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The question asks:

Let $\varphi$ be a linear fractional transformation which maps the upper half plane $\{z : \operatorname{Im}(z) > 0\}$ onto itself. Prove that if there exist distinct $z_1$ and $z_2$ having positive imaginary parts with $\varphi(z_1) = z_1$ and $\varphi(z2) = z_2$ then $\varphi(z) = z$ for all $z$.

There is an apparently related question that asks: 2.) Let $\varphi$ be a linear fractional transformation which maps the unit disk $\{z : |z| < 1\}$ onto itself. Prove that if there exist distinct $z_1$ and $z_2$ in the disk with $\varphi(z_1)=z_1$ and $\varphi(z_2)=z_2$ then $\varphi(z)=z$ for all $z$.

I was wondering if there is some sort of easy approach to this type of question that I'm missing? I don't believe they were intended to be particularly difficult, but after a significant amount of time I've run out of good ideas.

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Do you know what a generic LFT of the upper half plane looks like (I'm thinking matrices)? –  user641 Jul 6 '11 at 6:29
    
How about this approach: how many points can an LFT fix? And where do those points lie? –  John M Jul 6 '11 at 6:42
    
@Steve Well I was reading on Wikipedia, that the property of an LFT that maps the upper half plane to itself in terms of its matrix form is that ad - bc = 1, in the standard representation here. Is that what you're referring to? secure.wikimedia.org/wikipedia/en/wiki/… –  I Love Cake Jul 6 '11 at 6:47
    
@John Well I know that a LFT will have at most 2 fixed points if it's not the identity, but I wasn't aware of any restrictions on where the points lie. Edit: Do you think that I could use the process of determining the fixed points to somehow present that there are more than 2, maybe by setting up the quadratic using the two fixed points I have? I don't really know how to use the detail that both points I have are in the upper half plane though. secure.wikimedia.org/wikipedia/en/wiki/… –  I Love Cake Jul 6 '11 at 6:52
    
@I Love Cake : yes, and a,b,c,d are all real numbers. This is enough to show two fixed pts = identity (try finding a map which fixes i and infinity). –  user641 Jul 6 '11 at 7:13

2 Answers 2

up vote 4 down vote accepted

Think symmetry. Since $\varphi$ takes the upper half-plane into itself, $\varphi(\bar z)=\overline{\varphi(z)}$. Since $z_1$ and $z_2$ are fixed,so are $\bar z_1$ and $\bar z_2$.

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Answer to your second question:

Every Möbius transformation that preserves the unit disk is of the form

$m(z) = e^{i\theta} \frac{a-z } {1 - \bar{a}z }$, where $a$ is a point in the unit disk.

Proof (à la Julián Aguirre) : Take $a$ to be point that gets mapped to the origin, so immediately one has that its inverse is mapped to infinity. Then you require that some point on the unit circle, say $1$ be mapped to another point on the unit circle (The boundary of the unit disk). Done.

Can you deduce from here that if such a Möbius Transformation fixes two points, it is the identity transformation?

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I don't understand your Proof à la Julián. What do you mean by "it's inverse is mapped to infinity"? Easier argument (given we have the answer to 1 already): conjugate $\varphi$ with the Cayley transform. –  t.b. Jul 7 '11 at 5:10
    
@Theo Buehler Like I said, if $z=a$ is mapped to the origin, where $a$ is a point in the unit disk then its inverse $1/\bar{a}$ must be mapped to infinity. –  user38268 Jul 7 '11 at 8:22
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Ah, now I got it. Sorry didn't read properly. –  t.b. Jul 7 '11 at 8:27
    
In general, if $G$ is a group, $h$ is conjugate to $k$ if $k= ghg^{-1}$. Now let $h = \phi$ and let $g:\mathbb{D} \to \mathbb{H}$ be the Cayley transform. Then you get a Möbius transformation $k: \mathbb{H} \to \mathbb{H}$ having two fixed points in the upper half plane, and the problem is reduced to the one which is already settled. –  t.b. Jul 7 '11 at 8:37

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