Whether preserving inner products follows from preserving intrinsic distances
I know that if a map F between two surfaces preserves inner products of tangent vectors and is 1-1 and onto, then it must preserve intrinsic distances, but I’m not sure whether the inverse is true. Namely, is it true that any map F between two surfaces that is 1-1, onto, and preserves intrinsic distances will automatically preserve inner products of tangent vectors?
Thanks a lot!
geometry differential-geometry surfaces
asked Nov 25 at 7:05
David Petey Gao
454
add a comment |
I know that if a map F between two surfaces preserves inner products of tangent vectors and is 1-1 and onto, then it must preserve intrinsic distances, but I’m not sure whether the inverse is true. Namely, is it true that any map F between two surfaces that is 1-1, onto, and preserves intrinsic distances will automatically preserve inner products of tangent vectors?
Thanks a lot!
geometry differential-geometry surfaces
asked Nov 25 at 7:05
David Petey Gao
454
add a comment |
I know that if a map F between two surfaces preserves inner products of tangent vectors and is 1-1 and onto, then it must preserve intrinsic distances, but I’m not sure whether the inverse is true. Namely, is it true that any map F between two surfaces that is 1-1, onto, and preserves intrinsic distances will automatically preserve inner products of tangent vectors?
Thanks a lot!
geometry differential-geometry surfaces
asked Nov 25 at 7:05
David Petey Gao
454
I know that if a map F between two surfaces preserves inner products of tangent vectors and is 1-1 and onto, then it must preserve intrinsic distances, but I’m not sure whether the inverse is true. Namely, is it true that any map F between two surfaces that is 1-1, onto, and preserves intrinsic distances will automatically preserve inner products of tangent vectors?
Thanks a lot!
geometry differential-geometry surfaces
geometry differential-geometry surfaces
asked Nov 25 at 7:05
David Petey Gao
454
asked Nov 25 at 7:05
David Petey Gao
454
asked Nov 25 at 7:05
David Petey Gao
454
asked Nov 25 at 7:05
David Petey Gao
454
asked Nov 25 at 7:05
David Petey Gao
454
454
add a comment |
add a comment |
2 Answers
2
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Suppose $$F : (S, g) to (S', g')$$ is a metric isometry (homeomorphism that preserves the distance respective functions $d_g, d_{g'}$ of $g, g'$, that is, that satisfies $d_g = F^* d_{g'}$). Then, yes, $F$ is actually a (Riemannian) isometry, i.e., $F^* g' = g$.
To do this, it's certainly enough to give an explicit formula for $g$ in terms of $d_g$ (or even just $g(Z, Z)$, since we can recover $g$ by the Polarization Formula).
Hint Compare $g$ with the Euclidean metric in normal coordinates to establish that for any $p$ and $epsilon > 0$ there is a constant $C > 0$ for which the comparison
$$(1 - C|t|) |t X - t Y|_g leq d_g(exp t X, exp t Y) leq (1 + C|t|)|t X - t Y|_g$$ holds when $|t| leq 1$ for all $X, Y in B_{epsilon}(p)$.
Conclude that $$|X - Y|_g^2 = lim_{t to 0} frac{d_g(exp tX, exp tY)^2}{t^2} .$$
For a more detailed hint, see the erratum for p. 112, Problem 6-2 in the errata to Lee's Riemannian Manifolds.
answered Nov 25 at 8:41
Travis
59.4k767145
I've read the hint in the errata you mentioned, and it's of great help. Thanks a lot!
– David Petey Gao
Nov 26 at 19:01
However, the hint says the fact that metric isometries preserve geodesics needs to be proved first. I'm really new to this field and my understanding is a little bit messed up, but how exactly is this proved? Isn't it necessary to use the property of inner products preserving in proving this fact?
– David Petey Gao
Nov 26 at 19:08
You're welcome, I'm glad you've found it helpful. That question perhaps merits a question of its own, but: Geodesics can be defined in an invariant way in terms of the metric, namely, they are the solutions $gamma(t)$ to the differential equation $nabla^g_{gamma'} gamma' = 0$, so since the isometry maps one metric to the other, it maps the geodesics of the first to those of the second.
– Travis
Nov 26 at 19:16
If you prefer something more concrete, for any coordinate chart $(U, phi)$ on $S'$, $(F^{-1}(U), phi circ F)$ is a coordinate chart on $S$. Then, writing the geodesic equations in the two coordinate charts shows that if $gamma(t)$ is a geodesic of $g$ in $F^{-1}(U)$, then $(F circ gamma)(t)$ is a geodesic of $g'$ in $U$.
– Travis
Nov 26 at 19:19
I see. It's indeed true that if F preserves the metric tensor, then the geodesics would be preserved. However, the thing here is that we have only required F to preserve distances, not the metric tensor. The hint in the errata seems to say that in order to prove that such F would indeed preserve the metric tensor, we need to first prove that it will preserve geodesics. How can this be done? Thanks a lot!
– David Petey Gao
Nov 27 at 2:46
|
show 7 more comments
Yes, because $4acdot b=(a+b)^2-(a-b)^2$ or, in the complex case, $4acdot b=sum_{n=0}^3 i^{-n}Vert a+i^n bVert^2$.
answered Nov 25 at 7:12
J.G.
21.9k22034
add a comment |
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Suppose $$F : (S, g) to (S', g')$$ is a metric isometry (homeomorphism that preserves the distance respective functions $d_g, d_{g'}$ of $g, g'$, that is, that satisfies $d_g = F^* d_{g'}$). Then, yes, $F$ is actually a (Riemannian) isometry, i.e., $F^* g' = g$.
To do this, it's certainly enough to give an explicit formula for $g$ in terms of $d_g$ (or even just $g(Z, Z)$, since we can recover $g$ by the Polarization Formula).
Hint Compare $g$ with the Euclidean metric in normal coordinates to establish that for any $p$ and $epsilon > 0$ there is a constant $C > 0$ for which the comparison
$$(1 - C|t|) |t X - t Y|_g leq d_g(exp t X, exp t Y) leq (1 + C|t|)|t X - t Y|_g$$ holds when $|t| leq 1$ for all $X, Y in B_{epsilon}(p)$.
Conclude that $$|X - Y|_g^2 = lim_{t to 0} frac{d_g(exp tX, exp tY)^2}{t^2} .$$
For a more detailed hint, see the erratum for p. 112, Problem 6-2 in the errata to Lee's Riemannian Manifolds.
answered Nov 25 at 8:41
Travis
59.4k767145
I've read the hint in the errata you mentioned, and it's of great help. Thanks a lot!
– David Petey Gao
Nov 26 at 19:01
However, the hint says the fact that metric isometries preserve geodesics needs to be proved first. I'm really new to this field and my understanding is a little bit messed up, but how exactly is this proved? Isn't it necessary to use the property of inner products preserving in proving this fact?
– David Petey Gao
Nov 26 at 19:08
You're welcome, I'm glad you've found it helpful. That question perhaps merits a question of its own, but: Geodesics can be defined in an invariant way in terms of the metric, namely, they are the solutions $gamma(t)$ to the differential equation $nabla^g_{gamma'} gamma' = 0$, so since the isometry maps one metric to the other, it maps the geodesics of the first to those of the second.
– Travis
Nov 26 at 19:16
If you prefer something more concrete, for any coordinate chart $(U, phi)$ on $S'$, $(F^{-1}(U), phi circ F)$ is a coordinate chart on $S$. Then, writing the geodesic equations in the two coordinate charts shows that if $gamma(t)$ is a geodesic of $g$ in $F^{-1}(U)$, then $(F circ gamma)(t)$ is a geodesic of $g'$ in $U$.
– Travis
Nov 26 at 19:19
I see. It's indeed true that if F preserves the metric tensor, then the geodesics would be preserved. However, the thing here is that we have only required F to preserve distances, not the metric tensor. The hint in the errata seems to say that in order to prove that such F would indeed preserve the metric tensor, we need to first prove that it will preserve geodesics. How can this be done? Thanks a lot!
– David Petey Gao
Nov 27 at 2:46
|
show 7 more comments
Suppose $$F : (S, g) to (S', g')$$ is a metric isometry (homeomorphism that preserves the distance respective functions $d_g, d_{g'}$ of $g, g'$, that is, that satisfies $d_g = F^* d_{g'}$). Then, yes, $F$ is actually a (Riemannian) isometry, i.e., $F^* g' = g$.
To do this, it's certainly enough to give an explicit formula for $g$ in terms of $d_g$ (or even just $g(Z, Z)$, since we can recover $g$ by the Polarization Formula).
Hint Compare $g$ with the Euclidean metric in normal coordinates to establish that for any $p$ and $epsilon > 0$ there is a constant $C > 0$ for which the comparison
$$(1 - C|t|) |t X - t Y|_g leq d_g(exp t X, exp t Y) leq (1 + C|t|)|t X - t Y|_g$$ holds when $|t| leq 1$ for all $X, Y in B_{epsilon}(p)$.
Conclude that $$|X - Y|_g^2 = lim_{t to 0} frac{d_g(exp tX, exp tY)^2}{t^2} .$$
For a more detailed hint, see the erratum for p. 112, Problem 6-2 in the errata to Lee's Riemannian Manifolds.
answered Nov 25 at 8:41
Travis
59.4k767145
I've read the hint in the errata you mentioned, and it's of great help. Thanks a lot!
– David Petey Gao
Nov 26 at 19:01
However, the hint says the fact that metric isometries preserve geodesics needs to be proved first. I'm really new to this field and my understanding is a little bit messed up, but how exactly is this proved? Isn't it necessary to use the property of inner products preserving in proving this fact?
– David Petey Gao
Nov 26 at 19:08
You're welcome, I'm glad you've found it helpful. That question perhaps merits a question of its own, but: Geodesics can be defined in an invariant way in terms of the metric, namely, they are the solutions $gamma(t)$ to the differential equation $nabla^g_{gamma'} gamma' = 0$, so since the isometry maps one metric to the other, it maps the geodesics of the first to those of the second.
– Travis
Nov 26 at 19:16
If you prefer something more concrete, for any coordinate chart $(U, phi)$ on $S'$, $(F^{-1}(U), phi circ F)$ is a coordinate chart on $S$. Then, writing the geodesic equations in the two coordinate charts shows that if $gamma(t)$ is a geodesic of $g$ in $F^{-1}(U)$, then $(F circ gamma)(t)$ is a geodesic of $g'$ in $U$.
– Travis
Nov 26 at 19:19
I see. It's indeed true that if F preserves the metric tensor, then the geodesics would be preserved. However, the thing here is that we have only required F to preserve distances, not the metric tensor. The hint in the errata seems to say that in order to prove that such F would indeed preserve the metric tensor, we need to first prove that it will preserve geodesics. How can this be done? Thanks a lot!
– David Petey Gao
Nov 27 at 2:46
|
show 7 more comments
Suppose $$F : (S, g) to (S', g')$$ is a metric isometry (homeomorphism that preserves the distance respective functions $d_g, d_{g'}$ of $g, g'$, that is, that satisfies $d_g = F^* d_{g'}$). Then, yes, $F$ is actually a (Riemannian) isometry, i.e., $F^* g' = g$.
To do this, it's certainly enough to give an explicit formula for $g$ in terms of $d_g$ (or even just $g(Z, Z)$, since we can recover $g$ by the Polarization Formula).
Hint Compare $g$ with the Euclidean metric in normal coordinates to establish that for any $p$ and $epsilon > 0$ there is a constant $C > 0$ for which the comparison
$$(1 - C|t|) |t X - t Y|_g leq d_g(exp t X, exp t Y) leq (1 + C|t|)|t X - t Y|_g$$ holds when $|t| leq 1$ for all $X, Y in B_{epsilon}(p)$.
Conclude that $$|X - Y|_g^2 = lim_{t to 0} frac{d_g(exp tX, exp tY)^2}{t^2} .$$
For a more detailed hint, see the erratum for p. 112, Problem 6-2 in the errata to Lee's Riemannian Manifolds.
answered Nov 25 at 8:41
Travis
59.4k767145
Suppose $$F : (S, g) to (S', g')$$ is a metric isometry (homeomorphism that preserves the distance respective functions $d_g, d_{g'}$ of $g, g'$, that is, that satisfies $d_g = F^* d_{g'}$). Then, yes, $F$ is actually a (Riemannian) isometry, i.e., $F^* g' = g$.
To do this, it's certainly enough to give an explicit formula for $g$ in terms of $d_g$ (or even just $g(Z, Z)$, since we can recover $g$ by the Polarization Formula).
Hint Compare $g$ with the Euclidean metric in normal coordinates to establish that for any $p$ and $epsilon > 0$ there is a constant $C > 0$ for which the comparison
$$(1 - C|t|) |t X - t Y|_g leq d_g(exp t X, exp t Y) leq (1 + C|t|)|t X - t Y|_g$$ holds when $|t| leq 1$ for all $X, Y in B_{epsilon}(p)$.
Conclude that $$|X - Y|_g^2 = lim_{t to 0} frac{d_g(exp tX, exp tY)^2}{t^2} .$$
For a more detailed hint, see the erratum for p. 112, Problem 6-2 in the errata to Lee's Riemannian Manifolds.
answered Nov 25 at 8:41
Travis
59.4k767145
answered Nov 25 at 8:41
Travis
59.4k767145
answered Nov 25 at 8:41
Travis
59.4k767145
answered Nov 25 at 8:41
Travis
59.4k767145
59.4k767145
I've read the hint in the errata you mentioned, and it's of great help. Thanks a lot!
– David Petey Gao
Nov 26 at 19:01
However, the hint says the fact that metric isometries preserve geodesics needs to be proved first. I'm really new to this field and my understanding is a little bit messed up, but how exactly is this proved? Isn't it necessary to use the property of inner products preserving in proving this fact?
– David Petey Gao
Nov 26 at 19:08
You're welcome, I'm glad you've found it helpful. That question perhaps merits a question of its own, but: Geodesics can be defined in an invariant way in terms of the metric, namely, they are the solutions $gamma(t)$ to the differential equation $nabla^g_{gamma'} gamma' = 0$, so since the isometry maps one metric to the other, it maps the geodesics of the first to those of the second.
– Travis
Nov 26 at 19:16
If you prefer something more concrete, for any coordinate chart $(U, phi)$ on $S'$, $(F^{-1}(U), phi circ F)$ is a coordinate chart on $S$. Then, writing the geodesic equations in the two coordinate charts shows that if $gamma(t)$ is a geodesic of $g$ in $F^{-1}(U)$, then $(F circ gamma)(t)$ is a geodesic of $g'$ in $U$.
– Travis
Nov 26 at 19:19
I see. It's indeed true that if F preserves the metric tensor, then the geodesics would be preserved. However, the thing here is that we have only required F to preserve distances, not the metric tensor. The hint in the errata seems to say that in order to prove that such F would indeed preserve the metric tensor, we need to first prove that it will preserve geodesics. How can this be done? Thanks a lot!
– David Petey Gao
Nov 27 at 2:46
|
show 7 more comments
I've read the hint in the errata you mentioned, and it's of great help. Thanks a lot!
– David Petey Gao
Nov 26 at 19:01
However, the hint says the fact that metric isometries preserve geodesics needs to be proved first. I'm really new to this field and my understanding is a little bit messed up, but how exactly is this proved? Isn't it necessary to use the property of inner products preserving in proving this fact?
– David Petey Gao
Nov 26 at 19:08
You're welcome, I'm glad you've found it helpful. That question perhaps merits a question of its own, but: Geodesics can be defined in an invariant way in terms of the metric, namely, they are the solutions $gamma(t)$ to the differential equation $nabla^g_{gamma'} gamma' = 0$, so since the isometry maps one metric to the other, it maps the geodesics of the first to those of the second.
– Travis
Nov 26 at 19:16
If you prefer something more concrete, for any coordinate chart $(U, phi)$ on $S'$, $(F^{-1}(U), phi circ F)$ is a coordinate chart on $S$. Then, writing the geodesic equations in the two coordinate charts shows that if $gamma(t)$ is a geodesic of $g$ in $F^{-1}(U)$, then $(F circ gamma)(t)$ is a geodesic of $g'$ in $U$.
– Travis
Nov 26 at 19:19
I see. It's indeed true that if F preserves the metric tensor, then the geodesics would be preserved. However, the thing here is that we have only required F to preserve distances, not the metric tensor. The hint in the errata seems to say that in order to prove that such F would indeed preserve the metric tensor, we need to first prove that it will preserve geodesics. How can this be done? Thanks a lot!
– David Petey Gao
Nov 27 at 2:46
I've read the hint in the errata you mentioned, and it's of great help. Thanks a lot!
– David Petey Gao
Nov 26 at 19:01
I've read the hint in the errata you mentioned, and it's of great help. Thanks a lot!
– David Petey Gao
Nov 26 at 19:01
However, the hint says the fact that metric isometries preserve geodesics needs to be proved first. I'm really new to this field and my understanding is a little bit messed up, but how exactly is this proved? Isn't it necessary to use the property of inner products preserving in proving this fact?
– David Petey Gao
Nov 26 at 19:08
However, the hint says the fact that metric isometries preserve geodesics needs to be proved first. I'm really new to this field and my understanding is a little bit messed up, but how exactly is this proved? Isn't it necessary to use the property of inner products preserving in proving this fact?
– David Petey Gao
Nov 26 at 19:08
You're welcome, I'm glad you've found it helpful. That question perhaps merits a question of its own, but: Geodesics can be defined in an invariant way in terms of the metric, namely, they are the solutions $gamma(t)$ to the differential equation $nabla^g_{gamma'} gamma' = 0$, so since the isometry maps one metric to the other, it maps the geodesics of the first to those of the second.
– Travis
Nov 26 at 19:16
You're welcome, I'm glad you've found it helpful. That question perhaps merits a question of its own, but: Geodesics can be defined in an invariant way in terms of the metric, namely, they are the solutions $gamma(t)$ to the differential equation $nabla^g_{gamma'} gamma' = 0$, so since the isometry maps one metric to the other, it maps the geodesics of the first to those of the second.
– Travis
Nov 26 at 19:16
If you prefer something more concrete, for any coordinate chart $(U, phi)$ on $S'$, $(F^{-1}(U), phi circ F)$ is a coordinate chart on $S$. Then, writing the geodesic equations in the two coordinate charts shows that if $gamma(t)$ is a geodesic of $g$ in $F^{-1}(U)$, then $(F circ gamma)(t)$ is a geodesic of $g'$ in $U$.
– Travis
Nov 26 at 19:19
If you prefer something more concrete, for any coordinate chart $(U, phi)$ on $S'$, $(F^{-1}(U), phi circ F)$ is a coordinate chart on $S$. Then, writing the geodesic equations in the two coordinate charts shows that if $gamma(t)$ is a geodesic of $g$ in $F^{-1}(U)$, then $(F circ gamma)(t)$ is a geodesic of $g'$ in $U$.
– Travis
Nov 26 at 19:19
I see. It's indeed true that if F preserves the metric tensor, then the geodesics would be preserved. However, the thing here is that we have only required F to preserve distances, not the metric tensor. The hint in the errata seems to say that in order to prove that such F would indeed preserve the metric tensor, we need to first prove that it will preserve geodesics. How can this be done? Thanks a lot!
– David Petey Gao
Nov 27 at 2:46
I see. It's indeed true that if F preserves the metric tensor, then the geodesics would be preserved. However, the thing here is that we have only required F to preserve distances, not the metric tensor. The hint in the errata seems to say that in order to prove that such F would indeed preserve the metric tensor, we need to first prove that it will preserve geodesics. How can this be done? Thanks a lot!
– David Petey Gao
Nov 27 at 2:46
|
show 7 more comments
Yes, because $4acdot b=(a+b)^2-(a-b)^2$ or, in the complex case, $4acdot b=sum_{n=0}^3 i^{-n}Vert a+i^n bVert^2$.
answered Nov 25 at 7:12
J.G.
21.9k22034
add a comment |
Yes, because $4acdot b=(a+b)^2-(a-b)^2$ or, in the complex case, $4acdot b=sum_{n=0}^3 i^{-n}Vert a+i^n bVert^2$.
answered Nov 25 at 7:12
J.G.
21.9k22034
add a comment |
Yes, because $4acdot b=(a+b)^2-(a-b)^2$ or, in the complex case, $4acdot b=sum_{n=0}^3 i^{-n}Vert a+i^n bVert^2$.
answered Nov 25 at 7:12
J.G.
21.9k22034
Yes, because $4acdot b=(a+b)^2-(a-b)^2$ or, in the complex case, $4acdot b=sum_{n=0}^3 i^{-n}Vert a+i^n bVert^2$.
answered Nov 25 at 7:12
J.G.
21.9k22034
answered Nov 25 at 7:12
J.G.
21.9k22034
answered Nov 25 at 7:12
J.G.
21.9k22034
answered Nov 25 at 7:12
J.G.
21.9k22034
21.9k22034
add a comment |
add a comment |
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