# What is actually the geometry or analysis behind the fact that $Mob(\hat{\Bbb C})$ is simple?

Let, $$Mob(\hat{\Bbb C})$$ be the group of all Mobius transformations from the extended complex plane to itself i.e. from $$\hat{\Bbb C} \to \hat{\Bbb C}$$ .

I have been able to prove that (i) $$Mob(\hat{\Bbb C}) \cong PSL_{2}(\Bbb C)$$ where $$PSL_{2}(\Bbb C)$$ is defined to be , $$PSL_{2}(\Bbb C) := SL_{2}(\Bbb C)/ \{\pm I\}$$ and that,

(ii) $$SL_{2}(\Bbb C)$$ does not have a proper normal subgroup containing its center i.e. $$\{\pm I\}$$ and thus by correspondence theorem, $$PSL_{2}(\Bbb C) = SL_{2}(\Bbb C)/ \{\pm I\}$$ is simple.

Combining (i) and (ii) we get that $$Mob(\hat{\Bbb C})$$ is simple!

My question precisely is,

What is actually the geometry or analysis behind the fact that $$Mob(\hat{\Bbb C})$$ is simple ?

• Two thoughts - first, I suggest looking at the proof that $SL_2(\mathbb{C})$ is simple and use the isomorphism you constructed to pull the argument back to $Mob(\hat{\mathbb{C}})$. Second, you might find insight in the geometric interpretation of $Mob(\hat{\mathbb{C}})$ as isometries of $\mathbb{H}^3$. – Neal Jan 21 at 14:10
• @Neal I have studied the elements Mobius group as isomteries of $\Bbb H^3$, I want to picturize how this phenomena (being simple) affects the theory of fixed points under Mobius transformations etc. – Utsav Dewan Jan 21 at 14:13

Assume that $$N$$ is a normal subgroup of the Mobius group which is not the trivial subgroup. Then $$N$$ contains some non-identity element $$f$$. If $$f$$ is hyperbolic/loxodromic, i.e., if it has two fixed points $$z_0$$ and $$z_1$$, then you can pick a third point $$z_2$$ different from $$z_0$$ and $$z_1$$, and conjugate $$f$$ by a Mobius transformation $$h$$ which fixes $$z_0$$, and which maps $$z_1$$ to $$z_2$$, to obtain $$g = hfh^{-1} \in N$$ with fixed points $$z_0$$ and $$z_2$$. It is then easy to show that the commutator $$k = [f,g] = f^{-1}g^{-1}fg$$ fixes $$z_0$$ with $$k'(z_0)=1$$, but that it is not the identity, so it must be a parabolic Mobius transformation. This establishes that $$N$$ contains a parabolic Mobius transformation, and since any two parabolic Mobius transformations are conjugate, $$N$$ contains all of them. In particular $$N$$ contains $$F(z) = z+1$$ and $$G(z) = \frac{z}{1+az}$$ for any $$a \ne 0$$. Explicit calculations of the commutator of $$F$$ and $$G$$ shows that you get a hyperbolic/loxodromic element with trace $$2+a^2$$. (This calculation is probably easier with matrices than with explicit Mobius transformations.) This shows that $$N$$ contains hyperbolic/loxodromic elements with any given trace, so $$N$$ is the whole Mobius group.