I fell a bit behind on the continuation of the class field theory series because I was setting up a new laptop. Before I resume that, I want to talk about something very weird that I learned today.

Let {U \subset \mathbb{C}} be a set that omits at least two points. If {f: U \rightarrow U} holomorphic and is such that f(w)=w, {f'(w)=1} at one {w \in U}, then {f} is the identity.

This is a striking rigidity phenomenon!

But how do we prove it? The idea is to consider the sequence of iterates {f, f \circ f, \dots}. Suppose for simplicity {P=0}. Then in a neighborhood of {0}, we can write {f = z + cz^m + \dots }, where the {\dots} are omitted higher terms. If {f} is not identically the identity, then {c \neq 0}.

So, similarly, by direct computation, in some neighborhood of {P}, we have {f \circ f = z + 2c z^m + \dots}. Similarly, if we define {g_1 = f, g_2 = f \circ f, } for notational convenience, we have

\displaystyle g_k = z + kc z^m + \dots.

But the {g_k} are all holomorphic maps into {U}. Since {U} omits at least two points, the family {g_k} is normal by Montel’s theorem and consequently has a subsequence {g_{k_i}} that converges uniformly on compact sets.

Thus the derivatives {g^{(m)}_{k_i}(0) = m! k_i c} converge, which is impossible unless {c=0}.

Huh? I didn’t exactly see that coming. If {U} is the unit disk, then at least it looks familiar. A holomorphic map {f} of the unit disk into itself sending zero to zero must satisfy {|f'(0)| \leq 1}, and if equality holds {f} is a rotation. So perhaps this result should be thought of as a generalization of Schwarz’s lemma? (Nevertheless, the use of Montel’s theorem is quite a sledgehammer to prove something as elementary as Schwarz.)

I should say where I got this from: Krant’z Complex Analysis: The Geometric Viewpoint. Krantz didn’t prove exactly this, but the argument is the same.  Either this is standard fare that I missed when learning basic complex analysis, or I’m turning Climbing Mount Bourbaki into a comedy routine.