Two flat tori are isometric iff their lattices are isometric?
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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 10 hours ago
Juan Carlos Ortiz
825
add a comment |
up vote
2
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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 10 hours ago
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.
9 hours ago
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
9 hours ago
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
6 hours ago
Oh, you're right Mike, I'm being silly :)
– Juan Carlos Ortiz
6 hours ago
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
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 10 hours ago
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 10 hours ago
Juan Carlos Ortiz
825
asked 10 hours ago
Juan Carlos Ortiz
825
asked 10 hours ago
Juan Carlos Ortiz
825
asked 10 hours ago
Juan Carlos Ortiz
825
asked 10 hours ago
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.
9 hours ago
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
9 hours ago
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
6 hours ago
Oh, you're right Mike, I'm being silly :)
– Juan Carlos Ortiz
6 hours ago
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.
9 hours ago
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
9 hours ago
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
6 hours ago
Oh, you're right Mike, I'm being silly :)
– Juan Carlos Ortiz
6 hours ago
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.
9 hours ago
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.
9 hours ago
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
9 hours ago
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
9 hours ago
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
6 hours ago
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
6 hours ago
Oh, you're right Mike, I'm being silly :)
– Juan Carlos Ortiz
6 hours ago
Oh, you're right Mike, I'm being silly :)
– Juan Carlos Ortiz
6 hours ago
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.
9 hours ago
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
9 hours ago
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
6 hours ago
Oh, you're right Mike, I'm being silly :)
– Juan Carlos Ortiz
6 hours ago