Surjective differentiable map is an isometry












1














This is exercise 1.2 in Svetlana Katok's Fuchsian Groups.



$mathbb{H}$ is the upper half plane (with the hyperbolic metric), and $f:mathbb{H}rightarrowmathbb{H}$ is a surjective $C^1$ map. I want to show $f$ is an isometry (in terms of the hyperbolic metric) if and only if it preserves the Riemannian norm on the tangent bundle of $mathbb{H}$.



One direction I can do (isometry implies norm-preserving), but the other direction is giving me trouble. I've shown that if $f$ is norm-preserving, then it also preserves the length of curves, so that
$$ d(f(z),f(w))le d(z,w) $$



But I can't seem to show there's equality here. In particular, I can't show $f$ is injective. Am I missing something special about the upper half plane?






differential-geometry riemannian-geometry hyperbolic-geometry isometry







share|cite|improve this question


















  • 1




    Have you used the fact that $f$ must map a geodesic to a geodesic?
    – Ted Shifrin
    Nov 19 '18 at 6:06










  • @TedShifrin: yes that would solve it! How can I show that though? I don't even know that $f^{-1}$ exists yet.
    – Hempelicious
    Nov 19 '18 at 16:37






  • 1




    This follows from the fact that $f$ pulls back the Riemannian metric to itself. Geodesics are invariants of the metric.
    – Ted Shifrin
    Nov 19 '18 at 18:23












  • @TedShifrin: ok, thank you, that is more advanced than I was expecting. I was thinking of geodesics as being the shortest curves between two points, but I guess you have to use the Riemannian definition (I'm not really familiar here) to solve this problem.
    – Hempelicious
    Nov 19 '18 at 22:15










  • Yup, I would definitely use the Riemannian definition, myself.
    – Ted Shifrin
    Nov 20 '18 at 2:03
















1














This is exercise 1.2 in Svetlana Katok's Fuchsian Groups.



$mathbb{H}$ is the upper half plane (with the hyperbolic metric), and $f:mathbb{H}rightarrowmathbb{H}$ is a surjective $C^1$ map. I want to show $f$ is an isometry (in terms of the hyperbolic metric) if and only if it preserves the Riemannian norm on the tangent bundle of $mathbb{H}$.



One direction I can do (isometry implies norm-preserving), but the other direction is giving me trouble. I've shown that if $f$ is norm-preserving, then it also preserves the length of curves, so that
$$ d(f(z),f(w))le d(z,w) $$



But I can't seem to show there's equality here. In particular, I can't show $f$ is injective. Am I missing something special about the upper half plane?






differential-geometry riemannian-geometry hyperbolic-geometry isometry







share|cite|improve this question


















  • 1




    Have you used the fact that $f$ must map a geodesic to a geodesic?
    – Ted Shifrin
    Nov 19 '18 at 6:06










  • @TedShifrin: yes that would solve it! How can I show that though? I don't even know that $f^{-1}$ exists yet.
    – Hempelicious
    Nov 19 '18 at 16:37






  • 1




    This follows from the fact that $f$ pulls back the Riemannian metric to itself. Geodesics are invariants of the metric.
    – Ted Shifrin
    Nov 19 '18 at 18:23












  • @TedShifrin: ok, thank you, that is more advanced than I was expecting. I was thinking of geodesics as being the shortest curves between two points, but I guess you have to use the Riemannian definition (I'm not really familiar here) to solve this problem.
    – Hempelicious
    Nov 19 '18 at 22:15










  • Yup, I would definitely use the Riemannian definition, myself.
    – Ted Shifrin
    Nov 20 '18 at 2:03














1












1








1


2





This is exercise 1.2 in Svetlana Katok's Fuchsian Groups.



$mathbb{H}$ is the upper half plane (with the hyperbolic metric), and $f:mathbb{H}rightarrowmathbb{H}$ is a surjective $C^1$ map. I want to show $f$ is an isometry (in terms of the hyperbolic metric) if and only if it preserves the Riemannian norm on the tangent bundle of $mathbb{H}$.



One direction I can do (isometry implies norm-preserving), but the other direction is giving me trouble. I've shown that if $f$ is norm-preserving, then it also preserves the length of curves, so that
$$ d(f(z),f(w))le d(z,w) $$



But I can't seem to show there's equality here. In particular, I can't show $f$ is injective. Am I missing something special about the upper half plane?






differential-geometry riemannian-geometry hyperbolic-geometry isometry







share|cite|improve this question













This is exercise 1.2 in Svetlana Katok's Fuchsian Groups.



$mathbb{H}$ is the upper half plane (with the hyperbolic metric), and $f:mathbb{H}rightarrowmathbb{H}$ is a surjective $C^1$ map. I want to show $f$ is an isometry (in terms of the hyperbolic metric) if and only if it preserves the Riemannian norm on the tangent bundle of $mathbb{H}$.



One direction I can do (isometry implies norm-preserving), but the other direction is giving me trouble. I've shown that if $f$ is norm-preserving, then it also preserves the length of curves, so that
$$ d(f(z),f(w))le d(z,w) $$



But I can't seem to show there's equality here. In particular, I can't show $f$ is injective. Am I missing something special about the upper half plane?






differential-geometry riemannian-geometry hyperbolic-geometry isometry




differential-geometry riemannian-geometry hyperbolic-geometry isometry






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 19 '18 at 5:16









Hempelicious

127110




127110








  • 1




    Have you used the fact that $f$ must map a geodesic to a geodesic?
    – Ted Shifrin
    Nov 19 '18 at 6:06










  • @TedShifrin: yes that would solve it! How can I show that though? I don't even know that $f^{-1}$ exists yet.
    – Hempelicious
    Nov 19 '18 at 16:37






  • 1




    This follows from the fact that $f$ pulls back the Riemannian metric to itself. Geodesics are invariants of the metric.
    – Ted Shifrin
    Nov 19 '18 at 18:23












  • @TedShifrin: ok, thank you, that is more advanced than I was expecting. I was thinking of geodesics as being the shortest curves between two points, but I guess you have to use the Riemannian definition (I'm not really familiar here) to solve this problem.
    – Hempelicious
    Nov 19 '18 at 22:15










  • Yup, I would definitely use the Riemannian definition, myself.
    – Ted Shifrin
    Nov 20 '18 at 2:03














  • 1




    Have you used the fact that $f$ must map a geodesic to a geodesic?
    – Ted Shifrin
    Nov 19 '18 at 6:06










  • @TedShifrin: yes that would solve it! How can I show that though? I don't even know that $f^{-1}$ exists yet.
    – Hempelicious
    Nov 19 '18 at 16:37






  • 1




    This follows from the fact that $f$ pulls back the Riemannian metric to itself. Geodesics are invariants of the metric.
    – Ted Shifrin
    Nov 19 '18 at 18:23












  • @TedShifrin: ok, thank you, that is more advanced than I was expecting. I was thinking of geodesics as being the shortest curves between two points, but I guess you have to use the Riemannian definition (I'm not really familiar here) to solve this problem.
    – Hempelicious
    Nov 19 '18 at 22:15










  • Yup, I would definitely use the Riemannian definition, myself.
    – Ted Shifrin
    Nov 20 '18 at 2:03








1




1




Have you used the fact that $f$ must map a geodesic to a geodesic?
– Ted Shifrin
Nov 19 '18 at 6:06




Have you used the fact that $f$ must map a geodesic to a geodesic?
– Ted Shifrin
Nov 19 '18 at 6:06












@TedShifrin: yes that would solve it! How can I show that though? I don't even know that $f^{-1}$ exists yet.
– Hempelicious
Nov 19 '18 at 16:37




@TedShifrin: yes that would solve it! How can I show that though? I don't even know that $f^{-1}$ exists yet.
– Hempelicious
Nov 19 '18 at 16:37




1




1




This follows from the fact that $f$ pulls back the Riemannian metric to itself. Geodesics are invariants of the metric.
– Ted Shifrin
Nov 19 '18 at 18:23






This follows from the fact that $f$ pulls back the Riemannian metric to itself. Geodesics are invariants of the metric.
– Ted Shifrin
Nov 19 '18 at 18:23














@TedShifrin: ok, thank you, that is more advanced than I was expecting. I was thinking of geodesics as being the shortest curves between two points, but I guess you have to use the Riemannian definition (I'm not really familiar here) to solve this problem.
– Hempelicious
Nov 19 '18 at 22:15




@TedShifrin: ok, thank you, that is more advanced than I was expecting. I was thinking of geodesics as being the shortest curves between two points, but I guess you have to use the Riemannian definition (I'm not really familiar here) to solve this problem.
– Hempelicious
Nov 19 '18 at 22:15












Yup, I would definitely use the Riemannian definition, myself.
– Ted Shifrin
Nov 20 '18 at 2:03




Yup, I would definitely use the Riemannian definition, myself.
– Ted Shifrin
Nov 20 '18 at 2:03










1 Answer
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I think this works, based on comments by @TedShifrin.



We can characterize the geodesics as those curves $gamma$ with $nabla_{dot{gamma}_t}dot{gamma}_tequiv0$. Since $f$ preserves the Riemannian norm, it preserves the Riemannian metric (via the parallelogram rule), and so it preserves the covariant derivative. That is,
$$ nabla_{df(dot{gamma}_t)}df(dot{gamma}_t)equiv0 $$
So $fcircgamma$ is a geodesic.



I still don't understand why $f$ preserves the covariant derivative, but it seems to be true, and I've seen some messy formulas involving the Riemannian metric, claiming to prove it. Conceptually, I'm still unclear though.






share|cite|improve this answer





















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    I think this works, based on comments by @TedShifrin.



    We can characterize the geodesics as those curves $gamma$ with $nabla_{dot{gamma}_t}dot{gamma}_tequiv0$. Since $f$ preserves the Riemannian norm, it preserves the Riemannian metric (via the parallelogram rule), and so it preserves the covariant derivative. That is,
    $$ nabla_{df(dot{gamma}_t)}df(dot{gamma}_t)equiv0 $$
    So $fcircgamma$ is a geodesic.



    I still don't understand why $f$ preserves the covariant derivative, but it seems to be true, and I've seen some messy formulas involving the Riemannian metric, claiming to prove it. Conceptually, I'm still unclear though.






    share|cite|improve this answer


























      0














      I think this works, based on comments by @TedShifrin.



      We can characterize the geodesics as those curves $gamma$ with $nabla_{dot{gamma}_t}dot{gamma}_tequiv0$. Since $f$ preserves the Riemannian norm, it preserves the Riemannian metric (via the parallelogram rule), and so it preserves the covariant derivative. That is,
      $$ nabla_{df(dot{gamma}_t)}df(dot{gamma}_t)equiv0 $$
      So $fcircgamma$ is a geodesic.



      I still don't understand why $f$ preserves the covariant derivative, but it seems to be true, and I've seen some messy formulas involving the Riemannian metric, claiming to prove it. Conceptually, I'm still unclear though.






      share|cite|improve this answer
























        0












        0








        0






        I think this works, based on comments by @TedShifrin.



        We can characterize the geodesics as those curves $gamma$ with $nabla_{dot{gamma}_t}dot{gamma}_tequiv0$. Since $f$ preserves the Riemannian norm, it preserves the Riemannian metric (via the parallelogram rule), and so it preserves the covariant derivative. That is,
        $$ nabla_{df(dot{gamma}_t)}df(dot{gamma}_t)equiv0 $$
        So $fcircgamma$ is a geodesic.



        I still don't understand why $f$ preserves the covariant derivative, but it seems to be true, and I've seen some messy formulas involving the Riemannian metric, claiming to prove it. Conceptually, I'm still unclear though.






        share|cite|improve this answer












        I think this works, based on comments by @TedShifrin.



        We can characterize the geodesics as those curves $gamma$ with $nabla_{dot{gamma}_t}dot{gamma}_tequiv0$. Since $f$ preserves the Riemannian norm, it preserves the Riemannian metric (via the parallelogram rule), and so it preserves the covariant derivative. That is,
        $$ nabla_{df(dot{gamma}_t)}df(dot{gamma}_t)equiv0 $$
        So $fcircgamma$ is a geodesic.



        I still don't understand why $f$ preserves the covariant derivative, but it seems to be true, and I've seen some messy formulas involving the Riemannian metric, claiming to prove it. Conceptually, I'm still unclear though.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 27 '18 at 5:43









        Hempelicious

        127110




        127110






























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