The hypothesis $Kleq 0$ in the proof of Hadamard's Theorem












3














In chapter 7 from do Carmo's Riemannian Geometry, right after proving Hadamard's theorem, there is the following remark:





When he says "poles can exist in non-compact manifolds which have positive sectional curvature", I believe he is trying to justify the hypothesis "$Kleq 0$".



What I don't understand is: why did he have to talk about "non-compact manifolds"?



I thought he would say "poles can exist in complete manifolds with positive curvature", which makes perfect sense and justifies the hypothesis "$Kleq 0$".



I know I'm not crazy because exercise 13 (where $M$ is the parabolloid $z=x^2+y^2$) is precisely an example of a complete manifold with positive curvature having a pole (namely, the origin).



So what is this "non-compact" business?






riemannian-geometry smooth-manifolds curvature







share|cite|improve this question




















  • 1




    If $K>0$ and $M$ is compact then there are no poles. Of course, you can still have $K=0$, $M$ compact and no poles.
    – Moishe Cohen
    Jan 4 at 23:32










  • How can we prove that $K>0, M$ compact implies no poles?
    – rmdmc89
    2 days ago






  • 1




    Rauch Comparison Theorem.
    – Moishe Cohen
    2 days ago
















3














In chapter 7 from do Carmo's Riemannian Geometry, right after proving Hadamard's theorem, there is the following remark:





When he says "poles can exist in non-compact manifolds which have positive sectional curvature", I believe he is trying to justify the hypothesis "$Kleq 0$".



What I don't understand is: why did he have to talk about "non-compact manifolds"?



I thought he would say "poles can exist in complete manifolds with positive curvature", which makes perfect sense and justifies the hypothesis "$Kleq 0$".



I know I'm not crazy because exercise 13 (where $M$ is the parabolloid $z=x^2+y^2$) is precisely an example of a complete manifold with positive curvature having a pole (namely, the origin).



So what is this "non-compact" business?






riemannian-geometry smooth-manifolds curvature







share|cite|improve this question




















  • 1




    If $K>0$ and $M$ is compact then there are no poles. Of course, you can still have $K=0$, $M$ compact and no poles.
    – Moishe Cohen
    Jan 4 at 23:32










  • How can we prove that $K>0, M$ compact implies no poles?
    – rmdmc89
    2 days ago






  • 1




    Rauch Comparison Theorem.
    – Moishe Cohen
    2 days ago














3












3








3


3





In chapter 7 from do Carmo's Riemannian Geometry, right after proving Hadamard's theorem, there is the following remark:





When he says "poles can exist in non-compact manifolds which have positive sectional curvature", I believe he is trying to justify the hypothesis "$Kleq 0$".



What I don't understand is: why did he have to talk about "non-compact manifolds"?



I thought he would say "poles can exist in complete manifolds with positive curvature", which makes perfect sense and justifies the hypothesis "$Kleq 0$".



I know I'm not crazy because exercise 13 (where $M$ is the parabolloid $z=x^2+y^2$) is precisely an example of a complete manifold with positive curvature having a pole (namely, the origin).



So what is this "non-compact" business?






riemannian-geometry smooth-manifolds curvature







share|cite|improve this question















In chapter 7 from do Carmo's Riemannian Geometry, right after proving Hadamard's theorem, there is the following remark:





When he says "poles can exist in non-compact manifolds which have positive sectional curvature", I believe he is trying to justify the hypothesis "$Kleq 0$".



What I don't understand is: why did he have to talk about "non-compact manifolds"?



I thought he would say "poles can exist in complete manifolds with positive curvature", which makes perfect sense and justifies the hypothesis "$Kleq 0$".



I know I'm not crazy because exercise 13 (where $M$ is the parabolloid $z=x^2+y^2$) is precisely an example of a complete manifold with positive curvature having a pole (namely, the origin).



So what is this "non-compact" business?






riemannian-geometry smooth-manifolds curvature




riemannian-geometry smooth-manifolds curvature






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 12:53







rmdmc89

















asked Jan 4 at 12:45









rmdmc89rmdmc89

2,0931922




2,0931922








  • 1




    If $K>0$ and $M$ is compact then there are no poles. Of course, you can still have $K=0$, $M$ compact and no poles.
    – Moishe Cohen
    Jan 4 at 23:32










  • How can we prove that $K>0, M$ compact implies no poles?
    – rmdmc89
    2 days ago






  • 1




    Rauch Comparison Theorem.
    – Moishe Cohen
    2 days ago














  • 1




    If $K>0$ and $M$ is compact then there are no poles. Of course, you can still have $K=0$, $M$ compact and no poles.
    – Moishe Cohen
    Jan 4 at 23:32










  • How can we prove that $K>0, M$ compact implies no poles?
    – rmdmc89
    2 days ago






  • 1




    Rauch Comparison Theorem.
    – Moishe Cohen
    2 days ago








1




1




If $K>0$ and $M$ is compact then there are no poles. Of course, you can still have $K=0$, $M$ compact and no poles.
– Moishe Cohen
Jan 4 at 23:32




If $K>0$ and $M$ is compact then there are no poles. Of course, you can still have $K=0$, $M$ compact and no poles.
– Moishe Cohen
Jan 4 at 23:32












How can we prove that $K>0, M$ compact implies no poles?
– rmdmc89
2 days ago




How can we prove that $K>0, M$ compact implies no poles?
– rmdmc89
2 days ago




1




1




Rauch Comparison Theorem.
– Moishe Cohen
2 days ago




Rauch Comparison Theorem.
– Moishe Cohen
2 days ago










1 Answer
1






active

oldest

votes


















3














The reason Do Carmo restricts that comment to noncompact manifolds is because of some results that he will prove later, which show among other things that a manifold with sectional curvatures bounded below by a positive constant must have a conjugate point along every geodesic. In particular, as @MoisheCohen commented, if $M$ is compact and has strictly positive sectional curvature, then the curvature has a positive lower bound and thus $M$ has no poles.



But I don't understand your remark that the existence of poles in the positive curvature case would "justify the hypothesis $Kle 0$." My interpretation of that comment is sort of the opposite -- Hadamard's theorem shows that for complete manifolds, $Kle 0 $ is sufficient to guarantee the existence of poles, so one might wonder whether it's also necessary. Do Carmo's comment and the paraboloid example are meant to demonstrate that $Kle 0$ is far from necessary, in the sense that it's even possible to find a complete manifold with $K>0$ everywhere and that nonetheless has a pole.



It's worth noting, by the way, that Do Carmo's definition of a pole is nonstandard. Most Riemannian geometers define a pole to be a point $pin M$ such that the restricted exponential map $exp_pcolon T_pM to M$ is a global diffeomorphism. That implies that $p$ has no conjugate points, but it is much stronger. Of course, using the more usual definition of a pole, it's immediate that a manifold with a pole has to be diffeomorphic to $mathbb R^n$, so it can never be compact.






share|cite|improve this answer





















  • You're right about my comment, @ Jack Lee. I've confused the terms, but everything makes sense now. Thank you
    – rmdmc89
    2 days ago













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The reason Do Carmo restricts that comment to noncompact manifolds is because of some results that he will prove later, which show among other things that a manifold with sectional curvatures bounded below by a positive constant must have a conjugate point along every geodesic. In particular, as @MoisheCohen commented, if $M$ is compact and has strictly positive sectional curvature, then the curvature has a positive lower bound and thus $M$ has no poles.



But I don't understand your remark that the existence of poles in the positive curvature case would "justify the hypothesis $Kle 0$." My interpretation of that comment is sort of the opposite -- Hadamard's theorem shows that for complete manifolds, $Kle 0 $ is sufficient to guarantee the existence of poles, so one might wonder whether it's also necessary. Do Carmo's comment and the paraboloid example are meant to demonstrate that $Kle 0$ is far from necessary, in the sense that it's even possible to find a complete manifold with $K>0$ everywhere and that nonetheless has a pole.



It's worth noting, by the way, that Do Carmo's definition of a pole is nonstandard. Most Riemannian geometers define a pole to be a point $pin M$ such that the restricted exponential map $exp_pcolon T_pM to M$ is a global diffeomorphism. That implies that $p$ has no conjugate points, but it is much stronger. Of course, using the more usual definition of a pole, it's immediate that a manifold with a pole has to be diffeomorphic to $mathbb R^n$, so it can never be compact.






share|cite|improve this answer





















  • You're right about my comment, @ Jack Lee. I've confused the terms, but everything makes sense now. Thank you
    – rmdmc89
    2 days ago


















3














The reason Do Carmo restricts that comment to noncompact manifolds is because of some results that he will prove later, which show among other things that a manifold with sectional curvatures bounded below by a positive constant must have a conjugate point along every geodesic. In particular, as @MoisheCohen commented, if $M$ is compact and has strictly positive sectional curvature, then the curvature has a positive lower bound and thus $M$ has no poles.



But I don't understand your remark that the existence of poles in the positive curvature case would "justify the hypothesis $Kle 0$." My interpretation of that comment is sort of the opposite -- Hadamard's theorem shows that for complete manifolds, $Kle 0 $ is sufficient to guarantee the existence of poles, so one might wonder whether it's also necessary. Do Carmo's comment and the paraboloid example are meant to demonstrate that $Kle 0$ is far from necessary, in the sense that it's even possible to find a complete manifold with $K>0$ everywhere and that nonetheless has a pole.



It's worth noting, by the way, that Do Carmo's definition of a pole is nonstandard. Most Riemannian geometers define a pole to be a point $pin M$ such that the restricted exponential map $exp_pcolon T_pM to M$ is a global diffeomorphism. That implies that $p$ has no conjugate points, but it is much stronger. Of course, using the more usual definition of a pole, it's immediate that a manifold with a pole has to be diffeomorphic to $mathbb R^n$, so it can never be compact.






share|cite|improve this answer





















  • You're right about my comment, @ Jack Lee. I've confused the terms, but everything makes sense now. Thank you
    – rmdmc89
    2 days ago
















3












3








3






The reason Do Carmo restricts that comment to noncompact manifolds is because of some results that he will prove later, which show among other things that a manifold with sectional curvatures bounded below by a positive constant must have a conjugate point along every geodesic. In particular, as @MoisheCohen commented, if $M$ is compact and has strictly positive sectional curvature, then the curvature has a positive lower bound and thus $M$ has no poles.



But I don't understand your remark that the existence of poles in the positive curvature case would "justify the hypothesis $Kle 0$." My interpretation of that comment is sort of the opposite -- Hadamard's theorem shows that for complete manifolds, $Kle 0 $ is sufficient to guarantee the existence of poles, so one might wonder whether it's also necessary. Do Carmo's comment and the paraboloid example are meant to demonstrate that $Kle 0$ is far from necessary, in the sense that it's even possible to find a complete manifold with $K>0$ everywhere and that nonetheless has a pole.



It's worth noting, by the way, that Do Carmo's definition of a pole is nonstandard. Most Riemannian geometers define a pole to be a point $pin M$ such that the restricted exponential map $exp_pcolon T_pM to M$ is a global diffeomorphism. That implies that $p$ has no conjugate points, but it is much stronger. Of course, using the more usual definition of a pole, it's immediate that a manifold with a pole has to be diffeomorphic to $mathbb R^n$, so it can never be compact.






share|cite|improve this answer












The reason Do Carmo restricts that comment to noncompact manifolds is because of some results that he will prove later, which show among other things that a manifold with sectional curvatures bounded below by a positive constant must have a conjugate point along every geodesic. In particular, as @MoisheCohen commented, if $M$ is compact and has strictly positive sectional curvature, then the curvature has a positive lower bound and thus $M$ has no poles.



But I don't understand your remark that the existence of poles in the positive curvature case would "justify the hypothesis $Kle 0$." My interpretation of that comment is sort of the opposite -- Hadamard's theorem shows that for complete manifolds, $Kle 0 $ is sufficient to guarantee the existence of poles, so one might wonder whether it's also necessary. Do Carmo's comment and the paraboloid example are meant to demonstrate that $Kle 0$ is far from necessary, in the sense that it's even possible to find a complete manifold with $K>0$ everywhere and that nonetheless has a pole.



It's worth noting, by the way, that Do Carmo's definition of a pole is nonstandard. Most Riemannian geometers define a pole to be a point $pin M$ such that the restricted exponential map $exp_pcolon T_pM to M$ is a global diffeomorphism. That implies that $p$ has no conjugate points, but it is much stronger. Of course, using the more usual definition of a pole, it's immediate that a manifold with a pole has to be diffeomorphic to $mathbb R^n$, so it can never be compact.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 5 at 21:07









Jack LeeJack Lee

27k54565




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  • You're right about my comment, @ Jack Lee. I've confused the terms, but everything makes sense now. Thank you
    – rmdmc89
    2 days ago




















  • You're right about my comment, @ Jack Lee. I've confused the terms, but everything makes sense now. Thank you
    – rmdmc89
    2 days ago


















You're right about my comment, @ Jack Lee. I've confused the terms, but everything makes sense now. Thank you
– rmdmc89
2 days ago






You're right about my comment, @ Jack Lee. I've confused the terms, but everything makes sense now. Thank you
– rmdmc89
2 days ago




















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