## 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.

• 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.