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Find all Moebius Transformations preserving unit circle

Note: I am more interested if I got these computations right than the answer.

Approach-1

From page-124 of Needham, a general moebius transformation of form $f(z) = \frac{az+b}{cz+d}$ can be decomposed as:

  1. $z \to z + \frac{d}{c}$ (translate)

  2. $z \to \frac{1}{z}$ (inversion)

  3. $z \to -\frac{ad-bc}{c^2}z$ (rotation and dilation)

  4. $z \to z + \frac{a}{c}$ (translate)

It is clear to me that if it is to preserve the unit circle than 1. and 4. must be opposite , so I have:

$$ \frac{d}{c} = - \frac{a}{c} \implies d = -a \tag{1}$$

We also need that in 3. the magnitude should be preserved ( no other dilations), we have:

$$ c^2 = e^{it} ( a^2+bc) \implies c^2 e^{-it} -bc = a^2 \implies \pm \sqrt{c^2 e^{-it} -bc }=a\tag{2}$$

Putting the $d=-a$ in our function and then using (2), we have: $$ f(z) = \frac{\pm \sqrt{c^2 e^{-it} -bc }z+b}{cz-\pm \sqrt{c^2 e^{-it} -bc }}$$

Is this all the simplification possible or is it possible to kill any more variables?

Approach-2

This is based on a discussion with a friend. The idea is to begin with the symmetric principle i.e: symmetric points are mapped to symmetric point under moebius transformations, we have:

$$ w= f(0) = \frac{b}{d}$$

$$ \frac{1}{w} = \frac{a}{c}$$

This leads to the following constraint:

$$ w= \frac{b}{d} = \frac{c}{a}$$

Or,

$$ b = dw$$

and,

$$ c=aw$$

Leading to:

$$ f(z) = \frac{az+dw}{awz+d}= \frac {\frac{az}{d} + w}{\frac{a}{d} wz + 1}$$

Letting $\frac{a}{w} = \lambda$, we have:

$$ f(z) = \frac{\lambda z+w}{\lambda w z +1} \tag{3}$$

Now, without loss of generality let's suppose that point $1$ gets sent to some point $e^{it}$ on the unit circle, we have:

$$ \frac{\lambda + w}{ \lambda w +1} = e^{it}$$

$$ \lambda ( 1-e^{it} w)= e^{it} -w$$

$$ \lambda = \frac{e^{it} - w} {1- e^{it} w} \tag{4}$$

Can the expression received after plugging (3) into (4) be simplified any more?

Also, how exactly could I know how many independent variables there would be at them end of this problem?

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  • 1
    $\begingroup$ What do you mean by FLTS ? $\endgroup$
    – Jean Marie
    Jun 28 at 21:44
  • $\begingroup$ See Subgroups of the Möbius group, and also the questions listed under Related on the RHS. $\endgroup$
    – dxiv
    Jun 28 at 21:47
  • $\begingroup$ Edited, could you please check my work now @JeanMarie $\endgroup$ Jun 28 at 22:04
  • $\begingroup$ Nope. I edited it now. It should be clear why that quesiton is different than mine @Jean Marie $\endgroup$ Jun 28 at 22:17
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    $\begingroup$ @EthakkaappamwithChai Then you may want to change the first line of the question to make that more clear, and perhaps add the solution-verification tag, too. $\endgroup$
    – dxiv
    Jun 28 at 22:50

2 Answers 2

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Find all Moebius Transformations preserving unit circle

The general form of the tranformations preserving the unit circle is $\;\displaystyle f(z) = \omega\frac{z + \alpha}{\overline{\alpha}z + 1}\;$ with $\,|\omega|=1\,$ and $\,|\alpha| \ne 1\,$. Since $\,f(-\alpha)=0\,$ it follows that such transformations with $\,|\alpha| \lt 1\,$ preserve the unit disc, while those with $\,|\alpha| \gt 1\,$ map the interior of the unit disc to its exterior.

This is quoted in wikipedia under Subgroups of the Möbius group, and several proofs were posted on MSE under Möbius Transforms that preserve the unit disk and Can we characterize the Möbius transformations that maps the unit disk into itself?.

Approach-1

It is clear to me that if it is to preserve the unit circle than 1. and 4. must be opposite , so I have:

$$ \frac{d}{c} = - \frac{a}{c} \implies d = -a \tag{1}$$

Steps 2. and 3. do not preserve distances, so it is not obvious that the initial and final translations have to be opposite. In fact, with the notations from the first paragraph, this condition would be equivalent to $\,\omega = -1\,$, which is certainly not required.

Approach-2

This is based on a discussion with a friend. The idea is to begin with the symmetric principle i.e: symmetric points are mapped to symmetric point under moebius transformations, we have: $$ w= f(0) = \frac{b}{d}$$

$$ \frac{1}{w} = \frac{a}{c}$$

This is missing a conjugation. The symmetric of $\,w\,$ with respect to the unit circle is $\,w^* = \dfrac{1}{\overline w}\,$, and the symmetry principle gives $\,1 = \overline w w^* = \dfrac{\overline b}{\overline d} \cdot \dfrac{a}{c}\,$, or $\,a \overline b = c \overline d\,$.

The previous relation is not enough to define the transformation, but the idea can be made to work by writing the equality for an arbitrary pair of symmetric points $\,z, \dfrac{1}{\overline z}\,$, not just $\,0, \infty\,$:

$$ 1 = \overline{f(z)} \,f\left(\frac{1}{\overline z}\right) = \frac{\overline a \overline z + \overline b}{\overline c \overline z + \overline d} \,\frac{a + b \overline z}{c + d \overline z} \quad\implies\quad \begin{cases} \begin{align} \overline a b &= \overline c d \\ |a|^2+|b|^2 &= |c|^2 + |d|^2 \end{align} \end{cases} $$

This reduces the problem to the one in Sufficient conditions for a mobius transformation to map the unit circle to itself.

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  • $\begingroup$ Hello, but does this mean that if I specify the values of $\omega$ s.t. $|\omega|=1$ and $\alpha$ then I have a möbius transformation given by $$f(z)=\omega \frac{z+a}{\bar a z+1}$$ so an answer to this exercise would be $$f(z)=\omega \frac{z+a}{\bar a z+1}$$ such that $|\omega|=1$ or isn't this enough precise? $\endgroup$
    – Wave
    Jun 29 at 7:10
  • $\begingroup$ @Wave That's correct, with the additional condition that $\,|\alpha| \ne 1\,$. $\endgroup$
    – dxiv
    Jun 29 at 15:52
  • $\begingroup$ We should also add $f(z) = \omega/z$, which cannot be written as $\omega (z + \alpha)/(\overline \alpha z + 1)$. Approach 2 can be simplified by considering that there is a point $-\alpha$ (with $|\alpha| \neq 1$) which is mapped to $0$, then $-1/\overline \alpha$ is mapped to $\infty$. If $\alpha = \infty$, then $f(z) = \omega/z$, otherwise $f(z) = \omega (z + \alpha)/(\overline \alpha z + 1)$. $|f(1)| = 1$ implies $|\omega| = 1$, which gives the general form (any required transformation has to have this form, and any transformation of this form satisfies the requirements). $\endgroup$
    – Maxim
    Jun 30 at 14:08
  • $\begingroup$ @Maxim Right, $\,f(z)=\omega/z\,$ is also a valid answer, which corresponds to $\,\alpha=\infty\,$ in $\,\overline{\mathbb C}\,$. About Approach-2, I did not mean to imply that it's the easiest way to derive it, and different, arguably easier, alternatives can be found in the posted links. My point here was to show that OP's approach itself could be made to work, which it does. Incidentally, one of the forms that follows almost immediately from it is $\,f(z) = \omega \frac{az+b}{\bar b z + \bar a}\,$, which covers the case $\,f(z)=\omega/z\,$ explicitly. $\endgroup$
    – dxiv
    Jun 30 at 15:40
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Rather than explicit computation, I think it's more explanatory to use Schwarz' lemma to first show that injective maps of the open disk to itself that stabilize $0$ are of the form $z\to \mu z$ with $|\mu|=1$.

On the other hand, it is a reasonable and direct exercise to see that $U(1,1)=\{g\in GL_2(\mathbb C):g^*Sg=S\}$, with $S=\pmatrix{-1 & 0\cr 0 & 1}$ does stabilize the open or closed unit disk, and is transitive.

Combining these readily shows that your desired collection of linear fractional transformations is exactly $U(1,1)$.

EDIT: Oops, and forgot the inversion $z\to -1/z$! :) This normalizes $U(1,1)$, and together they generate an index-two over-group of $U(1,1)$. :) And, yes, as @moishekohan comments, since linear fractional transformations whose matrices differ by a scalar give the same "LFT", we might say $PU(1,1)$... and, still, I don't know a name for the larger group including the inversion.

EDIT2: actually, in terms of names for things, $GU(1,1)=\{g\in GL_2(\mathbb C):g^*Sg=\nu(g)\cdot S\}$ for some $\nu(g)\in \mathbb R^\times$ is a "unitary similitude group", and then $PGU(1,1)$, the projectivized unitary similitude group, is a way to name the group in question. :) But this terminology will not communicate clearly to many people! :)

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  • $\begingroup$ Your mathjax broke in second para $\endgroup$ Jun 28 at 22:17
  • $\begingroup$ $PU(1,1)$ to be more precise. $\endgroup$ Jun 28 at 22:18
  • $\begingroup$ I still don't understand the matrix formulation of it in projective coordinates. Also how can we apply Schwarz lemma here? How do we knw $|f(z)| \leq 1$ inside the disc? $\endgroup$ Jun 28 at 22:23
  • $\begingroup$ Since continuous maps preserve connectedness, an injective continuous map that maps the circle to itself must map the interior either to the interior, or to the exterior. If it mapped to the exterior, then composing with inversion would send it back to the interior. That kind of thing. $\endgroup$ Jun 28 at 22:24
  • $\begingroup$ ... using matrices and "projective coordinates" is really the only sane way to deal with LFT's... Small entrance fee, and then lots of things become vastly easier and clearer. Seriously. :) $\endgroup$ Jun 28 at 22:29

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