So here's my question:

Does there exist a minimal diffeomorphism of class at least $\mathcal{C^2}$ of a compact manifold X which is

  1. minimal

  2. uniquely ergodic with unique probability measure $\mu$

  3. not ergodic with respect to the Lebesgue measure ?

I don't really see why these requirements should contradict each other but I haven't been able to find an example. Note that the regularity hypothesis is necessary as (see R.W.'s answer below): there are $\mathcal{C}^1$ circle diffeomorphisms that satisfy those conditions, but one could argue that they are a bit artificial since as soon as the derivative is required to have bounded variation this can no longer be true.

I would also be happy with any example that is just a piecewise diffeomorphism!

  • $\begingroup$ Just take something like a shift on the circle, but use non-standard coordinates so that Lebesgue measure becomes something else in the new coordinates. (More to the point perhaps, what is "Lebesgue measure" anyway on a manifold?) $\endgroup$ Oct 12, 2018 at 16:37
  • $\begingroup$ Hi Christian. If you use "non-standard coordinates", then you loose the fact that your map is a diffeo. For any smooth manifold, the class of the Lebesgue measure is well-defined and it makes sense to say that a diffeo is ergodic (when the measure of any invariant set or its complement is zero for any representative of the class). It is standard that any sufficiently regular ($\mathcal{C}^2$) minimal circle diffeo is ergodic with respect to Lebesgue. $\endgroup$
    – Selim G
    Oct 12, 2018 at 16:45
  • $\begingroup$ By "non-standard coordinates," I meant a smooth change of variable, so the map stays $C^{\infty}$. $\endgroup$ Oct 12, 2018 at 18:08
  • 1
    $\begingroup$ So the definition of ergodic as I understand the OP is $T^{-1}A=A$ implies $\lambda(A)=A$. This makes sense for non-invariant measures. If one starts with a circle rotation and conjugated by something smooth, the new map preserves the push-forward of the original measure (which will generally not be Lebesgue). Nonetheless, Lebesgue measure is ergodic for the new map according to the definition that the OP is using. $\endgroup$ Oct 14, 2018 at 1:29
  • 1
    $\begingroup$ @Christian Remling As Anthony says, the definitions used by the OP are absolutely standard and the question is completely clear. The notion of ergodicity is routinely defined and used for quasi-invariant measures (and their classes) as well, and OP tacitly referred to the fact that the Lebesgue measure class (one shouldn't really talk about the Lebesgue measure on a smooth manifold) is quasi-invariant with respect to any diffeomorphism. $\endgroup$
    – R W
    Oct 14, 2018 at 22:30

2 Answers 2


I couldn't manage to find an online version, but this paper by Yoccoz provides an example of a diffeomorphism of the $2$-dimensional torus which is the product of two analytic circle diffeomorphisms, which is minimal, uniquely ergodic, and totally dissipative for Lebesgue measure (hence it cannot be ergodic).

  • $\begingroup$ Hi Andres, thank you for your answer. I have just read the abstract on Math Sci Net and it certainly answers the question. If anyone knows where to find the text I'd be super grateful... $\endgroup$
    – Selim G
    Oct 28, 2018 at 22:36

You haven't specified the smoothness, so that hopefully $C^1$ is OK. It was Denjoy who proved in 1932 that if a $C^1$ diffeomorphism $f$ of the circle has an irrational rotation number $\alpha$ and its derivative has bounded variation, then it is $C^0$-conjugate to the $\alpha$-rotation, and therefore is uniquely ergodic. Answering Denjoy's question, Herman (1979) and Katok (see Section 3.6 of Cornfeld-Fomin-Sinai) proved that if $f$ is $C^2$ and an irrational rotation number, then it is also ergodic with respect to the Lebesgue measure. Later the $C^2$ condition was replaced by Katok-Hasselblatt with Denjoy's condition ($C^1$ and bounded variation of the derivative).

Oliveira and da Rocha (2001) gave an example of a minimal non-ergodic (with respect to the Lebesgue measure) $C^1$ diffeomorphism $f$ of the circle which is $C^0$ conjugate to an irrational rotation (and therefore is uniquely ergodic). Finally, Kodama and Matsumoto (2013) showed that non-ergodicity in such examples can be made "the strongest possible", namely $f$ can be chosen to be completely dissipative with respect to the Lebesgue measure, i.e., such that its ergodic components are just orbits, or, equivalently, it admits a measurable "fundamental domain".

  • $\begingroup$ Hi RW, thank you for your answer. Actually what I had in mind was the problem in high regularity as I was aware of those examples (sorry for not being precise enough in my question). What actually initiated my interest in this question is the following problem: can you find minimal, uniquely ergodic $\mathcal{C}^{\infty}$ interval exchange transformations that are not Lebesgue ergodic? which would be the natural generalisation of the Herman-Katok result you are referring to... $\endgroup$
    – Selim G
    Oct 15, 2018 at 6:16
  • $\begingroup$ What do you mean by $C^\infty$ interval exchange transformations? $\endgroup$
    – R W
    Oct 15, 2018 at 7:16
  • $\begingroup$ A piecewise continuous bijection of the interval with finitely many discontinuity points which $\mathcal{C}^{\infty}$ where it is continuous. It is sometimes called 'generalised' interval exchange transformation. $\endgroup$
    – Selim G
    Oct 15, 2018 at 7:33
  • $\begingroup$ Is this the same as a usual exchange transformation combined with a $C^\infty$ time change? $\endgroup$
    – R W
    Oct 16, 2018 at 13:05
  • $\begingroup$ By global time change do you mean conjugated to via a $\mathcal{C}^{\infty}$? Because if it is the case the answer to your question is no. It is formally what I have said in my previous comment (see arxiv.org/pdf/1003.1191.pdf for precise definitions). I think a good way to say it is that $\mathcal{C}^{\infty}$ interval exchange transformation are to standard interval exchange transformations what circle diffeos are to rotations. $\endgroup$
    – Selim G
    Oct 16, 2018 at 14:49

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