
#1
Mar1308, 06:12 PM

P: 743

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 adbc=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 yaxis (imaginary axis) to the vertical line [itex] x=\frac{b}{d}, x= \frac{a}{c}[/itex] or a semicircle centred on the xaxis 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 adbc=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
Mar1308, 10:27 PM

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P: 25,165

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).




#3
Mar1308, 10:56 PM

P: 743

Yes, I've looked at the parametric solutions and b/d and a/c are fairly clear. But the semicircle case isn't entirely clear. Though I'm not too worried about it at this point. I'll just hand wave it.




#4
Mar1408, 05:26 AM

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Linear Fractional Transformation[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] = … 



#5
Mar1408, 08:51 AM

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P: 25,165




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