Gaussian systems

Examples of measure preserving systems with varied behaviours are vital in ergodic theory, to understand the general properties and to have counter examples to false statements. One classical method to craft examples with specific properties is the so-called Gaussian construction. In this post I define and give simple examples of applications of this construction. I took most of the material for this post from the book of Cornfeld, Fomin and Sinai.

— 1. Definition of Gaussian systems —

Definition 1

A probability measure ${\nu}$ on the Borel sets of ${{\mathbb R}}$ is called Gaussian if there exists a pair ${(\bar\nu,\sigma)\in{\mathbb R}\times{\mathbb R}^{\geq0}}$ such that
$\displaystyle \forall A\subset{\mathbb R}\text{ Borel }\qquad\nu(A)=\frac1{\sigma\sqrt{2\pi}}\int_A\exp\left(\frac{-(x-\bar\nu)^2}{2\sigma^2}\right)dx \ \ \ \ \ (1)$

Given a probability space ${(X,{\mathcal B},\mu)}$ and a Borel measurable function ${f:X\rightarrow{\mathbb R}}$ , we say that ${f}$ is a Gaussian function if the pushforward ${\nu:=f_*\mu}$ is a Gaussian measure.

Using probabilistic terminology, ${f}$ is a random variable, ${\bar\nu=\int_Xfd\mu}$ is the expectation (or mean or first moment) of ${f}$ and ${\sigma=\|f-\bar\nu\|_{L^2}}$ is the standard deviation. Since by subtracting a constant from ${f}$ we obtain ${\int fd\mu=\bar\nu=0}$ , we will from now on assume that every Gaussian function has mean ${0}$ . The following lemma is well known in probability theory and the proof will not be provided.

Lemma 2 If ${H\subset L^2(X)}$ is a (real) subspace and every ${f\in H}$ is a Gaussian function, then any ${f\in\overline{H}}$ in the ${L^2}$ closure of ${H}$ is also a Gaussian function.

Next we define Gaussian measure preserving systems.

Definition 3 Let ${(X,{\mathcal B},\mu,T)}$ be a measure preserving system and let ${f\in L^2(X)}$ .

We say that ${f}$ is Gauss element if any function in the linear span ${H_0}$ of ${\{T^nf:n\in{\mathbb Z}\}}$ is Gaussian.

We say that ${(X,{\mathcal B},\mu,T)}$ is a Gaussian system if there exists a Gauss element ${f}$ in ${L^2_0(X)}$ whose orbit generates the whole ${\sigma}$ -algebra, i.e.
$\displaystyle {\mathcal B}=\sigma\big(\big\{(T^nf)^{-1}(A):n\in{\mathbb Z},~A\subset{\mathbb R}\text{ is a Borel set}\big\}\big)$

Such ${f}$ is called a generating Gaussian element.

We want to stress the obvious fact that ${(X,{\mathcal B},\mu,T)}$ being a Gaussian system does not imply that every ${f\in L^2(X)}$ is a Gaussian function (consider for instance the indicator function of any measurable set) but rather that it contains a closed invariant subspace ${H}$ , generating the whole ${\sigma}$ -algebra, such that every ${f\in H}$ is a Gaussian function.

— 2. Positive definite sequences and Gaussian systems —

In the same way that a Gaussian measure (with mean ${0}$ ) is uniquely determined by its standard deviation, a Gaussian system is uniquely determined by the correlation sequence ${\gamma(n)=\langle T^nf,f\rangle}$ .

It is well known that for any measure preserving system ${(X,\mu,T)}$ and any ${f\in L^2(X)}$ , the sequence ${n\mapsto\langle T^nf,f\rangle}$ is positive definite:

Definition 4 A function ${\gamma:{\mathbb Z}\rightarrow{\mathbb R}}$ is called positive definite if for every ${f:{\mathbb Z}\rightarrow{\mathbb C}}$ with finite support

$\displaystyle \sum_{n,m\in{\mathbb Z}}f(n)\overline{f(m)}\gamma(n-m)\geq0$

In particular, the correlation sequence ${\gamma}$ of a Gaussian system is positive definite. It is not so obvious that for every positive definite sequence ${\gamma}$ there exist a Gaussian system with correlation sequence ${\gamma}$ , or indeed a measure preserving system ${(X,{\mathcal B},\mu,T)}$ and ${f\in L^2(X)}$ with ${\gamma(n)=\int_XT^nf\cdot fd\mu}$ .

Theorem 5 Every positive definite sequence ${\gamma}$ is the correlation sequence of a Gaussian system.

While so far we restricted our attention to ${{\mathbb Z}}$ -actions, Theorem 5 holds in far greater generality. Indeed we can replace the role of ${{\mathbb Z}}$ with any group (in fact the result goes even outside the scope of groups, by considering, instead of positive definite sequences, so-called reproducing kernels (which are essentially functions ${K}$ with domain in the cartesian square of any set ${S}$ satisfying certain properties. In the case of groups the reproducing kernel takes the shape of ${K(n,m)=\gamma(m^{-1}n)=\langle T_nf,T_mf\rangle}$ ).) The proof of Theorem 5 has two steps, both of which are non-trivial. The first step is to establish a halfway result (it can be found in this formulation as Theorem 5.20 in these notes of Paulsen):

Theorem 6 (Naimark’s Dilation Theorem) For every positive definite sequence ${\gamma:G\rightarrow{\mathbb R}}$ of a group ${G}$ there exists unitary representation of the group ${G}$ on a Hilbert space ${H}$ and a vector ${f\in H}$ such that ${\gamma(n)=\langle U_nf,f\rangle}$ for all ${n\in G}$ .

Proof: (sketch)

Let ${H_0}$ be the set of all functions ${f:G\rightarrow{\mathbb C}}$ with finite support. Define a inner product in ${H_0}$ by letting

$\displaystyle \langle f,g\rangle=\sum_{n,m\in G}f(n)\overline{g(m)}\gamma(m^{-1}n)$

Quotient out the subspace ${f:\langle f,f\rangle=0}$ and then let ${H}$ be the completion of ${H_0}$ with respect to the norm ${\|f\|^2=\langle f,f\rangle}$ ; the extension of the inner product turns ${H}$ into a Hilbert space. Define ${U_gf}$ for ${f\in U_0}$ as the function with finite support ${(U_gf)(n)=f(ng)}$ and extend it by continuity to ${H}$ ; this becomes a unitary representation of ${G}$ in ${H}$ . Finally take ${f=1_{1_G}}$ to be the indicator function of the singleton containing the identity of ${G}$ ; it is clear that ${\gamma(n)=\langle U_nf,f\rangle}$ . $\Box$

We can now prove Theorem 5:

Proof: Let ${X={\mathbb R}^G}$ , let ${\nu}$ be the Gaussian measure on ${{\mathbb R}}$ with mean ${\bar\nu=0}$ and standard deviation ${\sigma=1}$ and let ${\mu=\nu^G}$ be the product measure on the Borel sets of ${X}$ . Let ${H, (U_g)_{g\in G}}$ and ${f\in H}$ be the Hilbert space, unitary representation and vector given by Naimark’s dilation theorem. Let ${(e_n)_{n\in G}}$ be an orthonormal basis on ${H}$ and define ${\phi:H\rightarrow L^2(X)}$ by mapping ${e_n}$ to ${\pi_n}$ , where ${\pi_n:X\rightarrow{\mathbb R}}$ is the projection onto the ${n}$ -th coordinate. Extend ${\phi}$ to ${H}$ , it follows from Lemma 2 that every function ${h\in\phi(H)}$ is a Gaussian function.

Next let ${h=\phi(f)}$ , it is a Gaussian element. Since ${\phi}$ is an isometric isomorphism, it follows that ${\gamma(n)=\langle U_nf,f\rangle=\langle T_nh,h\rangle}$ , so ${\gamma}$ is the correlation function of the Gaussian system ${(X,{\mathcal B},\mu,T,h)}$ . $\Box$

— 3. Spectral properties of Gaussian systems —

In this section we return to ${{\mathbb Z}}$ -actions (in fact, most of this section applies to actions of locally compact abelian groups). Let ${(X,{\mathcal B},\mu,T)}$ be a Gaussian system and let ${f\in X}$ be a generating Gaussian element with correlation sequence ${\gamma}$ . Bochner-Herglotz theorem states that there exists a measure ${\rho}$ on ${{\mathbb T}:={\mathbb R}/{\mathbb Z}\cong\{z\in{\mathbb C}:|z|=1\}}$ such that ${\gamma(n)=\int_{\mathbb T} e_nd\rho}$ , where ${e_n:x\mapsto2^{2\pi i x}}$ . The measure ${\rho}$ is called the spectral measure of the Gaussian system. Since ${\gamma(n)=\gamma(-n)}$ it follows that ${\rho(A)=\rho(-A)}$ for any Borel ${A\subset{\mathbb T}}$ .

Let ${H\subset L^2(X)}$ denote the smallest closed invariant subspace containing ${f}$ . One can define the linear map ${\theta:H\rightarrow L^2({\mathbb T},\rho)}$ by sending ${T^kf}$ to ${\theta(T^kf):=e_k}$ and extending by linearity and continuity to ${H}$ . Observe that for a real valued function ${h\in H}$ , the image under ${\theta}$ satisfies ${(\theta h)(-t)=\overline{(\theta h)(t)}}$ .

Proposition 7 The map ${\theta:H\rightarrow L^2({\mathbb T},\rho)}$ is an isometric isomorphism such that

$\displaystyle \theta(T^nh)=e_n\cdot\theta(h)$

This proposition, whose proof is routine and will be omitted, already allows us to deduce the first non-trivial fact about Gaussian systems:

Proposition 8 If the spectral measure ${\rho}$ has atoms, then the Gaussian system ${(X,{\mathcal B},\mu,T)}$ is not ergodic.

Proof: Let ${x\in{\mathbb T}}$ be such that ${\rho(\{x\})>0}$ and let ${\varphi\in L^2({\mathbb T},\rho)}$ be the indicator function of the singleton ${\{x\}}$ . Observe that ${e_n\cdot\varphi=e_n(x)\varphi}$ (to be completely clear, the first member of the equation has a multiplication of functions, the second a multiplication of a scalar with a function). Let ${h=\theta^{-1}\varphi}$ , it follows from Proposition 7 that

$\displaystyle T^nh=T^n(\theta^{-1}\varphi)=\theta^{-1}(e_n\varphi)=\theta^{-1}(e_n(x)\varphi)=e_n(x)\theta^{-1}(h)=e_n(x)h$

Therefore ${|h|\in L^2(X)}$ is a ${T}$ -invariant function. Moreover, since ${h\in H}$ , both its real and imaginary part are Gaussian functions, therefore ${h}$ can not be a constant, and this shows that the Gaussian system ${(X,{\mathcal B},\mu,T)}$ is not ergodic. $\Box$

To show the converse (in a strong sense) to Proposition 8 we need a more fine understanding of the geometry of ${L^2(X)}$ . Denote by ${H_0\subset L^2(X)}$ the one dimensional space of constant functions and let ${H_1}$ be the space denote ${H}$ above. Then, for each ${m\geq2}$ we let ${H_m}$ be the orthogonal complement of ${H_0\oplus\cdots\oplus H_{m-1}}$ in the closed linear span of functions of the form ${f_1\cdot f_2\cdots f_m}$ for any ${f_j\in H\cup\{1\}}$ . Since ${H}$ generates the full ${\sigma}$ -algebra, we have the orthogonal decomposition

$\displaystyle L^2(X)=\bigoplus_{m=0}^\infty H_m\ \ \ \ \ (2)$

Observe that each ${H_m}$ is a closed invariant subspace of ${L^2(X)}$ . For any ${m\geq1}$ one can define a map ${\theta_m}$ analogous to the map ${\theta}$ defined above. Unfortunately, the construction is significantly more complicated , so I will just state the relevant properties without proof.

Theorem 9 For each ${m\in{\mathbb N}}$ there exists a map ${\theta_m:H_m\rightarrow L^2({\mathbb T}^m,\rho^m)}$ (where ${\rho^m}$ is just the product measure of ${\rho}$ with itself ${m}$ times) satisfying:

The image of ${\theta_m}$ is the set of functions ${\varphi\in L^2({\mathbb T}^m,\rho^m)}$ which are invariant under permutation of coordinates.

The map ${\theta_m}$ is an isometric isomorphism between ${H_m}$ and its image.

Let ${e\Delta:{\mathbb T}^m\rightarrow{\mathbb T}}$ denote the map ${{\bf t}\mapsto e^{2\pi i(t_1+\cdots+t_m)}}$ . Then for every ${h\in H_m}$
$\displaystyle \theta_m(Th)=e\Delta\cdot(\theta h)$

We can now prove the converse of Proposition 8.

Proposition 10 Let ${(X,{\mathcal B},\mu,T)}$ be a Gaussian system . If the spectral measure ${\rho}$ has no atoms, then the system ${(X,{\mathcal B},\mu,T)}$ is weak mixing.

Proof: Let ${h\in L^2(X)}$ be an eigenvector satisfying ${Th=\lambda h}$ ; we need to show that it is constant. Using (2), decompose ${h=\sum_{m=0}^\infty h_m}$ . Since the spaces ${H_m}$ are orthogonal and invariant, each ${h_m}$ is itself an eigenvector with ${Th_m=\lambda h_m}$ . Next, fix ${m\geq1}$ and let ${\varphi_m=\theta(h_m)}$ . Using Theorem 9 we deduce that ${e\Delta\cdot\varphi_m=\lambda\cdot\varphi_m}$ in ${L^2({\mathbb T}^m,\rho^m)}$ . More precisely

$\displaystyle \int_{{\mathbb T}^m}\left|e\Delta({\bf t})-\lambda\right|^2\cdot|\varphi_m({\bf t})|^2d\rho^m({\bf t})=0\ \ \ \ \ (3)$

Since ${\rho}$ is non-atomic, the codimension ${1}$ shifted subtorus ${S=\{{\bf t}\in{\mathbb T}^m:e\Delta({\bf t})=\lambda\}}$ has measure ${\rho^m(S)=0}$ . Hence it follows from (3) that ${\|\varphi\|_{L^2}=0}$ . Since ${m\geq1}$ was arbitrary, we conclude that ${h=h_0}$ , i.e. it is a constant as desired. $\Box$

Corollary 11 A Gaussian system is weak mixing if and only if it is ergodic.

Another property which can be easily deduced from the decomposition (2) characterizes strong mixing of a Gaussian system in terms of properties of the spectral measure ${\rho}$ .

Proposition 12 Let ${(X,{\mathcal B},\mu,T)}$ be a Gaussian system. The system is strongly mixing if and only if ${\lim_{|n|\rightarrow\infty}\hat\rho(n)=0}$ .

Proof: Let ${f\in L^2(X)}$ be a generating Gaussian element. If the system is strongly mixing then ${\lim_{|n|\rightarrow\infty}\hat\rho(n)=\lim_{|n|\rightarrow\infty}\langle T^nf,f\rangle=0}$ .

Now assume that ${\lim_{|n|\rightarrow\infty}\hat\rho(n)=0}$ and let ${h\in H_m}$ for some ${m\geq1}$ . Let ${{\mathcal A}}$ be the pullback ${\sigma}$ -algebra of the Borel ${\sigma}$ -algebra on ${{\mathbb T}}$ by the map ${{\bf t}\mapsto t_1+\cdots+t_m}$ . Let ${\varphi=\theta h}$ , let ${\rho^{*m}}$ be the convolution of ${\rho}$ with itself ${m}$ times (equivalently, ${\rho^{*m}}$ is the pushforward of ${\rho^m}$ through the map ${{\bf t}\mapsto t_1+\cdots+t_m}$ ) and let ${\psi=\mathop{\mathbb E}[|\varphi|^2\mid{\mathcal A}]\in L^2({\mathbb T},\rho^{*m})}$ be the conditional expectation of ${\varphi}$ on ${{\mathcal A}}$ . We have

$\displaystyle \langle T^nh,h\rangle=\int_{{\mathbb T}^m}(e\Delta)^n\cdot|\varphi|^2d\rho^m=\int_{\mathbb T} e_n\cdot\psi d\rho^{*m}$

Since ${\hat{\rho^{*m}}(n)=(\hat\rho(n)\big)^m}$ , we have that ${\lim_{|n|\rightarrow\infty}\hat{\rho^{*m}}(n)=0}$ . Therefore the generalized Riemann-Lebesgue lemma implies that ${\lim_{|n|\rightarrow\infty}\langle T^nh,h\rangle=0}$ , finishing the proof. $\Box$

It is a known fact that there exist continuous singular measures ${\rho}$ on ${{\mathbb T}}$ such that ${\hat\rho(n)}$ does not go to ${0}$ as ${|n|\rightarrow\infty}$ . This implies that there exist (Gaussian) systems which are weak mixing but not strongly mixing.

— 4. Rigid ${{\mathbb Z}^d}$ -action mixing of all shapes —

One interesting construction using Gaussian systems is of a rigid ${{\mathbb Z}^d}$ -action mixing of all shapes.

Definition 13 Let ${G}$ be an abelian group and ${F\subset G}$ be a finite set.

A probability preserving action ${(X,{\mathcal B},\mu,(T_g)_{g\in G})}$ of ${G}$ is ${F}$ -mixing if
$\displaystyle \lim_{k\rightarrow\infty}\mu\left(\bigcap_{n\in F}T_{kn}A_n\right)=\prod_{n\in F}\mu(A_n)$

for any ${A_n\in{\mathcal B}}$ for ${n\in F}$ .

The system ${(X,{\mathcal B},\mu,(T_g)_{g\in G})}$ is mixing of all shapes if it is ${F}$ -mixing for every finite ${F\subset G}$ .

The system ${(X,{\mathcal B},\mu,(T_g)_{g\in G})}$ is rigid if there exists a sequence ${(n_k)_{k\in{\mathbb N}}}$ in ${G}$ such that ${\lim_{k\rightarrow\infty}\mu(A\cap T_{n_k}A)=\mu(A)}$ for all ${A\in{\mathcal B}}$ .

Observe that if a system is mixing of order ${m}$ , then it is ${F}$ -mixing for every ${F\subset G}$ with ${|F|\leq m}$ . However, the converse is not necessarily true.

Theorem 14 (Ferenczi and Kamiński) There exists a probability preserving action of ${{\mathbb Z}^d}$ which is rigid and mixing of all shapes.

Proof: (Sketch)

In view of Theorem 5, it suffices to construct a positive definite sequence with certain properties. Indeed one can show that if ${\gamma:{\mathbb Z}^d\rightarrow{\mathbb R}}$ is a positive definite sequence such that ${\lim_{k\rightarrow\infty}\gamma(m+kn)=0}$ for all ${m,n\in{\mathbb Z}^d}$ with ${n\neq0}$ , then the Gaussian system induced by ${\gamma}$ is mixing of all shapes. Similarly, if ${\gamma({\bf 0})=1}$ and there exists a sequence ${(n_k)_{k\in{\mathbb N}}}$ in ${{\mathbb Z}^d}$ along which ${\lim_{k\rightarrow\infty}\gamma(n_k)=1}$ , then the induced Gaussian system is rigid (exactly along ${(n_k)}$ ). Taking ${1,\beta_1,\dots,\beta_d\in{\mathbb R}}$ linearly independent over ${{\mathbb Q}}$ , the sequence

$\displaystyle \gamma({\bf n})=\frac{\sin\big(2\pi(n_1\beta_1+\cdots+n_d\beta_d)\big)}{2\pi(n_1\beta_1+\cdots+n_d\beta_d)}$

satisfies the conditions. Indeed, it is clear to show that ${\lim_{k\rightarrow\infty}\gamma(m+kn)=0}$ for any ${m,n\in{\mathbb Z}^d}$ with ${n\neq0}$ . It is also not hard to find a sequence ${(n_k)_{k\in{\mathbb N}}}$ in ${{\mathbb Z}^d}$ such that ${\sum n_k^{(j)}\beta_j\rightarrow0}$ , which implies that ${\gamma(n_k)\rightarrow1}$ . $\Box$

Gaussian systems

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