Why is direct analytic continuation not transitive?

Question

Consider the following definitions and remarks which will be referenced in the question:

Defn( Direct analytic continuation) A function element in a domain $U$ is a pair $(f, D)$ where $D$ is a subdomain of $U$ and $f$ is an analytic function on $D$. Two function elements $(f,D)$ and $(g,E)$ are equivalent, write $(f,D) \sim (g,E)$ if $D \cap E \neq 0$ and $f = g$ on $D \cap E$. We say $(g,E)$ is a direct analytic continuation of $(f,D)$.

Why do we make such a definition? We know the power series $$\sum_{r \geq 0}z^k = \frac{1}{1-z}$$ is defined on $D(0,1)$ and cannot be extended to any larger domain due to natural boundary. However, $\frac{1}{1-z}$ is holomorphic on $\mathbb{C} \backslash \{1\}$ so sometimes the domain forced by the definition of a function is not the maximal possible. In other words, sometimes we are looking at the "correct" function with a "wrong" domain.

Defn(Analytic continuation along path) We say $( g, E )$ is an analytic continuation of $(f, D)$ along $\gamma$ if $\gamma : [0,1] \rightarrow 0$ and there exist function elements $(f_i, D_i), i \in \{ 0, \ldots , n \}$ and $0 = t_o < t_2 < \cdots < t_n = 1$ such that $$(f, D) = (f_0, D_0) \sim (f_1, D_1) \sim \cdots \sim (f_{n - 1}, D_{n - 1}) \sim (f_{n}, D_{n}) = (g, E)$$ and $\gamma \left( \left[ t_j, t_{j + 1} \right] \right) \subseteq D_j$ for $j \in \{ 0, \ldots, n - 1 \}$. Write $(f, D) \approx_{\gamma} (g, E)$.

Remark As $\mathbb{C}$ has a path-connected basis for the topology, domain are path-connected.

Defn(analytic continuation) We say $(g, E)$ is an analytic continuation of $(f,D)$ if there exists a path $\gamma$ such that $(f, D) \approx_{\gamma} (g, E)$. In this case we write $(f,D) \approx (g, E)$.

Remark

1. If $(f, D) \approx_{\gamma} (g, E)$ and $(f, D) \approx_{\gamma} (h, E)$ then $g = h$ by repeated application of the identity principle. In other words, $g$ is completely determined by $f$ and $\gamma$.
2. Analytics continuation is an equivalence relation (exercise), but direct analytic continuation is not transitive, even if we require pairwise intersections of the domains to be nonempty. If fact, that is the whole point of analytic continuation along path.

The author says that direct analytic continuation is not a transitive property. I am trying to understand why. The obvious thing to me that comes up to mind is that if $ A \cap B \neq \emptyset $ and $B \cap C \neq \emptyset$, then we could have $A \cap B = \emptyset$. Is this all there is or am I missing the larger picture?

Answer 1

Let $A, B, C$ be the open disks with radius $1$ and centers $1$, $e^{2\pi i/3}$, $e^{4\pi i/3}$ respectively (the third roots of unity). Let

• $f: A \to \Bbb C$ be the principal branch of the logarithm, i.e. $f(z) = \ln |z| + i \arg z$ with $-\pi/2 < \arg z < \pi/2$,
• $g: B \to \Bbb C$, $g(z) = \ln |z| + i \arg z$ with $2\pi i/3-\pi/2 < \arg z < 2\pi i/3+\pi/2$,
• $h: C \to \Bbb C$, $h(z) = \ln |z| + i \arg z$ with $4\pi i/3-\pi/2 < \arg z < 4\pi i/3+\pi/2$.

Then

• $f(z) = g(z)$ in $A\cap B$,
• $g(z) = h(z)$ in $B\cap C$,
• $h(z) = f(z)+ 2\pi i$ in $A\cap C$.

So $(A, f) \sim (B, g)$ and $(B, g) \sim (C, h)$, but not $(A, f) \sim (C, h)$.

The point is that three circles suffice to define an analytic continuation along a closed path, and the initial and the final function element of an analytic continuation along a closed path may not be equivalent.