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Thread: Regular polygons in the hyperbolic plane

  1. #1 Regular polygons in the hyperbolic plane 
    Forum Radioactive Isotope MagiMaster's Avatar
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    I've been working with the hyperbolic plane quite a bit lately, and I've run across a few difficult questions I can't find the answers for.

    First, I know that a regular polygon in the hyperbolic plane has to be constructed with a specific radius to be able to tessellate, and I've found a formula that works, but I haven't found any explaination for it.

    Second, I've found a way of describing a translation in the hyperbolic plane by using a specific pair of reflections (this should work in any space where perpendicular is well defined) but is there an easier way?

    I'll probably have some other questions later. Thanks.


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  3. #2  
    Forum Professor serpicojr's Avatar
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    Shouldn't you be able to tessellate so long as the angles of the polygon are of the form 2π/m? This would correspond to having m polygons meeting at each vertex. This is the same as tessellating by isosceles triangles with two angles of the form π/m and the last of the form π/n, where n is the number of sides, and 2/m + 1/n < 1. Then using hyperbolic trig, you should be able to get side lengths.

    I'm assuming that, by translations, you mean anything conjugate to the mappings z |-> z+r, r any real number. Equivalently, this is a mapping which fixes one point on the real axis or infinity, or a mapping corresponding to a linear fractional transformation with trace squared equal to 4. To me, the easiest way to describe these is by mapping the fixed point to infinity, describing it as a translation fixing infinity, and then mapping back to the original fixed point. I.e., if a is a real number, then your translations t(z) look like the composition of the maps:

    f(z) = -1/(z-a)
    g(z) = z+r, r any real number
    h(z) = f<sup>-1</sup>(z) = (az-1)/z

    So t(z) = hgf(z).


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  4. #3  
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    Hmm... I don't know why I didn't think of using hyperbolic trig, but now that I've done so, I get a different answer than the one I found before. I think that part of it is that one is giving a hyperbolic distance, and the other is giving a euclidean distance. After accounting for that, they're still different, so I'm not sure what's up. Anyway, here are the two equations.

    a and b are the two angles of a right triangle
    1) acosh((cos(a)*cos(b))/(sin(a)*sin(b)))
    2) sin(pi/2 - a - b)/(1 - sin(a)^2 - sin(b)*2)

    The second one seems to give the correct answer for the euclidean radius, but I don't know how to derive it. The first is what I got from the hyperbolic law of cosines, but it doesn't seem to give the right answer.

    Edit: I must have done something wrong, because they seem to be giving the same answer now.

    Anyway, about the translations. I don't understand your equations. First, is z the euclidean coordinates (expressed as x + iy) for the point being moved? Second, what is a and r in your equations? I think r is how far you want to move things, but I don't know what a means. Is it the angle things are moving?
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  5. #4  
    Forum Professor serpicojr's Avatar
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    I'm doing things via the upper half-plane model, so my coordinates are x+iy. I'm looking at the group of orientation-preserving isometries as being the fractional linear transformations in SL<sub>2</sub>(R), so the matrix A with rows [a,b] and [c,d], a, b, c, d real, acts via:

    Az = (az+b)/(cz+d)

    I'm not sure what you mean by a translation, so I took a guess. In the Euclidean plane, any orientation-preserving isometry is either a rotation or a translation--you either fix a point or you don't. In the hyperbolic plane, you can break down the non-rotations even further. You'll have either one or two fixed points on R U {∞}. If there's one fixed point, the map acts more like translation, at least in that if the fixed point is ∞, then it llterally is a (Euclidean) translation Az = z+b. If there are two fixed points, say 0 and ∞, then it acts more like a dilation, in that the map is then Az = az. (Technically, this should really be Az = az/d, ad = 1.) So were you talking about both of these cases or just the first case?

    In the case of one fixed point on the extended real line, I was letting a be the fixed point, and I was letting r be any arbitrary real number, and the map t(z) I constructed represents the general form of an isometry fixing a.
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  6. #5  
    Forum Radioactive Isotope MagiMaster's Avatar
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    I'm using the Poincare disc model, so Euclidean translations don't work at all. I suppose by translation I mean the most natural length-preserving transformation that takes a given point to a given destination. I don't think these transformations would fix any points other than ∞, but I could be wrong.
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  7. #6  
    Forum Professor serpicojr's Avatar
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    Okay, I just realized something. Rigid motions on the hyperbolic plane with no fixed points come in two varieties: they fix either one or two points on R U {∞}. In the former case, points move along horocycles of the fixed point--in the case of a point x on R, these are Euclidean circles tangent to the real line at the point x, and in the case of ∞, these are Euclidean straight lines parallel to the real line. None of these paths are hyperbolic geodesic. In the latter case, suppose x and y in R are the fixed points. Then points move along paths which are Euclidean circles passing through x and y. Exactly one of these is a hyperbolic geodesic--the one which meets x and y at right angles to the real line. If one of the fixed points is ∞ and the other is x, then the paths are all Euclidean straight lines going through x. Only the vertical line is a hyperbolic geodesic.

    So... which of these do we want to call translations? It seems natural to call the latter ones translations, since some points actually move along geodesics. However, there's something about the former ones that seems very translation-like to me--you're not taking the shortest path from A to B, but all points are moving along the same sorts of paths. And although horocycles are not geodesics, they play an integral role in the geometry of the hyperbolic plane, whereas I haven't seen much discussion of non-horocycle, non-geodesic paths.

    Which do you like better?
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  8. #7  
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    I never did figure out how the Poincare half-plane mapped onto the Poincare disc, so I'm having trouble visualizing the two motions. I think a translation would be anything that moves point A to point B along a geodesic and preserves length and angles. I know that reflecting first about the geodesic perpendicular to the geodesic AB through A, and then reflecting about the perpendicular bisector of AB does this, but that's not a very good solution when you're trying to move a lot of points around (and you can't simply concatenate several translations into one using that method).
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  9. #8  
    Forum Professor serpicojr's Avatar
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    There's a Mobius transformation that maps the unit disk to the upper-half plane, and this mapping is an isomorphism in every sense imaginable in this context, in particular as Riemannian manifolds. The map is:

    z |->i(z+1)/(-z+1)

    (Note that this map sends 0 to i, 1 to ∞, and -1 to 0--this actually implies that the interval (-1,1) is mapped to the vertical line iy, y > 0.) The inverse map is:

    z |-> (z-i)/(z+i)

    Okay, so let's go with translation as you defined it: something taking A to B along geodesic (i.e., so that all points on the geodesic passing through A and B remain on said geodesic after the translation is said and done) and which preserves angles and distances (which is what I've meant by isometry/rigid motion). You may be assuming this, too, but let's throw in that orientation is preserved.

    Given all of this, there is a unique translation mapping A to B. This can be proved, for example, by looking at the upper-half plane model and asking how we can map i to a point iy, y > 0. (You can always apply a Mobius transformation to get yourself into this scenario.) Well, the only way is multiplication by y: the geodesic between i and iy is the line from 0 to ∞, and you can show that the only transformations that fix 0 and ∞ are of the form z |-> yz, y > 0. You can come up with a general formula if you just apply Mobius transformations to get yourself to this setup.

    Okay, what is a real good way to do this? Well, first, come up with a way to map A to 0 in the disk model. One way of doing this is:

    1. Send A to the upper half-plane by the map I gave above, and if the image of A is x+iy, then subtract x and divide by y. Then map back to the disk. This composition of maps gives you a map on the disk which sends A to 0.

    2. Now see where B has gone under this composite map. The image of B is of the form re<sup>iT</sup>, 0 < r < 1. Multiply by e<sup>-iT</sup>, so that the image of B is now in the interval (0,1). Composing this map with the above, we have a map which sends A to 0 and B to (0,1).

    3. Map back to the upper half-plane by the map I gave. This maps A to i and B to iy, y > 1. Let's call the composition of all the maps we've done up until now F. F is a map which sends the disk to the upper-half plane, A to i, and B to iy for some y > 1. This is, in the upper half-plane, the translation from A to B.

    4. Let z be in the disk model. Then the translation from A to B is given by:

    T(z) = F<sup>-1</sup>(y*F(z))

    So you have to calculate F.
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  10. #9  
    Forum Radioactive Isotope MagiMaster's Avatar
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    Thanks. That may be just what I'm looking for, but it may take a while to solve. I have access to MATLAB, but it's pretty bad with complex numbers.

    The main problem I'm having with using the two reflections, besides any speed issues, is that I can't store the original position of a point and a composition of the movements it's made. This means that points can move close enough to the edge (fairly easily) to exceed double precision. Without solving the equations it's hard to tell, but hopefully, a direct transformation that can be easily concatenated can solve both problems.
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