Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 30 (Ergodic Theory)

A nonempty compact dynamical system $(X,T)$ is a minimal dynamical system if it has no proper nonempty closed forward-invariant subset. Equivalently, every forward orbit is dense: the orbit closure of any point is a nonempty closed forward-invariant set, and conversely a closed invariant set contains the orbit closure of each of its points.

Consider the collection of nonempty closed forward-invariant subsets of $X$, ordered by reverse inclusion. Every chain has an upper bound, its intersection: the chain has the finite-intersection property, compactness makes the intersection nonempty, and continuity is not even needed for its forward invariance. The Zorn lemma supplies a minimal member $Y$, so $(Y,T{|}_Y)$ is a minimal subsystem. Also $T(Y)$ is compact, nonempty and forward invariant, so minimality gives $T(Y)=Y$. Here the printed word “nontrivial” must mean nonempty; a singleton with its identity map shows that a subsystem with more than one point cannot always be required.

If $y\in Y$, every tail of its forward orbit is dense in $Y$, since it is the orbit of $T^{N}y\in Y$. Therefore every neighborhood of $y$ contains $T^{n}y$ for arbitrarily large $n$. Choosing these times successively for balls of radii $1/j$ gives $n_j\to \infty$ with $T^{n_j}y\to y$. This proves the Birkhoff recurrence theorem: every continuous map of a nonempty compact metric space has a recurrent point. In fact every point of the chosen minimal subsystem is recurrent.

Write the continuous circle extension as the skew productA sufficient condition, and in this setting also a necessary one, is the circle skew-product minimality criterion: for every integer $m\ne 0$ there is no continuous function $f:X\to {\mathbb{S}}^1$ such thatThe cocycle $\rho$ is assumed continuous; a merely measurable cocycle need not define this topological extension.

To prove sufficiency, choose a minimal subsystem $Z$ of the extension. Its projection is a nonempty closed forward-invariant subset of $X$, hence all of $X$. For $a\in \mathbb{R}/\mathbb{Z}$, vertical translation sends $Z$ to another minimal subsystem, since it commutes with $\tilde{T}$. Two such translates either are disjoint or coincide: a nonempty intersection is closed and forward invariant, so minimality forces equality. Define their common stabilizerThis is a closed subgroup of the circle group. Its closedness follows by taking limits of translations of points in the compact set $Z$, using both a translation and its inverse. Each fiber $Z_x$ is nonempty. If $y,z\in Z_x$, the sets $Z$ and $Z+(z-y)$ intersect at $(x,z)$, so $z-y\in H$. Conversely translating any fiber point by any member of $H$ stays in the fiber. Thus every fiber is exactly one coset of $H$.

If $Z$ is proper, $H$ is proper and therefore finite, say $H=\{0,1/m,\dots ,(m-1)/m\}$ for some positive integer $m$. The character $y\mapsto e^{2\pi imy}$ is constant on each fiber coset, giving a well-defined function $f(x)$ from $Z_x$. This function is continuous: its graph is the continuous image of $Z$ in $X\times {\mathbb{S}}^1$, and projection of that compact graph onto the Hausdorff space $X$ is a continuous bijection and hence a homeomorphism. The relation $\tilde{T}(Z)=Z$ gives $f(Tx)=e^{2\pi im\rho (x)}f(x)$, contrary to the criterion. Hence $Z=\tilde{X}$ and the extension is minimal.

Conversely, a solution of the displayed equation makes $F(x,y)=e^{2\pi imy}/f(x)$ a nonconstant continuous invariant function on the extension. Its nonempty level sets are proper closed invariant subsets, contradicting minimality. This also proves necessity of the circle skew-product minimality criterion.

Apply it with $\alpha =\sqrt{2}$, $X=\mathbb{R}/\mathbb{Z}$, $Tx=x+\alpha$, and $\rho (x)=2x+\alpha$. The base is a minimal dynamical system: the closure of the cyclic subgroup generated by an irrational angle is the whole circle group, and its positive forward orbit has the same closure. To see the subgroup assertion, a proper closed subgroup of the circle is finite; if it contained an irrational angle it would be infinite. The elementary classification follows because a closed subgroup with arbitrarily small nonzero elements is dense, while otherwise its least positive representative generates a finite subgroup.

There can be no forbidden $f$ for any $m\ne 0$. Indeed, the winding number of a continuous closed path of $x\mapsto f(x+\alpha )$ equals that of $x\mapsto f(x)$, by translation and homotopy invariance of winding number. Consequently their quotient has winding number zero. But the proposed quotient $e^{2\pi im(2x+\alpha )}$ has winding number $2m\ne 0$. Thus the skew product is minimal. Starting at $(0,0)$, induction givesbecause $n^2\alpha +2n\alpha +\alpha =(n+1{)}^2\alpha$. Its orbit is dense in the two-dimensional torus, so its second coordinate is dense in the circle. Removing the initial point does not change density, since every orbit tail in a minimal dynamical system is dense. Therefore

For a probability measure-preserving system, ergodicity means that every measurable set $A$ satisfying $T^{-1}A=A$ modulo null sets has measure zero or one. Equivalently, the only invariant functions are almost everywhere constant. The pointwise ergodic theorem states that, for $f\in L^1(\mu )$,where $\mathcal{I}$ is the invariant sigma-algebra; on a probability space the convergence also holds in $L^1$. For an ergodic transformation, the limit is the constant $\int fd\mu$.

For the Gauss continued-fraction map, the Gauss measure is the probability measureIts density integrates to one and is bounded above and below by positive constants, so it has the same null sets as Lebesgue measure. Apart from countably many endpoints, the inverse branches are $h_a(y)=1/(a+y)$, $a=1,2,\dots$. For a bounded measurable test function $F$, substitution on each branch givesThus $T$ is a measure-preserving transformation. Its definition at zero and the rational branch endpoints does not affect these identities.

For ergodicity, we give the full shrinking-cylinder argument. A digit cylinder with first $n$ partial quotients $a_1,\dots ,a_n$ is an interval $I=h((0,1))$, where $h=h_{a_1}\circ \cdots \circ h_{a_n}$ and $T^n\circ h$ is the identity. The continued fraction convergents giveHere $q_n\ge q_{n-1}>0$ after the initial step, and the recurrence $q_n=a_n{q}_{n-1}+q_{n-2}$ makes $q_n$ tend to infinity at least at the Fibonacci rate. Hence the nested cylinders of every irrational point shrink to that point. The bounded distortion of continued-fraction cylinders is uniform: the ratio of the largest to the smallest value of $|h^{\prime }|$ is at most four. Since $1/(1+h(y))$ ranges between $1/2$ and $1$, the densityhas maximum-to-minimum ratio at most eight. Consequently every measurable $E\subseteq (0,1)$ satisfiesThe last inequality uses $d\nu /dx\le 1/\log 2$.

Now suppose $T^{-1}A=A$ modulo null sets and $\nu (A)>0$. Choose an irrational density point $x\in A$, and let $I_n$ be its successive digit cylinders. The Lebesgue density theorem applies to these intervals even though they need not be centered: an interval of length $\ell$ containing $x$ lies in the centered interval of radius $\ell$, so its relative complement tends to zero. Comparability of the two densities then givesBut invariance gives $1_A(h(y))=1_A(y)$ almost everywhere for each inverse branch of $T^n$. Pulling the exceptional null sets through $h$ is legitimate since $h$ has a positive smooth Jacobian on this fixed branch. Thus the distortion estimate with $E=A^c$ givesLetting $n\to \infty$ forces $\nu (A^c)=0$. Therefore the Gauss continued-fraction map is ergodic for the Gauss measure.

For an irrational $x$, the successive continued fraction digits satisfy $a_j(x)=\lfloor 1/T^{j-1}x\rfloor$. The event that a digit equals two is the interval $(1/3,1/2]$, up to null endpoints. Apply the pointwise ergodic theorem to its indicator:This holds for almost every $x$ with respect to both the Gauss measure and Lebesgue measure, since they have the same null sets.

The discrete Bochner–Herglotz theorem says that a sequence $r:\mathbb{Z}\to \mathbb{C}$ is a positive-definite function, meaningif and only if there is a unique finite positive Borel measure $\sigma$ on $\mathbb{T}=\mathbb{R}/\mathbb{Z}$ such thatThe mass is $\sigma (\mathbb{T})=r(0)$. The necessity of positive definiteness follows by integrating $|{\sum }_j{c}_j{e}^{2\pi ijt}{|}^2$.

For sufficiency, positive definiteness makes the relevant Gram matrices Hermitian positive semidefinite, so $r(-n)=\overline{r(n)}$ and $|r(n)|\le r(0)$. DefineTaking $c_j=e^{-2\pi ijt}$ in the positive-definiteness condition gives $p_N(t)\ge 0$, while ${\int }_{\mathbb{T}}{p}_N(t)dt=r(0)$. Moreover,Thus the integrals against every trigonometric polynomial converge. These polynomials are uniformly dense in continuous functions on the circle: smooth periodic approximation followed by Fourier inversion for smooth functions gives this density. The uniform mass bound $r(0)$ then makes the integrals against every continuous $\phi$ converge as well. Their limit is a positive linear functional $L$ with $|L(\phi )|\le r(0)\Vert \phi {\Vert }_{\infty }$. The standard Riesz-Markov-Kakutani representation theorem gives a finite positive Borel measure $\sigma$ with $L(\phi )=\int \phi d\sigma$. The preceding coefficient computation proves its required Fourier coefficients. Finally, two such measures agree on all trigonometric polynomials, hence on all continuous functions by uniform density, and therefore are identical. This proves the Bochner–Herglotz theorem, including existence and uniqueness; if $r(0)=0$, the estimate $|r(n)|\le r(0)$ gives the zero measure directly.

Now let $U$ be the Koopman operator $Uf=f\circ T$ on the complex L2 space of the probability measure-preserving system. We use $⟨f,g⟩=\int f\overline{g}d\mu$, linear in the first entry. Since $U$ is an isometry, the sequence $r_f(n)=⟨U^{n}f,f⟩$ for $n\ge 0$, extended by $r_f(-n)=\overline{r_f(n)}$, is positive-definite: its matrix is the Gram matrix of $f,Uf,\dots$. The Bochner–Herglotz theorem gives its spectral measure of a Koopman observable ${\sigma }_f$.

We first prove exactly how an atom of a measure detects an eigenfunction. Fix $\zeta =e^{2\pi i{t}_0}$ and setThe mean ergodic theorem, applied to the isometry $V={\zeta }^{-1}U$, says these averages converge in L2 space to the orthogonal projection $P_{\zeta }f$ onto $\ker (U-\zeta I)$. Its usual Hilbert-space proof works without invertibility: the orthogonal complement of $\overline{\mathrm{ran}(I-V)}$ is $\ker (I-V^{\ast })=\ker (I-V)$, and the averages of $(I-V)h$ telescope to $(h-V^{N}h)/N\to 0$. The equality of the two kernels follows from $V$ being an isometry, by expanding $\Vert Vh-h{\Vert }^2$ when $V^{\ast }h=h$.

Expanding the averaged norm through the spectral measure of a Koopman observable givesThe integrand is at most one and converges to $1_{\{t_0\}}$. By the dominated convergence theorem,Thus a nonzero atom corresponds to a nonzero Koopman operator eigenfunction, with eigenvalue $\zeta$, in the cyclic spectral data of $f$.

We also need the Wiener Fourier coefficient lemma, which here has a short proof. For any finite positive circle measure $\sigma$ with coefficients $r(n)$,The bracket is bounded in absolute value by one and tends to the indicator of the diagonal $t=s$. Another application of the dominated convergence theorem givesIn particular the limit is zero for an atomless measure.

Suppose there are no nonconstant Koopman operator eigenfunctions. For every $f$ in the mean-zero L2 space, all its projections $P_{\zeta }f$ vanish: a nonzero projection would be a nonconstant eigenfunction, since it has mean zero. Hence ${\sigma }_f$ is atomless, and the Wiener Fourier coefficient lemma givesThe Cauchy-Schwarz inequality also gives vanishing averaged absolute autocorrelations. For two mean-zero observables $f,g$, the sesquilinear polarization identitythen gives vanishing averaged absolute cross-correlations. Taking $f=1_A-\mu (A)$ and $g=1_B-\mu (B)$ proveswhich is weak mixing.

Conversely, this set formulation of weak mixing extends to all pairs of mean-zero L2 space observables by finite-simple-function approximation and the uniform estimate $|⟨U^{n}f,g⟩|\le \Vert f{\Vert }_2\Vert g{\Vert }_2$. If $Uh=\zeta h$ and $h\ne 0$, the isometry property gives $|\zeta |=1$. If $\zeta \ne 1$, invariance of the integral gives $\int h=0$. If $\zeta =1$ and $h$ is nonconstant, replace it by $h-\int h$, a nonzero mean-zero eigenfunction. But in either casefor every $n$, contradicting weak mixing. We have proved weak mixing is equivalent to absence of nonconstant eigenfunctions, including for noninvertible transformations.

Finally, the doubling map $Tx=2x(\mathrm{mod}1)$ preserves normalized Lebesgue measure, as substitution on its two inverse branches shows. Let $f\in L^2(\mathbb{T})$ be an eigenfunction, $f(2x)=\zeta f(x)$, with Fourier coefficients $c_k$. The Fourier basis calculation givesSince $\zeta \ne 0$, all odd coefficients vanish. Every nonzero integer is $2^{r}m$ with $m$ odd, and repeated use of $\zeta c_{2^{r}m}=c_{2^{r-1}m}$ makes all nonzero coefficients vanish. Completeness of the Fourier basis then says $f$ is constant. Therefore the doubling map has no nonconstant eigenfunctions, and

Use the convention that level $k$ counts the $k$ nonzero shifts, hence an arithmetic progression of length $k+1$. The SZ property at level $k$ means that every measurable set $A$ of positive measure satisfiesThis is the averaged multiple-recurrence property, stronger than merely finding one positive intersection. Shifting the indexing convention from $k+1$ terms to $k$ terms does not change the all-levels assertion. We work with probability measure-preserving systems.

Here is the Furstenberg correspondence principle construction proving the Szemerédi theorem. Let $D\subseteq \mathbb{N}$ have positive upper density $\delta ={\mathrm{limsup}}_N|D\cap \{1,\dots ,N\}|/N$. Put $z\in \{0,1{\}}^{\mathbb{Z}}$ equal to the indicator of $D$ on the positive coordinates and zero elsewhere, and let $S$ be the left shift. The binary full shift $\Omega =\{0,1{\}}^{\mathbb{Z}}$ is a compact metric space in the product topology, and $S$ is a homeomorphism. Choose $N_i\to \infty$ attaining the upper density and define the empirical measuresWe use the standard weak compactness fact that every sequence of Borel probability measures on a compact metric space has a weakly convergent subsequence, where weak convergence means convergence of integrals of continuous functions. Pass to a subsequence with limit $\mu$. For every continuous $F$, telescoping givesPassing to the limit proves $S_{\ast }\mu =\mu$. The cylinder set $A=\{w:w(0)=1\}$ is clopen, so its indicator is continuous andIf every measure-preserving system has the SZ property at level $k$, there is some $n\ge 1$ for which $\mu (C_n)>0$, where $C_n={\bigcap }_{j=0}^k{S}^{-jn}A$. This is also a clopen cylinder set, so ${\mu }_i(C_n)\to \mu (C_n)>0$. For some sufficiently large $i$, at least one term of the empirical average is positive: $S^{m}z\in C_n$ for some $1\le m\le N_i$. Unwinding the coordinates givesTherefore $D$ contains an arithmetic progression of length $k+1$ with positive common difference. Having the SZ property at every level gives the infinite-density form of the Szemerédi theorem.

The finite uniform form follows from the same argument. If, for fixed $\delta >0$ and $k$, arbitrarily large $N_i$ admitted sets $D_i\subseteq \{1,\dots ,N_i\}$ with $|D_i|\ge \delta N_i$ and no such arithmetic progression, replace $z$ above by the indicator $z_i$ of $D_i$, extended by zero outside that interval. Their empirical measures still have an invariant weak limit with $\mu (A)\ge \delta$. For every fixed positive $n$, however, ${\mu }_i(C_n)=0$, since a point contributing to that cylinder would be an actual progression in $D_i$. Thus $\mu (C_n)=0$ for all positive $n$, contradicting the SZ property. Consequently a sufficiently large $N$ depending only on $\delta ,k$ forces the required progression.

A compact measure-preserving system is one for which every observable $f\in L^2(\mu )$ has relatively compact forward orbit $\{U^{n}f:n\ge 0\}$ under the Koopman operator $Uf=f\circ T$. Equivalently, each of these orbits is totally bounded in the global $L^2$ norm; compactness refers to observables, not just to the underlying topological space. For an invertible system it is usual to use all integer times. One may also define compactness by density of almost periodic observables: the two definitions coincide because $\Vert U^n(f-g){\Vert }_2=\Vert f-g{\Vert }_2$, so a finite net for an approximating observable is a finite net, with slightly larger radius, for $f$.

To prove the SZ property, fix $A$ with $a=\mu (A)>0$, put $f=1_A$, and chooseCompactness gives a finite $\epsilon$-net for its orbit with centers $U^{r_1}f,\dots ,U^{r_s}f$, where $r_i\ge 0$. Let $M={\max }_i{r}_i$. For every integer $n\ge M+1$, some center satisfies $\Vert U^{n}f-U^{r_i}f{\Vert }_2<\epsilon$. Since $U$ is an isometry, cancellation givesThus the return set $R=\{r\ge 1:\Vert U^{r}f-f{\Vert }_2<\epsilon \}$ meets every interval $\{n-M,\dots ,n\}$ for $n\ge M+1$. In particular, $R$ is a syndetic set with lower density at least $1/(M+1)$. This proves the needed syndetic near returns of an almost periodic observable without any assumption of ergodicity.

For $r\in R$, telescoping and the isometry property giveAs $f$ is an indicator, $\Vert U^{jr}f-f{\Vert }_2^2=\mu (A△T^{-jr}A)$. The union bound therefore givesAll other multiple-intersection measures are nonnegative. Averaging over the bounded-gap set $R$ yieldsThus every compact measure-preserving system has the SZ property at every level, including nonergodic systems and, under the forward-orbit definition, noninvertible transformations.