Rigorous proof that monodromy group of logarithm is $\mathbb{Z}$

Question

I have seen a lot of references on the topic but none of them really contained a proof from first principles that the monodromy group of logarithm is $\mathbb{Z}$. Here's the developed theory in my view.

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Consider the equation $f(x)=a$, where $f:X\to\mathbb{C}$ is a meromorphic non-constant function defined on a connected open set $X\subseteq\mathbb{C}$, and $a\in\mathbb{C}$ is a parameter. By the Identity Theorem in Complex Analysis, for any $a$ the set of roots is discrete.

Let $a$ draw a closed curve in the complex plane. Then the roots will also draw curves, as these are preimages of a path-connected set under a continuous map. However, these root curves are not necessarily closed, even though their union has to be a closed set, as it is a preimage of a closed set under a continuous map. We call these curves local branches. A critical point is a root of $f'(x) = 0$.

Denote the local branches by $\{x_1(a_0),\ldots,x_d(a_0)\}$ for some regular value $a_0$, i.e. not a critical point. It is possible that this set is infinite (but countable). For each $i=1,\ldots,d$ define $x_i^\gamma(a_0)$ as the set of values obtained by analytic continuation of $x_i(a_0)$ along a loop (closed curve) $\gamma$ around a critical point, starting at $a_0$. We require that the curve of $a$ must be such that any local branch does not contain a critical point.

Because no root collisions occur along $\gamma$ (a double root must be a root of the derivative, but we assumed $f'\neq 0$ on $\gamma$), each $x_i^\gamma$ must overlap with some $x_j(a_0)$. Thus, $\gamma$ induces a well-defined permutation $\sigma_\gamma\in S_d$ such that $$x_i^\gamma(a_0)=x_{\sigma_\gamma(i)}(a_0)\text{.}$$ (Intuitively, by using the continuation $\gamma$, we "move" from one local branch to another.) The image of the map $\rho:\pi_1(\mathbb{C}\setminus\{\text{critical points}\}, a_0)\to S_d$, given by $\rho(\gamma) = \sigma_\gamma$, is the monodromy group of $f$.

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Based on this, here is a proof that the monodromy of the exponential function is $\mathbb{Z}$. The image of $\exp$ is $\mathbb{C}^\times = \mathbb{C} \setminus \{0\}$. Suppose $a$ traverses a unit circle $\gamma$ around the origin, starting at $-1$. The local branches are $x_k(a) = \ln a + 2k\pi i$, $k\in \mathbb{Z}$. For $a = e^{i\theta}, \theta \in (-\pi, \pi)$ this becomes $x_k(a) = (2\pi k + \theta)i$. If we analytically continue along $\gamma$, then as $\theta$ runs through $(-\pi, \pi)$, for any $k$ the preimage $x_k(a)$ moves to $x_{k+1}(a)$. This corresponds to the permutation in which $k$ is mapped to $k+1$ for any $k$ and the subgroup generated by it is isomorphic to $\mathbb{Z}$.

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If I try to do this for the multivalued Log $z = \ln |z| + (\arg z + 2\pi n)i$ (or any single valued branch cut on it), I get a weird behaviour for $z$ - if $a = e^{i\theta}$ traverses a unit circle again, then $|z| = e^{\cos \theta}, \arg z + 2\pi n = \sin \theta$. I hope that there is a direct way to proceed from here or some other "elementary" manipulation of equations, rather than through some lengthy complicated equivalent theory. Any help appreciated!

Answer 1

I will be assuming here that the reader is familiar with basics of fundamental groups and covering spaces, as discussed, for instance, in Chapter 1 of Hatcher's "Algebraic Topology." (The same material can be found in many other algebraic topology textbooks.)

Consider the exponential map $\exp: \mathbb C\to \mathbb C^*$. One first verifies that this map is the (universal) cover with the group of covering (aka deck) transformations $G$ generated by $g: z\mapsto z+ 2\pi i$, and isomorphic to $\mathbb Z$. This group is isomorphic to the fundamental group $\pi_1(\mathbb C^*, 1)$. (The choice of a base-point $1$ is just for concreteness, you can use any base-point.) Concretely, this isomorphism is given by the path-lifting with respect to $\exp$: Take a loop $\gamma$ in $\mathbb C^*$ based at $1$. Then $\gamma$ has a unique lift $\tilde \gamma$, which is a path in $\mathbb C$ starting at $0$ and terminating at point $g^n(0)$, $n\in \mathbb Z$. The terminal point of $\tilde\gamma$ depends only on the (based) homotopy class of the loop $\gamma$. This is the isomorphism $$ \pi_1(\mathbb C^*, 1)\to G\to \mathbb Z, [\gamma]\mapsto g^n\mapsto n\in \mathbb Z. $$ As you check the definition of path-lifting in algebraic topology, it is exactly the same procedure as constructing the analytic continuation of $\log(z)$ defined in a small neighborhood of $1$ and normalized so that $\log(1)=0$. (There is nothing special here about $\exp$, the same will work for any regular covering map of any Riemann surface and even a complex manifold. See for instance Section 7 in Forster's "Lectures on Riemann Surfaces.") By the definition, the above map $\pi_1(\mathbb C^*, 1)\to Aut(\mathbb C)$ (given by the analytic continuation along based loops) is the monodromy homomorphism of $\log$.