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.

Conversely, suppose we have the condition on periodic orbits. Then we can choose a system of representatives ${x_{\alpha}, \alpha \in A}$ for each orbit by the axiom of choice. Then define ${f(T^i x_{\alpha}) = \sum_{j=0}^{i-1} f(T^j x_{\alpha})}$. This is seen to be well-defined in view of the vanishing of ${f}$ over periodic orbits, and it is equally easy to check that this solves the cohomological equation.

One motivation for the name “cohomology” comes from the fact that this can be interpreted as group cohomology. Indeed, the group ${\mathbb{Z}}$ acts on ${X}$ via ${T}$, and consequently on the set of functions ${f: X \rightarrow \mathbb{R}}$. So, this abelian group ${\mathrm{Fn}(X, \mathbb{R})}$ is actually a module over the group ring ${\mathbb{Z}[\mathbb{Z}]}$. A 1-cocycle corresponds to a function ${A: X \times \mathbb{Z} \rightarrow \mathbb{R}}$ such that

$\displaystyle A(x,n+m) = A( x, n) A (T^nx, m)$

which is evidently determined by the function ${A(\cdot, 1)}$ on ${X}$. To say that it is a coboundary is to say that we can solve the cohomological equation for ${A(\cdot, 1)}$.This interpretation via group cohomology is not (as far as I know) relevant to the problem that I’m working on, so I shall not spend much time on it, though. Readers should look at this post on Terence Tao’s blog for more in this vein.

In practice, in the theory of dynamical systems, we want something more: namely, we’ll have some structure, and we want ${f,g}$ to preserve this. For instance, if ${X}$ is a topological space and ${g}$ a continuous map, we’d like ${f}$ to be continuous as well. And the solution just given does not ensure that at all—in fact, ${f}$ needn’t even be Borel measurable!

Here is an example for when a cohomological equation as above arises in practice. Let ${M}$ be a compact, smooth oriented manifold and ${T}$ a diffeomorphism.

It is a basic theorem in ergodic theory (a corollary of weak* compactness of the unit ball in the dual space of continuous functions) that there exists a measure invariant with respect to ${T}$. However, this measure can have a rather trivial structure. For instance, if ${p \in M}$ is a fixed point, then the delta measure with respect to ${p}$ is invariant. What we’d want is a measure absolutely continuous with respect to the measure induced by ${\Omega}$. In particular, we want to find a positive function ${\rho: M \rightarrow \mathbb{R}}$ such that ${\rho \Omega}$ is invariant under ${T}$.

For this, note that the condition of an ${n}$-form ${\omega}$ being invariant under ${T}$ is precisely ${T^* \omega = \omega}$ by the change-of-variables formula. In particular, we are trying to find a positive ${\rho}$ such that

$\displaystyle T^*(\rho \Omega) = \rho \Omega .$

Now, for any function ${f: M \rightarrow M}$, define the Jacobian ${Jf}$ via ${f^* \Omega = (Jf) \Omega}$. Then this becomes the equation

$\displaystyle (\rho \circ T) (Jf) \Omega = \rho \Omega$

i.e.

$\displaystyle Jf = \frac{ \rho}{\rho \circ T}.$

This is just a multiplicative version of the cohomological equation. If we write ${\rho = e^{\psi}}$, this becomes precisely the additive one. It is now clear that a necessary condition for ${\rho}$ to exist is that

$\displaystyle Jf(x) Jf(Tx) \dots Jf(T^{n-1}x) = 1$

for ${x}$ satisfying ${T^n x = x}$. However, this is not sufficient if we want ${\rho}$ to be measurable (let alone continuous!), and this is why new techniques are necessary for the cohomological equation.

Measurability of solutions

Given the cohomological equation, there are not that many ways to get solutions. Basically, one has to define the function ${f}$ as sums over ${g}$ over orbits; the hard part is to show that the resulting ${f}$ has nice properties, for instance continuity.

Under certain weaker assumptions, we can at least deduce the measurability of solutions ${f}$ to the cohomological equation. Suppose that ${g}$ has vanishing periodic data and for every ${x\ in X}$, the sums

$\displaystyle S_n(x) = - \sum_{i=0}^{n-1} g(T^i x)$

are bounded. (Actually, the boundedness of these sums implies the vanishing of periodic data!)

Theorem 1 If the ${S_n(x)}$ are bounded for each ${x}$, then the cohomological equation has a measurable solution ${f}$.

The idea is that the ${S_n(x) = g(x) + S_{n-1}(Tx)}$, so that if the ${S_n(x)}$ have a limit ${S(x)}$ everywhere, then this limit would satisfy the cohomological equation ${S(x) = g(x) + S(Tx)}$. Also, the limit of measurable functions is always measurable. However, this is not necessarily the case.

Nevertheless, we can use a useful substitute for a limit of an arbitrary bounded sequence: namely, a Banach limit. Recall that a Banach limit is a continuous linear functional on ${l^{\infty}}$ that coincides with the usual limit on the subspace of sequences that do converge, which is translation-invariant. Let ${B}$ be one of them; they exist in view of (a slightly refined version of) the Hahn-Banach theorem.

Now if ${f_1, f_2, \dots, }$ is a sequence of continuous functions, the Banach limit ${B(f_1, \dots, )}$ exists because ${B}$ is continuous (though it may not be continuous). In detail, this follows because the inverse image of a Borel set in ${l^{\infty}}$ is a Borel set in ${X}$, so the map ${x \rightarrow (f_1, \dots, )}$ is at least Borel measurable.

So, define ${f(x) = B( \{ S_n(x)\}}$. Then it is clear that ${f(Tx) = B(\{S_{n}(Tx)\}) = g(x) + B(\{S_{n-1}(x)\}) = g(x) + f(x)}$. Thus, ${f}$ solves the cohomological equation and is measurable.

Side note: we didn’t actually have to have used Banach limits; sups would have been sufficient. Next time, we’ll discuss harder theorems and stronger methods for solving this.