Proof. Since star and plaquette operators commute, and for any \(s\in \mathop{\mathrm{{\mathbb{S}}}}_{\Lambda}\), we have \(e^{\beta A_s} = \cosh(\beta)\mathbb{1}+ \sinh(\beta)A_s\), since \(A_s^2=\mathbb{1}\) and analogously for plaquettes, we can expand the partition function: \[\require{physics} \begin{align} Z^\boxplus_{\Lambda} &= \Tr e^{-\beta H^\medwhitestar_{\Lambda}}e^{-\beta H^\square_{\Lambda}} = \Tr \left(\prod_{s\in \mathop{\mathrm{{\mathbb{S}}}}_{\Lambda}} (\cosh(\beta)\mathbb{1}+ \sinh(\beta)A_s) \prod_{p\in \mathop{\mathrm{{\mathbb{P}}}}_{\Lambda}} (\cosh(\beta)\mathbb{1}+ \sinh(\beta)B_p)\right) \notag\\ &= \cosh(\beta)^{|\mathop{\mathrm{{\mathbb{S}}}}_{\Lambda}| + |\mathop{\mathrm{{\mathbb{P}}}}_{\Lambda}|} \sum_{T\subseteq \mathop{\mathrm{{\mathbb{S}}}}_{\Lambda}} \tanh(\beta)^{|T|} \sum_{Q\subseteq \mathop{\mathrm{{\mathbb{P}}}}_{\Lambda}} \tanh(\beta)^{|Q|} \Tr\left(\bigotimes_{v\in dT} X_v \bigotimes_{v' \in \partial Q}Z_v\right) . \label{eq:HTexp} \end{align}\tag{20}\] The last line also used that \(\prod_{s\in T}A_s = \bigotimes_{v\in dT}X_v\) by definition of the coboundary \(dT\) of the star subset.

Since the Pauli operators are all traceless, the trace in the left side of 20 is non-zero if and only if the product \(\bigotimes_{v\in dT} X_v \bigotimes_{v' \in \partial Q}Z_{v'} = \mathbb{1}\). Expressing the trace in the \(X\)-basis, and using the fact that, in this basis, the diagonal elements of any product over Pauli \(Z\)-operators is independent of the \(X\)-state, we thus arrive at \[\require{physics} \Tr\left(\bigotimes_{v\in dT} X_v \bigotimes_{v' \in \partial Q}Z_v\right) = \Tr\left(\bigotimes_{v\in dT} X_v \right) \Tr\left( \bigotimes_{v' \in \partial Q}Z_v\right) 2^{-|\Lambda|}\] where \(2^{-|\Lambda|}\) corrects for the second \(\require{physics} \Tr \mathbb{1}\) term if both traces are non-zero. Plugging this back into the expansion of the partition function shows the factorization. The factorization of the Gibbs state follows directly from the factorization of the partition function in case \(R = \Lambda\). To decompose the reduced Gibbs state, we perform a similar expansion, replacing \(\require{physics} \Tr\) with \(\require{physics} \Tr_{R^c}\). We then use \[\require{physics} \Tr_{R^c}\left(\bigotimes_{v\in dT} X_v \bigotimes_{v' \in \partial Q}Z_v\right) = \Tr_{R^c}\left(\bigotimes_{v\in dT} X_v \right) \Tr_{R^c}\left( \bigotimes_{v' \in \partial Q}Z_v\right) 2^{-|R^c|} ,\] since the trace is only non-zero if both \(dT \cap R^c=\emptyset\) and \(\partial Q \cap R^c = \emptyset\). Together with the decomposition of the partition function, this yields the overall correction factor of \(2^{|\Lambda|-|R^c|}=2^{|R|}\) in ?? . Furthermore, considering only the star part, we find \[\require{physics} \Tr_{R^c}\left(\prod_{s\in T} A_s\right) = 1_{dT\cap R^c=\emptyset} 2^{|R^c|}\prod_{s\in T} A_s ,\] since \(\prod_{s\in T} A_s\) has no support on \(R^c\) iff \(dT\cap R^c=\emptyset\). Hence, the reduced star Gibbs state is still a polynomial of star operators \[\require{physics} \label{eq:traceoutGibbs} \Tr_{R^c}(e^{-\beta H^\medwhitestar_{\Lambda}}) = \cosh(\beta)^{|\mathop{\mathrm{{\mathbb{S}}}}_{\Lambda}|} 2^{|R^c|} \sum_{T\subseteq \mathop{\mathrm{{\mathbb{S}}}}_{\Lambda}} \tanh(\beta)^{|T|} 1_{dT\cap R^c=\emptyset} \prod_{s\in T} A_s ,\tag{21}\] and thus commutes with any plaquettes. Similarly, the marginal of any plaquette Gibbs state is a polynomial of plaquettes and commutes with any star. This yields ?? .

For the expectation of \(X\)- or \(Z\)-diagonal operators, the above expansion can be used on \(\require{physics} \Tr(e^{-\beta H^\boxplus_{\Lambda}} O_X)\), which, together with \[\require{physics} \Tr\left(\bigotimes_{v\in dT} X_v \bigotimes_{v' \in \partial Q}Z_v O_X\right) = \Tr\left(\bigotimes_{v\in dT} X_v O_X\right) \Tr\left( \bigotimes_{v' \in \partial Q}Z_v\right) 2^{-|\Lambda|}\] establishes the claim. The last identity is again most easily checked by expressing the trace in the \(X\)-basis, and using the fact that, in this basis, the diagonal elements of any product over Pauli \(Z\)-operators is independent of the \(X\)-state.

Finally, for the von Neumann entropy, we use the factorization of the Gibbs state and its expectation to find \[\require{physics} \begin{align} S(\rho^\boxplus_R) &= -\Tr ( \rho^\boxplus_R \ln ( \rho^\medwhitestar_R)) -\Tr ( \rho^\boxplus_R \ln ( \rho_R^\square)) - \ln 2^{|R|}\\ &= -\Tr (\rho^\medwhitestar_R \ln ( \rho_R^\medwhitestar)) -\Tr ( \rho_R^\square\ln ( \rho_R^\square)) - \ln 2^{|R|} , \end{align}\] since \(\ln \rho_R^\medwhitestar\) and \(\ln \rho_R^\square\) are diagonal in \(X\)- and \(Z\)-basis, respectively. ◻

Proof. We recall our conventions for the logical operators spelled above 2. For the logical \(X\)-operator we compute: \[\require{physics} \Tr(\rho^\boxplus_{\Lambda}X_L) = \Tr(\rho^\medwhitestar_{\Lambda} X_L) = \cosh(\beta)^{|\mathop{\mathrm{{\mathbb{S}}}}_{\Lambda}|}\sum_{T\subseteq \mathop{\mathrm{{\mathbb{S}}}}_{\Lambda}} \tanh(\beta)^{|T|} \Tr(X_L\prod_{s\in T}A_s) \;.\] This expression is \(0\), since \(X_L\prod_{s\in T}A_s\) is a product of Pauli \(X\)-operators. The trace of such a product is \(0\) unless it is the identity, i.e. unless \(X_L = \prod_{s\in T}A_s\). However, as a logical operator, \(X_L\) cannot be expressed as a product of star operators. Analogously \(\require{physics} \Tr(\rho^\boxplus_{\Lambda} Z_L) = \Tr(\rho^\square_{\Lambda} Z_L) =0\). For the expectation of \(Y_L=iX_L Z_L\) in \(\rho^\boxplus_{\Lambda}\), we expand both the star and plaquette Gibbs state, yielding a sum over some constants and traces of the form \[\require{physics} \Tr(X_L Z_L \prod_{s\in T}A_s \prod_{p\in Q}B_p)\] which is \(0\) for the same reasons. ◻

Proof. In this proof, we drop the \(\medwhitestar\) from the classical Gibbs measure. Employing 23 in case of \(R =U\) and \(R' = UVW\), respectively \(R'= UV\), and setting \(C \mathrel{\vcenter{:}}= (UVW)^c\), we arrive at 19 . Using the rules for the conditional probabilities on its right side, one may write: \[\frac{\mu_{UC}(\boldsymbol{x}_{U}|\boldsymbol{x}_{C})}{\mu_{UWC}(\boldsymbol{x}_{U}|\boldsymbol{x}_{W}\boldsymbol{x}_{C})} = \sum_{\boldsymbol{y}_W \in 2^{W}} \mu_{WC}(\boldsymbol{y}_W|\boldsymbol{x}_{C}) \frac{\mu_{UWC}(\boldsymbol{x}_{U}|\boldsymbol{y}_W\boldsymbol{x}_{C})}{\mu_{UWC}(\boldsymbol{x}_{U}|\boldsymbol{x}_{W}\boldsymbol{x}_{C})} \;.\] As is shown below, the Dobrushin and Shlosman condition IIId translated to the present situation implies that for some \(K^{\sharp}, \xi^{\sharp} \in (0,\infty)\): \[\label{eq:dobrushin95summed} \left| \frac{\mu_{UWC}(\boldsymbol{x}_{U}|\boldsymbol{y}_W\boldsymbol{x}_{C})}{\mu_{UWC}(\boldsymbol{x}_{U}|\boldsymbol{x}_{W}\boldsymbol{x}_{C})} - 1 \right| \leq K^{\sharp} e^{-\xi^{\sharp} \mathop{\mathrm{dist}}(U,W)}\tag{24}\] uniformly in the spin configurations. Hence \(\mathrm{DS}_\beta^{\sharp}(K^\sharp, \xi^\sharp)\) holds.

In the rest of the proof, we show that 24 follows indeed from [4]. To this end, we decompose \(W\) by distance to \(U\): \[W = \bigcupplus_{d=d_{min}}^{d_{max}} W_d,\quad W_d\mathrel{\vcenter{:}}= \{v\in W| \mathop{\mathrm{dist}}(v,U)=d\}\] where \(d_{min} = \min_{v\in W} \mathop{\mathrm{dist}}(v,U)\) and \(d_{max} = \max_{v\in W} \mathop{\mathrm{dist}}(v,U)\). For any \(W_d\), we pick an arbitrary enumeration of spins. We fix two spin configurations \(\boldsymbol{x}_W,\boldsymbol{y}_W\in 2^W\), and enumerate all spins in the symmetric difference \(\boldsymbol{x}_W\Delta\boldsymbol{y}_W\), that is, those with different configurations, thereby respecting distance and the enumeration on \(W_d\) we picked. Let \(F_{i}\) for \(i=1,\ldots, |\boldsymbol{x}_W\Delta\boldsymbol{y}_W|\) be the operation that flips spin \(i\). Let \(F_{\leq i} = \prod_{j\leq i}F_j\). Then, in particular \(\boldsymbol{y}_W= F_{\leq |\boldsymbol{x}_W\Delta\boldsymbol{y}_W|}\boldsymbol{x}_W\). We now use a standard telescopic expansion of the fraction in the left-hand side of 24 : \[\begin{align} \frac{\mu_{UWC}(\boldsymbol{x}_{U}|\boldsymbol{y}_W\boldsymbol{x}_{C})}{\mu_{UWC}(\boldsymbol{x}_{U}|\boldsymbol{x}_{W}\boldsymbol{x}_{C})} & = \prod_{i=1}^{|\boldsymbol{x}_W\Delta\boldsymbol{y}_W|} \frac{\mu_{UWC}(\boldsymbol{x}_{U}|F_{\leq i}(\boldsymbol{x}_W)\boldsymbol{x}_{C})}{\mu_{UWC}(\boldsymbol{x}_{U}|F_{\leq i-1}(\boldsymbol{x}_{W})\boldsymbol{x}_{C})} . \end{align}\] By assumption \(\mu\) satisfies the Dobrushin-Sholsmann condition IIId (4). For the current geometry this implies that there exist \(\hat{K}<\infty, \gamma>0\) such that for all \(U,V,W\), all \(\boldsymbol{x}_{U}, \boldsymbol{x}_{W}, \boldsymbol{x}_{C}\) and all \(i\): \[\left| \frac{\mu_{UWC}(\boldsymbol{x}_{U}|F_{\leq i}(\boldsymbol{x}_W)\boldsymbol{x}_{C})}{\mu_{UWC}(\boldsymbol{x}_{U}|F_{\leq i-1}(\boldsymbol{x}_{W})\boldsymbol{x}_{C})} -1\right| \leq \hat{K} e^{-\gamma \mathop{\mathrm{dist}}(i, U)} .\] Inserting this into the left side of 24 , we arrive at \[\begin{align} &\frac{\mu_{UWC}(\boldsymbol{x}_{U}|\boldsymbol{y}_W\boldsymbol{x}_{C})}{\mu_{UWC}(\boldsymbol{x}_{U}|\boldsymbol{x}_{W}\boldsymbol{x}_{C})} \leq \prod_{i=1}^{|\boldsymbol{x}_W\Delta\boldsymbol{y}_W|} \left( 1 + \hat{K} e^{-\gamma \mathop{\mathrm{dist}}(i, U)} \right) \leq \prod_{d=d_{min}}^{d_{max}} \left( 1 + \hat{K} e^{-\gamma d} \right)^{|W_d|}\\ &\leq \exp \left( \sum_{d=d_{min}}^{d_{max}} |W_d| \ln\left( 1 + \hat{K} e^{-\gamma d} \right)\right) \leq \exp \left( \max_{d} |W_d| \hat{K} e^{-\gamma d_{min}} \sum_{d=0}^{\infty} e^{-\gamma d} \right)\\ &\leq 1 + \frac{\max_{d} |W_d| \hat{K}}{\gamma} e^{-\gamma d_{min}} \exp \left( \frac{\max_{d} |W_d| \hat{K}}{\gamma} e^{-\gamma d_{min}} \right) . \end{align}\] If \(U,V,W\) are rectangles as assummed, and if \(\mathop{\mathrm{diam}}(W)\leq 2 \mathop{\mathrm{dist}}(U,W)^2\), then the size of \(W_d\) is bounded by \[|W_d|\leq |W|\leq 2^{D} \mathop{\mathrm{dist}}(U,W)^{2D} C_Q\] where \(C_Q\) is the number of qubits per unit cell. This yields 24 as an the upper bound with \(K^{\sharp} = \frac{2^{D} \hat{K} C_Q}{\gamma} f_1 e^{f_2}\) and \(\xi^{\sharp} = \gamma/2\), where \[f_1 \mathrel{\vcenter{:}}= \max_{d} d^{2D} e^{-\gamma/2 d} \quad f_2 \coloneq \max_{d} f(d) , \quad f(d) \mathrel{\vcenter{:}}= \frac{2^{D} d^{2D} C_Q \hat{K}}{\gamma} e^{-\gamma d} \;.\] Since this upper bound is independent of the spin configurations, by taking quotients, it also implies a lower bound on the left side in terms of \(1 - K^{\sharp} e^{-\xi^{\sharp} \mathop{\mathrm{dist}}(U,W)}\), which completes the argument for 24 . ◻

Proof. Using 4, it is sufficient to show that the model admits a trivial high-temperature expansion. We give the proof for the star case; the plaquette case is analogous. Similarly to 21 , we expand the marginal of the unnormalized Gibbs state: \[\require{physics} \Tr_R e^{-\beta H^\medwhitestar_R} = 2^{|R|} \cosh(\beta)^{|\mathop{\mathrm{{\mathbb{S}}}}_R|} \sum_{T\subseteq \mathop{\mathrm{{\mathbb{S}}}}_R} 1_{dT\cap R=\emptyset} \prod_{s\in T} \tanh(\beta) A_s \;.\] The condition \(dT \cap R = \emptyset\) holds only if \(T=\emptyset\) or \(T=\mathop{\mathrm{{\mathbb{S}}}}_R\). Assume \(dT \cap R = \emptyset\) but neither \(T=\emptyset\) nor \(T=\mathop{\mathrm{{\mathbb{S}}}}_R\). Then, there are \(s\in T, s'\in \mathop{\mathrm{{\mathbb{S}}}}_R\setminus T\). Since \(R\) is star-connected, there is a finite path \(s,v_1,s_2,v_2, \ldots, v_k, s'\) connecting \(s\) to \(s'\) with \(v_i \in R\) for all \(i\). Since \(s\in T, s'\notin T\) there exists some \(v_i\) with \(s_{i-1}\in T\) and \(s_{i}\notin T\). Since, by assumption, there are only two stars connected to \(v_i\), it is connected to an odd number of stars in \(T\) and hence \(v_i\in dT\cap R\), which is a contradiction. ◻

Proof of 4. Let \(U, V, W\subseteq \Lambda\) be as in 3 and let \(\beta<\infty\). Since, by assumption, any rectangle is star-connected, \(UV\), \(VW\), and \(UVW\) are star-connected. The rectangle \(V\) has either one or two connected components; the latter occurs if \(UVW\) has periodic boundary conditions (see 3). First, consider the case where \(V\) is star-connected. We use that \(H^\medwhitestar_{UVW}-H^\medwhitestar_{VW} = H^\medwhitestar_{UV}-H^\medwhitestar_{V}\) has no support on \(VW\) to rewrite the fraction in the DS-condition: \[\require{physics} \begin{align} &\frac{\Tr_{VW} \hat{\rho}^\medwhitestar_{UVW}}{\Tr_{V} \hat{\rho}^\medwhitestar_{UV}}= \frac{\Tr_{VW} e^{-\beta H^\medwhitestar_{VW}}}{\Tr_{UVW} e^{-\beta H^\medwhitestar_{UVW}}} \frac{\Tr_{UV} e^{-\beta H^\medwhitestar_{UV}}}{\Tr_{V} e^{-\beta H^\medwhitestar_{V}}} . \end{align}\] On each of the traces on the right side, we use the trivial high-temperature expansion 25 . To simplify notation let \(\tau =\tanh(\beta)\) and \(A^{\mathop{\mathrm{{\mathbb{S}}}}_R}\mathrel{\vcenter{:}}= 1_{d\mathop{\mathrm{{\mathbb{S}}}}_R \cap R=\emptyset} \prod_{s\in \mathop{\mathrm{{\mathbb{S}}}}_R}A_s\), then the above fraction equals: \[\begin{align} \frac{2^{|VW|} \cosh(\beta)^{|\mathop{\mathrm{{\mathbb{S}}}}_{VW}|}}{2^{|UVW|} \cosh(\beta)^{|\mathop{\mathrm{{\mathbb{S}}}}_{UVW}|}} \frac{\mathbb{1}+ \tau^{|\mathop{\mathrm{{\mathbb{S}}}}_{VW}|}A^{\mathop{\mathrm{{\mathbb{S}}}}_{VW}} }{\mathbb{1}+ \tau^{|\mathop{\mathrm{{\mathbb{S}}}}_{UVW}|}A^{\mathop{\mathrm{{\mathbb{S}}}}_{UVW}}} \cdot \frac{2^{|UV|} \cosh(\beta)^{|\mathop{\mathrm{{\mathbb{S}}}}_{UV}|}}{2^{|V|} \cosh(\beta)^{|\mathop{\mathrm{{\mathbb{S}}}}_{V}|}} \frac{\mathbb{1}+ \tau^{|\mathop{\mathrm{{\mathbb{S}}}}_{UV}|}A^{\mathop{\mathrm{{\mathbb{S}}}}_{UV}} }{\mathbb{1}+ \tau^{|\mathop{\mathrm{{\mathbb{S}}}}_{V}|}A^{\mathop{\mathrm{{\mathbb{S}}}}_{V}}} &\\ =\frac{\mathbb{1}+ \tau^{|\mathop{\mathrm{{\mathbb{S}}}}_{VW}|}A^{\mathop{\mathrm{{\mathbb{S}}}}_{VW}} }{\mathbb{1}+ \tau^{|\mathop{\mathrm{{\mathbb{S}}}}_{UVW}|}A^{\mathop{\mathrm{{\mathbb{S}}}}_{UVW}}} \frac{\mathbb{1}+ \tau^{|\mathop{\mathrm{{\mathbb{S}}}}_{UV}|}A^{\mathop{\mathrm{{\mathbb{S}}}}_{UV}} }{\mathbb{1}+ \tau^{|\mathop{\mathrm{{\mathbb{S}}}}_{V}|}A^{\mathop{\mathrm{{\mathbb{S}}}}_{V}}} & , \end{align}\] where the last step also used \(|\mathop{\mathrm{{\mathbb{S}}}}_{UV}| + |\mathop{\mathrm{{\mathbb{S}}}}_{VW}| = |\mathop{\mathrm{{\mathbb{S}}}}_{UVW}| + |\mathop{\mathrm{{\mathbb{S}}}}_{V}|\). This expression is diagonal in the \(X\) basis. Moreover, since the spectrum of \(A^{\mathop{\mathrm{{\mathbb{S}}}}_R}\) is one of \(\{\pm 1\}, \{1\}, \{0\}\) for any \(R\), we can bound the norm by replacing any \(A^{\mathop{\mathrm{{\mathbb{S}}}}_R}\) by either \(+1\) in the numerator and \(- 1\) in the denominator, hence taking the maximum over all such configurations. This yields \[\require{physics} \begin{align} \left\| \frac{\Tr_{VW} \hat{\rho}^\medwhitestar_{UVW}}{\Tr_{V} \hat{\rho}^\medwhitestar_{UV}} -1 \right\| & \leq \frac{\tau^{|\mathop{\mathrm{{\mathbb{S}}}}_{VW}|} + \tau^{|\mathop{\mathrm{{\mathbb{S}}}}_{UV}|} + \tau^{|\mathop{\mathrm{{\mathbb{S}}}}_{UV}|+|\mathop{\mathrm{{\mathbb{S}}}}_{VW}|} + \tau^{|\mathop{\mathrm{{\mathbb{S}}}}_{UVW}|} +\tau^{ |\mathop{\mathrm{{\mathbb{S}}}}_{V}|} + \tau^{|\mathop{\mathrm{{\mathbb{S}}}}_{UVW}| + |\mathop{\mathrm{{\mathbb{S}}}}_{V}|}}{\left( 1- \tau^{|\mathop{\mathrm{{\mathbb{S}}}}_{UVW}|} \right) \left(1-\tau^{|\mathop{\mathrm{{\mathbb{S}}}}_{V}|}\right)} \notag \\ & \leq \frac{ 6 \tau^{ |\mathop{\mathrm{{\mathbb{S}}}}_{V}|}}{ \left(1-\tau^{|\mathop{\mathrm{{\mathbb{S}}}}_{V}|}\right)^2} , \end{align}\] since \(\tau \in [0, 1)\). The last term is upper-bounded by \(6 \left(1-\tau\right)^{-2} e^{-|\ln(\tau)| |\mathop{\mathrm{{\mathbb{S}}}}_{V}|}\).

If \(V\) consists of two star-separated, connected components, \(V=V_1\uplus V_2\), the Gibbs state on \(V\) factorizes \(\require{physics} \Tr_V e^{-\beta H^{\medwhitestar}_V} = \Tr_{V_1} e^{-\beta H^{\medwhitestar}_{V_1}} \otimes \Tr_{V_2} e^{-\beta H^{\medwhitestar}_{V_2}}\). The other terms remain unchanged. Proceeding as above, we obtain three factors in the denominator and hence find \[\require{physics} \begin{align} \left\| \frac{\Tr_{VW} \hat{\rho}^\medwhitestar_{UVW}}{\Tr_{V} \hat{\rho}^\medwhitestar_{UV}} -1 \right\| &\leq \frac{ 6 }{ \left(1-\tau\right)^3} e^{-|\ln(\tau)| |\mathop{\mathrm{{\mathbb{S}}}}_{V}|} \;. \end{align}\] Taking the maximum over both cases yields \(K^\medwhitestar= 6 (1-\tau)^{-3}\) for ?? .

To obtain a decay rate \(\xi^\medwhitestar> 0\), we bound the number of stars on \(V\): \[|\mathop{\mathrm{{\mathbb{S}}}}_{V}| \geq \frac{|V|}{\max_{s\in \mathop{\mathrm{{\mathbb{S}}}}_V}|ds|}\geq \frac{\mathop{\mathrm{dist}}(U,W)}{C_Q \max_{s\in \mathop{\mathrm{{\mathbb{S}}}}_V}|ds|} ,\] where the maximal number of qubits in a star, \(\max_{s\in \mathop{\mathrm{{\mathbb{S}}}}_V}|ds|\) is bounded for translational invariant systems and where \(C_Q<\infty\) is the number of qubits per unit cell. Thus \[\xi^\medwhitestar= \frac{|\ln(\tau)|}{C_Q \max_{s\in \mathop{\mathrm{{\mathbb{S}}}}_V}|ds|}\] and the model satisfies \(DS^\medwhitestar(K^\medwhitestar, \xi^\medwhitestar)\) at any length scale. ◻

Proof. 1. The self-adjointness of \(\mathop{\mathrm{\mathcal{L}}}_v^\sharp\) with \(\sharp\in \{\medwhitestar, \square\}\) is immediate from ?? . The claim then follows by linearity.

. We use self-adjointness with respect to the GNS-inner product (\(s=1\)) and the unitality of the semigroup to conclude \[\require{physics} \begin{align} \Tr(O \mathop{\mathrm{\mathcal{L}}}^{\sharp, *}_R(\rho^{\boxplus}_{\Lambda})) = \Tr(\mathop{\mathrm{\mathcal{L}}}^{\sharp}_R(O)\rho^{\boxplus}_{\Lambda} \mathbb{1}) = -\Tr(O\rho^{\boxplus}_{\Lambda} \mathop{\mathrm{\mathcal{L}}}^{\sharp}_R(\mathbb{1}))=0 \end{align}\] for any \(O\in \mathop{\mathrm{\mathcal{B}}}(\mathop{\mathrm{\mathcal{H}}}_{\Lambda})\).

For the proof of the remaining assertions, we focus on the \(\medwhitestar\)-case, and recall from 3 that \(\rho^{\boxplus}_{\Lambda} = 2^{|\Lambda|} \rho^{\square}_{\Lambda}\rho^{\medwhitestar}_{\Lambda}\). Since the jump operators \(L_v^\medwhitestar(\omega)\) and their adjoints commute with the plaquette operators \(B_p\) and hence \(\rho^{\square}_{\Lambda}\), the second item implies that \[0 = \mathop{\mathrm{\mathcal{L}}}^{\medwhitestar, *}_R(\rho^{\boxplus}_{\Lambda}) = 2^{|\Lambda|} \rho^{\square}_{\Lambda} \mathop{\mathrm{\mathcal{L}}}^{\medwhitestar,*}_R(\rho_\Lambda^{\medwhitestar})\] and hence \(\mathop{\mathrm{\mathcal{L}}}^{\medwhitestar,*}_R(\rho_\Lambda^{\medwhitestar}) = 0\).

Finally, we may write \(e^{-\beta H_\Lambda^\medwhitestar} = e^{-\beta (H_{\Lambda}^\medwhitestar- H_{R'}^\medwhitestar)} e^{-\beta H_{R'}^\medwhitestar}\). The first factor commutes with the jump operators \(L_v(\omega)^\medwhitestar\) with \(v \in R\) and its adjoint. Since it is not supported on \(R\) it commutes with all \(Z_v\) for \(v\in R\) and, since they are built from stars, it commutes with all star projections \(P^\medwhitestar_{\partial v}(\omega)\). By a similar argument as above, one hence concludes \(\mathop{\mathrm{\mathcal{L}}}^{\medwhitestar,*}_R( e^{-\beta H_{R'}^\medwhitestar} ) = 0\), and hence the claim. ◻

Proof. It suffices to show that the single-site Lindbladians, \(\mathop{\mathrm{\mathcal{L}}}_v^\medwhitestar\) and \(\mathop{\mathrm{\mathcal{L}}}_{v'}^\square\), commute for any two sites \(v,v'\in \Lambda\). This is immediate from the representation 29 together with the fact that the jump operators either commute or anti-commute: \[\begin{align} \label{eq:commjump} L_{v'}^{\square}(\omega') L_v^{\medwhitestar}(\omega) \;& = X_{v'} P^\square_{dv'}(\omega') Z_v P^\medwhitestar_{\partial v}(\omega) = X_{v'} Z_v P^\square_{dv'}(\omega') P^\medwhitestar_{\partial v}(\omega) \notag \\ &= (-1)^{\delta_{v,v'}} L_v^{\medwhitestar} (\omega) L_{v'}^{\square}(\omega') \;. \end{align}\tag{30}\] This follows from the commutation rules of the constituents: as polynomials in star and plaquette operators (cf. 27 ), the projections \(P^\medwhitestar\) and \(P^\square\) commute. Moreover, \(P^\medwhitestar\) commutes with all \(X\)-operators, and \(P^\square\) commutes with all \(Z\)-operators. The same relation holds if we replace one or both jumps by their adjoints, using 28 . The jump operators thus share the commutation relation of \(Z_v\) and \(X_{v'}\). ◻

Proof. The kernel of a Lindbladian with a full-rank fixed point is known [73] to be given by the commutant of the set of jump operators and their adjoints. Since the Gibbs state \(\rho_\Lambda^{\boxplus}\) is a full rank fixed point (5) and the set of jump operators is self-adjoint by 28 , we have: \[\ker{\mathop{\mathrm{\mathcal{L}}}^{\medwhitestar}_R} = \left\{L_v(\omega)^\medwhitestar\right\}'_{v\in R, \omega} =\left\{Z_vP_{\partial v}^\medwhitestar(\omega) \right\}'_{v\in R, \omega} \;.\] We now use the fact that the commutant of a set is equal to the commutant of the algebra generated by this set to replace the jump operators with simpler operators. In a first step, we express \[\sum_{\omega} L^\medwhitestar_v(\omega) = Z_v \sum_{\omega} P_{\partial v}^\medwhitestar(\omega) = Z_v \;,\] since \(P_{\partial v}^\medwhitestar(\omega)\) are the spectral projections of \(-2\sum_{s\in\partial v} A_s\). This provides the first inclusion (\(\subseteq\)) of the equality \[\left\{L_v(\omega)^\medwhitestar\right\}'_{v\in R, \omega} =\left\{Z_v, P_{\partial v}^\medwhitestar(\omega) \right\}'_{v\in R, \omega} \;.\] The reverse inclusion (\(\supseteq\)) follows by multiplying the projection by \(Z_v\) from the left.

In a second step, we again use that \(P_{\partial v}^\medwhitestar(\omega)\) are spectral projections to obtain the sum of the star operators: \[\sum_{s\in\partial v} A_s = \sum_{\epsilon} \epsilon P_{\partial v}^\medwhitestar(-2\epsilon) \;.\] Here the sum is over all eigenvalues \(\epsilon\) of \(\sum_{s\in\partial v} A_s\). This yields the first inclusion (\(\subseteq\)) in the equality \[\left\{Z_v, P_{\partial v}^\medwhitestar(\omega) \right\}'_{v\in R, \omega} =\left\{Z_v, \sum_{s\in\partial v} A_s \right\}'_{v\in R}\;.\] The reverse inclusion (\(\supseteq\)) follows from the representation 27 of the spectral projection \(P_{\partial v}^\medwhitestar(\omega)\) as a polynomial. By the same argument, we also show the identity claimed for \(\ker\mathcal{L}_R^\square\). ◻

Proof of [item4]. in 11. Recall that \(\mathop{\mathrm{\mathbb{E}}}^\sharp_{R}\) and \(\mathop{\mathrm{\mathbb{E}}}^\sharp_{R'}\) are projections onto the kernels of \(\mathop{\mathrm{\mathcal{L}}}^\sharp_{R}\) and \(\mathop{\mathrm{\mathcal{L}}}^\sharp_{R'}\), respectively. By 8, the commutant of the set of jump operators gives these kernels. The kernel of the Lindbladian on the bigger set is hence contained in the kernel of the one on the smaller set: \(\ker \mathop{\mathrm{\mathcal{L}}}^\sharp_{R} \subseteq \ker \mathop{\mathrm{\mathcal{L}}}^\sharp_{R'}\). Consequently, the image of \(\mathop{\mathrm{\mathbb{E}}}^\sharp_{R}\) is contained in the image of \(\mathop{\mathrm{\mathbb{E}}}^\sharp_{R'}\). In particular, \(\mathop{\mathrm{\mathbb{E}}}^\sharp_{R'}\circ \mathop{\mathrm{\mathbb{E}}}^\sharp_{R} = \mathop{\mathrm{\mathbb{E}}}^\sharp_{R} = \mathop{\mathrm{\mathbb{E}}}^\sharp_{R}\circ \mathop{\mathrm{\mathbb{E}}}^\sharp_{R'}\), since these operators are orthogonal projections on nested spaces. This completes the proof. ◻

Proof.

1. Since \(\mathop{\mathrm{{\mathbb{S}}}}_R\) is the set of stars \(s\) with \(ds\cap R \neq \emptyset\), the representation of \(\mathop{\mathrm{{\mathbb{S}}}}_R\) as a union is immediate. The disjointness of this union follows from the fact that \(\mathop{\mathrm{{\mathbb{S}}}}_{r|R} \cap \mathop{\mathrm{{\mathbb{S}}}}_{w|R} = \emptyset\) if \(r\neq w\). This holds, since \(r=ds \cap R=w\) for any \(s\) in both sets.

2. Assume \(\mathop{\mathrm{{\mathbb{S}}}}_{r|R}\cap \mathop{\mathrm{{\mathbb{S}}}}_{r'|R'} \neq \emptyset\). Then \(r\cap R'=r'\), since for any \(s\in \mathop{\mathrm{{\mathbb{S}}}}_{r|R}\cap \mathop{\mathrm{{\mathbb{S}}}}_{r'|R'}\): \[r\cap R' = (ds\cap R)\cap R' = ds \cap R' = r' \;.\] Hence any \(s\in \mathop{\mathrm{{\mathbb{S}}}}_{r|R}\) has support in \(R'\) and thus is an element of \(\mathop{\mathrm{{\mathbb{S}}}}_{R'}\). Using these two facts and the definition, we may rewrite: \[\begin{align} \mathop{\mathrm{{\mathbb{S}}}}_{r|R} &= \left\{s\in \mathop{\mathrm{{\mathbb{S}}}}_{R}\middle| ds\cap R = r\right\}= \left\{s\in \mathop{\mathrm{{\mathbb{S}}}}_{R'}\middle| ds\cap R = r\right\}\\ &= \left\{s\in \mathop{\mathrm{{\mathbb{S}}}}_{R'}\middle| ds\cap R' = r' \;\mathrm{and} \; (ds\cap R)\setminus R' = r\setminus R'\right\} \subseteq \mathop{\mathrm{{\mathbb{S}}}}_{r'|R'} \;. \end{align}\]

3. By symmetry, we only need to prove the result in case \(i = 1\). We pick any \(v_0\in r\cap R^{-}_1\) and any \(s\in \mathop{\mathrm{{\mathbb{S}}}}_{r|R_1\cup R_2}\). Recall that the support of any star has diameter at most \(1\). Thus, the distance between \(v_0\) and any \(v\in r\) is at most one (cf.6), since they all lie in the support \(ds\) of \(s\). By assumption, the distance of \(v_0\) to \(R_2\setminus R_1\) is greater than \(2\). Hence the distance \(\mathop{\mathrm{dist}}(r,R_2\setminus R_1)\) is greater than \(1\) and \(r\subseteq R_1\). The latter also implies that \(\mathop{\mathrm{{\mathbb{S}}}}_{r|R_1\cup R_2}\subset \mathop{\mathrm{{\mathbb{S}}}}_{R_1}\) and, moreover, that \(\mathop{\mathrm{{\mathbb{S}}}}_{r|R_1}\) is well defined. Then, for every \(s\in \mathop{\mathrm{{\mathbb{S}}}}_{r|R_1\cup R_2}\) \[ds\cap R_1 = (ds\cap(R_1\cup R_2))\cap R_1 = r\cap R_1 = r \;.\] Thus \(\mathop{\mathrm{{\mathbb{S}}}}_{r|R_1\cup R_2}\subseteq \mathop{\mathrm{{\mathbb{S}}}}_{r|R_1}\). The other inclusion follows from the fact that the diameter of a star is at most \(1\). More precisely, since the distance from \(r\) to \(R_2\setminus R_1\) is greater than \(1\), no star with support in \(r\) can have support in \(R_2\setminus R_1\). In particular, for any \(s\in \mathop{\mathrm{{\mathbb{S}}}}_{r|R_1}\): \[r = ds \cap R_1 = (ds \cap R_1 ) \cup (ds \cap (R_2\setminus R_1)) = ds \cap (R_1\cup R_2) ,\] which concludes the proof of ?? for \(i = 1\).

This completes the proof. ◻

Proof. For any individual star \(s\in \mathop{\mathrm{{\mathbb{S}}}}_R\) the operator \(A_s\) either commutes or anti-commutes with any \(Z_v\): \(Z_v A_s Z_v = \pm A_s\), with \(-\) if and only if \(v\) is in the support of \(s\). Since all stars \(s \in \mathop{\mathrm{{\mathbb{S}}}}_{r|R}\) share the same sites \(v\) in their support when restricted to \(R\), any site \(v\) is either in the support of all of them or not in the support of any. ◻

Proof. Recall from Proposition 8 that \(\ker \mathop{\mathrm{\mathcal{L}}}_R^\medwhitestar= \left\{Z_v, \sum_{s\in \partial v} A_s\right\}_{v\in R}'\). We keep the \(Z_v\) as they are and replace the sums \(\sum_{s\in \partial v}A_s\) with those of the partition by support \(\Sigma_{r|R}^\medwhitestar\): \[\label{eq:equalcomm} \left\{Z_v, \sum_{s\in \partial v} A_s\right\}_{v\in R}' = \left\{Z_v, \sum_{s\in \mathop{\mathrm{{\mathbb{S}}}}_{r|R}} A_s\right\}_{v\in R, \;r\in J_R^\medwhitestar}' \;.\tag{32}\] The proof of this identity will be based on the fact that if two sets generate the same algebra, they also have the same commutant.

It is easy to see that the algebra generated by \(\sum_{s\in \partial v}A_s\) with \(v \in R\) is included in the one generated by \(\Sigma^\medwhitestar_{r|R}\) with \(r\in J^\medwhitestar_R\), since the former is the sum over all stars whose support intersects \(v\) while the latter is the sum over all stars whose support in \(R\) is exactly \(r\): \[\sum_{s\in \partial v}A_s = \sum_{r\in J^\medwhitestar_R\colon \;v\in r} \Sigma_{r|R}^\medwhitestar\;.\]

The other direction requires more work and involves the \(Z_v\) operators. Our goal is, given \(r\in J^\medwhitestar_R\), to construct \(\Sigma_{r|R}^\medwhitestar\). First, we pick any \(v_0\in r\) (recall that \(r\) is never empty). We then observe that for any other \(v\in r\), conjugating the sum \(\sum_{s\in \partial v_0}A_s\) with \(Z_v\) adds a negative sign to all stars supported on \(v\): \[\frac{1}{2}\left(\sum_{s\in \partial v_0} A_s - Z_{v}\sum_{s\in \partial v_0} A_s Z_{v}\right) = \sum_{s\in \partial v_0 \cap\partial v} A_s\] where \(s\in \partial v_0 \cap\partial v\) if and only if \(s\) is supported on both \(v\) and \(v_0\) (but not necessarily exactly). Repeating this step for all \(v\in r\), we can construct the sum of all stars with \(r\subseteq ds\). However, there may still be stars in this sum that have support on some \(v\in R\setminus r\). To exclude these, we again conjugate the sum with \(Z_v\), this time for \(v\in R\setminus r\) and with a plus sign: \[\frac{1}{2}\left(\sum_{s\in \partial v_0, \;r\subseteq ds} A_s + Z_{v}\sum_{s\in \partial v_0, \;r\subseteq ds} A_s Z_{v}\right) = \sum_{s\in \partial v_0, \;r\subseteq ds,\;s\notin \partial v} A_s \;.\] Repeating this for all \(v\in R\setminus r\) yields a sum over all stars \(s\) with \(r= ds\). This concludes the proof of 32 . ◻

Proof. We will use the fact that orthogonal projections are uniquely determined by the subspaces onto which they project. From 11 and 4, we recall that \(\mathop{\mathrm{\mathbb{E}}}^\medwhitestar_R\) is the self-adjoint projection onto \[\ker \mathop{\mathrm{\mathcal{L}}}^\medwhitestar_R = \left\{Z_v, \Sigma^\medwhitestar_{r|R}\right\}_{v\in R, \;r\in J_R^\medwhitestar}' \;.\] We will show that the expression in ?? is also a self-adjoint projection onto the same image. For a proof, we first note that by 5 below, the two pinchings \(\mathop{\mathrm{\mathcal{T}}}^\medwhitestar_R\) and \(\mathop{\mathrm{\mathcal{P}}}^\medwhitestar_R\) commute. Thus, any operator \(O\in \mathop{\mathrm{Im}}(\mathop{\mathrm{\mathcal{T}}}^\medwhitestar_R \circ \mathop{\mathrm{\mathcal{P}}}^\medwhitestar_R(\hat{\rho}_R \;\cdot \; ))\) commutes with any \(Z_v\) for any \(v\in R\), as well as with any \(\Sigma^\medwhitestar_{r|R}\) for any \(r\in J_R^\medwhitestar\). Hence \(O\in \ker \mathop{\mathrm{\mathcal{L}}}_R^\medwhitestar\), which establishes the inclusion \(\mathop{\mathrm{Im}}(\mathop{\mathrm{\mathcal{T}}}^\medwhitestar_R \circ \mathop{\mathrm{\mathcal{P}}}^\medwhitestar_R(\hat{\rho}_R^\medwhitestar\;\cdot \; )) \subseteq \ker \mathop{\mathrm{\mathcal{L}}}_R^\medwhitestar\). For the reverse inclusion, let \(O\in \ker \mathop{\mathrm{\mathcal{L}}}_R^\medwhitestar\). Then \(O\) commutes with any projection \(\ketbra{\mathbf{z}}{\mathbf{z}}\) onto any \(Z\)-basis state with \(\mathbf{z}\in 2^R\). It also commutes with the spectral projections \(\Pi_{r|R}^\medwhitestar(\omega)\) for any \(r\in J_R^\medwhitestar\) and any \(\omega \in \Omega_{r|R}^\medwhitestar\) justifying the first equality in \[\begin{align} \mathop{\mathrm{\mathcal{T}}}^\medwhitestar_R \circ \mathop{\mathrm{\mathcal{P}}}^\medwhitestar_R(\hat{\rho}^\medwhitestar_R O ) = \mathop{\mathrm{\mathcal{T}}}^\medwhitestar_R( \mathop{\mathrm{\mathcal{P}}}^\medwhitestar_R(\hat{\rho}^\medwhitestar_R) O ) = \mathop{\mathrm{\mathcal{T}}}^\medwhitestar_R \circ \mathop{\mathrm{\mathcal{P}}}^\medwhitestar_R(\hat{\rho}^\medwhitestar_R ) O = O . \end{align}\] Here, the last equality, which expresses the unitality of \(\mathop{\mathrm{\mathcal{T}}}^\medwhitestar_R \circ \mathop{\mathrm{\mathcal{P}}}^\medwhitestar_R(\hat{\rho}_R^\medwhitestar\;\cdot \; )\), follows from 6 below together with the definition of \(\hat{\rho}^\medwhitestar_R\), which imply \[\require{physics} \mathop{\mathrm{\mathcal{T}}}^\medwhitestar_R(\hat{\rho}^\medwhitestar_R) = \mathop{\mathrm{\mathcal{T}}}^\medwhitestar_R\left(\frac{e^{-\beta H^\medwhitestar_R}}{\Tr_R e^{-\beta H^\medwhitestar_R}}\right) = \frac{1}{\Tr_R e^{-\beta H^\medwhitestar_R}} \Tr_{R}(e^{-\beta H^\medwhitestar_R}) = \mathbb{1}\;.\]

Next, we show that \(\mathop{\mathrm{\mathcal{T}}}^\medwhitestar_R \circ \mathop{\mathrm{\mathcal{P}}}^\medwhitestar_R(\hat{\rho}^\medwhitestar_R \;\cdot\; )\) is a projection. Let \(O\in \mathop{\mathrm{\mathcal{B}}}(\mathop{\mathrm{\mathcal{H}}}_{\Lambda})\), then, by a repeated application 5 below and using \(\mathop{\mathrm{\mathcal{T}}}_R(\hat{\rho}_R^\medwhitestar)=\mathbb{1}\), we find \[\begin{align} \mathop{\mathrm{\mathcal{T}}}^\medwhitestar_R \circ \mathop{\mathrm{\mathcal{P}}}^\medwhitestar_R(\hat{\rho}^\medwhitestar_R \cdot \mathop{\mathrm{\mathcal{T}}}^\medwhitestar_R \circ \mathop{\mathrm{\mathcal{P}}}^\medwhitestar_R(\hat{\rho}^\medwhitestar_R O )) &= \mathop{\mathrm{\mathcal{P}}}^\medwhitestar_R \circ \mathop{\mathrm{\mathcal{T}}}^\medwhitestar_R(\hat{\rho}^\medwhitestar_R \cdot \mathop{\mathrm{\mathcal{T}}}^\medwhitestar_R \circ \mathop{\mathrm{\mathcal{P}}}^\medwhitestar_R(\hat{\rho}^\medwhitestar_R O ))\\ &= \mathop{\mathrm{\mathcal{P}}}^\medwhitestar_R( \mathop{\mathrm{\mathcal{T}}}^\medwhitestar_R(\hat{\rho}^\medwhitestar_R) \cdot \mathop{\mathrm{\mathcal{T}}}^\medwhitestar_R \circ\mathop{\mathrm{\mathcal{P}}}^\medwhitestar_R(\hat{\rho}^\medwhitestar_R O ))\\ &= \mathop{\mathrm{\mathcal{P}}}^\medwhitestar_R(\mathbb{1}\cdot \mathop{\mathrm{\mathcal{T}}}^\medwhitestar_R \circ\mathop{\mathrm{\mathcal{P}}}^\medwhitestar_R(\hat{\rho}^\medwhitestar_R O ))\\ &= \mathop{\mathrm{\mathcal{T}}}^\medwhitestar_R \circ\mathop{\mathrm{\mathcal{P}}}^\medwhitestar_R(\hat{\rho}^\medwhitestar_R O ) . \end{align}\] In the second line we also used that \([\mathop{\mathrm{\mathcal{T}}}^\medwhitestar_R(O),\ketbra{\mathbf{z}}{\mathbf{z}}] = 0\) for any \(O \in \mathop{\mathrm{\mathcal{B}}}(\mathop{\mathrm{\mathcal{H}}}_{\Lambda})\) and any \(\mathbf{z}\in 2^R\).

To show self-adjointness with respect to the scalar product 26 with \(s \in [0,1]\) arbitrary, we note that for any \(O\in \mathop{\mathrm{\mathcal{B}}}(\mathop{\mathrm{\mathcal{H}}}_{\Lambda})\) \[\label{eq:pinchingsonrho} \mathop{\mathrm{\mathcal{P}}}^\medwhitestar_R(\hat{\rho}^\medwhitestar_R O )= \mathop{\mathrm{\mathcal{P}}}^\medwhitestar_R((\hat{\rho}^\medwhitestar_R)^{1-s} O (\hat{\rho}^\medwhitestar_R)^s)\tag{35}\] since by the explicit form of the states \(\hat{\rho}^\medwhitestar_R\) and the fact that \(H^\medwhitestar_R = - \sum_{r \in J_R^\medwhitestar} \Sigma^\medwhitestar_{r|R}\) one may let the star-pinching act on the conjugating operators. We now use the definition 26 of the scalar product to conclude that for all \(O_1, O_2 \in \mathop{\mathrm{\mathcal{B}}}(\mathop{\mathrm{\mathcal{H}}}_{\Lambda})\): \[\require{physics} \begin{align} & \mkern-50mu \langle O_1, \mathop{\mathrm{\mathcal{T}}}^\medwhitestar_R \circ \mathop{\mathrm{\mathcal{P}}}^\medwhitestar_R(\hat{\rho}^\medwhitestar_R O_2 )\rangle_{\rho_\Lambda^\boxplus,s} = \Tr \left((\rho_\Lambda^\boxplus)^{s} O_1^\dagger (\rho_\Lambda^\boxplus)^{1-s} \mathop{\mathrm{\mathcal{T}}}^\medwhitestar_R \circ \mathop{\mathrm{\mathcal{P}}}^\medwhitestar_R((\hat{\rho}^\medwhitestar_R)^{1-s} O_2 (\hat{\rho}^\medwhitestar_R)^s) \right)\\ &= \Tr \left( \frac{e^{- s\beta H^\medwhitestar_R}}{(\Tr_{R} e^{-\beta H^\medwhitestar_R})^{s}} \mathop{\mathrm{\mathcal{T}}}^\medwhitestar_{R}\circ \mathop{\mathrm{\mathcal{P}}}^\medwhitestar_{R}\left( \frac{e^{-s \beta H_\Lambda^\boxplus}}{(Z_\Lambda^\boxplus)^s} O_1^\dagger \frac{e^{- (1-s)\beta H_\Lambda^\boxplus}}{(Z_\Lambda^\boxplus)^{1-s}}\right) \frac{e^{- (1-s)\beta H^\medwhitestar_R}}{(\Tr_{R} e^{-\beta H^\medwhitestar_R})^{1-s}} O_2 \right)\\ &= \Tr \left( \frac{e^{-s \beta H_\Lambda^\boxplus}}{(Z_\Lambda^\boxplus)^s} \mathop{\mathrm{\mathcal{T}}}^\medwhitestar_{R}\circ \mathop{\mathrm{\mathcal{P}}}^\medwhitestar_{R}\left( \frac{e^{- s\beta H^\medwhitestar_R}}{(\Tr_{R} e^{-\beta H^\medwhitestar_R})^{s}} O_1^\dagger \frac{e^{- (1-s)\beta H^\medwhitestar_R}}{(\Tr_{R} e^{-\beta H^\medwhitestar_R})^{1-s}} \right) \frac{e^{- (1-s)\beta H_\Lambda^\boxplus}}{(Z_\Lambda^\boxplus)^{1-s}} O_2 \right)\\ &= \langle \mathop{\mathrm{\mathcal{T}}}^\medwhitestar_R \circ \mathop{\mathrm{\mathcal{P}}}^\medwhitestar_R(\hat{\rho}^\medwhitestar_R O_1 ), O_2 \rangle_{\rho_\Lambda^\boxplus,s} . \end{align}\] In the second equality, we commuted the pinchings using 5, and let them act as their Hilbert-Schmidt duals. The next line relies on the fact that \(\require{physics} (\Tr_{R} e^{-\beta H^\medwhitestar_R})^{-1}\) and \(e^{-\beta (H^\medwhitestar_{\Lambda}-H^\medwhitestar_R)}\) have no support on \(R\), and thus commute with any projection \(\ketbra{\mathbf{z}}{\mathbf{z}}\) supported on \(R\). Thus, they can be pulled in and out of \(\mathop{\mathrm{\mathcal{T}}}^\medwhitestar_R\). Since they also commute with any star operator, they furthermore can be pulled in and out of \(\mathop{\mathrm{\mathcal{P}}}^\medwhitestar_R\). Moreover, for any \(O\in \mathop{\mathrm{\mathcal{B}}}(\mathop{\mathrm{\mathcal{H}}}_{\Lambda})\) \[\mathop{\mathrm{\mathcal{T}}}^\medwhitestar_R \circ \mathop{\mathrm{\mathcal{P}}}^\medwhitestar_R(e^{-s\beta H^\square} O e^{-(1-s) \beta H^\square} )= e^{-s\beta H^\square} \mathop{\mathrm{\mathcal{T}}}^\medwhitestar_R \circ \mathop{\mathrm{\mathcal{P}}}^\medwhitestar_R(O ) e^{-\beta (1-s) H^\square} ,\] since every plaquette operator \(B_p\) commutes with every star operator \(A_s\) and every \(Z_v\). The final step in the above chain of equalities is again based on 35 and the fact that both pinchings are \(\dagger\)-homomorphisms. This concludes the proof. ◻

Proof. Since star operators are diagonal in \(X\), plaquettes are diagonal in \(Z\) and stars and plaquettes commute, the only non-trivial pairings are \([\mathop{\mathrm{\mathcal{T}}}_{R_1}^\medwhitestar, \mathop{\mathrm{\mathcal{T}}}_{R_3}^\square]\), \([\mathop{\mathrm{\mathcal{T}}}_{R_1}^\medwhitestar, \mathop{\mathrm{\mathcal{P}}}_{R_2}^\medwhitestar]\) and \([\mathop{\mathrm{\mathcal{T}}}_{R_3}^\square, \mathop{\mathrm{\mathcal{P}}}_{R_4}^\square]\) and the latter two are equivalent. The fact that \(X\)- and \(Z\)-pinchings commute is easy to check using the fact that \(|\braket{\mathbf{x}|\mathbf{z}}|^2\) is independent of \(\mathbf{x} \in 2^R\) and \(\mathbf{z}\in 2^R\). It thus remains to prove the last case.

From 3 we recall that \(Z_v \Sigma_{r|R_{2}}^\medwhitestar Z_v=\pm \Sigma_{r|R_{2}}^\medwhitestar\) for any \(r\in J^\medwhitestar_{R_2}\) and \(v\in R_1\). Conjugating the corresponding spectral projections may hence only flip the sign of the eigenvalue \(\omega \in \Omega_{r|R}^\medwhitestar\): \[Z_v \Pi_{r|R}^\medwhitestar(\omega) Z_v=\Pi_{r|R}^\medwhitestar(\pm\omega).\] For the rest of the proof, we abbreviate \(\mathbf{\Pi}_R(\pm_v\boldsymbol{\omega}) \mathrel{\vcenter{:}}= Z_v\mathbf{\Pi}_R^\medwhitestar(\boldsymbol{\omega})Z_v\).

We now compute the action of \(\mathop{\mathrm{\mathcal{T}}}_{R_1}^\medwhitestar\circ \mathop{\mathrm{\mathcal{P}}}_{R_2}^\medwhitestar\) in the \(Z\)-basis, that is, for any \(\mathbf{z}_1, \mathbf{z}_2, \mathbf{z}_3, \mathbf{z}_4 \in 2^{\Lambda}\): \[\begin{align} &2^{-|R_1|} \bra{\mathbf{z}_1{}}\mathop{\mathrm{\mathcal{T}}}_{R_1}^\medwhitestar\circ \mathop{\mathrm{\mathcal{P}}}_{R_2}^\medwhitestar(\ketbra{\mathbf{z}_2{}}{\mathbf{z}_3{}})\ket{\mathbf{z}_4{}} \\ &= 1[\mathbf{z}_1|_{R_1} = \mathbf{z}_4|_{R_1}] \;\bra{\mathbf{z}_1{}} \mathop{\mathrm{\mathcal{P}}}_{R_2}^\medwhitestar(\ketbra{\mathbf{z}_2{}}{\mathbf{z}_3{}})\ket{\mathbf{z}_4{}}\\ &= 1[\mathbf{z}_1|_{R_1} = \mathbf{z}_4|_{R_1}] (\mathbf{z}_1)_v (\mathbf{z}_2)_v (\mathbf{z}_3)_v (\mathbf{z}_4)_v \bra{\mathbf{z}_1{}} Z_v \mathop{\mathrm{\mathcal{P}}}_{R_2}^\medwhitestar(Z_v \ketbra{\mathbf{z}_2{}}{\mathbf{z}_3{}} Z_v)Z_v\ket{\mathbf{z}_4{}}\\ &= 1[\mathbf{z}_1|_{R_1} = \mathbf{z}_4|_{R_1}] (\mathbf{z}_2)_v (\mathbf{z}_3)_v \sum_{\boldsymbol{\omega}\in \Omega^\medwhitestar_{R_2}} \bra{\mathbf{z}_1{}} \mathbf{\Pi}_{R_2}(\pm_v\boldsymbol{\omega}) \ketbra{\mathbf{z}_2{}}{\mathbf{z}_3{}} \mathbf{\Pi}_{R_2}(\pm_v\boldsymbol{\omega}) \ket{\mathbf{z}_4{}}\\ &= 1[\mathbf{z}_1|_{R_1} = \mathbf{z}_4|_{R_1}] (\mathbf{z}_2)_v (\mathbf{z}_3)_v \bra{\mathbf{z}_1{}} \mathop{\mathrm{\mathcal{P}}}_{R_2}^\medwhitestar(\ketbra{\mathbf{z}_2{}}{\mathbf{z}_3{}})\ket{\mathbf{z}_4{}} . \end{align}\] Here \(1[\dots]\) denotes the indicator function of the set in brackets. In the first step, the indicator function resulted from the \(Z\)-pinching. For the second step, we used that for \(v\in R_i\) the values \((\mathbf{z}_i)_v \in \{\pm 1\}\) are the eigenvalues of \(Z_v\) on the vector \(\ket{\mathbf{z}_i}\). In the last step, we absorbed the sign \(\pm_v\) into the sum since the spectra \(\Omega_{r|R}^\medwhitestar\) are symmetric. To proceed, we compare the second with the last line in the equation above, finding \[\begin{align} 1[\mathbf{z}_1|_{R_1} = \mathbf{z}_4|_{R_1}]\bra{\mathbf{z}_1{}} \mathop{\mathrm{\mathcal{P}}}_{R_2}^\medwhitestar(\ketbra{\mathbf{z}_2{}}{\mathbf{z}_3{}})\ket{\mathbf{z}_4{}} = (\mathbf{z}_2)_v (\mathbf{z}_3)_v 1[\mathbf{z}_1|_{R_1} = \mathbf{z}_4|_{R_1}]\bra{\mathbf{z}_1{}} \mathop{\mathrm{\mathcal{P}}}_{R_2}^\medwhitestar(\ketbra{\mathbf{z}_2{}}{\mathbf{z}_3{}})\ket{\mathbf{z}_4{}} \end{align}\] and hence \(1[\mathbf{z}_1|_{R_1} = \mathbf{z}_4|_{R_1}] \bra{\mathbf{z}_1{}} \mathop{\mathrm{\mathcal{P}}}_{R_2}^\medwhitestar(\ketbra{\mathbf{z}_2{}}{\mathbf{z}_3{}})\ket{\mathbf{z}_4{}} = 0\) unless \((\mathbf{z}_2)_v (\mathbf{z}_3)_v = 1\). Since \(v \in R_1\) was arbitrary, we conclude \[\begin{gather} 2^{-|R_1|} \bra{\mathbf{z}_1{}}\mathop{\mathrm{\mathcal{T}}}_{R_1}^\medwhitestar\circ \mathop{\mathrm{\mathcal{P}}}_{R_2}^\medwhitestar(\ketbra{\mathbf{z}_2{}}{\mathbf{z}_3{}})\ket{\mathbf{z}_4{}} \\ = 1[\mathbf{z}_1|_{R_1} = \mathbf{z}_4|_{R_1}] 1[\mathbf{z}_2|_{R_1} = \mathbf{z}_3|_{R_1}] \bra{\mathbf{z}_1{}} \mathop{\mathrm{\mathcal{P}}}_{R_2}^\medwhitestar(\ketbra{\mathbf{z}_2{}}{\mathbf{z}_3{}})\ket{\mathbf{z}_4{}} . \end{gather}\] Repeating the same argument for \(\mathop{\mathrm{\mathcal{P}}}_{R_2}^\medwhitestar\circ\mathop{\mathrm{\mathcal{T}}}_{R_1}^\medwhitestar\) gives an indicator function on \(\mathbf{z}_2, \mathbf{z}_3\) from the \(Z\)-pinching and another indicator on \(\mathbf{z}_1,\mathbf{z}_4\) from inserting \(Z_v\). Thus, the two expressions are identical, as claimed. ◻

Proof. Let \(\Xi_{\boldsymbol{r}}^\medwhitestar\mathrel{\vcenter{:}}=\Sigma_{r_1|R_2}^\medwhitestar\cdot \Sigma_{r_2|R_2}^\medwhitestar\cdot \ldots\) be a product of \(\Sigma_{r_i|R_2}^{\medwhitestar}\) with \(r_i\in J^\medwhitestar_{R_2}\). Then the operator \(\bra{\mathbf{z}} \Xi_{\boldsymbol{r}}^\medwhitestar\ket{\mathbf{z}} \in \mathop{\mathrm{\mathcal{B}}}(\mathop{\mathrm{\mathcal{H}}}_{R_1^c})\) is independent of \(\mathbf{z} \in 2^{R_1}\), which follows from the fact that any \(\mathbf{z}' \in 2^{R_1}\) may be reached by spin flips, \(\ket{\mathbf{z}'} = \bigotimes_{v \in \mathbf{z}\Delta \mathbf{z}'} X_v \ket{\mathbf{z}}\). Since \([\Xi_{\boldsymbol{r}}^\medwhitestar,X_v]=0\) for any \(v\), this implies \[\bra{\mathbf{z}} \Xi_{\boldsymbol{r}}^\medwhitestar\ket{\mathbf{z}} = \bra{\mathbf{z}} \left(\bigotimes_{v \in \mathbf{z}\Delta \mathbf{z}'} X_v\right)^2 \Xi_{\boldsymbol{r}}^\medwhitestar\ket{\mathbf{z}} = \bra{\mathbf{z}'} \Xi_{\boldsymbol{r}}^\medwhitestar\ket{\mathbf{z}'}\;.\] Now, any operator \(O\in \mathop{\mathrm{\mathcal{A}}}_{R_2}\) can be written as a finite linear combination \(O = \sum_{\boldsymbol{r}} o_{\boldsymbol{r}}\Xi^\medwhitestar_{\boldsymbol{r}}\) with some \(o_{\boldsymbol{r}}\in \mathop{\mathrm{\mathbb{C}}}\), which implies: \[\require{physics} \begin{align} \mathop{\mathrm{\mathcal{T}}}^\medwhitestar_{R_1}(O) &= 2^{|R_1|}\sum_{\mathbf{z}\in 2^{R_1}}\ketbra{\mathbf{z}}{\mathbf{z}} \sum_{\boldsymbol{r}} o_{\boldsymbol{r}} \Xi_{\boldsymbol{r}}^\medwhitestar \ketbra{\mathbf{z}}{\mathbf{z}} \\ &= \sum_{\mathbf{z}\in 2^{R_1}}\sum_{\boldsymbol{r}} o_{\boldsymbol{r}}\sum_{\mathbf{z}'\in 2^{R_1}} \ketbra{\mathbf{z}}{\mathbf{z'}} \Xi_{\boldsymbol{r}}^\medwhitestar \ketbra{\mathbf{z}'}{\mathbf{z}} \\ &= \mathbb{1}_{R_1}\otimes\sum_{\boldsymbol{r}} o_{\boldsymbol{r}} \Tr_{R_1}\Xi_{\boldsymbol{r}}^\medwhitestar = \Tr_{R_1} O \;. \end{align}\] To show that \(\require{physics} \Tr_{R_1}(O)\in\mathop{\mathrm{\mathcal{A}}}_{R_2}^\medwhitestar\), we recall from 3 that \(Z_v \Sigma^\medwhitestar_{r|R_2} = \pm\Sigma^\medwhitestar_{r|R_2} Z_v\) for any \(v\in R_1\) and any \(r\in J^\medwhitestar_{R_2}\), and thus also \(Z_v\Xi^\medwhitestar_{\boldsymbol{r}} = \pm \Xi^\medwhitestar_{\boldsymbol{r}} Z_v\). We now consider the \(Z\)-pinching of \(\Xi^\medwhitestar_{\boldsymbol{r}}\) on a single qubit \(v\in R_1\): \[\begin{align} \mathop{\mathrm{\mathcal{T}}}^\medwhitestar_{v}(\Xi^\medwhitestar_{\boldsymbol{r}}) &= 2\left(\frac{\mathbb{1}+ Z_v}{2} \Xi^\medwhitestar_{\boldsymbol{r}} \frac{\mathbb{1}+ Z_v}{2} + \frac{\mathbb{1}- Z_v}{2} \Xi^\medwhitestar_{\boldsymbol{r}}\frac{\mathbb{1}- Z_v}{2}\right)\\ &= 2\, \Xi^\medwhitestar_{\boldsymbol{r}} \frac{\mathbb{1}\pm Z_v}{2}\frac{\mathbb{1}+ Z_v}{2} + 2\, \Xi^\medwhitestar_{\boldsymbol{r}} \frac{\mathbb{1}\mp Z_v}{2}\frac{\mathbb{1}- Z_v}{2}\\ &= \begin{cases}2\, \Xi^\medwhitestar_{\boldsymbol{r}} &+\\ 0 & - \end{cases} \;. \end{align}\] Thus, \(\require{physics} \Tr_{R_1} O\) is indeed an element of \(\mathop{\mathrm{\mathcal{A}}}_{R_2}^\medwhitestar\).

For a proof of the last claim, note that \(H^\medwhitestar_R = - \sum_{r \in J^\medwhitestar_R}\Sigma^\medwhitestar_{r|R}\) and thus \(\exp\left(-\beta H^\medwhitestar_R \right) \in \mathop{\mathrm{\mathcal{A}}}^\medwhitestar_R\). By the above argument, \(\require{physics} \Tr_R(e^{-\beta H^\medwhitestar_R})\in \mathop{\mathrm{\mathcal{A}}}^\medwhitestar_{R}\), and thus \(\hat{\rho}^\medwhitestar_R\in \mathop{\mathrm{\mathcal{A}}}^\medwhitestar_{R}\) as well. ◻

Proof. The proof is based on 2. Since any \(r\in J^\medwhitestar_{R_1\cup R_2}\) intersects at least one of \(R_1^{-}\) or \(R_2^{-}\) by assumption, the partition of the stars of \(R_1\cup R_2\) is a subset of the partitions of \(R_1\) and \(R_2\): \[\label{eq:pinchingoverlapproof1} \{\mathop{\mathrm{{\mathbb{S}}}}_{r|R_1\cup R_2}\}_{r\in J_{R_1 \cup R_2}^\medwhitestar} \subseteq \{\mathop{\mathrm{{\mathbb{S}}}}_{r|R_1}\}_{r\in J_{R_1}^\medwhitestar} \cup \{\mathop{\mathrm{{\mathbb{S}}}}_{r|R_2}\}_{r\in J_{ R_2}^\medwhitestar} \;.\tag{36}\] Denoting the remaining terms in the union of the partitions by \(\tilde{r}_i \in \tilde{J}^\medwhitestar_{R_i}\) and the corresponding projections by \(\mathop{\mathrm{\mathcal{P}}}^\medwhitestar_{\tilde{J}^\medwhitestar_{R_i}}\), we may rewrite: \[\label{eq:pinchingoverlapproof2} \mathop{\mathrm{\mathcal{P}}}^\medwhitestar_{R_1}\circ \mathop{\mathrm{\mathcal{P}}}^\medwhitestar_{R_2} = \mathop{\mathrm{\mathcal{P}}}^\medwhitestar_{R_1\cup R_2} \circ \mathop{\mathrm{\mathcal{P}}}^\medwhitestar_{\tilde{J}^\medwhitestar_{R_1}} \circ \mathop{\mathrm{\mathcal{P}}}^\medwhitestar_{\tilde{J}^\medwhitestar_{R_2}} \;.\tag{37}\] To prove that the right-hand side of 37 equals \(\mathop{\mathrm{\mathcal{P}}}^\medwhitestar_{R_1\cup R_2}\), it suffices to show that for a single \(\tilde{r}_1 \in \tilde{J}_{R_1}^\medwhitestar\), its pinching \[\mathop{\mathrm{\mathcal{P}}}^\medwhitestar_{\tilde{r}_1} (O)= \sum_{\omega\in \Omega_{\tilde{r}_1|R_1}^\medwhitestar} \Pi_{\tilde{r}_1|R_1}^\medwhitestar(\omega) \;O\; \Pi_{\tilde{r}_1|R_1}^\medwhitestar(\omega)\] satisfies \(\mathop{\mathrm{\mathcal{P}}}^\medwhitestar_{R_1\cup R_2}\circ \mathop{\mathrm{\mathcal{P}}}^\medwhitestar_{\tilde{r}_1} = \mathop{\mathrm{\mathcal{P}}}^\medwhitestar_{R_1\cup R_2}\). In turn, by 2 we know that any \(\mathop{\mathrm{{\mathbb{S}}}}_{\tilde{r}_1|R_1}\) is a disjoint union \[\mathop{\mathrm{{\mathbb{S}}}}_{\tilde{r}_1|R_1} = \bigcupplus_{k} \mathop{\mathrm{{\mathbb{S}}}}_{r_k|R_1\cup R_2}\] for a finite set \(\{r_k\}\subset J^\medwhitestar_{R_1\cup R_2}\). Thus, the spectral projection \(\Pi_{\tilde{r}_1|R_1}^\medwhitestar(\omega)\) projects onto the sum of the eigenvalues of the \(\{\Sigma^\medwhitestar_{r_k|R_1\cup R_2}\}\). In other words, for any \(\boldsymbol{\omega} \in \Omega_{R_1\cup R_2}^\medwhitestar\) and \(\omega\in\Omega_{\tilde{r}_i}^\medwhitestar\) \[\prod_{r\in J_{R_1\cup R_2}^\medwhitestar} \Pi_{r|R_1\cup R_2}^\medwhitestar(\omega_r) \Pi_{\tilde{r}_1|R_1}^\medwhitestar(\omega)= \prod_{r\in J_{R_1\cup R_2}^\medwhitestar} \Pi_{r|R_1\cup R_2}^\medwhitestar(\omega_r) 1[\Sigma_k\omega_{r_k}=\omega] \;.\] The sum over \(\omega\in \Omega_{\tilde{r}_i}^\medwhitestar\) removes the indicator function in the pinching. Thus \[\mathop{\mathrm{\mathcal{P}}}^\medwhitestar_{R_1\cup R_2} \circ \mathop{\mathrm{\mathcal{P}}}_{\tilde{r}_1}^\medwhitestar= \mathop{\mathrm{\mathcal{P}}}^\medwhitestar_{R_1\cup R_2} \;\] as claimed. ◻

Proof. The proof of ?? is in [66]. The bound ?? follows from elementary estimates: setting \(k\mathrel{\vcenter{:}}=2\ln2-1\), we have \(1/k\leq 3\), \(\varepsilon^2\leq\varepsilon\leq \frac{1}{14}\) and \[\left(\frac{1-\varepsilon}{1+\varepsilon} - \frac{\varepsilon}{(1-\varepsilon)k}\right)^{-1} = \frac{1-\varepsilon^2}{1-\varepsilon(\frac{1}{k}+2)-\varepsilon^2(\frac{1}{k}-1)} \leq \frac{1}{1-7\varepsilon} \leq 1+14 \varepsilon .\] ◻

Proof. The proof is spelled out for the star case, and, to make things more readable, we will drop the superscript \(\medwhitestar\) from \(\mathop{\mathrm{\mathbb{E}}}, \mathop{\mathrm{\mathcal{T}}}, \mathop{\mathrm{\mathcal{P}}}, \mathop{\mathrm{\mathcal{A}}}, H, \Sigma,J\) and \(\hat{\rho}\) for the rest of this proof. We will use the explicit expressions from 5 and the following facts:

1. By 6 the partial traces \(\require{physics} \Tr_{UVW}e^{-\beta H_{UVW}}\) and \(\require{physics} \Tr_{VW}e^{-\beta H_{UVW}}\), which as partial traces of Gibbs states are also invertible, are elements of \(\mathop{\mathrm{\mathcal{A}}}_{UVW}\), the algebra generated by \(\{\Sigma_{r|UVW}\}_{r\in J_{UWV}}\). They are hence polynomials in the star sums \(\Sigma_{r|UVW}\) with \(r\in J_{UVW}\), and commute with each other and with any spectral projector of the star sums. In particular, they can be pulled in and out of the star pinching \(\mathop{\mathrm{\mathcal{P}}}_{UVW}\). In addition, both partial traces have no support on \(VW\) and can thus be pulled in and out of any \(Z\)-pinching \(\mathop{\mathrm{\mathcal{T}}}_{VW}\).

2. By 5: \([\mathop{\mathrm{\mathcal{T}}}_{VW}, \mathop{\mathrm{\mathcal{P}}}_{UVW}]=0\) and \([\mathop{\mathrm{\mathcal{T}}}_{V}, \mathop{\mathrm{\mathcal{P}}}_{UVW}]=0\).

3. \(\mathop{\mathrm{\mathcal{P}}}_{UVW} = \mathop{\mathrm{\mathcal{P}}}_{UV}\circ \mathop{\mathrm{\mathcal{P}}}_{VW} = \mathop{\mathrm{\mathcal{P}}}_{UVW} \circ \mathop{\mathrm{\mathcal{P}}}_{VW}\) by 7, as the distance between \(U\) and \(W\) is larger than \(4\) and thus, any \(v\in V\) is either in the interior of \(UV\) or in the interior of \(VW\).

Applying \(\mathop{\mathrm{\mathbb{E}}}_{UVW}\) to \(O \in \mathop{\mathrm{\mathcal{B}}}(\mathop{\mathrm{\mathcal{H}}}_{\Lambda})\) and using Fact [item1] in the second and forth subsequent equality, and Facts [item2]-[item3] for the fourth and fifth equality, we obtain: \[\require{physics} \begin{align} \label{eq:cond3} \mathop{\mathrm{\mathbb{E}}}_{UVW}(O) &= \mathop{\mathrm{\mathcal{T}}}_{UVW}\left(\frac{\Tr_{VW}e^{-\beta H_{UVW}}}{\Tr_{VW}e^{-\beta H_{UVW}}} \mathop{\mathrm{\mathcal{P}}}_{UVW}\left(\frac{e^{-\beta H_{UVW}}}{\Tr_{UVW}e^{-\beta H_{UVW}}} O \right)\right) \notag\\ &= \mathop{\mathrm{\mathcal{T}}}_{UVW}\left(\frac{\Tr_{VW}e^{-\beta H_{UVW}}}{\Tr_{UVW}e^{-\beta H_{UVW}}} \mathop{\mathrm{\mathcal{P}}}_{UVW}\left(\frac{e^{-\beta H_{UVW}}}{\Tr_{VW}e^{-\beta H_{UVW}}} O \right)\right) \notag\\ &= \mathop{\mathrm{\mathcal{T}}}_{U}\circ \mathop{\mathrm{\mathcal{T}}}_{VW}\left(\Tr_{VW}(\hat{\rho}_{UVW}) \mathop{\mathrm{\mathcal{P}}}_{UVW}\left(\hat{\rho}_{VW} O \right)\right) \notag \\ &= \mathop{\mathrm{\mathcal{T}}}_{U} \left(\Tr_{VW}(\hat{\rho}_{UVW}) \mathop{\mathrm{\mathcal{P}}}_{UVW}\circ \mathop{\mathrm{\mathcal{T}}}_{VW}\left(\hat{\rho}_{VW} O \right)\right) \notag\\ &= \mathop{\mathrm{\mathcal{T}}}_{U} \left(\Tr_{VW}(\hat{\rho}_{UVW}) \mathop{\mathrm{\mathcal{P}}}_{UVW}\circ \mathop{\mathrm{\mathcal{P}}}_{VW}\circ \mathop{\mathrm{\mathcal{T}}}_{VW}\left(\hat{\rho}_{VW} O \right)\right) \notag \\ &= \mathop{\mathrm{\mathcal{T}}}_{U} \left(\Tr_{VW}(\hat{\rho}_{UVW}) \mathop{\mathrm{\mathcal{P}}}_{UVW}\circ \mathop{\mathrm{\mathbb{E}}}_{VW}(O)\right) . \end{align}\tag{38}\] The third equality is also based on the observation that \[\require{physics} \hat{\rho}_{VW} = e^{-\beta H_{VW}}\left(\Tr_{VW}e^{-\beta H_{VW}}\right)^{-1} = e^{-\beta H_{UVW}}\left(\Tr_{VW}e^{-\beta H_{UVW}}\right)^{-1}\] since \(H_{UVW}-H_{VW}\) has no support on \(VW\).

Applying \(\mathop{\mathrm{\mathbb{E}}}_{UV}\circ \mathop{\mathrm{\mathbb{E}}}_{VW}\) to \(O\in \mathop{\mathrm{\mathcal{B}}}(\mathop{\mathrm{\mathcal{H}}}_{\Lambda})\) and using the facts above, we similarly obtain: \[\require{physics} \begin{align} \label{eq:conditer} \mathop{\mathrm{\mathbb{E}}}_{UV}\circ \mathop{\mathrm{\mathbb{E}}}_{VW}(O) &= \mathop{\mathrm{\mathcal{T}}}_{UV}\circ \mathop{\mathrm{\mathcal{P}}}_{UV}\left(\hat{\rho}_{UV} \mathop{\mathrm{\mathbb{E}}}_{VW}(O) \right)\notag\\ &= \mathop{\mathrm{\mathcal{T}}}_{UV}\left(\hat{\rho}_{UV} \mathop{\mathrm{\mathcal{P}}}_{UV}\circ\mathop{\mathrm{\mathbb{E}}}_{VW}(O) \right) \notag\\ &= \mathop{\mathrm{\mathcal{T}}}_{U}\circ \mathop{\mathrm{\mathcal{T}}}_{V}\left(\hat{\rho}_{UV} \mathop{\mathrm{\mathcal{P}}}_{UVW}\circ(2^{-|V|}\mathop{\mathrm{\mathcal{T}}}_{V})\circ\mathop{\mathrm{\mathbb{E}}}_{VW}(O)\right)\notag\\ &= \mathop{\mathrm{\mathcal{T}}}_{U}\circ \mathop{\mathrm{\mathcal{T}}}_{V}\left(\hat{\rho}_{UV} (2^{-|V|}\mathop{\mathrm{\mathcal{T}}}_{V})\circ\mathop{\mathrm{\mathcal{P}}}_{UVW}\circ\mathop{\mathrm{\mathbb{E}}}_{VW}(O)\right)\notag\\ &= \mathop{\mathrm{\mathcal{T}}}_{U}\left(\mathop{\mathrm{\mathcal{T}}}_{V}(\hat{\rho}_{UV}) \mathop{\mathrm{\mathcal{P}}}_{UVW}\circ\mathop{\mathrm{\mathbb{E}}}_{VW}(O)\right) \notag\\ &= \mathop{\mathrm{\mathcal{T}}}_{U}\left(\Tr_V(\hat{\rho}_{UV}) \mathop{\mathrm{\mathcal{P}}}_{UVW}\circ\mathop{\mathrm{\mathbb{E}}}_{VW}(O)\right) . \end{align}\tag{39}\] The factor \(2^{|V|}\) on lines 3 and 4 normalizes \(\mathop{\mathrm{\mathcal{T}}}_{V}\) into a \(Z\)-pinching. In particular, we use that \(\mathop{\mathrm{\mathcal{T}}}_{V}\circ\mathop{\mathrm{\mathcal{T}}}_{V} = 2^{|V|}\mathop{\mathrm{\mathcal{T}}}_{V}\) to pull a \(Z\)-pinching out of \(\mathop{\mathrm{\mathbb{E}}}_{VW}(O)\) on line 3 and absorb it again on line 5. In order to compare the expressions in 38 and 39 in the sense of positivity order for non-negative \(O \geq 0\), we note that the following objects all commute:

1. \(\require{physics} [\Tr_{VW}(\hat{\rho}_{UVW}),\Tr_V(\hat{\rho}_{UV})]=0\), since both are contained in \(\mathop{\mathrm{\mathcal{A}}}_{UVW}\) by 6.

2. \(\require{physics} [\Tr_{VW}(\hat{\rho}_{UVW}), \mathop{\mathrm{\mathcal{P}}}_{UVW}\circ \mathop{\mathrm{\mathbb{E}}}_{VW}(O)]=0\), since \(\require{physics} \Tr_{VW}(\hat{\rho}_{UVW})\) is an element of \(\mathop{\mathrm{\mathcal{A}}}_{UVW}\) and \(\mathop{\mathrm{\mathcal{P}}}_{UVW}\) projects onto the commutant \((\mathop{\mathrm{\mathcal{A}}}_{UVW})'\).

3. \(\require{physics} [\Tr_V(\hat{\rho}_{UV}),\mathop{\mathrm{\mathcal{P}}}_{UVW}\circ \mathop{\mathrm{\mathbb{E}}}_{VW}(O)]=0\) for the same reason.

Thus, if \[\require{physics} (1-\varepsilon) \Tr_V(\hat{\rho}_{UV}) \leq \Tr_{VW}(\hat{\rho}_{UVW}) \leq (1+\varepsilon) \Tr_V(\hat{\rho}_{UV})\] and since \(\mathop{\mathrm{\mathcal{P}}}_{UVW}\circ \mathop{\mathrm{\mathbb{E}}}_{VW}(O)\) is non-negative by assumption on \(O\geq 0\) and commutes with the partial traces inside the quantum channel \(\mathop{\mathrm{\mathcal{T}}}_{U}\) featuring on the outside of 38 and 39 , we arrive at the bounds \[(1-\varepsilon) \mathop{\mathrm{\mathbb{E}}}_{UV}\circ \mathop{\mathrm{\mathbb{E}}}_{VW}(O) \leq \mathop{\mathrm{\mathbb{E}}}_{UVW}(O) \leq (1+\varepsilon) \mathop{\mathrm{\mathbb{E}}}_{UV}\circ \mathop{\mathrm{\mathbb{E}}}_{VW}(O) .\] To establish the claimed bound, we note that \(\varepsilon\leq \frac{1}{2}\) implies the bounds \(\frac{1}{1-\varepsilon}\leq 1+2\varepsilon\) and \(\frac{1}{1+\varepsilon}\geq 1-2\varepsilon\). This finishes the proof. ◻

Proof of 6. Combining 12 and 8 yields the bound: \[D(\sigma \|\mathop{\mathrm{\mathbb{E}}}^{\medwhitestar,*}_{UVW}(\sigma))\leq (1+28\varepsilon) D(\sigma \|\mathop{\mathrm{\mathbb{E}}}^{\medwhitestar,*}_{VW}\circ\mathop{\mathrm{\mathbb{E}}}^{\medwhitestar,*}_{UV}(\sigma)) \;.\] We now use the fact that \(\ln\mathop{\mathrm{\mathbb{E}}}^{\medwhitestar,*}_{VW}(\sigma) - \ln \mathop{\mathrm{\mathbb{E}}}^{\medwhitestar,*}_{VW}\circ\mathop{\mathrm{\mathbb{E}}}^{\medwhitestar,*}_{UV}(\sigma)\) is an element of the von Neumann subalgebra of \(\mathop{\mathrm{\mathbb{E}}}^{\medwhitestar}_{VW}\) (cf.10) to rewrite the relative entropy for full-rank states \(\sigma\) (cf. 9): \[\require{physics} \begin{align} D(\sigma \|\mathop{\mathrm{\mathbb{E}}}^{\medwhitestar,*}_{VW}\circ\mathop{\mathrm{\mathbb{E}}}^{\medwhitestar,*}_{UV}(\sigma)) &= \Tr(\sigma (\ln\sigma -\ln \mathop{\mathrm{\mathbb{E}}}^{\medwhitestar,*}_{VW}\circ\mathop{\mathrm{\mathbb{E}}}^{\medwhitestar,*}_{UV}(\sigma)))\\ &= \Tr(\sigma (\ln\sigma -\ln\mathop{\mathrm{\mathbb{E}}}^{\medwhitestar,*}_{VW}(\sigma)))\\ &\phantom{=} + \Tr(\sigma(\ln\mathop{\mathrm{\mathbb{E}}}^{\medwhitestar,*}_{VW}(\sigma) - \ln \mathop{\mathrm{\mathbb{E}}}^{\medwhitestar,*}_{VW}\circ\mathop{\mathrm{\mathbb{E}}}^{\medwhitestar,*}_{UV}(\sigma)))\\ &=D(\sigma \|\mathop{\mathrm{\mathbb{E}}}^{\medwhitestar,*}_{VW}(\sigma)) + D(\mathop{\mathrm{\mathbb{E}}}^{\medwhitestar,*}_{VW}(\sigma) \|\mathop{\mathrm{\mathbb{E}}}^{\medwhitestar,*}_{VW}\circ\mathop{\mathrm{\mathbb{E}}}^{\medwhitestar,*}_{UV}(\sigma))\\ &\leq D(\sigma \|\mathop{\mathrm{\mathbb{E}}}^{\medwhitestar,*}_{VW}(\sigma)) + D(\sigma \|\mathop{\mathrm{\mathbb{E}}}^{\medwhitestar,*}_{UV}(\sigma)) . \end{align}\] The equality from the first to the fourth line is sometimes refered to as the chain rule. The last inequality is the data processing inequality for the relative entropy. ◻

Proof of 2. By 13 below and assuming ?? , there exists a length scale \(L^\medwhitestar_0\), which is larger or equal to the scale in 3 (and, by a slight abuse of notation, denoted by the same symbol), such that for \(J\) such that \(R\in \mathop{\mathrm{\mathcal{R}}}_{2^J L_0}\): \[\alpha^\medwhitestar(R) \geq \alpha^\medwhitestar(2^J L_0^{\medwhitestar}) \geq \prod_{j=0}^{J-1} \left(1+ \frac{12 D}{(L_0^\medwhitestar)^{1/3} 2^{j/3}} \right)^{-1} \alpha^\medwhitestar(L^\medwhitestar_0)\] where we iterated ?? below exactly \(J\) times. Next, we bound the product in the right side using \(\ln(1+x)\leq x\) to arrive at: \[\label{eq:MLSIlb} \alpha^\medwhitestar(R) \geq \exp\left[-\frac{12 D}{(L_0^\medwhitestar)^{1/3}} \sum_{j=0}^{\infty}2^{-j/3}\right] \alpha^\medwhitestar(L_0^\medwhitestar)\geq \exp\left[\frac{-60 D}{(L_0^\medwhitestar)^{1/3}} \right]\alpha^\medwhitestar(L^\medwhitestar_0) \;.\tag{41}\] Note that this bound is independent of the system size and \(J\).

It remains to establish a uniform lower bound on \(\alpha^\medwhitestar(L^\medwhitestar_0)\) for the initial length scale. To do so, we use the lower bound from [33] for any \(R \in \mathcal{R}_{L^\medwhitestar_0}\): \[\label{eq:gaorouze} \alpha^\medwhitestar(R) \geq \frac{\mathop{\mathrm{gap}}\mathop{\mathrm{\mathcal{L}}}_{R}^\medwhitestar}{C_{cb}(\mathop{\mathrm{\mathbb{E}}}^{\medwhitestar}_{R})} ,\tag{42}\] which involve a spectral gap and the completely bounded Pimsner-Popa index \(C_{cb}(\mathop{\mathrm{\mathbb{E}}}^{\medwhitestar}_{R})\) associated with the conditional expectation. By [33] the latter is bounded from above \[C_{cb}(\mathop{\mathrm{\mathbb{E}}}^{\medwhitestar}_{R}) \leq (\dim \mathop{\mathrm{\mathcal{H}}}_{R'})^2 (\hat{\rho}^\medwhitestar_{R})_{\textrm{min}}^{-1} ,\] \(R'\) being the set of qubits at distance at most \(1\) from \(R\), and since 5 ensures that the \(R\)-local Gibbs state is invariant, \(\hat{\rho}^\medwhitestar_{R} =\mathop{\mathrm{\mathbb{E}}}^{\medwhitestar,*}_{R}(\hat{\rho}_{R}^\medwhitestar)\). Its minimal eigenvalue is lower bounded according to \[\require{physics} (\hat{\rho}^\medwhitestar_{R})_{\textrm{min}}^{-1} \leq \left\| e^{\beta H_R^\medwhitestar} \Tr_R e^{-\beta H_R^\medwhitestar} \right\| \leq 2^{|R|} \exp\left( 2 \beta \| H_{R}^\medwhitestar\|\right) .\] Since the Hamiltonian is a sum of bounded local terms, \(\| H_{R}^\medwhitestar\|\) is estimated by a polynomial in the size \(| R' |\), which is uniformly bounded in terms of \(L_0^\medwhitestar\).

Next, we recall from [33] the details concerning the spectral gap featuring in 42 and provide a lower bound, which is uniformly positive. Since we may exchange [33] the invariant state \(\mathop{\mathrm{\mathbb{E}}}^{\medwhitestar,*}_{R}(\sigma)\) by the full Gibbs state \(\rho^\boxplus_\Lambda\), the notion of spectral gap from [33] agrees with the lowest non-zero eigenvalue associated with the Dirichlet form ?? averaged over \(s \in [0,1]\): \[\mathop{\mathrm{gap}}\mathop{\mathrm{\mathcal{L}}}_{R}^\medwhitestar\mathrel{\vcenter{:}}= \inf_{O \in \mathop{\mathrm{\mathcal{B}}}(\mathop{\mathrm{\mathcal{H}}}_R)} \int_0^1 \frac{- \langle O, \mathop{\mathrm{\mathcal{L}}}^\medwhitestar_R(O) \rangle_{\rho_\Lambda^\boxplus,s}}{\| O -\mathop{\mathrm{\mathbb{E}}}^{\medwhitestar}_{R}(\sigma)(O) \|_{\textrm{BKM}} } ds .\] The average corresponds the Bogoliubov-Kubo-Mori scalar product whose norm is abbreviated by \(\| \cdot \|_{\textrm{BKM}}\). By self-adjointness of \(\mathop{\mathrm{\mathcal{L}}}^\medwhitestar_R\) (5), the above is indeed the second largest eigenvalue of the linear operator \(\mathop{\mathrm{\mathcal{L}}}^\medwhitestar_R\) on the finite-dimensional Hilbert space \(\mathop{\mathrm{\mathcal{B}}}(\mathop{\mathrm{\mathcal{H}}}_R)\).

By the assumption on the jump rates (1) and using the representation ?? , the Dirichlet form \(- \langle O, \mathop{\mathrm{\mathcal{L}}}^\medwhitestar_R(O) \rangle_{\rho^\boxplus_\Lambda,s}\) is lower bounded in terms of a Dirichlet form, in which the jump rates are set to \(\widetilde{h}^\medwhitestar(\omega) \mathrel{\vcenter{:}}= g^\medwhitestar e^{\beta\omega/2}\). Since these jump rates still satisfy detailed balance, they correspond to a Davies Lindbladian \(\widetilde{\mathop{\mathrm{\mathcal{L}}}}_{R}^\medwhitestar\), whose jump rates are independent of \(v\), and for which, by the above reasoning: \[\mathop{\mathrm{gap}}\mathop{\mathrm{\mathcal{L}}}_{R}^\medwhitestar\geq \mathop{\mathrm{gap}}\widetilde{\mathop{\mathrm{\mathcal{L}}}}_{R}^\medwhitestar.\] The right side is bounded uniformly in all \(R \in \mathcal{R}_{L^\medwhitestar_0}\), since by translation invariance of \(\widetilde{\mathop{\mathrm{\mathcal{L}}}}_{R}^\medwhitestar\), one only needs to consider finitely many rectangles of the given length scale, and on which the second largest eigenvalue of \(\widetilde{\mathop{\mathrm{\mathcal{L}}}}_{R}^\medwhitestar\) does not vanish. This completes the proof of \(\alpha^\medwhitestar(L^\medwhitestar_0) > 0\). ◻

Proof of 1. By assumption, both \(\mathop{\mathrm{\mathcal{L}}}_\Lambda^\medwhitestar\) and \(\mathop{\mathrm{\mathcal{L}}}^\square_\Lambda\) satisfy an MLSI. Since they commute by 1, we have \(\mathop{\mathrm{\mathbb{E}}}^{\boxplus,*}_R = \mathop{\mathrm{\mathbb{E}}}^{\medwhitestar,*}_R\circ \mathop{\mathrm{\mathbb{E}}}^{\square,*}_R\) together with 10 for any full rank state \(\sigma\): \[\begin{align} D(\sigma \| \mathop{\mathrm{\mathbb{E}}}^{\boxplus,*}_R(\sigma) )&\leq D(\sigma \| \mathop{\mathrm{\mathbb{E}}}^{\medwhitestar,*}_R(\sigma))+D(\sigma \| \mathop{\mathrm{\mathbb{E}}}^{\square,*}_R(\sigma))\\ &\leq \frac{EP_R^\medwhitestar(\sigma)}{2\alpha^\medwhitestar(R)} + \frac{EP_R^\square(\sigma)}{2\alpha^\square(R)}\\ &\leq \frac{1}{2\min\{\alpha^\medwhitestar(R), \alpha^\square(R)\}}EP_R^\boxplus(\sigma) . \end{align}\] In the last step, we used 15, which guaranteed that the entropy production of \(\mathop{\mathrm{\mathcal{L}}}_R^\medwhitestar\) and \(\mathop{\mathrm{\mathcal{L}}}^\square_R\) is additive. ◻

Proof. We fix any rectangle \(R\in \mathop{\mathrm{\mathcal{R}}}_{2L}\). If no side of \(R\) is longer than \(L\), \(R\in\mathop{\mathrm{\mathcal{R}}}_{L}\), and thus \(\alpha^\medwhitestar(R) \geq \alpha^\medwhitestar(L)\) by definition. For the rest of the proof, we hence assume that one side of \(R\), which we will call \(l_1\), is longer than \(L\), that is \(L<l_1\leq 2L\). To simplify notation, we abbreviate in this proof for any \(R\): \[D_R \mathrel{\vcenter{:}}= D(\sigma \| \mathop{\mathrm{\mathbb{E}}}^{\medwhitestar,*}_{R}(\sigma)).\]

We partition \(R = U\uplus V\uplus W\) into overlapping rectangles \(UV\) and \(VW\) along the direction of \(l_1\). If the boundary is periodic in the direction of \(l_1\), \(R\) might have a winding number in this direction. In this case, \(V\) contains two components \(V=V_1\uplus V_2\), see 3. Assume furthermore that the length of the overlap \(V\), that is, the distance \(\mathop{\mathrm{dist}}(U,W)\), is at least \(\sqrt{L}\) and larger than \(4\). Then the diameter of \(W\) is shorter than \(2L\leq 2 \mathop{\mathrm{dist}}(U,W)^2\) and the DS bound ?? holds. We now take \(L_0^{\medwhitestar}\) large enough such that the DS bound ?? satisfies for all \(L\geq L_0^{\medwhitestar}\): \[K^\medwhitestar e^{-\xi^\medwhitestar\sqrt{L}}\leq \frac{1}{28} .\] By approximate tensorization, 6 above, we may then bound the relative entropy with \(K'=28K^\medwhitestar\): \[\begin{align} D_R &\leq (1 + K' e^{-\xi^{\medwhitestar} \sqrt{L}}) (D_{UV} + D_{VW})\\ &\leq (1 + K' e^{-\xi^{\medwhitestar} \sqrt{L}}) \left(\frac{EP^\medwhitestar_{UV}}{2 \alpha^\medwhitestar(UV)} + \frac{EP^\medwhitestar_{VW}}{2 \alpha^\medwhitestar(UV)} \right)\\ &\leq (1 + K' e^{-\xi^{\medwhitestar} \sqrt{L}}) \frac{1}{2 \min\{\alpha^\medwhitestar(UV), \alpha^\medwhitestar(VW)\}}\left(EP^\medwhitestar_{UV} + EP^\medwhitestar_{VW} \right) . \end{align}\] Here we dropped the argument \(\sigma\) not only in the relative entropy, but also in the entropy production.

Our next aim is to use the subadditivity of the entropy production (15). However, due to the overlap \(V\), this is not possible for \(EP^\medwhitestar_{UV} + EP^\medwhitestar_{VW}\). One way to get around this is to average over many partitions as was done in [43].

Consider a family \((U_i, V_i, W_i)_{i=1}^I\) of partitions with \(I=\lfloor\sqrt[3]{L}\rfloor\) and the property that for all \(i=1,\ldots, I-1\) the regions \(U_iV_i\) and \(V_{i+1}W_{i+1}\) do not overlap (cf. 7). By 15 the sum of the entropy productions of disjoint regions can be bounded by the entropy production of their union: \[EP^\medwhitestar_{U_i V_i} + EP^\medwhitestar_{V_{i+1} W_{i+1}} \leq EP^\medwhitestar_{R} \;.\] Averaging over all \(i\) and pairwise combining entropy productions of \(i\) and \(i+1\) except \(EP^\medwhitestar_{V_1W_1}\) and \(EP^\medwhitestar_{U_IV_I}\), which we both bound individually by \(EP^\medwhitestar_{R}\), we find: \[\begin{align} \label{eq:afterav} D_R \leq (1 + K' e^{-\xi^{\medwhitestar} \sqrt{L}}) \frac{1}{2 \min\{\alpha^\medwhitestar(U_iV_i), \alpha^\medwhitestar(V_iW_i)\}_{i=1}^{I}} \left(1+L^{-1/3}\right) EP^\medwhitestar_R \;. \end{align}\tag{43}\]

Next, we let \(R' \in\{U_iV_i, V_iW_i\}_{i=1}^I\) be the rectangle minimizing \(\alpha^\medwhitestar\). We will argue below that for \(L\) large enough, the width of \(R'\) in the coordinate direction of \(l_1\) can be taken at most \(3/2 L\). Then, we distinguish two cases:

1. the size of \(R'\) in all other directions is less than \(\frac{3}{2}L\) and hence \(R'\in \mathop{\mathrm{\mathcal{R}}}_{\frac{3}{2}L}\),

2. \(R'\) is longer than \(\frac{3}{2}L\) in at least one direction.

In the second case, we repeat the above procedure a maximum of \(D-1\) times, since every coordinate direction in which we cut will then be shorter than \(\frac{3}{2}L\). We then conclude from 43 : \[\label{eq:recursionthird} \alpha^\medwhitestar(3/2L) \leq (1 + K' e^{-\xi^{\medwhitestar} \sqrt{L}})^D \left(1+L^{-1/3}\right)^D \alpha^\medwhitestar(2L).\tag{44}\] Setting \(L_0^{\medwhitestar}\) such that \(K' e^{-\xi^{\medwhitestar} \sqrt{L}} \leq L^{-1/3}\leq \frac{\ln 2}{6D}\) for all \(L\geq 3/2 L_0^{\medwhitestar}\) and iterating 44 three times, we then arrive at \[\begin{align} \alpha^\medwhitestar(L) &\leq (1 + K' e^{-\xi^{\medwhitestar} \sqrt{\frac{2}{3} L}})^{3D}\left(1+\left(\frac{2}{3} L\right)^{-1/3}\right)^{3D} \alpha^\medwhitestar(64/27 L)\\ &\leq \left(1+ L^{-1/3}\right)^{6D}\alpha^\medwhitestar(2 L) \leq \left(1+12 D L^{-1/3}\right)\alpha^\medwhitestar(2 L). \end{align}\]

It remains to be shown that such a family of partitions exists. There are two cases. If \(R\) does not wrap around the torus along \(l_1\), \(V\) only has one connected component, and we choose the following partition: Let the length of \(V_i\) be \(\lceil \sqrt{L}\rceil\), let the length of \(U_i\) be \(\lfloor l_{0} \rfloor + i \lceil\sqrt{L}\rceil\) and let the length of \(W_i\) be \(\lceil l_0 \rceil + (I-i+1) \lceil\sqrt{L}\rceil\), where \[l_0 \mathrel{\vcenter{:}}= \frac{l_1- (I+2)\lceil\sqrt{L}\rceil}{2}\;.\] This defines a valid partition if \(\lceil \sqrt L \rceil >4\) and if the lengths of \(U_1\) and \(W_I\) are at least \(1\). The latter condition is implied (using \(l_1\geq L\)) by 43 : \(L - I \lceil\sqrt{L}\rceil\geq 4\). Both conditions are satisfied for all \(L\geq L_0^{\medwhitestar}\) if we pick \(L_0^{\medwhitestar}\geq16\).

In case \(R\) wraps around the torus along \(l_1\), then \(V\) contains two connected components, \(V= V^1 \sqcup V^2\) (see 3). The length of \(V^1\) and \(V^2\) are taken both as \(\lceil\sqrt{L}\rceil\) which also agrees with \(\mathop{\mathrm{dist}}(U,W)\). Similarly to the first case, we define a partition with a growing \(U_i\) and shrinking \(W_i\). However, in this case, \(U_i\) grows on both ends and \(W_i\) shrinks accordingly. That is, the length of \(U_i\) is \(\lfloor l_{0}' \rfloor + 2i \lceil\sqrt{L}\rceil\) and the length of \(W_i\) is \(\lceil l_0' \rceil + 2(I-i+1) \lceil\sqrt{L}\rceil\) with \[l_0' \mathrel{\vcenter{:}}= \frac{l_1- 2(I+2)\lceil\sqrt{L}\rceil}{2}\;.\] Requiring that the lengths of \(U_i\) and \(W_i\) are at least \(1\) leads to the following condition: \(L - 2 I \lceil\sqrt{L}\rceil\geq 4\). This holds for \(L_0^{\medwhitestar}\geq 125\).

Finally, the longest component in both cases is \(W_1\). Its length is at most \(L+\frac{\sqrt[3]{L}(\sqrt{L}+1)}{2}+1\) in the first case or \(L+\sqrt[3]{L}(\sqrt{L}+1)+1\) in the second case. This length is shorter than \(3/2 L\) if \(L_0^{\medwhitestar}\geq 1000\). ◻

Proof of 2. By an approximation argument, it suffices to establish the bound for any full-rank state \(\sigma \in \mathop{\mathrm{\mathcal{S}}}(\mathop{\mathrm{\mathcal{H}}}_{\Lambda})\). Since \(\mathop{\mathrm{\mathcal{L}}}^\medwhitestar_{\Lambda}\) satisfies an MLSI: \[\|e^{t \mathop{\mathrm{\mathcal{L}}}^{\medwhitestar, *}_{\Lambda}}(\sigma) - \mathop{\mathrm{\mathbb{E}}}^{\medwhitestar,*}_{\Lambda}(\sigma) \|_1 \leq e^{-\alpha^{\medwhitestar} t} \sqrt{ 2 D\left(\sigma\| \mathop{\mathrm{\mathbb{E}}}^{\medwhitestar,*}_{\Lambda}(\sigma)\right)} \;.\] Using the fact that \(\ln \mathop{\mathrm{\mathbb{E}}}^{\medwhitestar,*}_{\Lambda}(\sigma) - \ln \rho^\boxplus_{\Lambda}\) is an element of the subalgebra associated with \(\mathop{\mathrm{\mathbb{E}}}^{\medwhitestar}_{\Lambda}\) (10), we rewrite the relative entropy as a difference and drop the non-positive term: \[\begin{align} D\left(\sigma\| \mathop{\mathrm{\mathbb{E}}}^{\medwhitestar,*}_{\Lambda}(\sigma)\right) &= D\left(\sigma\| \rho^\boxplus_{\Lambda}\right) - D\left(\mathop{\mathrm{\mathbb{E}}}^{\medwhitestar,*}_{\Lambda}(\sigma)\| \rho^\boxplus_{\Lambda}\right) \leq D\left(\sigma\| \rho^\boxplus_{\Lambda} \right) \;. \end{align}\] By the standard bound \[D\left( \sigma \| \rho^{\boxplus}_{\Lambda} \right) \leq \ln((\rho^{\boxplus}_{\Lambda})^{-1}_{\min}) = \mathrm{poly}(|\Lambda|) ,\] we thus conclude that the star part mixes rapidly to the kernel of \(\mathop{\mathrm{\mathcal{L}}}^{\medwhitestar}_{\Lambda}\). By construction, the logical operators \(X_L\) and \(Y_L=iX_LZ_L\) commute with all stars and plaquettes, and are annihilated by the star dynamics: \[\require{physics} \begin{align} \mathop{\mathrm{\mathbb{E}}}^\medwhitestar_{\Lambda}(X_L) &= \mathop{\mathrm{\mathcal{P}}}^\medwhitestar_{\Lambda} \circ \mathop{\mathrm{\mathcal{T}}}^\medwhitestar_{\Lambda}(\hat{\rho}^{\medwhitestar}_{\Lambda}X_L) =\mathop{\mathrm{\mathcal{P}}}^\medwhitestar_{\Lambda}\left(\Tr(\hat{\rho}^{\medwhitestar}_{\Lambda}X_L)\right) =0 ,\\ \mathop{\mathrm{\mathbb{E}}}^\medwhitestar_{\Lambda}(Y_L) &= i \mathop{\mathrm{\mathcal{P}}}^\medwhitestar_{\Lambda} \left( \mathop{\mathrm{\mathcal{T}}}^\medwhitestar_{\Lambda}(\hat{\rho}^{\medwhitestar}_{\Lambda}X_L)Z_L\right) =i \mathop{\mathrm{\mathcal{P}}}^\medwhitestar_{\Lambda} \left( \Tr(\hat{\rho}^{\medwhitestar}_{\Lambda}X_L)Z_L\right) =0 . \end{align}\] This follows from the fact that \(\mathop{\mathrm{\mathcal{T}}}^\medwhitestar_{\Lambda}\) is the \(Z\)-pinching on the full lattice and \(\hat{\rho}^{\medwhitestar}_{\Lambda}X_L\) is diagonal in the \(X\)-basis. The above trace is zero by 3.

The above facts then allow us to conclude that for both \(O_L\in \{X_L, Y_L\}\): \[\require{physics} \begin{align} \left|\Tr( e^{t\mathop{\mathrm{\mathcal{L}}}^{\boxplus, *}_{\Lambda}}(\sigma)O_L )\right| &= \left|\Tr( \left(e^{t\mathop{\mathrm{\mathcal{L}}}^{\boxplus,*}_{\Lambda}}(\sigma)- \mathop{\mathrm{\mathbb{E}}}^{\medwhitestar,*}_{\Lambda}\circ e^{t\mathop{\mathrm{\mathcal{L}}}^{\square,*}_{\Lambda}}(\sigma)\right)O_L)\right|\\ &= \left|\Tr( e^{t\mathop{\mathrm{\mathcal{L}}}^{\square,*}_{\Lambda}}\left(e^{t\mathop{\mathrm{\mathcal{L}}}^{\medwhitestar,*}_{\Lambda}}(\sigma)- \mathop{\mathrm{\mathbb{E}}}^{\medwhitestar,*}_{\Lambda}(\sigma)\right)O_L)\right|\\ &\leq \left\| e^{t\mathop{\mathrm{\mathcal{L}}}^{\square,*}_{\Lambda}}\left(e^{t\mathop{\mathrm{\mathcal{L}}}^{\medwhitestar,*}_{\Lambda}}(\sigma)- \mathop{\mathrm{\mathbb{E}}}^{\medwhitestar,*}_{\Lambda}(\sigma)\right) \right\|_1\\ &\leq \left\| e^{t\mathop{\mathrm{\mathcal{L}}}^{\medwhitestar,*}_{\Lambda}}(\sigma)- \mathop{\mathrm{\mathbb{E}}}^{\medwhitestar,*}_{\Lambda}(\sigma) \right\|_1 \leq \mathrm{poly}(|\Lambda|)e^{-\alpha^{\medwhitestar} t}. \end{align}\] Here, the second line follows from the fact that the star and plaquette parts of the Lindbladian commute (1). The second inequality stems from the contractivity of any Lindbladian evolution. ◻

Proof. As \(\mathop{\mathrm{\mathcal{L}}}^*\) has a full rank fixed point, there exists a decomposition of the Hilbert space \[\mathop{\mathrm{\mathcal{H}}}= \bigoplus_{i\in I} \mathop{\mathrm{\mathcal{H}}}_i \otimes \mathop{\mathrm{\mathcal{K}}}_i\] such that the conditional expectation has the following decomposition [80]: \[\require{physics} \mathop{\mathrm{\mathbb{E}}}^*_{\mathop{\mathrm{\mathcal{L}}}} (\sigma) = \sum_{i\in I} \Tr_{\mathop{\mathrm{\mathcal{K}}}_i}(P_i \sigma P_i)\otimes \tau_i\] where \(P_i\) is the projection onto the \(i\)-th block \(\mathop{\mathrm{\mathcal{H}}}_i\otimes \mathop{\mathrm{\mathcal{K}}}_i\) and where \(\tau_i\) are full-rank states on \(\mathop{\mathrm{\mathcal{K}}}_i\). First, since \(\mathop{\mathrm{\mathbb{E}}}_{\mathop{\mathrm{\mathcal{L}}}}^*(\sigma)\) is block-diagonal, it’s kernel is \[\require{physics} \ker(\mathop{\mathrm{\mathbb{E}}}^*_{\mathop{\mathrm{\mathcal{L}}}}(\sigma)) = \bigoplus_{i\in I} \ker\left(\Tr_{\mathop{\mathrm{\mathcal{K}}}_i}(P_i \sigma P_i)\otimes \tau_i\right) \;.\] Without loss of generality, fix some \(i\in I\). Since \(\tau_i\) is full-rank, its kernel is empty and thus \[\require{physics} \ker\left(\Tr_{\mathop{\mathrm{\mathcal{K}}}_i}(P_i \sigma P_i)\otimes \tau_i\right) = \ker\left(\Tr_{\mathop{\mathrm{\mathcal{K}}}_i}(P_i \sigma P_i)\right) \otimes \mathop{\mathrm{\mathcal{K}}}_i \;.\] Let \(\ket{\psi}\in \mathop{\mathrm{\mathcal{H}}}_i\) be an element of the kernel of \(\require{physics} \Tr_{\mathop{\mathrm{\mathcal{K}}}_i}(P_i \sigma P_i)\). Then \[\require{physics} 0 = \Tr_{\mathop{\mathrm{\mathcal{H}}}_i}(\Tr_{\mathop{\mathrm{\mathcal{K}}}_i}(P_i\sigma P_i) \ketbra{\psi}{\psi}) = \Tr_{\mathop{\mathrm{\mathcal{H}}}_i\otimes\mathop{\mathrm{\mathcal{K}}}_i}(\ketbra{\psi}{\psi}P_i\sigma P_i\ketbra{\psi}{\psi})\] and, since \(\ketbra{\psi}{\psi}P_i\sigma P_i\ketbra{\psi}{\psi}=O^\dagger O\), \(0=O=\sqrt{\sigma}P_i\ketbra{\psi}{\psi}\) and thus also \(\sigma P_i\ketbra{\psi}{\psi} = \sigma \ketbra{\psi}{\psi} =0\). Where we embedded \(\ket{\psi}\) into \(\mathop{\mathrm{\mathcal{H}}}\) in the last step. Thus, for any element \(\require{physics} \sum_{k}\ket{\psi_k}\otimes \ket{\phi_k}\in \ker\left(\Tr_{\mathop{\mathrm{\mathcal{K}}}_i}(P_i \sigma P_i)\right) \otimes \mathop{\mathrm{\mathcal{K}}}_i\) it holds that \[\sigma \sum_k \ket{\psi_k}\otimes \ket{\phi_k} = \sum_k \sigma \ketbra{\psi_k}{\psi_k}\ket{\psi_k}\otimes \ket{\phi_k} =0 ,\] and hence \(\sum_{k}\ket{\psi_k}\otimes \ket{\phi_k} \in \ker \sigma\). ◻

Proof. By 9, \(\ker\mathop{\mathrm{\mathbb{E}}}_{\mathop{\mathrm{\mathcal{L}}}}^*(\sigma) \subseteq \ker\sigma\) for any \(\sigma\). In particular, \[\ker(\mathop{\mathrm{\mathbb{E}}}_{\mathop{\mathrm{\mathcal{L}}}}^*( \sigma)) = \ker(\mathop{\mathrm{\mathbb{E}}}_{\mathop{\mathrm{\mathcal{L}}}}^* \circ e^{t\mathop{\mathrm{\mathcal{L}}}}(\sigma)) \subseteq \ker (e^{t\mathop{\mathrm{\mathcal{L}}}} \sigma) .\] This completes the proof, since the relative entropy is infinite only if the kernel of the second argument is not contained in the kernel of the first. ◻

Proof. As is shown in [9] positivity follows from the monotonicity of \(t \mapsto D(e^{t\mathop{\mathrm{\mathcal{L}}}^*_{R}}(\sigma)\|\mathop{\mathrm{\mathbb{E}}}^*_{R}(\sigma))\), which is finite by 6. The monotonicity in the region follows directly from linearity and positivity. To show linearity, consider the explicit expression from [9]: \[\require{physics} EP_{R_i}(\sigma) = -\Tr\left(\mathop{\mathrm{\mathcal{L}}}^*_{R_i}(\sigma)\left(\ln(\sigma) - \ln(\mathop{\mathrm{\mathbb{E}}}^*_{R_i}(\sigma))\right)\right)\;,\] where \(\mathop{\mathrm{\mathbb{E}}}^*_{R_i}:=\mathop{\mathrm{\mathbb{E}}}^*_{\mathop{\mathrm{\mathcal{L}}}_{R_i}}\) is the conditional expectation for the region \(R_i\). Since the Lindbladian is frustration-free, \(\rho\) is a full rank fixed point of \(\mathop{\mathrm{\mathcal{L}}}^*_{R_1}\), \(\mathop{\mathrm{\mathcal{L}}}^*_{R_2}\) and \(\mathop{\mathrm{\mathcal{L}}}^*_{R_1 \cup R_2}\). Using 10 below for each of the three Lindbladians, we can swap the fixed points to find: \[\require{physics} \begin{align} EP_{R_1}(\sigma) + EP_{R_2}(\sigma) &= -\Tr\left((\mathop{\mathrm{\mathcal{L}}}^*_{R_1}(\sigma) +\mathop{\mathrm{\mathcal{L}}}^*_{R_2}(\sigma) )\left(\ln(\sigma) - \ln(\rho)\right)\right) \\ & = -\Tr\left(\mathop{\mathrm{\mathcal{L}}}^*_{R_1\cup R_2}(\sigma)\left(\ln(\sigma) - \ln(\rho)\right)\right) \\ &= -\Tr\left(\mathop{\mathrm{\mathcal{L}}}^*_{R_1 \cup R_2}(\sigma) \left(\ln(\sigma) - \ln(\mathop{\mathrm{\mathbb{E}}}^*_{R_1\cup R_2}(\sigma))\right)\right) \\ &= EP_{R_1 \cup R_2}(\sigma)\;. \end{align}\] On the second line, we used that \(R_1\) and \(R_2\) are disjoint. In the last step, we also used 9, which ensures that \(\mathop{\mathrm{\mathbb{E}}}^*_{R_1\cup R_2}(\sigma)\) is full rank. ◻

Proof. As \(\mathop{\mathrm{\mathcal{L}}}^*\) has a full-rank fixed point, there exists a decomposition of the Hilbert space \[\mathop{\mathrm{\mathcal{H}}}= \bigoplus_{i\in I} \mathop{\mathrm{\mathcal{H}}}_i \otimes \mathop{\mathrm{\mathcal{K}}}_i\] such that the conditional expectation has the following decomposition [80]: \[\require{physics} \mathop{\mathrm{\mathbb{E}}}^*_{\mathop{\mathrm{\mathcal{L}}}} (\sigma) = \sum_{i\in I} \Tr_{\mathop{\mathrm{\mathcal{K}}}_i}(P_i \sigma P_i)\otimes \tau_i\] where \(P_i\) is the projection onto the \(i\)-th block \(\mathop{\mathrm{\mathcal{H}}}_i\otimes \mathop{\mathrm{\mathcal{K}}}_i\) and where \(\tau_i\) are full-rank states on \(\mathop{\mathrm{\mathcal{K}}}_i\). Furthermore, in this decomposition, the kernel of \(\mathop{\mathrm{\mathcal{L}}}\) is given by \(\bigoplus_{i\in I} \mathop{\mathrm{\mathcal{B}}}(\mathop{\mathrm{\mathcal{H}}}_i)\otimes \mathbb{1}_{\mathop{\mathrm{\mathcal{K}}}_i}\). Since \(\rho_1\) and \(\rho_2\) have full rank, we can directly evaluate the difference of logarithms, using that \(\rho_j = \mathop{\mathrm{\mathbb{E}}}^*_{\mathop{\mathrm{\mathcal{L}}}}(\rho_j)\) for \(j=1,2\): \[\require{physics} \ln(\rho_1) - \ln(\rho_2) = \sum_{i\in I}\left(\ln(\Tr_{\mathop{\mathrm{\mathcal{K}}}_i}(P_i \rho_1 P_i))-\ln(\Tr_{\mathop{\mathrm{\mathcal{K}}}_i}(P_i \rho_2 P_i))\right)\otimes\mathbb{1}_{\mathop{\mathrm{\mathcal{K}}}_i}\] which is an element of the kernel of \(\mathop{\mathrm{\mathcal{L}}}\). Thus, \[\require{physics} \Tr\left(\mathop{\mathrm{\mathcal{L}}}^*(\sigma)\left(\ln(\rho_1) - \ln(\rho_2)\right)\right) = \Tr\left(\sigma \mathop{\mathrm{\mathcal{L}}}\left(\ln(\rho_1) - \ln(\rho_2)\right)\right) = 0 .\] ◻

Proof. By [81] the relative entropy of any two states \(\sigma, \sigma'\in \mathop{\mathrm{\mathcal{S}}}(\mathop{\mathrm{\mathcal{H}}})\) can be approximated by \[\label{eq:approxentr} D(\sigma\|\sigma') = \lim_{n\to\infty} D(\sigma + \xi_n\|\sigma'+\xi_n)\tag{45}\] for any sequence \(\xi_n\) of positive, bounded operators with \(\lim_{n\to\infty}\xi_n=0\) and as long as \(\sigma+\xi_n>0\) and \(\sigma'+\xi_n>0\) for all \(n\). Recall that \(\rho>0\) is a fixed point of \(e^{t\mathop{\mathrm{\mathcal{L}}}^*}\) and thus also of \(\mathop{\mathrm{\mathbb{E}}}^*_{\mathop{\mathrm{\mathcal{L}}}}\). Consider \(\xi_n=n^{-1}\rho>0\) and note that \(\sigma+\frac{1}{n}\rho\) is of full rank for any state \(\sigma\).Thus, for any (not necessarily full rank) state \(\sigma\), we find \[\begin{align} D(e^{t\mathop{\mathrm{\mathcal{L}}}^*}\sigma \| \mathop{\mathrm{\mathbb{E}}}^*_{\mathop{\mathrm{\mathcal{L}}}}\sigma) &= \lim_{n\to\infty} D(e^{t\mathop{\mathrm{\mathcal{L}}}^*}(\sigma) + n^{-1}\rho\| \mathop{\mathrm{\mathbb{E}}}^*_{\mathop{\mathrm{\mathcal{L}}}}(\sigma)+n^{-1}\rho)\\ &= \lim_{n\to\infty} D(e^{t\mathop{\mathrm{\mathcal{L}}}^*}(\sigma + n^{-1}\rho)\| \mathop{\mathrm{\mathbb{E}}}^*_{\mathop{\mathrm{\mathcal{L}}}}(\sigma+n^{-1}\rho))\\ &\leq f(t) \lim_{n\to\infty} D(\sigma + n^{-1}\rho\| \mathop{\mathrm{\mathbb{E}}}^*_{\mathop{\mathrm{\mathcal{L}}}}(\sigma+n^{-1}\rho))\\ &= f(t) D(\sigma \| \mathop{\mathrm{\mathbb{E}}}^*_{\mathop{\mathrm{\mathcal{L}}}}\sigma) \;, \end{align}\] where the last step is based on the linearity of the conditional expectation and 45 . Note that we assume the latter only for normalized states but applied it to non-normalized states. However, for any two normalized states \(\sigma_1,\sigma_2\) and any positive number \(c\), it holds that \[\require{physics} D(c\sigma_1\|c\sigma_2) = \Tr\left(c\sigma_1 (\ln\sigma_1+\ln c -\ln\sigma_2 -\ln c)\right) = c D(\sigma_1\|\sigma_2) \;.\] Thus, 45 can be extended to non-normalized states \(\sigma\). ◻