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?



Existence conditions for twisted cohomological equations?
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.


