In my previous question it was pointed out to me that an "automorphism" of the projective line/Riemann sphere (=that is a Möbius transform) is a bijection that is meromorphic in the local coordinate z.
Since I didn't know Möbius transforms I tried to find information about them and then on Wikipedia about Möbus transofmrs it is stated that Möbius transformations are exactly the bijective conformal maps from the Riemann sphere to itself.
But it is also stated that it is a bijective holomorphic function from the Riemann sphere to the Riemann sphere.
I understand that meromorphic is weaker than holomorphic but it's not clear to me where conformal fits in. But mostly it's not clear to me which of these is right.
While I was trying to find out which of these 3 descriptions is correct I found out that every Möbius transofrm can be composed into translations, scaling and $z \mapsto 1/z$. Then maybe using this composition, if $1/z$ is holomorphic then so is the Möbius transform?
Is a Möbius transform holomorphic (or just meromorphic or conformal) on the Riemann sphere?
and
Is $1/z$ holomorphic on the Riemann sphere?
Best Answer
A generic Möbius transformation is a meromorphic function on $\mathbb{C}$, and a holomorphic map of the Riemann sphere onto itself. There is a difference between "function" and "map" here.
When the Riemann sphere is equipped with the structure of a complex manifold, a neighborhood of $\infty$ is covered by the coordinate patch $z\mapsto 1/z$. A function with a pole qualifies as a holomorphic map into the sphere, because at a point $a$ where $f(a)=\infty$, the composition with the above chart map is $1/f$ (which is holomorphic).