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Linear Fractional Transformation

  1. Mar 13, 2008 #1
    1. The problem statement, all variables and given/known data
    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...
     
  2. jcsd
  3. Mar 13, 2008 #2

    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).
     
  4. Mar 13, 2008 #3
    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.
     
  5. Mar 14, 2008 #4

    tiny-tim

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    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:
     
  6. Mar 14, 2008 #5

    Dick

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    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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