Minimal but non-ergodic volume-preserving systems

In this post we will describe the example constructed by Furstenberg, a volume-preserving diffeomorphism f\in\mathrm{Diff}^{\omega}_m(\mathbb{T}^2) which is minimal, but not ergodic. See also Parry’s book Topics in Ergodic Theory.

Let \alpha be an irrational number and R_\alpha:\mathbb{T}\to\mathbb{T} be the irrational rotation. Let r:\mathbb{T}\to\mathbb{R} be a smooth function, which induces a skew-product f:\mathbb{T}^2\to\mathbb{T}^2, (x,y)\mapsto (x+\alpha,y+r(x)). Consider the following cohomological equation:

(*)      \phi(x+\alpha)=e^{2\pi ik\cdot r(x)}\cdot\phi(x),

(⋆)      \phi(x+\alpha)=k\cdot r(x)+\phi(x).

Remark 1. Viewed (⋆) as a real-valued equation, there is an obstruction for it to admit any solution since r_0=\int r(x)dx >0. But viewed as a \mathbb{T}-valued function, the obstruction is trivial for k if k\cdot r_0 is an integer. (also nontrivial if r_0 is an irrational..)

Proposition 1. If f is not minimal, then the equation (*) has a continuous S^1-valued solution for some k\ge1.

Proposition 2. If the equation (*) has some measurable solution for some k\neq 0, then f is not ergodic.

Remark 2. If r(x)\equiv0.5, then such f is far from minimal and (⋆) has no real-valued solution for all d\neq0. But k\cdot r_0 is an integer for even number k and the equation (*) for such k admits (trivial) constant solutions.

Theorem. By a suitable choice of \alpha and r, the above equation has a L^2-solution, but no continuous solution. In particular the corresponding system is minimal but not ergodic.

Remark 3. For circle homeomorphism f on \mathbb{T}, it has only one rotation number \rho. If \rho is irrational, then f is semiconjugate to a rotation. If $f$ is C^2, then it is transitive and is actually conjugate to a rotation. A natural question is whether Denjoy’s theory admits a reasonable generalization in higher dimensions. A homeomorphism of a higher dimensional torus does not, in general, have a unique rotation vector. Even if this is the case, the minimality of the corresponding translation does not imply the
existence of a continuous semi-conjugacy to it.

Proof of Prop 1. If f is not minimal, then pick a proper minimal subset K\subset \mathbb{T}^2. Inductively we see f^n(x,y)=(x+n\alpha,y+r_n(x,\alpha)), where r_n(x,\alpha)=r(x)+r(x+\alpha)+\cdots+r(x+(n-1)\alpha). So the first component projection of K covers \mathbb{T}.

Consider the translation of the second component L_\beta:(x,y)\mapsto (x,y+\beta). Then we have

L_\beta K (for different \beta‘s) are either coincide, or disjoint;

– the union of all L_\beta K covers \mathbb{T}^2.

Moreover the set G=\{\beta: L_\beta K=K\} forms a nonempty, proper, closed subgroup of \mathbb{T}, and hence a finite group. Let d be the order of G, and define a map \phi:x\in\mathbb{T}\mapsto (x,k_x)\in K\mapsto d\cdot k_x\in\mathbb{T}. Note that \phi(x) is independent of the choices of k_x in the fiber of K over x (hence a well-defined \mathbb{T}-valued function). Also \phi is continuous, since K is closed. Note that (x+\alpha,k_x+r(x))=f(x,k_x)\in fK=K. Therefore \phi(x+\alpha)=d\cdot (k_{x}+r(x))=\phi(x)+d\cdot r(x) for all x\in \mathbb{T}, and e^{2\pi i\phi} serves as a continuous solution of equation (*). Q.E.D.

Proof of Prop 2. Let \phi be a measurable solution of (*) for some k\neq0. Then consider the following function \Phi:(x,y)\in\mathbb{T}^2\mapsto \phi(x)\cdot e^{-2\pi ky}. Then \Phi(f(x,y))=\Phi(x+\alpha,y+r(x))=\phi(x+\alpha)e^{-2\pi ik(y+r(x)}=\phi(x)\cdot e^{-2\pi ky}=\Phi(x,y). So \Phi is f-invariant (and obviously not constant if we keep x fixed and vary y). So f is not ergodic. Q.E.D.

Proof of Theorem. Consider the Fourier series \phi(x)=\sum_n \phi_n e^{2\pi inx} and r(x)=\sum_n r_n e^{2\pi inx}. Then equation (⋆) turns out to be

\phi_n\cdot e^{2\pi i n\alpha}=\phi_n+k\cdot r_n, or equivalently, \displaystyle \phi_n=\frac{k\cdot r_n}{e^{2\pi i n\alpha}-1}, for all n\neq 0.

By a suitable choice of \alpha (say some Liouville number) and r with kr_0\in\mathbb{Z} (can be analytic), we can ensure that the series \sum_n \phi_n e^{2\pi i nx} converges at every point, but discontinues at some points. This completes the proof. Q.E.D.

Define v_k inductively by setting v_1=1, v_{k+1}=2^{v_k}+v_k+1. Let n_k=2^{n_k} \alpha=\sum_{k\ge 1}1/n_k. Note that \displaystyle |n_k\alpha-[n_k\alpha]|=\sum_{i>k} 2^{v_k-n_i}<\sum_{i>k} 2^{v_k-v_{k+1}+1-i}\le 2\cdot 2^{v_k-v_{k+1}}=2^{-2^{v_k}}=2^{-n_k}.

Let n_{-k}=-n_k and \displaystyle \phi(x)=\sum_{k\neq 0}\frac{1}{|k|}e^{2\pi in_k x} and r(x)=\phi(x+\alpha)-\phi(x)=\sum_{k\neq 0}\frac{e^{2\pi i n_k \alpha}-1}{|k|}\cdot e^{2\pi in_k x}. Then we see \phi does not converge at x=0, but r is analytic. The map (x,z)\mapsto (x+\alpha,z+r(x)) is minimal but not ergodic.

In the following let’s copy some materials of Furstenberg’s paper. A system (X,T) is said to be uniquely ergodic if there is only one ergodic probability. In general a measure-preserving triple (X,T,\mu) is referred as a process. For a fixed \phi\in C(X), the sequence T^n\phi=\phi\circ T^n forms a stationary stochastic process. We say the process is ergodic when (X,T,\mu) is ergodic. We say the process is uniquely ergodic when (X,T) is uniquely ergodic.

The system (X,T) is said to be almost periodic, if T is minimal and \{T^n\} 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 \mu\in \mathcal{M}(f) and \phi\in C(X). Then \phi\circ T^n are uniformly continuous, so is \displaystyle A_n\phi=\frac{1}{n}\sum_{k=0}^{n-1}\phi\circ T^k. Then a subsequence of A_n converge uniformly to some limit g_\phi\in C(X). Clearly g_\phi is T-invariant and hence a constant (since T is minimal). Note that \mu(A_n\phi)=\mu(\phi) for all n. So g_\phi=\mu(\phi) and the original sequence actually has only one limit g_\phi. This procedure can be applied to any other \nu\in\mathcal{M}(f) and lead to \nu=\mu. So (X,T) is uniquely ergodic.

Theorem 1.3. Let f:\mathbb{T}\to\mathbb{T} be a circle diffeomorphism.
If f is periodic free, then it is uniquely ergodic.
More generally, every point on \mathbb{T} is a generic point for some ergodic measures, (even non-uniquely ergodic case).

Let (X,f) be a uniquely ergodic system with \mu be its ergodic measure.
Consider a continuous map r:X\to\mathbb{T} and the induced extension F:X\times\mathbb{T}\to X\times \mathbb{T},
(x,z)\mapsto (fx,z+r(x)). Consider the following equation:

(k)      \phi(fx)=e^{2\pi i k r(z)}\phi(x)

Lemma 2.1. (X\times\mathbb{T},F) is uniquely ergodic iff (X\times\mathbb{T},F,\mu\times m) is ergodic,
if and only if the equation (1) has no measurable solution for all k\neq0.

Proof. Clearly \mu\times m is F-invariant. Moreover it is L_\beta-invariant.
It implies that if a point (x,z) is generic with respect to \mu\times m, so is (x,z+\beta).
So either \mu\times m has no generic point, or every point is its generic point. This proves the first part.

Let’s consider a function \Phi(x,z)\in\mathrm{L}^2(\mu\times m). Being a product measure, we can write
\Phi(x,z)\sim \sum_n \phi_n(x)\cdot e^{2n\pi z}. If \Phi is F-invariant, then \phi_n(fx)e^{2\pi in r(z)}=\phi_n(x) for all n.
Since (X,f,\mu) is ergodic, \phi_n\neq 0 for some n\neq0. In fact \phi_n is nonzero almost everywhere since f is ergodic. So its inverse \phi_n^{-1} provides the solution
of the cohomological equation.
On the other hand, for a solution of (k), the induced function \Phi(x,z)=\phi(x)e^{-2\pi ikz} is F-invariant. Q.E.D.

Note that for a nonzero solution \phi of (k), we know |\phi| is f-invariant and hence a nonzero constant. We may assume |\phi|=1.

Lemma 2.2. Suppose \mu\in\mathcal{M}(X,f), \mu_1=\mu\times m is ergodic with respect to F:(x,z)\in X\times\mathbb{T}\mapsto (fx,z+r(x)). Consider a \mathbb{T}-valued map g(x,z) such that g(x,\cdot) has nonzero degree for each fixed x and uniform Lipschitz. Then the following equation:

(k)      \Phi\circ F=e^{2\pi i kg}\cdot \Phi

has no measurable solution.

In particular F_2=(F,g) is uniquely ergodic if and only if f is.

Consider the suspension T_t over \mathbb{T}^2_f. Let \mathcal{O}_f be the orbit foliation of T_t on \mathbb{T}^2_f. Then \mathcal{O}_f is a minimal but non-ergodic foliation.

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  • silver account  On December 9, 2012 at 4:58 am

    concerning Exercise 12 there seems to be a small problem (hope I understood the definitions and details): Suppose T is a homeomorphism of the circle K with exactly one fixed point, say 0, whose complement K\backslash\{0\} is wandering. E.g. we may take the map z=\exp(2\pi ix) \mapsto T(z)=exp(2\pi ix^2) where x is in [0,1]. Then T is uniquely ergodic as every invariant measure is supported in the non-wandering set, thus there is only one invariant measure, the Dirac mass in 0. Nevertheless T is not minimal as is has a fixed point.

    • Pengfei  On December 12, 2012 at 4:11 am

      Yes your example is uniquely ergodic and non-minimal. Furstenberg’s example goes the other way: minimal and non-uniquely ergodic. In fact the special measure– the volume, is preserved but not ergodic.

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