(see, [*e.g.*]{}, [@Kato]) entails that $A \Omega = F_0 A \Omega$, where $F_0$ is the projection onto the subspace of vectors in $\Hs$ each left invariant under $U({\mbox{SO$_0(2,n\!-\!1)$}})$, using Lemma \[invariance\]. Since the mixing property in condition (iii) entails that the rank of $F_0$ is 1 and since $\Omega$ is separating for $\As(W_R)$, it follows that $A$ is a multiple of the identity. But this entails $$\CC 1 = J_W (\As(W) \cap \As(W')) J_W =
J_W \As(W) J_W \cap J_W \As(W') J_W =
\As(W)' \cap \As(W')' \, ,$$ using (\[Js\]), so that $\As(W) \bigvee \As(W') = \Bs(\Hs)$, for every $W \in \Ws$.

Before we close this section, we have a final proposition to prove.

\[unequal\] Let Standing Assumptions (i)–(iv) hold, $n \geq 3$ and $W_1,W_2 \in \Ws$. If $W_2 \neq \pm W_1$, then $\As(W_1) \neq \As(W_2)$.

Since two unequal wedges $W_1,W_2 \in \Ws$ have unequal reflections about their edges, unless $W_2$ coincides with $W_1{}', \widetilde{W}_1$ or $\widetilde{W}_1{}'$, Lemma \[TCPgroup\] entails $J_{W_1} \neq J_{W_2}$ and, thus, $\As(W_1) \neq \As(W_2)$.

If $\As(W_1) \subset \As(W_1{}')$, then Theorem \[weaklocality1\] entails that the restriction of $\omega$ to $\Rs(W_1)$ is a trace, which is excluded by assumptions (i) and (iii). Similarly Theorem \[weaklocality2\] yields $\As(W_1) \neq
