Is there any non-complete metric on $Bbb R^2$?
I've read several books with this content:
Let $g = dr^2 + f^2(r)dtheta^2$ be a smooth metric on $Bbb R^2$ expressed in
polar coordinates. This metric is complete and the volume can be finite even though this manifold is not-compact.
I thought that $Bbb R^2$ always have infinite volume.
Question: Why authors emphasize that this metric is complete while we know that $Bbb R^2$ is complete topological space. Is there any non-complete metric on $Bbb R^2$?
differential-geometry riemannian-geometry
|
show 1 more comment
I've read several books with this content:
Let $g = dr^2 + f^2(r)dtheta^2$ be a smooth metric on $Bbb R^2$ expressed in
polar coordinates. This metric is complete and the volume can be finite even though this manifold is not-compact.
I thought that $Bbb R^2$ always have infinite volume.
Question: Why authors emphasize that this metric is complete while we know that $Bbb R^2$ is complete topological space. Is there any non-complete metric on $Bbb R^2$?
differential-geometry riemannian-geometry
Does the author mean Cauchy complete or geodesic complete?
– edm
Nov 28 '18 at 5:36
is there difference between them? I think geo. complete.
– C.F.G
Nov 28 '18 at 5:39
1
Completeness is not a property of topological spaces so I don't know what you mean by "$mathbb{R}^2$ is complete topological space."
– Qiaochu Yuan
Nov 28 '18 at 5:47
1
@edm According to Hopf–Rinow, those are equivalent.
– Akiva Weinberger
Nov 28 '18 at 5:59
@QiaochuYuan, Thank you for your remark.
– C.F.G
Nov 28 '18 at 6:07
|
show 1 more comment
I've read several books with this content:
Let $g = dr^2 + f^2(r)dtheta^2$ be a smooth metric on $Bbb R^2$ expressed in
polar coordinates. This metric is complete and the volume can be finite even though this manifold is not-compact.
I thought that $Bbb R^2$ always have infinite volume.
Question: Why authors emphasize that this metric is complete while we know that $Bbb R^2$ is complete topological space. Is there any non-complete metric on $Bbb R^2$?
differential-geometry riemannian-geometry
I've read several books with this content:
Let $g = dr^2 + f^2(r)dtheta^2$ be a smooth metric on $Bbb R^2$ expressed in
polar coordinates. This metric is complete and the volume can be finite even though this manifold is not-compact.
I thought that $Bbb R^2$ always have infinite volume.
Question: Why authors emphasize that this metric is complete while we know that $Bbb R^2$ is complete topological space. Is there any non-complete metric on $Bbb R^2$?
differential-geometry riemannian-geometry
differential-geometry riemannian-geometry
edited Nov 28 '18 at 5:47
C.F.G
asked Nov 28 '18 at 5:30
C.F.GC.F.G
1,4281821
1,4281821
Does the author mean Cauchy complete or geodesic complete?
– edm
Nov 28 '18 at 5:36
is there difference between them? I think geo. complete.
– C.F.G
Nov 28 '18 at 5:39
1
Completeness is not a property of topological spaces so I don't know what you mean by "$mathbb{R}^2$ is complete topological space."
– Qiaochu Yuan
Nov 28 '18 at 5:47
1
@edm According to Hopf–Rinow, those are equivalent.
– Akiva Weinberger
Nov 28 '18 at 5:59
@QiaochuYuan, Thank you for your remark.
– C.F.G
Nov 28 '18 at 6:07
|
show 1 more comment
Does the author mean Cauchy complete or geodesic complete?
– edm
Nov 28 '18 at 5:36
is there difference between them? I think geo. complete.
– C.F.G
Nov 28 '18 at 5:39
1
Completeness is not a property of topological spaces so I don't know what you mean by "$mathbb{R}^2$ is complete topological space."
– Qiaochu Yuan
Nov 28 '18 at 5:47
1
@edm According to Hopf–Rinow, those are equivalent.
– Akiva Weinberger
Nov 28 '18 at 5:59
@QiaochuYuan, Thank you for your remark.
– C.F.G
Nov 28 '18 at 6:07
Does the author mean Cauchy complete or geodesic complete?
– edm
Nov 28 '18 at 5:36
Does the author mean Cauchy complete or geodesic complete?
– edm
Nov 28 '18 at 5:36
is there difference between them? I think geo. complete.
– C.F.G
Nov 28 '18 at 5:39
is there difference between them? I think geo. complete.
– C.F.G
Nov 28 '18 at 5:39
1
1
Completeness is not a property of topological spaces so I don't know what you mean by "$mathbb{R}^2$ is complete topological space."
– Qiaochu Yuan
Nov 28 '18 at 5:47
Completeness is not a property of topological spaces so I don't know what you mean by "$mathbb{R}^2$ is complete topological space."
– Qiaochu Yuan
Nov 28 '18 at 5:47
1
1
@edm According to Hopf–Rinow, those are equivalent.
– Akiva Weinberger
Nov 28 '18 at 5:59
@edm According to Hopf–Rinow, those are equivalent.
– Akiva Weinberger
Nov 28 '18 at 5:59
@QiaochuYuan, Thank you for your remark.
– C.F.G
Nov 28 '18 at 6:07
@QiaochuYuan, Thank you for your remark.
– C.F.G
Nov 28 '18 at 6:07
|
show 1 more comment
1 Answer
1
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There exists a diffeomorphism from $Bbb R^2$ to the open unit disc. One such function, given in polar form, is $(r,theta)mapsto(tanh r,theta)$. You can replace $tanh$ with your favorite sigmoid function. Give the open unit disc the standard metric, and then give $Bbb R^2$ the metric it inherits from the diffeomorphism. This metric is not complete.
This doesn't contradict the author's statement, because metrics of the form $dr^2+f^2(r )dtheta^2$ do not change the lengths of rays from the origin, and so the metric I described above is not of this form.
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1 Answer
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1 Answer
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There exists a diffeomorphism from $Bbb R^2$ to the open unit disc. One such function, given in polar form, is $(r,theta)mapsto(tanh r,theta)$. You can replace $tanh$ with your favorite sigmoid function. Give the open unit disc the standard metric, and then give $Bbb R^2$ the metric it inherits from the diffeomorphism. This metric is not complete.
This doesn't contradict the author's statement, because metrics of the form $dr^2+f^2(r )dtheta^2$ do not change the lengths of rays from the origin, and so the metric I described above is not of this form.
add a comment |
There exists a diffeomorphism from $Bbb R^2$ to the open unit disc. One such function, given in polar form, is $(r,theta)mapsto(tanh r,theta)$. You can replace $tanh$ with your favorite sigmoid function. Give the open unit disc the standard metric, and then give $Bbb R^2$ the metric it inherits from the diffeomorphism. This metric is not complete.
This doesn't contradict the author's statement, because metrics of the form $dr^2+f^2(r )dtheta^2$ do not change the lengths of rays from the origin, and so the metric I described above is not of this form.
add a comment |
There exists a diffeomorphism from $Bbb R^2$ to the open unit disc. One such function, given in polar form, is $(r,theta)mapsto(tanh r,theta)$. You can replace $tanh$ with your favorite sigmoid function. Give the open unit disc the standard metric, and then give $Bbb R^2$ the metric it inherits from the diffeomorphism. This metric is not complete.
This doesn't contradict the author's statement, because metrics of the form $dr^2+f^2(r )dtheta^2$ do not change the lengths of rays from the origin, and so the metric I described above is not of this form.
There exists a diffeomorphism from $Bbb R^2$ to the open unit disc. One such function, given in polar form, is $(r,theta)mapsto(tanh r,theta)$. You can replace $tanh$ with your favorite sigmoid function. Give the open unit disc the standard metric, and then give $Bbb R^2$ the metric it inherits from the diffeomorphism. This metric is not complete.
This doesn't contradict the author's statement, because metrics of the form $dr^2+f^2(r )dtheta^2$ do not change the lengths of rays from the origin, and so the metric I described above is not of this form.
answered Nov 28 '18 at 6:03
Akiva WeinbergerAkiva Weinberger
13.8k12167
13.8k12167
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Does the author mean Cauchy complete or geodesic complete?
– edm
Nov 28 '18 at 5:36
is there difference between them? I think geo. complete.
– C.F.G
Nov 28 '18 at 5:39
1
Completeness is not a property of topological spaces so I don't know what you mean by "$mathbb{R}^2$ is complete topological space."
– Qiaochu Yuan
Nov 28 '18 at 5:47
1
@edm According to Hopf–Rinow, those are equivalent.
– Akiva Weinberger
Nov 28 '18 at 5:59
@QiaochuYuan, Thank you for your remark.
– C.F.G
Nov 28 '18 at 6:07