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Linear Fractional Transformation

  • Thread starter Kreizhn
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1. Homework Statement
Let [itex]\mathbb{H}^2 = \{z=x+iy\in \mathbb{C} | y>0 \}. [/itex] For [itex]a,b,c,d\in \mathbb{R}[/itex] satisfying ad-bc=1 define [itex] T: \mathbb{H}^2 \rightarrow \mathbb{H}^2 [/itex] by
[tex] T(z) = \frac{az+b}{cz+d} [/tex]

Show that T maps that positive y-axis (imaginary axis) to the vertical line [itex] x=\frac{b}{d}, x= \frac{a}{c}[/itex] or a semicircle centred on the x-axis containing both [itex] (\frac{b}{d},0) \text{ and } (\frac{a}{c},0)[/itex]


3. The Attempt at a Solution

This seems like it should be fairly easy, but the answer has been eluding me. I began by proceeding as we would in finding the isotropy group, by taking the linear fractional map as a change of variables. Doing this we can set [itex] iy = \frac{aiy+b}{ciy+d} [/itex] and conclude that in general [itex] a=d \text{ and } b=-cy^2 [/itex]. Then using ad-bc=1 we get that [itex] a^2+c^2y^2=1[/itex].

I'm wondering if we need to use anything special about the fact that we can express this mapping as
[tex] \begin{pmatrix} a&b\\ c&d \end{pmatrix} \in SL_2(\mathbb{R})[/tex]

I can't quite seem to figure out where to go from there...
 

Answers and Replies

Dick
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You are kind of off on the wrong track. Put z=i*t, t>=0. Then set Re(T(it))=x(t) and Im(T(it))=y(t). Now you have a parametric form of the curve (x(t),y(t)). It should be pretty clear that x=b/d corresponds to the c=0 case and x=a/c corresponds to the d=0 case. For the semicircle case you should be able to do brute force and show (x(t),y(t)) satisfies the equation of the given circle. I'm not sure if there is a cleverer way, but you can at least note (x(0),y(0))=(b/d,0) and limit t->infinity of (x(t),y(t))=(a/c,0).
 
743
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Yes, I've looked at the parametric solutions and b/d and a/c are fairly clear. But the semi-circle case isn't entirely clear. Though I'm not too worried about it at this point. I'll just hand wave it.
 
tiny-tim
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For [itex]a,b,c,d\in \mathbb{R}[/itex] satisfying ad-bc=1 define [itex] T: \mathbb{H}^2 \rightarrow \mathbb{H}^2 [/itex] by
[tex] T(z) = \frac{az+b}{cz+d} [/tex]

Show that T maps that positive y-axis (imaginary axis) to the vertical line [itex] x=\frac{b}{d}, x= \frac{a}{c}[/itex] or a semicircle centred on the x-axis containing both [itex] (\frac{b}{d},0) \text{ and } (\frac{a}{c},0)[/itex]
How about using complex conjugates:

[tex] (T(iy) - (q,0))^2 = (\frac{aiy\,+\,b}{ciy\,+\,d}\,-\,q)^2[/tex]

[tex] =\,\frac{(aiy\,+\,b)\,-\,q(ciy\,+\,d)}{ciy\,+\,d}\,\frac{(-aiy\,+\,b)\,-\,q(-ciy\,+\,d)}{-ciy\,+\,d}[/tex]

= … :smile:
 
Dick
Science Advisor
Homework Helper
26,258
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Yes, I've looked at the parametric solutions and b/d and a/c are fairly clear. But the semi-circle case isn't entirely clear. Though I'm not too worried about it at this point. I'll just hand wave it.
Here's an argument assuming you know these transformations map lines and circles to lines and circles AND are angle preserving. T maps the x-axis to the x-axis. For T(t)=(x(t),y(t)), T(0)=(b/d,0), T(infinity)=(a/c,0). So T(t) is a curve passing through T(0) and T(infinity). But since the y-axis is orthogonal to the x-axis, the curve T(t) is also orthogonal to the x-axis at T(0). That's enough geometric facts to force it to be the circle you described.
 

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