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 '19 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 '19 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 '19 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 '19 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 '19 at 13:35

Please note that it does not read $p_\zeta(g)$ but $\rho_\zeta(g)$ (rho), the germ of $g$ at the point $\zeta$.

The path chosen in $Z$ from $c$ to $\zeta$ is not decisive. Important is the germ $\rho_\zeta(g) \in \mathcal{O}_{Z,\zeta}$. Apparently it is independent from the path. It projects to the germ $$\eta:=q_*(\rho_\zeta(g)) \in \mathcal{O}_{X,x}$$ The latter is obtained as analytic continuation of $\phi$ along a curve - here you need the existence of at least one curve. Then by definition $\eta$ is a point of $Y$ and one defines $$F(\zeta):=\eta.$$


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.