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Complex analysis/linear fractional transformation

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arthurhenry
#1
Mar8-12, 10:17 PM
P: 43
In the text by Joseph Bak,

He is trying to determine all automorphisms of the unit disk such that f(a)=0.
He says "let us suppose that this automorphism is a linear fractional transformation. Then it must map the unit circle onto the unit circle.

I am asking for help in understanding this deduction/conclusion.

Thank you
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morphism
#2
Mar8-12, 10:45 PM
Sci Advisor
HW Helper
P: 2,020
A linear fractional transformation (lft) takes circles to circles. Since the lft under consideration here is assumed to be an automorphism of the unit disk, it must take points inside the disk to points inside the disk (i.e. inversion in any circle inside the disk is ruled out), so it must take the unit circle to itself.

Alternatively, you can go the long route: starting from the formula $$z\mapsto f(z) = \frac{az+b}{cz+d}$$ (with ##ad-bc=1##, wlog), show that the stipulation ##|z|\leq1 \implies |f(z)|\leq1## puts some severe restrictions on what a,b,c,d could be. And then conclude that points with ##|z|=1## get mapped to points with ##|f(z)|=1##. This will be fairly messy though.
arthurhenry
#3
Mar9-12, 01:14 AM
P: 43
I think I understand it now.
I think you are saying:
suppose p is a point inside the disk, i.e. an interior point. Take nbhd around p that is contained in the unit disk still. Then by Open Mapping Theorem, the image of this disk is open, i.e., the f(p) is also contained in a nbhd that is also inside the unit circle, so f(p) cannot be a boundary point. Since the LFT is injective, all interior points is taken as the images of interior points and the only place for a boundary point to be sent is the boundary.
Hope I am right...in the sense that I am not able conclude this without the Open mapping theorem.

morphism
#4
Mar9-12, 01:35 AM
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P: 2,020
Complex analysis/linear fractional transformation

Your argument doesn't work because it doesn't explain why f takes the open disk onto itself.

I was just using the following facts: An lft takes a circle C_1 to a circle C_2, and takes the region inside of C_1 to either the region inside of C_2 or to the region outside of C_2. If it takes the interior of C_1 to the interior of C_2, it will take the exterior of C_1 to the exterior of C_2. A similar comment applies in the other case.

Now think about the situation in your proof: say f takes the unit circle C_1 to some circle C_2. Since f maps the interior of the unit circle (i.e. the open unit disk) onto itself, C_2 had better be the unit circle.
arthurhenry
#5
Mar9-12, 06:15 PM
P: 43
Thank you, that has cleared things very nicely for me.


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