What do you want to prove?
– Anubhav Mukherjee
Jul 26 at 20:39
What happened if you remove a point from $mathbb H^3$?
– Anubhav Mukherjee
Jul 26 at 20:41
I'm trying to show that there can be at most one geodesic connecting two points on $M$ if $M$ is a manifold that is simply connected, has nonpositive curvatures, but is not necessarily complete.
– NaotoK
Jul 26 at 21:23
Also the statement still holds for $mathbbH^3 setminus textorigin$ if you were thinking about two semicircles connecting, say $(1,0,0)$ and $(-1,0,0)$. The geodesic in that case simply doesn't exist.
– NaotoK
Jul 26 at 21:43