Forster defines analytic continuation of a germ of a holomorphic function at a point on a Riemann surface as follows.

Suppose $ X$ is a Riemann surface, $a\in X$ and $\phi\in\mathcal{O}_a$ is a function germ. The quadrupel $(Y,p,f,b)$ is called an analytic continuation of $\phi$ if:

a) $Y$ is a riemann surface and $p:Y\longrightarrow X$ is an unbranched holomorphic map.

b)$f$ is a homolomprhic function on $Y$.

c) $b$ is a point of $Y$ such that $p(b)=a$ and $p_*(\rho_b(f))=\phi$, where $\rho_b(f)$ is the germ of $f$ at $b$ as an element of $\mathcal{O}_b$. Here $p_*:\mathcal{O}_{Y,y}\longrightarrow \mathcal{O}_{X,p(y)}$ is the inverse to the natural isomorphism $\mathcal{O}_{X,p(y)}\longrightarrow\mathcal{O}_{Y,y}$.

He later proves the following : if $X$ is a Riemann surface, $a\in X$ and $\phi\in\mathcal{O}_a$, then there exists a maximal analytic continuation $(Y,p,f,b)$ of $\phi$.

I have a doubt in the proof which goes as follows:

Let $Y$ be the connected component of $|O| = \amalg\mathcal{O}_p$ (with the natural topology) which contains $\phi$. And $p$ be the restriction of the projection map $|\mathcal{O}|\longrightarrow X$ to $Y$. Then we know by previous results that $p$ is a local homemorphism, and that we can give a Riemann surface structure on $Y$ such that $p$ becomes holomorphic.

Now we define a holomorphic function $f:Y\longrightarrow\mathbb{C}$ as follows : By definition, every $\sigma\in Y$ is a function germ at the point $p(\sigma)\in X$. Set $f(\sigma)=\sigma(p(\sigma))$ for all $\sigma\in Y$.

My question is why is this function $f$ holomorphic? Any help will be appreciated!


I will follow Forster's notaion. It suffices to show that the restriction $f|[U,g]$ is holomorphic for every $g\in\mathcal{O}(U)$. Without loss of generality, we may assume that $(U,\varphi)$ is a chart. Since $p|[U,g]\colon[U,g]\to U$ is a homeomorphism, $([U,g],\varphi\circ p|[U,g])$ is a chart on $|\mathcal{O}|$. It follows that $$\begin{align} f(\varphi\circ p|[U,g])^{-1}(z) &=f(p|[U,g])^{-1}\varphi^{-1}(z)\\ &=f(\rho_{\varphi^{-1}(z)}(g))\\ &=\rho_{\varphi^{-1}(z)}(g)(\varphi^{-1}(z))\\ &=g\varphi^{-1}(z) \end{align}$$ for every $z\in\varphi(U)$. Hence, $f$ is holomorphic in $[U,g]$.


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.