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Linear Fractional Transformations

Linear Fractional Transformations

We can express a coordinate in 2-space as a complex number

(x, y) = x + iy = z

Here we write down several properties of complex numbers for those who are not already familiar with them.

(Real part) Re(z) = x           (Imaginary part) Im(z) = y
z(t) = x(t) + iy(t)               z(t) = x(t) - iy(t)
              z'(t) = x'(t) + iy'(t)             (z(t))' = x'(t) - iy'(t) = z'(t)
|z(t)| = (z(t)z(t)) = (x(t)2 + y(t)2)
length = ab(x'(t)2 + y'(t)2)dt = ab(z'(t)(z(t))')dt

A linear fraction transformation is defined by the map

f(z) = (az + b) / (cz + d)

where a, b, c, d are real numbers such that ad - bc = 1. With this definition, suppose we want to apply a linear fractional transformation to a point (x, y) in the upper half-plane. We can write this point as a complex number z = x + iy. The image of z is f(z) = w, which corresponds to a point (Re(w), Im(w)).

Claim: f sends the UPH to the UPH.
Proof: To show that linear fractional transformations map the UPH to the UPH, we need to prove that Im(z) > 0 implies Im(w) > 0.
Assume that Im(z) > 0. Then, (1/2i)(z - z) > 0.

    Im(w) = (1/2i)(w - w) = [(az + b)/(cz + d) - (az + b)/(cz + d)]/(2i)
              = [(az + b)(cz + d) - (cz + d)(az + b)]/[(2i)(c2zz + cdz + cdz + d2)]
              = [aczz + bcz + adz + bd - aczz - adzz - bcz - bd]/[(2i)(c2zz + cd(z + z) + d2)]
              = [bc(z-z) + ad(z-z)]/[(2i)(c2|z|2 + 2cdx + d2)]
              = [ad(2iy) - bc(2iy)]/[(2i)(c2|z|2 + 2cdx + d2)]
              = (ad - bc)y/[(cx + d)2 + y2]
              = y/[(cx + d)2 + y2]
Since y = Im(z), y > 0 from our original assumption. This, in turn, means that Im(w) > 0, so the claim is true.

Claim: Linear fractional transformations are isometries of the UPH.
Proof: We want to show that (1/Im(w))((dw/dt) · (dw/dt)) = (1/Im(z))((dz/dt) · (dz/dt)).

    w(t) = (az(t) + b)/(cz(t) + d), so
    w'(t) = [(cz(t)+d)az'(t) - (az(t)+b)(cz'(t))]/(cz(t)+d)2
              = (acz(t)z'(t) + adz'(t) - acz(t)z'(t) - bcz'(t))/(cz(t)+d)2
              = [(ad - bc)z'(t)]/(cz(t) + d)2, which means that
    (dw/dt) = (1/(cz + d)2)(dz/dt)     and     (dw/dt) = (1/(cz + d)2)(dz/dt), and
    (dw/dt)(dw/dt) = (1/(cz + d)4)(dz/dt)(dz/dt)
But, Im(w) = (1/(cz+d)2)Im(z), so
    ((dw/dt) · (dww/dt)) = (Im(w)/Im(z))((dz/dt) · (dz/dt)).
Demonstration

The purpose of this demonstration is to illustrate how linear fractional transformations work. In one window, we draw a triangle and a circle in the upper half-plane and and in a second window, we show the image of these curves under a linear fractional transformation. In the control panel the coefficients of the LFT, which are labeled a, b, c, and d, can be entered. Note that any LFT will take the UHP to the UHP. In a third window, we map the UHP to the unit disc.

Exercises

    1. Show that the left inverse of a Linear Fractional Transformation (LFT) is another LFT.

    2. Show that the points on the curve z(t) = [icos(t/2)ea + sin(t/2)]/[-isin(t/2)ea + cos(t/2)] lie on a Euclidean circle centered at cosh(a)i with radius sinh(a).

    3. Show that the Gauss-Bonnet theorem holds for a geodesic quadrilateral in the Upper Half Plane.

    4. Show that the G-B Thm. holds for a simple closed polygon in the UHP. You may assume that the polygonal region bounded by the polygon can be decomposed into triangles by diagonals lying in the interior of the region.

    5. Show that the vertices iea, -tanh(a) + isech(a), ie(-a), tanh(a) + isech(a) are the vertices of a quadrilateral with equal sides and equal angles. Explain why there is a value a such that the angles of this quadrilateral are all /4.