Proof. Let $({\mathcal H}_{\omega },\pi _{\omega },\Omega _{\omega }), \left ({\mathcal H}_{\varphi _{{\mathfrak A}}},\pi _{\varphi _{{\mathfrak A}}},\Omega _{\varphi _{{\mathfrak A}}}\right ), \left ({\mathcal H}_{\varphi _{{\mathfrak B}}},\pi _{\varphi _{{\mathfrak B}}},\Omega _{\varphi _{{\mathfrak B}}}\right )$ be GNS triples of $\omega , \varphi _{{\mathfrak A}}, \varphi _{{\mathfrak B}}$, respectively. Then $\left ({\mathcal H}_{\varphi _{{\mathfrak A}}}\otimes {\mathcal H}_{\varphi _{{\mathfrak B}}},\pi _{\varphi _{{\mathfrak A}}}\otimes \pi _{\varphi _{{\mathfrak B}}},\Omega _{\varphi _{{\mathfrak A}}}\otimes \Omega _{\varphi _{{\mathfrak B}}}\right )$ is a GNS triple of $\varphi _{{\mathfrak A}}\otimes \varphi _{{\mathfrak B}}$. As $\omega $ is quasiequivalent to $\varphi _{{\mathfrak A}}\otimes \varphi _{{\mathfrak B}}$, there is a $*$-isomorphism $\tau :\pi _{\omega }\left ({\mathfrak A}\otimes {\mathfrak B}\right )''\to \pi _{\varphi _{{\mathfrak A}}}({\mathfrak A})''\otimes \pi _{\varphi _{{\mathfrak B}}}({\mathfrak B})''$ such that $\tau \circ \pi _{\omega }=\pi _{\varphi _{{\mathfrak A}}}\otimes \pi _{\varphi _{{\mathfrak B}}}$. Because $\omega $ is pure, we have $\pi _{\omega }\left ({\mathfrak A}\otimes {\mathfrak B}\right )''={\mathcal B}({\mathcal H}_{\omega })$, and from the isomorphism $\tau $, we have that $\pi _{\varphi _{{\mathfrak A}}}({\mathfrak A})''\otimes \pi _{\varphi _{{\mathfrak B}}}({\mathfrak B})''$ is also a type I factor. Then from [Reference TakesakiT, Theorem 2.30V], both $\pi _{\varphi _{{\mathfrak A}}}({\mathfrak A})''$ and $\pi _{\varphi _{{\mathfrak B}}}({\mathfrak B})''$ are type I factors. The restriction of $\tau $ to $\pi _{\omega }\left ({\mathfrak A}\otimes {\mathbb C}\mathbb I_{{\mathfrak B}}\right )''$ implies a $*$-isomorphism from $\pi _{\omega }\left ({\mathfrak A}\otimes {\mathbb C}\mathbb I_{{\mathfrak B}}\right )''$ onto the type I factor $\pi _{\varphi _{{\mathfrak A}}}({\mathfrak A})''$. Hence we see that $\pi _{\omega }\left ({\mathfrak A}\otimes {\mathbb C}\mathbb I_{{\mathfrak B}}\right )''$ is a type I factor. Therefore, from [Reference TakesakiT, Theorem 1.31V], there are Hilbert spaces ${\mathcal K}_{{\mathfrak A}},{\mathcal K}_{{\mathfrak B}}$ and a unitary $W: {\mathcal H}_\omega \to {\mathcal K}_{{\mathfrak A}}\otimes {\mathcal K}_{{\mathfrak B}}$ such that $\mathop {\mathrm {Ad}}\nolimits \left ( W\right )\left ( \pi _{\omega }\left ({\mathfrak A}\otimes {\mathbb C}\mathbb I_{{\mathfrak B}}\right )''\right )= {\mathcal B}\left ( {\mathcal K}_{{\mathfrak A}}\right )\otimes {\mathbb C}\mathbb I_{{\mathcal K}_{{\mathfrak B}}} $. Because $\omega $ is pure, we also have $\mathop {\mathrm {Ad}}\nolimits \left ( W\right )\left ( \pi _{\omega }\left ({\mathbb C}\mathbb I_{{\mathfrak A}} \otimes {\mathfrak B}\right )''\right )={\mathbb C}\mathbb I_{{\mathcal K}_{{\mathfrak A}}}\otimes {\mathcal B}({\mathcal K}_{{\mathfrak B}})$. From this, we see that there are irreducible representations $\rho _{{\mathfrak A}}, \rho _{{\mathfrak B}}$ of ${\mathfrak A}$ and ${\mathfrak B}$ on ${\mathcal K}_{{\mathfrak A}}, {\mathcal K}_{{\mathfrak B}}$ such that $\mathop {\mathrm {Ad}}\nolimits (W)\circ \pi _{\omega }=\rho _{{\mathfrak A}}\otimes \rho _{{\mathfrak B}}$. Fix some unit vectors $\xi _{{\mathfrak A}}\in {\mathcal K}_{{\mathfrak A}}, \xi _{{\mathfrak B}}\in {\mathcal K}_{{\mathfrak B}}$. Then because of the irreducibility of $\rho _{{\mathfrak B}}$ and $\rho _{{\mathfrak B}}$, we have that $\omega _{{\mathfrak A}}:=\left \langle \xi _{{\mathfrak A}},\rho _{{\mathfrak A}}\left (\cdot \right )\xi _{{\mathfrak A}}\right \rangle $ and $\omega _{{\mathfrak B}}:=\left \langle \xi _{{\mathfrak B}},\rho _{{\mathfrak B}}\left (\cdot \right )\xi _{{\mathfrak B}}\right \rangle $ are pure states on ${\mathfrak A}, {\mathfrak B}$. By [Reference Kishimoto, Ozawa and SakaiKOS, Theorem 1.1] (originally proved by Powers [Reference PowersP] for UHF-algebras), for any pure states $\psi _{{\mathfrak A}}, \psi _{{\mathfrak B}}$ on ${\mathfrak A}, {\mathfrak B}$, there exist automorphisms $\gamma _{{\mathfrak A}}\in \mathop {\mathrm {Aut}}\nolimits ({\mathfrak A}), \gamma _{{\mathfrak B}}\in \mathop {\mathrm {Aut}}\nolimits ({\mathfrak B})$ such that $\omega _{{\mathfrak A}}=\psi _{{\mathfrak A}}\circ \gamma _{{\mathfrak A}}$ $\omega _{{\mathfrak B}}=\psi _{{\mathfrak B}}\circ \gamma _{{\mathfrak B}}$. Now for unit vectors $W^{*}\left (\xi _{{\mathfrak A}}\otimes \xi _{{\mathfrak B}}\right ),\Omega _{\omega }\in {\mathcal H}_{\omega }$, by Kadison’s transitivity theorem and the irreducibility of $\pi _\omega $ there exists a unitary $u\in {\mathcal U}\left ( {\mathfrak A}\otimes {\mathfrak B}\right )$ such that $\pi _{\omega }(u)\Omega _{\omega }=W^{*}\left (\xi _{{\mathfrak A}}\otimes \xi _{{\mathfrak B}}\right )$. Substituting this, we obtain

(3.4)$$ \begin{align} \omega&=\left\langle\Omega_{\omega},\pi_{\omega}\left (\cdot\right ) \Omega_{\omega}\right\rangle =\left\langle\pi_{\omega}(u^{*})W^{*}\left (\xi_{{\mathfrak A}}\otimes \xi_{{\mathfrak B}}\right ),\pi_{\omega}\left (\cdot\right )\pi_{\omega}(u^{*})W^{*}\left (\xi_{{\mathfrak A}}\otimes \xi_{{\mathfrak B}}\right )\right\rangle\notag\\ &=\left\langle W^{*}\left (\xi_{{\mathfrak A}}\otimes \xi_{{\mathfrak B}}\right ),\pi_{\omega}\circ\mathop{\mathrm{Ad}}\nolimits(u)\left (\cdot\right ) W^{*}\left (\xi_{{\mathfrak A}}\otimes \xi_{{\mathfrak B}}\right )\right\rangle\notag\\ &=\left\langle\left (\xi_{{\mathfrak A}}\otimes \xi_{{\mathfrak B}}\right ),\left (\rho_{{\mathfrak A}}\otimes \rho_{{\mathfrak B}}\right )\circ\mathop{\mathrm{Ad}}\nolimits(u)\left (\cdot\right ) \left (\xi_{{\mathfrak A}}\otimes \xi_{{\mathfrak B}}\right )\right\rangle \notag \\ &=\left ( \omega_{{\mathfrak A}}\otimes\omega_{{\mathfrak B}}\right )\circ \mathop{\mathrm{Ad}}\nolimits(u) =\left ( \psi_{{\mathfrak A}}\circ \gamma_{{\mathfrak A}}\otimes \psi_{{\mathfrak B}}\circ \gamma_{{\mathfrak B}}\right ) \circ \mathop{\mathrm{Ad}}\nolimits(u). \end{align} $$

Now assume that $\psi _{{\mathfrak A}}$ and $\varphi _{{\mathfrak A}}$ are quasiequivalent – that is, the GNS representations of $\psi _{{\mathfrak A}}, \varphi _{{\mathfrak A}}$, denoted by $\pi _{\psi _{{\mathfrak A}}}$ and $\pi _{\varphi _{{\mathfrak A}}}$ are quasiequivalent. From the foregoing argument, $\pi _{\omega }\rvert _{{\mathfrak A}}$ and $\pi _{\varphi _{{\mathfrak A}}}$ are quasiequivalent. At the same time, $\pi _{\omega }\rvert _{{\mathfrak A}}$ and $\rho _{{\mathfrak A}}$ are quasiequivalent. Therefore, $\pi _{\psi _{{\mathfrak A}}}$ and $\rho _{{\mathfrak A}}$ are quasiequivalent. Because both of them are irreducible, we see that a pure state ${\psi _{{\mathfrak A}}}$ can be represented by a unit vector $\zeta \in {\mathcal K}_{{\mathfrak A}}$, as ${\psi _{{\mathfrak A}}}=\left \langle \zeta ,\rho _{{\mathfrak A}}\left (\cdot \right ) \zeta \right \rangle $. Because $\rho _{{\mathfrak A}}$ is irreducible, by Kadison’s transitivity theorem there exists a unitary $w\in {\mathcal U}\left ( {\mathfrak A}\right )$ such that $\rho _{{\mathfrak A}}(w^{*})\zeta =\xi _{{\mathfrak A}}$. Hence we obtain ${\psi _{{\mathfrak A}}}\circ \mathop {\mathrm {Ad}}\nolimits (w)=\omega _{{\mathfrak A}}$. Substituting this instead of $\omega _{{\mathfrak A}}=\psi _{{\mathfrak A}}\circ \gamma _{{\mathfrak A}}$ in equation (3.4), we obtain

(3.5)$$ \begin{align} \omega= \left ( \psi_{{\mathfrak A}}\otimes \psi_{{\mathfrak B}}\circ \gamma_{{\mathfrak B}}\right ) \circ \mathop{\mathrm{Ad}}\nolimits\left ( \left ( w\otimes \mathop{\mathrm{id}}\nolimits_{{\mathfrak B}}\right ) u\right ), \end{align} $$

proving the last claim.

Proof. Because the pure state $\omega $ is quasiequivalent to $\left .\left ( \psi \otimes \varphi _{L}^{\left (1,2\right )}\otimes \varphi _{R}^{\left (1,2\right )} \right )\right \rvert _{{\mathfrak B}\otimes {\mathfrak A}_{1}} =\psi \otimes \left .\left ( \varphi _{L}^{\left (1,2\right )}\otimes \varphi _{R}^{\left (1,2\right )} \right )\right \rvert _{{\mathfrak A}_{1}}$, applying Lemma 3.2 means that for any pure states $\varphi _{L}^{(1)}, \varphi _{R}^{(1)}$ on ${\mathfrak A}_{1,L}, {\mathfrak A}_{1,R}$, there exist an automorphism $S\in \mathop {\mathrm {Aut}}\nolimits {\mathfrak A}_1$ and a unitary $v\in {\mathcal U}\left ( {\mathfrak B}\otimes {\mathfrak A}_1\right )$ such that

(3.7)$$ \begin{align} \omega=\left ( \psi\otimes\left ( \left ( \varphi_{L}^{(1)} \otimes \varphi_{R}^{(1)}\right )\circ S\right ) \right )\circ \mathop{\mathrm{Ad}}\nolimits v. \end{align} $$

From equation (3.7) and $\omega \sim _{\text {q.e.}}\left .\left ( \psi \otimes \varphi _{L}^{\left (1,2\right )}\otimes \varphi _{R}^{\left (1,2\right )} \right )\right \rvert _{{\mathfrak B}\otimes {\mathfrak A}_{1}}$, we get $ \left ( \psi \otimes \left ( \left ( \varphi _{L}^{(1)} \otimes \varphi _{R}^{(1)}\right )\circ S\right ) \right )\sim _{\text {q.e.}} \left .\left ( \psi \otimes \varphi _{L}^{\left (1,2\right )}\otimes \varphi _{R}^{\left (1,2\right )} \right )\right \rvert _{{\mathfrak B}\otimes {\mathfrak A}_{1}}$, which implies

(3.8)$$ \begin{align} \left ( \varphi_{L}^{(1)} \otimes \varphi_{R}^{(1)}\right )\circ S\sim_{\text{q.e.}} \left.\left ( \varphi_{L}^{\left(1,2\right)}\otimes \varphi_{R}^{\left(1,2\right)} \right )\right\rvert_{{\mathfrak A}_{1}}. \end{align} $$

Applying Lemma 3.2 to formula (3.8), there are automorphisms $\gamma _{L}^{(1)}\in \mathop {\mathrm {Aut}}\nolimits \left ( {\mathfrak A}_{1,L}\right ), \gamma _{R}^{(1)}\in \mathop {\mathrm {Aut}}\nolimits \left ( {\mathfrak A}_{1,R}\right )$ and a unitary $w\in {\mathcal U}\left ( {\mathfrak A}_{1}\right )$ such that

(3.9)$$ \begin{align} \left ( \varphi_{L}^{(1)} \otimes \varphi_{R}^{(1)}\right )\circ S=\left (\left ( \varphi_{L}^{(1)}\circ\gamma_{L}^{(1)}\right ) \otimes \left ( \varphi_{R}^{(1)}\circ\gamma_{R}^{(1)}\right ) \right )\circ\mathop{\mathrm{Ad}}\nolimits w. \end{align} $$

Substituting this into equation (3.7), we obtain equation (3.6).

Proof. First we claim that there are automorphisms $\theta _{\mathfrak B}^{LU}\in \mathop {\mathrm {Aut}}\nolimits {{\mathfrak B}_{LU}}, \theta _{\mathfrak B}^{RU}\in \mathop {\mathrm {Aut}}\nolimits {\mathfrak B}_{RU}$ and a unitary $u\in {\mathcal U}\left ( {\mathfrak B}^U\otimes {\mathfrak C}^U\right )$ such that

(3.14)$$ \begin{align} \omega_{{\mathfrak B}{\mathfrak C}}^U\circ\rho_{{\mathfrak B}{\mathfrak C}}^U =\omega_{{\mathfrak B}{\mathfrak C}}^U\circ \left ( \theta_{{\mathfrak B}}^{LU}\otimes \theta_{{\mathfrak B}}^{RU}\otimes\mathop{\mathrm{id}}\nolimits_{{\mathfrak C}^U}\right ) \circ\mathop{\mathrm{Ad}}\nolimits\left ( u\right ). \end{align} $$

To prove this, we first note that from $\omega _0\circ \hat \alpha =\omega _0$ and the decomposition (3.12), we have

(3.15)$$ \begin{align} \omega_{{\mathfrak B}{\mathfrak C}}^U\circ\rho_{{\mathfrak B}{\mathfrak C}}^U\otimes \omega_{{\mathfrak A}} \otimes \omega_{{\mathfrak B}{\mathfrak C}}^D\circ\rho_{{\mathfrak B}{\mathfrak C}}^D \sim_{\text{q.e.}} \omega_{{\mathfrak C}_U} \otimes\omega_{{\mathfrak A}{\mathfrak B}}^L\circ\left (\widehat{ \gamma_{{\mathfrak A}{\mathfrak B}}^L}\right )^{-1} \otimes \omega_{{\mathfrak A}{\mathfrak B}}^R\circ\left ( \widehat{\gamma_{{\mathfrak A}{\mathfrak B}}^R}\right )^{-1} \otimes\omega_{{\mathfrak C}_D}. \end{align} $$

From this, because both states are pure (hence the restrictions of their GNS representations onto ${\mathfrak C}_U\otimes {\mathfrak B}_U$ are factors), we have

(3.16)$$ \begin{align} \omega_{{\mathfrak B}{\mathfrak C}}^U\circ\rho_{{\mathfrak B}{\mathfrak C}}^U&= \left.\left ( \omega_{{\mathfrak B}{\mathfrak C}}^U\circ\rho_{{\mathfrak B}{\mathfrak C}}^U\otimes \omega_{{\mathfrak A}} \otimes \omega_{{\mathfrak B}{\mathfrak C}}^D\circ\rho_{{\mathfrak B}{\mathfrak C}}^D \right ) \right\rvert_{{\mathfrak C}_U\otimes {\mathfrak B}_U}\notag\\ &\sim_{\text{q.e.}} \omega_{{\mathfrak C}_U} \otimes\left.\left ( \omega_{{\mathfrak A}{\mathfrak B}}^L\circ\left ( \widehat{\gamma_{{\mathfrak A}{\mathfrak B}}^L}\right )^{-1} \otimes \omega_{{\mathfrak A}{\mathfrak B}}^R\circ\left ( \widehat{\gamma_{{\mathfrak A}{\mathfrak B}}^R}\right )^{-1}\right )\right\rvert_{{\mathfrak B}_U}. \end{align} $$

We apply Lemma 3.3 for ${\mathfrak B}, {\mathfrak A}_{1L}, {\mathfrak A}_{1R}, {\mathfrak A}_{2L}, {\mathfrak A}_{2R}, \omega , \varphi _L^{(1,2)}, \varphi _R^{(1,2)}, \psi $, replaced by ${\mathfrak C}_U, {\mathfrak B}_{LU}, {\mathfrak B}_{RU}, {\mathfrak A}_{L}\otimes {\mathfrak B}_{LD}, {\mathfrak A}_R\otimes {\mathfrak B}_{RD}, \omega _{{\mathfrak B}{\mathfrak C}}^U\circ \rho _{{\mathfrak B}{\mathfrak C}}^U, \omega _{{\mathfrak A}{\mathfrak B}}^L\circ \left (\widehat {\gamma _{{\mathfrak A}{\mathfrak B}}^L}\right )^{-1}, \omega _{{\mathfrak A}{\mathfrak B}}^R\circ \left ( \widehat {\gamma _{{\mathfrak A}{\mathfrak B}}^R}\right )^{-1}, \omega _{{\mathfrak C}_U}$, respectively. From equation (3.16), they satisfy the conditions in Lemma 3.3. Applying Lemma 3.3 – for pure states $\varphi _L^{(1)}=\omega _{{\mathfrak B}_{LU}}$ and $\varphi _R^{(1)}=\omega _{{\mathfrak B}_{RU}}$ – we obtain automorphisms $\theta _{{\mathfrak B}}^{LU}\in \mathop {\mathrm {Aut}}\nolimits \left ({\mathfrak B}_{LU}\right ), \theta _{{\mathfrak B}}^{RU}\in \mathop {\mathrm {Aut}}\nolimits \left ({\mathfrak B}_{RU}\right )$ and a unitary $u\in {\mathcal U}\left ({\mathfrak B}_U\otimes {\mathfrak C}_U\right )$ satisfying equation (3.14).

We set

(3.17)$$ \begin{align} \begin{split} \eta_L&:=\left ( \theta^{LU}_{{\mathfrak B}}\otimes \mathop{\mathrm{id}}\nolimits_{{\mathfrak A}_L}\otimes \mathop{\mathrm{id}}\nolimits_{{\mathfrak B}_{LD}}\right ) \circ\gamma_{{\mathfrak A}{\mathfrak B}}^L \in \mathop{\mathrm{Aut}}\nolimits\left ({\mathfrak B}_{LU}\otimes {\mathfrak A}_L\otimes{\mathfrak B}_{LD}\right )\\ \eta_R&:=\left ( \theta^{RU}_{{\mathfrak B}}\otimes \mathop{\mathrm{id}}\nolimits_{{\mathfrak A}_R}\otimes \mathop{\mathrm{id}}\nolimits_{{\mathfrak B}_{RD}}\right ) \circ\gamma_{{\mathfrak A}{\mathfrak B}}^R \in \mathop{\mathrm{Aut}}\nolimits\left ( {\mathfrak B}_{RU}\otimes {\mathfrak A}_R\otimes{\mathfrak B}_{RD}\right ). \end{split} \end{align} $$

Then we have

(3.18)$$ \begin{align} \begin{split} \omega_0\circ\alpha&= \left ( \omega_{{\mathfrak A}_L}\otimes\omega_{{\mathfrak A}_R} \otimes \omega_{{\mathfrak B}{\mathfrak C}}^U\otimes \omega_{{\mathfrak B}{\mathfrak C}}^D \right )\circ\alpha\\ &\sim_{\text{q.e.}} \left ( \omega_{{\mathfrak A}_L}\otimes\omega_{{\mathfrak A}_R} \otimes \omega_{{\mathfrak B}{\mathfrak C}}^U\circ\rho_{{\mathfrak B}{\mathfrak C}}^U \otimes \omega_{{\mathfrak B}{\mathfrak C}}^D \right ) \circ \left ( \mathop{\mathrm{id}}\nolimits_{{\mathfrak C}_U}\otimes \gamma_{{\mathfrak A}{\mathfrak B}}^L\otimes \gamma_{{\mathfrak A}{\mathfrak B}}^R \otimes \mathop{\mathrm{id}}\nolimits_{{\mathfrak C}_D} \right )\\ &\sim_{\text{q.e.}} \left ( \omega_{{\mathfrak A}_L}\otimes\omega_{{\mathfrak A}_R} \otimes \omega_{{\mathfrak B}{\mathfrak C}}^U \otimes \omega_{{\mathfrak B}{\mathfrak C}}^D \right )\\ &\quad \circ \left ( \mathop{\mathrm{id}}\nolimits_{{\mathfrak C}_U}\otimes \left ( \left ( \theta_{{\mathfrak B}}^{LU}\otimes \mathop{\mathrm{id}}\nolimits_{{\mathfrak A}_L}\otimes\mathop{\mathrm{id}}\nolimits_{{\mathfrak B}_{LD}} \right )\circ\gamma_{{\mathfrak A}{\mathfrak B}}^L\right ) \otimes \left ( \left ( \theta_{{\mathfrak B}}^{RU}\otimes \mathop{\mathrm{id}}\nolimits_{{\mathfrak A}_R}\otimes\mathop{\mathrm{id}}\nolimits_{{\mathfrak B}_{RD}} \right )\circ\gamma_{{\mathfrak A}{\mathfrak B}}^R\right ) \otimes\mathop{\mathrm{id}}\nolimits_{{\mathfrak C}_D} \right )\\ &=\omega_0\circ\left (\mathop{\mathrm{id}}\nolimits_{{\mathfrak C}_U}\otimes\eta_L\otimes\eta_R\otimes\mathop{\mathrm{id}}\nolimits_{{\mathfrak C}_D}\right ). \end{split} \end{align} $$

This completes the proof.

Proof of Theorem 3.1.

Set $0<\theta <\frac \pi 2$ and $\alpha \in \mathop {\mathrm {SQAut}}\nolimits ({\mathcal A})$ satisfying $ \omega _0\circ \alpha \circ \beta _g=\omega _0\circ \alpha $ for all $g\in G$. We would like to show that $\mathop {\mathrm {IG}}\nolimits (\omega _0\circ \alpha ,\theta )$ is not empty.

Let us set $\theta _{2.2}:=\theta $ and consider $\theta _{0.8}, \theta _1, \theta _{1.2}, \theta _{1.8}, \theta _2, \theta _{2.8}, \theta _3, \theta _{3.2}$ satisfying formula (2.11) for this $\theta _{2.2}$. Because $\alpha \in \mathop {\mathrm {SQAut}}\nolimits ({\mathcal A})$, there is a decomposition given by formulas (2.12), (2.13) and (2.14). Using this decomposition, set

(3.19)$$ \begin{align} \begin{split} \alpha_1&:=\alpha_{1D}\otimes\alpha_{1U},\quad \text{where}\\ \alpha_{1\zeta} &:= \left ( \alpha_{\left(\theta_1,\theta_2\right],\zeta} \otimes \alpha_{\left(\theta_2,\theta_3\right],\zeta}\otimes \alpha_{\left(\theta_3,\frac\pi 2\right],\zeta} \right ) \\ &\quad \circ \left ( \alpha_{\left(\theta_{0.8}, \theta_{1.2}\right],\zeta}\otimes \alpha_{\left(\theta_{1.8},\theta_{2.2}\right],\zeta} \otimes \alpha_{\left(\theta_{2.8},\theta_{3.2}\right],\zeta} \right )\in\mathop{\mathrm{Aut}}\nolimits\left ( {\mathcal A}_{\left ( \left ( C_{\theta_{0.8}}\right )^c\right )_\zeta} \right ) ,\quad\zeta=U,D,\\ \alpha_2&:=\alpha_{\left[0,\theta_1\right]} \in\mathop{\mathrm{Aut}}\nolimits\left ({\mathcal A}_{C_{\theta_1}}\right ). \end{split} \end{align} $$

We have $\alpha =({\textrm {{inner}}})\circ \alpha _2\circ \alpha _1$.

We would like to show that $\left (\alpha \circ \beta _g^U\circ \alpha ^{-1}, \alpha \circ \beta _g\circ \alpha ^{-1}\right )$ satisfy the conditions of $(\alpha ,\hat \alpha )$ in Lemma 3.4. We first show that they satisfy a decomposition corresponding to equations (3.12) and (3.13). For $\Gamma ={\mathbb Z}^2, H_U$, we have

(3.20)$$ \begin{align} \left ( \beta_g^\Gamma\right )^{-1}\alpha\circ\beta_g^\Gamma\circ\alpha^{-1} =({\textrm{{inner}}})\circ\left ( \beta_g^\Gamma\right )^{-1}\circ\left ( \alpha_1\beta_g^\Gamma\alpha_1^{-1}\right ) \left ( \alpha_1\beta_g^\Gamma\alpha_1^{-1}\right )^{-1} \alpha_2\alpha_1\beta_g^{\Gamma}\alpha_1^{-1}\alpha_2^{-1}. \end{align} $$

The latter part, $\left ( \alpha _1\beta _g^\Gamma \alpha _1^{-1}\right )^{-1} \alpha _2\alpha _1\beta _g^{\Gamma }\alpha _1^{-1}\alpha _2^{-1}$, decomposes to left and right. To see this, first note that

(3.21)$$ \begin{align} \alpha_1^{-1}\alpha_2\alpha_1 =\alpha_{\left(\theta_{0.8}, \theta_{1.2}\right]}^{-1}\alpha_{\left[0,\theta_1\right]}\alpha_{\left(\theta_{0.8}, \theta_{1.2}\right]} \in \mathop{\mathrm{Aut}}\nolimits\left ( {\mathcal A}_{C_{\theta_{1.2}}} \right ). \end{align} $$

Because the conjugation $\left ( \beta _g^\Gamma \right )^{-1}\cdot \beta _g^{\Gamma }$ does not change the support of an automorphism, $\left ( \beta _g^\Gamma \right )^{-1}\left ( \alpha _1^{-1}\alpha _2\alpha _1\right ) \beta _g^{\Gamma }$ is also supported on ${C_{\theta _{1.2}}}$. Therefore, we have

(3.22)$$ \begin{align} &\alpha_1\left ( \left ( \beta_g^\Gamma\right )^{-1}\left ( \alpha_1^{-1}\alpha_2\alpha_1\right ) \beta_g^{\Gamma}\right ) \alpha_1^{-1} \notag \\ &=\alpha_{\left(\theta_1,\theta_2\right]}\alpha_{\left(\theta_{0.8}, \theta_{1.2}\right]} \left ( \beta_g^\Gamma\right )^{-1} \alpha_{\left(\theta_{0.8}, \theta_{1.2}\right]}^{-1}\alpha_{\left[0,\theta_1\right]}\alpha_{\left(\theta_{0.8}, \theta_{1.2}\right]} \beta_g^\Gamma\alpha_{\left(\theta_{0.8}, \theta_{1.2}\right]}^{-1}\alpha_{\left(\theta_1,\theta_2\right]}^{-1}. \end{align} $$

Hence we get the left-right decomposition

(3.23)$$ \begin{align} \begin{split} \left ( \alpha_1\beta_g^\Gamma\alpha_1^{-1}\right )^{-1} &\alpha_2\alpha_1\beta_g^{\Gamma}\alpha_1^{-1}\alpha_2^{-1} =\alpha_1\left ( \left ( \beta_g^\Gamma\right )^{-1}\left ( \alpha_1^{-1}\alpha_2\alpha_1\right ) \beta_g^{\Gamma}\right ) \alpha_1^{-1}\alpha_2^{-1}\\ &=\alpha_{\left(\theta_1,\theta_2\right]}\alpha_{\left(\theta_{0.8}, \theta_{1.2}\right]} \left ( \beta_g^\Gamma\right )^{-1} \alpha_{\left(\theta_{0.8}, \theta_{1.2}\right]}^{-1}\alpha_{\left[0,\theta_1\right]}\alpha_{\left(\theta_{0.8}, \theta_{1.2}\right]} \beta_g^\Gamma\alpha_{\left(\theta_{0.8}, \theta_{1.2}\right]}^{-1}\alpha_{\left(\theta_1,\theta_2\right]}^{-1}\circ\alpha_{\left[0,\theta_1\right]}^{-1}\\ &=:\bigotimes_{\sigma=L,R}\Xi_{\Gamma,g,\sigma}. \end{split} \end{align} $$

Here we set

(3.24)$$ \begin{align} &\Xi_{\Gamma,g,\sigma}:=\left ( \alpha_{\left(\theta_1,\theta_2\right],\sigma}\alpha_{\left(\theta_{0.8}, \theta_{1.2}\right],\sigma} \left ( \beta_g^{\Gamma_\sigma}\right )^{-1} \right. \notag\\ &\quad \left.\alpha_{\left(\theta_{0.8}, \theta_{1.2}\right],\sigma}^{-1}\alpha_{\left[0,\theta_1\right],\sigma }\alpha_{\left(\theta_{0.8}, \theta_{1.2}\right],\sigma} \beta_g^{\Gamma_\sigma}\alpha_{\left(\theta_{0.8}, \theta_{1.2}\right],\sigma}^{-1}\alpha_{\left(\theta_1,\theta_2\right],\sigma}^{-1}\circ\alpha_{\left[0,\theta_1\right],\sigma}^{-1} \right )\quad \in\mathop{\mathrm{Aut}}\nolimits\left ({\mathcal A}_{\left ( C_{\theta_2}\right )_\sigma}\right ). \end{align} $$

On the other hand, the first part of equation (3.20) with $\Gamma ={\mathbb Z}^2,H_U$ satisfies

(3.25)$$ \begin{align} \begin{split} \beta_g^{-1}\alpha_1\beta_g\alpha_1^{-1} = \xi_D\otimes\xi_U, \qquad \left ( \beta_g^{U}\right )^{-1}\alpha_1\beta_g^U\alpha_1^{-1} = \mathop{\mathrm{id}}\nolimits_{{\mathcal A}_{H_D}}\otimes \xi_U, \end{split} \end{align} $$

where

(3.26)$$ \begin{align} \xi_\zeta:=\left ( \beta_g^{\zeta}\right )^{-1}\alpha_{1,\zeta}\beta_g^{\zeta}\alpha_{1,\zeta}^{-1}\in \mathop{\mathrm{Aut}}\nolimits\left ( {\mathcal A}_{\left ( \left ( C_{\theta_{0.8}}\right )^c\right )_\zeta} \right ),\quad \zeta=U,D. \end{align} $$

Hence we obtain decompositions

(3.27)$$ \begin{align} \begin{split} \left ( \beta_g^U\right )^{-1}\circ\alpha\circ \beta_g^U\circ \alpha^{-1} &=({\textrm{{inner}}})\circ \left ( \mathop{\mathrm{id}}\nolimits_{{\mathcal A}_{H_D}}\otimes \xi_U \right )\circ \left ( \Xi_{H_U,g,L}\otimes \Xi_{H_U,g,R}\right ),\\ \left ( \beta_g\right )^{-1}\circ\alpha\circ \beta_g\circ \alpha^{-1} &=({\textrm{{inner}}})\circ \left ( \xi_D\otimes\xi_U\right ) \circ \left ( \Xi_{{\mathbb Z}^2,g,L}\otimes \Xi_{{\mathbb Z}^2,g,R}\right ). \end{split} \end{align} $$

Because $\xi _{\zeta }\in \mathop {\mathrm {Aut}}\nolimits \left ( {\mathcal A}_{\left ( \left ( C_{\theta _{0.8}}\right )^c\right )_\zeta } \right )$ commutes with $\beta _{g}^{C_{\left [0,\theta _{0.8}\right ]}}$ and $\beta _{g}^{C_{\left [0,\theta _{0.8}\right ],U}}$, we get

(3.28)$$ \begin{align} \alpha\circ \beta_g^U\circ \alpha^{-1} &=({\textrm{{inner}}})\circ \left ( \mathop{\mathrm{id}}\nolimits_{{\mathcal A}_{H_D}}\otimes \beta_{g}^{C_{\left(\theta_{0.8},\frac\pi 2\right],U}}\xi_U \right )\circ \left ( \beta_{g}^{C_{\left[0,\theta_{0.8}\right],L,U}}\Xi_{H_U,g,L}\otimes \beta_{g}^{C_{\left[0,\theta_{0.8}\right],R,U}} \Xi_{H_U,g,R}\right ), \notag\\ \alpha\circ \beta_g\circ \alpha^{-1} &=({\textrm{{inner}}})\circ \left ( \beta_{g}^{C_{(\theta_{0.8},\frac\pi 2],D}} \xi_D\otimes\beta_{g}^{C_{(\theta_{0.8},\frac\pi 2],U}}\xi_U \!\right ) \circ \left ( \beta_{g}^{C_{[0,\theta_{0.8}],L}}\Xi_{{\mathbb Z}^2,g,L}\otimes \beta_{g}^{C_{[0,\theta_{0.8}],R}}\Xi_{{\mathbb Z}^2,g,R} \!\right ). \end{align} $$

Furthermore, from the $\beta _g$-invariance of $\omega _0\circ \alpha $, we have

(3.29)$$ \begin{align} \omega_0\circ \alpha\circ \beta_g\circ \alpha^{-1} =\omega_0. \end{align} $$

Now we apply Lemma 3.4 for ${\mathfrak A}_\sigma , {\mathfrak B}_{\sigma \zeta }, {\mathfrak C}_\zeta $ replaced by ${\mathcal A}_{\left ( C_{\left [0,\theta _{0.8}\right ]}\right )_\sigma }, {\mathcal A}_{\left ( C_{\left (\theta _{0.8},\theta _{2}\right ]}\right )_{\sigma ,\zeta }}, {\mathcal A}_{\left ( C_{\left (\theta _2,\frac \pi 2\right ]}\right )_\zeta }$, for $\sigma =L,R$, $\zeta =D,U$. By equations (3.29) and (3.28), $\left (\alpha \circ \beta _g^U\circ \alpha ^{-1}, \alpha \circ \beta _g\circ \alpha ^{-1}\right )$ satisfy the conditions of $(\alpha ,\hat \alpha )$ in Lemma 3.4, for $\omega _0$ and its restrictions. Applying Lemma 3.4, there are $\tilde \eta _{\sigma ,g}\in \mathop {\mathrm {Aut}}\nolimits \left ( {\mathcal A}_{\left ( C_{\theta _2}\right )_\sigma }\right )$, $g\in G$, $\sigma =L,R$, such that

(3.30)$$ \begin{align} \omega_0\circ\alpha\circ \beta_g^U\circ \alpha^{-1} \sim_{\text{q.e.}} \omega_0\circ \left ( \tilde\eta_{Lg}\otimes \tilde\eta_{Rg} \right ),\quad g\in G. \end{align} $$

Because both $\omega _0\circ \alpha \circ \beta _g^U\circ \alpha ^{-1}$ and $\omega _0\circ \left ( \tilde \eta _{Lg}\otimes \tilde \eta _{Rg} \right )$ are pure, by Kadison’s transitivity theorem there exists a unitary $\tilde v_g\in {\mathcal U}({\mathcal A})$ such that

(3.31)$$ \begin{align} \omega_0\circ\alpha\circ \beta_g^U\circ \alpha^{-1} = \omega_0\circ\mathop{\mathrm{Ad}}\nolimits_{\tilde v_g}\circ \left ( \tilde\eta_{Lg}\otimes \tilde\eta_{Rg} \right ),\quad g\in G. \end{align} $$

We define

(3.32)$$ \begin{align} \tilde \beta_g:= \mathop{\mathrm{Ad}}\nolimits\left ( \alpha^{-1}\left (\tilde v_{g^{-1}}\right )\right ) \circ\alpha^{-1} \circ\left ( \tilde\eta_{Lg^{-1}}\otimes \tilde\eta_{Rg^{-1}} \right )\circ\alpha\circ\beta_g^U,\quad g\in G. \end{align} $$

It suffices to show that $\left (\tilde \beta _g\right )\in \mathop {\mathrm {IG}}\nolimits (\omega _0\circ \alpha ,\theta )=\mathop {\mathrm {IG}}\nolimits \left (\omega _0\circ \alpha ,\theta _{2.2}\right )$. By equation (3.31), we have $\omega _0\circ \alpha \circ \tilde \beta _g=\omega _0\circ \alpha $. Therefore, what is left to be proven is that there are $\eta _{g}^\sigma \in \mathop {\mathrm {Aut}}\nolimits \left ( \left ( C_\theta \right )_\sigma \right )$, $g\in G$, $\sigma =L,R$, such that

(3.33)$$ \begin{align} \tilde\beta_g=({\textrm{{inner}}})\circ\left ( \eta_g^L\otimes\eta_g^R\right )\circ\beta_g^U,\quad \text{for all } g\in G. \end{align} $$

By the decomposition (2.12) and the fact that $\tilde \eta _{Lg^{-1}}\otimes \tilde \eta _{Rg^{-1}}$ has support in $C_{\theta _2}$, we have

(3.34)$$ \begin{align} \begin{split} \alpha^{-1} &\circ\left ( \tilde\eta_{Lg^{-1}}\otimes \tilde\eta_{Rg^{-1}} \right )\circ\alpha\\ &=({\textrm{{inner}}})\circ \left ( \alpha_{\left(\theta_{0.8}, \theta_{1.2}\right]}\otimes \alpha_{\left(\theta_{1.8},\theta_{2.2}\right]} \right )^{-1} \left ( \alpha_{\left[0,\theta_1\right]}\otimes\alpha_{\left(\theta_1,\theta_2\right]} \right )^{-1} \left ( \tilde\eta_{Lg^{-1}}\otimes \tilde\eta_{Rg^{-1}} \right ) \left ( \alpha_{\left[0,\theta_1\right]}\otimes\alpha_{\left(\theta_1,\theta_2\right]} \right )\\ & \quad \circ \left ( \alpha_{\left(\theta_{0.8}, \theta_{1.2}\right]}\otimes \alpha_{\left(\theta_{1.8},\theta_{2.2}\right]} \right )\\ &=({\textrm{{inner}}})\circ\left ( \eta_g^L\otimes\eta_g^R\right ), \end{split} \end{align} $$

where

(3.35)$$ \begin{align} \begin{split} \eta_{g}^\sigma&=\left ( \alpha_{\left(\theta_{0.8}, \theta_{1.2}\right],\sigma}\otimes \alpha_{\left(\theta_{1.8},\theta_{2.2}\right],\sigma} \right )^{-1} \left ( \alpha_{\left[0,\theta_1\right],\sigma}\otimes\alpha_{\left(\theta_1,\theta_2\right],\sigma} \right )^{-1} \left ( \tilde\eta_{\sigma g^{-1}} \right ) \left ( \alpha_{\left[0,\theta_1\right],\sigma}\otimes\alpha_{\left(\theta_1,\theta_2\right],\sigma} \right ) \\ & \quad \circ \left ( \alpha_{\left(\theta_{0.8}, \theta_{1.2}\right],\sigma}\otimes \alpha_{\left(\theta_{1.8},\theta_{2.2}\right], \sigma} \right )\\ &\quad \in \mathop{\mathrm{Aut}}\nolimits\left ( \left ( C_{\theta_{2.2}}\right )_\sigma\right ),\quad \sigma=L,R. \end{split} \end{align} $$

Substituting this into formula (3.32), we obtain equation (3.33). This completes the proof.

Proof. The point of the proof is that we can derive $\left (\hat \alpha _{L},\hat \alpha _{R},\hat \Theta \right )\in {\mathcal D}_{\alpha \gamma }^{\theta _{1.2}}$ (formulas (4.10) and (4.11)) and $\left ( \gamma ^{-1}\tilde \beta _{g}\gamma \right )\in \mathop {\mathrm {IG}}\nolimits (\omega \circ \gamma ,\theta _{1.2}), \left (\hat \eta _{g}^{\sigma }\right )\in {\mathcal T}\left (\theta _{1.2}, \left ( \gamma ^{-1}\tilde \beta _{g}\gamma \right )\right )$ (formula (4.16)) from the corresponding objects for $\alpha $, using the factorisation property of $\alpha ,\gamma $. And it is straightforward to see that the $\beta _{g}^{U}$-invariance of $\gamma _{C}$ results in $\mathop {\mathrm {IP}}\nolimits \left ( \omega , \alpha , \theta _{2}, \left (\tilde \beta _g\right ), (\eta _{g}^\sigma ), (\alpha _L,\alpha _R,\Theta ) \right ) =\mathop {\mathrm {IP}}\nolimits \left ( \omega \circ \gamma , \alpha \circ \gamma , \theta _{1.2}, \left (\gamma ^{-1}\tilde \beta _g\gamma \right ), \left (\hat \eta _{g}^\sigma \right ), \left (\hat \alpha _L,\hat \alpha _R,\hat \Theta \right ) \right )$, which immediately implies the Theorem.

Step 1. From $\omega \in \mathop {\mathcal {SL}}\nolimits $, there is an $ \alpha \in \mathop {\mathrm {EAut}}\nolimits (\omega )$. For any $0<\theta <\frac \pi 2$ fixed, we show that ${\mathcal D}^{\theta }_{\alpha \circ \gamma }\neq \emptyset $, hence $\alpha \circ \gamma \in \mathop {\mathrm {QAut}}\nolimits ({\mathcal A})$ and $\omega \circ \gamma =\omega _{0}\circ \alpha \gamma \in \mathop {\mathcal {SL}}\nolimits $. Set $\theta _{1.2}:=\theta $ and choose

(4.2)$$ \begin{align} 0<\theta_{0}<\theta_{0.8}<\theta_1<\theta_{1.2}:=\theta<\theta_{1.8}<\theta_2<\theta_{2.2}< \theta_{2.8}<\theta_3<\theta_{3.2}<\frac\pi 2. \end{align} $$

Because $\alpha \in \mathop {\mathrm {QAut}}\nolimits ({\mathcal A})$, there exists some $(\alpha _L,\alpha _R,\Theta )\in {\mathcal D}_{\alpha }^{\theta _{2}}$. Setting $\alpha _0:=\alpha _L\otimes \alpha _R$, we have $\alpha =({\textrm {{inner}}})\circ \alpha _0\circ \Theta $. Because $\gamma \in \mathop {\mathrm {GUQAut}}\nolimits \left ({\mathcal A}\right )$, there are $\gamma _{H}\in \mathop {\mathrm {HAut}}\nolimits ({\mathcal A})$ and $\gamma _{C}\in \mathop {\mathrm {GSQAut}}\nolimits ({\mathcal A})$ such that

(4.3)$$ \begin{align} \gamma=\gamma_{C}\circ\gamma_{H}. \end{align} $$

Because $\gamma _{H}\in \mathop {\mathrm {HAut}}\nolimits ({\mathcal A})$, we may decompose $\gamma _{H}$ as

(4.4)$$ \begin{align} \gamma_{H}=({\textrm{{inner}}})\circ\left ( \gamma_{H, {L}}\otimes \gamma_{H,R}\right ) =({\textrm{{inner}}})\circ\gamma_{0}, \end{align} $$

with some $\gamma _{H, {\sigma }}\in \mathop {\mathrm {Aut}}\nolimits \left ( {\mathcal A}_{{\left ( C_{\theta _{0}}\right )_\sigma }}\right )$, $\sigma =L,R$. We set $\gamma _{0}:=\gamma _{H, {L}}\otimes \gamma _{H,R}\in \mathop {\mathrm {Aut}}\nolimits \left ( {\mathcal A}_{{ C_{\theta _{0}}}}\right )$. By definition, $\gamma _{C}\in \mathop {\mathrm {GSQAut}}\nolimits ({\mathcal A})$ allows a decomposition

(4.5)$$ \begin{align} \gamma_{C}&=({\textrm{{inner}}})\circ\gamma_{CS},\notag\\ \gamma_{CS}&=\left ( \gamma_{\left[0,\theta_1\right]}\otimes\gamma_{\left(\theta_1,\theta_2\right]} \otimes \gamma_{\left(\theta_2,\theta_3\right]}\otimes \gamma_{\left(\theta_3,\frac\pi 2\right]} \right ) \circ \left ( \gamma_{\left(\theta_{0.8}, \theta_{1.2}\right]}\otimes \gamma_{\left(\theta_{1.8},\theta_{2.2}\right]} \otimes \gamma_{\left(\theta_{2.8},\theta_{3.2}\right]} \right ), \end{align} $$

with

(4.6)$$ \begin{align} \gamma_X&:=\bigotimes_{\sigma=L,R, \, \zeta=D,U} \gamma_{X,\sigma,\zeta},& \gamma_{\left[0,\theta_{1}\right]}&:=\bigotimes_{\sigma=L,R}\gamma_{\left[0,\theta_{1}\right],\sigma},& \gamma_{\left(\theta_3,\frac\pi 2\right]}&:=\bigotimes_{\zeta=D,U} \gamma_{\left(\theta_3,\frac\pi 2\right],\zeta}, \nonumber\\ \gamma_{X,\sigma,\zeta}&\in \mathop{\mathrm{Aut}}\nolimits\left ({\mathcal A}_{C_{X,\sigma,\zeta}}\right ),& \gamma_{X,\sigma}&:=\bigotimes_{\zeta=U,D}\gamma_{X,\sigma,\zeta},& \gamma_{X,\zeta}&:=\bigotimes_{\sigma=L,R}\gamma_{X,\sigma,\zeta},\\ \gamma_{\left[0,\theta_{1}\right],\sigma}&\in \mathop{\mathrm{Aut}}\nolimits\left ({\mathcal A}_{C_{\left[0,\theta_{1}\right],\sigma}}\right ),& \gamma_{\left(\theta_3,\frac\pi 2\right],\zeta}&\in \mathop{\mathrm{Aut}}\nolimits\left ({\mathcal A}_{C_{\left(\theta_3,\frac\pi 2\right],\zeta}}\right ), \nonumber \end{align} $$

for

(4.7)$$ \begin{align} X=(\theta_1,\theta_2], (\theta_2,\theta_3], (\theta_{0.8},\theta_{1.2}], (\theta_{1.8},\theta_{2.2}], (\theta_{2.8},\theta_{3.2}],\quad \sigma=L,R,\ \zeta=D,U. \end{align} $$

Here we have

(4.8)$$ \begin{align} \gamma_{I}\circ\beta_g^{U}=\beta_g^{U}\circ\gamma_{I}\quad\text{for all } g\in G, \end{align} $$

for any

(4.9)$$ \begin{align} I=[0,\theta_1],(\theta_1,\theta_2], (\theta_2,\theta_3], \left(\theta_3,\frac\pi 2\right], (\theta_{0.8}, \theta_{1.2}], (\theta_{1.8},\theta_{2.2}], (\theta_{2.8},\theta_{3.2}]. \end{align} $$

Set

(4.10)$$ \begin{align} \hat\Theta:=\Theta\circ \left ( \gamma_{\left(\theta_2,\theta_3\right]}\otimes \gamma_{\left(\theta_3,\frac\pi 2\right]} \right ) \circ \left ( \gamma_{\left(\theta_{1.8},\theta_{2.2}\right]} \otimes \gamma_{\left(\theta_{2.8},\theta_{3.2}\right]} \right ) \in\mathop{\mathrm{Aut}}\nolimits\left ( {\mathcal A}_{C_{\theta_{1.8}}^{c}}\right ) \subset \mathop{\mathrm{Aut}}\nolimits\left ( {\mathcal A}_{C_{\theta_{1.2}}^{c}}\right ) \end{align} $$

and

(4.11)$$ \begin{align} \hat\alpha_{\sigma}:= \alpha_{\sigma}\circ \left ( \gamma_{\left[0,\theta_1\right],\sigma}\otimes\gamma_{\left(\theta_1,\theta_2\right],\sigma} \right ) \circ \gamma_{\left(\theta_{0.8}, \theta_{1.2}\right],\sigma} \circ\gamma_{H,\sigma}\in\mathop{\mathrm{Aut}}\nolimits\left({\mathcal A}_{H_{\sigma}}\right),\quad\sigma=L,R. \end{align} $$

We claim

(4.12)$$ \begin{align} \alpha\circ\gamma=({\textrm{{inner}}})\circ\left (\hat\alpha_{L}\otimes\hat \alpha_{R}\right )\circ\hat\Theta. \end{align} $$

This means $(\hat \alpha _{L},\hat \alpha _{R},\hat \Theta )\in {\mathcal D}_{\alpha \gamma }^{\theta _{1.2}}$, hence ${\mathcal D}_{\alpha \gamma }^{\theta }={\mathcal D}_{\alpha \gamma }^{\theta _{1.2}}\neq \emptyset $. The claim (4.12) can be checked as follows. Note that $\gamma _{\left (\theta _2,\theta _3\right ]}\otimes \gamma _{\left (\theta _3,\frac \pi 2\right ]}$ and $\gamma _{\left (\theta _{0.8}, \theta _{1.2}\right ]}$ commute because of their disjoint supports. Because $\Theta \in \mathop {\mathrm {Aut}}\nolimits ({\mathcal A}_{C_{\theta _{2}}^{c}})$, it commutes with $\gamma _{\left [0,\theta _1\right ]}\otimes \gamma _{\left (\theta _1,\theta _2\right ]}$ and $\gamma _{\left (\theta _{0.8}, \theta _{1.2}\right ]}$. Therefore, we have

(4.13)$$ \begin{align} \alpha\circ\gamma&=({\textrm{{inner}}})\circ\alpha_{0}\circ \Theta\circ\left ( \gamma_{\left[0,\theta_1\right]}\otimes\gamma_{\left(\theta_1,\theta_2\right]} \otimes \gamma_{\left(\theta_2,\theta_3\right]}\otimes \gamma_{\left(\theta_3,\frac\pi 2\right]} \right ) \notag\\ &\quad \circ \left ( \gamma_{\left(\theta_{0.8}, \theta_{1.2}\right]}\otimes \gamma_{\left(\theta_{1.8},\theta_{2.2}\right]} \otimes \gamma_{\left(\theta_{2.8},\theta_{3.2}\right]} \right )\circ\gamma_{0}\notag\\ &=({\textrm{{inner}}})\circ\alpha_{0}\circ\left ( \gamma_{\left[0,\theta_1\right]}\otimes\gamma_{\left(\theta_1,\theta_2\right]} \right )\circ\gamma_{\left(\theta_{0.8}, \theta_{1.2}\right]}\circ \Theta\circ\left ( \gamma_{\left(\theta_2,\theta_3\right]}\otimes \gamma_{\left(\theta_3,\frac\pi 2\right]}\right ) \notag\\ &\quad \circ\left ( \gamma_{\left(\theta_{1.8},\theta_{2.2}\right]} \otimes \gamma_{\left(\theta_{2.8},\theta_{3.2}\right]} \right )\circ\gamma_{0}\notag\\ &=({\textrm{{inner}}})\circ\alpha_{0}\circ\left ( \gamma_{\left[0,\theta_1\right]}\otimes\gamma_{\left(\theta_1,\theta_2\right]} \right )\circ\gamma_{\left(\theta_{0.8}, \theta_{1.2}\right]}\circ\hat\Theta\circ\gamma_{0}. \end{align} $$

Because $\gamma _{0}\in \mathop {\mathrm {Aut}}\nolimits \left ( {\mathcal A}_{{ C_{\theta _{0}}}}\right )$ and $\hat \Theta \in \mathop {\mathrm {Aut}}\nolimits \left ( {\mathcal A}_{C_{\theta _{1.8}}^{c}}\right )$ commute, we have

(4.14)$$ \begin{align} \alpha\circ\gamma&=\text{equation }(4.13)= ({\textrm{{inner}}})\circ\alpha_{0}\circ\left ( \gamma_{\left[0,\theta_1\right]}\otimes\gamma_{\left(\theta_1,\theta_2\right]} \right )\circ\gamma_{\left(\theta_{0.8}, \theta_{1.2}\right]}\circ\gamma_{0}\circ\hat\Theta \notag\\ &=({\textrm{{inner}}})\circ\left (\hat\alpha_{L}\otimes\hat \alpha_{R}\right )\circ\hat\Theta, \end{align} $$

proving equation (4.12).

Step 2. From $\mathop {\mathrm {IG}}\nolimits (\omega )\neq \emptyset $, we fix a $0<\theta _{0}<\frac \pi 2$ such that $\mathop {\mathrm {IG}}\nolimits (\omega ,\theta _{0})\neq \emptyset $. We choose $\theta _{0.8},\theta _1,\theta _{1.2},\theta _{1.8},\theta _2,\theta _{2.2}, \theta _{2.8},\theta _3,\theta _{3.2}$ such that

(4.15)$$ \begin{align} 0<\theta_{0}<\theta_{0.8}<\theta_1<\theta_{1.2}<\theta_{1.8}<\theta_2<\theta_{2.2}< \theta_{2.8}<\theta_3<\theta_{3.2}<\frac\pi 2. \end{align} $$

For these $\theta $s, we associate the decomposition of $\gamma $ in step 1. Fix $\left ( \tilde \beta _{g}\right )\in \mathop {\mathrm {IG}}\nolimits (\omega ,\theta _{0})$ and $ \left (\eta _{g}^{\sigma }\right )\in {\mathcal T}\left (\theta _{0}, \left (\tilde \beta _g\right )\right )$. Set $\eta _{g}:=\eta _{g}^{L}\otimes \eta _{g}^{R}$. Note that $\left (\eta _{g}^{\sigma }\right )$ also belongs to ${\mathcal T}\left (\theta _{2}, \left (\tilde \beta _g\right )\right )$. Set

(4.16)$$ \begin{align} \hat\eta_{g}^{\sigma} &:= \left ( \gamma_{\left[0,\theta_1\right],\sigma}\gamma_{\left(\theta_{0.8}, \theta_{1.2}\right],\sigma}\gamma_{H,\sigma}\right )^{-1} \eta_{g}^{\sigma}\left ( \beta_{g}^{\sigma U} \gamma_{\left[0,\theta_1\right],\sigma}\gamma_{\left(\theta_{0.8}, \theta_{1.2}\right],\sigma}\gamma_{H,\sigma} \left ( \beta_{g}^{\sigma U} \right )^{-1}\right )\notag\\ &\quad \in\mathop{\mathrm{Aut}}\nolimits\left ( {\mathcal A}_{\left ( C_{\theta_{1.2}}\right )_{\sigma}}\right ), \end{align} $$

for $\sigma =L,R$. We also set $\hat \eta _{g}:=\hat \eta _{g}^{L}\otimes \hat \eta _{g}^{R}$. We claim that $\left ( \gamma ^{-1}\tilde \beta _{g}\gamma \right )\in \mathop {\mathrm {IG}}\nolimits (\omega \circ \gamma ,\theta _{1.2})$ with $\left (\hat \eta _{g}^{\sigma }\right )\in {\mathcal T}\left (\theta _{1.2}, \left ( \gamma ^{-1}\tilde \beta _{g}\gamma \right )\right )$. Clearly we have

(4.17)$$ \begin{align} \omega\circ\gamma\circ\left ( \gamma^{-1}\tilde \beta_{g}\gamma\right ) =\omega\circ\tilde \beta_{g}\circ\gamma =\omega\circ\gamma. \end{align} $$

Therefore, what remains to be shown is

(4.18)$$ \begin{align} \gamma^{-1}\tilde \beta_{g}\gamma =({\textrm{{inner}}}) \circ\left (\hat\eta_{g}^L\otimes \hat\eta_{g}^R\right )\circ \beta_{g}^{U}. \end{align} $$

To see this, we first have

(4.19)$$ \begin{align} &\gamma^{-1}\circ\eta_{g}\circ\gamma=({\textrm{{inner}}})\circ\gamma_{0}^{-1}\circ \left ( \gamma_{\left(\theta_{0.8}, \theta_{1.2}\right]}\otimes \gamma_{\left(\theta_{1.8},\theta_{2.2}\right]} \otimes \gamma_{\left(\theta_{2.8},\theta_{3.2}\right]} \right )^{-1}\notag\\ &\quad \circ \left ( \gamma_{\left[0,\theta_1\right]}\otimes\gamma_{\left(\theta_1,\theta_2\right]} \otimes \gamma_{\left(\theta_2,\theta_3\right]}\otimes \gamma_{\left(\theta_3,\frac\pi 2\right]} \right )^{-1}\notag\\ & \quad \circ\eta_{g}\circ\left ( \gamma_{\left[0,\theta_1\right]}\otimes\gamma_{\left(\theta_1,\theta_2\right]} \otimes \gamma_{\left(\theta_2,\theta_3\right]}\otimes \gamma_{\left(\theta_3,\frac\pi 2\right]} \right ) \circ \left ( \gamma_{\left(\theta_{0.8}, \theta_{1.2}\right]}\otimes \gamma_{\left(\theta_{1.8},\theta_{2.2}\right]} \otimes \gamma_{\left(\theta_{2.8},\theta_{3.2}\right]} \right ) \gamma_{0} \end{align} $$

from the decomposition of equations (4.3), (4.4) and (4.5). Because $\gamma _{\left (\theta _1,\theta _2\right ]} \otimes \gamma _{\left (\theta _2,\theta _3\right ]}\otimes \gamma _{\left (\theta _3,\frac \pi 2\right ]}$ commutes with $\eta _{g}\in \mathop {\mathrm {Aut}}\nolimits ( {\mathcal A}_{{ C_{\theta _{0}}}})$ and $\gamma _{\left (\theta _{1.8},\theta _{2.2}\right ]} \otimes \gamma _{\left (\theta _{2.8},\theta _{3.2}\right ]}$ commutes with $\left ( \gamma _{\left [0,\theta _1\right ]}\right )^{-1}\eta _{g}\gamma _{[0,\theta _1]}\in \mathop {\mathrm {Aut}}\nolimits ( {\mathcal A}_{{ C_{\theta _{1}}}})$, we have

(4.20)$$ \begin{align} \gamma^{-1}&\circ\eta_{g}\circ\gamma\notag\\ &=\text{equation }(4.19) = ({\textrm{{inner}}})\circ\gamma_{0}^{-1}\circ \left ( \gamma_{\left(\theta_{0.8}, \theta_{1.2}\right]}\otimes \gamma_{\left(\theta_{1.8},\theta_{2.2}\right]} \otimes \gamma_{\left(\theta_{2.8},\theta_{3.2}\right]} \right )^{-1}\notag\\ &\quad\circ \left ( \gamma_{\left[0,\theta_1\right]} \right )^{-1}\circ\eta_{g} \circ\left ( \gamma_{\left[0,\theta_1\right]} \right ) \circ \left ( \gamma_{\left(\theta_{0.8}, \theta_{1.2}\right]}\otimes \gamma_{\left(\theta_{1.8},\theta_{2.2}\right]} \otimes \gamma_{\left(\theta_{2.8},\theta_{3.2}\right]} \right ) \gamma_{0}\notag\\ &=({\textrm{{inner}}})\circ\gamma_{0}^{-1}\circ \left ( \gamma_{\left(\theta_{0.8}, \theta_{1.2}\right]} \right )^{-1}\circ \left ( \gamma_{\left[0,\theta_1\right]} \right )^{-1}\circ\eta_{g} \circ\left ( \gamma_{\left[0,\theta_1\right]} \right ) \circ \left ( \gamma_{\left(\theta_{0.8}, \theta_{1.2}\right]} \right ) \gamma_{0}. \end{align} $$

On the other hand, because $\gamma _{CS}$ and $\beta _{g}^{U}$ commute, we have

(4.21)$$ \begin{align} \gamma^{-1}\circ\beta_{g}^{U}\circ\gamma=({\textrm{{inner}}})\gamma_{0}^{-1}\circ\gamma_{CS}^{-1}\beta_{g}^{U}\gamma_{CS}\gamma_{0} =({\textrm{{inner}}})\gamma_{0}^{-1}\circ\beta_{g}^{U}\gamma_{0}. \end{align} $$

Combining equations (4.20) and (4.21), we obtain

(4.22)$$ \begin{align} \gamma^{-1}\tilde \beta_{g}\gamma &=({\textrm{{inner}}})\circ \gamma_{0}^{-1} \left ( \gamma_{\left(\theta_{0.8}, \theta_{1.2}\right]} \right )^{-1}\circ \left ( \gamma_{\left[0,\theta_1\right]} \right )^{-1}\circ\eta_{g} \circ\left ( \gamma_{\left[0,\theta_1\right]} \right ) \circ \left ( \gamma_{\left(\theta_{0.8}, \theta_{1.2}\right]} \right ) \gamma_{0} \circ \gamma_{0}^{-1}\circ\beta_{g}^{U}\gamma_{0}\notag\\ &=({\textrm{{inner}}})\circ \gamma_{0}^{-1} \left ( \gamma_{\left(\theta_{0.8}, \theta_{1.2}\right]} \right )^{-1}\circ \left ( \gamma_{\left[0,\theta_1\right]} \right )^{-1}\circ\eta_{g}\beta_{g}^{U} \circ\left ( \gamma_{\left[0,\theta_1\right]} \right ) \circ \left ( \gamma_{\left(\theta_{0.8}, \theta_{1.2}\right]} \right ) \circ\gamma_{0}\notag\\ &=({\textrm{{inner}}}) \circ\left (\hat\eta_{g}^L\otimes \hat\eta_{g}^R\right )\circ \beta_{g}^{U}. \end{align} $$

In the second equality, we used the fact that $ \gamma _{[0,\theta _1]} \gamma _{(\theta _{0.8}, \theta _{1.2}]} $ and $\beta _{g}^{U}$ commute. This completes the proof of the claim.

Step 3. We use the setting and notation of steps 1 and 2 (with $\theta _{0}$ chosen in step 2). By Lemma 2.1, there exists

(4.23)$$ \begin{align} ( (W_g), (u_\sigma(g,h)))\in \mathop{\mathrm{IP}}\nolimits\left ( \omega, \alpha, \theta_{2}, (\tilde\beta_g ), (\eta_{g}^\sigma ), (\alpha_L,\alpha_R,\Theta) \right ). \end{align} $$

Now we have

(4.24)$$ \begin{align} \begin{split} &\omega\circ\gamma\in \mathop{\mathcal{SL}}\nolimits,\qquad \alpha\circ\gamma\in \mathop{\mathrm{EAut}}\nolimits(\omega\circ\gamma),\qquad \left ( \gamma^{-1}\circ\tilde \beta_{g}\circ\gamma\right )\in \mathop{\mathrm{IG}}\nolimits(\omega\circ\gamma,\theta_{1.2}),\\ &\left(\hat\eta_{g}^{\sigma}\right)\in {\mathcal T}\left (\theta_{1.2}, \left ( \gamma^{-1}\tilde \beta_{g}\gamma\right )\right ),\qquad \left(\hat \alpha_L,\hat \alpha_R,\hat\Theta\right)\in{\mathcal D}_{\alpha\gamma}^{\theta_{1.2}}. \end{split} \end{align} $$

We claim

(4.25)$$ \begin{align} ( (W_g), (u_\sigma(g,h)))\in \mathop{\mathrm{IP}}\nolimits\left ( \omega\circ\gamma, \alpha\circ\gamma, \theta_{1.2}, \left(\gamma^{-1}\tilde\beta_g\gamma\right), \left(\hat \eta_{g}^\sigma\right), \left(\hat \alpha_L,\hat \alpha_R,\hat \Theta\right) \right ). \end{align} $$

This immediately implies $h(\omega )=h(\omega \circ \gamma )$. To prove the claim, we first see from formulas (4.10) and (4.11) that

(4.26)$$ \begin{align} &\left ( \hat \alpha_L\otimes\hat \alpha_R\right )\circ \hat\Theta\circ \gamma_{0}^{-1} \left ( \gamma_{\left(\theta_{0.8}, \theta_{1.2}\right]} \right )^{-1}\circ \left ( \gamma_{\left[0,\theta_1\right]} \right )^{-1}\notag\\ &=\alpha_{0}\circ \left ( \gamma_{\left[0,\theta_1\right]}\otimes\gamma_{\left(\theta_1,\theta_2\right]} \right ) \circ \gamma_{\left(\theta_{0.8}, \theta_{1.2}\right]} \circ\gamma_{0} \circ\Theta\circ \left ( \gamma_{\left(\theta_2,\theta_3\right]}\otimes \gamma_{\left(\theta_3,\frac\pi 2\right]} \right ) \circ \left ( \gamma_{\left(\theta_{1.8},\theta_{2.2}\right]} \otimes \gamma_{\left(\theta_{2.8},\theta_{3.2}\right]} \right )\notag\\ &\quad \circ \gamma_{0}^{-1} \left ( \gamma_{\left(\theta_{0.8}, \theta_{1.2}\right]} \right )^{-1}\circ \left ( \gamma_{\left[0,\theta_1\right]} \right )^{-1}\notag\\ &=\alpha_{0}\circ \left ( \gamma_{\left[0,\theta_1\right]}\otimes\gamma_{\left(\theta_1,\theta_2\right]} \right ) \circ \Theta\circ \left ( \gamma_{\left(\theta_2,\theta_3\right]}\otimes \gamma_{\left(\theta_3,\frac\pi 2\right]} \right ) \circ \left ( \gamma_{\left(\theta_{1.8},\theta_{2.2}\right]} \otimes \gamma_{\left(\theta_{2.8},\theta_{3.2}\right]} \right )\circ \left ( \gamma_{\left[0,\theta_1\right]} \right )^{-1}, \end{align} $$

because $\gamma _{\left (\theta _{0.8}, \theta _{1.2}\right ]}\circ \gamma _{0}\in \mathop {\mathrm {Aut}}\nolimits ( {\mathcal A}_{{ C_{\theta _{1.2}}}})$ and $\Theta \circ ( \gamma _{(\theta _2,\theta _3 ]}\otimes \gamma _{\left (\theta _3,\frac \pi 2\right ]} ) \circ ( \gamma _{\left (\theta _{1.8},\theta _{2.2}\right ]} \otimes \gamma _{\left (\theta _{2.8},\theta _{3.2}\right ]} )\in \mathop {\mathrm {Aut}}\nolimits ( {\mathcal A}_{{ C_{\theta _{1.8}}}^{c}})$ commute. Furthermore, because $\gamma _{\left [0,\theta _1\right ]}$ and $\Theta \circ ( \gamma _{(\theta _2,\theta _3 ]}\otimes \gamma _{\left (\theta _3,\frac \pi 2\right ]} ) \circ \left ( \gamma _{\left (\theta _{1.8},\theta _{2.2}\right ]} \otimes \gamma _{\left (\theta _{2.8},\theta _{3.2}\right ]} \right ) \in \mathop {\mathrm {Aut}}\nolimits ( {\mathcal A}_{{ C_{\theta _{1.8}}}^{c}})$ commute and $\gamma _{\left (\theta _1,\theta _2\right ]}$ and $\Theta \in \mathop {\mathrm {Aut}}\nolimits ( {\mathcal A}_{{ C_{\theta _{2}}}^{c}} )$ commute, we have

(4.27)$$ \begin{align} \left ( \hat \alpha_L\otimes\hat \alpha_R\right )&\circ \hat\Theta\circ \gamma_{0}^{-1} \left ( \gamma_{\left(\theta_{0.8}, \theta_{1.2}\right]} \right )^{-1}\circ \left ( \gamma_{\left[0,\theta_1\right]} \right )^{-1} =\text{equation }(4.26) \notag\\ & =\alpha_{0}\circ \gamma_{\left(\theta_1,\theta_2\right]} \circ \Theta\circ \left ( \gamma_{\left(\theta_2,\theta_3\right]}\otimes \gamma_{\left(\theta_3,\frac\pi 2\right]} \right ) \circ \left ( \gamma_{\left(\theta_{1.8},\theta_{2.2}\right]} \otimes \gamma_{\left(\theta_{2.8},\theta_{3.2}\right]} \right )\notag\\ &=\alpha_{0}\circ \Theta\circ\gamma_{\left(\theta_1,\theta_2\right]} \circ\left ( \gamma_{\left(\theta_2,\theta_3\right]}\otimes \gamma_{\left(\theta_3,\frac\pi 2\right]} \right ) \circ \left ( \gamma_{\left(\theta_{1.8},\theta_{2.2}\right]} \otimes \gamma_{\left(\theta_{2.8},\theta_{3.2}\right]} \right ) =\alpha_{0}\circ \Theta \circ\hat\gamma. \end{align} $$

Here $\hat \gamma :=\gamma _{(\theta _1,\theta _2 ]} \circ ( \gamma _{ (\theta _2,\theta _3 ]}\otimes \gamma _{\left (\theta _3,\frac \pi 2\right ]} ) \circ \left ( \gamma _{\left (\theta _{1.8},\theta _{2.2}\right ]} \otimes \gamma _{\left (\theta _{2.8},\theta _{3.2}\right ]} \right )\in \mathop {\mathrm {Aut}}\nolimits ( {\mathcal A}_{{ C_{\theta _{1}}}^{c}} )$ commutes with $\beta _{g}^{U}$. Combining this and

(4.28)$$ \begin{align} \hat \eta_{g}\beta_{g}^{U}= \left ( \gamma_{\left[0,\theta_1\right]}\gamma_{\left(\theta_{0.8}, \theta_{1.2}\right]}\gamma_{0}\right )^{-1} \eta_{g} \beta_{g}^{ U} \gamma_{\left[0,\theta_1\right]}\gamma_{\left(\theta_{0.8}, \theta_{1.2}\right]}\gamma_{0}, \end{align} $$

we obtain

(4.29)$$ \begin{align} \pi_{0}\circ \left ( \hat \alpha_L\otimes\hat \alpha_R\right )\circ \hat\Theta\circ\hat\eta_{g}\beta_{g}^{U} ( \hat\Theta )^{-1} \left ( \hat \alpha_L\otimes\hat \alpha_R\right )^{-1} =\pi_{0}\circ\alpha_{0}\circ \Theta \circ\hat\gamma\circ \eta_{g} \beta_{g}^{ U}\circ \hat\gamma^{-1}\circ\Theta^{-1}\circ \alpha_{0}^{-1}. \end{align} $$

Because $\hat \gamma $ commutes with $\beta _{g}^{U}$ and $ \eta _{g}\in \mathop {\mathrm {Aut}}\nolimits ( {\mathcal A}_{{ C_{\theta _{0}}}} )$ commutes with $\hat \gamma \in \mathop {\mathrm {Aut}}\nolimits ( {\mathcal A}_{{ C_{\theta _{1}}}^{c}} )$, we have

(4.30)$$ \begin{align} &\pi_{0}\circ \left ( \hat \alpha_L\otimes\hat \alpha_R\right )\circ \hat\Theta\circ\hat\eta_{g}\beta_{g}^{U} ( \hat\Theta )^{-1} \left ( \hat \alpha_L\otimes\hat \alpha_R\right )^{-1} \notag \\&\quad =\text{equation }(4.29) =\pi_{0}\circ\alpha_{0}\circ \Theta \circ \eta_{g} \beta_{g}^{ U} \circ\Theta^{-1}\circ \alpha_{0}^{-1} =\mathop{\mathrm{Ad}}\nolimits( W_{g} )\circ\pi_{0}. \end{align} $$

Hence the condition for $W_{g}$ in formula (4.25) is checked. On the other hand, substituting formulas (4.11) and (4.16), we get

(4.31)$$ \begin{align} \pi_R&\circ\hat \alpha_R\circ\hat \eta_g^R\beta_g^{R U} \hat \eta_h^R\left (\beta_g^{R U}\right )^{-1}\left ( \hat \eta_{gh}^R\right )^{-1}\hat\alpha_{R}^{-1}\notag\\ &= \pi_{R}\circ\alpha_R\circ \left ( \gamma_{\left[0,\theta_1\right],R}\otimes\gamma_{\left(\theta_1,\theta_2\right],R} \right ) \circ \gamma_{\left(\theta_{0.8}, \theta_{1.2}\right],R} \circ\gamma_{H,R} \circ \left ( \gamma_{\left[0,\theta_1\right],R}\circ\gamma_{\left(\theta_{0.8}, \theta_{1.2}\right],R}\circ\gamma_{H,R}\right )^{-1}\notag\\ & \quad \eta_g^R\beta_g^{RU} \eta_h^R\left (\beta_g^{R U}\right )^{-1}\left ( \eta_{gh}^R\right )^{-1}\circ \gamma_{\left[0,\theta_1\right],R}\circ\gamma_{\left(\theta_{0.8}, \theta_{1.2}\right],R} \notag\\ &\quad\circ\gamma_{H,R} \circ \left ( \left ( \gamma_{\left[0,\theta_1\right],R}\otimes\gamma_{\left(\theta_1,\theta_2\right],R} \right ) \circ \gamma_{\left(\theta_{0.8}, \theta_{1.2}\right],R} \circ\gamma_{H,R} \right )^{-1}\alpha_R^{-1}\notag\\ &= \pi_{R}\circ\alpha_R\circ \gamma_{\left(\theta_1,\theta_2\right],R} \circ \eta_g^R\beta_g^{RU} \eta_h^R\left (\beta_g^{R U}\right )^{-1}\left ( \eta_{gh}^R\right )^{-1}\circ \left ( \gamma_{\left(\theta_1,\theta_2\right],R} \right )^{-1}\circ\alpha_R^{-1}. \end{align} $$

Because $ \eta _g^R\beta _g^{RU} \eta _h^R(\beta _g^{R U})^{-1}( \eta _{gh}^R)^{-1}\in \mathop {\mathrm {Aut}}\nolimits ( {\mathcal A}_{{ C_{\theta _{0}}}})$ commutes with $\gamma _{\left (\theta _1,\theta _2\right ],R}$, we obtain

(4.32)$$ \begin{align} \pi_R&\circ\hat \alpha_R\circ\hat \eta_g^R\beta_g^{R U} \hat \eta_h^R\left (\beta_g^{R U}\right )^{-1}\left ( \hat \eta_{gh}^R\right )^{-1}\hat\alpha_{R}^{-1} =\text{equation }(4.31)\notag\\ &=\pi_{R}\circ\alpha_R\circ \eta_g^R\beta_g^{RU} \eta_h^R\left (\beta_g^{R U}\right )^{-1}\left ( \eta_{gh}^R\right )^{-1} \alpha_R^{-1} =\mathop{\mathrm{Ad}}\nolimits\left ( u_{R}(g,h)\right )\circ\pi_{R}. \end{align} $$

An analogous statement for $\sigma =L$ also holds. This completes the proof of formula (4.25). Hence the statement of the theorem is proven.

Proof. Fix arbitrary

(5.1)$$ \begin{align} 0<\theta_{0.8}<\theta_1<\theta_{1.2}<\theta_{1.8}<\theta_2<\theta_{2.2}< \theta_{2.8}<\theta_3<\theta_{3.2}<\frac\pi 2. \end{align} $$

We show the existence of the decomposition

(5.2)$$ \begin{align} \begin{split} \tau_{1,0}^\Psi=&\mathop{\mathrm{Ad}}\nolimits(u)\circ \left ( \alpha_{\left(0,\theta_1\right]}\otimes \alpha_{\left(\theta_1,\theta_2\right]}\otimes \alpha_{\left(\theta_2,\theta_3\right]} \otimes \alpha_{\left(\theta_3,\frac\pi 2\right]} \right )\\ &\circ \left ( \alpha_{\left(\theta_{0.8},\theta_{1.2}\right]} \alpha_{\left(\theta_{1.8},\theta_{2.2}\right]}\otimes \alpha_{\left(\theta_{2.8},\theta_{3.2}\right]} \right ), \end{split} \end{align} $$

with $\alpha $s of the forms in formulas (2.13) and (2.14). We follow the strategy of [Reference Naaijkens and OgataNO].

Step 1. Fix some $0<\theta '<\theta $ and set

(5.3)$$ \begin{align} \tilde F(r):=\frac{\exp\left ( {-r^{\theta'}}\right )}{(1+r)^{4}}. \end{align} $$

With a suitably chosen constant $c_{1}>0$, we have

(5.4)$$ \begin{align} \max\left\{ F\left (\frac r 3\right ), \left ( F\left ( \left [ \frac r 3 \right ] \right )\right )^{\frac 12}\right\}\le c_{1}\tilde F(r),\quad r\ge 0. \end{align} $$

Namely, $c_{1}\tilde F(r)$ satisfy the condition on $\tilde F_{\theta }$ in Definition C.2(ii) for our $F=\frac {\exp \left ( {-cr^{\theta }}\right )}{(1+r)^{4}}$ and $\theta =\frac 12$. Set

(5.5)$$ \begin{align} {\mathcal C}_{0}&:=\left\{ \begin{gathered} C_{\left[0,\theta_1\right],\sigma}, \ C_{\left(\theta_1,\theta_2\right], \sigma,\zeta}, \ C_{\left(\theta_2,\theta_3\right], \sigma,\zeta}, \ C_{\left(\theta_3,\frac\pi 2\right], \zeta},\\ \sigma=L,R,\ \zeta=D,U \end{gathered} \right\}, \end{align} $$

(5.6)$$ \begin{align} {\mathcal C}_{1}&:=\left\{ \begin{gathered} C_{\left(\theta_{0.8},\theta_{1.2}\right),\sigma,\zeta}, \ C_{\left(\theta_{1.8},\theta_{2.2}\right), \sigma,\zeta}, \ C_{\left(\theta_{2.8},\theta_{3.2}\right), \sigma,\zeta},\\ \sigma=L,R,\ \zeta=D,U \end{gathered} \right\}. \end{align} $$

Define $\Psi ^{(0)}, \Psi ^{(1)}\in \hat {\mathcal B}_{F}([0,1])$ by

(5.7)$$ \begin{align} \begin{split} \Psi^{(0)}\left ( X; t\right )&:= \begin{cases} \Psi\left ( X; t\right ) & \text{if there exists a } C\in{\mathcal C}_0 \text{ such that } X\subset C, \\ 0 & \text{otherwise}, \end{cases} \\ \Psi^{(1)}\left ( X; t\right )&:=\Psi^{(0)}\left ( X; t\right )-\Psi\left ( X; t\right ), \end{split} \end{align} $$

for each $X\in {\mathfrak S}_{{\mathbb Z}^2}$, $t\in [0,1]$.

First we would like to represent $ ( \tau _{1,0}^{\Psi ^{(0)}})^{-1}\circ \tau _{1,0}^{ \Psi }$ as some quasilocal automorphism. Set $t,s\in [0,1]$. We apply Proposition D.6 for $\Psi $ replaced by $\Psi ^{(1)}$ and $\tilde \Psi $ by $\Psi $. Hence we set

(5.8)$$ \begin{align} \Xi^{(s)}\left ( Z, t \right ):= \sum_{m\ge 0} \sum_{X\subset Z,\, X(m)=Z} \Delta_{X(m)}\left ( \tau_{t,s}^{\Psi}\left ( \Psi^{(1)}\left ( X; t\right ) \right ) \right ) \end{align} $$

and

(5.9)$$ \begin{align} \Xi^{(n)(s)}\left ( Z, t \right ):= \sum_{m\ge 0} \sum_{X\subset Z, \, X(m)\cap\Lambda_{n}=Z} \Delta_{X(m)}\left ( \tau_{t,s}^{\left(\Lambda_n\right)\Psi}\left ( \Psi^{{(1)}}\left ( X; t\right ) \right ) \right ). \end{align} $$

Corresponding to equation (D.31), we obtain

(5.10)$$ \begin{align} \tau_{t,s}^{\left(\Lambda_n\right),\Psi} \left ( H_{\Lambda_n,\Psi^{(1)}}(t) \right ) = H_{\Lambda_n,\Xi^{(n)(s)}}(t). \end{align} $$

Applying Proposition D.6. we have $\Xi ^{(n)(s)}, \Xi ^{(s)}\in \hat {\mathcal B}_{\tilde F}([0,1])$, and

(5.11)$$ \begin{align} \lim_{n\to\infty}\left \lVert \tau_{t,u}^{\Xi^{(n)(s)}}\left ( A\right ) -\tau_{t,u}^{\Xi^{(s)}}\left ( A\right ) \right \rVert=0,\quad A\in{\mathcal A},\ t,u\in [0,1], \end{align} $$

holds. Two functions $\hat \tau _{t,s}^{(\Lambda _n), \Xi ^{(n)(s)}}(A)$ and $ \tau _{t,s}^{(\Lambda _n ), \Psi }\circ ( \tau _{t,s}^{(\Lambda _n ),\Psi ^{(0)}} )^{-1}(A)$ satisfy the same differential equation and initial condition. Therefore we obtain

(5.12)$$ \begin{align} \hat\tau_{t,s}^{\left(\Lambda_n\right), \Xi^{(n)(s)}}(A)= \tau_{t,s}^{\left(\Lambda_n\right), \Psi}\circ \left ( \tau_{t,s}^{\left(\Lambda_n\right),\Psi^{(0)}}\right )^{-1}(A),\quad t\in [0,1],\ A\in{\mathcal A}. \end{align} $$

From the fact that $ \hat \tau _{t,u}^{\left (\Lambda _n\right ), \Xi ^{(n)(s)}}=\tau _{u,t}^{\left (\Lambda _n\right ), \Xi ^{(n)(s)}}=\tau _{u,t}^{\Xi ^{(n)(s)}}$ converges strongly to an automorphism $\tau _{u,t}^{\Xi ^{(s)}}$ on ${\mathcal A}$ (equation (5.11)), we have

(5.13)$$ \begin{align} \lim_{n\to\infty}\left \lVert \hat\tau_{t,s}^{\left(\Lambda_{n}\right)\Xi^{(n)(s)}}\left ( A\right ) -\tau_{s,t}^{\Xi^{(s)}}\left ( A\right ) \right \rVert=0,\quad A\in{\mathcal A}. \end{align} $$

On the other hand, by Theorem D.3 we have, for $t \in [0,1]$ and $A \in {\mathcal A}$,

(5.14)$$ \begin{align} \lim_{n\to\infty}\left \lVert \tau_{t,s}^{\left(\Lambda_n\right), \Psi}\circ \left ( \tau_{t,s}^{\left(\Lambda_n\right),\Psi^{(0)}}\right )^{-1}(A) -\tau_{t,s}^{ \Psi}\circ \left ( \tau_{t,s}^{\Psi^{(0)}}\right )^{-1}(A) \right \rVert=0. \end{align} $$

Therefore, taking the $n\to \infty $ limit in equation (5.12), we obtain

(5.15)$$ \begin{align} \tau_{s,t}^{ \Xi^{(s)}}(A)= \tau_{t,s}^{\Psi}\circ \left ( \tau_{t,s}^{\Psi^{(0)}}\right )^{-1}(A),\quad t,s\in [0,1],\ A\in{\mathcal A}. \end{align} $$

Hence we have

(5.16)$$ \begin{align} \tau_{s,t}^{\Psi}=\left ( \tau_{t,s}^{\Psi}\right )^{-1} =\left ( \tau_{t,s}^{\Psi^{(0)}}\right )^{-1}\left (\tau_{s,t}^{ \Xi^{(s)}}\right )^{-1}= \tau_{s,t}^{\Psi^{(0)}}\tau_{t,s}^{ \Xi^{(s)}}. \end{align} $$

In particular, we get

(5.17)$$ \begin{align} \tau_{1,0}^{\Psi}= \tau_{1,0}^{\Psi^{(0)}}\tau_{0,1}^{ \Xi^{(1)}}. \end{align} $$

Step 2. We show

(5.18)$$ \begin{align} \sum_{\substack{Z\in{\mathfrak S}\left({\mathbb Z}^2\right), \\\not\exists C\in{\mathcal C}_{1}\text{ s.t. }Z\subset C}} \sup_{t\in \left[0,1\right]}\left \lVert \Xi^{(1)}\left ( Z,t\right )\right \rVert<\infty. \end{align} $$

From this,

(5.19)$$ \begin{align} V(t):=\sum_{\substack{Z\in{\mathfrak S}\left({\mathbb Z}^2\right), \\\not\exists C\in{\mathcal C}_{1}\text{ s.t. }Z\subset C}} \Xi^{(1)}\left ( Z,t\right ) \in{\mathcal A} \end{align} $$

converges absolutely in the norm topology and defines an element in ${\mathcal A}$. Furthermore, for

(5.20)$$ \begin{align} V_{n}(t):=\sum_{\substack{Z\in{\mathfrak S}\left({\mathbb Z}^2\right),\, Z\subset \Lambda_{n} \\\not\exists C\in{\mathcal C}_{1}\text{ s.t. }Z\subset C} } \Xi^{(1)}\left ( Z,t\right ) \in{\mathcal A}_{\Lambda_{n}},\quad n\in\mathbb{N}, \end{align} $$

we get

(5.21)$$ \begin{align} \lim_{n\to\infty}\sup_{t\in\left[0,1\right]}\left \lVert V_{n}(t)-V(t)\right \rVert=0 \end{align} $$

from formula (5.18).

To prove formula (5.18), we first bound

(5.22)$$ \begin{align} \begin{split} \sum_{\substack{Z\in{\mathfrak S}\left({\mathbb Z}^2\right) \\ \not\exists C\in{\mathcal C}_{1}\text{ s.t. }Z\subset C } } &\sup_{t\in\left[0,1\right]} \left \lVert \Xi^{(1)}\left ( Z,t\right ) \right \rVert \\ &\le \sum_{\substack{Z\in{\mathfrak S}\left({\mathbb Z}^2\right), \\ \not\exists C\in{\mathcal C}_{1}\text{ s.t. }Z\subset C } } \sum_{m\ge 0} \sum_{\substack{X: X\subset Z\\ X(m)=Z}} \left[ \sup_{t\in\left[0,1\right]}\left \lVert \Delta_{X(m)}\left ( \tau_{t,1}^{\Psi}\left ( \Psi^{(1)}\left ( X; t\right ) \right ) \right ) \right \rVert \right]\\ &\le \sum_{m\ge 0} \sum_{\substack{X: \\ \not\exists C\in{\mathcal C}_{1}\text{ s.t. }X(m)\subset C}} \sup_{t\in\left[0,1\right]}\left \lVert \Delta_{X(m)}\left ( \tau_{t,1}^{\Psi}\left ( \Psi^{(1)}\left ( X; t\right ) \right ) \right ) \right \rVert\\ & \le \sum_{m\ge 0} \sum_{\substack{X: \\ \not\exists C\in{\mathcal C}_{1}\text{ s.t. }X(m)\subset C}} \left[ \sup_{t\in\left[0,1\right]} \frac{8\left \lVert \Psi^{(1)}\left ( X; t\right ) \right \rVert}{C_{F}}\left ( e^{2I_{F}(\Psi)}-1\right )\left \lvert X\right \rvert G_{F}\left ( m\right ) \right] \\ &=\frac{8}{C_{F}}\left ( e^{2I_F(\Psi)}-1\right ) \sum_{m\ge 0} \sum_{\substack{X: \\ \not\exists C\in{\mathcal C}_{1}\text{ s.t. }X(m)\subset C}} \left[ \sup_{t\in\left[0,1\right]}\left ( \left \lVert \Psi^{(1)}\left ( X; t\right ) \right \rVert\right ) \left \lvert X\right \rvert G_{F}\left ( m\right ) \right]. \end{split} \end{align} $$

For the third inequality, we used Theorem D.3 3. For any cone $C_1,C_2$ of ${\mathbb Z}^2$ with its apex at the origin, we set

(5.23)$$ \begin{align} M(C_1,C_2):= \sum_{m\ge 0} \sum_{\substack{X: \\ \forall C\in{\mathcal C}_{1}, \, X\cap \left ( \left ( C^{c}\right )(m)\right )\neq\emptyset,\\ X\cap C_{1}\neq\emptyset,\, X\cap C_{2}\neq\emptyset }} \left[ \sup_{t\in\left[0,1\right]}\left ( \left \lVert \Psi^{(1)}\left ( X; t\right ) \right \rVert\right ) \left \lvert X\right \rvert G_{F}\left ( m\right ) \right]. \end{align} $$

From the definition of $ \Psi ^{(1)}$, we have $ \Psi ^{(1)}\left ( X; t\right ) =0, $ unless X has a nonempty intersection with at least two elements in ${\mathcal C}_{0}$. Therefore, if X gives a nonzero contribution in formula (5.22), then it has to satisfy

$$ \begin{gather*} X\cap \left ( \left ( C^{c}\right )(m)\right )\neq\emptyset,\quad \text{for all } C\in{\mathcal C}_{1},\\ \exists C_{1},C_{2}\in{\mathcal C}_{0}\text{ such that } C_1\neq C_2, \ X\cap C_{1}\neq\emptyset,\ X\cap C_{2}\neq\emptyset. \end{gather*} $$

Hence we have

(5.24)$$ \begin{align} \begin{split} \text{formula }(5.22)\le \frac{8}{C_{F}}\left ( e^{2I_F(\Psi)}-1\right ) \sum_{\substack{C_{1},C_{2}\in{\mathcal C}_{0}\\C_1\neq C_2 } } M({C_1,C_2}). \end{split} \end{align} $$

Hence it suffices to show that $M({C_1,C_2})<\infty $ for all $C_{1},C_{2}\in {\mathcal C}_{0}$ with $C_1\neq C_2$.

In order to proceed, we prepare two estimates. We will freely identify ${\mathbb C}$ and ${\mathbb R}^{2}$ in an obvious manner. In particular, $\arg z$ of $z\in {\mathbb Z}^{2}\subset {\mathbb R}^{2}$ in the following definition is considered with this identification: For $\varphi _{1}<\varphi _{2}$, we set

(5.25)$$ \begin{align} \check C_{\left[\varphi_{1},\varphi_{2}\right]}:= \left\{ z\in {\mathbb Z}^{2}\mid \arg z\in [\varphi_{1},\varphi_{2}] \right\}. \end{align} $$

We define $\check C_{\left (\varphi _{1},\varphi _{2}\right )}$ and so on analogously. Set

(5.26)$$ \begin{align} {c^{(0)}}_{\zeta_1,\zeta_2,\zeta_3,\zeta_4} :=\sqrt{1-\max\left\{\cos (\zeta_3-\zeta_2),\cos (\zeta_4-\zeta_1),0\right\}},\quad \zeta_1,\zeta_2,\zeta_3,\zeta_4 \in {\mathbb R}. \end{align} $$

Lemma 5.3. Set $\varphi _1<\varphi _2<\varphi _3<\varphi _4$ with $\varphi _4-\varphi _1<2\pi $. Then

$$ \begin{align*} \begin{split} b_0(\varphi_1,\varphi_2,\varphi_3,\varphi_4) &:= \sum_{m\ge 0} \sum_{\substack{X: \\ X\cap \check C_{\left[\varphi_1,\varphi_2\right]}\neq\emptyset\\ X\cap \check C_{\left[\varphi_3,\varphi_4\right]}\neq\emptyset }} \left[ \sup_{t\in\left[0,1\right]}\left (\left \lVert \Psi\left ( X; t\right )\right \rVert\right ) \left \lvert X\right \rvert G_{F}\left ( m\right ) \right] \notag\\ &\le (64)^3 \frac {3^{4}\kappa_{1, 4, F}}{\left ( c^{(0)}_{\varphi_1,\varphi_2,\varphi_3,\varphi_4}\right ) ^{4}} \left ( \left \lVert\left \lvert \Psi_1\right \rVert\right \rvert_{F}\right ) \left ( \sum_{m\ge 0}G_{F}\left ( m\right )\right )<\infty. \end{split} \end{align*} $$

Proof. Substituting Lemma C.4, we obtain

(5.27)$$ \begin{align} b_0(\varphi_1,\varphi_2,\varphi_3,\varphi_4) &:= \sum_{m\ge 0} \sum_{\substack{X: \\ X\cap \check {C}_{\left[\varphi_1,\varphi_2\right]}\neq\emptyset\\ X\cap \check {C}_{\left[\varphi_3,\varphi_4\right]}\neq\emptyset }} \left[ \sup_{t\in\left[0,1\right]}\left ( \left \lVert \Psi\left ( X; t\right ) \right \rVert\right ) \left \lvert X\right \rvert G_{F}\left ( m\right ) \right] \notag\\ &\le \sum_{m\ge 0} \sum_{\substack{x\in \check {C}_{\left[\varphi_1,\varphi_2\right]} \\y\in \check {C}_{\left[\varphi_3,\varphi_4\right]}}} \sum_{X\ni x,y}\left[ \sup_{t\in\left[0,1\right]}\left ( \left \lVert \Psi\left ( X; t\right ) \right \rVert\right ) \left \lvert X\right \rvert G_{F}\left ( m\right ) \right]\notag\\ &\le \left ( \left \lVert\left \lvert \Psi_1\right \rVert\right \rvert_{F}\right ) \sum_{\substack{x\in \check {C}_{\left[\varphi_1,\varphi_2\right]}\\ y\in \check {C}_{\left[\varphi_3,\varphi_4\right]}}} F\left ( {\mathrm{d}}(x,y)\right ) \left ( \sum_{m\ge 0}G_{F}\left ( m\right )\right )\notag\\ &\le (64)^3 \frac {3^{4}\kappa_{1, 4, F}}{\left ( c^{(0)}_{\varphi_1,\varphi_2,\varphi_3,\varphi_4}\right ) ^{4}} \left ( \left \lVert\left \lvert \Psi_1\right \rVert\right \rvert_{F}\right ) \left ( \sum_{m\ge 0}G_{F}\left ( m\right )\right )<\infty. \end{align} $$

We used Lemma C.4 in the last inequality. The last value is finite by equation (C.14) for our $F\in {\mathcal F}_{a}$.

Set

(5.28)$$ \begin{align} {c^{(1)}}_{\zeta_1,\zeta_2,\zeta_3} :=\sqrt{1-\max\left\{\cos (\zeta_1-\zeta_2),\cos (\zeta_1-\zeta_3)\right\}},\quad \zeta_1,\zeta_2,\zeta_3\in [0,2\pi). \end{align} $$

Lemma 5.4. For $\varphi _1<\varphi _2<\varphi _3$ with $\varphi _3-\varphi _1<\frac \pi 2$, we have

(5.29)$$ \begin{align} \begin{split} b_1(\varphi_1,\varphi_2,\varphi_3) &:=\sum_{m\ge 0} \sum_{\substack{X: \\ X\subset \check {C}_{\left[\varphi_1,\varphi_3\right]}\\ X\cap \check {C}_{\left[\varphi_1,\varphi_2\right]}\neq\emptyset\\ X\cap \check C_{\left[\varphi_2,\varphi_3\right]}\neq\emptyset\\ X\cap \left ( \left ( \left ( \check C_{\left(\varphi_1,\varphi_3\right)}\right )^c\right )(m)\right )\neq \emptyset }} \left[ \sup_{t\in\left[0,1\right]}\left ( \left \lVert \Psi\left ( X; t\right ) \right \rVert\right ) \left \lvert X\right \rvert G_{F}\left ( m\right ) \right]\\ &\le64\cdot 144\cdot 24\cdot \left ( \pi \kappa_{1,2,F}+F(0) \right ) \left ( \left \lVert\left \lvert \Psi_1\right \rVert\right \rvert_F\right ) \left ( \sum_{m\ge 0} (m+1)^{4}G_{F}\left ( m\right )\right ) \\ &\quad \left ( \left ( {c^{(1)}}_{\varphi_{1},\varphi_{2},\varphi_{3}}\right )^{-4} +\left ( {c^{(1)}}_{\varphi_{3},\varphi_{1},\varphi_2}\right )^{-4} \right ) <\infty. \end{split} \end{align} $$

Proof. Set

(5.30)$$ \begin{align} L_\varphi:= \left\{ z\in{\mathbb R}^2\mid \arg z=\varphi \right\},\quad \varphi\in [0,2\pi). \end{align} $$

Note that if $X\in {\mathfrak S}_{{\mathbb Z}^2}$ satisfies $X\subset \check {C}_{\left [\varphi _1,\varphi _3\right ]}$ and $X\cap ( ( ( \check C_{ (\varphi _1,\varphi _3 )} )^c )(m) )\neq \emptyset $, then we have

(5.31)$$ \begin{align} d(X, L_{\varphi_1})\le m\quad\text{or}\quad d(X, L_{\varphi_3})\le m. \end{align} $$

Therefore, we have

(5.32)$$ \begin{align}\sum_{m\ge 0}& \sum_{\substack{X: \\ X\subset \check {C}_{\left[\varphi_1,\varphi_3\right]}\\ X\cap \check {C}_{\left[\varphi_1,\varphi_2\right]}\neq\emptyset\\ X\cap \check C_{\left[\varphi_2,\varphi_3\right]}\neq\emptyset\\ X\cap \left ( \left ( \left ( \check C_{\left(\varphi_1,\varphi_3\right)}\right )^c\right )(m)\right )\neq \emptyset }} \left[ \sup_{t\in\left[0,1\right]}\left ( \left \lVert \Psi\left ( X; t\right ) \right \rVert\right ) \left \lvert X\right \rvert G_{F}\left ( m\right ) \right] \nonumber \\ &\le\sum_{m\ge 0} G_{F}\left ( m\right ) \left ( \sum_{\substack{X: \\ X\cap \check C_{\left[\varphi_2,\varphi_3\right]}\neq\emptyset\\ d\left(X, L_{\varphi_1}\right)\le m }} +\sum_{\substack{X: \\ X\cap \check{C}_{\left[\varphi_1,\varphi_2\right]}\neq\emptyset\\ d\left(X, L_{\varphi_3}\right)\le m }} \right ) \left[ \sup_{t\in\left[0,1\right]}\left ( \left \lVert \Psi\left ( X; t\right ) \right \rVert\right ) \left \lvert X\right \rvert \right] \nonumber\\ &\le \sum_{m\ge 0} G_{F}\left ( m\right ) \left ( \sum_{\substack{x\in \check C_{\left[\varphi_2,\varphi_3\right]}\\ y\in L_{\varphi_1}(m) }} +\sum_{\substack{x\in \check {C}_{\left[\varphi_1,\varphi_2\right]}\\ y\in L_{\varphi_3}(m) }} \right ) \sum_{\substack{X: X\ni x,y }} \left[ \sup_{t\in\left[0,1\right]}\left ( \left \lVert \Psi\left ( X; t\right ) \right \rVert\right ) \left \lvert X\right \rvert \right] \nonumber\\ &\le \left ( \left \lVert\left \lvert \Psi_1\right \rVert\right \rvert_F\right ) \sum_{m\ge 0} G_{F}\left ( m\right ) \left ( \sum_{\substack{x\in \check C_{\left[\varphi_2,\varphi_3\right]}\\ y\in L_{\varphi_1}(m) }} +\sum_{\substack{x\in \check {C}_{\left[\varphi_1,\varphi_2\right]}\\ y\in L_{\varphi_3}(m) }} \right ) F\left ( {\mathrm{d}}(x,y)\right )\nonumber\\ &\le 64\cdot 144\cdot 24\cdot \left ( \pi \kappa_{1,2,F}+F(0) \right ) \left ( \left \lVert\left \lvert \Psi_1\right \rVert\right \rvert_F\right )\nonumber \\ &\quad \left ( \sum_{m\ge 0} (m+1)^{4}G_{F}\left ( m\right )\right ) \left ( \left ( {c^{(1)}}_{\varphi_{1},\varphi_{2},\varphi_{3}}\right )^{-4} +\left ( {c^{(1)}}_{\varphi_{3},\varphi_{1},\varphi_2}\right )^{-4} \right ). \end{align} $$

In the last inequality, we used Lemma C.5 with $\varphi _3-\varphi _1<\frac \pi 2$. Because $\varphi _3-\varphi _1<\frac \pi 2$ and because of formula (C.14), the last value is finite.

Now let us go back to the estimate of formula (5.23). If $C_1,C_2\in {\mathcal C}_0$ are $C_1=\check {C}_{\left [\varphi _1,\varphi _2\right ]}, C_2=\check {C}_{\left [\varphi _3,\varphi _4\right ]}$ with $\varphi _1<\varphi _2<\varphi _3<\varphi _4, \varphi _4-\varphi _1<2\pi $, then from Lemma 5.3, we have

(5.33)$$ \begin{align} \begin{split} & M(C_1,C_2) \le b_0(\varphi_1,\varphi_2,\varphi_3,\varphi_4) <\infty. \end{split} \end{align} $$

Now suppose that $C_1,C_2\in {\mathcal C}_0$ are $C_1=\check {C}_{\left [\varphi _1,\varphi _2\right ]}, C_2=\check C_{\left [\varphi _2,\varphi _3\right ]}$ with $\varphi _1<\varphi _2<\varphi _3, \varphi _3-\varphi _1<2\pi $. (Recall definition (5.5).) By the definition of ${\mathcal C}_0$ and ${\mathcal C}_1$, there is some $C=C_{\left (\zeta _1,\zeta _2\right )}\in {\mathcal C}_1$ such that $\varphi _1<\zeta _1<\varphi _2<\zeta _2<\varphi _3$ and $\zeta _2-\zeta _1<\frac \pi 2$. For $X\in {\mathfrak S}_{{\mathbb Z}^2}$ to give a nonzero contribution in formula (5.23), it has to satisfy

(5.34)$$ \begin{align} X(m)\cap ( \check C_{\left[\zeta_1,\zeta_2\right]})^c\neq\emptyset,\qquad X\cap \check {C}_{\left[\varphi_1,\varphi_2\right]}\neq\emptyset,\qquad X\cap \check C_{\left[\varphi_2,\varphi_3\right]}\neq\emptyset. \end{align} $$

For such an X, one of the following occurs:

1. (i) $X\cap \check C_{\left [\zeta _2,\varphi _3\right ]}\neq \emptyset $ and $X\cap \check {C}_{\left [\varphi _1,\varphi _2\right ]}\neq \emptyset $.

2. (ii) $X\cap \check C_{\left [\varphi _1,\zeta _1\right ]}\neq \emptyset $ and $X\cap \check C_{\left [\varphi _2,\varphi _3\right ]}\neq \emptyset $.

3. (iii) $X\cap \check C_{\left [\varphi _2, \zeta _2\right ]}\neq \emptyset (\text {and } X\cap \check C_{\left [\zeta _1,\varphi _2\right ]}\neq \emptyset )$ and $X\cap \check C_{\left [\varphi _3,\varphi _1+2\pi \right ]}\neq \emptyset $.

4. (iv) $X\subset \check C_{\zeta _1,\zeta _2}$, $X\cap ( ( \check C_{\zeta _1,\zeta _2} )^c )(m)\neq \emptyset $, $X\cap \check C_{\left [\varphi _2, \zeta _2\right ]}\neq \emptyset $ and $X\cap \check C_{\left [\zeta _1,\varphi _2\right ]}\neq \emptyset $.

Hence we get

(5.35)$$ \begin{align} \begin{split} M({C_1,C_2}) &\le b_0(\varphi_1,\varphi_2, \zeta_2,\varphi_3) +b_0(\varphi_1,\zeta_1, \varphi_2,\varphi_3) +b_0(\varphi_2, \zeta_2, \varphi_3,\varphi_1+2\pi) +b_1(\zeta_1,\varphi_2,\zeta_2)\\ &<\infty. \end{split} \end{align} $$

Hence we have proven the claim of step 2.

Step 3. Next we set

(5.36)$$ \begin{align} \tilde\Xi(Z,t):= \begin{cases} \Xi^{(1)}(Z,t) & \text{if }\exists C\in{\mathcal C}_{1}\text{ s.t. } Z\subset C,\\ 0&\text{otherwise}. \end{cases} \end{align} $$

Clearly, we have $\tilde \Xi \in \hat {\mathcal B}_{\tilde F}([0,1])$. Note that

(5.37)$$ \begin{align} H_{\Lambda_{n}, \tilde\Xi}(t)+V_{n}(t) = H_{\Lambda_{n}, \Xi^{(1)}}(t). \end{align} $$

As a uniform limit of $[0,1]\ni t\mapsto V_{n}(t)\in {\mathcal A}$ (equation (5.21)), $[0,1]\ni t\mapsto V(t)\in {\mathcal A}$ is norm-continuous. Because $\tilde \Xi \in \hat {\mathcal B}_{\tilde F}([0,1])$, $[0,1]\ni t\mapsto \tau _{t,s}^{\tilde \Xi }\left ( V(t)\right )\in {\mathcal A}$ is also norm-continuous, for each $s\in [0,1]$. Therefore, for each $s\in [0,1]$, there is a unique norm-differentiable map $[0,1]\ni t \mapsto W^{(s)}(t) \in {\mathcal U}\left ( {\mathcal A}\right )$ such that

(5.38)$$ \begin{align} \frac{d}{dt} W^{(s)}(t)=-i \tau_{t,s}^{\tilde\Xi}\left ( V(t)\right ) W^{(s)}(t),\quad W^{(s)}(s)=\mathbb I. \end{align} $$

It is given by

(5.39)$$ \begin{align} W^{(s)}(t) :=\sum_{k=0}^{\infty }(-i)^{k} \int_{s}^{t}ds_{1}\int_{s}^{s_{1}}ds_{2}\dotsi \int_{s}^{s_{k-1}}ds_{k} \tau_{s_{1},s}^{\tilde\Xi}\left ( V(s_{1})\right ) \dotsm \tau_{s_{k},s}^{\tilde\Xi}\left ( V(s_{k})\right ). \end{align} $$

Analogously, for each $s\in [0,1]$ and $n\in \mathbb {N}$, we define a unique norm-differentiable map from $[0,1]$ to $ {\mathcal U}\left ( {\mathcal A}\right )$ such that

(5.40)$$ \begin{align} \frac{d}{dt} W_{n}^{(s)}(t)=-i \tau_{t,s}^{(\Lambda_{n})\tilde\Xi}\left ( V_{n}(t)\right ) W_{n}^{(s)}(t),\quad W_{n}^{(s)}(s)=\mathbb I. \end{align} $$

It is given by

(5.41)$$ \begin{align} W_{n}^{(s)}(t) :=\sum_{k=0}^{\infty }(-i)^{k} \int_{s}^{t}ds_{1}\int_{s}^{s_{1}}ds_{2}\dotsi \int_{s}^{s_{k-1}}ds_{k} \tau_{s_{1},s}^{\left(\Lambda_{n}\right)\tilde\Xi}\left ( V_{n}(s_{1})\right ) \dotsm \tau_{s_{k},s}^{\left(\Lambda_{n}\right)\tilde\Xi}\left ( V_{n}(s_{k})\right ). \end{align} $$

By the uniform convergence (5.21) and Lemma D.3, we have

(5.42)$$ \begin{align}\lim_{n\to\infty} \sup_{t\in\left[0,1\right]}\left \lVert \tau_{t,s}^{\left(\Lambda_{n}\right)\tilde\Xi}\left ( V_{n}(t)\right ) -\tau_{t,s}^{\tilde\Xi}\left ( V(t)\right )\right \rVert=0. \end{align} $$

From this and formulas (5.39) and (5.41), we obtain

(5.43)$$ \begin{align} \lim_{n\to\infty}\sup_{t\in\left[0,1\right]}\left \lVert W_{n}^{(s)}(t)- W^{(s)}(t) \right \rVert =0. \end{align} $$

This and Theorem D.3 4 for $\Xi ^{(1)}, \tilde \Xi \in {\mathcal B}_{\tilde F}([0,1])$ imply

(5.44)$$ \begin{align} \begin{split} \lim_{n\to\infty} \tau_{s,t}^{\left(\Lambda_{n}\right), \tilde\Xi}\circ \mathop{\mathrm{Ad}}\nolimits \left ( W_{n}^{(s)}(t)\right ) (A)&= \tau_{s,t}^{ \tilde\Xi}\circ \mathop{\mathrm{Ad}}\nolimits \left ( W^{(s)}(t)\right ) (A),\\ \lim_{n\to\infty} \tau_{s,t}^{\left(\Lambda_{n}\right), \Xi^{(1)}}(A)&= \tau_{s,t}^{ \Xi^{(1)}}(A), \end{split} \end{align} $$

for any $A\in {\mathcal A}$.

Note that for any $A\in {\mathcal A}$,

$$ \begin{align*} \begin{split} \frac{d}{dt} \tau_{s,t}^{\left(\Lambda_{n}\right), \tilde\Xi}\circ \mathop{\mathrm{Ad}}\nolimits \left ( W_{n}^{(s)}(t)\right ) (A) &=-i\left[ H_{\Lambda_{n}, \tilde\Xi}(t), \tau_{s,t}^{\left(\Lambda_{n}\right), \tilde\Xi}\circ \mathop{\mathrm{Ad}}\nolimits \left ( W_{n}^{(s)}(t)\right ) (A) \right] \\ &\quad -i\tau_{s,t}^{\left(\Lambda_{n}\right), \tilde\Xi}\left ( \left[ \tau_{t,s}^{\left(\Lambda_{n}\right), \tilde\Xi}\left ( V_{n}(t)\right ), \mathop{\mathrm{Ad}}\nolimits \left ( W_{n}^{(s)}(t)\right ) (A) \right] \right ) \\ &=-i\left[ H_{\Lambda_{n}, \tilde\Xi}(t)+V_{n}(t), \tau_{s,t}^{\left(\Lambda_{n}\right), \tilde\Xi}\circ \mathop{\mathrm{Ad}}\nolimits \left ( W_{n}^{(s)}(t)\right ) (A) \right]\nonumber\\ &=-i\left[ H_{\Lambda_{n}, \Xi^{(1)}}(t), \tau_{s,t}^{\left(\Lambda_{n}\right), \tilde\Xi}\circ \mathop{\mathrm{Ad}}\nolimits \left ( W_{n}^{(s)}(t)\right ) (A) \right]. \end{split} \end{align*} $$

We used equation (D.10) for the second equality and equation (5.37) for the third. On the other hand, for any $A\in {\mathcal A}$, we have

(5.45)$$ \begin{align} \frac{d}{dt} \tau_{s,t}^{\left(\Lambda_{n}\right), \Xi^{(1)}} (A)=-i\left[ H_{\Lambda_{n}, \Xi^{(1)}}(t), \tau_{s,t}^{\left(\Lambda_{n}\right), \Xi^{(1)}} (A) \right]. \end{align} $$

Therefore, $\tau _{s,t}^{\left (\Lambda _{n}\right ), \tilde \Xi }\circ \mathop {\mathrm {Ad}}\nolimits ( W_{n}^{(s)}(t) ) (A)$ and $ \tau _{s,t}^{\left (\Lambda _{n}\right ), \Xi ^{(1)}} (A)$ satisfy the same differential equation. Also note that we have $\tau _{s,s}^{\left (\Lambda _{n}\right ), \tilde \Xi }\circ \mathop {\mathrm {Ad}}\nolimits ( W_{n}^{(s)}(s) ) (A)= \tau _{s,s}^{\left (\Lambda _{n}\right ), \Xi ^{(1)}} (A)=A$. Therefore, we get

(5.46)$$ \begin{align} \tau_{s,t}^{\left(\Lambda_{n}\right), \tilde\Xi}\circ \mathop{\mathrm{Ad}}\nolimits \left ( W_{n}^{(s)}(t)\right ) (A) =\tau_{s,t}^{\left(\Lambda_{n}\right), \Xi^{(1)}} (A). \end{align} $$

By equation (5.44), we obtain

(5.47)$$ \begin{align} \tau_{s,t}^{\tilde\Xi}\circ \mathop{\mathrm{Ad}}\nolimits \left ( W^{(s)}(t)\right ) (A) =\tau_{s,t}^{\Xi^{(1)}} (A),\quad A\in{\mathcal A}, \ t,s\in[0,1]. \end{align} $$

Taking the inverse, we get

(5.48)$$ \begin{align} \mathop{\mathrm{Ad}}\nolimits \left ( W^{(s)^{*}}(t)\right )\circ\tau_{t,s }^{\tilde\Xi} =\tau_{t,s}^{\Xi^{(1)}},\quad t,s\in[0,1]. \end{align} $$

Step 4. Combining equations (5.17) and (5.48), we have

(5.49)$$ \begin{align} \tau_{1,0}^{\Psi}= \tau_{1,0}^{\Psi^{(0)}}\tau_{0,1}^{ \Xi^{(1)}} =\tau_{1,0}^{\Psi^{(0)}}\circ\mathop{\mathrm{Ad}}\nolimits \left ( \left ( W^{(1)}(0)\right )^{*}\right )\circ\tau_{0,1 }^{\tilde\Xi}. \end{align} $$

By the definitions of $\Psi ^{(0)}$ and $\tilde \Xi $, we obtain decompositions

(5.50)$$ \begin{align} \begin{split} \tau_{1,0}^{\Psi^{(0)}} &=\alpha_{\left[0,\theta_{1}\right]}\otimes \alpha_{\left(\theta_1,\theta_2\right]}\otimes \alpha_{\left(\theta_2,\theta_3\right]} \otimes \alpha_{\left(\theta_3,\frac\pi 2\right]},\\ \tau_{0,1}^{\tilde\Xi} &= \alpha_{\left(\theta_{0.8},\theta_{1.2}\right]}\otimes \alpha_{\left(\theta_{1.8},\theta_{2.2}\right]}\otimes \alpha_{\left(\theta_{2.8},\theta_{3.2}\right]}, \end{split} \end{align} $$

with $\alpha $s in the form of formulas (2.13) and (2.14). This completes the proof of the first part.

Step 5. Suppose that $\beta _{g}^{U}\left ( \Psi (X;t)\right )=\Psi (X;t)$ for any $X\in {\mathfrak S}_{{\mathbb Z}^2}$, $t\in [0,1]$ and $g\in G$. Then clearly we have $\beta _{g}^{U}\left ( \Psi ^{(0)}(X;t)\right )=\Psi ^{(0)}(X;t)$ for any $X\in {\mathfrak S}_{{\mathbb Z}^2}$, $t\in [0,1]$ and $g\in G$. By Theorem D.3 5, this implies $\tau _{1,0}^{\Psi ^{(0)}}{\beta _g^U}={\beta _g^U}\tau _{1,0}^{\Psi ^{(0)}}$. From the decomposition (5.50), this means that all of $\alpha _{\left [0,\theta _{1}\right ],\sigma }, \alpha _{\left (\theta _1,\theta _2\right ],\sigma ,\zeta }, \alpha _{\left (\theta _2,\theta _3\right ],\sigma ,\zeta }, \alpha _{\left (\theta _3,\frac \pi 2\right ],\zeta }$, $\sigma =L,R, \zeta =U,D$, commute with ${\beta _g^U}$. Because $\Pi _{X}$ commutes with $\beta _{g}^{U}$, $\tau _{t,s}^{\Psi }$ commutes with $\beta _{g}^{U}$ (Theorem D.3 5), and $\Psi ^{(1)}$ and $\Xi ^{(s)}$ are $\beta _{g}^{U}$-invariant from the definition (5.8). Therefore, from the definition (5.36), $\tilde \Xi $ is also $\beta _{g}^{U}$-invariant. Hence by Theorem D.3 5, $\tau _{0,1}^{ \tilde \Xi }$ commutes with ${\beta _g^U}$. The decomposition (5.50) then implies that $\alpha _{\left (\theta _{0.8},\theta _{1.2}\right ],\sigma ,\zeta }, \alpha _{\left (\theta _{1.8},\theta _{2.2}\right ],\sigma ,\zeta }, \alpha _{\left (\theta _{2.8},\theta _{3.2}\right ],\sigma ,\zeta }$, $\sigma =L,R, \ \zeta =U,D$, commute with ${\beta _g^U}$.

Proof. Define $\tilde F$ as in formula (5.3) with some $0<\theta '<\theta $. The same argument as in Theorem 5.2, step 2, implies that there exists $\Xi ^{(1)}\in \hat {\mathcal B}_{\tilde F}[0,1]$ with $\tilde F\in {\mathcal F}_a$, such that

(5.52)$$ \begin{align} \tau_{1,0}^{\Psi}= \tau_{1,0}^{\Psi^{(0)}}\tau_{0,1}^{ \Xi^{(1)}}. \end{align} $$

This $\Xi ^{(1)}$ is given by formula (5.8) for current $\Psi $ and $\Psi ^{(1)}\left ( X; t\right ):=\Psi ^{(0)}\left ( X; t\right )-\Psi \left ( X; t\right )$. To prove the theorem, it suffices to show that $\tau _{0,1}^{ \Xi ^{(1)}}$ belongs to $\mathop {\mathrm {HAut}}\nolimits ({\mathcal A})$. Indeed, for any $0<\theta _0<\frac \pi 4$, as in Theorem 5.2, step 2, we have

(5.53)$$ \begin{align} &\sum_{\substack{Z: Z\nsubseteq C_{\left[0,\theta_0\right],L} \\Z\nsubseteq C_{\left[0,\theta_0\right],R} }} \sup_{t\in \left[0,1\right]}\left \lVert \Xi^{(1)}\left ( Z,t\right )\right \rVert\nonumber\\ &\qquad\qquad\le \frac{8}{C_{F}}\left ( e^{2I_F(\Psi)}-1\right ) \sum_{m\ge 0} \sum_{\substack{X: X(m)\nsubseteq C_{\left[0,\theta_0\right],L} \\X(m)\nsubseteq C_{\left[0,\theta_0\right],R} } } \left[ \sup_{t\in\left[0,1\right]}\left ( \left \lVert \Psi^{(1)}\left ( X; t\right ) \right \rVert\right ) \left \lvert X\right \rvert G_{F}\left ( m\right ) \right] <\infty. \end{align} $$

To see this, note that if X in the last line has a nonzero contribution to the sum, then at least one of the following occurs:

1. (i) $X\cap C_{\left [\theta _{0}, \frac \pi 2\right ],U}\neq \emptyset $ and $X\cap H_{D}\neq \emptyset $.

2. (ii) $X\cap C_{\left [\theta _{0}, \frac \pi 2\right ],D}\neq \emptyset $ and $X\cap H_{U}\neq \emptyset $.

3. (iii) $X\subset C_{\left [0,\theta _{0}\right ]}$ and

   1. (1) $X\cap C_{\left [0,\theta _{0}\right ],L}\neq \emptyset $ and $X\cap C_{\left [0,\theta _{0}\right ],R}\neq \emptyset $, or

   2. (2) $X\subset C_{\left [0,\theta _{0}\right ], R}$, $X\cap \check C_{\left [0,\theta _{0}\right ]}\neq \emptyset $, $X\cap \check C_{\left [-\theta _{0},0\right ]}\neq \emptyset $ and $X{(m)}\cap \left ( C_{\left [0,\theta _{0}\right ], R}\right )^{c}\neq \emptyset $, or

   3. (3) $X\subset C_{\left [0,\theta _{0}\right ], L}$, $X\cap \check C_{\left [\pi -\theta _{0},\pi \right ]}\neq \emptyset $, $X\cap \check C_{\left [\pi , \pi +\theta _{0}\right ]}\neq \emptyset $ and $X{(m)}\cap \left ( C_{\left [0,\theta _{0}\right ], L}\right )^{c}\neq \emptyset $.

Therefore, the summation in the second line of formula (5.53) is bounded by

$$ \begin{align*} \frac{8}{C_{F}}\left ( e^{2I_F(\Psi)}-1\right ) \left ( \begin{gathered} b_{0}(\theta_{0}, \pi-\theta_{0}, \pi, 2\pi) +b_{0}(0,\pi, \pi+\theta_{0}, 2\pi-\theta_{0}) +b_{0}(-\theta_{0}, \theta_{0}, \pi-\theta_{0}, \pi+\theta_{0})\\ +b_{1}(-\theta_{0}, 0,\theta_{0}) +b_{1}(\pi-\theta_{0}, \pi, \pi+\theta_{0}) \end{gathered} \right )<\infty, \end{align*} $$

from Lemmas 5.3 and 5.4, proving formula (5.53).

Therefore, as in step 3 of Theorem 5.2, setting

(5.54)$$ \begin{align} \tilde\Xi(Z,t):= \begin{cases} \Xi^{(1)}(Z,t)&\text{if } Z\subseteq C_{\left[0,\theta_0\right],L} \text{ or }Z\subseteq C_{\left[0,\theta_0\right],R}, \\ 0&\text{otherwise}, \end{cases} \end{align} $$

we obtain $\tau _{0,1}^{ \Xi ^{(1)}}=({\textrm {{inner}}})\circ \tau _{0,1}^{ \tilde \Xi }$. By the definition, $\tau _{0,1}^{ \tilde \Xi }$ decomposes as $\tau _{0,1}^{ \tilde \Xi }= \zeta _L\otimes \zeta _R$, with some $\zeta _\sigma \in \mathop {\mathrm {Aut}}\nolimits ({\mathcal A}_{C_{\left [0,\theta _0\right ],\sigma }} )$, $\sigma =L,R$. As this holds for any $0<\theta _0<\frac \pi 4$, we conclude $\tau _{0,1}^{ \Xi ^{(1)}}\in \mathop {\mathrm {HAut}}\nolimits ({\mathcal A})$.