Is there any non-complete metric on $Bbb R^2$?












0














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







share|cite|improve this question
























  • 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
















0














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







share|cite|improve this question
























  • 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














0












0








0







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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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


















  • 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










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.






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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.






    share|cite|improve this answer


























      2














      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.






      share|cite|improve this answer
























        2












        2








        2






        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.






        share|cite|improve this answer












        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 28 '18 at 6:03









        Akiva WeinbergerAkiva Weinberger

        13.8k12167




        13.8k12167






























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