3
$\begingroup$

I have a problem in understanding why a 1-1, onto analytic map which sends upper half plane to unit disk is of the form $f(z)=\frac {az+b}{cz+d}$ where $ad-bc \neq 0$. (I know the map can be refined further but only have problems in understanding this part.)

So please explain why we have to start our work with this form.

$\endgroup$
  • 1
    $\begingroup$ Have you studied cross ratio, orientation and symmetry principle? $\endgroup$ – Swapnil Tripathi Nov 8 '14 at 23:45
  • $\begingroup$ What does $f(x)=\dfrac1x$ do ? Now, imagine rotating the real axis around the origin. What does x become, and what does $f(x)=\dfrac1x$ become ? Now, how can we modify $f(x)=\dfrac1x$ so that instead of mapping $(1,\infty)$ to $(0,1)$, it maps $(0,\infty)$ to $(0,1)$ ? Now, as we again rotate the real axis around the origin, what does this new function become ? $\endgroup$ – Lucian Nov 9 '14 at 1:02
2
$\begingroup$

Claim 1. Every conformal map $f$ of $\mathbb D$ onto itself such that $f(0)=0$ is the rotation, $f(z)=e^{i\theta} z$.

Proof: Apply the Schwarz lemma to $f$ and to $f^{-1}$; conclude that equality holds, use the equality statement in the lemma.

Claim 2. For every $w\in \mathbb H$ (the upper halfplane) the map $\phi(z) = \dfrac{z-w}{z-\overline w}$ transforms $\mathbb H$ onto $\mathbb D$.

Proof. It's invertible, and the real axis goes to the unit circle.

Claim 3. Every conformal map $f$ of $\mathbb H$ onto $\mathbb D$ is a fractional linear transformation.

Proof: Apply Claim 1 to $$g(z) = \frac{f(z)-f(0)}{f(z)-\overline{f(0)}}$$ which maps $\mathbb D$ onto $\mathbb D$ by virtue of Claim 2.

Then use the fact that fractional linear transformations form a group (in particular, the composition of two such transformations is again of the same form).


Answer based on Which conformal maps UHP$\to$UHP extend continuously to the closure?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.