Helllo friends!
I want to prove that polynomial function $p(z)=z^2-4$ extends to a holomorphic mapping from the Riemann Sphere to itself. I know that four possible combinations of holomorphic charts in domain and codomain must be holomorphic but I do not know how to define these charts to show that the following 4 possible compositions are hol.:
$\psi_1\ o\ f\ o\ \psi_1^{-1}$,
$\psi_2\ o\ f\ o\ \psi_1^{-1}$,
$\psi_1\ o\ f\ o\ \psi_2^{-1}$,
$\psi_2\ o\ f\ o\ \psi_2^{-1}$
I appreciate any help in this problem.
Best Answer
First, consider the extension defined at infinity as $p (\infty) = \infty$. By definition, $p$ is holomorphic at a point $x $ in the sphere iff there exists charts $\phi$ and $\psi$ on neighbourhoods of $x $ and $p (x) $ respectively, such that the function $f = \psi \circ p \circ \phi^{-1}$ is holomorphic (defined in an open set the complex plane) at the point $\phi(x)$. If $z \neq \infty$, the charts are the identity charts (you are in the complex plane). Then, you simply have to check for the infinity point. Taking the chart of infinity at domain and codomain: $\phi = \psi = 1/z$, you can calculate:
$$ f (z) = \psi \circ p \circ \phi^{-1} = \frac{1}{(\frac{1}{z})^2 - 4} = \frac{z}{1 - 4z^2} $$
This new function is clearly holomorphic at $\phi(\infty) = 0$.
Edit: Denoting the sphere with $S$, open sets with $U = S \setminus \{ \infty \}$ and $V = S \setminus \{ 0\}$ and define the charts $f_1(z) = z$ in $U $ and $f_2 (z) = 1/z$ (this one is the chart of infinity) in V. In the intersection of those you can use either of them.