If two Riemannian manifolds are isometric as metric spaces, must they be isometric as Riemannian manifolds?
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In other words, is it possible that distinct Riemannian metrics can induce the same metric space structure on a smooth manifold?
differential-topology riemannian-geometry
asked Nov 24 at 5:09
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In other words, is it possible that distinct Riemannian metrics can induce the same metric space structure on a smooth manifold?
differential-topology riemannian-geometry
asked Nov 24 at 5:09
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I believe you're looking for the Meyers-Steenrod Theorem: If $f:Mto N$ is distance-preserving (i.e., $d_N(f(p),f(q))=d_M(p,q)$ for all $p,qin M$) and surjective, then $f$ is a Riemannian isometry. I can't think of a reference off hand, but I believe it's typically an exercise in most texts (I seem to recall it as an exercise in Lee's Riemannian text, but I could be mistaken).
– Matt
Nov 24 at 8:46
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up vote
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down vote
favorite
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In other words, is it possible that distinct Riemannian metrics can induce the same metric space structure on a smooth manifold?
differential-topology riemannian-geometry
asked Nov 24 at 5:09
SolveIt
424514
In other words, is it possible that distinct Riemannian metrics can induce the same metric space structure on a smooth manifold?
differential-topology riemannian-geometry
differential-topology riemannian-geometry
asked Nov 24 at 5:09
SolveIt
424514
asked Nov 24 at 5:09
SolveIt
424514
asked Nov 24 at 5:09
SolveIt
424514
asked Nov 24 at 5:09
SolveIt
424514
asked Nov 24 at 5:09
SolveIt
424514
424514
5
I believe you're looking for the Meyers-Steenrod Theorem: If $f:Mto N$ is distance-preserving (i.e., $d_N(f(p),f(q))=d_M(p,q)$ for all $p,qin M$) and surjective, then $f$ is a Riemannian isometry. I can't think of a reference off hand, but I believe it's typically an exercise in most texts (I seem to recall it as an exercise in Lee's Riemannian text, but I could be mistaken).
– Matt
Nov 24 at 8:46
add a comment |
5
I believe you're looking for the Meyers-Steenrod Theorem: If $f:Mto N$ is distance-preserving (i.e., $d_N(f(p),f(q))=d_M(p,q)$ for all $p,qin M$) and surjective, then $f$ is a Riemannian isometry. I can't think of a reference off hand, but I believe it's typically an exercise in most texts (I seem to recall it as an exercise in Lee's Riemannian text, but I could be mistaken).
– Matt
Nov 24 at 8:46
5
5
I believe you're looking for the Meyers-Steenrod Theorem: If $f:Mto N$ is distance-preserving (i.e., $d_N(f(p),f(q))=d_M(p,q)$ for all $p,qin M$) and surjective, then $f$ is a Riemannian isometry. I can't think of a reference off hand, but I believe it's typically an exercise in most texts (I seem to recall it as an exercise in Lee's Riemannian text, but I could be mistaken).
– Matt
Nov 24 at 8:46
I believe you're looking for the Meyers-Steenrod Theorem: If $f:Mto N$ is distance-preserving (i.e., $d_N(f(p),f(q))=d_M(p,q)$ for all $p,qin M$) and surjective, then $f$ is a Riemannian isometry. I can't think of a reference off hand, but I believe it's typically an exercise in most texts (I seem to recall it as an exercise in Lee's Riemannian text, but I could be mistaken).
– Matt
Nov 24 at 8:46
add a comment |
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I believe you're looking for the Meyers-Steenrod Theorem: If $f:Mto N$ is distance-preserving (i.e., $d_N(f(p),f(q))=d_M(p,q)$ for all $p,qin M$) and surjective, then $f$ is a Riemannian isometry. I can't think of a reference off hand, but I believe it's typically an exercise in most texts (I seem to recall it as an exercise in Lee's Riemannian text, but I could be mistaken).
– Matt
Nov 24 at 8:46