# Möbius mappings

1. Sep 20, 2007

### jostpuur

I have an exercise where the set of möbius mappings is defined like this

$$\textrm{Mob} = \{ f_A:\mathbb{C}\to\mathbb{C}\;|\; f_A(z)=\frac{az+b}{cz+d},\; A=\left[\begin{array}{cc}a & b \\ c & d \\ \end{array}\right]\in \textrm{SL}(2,\mathbb{C})\}$$

Is it probable, that there is a mistake and the the special linear group should be replaced with $\textrm{GL}(2,\mathbb{C})$?

The exercise uses wording "let us consider the set of möbius mappings", and I started thinking, that could that "set" be "subset". Is "Mob" common name for the full set of Möbius mappings?

Last edited: Sep 20, 2007
2. Sep 20, 2007

### cristo

Staff Emeritus
If it's really the special linear group, then yes, it is a subset of all the Moebius transformations, since a Moebius transformation is defined as a mapping from C to C of the form (az+b)/(cz+d) where ad-bc!=0

3. Sep 20, 2007

### jostpuur

So there is a mistake in the exercise, but I cannot logically conclude what it is, because there could be at least two different kind of mistakes. I think I'll assume that the Mob is a subset of the set of Möbius mappings, because this way the exercise is easier.

hmhm... or no. It's not really mistake, only very confusing. "set of möbius mappings" can mean any set, whose members are möbius mappings, and thus "set of möbius mappings" is now some "subset of the set of all möbius mappings"...

Last edited: Sep 20, 2007
4. Sep 20, 2007

### Chris Hillman

Hmm... the exercise looks fine to me. I think Christo forgot a technical point.

The notation $Mob$ is not exactly standard but it is reasonable enough. There is no universally standard notation for the group of Moebius transformations, but the definition as the group of mappings $z \mapsto \frac{a \, z + b}{c \, z + d}, \; a \, d - b \, c \neq 0$ is standard. The bit you have quite grokked yet is the relationship between this group and $SL(2,{\bf Z})$.

Do you know what $PSL(2,{\bf Z})$ is? How about $PGL(2,{\bf Z})$? How are these related? (Hint: might be a trick questions Another hint: would your "correction" really change the definition offered in the exercise?)

I take it this is a homework exercise so I don't want to just tell you the answer, but if you really get stuck, see Theorem 2.1.3 in Jones and Singerman, Complex Functions, Cambridge University Press, 1987.

Last edited: Sep 20, 2007
5. Sep 20, 2007

### jostpuur

I knew there was something tricky going on, because I'm now fighting with the equivalence classes of these matrices.

6. Sep 20, 2007

### Chris Hillman

Exactly.

7. Sep 21, 2007

### ZioX

Incidentally these mobius transformations define the automorphisms of conformal mappings on the upper half plane.

For fun, prove this! ;0

I recently wrote an exposition on hyperbolic geometry utilizing mobius transformations. One of my sources used Mob(H) to be the set of mobius transformation as you defined.

Last edited: Sep 21, 2007
8. Sep 21, 2007

### Chris Hillman

I think you mean $PSL(2, {\bf R})$ (Moebius transformations of the upper half plane), which is isomorphic to $PSU(1,1)$ (Moebius transformations of the unit disk). Both have three real parameters, while $PSL(2,{\bf C})$ has six (like the Lorentz group, not a coincidence). In fact, $PSL(2, {\bf R})$ is the stabilizer of the upper half plane (in fact, of its boundary, the real line--- which is a circle on the Riemann sphere) while $PSU(1,1)$ is the stabilizer of the unit disk (in fact, of its boundary, the unit circle). So the first two are conjugate subgroups of the Moebius group itself (the third group).

Last edited: Sep 21, 2007