Hyperbolic isometry and line segments The 2019 Stack Overflow Developer Survey Results Are InDifference between a hyperbolic line and a geodesicHyperbolic quadrilaterals with two adjacent right anglesTriangular tiling of hyperbolic planeCan an isometry of the hyperbolic plane that maps a circle to a disjoint circle have a fixed point?“Square,” line-preserving models of the hyperbolic planeHyperbolic segment from $(0,0)$ to $(0,0)$Hyperbolic polygons as fundamental regions for groups $Gammasubset textIso(mathbbH^2)$Why are hyperbolic circles in the upper half plane also Euclidean circles?Hyperbolic CrownInversion as hyperbolic isometry (Poincaré disk model).
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Hyperbolic isometry and line segments
The 2019 Stack Overflow Developer Survey Results Are InDifference between a hyperbolic line and a geodesicHyperbolic quadrilaterals with two adjacent right anglesTriangular tiling of hyperbolic planeCan an isometry of the hyperbolic plane that maps a circle to a disjoint circle have a fixed point?“Square,” line-preserving models of the hyperbolic planeHyperbolic segment from $(0,0)$ to $(0,0)$Hyperbolic polygons as fundamental regions for groups $Gammasubset textIso(mathbbH^2)$Why are hyperbolic circles in the upper half plane also Euclidean circles?Hyperbolic CrownInversion as hyperbolic isometry (Poincaré disk model).
$begingroup$
I was trying to apply Poincare's Polygon theorem, for that I had to give a pairing of sides, i.e., to have an isometry of the hyperbolic plane that will take a side of a polygon to another side (of the same length). So my question is that -
Given any two hyperbolic line segments of equal hyperbolic length, is there an isometry of the hyperbolic plane that sends one of the line segments to the other?
Its is easy to see that there will be mobius transformations that will take one line segment to the other, but I couldn't see why will there be one such transformation that will also fix the upper half-plane (i.e. will be an isometry of the hyperbolic plane).
A proof of existence will be enough for me.
hyperbolic-geometry
asked Mar 24 at 5:03
Agneedh BasuAgneedh Basu
61
$endgroup$
add a comment |
$begingroup$
I was trying to apply Poincare's Polygon theorem, for that I had to give a pairing of sides, i.e., to have an isometry of the hyperbolic plane that will take a side of a polygon to another side (of the same length). So my question is that -
Given any two hyperbolic line segments of equal hyperbolic length, is there an isometry of the hyperbolic plane that sends one of the line segments to the other?
Its is easy to see that there will be mobius transformations that will take one line segment to the other, but I couldn't see why will there be one such transformation that will also fix the upper half-plane (i.e. will be an isometry of the hyperbolic plane).
A proof of existence will be enough for me.
hyperbolic-geometry
asked Mar 24 at 5:03
Agneedh BasuAgneedh Basu
61
$endgroup$
1
$begingroup$
Hi and welcome to MSE! I want to remind you that it is generally preferred you include context when asking a question here (which can include: where this problem came from, your own attempts, and a specific idea as to where you're stuck) - it also lets us help you better! As is, your question is little more than an isolated problem, and thus likely to get a lot of downvotes and closed. Feel free to edit the context into your post though! Here's some useful links: asking a good question and a MathJax reference to format your work.
$endgroup$
– Eevee Trainer
Mar 24 at 5:06
$begingroup$
To amplify Eevee's comment ... Are you looking for an "abstract" proof that such an isometry exists? (What fundamental notions of isometries do you know?) Are you looking at a particular model —Poincaré disk? Upper Half-Plane?— and wanting to determine an explicit transformation function? (What do you know about such functions?) Explaining exactly what you know and/or want helps answerers tailor their responses, without wasting time (yours or theirs) telling you things you already know or using approaches you haven't seen.
$endgroup$
– Blue
Mar 24 at 5:31
$begingroup$
I would use the hyperboloid model instead. Then the existence of such isometry is as obvious as on a sphere.
$endgroup$
– Zeno Rogue
Mar 24 at 23:50
add a comment |
$begingroup$
I was trying to apply Poincare's Polygon theorem, for that I had to give a pairing of sides, i.e., to have an isometry of the hyperbolic plane that will take a side of a polygon to another side (of the same length). So my question is that -
Given any two hyperbolic line segments of equal hyperbolic length, is there an isometry of the hyperbolic plane that sends one of the line segments to the other?
Its is easy to see that there will be mobius transformations that will take one line segment to the other, but I couldn't see why will there be one such transformation that will also fix the upper half-plane (i.e. will be an isometry of the hyperbolic plane).
A proof of existence will be enough for me.
hyperbolic-geometry
asked Mar 24 at 5:03
Agneedh BasuAgneedh Basu
61
$endgroup$
I was trying to apply Poincare's Polygon theorem, for that I had to give a pairing of sides, i.e., to have an isometry of the hyperbolic plane that will take a side of a polygon to another side (of the same length). So my question is that -
Given any two hyperbolic line segments of equal hyperbolic length, is there an isometry of the hyperbolic plane that sends one of the line segments to the other?
Its is easy to see that there will be mobius transformations that will take one line segment to the other, but I couldn't see why will there be one such transformation that will also fix the upper half-plane (i.e. will be an isometry of the hyperbolic plane).
A proof of existence will be enough for me.
hyperbolic-geometry
hyperbolic-geometry
asked Mar 24 at 5:03
Agneedh BasuAgneedh Basu
61
asked Mar 24 at 5:03
Agneedh BasuAgneedh Basu
61
edited Mar 24 at 5:32
Agneedh Basu
asked Mar 24 at 5:03
Agneedh BasuAgneedh Basu
61
asked Mar 24 at 5:03
Agneedh BasuAgneedh Basu
61
asked Mar 24 at 5:03
Agneedh BasuAgneedh Basu
61
61
1
$begingroup$
Hi and welcome to MSE! I want to remind you that it is generally preferred you include context when asking a question here (which can include: where this problem came from, your own attempts, and a specific idea as to where you're stuck) - it also lets us help you better! As is, your question is little more than an isolated problem, and thus likely to get a lot of downvotes and closed. Feel free to edit the context into your post though! Here's some useful links: asking a good question and a MathJax reference to format your work.
$endgroup$
– Eevee Trainer
Mar 24 at 5:06
$begingroup$
To amplify Eevee's comment ... Are you looking for an "abstract" proof that such an isometry exists? (What fundamental notions of isometries do you know?) Are you looking at a particular model —Poincaré disk? Upper Half-Plane?— and wanting to determine an explicit transformation function? (What do you know about such functions?) Explaining exactly what you know and/or want helps answerers tailor their responses, without wasting time (yours or theirs) telling you things you already know or using approaches you haven't seen.
$endgroup$
– Blue
Mar 24 at 5:31
$begingroup$
I would use the hyperboloid model instead. Then the existence of such isometry is as obvious as on a sphere.
$endgroup$
– Zeno Rogue
Mar 24 at 23:50
add a comment |
1
$begingroup$
Hi and welcome to MSE! I want to remind you that it is generally preferred you include context when asking a question here (which can include: where this problem came from, your own attempts, and a specific idea as to where you're stuck) - it also lets us help you better! As is, your question is little more than an isolated problem, and thus likely to get a lot of downvotes and closed. Feel free to edit the context into your post though! Here's some useful links: asking a good question and a MathJax reference to format your work.
$endgroup$
– Eevee Trainer
Mar 24 at 5:06
$begingroup$
To amplify Eevee's comment ... Are you looking for an "abstract" proof that such an isometry exists? (What fundamental notions of isometries do you know?) Are you looking at a particular model —Poincaré disk? Upper Half-Plane?— and wanting to determine an explicit transformation function? (What do you know about such functions?) Explaining exactly what you know and/or want helps answerers tailor their responses, without wasting time (yours or theirs) telling you things you already know or using approaches you haven't seen.
$endgroup$
– Blue
Mar 24 at 5:31
$begingroup$
I would use the hyperboloid model instead. Then the existence of such isometry is as obvious as on a sphere.
$endgroup$
– Zeno Rogue
Mar 24 at 23:50
1
1
$begingroup$
Hi and welcome to MSE! I want to remind you that it is generally preferred you include context when asking a question here (which can include: where this problem came from, your own attempts, and a specific idea as to where you're stuck) - it also lets us help you better! As is, your question is little more than an isolated problem, and thus likely to get a lot of downvotes and closed. Feel free to edit the context into your post though! Here's some useful links: asking a good question and a MathJax reference to format your work.
$endgroup$
– Eevee Trainer
Mar 24 at 5:06
$begingroup$
Hi and welcome to MSE! I want to remind you that it is generally preferred you include context when asking a question here (which can include: where this problem came from, your own attempts, and a specific idea as to where you're stuck) - it also lets us help you better! As is, your question is little more than an isolated problem, and thus likely to get a lot of downvotes and closed. Feel free to edit the context into your post though! Here's some useful links: asking a good question and a MathJax reference to format your work.
$endgroup$
– Eevee Trainer
Mar 24 at 5:06
$begingroup$
To amplify Eevee's comment ... Are you looking for an "abstract" proof that such an isometry exists? (What fundamental notions of isometries do you know?) Are you looking at a particular model —Poincaré disk? Upper Half-Plane?— and wanting to determine an explicit transformation function? (What do you know about such functions?) Explaining exactly what you know and/or want helps answerers tailor their responses, without wasting time (yours or theirs) telling you things you already know or using approaches you haven't seen.
$endgroup$
– Blue
Mar 24 at 5:31
$begingroup$
To amplify Eevee's comment ... Are you looking for an "abstract" proof that such an isometry exists? (What fundamental notions of isometries do you know?) Are you looking at a particular model —Poincaré disk? Upper Half-Plane?— and wanting to determine an explicit transformation function? (What do you know about such functions?) Explaining exactly what you know and/or want helps answerers tailor their responses, without wasting time (yours or theirs) telling you things you already know or using approaches you haven't seen.
$endgroup$
– Blue
Mar 24 at 5:31
$begingroup$
I would use the hyperboloid model instead. Then the existence of such isometry is as obvious as on a sphere.
$endgroup$
– Zeno Rogue
Mar 24 at 23:50
$begingroup$
I would use the hyperboloid model instead. Then the existence of such isometry is as obvious as on a sphere.
$endgroup$
– Zeno Rogue
Mar 24 at 23:50
add a comment |
1 Answer
1
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oldest
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$begingroup$
A Möbius transformation is uniquely defined by 3 points and their images. If you have $z_1mapsto z_1'$ and $z_2mapsto z_2'$ mapping the endpoints of the line segments, then add $overlinez_1mapsto overlinez_1'$, i.e. map the complex conjugates for one point and its image. If the segments $(z_1,z_2)$ and $(z_1',z_2')$ are indeed of equal length, then the map defined by these three points will also map $overlinez_2mapsto overlinez_2'$ and it will have a representation using real coefficients only, so that it preserves the real axis.
If some other reader wants the same for the Poincaré disk, use inversion in the unit circle instead of complex conjugate i.e. reflection in the real axis. The idea is that in a way, the upper and the lower half plane in the half plane model, or the inside and the outside (including the point at infinity) of the disk in the disk model, are algebraically pretty much equivalent. It makes sense to think of a hyperbolic point in the half plane model not as a single point in the upper half plane, but as a pair of points reflected in the real axis. Using this helps adding additional constraints for the Möbius transformation.
answered Mar 26 at 21:32
MvGMvG
31.2k450107
$endgroup$
add a comment |
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$begingroup$
A Möbius transformation is uniquely defined by 3 points and their images. If you have $z_1mapsto z_1'$ and $z_2mapsto z_2'$ mapping the endpoints of the line segments, then add $overlinez_1mapsto overlinez_1'$, i.e. map the complex conjugates for one point and its image. If the segments $(z_1,z_2)$ and $(z_1',z_2')$ are indeed of equal length, then the map defined by these three points will also map $overlinez_2mapsto overlinez_2'$ and it will have a representation using real coefficients only, so that it preserves the real axis.
If some other reader wants the same for the Poincaré disk, use inversion in the unit circle instead of complex conjugate i.e. reflection in the real axis. The idea is that in a way, the upper and the lower half plane in the half plane model, or the inside and the outside (including the point at infinity) of the disk in the disk model, are algebraically pretty much equivalent. It makes sense to think of a hyperbolic point in the half plane model not as a single point in the upper half plane, but as a pair of points reflected in the real axis. Using this helps adding additional constraints for the Möbius transformation.
answered Mar 26 at 21:32
MvGMvG
31.2k450107
$endgroup$
add a comment |
$begingroup$
A Möbius transformation is uniquely defined by 3 points and their images. If you have $z_1mapsto z_1'$ and $z_2mapsto z_2'$ mapping the endpoints of the line segments, then add $overlinez_1mapsto overlinez_1'$, i.e. map the complex conjugates for one point and its image. If the segments $(z_1,z_2)$ and $(z_1',z_2')$ are indeed of equal length, then the map defined by these three points will also map $overlinez_2mapsto overlinez_2'$ and it will have a representation using real coefficients only, so that it preserves the real axis.
If some other reader wants the same for the Poincaré disk, use inversion in the unit circle instead of complex conjugate i.e. reflection in the real axis. The idea is that in a way, the upper and the lower half plane in the half plane model, or the inside and the outside (including the point at infinity) of the disk in the disk model, are algebraically pretty much equivalent. It makes sense to think of a hyperbolic point in the half plane model not as a single point in the upper half plane, but as a pair of points reflected in the real axis. Using this helps adding additional constraints for the Möbius transformation.
answered Mar 26 at 21:32
MvGMvG
31.2k450107
$endgroup$
add a comment |
$begingroup$
A Möbius transformation is uniquely defined by 3 points and their images. If you have $z_1mapsto z_1'$ and $z_2mapsto z_2'$ mapping the endpoints of the line segments, then add $overlinez_1mapsto overlinez_1'$, i.e. map the complex conjugates for one point and its image. If the segments $(z_1,z_2)$ and $(z_1',z_2')$ are indeed of equal length, then the map defined by these three points will also map $overlinez_2mapsto overlinez_2'$ and it will have a representation using real coefficients only, so that it preserves the real axis.
If some other reader wants the same for the Poincaré disk, use inversion in the unit circle instead of complex conjugate i.e. reflection in the real axis. The idea is that in a way, the upper and the lower half plane in the half plane model, or the inside and the outside (including the point at infinity) of the disk in the disk model, are algebraically pretty much equivalent. It makes sense to think of a hyperbolic point in the half plane model not as a single point in the upper half plane, but as a pair of points reflected in the real axis. Using this helps adding additional constraints for the Möbius transformation.
answered Mar 26 at 21:32
MvGMvG
31.2k450107
$endgroup$
A Möbius transformation is uniquely defined by 3 points and their images. If you have $z_1mapsto z_1'$ and $z_2mapsto z_2'$ mapping the endpoints of the line segments, then add $overlinez_1mapsto overlinez_1'$, i.e. map the complex conjugates for one point and its image. If the segments $(z_1,z_2)$ and $(z_1',z_2')$ are indeed of equal length, then the map defined by these three points will also map $overlinez_2mapsto overlinez_2'$ and it will have a representation using real coefficients only, so that it preserves the real axis.
If some other reader wants the same for the Poincaré disk, use inversion in the unit circle instead of complex conjugate i.e. reflection in the real axis. The idea is that in a way, the upper and the lower half plane in the half plane model, or the inside and the outside (including the point at infinity) of the disk in the disk model, are algebraically pretty much equivalent. It makes sense to think of a hyperbolic point in the half plane model not as a single point in the upper half plane, but as a pair of points reflected in the real axis. Using this helps adding additional constraints for the Möbius transformation.
answered Mar 26 at 21:32
MvGMvG
31.2k450107
answered Mar 26 at 21:32
MvGMvG
31.2k450107
answered Mar 26 at 21:32
MvGMvG
31.2k450107
answered Mar 26 at 21:32
MvGMvG
31.2k450107
31.2k450107
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$begingroup$
Hi and welcome to MSE! I want to remind you that it is generally preferred you include context when asking a question here (which can include: where this problem came from, your own attempts, and a specific idea as to where you're stuck) - it also lets us help you better! As is, your question is little more than an isolated problem, and thus likely to get a lot of downvotes and closed. Feel free to edit the context into your post though! Here's some useful links: asking a good question and a MathJax reference to format your work.
$endgroup$
– Eevee Trainer
Mar 24 at 5:06
$begingroup$
To amplify Eevee's comment ... Are you looking for an "abstract" proof that such an isometry exists? (What fundamental notions of isometries do you know?) Are you looking at a particular model —Poincaré disk? Upper Half-Plane?— and wanting to determine an explicit transformation function? (What do you know about such functions?) Explaining exactly what you know and/or want helps answerers tailor their responses, without wasting time (yours or theirs) telling you things you already know or using approaches you haven't seen.
$endgroup$
– Blue
Mar 24 at 5:31
$begingroup$
I would use the hyperboloid model instead. Then the existence of such isometry is as obvious as on a sphere.
$endgroup$
– Zeno Rogue
Mar 24 at 23:50