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Polynomial rigidity and the spectra of Sidon automorphisms V. V. Ryzhikov Faculty of Mechanics and Mathematics, Lomonosov Moscow State University, Moscow, Russia
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     § 1. IntroductionA generalization of the rigidity property of an operatorLet $T\colon H\to H$ be a bounded operator in a normed space $H$ that has the following property: for some sequence $n_j\to +\infty$ and a dense subspace $F$ of $H$ there exists a sequence of polynomials $ P_j(T)=\sum_{n,\,|n|>n_j} a_n T^n$ such that for each $f\in F$ Then we say that the operator $T$ has the property of $P_j$-rigidity. We can also say that this operator possesses generalized (polynomial) rigidity. Recall that an operator $T$ is said to be rigid if $T^{n_j}\to_w I$ along some sequence $n_i\to+\infty$. For a unitary operator $T$ this implies that the spectral measure of $T$ is singular.Polynomial rigidity of operators in Hilbert spaces is closely connected with linear determinism in the sense of Kolmogorov, that is, with our ability to recover a vector as a linear combination of its far translations (see [1]; also see [2] and [3] on the development of this topic). Polynomial mixingWe say that a sequence of polynomials of the form
$$
\begin{equation*}
P_j(T)=\sum_{n,\,|n|\leqslant n_j} a_{j,n} T^n, \qquad a_{j,n}\geqslant 0, \quad n_j\to+\infty,
\end{equation*}
\notag
$$
is mixing if $\|P_j(T)v\|\to 0$ for each $v$ in some dense subspace $V$ of $ H$. This terminology comes from ergodic theory: the mixing property of an automorphism $T$ of a probability space is equivalent to the weak convergence to 0 of powers of the induced unitary operator in the space orthogonal to the constants. We consider the case when $H=L_2(\mu)$, where $\mu$ is a sigma-finite measure, so nontrivial constants do not belong to $L_2(\mu)$. Disjointness of operatorsThe spectral measures of a rigid operator $T$ along a sequence $n_j$ and those of a mixing operator $\widetilde T$ along the same sequence are mutually singular. This property is equivalent to these operators being disjoint: for a bounded operator $J$ it follows from the equality $TJ=J\widetilde T$ that $J=0$. In fact, taking the weak limits in the equality в $T^{n_j}J=J\widetilde T^{n_j}$ we obtain $J=0$. A modification of this idea enables us to distinguish the spectra of two mixing operators $T$ and $\widetilde T$. If there exists a sequence of polynomials $P_j$ such that $P_j(T)$ is a rigid sequence of operators and the sequence $P_j(\widetilde T)$ is mixing, then $T$ and $\widetilde T$ are disjoint. The lack of the group property of the spectrumIn the 1960s Kolmogorov stated a conjecture that the spectrum of an automorphism has the group property, which, on being solved in the negative (see [4]), prompted the development of new approaches to the spectra of ergodic actions. In this paper we consider so-called simple Sidon automorphisms $T$, which preserve a sigma-finite measure. We show that their spectra are singular and the convolution squares of their spectra are equivalent to Lebesgue measure under certain additional assumptions. Thus we produce new examples of automorphisms $T$ for which the convolution $\sigma_T\ast\sigma_T$ is mutually singular with the spectral measure $\sigma_T$. Tensor roots of a planar translationIn connection with the Thouvenot problem of nonisomorphic tensor roots of the tensor square of an automorphism (see [5]) we show that a nontrivial translation $S\colon \mathbb R^2\to\mathbb R^2$, regarded as a transformation preserving the planar Lebesgue measure, has spectrally nonisomorphic tensor roots. We show that for continuum many spectrally disjoint Sidon automorphisms $T\colon \mathbb R\to\mathbb R$ their products $T\times T$ are conjugate to the translation $S$ by means of some automorphisms $\Phi_T$ (that is, $T\times T=\Phi_T S\Phi^{-1}_T$). Sidon constructions were considered in [6] as examples of metrically nonisomorphic tensor roots from the transformation $S$. Applications to Poisson and Gaussian dynamical systemsTwo classical constructions of automorphisms of a probability space correspond to an automorphism $S$ of the standard space with a sigma-finite measure: the Gaussian automorphism $G(S)$ and the Poisson suspension $P(S)$, which, apart from ergodic theory, are also well known in group representation theory (for instance, see [7]). In [8] various spectral multiplicities of the form $M\cup\{\infty\}$ were realized for nonmixing Gaussian systems and Poisson suspensions. Our results in this paper lead to a similar realization in the class of mixing systems. § 2. Auxiliary results. Constructions of automorphisms of rank $1$The fact that Sidon automorphisms (see the definition in § 3) have singular spectra can be shown using the method and results from [9]. Note that for other classes of automorphisms such singularity was proved in [10] and [11]. We do not use generalized Riesz products, but in place of products we use sums of polynomials which ensure generalized rigidity. For a simple Sidon automorphism we show that its powers are disjoint and, as a consequence, its spectrum is singular. The approach put forward is based on the following results. Lemma 2.1. Let $S$ and $T$ be unitary operators such that the sequence $P_j(T)$ is rigid and the sequence $P_j(S^\ast)$ is mixing. Then the operators $S$ and $T$ are disjoint: for each bounded operator $J$ it follows from the equality $SJ=JT$ that $J=0$. This is a special case of the following result. Theorem 2.2. Given unitary operators $S$ and $T$ in the space $H$, let there exist dense subspaces $F$ and $ G$ of $H$ such that the sequence $P_j(T)f$ converges to $P(T)f$ for $f\in F$, and $P_j(S^\ast)g$ converges to $P'(S^\ast)g$ for $g\in G$. If the polynomials $P$ and $P'$ are different and $T$ has continuous spectrum, then $T$ and $S$ are disjoint. Proof. Let $SJ=JT$. Then we have Therefore, Since $T$ has continuous spectrum, the vectors of the form $P(T)f-P'(T)f$, $f\in F$, are dense in $H$. However, then $J^\ast=0$.
The proof is complete. Lemma 2.3. Given a unitary operator $T$, let the sequence $P_j(T)$ be rigid and the sequences $P_j(T^p)$ be mixing for all $p>1$, where all polynomials have real coefficients. Then $T$ has singular spectrum. Proof. For $T$ we have (this is obvious when the polynomial $P_j$ has real coefficients). Hence it follows from Lemma 2.1 that $T$ is disjoint from its powers $T^p$ for $p>1$. When $T$ has an absolutely continuous component $\sigma_{1}$, this component cannot be mutually singular with the absolutely continuous component $\sigma_p$ of the operator $T^p$ for large $p$ (the measure $\sigma_p$ is the image of $\sigma_1$ under the map $z\to z^p$). In fact, assume that $\sigma_1$ has a Radon–Nikodym density $\rho_1(z)$, while the relevant density $\rho_p(z)$ for $T^p$ is a sum $\sum_{w,\,w^p=z} \rho_1(w)/p$. The measure of the points $z$ such that $\rho_p(z)=0$ tends to 0 as $p\to\infty$. Hence the set of points $z$ such that $\rho_1(z)>0$ and $\rho_p(z)>0$ has a positive measure for large $p$.
The proof is complete. Constructions of automorphisms of rank $1$Recall the definition of a rank-one automorphism, which we need for what follows. Fix an integer $h_1\geqslant 1$ (the hight of the tower at step $j=1$), a sequence $r_j\to\infty$ (the parameter $r_j$ is equal to the number of columns into which the tower at step $j$ is virtually cut) and a sequence of integer vectors (the parameters of spacers)
$$
\begin{equation*}
\overline s_j=(s_j(1), s_j(2),\dots, s_j(r_j-1),s_j(r_j)).
\end{equation*}
\notag
$$
We turn to the description of the construction of a measure-preserving transformation, which is fully determined by the parameters $h_1$, $r_j$ and $\overline s_j$. At step $j=1$ we fix a half-open interval $E_1$. At step $j$ we define a system of disjoint half-open intervals such that on $E_j, TE_j, \dots, T^{h_j-2}E_j$ the transformation $T$ is a parallel translation. Such a system of half-open intervals is called the tower of step $j$; their union is denoted by $X_j$ and also called a tower.We represent $E_j$ as a disjoint union of $r_j$ half-open intervals of equal length. For $i=1,2,\dots, r_j$ consider the so-called column To a column with index $i$ we add $s_j(i)$ disjoint half-open intervals (levels) of length equal to that of $E_j^i$. We call the resulting systems of intervals for fixed $i$ and $j$ columns with spacers $X_{i,j}$. Note that for fixed $j$ the columns $X_{i,j}$ are disjoint by construction. Now using parallel translations of intervals we extends the definition of $T$ so that the columns $X_{i,j}$ have the form
$$
\begin{equation*}
E_j^i, TE_j^i, T^2 E_j^i,\dots, T^{h_j-1}E_j^i, T^{h_j}E_j^i, T^{h_j+1}E_j^i, \dots, T^{h_j+s_j(i)-1}E_j^i,
\end{equation*}
\notag
$$
and $T$ maps the top levels $T^{h_j+s_j(i)-1}E_j^i$ of the columns $X_{i,j}$ ($i<r_j$) to the bottom levels of the $X_{i+1,j}$ by parallel translations: Setting $E_{j+1}= E^1_j$ we notice that in the new notation all levels of the columns with spacers which we mentioned above have the form
$$
\begin{equation*}
E_{j+1}, TE_{j+1}, T^2 E_{j+1},\dots, T^{h_{j+1}-1}E_{j+1},
\end{equation*}
\notag
$$
and form a tower of step $j+1$, which has height Thus, we have defined the transformation $T$ on all but the last levels of the tower of step $j+1$. This partial definition of $T$ at step $j$ is preserved at the steps that follow. As a result, we obtain a space $X=\bigcup_j X_j$ and an invertible transformation $T\colon X\to X$ preserving the standard Lebesgue measure $\mu$ on $X$. If
$$
\begin{equation*}
\sum_j \frac{s_j(1)+s_j(2)+\dots+s_j(r_j)}{h_jr_j}=\infty,
\end{equation*}
\notag
$$
then $X$ has an infinite measure, and this is just the case we consider in our paper. Transformations of rank $1$ are ergodic, and in the case when the phase space has an infinite measure, they have continuous spectra. Moreover, it is well known that linear combinations of the indicators of intervals involved in the construction of $T$ are cyclic vectors of the operator $T$ (throughout, we use the same notation for a transformation and the induced unitary operator). Various results on rank-one automorphisms were discussed in [6].
§ 3. Sidon automorphisms and dissipativity of the tensor squareAssume that the construction of an automorphism $T$ of rank 1 has the following property: for all large $j$, provided that $h_{j}<m\leqslant h_{j+1}$, the intersection $X_j \cap T^mX_j$ can only lie in a single column $X_{i,j}$ of the tower $X_j$. This is called a Sidon construction in view of the following analogy: a subset $M$ of the set of positive integers is said to be Sidon if $|M\cap M+n|\leqslant 1$ for all $n>0$. Note that for such $T$ and the set $A$ equal to the union of the levels of the tower of step $j_0$ we have for all $m\in [h_j, h_{j+1}]$, $j\geqslant j_0$. Hence if $r_j\to\infty$, then the Sidon construction has the mixing property. Recall that an automorphism $T$ of a space with a sigma-finite measure is mixing if for all sets $A$ and $B$ of finite measure we haveBelow we focus on so-called simple Sidon constructions, whose parameters ensure the above properties of the intersections $X_j\cap T^mX_j$. In the case when $m$ is comparable with $h_{j+1}$, for these constructions we have $X_j\cap T^mX_j\subset X_{1,j}$. An example of a Sidon constructionSet
$$
\begin{equation*}
s_j(i)=10^ih_j - h_j, \qquad 1\leqslant i\leqslant r_j.
\end{equation*}
\notag
$$
Note that for $h_{j}<m\leqslant h_{j+1}$ the intersection $ X_j\cap T^mX_j$ is nonempty only when $m$ is comparable with $(10^k -10^i)h_j$ for some $i$, $k$; moreover, for them
$$
\begin{equation}
T^m X_j\cap X_j= T^mX_{i,j}\cap X_{k,j}, \qquad 1\leqslant i< k<r_j.
\end{equation}
\tag{*}
$$
Here we see an analogy with the Sidon set $M=\{10^q\colon 1\leqslant q\leqslant r\}$, for which $(M\cap M+m)\neq \varnothing $ only for $m=10^k-10^i$. It must be mentioned that, because of spacers of step $j+1$, we have $X_j\cap T^mX_j=\varnothing$ for $0.12h_{j+1}<m< 9 h_{j+1}$. The nonempty intersections $T^m X_j\cap X_j$ for $h_{j}<m\leqslant h_{j+1}$ satisfy (*), which means that our construction is Sidon. Looking ahead we notice that, for example, the construction with parameters
$$
\begin{equation*}
s_j(i)=(10+j)^ih_j, \qquad 1\leqslant i\leqslant r_j,
\end{equation*}
\notag
$$
has the following property: for fixed $p$ and sufficiently large $j$, if $T^m X_j \cap X_j\neq \varnothing$ for $m>h_j$, then $T^{pm} X_j\cap X_j= \varnothing$. We show below that this property implies that powers of the automorphism $T$ have disjoint spectra. Given a Sidon construction, if the sequence $r_j$ increases sufficiently rapidly, then the product $T\times T$ is dissipative (for this reason the operator $T\otimes T$ has Lebesgue spectrum). Theorem 3.1. Let $T$ be a Sidon construction such that Then $T\times T$ is a dissipative transformation with wandering set of infinite measure. Proof. Let $S=T\times T$. Our aim is to show that for some (wandering) set $W$ of infinite measure. First we find a wandering set of finite measure. Fix a positive integer $p$. Throughout what follows we use the notation $\overline \mu=\mu\otimes\mu$. For time between $h_j$ and $h_{j+1}$ the measure of the set of recurrence in $C_p=X_p\times X_p$, where $j\geqslant p$, is at most $\overline \mu(C_p)/r_j$. Indeed, the condition $S^iC_p\cap C_p\neq\varnothing$ if equivalent to $T^iX_p\cap X_p\neq\varnothing$, and we have
$$
\begin{equation*}
S^iC_p\cap C_p\subset (T^hX_p\cap X_p)\times (T^hX_p\cap X_p).
\end{equation*}
\notag
$$
For $h_j\leqslant h < h_{j+1}$, in a Sidon construction the intersection $T^hX_p \cap X_p$ lies in one of the columns $ X_{i,j}$, so that for some $i=i(h)$, $1<i\leqslant r_j$. Since
$$
\begin{equation*}
\overline \mu\biggl(\bigcup_{i=1}^{r_j}X_{i,j}\times X_{i,j}\biggr)=\frac {\overline \mu(C_p)} {r_j},
\end{equation*}
\notag
$$
we see that the set returning to $C_p$ since the moment of time $h_p$ has measure at most $\sum_{j=p}^{\infty }{\overline\mu(C_p)}/ {r_j}$. As the series $\sum_{i=1}^{\infty }r_j^{-1}$ is convergent, it follows from the above that the measure of a maximal wandering set $W$ is at least
$$
\begin{equation*}
(1-\varepsilon_p)\overline\mu(C_p)\to\infty, \quad\text{where} \quad \varepsilon_p\to 0.
\end{equation*}
\notag
$$
Some explanations are in order here. By the above we have If the maximal wandering set $W$ has a finite measure, then for each $Y$ such that $\overline \mu(Y)<\infty$ we have Hence the measure of $W$ is infinite in our case.
Theorem 3.1 is proved. In [12] the reader can find examples of automorphisms $T$ that have a conservative (and even ergodic) power $ (T\times\dots\times T)$ of an arbitrary order such that the product $ (T\times\dots\times T)\times T$ is dissipative. Comments on conservativity and the spectra of tensor powersThe product $T\times T$ is conservative in the case of Sidon automorphisms satisfying the conditions $ r_j\to\infty$ and $ \sum_{i=1}^{\infty }r_j^{-1} = \infty$. In this class of automorphisms, for each positive integer $m$ there exists $T$ such that $T^{\otimes m}$ has singular spectrum but $T^{\otimes n}$ is dissipative for some $n>m$. Using such automorphisms we can obtain new spectral types of ergodic actions on a probability space. For example, for each $n>1$ there exist a mixing Poisson suspension and a mixing Gaussian automorphism such that the corresponding spectral type has the form $\sigma +\sigma^{\ast 2}+\dots + \sigma^{\ast n}+\lambda$, where the convolution power $\sigma^{\ast n}$ is singular and $\lambda$ is the Lebesgue measure. The author plans to present these result in a separate paper. § 4. Polynomial rigidity and mixing for simple Sidon automorphismsLet $\psi>3$, and assume that for the parameters of a construction of rank 1 at $j$th step, starting with some $j$, we have
$$
\begin{equation*}
s_j(1)>\psi h_j, \quad s_j(i+1)> \psi s_j(i), \qquad 1\leqslant i <r_j.
\end{equation*}
\notag
$$
It is easy to verify that such a construction is Sidon. By a simple Sidon construction we mean one with parameters satisfying
$$
\begin{equation*}
h_j\ll s_j(1)\ll s_j(2)\ll \dots\ll s_j(r_j-1)\ll s_j(r_j), \qquad r_j\to\infty,
\end{equation*}
\notag
$$
where expressions of the form $a_j\ll b_j$ indicate that $b_j>\psi(j)a_j$ for some fixed sequence $\psi(j)\to +\infty$. Note one property of Sidon constructions, which is important for what follows. If
$$
\begin{equation*}
T^nX_1\cap X_1\neq\varnothing \quad\text{for } h_j< n\leqslant 2s_j(r_j-1),
\end{equation*}
\notag
$$
then for some $i<r_j$ this intersection $T^nX_1\cap X_1$ lies in the column $X_{i+1,j}$ and
$$
\begin{equation*}
T^nX_1\cap X_1= T^n(X_1\cap X_{i,j})\cap (X_1\cap X_{i+1,j}).
\end{equation*}
\notag
$$
In this case Since $h_j\ll s_j(i)\ll s_j(i+1)$, the set $T^{pn}(X_1\cap X_{i,j})$ occurs in the spacer over $X_{i+1,j}$ and is disjoint from $X_1\subset X_j$. Moreover, the intersections of the set $X_1$ with all columns with indices at most $i$ occurs in the same spacer under the action of $T^{pn}$. This observation (for fixed $p$, for all sufficiently large $j$) follows from the inequality
$$
\begin{equation*}
s_j(1)+h_j + s_j(2)+h_j +\dots +s_j(i)+h_j < pn< s_j(i+1).
\end{equation*}
\notag
$$
Thus, We will see that, based on this property, for suitable polynomials $P_j$ we will obtain the properties of $P_j(T)$-rigidity and $P_j(T^p)$-mixing at the same time. For simple Sidon constructions $T$ we will find a sequence of polynomials $ P_j$ such that for some cyclic vector $f$ we have $ \|P_j(T)f -f\|\to 0$, but at the same time ${\|P_j(T^p)f\|\,{\to}\, 0}$ for each $p>1$. Hence the spectrum of $T$ is singular. Theorem 4.1. Let $T$ be a simple Sidon automorphism with parameters $s_j(i)$ and $r_j$. Set $k(1,j)=0$, and for some sequence $m(j)\to\infty$. Then the sequence $P_j(T)$ is rigid and for $p>1$ the sequence $P_j(T^p)$ is mixing. Proof of Theorem 4.1. Let $f$ denote the indicator of the tower $X_1$. We show that
(i) $P_j(T^p)f\to 0$ for $p>1$; (ii) $P_j(T)f\to f$. Since the spectrum of $T$ is continuous, for a rank-one automorphism such a function $f$ is a cyclic vector (which is well known). Thus, once we will have established (i) and (ii) for $f$, we will have also proved (i) and (ii) for a dense subspace of functions in $L_2(X,\mu)$. This will prove the theorem. Proof of (i). For all sufficiently large $j$ the intersection $T^{pk(i,j)-pk(i',j)}X_1\cap X_1$ is empty for $1\leqslant i\neq i'\leqslant r_j$. This is because otherwise $pk(i,j)-pk(i',j)/k(m,j)\approx 1$ for some $m$. However, this is impossible for $p>1$ by the relation which is obvious. In a similar way we verify that we only have
$$
\begin{equation*}
T^{pk(i,j)-pk(i',j)-pk(m,j)+pk(m',j)}X_1\cap X_1\neq \varnothing
\end{equation*}
\notag
$$
for $i=i'$ and $m=m'$, so that the intersection coincides with $X_1$.
In using sum signs below we assume implicitly that $1\leqslant i,i',m,m'\leqslant r_j$. For example, $\sum_{i\neq i'} $ means actually $\sum_{1\leqslant i\neq i'\leqslant r_j}$. For $f=\chi_{X_1}$ we obtain
$$
\begin{equation*}
\begin{aligned} \, \|Q_j(T^p)f\|^2 &=\frac 1{r_j^2}\sum_{i\neq i'} \sum_{m\neq m'}(T^{pk(i,j)-pk(i',j)-pk(m,j)+pk(m',j)}f,f) \\ &=\frac 1{r_j^2}\sum_{i=m\neq i'=m'} \mu(X_1)=\frac{(r_j-1)\|f\|^2}{r_j}. \end{aligned}
\end{equation*}
\notag
$$
Note that the set $T^{pk(i,j)-pk(i',j)}X_1$ lies in $X_{j+1}\setminus X_j$ (for all large $j$). Hence the functions $Q_{j+l}(T^p)f$, $l=1,2,\dots, m(j)$, have disjoint supports. Therefore, these functions are pairwise orthogonal, so that
$$
\begin{equation*}
\|P_j(T^p)f\|< \frac{\|f\|}{\sqrt{m(j)}}\to 0, \qquad m(j)\to\infty.
\end{equation*}
\notag
$$
Proof of (ii). It is characteristic for Sidon constructions that the map $T^{k(i,j)-k(i',j)}$ takes the $i'$th column $X_{i',j}$ to the $i$th column $X_{i,j}$, while the images of the other columns for $i\neq i'$ occur in $X_{j+1}\setminus X_j$. For us it is important to know that
$$
\begin{equation*}
\mu(T^{k(i,j)-k(i',j)-k(m,j)+k(m',j)}X_1\cap X_1)\in \biggl\{0,\frac{\mu(X_1)}{r_j},\mu(X_1)\biggr\}.
\end{equation*}
\notag
$$
The case of a nonzero measure is possible when $k(i,j)-k(i',j)-k(m,j)+k(m',j)$ is different from a quantity of the form $k(n,j)-k(n',j)$ by at most $h_j$. From $h_j\ll k(1,j)\ll k(2,j)\ll \dots \ll k(r_j,j)$ we see that this condition is satisfied only when two integers from $i$, $i'$, $m$ and $m'$ coincide. However, this implies that two numbers in the set $i,i',m,m'$ are equal. For instance, let $i=m$ and $i'\neq m'$. Note that the intersection $T^{k(i,j)-k(i',j)}X_1 \cap T^{k(i,j)-k(m',j)}X_1$ contains $X_1\cap X_{i,j}$, which is straightforward to verify. Then the set
$$
\begin{equation*}
T^{k(i,j)-k(i',j)- k(i,j)+k(m',j)}X_1\cap X_1 =T^{k(m',j)-k(i',j)}X_1\cap X_1
\end{equation*}
\notag
$$
contains $X_1\cap X_{m',j}$ and, in fact, coincides with it because this intersection can only lie in one of the columns $X_{i'',j}$. We have thus verified that for $i=m$ and $i'\neq m'$ the set $T^{k(i,j)-k(i',j)}X_1 \cap T^{k(i,j)-k(m',j)}X_1$ has measure $\mu(X_1)/r_j$. If $i=m$ and $i'= m'$, then it is obvious that the measure of the intersection is $\mu(X_1)$.
We calculate $\|Q_j(T)f\|$ for $f=\chi_{X_1}$. We have
$$
\begin{equation*}
\begin{aligned} \, \|Q_j(T)f\|^2 &=\frac 1{r_j^2}\sum_{i\neq i',\,m\neq m'}(T^{k(i,j)-k(i',j)-k(m,j)+k(m',j)}f,f) \\ & =\frac 1{r_j^2}\biggl(\sum_{i=m\neq i'= m'} \mu(X_1)+\sum_{i=m,\,i'\neq m'} \frac {\mu(X_1)}{r_j}+\sum_{i\neq m,\,i'=m'} \frac {\mu(X_1)}{r_j}\biggr) \\ &= \frac{3(r_j-1)\|f\|^2}{r_j}. \end{aligned}
\end{equation*}
\notag
$$
Set
For distinct $j$ the functions $\Delta_j$ are pairwise orthogonal because their supports are disjoint (they lie in the corresponding differences $X_{j+1}\setminus X_j$). Thus, and
$$
\begin{equation*}
\biggl\|P_j(T)f-\biggl(1-\frac 1{r_j}\biggr)f\biggr\|^2 <\frac {2\|f\|^2} {m(j)}.
\end{equation*}
\notag
$$
Since $r_j\to\infty$, we obtain
The convergences (i) and (ii) that we have established yield the assertion of Theorem 4.1. Owing to this theorem, from Lemmas 2.1 and 2.3 we see that the automorphism $T$ has singular spectrum. Remark 4.3. Setting for $P_j^+=(Q_{j+1}^++\dots+Q_{j+m(j)}^+)/m(j)$ we again obtain an approximation: although this must be substantiated additionally. In this way we achieve polynomial rigidity, under the assumption that the polynomials $P_j^+$ have positive powers and positive coefficients. The following question arises in this connection: does $P_j(T)$-rigidity under this assumption mean that the automorphism $T$ has singular spectrum? § 5. Disjoint Sidon automorphisms, tensor roots and Poisson suspensionsTheorem 5.1. There exists a family of Sidon automorphisms of the cardinality of a continuum that have mutually singular multiplicity-one spectra such that all of their various tensor products are dissipative (have Lebesgue spectra). There exists a dissipative transformation isomorphic to a planar translation that has continuum many spectrally nonisomorphic tensor roots. Remark 5.2. Given a family of automorphisms with mutually singular spectral measures that has the cardinality of a continuum, the spectra of at most countably many of them contain absolutely continuous components. Almost all (in the sense of cardinality) automorphisms in this family must have singular spectra. Proof of Theorem 5.1. Fix some simple Sidon construction $T$. Recall that its parameters satisfy where $h_j$ is the height of the tower at step $j$ of the construction. We also assume that $\sum_{i=1}^{\infty }{r_j^{-1}}<\infty$, which implies that $T\times T$ is dissipative.
Fictitious steps for $r_j=1$. We turn to the description of a family of Sidon automorphisms $\widetilde T$ for which we will show that their spectra are mutually disjoint. In the definition of the construction of rank $1$ we impose the natural condition ${r_j\geqslant 2}$. Here we also let $r_j=1$ for convenience, but we assume that $r_j\geqslant 2$ for an infinite number of steps with indices $j$. We call steps $j$ with $r_j=1$ fictitious. We increase merely the parameter $s_{j-1}(r_j)$ of step $j-1$ and call this the $j$th step. We increase this parameter so that for the heights of towers in the construction of $\widetilde T$ we have ${\widetilde h_{j}=h_{j}}$. Set $G_n = \{n^2,n^2+1,\dots, n^2+2n\}$. Let $\gamma$ be an infinite set of positive integers. We define the parameters of the construction of $\widetilde T=T_\gamma$ as follows:
We consider the polynomials involved in Theorem 4.1 by setting
$$
\begin{equation*}
\begin{gathered} \, k(i,j)=(i-1)h_j +s_j(1)+s_j(2)+\dots +s_j(i), \qquad s_j(0)=0, \\ Q_j(T)=\frac 1{r_j}\sum_{1\leqslant i\neq i'\leqslant r_j} T^{k(i,j)-k(i',j)} \end{gathered}
\end{equation*}
\notag
$$
and
Repeating the arguments in the proof of Theorem 4.1 we obtain that for $j_n\to\infty$, $j_n\in \gamma$, the sequence $P_{j_n}(T_\gamma)$ is rigid. For $j_n\to\infty$, $j_n\notin \gamma$, the sequence $P_{j_n}(T_{\gamma'})$ is mixing. We fix a family $\Gamma$ of the cardinality of a continuum consisting of infinite sets of positive integers such that all intersections of distinct elements of $\Gamma$ have a finite cardinality. For example, we can construct it as an analogue of continuum many ‘clearings in the forest’ $\mathbb Z^2$, when the intersection of two distinct ‘clearings’ is the set of ‘trees’ inside a parallelogram. Then the family of automorphisms $T_\gamma$, $\gamma\in \Gamma$, is as required. If $\gamma\neq\gamma'$, then for $j_n\in \gamma$ the sequence $P_{j_n}(T_\gamma)$ is rigid but $P_{j_n}(T_{\gamma'})$ is mixing. Then it follows from Lemma 2.1 that the automorphisms $T_\gamma$ and $ T_{\gamma'}$ are disjoint. It remains to observe that the products $T_\gamma\times T_{\gamma'}$ are dissipative. If $\gamma\neq \gamma'$, then for large $k$ the tower $X_1$ has the following property: if $T_\gamma^kX_1\cap X_1\neq\varnothing$, then $T_{\gamma'}^kX_1\cap X_1=\varnothing$. In turn this means that the product $T_\gamma\times T_{\gamma'}$ is dissipative. The product $T_\gamma\times T_{\gamma}$ is dissipative for the same reasons as $T\times T$ in the proof of Theorem 3.1 is. In all cases the wandering set has an infinite measure, so all products $T_\gamma\times T_{\gamma'}$ are isomorphic to a planar translation. Theorem 5.1 is proved. The spectral multiplicities of mixing Gaussian and Poisson suspensionsIn [8] we produced spectral multiplicities of the form $M \cup \{\infty\}$ for nonmixing Gaussian ($G(S)$) and Poisson ($P(S)$) suspensions over an automorphism $S$ of a space with a sigma-finite measure. The following result is a consequence of Theorem 5.1 and the well-known fact that $G(S)$ and $P(S)$ are unitarily equivalent to the operator
$$
\begin{equation*}
\exp(S)=1\oplus S\oplus S^{\odot 2}\oplus S^{\odot 3}\oplus \dotsb.
\end{equation*}
\notag
$$
Theorem 5.3. For each set $M$ of positive integers there exist a mixing Poisson suspension and a mixing Gaussian automorphism such that their set of spectral multiplicities is $M\cup\{\infty\}$. Furthermore, the infinite multiplicity is characteristic for the Lebesgue component. In fact, $M\cup\{\infty\}$ is the set of spectral multiplicities of the operator
$$
\begin{equation*}
P =\exp\biggl(\bigoplus_{m\in M} \bigoplus_{i=1}^m T_m\biggr),
\end{equation*}
\notag
$$
provided that the singular spectra of the operators $T_m$ have multiplicity one and are pairwise disjoint, and all products of the form $T_m\otimes T_{m'}$ have Lebesgue spectrum. The Poisson suspension over a direct sums $\bigoplus_{m\in M} \bigoplus_{i=1}^m T_m$ is a dynamical realization of the operator $P$ (and similarly for a Gaussian automorphism). A suitable set of automorphisms $T_m$ can be found using Theorem 5.1. § 6. Remarks on the deterministic nature of the shift in $l_p( {\mathbb Z})$The property of $P_j(S)$-rigidity is an invariant of the operator $S$, so it can be of independent interest. For example, we show how we can establish this property for the shift operator $S$ in the Banach space $l_p(\mathbb Z)$ for $p>2$. Proof. We will define a two-sided sequence $v\colon \mathbb Z\to \mathbb R$ such that the quantities $v(i)$ are distinct from zero only for $i=-10,- 10^2, \dots, -10^n, \dots$ . An additional assumption: all nonzero elements $v(i)$ have the form $2^{-j}$, where $j=1,2,\dots$, and each quantity $2^{-j}$ occurs $q(j)= [2^{cj}]$ times among the $v(i)$, where $c$ satisfying $2<c<p$ is fixed. We also assume for convenience that the nonzero values of components of the vector $v$ are decreasing from left to right. Then there exists a sequence $h_j\to\infty$ such that for $i\in [h_j, h_{j+1})$ we have $v(i)=2^{-j}$ or $v(i)=0$, and for all $i\notin[h_j, h_{j+1})$ we have $v(i)\neq 2^{-j}$. Since $c<p$, we obtain
$$
\begin{equation*}
\|v\|_p^p = \sum_{j=1}^\infty q(j)2^{-pj}<\sum 2^{(c-p)j}<\infty.
\end{equation*}
\notag
$$
Let $e_i$ denote the basis vector such that $e_i(i)=1$ and its other coordinates are zero. Set then
$$
\begin{equation*}
Q_j(S)v=\sum_{n,\,10^n\in [h_j, h_{j+1})} 2^{-j} S^{10^n}v = q(j)2^{-j}e_0 + \Delta_j,
\end{equation*}
\notag
$$
where $\Delta_j$ is a sum of $q(j)$ vectors with norms less than $\|v\|$ and disjoint supports. Then we have
$$
\begin{equation*}
\|\Delta_j\|_p< q(j)^{1/p} \|v\|_p \leqslant 2^{cj/p} \|v\|_p.
\end{equation*}
\notag
$$
Taking the relations $q(j)= [2^{cj}]$ and $2<c<p$ into account we obtain Then
Since $S^iR_j(S)v$ tends to $ e_i$ each vector $\sum_i a_ie_i=\sum_i a_iS^ie_0$ with finite support can be approximated by the vectors $R_j(S)\sum_i a_iS^iv$. The vectors with finite support are dense in $l_p$, so some appropriate sequence of polynomials $P_j(S)$ is rigid, that is, $P_j(S)v\to v$, the polynomials $P_{j}$ have nonnegative coefficients, and the powers of $S$ involved in $P_{j}(S)$ with positive coefficients tend to $+\infty$ with growth of $j$. Proposition 6.1 is proved. The shift in $l_2( {\mathbb Z})$The shift operator in $l_2(\mathbb Z)$ is isomorphic to the operator $U$ of multiplication by $z$ in the space $L_2(\mathbb T,\lambda)$, $|z|=1$, where $\mathbb T$ is the unit circle in the complex plane with Lebesgue measure $\lambda$. We prove polynomial rigidity for $U$. By the Kolmogorov–Szegő theorem (for instance, see [3]) there exists a bounded positive weight function $w(z)$ on $\mathbb T$ such that for each $j$ the system $\{z^n\colon n>j \}$ is complete in $L_2(\mathbb T,w)$. Since these systems are complete, there exist polynomials $P_j$, which are linear combinations of the functions $z^n$ for $n>j$, such that $P_j\to 1$ in $L_2(\mathbb T,w)$. This means that which implies that
$$
\begin{equation*}
\int_\mathbb T |P_j(z)w-w|^2\,d\lambda(z)= \int_\mathbb T |P_j(z)-1|^2w(z)^2\,d\lambda(z) \to 0.
\end{equation*}
\notag
$$
Thus, $P_jw\to w$ in $L_2(\mathbb T,\lambda)$. Since the function $w$ is distinct from zero almost everywhere, it is a cyclic vector of the operator $U$. As a dense subset of $F$ we take the linear span of the vectors $z^nw$, $n\in \mathbb Z$; then $P_jf\to f$ in $L_2(\mathbb T,\lambda)$ for each $f\in F$. This means that the sequence of polynomial $P_j$ is rigid. By contrast with the previous examples, we do not claim here that the polynomials $P_j$ have positive coefficients. AcknowledgementsThe author is obliged to J.-P. Thouvenot and the participants of the seminar on the theory of functions at the Faculty of Mechanics and Mathematics at Moscow State University for useful discussions. The author is grateful to the referee for their comments. 
Citation in format AMSBIB
\Bibitem{Ryz24} |
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