Possible paths of analytic continuation

Question

Let $U\subset \mathbb{C}$ and $U$ open. Let $f:U\rightarrow \mathbb{C}$ and $f$ holomorphic. Let $p\in U$. Let $\gamma$ be path starting from $p$.

definition $f$ can be analytically continued along path $\gamma$ iff there exists an indexed finite set of tuples $(D_i,f_i)_{i\in I}$ where $D_i$ is a an open disc and $f_i$ is a holomorphic function on it such that for all $i,j \in I$ $f_i|_{D_i\cap D_j}=f_j|_{D_i\cap D_j}$ and for all $i\in I$ $f_i|_{D_i\cap U}=f$.

$V(f):=\{q\in \mathbb{C}: \ \exists $ a path $ \gamma $ starting at $p$ and ending at $q $ such that $f$ can be continued along it$\}$

The question is: can we perform analytic continuation along any path in $V$ starting from a point $p$?

So far, I have only encountered functions for which this is possible. For example: meromorphic functions, sections of functions as $z\rightarrow z^n$ and solutions of linear ode's on $\mathbb{C}$.

(edit: I want to stress that the resulting extension of $f$ could be multivalued, so that is not the issue.)

Answer 1

Okay, I found a counterexample myself. Define $z\mapsto \log(z)$ as a local inverse around 1 of $z\mapsto e^z$ such that $\log (1)=0$. Define $$ g(z)=\frac{1}{z-2 \pi i}. $$ The set of points for which there exists a path starting at 1 along which we can analytically continue the function $g(\log(z))$ is $\mathbb{C}\backslash \{0\}$. However, we cannot analytically continue this along the path $t\mapsto e^{2t\pi i}$, since there $\log(z)$ reaches $2\pi i$ and $g$ explodes on that value.

Answer 2

The question is: can we perform analytic continuation along any path in $V$ starting from a point $p$?

This question looks clear enough for me to answer it.

Yes, we can perform analytic continuation along any path having enough space around it to allow it. The resulting functions might however differ. These different functions are called branches.

Answer 3

I don't have my copy of Lang's "Complex Analysis" handy to see exactly what he wrote, but I suspect that there is a small underlying misunderstanding that makes it harder to address this precisely.

Also, there is a not-so-trivial theorem (perhaps attributable to Weierstrass?) about a condition under which analytic continuation is guaranteed... Probably some version of it is in Lang's book. I believe also in Ahlfors'.

First, I have a suspicion that the desired hypothesis is something in the following direction: for some non-empty open subset $U$ of $\mathbb C$, for every $z_o\in U$ there is an open neighborhood $U_{z_o}\subset U$ of $z_o$ and a holomorphic function $f_{z_o}$ on $U_{z_o}$ (possibly a set of them, or a vector space, or...) ... with some sort of compatibility conditions on non-empty overlaps $U_{z_1}\cap U_{z_2}$.

(Typical instances are that the local bits $f_{z_o}$ are local solutions of an algebraic equation, or of a linear differential equation. But the set-up does admit some abstractions/generalizations.)

Then the theorem asserts the existence of an analytic continuation along any path inside $U$.

No, it does not assert that the analytically continued pointwise value(s) match the starting value, in case the path returns to the beginning point.

In fact, the Monodromy Theorem asserts that the values obtained by analytic continuation depend only on the homotopy class of a closed path (and base point, if the open $U$ is not connected).

Thus, analytic continuation (with a connected open) gives a group homomorphism from the first homotopy group to a Galois group, in the case of an algebraic equation, and to a group of automorphisms of the vector space of solutions to a homogeneous linear differential equation (with analytic coefficients...)