Whether preserving inner products follows from preserving intrinsic distances












1














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







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    1














    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







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      1












      1








      1







      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







      share|cite|improve this question













      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






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      share|cite|improve this question











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      asked Nov 25 at 7:05









      David Petey Gao

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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.






          share|cite|improve this answer





















          • 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



















          0














          Yes, because $4acdot b=(a+b)^2-(a-b)^2$ or, in the complex case, $4acdot b=sum_{n=0}^3 i^{-n}Vert a+i^n bVert^2$.






          share|cite|improve this answer





















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            2














            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.






            share|cite|improve this answer





















            • 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
















            2














            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.






            share|cite|improve this answer





















            • 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














            2












            2








            2






            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.






            share|cite|improve this answer












            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.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 25 at 8:41









            Travis

            59.5k767146




            59.5k767146












            • 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










            • 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











            0














            Yes, because $4acdot b=(a+b)^2-(a-b)^2$ or, in the complex case, $4acdot b=sum_{n=0}^3 i^{-n}Vert a+i^n bVert^2$.






            share|cite|improve this answer


























              0














              Yes, because $4acdot b=(a+b)^2-(a-b)^2$ or, in the complex case, $4acdot b=sum_{n=0}^3 i^{-n}Vert a+i^n bVert^2$.






              share|cite|improve this answer
























                0












                0








                0






                Yes, because $4acdot b=(a+b)^2-(a-b)^2$ or, in the complex case, $4acdot b=sum_{n=0}^3 i^{-n}Vert a+i^n bVert^2$.






                share|cite|improve this answer












                Yes, because $4acdot b=(a+b)^2-(a-b)^2$ or, in the complex case, $4acdot b=sum_{n=0}^3 i^{-n}Vert a+i^n bVert^2$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 25 at 7:12









                J.G.

                22.6k22136




                22.6k22136






























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                    Musa

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                    Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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