In this post we will describe the example constructed by Furstenberg, a volume-preserving diffeomorphism which is minimal, but not ergodic. See also Parry’s book Topics in Ergodic Theory.
Let be an irrational number and be the irrational rotation. Let be a smooth function, which induces a skew-product , . Consider the following cohomological equation:
Remark 1. Viewed (@) as a real-valued equation, there is an obstruction for it to admit any solution since . But viewed as a -valued function, the obstruction is trivial for if is an integer. (also nontrivial if is an irrational..)
Remark 2. If , then such is far from minimal and (@) has no real-valued solution for all . But is an integer for even number and the equation (*) for such admits (trivial) constant solutions.
Proof of Prop 1. If is not minimal, then pick a proper minimal subset . Inductively we see , where . So the first component projection of covers .
Consider the translation of the second component . Then we have
– (for different ‘s) are either coincide, or disjoint;
– the union of all covers .
Moreover the set forms a nonempty, proper, closed subgroup of , and hence a finite group. Let be the order of , and define a map . Note that is independent of the choices of in the fiber of over (hence a well-defined -valued function). Also is continuous, since is closed. Note that . Therefore for all , and serves as a continuous solution of equation (*). Q.E.D.
Proof of Prop 2. Let be a measurable solution of (*) for some . Then consider the following function . Then . So is -invariant (and obviously not constant if we keep fixed and vary ). So is not ergodic. Q.E.D.
Proof of Theorem. Consider the Fourier series and . Then equation (@) turns out to be
By a suitable choice of (say some Liouville number) and with (can be analytic), we can ensure that the series converges at every point, but discontinues at some points. This completes the proof. Q.E.D.
Define inductively by setting , . Let . Note that .
Let and and . Then we see does not converge at , but is analytic. The map is minimal but not ergodic.
In the following let’s copy some materials of Furstenberg’s paper. A system is said to be uniquely ergodic if there is only one ergodic probability. In general a measure-preserving triple is referred as a process. For a fixed , the sequence forms a stationary stochastic process. We say the process is ergodic when is ergodic. We say the process is uniquely ergodic when is uniquely ergodic.
The system is said to be almost periodic, if is minimal and forms a equicontinuous family of transformations. Such a process is called an almost periodic process.
Theorem 1.2. Every almost periodic transformation is uniquely ergodic.
Proof. Let and . Then are uniformly continuous, so is . Then a subsequence of converge uniformly to some limit . Clearly is -invariant and hence a constant (since is minimal). Note that for all . So and the original sequence actually has only one limit . This procedure can be applied to any other and lead to . So is uniquely ergodic.
Theorem 1.3. Let be a circle diffeomorphism.
If is periodic free, then it is uniquely ergodic.
More generally, every point on is a generic point for some ergodic measures, (even non-uniquely ergodic case).
Let be a uniquely ergodic system with be its ergodic measure.
Consider a continuous map and the induced extension ,
. Consider the following equation:
Lemma 2.1. is uniquely ergodic iff is ergodic,
if and only if the equation (1) has no measurable solution for all .
Proof. Clearly is -invariant. Moreover it is -invariant.
It implies that if a point is generic with respect to , so is .
So either has no generic point, or every point is its generic point. This proves the first part.
Let’s consider a function . Being a product measure, we can write
. If is -invariant, then for all .
Since is ergodic, for some . In fact is nonzero almost everywhere since is ergodic. So its inverse provides the solution
of the cohomological equation.
On the other hand, for a solution of (k), the induced function is -invariant. Q.E.D.
Note that for a nonzero solution of (k), we know is -invariant and hence a nonzero constant. We may assume .
Lemma 2.2. Suppose , is ergodic with respect to . Consider a -valued map such that has nonzero degree for each fixed and uniform Lipschitz. Then the following equation:
has no measurable solution.
In particular is uniquely ergodic if and only if is.