We continue the discussion of the cohomological equation started yesterday and explain a situation where it can be solved.﻿

Given a compact space ${X}$, a continuous map ${T: X \rightarrow X}$, and a continuous ${g: X \rightarrow \mathbb{R}}$ with vanishing periodic data as in yesterday’s post, there are not many ways to construct a solution ${f}$ of the cohomological equation

$\displaystyle g = f \circ T - f.$

The basic thing to note that if ${f(x)}$ is known, then recursively we can determine ${f}$ on the entire orbit of ${x}$ in terms of ${g}$. In case the map ${T}$ is topologically transitive, say with a dense orbit generated by ${x_0}$, then by continuity the entire map ${f}$ is determined by its value on ${x_0}$.

This also provides the method for obtaining ${f}$ in the topologically transitive case. Namely, one picks ${f(x_0)}$ aribtrarily, defines ${f(T^ix_0)}$ in the only way possible by the cohomological equation. In this way one has ${f}$ defined on the entire orbit ${T^{\mathbb{Z}}(x_0)}$ such that on this orbit, the equation is satisfied. If one can show that ${f}$ is uinformly continuous on ${T^{\mathbb{Z}}(x_0)}$, then it extends to the whole space and must by continuity satisfy the cohomological equation there too.

This is the strategy behind the proof of the theorem of Livsic from the seventies, whose proof we shall sketch:

Theorem 1 (Livsic) Let ${M}$ be a compact Riemannian manifold, ${T: M \rightarrow M}$ a topologically transitive Anosov diffeomorphism. If ${g: M \rightarrow \mathbb{R}}$ is an ${\alpha}$-Holder function such that ${T^n p =p}$ implies ${\sum_{i=0}^{n-1} g(T^i p) = 0}$, then there exists an ${\alpha}$-Holder ${f: M \rightarrow \mathbb{R}}$ such that

$\displaystyle g = f \circ T -f .$

We start by considering a very simple problem. Let ${X}$ be a set, ${T: X \rightarrow X}$ be a bijection function, and ${g: X \rightarrow \mathbb{R}}$ a function. We want to know when the cohomological equation

$\displaystyle g = f \circ T - f$

can be solved for some ${f: X \rightarrow \mathbb{R}}$.

It turns out that this very simple question has an equally simple answer. The answer is that the equation can be solved if and only if for every finite (i.e., periodic) orbit ${O \subset X}$, we have ${\sum_{x \in O} g(x) = 0}$. The necessity of this is evident, because if we have such a solution, then

$\displaystyle \sum_O g(x) = \sum_0 f \circ T(x) - f(x) = \sum_O f(x) - \sum_O f(x) = 0$

because ${T}$ induces a bijection of ${O}$ with itself. This condition is called the vanishing of the periodic obstruction.