MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $T: X \to X$ be an Anosov diffeomorphism. Suppose $f: X \to \mathbb{R}$ is Holder continuous (say with exponent $\alpha$). The question arises as to when $f$ can be written as $g \circ T - g $ for some $\alpha$-Hölder $g: X \to \mathbb{R}$. It is easily checked that a necessary condition is that the periodic data vanishes, i.e. the sum of $f$ over every periodic orbit is zero. It is also a (nontrivial) theorem of Livsic that the converse is true, namely vanishing periodic data implies that this cohomological equation is solvable.

I'm interested in the following variant of the cohomological equation. Let $A: \mathbb{R}^k \to \mathbb{R}^k$ be an isomorphism. Suppose $f: X \to \mathbb{R}^k$ is $\alpha$-Hölder. When is the equation $f = A( g\circ T) - g$ solvable in $g$?

There seem to be sufficient conditions known for the case of compact groups. Are there conditions known in the noncompact but abelian case?

share|cite|improve this question
Akhil -- out of curiosity: why do you call $f=A(g\circ T)-g$ a cohomological equation? – algori Jul 28 '10 at 22:42
I think it's called a "twisted" cohomological equation. The equation $f = g\circ T -g$ comes from group cohomology (it says that the cocycle generated by $f$ is a coboundary). – Akhil Mathew Jul 28 '10 at 22:55
up vote 6 down vote accepted

Part I: The answer is yes under additional conditions:

  1. Periodic data conditions are satisfied. That is, for any periodic point $p$ $$ \sum_{x\in O(p)}f(x)=0. $$
  2. Exponent $\alpha$ is sufficiently close to 1.
  3. Transformation $A$ is dominated by $T$. That is, the map $(x,v)\mapsto(Tx, Av)$ is partially hyperbolic.

Then Walkden's paper "Solutions to the twisted cocycle equation over hyperbolic systems" proves that there exists an $\alpha$-Holder solution $g$. The result is more general: the target group is any Lie group with a bi-invariant metric and the equation is the cohomological equation for two cocycles rather than just coboundary equation.

Part II: Notice however that if $A\neq Id$ then the periodic conditions may be no longer necessary. Let's restrict to the case when $k=1$ then our equation takes form $$ f=\lambda g\circ T-g, $$ where $\lambda<1$. Direct computation shows that $$ g=-\sum_{i\ge 0} \lambda^if\circ T^i $$ is a solution. It is also clear that $g$ is Holder continuous. Moreover, in this case uniqueness is clear as well since the above formula for $g$ is obtained recurrently from $$ g=-f+\lambda g\circ T. $$ It seems that this generalizes rather straightforwardly to the case when $A$ is hyperbolic. And I think it's worthwhile to see if anything interesting happens in the case then $A$ has some eigenvalues on the unit circle and some off.

share|cite|improve this answer
And of course periodic data conditions must be sutisfied! – Andrey Gogolev Jul 28 '10 at 22:54
Thanks! $ $ – Akhil Mathew Jul 28 '10 at 22:56
See an update. – Andrey Gogolev Jul 29 '10 at 5:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.