The following is from Lectures on Riemann Surfaces by O. Forster:

7.8. Theorem. Suppose $X$ is a Riemann surface, $a\in X$ and $\varphi\in\mathcal{O}_a$ is a holomorphic function germ at the point $a$. Then there exists a maximal analytic continuation $(Y, p,f, b)$ of $\varphi$.

Pʀᴏᴏғ. Let $Y$ be the connected component of $|\mathcal{O}|$ containing $\varphi$. Let $p$ also denote the restriction of the mapping $p: |\mathcal{O}| \to X$ to $Y$. Then $p: Y \to X$ is a local homeomorphism. By Theorem (4.6) there is a complex structure on $Y$ so that it becomes a Riemann surface and the mapping $p: Y \to X$ is holomorphic. Now define a holomorphic function $f: Y \to \mathbb{C}$ as follows. By definition every $\eta\in Y$ is a function germ at the point $p(\eta)$. Set $f(\eta) := \eta(p(\eta))$. One easily sees that $f$ is holomorphic and $p_*(p_\eta(f)) = \eta$ for every $\eta\in y$. Thus if one lets $b := \varphi$, then $(Y, p,f, b)$ is an analytic continuation of $\varphi$.

Now we will show that $(Y, p,f, b)$ is a maximal analytic continuation of $\varphi$. Suppose $(Z, q, g, c)$ is another analytic continuation of $\varphi$. Define the map $F: Z \to Y$ as follows. Suppose $\zeta \in Z$ and $q(\zeta) =: x$. By Lemma (7.7) the function germ $q_*(p_\zeta(g))\in\mathcal{O}_x$ arises by analytic continuation along a curve from $a$ to $x$ from the function germ $\varphi$. By Lemma (7.2) $Y$ consists of all function germs which are obtained by the analytic continuation of $\varphi$ along curves. Hence there exists exactly one $\eta\in Y$ such that $q_*(p_\zeta(g))=\eta$. Let $F(\zeta) = \eta$. It is easy to check that $F: Z \to Y$ is a fiber-preserving holomorphic map such that $F(c)=b$ and $F^*(f)=g$. $\square$

I don't understand the second paragraph of the proof. Why is the map $F:Z\to Y$ well-defined? If we choose a different curve from $a$ to $x$, we might get a different $\eta$...

For the relevant definitions:

Suppose $X$ and $Y$ are Riemann surfaces and $\mathcal{O}_X$ and $\mathcal{O}_Y$ are the sheaves of holomorphic functions on them. Suppose $p: Y \to X$ is an unbranched holomorphic map. Since $p$ is locally biholomorphic, for each $y\in Y$ it induces an isomorphism $p^*: \mathcal{O}_{X, p(y)} \to \mathcal{O}_{Y, y}$. Let

$$ p_*: \mathcal{O}_{Y, y} \to \mathcal{O}_{X, p(y)} $$

be the inverse of $p^*$.

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

  1. $Y$ is a Riemann surface and $p:Y \to X$ is an unbranched holomorphic map.
  2. $f$ is a holomorphic function on $Y$.
  3. $b$ is a point of $Y$ such that $p(b) = a$ and $$ p_*(p_b(f)) = \varphi.$$

An analytic continuation $(Y, p,f, b)$ of $\varphi$ is said to be maximal if it has the following universal property. If $(Z, q, g, c)$ is any other analytic continuation of $\varphi$, then there exists a fiber-preserving holomorphic mapping $F: Z\to Y$ such that $F(c) = b$ and $F^*(f) = g$.

  • $\begingroup$ That's the whole point of defining $Y$ so different (homotopically) curves in $X$ lead to different points in $Y$ or conversely given a path in $X$ there is a unique point in $Y$ corresponding to the homotopy class of the path in $X$ (and the continuation of $\varphi$ on that path) $\endgroup$ – Conrad Apr 21 at 17:11
  • $\begingroup$ How to make simple things very abstract. For a function $f$ analytic on a small disk around $a$ you are considering the set of curves $\gamma : \gamma(0)=a \to \gamma(1)$ such that $f$ can be analytically continued along $\gamma$, which gives a new function $f_\gamma$ analytic around $\gamma(1)$ (a power series), the full analytic continuation of $f$ is the set of those $(\gamma(1),f_\gamma)$ which is a Riemann surface $Y$. You can define homotopy of two such curves in term of if $f_{\gamma_r}$ is well-defined, which is the same as homotopy on $Y$. $\endgroup$ – reuns Apr 21 at 17:43
  • $\begingroup$ @Conrad In general it is not the homotopy in $X$, to do so you need to assume for any $b \in Y$ and simply connected set $U \ni f(b)\subset X$ then $U = f(V)$ with $V\subset Y$ simply connected. $\endgroup$ – reuns Apr 21 at 18:50
  • $\begingroup$ @Conrad Sorry for the late response. I think the reasoning is this: For $\zeta\in Z$ find curve $\gamma:[0,1]\to Z$ such that $\gamma(0)=c$, $\gamma(1)=\zeta$. Then $q_*(\rho_\zeta(g))$ is the analytic continuation of the function germ $\varphi$ along $q\circ\gamma$. Now there is a lifting $\Gamma:[0,1]\to Y\subset|\mathcal{O}|$ of $q\circ\gamma$ along $p$. Then we define $F(\zeta)=\Gamma(1)$. Why is it independent of $\gamma$? $\endgroup$ – Colescu May 2 at 13:28
  • $\begingroup$ Hmmm I seem to understand it now. So $F(\zeta)$ is just $q_*(\rho_\zeta(g))$ and the lifting $\Gamma$ is only used to show that $q_*(\rho_\zeta(g))$ lies in the connected component $Y$? Am I right? $\endgroup$ – Colescu May 2 at 13:35

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