Moebius transformations preserving unit circle

Question

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?

Answer 1

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.

Answer 2

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! :)