There are no topologically transitive operators in the noncommutative Schwartz space

Abstract

The aim of this note is to prove that there are no topologically transitive operators in the noncommutative Schwartz space.

1 Introduction

Let s and $s'$ be the spaces of rapidly decreasing and slowly increasing sequences, respectively, equipped with their natural locally convex topologies. The so-called noncommutative Schwartz space is the space ${\mathcal {S}}:=L(s',s)$ of all bounded linear operators acting from $s'$ into s. This space becomes a Fréchet algebra in a natural way: if $T_1$ and $T_2$ are in ${\mathcal {S}}$, then the product $T_1T_2$ is defined by the formula $T_1T_2=T_1\circ \iota \circ T_2$, where $\iota$ is the natural embedding of s into $s'$. In fact, ${\mathcal {S}}$ embeds algebraically into the $C^*$-algebra ${\mathcal {B}}(\ell _2)$ of all bounded and linear operators on the Hilbert space $\ell _2$.

The noncommutative Schwartz space is isomorphic (as a Fréchet $^*$-algebra) to a number of other natural objects of analysis, e.g., ${\mathcal {S}}\simeq {\mathcal {S}}({\mathbb {R}}^2)$—the Schwartz space of rapidly decreasing functions on ${\mathbb {R}}^2$ equipped with the Volterra multiplication $(f\cdot g)(x,y):=\int _{{\mathbb {R}}}f(x,z)g(z,y)dz$ and involution $f^*(x,y):=\overline{f(y,x)}$. It plays an important role, e.g., in K-theory—see [4, 12], cyclic cohomology for crossed products—see [9, 14], noncommutative geometry—see [3], operator spaces—see [7, 8]. Another motivation comes from quantum mechanics where ${\mathcal {S}}$ is called the space of physical states and its dual is the so-called space of observables—see [6] for details.

Since ${\mathcal {S}}$ is a Fréchet algebra of operators, it is natural to ask about the dynamical properties of the elements of ${\mathcal {S}}$. Recall that an operator $T:s'\rightarrow s$ is topologically transitive if for every two non-empty and open sets $U\subset s'$, $V\subset s$, there exists $n\ge 0$ such that $T^n(U)\cap V\ne \emptyset$ and hypercyclic if there exists $x\in s'$ such that the set $\{T^{n}x:n\in {\mathbb {N}}\}$ is dense in s. It is clear that hypercyclicity of T implies that it is topologically transitive. Formally it could happen that the latter is a weaker property since $s'$ is not a metric space.

The main goal of this note is to show that there are no topologically transitive operators in ${\mathcal {S}}$. The main difficulty of the paper is to understand the spectral properties of operators from ${\mathcal {S}}$, those are investigated in Sect. 2.

We refer the reader to [1, 10, 11] for unexplained details from linear dynamics and functional analysis, respectively.

2 Notation and terminology

Recall that

$$\begin{aligned} s=\left\{ \xi =(\xi _j)_{j\in {\mathbb {N}}}\subset {\mathbb {C}}^{{\mathbb {N}}}:\,\,|\xi |_t^2:=\sum _{j=1}^{+\infty }|\xi _j|^2j^{2t}<+\infty \,\,\text {for all}\,\,t\geqslant 0\right\} \end{aligned}$$

and its topological dual

$$\begin{aligned} s'=\left\{ \eta =(\eta _j)_{j\in {\mathbb {N}}}\subset {\mathbb {C}}^{{\mathbb {N}}}:\,\,(|\eta |_t')^2:=\sum _{j=1}^{+\infty }|\eta _j|^2j^{-2t}<+\infty \,\,\text {for some}\,\,t\geqslant 0\right\} \end{aligned}$$

are the so-called spaces of rapidly decreasing and slowly increasing sequences, respectively.

Furthermore we consider the space ${\mathcal {S}}:=L(s',s)$ of all linear and continuous operators from $s'$ into s, equipped with the topology of uniform convergence on bounded sets. Consequently, the topology of ${\mathcal {S}}$ is given by the scale $(\Vert \cdot \Vert _t)_{t\geqslant 0}$ of norms, defined as

$$\begin{aligned} \Vert T\Vert _t:=\sup \{|T\eta |_t:\,\,|\eta |'_t\leqslant 1\}{}(T\in {\mathcal {S}},\,t\geqslant 0). \end{aligned}$$

If we denote $H_t:=\ell _2((j^t)_{j\in {\mathbb {N}}}),\,t\in {\mathbb {R}}$, then $H_t'\cong H_{-t}$ and every $T\in {\mathcal {S}}$ is a Hilbert space operator in the sense that $T:H_t'\rightarrow H_t$ and

$$\begin{aligned} \Vert T\Vert _t=\Vert T\Vert _{H_t'\rightarrow H_t}\qquad (t\geqslant 0). \end{aligned}$$

In other words, if we denote by $D_t:={{\,\mathrm{diag}\,}}(j^t),\,t\in {\mathbb {R}}$, an infinite diagonal matrix, then $D_t$ becomes simultaneously an isometry $D_t:H_t\rightarrow \ell _2$ and $D_t:\ell _2\rightarrow H_t'$ and

$$\begin{aligned} \Vert T\Vert _t=\Vert D_tTD_t\Vert _{{\mathcal {B}}(\ell _2)}\qquad (T\in {\mathcal {S}},\,t\geqslant 0). \end{aligned}$$

(1)

In particular, ${\mathcal {S}}={{\,\mathrm{proj}\,}}_{t\geqslant 0}{\mathcal {B}}(H_t',H_t)={{\,\mathrm{proj}\,}}_{k\in {\mathbb {N}}}{\mathcal {B}}(H_k',H_k)$. We will be using these properties interchangably.

Since $s\hookrightarrow \ s'$, we can define multiplication in ${\mathcal {S}}$ as

$$\begin{aligned} T_1T_2:=T_1\circ \iota \circ T_2{}(T_1,T_2\in {\mathcal {S}}), \end{aligned}$$

where $\iota :s\rightarrow s',\,\iota (\xi ):=\xi$ is the formal embedding. Altogether it turns ${\mathcal {S}}$ into an m-convex Fréchet algebra. It comes endowed also with the involution (or the adjoint map) given as

$$\begin{aligned} \langle T^*\xi ,\eta \rangle :=\langle \xi ,T\eta \rangle {}(\xi ,\eta \in s',\,T\in {\mathcal {S}}). \end{aligned}$$

It is worth noting that $s\hookrightarrow \ell _2$ and, by dualization, also $\ell _2\hookrightarrow s'$ therefore ${\mathcal {S}}$ is algebraically contained in the $C^*$-algebra ${\mathcal {B}}(\ell _2)$ of all bounded and linear operators on the Hilbert space $\ell _2$. Therefore multiplication in ${\mathcal {S}}$ is essentially the multiplication in ${\mathcal {B}}(\ell _2)$ with the additional property that the resulting operator belongs to ${\mathcal {S}}$. The same applies to involution.

The unitization of ${\mathcal {S}}$ will be denoted by ${\mathcal {S}}_1$. Clearly, the unit in ${\mathcal {S}}_1$ is the identity operator on $\ell _2$ denoted by $\mathbb {1}$. The algebra ${\mathcal {S}}$ is called the noncommutative Schwartz space and the elements of ${\mathcal {S}}$ are called smooth operators. We refer the reader to [2, 13] for more information on the properties of this algebra.

3. Spectral properties of operators in $\mathcal {S}$. We start by showing some spectral properties of smooth operators.

Proposition 3.1

([5, Proposition 3.1 and Theorem 3.3]). Every smooth operator is compact, i.e., ${\mathcal {S}}\hookrightarrow {\mathcal {K}}(\ell _2)$ and

$$\begin{aligned} \sigma _{{\mathcal {S}}_1}(T)=\sigma _{{\mathcal {B}}(\ell _2)}(T)\qquad (T\in {\mathcal {S}}). \end{aligned}$$

In particular, the spectrum of every smooth operator consists of zero and a (possibly) null sequence of eigenvalues.

Lemma 3.2

For any $t\in {\mathbb {R}}$ and every smooth operator $T\in {\mathcal {S}}$, we have

$$\begin{aligned} \sigma _{{\mathcal {B}}(\ell _2)}(D_tTD_{-t})\subset \sigma _{{\mathcal {S}}_1}(T). \end{aligned}$$

Proof

Let $t\in {\mathbb {R}}$ and $T\in {\mathcal {S}}$ be fixed. Suppose that $\lambda \in \rho _{{\mathcal {S}}_1}(T)$, i.e., there is $S\in {\mathcal {S}}$ such that

$$\begin{aligned} \left( S-\frac{1}{\lambda }\mathbb {1}\right) (T-\lambda \mathbb {1})=(T-\lambda \mathbb {1})\left( S-\frac{1}{\lambda }\mathbb {1}\right) =\mathbb {1}, \end{aligned}$$

where $\mathbb {1}$ is the identity operator on $\ell _2$. Then

$$\begin{aligned} \ell _2\xrightarrow {D_{-t}}H_t\hookrightarrow s'\xrightarrow {S}s\hookrightarrow H_t\xrightarrow {D_t}\ell _2 \end{aligned}$$

and

$$\begin{aligned} \left( D_tSD_{-t}-\frac{1}{\lambda }\mathbb {1}\right) (D_tTD_{-t}-\lambda \mathbb {1})=(D_tTD_{-t}-\lambda \mathbb {1})\left( D_tSD_{-t}-\frac{1}{\lambda }\mathbb {1}\right) =\mathbb {1}. \end{aligned}$$

Consequently, $\lambda \in \rho _{{\mathcal {B}}(\ell _2)}(D_tTD_{-t})$ and the proof is thereby complete. $\square$

Corollary 3.3

If a smooth operator $T\in {\mathcal {S}}$ satisfies

$$\begin{aligned} \sigma _{{\mathcal {S}}_1}(T)\subset {\mathbb {D}}, \end{aligned}$$

(2)

then the sequence $(T^n)_{n\in {\mathbb {N}}}$ is bounded in ${\mathcal {S}}$.

Proof

Let the smooth operator $T\in {\mathcal {S}}$ satisfy (2). Since $D_t^{-1}=D_{-t}$, we obtain that for every $n\in {\mathbb {N}}$ and every $t\geqslant 0$,

$$\begin{aligned} \Vert T^n\Vert _t&=\Vert D_tT^nD_t\Vert _{{\mathcal {B}}(\ell _2)} \\&=\Vert \underbrace{D_tTD_{-t}D_tTD_{-t}\cdots D_tTD_{-t}}_{n-1}D_tTD_t\Vert _{{\mathcal {B}}(\ell _2)} \\&\leqslant \Vert (D_tTD_{-t})^{n-1}\Vert _{{\mathcal {B}}(\ell _2)}\Vert T\Vert _t. \end{aligned}$$

From Lemma 3.2, the spectral radius formula, and compactness of the spectrum, it now follows that there is ${\varepsilon }>0$ such that

$$\begin{aligned} \nu (D_tTD_{-t})=\lim _{n\rightarrow \infty }\Vert (D_tTD_{-t})^n\Vert _{{\mathcal {B}}(\ell _2)}^{1/n}\leqslant 1-{\varepsilon }. \end{aligned}$$

Hence there is $N\in {\mathbb {N}}$ such that for every $n\geqslant N$, we have

$$\begin{aligned} \Vert (D_tTD_{-t})^n\Vert _{{\mathcal {B}}(\ell _2)}\leqslant 1. \end{aligned}$$

If we now define $C_t:=\max \{\Vert T\Vert _t,\Vert T^2\Vert _t,\ldots ,\Vert T^N\Vert _t,1\}\cdot \Vert T\Vert _t$, then

$$\begin{aligned}\sup _{n\in {\mathbb {N}}}\Vert T^n\Vert _t\leqslant C_t<\infty .\end{aligned}$$

Consequently, $(T^n)_{n\in {\mathbb {N}}}$ is a bounded sequence in the noncommutative Schwartz space. $\square$

3 Main result

Theorem 3.1

There are no topologically transitive operators in ${\mathcal {S}}$. In particular, the operators in ${\mathcal {S}}$ are not hypercyclic.

Proof

Let $T \in {\mathcal {S}}$ be arbitrary. There are two possible cases: either $\sigma _{{\mathcal {S}}_1}(T)=\{0\}$ or there exists $0\not =\lambda \in \sigma _{{\mathcal {S}}_1}(T)$.

If $\sigma _{{\mathcal {S}}_1}(T)=\{0\}$, then from Corollary 3.3, the sequence $(T^n)_{n\in {\mathbb {N}}}$ is bounded in ${\mathcal {S}}$ and therefore it is equicontinuous. In particular, for every zero neighbourhood $U \subset s'$, there is a zero neighbourhood $V\subset s$ such that

$$\begin{aligned} T^n(U)\subset \frac{1}{2}V\qquad (n\in {\mathbb {N}}). \end{aligned}$$

We choose now $\xi \in s\setminus V$ and suppose that there is $n\in {\mathbb {N}}$ and $\eta \in s$ such that

$$\begin{aligned} \eta \in T^n(U)\cap \left( \xi +\frac{1}{2}V\right) . \end{aligned}$$

This implies that for some $\zeta \in \frac{1}{2}V$, we have

$$\begin{aligned} \xi =\eta -\zeta \in \frac{1}{2}V+\frac{1}{2}V=V. \end{aligned}$$

This contradicts the choice of $\xi \in s$ and shows that in this case T is not topologically transitive.

If there exists $0\not =\lambda \in \sigma _{{\mathcal {S}}_1}(T)$, then from Proposition 3.1, it follows that $\lambda$ is an eigenvalue and we let f be a holomorphic function on a neighbourhood of $\sigma _{{\mathcal {S}}_1}(T)$ such that $f(\lambda )=1$ and $f(z)=0$ for $z\in \sigma _{{\mathcal {S}}_1}(T)\setminus \{\lambda \}$. Using the holomorphic functional calculus (which is available in ${\mathcal {S}}$ by [12, Lemma 1.3]), we can now consider the operator $f(T)\in {\mathcal {S}}$. Let $M={\text {Im}}(f(T))$. It is clear that M is a non-trivial and finite dimensional subspace of s (every non-zero element of M is an eigenvector of the compact operator f(T)). The properties of the functional calculus imply that the diagram

commutes and one can easily verify that topological transitivity of T would imply topological transitivity of $T_{|M}$. Since M is finite dimensional this implies that T is not topologically transitive. $\square$