Proof. The only point that requires explanation is the surjectivity of the map. Let $u:=\{u_x\}_{x \in P}$ be a unit of $\alpha$. Set

\[ Q:=\{x \in P: \textrm{there exist }\lambda_x \in \mathbb{C}^{{\times}}\text{ and }\xi_x \in Ker(V_x^{*})\text{ such that }u_x=\lambda_xT^{(x)}_{e(\xi_x)}\}. \]

Note that $Q$ is a closed subsemigroup of $P$.

Fix $X \in L(P)$. For $t \geq 0$, let $\alpha _t:=\alpha _{\exp (tX)}$. Then, $\{\alpha _t\}_{t \geq 0}$ is the one parameter CCR flow associated with the isometric representation $\{V_{\exp (tX)}\}_{t \geq 0}$. Making use of Wold decomposition and Theorem 2.6.4 of [Reference Arveson4], it is routine to see that $\exp (tX) \in Q$ for every $t \geq 0$. Hence, $Q$ contains the semigroup generated by $\exp (L(P))$. But the semigroup generated by $\exp (L(P))$ is dense in $P$. Therefore $Q=P$.

Thus, for every $x \in P$, there exists $\lambda _x \in \mathbb {C}^{\times }$ and $\xi _x \in Ker(V_x^{*})$ such that

\[ u_x=\lambda_xT^{(x)}_{e(\xi_x)}. \]

The rest of the proof is similar to Theorem 5.10 of [Reference Arjunan, Srinivasan and Sundar1].

Proof. Observe that for units $u,\,v \in \mathcal {U}_{\alpha }$, the map $P \ni x\to v_x^{*}u_x \in \mathbb {C}^{\times }$ is multiplicative. Let $a,\,b \in \varOmega$ be given. Denote the seminorms corresponding to the semi-definite inner products $\langle ~|~\rangle _a$ and $\langle ~|~\rangle _b$ by $||~||_a$ and $||~||_b$ respectively.

Let $b \in \varOmega$. By the Archimedean property, there exists $n \in \mathbb {N}$ such that $b< na$, i.e. $na-b \in \varOmega$. So, $na=b+c$, for some $c \in \varOmega$. Let $u,\,v \in \mathcal {U}_{\alpha }$ be given. Then,

\[ e^{tc_{na}(u,v)}=e^{ntc_{a}(u,v)}, \quad e^{tc_{b+c}(u,v)}=e^{t(c_{b}(u,v)+c_{c}(u,v))} \]

which gives us the relation $nc_{a}(u,\,v)=c_{b}(u,\,v)+c_{c}(u,\,v)$. Similarly, there exists $m \in \mathbb {N}$ such that $mb=a+d$, for some $d \in \varOmega$, and $mc_{b}(u,\,v)=c_{a}(u,\,v)+c_{d}(u,\,v)$. Combining the two, we have for $f \in C_0(\mathcal {U}_\alpha )$, $n||f||_{a}^{2}=||f||_{b}^{2}+||f||_{c}^{2}$ and $m||f||_{b}^{2}=||f||_{a}^{2}+||f||_{d}^{2}$. Therefore, for $f \in C_0(\mathcal {U}_\alpha )$,

\[ ||f||_{a} \leq \sqrt{m}||f||_{b} \quad \textrm{and} \quad ||f||_{b} \leq \sqrt{n} ||f||_{a}. \]

Thus, both the seminorms $||~||_a$ and $||~||_b$ are equivalent. Consequently, the dimension of $\mathcal {H}(\mathcal {U}_{\alpha })$ is independent of the chosen interior point.

Proof. Note that for $\xi \in \mathcal {A}(V)$, the map $P \ni x \to \langle \xi _x|\xi _x \rangle \in [0,\,\infty )$ is additive. Let $\xi \in \mathcal {A}(V)$ be given. Let $x \in P$ and let $n$ be a natural number such that $x \leq na$. Then, $na=x+y$ for some $y \in P$. Note that $\xi _{na}=\xi _{x}+V_{x}\xi _{y}$, and consequently

\[ n ||\xi_{a}||^{2}=\langle \xi_{na}|\xi_{na}\rangle=\langle \xi_x| \xi_x\rangle+\langle \xi_y|\xi_y \rangle. \]

This implies

(2.2)\begin{equation} ||\xi_x||^{2} \leq n ||\xi_a||^{2}. \end{equation}

By the Archimedean property, given any $x \in P$, we have an $n \in \mathbb {N}$ such that $na=x+y$, for some $y \in P$. By Equation (2.2), if for some $\xi$, $\langle \xi | \xi \rangle _a= \langle \xi _a| \xi _a \rangle =0$, then $||\xi _x||=0$, for all $x \in P$, and $\xi =0$ in $\mathcal {A}(V)$. This proves that the sesquilinear form is indeed an inner product.

Given $b \in \varOmega$, there exist natural numbers $k$ and $m$ such that $ka-b \in P$ and $mb-a \in P$. By Equation (2.2), $\frac {1}{k}||\xi _b||^{2} \leq ||\xi _a||^{2} \leq m ||\xi _b||^{2}$. Thus, the norms induced by any two interior points $a$ and $b$ are equivalent.

To prove that $\mathcal {A}(V)$ is a Hilbert space, let $\left \lbrace \xi ^{n} \right \rbrace$ be a Cauchy sequence in $\mathcal {A}(V)$. Let $x \in P$ be given. Choose a natural number $k$ such that $ka>x$. Then, by Equation (2.2), for $n,\,m \geq 1$,

\[ ||\xi^{n}_x-\xi^{m}_x|| \leq k ||\xi^{n}_a-\xi^{m}_a||. \]

Hence, for all $x\in P$, $\left \lbrace \xi ^{n}_{x}\right \rbrace$ is Cauchy in $Ker(V_x^{*})$ and hence converges to, say, $\eta _{x}$ in $Ker(V_{x}^{*})$.

Fix $x \in P$, $0< \delta < 1$ and $a \in \varOmega$. By the Archimedean Property, given $y \in B(x,\, \delta ) \cap P$, there exists $k \in \mathbb {N}$ such that $y < ka$. Now, $P=\displaystyle \bigcup _{n \geq 1} \left \lbrace y \in P | y < na\right \rbrace$ and $B(x,\,\delta ) \cap P$ has compact closure. So, there exists $k \in \mathbb {N}$ such that $B(x,\, \delta ) \cap P \subset \left \lbrace y \in P| y< ka\right \rbrace$. Since $y < ka$, for all $y \in B(x,\, \delta ) \cap P$, we have by Equation (2.2) that for all $n,\,m$,

\[ ||\xi^{n}_y-\xi_y^{m}|| \leq k ||\xi^{n}_a-\xi^{m}_a||. \]

This shows that the sequence $\left \lbrace \xi ^{n} \right \rbrace$ is locally uniformly Cauchy. Consequently, the map $\eta = \left \lbrace \eta _x \right \rbrace _{x \in P}$ is continuous. Also, $\eta _{x+y}$ being the limit of $\xi ^{n}_{x+y}=\xi ^{n}_{x}+V_{x} \xi ^{n}_{y}$, we have for $x,\,y \in P$, $\eta _{x+y}=\eta _{x}+V_{x}\eta _{y}$. Thus, $\eta$ belongs to $\mathcal {A}(V)$. This completes the proof.

Proof. Fix $a \in \varOmega$. For two additive cocycles $\xi,\, \eta \in \mathcal {A}(V)$ and for any $\lambda,\, \mu \in \mathbb {C}^{d}$, note that

\[ c_{a}(u^{(\lambda,\xi)},u^{(\mu,\eta)}) = \langle \lambda|a \rangle + \overline{\langle \mu|a \rangle} + \langle \xi|\eta \rangle_a. \]

Calculate as follows to observe that

\begin{align*} \left\langle [u^{(0,\xi)}]-[u^{(0,0)}] \Big| [u^{(0,\eta)}]-[u^{(0,0)}] \right\rangle & = c_{a}(u^{(0,\xi)},u^{(0,\eta)}) - c_{a}(u^{(0,\xi)},u^{(0,0)}) \\ & \quad - c_{a}(u^{(0,0)},u^{(0,\eta)}) + c_{a}(u^{(0,0)},u^{(0,0)}) \\ & = \langle \xi| \eta \rangle_a - \langle \xi |0 \rangle_a - \langle 0 | \eta \rangle_a + \langle 0|0 \rangle_a \\ & = \langle \xi | \eta \rangle_a. \end{align*}

Thus, the prescribed map preserves the inner product. A routine calculation shows that

\[ \langle [u^{(\lambda,\xi)}]-[u^{(0,\xi)}] | [u^{(\lambda,\xi)}]-[u^{(0,\xi)}] \rangle =0. \]

Thus, $[u^{(\lambda,\xi )}]-[u^{(0,0)}]=[u^{(0,\xi )}]-[u^{(0,0)}]$. Also, the set $\left \lbrace [u]-[v]| u,\, v \in \mathcal {U}_{\alpha } \right \rbrace$ is total in $\mathcal {H}(\mathcal {U}_\alpha )$ and

\[ [u^{(\lambda,\xi)}]-[u^{(\mu,\eta)}]=([u^{(0,\xi)}]-[u^{(0,0)}])-([u^{(0,\eta)}]-[u^{(0,0)}]). \]

Consequently, the set $\left \lbrace [u^{(0,\xi )}]-[u^{(0,0)}] | \xi \in \mathcal {A}(V) \right \rbrace$ is total in $\mathcal {H}(\mathcal {U}_\alpha )$. Hence, the prescribed map is a unitary, and $Ind(\alpha )=$dim $\mathcal {A}(V)$.

Proof. By considering a right translate of $A$, we can without loss of generality assume that $e \in A$, and consequently $P \subset A$. Fix a scalar $\lambda \in \mathbb {C}$. Let us assume that $A \backslash xA$ has a finite measure for all $x \in P$. For $x \in P$, let $\xi _x:=\lambda 1_{A\backslash xA}$. Then, $\xi _x \in Ker(V_x^{*})$. Observe that for $x,\,y \in P$, $(A \backslash xyA) = (A \backslash xA) \amalg x(A \backslash yA)$. Hence, for $x,\,y \in P$, we have $\xi _{xy} = \xi _x + V_x \xi _y$.

Let $(x_n)$ be a sequence in $P$ such that $(x_n)$ converges to $x$. Since $A$ is closed, it follows that $1_{A\backslash x_nA}(y) \to 1_{A\backslash xA}(y)$ whenever $y \notin x \partial A$. Thanks to Lemma II.12 of [Reference Hilgert and Neeb9], it follows that $x\partial A$ has a measure zero. Therefore, $(1_{A\backslash x_nA}) \to 1_{A \backslash xA}$ a.e. Consider an increasing cofinal sequence, say $(a_n)$, in $\varOmega$. Then, $x < a_N$, for some $N$; $x^{-1}a_N \in \varOmega$. Thus, there exists $k$ such that $x_n^{-1}a_N \in \varOmega$, for all $n \geq k$. For $n \geq k$, $(A \backslash x_n A) \subset (A \backslash a_NA)$ and hence, $1_{A \backslash x_n A}$ is bounded above by $1_{A \backslash a_NA}$. By DCT, the sequence $(1_{A \backslash x_n A})$ converges to $1_{A \backslash xA}$ in $L^{2}(A ,\,\mu )$. Thus, the map $P \ni x \mapsto \xi _x \in L^{2}(A,\, \mu )$ is continuous. Therefore, $\xi = \left \lbrace \lambda (1_{A \backslash xA}) \right \rbrace _{x \in P}$ is an additive cocycle of $V$.

Now, let $\xi =\left \lbrace \xi _x \right \rbrace _{x \in P}$ be an additive cocycle of $V$. Each $\xi _x$ belongs to $L^{2} (A \backslash xA)$. Let $(a_n)$ be an increasing cofinal sequence in $\varOmega$ and set $a_0=e$. Then, $A=\displaystyle \amalg _{n \geq 1}( a_{n-1}A \backslash a_n A)$. Define $f: A \longrightarrow \mathbb {C}$ by,

\[ f(x):= \xi_{a_n}(x), \quad x \in a_{n-1} A \backslash a_n A. \]

The map $f$ is well defined, and is also measurable by definition.

1. (1) For all $n$, $f = \xi _{a_n}$ a.e. on $A \backslash a_n A$. Fix $n$. Note that $A \backslash a_n A = \bigcup _{1 \leq k \leq n} ( a_{k-1} A \backslash a_k A)$. For $m< n$ and $x \in a_{m-1}A \backslash a_m A$, $a_m^{-1}x \notin A$, so $V_{a_m} \xi _{a_m^{-1}a_n}(x)=0$. Thus, for $m< n$

   \[ \xi_{a_n}(x) = \xi_{a_m}(x) + V_{a_m} \xi_{a_m^{{-}1}a_n}(x) = \xi_{a_m}(x) = f(x) \]
   for almost all $x \in a_{m-1}A \backslash a_m A$. Hence, $f(x) = \xi _{a_n}(x)$ a.e. on $A \backslash a_n A$.

2. (2) Let $a \in P$ be given. Then $f(x) = \xi _a(x)$ a.e. on $A \backslash aA$. Fix $a$ in $P$. Choose $n$ for which $a < a_n$, so $(A \backslash aA) \subset (A \backslash a_n A)$. But, for $x$ in $A \backslash aA$, $a^{-1}x \notin A$, so $V_a \xi _{a^{-1}a_n}(x)=0$. Hence, for almost every $x \in (A \backslash aA)$,

   \[ \xi_{a_n}(x)= \xi_a(x) + V_a \xi_{a^{{-}1}a_n}(x)= \xi_a(x). \]
   Thus, $f (x)= \xi _a(x)$ for almost all $x \in A \backslash aA$.

3. (3) For every $a \in P$, $f(ax)=f(x)$, a.e. on $A$. Fix $a$ in $P$. Note that, for all $x \in A$, $\xi _a (ax)=0$. Thus, for each $n$,

   \[ \xi_{aa_n}(ax) = \xi_a(ax) + V_a \xi_{a_n}(ax) = \xi_{a_n}(x) \]
   for a.e. $x \in A$. But by $(2)$, $\xi _{a_n}(x) = f(x)$ and $\xi _{aa_n}(ax)=f(ax)$ for a.e $x$ in $A \backslash a_n A$. Since $A= \bigcup _n (A \backslash a_n A)$, $f(ax)=f(x)$ for almost every $x \in A$.

We define another order on $G$ as follows. For $g,\, h\in G$, we say $g \leq _r h$ if $g^{-1}h \in P^{-1}$, or equivalently $h^{-1}g \in P$. With this order, $\varOmega ^{-1}$ has an increasing cofinal sequence, say $(b_n)$, and we set $b_0=e$. For $g \in G$, $g \leq _r b_n$, for some $n$. Hence, $g \in b_n P \subset b_n A$. Therefore, $G=\bigcup _{n \geq 0}b_nA$. Also, for $m < n$, we have $b_m A \subset b_n A$. We extend $f$ to $G$ by:

\begin{align*} \widetilde{f}|_{(b_0 A =A)}(x)& =f(x),\\ \widetilde{f}|_{(b_n A \backslash b_{n-1}A)}(x)& = f(b_n^{{-}1}x). \end{align*}

Then, $\widetilde {f}$ is well defined, measurable, and is an extension of $f$ to $G$.

1. (1) For all $n$, $\widetilde {f}(x) = f(b_n ^{-1} x)$, for almost all $x \in b_n A$. Fix a natural number $n$. Observe that $b_n A = \displaystyle A \amalg _{1 \leq k \leq n} (b_k A \backslash b_{k-1}A)$. By (3),

   \[ f(b_n^{{-}1}x) = f(b_n^{{-}1} b_k b_k^{{-}1}x) = f(b_k^{{-}1}x) = \widetilde{f}(x) \]
   for almost every $x \in (b_k A \backslash b_{k-1} A)$ and for all $1 \leq k \leq n$. Thus, for every $n$, $\widetilde {f}|_{b_n A}(x)=f(b_n^{-1}x)$ for almost every $x \in b_n A$.

2. (2) Let $b \in P^{-1}$ be given. We claim that $\widetilde {f}(x)=f(b^{-1}x)$ for almost every $x \in bA$. Since $(b_n)$ is cofinal, it follows that for some $n$, $b \leq _r b_n$, i.e. $b_n^{-1}b \in P$. Then, by (3) and (4), for almost every $x \in bA \subset b_nA$,

   \[ \widetilde{f}(x)=f(b_n^{{-}1}x)=f(b_n^{{-}1}bb^{{-}1}x)=f(b^{{-}1}x). \]

Fix $b \in P^{-1}$. For almost all $x \in b_nA$,

\[ \widetilde{f}|_{bb_nA}(bx) = f((bb_n)^{{-}1}bx) = f(b_n^{{-}1}x) = \widetilde{f}(x), \]

for all $n$. But $G = \displaystyle \bigcup _n b_nA$, and hence $\widetilde {f}(bx) = \widetilde {f}(x)$ for almost all $x \in G$. Since $P^{-1}$ generates $G$, it follows that for every $g \in G$, $\widetilde {f}(gx)=\widetilde {f}(x)$ for almost all $x \in G$. But the left translation action of $G$ on $G$ is ergodic. Therefore, there exists a complex number $\lambda$ such that $\widetilde {f}=\lambda$ a.e. By (2), for $x$ in $P$, $\xi _x = f|_{(A \backslash xA)} = \lambda 1_{(A \backslash xA)}$. The proof is now complete.

Proof. Let $b,\,c \in (e,\,a)$ be given. Choose a sequence $(s_{n}) \in \varOmega$ such that $(s_{n})$ converges to $e$. Since $b \in \varOmega$ and $(s_{n}^{-1}b)\to b \in \varOmega$, we have $s_n^{-1}b \in \varOmega$ for large $n$. Similarly $s_n^{-1}c \in \varOmega$ for large $n$. Choose $N$ such that $s_N^{-1}b \in \varOmega$ and $s_{N}^{-1}c \in \varOmega$. Then, $e< s_{N}< b< a$. By Remark 3.1, there exists a monotone path $\sigma :[0,\,1]\longrightarrow P$ joining $e$ and $s_{N}^{-1}b$. Then, $s_{N}\sigma$ is a monotone path in $(e,\,a)$ joining $s_{N}$ and $b$. Similarly, there exists a path in $(e,\,a)$ joining $s_{N}$ and $c$. Thus, we have a path in $(e,\,a)$ joining $b$ and $c$. Therefore, $(e,\,a)$ is path-connected. This proves (1).

Next, to see that the map defined in $(2)$ is well defined, let $b \in (e,\,a) \subset \varOmega$ and $x \in \partial (A) \subset A$ be given. Then, $bx \in Int(A)$. If $bx \in aA$, then $bx=ak$ for some $k \in A$ and $x=b^{-1}ak$. Since $b \in (e,\,a)$, we have $b^{-1}a \in \varOmega$. But $\varOmega A \subset Int(A)$. Therefore, $x \in Int(A)$, which is a contradiction. Hence $bx \in Int (A) \smallsetminus aA$ and our map is well defined which we call as $\psi$.

To prove the surjectivity of $\psi$, let $b \in Int (A) \smallsetminus aA$ be given. Consider the map $\phi : (e,\,a) \longrightarrow G$ defined by $\phi (s)=s^{-1}b$. We claim that $\phi ((e,\,a)) \cap A^{c}$ is non-empty. Thanks to Remark 3.1, there exists a sequence $(s_n)$ in $(e,\,a)$ such that $s_n \to a$. Note that $s_n^{-1}b \to a^{-1}b$, $a^{-1}b \in A^{c}$ and $A^{c}$ are open. Thus, there exists $N \geq 1$ such that $s_{n}^{-1}b \in A^{c}$ for all $n\geq N$. This proves that $\phi ((e,\,a)) \cap A^{c}$ is non-empty.

Let $(t_{n})$ be a sequence in $(e,\,a)$ that converges to $e$. Since $b \in Int(A)$, $t_{k}^{-1}b \in Int (A)$, for large $k$. Hence, $\phi ((e,\,a))\cap Int (A)$ is non-empty. However, $\phi ((e,\,a))$ is connected as $(e,\,a)$ is connected. Therefore, $\phi ((e,\,a)) \cap \partial A$ is non-empty. This implies that there exists $s \in (e,\,a)$ such that $s^{-1}b \in \partial A$ and $\psi (s^{-1}b,\,s)=ss^{-1}b=b$. Consequently, the map $\psi$ is onto. This completes the proof.

Proof. Assume that $(2)$ holds. Suppose $\partial A$ is not compact. Fix $a \in \varOmega$. Choose a compact set $E \subset (e,\,a)$ with non-empty interior. Since $\partial A$ is not compact, there exists a sequence $(x_{n}) \in \partial A$ which has no convergent subsequence. We claim that there exists $k_1 >1$ such that $Ex_{k_1} \cap Ex_1=\emptyset$. Because if not, then $Ex_{n} \cap Ex_{1} \neq \emptyset$, for all $n$. Choose $y_{n} \in Ex_{n} \cap Ex_{1}$. Write $y_{n}=h_{n}x_{n}=g_{n}x_{1}$ for some $h_{n},\, g_{n} \in E$. Then, $x_n x_1^{-1}=h_{n}^{-1}g_n$. But $h_{n}^{-1}g_n \in E^{-1}E$, which is compact. Thus, $(x_{n}x_{1}^{-1})$, and consequently $(x_{n})$, will have a convergent subsequence, which is a contradiction. Hence, there exists $k_{1}$ such that $Ex_{k_{1}} \cap Ex_{1}$ is the null set. Let

\[ I_{k_{1}}= \left\lbrace k \in \mathbb{N} | k \geq k_{1}, Ex_{k} \cap Ex_{k_{1}} = \emptyset, Ex_{k} \cap Ex_{1} = \emptyset \right\rbrace . \]

By a very similar argument, we see that $I_{k_{1}}$ is non-empty. So, there exists $k_{2} >k_1 >1$ such that $Ex_{1} \cap Ex_{k_{2}} = \emptyset$ and $Ex_{k_{1}} \cap Ex_{k_{2}} = \emptyset$. We continue this process to get a subsequence $(x_{n_{k}})$ such that $Ex_{n_{k}} \cap Ex_{n_{m}} = \emptyset$, for all $k \neq m$. Since $G$ is unimodular, $\mu (Ex_{n_k})=\mu (E)$. As verified in Lemma 3.3, $Ex_{n_{k}} \subset Int~A \smallsetminus aA$ for all $k$. Consequently, $Int(A)\backslash aA$ contains the disjoint union $\displaystyle \amalg _{k=1}Ex_{n_k}$ and the latter set has an infinite measure. Therefore, $\mu (Int A \smallsetminus aA) = \infty$. This completes the proof of the implication $(2) \implies (3)$.

Suppose that $(3)$ holds. Set $H:=P \cap P^{-1}$. We claim that $H$ is compact. Since $PA \subset A$, it follows that for $h \in H$, $hA \subset A$ and $h^{-1}A \subset A$. In other words, $hA=A$ for every $h \in H$. Consequently, $h \partial A=\partial A$ for every $h \in H$. Fix $x_0 \in \partial A$. Then, the map

\[ H \ni h \to hx_0 \in \partial A \]

is a topological embedding. Since $\partial A$ is compact, it follows that $H$ is compact. This proves the claim.

Let $\widetilde {G}=G / H$ be the homogeneous space of left cosets of $H$. For $x \in G$, we denote the left coset $xH$ by $\widetilde {x}$. The map $G \ni x \to \widetilde {x} \in \widetilde {G}$ will be denoted by $\pi$. The preorder $\leq$ descends to a closed partial order on $\widetilde {G}$. That is, for $x,\,y\in G$, $\widetilde {x} \leq \widetilde {y}$ if $x \leq y$. It is easily verifiable, using Remark 3.1, that $\widetilde {G}$ has the chain approximation property defined in Page 116, Chapter 4 of [Reference Hilgert and Neeb8]. Making use of Prop. 4.4 of [Reference Hilgert and Neeb8], choose an open set $\widetilde {U}$ containing $\widetilde {e}$ such that for all $\widetilde {a} \in \widetilde {U}$, $[\widetilde {e},\, \widetilde {a}]$ is compact whenever $\widetilde {a} \geq \widetilde {e}$. Let $U:=\pi ^{-1}(\widetilde {U})$. Note that $U$ contains $e$. Since $H$ is compact, it follows that $\pi$ is proper, i.e. the inverse image of a compact set is compact. Therefore, for $a \in U$, $[e,\,a]=\pi ^{-1}([\widetilde {e},\,\widetilde {a}])$ is compact.

Claim: For $a \in U \cap P$, $Int(A)\smallsetminus aA$ has a finite measure.

Let $a \in U \cap P$ be given.

Case 1: Suppose $a \in \varOmega$. Consider the map $\psi :\partial (A) \times [e,\,a] \to vx \in G$ defined by $\psi (x,\,v)=vx$. Then, $\psi$ is a continuous map and has a compact image. Thanks to Lemma 3.3, the image of $\psi$ contains $Int(A)\backslash aA$. Hence, $Int(A) \smallsetminus aA$ has a finite measure.

Case 2: Suppose $a \in \partial P$. Since $e \in U$, there exists a sequence $(s_n)$ in $U \cap \varOmega$ that converges to $e$. Thus, $(as_n)$ converges to $a \in U$. Hence, there exists $N$ for which $as_N \in U$. Clearly $as_N \in \varOmega$. By Case 1, $Int(A)\backslash as_N A$ has a finite measure. Note that

\[ Int(A)\backslash as_{N} A=(Int(A) \backslash aA) \coprod a(A \backslash s_N A). \]

Hence $Int(A)\backslash aA$ has a finite measure.

This proves our claim.

Note that by Lemma II.12 of [Reference Hilgert and Neeb9], $\partial A$ has measure zero. Therefore, for $a \in U \cap P$, $A\backslash aA$ has a finite measure. This is because, up to a set of measure zero, $A \backslash aA=Int(A)\backslash aA$.

Let $a \in \varOmega$ be given. Write $a=expX_{1}expX_{2}\ldots expX_{n}$, for some $X_{1},\,\ldots X_{n} \in L(P)$. Since $U$ is open and contains $e$, there exists a natural number $N$ such that for every $i$, $\exp (\tfrac {X_i}{N}) \in U$. By what we have proved above, $\mu (A\backslash \exp (\tfrac {X_i}{N}))<\infty$.

But $A \backslash \exp (X_i)A$ is the disjoint union of the sets $A \backslash \exp (\tfrac {X_i}{N})A$, $\exp (\tfrac {X_i}{N})(A \backslash \exp (\tfrac {X_i}{N})A), \cdots$ $\exp (\tfrac {(N-1)X_i}{N})(A \backslash \exp (\tfrac {X_i}{N})A)$ each having finite measure. Therefore, for each $i$, $A \backslash \exp (X_i)A$ has a finite measure. Note that $A \backslash aA$ is the disjoint union of the sets $A \backslash \exp (X_1)A$, $\exp (X_1)(A \backslash \exp (X_2)A)$, $\cdots$, $\exp (X_1)\exp (\!X_2\!)\cdots \exp (X_{n-1})(A \backslash \exp (\!X_n\!)A)$. Hence $A \backslash aA$ has a finite measure for every $a \in \varOmega$.

Let $x \in P$ be given. Choose $s \in \varOmega$. Then $A \backslash xA \subset A \backslash xs A$. But $xs \in \varOmega$ and $A \backslash xs A$ have finite measure. Therefore, $A \backslash xA$ has a finite measure for every $x \in P$. This completes the proof of the implication $(3) \implies (2)$.

Proof. This is immediate from Propositions 3.2 and 3.4.

Proof. There is no loss of generality in assuming that $A$ contains $P$. By the Malcev–Iwasawa Theorem, we have $G= K \times \mathbb {R}^{n}$, up to a homeomorphism, for some $n$. Let the boundary of $A$, $\partial A$, be compact. Then $\partial A \subset K \times B[0,\,R]$, for some large $R>0$, where $B[0,\,R]$ is the closed ball of radius $R$ in $\mathbb {R}^{n}$. Now,

\[ G \backslash (K \times B[0,R]) = (Int(A) \backslash (K \times B[0,R])) \cup (A^{c} \backslash (K \times B[0,R])). \]

Observe that $Int(A)$ contains $\varOmega$, which does not have compact closure. Note that $P^{-1}A^{c} \subset A^{c}$. Therefore, $A^{c}$ contains a translate of $\varOmega ^{-1}$ which again does not have compact closure. So, both $Int(A) \backslash (K \times B[0,\,R])$ and $A^{c} \backslash (K \times B[0,\,R])$ are non-empty open sets, which do not intersect. Consequently, $G \backslash (K \times B[0,\,R])$ is not a connected set. This implies that $n=dim(G / K) = 1$. Hence the result.

Proof. Let $x=(x_{1},\,\ldots,\,x_{d})$ in $P_{1}$ and $(y_{1},\,\ldots,\,y_{r})$ in $S(x)$. Since $L(P)$ is a cone, it follows that for $t \geq 0$, $t(x_{1},\,\ldots,\,x_{d},\,y_{1},\,\ldots,\,y_{r}) \in L(P)$. Thus, $t(y_{1},\,\ldots,\,y_{r})$ belongs to $S(tx)$, for all $t\geq 0$, and we get $tS(x)\subset S(tx)$. This proves (1).

To prove (2), let $a=(a_{1},\,\ldots,\,a_{d})\in A_{1}$ and $x=(x_{1},\,\ldots,\,x_{d})\in P_{1}$ be given. Let $(b_{1},\,\ldots,\,b_{r})\in T(a)$ and $(y_{1},\,\ldots,\,y_{r})\in S(x)$ be given. Then, $(a_{1},\,\ldots a_{d},\,e^{2\pi i b_{1}},\,\ldots,\,e^{2\pi i b_{r}}) \in A$ and $(x_{1},\,\ldots,\,x_{d},\,e^{2\pi iy_{1}},\,\ldots,\,e^{2\pi i y_{r}})\in P$. Since $A+P \subset A$, it follows that

\[ (a_{1}+x_{1},\ldots ,a_{d}+x_{d},e^{2\pi i (b_{1}+y_{1})},\ldots ,e^{2\pi i (b_{r}+y_{r})})\in A. \]

Thus, $(b_{1}+y_{1},\,\ldots,\,b_{r}+y_{r})=(b_{1},\,\ldots,\,b_{r})+(y_{1},\,\ldots,\,y_{r}) \in T(a+x)$. Consequently, the inclusion $T(a)+S(x)\subset T(a+x)$ holds.

For proving (3), let $x=(x_{1},\,\ldots,\,x_{d})$ and $y=(y_{1},\,\ldots,\,y_{r})$ be such that $(x;y)\in Int ~L(P)$. Then there exists $R>0$ such that $(x;y) \in B(x,\,R) \times B(y,\,R) \subset Int ~ L(P)$ , in $\mathbb {R}^{d+r}$, where $B(x,\,R)$ and $B(y,\,R)$ are open balls of radius $R$ centred at $x$ and $y$ in $\mathbb {R}^{d}$ and $\mathbb {R}^{r}$ respectively. So, $x \in \pi (B(x,\,R) \times B(y,\,R)) \subset \pi (Int ~ L(P))$. Also, for $z \in B(y,\,R)$, $(x;z) \in B(x,\,R) \times B(y,\,R) \subset L(P)$. Thus, $y \in B(y,\,R) \subset S(x)$, i.e. $y \in Int ~ S(x)$. This proves (3).

Proof. Let $x \in \pi (Int ~ L(P))$. From (3) of Lemma 4.2, we see that $Int ~S(x)$ is non-empty. Hence, $S(x)$ contains a r-dimensional hypercube of side $l$, say. Using (1) from Lemma 4.2, we can thus obtain a $t_{x}>0$ such that $S(tx)$ contains a r-dimensional hypercube of side $1$ in $\mathbb {R}^{r}$, for all $t\geq t_{x}$.

Now, let $a \in A_{1}$ and $t\geq t_{x}$. We can write $T(a)+S(tx)=\bigcup _{b\in T(a)}(b+S(tx))$. Since $S(tx)$ contains a r-dimensional hypercube of side $1$, each set $b+tS(x)$ contains a r-dimensional hypercube of side $1$ in $\mathbb {R}^{r}$. But $T(a)+S(tx)$ is the union of such sets, hence it also contains a r-dimensional hypercube of side $1$. Using (2) from Lemma 4.2, we get that $T(a+tx)$ contains a r-dimensional hypercube of side $1$ in $\mathbb {R}^{r}$. Therefore, under the exponential map,

\[ \exp(\left\lbrace a+tx\right\rbrace \times T(a+tx))=\left\lbrace a+tx\right\rbrace \times \mathbb{T}^{r}, \quad \forall ~ t\geq t_{x}. \]

This completes the proof.

Proof. Let $\partial A$ be compact. Since $G = \mathbb {R}^{d} \times \mathbb {T}^{r}$, the maximal compact subgroup of $G$ is $\mathbb {T}^{r}$. Hence, by Theorem 4.1, $d=1$.

To prove the converse, let us assume $d=1$. Then, $P_{1}=[ 0,\,\infty )$ or $P_{1}=(-\infty,\,0]$. Consider the case when $P_{1}=[0,\, \infty )$. Choose $x_{0}$ from $\pi (Int ~L(P))$. Let $t_0:=t_{x_0}$ be as in Lemma 4.3. Set $x:=t_0x_0$.

Since $A_{1}$ is not entire $\mathbb {R}$ and $A_1$ is a $P_1$-space, $A_{1}=[k,\, \infty )$ for some $k \leq 0$. By Lemma 4.3, for all $a\in A_{1}$, $\exp (\left \lbrace a \right \rbrace \times T(a))= \left \lbrace a \right \rbrace \times \mathbb {T}^{r}$, whenever $a\geq (k+x)$, and we get

\[ [k+x, \infty) \times \mathbb{T}^{r} \subset A \subset [ k, \infty) \times \mathbb{T}^{r} . \]

Thus, $\partial A$ is contained in the compact set $[ k,\, k+x] \times \mathbb {T}^{r}$, and hence is compact. The case $P_{1}=(-\infty,\, 0]$ is similar.

Proof. Without loss of generality, we can assume that $V$ and $W$ are pure. From the Wold decomposition of a $1$-parameter semigroup of isometries and from Arveson's index computation for $1$-parameter CCR flows (Theorem 2.6.4 of [Reference Arveson4]), we can conclude that there exists a unitary $X:\mathcal {H} \to \mathcal {K}$ such that $XV_tX^{*}=W_t$ for every $t \geq 0$.

Define $\widetilde {\theta }_t:E(t) \to F(t)$ by

\[ \widetilde{\theta}_t(e(\xi))=e(X\xi). \]

Then, $\widetilde {\theta }:=\{\widetilde {\theta }_t\}_{t \geq 0}:E \to F$ is an isomorphism.

Define $\theta ^{'}:=\theta \circ \widetilde {\theta }^{-1}$. Then, $\theta ^{'}$ is an automorphism of $F$. By Remark 5.2, there exist an additive cocycle $\eta :=\{\eta _t\}_{t \geq 0}$ of $W$, a unitary $U \in B(\mathcal {K})$ which lies in the commutant of $\{W_t,\,W_{t}^{*}:t \geq 0\}$ and a scalar $\lambda \in \mathbb {R}$ such that

\[ \theta^{'}_{t}(e(\eta))=e^{i\lambda t}W(\eta_t)\varGamma(U)e(\eta)=e^{i \lambda t}e^{-\frac{||\eta_t||^{2}}{2}}e^{-\langle U\eta|\eta_t\rangle}e(\eta_t+U\eta). \]

Let $t \geq 0$ and $\xi \in Ker(V_t^{*})$ be given. Calculate as follows to observe

\begin{align*} \theta_{t}(e(\xi))& =(\theta^{'}_{t} \circ \widetilde{\theta}_{t})(e(\xi))\\ & =\theta^{'}_{t}(e(X\xi))\\ & =e^{i\lambda t}e^{-\frac{||\eta_t||^{2}}{2}}e^{-\langle UX\xi|\eta_t\rangle}e(\eta_t+UX\xi). \end{align*}

Set $c_{t,\xi }=e^{i\lambda t}e^{-\frac {||\eta _t||^{2}}{2}}e^{-\langle UX\xi |\eta _t\rangle }$ and $\widetilde {\xi }_{t}:=\eta _t+UX\xi$. Then, $\theta _t(e(\xi ))=c_{t,\xi }e(\widetilde {\xi }_t)$. This proves the existence part and the uniqueness part are obvious. Hence the proof.

Proof. Let $\alpha$ and $\beta$ be the CCR flows associated with $V$ and $W$ respectively. Assume that $\alpha$ and $\beta$ are cocycle conjugate. Denote the product systems of $\alpha$ and $\beta$ by $\mathcal {E}^{\alpha } = \displaystyle \amalg _{x\in P} E^{\alpha }(x)$ and $\mathcal {E}^{\beta } = \amalg _{x \in P} E^{\beta }(x)$ respectively. Then, there exists a Borel isomorphism $\theta : \mathcal {E}^{\alpha } \longrightarrow \mathcal {E}^{\beta }$ such that $\theta (uv)= \theta (u) \theta (v)$ for $u \in E^{\alpha }(x)$ and $v \in E^{\alpha }(y)$ and

\[ \theta_x : E^{\alpha}(x) \longrightarrow E^{\beta}(x), \quad \theta_x=\theta|_{E^{\alpha}(x)}, \]

is a unitary for all $x\in P$.

We claim that for $a \in \varOmega$ and $\xi \in Ker(V_a^{*})$, there exist a unique non-zero complex number $c_{a,\xi }$ and a unique vector $\widetilde {\xi }_{a} \in Ker(W_a^{*})$ such that

\[ \theta_{a}(e(\xi))=c_{a,\xi}e(\widetilde{\xi}_{a}). \]

Let $X \in L(P)$ be given. Restricting the product systems to the ray $\{\exp (tX): t \geq 0\}$ and applying Lemma 5.3, we see that given $\xi \in Ker(V_{\exp (X)}^{*})$, there exist a non-zero complex number $c_{X,\xi }$ and a vector $\widetilde {\xi }_{X} \in Ker(W_{\exp (X)}^{*})$ such that

\[ \theta_{\exp(X)}e(\xi)=c_{X,\xi}e(\widetilde{\xi}_{X}). \]

Fix $a \in \varOmega$. Since $\varOmega$ is contained in the semigroup generated by $\exp (L(P))$, there exist $X_1,\, X_2,\,\ldots,\, X_n$ in $L(P)$ such that $a=\exp (X_1)\exp (X_2)\ldots \exp (X_n)$. We can expand $Ker(V_a^{*})$ as

\begin{align*} Ker( V_a^{*}) & = Ker(V_{\exp(X_1)}^{*}) \oplus V_{\exp(X_1)}Ker(V_{\exp(X_2)}^{*}) \oplus \cdots \oplus V_{\exp(X_1) \exp(X_2)\ldots \exp(X_{n-1})} \\ & \quad Ker(V_{\exp(X_n)}^{*}). \end{align*}

Thus, for any $\xi \in Ker(V_a^{*})$, there exist $\xi _i \in Ker( V_{\exp (X_i)}^{*})$, $1 \leq i \leq n$, such that

\[ \xi = \xi_1 + V_{\exp(X_1)} \xi_2 +\cdots+V_{\exp(X_1)\cdots\exp(X_{n-1})} \xi_n. \]

Note that $e(\xi )=e(\xi _1)e(\xi _2)\cdots e(\xi _n)$. Since $\theta$ is multiplicative, it follows that

\begin{align*} \theta_a e(\xi) & = \theta_{\exp(X_1)} e(\xi_1) \theta_{\exp(X_2)} e(\xi_2)\ldots\theta_{\exp(X_n)} e(\xi_n) \\ & = c_{X_1, \xi_1}e(\widetilde{\xi}_1) c_{X_2, \xi_2}e(\widetilde{\xi}_2) \ldots c_{X_n, \xi_n}e(\widetilde{\xi}_n) \\ & = c_{X_1, \xi_1} c_{X_2, \xi_2}\cdots c_{X_n, \xi_n}e( \widetilde{\xi}_1 + W_{\exp(X_1)} \widetilde{\xi}_2 + \cdots + W_{\exp(X_1)\cdots\exp(X_{n-1})} \widetilde{\xi}_n) \\ & = c_{a,\xi} e (\widetilde{\xi}_a) \end{align*}

where

\[ \widetilde{\xi}_a:=\widetilde{\xi}_{1}+W_{\exp(X_1)}\widetilde{\xi}_{2}+\cdots +W_{\exp(X_1)\cdots \exp(X_n)}\widetilde{\xi}_n \in Ker(W_{a}^{*}) \]

and $c_{a, \xi } := c_{X_1, \xi _1} c_{X_2, \xi _2}\cdots c_{X_n, \xi _n}$. Uniqueness of $c_{a,\xi }$ and $\widetilde {\xi }_{a}$ is clear. This proves the claim.

Now, for any $a,\, b \in \varOmega$,

\begin{align*} \theta_{ab} e(0) & = \theta_a e(0) \theta_b e(0) \\ & = c_{a, 0} e(\widetilde{0}_a) c_{b, 0} e (\widetilde{0}_b)\\ & = c_{a,0}c_{b,0} e(\widetilde{0}_{a} + W_a \widetilde{0}_b). \end{align*}

Thus, for all $a,\,b \in \varOmega$, $\widetilde {0}_{ab} = \widetilde {0}_a + W_a \widetilde {0}_b$. For $a \in \varOmega$, let $\eta _a := - \widetilde {0}_a$. By Remark 5.2, $\{W(\eta _a)\}_{a \in \varOmega }$ is an automorphism of $\mathcal {E}^{\beta }$. Note that $\left \lbrace W(\eta _a) \theta _a e(0) = c_{a, 0} e^{\frac {||\eta _a||^{2}}{2}} e(0)\right \rbrace _{a \in \varOmega }$ is a unit of $\beta$. By Proposition 2.3, there exists $\chi \in Hom(G,\,\mathbb {C}^{\times })$ such that

\[ \chi(a) W(\eta_a) \theta_a e(0) = e(0). \]

Taking norm in the above equality, we observe that $\chi \in Hom(G,\,\mathbb {T})$. Observe that $\{\chi (a)W(\eta _a)\}_{a \in \varOmega }$ is an automorphism of $\mathcal {E}^{\beta }$.

For $\xi \in Ker(V_{a}^{*})$, note that

\begin{align*} 1 & = \langle e(\xi) | e(0) \rangle \\ & = \langle \chi(a) W(\eta_a) \theta_a e(\xi) | \chi(a) W(\eta_a) \theta_a e(0) \rangle \\ & = \langle \chi(a) c_{a, \xi} e^{ - \frac{|| \eta_a||^{2}}{2} - \langle \widetilde{\xi}_a | \eta_a \rangle} e(\widetilde{\xi}_a + \eta_a) | e(0) \rangle \\ & = \chi(a) c_{a, \xi} e^{ - \frac{|| \eta_a||^{2}}{2} - \langle \widetilde{\xi}_a | \eta_a \rangle} \langle e(\widetilde{\xi}_a + \eta_a) | e(0) \rangle \\ & = \chi(a) c_{a, \xi} e^{ - \frac{|| \eta_a||^{2}}{2} - \langle \widetilde{\xi}_a | \eta_a \rangle}. \end{align*}

Thus, $\chi (a) W(\eta _a) \theta _a e(\xi ) = e(\widetilde {\xi }_a + \eta _a).$

Suppose $\xi \in Ker(V_a^{*}) \cap Ker(V_b^{*})$ with $a,\,b \in \varOmega$. Choose $x \in \varOmega$ such that $a < x$ and $b < x$. This is possible since $a\varOmega \cap b\varOmega$ is non-empty. Then, $\xi \in Ker(V_a^{*}) \subset Ker(V_x^{*})$, and

\begin{align*} e(\eta_x + \widetilde{\xi}_x) & = \chi(x)W(\eta_x) \theta_x e(\xi) \\ & = \chi(a)W(\eta_a)\theta_a e(\xi) \chi(a^{{-}1}x)W(\eta_{a^{{-}1}x}) \theta_{a^{{-}1}x} e(0) \\ & = e(\eta_a + \widetilde{\xi}_a)e(0) = e(\eta_a + \widetilde{\xi}_a). \end{align*}

Hence, $\eta _x + \widetilde {\xi }_x = \eta _a + \widetilde {\xi }_a$. Similarly, $\eta _x + \widetilde {\xi }_x = \eta _b + \widetilde {\xi }_b$. Hence, $\eta _a + \widetilde {\xi }_a = \eta _b + \widetilde {\xi }_b$, whenever $\xi \in Ker(V_a^{*}) \cap Ker(V_b^{*})$.

Let $D=\bigcup _{a \in \varOmega } Ker(V_a^{*})$. Since $V$ is pure, $D$ is dense in $\mathcal {H}$. Define $U: D \longrightarrow \mathcal {K}$ by

\[ U (\xi) := \eta_a + \widetilde{\xi}_a \]

for $\xi \in Ker(V_a^{*})$. This map is well defined, as proved earlier. For $a \in \varOmega$, the restriction $U|_{Ker(V_a^{*})}$ is continuous.

Fix $\xi,\, \zeta$ in $D$. Choose $a \in \varOmega$ such that $\xi,\,\zeta \in Ker(V_a^{*})$. Calculate as follows to observe that