Proof From [Reference Arveson8, Corollary 6.8.3], $D_{\Omega }^{E}(t)$ is total in $E(t)$ , and $D_{\Omega }^{F}(t)$ is total in $F(t)$ for $t\geq 0$ . Also, by Proposition 2.1,

$$\begin{align*}\langle Exp(\eta)|Exp(\xi)\rangle=e^{\langle \eta|\xi \rangle}=\langle e(\eta)|e(\xi)\rangle\end{align*}$$

for $\eta ,\xi \in Ker(V_{t}^{*})$ for $t\geq 0$ . Hence, $\Psi _{t}$ can be extended to a unitary operator, again denoted $\Psi _t$ , $\Psi _{t}:E(t)\to F(t)$ , for $t\geq 0$ . It follows from (2.4) and (3.1) that $\Psi :=\{\Psi _t\}_{t \geq 0}:E \to F$ is an isomorphism. Hence the proof.

Proof We denote by $F_{1}$ and $F_{2}$ the product systems of the CCR flows associated with $V^{(1)}$ and $V^{(2)}$ respectively. For $i=1,2$ , let $\Theta _{i}=\{\Theta _{i}(t)\}_{t \geq 0}$ be the isomorphism from $E_{i}$ onto $F_{i}$ given by

$$\begin{align*}\Theta_{i}(t)Exp(\eta)=e(\eta)\end{align*}$$

for $\eta \in Ker(V^{(i)*}_{t})$ and $t\geq 0$ . The isomorphism $\Theta _i$ for $i=1,2$ is guaranteed by Lemma 3.1. Then, $\Delta =\Theta _{2}\circ \Psi \circ \Theta _{1}^{-1}$ is an isomorphism from $F_{1}$ onto $F_{2}$ .

It follows from [Reference Arveson8, Corollary 2.6.10] that $V^{(1)}$ and $V^{(2)}$ are unitarily equivalent. Suppose $W:\mathcal {H}\to \mathcal {K}$ is a unitary such that $WV^{(1)}_{t}=V^{(2)}_{t}W$ for $t\geq 0$ . For $t\geq 0$ , let $\Lambda _t:F_1(t) \to F_2(t)$ be the unitary operator such that

$$\begin{align*}\Lambda_{t} exp(\xi)=exp(W\xi)\end{align*}$$

for $\xi \in Ker(V^{(1)*}_t)$ . Clearly, $\Lambda :=\{\Lambda _{t}\}_{t\geq 0}:F_{1}\to F_{2}$ is an isomorphism.

Now consider the map $T:=\Lambda ^{-1}\circ \Delta $ . The map T is an automorphism of $F_{1}$ . By [Reference Arveson8, Theorem 3.8.4], there exists $U \in \mathscr {U}(V^{(1)},V^{(1)})$ , an additive cocycle $\xi =\{\xi _{t}\}_{t\geq 0}$ for $V^{(1)}$ and $\lambda \in \mathbb {R}$ such that

$$\begin{align*}T_{t} e(\eta)=e^{i\lambda t}e^{-\frac{||\xi_{t}||^{2}}{2}-\langle U\eta|\xi_{t}\rangle}e(U\eta+\xi_{t})\end{align*}$$

for $\eta \in Ker(V_t^{(1)*})$ . For $t \geq 0$ , set $\widetilde {\xi }_t:=W\xi _t$ and $\widetilde {U}:=WU$ . Then, $\widetilde {U} \in \mathscr {U}(V^{(1)},V^{(2)})$ and $\widetilde {\xi }:=\{\widetilde {\xi }_t\}_{t\geq 0}$ is an additive cocycle for $V^{(2)}$ .

Since $\Delta :=\Lambda \circ T$ , it follows that for for $t\geq 0$ and $\eta \in Ker(V^{(1)*}_{t})$ ,

$$ \begin{align*} \Delta_{t}e(\eta)&=e^{i\lambda t}e^{-\frac{||\xi_{t}||^{2}}{2}-\langle U\eta|\xi_{t}\rangle}e(WU\eta+W\xi_{t})\\ &=e^{i\lambda t}e^{-\frac{||W\xi_{t}||^{2}}{2}-\langle WU\eta|W\xi_{t}\rangle}e(WU\eta+\widetilde{\xi}_{t})\\ &=e^{i \lambda t}e^{-\frac{||\widetilde{\xi}_t||^{2}}{2}-\langle \widetilde{U}\eta|\widetilde{\xi}_t\rangle}e(\widetilde{U}\eta+\widetilde{\xi}_t). \end{align*} $$

Since $\Psi =\Theta _2^{-1} \circ \Delta \circ \Theta _1$ , it is immediate that

$$\begin{align*}\Psi_{t} Exp(\eta)=e^{i\lambda t}e^{\frac{-||\widetilde{\xi}_{t}||^{2}}{2}-\langle \widetilde{U}\eta|\widetilde{\xi}_{t}\rangle }Exp(\widetilde{U}\eta+\widetilde{\xi}_{t})\end{align*}$$

for $\eta \in Ker(V^{(1)*}_{t})$ and $t\geq 0$ . Hence the proof.

Proof Suppose $E^{(1)}$ and $E^{(2)}$ are isomorphic. Let $\Psi =\{\Psi _{a}\}_{a\in P}:E^{(1)}\to E^{(2)}$ be an isomorphism. Let $a\in Int(P)$ . Consider the one parameter product system $E_{a}^{(1)}:= \{E_{a}^{(1)} (t)\}_{t \geq 0}$ , where for $t \geq 0$ , $E_{a}^{(1)}(t):=E^{(1)}(ta)$ . Similarly, consider the one parameter product system $E_{a}^{(2)}:= \{E_{a}^{(2)}(t)\}_{t \geq 0}$ , where $E_{a}^{(2)}(t):=E^{(2)}(ta)$ for $t\geq 0$ . Then, $\{\Psi _{ta}\}_{t \geq 0}:E_a^{(1)} \to E_a^{(2)}$ is an isomorphism.

By Lemma 3.2, there exists a unitary $U_{a}:\mathcal {H}_1\to \mathcal {H}_2$ intertwining $\{V^{(1)}_{ta}\}_{t\geq 0}$ and $\{V^{(2)}_{ta}\}_{t\geq 0}$ , an additive cocycle $ \{\xi ^{a}_{t}\}_{t\geq 0}$ of $\{V^{(2)}_{ta}\}_{t\geq 0}$ and $\lambda _{a}\in \mathbb {R}$ such that $\Psi _{ta}$ is of the form

(3.2) $$ \begin{align}\Psi_{ta}Exp_{a}(\xi)=e^{i\lambda_{a}t} e^{-\frac{||\xi^{a}_{t}||^{2}}{2}}e^{-\langle U_{a}\xi|\xi^{a}_{t}\rangle}Exp_{a}(U_{a}\xi+\xi^{a}_{t}), \end{align} $$

for $\xi \in Ker(V^{(1)*}_{ta}) $ and $t>0$ . We will denote $\xi ^{a}_{1}$ by $\xi _{a}$ . Hence,

(3.3) $$ \begin{align} \Psi_{a}Exp_{a}(\xi)=e^{i\lambda_{a}} e^{-\frac{||\xi_s||^{2}}{2}}e^{-\langle U_{a}\xi|\xi_a\rangle}Exp_{a}(U_{a}\xi+\xi_a) \end{align} $$

for $a \in Int(P)$ and $\xi \in Ker(V^{(1)*}_a)$ .

Let $a, b\in Int(P)$ . Since $\Psi $ is an isomorphism, we have

$$ \begin{align*}\Psi_{a}Exp_{a}(\xi). \Psi_{b}Exp_{b}(\eta)=\Psi_{a+b}(Exp_{a}(\xi).Exp_{b}(\eta)), \end{align*} $$

for $\xi \in Ker(V^{(1)*}_{a}) $ and $\eta \in Ker(V^{(1)*}_{b}) $ .

In particular,

$$ \begin{align*}\Psi_{a+b}(Exp_{a}(0).Exp_{b}(0))=\Psi_{a}Exp_{a}(0).\Psi_{b}Exp_{b}(0), \end{align*} $$

i.e.,

$$\begin{align*}\Psi_{a+b}Exp_{a+b}(0)=e^{i(\lambda_a+\lambda_b)}e^{-\frac{(||\xi_{a}||^{2}+||\xi_{b}||^{2})}{2}}Exp_{a}(\xi_{a}).Exp_{b}(\xi_{b})\end{align*}$$

(3.4) $$ \begin{align} e^{i\lambda_{a+b}}e^{-\frac{||\xi_{a+b}||^{2}}{2}}Exp_{a+b}(\xi_{a+b})=e^{i(\lambda_{a}+\lambda_{b})}e^{-\frac{(||\xi_{a}||^{2}+||\xi_{b}||^{2})}{2}}Exp_{a}(\xi_{a}).Exp_{b}(\xi_{b}). \end{align} $$

Comparing the $0$ -particle vectors in LHS and RHS of (3.4), we have

(3.5) $$ \begin{align} e^{i\lambda_{a+b}-\frac{||\xi_{a+b}||^{2}}{2}}= e^{i(\lambda_{a}+\lambda_{b})-\frac{(||\xi_{a}||^{2}+||\xi_{b}||^{2})}{2}}. \end{align} $$

Now, comparing the $1$ -particle vectors of LHS and RHS in (3.4), we have

(3.6) $$ \begin{align} \xi_{a+b}=\xi_{a}+V^{(2)}_{a}\xi_{b}. \end{align} $$

Let $\xi \in Ker(V^{(1)*}_{a})$ . Consider the equation

(3.7) $$ \begin{align}\Psi_{a+b}\xi =\Psi_{a}\xi. \Psi_{b}Exp_{b}(0). \end{align} $$

Considering the exponential map in the one parameter product system $E^{(1)}_a$ , we have

$$\begin{align*}Exp_{a}(\xi)=\sum_{n=0}^{\infty}x_{1}^{(n)}\end{align*}$$

where for $r \in (0,1]$ , $x_r^{(0)}=\Omega _{ra}$ , $x_r^{(1)}= (1-V_{ra}^{(1)}V_{ra}^{(1)*})\xi $ and $x_r^{(n)}=\int _{0}^{r}x_t^{(n-1)}d\xi _{t}$ . Here, for $t \in (0,1]$ , $\xi _{t}=(1-V^{(1)}_{ta}V^{(1)*}_{ta})\xi $ .

Let $s \in \mathbb {R}$ . Suppose

$$\begin{align*}Exp_{a}(s\xi)=\sum_{n=0}^{\infty}y_1^{(n)},\end{align*}$$

where the summands are obtained via Itô integration. A moment’s reflection on the definition of the Itô integral reveals that

$$\begin{align*}y_{1}^{(n)}=s^{n}x_1^{(n)}\end{align*}$$

for $n=0,1,2,\cdots $ .

Note that

(3.8) $$ \begin{align}\Psi_a Exp_{a}(t\xi)=\sum_{n=0}^{\infty}\Psi_a y_1^{(n)}=\sum_{n=0}^{\infty}t^{n}\Psi_a x_1^{(n)}\end{align} $$

for $t \in \mathbb {R}$ . Since the power series in (3.8) is norm convergent for every $t \in \mathbb {R}$ , it may be differentiated term by term, and the derivative is given by

$$\begin{align*}\frac{d}{dt}\Psi_a Exp_{a}(t\xi)=\sum_{n=1}^{\infty}nt^{n-1}\Psi_a( x_1^{(n)}).\end{align*}$$

Hence,

(3.9) $$ \begin{align} \Psi_{a}(\xi)=\frac{d}{dt}\Big|_{t=0}\Psi_{a}Exp_{a}(t\xi). \end{align} $$

From (3.3) we have,

$$ \begin{align*} & \frac{d}{dt}\Psi_{a}Exp_{a}(t\xi) \\ &\quad =\frac{d}{dt} e^{i\lambda_{a}}e^{-\frac{||\xi_{a}||^{2}}{2}}e^{-t\langle U_{a}\xi|\xi_{a}\rangle}Exp_{a}(tU_{a}\xi+\xi_{a})\\ &\quad =e^{i\lambda_{a}}e^{-\frac{||\xi_{a}||^{2}}{2}}\Big(Exp_{a}(tU_{a}\xi+\xi_{a})\frac{d}{dt}e^{-t\langle U_{a}\xi|\xi_{a}\rangle}+ e^{-t\langle U_{a}\xi|\xi_{a}\rangle}\frac{d}{dt}Exp_{a}(tU_{a}\xi+\xi_{a})\Big)\\ &\quad =e^{i\lambda_{a}-\frac{||\xi_{a}||^{2}}{2}}e^{-t\langle U_{a}\xi|\xi_{a}\rangle}\Big(-\langle U_{a}\xi|\xi_{a}\rangle Exp_{a}(tU_{a}\xi+\xi_{a})+U_{a}\xi+q_{a}(t,\xi)\Big) \end{align*} $$

where the projection of $q_{a}(t,\xi )$ onto $0$ -particle and $1$ -particle space is zero. By (3.9),

$$\begin{align*}\Psi_{a}(\xi)=\frac{d}{dt}\Big|_{t=0}\Psi_{a}Exp_{a}(t\xi)=e^{i\lambda_{a}-\frac{||\xi_{a}||^{2}}{2}}(-\langle U_{a}\xi|\xi_{a}\rangle Exp_{a}(\xi_{a})+U_{a}\xi+q_{a}(0,\xi)).\end{align*}$$

Similarly,

$$ \begin{align*}\Psi_{a+b}(\xi)=e^{i\lambda_{a+b}-\frac{||\xi_{a+b}||^{2}}{2}}(-\langle U_{a+b}\xi|\xi_{a+b}\rangle Exp_{a+b}(\xi_{a+b})+U_{a+b}\xi+q_{a+b}(0,\xi)) \end{align*} $$

where the projection of $q_{a+b}(0,\xi )$ onto $0$ -particle and $1$ -particle space is zero.

Therefore, (3.7) implies,

$$ \begin{align*} &e^{i(\lambda_{a}+\lambda_{b})-\frac{(||\xi_{a}||^{2}+||\xi_{b}||^{2})}{2}}\Big(\!\!-\langle U_{a}\xi|\xi_{a}\rangle Exp_{a}(\xi_{a})+U_{a}\xi+q_{a}(0,\xi)\Big).Exp_{b}(\xi_{b})\\ & \quad =e^{i\lambda_{a+b}-\frac{||\xi_{a+b}||^{2}}{2}}\Big(\!\!-\langle U_{a+b}\xi|\xi_{a+b}\rangle Exp_{a+b}(\xi_{a+b})+U_{a+b}\xi+q_{a+b}(0,\xi)\Big). \end{align*} $$

Using (3.5), the above equation reduces to

(3.10) $$ \begin{align} &\big(\!\!-\langle U_{a}\xi|\xi_{a}\rangle Exp_{a}(\xi_{a})+U_{a}\xi+q_{a}(0,\xi)\Big).Exp_{b}(\xi_{b}) \nonumber \\ &\quad =-\langle U_{a+b}\xi|\xi_{a+b}\rangle Exp_{a+b}(\xi_{a+b})+U_{a+b}\xi+q_{a+b}(0,\xi). \end{align} $$

By equating the $0$ -particle vectors in the above equation, we see that

(3.11) $$ \begin{align}\langle U_{a}\xi|\xi_{a}\rangle=\langle U_{a+b}\xi|\xi_{a+b}\rangle. \end{align} $$

Equating the $1$ -particle vectors in (3.10), we have

$$ \begin{align*} -\langle U_{a}\xi|\xi_{a}\rangle (\xi_a+V^{(2)}_a\xi_b)+U_{a}\xi=-\langle U_{a+b}\xi|\xi_{a+b}\rangle\xi_{a+b}+U_{a+b}\xi. \end{align*} $$

Hence, by (3.6) and (3.11),

$$ \begin{align*}U_{a}\xi=U_{a+b}\xi. \end{align*} $$

Thus, if $\xi \in Ker( V^{(1)*}_{a})$ and $\xi \in Ker( V^{(1)*}_{b})$ for $a,b\in Int(P)$ , $U_{a}\xi =U_{a+b}\xi =U_{b}\xi $ . Since $ \bigcup _{a \in Int(P)}Ker(V_a^{(1)*})$ is dense in $\mathcal {H}_1$ , it follows that there exists a unitary operator $U: \mathcal {H}_1\to \mathcal {H}_2$ such that

$$\begin{align*}U\xi=U_a\xi \end{align*}$$

if $\xi \in Ker(V^{(1)*}_{a})$ .

Similarly, considering the equation

$$ \begin{align*}\Psi_{a+b}(Exp_{a}(0).\eta)=\Psi_{a}(Exp_{a}(0)).\Psi_{b}\eta \end{align*} $$

for $\eta \in Ker(V^{(1)*}_{b})$ and comparing $0$ and $1$ -particle vectors as before, it is not difficult to see that for $a,b \in Int(P)$ and $\eta \in Ker(V_b^{(1)*})$

$$ \begin{align*} UV^{(1)}_{a}\eta=V^{(2)}_{a}U\eta. \end{align*} $$

Since $\displaystyle \bigcup _{b \in Int(P)}Ker(V_b^{(1)*})$ is dense in $\mathcal {H}_1$ , it follows that $UV^{(1)}_a=V^{(2)}_aU$ for every $a \in Int(P)$ . As $Int(P)$ is dense in P, U intertwines $V^{(1)}$ and $V^{(2)}$ . This completes the proof.

Proof We first show that $V^{\mu }$ is unitarily equivalent to $S^{\lambda }$ . The reader may verify that the map $\mathcal {F}: Y^{\lambda }_{u}\to \mathbb {R}$ defined by

$$\begin{align*}\mathcal{F}(H^{\lambda}+t\lambda)=t\end{align*}$$

is a homeomorphism, and $\mathcal {F}(X_u^{\lambda })=[0,\infty )$ . Also,

$$\begin{align*}\mathcal{F}((H_\lambda+t\lambda)+a)=t+\langle \lambda|a \rangle\end{align*}$$

for $t \in \mathbb {R}$ and $a \in \mathbb {R}^d$ . Since $\mu $ is an invariant measure, the push forward measure $\mathcal {F}_{*}\mu $ is the Lebesgue measure, denoted by m. Define a unitary operator $U: L^{2}([0,\infty ),m)\to L^{2}(X_{u}^{\lambda },\mu )$ by

$$\begin{align*}Uf=f \circ \mathcal{F}.\end{align*}$$

It is routine to verify that U intertwines $S^{\lambda }$ and $V^{\mu }$ . Note that $U^{-1}(1_{X_{u}\setminus X_{u}+a})=1_{(0,\langle \lambda |a\rangle )}$ for $a\in P$ . By abusing notation, we may assume that $V^{\mu }=S^{\lambda }$ and $\eta _{a}=1_{(0,\langle \lambda |a\rangle )}$ for $a\in P$ .

Let $Exp$ denote the exponential map of the one parameter product system $\widetilde {E} $ of the CAR flow associated with $\{S_{t}\}_{t\geq 0}$ . Let $u_{t}=Exp(1_{(0,t)})$ for $t\geq 0$ . It follows from Remark 2.2 and Propostion 2.1 that $u=\{u_t\}_{t \geq 0}$ is a unit for $\widetilde {E}$ . Note that $Exp_{a}(\eta _{a})=u_{\langle \lambda |a\rangle }$ for $a\in P$ . Observe that

$$\begin{align*}Exp_{a}(\eta_{a}).Exp_{b}(\eta_{b})=u_{\langle\lambda|a\rangle}.u_{\langle\lambda|b\rangle}=u_{\langle\lambda|a+b\rangle}=Exp_{a+b}(\eta_{a+b}).\end{align*}$$

for $a, b\in P$ , i.e., $Exp(\eta )$ is a unit.

Proof Uniqueness is clear. Without loss of generality, we assume $i=d$ and write $Q_{d}= Q$ . For $x \in Q$ , let

$$\begin{align*}A_x:=\{t \in \mathbb{R}: x+tv_d \in A\}.\end{align*}$$

Since $\mathbb {R}^d \cong Q \times \mathbb {R}$ , $A=\{x+tv_d: x \in Q, t \in A_x\}$ . Since A is closed, $A_x$ is closed for each $x \in Q$ .

Claim: For each $x \in Q$ , $A_x \neq \emptyset $ and $A_x \neq \mathbb {R}$ .

Note that since $-P+A \subset A$ , $-Int(P)+A \subset Int(A)$ and $-P+Int(A) \subset Int(A)$ . Let $B= \mathbb {R}^{d}\setminus Int(A)$ . Then, B is a proper closed subset of $\mathbb {R}^{d}$ such that $P+B\subset B$ . Also, $Int(B)=\mathbb {R}^{d}\setminus A$ . It is clear that $\mathbb {R}^d \setminus A \subset Int(B)$ as A is closed. If $z \in \partial A$ , then every neighbourhood of z intersects $Int(A)$ . This is because $-Int(P)+A \subset Int(A)$ which in turn implies that $-tv_d+z \in Int(A)$ as $t \to 0+$ . Thus, $Int(B)=\mathbb {R}^d\backslash A$ .

Let $x \in Q$ be given. For $y\in \mathbb {R}^{d}$ , there exists $n_{0} \in \mathbb {N}$ such that $y-n_{0}v_{d}\in -Int(P)$ . For, $-Int(P)$ is an open cone, $v_d \in Int(P)$ and $\frac {y}{n}-v_{d}\in -Int(P)$ for large $n\in \mathbb {N}$ . Fix $z_0\in A$ . Since $-Int(P) +z_0 \subset Int(A)$ , given $y\in \mathbb {R}^{d}$ , there exists a natural number $n_{0}\in \mathbb {N}$ such that $y-n_{0}v_{d}\in -Int(P) \subset Int(A)-z_0$ . Setting $y=x-z_0$ , we see that there exists $n_{0}\in \mathbb {N}$ such that $x-n_{0}v_{d}\in Int(A)$ . Replacing A by B, P by $-P$ and $v_{d}$ by $-v_{d}$ and arguing similarly, we see that there exists $n_{1}\in \mathbb {N}$ such that $x+n_{1}v_{d}\in Int(B)= \mathbb {R}^{d}\setminus A$ . In particular, there exists $m,n\in \mathbb {Z}$ such that $m\in A_{x}$ and $n\in \mathbb {R}\setminus A_{x}$ . This proves the claim.

As $-P+A \subset A$ , note that if $x+tv_{d}\in A$ for $t\in \mathbb {R}$ , then $x+sv_{d}\in A$ for every $s<t$ . Thus, for each $x \in Q$ , $A_x$ is a closed interval which is not bounded below. Since $A_x \neq \mathbb {R}$ , $A_x=(-\infty ,f(x)]$ , where

$$\begin{align*}f(x)=\sup A_x=\sup\{t \in \mathbb{R}: x+tv_d \in A\} <\infty.\end{align*}$$

Hence, $A=A_f$ .

Next, we prove that f is continuous. Let $x\in Q$ and let $\epsilon>0$ be given. Let $s_{0},s_{1}\in \mathbb {R}$ be such that $f(x)>s_{0}>f(x)-\epsilon $ and $f(x)+\epsilon>s_{1}>f(x)$ . Since $x+f(x)v_d \in A$ and $-Int(P)+A \subset Int(A)$ , $x+s_0v_d \in Int(A)$ . Also, $x+s_{1}v_{d}\in \mathbb {R}^d \setminus A$ . Let $\delta>0$ be such that the open ball $B(x+s_{0}v_{d},\delta )\subset Int(A)$ and $B(x+s_{1}v_{d},\delta )\subset \mathbb {R}^d \setminus A$ , respectively. Let $y\in Q$ be such that $|x-y|<\delta $ . Since $|(x+s_{0}v_{d})-(y+s_{0}v_{d})|=|x-y|<\delta $ , $y+s_{0}v_{d}\in A$ . Hence,

$$\begin{align*}f(y)\geq s_{0}> f(x)-\epsilon\end{align*}$$

Similarly, since $|(x+s_{1}v_{d})-(y+s_{1}v_{d})|=|x-y|<\delta $ , $y+s_{1}v_{d}\in \mathbb {R}^d\setminus A $ , i.e.,

$$\begin{align*}f(y)< s_{1}<f(x)+\epsilon.\end{align*}$$

This implies $|f(x)-f(y)|<\epsilon $ . Hence, f is continuous. This completes the proof.

Proof Fix $i\in \{1,2,\ldots ,d\}$ . Suppose $(A_{n})$ is a sequence in $Y_{u}$ converging to $A\in Y_{u}$ . For simplicity, we denote $f_i^{A_{n}}$ and $f_i^{{A}}$ by $f_{n}$ and f respectively. From the proof of [Reference Hilgert and Neeb14, Proposition II.13], $1_{A_{n}}(a)\to 1_{A}(a)$ pointwise for every $a \in \mathbb {R}^{d}\setminus \partial A$ , where $\partial A$ is the boundary of A. Recall that

$$\begin{align*}A=\left\{\sum_{i=1}^{d}x_{j}v_{j}\in \mathbb{R}^{d}: { x_{i}\leq f\left(\sum_{k\neq i}x_{k}v_{k}\right)}\right\}.\end{align*}$$

Let $x\in Q_{i}$ and $t>0$ . Let $y=x+(f(x)-t)v_{i}$ . Since $y\in Int(A)$ , $y\in A_{n}$ eventually, i.e., for $n\in \mathbb {N}$ sufficiently large, there exists $s_{n} \geq 0$ and $x_n \in Q_i$ such that $y=x_{n}+(f_{n}(x)-s_{n})v_{i}$ . Now,

$$\begin{align*}f(x)-t=f_{n}(x)-s_{n}.\end{align*}$$

for sufficiently large $n\in \mathbb {N}$ . In particular,

$$\begin{align*}f_{n}(x)-f(x)\geq - t\end{align*}$$

for large $n\in \mathbb {N}$ .

Let $z= x+(f(x)+t)v_{i}$ . Then, $z\in \mathbb {R}^{d}\setminus A$ . Therefore, $z\in \mathbb {R}^{d}\setminus A_{n}$ eventually, i.e., for sufficiently large $n\in \mathbb {N}$ , there exists $t_{n}>0$ and $z_n \in Q_i$ such that $z=z_n+(f_{n}(x)+t_{n})v_{i}$ . Now,

$$\begin{align*}f(x)+t=f_{n}(x)+t_{n}\end{align*}$$

for large $n\in \mathbb {N}$ . Therefore, for large n,

(4.1) $$ \begin{align} f_{n}(x)-f(x)\leq t. \end{align} $$

Since $t>0$ is arbitrary, we conclude that $f_{n}(x)\to f(x)$ . This completes the proof.

Proof Fix $i\in \{1,2,\ldots , d\}$ . Let $\nu =(\Psi _{i}^{-1})_{*}\mu $ . Since $\mu $ is invariant, it follows from (4.2) that $\nu $ is invariant under the action of $\mathbb {R}$ on $Y^{i}_{u} \times \mathbb {R}$ given by $s.(A,t)=(A,s+t)$ . Hence, $\nu $ is a product measure of the form $\nu _0 \times m$ where m is the Lebesgue measure on $\mathbb {R}$ . Now, $\mu (Y^{i}_{u})=\nu (Y^{i}_{u}\times \{0\})=0$ . This proves $(i)$ .

Fix $B\in Y_{u}$ , and let $t:=f^{B}_{i}(0)$ . Then, $1_{\mathcal {N}}(A,B)=1$ if and only if $f^{A}_{i}(0)=f^{B}_{i}(0)=t$ . But $\{A \in Y_u: {f_i^{A}(0)=t}\}=Y^{i}_{u}+tv_{i}$ is a null set. Now $(ii)$ follows from Fubini’s theorem.

Proof The computation is similar to the computation done in [Reference Arjunan1, Proposition 5]. Fix $t>0$ and $i\in \{1,2,\ldots , d\}$ . For simplicity, write $\eta _{sv_{i}}=\eta _{s}$ for $s\geq 0$ . Let $x_{s}^{(1)}=\eta _{s}$ .

By definition,

(4.3) $$ \begin{align} P_{2}\Big(Exp_{i}(\eta_{tv_{i}})\Big)&=\int_{0}^{t}x_{s}^{(1)}d\eta_{s} \hspace{6pt}\qquad\qquad\qquad\qquad \qquad\qquad \end{align} $$

(4.4) $$ \begin{align} &=\lim_{n\to\infty} \sum_{j=0}^{n-1}\eta_{\frac{jt}{n}}.\eta_{\frac{t}{n}}\nonumber\\ &=\lim_{n\to\infty} \sum_{j=0}^{n-1}V_{\frac{jtv_{i}}{n}}\eta_{\frac{t}{n}}\wedge \eta_{\frac{jt}{n}}\nonumber\\ &=\lim_{n\to\infty} \sum_{j=0}^{n-1}(\eta_{\frac{(j+1)t}{n}}-\eta_{\frac{jt}{n}})\wedge \eta_{\frac{jt}{n}}\nonumber\\ &=\lim_{n\to\infty} \sum_{j=0}^{n-1}\eta_{\frac{(j+1)t}{n}}\wedge \eta_{\frac{jt}{n}}.\end{align} $$

Let $s_{n}= \displaystyle \sum _{j=0}^{n-1}\eta _{\frac {(j+1)t}{n}}\wedge \eta _{\frac {jt}{n}}$ for $n\in \mathbb {N}$ . The proof will be over if we show that

$$\begin{align*}\lim_{n\to\infty }s_{n}(A,B)=\frac{1}{\sqrt{2}}\epsilon_i(A,B)\eta_{tv_i}(A)\eta_{tv_i}(B)\end{align*}$$

for almost every $(A,B)\in X_{u}\setminus (X_u+tv_i) \times X_u\setminus (X_u+tv_i)$ .

For $n\in \mathbb {N}$ , suppose $A,B\in X_{u}$ are such that $f^{A}_{i}(0)<\frac {jt}{n}$ , $\frac {jt}{n} \leq f^{B}_{i}(0)< \frac {(j+1)t}{m}$ for some $j\in \{1,2,\ldots ,n-1\}$ . Then,

$$ \begin{align*}s_{n}(A,B) = 1_{X_{u}\setminus x_{u}+(\frac{(j+1)tv_{i}}{n})}\wedge 1_{X_{u}\setminus x_{u}+(\frac{jtv_{i}}{n})}(A,B). \end{align*} $$

Since $A \in X_u\setminus (X_u+\frac {jtv_i}{n})$ and $B \in (X_u+\frac {jtv_i}{n})\setminus (X_u+\frac {(j+1)tv_i}{n})$ ,

(4.5) $$ \begin{align} s_n(A,B)=1_{X_{u}\setminus x_{u}+(\frac{(j+1)tv_{i}}{n})}\wedge 1_{X_{u}\setminus x_{u}+(\frac{jtv_{i}}{n})}(A,B)=-\frac{1}{\sqrt{2}}. \end{align} $$

Let $A,B\in X_{u}\setminus X_{u}+tv_{i}$ . Suppose $f^{A}_{i}(0)<f^{B}_{i}(0)$ . For sufficiently large $n\in \mathbb {N}$ , there exists a unique $j_{n}\in \{1,2,\ldots , n-1\}$ such that $f^{A}_{i}(0)< \frac {j_{n}t}{n}$ , $\frac {j_{n}t}{n}\leq f^{B}_{i}(0)< \frac {(j_{n}+1)t}{n}$ . By (4.5), for sufficiently large n,

$$ \begin{align*} s_n(A,B)&=-\frac{1}{\sqrt{2}} \\ &=\frac{1}{\sqrt{2}}\Big(\epsilon_{i}(A,B)\eta_{tv_{i}}(A) \eta_{tv_{i}}(B)\Big). \end{align*} $$

Similarly, it may be proved that, when $f^A_i(0)>f_i^B(0)$ , for n sufficiently large,

$$\begin{align*}s_n(A,B)=\frac{1}{\sqrt{2}}\epsilon_i(A,B)\eta_{tv_i}(A)\eta_{tv_i}(B). \end{align*}$$

The result follows from Lemma 4.8. The proof is complete.

Proof Write $V^{\mu }=V$ , and let $\eta _{a}=1_{X_{u}\setminus (X_{u}+a)}$ for $a\in P$ . Let $i,j \in \{1,2,\ldots ,d\}$ be given. Since $Exp(\eta )$ is a unit,

(4.6) $$ \begin{align}Exp_{i}(\eta_{sv_{i}}).Exp_{j}(\eta_{tv_{j}})=Exp_{j}(\eta_{tv_{j}}).Exp_{i}(\eta_{sv_{i}})\end{align} $$

for $i,j\in \{1,2,\ldots , d\}$ and $s,t> 0$ .

Hence,

(4.7) $$ \begin{align}P_{2}\Big(Exp_{i}(\eta_{sv_{i}}).Exp_{j}(\eta_{tv_{j}})\Big)=P_{2}\Big(Exp_{j}(\eta_{tv_{j}}).Exp_{i}(\eta_{sv_{i}})\Big)\end{align} $$

for $i,j\in \{1,2,\ldots , d\}$ and $s,t> 0$ .

Thanks to Lemma 4.9,

$$ \begin{align*}P_{2}\Big(Exp_{i}(\eta_{sv_{i}}).Exp_{j}(\eta_{tv_{j}})\Big)(A,B)&=\epsilon_i(A,B) \eta_{sv_{i}}(A) \eta_{sv_{i}}(B)+V_{sv_{i}}\eta_{tv_{j}}\wedge \eta_{sv_{i}}(A,B)\\ &\quad +\epsilon_{j}(A-sv_i,B-sv_i)V_{sv_{i}}\eta_{tv_{j}}(A) V_{sv_{i}}\eta_{tv_{j}}(B)\end{align*} $$

for almost every $(A,B)\in X_{u}\times X_{u}$ . Similarly,

$$ \begin{align*}P_{2}\Big(Exp_{j}(\eta_{tv_{j}}).Exp_{i}(\eta_{sv_{i}})\Big)(A,B)&=\epsilon_j(A,B) \eta_{tv_{j}}(A) \eta_{tv_{j}}(B)+V_{tv_{j}}\eta_{sv_{i}}\wedge \eta_{tv_{j}}(A,B)\\ & \quad +\epsilon_{i}(A-tv_j,B-tv_j)V_{tv_{j}}\eta_{sv_{i}}(A) V_{tv_{j}}\eta_{sv_{i}}(B)\end{align*} $$

for almost every $(A,B)\in X_{u}\times X_{u}$ .

For $s>0$ and $i\in \{1,2,\ldots , d\}$ , let $L_{i,s}:=(1_{X_{u}}\setminus X_{u}+sv_{i})\times (1_{X_{u}}\setminus X_{u}+sv_{i})$ . Let $s,t>0$ . For almost all $(A,B)\in L_{i,s}\cap L_{j,t}$ ,

$$\begin{align*}P_{2}\Big(Exp_{i}(\eta_{sv_{i}}).Exp_{j}(\eta_{tv_{j}})\Big)(A,B)=\epsilon_{i}(A,B)\end{align*}$$

and

$$\begin{align*}P_{2}\Big(Exp_{j}(\eta_{tv_{j}}).Exp_{i}(\eta_{sv_{i}})\Big)(A,B)=\epsilon_{j}(A,B).\end{align*}$$

Therefore, (4.7) implies,

$$\begin{align*}\epsilon_{i}(A,B)=\epsilon_{j}(A,B)\end{align*}$$

for almost every $(A,B)\in L_{i,s}\cap L_{j,t}$ . Now observe that $X_{u}\times X_{u}=\displaystyle {\bigcup _{m,n\in \mathbb {N}}}L_{i,m}\cap L_{j,n}$ . Hence the proof follows.

Notation: Let $X:=supp(\mu ) \cap X_u$ . Recall that $Y_u\setminus supp(\mu )$ is the largest open set that has $\mu $ measure zero. Since $\mu $ is invariant, $supp(\mu )$ is $\mathbb {R}^d$ -invariant. It follows from Remark 4.7 that $\mu |_{X}$ has full support.

Proof Write $\eta _{a}=1_{X_{u}\setminus X_{u}+a}$ for $a\in P$ . Assume that $Exp(\eta )$ is a unit for $E^{\mu }$ . For $A,B\in Y_{u}$ and $i\in \{1,2,\ldots ,d\}$ , note that $A\subset B$ if and only if $f_{i}^{A}(x)\leq f^{B}_{i}(x)$ for each $x\in Q_{i}$ .

Let $A,B\in X$ , and $i\in \{1,2,\ldots ,d\}$ . Suppose $f^{A}_{i}(0)\leq f^{B}_{i}(0)$ . Then, we claim the following.

1. (i) $f^{A}_{j}(0)\leq f^{B}_{j}(0)$ for each $j\in \{1,2,\ldots ,d\}$ , and

2. (ii) if $t_{j}>0$ for $j\in \{1,2,\ldots ,d-1\}$ , then

   $$\begin{align*}f^{A}_{d} \left(-\sum_{j=1}^{d-1}t_{j}v_{j}\right)\leq f^{B}_{d} \left(-\sum_{j=1}^{d-1}t_{j}v_{j}\right).\end{align*}$$

Observe that X is a measurable subset of $X_{u}$ such that $X+P \subset X$ . Consider the set

$$\begin{align*}N_{i,j}=\{(A^{'},B^{'})\in X\times X: f^{A^{'}}_{i}(0)<f^{B^{'}}_{i}(0), f^{A^{'}}_{j}(0)>f^{B^{'}}_{j}(0)\}\end{align*}$$

for $i,j\in \{1,2,\ldots , d\}$ . By Proposition 4.6, $N_{i,j}$ is an open set, and by Lemma 4.10, $N_{i,j}$ is a null set. Therefore, $N_{i,j}$ is empty. Similarly, consider the set

$$\begin{align*}M_{i,j}=\{(A^{'},B^{'}) \in X \times X: { f^{A^{'}}_{i}(0)=f^{B^{'}}_{i}(0), f^{A^{'}}_{j}(0)>f^{B^{'}}_{j}(0)}\}\end{align*}$$

for $i,j\in \{1,2,\ldots ,d\}$ . Fix $i,j\in \{1,2,\ldots ,d\}$ with $i \neq j$ . Suppose $(A^{'},B^{'})\in M_{i,j}$ . Let $s,t>0$ be sufficiently small such that $s<t$ and $f^{A^{'}}_{j}(0)>f^{B^{'}}_{j}(-tv_{i})$ . Then,

$$\begin{align*}f_i^{A^{'}+sv_i}=f^{A^{'}}_{i}(0)+s<f^{B^{'}}_{i}(0)+t=f_i^{B^{'}+sv_i}.\end{align*}$$

Since $f^{A^{'}}_{j} \in \mathcal {F}_j$ ,

$$\begin{align*}f_{j}^{A^{'}+sv_i}(0)=f^{A^{'}}_{j}(-sv_{i})\geq f^{A^{'}}_{j}(0)>f^{B^{'}}_{j}(-tv_{i})=f_j^{B^{'}+tv_i}(0).\end{align*}$$

This implies $(A^{'}+sv_{i},B^{'}+tv_{i})\in N_{i,j}$ , which is a contradiction. Hence, $M_{i,j}=\emptyset $ . Therefore, for $A^{'},B^{'}\in X$ and $i,j\in \{1,2,\ldots d\}$ , $f^{A^{'}}_{i}(0)\leq f^{B^{'}}_{i}(0)$ if and only if $f^{A^{'}}_{j}(0)\leq f^{B^{'}}_{j}(0)$ , which proves $(i)$ .

For each $j \in \{1,2,\ldots ,d-1\}$ , let $t_{j}>0$ . Let $A,B \in X$ be such that $f_i^A(0) \leq f_i^B(0)$ for some i. Since $f_{i}^{A}(0)\leq f_{i}^{B}(0)$ , by $(i)$ , $f_{1}^{A}(0)\leq f_{1}^{B}(0)$ . Since $f^{A+t_{1}v_{1}}_{1}(0)\leq f^{B+t_{1}v_{1}}_{1}(0)$ , by $(i)$ , $f^{A+t_{1}v_{1}}_{2}(0)\leq f^{B+t_{1}v_{1}}_{2}(0)$ , i.e.,

$$\begin{align*}f^{A}_{2}(-t_{1}v_{1})\leq f^{B}_{2}(-t_{1}v_{1}).\end{align*}$$

Since $f_{2}^{A+t_{1}v_{1}}(0)+t_{2}\leq f^{B+t_{1}v_{1}}_{2}(0)+t_{2}$ ,

$$\begin{align*}f^{A+t_{1}v_{1}+t_{2}v_{2}}_{2}(0)\leq f^{B+t_{1}v_{1}+t_{2}v_{2}}_{2}(0).\end{align*}$$

Once again by $(i)$ ,

$$\begin{align*}f^{A}_{3}(-t_{1}v_{1}-t_{2}v_{2})=f_3^{A+t_1v_1+t_2v_2}(0)\leq f_3^{B+t_1v_1+t_2v_2}(0) = f^{B}_{3}(-t_{1}v_{1}-t_{2}v_{2}).\end{align*}$$

Inductively,

(4.8) $$ \begin{align}f^{A}_{d}\left(-\sum_{j=1}^{d-1}t_{j}v_{j}\right)\leq f^{B}_{d}\left(-\sum_{j=1}^{d-1}t_{j}v_{j}\right).\end{align} $$

We have proved $(ii)$ .

It follows from $(i)$ that for $i \in \{1,2,\ldots ,d\}$ and $A,B \in X$ , $f_i^A(0)>f_i^B(0)$ if and only if $f_j^A(0)>f_j^B(0)$ for every $j \in \{1,2,\ldots ,d\}$ . Suppose $f_i^A(0)>f_i^B(0)$ for $A,B \in X$ and $i \in \{1,2,\ldots ,d\}$ . Arguing as before, we see that

(4.9) $$ \begin{align} f^{A}_{d}\left(-\sum_{j=1}^{d-1}t_{j}v_{j}\right)>f^{B}_{d}\left(-\sum_{j=1}^{d-1}t_{j}v_{j}\right) \end{align} $$

whenever $t_{j}>0$ for $j\in \{1,2,\ldots ,d-1\}$ .

Let $A,B\in X$ . Without loss of generality, we can assume that $f^{A}_{d}(0)\leq f^{B}_{d}(0)$ . Then,

(4.10) $$ \begin{align} f^{A}_{d}\left(-\sum_{j=1}^{d-1}t_{j}v_{j}\right) \leq f^{B}_{d}\left(-\sum_{j=1}^{d-1}t_{j}v_{j}\right) \end{align} $$

whenever $t_{j}>0$ for $j\in \{1,2,\ldots ,d-1\}$ .

Let $x \in Q_d$ . Choose $t>0$ such that $A-x+tv_{d}\in X$ and $B-x+tv_{d}\in X$ , i.e.,

$$\begin{align*}f_{d}^{A-x+tv_{d}}(0)=f_d^{A}(x)+t\geq 0\end{align*}$$

and

$$\begin{align*}f_{d}^{B-x+tv_{d}}(0)=f_d^B(x)+t\geq 0.\end{align*}$$

Let $\widetilde {A}=A-x+tv_{d}$ and $\widetilde {B}=B-x+tv_{d}$ . Suppose $f^{A}_{d}(x)>f^{B}_{d}(x)$ . Then,

$$\begin{align*}f^{\widetilde{A}}_{d}(0)=f^{A}_{d}(x)+t> f^{B}_{d}(x)+t=f^{\widetilde{B}}_{d}(0).\end{align*}$$

By (4.9),

$$\begin{align*}f^{\widetilde{A}}_{d}\left(-\sum_{j=1}^{d-1}r_{j}v_{j}\right)>f^{\widetilde{B}}_{d}\left(-\sum_{j=1}^{d-1}r_{j}v_{j}\right)\end{align*}$$

whenever $r_{j}>0$ for $j=1,2,\ldots , d-1$ . This implies

$$\begin{align*}f^{A}_{d}\left(\sum_{j=1}^{d-1}(x_{j}-r_{j})v_{j}\right)>f^{B}_{d} \left(\sum_{j=1}^{d-1}(x_{j}-r_{j})v_{j}\right)\end{align*}$$

whenever $r_{j}>0$ for $j=1,2,\ldots , d-1$ . This contradicts (4.10). Hence, $f^{A}_{d}(x)\leq f^{B}_{d}(x)$ for every $x\in Q_{d}$ , i.e., $A\subset B$ .

Now suppose $A,B\in supp(\mu )$ . Then, there exists $t>0$ such that $A+tv_{1}, B+tv_{1}\in X$ . Without loss of generality, assume that $A+tv_{1}\subset B+tv_{1}$ . Then, $A\subset B$ . Therefore, if $A,B\in supp(\mu )$ , $A\subset B$ or $B\subset A$ . The proof is complete.

Proof By Lemma 4.2, if $supp(\mu )=Y^{\lambda }_{u}$ for some $\lambda \in S(P^{*})$ , then $\{Exp(1_{X_{u}\setminus X_{u}+a})\}_{a\in P}$ is a unit for $E^{\mu }$ .

To prove the converse, let $A\in supp(\mu )$ . Since $\mu $ is invariant, $supp(\mu )$ is $\mathbb {R}^d$ -invariant. By Lemma 4.11, for $x\in \mathbb {R}^{d}$ , either $A+x\subset A$ or $A\subset A+x$ . Define $Q_{A}:=\{x\in \mathbb {R}^{d}: {A+x\subset A}\}$ . Then, $Q_{A}\cup -Q_{A}=\mathbb {R}^{d}$ . Note that $-P\subset Q_{A}$ and $-P+Q_A \subset Q_A$ . Let $Q:=Q_{A}\cap -Q_{A}$ . We claim the following:

1. (i) $\partial Q_{A}=Q$ , where $\partial Q_A$ is the boundary of $Q_{A}$ , and

2. (ii) Q is a vector subspace of $\mathbb {R}^{d}$ .

We identify $span\{v_{1},v_{2},\ldots v_{d-1}\}$ with $\mathbb {R}^{d-1}$ . Let $f:\mathbb {R}^{d-1}\to \mathbb {R}$ be a continuous function such that $A=\{(x,t): {t\leq f(x)}\}$ . Note that

$$\begin{align*}Q_{A} =\{(y,t) \in \mathbb{R}^{d-1}\times \mathbb{R}: f(x)\leq f(x+y)-t \text{ for all } x\in\mathbb{R}^{d-1} \},\end{align*}$$

and

$$\begin{align*}-Q_{A}=\{(y,t)\in\mathbb{R}^{d-1}\times\mathbb{R}: {f(x)\geq f(x+y)-t \text{ for all } x\in\mathbb{R}^{d-1} }\}.\end{align*}$$

It is not difficult to deduce using the fact that $Q_A \cup -Q_A=\mathbb {R}^d$ that

$$\begin{align*}\partial Q_{A}\subset \{(y,t)\in\mathbb{R}^{d-1}\times\mathbb{R}: {f(x)= f(x+y)-t \text{ for all } x\in\mathbb{R}^{d-1} }\}=Q.\end{align*}$$

Suppose $(y,t) \in Q$ . Set $t_n:=t-\frac {1}{n}$ and $s_n:=t+\frac {1}{n}$ . Then, $(y,t_n) \in Q_A$ and $(y,t_n) \to (y,t)$ . Also, $(y,s_n) \notin Q_A$ and $(y,s_n) \to (y,t)$ . Therefore, $(y,t) \in \partial Q_A$ . Hence, $\partial Q_A=Q$ .

Let $g:\mathbb {R}^{d-1}\to \mathbb {R}$ be a continuous function such that $Q_{A}=\{(x,t): {t\leq g(x)}\}$ . Clearly, $Q=\{x \in \mathbb {R}^d:A +x=A\}$ is a closed subgroup of $\mathbb {R}^{d}$ . Since $Q=\partial Q_A$ is the graph of the continuous function g, it is connected. Therefore, Q is a vector space and consequently, g is linear. Hence, there exists $\lambda \in \mathbb {R}^d$ , $||\lambda ||=1$ such that

$$\begin{align*}Q_{A}=\{y\in\mathbb{R}^{d}: {\langle\lambda|y\rangle\leq 0}\}.\end{align*}$$

Since $-P\subset Q_{A}$ , $\lambda \in S(P^{*})$ .

Since $Q_{A}+A\subset A$ , there exists $y\in \mathbb {R}^{d}$ such that $A=y+Q_{A}$ , i.e., $A \in Y_u^{\lambda }$ . Since $supp(\mu ) \subset Y_{u}$ is invariant, $x+A=x+y+Q_A\in supp(\mu )$ for each $x\in \mathbb {R}^{d}$ , i.e., $Y^{\lambda }_{u}\subset supp(\mu )$ . To stress the dependence of $\lambda $ on A, we write $\lambda =\lambda _{A}$ . We have proved that

$$\begin{align*}supp(\mu)=\displaystyle \bigcup_{A\in supp(\mu)}Y^{\lambda_{A}}_{u}.\end{align*}$$

Suppose $A, B\in supp(\mu )$ . Note that $Q_{A}\in Y^{\lambda _{A}}_{u}\subset supp(\mu )$ and $Q_{B}\in Y^{\lambda _{B}}_{u}\subset supp(\mu )$ . Then, by Proposition 4.11, either $Q_{A}\subset Q_{B}$ or $Q_{B}\subset Q_{A}$ . But this can happen only if $\lambda _{A}=\lambda _{B}$ . In that case, $Y^{\lambda _{A}}_{u}=Y^{\lambda _{B}}_{u}$ . Therefore, there exists $\lambda \in S(P^{*})$ such that $supp(\mu )=Y^{\lambda }_{u}$ . Now the proof is complete.

Notation and Convention: Let V be an isometric representation of P on a Hilbert space $\mathcal {H}$ . Let $\xi =\{\xi _{a}\}_{a\in P}\in \mathcal {A}(V)$ . We define $\mathcal {H}^{\xi }$ to be the smallest, closed, reducing subspace of $\mathcal {H}$ containing $\{\xi _{a}:{ a\in P}\}$ . (Hereafter, all reducing subspaces will be assumed to be closed.) If W is a direct summand of V, we view the product system $E^W$ as a subsystem of $E^V$ . Similarly, we view $\mathcal {A}(W)$ as a subspace of $\mathcal {A}(V)$ .

Proof For the proof of $(1)$ , we refer the reader to [Reference Sundar34, Theorem 3.16 and Remark 3.17]. From $(1)$ , $V^{\xi }:=V|_{\mathcal {H}^{\xi }}$ is unitarily equivalent to $V^{\mu }$ . We can see from the proof of [Reference Sundar34, Theorem 3.16] that the unitary U intertwining $V^{\xi }$ and $V^{\mu }$ can be chosen such that $U\xi _{a}=1_{X_{u}\setminus X_{u}+a}$ for $a\in P$ . Suppose $Exp(\xi )$ is a unit for $E^{V}$ . Note that the subsystem $E^{V^\xi }$ contains $Exp(\xi )$ . Hence, $\{Exp(1_{X_{u}\setminus X_{u}+a})\}_{a\in P}$ is a unit for $E^{V^\mu }$ . It follows from Theorem 4.12 and Lemma 4.2 that $V^{\mu }$ is unitarily equivalent to $S^{\lambda }$ for some $\lambda \in S(P^{*})$ . This proves $(2)$ .