If $\alpha \in \mathbb{R}\setminus \mathbb{Q}$ is an irrational number, then the rotation $X = (S^1, +\alpha)$ has "trivial" cohomology i.e. $$H^1(X) := C(X,\mathbb{C})/\beta C(X,\mathbb{C})$$ consists only of scalar multiples of the class $[1]$ of the constant function (where $C(X,\mathbb{C})$ is the additive group of complex-valued continuous functions on $X$ and $\beta g(x) = g(x+\alpha)-g(x)$ is the coboundary map). My question is, does this property characterizes irrational rotations? More precisely: let $X$ be compact metrizable and $T\colon X\to X$ an homeomorphism such that $(X,T)$ is minimal and $H^1(X,T)$ trivial (in the sense of above). Then, is $(X,T)$ conjugated, in some sense, to $(S^1,+\alpha)$, for some $\alpha$? or at least to a more general group rotation?

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    $\begingroup$ "other minimal group rotations"? but the minimal group rotations are precisely the irrational rotations. What is the question? $\endgroup$ – YCor Feb 7 at 13:30
  • $\begingroup$ If $X = (G,\cdot a)$ is a group rotatation such that the orbit $\{a^n:n\in\mathbb{Z}\}$ is dense in $G$, isn't it $G$ minimal? $\endgroup$ – Veridian Dynamics Feb 7 at 13:34
  • $\begingroup$ My question is: if $(X,T)$ is a topological dynamical system such that $H^1(X)$ is trivial then, is $(X,T)$ conjugated in some sense to $(S^1,+\alpha)$, for some $\alpha$? $\endgroup$ – Veridian Dynamics Feb 7 at 13:36
  • $\begingroup$ Well, can you edit your question so as to present the setting. I have no idea if you work on the circle or more general spaces/ compact group, and which which restrictions on the homeomorphism. The $H^1$ you define works for an arbitrary self-homeomorphism of a compact space, then can be specified to rotations of the circle. (By the way, $H^1(X)=\mathbf{C}$ is not trivial, it's 1-dimensional. For more general group actions, the space of co-invariants might be 0-dimensional.) $\endgroup$ – YCor Feb 7 at 13:37
  • $\begingroup$ Aren't you precisely asking which systems $(X,T)$ are uniquely ergodic? Uniquely ergodic means that the only positive invariant linear forms on $C(X)$ are positive scalar multiples of "taking the integral" with respect to a given invariant measure. So clearly your condition implies uniquely ergodic, but I was wondering whether the converse also holds. It seems so: if one has a invariant signed measure, its positive and negative mutually singular parts are invariant too. So if this is correct there is plethora of examples beyond irrational rotations. $\endgroup$ – YCor Feb 7 at 14:33

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