Coordinate system with easy linear point transformations in hyperbolic plane

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
1
down vote

favorite












I want to do some analytic geometry in the hyperbolic plane. I haven't chosen a coordinate system yet since I don't know which would make the math easiest. It's a lot to ask for, but I want the description of the coordinate system (map and inverse map for these coordinates to hyperboloid model coordinates) and closed form expressions for as many of these as possible (intermediate variables is okay if it reduces the work needed to calculate these). Also avoid unsolveable integrals. The rest I'll try to fill in myself. I just don't want to be doing the math for all the different options myself when someone else has probably already done it before.



In my descriptions I treat points as interchangeable with vectors, though I don't assume operations that work in the Euclidean plane also work in the hyperbolic plane.



Geometry requires the origin be an actual point, though usually we assume it to be $O(0,0)$ or equivalent to simplify math.




Point Magnitude



Given origin point $O$ and point $A$. $|A|$ is the length of the shortest path from $O$ to $A$.



Point Negation



Given origin point $O$ and point $A$. $-A$ is $A$ rotated $180^circ$ about $O$, or $A$ scaled by $-1$ relative to $O$.



This should be easy to calculate, if it isn't just something like $(x,y)to (-x,-y)$.



Point Axis Reflection



Given origin point $O$ and point $A$. $overline A$ is $A$ reflected in the $x$ axis for $O$.



This should also be easy to calculate, if it's not just something like $(x,y)to (x,-y)$.



Point Translation



Given origin point $O$, point $A$, and translation vector $B$. The translated point $C$ is whatever $A$ would be called if $-B$ was the origin, setting the rotation so that $O$ to the new origin $-B$ looks the same as $B$ to the old origin $O$.



This translation is the basis of the new non-commutative addition and subtraction operators: $C=A+B,C-B=A$. Since the side-angle-side congruence still holds, we have an alternative choice for the geometrical interpretation:



Treating A as the new origin, B is drawn, which is C to the origin. Treating -B as the new origin, C is drawn, which is A to the origin.



Don't mind the straight lines. $A+Bne B+A$ because you can't have parallelograms without parallel lines. Other working geometric interpretations reduce to this or produce the same triangle.



Because this is the hyperbolic plane, translations result in rotations. This is okay.



Using this and magnitude, the equation for line length $|A-B|$ can be derived.



Point Rotation



Given origin point $O$, point $A$, and angle $theta$. The rotated point $B$ is $A$ rotated by $theta$ about the origin, satisfying $|A|=|B|,angle BOA=theta$.



Since translations also rotate, this can be derived the hard way from a series of translations.



Point Scale



Given origin point $O$, point $A$, and scale factor $c$. The scaled point $B$ is $A$ scaled by $c$ relative to the origin, or $B=cA$ in normal terms, satisfying $hat B=hat chat A,|B|=|c||A|$, using $hat x$ to denote the direction/sign of $x$. Negative scale can alternatively be calculated as $B=(-c)(-A)$.



Rectangles don't work like in Euclidean space, so it's difficult to define non-uniform scale, hence I only ask for uniform.



Using this and translation, we can use linear interpolation to get a line equation $C=(1-t)A+tB=A+t(B-A)$.




Polar coordinates seem like a pretty good candidate except for translation:



  • Mapping: $(r,theta)to (kcostheta,ksintheta,z)$ using $z=cosh(r),k=sinh(r)=sqrtz^2-1$

  • Inverse mapping: $(x,y,z)to (textacosh(z),textatan_2(y,x))$ using atan2

  • Magnitude: $(r,theta)to r$

  • Negation: $(r,theta)to (r,theta+pi)$

  • Reflection: $(r,theta)to (r,-theta)$

  • Translation: $(r_1,theta_1),(r_2,theta_2)totextprobably something nasty$

  • Rotation: $(r,theta),alphato (r,theta+alpha)$

  • Scale: $(r,theta),cto (rc,theta)text or (-rc,theta+pi)$

There's still other obscure coordinate systems that might be easier to work with, being a little more difficult for the others but having easier translation. So I'm not settling on polar yet.







share|cite|improve this question


















  • 2




    Are you familiar with the unit disk and the upper half plane models of the hyperbolic plane?
    – Moishe Cohen
    Mar 5 at 1:29










  • I concur most strongly with the implied suggestion of @MoisheCohen that you familiarize yourself with the unit disk or upper half plane representation of the hyperbolic plane.
    – Lubin
    Mar 5 at 4:02






  • 1




    In my experience the hyperboloid model is the easiest to work with, much easier than disk/half plane/polar. My formulas are here, I think they cover most of what you want. Some of the formulas can be simplified. I use rgpushxto0 as the translation matrix.
    – Zeno Rogue
    Mar 6 at 11:22






  • 1




    I didn't notice before that linear transforms can be done in the hyperboloid model as linear transforms. That's very useful. Will give it a try, thanks!
    – EPICI
    Mar 6 at 19:09














up vote
1
down vote

favorite












I want to do some analytic geometry in the hyperbolic plane. I haven't chosen a coordinate system yet since I don't know which would make the math easiest. It's a lot to ask for, but I want the description of the coordinate system (map and inverse map for these coordinates to hyperboloid model coordinates) and closed form expressions for as many of these as possible (intermediate variables is okay if it reduces the work needed to calculate these). Also avoid unsolveable integrals. The rest I'll try to fill in myself. I just don't want to be doing the math for all the different options myself when someone else has probably already done it before.



In my descriptions I treat points as interchangeable with vectors, though I don't assume operations that work in the Euclidean plane also work in the hyperbolic plane.



Geometry requires the origin be an actual point, though usually we assume it to be $O(0,0)$ or equivalent to simplify math.




Point Magnitude



Given origin point $O$ and point $A$. $|A|$ is the length of the shortest path from $O$ to $A$.



Point Negation



Given origin point $O$ and point $A$. $-A$ is $A$ rotated $180^circ$ about $O$, or $A$ scaled by $-1$ relative to $O$.



This should be easy to calculate, if it isn't just something like $(x,y)to (-x,-y)$.



Point Axis Reflection



Given origin point $O$ and point $A$. $overline A$ is $A$ reflected in the $x$ axis for $O$.



This should also be easy to calculate, if it's not just something like $(x,y)to (x,-y)$.



Point Translation



Given origin point $O$, point $A$, and translation vector $B$. The translated point $C$ is whatever $A$ would be called if $-B$ was the origin, setting the rotation so that $O$ to the new origin $-B$ looks the same as $B$ to the old origin $O$.



This translation is the basis of the new non-commutative addition and subtraction operators: $C=A+B,C-B=A$. Since the side-angle-side congruence still holds, we have an alternative choice for the geometrical interpretation:



Treating A as the new origin, B is drawn, which is C to the origin. Treating -B as the new origin, C is drawn, which is A to the origin.



Don't mind the straight lines. $A+Bne B+A$ because you can't have parallelograms without parallel lines. Other working geometric interpretations reduce to this or produce the same triangle.



Because this is the hyperbolic plane, translations result in rotations. This is okay.



Using this and magnitude, the equation for line length $|A-B|$ can be derived.



Point Rotation



Given origin point $O$, point $A$, and angle $theta$. The rotated point $B$ is $A$ rotated by $theta$ about the origin, satisfying $|A|=|B|,angle BOA=theta$.



Since translations also rotate, this can be derived the hard way from a series of translations.



Point Scale



Given origin point $O$, point $A$, and scale factor $c$. The scaled point $B$ is $A$ scaled by $c$ relative to the origin, or $B=cA$ in normal terms, satisfying $hat B=hat chat A,|B|=|c||A|$, using $hat x$ to denote the direction/sign of $x$. Negative scale can alternatively be calculated as $B=(-c)(-A)$.



Rectangles don't work like in Euclidean space, so it's difficult to define non-uniform scale, hence I only ask for uniform.



Using this and translation, we can use linear interpolation to get a line equation $C=(1-t)A+tB=A+t(B-A)$.




Polar coordinates seem like a pretty good candidate except for translation:



  • Mapping: $(r,theta)to (kcostheta,ksintheta,z)$ using $z=cosh(r),k=sinh(r)=sqrtz^2-1$

  • Inverse mapping: $(x,y,z)to (textacosh(z),textatan_2(y,x))$ using atan2

  • Magnitude: $(r,theta)to r$

  • Negation: $(r,theta)to (r,theta+pi)$

  • Reflection: $(r,theta)to (r,-theta)$

  • Translation: $(r_1,theta_1),(r_2,theta_2)totextprobably something nasty$

  • Rotation: $(r,theta),alphato (r,theta+alpha)$

  • Scale: $(r,theta),cto (rc,theta)text or (-rc,theta+pi)$

There's still other obscure coordinate systems that might be easier to work with, being a little more difficult for the others but having easier translation. So I'm not settling on polar yet.







share|cite|improve this question


















  • 2




    Are you familiar with the unit disk and the upper half plane models of the hyperbolic plane?
    – Moishe Cohen
    Mar 5 at 1:29










  • I concur most strongly with the implied suggestion of @MoisheCohen that you familiarize yourself with the unit disk or upper half plane representation of the hyperbolic plane.
    – Lubin
    Mar 5 at 4:02






  • 1




    In my experience the hyperboloid model is the easiest to work with, much easier than disk/half plane/polar. My formulas are here, I think they cover most of what you want. Some of the formulas can be simplified. I use rgpushxto0 as the translation matrix.
    – Zeno Rogue
    Mar 6 at 11:22






  • 1




    I didn't notice before that linear transforms can be done in the hyperboloid model as linear transforms. That's very useful. Will give it a try, thanks!
    – EPICI
    Mar 6 at 19:09












up vote
1
down vote

favorite









up vote
1
down vote

favorite











I want to do some analytic geometry in the hyperbolic plane. I haven't chosen a coordinate system yet since I don't know which would make the math easiest. It's a lot to ask for, but I want the description of the coordinate system (map and inverse map for these coordinates to hyperboloid model coordinates) and closed form expressions for as many of these as possible (intermediate variables is okay if it reduces the work needed to calculate these). Also avoid unsolveable integrals. The rest I'll try to fill in myself. I just don't want to be doing the math for all the different options myself when someone else has probably already done it before.



In my descriptions I treat points as interchangeable with vectors, though I don't assume operations that work in the Euclidean plane also work in the hyperbolic plane.



Geometry requires the origin be an actual point, though usually we assume it to be $O(0,0)$ or equivalent to simplify math.




Point Magnitude



Given origin point $O$ and point $A$. $|A|$ is the length of the shortest path from $O$ to $A$.



Point Negation



Given origin point $O$ and point $A$. $-A$ is $A$ rotated $180^circ$ about $O$, or $A$ scaled by $-1$ relative to $O$.



This should be easy to calculate, if it isn't just something like $(x,y)to (-x,-y)$.



Point Axis Reflection



Given origin point $O$ and point $A$. $overline A$ is $A$ reflected in the $x$ axis for $O$.



This should also be easy to calculate, if it's not just something like $(x,y)to (x,-y)$.



Point Translation



Given origin point $O$, point $A$, and translation vector $B$. The translated point $C$ is whatever $A$ would be called if $-B$ was the origin, setting the rotation so that $O$ to the new origin $-B$ looks the same as $B$ to the old origin $O$.



This translation is the basis of the new non-commutative addition and subtraction operators: $C=A+B,C-B=A$. Since the side-angle-side congruence still holds, we have an alternative choice for the geometrical interpretation:



Treating A as the new origin, B is drawn, which is C to the origin. Treating -B as the new origin, C is drawn, which is A to the origin.



Don't mind the straight lines. $A+Bne B+A$ because you can't have parallelograms without parallel lines. Other working geometric interpretations reduce to this or produce the same triangle.



Because this is the hyperbolic plane, translations result in rotations. This is okay.



Using this and magnitude, the equation for line length $|A-B|$ can be derived.



Point Rotation



Given origin point $O$, point $A$, and angle $theta$. The rotated point $B$ is $A$ rotated by $theta$ about the origin, satisfying $|A|=|B|,angle BOA=theta$.



Since translations also rotate, this can be derived the hard way from a series of translations.



Point Scale



Given origin point $O$, point $A$, and scale factor $c$. The scaled point $B$ is $A$ scaled by $c$ relative to the origin, or $B=cA$ in normal terms, satisfying $hat B=hat chat A,|B|=|c||A|$, using $hat x$ to denote the direction/sign of $x$. Negative scale can alternatively be calculated as $B=(-c)(-A)$.



Rectangles don't work like in Euclidean space, so it's difficult to define non-uniform scale, hence I only ask for uniform.



Using this and translation, we can use linear interpolation to get a line equation $C=(1-t)A+tB=A+t(B-A)$.




Polar coordinates seem like a pretty good candidate except for translation:



  • Mapping: $(r,theta)to (kcostheta,ksintheta,z)$ using $z=cosh(r),k=sinh(r)=sqrtz^2-1$

  • Inverse mapping: $(x,y,z)to (textacosh(z),textatan_2(y,x))$ using atan2

  • Magnitude: $(r,theta)to r$

  • Negation: $(r,theta)to (r,theta+pi)$

  • Reflection: $(r,theta)to (r,-theta)$

  • Translation: $(r_1,theta_1),(r_2,theta_2)totextprobably something nasty$

  • Rotation: $(r,theta),alphato (r,theta+alpha)$

  • Scale: $(r,theta),cto (rc,theta)text or (-rc,theta+pi)$

There's still other obscure coordinate systems that might be easier to work with, being a little more difficult for the others but having easier translation. So I'm not settling on polar yet.







share|cite|improve this question














I want to do some analytic geometry in the hyperbolic plane. I haven't chosen a coordinate system yet since I don't know which would make the math easiest. It's a lot to ask for, but I want the description of the coordinate system (map and inverse map for these coordinates to hyperboloid model coordinates) and closed form expressions for as many of these as possible (intermediate variables is okay if it reduces the work needed to calculate these). Also avoid unsolveable integrals. The rest I'll try to fill in myself. I just don't want to be doing the math for all the different options myself when someone else has probably already done it before.



In my descriptions I treat points as interchangeable with vectors, though I don't assume operations that work in the Euclidean plane also work in the hyperbolic plane.



Geometry requires the origin be an actual point, though usually we assume it to be $O(0,0)$ or equivalent to simplify math.




Point Magnitude



Given origin point $O$ and point $A$. $|A|$ is the length of the shortest path from $O$ to $A$.



Point Negation



Given origin point $O$ and point $A$. $-A$ is $A$ rotated $180^circ$ about $O$, or $A$ scaled by $-1$ relative to $O$.



This should be easy to calculate, if it isn't just something like $(x,y)to (-x,-y)$.



Point Axis Reflection



Given origin point $O$ and point $A$. $overline A$ is $A$ reflected in the $x$ axis for $O$.



This should also be easy to calculate, if it's not just something like $(x,y)to (x,-y)$.



Point Translation



Given origin point $O$, point $A$, and translation vector $B$. The translated point $C$ is whatever $A$ would be called if $-B$ was the origin, setting the rotation so that $O$ to the new origin $-B$ looks the same as $B$ to the old origin $O$.



This translation is the basis of the new non-commutative addition and subtraction operators: $C=A+B,C-B=A$. Since the side-angle-side congruence still holds, we have an alternative choice for the geometrical interpretation:



Treating A as the new origin, B is drawn, which is C to the origin. Treating -B as the new origin, C is drawn, which is A to the origin.



Don't mind the straight lines. $A+Bne B+A$ because you can't have parallelograms without parallel lines. Other working geometric interpretations reduce to this or produce the same triangle.



Because this is the hyperbolic plane, translations result in rotations. This is okay.



Using this and magnitude, the equation for line length $|A-B|$ can be derived.



Point Rotation



Given origin point $O$, point $A$, and angle $theta$. The rotated point $B$ is $A$ rotated by $theta$ about the origin, satisfying $|A|=|B|,angle BOA=theta$.



Since translations also rotate, this can be derived the hard way from a series of translations.



Point Scale



Given origin point $O$, point $A$, and scale factor $c$. The scaled point $B$ is $A$ scaled by $c$ relative to the origin, or $B=cA$ in normal terms, satisfying $hat B=hat chat A,|B|=|c||A|$, using $hat x$ to denote the direction/sign of $x$. Negative scale can alternatively be calculated as $B=(-c)(-A)$.



Rectangles don't work like in Euclidean space, so it's difficult to define non-uniform scale, hence I only ask for uniform.



Using this and translation, we can use linear interpolation to get a line equation $C=(1-t)A+tB=A+t(B-A)$.




Polar coordinates seem like a pretty good candidate except for translation:



  • Mapping: $(r,theta)to (kcostheta,ksintheta,z)$ using $z=cosh(r),k=sinh(r)=sqrtz^2-1$

  • Inverse mapping: $(x,y,z)to (textacosh(z),textatan_2(y,x))$ using atan2

  • Magnitude: $(r,theta)to r$

  • Negation: $(r,theta)to (r,theta+pi)$

  • Reflection: $(r,theta)to (r,-theta)$

  • Translation: $(r_1,theta_1),(r_2,theta_2)totextprobably something nasty$

  • Rotation: $(r,theta),alphato (r,theta+alpha)$

  • Scale: $(r,theta),cto (rc,theta)text or (-rc,theta+pi)$

There's still other obscure coordinate systems that might be easier to work with, being a little more difficult for the others but having easier translation. So I'm not settling on polar yet.









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 5 at 6:28

























asked Mar 4 at 23:21









EPICI

757




757







  • 2




    Are you familiar with the unit disk and the upper half plane models of the hyperbolic plane?
    – Moishe Cohen
    Mar 5 at 1:29










  • I concur most strongly with the implied suggestion of @MoisheCohen that you familiarize yourself with the unit disk or upper half plane representation of the hyperbolic plane.
    – Lubin
    Mar 5 at 4:02






  • 1




    In my experience the hyperboloid model is the easiest to work with, much easier than disk/half plane/polar. My formulas are here, I think they cover most of what you want. Some of the formulas can be simplified. I use rgpushxto0 as the translation matrix.
    – Zeno Rogue
    Mar 6 at 11:22






  • 1




    I didn't notice before that linear transforms can be done in the hyperboloid model as linear transforms. That's very useful. Will give it a try, thanks!
    – EPICI
    Mar 6 at 19:09












  • 2




    Are you familiar with the unit disk and the upper half plane models of the hyperbolic plane?
    – Moishe Cohen
    Mar 5 at 1:29










  • I concur most strongly with the implied suggestion of @MoisheCohen that you familiarize yourself with the unit disk or upper half plane representation of the hyperbolic plane.
    – Lubin
    Mar 5 at 4:02






  • 1




    In my experience the hyperboloid model is the easiest to work with, much easier than disk/half plane/polar. My formulas are here, I think they cover most of what you want. Some of the formulas can be simplified. I use rgpushxto0 as the translation matrix.
    – Zeno Rogue
    Mar 6 at 11:22






  • 1




    I didn't notice before that linear transforms can be done in the hyperboloid model as linear transforms. That's very useful. Will give it a try, thanks!
    – EPICI
    Mar 6 at 19:09







2




2




Are you familiar with the unit disk and the upper half plane models of the hyperbolic plane?
– Moishe Cohen
Mar 5 at 1:29




Are you familiar with the unit disk and the upper half plane models of the hyperbolic plane?
– Moishe Cohen
Mar 5 at 1:29












I concur most strongly with the implied suggestion of @MoisheCohen that you familiarize yourself with the unit disk or upper half plane representation of the hyperbolic plane.
– Lubin
Mar 5 at 4:02




I concur most strongly with the implied suggestion of @MoisheCohen that you familiarize yourself with the unit disk or upper half plane representation of the hyperbolic plane.
– Lubin
Mar 5 at 4:02




1




1




In my experience the hyperboloid model is the easiest to work with, much easier than disk/half plane/polar. My formulas are here, I think they cover most of what you want. Some of the formulas can be simplified. I use rgpushxto0 as the translation matrix.
– Zeno Rogue
Mar 6 at 11:22




In my experience the hyperboloid model is the easiest to work with, much easier than disk/half plane/polar. My formulas are here, I think they cover most of what you want. Some of the formulas can be simplified. I use rgpushxto0 as the translation matrix.
– Zeno Rogue
Mar 6 at 11:22




1




1




I didn't notice before that linear transforms can be done in the hyperboloid model as linear transforms. That's very useful. Will give it a try, thanks!
– EPICI
Mar 6 at 19:09




I didn't notice before that linear transforms can be done in the hyperboloid model as linear transforms. That's very useful. Will give it a try, thanks!
– EPICI
Mar 6 at 19:09










1 Answer
1






active

oldest

votes

















up vote
1
down vote



accepted










Here are six simple coordinate systems, using the hyperboloid model for comparison. The metric is determined by $ds^2=dx^2+dy^2-dz^2$.




Polar (circular):



$$beginalign &x=sinhvarphicostheta \ &y=sinhvarphisintheta \ &z=coshvarphi endalign$$
$$ds^2=dvarphi^2+sinh^2varphi,dtheta^2$$



Rotation by angle $alpha$ around the point $varphi=0$ : $quadthetamapstotheta+alpha$



Reflection across the line $theta=alpha$ : $quadthetamapsto2alpha-theta$




Isothermal polar:



$$beginalign &x=textcsch,rhocostheta \ &y=textcsch,rhosintheta \ &z=coth,rho endalign$$
$$ds^2=textcsch^2rho,(drho^2+dtheta^2)$$




Fermi coordinates (hypercyclic):



$$beginalign &x=sinh ucosh v \ &y=sinh v \ &z=cosh ucosh v endalign$$
$$ds^2=du^2cosh^2v+dv^2$$



Translation by distance $alpha$ along the line $v=0$ : $quad umapsto u+alpha$



(Points not on this line will move a greater distance than $alpha$, but will maintain their distance $v$ from the line.)



Reflection across the line $u=alpha$ : $quad umapsto2alpha-u$




Isothermal Fermi:



$$beginalign &x=sinh usecomega \ &y=tanomega \ &z=cosh usecomega endalign$$
$$ds^2=sec^2omega,(du^2+domega^2)$$




Horocyclic:



$$beginalign &x=Ue^-V \ &y=tfrac12U^2e^-V+sinh V \ &z=tfrac12U^2e^-V+cosh V endalign$$
$$ds^2=e^-2VdU^2+dV^2$$



"Rotation"/"Translation" by length $alpha$ along the horocycle $V=0$ : $quad Umapsto U+alpha$



(Points not on this horocycle will maintain their distance $V$ from it.)



Reflection across the line $U=alpha$ : $quad Umapsto2alpha-U$




Isothermal horocyclic (half-plane):



$$beginalign &x=(2U)/(2W) \ &y=(U^2+W^2-1)/(2W) \ &z=(U^2+W^2+1)/(2W) endalign$$
$$ds^2=tfrac1W^2(dU^2+dW^2)$$




I agree with ZenoRogue that the hyperboloid model is the easiest and most useful. Isometries of the hyperbolic plane correspond to pseudo-Euclidean isometries of the hyperboloid. Such a transformation can be represented not only by a matrix, but also by multiplication of vectors.



For example, the reflection of a point $vec p$ across a plane (in psEuc space) with normal vector $vec n$ is



$$pmapsto-n^-1pn$$



The inverse is $n^-1=frac1n^2n=frac1ncdot nn$. This points in the same direction as $n$, but has reciprocal magnitude. (Note that a psEuc vector may square to a negative number, so its inverse points in the opposite direction.)



Any isometry can be composed from reflections. Two consecutive reflections across planes intersecting at an angle $alpha$ make a rotation by angle $2alpha$. If the two planes' normals are $n_1$ and $n_2$, then the rotation is



$$pmapsto n_2^-1n_1^-1pn_1n_2=(n_1n_2)^-1p(n_1n_2)$$






share|cite|improve this answer




















    Your Answer




    StackExchange.ifUsing("editor", function ()
    return StackExchange.using("mathjaxEditing", function ()
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    );
    );
    , "mathjax-editing");

    StackExchange.ready(function()
    var channelOptions =
    tags: "".split(" "),
    id: "69"
    ;
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function()
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled)
    StackExchange.using("snippets", function()
    createEditor();
    );

    else
    createEditor();

    );

    function createEditor()
    StackExchange.prepareEditor(
    heartbeatType: 'answer',
    convertImagesToLinks: true,
    noModals: false,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    );



    );













     

    draft saved


    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2677006%2fcoordinate-system-with-easy-linear-point-transformations-in-hyperbolic-plane%23new-answer', 'question_page');

    );

    Post as a guest






























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    Here are six simple coordinate systems, using the hyperboloid model for comparison. The metric is determined by $ds^2=dx^2+dy^2-dz^2$.




    Polar (circular):



    $$beginalign &x=sinhvarphicostheta \ &y=sinhvarphisintheta \ &z=coshvarphi endalign$$
    $$ds^2=dvarphi^2+sinh^2varphi,dtheta^2$$



    Rotation by angle $alpha$ around the point $varphi=0$ : $quadthetamapstotheta+alpha$



    Reflection across the line $theta=alpha$ : $quadthetamapsto2alpha-theta$




    Isothermal polar:



    $$beginalign &x=textcsch,rhocostheta \ &y=textcsch,rhosintheta \ &z=coth,rho endalign$$
    $$ds^2=textcsch^2rho,(drho^2+dtheta^2)$$




    Fermi coordinates (hypercyclic):



    $$beginalign &x=sinh ucosh v \ &y=sinh v \ &z=cosh ucosh v endalign$$
    $$ds^2=du^2cosh^2v+dv^2$$



    Translation by distance $alpha$ along the line $v=0$ : $quad umapsto u+alpha$



    (Points not on this line will move a greater distance than $alpha$, but will maintain their distance $v$ from the line.)



    Reflection across the line $u=alpha$ : $quad umapsto2alpha-u$




    Isothermal Fermi:



    $$beginalign &x=sinh usecomega \ &y=tanomega \ &z=cosh usecomega endalign$$
    $$ds^2=sec^2omega,(du^2+domega^2)$$




    Horocyclic:



    $$beginalign &x=Ue^-V \ &y=tfrac12U^2e^-V+sinh V \ &z=tfrac12U^2e^-V+cosh V endalign$$
    $$ds^2=e^-2VdU^2+dV^2$$



    "Rotation"/"Translation" by length $alpha$ along the horocycle $V=0$ : $quad Umapsto U+alpha$



    (Points not on this horocycle will maintain their distance $V$ from it.)



    Reflection across the line $U=alpha$ : $quad Umapsto2alpha-U$




    Isothermal horocyclic (half-plane):



    $$beginalign &x=(2U)/(2W) \ &y=(U^2+W^2-1)/(2W) \ &z=(U^2+W^2+1)/(2W) endalign$$
    $$ds^2=tfrac1W^2(dU^2+dW^2)$$




    I agree with ZenoRogue that the hyperboloid model is the easiest and most useful. Isometries of the hyperbolic plane correspond to pseudo-Euclidean isometries of the hyperboloid. Such a transformation can be represented not only by a matrix, but also by multiplication of vectors.



    For example, the reflection of a point $vec p$ across a plane (in psEuc space) with normal vector $vec n$ is



    $$pmapsto-n^-1pn$$



    The inverse is $n^-1=frac1n^2n=frac1ncdot nn$. This points in the same direction as $n$, but has reciprocal magnitude. (Note that a psEuc vector may square to a negative number, so its inverse points in the opposite direction.)



    Any isometry can be composed from reflections. Two consecutive reflections across planes intersecting at an angle $alpha$ make a rotation by angle $2alpha$. If the two planes' normals are $n_1$ and $n_2$, then the rotation is



    $$pmapsto n_2^-1n_1^-1pn_1n_2=(n_1n_2)^-1p(n_1n_2)$$






    share|cite|improve this answer
























      up vote
      1
      down vote



      accepted










      Here are six simple coordinate systems, using the hyperboloid model for comparison. The metric is determined by $ds^2=dx^2+dy^2-dz^2$.




      Polar (circular):



      $$beginalign &x=sinhvarphicostheta \ &y=sinhvarphisintheta \ &z=coshvarphi endalign$$
      $$ds^2=dvarphi^2+sinh^2varphi,dtheta^2$$



      Rotation by angle $alpha$ around the point $varphi=0$ : $quadthetamapstotheta+alpha$



      Reflection across the line $theta=alpha$ : $quadthetamapsto2alpha-theta$




      Isothermal polar:



      $$beginalign &x=textcsch,rhocostheta \ &y=textcsch,rhosintheta \ &z=coth,rho endalign$$
      $$ds^2=textcsch^2rho,(drho^2+dtheta^2)$$




      Fermi coordinates (hypercyclic):



      $$beginalign &x=sinh ucosh v \ &y=sinh v \ &z=cosh ucosh v endalign$$
      $$ds^2=du^2cosh^2v+dv^2$$



      Translation by distance $alpha$ along the line $v=0$ : $quad umapsto u+alpha$



      (Points not on this line will move a greater distance than $alpha$, but will maintain their distance $v$ from the line.)



      Reflection across the line $u=alpha$ : $quad umapsto2alpha-u$




      Isothermal Fermi:



      $$beginalign &x=sinh usecomega \ &y=tanomega \ &z=cosh usecomega endalign$$
      $$ds^2=sec^2omega,(du^2+domega^2)$$




      Horocyclic:



      $$beginalign &x=Ue^-V \ &y=tfrac12U^2e^-V+sinh V \ &z=tfrac12U^2e^-V+cosh V endalign$$
      $$ds^2=e^-2VdU^2+dV^2$$



      "Rotation"/"Translation" by length $alpha$ along the horocycle $V=0$ : $quad Umapsto U+alpha$



      (Points not on this horocycle will maintain their distance $V$ from it.)



      Reflection across the line $U=alpha$ : $quad Umapsto2alpha-U$




      Isothermal horocyclic (half-plane):



      $$beginalign &x=(2U)/(2W) \ &y=(U^2+W^2-1)/(2W) \ &z=(U^2+W^2+1)/(2W) endalign$$
      $$ds^2=tfrac1W^2(dU^2+dW^2)$$




      I agree with ZenoRogue that the hyperboloid model is the easiest and most useful. Isometries of the hyperbolic plane correspond to pseudo-Euclidean isometries of the hyperboloid. Such a transformation can be represented not only by a matrix, but also by multiplication of vectors.



      For example, the reflection of a point $vec p$ across a plane (in psEuc space) with normal vector $vec n$ is



      $$pmapsto-n^-1pn$$



      The inverse is $n^-1=frac1n^2n=frac1ncdot nn$. This points in the same direction as $n$, but has reciprocal magnitude. (Note that a psEuc vector may square to a negative number, so its inverse points in the opposite direction.)



      Any isometry can be composed from reflections. Two consecutive reflections across planes intersecting at an angle $alpha$ make a rotation by angle $2alpha$. If the two planes' normals are $n_1$ and $n_2$, then the rotation is



      $$pmapsto n_2^-1n_1^-1pn_1n_2=(n_1n_2)^-1p(n_1n_2)$$






      share|cite|improve this answer






















        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        Here are six simple coordinate systems, using the hyperboloid model for comparison. The metric is determined by $ds^2=dx^2+dy^2-dz^2$.




        Polar (circular):



        $$beginalign &x=sinhvarphicostheta \ &y=sinhvarphisintheta \ &z=coshvarphi endalign$$
        $$ds^2=dvarphi^2+sinh^2varphi,dtheta^2$$



        Rotation by angle $alpha$ around the point $varphi=0$ : $quadthetamapstotheta+alpha$



        Reflection across the line $theta=alpha$ : $quadthetamapsto2alpha-theta$




        Isothermal polar:



        $$beginalign &x=textcsch,rhocostheta \ &y=textcsch,rhosintheta \ &z=coth,rho endalign$$
        $$ds^2=textcsch^2rho,(drho^2+dtheta^2)$$




        Fermi coordinates (hypercyclic):



        $$beginalign &x=sinh ucosh v \ &y=sinh v \ &z=cosh ucosh v endalign$$
        $$ds^2=du^2cosh^2v+dv^2$$



        Translation by distance $alpha$ along the line $v=0$ : $quad umapsto u+alpha$



        (Points not on this line will move a greater distance than $alpha$, but will maintain their distance $v$ from the line.)



        Reflection across the line $u=alpha$ : $quad umapsto2alpha-u$




        Isothermal Fermi:



        $$beginalign &x=sinh usecomega \ &y=tanomega \ &z=cosh usecomega endalign$$
        $$ds^2=sec^2omega,(du^2+domega^2)$$




        Horocyclic:



        $$beginalign &x=Ue^-V \ &y=tfrac12U^2e^-V+sinh V \ &z=tfrac12U^2e^-V+cosh V endalign$$
        $$ds^2=e^-2VdU^2+dV^2$$



        "Rotation"/"Translation" by length $alpha$ along the horocycle $V=0$ : $quad Umapsto U+alpha$



        (Points not on this horocycle will maintain their distance $V$ from it.)



        Reflection across the line $U=alpha$ : $quad Umapsto2alpha-U$




        Isothermal horocyclic (half-plane):



        $$beginalign &x=(2U)/(2W) \ &y=(U^2+W^2-1)/(2W) \ &z=(U^2+W^2+1)/(2W) endalign$$
        $$ds^2=tfrac1W^2(dU^2+dW^2)$$




        I agree with ZenoRogue that the hyperboloid model is the easiest and most useful. Isometries of the hyperbolic plane correspond to pseudo-Euclidean isometries of the hyperboloid. Such a transformation can be represented not only by a matrix, but also by multiplication of vectors.



        For example, the reflection of a point $vec p$ across a plane (in psEuc space) with normal vector $vec n$ is



        $$pmapsto-n^-1pn$$



        The inverse is $n^-1=frac1n^2n=frac1ncdot nn$. This points in the same direction as $n$, but has reciprocal magnitude. (Note that a psEuc vector may square to a negative number, so its inverse points in the opposite direction.)



        Any isometry can be composed from reflections. Two consecutive reflections across planes intersecting at an angle $alpha$ make a rotation by angle $2alpha$. If the two planes' normals are $n_1$ and $n_2$, then the rotation is



        $$pmapsto n_2^-1n_1^-1pn_1n_2=(n_1n_2)^-1p(n_1n_2)$$






        share|cite|improve this answer












        Here are six simple coordinate systems, using the hyperboloid model for comparison. The metric is determined by $ds^2=dx^2+dy^2-dz^2$.




        Polar (circular):



        $$beginalign &x=sinhvarphicostheta \ &y=sinhvarphisintheta \ &z=coshvarphi endalign$$
        $$ds^2=dvarphi^2+sinh^2varphi,dtheta^2$$



        Rotation by angle $alpha$ around the point $varphi=0$ : $quadthetamapstotheta+alpha$



        Reflection across the line $theta=alpha$ : $quadthetamapsto2alpha-theta$




        Isothermal polar:



        $$beginalign &x=textcsch,rhocostheta \ &y=textcsch,rhosintheta \ &z=coth,rho endalign$$
        $$ds^2=textcsch^2rho,(drho^2+dtheta^2)$$




        Fermi coordinates (hypercyclic):



        $$beginalign &x=sinh ucosh v \ &y=sinh v \ &z=cosh ucosh v endalign$$
        $$ds^2=du^2cosh^2v+dv^2$$



        Translation by distance $alpha$ along the line $v=0$ : $quad umapsto u+alpha$



        (Points not on this line will move a greater distance than $alpha$, but will maintain their distance $v$ from the line.)



        Reflection across the line $u=alpha$ : $quad umapsto2alpha-u$




        Isothermal Fermi:



        $$beginalign &x=sinh usecomega \ &y=tanomega \ &z=cosh usecomega endalign$$
        $$ds^2=sec^2omega,(du^2+domega^2)$$




        Horocyclic:



        $$beginalign &x=Ue^-V \ &y=tfrac12U^2e^-V+sinh V \ &z=tfrac12U^2e^-V+cosh V endalign$$
        $$ds^2=e^-2VdU^2+dV^2$$



        "Rotation"/"Translation" by length $alpha$ along the horocycle $V=0$ : $quad Umapsto U+alpha$



        (Points not on this horocycle will maintain their distance $V$ from it.)



        Reflection across the line $U=alpha$ : $quad Umapsto2alpha-U$




        Isothermal horocyclic (half-plane):



        $$beginalign &x=(2U)/(2W) \ &y=(U^2+W^2-1)/(2W) \ &z=(U^2+W^2+1)/(2W) endalign$$
        $$ds^2=tfrac1W^2(dU^2+dW^2)$$




        I agree with ZenoRogue that the hyperboloid model is the easiest and most useful. Isometries of the hyperbolic plane correspond to pseudo-Euclidean isometries of the hyperboloid. Such a transformation can be represented not only by a matrix, but also by multiplication of vectors.



        For example, the reflection of a point $vec p$ across a plane (in psEuc space) with normal vector $vec n$ is



        $$pmapsto-n^-1pn$$



        The inverse is $n^-1=frac1n^2n=frac1ncdot nn$. This points in the same direction as $n$, but has reciprocal magnitude. (Note that a psEuc vector may square to a negative number, so its inverse points in the opposite direction.)



        Any isometry can be composed from reflections. Two consecutive reflections across planes intersecting at an angle $alpha$ make a rotation by angle $2alpha$. If the two planes' normals are $n_1$ and $n_2$, then the rotation is



        $$pmapsto n_2^-1n_1^-1pn_1n_2=(n_1n_2)^-1p(n_1n_2)$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 26 at 20:44









        mr_e_man

        1,049422




        1,049422



























             

            draft saved


            draft discarded















































             


            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2677006%2fcoordinate-system-with-easy-linear-point-transformations-in-hyperbolic-plane%23new-answer', 'question_page');

            );

            Post as a guest













































































            這個網誌中的熱門文章

            How to combine Bézier curves to a surface?

            Why am i infinitely getting the same tweet with the Twitter Search API?

            Carbon dioxide