Surjectivity and Invertibility of the exponential map












3












$begingroup$


Let $(mathcal{M},g)$ be a geodesically complete Riemannian manifold. So $forall x in mathcal{M}$, the exponential map $exp_x$ is defined on all of $T_x mathcal{M}$. I have two related questions:



(1) Is $exp_x : T_x mathcal{M} rightarrow mathcal{M}$ surjective for all $x in mathcal{M}$? If not, under what conditions on $mathcal{M}$ is $exp_x$ surjective (e.g., assuming $mathcal{M}$ is compact)?



(2) If $exp_x : T_x mathcal{M} rightarrow mathcal{M}$ is surjective, is there a subset of $S_x subset T_x mathcal{M}$ such that $exp_x$ is invertible on $S_x$ and the image of $S_x$ is almost all of $mathcal{M}$? I.e., $exp_x(S_x) = mathcal{M} setminus U_x$ for some small (e.g., 'measure zero') set $U_x$. For example, consider the unit 2-sphere: $mathcal{M} = S^2 subset mathbb{R}^3$. Then $exp_x : B_{pi} rightarrow S^2 setminus {-x}$ is invertible.






differential-geometry riemannian-geometry







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$endgroup$












  • $begingroup$
    i'm pretty sure the answer to (1) is yes. do you have Lee's book on riemannian manifolds?
    $endgroup$
    – Tim kinsella
    Dec 31 '18 at 20:05












  • $begingroup$
    You forgot to assume that $M$ is connected.
    $endgroup$
    – Moishe Kohan
    Dec 31 '18 at 20:29










  • $begingroup$
    I think at least (1) is proved in Milnors Morse Theory. You might get (2) using Sard's theorem and the fact that $exp$ fails to be of maximal rank at conjugate points; my memory is rusty though.
    $endgroup$
    – Robert Lewis
    Dec 31 '18 at 21:04


















3












$begingroup$


Let $(mathcal{M},g)$ be a geodesically complete Riemannian manifold. So $forall x in mathcal{M}$, the exponential map $exp_x$ is defined on all of $T_x mathcal{M}$. I have two related questions:



(1) Is $exp_x : T_x mathcal{M} rightarrow mathcal{M}$ surjective for all $x in mathcal{M}$? If not, under what conditions on $mathcal{M}$ is $exp_x$ surjective (e.g., assuming $mathcal{M}$ is compact)?



(2) If $exp_x : T_x mathcal{M} rightarrow mathcal{M}$ is surjective, is there a subset of $S_x subset T_x mathcal{M}$ such that $exp_x$ is invertible on $S_x$ and the image of $S_x$ is almost all of $mathcal{M}$? I.e., $exp_x(S_x) = mathcal{M} setminus U_x$ for some small (e.g., 'measure zero') set $U_x$. For example, consider the unit 2-sphere: $mathcal{M} = S^2 subset mathbb{R}^3$. Then $exp_x : B_{pi} rightarrow S^2 setminus {-x}$ is invertible.






differential-geometry riemannian-geometry







share|cite|improve this question









$endgroup$












  • $begingroup$
    i'm pretty sure the answer to (1) is yes. do you have Lee's book on riemannian manifolds?
    $endgroup$
    – Tim kinsella
    Dec 31 '18 at 20:05












  • $begingroup$
    You forgot to assume that $M$ is connected.
    $endgroup$
    – Moishe Kohan
    Dec 31 '18 at 20:29










  • $begingroup$
    I think at least (1) is proved in Milnors Morse Theory. You might get (2) using Sard's theorem and the fact that $exp$ fails to be of maximal rank at conjugate points; my memory is rusty though.
    $endgroup$
    – Robert Lewis
    Dec 31 '18 at 21:04
















3












3








3


1



$begingroup$


Let $(mathcal{M},g)$ be a geodesically complete Riemannian manifold. So $forall x in mathcal{M}$, the exponential map $exp_x$ is defined on all of $T_x mathcal{M}$. I have two related questions:



(1) Is $exp_x : T_x mathcal{M} rightarrow mathcal{M}$ surjective for all $x in mathcal{M}$? If not, under what conditions on $mathcal{M}$ is $exp_x$ surjective (e.g., assuming $mathcal{M}$ is compact)?



(2) If $exp_x : T_x mathcal{M} rightarrow mathcal{M}$ is surjective, is there a subset of $S_x subset T_x mathcal{M}$ such that $exp_x$ is invertible on $S_x$ and the image of $S_x$ is almost all of $mathcal{M}$? I.e., $exp_x(S_x) = mathcal{M} setminus U_x$ for some small (e.g., 'measure zero') set $U_x$. For example, consider the unit 2-sphere: $mathcal{M} = S^2 subset mathbb{R}^3$. Then $exp_x : B_{pi} rightarrow S^2 setminus {-x}$ is invertible.






differential-geometry riemannian-geometry







share|cite|improve this question









$endgroup$




Let $(mathcal{M},g)$ be a geodesically complete Riemannian manifold. So $forall x in mathcal{M}$, the exponential map $exp_x$ is defined on all of $T_x mathcal{M}$. I have two related questions:



(1) Is $exp_x : T_x mathcal{M} rightarrow mathcal{M}$ surjective for all $x in mathcal{M}$? If not, under what conditions on $mathcal{M}$ is $exp_x$ surjective (e.g., assuming $mathcal{M}$ is compact)?



(2) If $exp_x : T_x mathcal{M} rightarrow mathcal{M}$ is surjective, is there a subset of $S_x subset T_x mathcal{M}$ such that $exp_x$ is invertible on $S_x$ and the image of $S_x$ is almost all of $mathcal{M}$? I.e., $exp_x(S_x) = mathcal{M} setminus U_x$ for some small (e.g., 'measure zero') set $U_x$. For example, consider the unit 2-sphere: $mathcal{M} = S^2 subset mathbb{R}^3$. Then $exp_x : B_{pi} rightarrow S^2 setminus {-x}$ is invertible.






differential-geometry riemannian-geometry




differential-geometry riemannian-geometry






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asked Dec 31 '18 at 19:02









DoggyyDoggyy

213




213












  • $begingroup$
    i'm pretty sure the answer to (1) is yes. do you have Lee's book on riemannian manifolds?
    $endgroup$
    – Tim kinsella
    Dec 31 '18 at 20:05












  • $begingroup$
    You forgot to assume that $M$ is connected.
    $endgroup$
    – Moishe Kohan
    Dec 31 '18 at 20:29










  • $begingroup$
    I think at least (1) is proved in Milnors Morse Theory. You might get (2) using Sard's theorem and the fact that $exp$ fails to be of maximal rank at conjugate points; my memory is rusty though.
    $endgroup$
    – Robert Lewis
    Dec 31 '18 at 21:04




















  • $begingroup$
    i'm pretty sure the answer to (1) is yes. do you have Lee's book on riemannian manifolds?
    $endgroup$
    – Tim kinsella
    Dec 31 '18 at 20:05












  • $begingroup$
    You forgot to assume that $M$ is connected.
    $endgroup$
    – Moishe Kohan
    Dec 31 '18 at 20:29










  • $begingroup$
    I think at least (1) is proved in Milnors Morse Theory. You might get (2) using Sard's theorem and the fact that $exp$ fails to be of maximal rank at conjugate points; my memory is rusty though.
    $endgroup$
    – Robert Lewis
    Dec 31 '18 at 21:04


















$begingroup$
i'm pretty sure the answer to (1) is yes. do you have Lee's book on riemannian manifolds?
$endgroup$
– Tim kinsella
Dec 31 '18 at 20:05






$begingroup$
i'm pretty sure the answer to (1) is yes. do you have Lee's book on riemannian manifolds?
$endgroup$
– Tim kinsella
Dec 31 '18 at 20:05














$begingroup$
You forgot to assume that $M$ is connected.
$endgroup$
– Moishe Kohan
Dec 31 '18 at 20:29




$begingroup$
You forgot to assume that $M$ is connected.
$endgroup$
– Moishe Kohan
Dec 31 '18 at 20:29












$begingroup$
I think at least (1) is proved in Milnors Morse Theory. You might get (2) using Sard's theorem and the fact that $exp$ fails to be of maximal rank at conjugate points; my memory is rusty though.
$endgroup$
– Robert Lewis
Dec 31 '18 at 21:04






$begingroup$
I think at least (1) is proved in Milnors Morse Theory. You might get (2) using Sard's theorem and the fact that $exp$ fails to be of maximal rank at conjugate points; my memory is rusty though.
$endgroup$
– Robert Lewis
Dec 31 '18 at 21:04












1 Answer
1






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2












$begingroup$

1 is is a corrolary of the Hopf Rinow theorem:



See the third condition in the statement of the theorem.



https://en.wikipedia.org/wiki/Hopf%E2%80%93Rinow_theorem






share|cite|improve this answer









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    1 Answer
    1






    active

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    active

    oldest

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    active

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    2












    $begingroup$

    1 is is a corrolary of the Hopf Rinow theorem:



    See the third condition in the statement of the theorem.



    https://en.wikipedia.org/wiki/Hopf%E2%80%93Rinow_theorem






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      1 is is a corrolary of the Hopf Rinow theorem:



      See the third condition in the statement of the theorem.



      https://en.wikipedia.org/wiki/Hopf%E2%80%93Rinow_theorem






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        1 is is a corrolary of the Hopf Rinow theorem:



        See the third condition in the statement of the theorem.



        https://en.wikipedia.org/wiki/Hopf%E2%80%93Rinow_theorem






        share|cite|improve this answer









        $endgroup$



        1 is is a corrolary of the Hopf Rinow theorem:



        See the third condition in the statement of the theorem.



        https://en.wikipedia.org/wiki/Hopf%E2%80%93Rinow_theorem







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 31 '18 at 20:21









        Tsemo AristideTsemo Aristide

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