Systems with trivial cohomology

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?

• "other minimal group rotations"? but the minimal group rotations are precisely the irrational rotations. What is the question? – YCor Feb 7 at 13:30
• 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? – Veridian Dynamics Feb 7 at 13:34
• 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 $\alpha$? – Veridian Dynamics Feb 7 at 13:36
• 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.) – YCor Feb 7 at 13:37
• 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. – YCor Feb 7 at 14:33