Let a,b,c,da, b, c, d be complex numbers with ad−bc≠0ad – bc \ne 0. Then

f(z)=az+bcz+d f(z) = \frac{az + b}{cz + d}

is called the Mobius transformation.

If H:={z∈C:ℑ(z)>0}H := \{ z \in \mathbb{C} : \Im(z) > 0 \} is the open upper half plane, show that any Mobius transformation from HH onto itself can be written with real coefficients a,b,c,da, b, c, d with ad−bc=1ad – bc = 1.

Can someone please show me how to do this problem?

I already showed that the Cayley transform C(z)=z−iz+iC(z) = \frac{z-i}{z+i} is a biholomorphic map (analytic, injective, and onto) from HH onto the unit disk B(0,1)B(0,1). I also showed that any biholomorphic map from HH onto HH is a Mobius transformation. How can I go further and use this?

There is also another hint telling me that such transformation maps the real axis into the real axis, but I don’t see how that would help me.

Thank you.

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2 Answers

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Here is a sketch of a possible proof.

Suppose ff is a Mأ¶bius transformation taking R∞\mathbb{R}_{\infty} to R∞\mathbb{R}_{\infty}. Then I claim the coefficients a,b,c,da,b,c,d of ff are all real. suppose a1,a2,a3∈R∞a_1,a_2,a_3\in\mathbb{R_{\infty}} are three distinct points that are mapped to b1,b2,b3∈R∞b_1,b_2,b_3\in\mathbb{R_{\infty}}. Then by invariance of cross-ratio we have for all z∈Hz\in H

zâˆ′a1a3âˆ′a1:zâˆ′a2a3âˆ′a1=f(z)âˆ′b1b3âˆ′b1:f(z)âˆ′b2b3âˆ′b2\frac{z âˆ’ a_1}{a_3 âˆ’ a_1}:\frac{z âˆ’ a_2}{a_3 âˆ’ a_1}=\frac{f(z) âˆ’ b_1}{b_3 âˆ’ b_1}:\frac{f(z) âˆ’ b_2}{b_3 âˆ’ b_2}

with the appropriate conventions on ∞\infty.

It is easy to see this implies that ff can be written in the desired form.

Next, we have

ℑ(f(z))=ad−bc|cz+d|2⋅ℑ(z).\Im(f(z))=\frac{ad-bc}{|cz+d|^2}\cdot\Im(z). This implies ad−bc>0ad-bc>0, which is what we wanted.

Thank you for your answer. I used a different approach but were able to show that ad−bc>0ad – bc > 0, but then how can I show it is, in fact, equal to 11?

– dh16

yesterday

1

Since ad−bcad-bc is nonzero you can just normalize by dividing all the coefficients by √ad−bd\sqrt{ad-bd}

– ArtW

yesterday

w=az+bcz+dw=\frac{az+b}{cz+d} gives z=b−wdcw−az=\frac{b-wd}{cw-a}. The upper half plane is y≥0y\geq 0 or z−¯z≥0z-\overline z\geq 0

⟹b−wdcw−a−b−¯wdc¯w−a≥0\implies \frac{b-wd}{cw-a}-\frac{b-\overline wd}{c\overline w-a}\geq 0

⟹(ad−bc)(w−¯w)≥0\implies (ad-bc)(w-\overline w)\geq 0 (On simplifying)

You need ad−bc=1ad-bc=1 so that w−¯w≥0w-\overline w\geq 0 is the upper half ww-plane.