$f:Mrightarrow M$ is an isometry and $ker(Id - d_{gamma(t)}f) = mathbb{R} dot{gamma}(t)$, show that $gamma$...
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Let M be a Riemannian manifold, $gamma: I rightarrow M$ be a curve which is parameterized with constant speed and $f:Mrightarrow M$ be an isometry with $f circ gamma = gamma$. Furthermore $ker(Id - d_{gamma(t)}f) = mathbb{R} dot{gamma}(t), forall t$. Show that $gamma$ is a geodesic then.
I need some help with this. I don't know how f plays into this. We know that $d_{gamma(t)}f(adot{gamma}(t)) = adot{gamma}(t)$ for any $a in mathbb{R}$ , then for $ mathbb{R} ni c = langle dot{gamma(t)}, dot{gamma(t)} rangle = langle d_{gamma(t)}f(dot{gamma}(t)), d_{gamma(t)}f(dot{gamma}(t)) rangle$.
I don't see how any of this is supposed to be related to $nabla_{dot{gamma}}dot{gamma}$.
Why is it important that f is an isometry?
differential-geometry
asked Jan 7 at 13:26
eager2learneager2learn
1,23711430
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add a comment |
$begingroup$
Let M be a Riemannian manifold, $gamma: I rightarrow M$ be a curve which is parameterized with constant speed and $f:Mrightarrow M$ be an isometry with $f circ gamma = gamma$. Furthermore $ker(Id - d_{gamma(t)}f) = mathbb{R} dot{gamma}(t), forall t$. Show that $gamma$ is a geodesic then.
I need some help with this. I don't know how f plays into this. We know that $d_{gamma(t)}f(adot{gamma}(t)) = adot{gamma}(t)$ for any $a in mathbb{R}$ , then for $ mathbb{R} ni c = langle dot{gamma(t)}, dot{gamma(t)} rangle = langle d_{gamma(t)}f(dot{gamma}(t)), d_{gamma(t)}f(dot{gamma}(t)) rangle$.
I don't see how any of this is supposed to be related to $nabla_{dot{gamma}}dot{gamma}$.
Why is it important that f is an isometry?
differential-geometry
asked Jan 7 at 13:26
eager2learneager2learn
1,23711430
$endgroup$
add a comment |
$begingroup$
Let M be a Riemannian manifold, $gamma: I rightarrow M$ be a curve which is parameterized with constant speed and $f:Mrightarrow M$ be an isometry with $f circ gamma = gamma$. Furthermore $ker(Id - d_{gamma(t)}f) = mathbb{R} dot{gamma}(t), forall t$. Show that $gamma$ is a geodesic then.
I need some help with this. I don't know how f plays into this. We know that $d_{gamma(t)}f(adot{gamma}(t)) = adot{gamma}(t)$ for any $a in mathbb{R}$ , then for $ mathbb{R} ni c = langle dot{gamma(t)}, dot{gamma(t)} rangle = langle d_{gamma(t)}f(dot{gamma}(t)), d_{gamma(t)}f(dot{gamma}(t)) rangle$.
I don't see how any of this is supposed to be related to $nabla_{dot{gamma}}dot{gamma}$.
Why is it important that f is an isometry?
differential-geometry
asked Jan 7 at 13:26
eager2learneager2learn
1,23711430
$endgroup$
Let M be a Riemannian manifold, $gamma: I rightarrow M$ be a curve which is parameterized with constant speed and $f:Mrightarrow M$ be an isometry with $f circ gamma = gamma$. Furthermore $ker(Id - d_{gamma(t)}f) = mathbb{R} dot{gamma}(t), forall t$. Show that $gamma$ is a geodesic then.
I need some help with this. I don't know how f plays into this. We know that $d_{gamma(t)}f(adot{gamma}(t)) = adot{gamma}(t)$ for any $a in mathbb{R}$ , then for $ mathbb{R} ni c = langle dot{gamma(t)}, dot{gamma(t)} rangle = langle d_{gamma(t)}f(dot{gamma}(t)), d_{gamma(t)}f(dot{gamma}(t)) rangle$.
I don't see how any of this is supposed to be related to $nabla_{dot{gamma}}dot{gamma}$.
Why is it important that f is an isometry?
differential-geometry
differential-geometry
asked Jan 7 at 13:26
eager2learneager2learn
1,23711430
asked Jan 7 at 13:26
eager2learneager2learn
1,23711430
asked Jan 7 at 13:26
eager2learneager2learn
1,23711430
asked Jan 7 at 13:26
eager2learneager2learn
1,23711430
asked Jan 7 at 13:26
eager2learneager2learn
1,23711430
1,23711430
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add a comment |
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This is a consequence of the fact that, under your given assumptions, $gamma$ constitutes a local parameterization of an immersed submanifold of $M$ which is fixed by an isometry. But such immersed submanifolds are totally-geodesic. In this instance, the immersed submanifold is of dimension 1 and an immersed, totally-geodesic submanifold of dimension 1 is just a geodesic.
answered 2 days ago
Or EisenbergOr Eisenberg
1346
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1 Answer
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$begingroup$
This is a consequence of the fact that, under your given assumptions, $gamma$ constitutes a local parameterization of an immersed submanifold of $M$ which is fixed by an isometry. But such immersed submanifolds are totally-geodesic. In this instance, the immersed submanifold is of dimension 1 and an immersed, totally-geodesic submanifold of dimension 1 is just a geodesic.
answered 2 days ago
Or EisenbergOr Eisenberg
1346
$endgroup$
add a comment |
$begingroup$
This is a consequence of the fact that, under your given assumptions, $gamma$ constitutes a local parameterization of an immersed submanifold of $M$ which is fixed by an isometry. But such immersed submanifolds are totally-geodesic. In this instance, the immersed submanifold is of dimension 1 and an immersed, totally-geodesic submanifold of dimension 1 is just a geodesic.
answered 2 days ago
Or EisenbergOr Eisenberg
1346
$endgroup$
add a comment |
$begingroup$
This is a consequence of the fact that, under your given assumptions, $gamma$ constitutes a local parameterization of an immersed submanifold of $M$ which is fixed by an isometry. But such immersed submanifolds are totally-geodesic. In this instance, the immersed submanifold is of dimension 1 and an immersed, totally-geodesic submanifold of dimension 1 is just a geodesic.
answered 2 days ago
Or EisenbergOr Eisenberg
1346
$endgroup$
This is a consequence of the fact that, under your given assumptions, $gamma$ constitutes a local parameterization of an immersed submanifold of $M$ which is fixed by an isometry. But such immersed submanifolds are totally-geodesic. In this instance, the immersed submanifold is of dimension 1 and an immersed, totally-geodesic submanifold of dimension 1 is just a geodesic.
answered 2 days ago
Or EisenbergOr Eisenberg
1346
answered 2 days ago
Or EisenbergOr Eisenberg
1346
answered 2 days ago
Or EisenbergOr Eisenberg
1346
answered 2 days ago
Or EisenbergOr Eisenberg
1346
1346
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