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I'm currently reading about the concept of a “maximal analytic continuation” from Forster's book Lectures on Riemann Surfaces (see Section 7). There are a bunch of definitions to unpack before I can ask my question, so I'll start with those.


Definitions

Let $\mathcal{O}$ be the sheaf of germs of holomorphic functions over a Riemann surface $X$. Given a germ $\varphi \in \mathcal{O}_x$ ($\mathcal{O}_x$ denotes the stalk of $\mathcal{O}$ over the point $x \in X$), we define an analytic continuation of $\varphi$ as a quadruplet $(Y, p, f, y)$ satisfying:

  1. $Y$ is a Riemann surface, and $p : Y \to X$ is an unbranched holomorphic map (and thus a local diffeomorphism).
  2. $f : Y \to \mathbb{C}$ is a holomorphic map on $Y$.
  3. $p(y) = x$.
  4. $[f \circ (p|_{U_y})^{-1}]_x = \varphi$, where $U_y$ is a small neighborhood about $y$ such that the restriction $p|_{U_y}$ is a diffeomorphism.

Here's an example of an analytic continuation to provide some intuition. Suppose $X = \mathbb{C} \setminus \{0\}$, and suppose $\varphi$ is a germ associated to the complex logarithm at a point $x \in X$. Let $Y$ be the “graph” $$ Y := \{(z, w) : z \in \mathbb{C} \setminus \{0\}, \exp(w) = z\}. $$ ($Y$ is just the set of ordered pairs $(z, \log z)$ over all branches of the logarithm.) Let $p : (z, w) \mapsto z$ and $f : (z, w) \mapsto w$ be the associated projection maps. Setting $y$ to be the value of the germ $\varphi$ at $x$, one can show that $(Y, p, f, y)$ is an analytic continuation of $\varphi$. Viewing $Y$ as a subset of $\mathbb{R}^4$ and projecting it onto its first, second, and fourth coordinates, we see that $Y$ is isomorphic to a helicoid: the Riemann surface that the complex logarithm has classically been defined over.

Forster further defines the notion of the maximal analytic continuation of a germ, which he makes rigorous by defining via a universal property. We can always construct the maximal analytic continuation of a germ $\varphi$ as follows. We define the étalé space $E(\mathcal{O})$ associated to $\mathcal{O}$ as the disjoint union of all stalks of $\mathcal{O}$ (i.e., the set of all germs). The étalé space is a Hausdorff topological space, and has the structure of a Riemann surface via the projection map $\pi : [g]_x \mapsto x$. Let $Y$ be the connected component of the étalé space $E(\mathcal{O})$ that contains the germ $\varphi$. Setting $f([g]_x) := g(x)$, one can show that $(Y, \pi, f, \varphi)$ is the maximal analytic continuation of $\varphi$.


Question

My issue is that this definition does not seem to be the “correct” definition of analytic continuation. In particular, the maximal analytic continuation of a germ $\varphi$ depends on the choice of the space $X$. This is easily seen from Forster's construction of a maximal analytic continuation. If $\varphi$ is a germ about a point $x \in X$, we can repeat the construction of the maximal analytic continuation with the space $X$ replaced by another space $X' \subseteq X$ that contains the point $x$. However, the sheaf $\mathcal{O}_{X'}$ associated to $X'$ may not be the same as the sheaf $\mathcal{O}_X$ associated to $X$, so this may yield two different maximal analytic continuations! Meanwhile, the germ $\varphi$ is a local property, so it seems nonsensical for its analytic continuation to depend on the space $X$ it lives in.

Is there some purpose to Forster's definition that I'm not understanding? Perhaps, is there some alternative notion of a “maximal analytic continuation” that is independent of the choice of $X$?

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  • $\begingroup$ Why do you expect uniqueness? I guess that any definition of "analytic continuation" of a germ of holomorphic function at $x$ will be given up to local biholomorphisms around $x$. And any unbranched holomorphic map $Y \to X$ is a local biholomorphism (by the Inverse Function Theorem). $\endgroup$ Commented Mar 11, 2022 at 8:49
  • $\begingroup$ @FrancescoPolizzi Yeah, it makes sense why analytic continuations are defined up to local biholomorphisms. My issue is that the definition of what an analytic continuations is (and what ends up being the maximal analytic continuation) is dependent on the space $X$. It seems that swapping $X$ for another surface may yield a different set of analytic continuations (in a non-isomorphic sense). $\endgroup$
    – Frank
    Commented Mar 11, 2022 at 9:04

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I think you're certainly right that one can define the category of quadruples $(Y, p, f, y)$ where

  • $Y$ is a Riemann surface and $y \in Y$ is a point,
  • $p$ is an equivalence class of open embeddings $U \hookrightarrow Y$ that are biholomorphisms onto the image, where $U$ is a small open neighborhood of $x \in X$ and $p(x) = y$,
  • $f$ is a holomorphic function on $Y$ extending the germ in $\mathcal{O}_y$ corresponding to $\varphi \in \mathcal{O}_x$.

Then we can ask whether there exists a final object in this category, and define that to be the maximal analytic continuation of $\varphi \in \mathcal{O}_x$.

It's just that this is completely uninteresting. Let's take the example of $X = \mathbb{C} \setminus \{0\}$ and the germ of the logarithm function at $x = 1$. The final object simply turns out to be $Y = \mathbb{C}$ with $f(x) = x$, where $p$ is the logarithm function around a neighborhood of $1$. Indeed, if $(Y^\prime, p^\prime, f^\prime, y^\prime)$ is any analytic continuation, then the function $f^\prime$ defines the morphism $Y^\prime \to \mathbb{C} = Y$.

What's happening is that equivalence classes of germs $(X, x, \varphi)$ are classified precisely by the value $\varphi(x) \in \mathbb{C}$ and the degree of the first nonzero non-constant term in the Taylor expansion of $\varphi$. When $\varphi^\prime(x) \neq 0$, the terminal object will always be $Y = \mathbb{C}$. It's probably true that when $\varphi^\prime(x) = 0$ there is no terminal object in the category of analytic continuations.

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  • $\begingroup$ Thanks for answering this old question! I'm guessing you're defining morphisms in this category to be a local diffeomorphism at $x$? Otherwise, I believe the final object should still be $\mathbb{C}$ by the same argument. $\endgroup$
    – Frank
    Commented Nov 4 at 15:56

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