Unique geodesic between two points
1
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Suppose $(X,g)$ is a simply connected complete Riemannian manifold, and $X$ has negative sectional curvature everywhere. Is it true in such a case, for any two points in this space, namely $A$ and $B$, there exists a unique geodesic between them?
I know it's true for $dim(X) = 2$ since we can use Gauss Bonnet. How should I approach for higher dimension? If it's not true, are there some counterexamples?
riemannian-geometry
asked Jan 13 at 18:01
DaiDai
316210
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|
show 3 more comments
1
$begingroup$
Suppose $(X,g)$ is a simply connected complete Riemannian manifold, and $X$ has negative sectional curvature everywhere. Is it true in such a case, for any two points in this space, namely $A$ and $B$, there exists a unique geodesic between them?
I know it's true for $dim(X) = 2$ since we can use Gauss Bonnet. How should I approach for higher dimension? If it's not true, are there some counterexamples?
riemannian-geometry
asked Jan 13 at 18:01
DaiDai
316210
$endgroup$
$begingroup$
First of all, what do you mean by "trivial topology"? Simply connected? Contractible? Diffeomorphic to $mathbb R^n$?
$endgroup$
– Jack Lee
Jan 13 at 22:38
$begingroup$
Second, how do you propose to use Gauss-Bonnet to answer this question in two dimensions? Gauss-Bonnet only applies to compact manifolds, which probably don't have "trivial topology" (unless you just mean simply connected, in which case there's only the sphere, which has no metric of negative curvature).
$endgroup$
– Jack Lee
Jan 13 at 22:39
$begingroup$
In any case, the answer is certainly no if you don't assume that the metric is complete. For example, if you remove a ray from hyperbolic space, you get a manifold diffeomorphic to $mathbb R^n$ with negative curvature, but points on opposite sides of the removed ray will have no minimizing geodesic between them.
$endgroup$
– Jack Lee
Jan 13 at 22:41
$begingroup$
Sorry, I should have said "points on opposite sides of the removed ray will have no geodesic between them."
$endgroup$
– Jack Lee
Jan 13 at 23:18
2
$begingroup$
I think math.stackexchange.com/questions/170964/… answers you question affirmatively.
$endgroup$
– Jason DeVito
Jan 14 at 18:00
|
show 3 more comments
1
1
1
$begingroup$
Suppose $(X,g)$ is a simply connected complete Riemannian manifold, and $X$ has negative sectional curvature everywhere. Is it true in such a case, for any two points in this space, namely $A$ and $B$, there exists a unique geodesic between them?
I know it's true for $dim(X) = 2$ since we can use Gauss Bonnet. How should I approach for higher dimension? If it's not true, are there some counterexamples?
riemannian-geometry
asked Jan 13 at 18:01
DaiDai
316210
$endgroup$
Suppose $(X,g)$ is a simply connected complete Riemannian manifold, and $X$ has negative sectional curvature everywhere. Is it true in such a case, for any two points in this space, namely $A$ and $B$, there exists a unique geodesic between them?
I know it's true for $dim(X) = 2$ since we can use Gauss Bonnet. How should I approach for higher dimension? If it's not true, are there some counterexamples?
riemannian-geometry
riemannian-geometry
asked Jan 13 at 18:01
DaiDai
316210
asked Jan 13 at 18:01
DaiDai
316210
edited Jan 14 at 17:09
Dai
asked Jan 13 at 18:01
DaiDai
316210
asked Jan 13 at 18:01
DaiDai
316210
asked Jan 13 at 18:01
DaiDai
316210
316210
$begingroup$
First of all, what do you mean by "trivial topology"? Simply connected? Contractible? Diffeomorphic to $mathbb R^n$?
$endgroup$
– Jack Lee
Jan 13 at 22:38
$begingroup$
Second, how do you propose to use Gauss-Bonnet to answer this question in two dimensions? Gauss-Bonnet only applies to compact manifolds, which probably don't have "trivial topology" (unless you just mean simply connected, in which case there's only the sphere, which has no metric of negative curvature).
$endgroup$
– Jack Lee
Jan 13 at 22:39
$begingroup$
In any case, the answer is certainly no if you don't assume that the metric is complete. For example, if you remove a ray from hyperbolic space, you get a manifold diffeomorphic to $mathbb R^n$ with negative curvature, but points on opposite sides of the removed ray will have no minimizing geodesic between them.
$endgroup$
– Jack Lee
Jan 13 at 22:41
$begingroup$
Sorry, I should have said "points on opposite sides of the removed ray will have no geodesic between them."
$endgroup$
– Jack Lee
Jan 13 at 23:18
2
$begingroup$
I think math.stackexchange.com/questions/170964/… answers you question affirmatively.
$endgroup$
– Jason DeVito
Jan 14 at 18:00
|
show 3 more comments
$begingroup$
First of all, what do you mean by "trivial topology"? Simply connected? Contractible? Diffeomorphic to $mathbb R^n$?
$endgroup$
– Jack Lee
Jan 13 at 22:38
$begingroup$
Second, how do you propose to use Gauss-Bonnet to answer this question in two dimensions? Gauss-Bonnet only applies to compact manifolds, which probably don't have "trivial topology" (unless you just mean simply connected, in which case there's only the sphere, which has no metric of negative curvature).
$endgroup$
– Jack Lee
Jan 13 at 22:39
$begingroup$
In any case, the answer is certainly no if you don't assume that the metric is complete. For example, if you remove a ray from hyperbolic space, you get a manifold diffeomorphic to $mathbb R^n$ with negative curvature, but points on opposite sides of the removed ray will have no minimizing geodesic between them.
$endgroup$
– Jack Lee
Jan 13 at 22:41
$begingroup$
Sorry, I should have said "points on opposite sides of the removed ray will have no geodesic between them."
$endgroup$
– Jack Lee
Jan 13 at 23:18
2
$begingroup$
I think math.stackexchange.com/questions/170964/… answers you question affirmatively.
$endgroup$
– Jason DeVito
Jan 14 at 18:00
$begingroup$
First of all, what do you mean by "trivial topology"? Simply connected? Contractible? Diffeomorphic to $mathbb R^n$?
$endgroup$
– Jack Lee
Jan 13 at 22:38
$begingroup$
First of all, what do you mean by "trivial topology"? Simply connected? Contractible? Diffeomorphic to $mathbb R^n$?
$endgroup$
– Jack Lee
Jan 13 at 22:38
$begingroup$
Second, how do you propose to use Gauss-Bonnet to answer this question in two dimensions? Gauss-Bonnet only applies to compact manifolds, which probably don't have "trivial topology" (unless you just mean simply connected, in which case there's only the sphere, which has no metric of negative curvature).
$endgroup$
– Jack Lee
Jan 13 at 22:39
$begingroup$
Second, how do you propose to use Gauss-Bonnet to answer this question in two dimensions? Gauss-Bonnet only applies to compact manifolds, which probably don't have "trivial topology" (unless you just mean simply connected, in which case there's only the sphere, which has no metric of negative curvature).
$endgroup$
– Jack Lee
Jan 13 at 22:39
$begingroup$
In any case, the answer is certainly no if you don't assume that the metric is complete. For example, if you remove a ray from hyperbolic space, you get a manifold diffeomorphic to $mathbb R^n$ with negative curvature, but points on opposite sides of the removed ray will have no minimizing geodesic between them.
$endgroup$
– Jack Lee
Jan 13 at 22:41
$begingroup$
In any case, the answer is certainly no if you don't assume that the metric is complete. For example, if you remove a ray from hyperbolic space, you get a manifold diffeomorphic to $mathbb R^n$ with negative curvature, but points on opposite sides of the removed ray will have no minimizing geodesic between them.
$endgroup$
– Jack Lee
Jan 13 at 22:41
$begingroup$
Sorry, I should have said "points on opposite sides of the removed ray will have no geodesic between them."
$endgroup$
– Jack Lee
Jan 13 at 23:18
$begingroup$
Sorry, I should have said "points on opposite sides of the removed ray will have no geodesic between them."
$endgroup$
– Jack Lee
Jan 13 at 23:18
2
2
$begingroup$
I think math.stackexchange.com/questions/170964/… answers you question affirmatively.
$endgroup$
– Jason DeVito
Jan 14 at 18:00
$begingroup$
I think math.stackexchange.com/questions/170964/… answers you question affirmatively.
$endgroup$
– Jason DeVito
Jan 14 at 18:00
|
show 3 more comments
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$begingroup$
First of all, what do you mean by "trivial topology"? Simply connected? Contractible? Diffeomorphic to $mathbb R^n$?
$endgroup$
– Jack Lee
Jan 13 at 22:38
$begingroup$
Second, how do you propose to use Gauss-Bonnet to answer this question in two dimensions? Gauss-Bonnet only applies to compact manifolds, which probably don't have "trivial topology" (unless you just mean simply connected, in which case there's only the sphere, which has no metric of negative curvature).
$endgroup$
– Jack Lee
Jan 13 at 22:39
$begingroup$
In any case, the answer is certainly no if you don't assume that the metric is complete. For example, if you remove a ray from hyperbolic space, you get a manifold diffeomorphic to $mathbb R^n$ with negative curvature, but points on opposite sides of the removed ray will have no minimizing geodesic between them.
$endgroup$
– Jack Lee
Jan 13 at 22:41
$begingroup$
Sorry, I should have said "points on opposite sides of the removed ray will have no geodesic between them."
$endgroup$
– Jack Lee
Jan 13 at 23:18
2
$begingroup$
I think math.stackexchange.com/questions/170964/… answers you question affirmatively.
$endgroup$
– Jason DeVito
Jan 14 at 18:00