Surjectivity and Invertibility of the exponential map
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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
asked Dec 31 '18 at 19:02
DoggyyDoggyy
213
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add a comment |
$begingroup$
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
asked Dec 31 '18 at 19:02
DoggyyDoggyy
213
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i'm pretty sure the answer to (1) is yes. do you have Lee's book on riemannian manifolds?
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– Tim kinsella
Dec 31 '18 at 20:05
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You forgot to assume that $M$ is connected.
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– Moishe Kohan
Dec 31 '18 at 20:29
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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
add a comment |
$begingroup$
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
asked Dec 31 '18 at 19:02
DoggyyDoggyy
213
$endgroup$
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
asked Dec 31 '18 at 19:02
DoggyyDoggyy
213
asked Dec 31 '18 at 19:02
DoggyyDoggyy
213
asked Dec 31 '18 at 19:02
DoggyyDoggyy
213
asked Dec 31 '18 at 19:02
DoggyyDoggyy
213
asked Dec 31 '18 at 19:02
DoggyyDoggyy
213
213
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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
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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
add a comment |
$begingroup$
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
add a comment |
1 Answer
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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
answered Dec 31 '18 at 20:21
Tsemo AristideTsemo Aristide
60.9k11446
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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
answered Dec 31 '18 at 20:21
Tsemo AristideTsemo Aristide
60.9k11446
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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
answered Dec 31 '18 at 20:21
Tsemo AristideTsemo Aristide
60.9k11446
$endgroup$
add a comment |
$begingroup$
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
answered Dec 31 '18 at 20:21
Tsemo AristideTsemo Aristide
60.9k11446
$endgroup$
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
answered Dec 31 '18 at 20:21
Tsemo AristideTsemo Aristide
60.9k11446
answered Dec 31 '18 at 20:21
Tsemo AristideTsemo Aristide
60.9k11446
answered Dec 31 '18 at 20:21
Tsemo AristideTsemo Aristide
60.9k11446
answered Dec 31 '18 at 20:21
Tsemo AristideTsemo Aristide
60.9k11446
60.9k11446
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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