Product systems associated to compound Poisson processes

Let $P$ be a closed convex cone in a Euclidean space $\mathbb{R}^{d}$. We assume that $P$ spans $\mathbb{R}^{d}$ and $P$ contains no line. An $E_{0}$-semigroup over $P$ is a semigroup of normal unital $*$-endomorphisms, indexed by $P$, of the algebra of bounded operators of an infinite dimensional separable Hilbert space satisfying a natural continuity assumption. It was first established by Arveson ([1], [3]), in the one parameter context, that $E_{0}$-semigroups are in one-one correspondence with product systems. This was recently extended to the multiparameter case in [9]. Roughly speaking, a product system over $P$ is a measurable field of separable Hilbert spaces carrying an associative multiplication rule.

In the one parameter setting, examples of product systems, which are in fact exotic, were constructed by Tsirelson in ([13], [12], [14]) and by Liebscher in [7] by making use of probabilistic models. The remarkable works of Tsirelson and that of Liebscher have amply demonstrated the rich interaction that exists between probability theory and the theory of $E_{0}$-semigroups. We expect similar things to take place in the multiparameter setting too. We explore a simple test case here.

The simplest probabilistic models that give rise to product systems in the multiparameter setting are stationary compound Poisson processes. In this paper, we define a product system corresponding to a stationary compound Poisson process. This is akin to associating product systems to Lévy processes ([8]) in the one parameter setting. We show that in the case of a stationary Poisson process and in the case of a stationary compound Poisson process, the resulting $E_{0}$-semigroups are CCR flows.

In the Poisson case, this is expected due to the Wiener-Itô Chaos decomposition (See Chapter 18, [6]). However it is quite easy to exhibit an explicit isomorphism in the Poisson case and it does not require sophisticated knowledge of Poisson processes. We then use the results of the Poisson case and the computation of the local projective cocycles carried out in [10] to identify the $E_{0}$-semigroup associated to a stationary compound Poisson process.

To keep the prerequistes on point processes to a minimum, in Section 2, we have collected the preliminaries on Poisson processes. For completeness, we have included proofs of some results. Our exposition is based on [5] and [6]. We make the following assumptions throughout this paper.

1. (1)

   The probability triples $(\Omega,\mathcal{F},\mathbb{P})$ that we consider are assumed to be complete.

2. (2)

   Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space. All sub $\sigma$-algebras of $\mathcal{F}$ are assumed to be complete. Let $\Sigma\subset\mathcal{F}$ be a sub $\sigma$-algebra. Recall that $\Sigma$ is said to be complete if it satisfies the following. If $A\subset\Omega$ equals a set in $\Sigma$ up to measure zero then $A\in\Sigma$.

3. (3)

   For random variables $X_{i}$, by the smallest $\sigma$-algebra generated by $X_{i}$, we mean the smallest complete $\sigma$-algebra which makes $X_{i}$’s measurable.

The author would like to thank Prof. Partha Sarathi Chakraborty for his suggestion to investigate the relation between point processes and $E_{0}$-semigroups.

Let $(\mathbb{X},\mathcal{B})$ be a measurable space. We assume that $\mathcal{B}$ is countably generated. Let $\overline{\mathbb{N}}=\mathbb{N}\cup\{\infty\}$. Denote the set of $\overline{\mathbb{N}}$-valued $\sigma$-finite measures by $\mathcal{N}(\mathbb{X})$. We endow $\mathcal{N}(\mathbb{X})$ with the smallest $\sigma$-algebra which makes the map

$$\mathcal{N}(\mathbb{X})\ni\mu\to\mu(B)\in\overline{\mathbb{N}}$$

measurable for every $B\in\mathcal{B}$.

A point process on $\mathbb{X}$ is a random element of $\mathcal{N}(\mathbb{X})$, i.e. a measurable mapping $\eta:\Omega\to\mathcal{N}(\mathbb{X})$ where $(\Omega,\mathcal{F},\mathbb{P})$ is an underlying probability space. Let $\eta$ be a point process on $\mathbb{X}$. For $\omega\in\Omega$, $\eta(\omega)$ is a measure on $\mathbb{X}$ and for $B\in\mathcal{B}$, the map $\omega\to\eta(\omega)(B)$ is a random variable. We denote the random variable $\omega\to\eta(\omega)(B)$ by $\eta(B)$.

Let $\lambda$ be a $\sigma$-finite measure on $\mathbb{X}$. A point process $\eta$ on $\mathbb{X}$ is called a Poisson process with intensity measure $\lambda$ if the following two conditions are satisfied.

1. (1)

   For $B\in\mathcal{B}$, $\eta(B)$ is a Poisson random variable with parameter $\lambda(B)$, i.e. for $k\in\mathbb{N}$, $\mathbb{P}(\eta(B)=k)=\frac{(\lambda(B))^{k}}{k!}e^{-\lambda(B)}$.

2. (2)

   The random variables $\eta(B_{1}),\eta(B_{2}),\cdots,\eta(B_{n})$ are independent whenever $(B_{i})_{i=1}^{n}$ is a disjoint family of measurable subsets of $\mathbb{X}$.

For a $\sigma$-finite measure $\lambda$, there exists a unique (up to equality in distribution) Poisson process with intensity measure $\lambda$. For a measure $\nu$ on $\mathbb{X}$ and an integrable complex valued function $u$ on $\mathbb{X}$, we denote $\int u(x)\nu(dx)$ by $\nu(u)$.

Proof. Write $\displaystyle u=\sum_{i=1}^{n}a_{i}1_{B_{i}}$ with $B_{i}$’s disjoint and $a_{i}\in\mathbb{C}\backslash\{0\}$. Note that $\lambda(B_{i})<\infty$ for every $i$. Then $\displaystyle e^{\eta(u)}=\prod_{i=1}^{n}e^{a_{i}\eta(B_{i})}$. Making use of the independence of $\eta(B_{i})$ and the fact that $\eta(B_{i})$ is a Poisson random variable with parameter $\lambda(B_{i})$, calculate as follows to observe that

	$\displaystyle\mathbb{E}(e^{\eta(u)})$	$\displaystyle=\prod_{i=1}^{n}\mathbb{E}(e^{a_{i}\eta(B_{i})})$
		$\displaystyle=\prod_{i=1}^{n}exp(\lambda(B_{i})(e^{a_{i}}-1))$
		$\displaystyle=exp(\sum_{i=1}^{n}(e^{a_{i}}-1)\lambda(B_{i}))$
		$\displaystyle=exp\Big{(}\int(e^{u(x)}-1)\lambda(dx)\Big{)}.$

The proof is now complete. $\Box$

Let $(\Omega,\mathcal{F},\mathbb{P})$ be an underlying probability space realising the Poisson process $\eta$ with intensity measure $\lambda$. We can and will assume that $\mathcal{F}$ is the smallest $\sigma$-algebra which makes the family of random variables $\eta(B)$, $B\in\mathcal{B}$, measurable.

See Lemma 18.4 of [6] for a proof.

Proof. Let $\mathcal{A}\subset\mathcal{B}$ be an algebra such that $\mathcal{A}$ is countable and $\mathcal{A}$ generates $\mathcal{B}$. Enumerate the elements of $\mathcal{A}$ as $A_{1},A_{2},\cdots$. Choose $B_{n}\in\mathcal{B}$ such that $\lambda(B_{n})<\infty$ and $\coprod_{n=1}^{\infty}B_{n}=\mathbb{X}$. It suffices to prove that $\mathcal{F}$ is the smallest $\sigma$-algebra which makes the random variables $\eta(A_{m}\cap B_{n})$ measurable. To that end, let $\mathcal{G}$ be the smallest $\sigma$-algebra which makes the random variables $\eta(A_{m}\cap B_{n})$, $m,n\in\mathbb{N}$, measurable.

For $n\in\mathbb{N}$, let $\mathcal{G}_{n}:=\{B\in\mathcal{B}:B\subset B_{n},\eta(B)\textrm{ is $\mathcal{G}$ measurable}\}$. It is immediate that $\mathcal{G}_{n}$ is a monotone class and contains the algebra $\mathcal{A}\cap B_{n}$. Hence $\mathcal{G}_{n}$ coincides with the $\sigma$-algebra of subsets of $B_{n}$ generated by $\mathcal{A}\cap B_{n}$. Consequently $\mathcal{G}_{n}=\mathcal{B}\cap B_{n}$. Now for $B\in\mathcal{B}$, $\eta(B)=\sum_{n=1}^{\infty}\eta(B\cap B_{n})$. Hence $\eta(B)$ is $\mathcal{G}$-measurable for every $B\in\mathcal{B}$. Hence $\mathcal{G}=\mathcal{F}$. This completes the proof. $\Box$

In this section, we associate a product system to a stationary Poisson process. Let $P$ be a closed convex cone in $\mathbb{R}^{d}$ which we assume is spanning and pointed, i.e. $P-P=\mathbb{R}^{d}$ and $P\cap-P=\{0\}$. Denote the interior of $P$ by $Int(P)$. For $x,y\in\mathbb{R}^{d}$, we write $x\leq y$ if $y-x\in P$ and we write $x<y$ if $y-x\in Int(P)$. The cone $P$ is fixed for the reminder of this paper. The setting that we consider is as follows.

Let $(Y,\mathcal{B})$ be a measurable space on which $\mathbb{R}^{d}$ acts in a measurable fashion. We use additive notation for the action. Assume that $\mathcal{B}$ is countably generated. Let $\lambda$ be a $\sigma$-finite measure on $Y$ which is $\mathbb{R}^{d}$ invariant. Consider a Poisson process $\eta$ on $Y$ with intensity measure $\lambda$. Then $\eta$ is stationary, i.e. for measurable subsets $B_{1},B_{2},\cdots,B_{n}$ of $Y$ and $x\in\mathbb{R}^{d}$, $(\eta(B_{1}),\cdots,\eta(B_{n}))=(\eta(B_{1}+x),\eta(B_{2}+x),\cdots,\eta(B_{n}+x))$ in distribution.

Suppose that $X\in\mathcal{B}$ and is $P$-invariant, i.e. $x+a\in X$ whenever $x\in X$ and $a\in P$. We assume that the action of $P$ on $X$ is pure, i.e. $\displaystyle\bigcap_{a\in P}(X+a)=\emptyset$. Note that for $a,b\in P$ with $a\leq b$, $X+b\subset X+a$. Let $(\Omega,\mathcal{F},\mathbb{P})$ be an underlying probability space realising the Poisson procees $\eta$. For $a,b\in P$ with $a\leq b$, let $\mathcal{F}_{a,b}$ be the $\sigma$-algebra generated by $\{\eta(B):B\subset(X+a)\backslash(X+b),B\in\mathcal{B}\}$. For $a\in P$, set $\mathcal{F}_{a}:=\mathcal{F}_{0,a}$.

Note that for $a\leq b\leq c$, $(X+a)\backslash(X+c)=((X+a)\backslash(X+b))\coprod((X+b)\backslash(X+c))$. This together with the complete independence of the Poisson process implies that for $a\leq b\leq c$, $\mathcal{F}_{a,b}$ and $\mathcal{F}_{b,c}$ are independent and $\mathcal{F}_{a,c}$ is generated by $\{\mathcal{F}_{a,b},\mathcal{F}_{b,c}\}$.

The stationarity of the Poisson process implies that for every $c\in P$, there exists a unitary $S_{c}:L^{2}(\Omega,\mathcal{F}_{a,b},\mathbb{P})\to L^{2}(\Omega,\mathcal{F}_{a+c,b+c},\mathbb{P})$ such that

$$S_{c}(f(\eta(B_{1}),\eta(B_{2}),\cdots,\eta(B_{n})))=f(\eta(B_{1}+c),\eta(B_{2}+c),\cdots,\eta(B_{n}+c)).$$

We are now in a position to define the product system given the above data. Let

$$E:=\{(a,f):a\in P,f\in L^{2}(\Omega,\mathcal{F}_{a},\mathbb{P})\}.$$

(3.1)

The first projection of $E$ onto $P$ is denoted by $p$. Note that by Lemma 2.3, the fibres $E(a)$, for $a\in P$, are separable. The product structure on $E$ is defined as follows

$$(a,f)(b,g)=(a+b,fS_{a}(g)).$$

It is routine to verify from discussions above that the algebraic requirements for $E$ to be a product system are met. We proceed towards the proof of the fact that $E$ is a product system in the sense of Definition 9.2 of [11]. A careful look at the axioms of Definition 9.2 of [11] reveals that it suffices to prove the following.

1. (1)

   The set $E$ is a Borel subset of $P\times L^{2}(\Omega,\mathcal{F},\mathbb{P})$.

2. (2)

   The multiplication $E\times E\ni(a,f)\times(b,g)\to(a+b,fS_{a}(g))$ is measurable.

3. (3)

   The pair $(E,\Gamma)$ forms a measurable field of Hilbert spaces where

   $$\Gamma:=\{s:P\to E:\textit{$s$ is measurable and}~{}s(a)\in E(a),\forall a\in P\}.$$

Set $\mathcal{H}:=L^{2}(\Omega,\mathcal{F},\mathbb{P})$. For $a\in P$, let $Q(a)$ be the projection on $\mathcal{H}$ that corresponds to the subspace $L^{2}(\Omega,\mathcal{F}_{a},\mathbb{P})$. Then for $a\in P$ and $f\in\mathcal{H}$, $Q(a)f=\mathbb{E}\{f|\mathcal{F}_{a}\}$. Note that

$$E:=\{(a,f)\in P\times L^{2}(\Omega,\mathcal{F},\mathbb{P}):Q(a)f=f\}.$$

Thus to prove $(1)$ and $(3)$, it suffices to show that $\{Q(a)\}_{a\in P}$ is a weakly measurable family of projections, i.e. for $f,g\in\mathcal{H}$, the map $P\ni a\to\langle Q(a)f|g\rangle$ is Borel measurable.

Proof. It suffices to prove that for $f,g$ in a total set, the map $P\ni a\to\langle Q(a)f|g\rangle\in\mathbb{C}$ is measurable. In view of Prop.2.2, it is enough to show that for simple $L^{2}$-functions $u,v$ on $Y$, the map $P\ni a\to\langle Q(a)e^{\eta(u)}|e^{\eta(v)}\rangle\in\mathbb{C}$ is measurable. For $a\in P$, let $L_{a}:=X\backslash(X+a)$.

Write $u:=\sum_{i=1}^{n}a_{i}1_{B_{i}}$ and $v:=\sum_{i=1}^{n}b_{i}1_{B_{i}}$ with $\displaystyle\coprod_{i=1}^{n}B_{i}=Y$ and $\lambda(B_{i})<\infty$. Calculate as follows to observe that

	$\displaystyle Q(a)e^{\eta(u)}$	$\displaystyle=\mathbb{E}\{\prod_{i=1}e^{a_{i}\eta(B_{i})}|\mathcal{F}_{a}\}$
		$\displaystyle=\mathbb{E}\{\prod_{i=1}^{n}e^{a_{i}\eta(B_{i}\cap L_{a})}e^{a_{i}\eta(B_{i}\cap L_{a}^{c})}|\mathcal{F}_{a}\}$
		$\displaystyle=\prod_{i=1}^{n}e^{a_{i}\eta(B_{i}\cap L_{a})}\prod_{i=1}^{n}\mathbb{E}\big{(}e^{a_{i}\eta(B_{i}\cap L_{a}^{c})}\big{)}~{}~{}(\textrm{by the complete independence of $\eta$}).$

Similarly, $\displaystyle Q(a)e^{\eta(v)}=\prod_{i=1}^{n}e^{b_{i}\eta(B_{i}\cap L_{a})}\prod_{i=1}^{n}\mathbb{E}\big{(}e^{b_{i}\eta(B_{i}\cap L_{a}^{c})}\big{)}$. Using the complete independence of the Poisson process , we arrive at the following equation

$$\langle Q(a)e^{\eta(u)}|Q(a)e^{\eta(v)}\rangle=\prod_{i=1}^{n}\mathbb{E}\big{(}e^{(a_{i}+\overline{b_{i}})\eta(B_{i}\cap L_{a})}\big{)}\prod_{i=1}^{n}\mathbb{E}\big{(}e^{a_{i}\eta(B_{i}\cap L_{a}^{c})}\big{)}\prod_{i=1}^{n}\mathbb{E}\big{(}e^{\overline{b_{i}}\eta(B_{i}\cap L_{a}^{c})}\big{)}$$

It is now sufficient to prove that for a measurable set $B$ of finite measure and a complex number $z$, the maps $P\ni a\to\mathbb{E}(e^{z\eta(B\cap L_{a})})\in\mathbb{C}$ and $P\ni a\to\mathbb{E}(e^{z\eta(B\cap L_{a}^{c})})$ are measurable. Note that

$$\mathbb{E}(e^{z\eta(B\cap L_{a})})=exp(\lambda(B\cap L_{a})(e^{z}-1))$$

and

$$\mathbb{E}(e^{z\eta(B\cap L_{a}^{c})})=exp(\lambda(B\cap L_{a}^{c})(e^{z}-1)).$$

By Tonelli’s theorem, the map

$$P\ni a\to\int 1_{B}(y)1_{X+a}(y)\lambda(dy)\in\mathbb{C}$$

is measurable. Consequently, for a measurable subset $B$ of finite measure, the maps $P\ni a\to\lambda(B\cap L_{a})\in\mathbb{C}$ and $P\ni a\to\lambda(B\cap L_{a}^{c})\in\mathbb{C}$ are measurable. The conclusion is now immediate and the proof is complete. $\Box$

Let us fix a few notation. For $a\in P$, let $L_{-\infty,a}=Y\backslash(X+a)$ and let $L_{a,\infty}=X+a$. For $a,b\in P$ with $a\leq b$, let $L_{a,b}=(X+a)\backslash(X+b)$. Note that for $a\leq b\leq c$,

	$\displaystyle L_{a,c}$	$\displaystyle=L_{a,b}\sqcup L_{b,c}$
	$\displaystyle Y$	$\displaystyle=L_{-\infty,a}\sqcup L_{a,b}\sqcup L_{b,\infty}.$

For $a\in P$, we write $L_{a}=L_{0,a}$.

Proof. Let $B$ be the complement of $\coprod_{i=1}^{n}B_{i}$ in $Y$. To prove that the first map is measurable, It suffices to prove that for a simple $L^{2}$-function $u$, the map

$$P\ni a\to\mathbb{E}\Big{(}e^{\sum_{i=1}^{n}z_{i}\eta(B_{i}\cap L_{a})}e^{\eta(u)}\Big{)}\in\mathbb{C}$$

is measurable. Let $u$ be a simple $L^{2}$-function. Write $u=\sum_{j=1}^{m}w_{j}1_{C_{j}}$ with $\coprod_{j=1}^{m}C_{j}=Y$ and $\lambda(C_{j})<\infty$. Set $A_{ij}=B_{i}\cap C_{j}$.

A routine calculation using the complete independence of the Poisson process $\eta$ reveals that $\mathbb{E}\Big{(}e^{\sum_{i=1}^{n}z_{i}\eta(B_{i}\cap L_{a})}e^{\sum_{j=1}^{m}w_{j}\eta(C_{j})}\Big{)}$ is the product of $\prod_{j=1}^{m}exp(\lambda(C_{j}\cap B\cap L_{a})(e^{w_{j}}-1))$ and

$$\prod_{i,j}exp(\lambda(A_{ij}\cap L_{a})(e^{z_{i}+w_{j}}-1))\prod_{j=1}^{m}exp(\lambda(C_{j}\cap L_{-\infty,0})(e^{w_{j}}-1))\prod_{j=1}^{m}exp(\lambda(C_{j}\cap L_{a,\infty})(e^{w_{j}}-1)).$$

The measurability of each term follows from the fact that for a measurable set $B$ of finite measure the maps $P\ni a\to\lambda(B\cap L_{a})$ and $P\ni a\to\lambda(B\cap L_{a,\infty})$ are measurable. The proofs of other assertions are similar and we omit the proof. $\Box$

Proof. It suffices to show that for a simple function $L^{2}$-function $u$,

$$((a,f),(b,g))\to\mathbb{E}(fS_{a}ge^{\eta(u)})$$

is measurable. Write $u=\sum_{i=1}^{n}z_{i}1_{B_{i}}$ with $\coprod_{i=1}^{n}B_{i}=Y$ and $\lambda(B_{i})<\infty$. Using the complete independence and the stationarity of the Poisson process $\eta$, note that $\mathbb{E}(fS_{a}ge^{\eta(u)})$ is

$$\mathbb{E}\Big{(}e^{\sum_{i}z_{i}\eta(B_{i}\cap L_{-\infty,0})}\Big{)}\mathbb{E}\Big{(}fe^{\sum_{i}z_{i}\eta(B_{i}\cap L_{a})}\Big{)}\mathbb{E}\Big{(}ge^{\sum_{i}z_{i}\eta((B_{i}-a)\cap L_{b})}\Big{)}\mathbb{E}\Big{(}e^{\sum_{i}z_{i}\eta(B_{i}\cap L_{a+b,\infty})}\Big{)}.$$

The required measurability conclusion follows from Lemma 3.2. This completes the proof. $\Box$

In short, we have the following theorem.

In this section, we define a product system associated to a stationary compound Poisson process and identify it explicitly. First we consider the Poisson case. We use the notation introduced in Section 3. For $a\in P$, let $\mathbbm{1}_{a}\in L^{2}(\Omega,\mathcal{F}_{a},\mathbb{P})$ be the constant function $1$. Note that $(\mathbbm{1}_{a})_{a\in P}$ is a unit of $E$ (i.e. it is a non-zero multiplicative measurable cross section of $E$). For the rest of this paper, we fix a measurable logarithm, i.e. a measurable map $\ell:\mathbb{C}\backslash\{0\}\to\mathbb{C}$ such that $e^{\ell(z)}=z$ for every $z\in\mathbb{C}\backslash\{0\}$ and $\ell(x)=\log(x)$ if $x>0$.

Let $\mathcal{S}$ denote the set of all complex valued simple functions $f$ on $X$ which are square integrable (which is the same as integrable) such that $f(x)\neq-1$ for every $x\in X$. For $f\in\mathcal{S}$, set

$$u_{f}(x):=\ell(1+f(x)).$$

(4.2)

Note that $u_{f}$ is simple and square integrable. For $f\in\mathcal{S}$, let $\Sigma_{f}$ be the random variable defined by

$$\Sigma_{f}:=\frac{e^{\eta(u_{f})}}{\mathbb{E}\big{(}e^{\eta(u_{f})}\big{)}}.$$

Observe that $\mathcal{S}$ is a dense subset of $L^{2}(X,\lambda)$. A simple calculation using Lemma 2.1 reveals that for $f,g\in\mathcal{S}$,

$$\mathbb{E}(\Sigma_{f}\overline{\Sigma_{g}})=exp\Big{(}\int f(x)\overline{g(x)}\lambda(dx)\Big{)}.$$

(4.3)

Hence for $f,g\in\mathcal{S}$,

$$\mathbb{E}\Big{(}(\Sigma_{f}-\Sigma_{g})\overline{(\Sigma_{f}-\Sigma_{g})}\Big{)}=exp(\langle f|f\rangle)+exp(\langle g|g\rangle)-2Re~{}exp(\langle f|g\rangle).$$

(4.4)

Let $f\in L^{2}(X,\lambda)$ be given. Choose a sequence $(f_{n})\in\mathcal{S}$ such that $f_{n}\to f$ in $L^{2}(X,\lambda)$. Equation 4.4 implies that $\Sigma_{f_{n}}$ converges in $L^{2}(\Omega,\mathcal{F},\mathbb{P})$. Making use of Equation 4.4 again, it follows that the limit $\displaystyle\lim_{n\to\infty}\Sigma_{f_{n}}$ is independent of the chosen sequence $(f_{n})$. Define

$$\Sigma_{f}=\lim_{n\to\infty}\Sigma_{f_{n}}.$$

Note that Equation 4.3 is valid for $f,g\in L^{2}(X,\lambda)$. Observe that for $a\in P$, if $f\in L^{2}(X\backslash(X+a))$ then $\Sigma_{f}\in L^{2}(\Omega,\mathcal{F}_{a},\mathbb{P})$.

Consider the Hilbert space $H:=L^{2}(X,\lambda)$. For $a\in P$, let $V_{a}$ be the isometry on $H$ defined by the equation

$$V_{a}(f)(y):=\begin{cases}f(y-a)&\mbox{ if }y-a\in X,\cr&\cr 0&\mbox{ if }y-a\notin X.\end{cases}$$

(4.5)

Then $(V_{a})_{a\in P}$ is a weakly measurable semigroup of isometries and hence is strongly continuous.

Proof. Let $a\in P$ be given. For $f,g\in L^{2}(X\backslash(X+a))$,

$$\mathbb{E}(\Sigma_{f}\overline{\Sigma_{g}})=exp(\langle f|g\rangle)=\langle e(f)|e(g)\rangle.$$

Moreover the set $\{\Sigma_{f}:f\in L^{2}(X\backslash(X+a))\}$ is total in $E(a)$. This is because if $u$ is a non-negative simple function and if we set $f(x)=e^{u(x)}-1$ then $u_{f}=u$. Also the set $\{e(f):f\in L^{2}(X\backslash(X+a))\}$ is total in $\Gamma(L^{2}(X\backslash(X+a)))$. Consequently, there exists a unitary $\theta_{a}:\Gamma(L^{2}(X\backslash(X+a)))\to E(a)$ such that $\theta_{a}(e(f))=\Sigma_{f}$. Set $\displaystyle\theta:=\coprod_{a\in P}\theta_{a}$.

Let $a,b\in P$ and let $f\in L^{2}(X\backslash(X+a))$, $g\in L^{2}(X\backslash(X+b))$ be given. Observe that

$$(a,\Sigma_{f})(b,\Sigma_{g})=(a+b,\Sigma_{h})$$

where $h=f+V_{a}g$. Thus $\theta$ preserves the product structure. We leave the verification that $\theta$ is measurable to the reader. This completes the proof. $\Box$

The following is an immediate corollary of Prop. 4.1 and Prop. 2.1 of [10].

Next we associate a product system to a stationary compound Poisson process. First we collect a few preliminaries on compound Poisson processes from [6]. Let $(\mathbb{X},\mathcal{B})$ be a measurable space. We assume that $\mathcal{B}$ is countably generated. Denote the set of all $\sigma$-finite measures on $\mathbb{X}$ by $M(\mathbb{X})$. We endow $M(\mathbb{X})$ with the smallest $\sigma$-algebra which makes the map $\mu\to\mu(B)$ measurable for every $B\in\mathcal{B}$. A measurable mapping $\omega\to\xi(\omega)\in M(\mathbb{X})$ is called a random measure on $\mathbb{X}$ where $(\Omega,\mathcal{F},\mathbb{P})$ is an underlying probability space. Just like in the case of point processes, for every $\omega$, $\xi(\omega)$ is a measure and for every $B\in\mathcal{B}$, $\xi(B)$ is a random variable.

The random measure that we will be interested in are compound Poisson processes. Let $(Y,\mathcal{B}_{Y})$ be a measurable space and assume that $B_{Y}$ is countably generated. Suppose $\rho_{0}$ is a $\sigma$-finite measure on $Y$. Let $\nu$ be a “Lévy measure” on $(0,\infty)$, i.e. the integral $\int(r\wedge 1)\nu(dr)<\infty$ which is also equivalent to the fact that $\int(1-e^{-tr})\nu(dr)<\infty$ for every $t>0$. Let $\eta$ be a Poisson process on $Y\times(0,\infty)$ with intensity measure $\lambda:=\rho_{0}\otimes\nu$.

For $B\in B_{Y}$, let

$$\xi(B)=\eta(r1_{B}(y))=\int r1_{B}(y)\eta(d(y,r)).$$

Then $\xi$ is a random measure on $Y$. The random measure $\xi$ is called the $\rho_{0}$-symmetric compound Poisson process on $Y$ with Lévy measure $\nu$. The following are the basic properties of the compound Poisson process $\xi$.

1. (1)

   The compound Poisson process $\xi$ is completely independent, i.e. for disjoint measurable subsets $B_{1},B_{2},\cdots,B_{n}$, the random variables $\xi(B_{1}),\xi(B_{2}),\cdots,\xi(B_{n})$ are independent.

2. (2)

   For $B\in B_{Y}$, the Laplace transform of the random variable $\xi(B)$ is given by

   $$\mathbb{E}(e^{-t\xi(B)})=exp\Big{(}-\rho_{0}(B)\int(1-e^{-tr})\nu(dr)\Big{)}.$$

Just like in the Poisson case, we can associate a product system to a stationary compound Poisson process. The setting is as before. Let $(Y,\mathcal{B}_{Y})$ be a measurable space on which $\mathbb{R}^{d}$ acts and let $\rho_{0}$ be a $\sigma$-finite measure on $Y$ which is $\mathbb{R}^{d}$ invariant. Suppose $X\subset Y$ is a measurable subset which is $P$-invariant. We also assume that the action of $P$ on $X$ is pure. Let $\nu$ be a Lévy measure on $(0,\infty)$ and $\eta$ be the Poisson process on $Y\times(0,\infty)$ with intensity measure $\lambda:=\rho_{0}\otimes\nu$. Let $\xi$ be the $\rho_{0}$-symmetric compound Poisson process on $Y$ with Lévy measure $\nu$. Since $\rho_{0}$ is $\mathbb{R}^{d}$-invariant, the random measure $\xi$ is stationary, i.e. for a measurable subset $B$ of $Y$, $\xi(B+x)$ and $\xi(B)$ have the same distribution for every $x\in\mathbb{R}^{d}$.

The action of $\mathbb{R}^{d}$ on $Y$ induces an action of $\mathbb{R}^{d}$ on $\widetilde{Y}:=Y\times(0,\infty)$ where the action is on the first coordinate. Let $X$ be a $P$-invariant measurable subset of $Y$ and set $\widetilde{X}:=X\times(0,\infty)$. Consider a probability space $(\Omega,\mathcal{F},\mathbb{P})$ which realises the Poisson process $\eta$. The $\sigma$-algebras $\mathcal{F}_{a}$, $\mathcal{F}_{a,b}$ etc.. are defined as in the Poisson case with $\eta$ replaced by $\xi$. The $\sigma$-algebras corresponding to the Poisson process $\eta$ associated to the data $(\widetilde{Y},\widetilde{X},\lambda)$ are denoted by $\widetilde{\mathcal{F}_{a}}$, $\widetilde{\mathcal{F}_{a,b}}$ etc…

Set

$$E:=\{(a,f):a\in P,f\in L^{2}(\Omega,\mathcal{F}_{a},\mathbb{P})\}.$$

The product rule on $E$ is defined exactly as in the Poisson case. Using the complete independence of $\xi$ and the stationarity of $\xi$, it is quite routine to check that the algebraic requirements for $E$ to be a product system is satisfied. It is possible to prove as in the Poisson case that $E$ satisfies the measurability requirements. Howeover, it is automatic that the measurability requirements are met given that the Poisson case is already verified. We explain this below.

Let $(E,p)$ be a product system over $P$. Let $F\subset E$ be given. For $x\in P$, let $F(x)=E(x)\cap F$. We say that $F$ is a subsystem of $E$ if

1. (1)

   for every $x\in P$, $F(x)$ is a non-zero closed subspace of $E(x)$, and

2. (2)

   for $x,y\in P$, $F(x)F(y)\subset F(x+y)$ and is total in $F(x+y)$.

Suppose $E$ is a product system and $F\subset E$ is a subsystem. We prove that $F$ is a measurable subset of $E$ and with the measurable structure induced from $E$, $F$ is a product system on its own right. Realise $E$ as a product system of an $E_{0}$-semigroup, say $\alpha:=\{\alpha_{x}\}_{x\in P}$ on $B(\mathcal{H})$.

For $x\in P$, let $\theta_{x}:E(x)\to E(x)$ be the orthogonal projection onto $F(x)$. Let $p_{x}\in\alpha_{x}(B(\mathcal{H}))^{{}^{\prime}}$ be such that $\theta_{x}(T)=p_{x}T$. Then $p_{x}$ is a non-zero projection for every $x\in P$. Note that for $u\in E(x)$ and $v\in E(y)$, $\theta_{x+y}(uv)=\theta_{x}(u)\theta_{y}(v)$. This translates to the fact that $p_{x}\alpha_{x}(p_{y})=p_{x+y}$ for $x,y\in P$. Then

$$F=\{(x,T):x\in P,T\in E(x),p_{x}T=T\}.$$

The required conclusion is immediate provided $(p_{x})_{x\in P}$ is a weakly continuous family of projections. The latter assertion is proved as in Prop. 8.9.9 of [2] with the aid of Theorem 10.8.1 of [4].