Why is Mobius group denoted by $Aut(\hat{\mathbb{C}})$ In some articles, Möbius group is denoted by $Aut(\hat{\mathbb{C}})$, even in wikipedia.
I'm wondering why. Can we gather all the Möbius transformation under some properties of functions on $\hat{\mathbb{C}}$?
For example, in group-theory, $Aut(G)$ is a automorphism group which is the collection of all the group-homomorphisms on $G$. Is the notation $Aut(\hat{\mathbb{C}})$ somewhat consistent with this?
 A: $\hat{\mathbf{C}}$ is a Riemann surface, and $\text{Aut}(\hat{\mathbf{C}})$ is its automorphism group: meromorphic bijections.
A: Because $\hat{\mathbb{C}}$ is compact, meromorphic functions on it can only have finitely many poles. Therefore, by the definition of meromorphic functions (isolated sigularities and holomorphicity), any meromorphic function can be written as $f=\frac{p}{q}$ where $p,q$ are polynomials and $q\neq0$. Since $\operatorname{Aut}(\hat{\mathbb{C}})$ is the group of meromorphic bijections, functions in it must have the form $f=\frac{az+b}{cz+d}$.  By Liouville’s theorem, one can map $\left(GL(2,\mathbb{C}),\cdot\right)\to\left(\operatorname{Aut}(\hat{\mathbb{C}}),\circ\right)$ surjectively by the homomorphism
$$f:\begin{pmatrix}
a & b\\
c & d
\end{pmatrix}\to\left\{f(z)=\frac{az+b}{cz+d}\right\}$$
and this homomorphism has a kernel $$\ker f=\left\{\begin{pmatrix}
a & 0\\
0 & a
\end{pmatrix}: a\in\mathbb{C}^*\right\}$$
Therefore, you can see that
$$\operatorname{Aut}(\hat{\mathbb{C}})=f(GL(2,\mathbb{C}))\cong GL(2,\mathbb{C})/\ker f=PSL(2,\mathbb{C})$$
And that's why in some articles the Mobius group is denoted by $\operatorname{Aut}(\hat{\mathbb{C}})$.
