Intuitively speaking why use geodesics to capture the idea of completeness?
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Let $(M,g)$ be a Riemannian manifold. We would like to define one notion of completeness which captures the idea of "missing points". For example $mathbbR^nsetminus 0$ should be incomplete in this sense.
Now, in the Riemmanian case, the metrig $g$ gives rise to a distance function $d_g : Mtimes Mto [0,infty)$ and $M$ is a metric space in the usual sense. This allows us to use the familiar idea of completeness of a metric space. We hence say that $(M,g)$ is $m$-complete (or metrically complete) if the metric space $(M,d_g)$ is complete in the sense that every Cauchy sequence converges.
Now forget the metric space structure and focus just on the Riemmanian manifold structure. The usual thing to do is to say that $(M,g)$ is $g$-complete (or geodesically complete) if every inextendible geodesic is defined on the whole $mathbbR$.
The Hopf-Rinow theorem them says that $(M,g)$ is geodesically complete if and only if it is metrically complete.
Anyway, the question is about geodesic completeness. What we want is to capture the idea that some point is missing. Well it is then natural to pick a curve and ask whether we can continuously extend it to the whole $mathbbR$ or not. If we can't intuitively speaking there's a point missing.
Take $mathbbR^2setminus (1,0)$ then the curve $(costheta,sintheta)$ with $theta in (0,2pi)$ cannot be extended to $mathbbR$ continuously. Indeed whatever the value we set for it at $0,2pi$ the curve will be discontinuous there. In a sense: it was going in the direction of a missing point and to extend it, it must jump.
The question is: why do we need geodesics to define this? Why can't we say:
$M$ is complete if every continuous inextendible curve is defined on the whole $mathbbR$.
The point is that I don't get why do we need the metric structure encoded in the requirement that the definition is made with geodesics to capture the idea of missing points.
What is the intuition that we need geodesics/metric structure to capture the idea that there are points missing in the manifold?
general-topology differential-geometry definition intuition motivation
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up vote
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Let $(M,g)$ be a Riemannian manifold. We would like to define one notion of completeness which captures the idea of "missing points". For example $mathbbR^nsetminus 0$ should be incomplete in this sense.
Now, in the Riemmanian case, the metrig $g$ gives rise to a distance function $d_g : Mtimes Mto [0,infty)$ and $M$ is a metric space in the usual sense. This allows us to use the familiar idea of completeness of a metric space. We hence say that $(M,g)$ is $m$-complete (or metrically complete) if the metric space $(M,d_g)$ is complete in the sense that every Cauchy sequence converges.
Now forget the metric space structure and focus just on the Riemmanian manifold structure. The usual thing to do is to say that $(M,g)$ is $g$-complete (or geodesically complete) if every inextendible geodesic is defined on the whole $mathbbR$.
The Hopf-Rinow theorem them says that $(M,g)$ is geodesically complete if and only if it is metrically complete.
Anyway, the question is about geodesic completeness. What we want is to capture the idea that some point is missing. Well it is then natural to pick a curve and ask whether we can continuously extend it to the whole $mathbbR$ or not. If we can't intuitively speaking there's a point missing.
Take $mathbbR^2setminus (1,0)$ then the curve $(costheta,sintheta)$ with $theta in (0,2pi)$ cannot be extended to $mathbbR$ continuously. Indeed whatever the value we set for it at $0,2pi$ the curve will be discontinuous there. In a sense: it was going in the direction of a missing point and to extend it, it must jump.
The question is: why do we need geodesics to define this? Why can't we say:
$M$ is complete if every continuous inextendible curve is defined on the whole $mathbbR$.
The point is that I don't get why do we need the metric structure encoded in the requirement that the definition is made with geodesics to capture the idea of missing points.
What is the intuition that we need geodesics/metric structure to capture the idea that there are points missing in the manifold?
general-topology differential-geometry definition intuition motivation
2
You're missing an important part: The curves should be parametrized by arc length. The curve $f$ on the complete manifold $Bbb R$ with the standard metric, given by $f(t) = tan t$ is defined for $tin (-pi/2, pi/2)$. It cannot be extended, but doesn't have the whole real line as domain.
â Arthur
Aug 28 at 16:59
1
If you didn't need the metric structure, $Bbb R^2 cong (S^2 - x)$ could not possibly be complete.
â Mike Miller
Aug 28 at 17:26
@Arthur I get one part of your point. I mean, we can have a curve which is inextendible and not defined on the whole of $mathbbR$ because of a bad parametrization. But still, I confess I don't get intuitively why we need the metric structure, in other words, why do we need to parametrize by arc lenght to analyze incompleteness. I mean, as a definition one can just say "fine, let's work with it", but I want to understand why it is done the way it is done.
â user1620696
Aug 28 at 17:36
add a comment |Â
up vote
3
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favorite
up vote
3
down vote
favorite
Let $(M,g)$ be a Riemannian manifold. We would like to define one notion of completeness which captures the idea of "missing points". For example $mathbbR^nsetminus 0$ should be incomplete in this sense.
Now, in the Riemmanian case, the metrig $g$ gives rise to a distance function $d_g : Mtimes Mto [0,infty)$ and $M$ is a metric space in the usual sense. This allows us to use the familiar idea of completeness of a metric space. We hence say that $(M,g)$ is $m$-complete (or metrically complete) if the metric space $(M,d_g)$ is complete in the sense that every Cauchy sequence converges.
Now forget the metric space structure and focus just on the Riemmanian manifold structure. The usual thing to do is to say that $(M,g)$ is $g$-complete (or geodesically complete) if every inextendible geodesic is defined on the whole $mathbbR$.
The Hopf-Rinow theorem them says that $(M,g)$ is geodesically complete if and only if it is metrically complete.
Anyway, the question is about geodesic completeness. What we want is to capture the idea that some point is missing. Well it is then natural to pick a curve and ask whether we can continuously extend it to the whole $mathbbR$ or not. If we can't intuitively speaking there's a point missing.
Take $mathbbR^2setminus (1,0)$ then the curve $(costheta,sintheta)$ with $theta in (0,2pi)$ cannot be extended to $mathbbR$ continuously. Indeed whatever the value we set for it at $0,2pi$ the curve will be discontinuous there. In a sense: it was going in the direction of a missing point and to extend it, it must jump.
The question is: why do we need geodesics to define this? Why can't we say:
$M$ is complete if every continuous inextendible curve is defined on the whole $mathbbR$.
The point is that I don't get why do we need the metric structure encoded in the requirement that the definition is made with geodesics to capture the idea of missing points.
What is the intuition that we need geodesics/metric structure to capture the idea that there are points missing in the manifold?
general-topology differential-geometry definition intuition motivation
Let $(M,g)$ be a Riemannian manifold. We would like to define one notion of completeness which captures the idea of "missing points". For example $mathbbR^nsetminus 0$ should be incomplete in this sense.
Now, in the Riemmanian case, the metrig $g$ gives rise to a distance function $d_g : Mtimes Mto [0,infty)$ and $M$ is a metric space in the usual sense. This allows us to use the familiar idea of completeness of a metric space. We hence say that $(M,g)$ is $m$-complete (or metrically complete) if the metric space $(M,d_g)$ is complete in the sense that every Cauchy sequence converges.
Now forget the metric space structure and focus just on the Riemmanian manifold structure. The usual thing to do is to say that $(M,g)$ is $g$-complete (or geodesically complete) if every inextendible geodesic is defined on the whole $mathbbR$.
The Hopf-Rinow theorem them says that $(M,g)$ is geodesically complete if and only if it is metrically complete.
Anyway, the question is about geodesic completeness. What we want is to capture the idea that some point is missing. Well it is then natural to pick a curve and ask whether we can continuously extend it to the whole $mathbbR$ or not. If we can't intuitively speaking there's a point missing.
Take $mathbbR^2setminus (1,0)$ then the curve $(costheta,sintheta)$ with $theta in (0,2pi)$ cannot be extended to $mathbbR$ continuously. Indeed whatever the value we set for it at $0,2pi$ the curve will be discontinuous there. In a sense: it was going in the direction of a missing point and to extend it, it must jump.
The question is: why do we need geodesics to define this? Why can't we say:
$M$ is complete if every continuous inextendible curve is defined on the whole $mathbbR$.
The point is that I don't get why do we need the metric structure encoded in the requirement that the definition is made with geodesics to capture the idea of missing points.
What is the intuition that we need geodesics/metric structure to capture the idea that there are points missing in the manifold?
general-topology differential-geometry definition intuition motivation
asked Aug 28 at 16:55
user1620696
11k339104
11k339104
2
You're missing an important part: The curves should be parametrized by arc length. The curve $f$ on the complete manifold $Bbb R$ with the standard metric, given by $f(t) = tan t$ is defined for $tin (-pi/2, pi/2)$. It cannot be extended, but doesn't have the whole real line as domain.
â Arthur
Aug 28 at 16:59
1
If you didn't need the metric structure, $Bbb R^2 cong (S^2 - x)$ could not possibly be complete.
â Mike Miller
Aug 28 at 17:26
@Arthur I get one part of your point. I mean, we can have a curve which is inextendible and not defined on the whole of $mathbbR$ because of a bad parametrization. But still, I confess I don't get intuitively why we need the metric structure, in other words, why do we need to parametrize by arc lenght to analyze incompleteness. I mean, as a definition one can just say "fine, let's work with it", but I want to understand why it is done the way it is done.
â user1620696
Aug 28 at 17:36
add a comment |Â
2
You're missing an important part: The curves should be parametrized by arc length. The curve $f$ on the complete manifold $Bbb R$ with the standard metric, given by $f(t) = tan t$ is defined for $tin (-pi/2, pi/2)$. It cannot be extended, but doesn't have the whole real line as domain.
â Arthur
Aug 28 at 16:59
1
If you didn't need the metric structure, $Bbb R^2 cong (S^2 - x)$ could not possibly be complete.
â Mike Miller
Aug 28 at 17:26
@Arthur I get one part of your point. I mean, we can have a curve which is inextendible and not defined on the whole of $mathbbR$ because of a bad parametrization. But still, I confess I don't get intuitively why we need the metric structure, in other words, why do we need to parametrize by arc lenght to analyze incompleteness. I mean, as a definition one can just say "fine, let's work with it", but I want to understand why it is done the way it is done.
â user1620696
Aug 28 at 17:36
2
2
You're missing an important part: The curves should be parametrized by arc length. The curve $f$ on the complete manifold $Bbb R$ with the standard metric, given by $f(t) = tan t$ is defined for $tin (-pi/2, pi/2)$. It cannot be extended, but doesn't have the whole real line as domain.
â Arthur
Aug 28 at 16:59
You're missing an important part: The curves should be parametrized by arc length. The curve $f$ on the complete manifold $Bbb R$ with the standard metric, given by $f(t) = tan t$ is defined for $tin (-pi/2, pi/2)$. It cannot be extended, but doesn't have the whole real line as domain.
â Arthur
Aug 28 at 16:59
1
1
If you didn't need the metric structure, $Bbb R^2 cong (S^2 - x)$ could not possibly be complete.
â Mike Miller
Aug 28 at 17:26
If you didn't need the metric structure, $Bbb R^2 cong (S^2 - x)$ could not possibly be complete.
â Mike Miller
Aug 28 at 17:26
@Arthur I get one part of your point. I mean, we can have a curve which is inextendible and not defined on the whole of $mathbbR$ because of a bad parametrization. But still, I confess I don't get intuitively why we need the metric structure, in other words, why do we need to parametrize by arc lenght to analyze incompleteness. I mean, as a definition one can just say "fine, let's work with it", but I want to understand why it is done the way it is done.
â user1620696
Aug 28 at 17:36
@Arthur I get one part of your point. I mean, we can have a curve which is inextendible and not defined on the whole of $mathbbR$ because of a bad parametrization. But still, I confess I don't get intuitively why we need the metric structure, in other words, why do we need to parametrize by arc lenght to analyze incompleteness. I mean, as a definition one can just say "fine, let's work with it", but I want to understand why it is done the way it is done.
â user1620696
Aug 28 at 17:36
add a comment |Â
2 Answers
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Two examples in $mathbb R^3 ; :$ (A) take the unit sphere $x^2 + y^2 + z^2 = 1$ and delete the north and south poles; (B) the infinite cylinder $x^2 + y^2 = 1$ with $-infty <z < infty.$
Examples (A) and (B) are diffeomorphic, by a projection around the origin. (A) is incomplete as a metric space, (B) is complete.
This addresses why parametrisation by arc length is important, but not why geodesics are important.
â Arthur
Aug 28 at 17:45
add a comment |Â
up vote
2
down vote
As others have pointed out, you certainly need the metric structure to talk about completeness, because the same manifold can be complete with one metric and incomplete with another.
But let me address the more focused question that you posed in the comments: why are geodesics important?
I think probably you're misinterpreting the reason why people focus on geodesic completeness. It's not that we need geodesics to detect completeness -- it's certainly possible to give lots of characterizations of complete Riemannian manifolds that don't refer to geodesics. For example, a Riemannian manifold is (metrically) complete if and only if any of the following conditions is satisfied:
- Every Cauchy sequence converges.
- Every closed and bounded subset is compact.
- Every piecewise $C^1$ unit-speed curve defined on a proper subinterval of $mathbb R$ can be extended to a piecewise $C^1$ unit-speed curve on a strictly larger interval.
- Every piecewise $C^1$ curve that diverges to infinity (meaning that it escapes from every compact set) has infinite length.
The real significance of the Hopf-Rinow theorem is that it tells us something about geodesics, not about metric completeness. For example, as a consequence of the Hopf-Rinow theorem, we know that
- If $(M,g)$ is metrically complete, every geodesic can be extended to all of $mathbb R$ as a geodesic (not just as a piecewise $C^1$ curve).
- If there is a point $pin M$ such that every geodesic starting at $p$ can be extended to all of $mathbb R$, then $M$ is metrically complete. (Technically, this is a consequence of one step in the proof of Hopf-Rinow, not of the theorem itself.)
These results have lots of important consequences in the study of Riemannian geometry, far beyond just detecting whether a manifold is metrically complete or not.
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Two examples in $mathbb R^3 ; :$ (A) take the unit sphere $x^2 + y^2 + z^2 = 1$ and delete the north and south poles; (B) the infinite cylinder $x^2 + y^2 = 1$ with $-infty <z < infty.$
Examples (A) and (B) are diffeomorphic, by a projection around the origin. (A) is incomplete as a metric space, (B) is complete.
This addresses why parametrisation by arc length is important, but not why geodesics are important.
â Arthur
Aug 28 at 17:45
add a comment |Â
up vote
2
down vote
Two examples in $mathbb R^3 ; :$ (A) take the unit sphere $x^2 + y^2 + z^2 = 1$ and delete the north and south poles; (B) the infinite cylinder $x^2 + y^2 = 1$ with $-infty <z < infty.$
Examples (A) and (B) are diffeomorphic, by a projection around the origin. (A) is incomplete as a metric space, (B) is complete.
This addresses why parametrisation by arc length is important, but not why geodesics are important.
â Arthur
Aug 28 at 17:45
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Two examples in $mathbb R^3 ; :$ (A) take the unit sphere $x^2 + y^2 + z^2 = 1$ and delete the north and south poles; (B) the infinite cylinder $x^2 + y^2 = 1$ with $-infty <z < infty.$
Examples (A) and (B) are diffeomorphic, by a projection around the origin. (A) is incomplete as a metric space, (B) is complete.
Two examples in $mathbb R^3 ; :$ (A) take the unit sphere $x^2 + y^2 + z^2 = 1$ and delete the north and south poles; (B) the infinite cylinder $x^2 + y^2 = 1$ with $-infty <z < infty.$
Examples (A) and (B) are diffeomorphic, by a projection around the origin. (A) is incomplete as a metric space, (B) is complete.
answered Aug 28 at 17:27
Will Jagy
97.8k595196
97.8k595196
This addresses why parametrisation by arc length is important, but not why geodesics are important.
â Arthur
Aug 28 at 17:45
add a comment |Â
This addresses why parametrisation by arc length is important, but not why geodesics are important.
â Arthur
Aug 28 at 17:45
This addresses why parametrisation by arc length is important, but not why geodesics are important.
â Arthur
Aug 28 at 17:45
This addresses why parametrisation by arc length is important, but not why geodesics are important.
â Arthur
Aug 28 at 17:45
add a comment |Â
up vote
2
down vote
As others have pointed out, you certainly need the metric structure to talk about completeness, because the same manifold can be complete with one metric and incomplete with another.
But let me address the more focused question that you posed in the comments: why are geodesics important?
I think probably you're misinterpreting the reason why people focus on geodesic completeness. It's not that we need geodesics to detect completeness -- it's certainly possible to give lots of characterizations of complete Riemannian manifolds that don't refer to geodesics. For example, a Riemannian manifold is (metrically) complete if and only if any of the following conditions is satisfied:
- Every Cauchy sequence converges.
- Every closed and bounded subset is compact.
- Every piecewise $C^1$ unit-speed curve defined on a proper subinterval of $mathbb R$ can be extended to a piecewise $C^1$ unit-speed curve on a strictly larger interval.
- Every piecewise $C^1$ curve that diverges to infinity (meaning that it escapes from every compact set) has infinite length.
The real significance of the Hopf-Rinow theorem is that it tells us something about geodesics, not about metric completeness. For example, as a consequence of the Hopf-Rinow theorem, we know that
- If $(M,g)$ is metrically complete, every geodesic can be extended to all of $mathbb R$ as a geodesic (not just as a piecewise $C^1$ curve).
- If there is a point $pin M$ such that every geodesic starting at $p$ can be extended to all of $mathbb R$, then $M$ is metrically complete. (Technically, this is a consequence of one step in the proof of Hopf-Rinow, not of the theorem itself.)
These results have lots of important consequences in the study of Riemannian geometry, far beyond just detecting whether a manifold is metrically complete or not.
add a comment |Â
up vote
2
down vote
As others have pointed out, you certainly need the metric structure to talk about completeness, because the same manifold can be complete with one metric and incomplete with another.
But let me address the more focused question that you posed in the comments: why are geodesics important?
I think probably you're misinterpreting the reason why people focus on geodesic completeness. It's not that we need geodesics to detect completeness -- it's certainly possible to give lots of characterizations of complete Riemannian manifolds that don't refer to geodesics. For example, a Riemannian manifold is (metrically) complete if and only if any of the following conditions is satisfied:
- Every Cauchy sequence converges.
- Every closed and bounded subset is compact.
- Every piecewise $C^1$ unit-speed curve defined on a proper subinterval of $mathbb R$ can be extended to a piecewise $C^1$ unit-speed curve on a strictly larger interval.
- Every piecewise $C^1$ curve that diverges to infinity (meaning that it escapes from every compact set) has infinite length.
The real significance of the Hopf-Rinow theorem is that it tells us something about geodesics, not about metric completeness. For example, as a consequence of the Hopf-Rinow theorem, we know that
- If $(M,g)$ is metrically complete, every geodesic can be extended to all of $mathbb R$ as a geodesic (not just as a piecewise $C^1$ curve).
- If there is a point $pin M$ such that every geodesic starting at $p$ can be extended to all of $mathbb R$, then $M$ is metrically complete. (Technically, this is a consequence of one step in the proof of Hopf-Rinow, not of the theorem itself.)
These results have lots of important consequences in the study of Riemannian geometry, far beyond just detecting whether a manifold is metrically complete or not.
add a comment |Â
up vote
2
down vote
up vote
2
down vote
As others have pointed out, you certainly need the metric structure to talk about completeness, because the same manifold can be complete with one metric and incomplete with another.
But let me address the more focused question that you posed in the comments: why are geodesics important?
I think probably you're misinterpreting the reason why people focus on geodesic completeness. It's not that we need geodesics to detect completeness -- it's certainly possible to give lots of characterizations of complete Riemannian manifolds that don't refer to geodesics. For example, a Riemannian manifold is (metrically) complete if and only if any of the following conditions is satisfied:
- Every Cauchy sequence converges.
- Every closed and bounded subset is compact.
- Every piecewise $C^1$ unit-speed curve defined on a proper subinterval of $mathbb R$ can be extended to a piecewise $C^1$ unit-speed curve on a strictly larger interval.
- Every piecewise $C^1$ curve that diverges to infinity (meaning that it escapes from every compact set) has infinite length.
The real significance of the Hopf-Rinow theorem is that it tells us something about geodesics, not about metric completeness. For example, as a consequence of the Hopf-Rinow theorem, we know that
- If $(M,g)$ is metrically complete, every geodesic can be extended to all of $mathbb R$ as a geodesic (not just as a piecewise $C^1$ curve).
- If there is a point $pin M$ such that every geodesic starting at $p$ can be extended to all of $mathbb R$, then $M$ is metrically complete. (Technically, this is a consequence of one step in the proof of Hopf-Rinow, not of the theorem itself.)
These results have lots of important consequences in the study of Riemannian geometry, far beyond just detecting whether a manifold is metrically complete or not.
As others have pointed out, you certainly need the metric structure to talk about completeness, because the same manifold can be complete with one metric and incomplete with another.
But let me address the more focused question that you posed in the comments: why are geodesics important?
I think probably you're misinterpreting the reason why people focus on geodesic completeness. It's not that we need geodesics to detect completeness -- it's certainly possible to give lots of characterizations of complete Riemannian manifolds that don't refer to geodesics. For example, a Riemannian manifold is (metrically) complete if and only if any of the following conditions is satisfied:
- Every Cauchy sequence converges.
- Every closed and bounded subset is compact.
- Every piecewise $C^1$ unit-speed curve defined on a proper subinterval of $mathbb R$ can be extended to a piecewise $C^1$ unit-speed curve on a strictly larger interval.
- Every piecewise $C^1$ curve that diverges to infinity (meaning that it escapes from every compact set) has infinite length.
The real significance of the Hopf-Rinow theorem is that it tells us something about geodesics, not about metric completeness. For example, as a consequence of the Hopf-Rinow theorem, we know that
- If $(M,g)$ is metrically complete, every geodesic can be extended to all of $mathbb R$ as a geodesic (not just as a piecewise $C^1$ curve).
- If there is a point $pin M$ such that every geodesic starting at $p$ can be extended to all of $mathbb R$, then $M$ is metrically complete. (Technically, this is a consequence of one step in the proof of Hopf-Rinow, not of the theorem itself.)
These results have lots of important consequences in the study of Riemannian geometry, far beyond just detecting whether a manifold is metrically complete or not.
answered Aug 28 at 21:41
Jack Lee
25.5k44362
25.5k44362
add a comment |Â
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2
You're missing an important part: The curves should be parametrized by arc length. The curve $f$ on the complete manifold $Bbb R$ with the standard metric, given by $f(t) = tan t$ is defined for $tin (-pi/2, pi/2)$. It cannot be extended, but doesn't have the whole real line as domain.
â Arthur
Aug 28 at 16:59
1
If you didn't need the metric structure, $Bbb R^2 cong (S^2 - x)$ could not possibly be complete.
â Mike Miller
Aug 28 at 17:26
@Arthur I get one part of your point. I mean, we can have a curve which is inextendible and not defined on the whole of $mathbbR$ because of a bad parametrization. But still, I confess I don't get intuitively why we need the metric structure, in other words, why do we need to parametrize by arc lenght to analyze incompleteness. I mean, as a definition one can just say "fine, let's work with it", but I want to understand why it is done the way it is done.
â user1620696
Aug 28 at 17:36