Surjective differentiable map is an isometry
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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
asked Nov 19 at 5:16
Hempelicious
7510
|
show 2 more comments
up vote
1
down vote
favorite
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
asked Nov 19 at 5:16
Hempelicious
7510
1
Have you used the fact that $f$ must map a geodesic to a geodesic?
– Ted Shifrin
Nov 19 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 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 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 at 22:15
Yup, I would definitely use the Riemannian definition, myself.
– Ted Shifrin
Nov 20 at 2:03
|
show 2 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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
asked Nov 19 at 5:16
Hempelicious
7510
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
asked Nov 19 at 5:16
Hempelicious
7510
asked Nov 19 at 5:16
Hempelicious
7510
asked Nov 19 at 5:16
Hempelicious
7510
asked Nov 19 at 5:16
Hempelicious
7510
asked Nov 19 at 5:16
Hempelicious
7510
7510
1
Have you used the fact that $f$ must map a geodesic to a geodesic?
– Ted Shifrin
Nov 19 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 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 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 at 22:15
Yup, I would definitely use the Riemannian definition, myself.
– Ted Shifrin
Nov 20 at 2:03
|
show 2 more comments
1
Have you used the fact that $f$ must map a geodesic to a geodesic?
– Ted Shifrin
Nov 19 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 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 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 at 22:15
Yup, I would definitely use the Riemannian definition, myself.
– Ted Shifrin
Nov 20 at 2:03
1
1
Have you used the fact that $f$ must map a geodesic to a geodesic?
– Ted Shifrin
Nov 19 at 6:06
Have you used the fact that $f$ must map a geodesic to a geodesic?
– Ted Shifrin
Nov 19 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 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 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 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 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 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 at 22:15
Yup, I would definitely use the Riemannian definition, myself.
– Ted Shifrin
Nov 20 at 2:03
Yup, I would definitely use the Riemannian definition, myself.
– Ted Shifrin
Nov 20 at 2:03
|
show 2 more comments
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.
answered Nov 27 at 5:43
Hempelicious
7510
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
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.
answered Nov 27 at 5:43
Hempelicious
7510
add a comment |
up vote
0
down vote
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.
answered Nov 27 at 5:43
Hempelicious
7510
add a comment |
up vote
0
down vote
up vote
0
down vote
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.
answered Nov 27 at 5:43
Hempelicious
7510
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.
answered Nov 27 at 5:43
Hempelicious
7510
answered Nov 27 at 5:43
Hempelicious
7510
answered Nov 27 at 5:43
Hempelicious
7510
answered Nov 27 at 5:43
Hempelicious
7510
7510
add a comment |
add a comment |
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Have you used the fact that $f$ must map a geodesic to a geodesic?
– Ted Shifrin
Nov 19 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 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 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 at 22:15
Yup, I would definitely use the Riemannian definition, myself.
– Ted Shifrin
Nov 20 at 2:03