3
$\begingroup$

I am reading the book on Riemann surfaces by Miranda as a part of my reading course. I am facing this (maybe elementary) confusion.

Let $S^2$ denote the unit sphere in $\mathbb{R}^3$. This is given a Riemann surface structure by means of charts: $$\phi:S^2-\{(0,0,1)\}\rightarrow \mathbb{C} : \phi(x,y,z)=(x+iy)/(1-w)\text{ and}$$ $$\psi:S^2-\{(0,0,-1)\}\rightarrow \mathbb{C} : \psi(x,y,z)=(x-i y)/(1+w).$$ Here, the point $(0,0,1)$ is identified with the point at $\infty$. These charts are compatible because $\phi\circ\psi^{-1}(z)=1/z=\psi\circ\phi^{-1}(z)$, which is holomophic in the domain of definition.

Now, a complex valued function $f$ defined in a neighbourhood of a point $p$ of a Riemann surface $X$ is defined to be holomorphic at $p$ if there is a chart $\tau:U\rightarrow V$ such that $f\circ\tau^{-1}$ is holomorphic at $\tau(p)$.

Now he states the following in which I have a doubt.

Let $f$ be a complex valued function defined in a neighbourhood of $\infty$ in the Riemann sphere. The claim is that $f$ is holomorphic at $\infty$ iff $f(1/z)$ is holomorphic at $z=0$.

I am trying to write this out using definitions and getting stuck. Since $(0,0,1)$ is identified with $\infty$, the chart about $\infty$ is $\psi$ above.

Now $ f$ is holomorphic at $(0,0,1)\ \iff\ f\circ\psi^{-1}(z)$ is holomorphic at $\psi(0,0,1)=0$. How do I get $f(1/z)$ from this. Also technically $1/z$ is function which is defined on the punctured complex plane right. So what is $f(1/z)$?

The closest I can think of is $f\circ\psi^{-1}(z)=f\circ\phi^{-1}\circ\phi\circ\psi^{-1}(z)=f\circ\phi^{-1}(1/z)$ on $\mathbb{C}^*$. Is this what he means? But this function is not even defined at zero.

I think I am getting caught in pedantic details. But it would be great if some can help by indicating a proof of the above claim.

$\endgroup$
  • $\begingroup$ I suppose that there's a typo in the definitions of $\phi$ and $\psi$. $\endgroup$ – José Carlos Santos Feb 10 '18 at 14:28
  • $\begingroup$ It is not a "claim" that "$f(z)$ is holomorphic at $\infty$ iff $f(1/z)$ is holomorphic at $0$", it is simply the definition of being holomorphic near $\infty$. If you want to be more formal, your $f$ is defined on the unit sphere in $R^3$, so $f(\infty)$ does not make sense. You can also define the Riemann sphere without ever mentioning the unit sphere in $R^3$, you simply let $S$ be the 1-point compactification of the complex plane (or, as some would say, the complex line). Then you define a holomorphic atlas on $S$, etc. $\endgroup$ – Moishe Kohan Feb 10 '18 at 17:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.