Mixing
Two flat tori are isometric iff their lattices are isometric?
Say $T_1,T_2$ are flat tori defined as quotients of $mathbb{R}^n$ by the lattices $Gamma_1,Gamma_2$, resp. It is easy to prove that an isometry $mathbb{R}^ntomathbb{R}^n$ sending $Gamma_1toGamma_2$ must descend to an isometry $T_1to T_2$. How can one prove the converse: that if there exists an isometry $T_1to T_2$, one can lift it to an isometry $mathbb{R}^ntomathbb{R}^n$ (which sends $Gamma_1mapstoGamma_2$)?
differential-geometry riemannian-geometry
asked Nov 20 '18 at 19:01
Juan Carlos Ortiz
825
add a comment |
Say $T_1,T_2$ are flat tori defined as quotients of $mathbb{R}^n$ by the lattices $Gamma_1,Gamma_2$, resp. It is easy to prove that an isometry $mathbb{R}^ntomathbb{R}^n$ sending $Gamma_1toGamma_2$ must descend to an isometry $T_1to T_2$. How can one prove the converse: that if there exists an isometry $T_1to T_2$, one can lift it to an isometry $mathbb{R}^ntomathbb{R}^n$ (which sends $Gamma_1mapstoGamma_2$)?
differential-geometry riemannian-geometry
asked Nov 20 '18 at 19:01
Juan Carlos Ortiz
825
2
Intuitively, it seems pretty straightforward to me: your map $T_1 to T_2$ gives you a family of maps between any two fundamental domains for $T_1$ and $T_2$ in $Bbb{R}^n$; then it's just a matter of choosing one and stitching the rest together following the relevant boundary conditions. I don't think that the fact that the map is an isometry plays any role here, but I may be wrong.
– A.P.
Nov 20 '18 at 20:10
3
An isometry of metric spaces induces an isometry of their universal covers once you choose where to send a single point in the universal cover.
– Mike Miller
Nov 20 '18 at 20:10
A.P., yeah I reckon you're right, thet fact that $T_1to T_2$ is an isometry is only used to prove that $mathbb{R}^ntomathbb{R}^n$ is an isometry.
– Juan Carlos Ortiz
Nov 20 '18 at 22:55
Oh, you're right Mike, I'm being silly :)
– Juan Carlos Ortiz
Nov 20 '18 at 23:04
add a comment |
Say $T_1,T_2$ are flat tori defined as quotients of $mathbb{R}^n$ by the lattices $Gamma_1,Gamma_2$, resp. It is easy to prove that an isometry $mathbb{R}^ntomathbb{R}^n$ sending $Gamma_1toGamma_2$ must descend to an isometry $T_1to T_2$. How can one prove the converse: that if there exists an isometry $T_1to T_2$, one can lift it to an isometry $mathbb{R}^ntomathbb{R}^n$ (which sends $Gamma_1mapstoGamma_2$)?
differential-geometry riemannian-geometry
asked Nov 20 '18 at 19:01
Juan Carlos Ortiz
825
Say $T_1,T_2$ are flat tori defined as quotients of $mathbb{R}^n$ by the lattices $Gamma_1,Gamma_2$, resp. It is easy to prove that an isometry $mathbb{R}^ntomathbb{R}^n$ sending $Gamma_1toGamma_2$ must descend to an isometry $T_1to T_2$. How can one prove the converse: that if there exists an isometry $T_1to T_2$, one can lift it to an isometry $mathbb{R}^ntomathbb{R}^n$ (which sends $Gamma_1mapstoGamma_2$)?
differential-geometry riemannian-geometry
differential-geometry riemannian-geometry
asked Nov 20 '18 at 19:01
Juan Carlos Ortiz
825
asked Nov 20 '18 at 19:01
Juan Carlos Ortiz
825
asked Nov 20 '18 at 19:01
Juan Carlos Ortiz
825
asked Nov 20 '18 at 19:01
Juan Carlos Ortiz
825
asked Nov 20 '18 at 19:01
Juan Carlos Ortiz
825
825
2
Intuitively, it seems pretty straightforward to me: your map $T_1 to T_2$ gives you a family of maps between any two fundamental domains for $T_1$ and $T_2$ in $Bbb{R}^n$; then it's just a matter of choosing one and stitching the rest together following the relevant boundary conditions. I don't think that the fact that the map is an isometry plays any role here, but I may be wrong.
– A.P.
Nov 20 '18 at 20:10
3
An isometry of metric spaces induces an isometry of their universal covers once you choose where to send a single point in the universal cover.
– Mike Miller
Nov 20 '18 at 20:10
A.P., yeah I reckon you're right, thet fact that $T_1to T_2$ is an isometry is only used to prove that $mathbb{R}^ntomathbb{R}^n$ is an isometry.
– Juan Carlos Ortiz
Nov 20 '18 at 22:55
Oh, you're right Mike, I'm being silly :)
– Juan Carlos Ortiz
Nov 20 '18 at 23:04
add a comment |
2
Intuitively, it seems pretty straightforward to me: your map $T_1 to T_2$ gives you a family of maps between any two fundamental domains for $T_1$ and $T_2$ in $Bbb{R}^n$; then it's just a matter of choosing one and stitching the rest together following the relevant boundary conditions. I don't think that the fact that the map is an isometry plays any role here, but I may be wrong.
– A.P.
Nov 20 '18 at 20:10
3
An isometry of metric spaces induces an isometry of their universal covers once you choose where to send a single point in the universal cover.
– Mike Miller
Nov 20 '18 at 20:10
A.P., yeah I reckon you're right, thet fact that $T_1to T_2$ is an isometry is only used to prove that $mathbb{R}^ntomathbb{R}^n$ is an isometry.
– Juan Carlos Ortiz
Nov 20 '18 at 22:55
Oh, you're right Mike, I'm being silly :)
– Juan Carlos Ortiz
Nov 20 '18 at 23:04
2
2
Intuitively, it seems pretty straightforward to me: your map $T_1 to T_2$ gives you a family of maps between any two fundamental domains for $T_1$ and $T_2$ in $Bbb{R}^n$; then it's just a matter of choosing one and stitching the rest together following the relevant boundary conditions. I don't think that the fact that the map is an isometry plays any role here, but I may be wrong.
– A.P.
Nov 20 '18 at 20:10
Intuitively, it seems pretty straightforward to me: your map $T_1 to T_2$ gives you a family of maps between any two fundamental domains for $T_1$ and $T_2$ in $Bbb{R}^n$; then it's just a matter of choosing one and stitching the rest together following the relevant boundary conditions. I don't think that the fact that the map is an isometry plays any role here, but I may be wrong.
– A.P.
Nov 20 '18 at 20:10
3
3
An isometry of metric spaces induces an isometry of their universal covers once you choose where to send a single point in the universal cover.
– Mike Miller
Nov 20 '18 at 20:10
An isometry of metric spaces induces an isometry of their universal covers once you choose where to send a single point in the universal cover.
– Mike Miller
Nov 20 '18 at 20:10
A.P., yeah I reckon you're right, thet fact that $T_1to T_2$ is an isometry is only used to prove that $mathbb{R}^ntomathbb{R}^n$ is an isometry.
– Juan Carlos Ortiz
Nov 20 '18 at 22:55
A.P., yeah I reckon you're right, thet fact that $T_1to T_2$ is an isometry is only used to prove that $mathbb{R}^ntomathbb{R}^n$ is an isometry.
– Juan Carlos Ortiz
Nov 20 '18 at 22:55
Oh, you're right Mike, I'm being silly :)
– Juan Carlos Ortiz
Nov 20 '18 at 23:04
Oh, you're right Mike, I'm being silly :)
– Juan Carlos Ortiz
Nov 20 '18 at 23:04
add a comment |
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2
Intuitively, it seems pretty straightforward to me: your map $T_1 to T_2$ gives you a family of maps between any two fundamental domains for $T_1$ and $T_2$ in $Bbb{R}^n$; then it's just a matter of choosing one and stitching the rest together following the relevant boundary conditions. I don't think that the fact that the map is an isometry plays any role here, but I may be wrong.
– A.P.
Nov 20 '18 at 20:10
3
An isometry of metric spaces induces an isometry of their universal covers once you choose where to send a single point in the universal cover.
– Mike Miller
Nov 20 '18 at 20:10
A.P., yeah I reckon you're right, thet fact that $T_1to T_2$ is an isometry is only used to prove that $mathbb{R}^ntomathbb{R}^n$ is an isometry.
– Juan Carlos Ortiz
Nov 20 '18 at 22:55
Oh, you're right Mike, I'm being silly :)
– Juan Carlos Ortiz
Nov 20 '18 at 23:04