Proof. [it:tau1] Since \(\chi_L\) is finite-rank operator, then \(\chi_L A \chi_L\) is trace class. [32] implies the statement.
[it:tau2] It follows by [32] together with [32].
[it:tau3] [32] proves the statement.
[it:tau4] We note that for finite \(L\) by linearity and by exploiting the inequality \(\mathop{\mathrm{Tr}}(\left\lvert BC\right\rvert)\leq \mathop{\mathrm{Tr}}(\left\lvert B\right\rvert)\left\| C \right\|\) for any trace class operator \(B\) and and any bounded operator \(C\), we have \[\frac{1}{\left\lvert C_L\right\rvert}\mathop{\mathrm{Tr}}\chi_{L} \int_{\mathbb{R}} A(t) \mathrm{d}t \chi_{L} = \frac{1}{\left\lvert C_L\right\rvert}\int_{\mathbb{R}} \mathop{\mathrm{Tr}}\chi_{L} A(t) \chi_{L} \mathrm{d}t \;,\] and thus, since \(|C_{L}|^{-1}\mathop{\mathrm{Tr}}\chi_{L} A(t) \chi_{L} \leq c \| A(t) \|\) for some uniform constant \(c\), we can take the limit \(L \to \infty\) inside of the integral by dominated convergence and obtain the claim.
[it:tau5] The proof can be found e. g. [41].
[it:tau6] Observe that in view of Lemma 1[it:tau1], decomposition ?? and the cyclicity of the trace, we have that \(\left\lvert C_1\right\rvert\tau(A)=\mathop{\mathrm{Tr}}( A \chi_{\mathcal{P}_{\gamma_1}})+\mathop{\mathrm{Tr}}( A\chi_{\mathcal{P}_{\gamma_2}})=0\). Thus, by using [32], we obtain that \[\label{eqn:orsymnotused} \frac{1}{\left\lvert C_L\right\rvert}\mathop{\mathrm{Tr}}(\chi_L X_j A \chi_L)=\frac{1}{\left\lvert C_1\right\rvert}\mathop{\mathrm{Tr}}( \chi_1 X_j A \chi_1)+\frac{\tau(A)}{L^2} \left(\sum_{\gamma\in \Gamma\cap C_L}\gamma_j\right)=\frac{1}{\left\lvert C_1\right\rvert}\mathop{\mathrm{Tr}}( \chi_1 X_j A \chi_1),\tag{17}\] where the right-hand side term does not depend on the side \(L\). Hence, by employing again decomposition ?? we get that \[\begin{align} \tau\left( X_j A\right)&=\frac{1}{\left\lvert C_1\right\rvert}\left[ \mathop{\mathrm{Tr}}(\chi_{\mathcal{P}_{\gamma_1}} X_j A \chi_{\mathcal{P}_{\gamma_1}}) + \mathop{\mathrm{Tr}}(\chi_{\mathcal{P}_{\gamma_2}} X_j A \chi_{\mathcal{P}_{\gamma_2}}) \right]\\ &=\frac{1}{\left\lvert C_1\right\rvert}\left[ \lambda_{1j}\mathop{\mathrm{Tr}}( A \chi_{\mathcal{P}_{\gamma_1}}) + \lambda_{2j}\mathop{\mathrm{Tr}}( A \chi_{\mathcal{P}_{\gamma_2}}) \right]=0, \end{align}\] where we have used that \(\chi_{\mathcal{P}_{\gamma_i}} X_j=\lambda_{ij}\chi_{\mathcal{P}_{\gamma_i}}\). ◻

Proof. The first claim in [it:i-P0] follows by the triangle inequality. To prove the second claim in [it:i-P0], we note that by Cauchy integral formula we can write \[f(H) = \frac{\mathrm{i}}{2 \pi} \oint_{\mathcal{C}} \mathrm{d}z f(z) (H-z)^{-1} \;,\] where \(\mathcal{C}\) is a positively-oriented complex contour within the strip of analyticity of \(f\) and crossing the real axis away from the spectrum of \(H\). The claim then follows by Combes–Thomas estimates for the resolvent \((H-z)^{-1}\), which give exponential decay for such \(z \in \mathcal{C}\).

We now prove [it:ii-P0]. Recalling the dimerization notation in 5 , we obtain that \[\label{eqn:dimpos} {\left(X_i\right)}^{(j)}_{n,m}=m_i\delta_{n,m} \mathbb{1}_{\mathbb{C}^4}+ d_{j,i}\delta_{n,m}\begin{pmatrix} E_4 & 0\\ 0 & E_4 \end{pmatrix},\tag{21}\] where \(d_{j,i}\) denotes the \(i\)-th component of the vector \(d_j\) and \(E_4:=\begin{pmatrix}0 & 0\\ 0 & 1\end{pmatrix}\). Therefore, one has that \[{\left([A,X_i]\right)}^{(j)}_{n,m}=(m_i-n_i){\left(A\right)}^{(j)}_{n,m}+d_{j,i}\left[{\left(A\right)}^{(j)}_{n,m},\begin{pmatrix} E_4 &0\\0 & E_4 \end{pmatrix}\right],\] thus \([A,X_i] \in \mathcal{B}_{0}(\mathcal{H})\). Finally, if \(A\) is periodic, by the Jacobi identity for commutators we get \[\begin{align} [[A,X_i],T_\gamma]&= -[[T_\gamma,A],X_i]-[[X_i,T_\gamma],A] =\gamma_i [ T_\gamma,A]=0\, , \end{align}\] where we have used that \([X_i,T_\gamma]=\gamma_i T_\gamma\). ◻

Proof. In order to prove both statements [it:IA] and [it:etaI], we first notice that both integrals can be fibered via the Bloch–Floquet transform (see 11 ) since all the involved operators are in \(\mathcal{P}(\mathcal{H})\). Let us restrict to the case in which \(A=PAP^\perp\), as the other term \(P^\perp AP\) can be treated in a similar way. By using the spectral decomposition \(H(k)=\sum_{n=1}^4E_n(k)P_n(k)\), where \(P_n(k)\) is the eigenprojector associated with the eigenvalue \(E_n(k)\), we have that \[\mathrm{e}^{\mathrm{i}s H(k)}P(k)=\sum_{n:\, E_n(k)\leq \mu}\mathrm{e}^{\mathrm{i}s E_n(k)}P_n(k),\qquad \mathrm{e}^{-\mathrm{i}s H(k)}P^\perp(k)=\sum_{m:\, E_m(k)> \mu}\mathrm{e}^{-\mathrm{i}s E_m(k)}P_m(k).\] Thus, we get that \[\label{eqn:Apart} \left(\mathcal{F}_i \,\mathrm{e}^{\mathrm{i}s H}P A P^\perp \mathrm{e}^{-\mathrm{i}s H} \, \mathcal{F}_i ^{*}\right)(k)= \sum_{\substack{n:\, E_n(k)\leq \mu\\m:\, E_m(k)> \mu } } \mathrm{e}^{\mathrm{i}s (E_n(k)-E_m(k))}\,P_n(k)A(k)P_m(k).\tag{24}\]

[it:IA] By using 24 , we deduce that \[\label{eqn:repr-P1} \begin{align} \left(\mathcal{F}_i \, I(P A P^\perp )\,\mathcal{F}_i ^{*}\right)(k)&=\lim_{\eta\to 0^+}\mathrm{i}\sum_{\substack{n:\, E_n(k)\leq \mu\\m:\, E_m(k)> \mu } }P_n(k)A(k)P_m(k)\int_{-\infty}^{0}\mathrm{d}s\, \mathrm{e}^{ s (\eta +\mathrm{i}(E_n(k)-E_m(k)))}\\ &=\lim_{\eta\to 0^+}\mathrm{i}\sum_{\substack{n:\, E_n(k)\leq \mu\\m:\, E_m(k)> \mu } }P_n(k)A(k)P_m(k)\frac{1}{\eta +\mathrm{i}(E_n(k)-E_m(k))}\\ &=\sum_{\substack{n:\, E_n(k)\leq \mu\\m:\, E_m(k)> \mu } }\frac{P_n(k)A(k)P_m(k)}{ E_n(k)-E_m(k)}, \end{align}\tag{25}\] in the last equality we have used that \(E_n(k)-E_m(k)\neq 0\) for all \(k\in \mathbb{T}^2_*\) due to the hypothesis on \(\mu\) in a spectral gap [47] and the continuity of the eigenvalues \(E_n(k)\) for all \(1\leq n\leq 4\). From this explicit computation in \(k\)-space, we have that \[I(PAP^\perp)=\mathcal{L}_H^{-1}\left( PAP^\perp\right).\] [it:etaI] By using equality 24 , we have that \[\begin{align} &\lim _{\eta \to 0^{+}}\mathrm{i}\int_{-\infty}^{0}\mathrm{d}s\, \eta \mathrm{e}^{\eta s} B(k)\mathrm{e}^{\mathrm{i}s H(k)}A(k)\mathrm{e}^{-\mathrm{i}s H(k)}\\ &\qquad=\lim _{\eta \to 0^{+}}\sum_{\substack{n:\, E_n(k)\leq \mu\\m:\, E_m(k)> \mu } } \,B(k)P_n(k)A(k)P_m(k)\int_{-\infty}^{0}\mathrm{d}s\, \eta\mathrm{e}^{ s (\eta +\mathrm{i}(E_n(k)-E_m(k)))}\\ &\qquad=\lim _{\eta \to 0^{+}}\sum_{\substack{n:\, E_n(k)\leq \mu\\m:\, E_m(k)> \mu } } \,B(k)P_n(k)A(k)P_m(k)\frac{\eta}{\eta +\mathrm{i}(E_n(k)-E_m(k))}=0, \end{align}\] by exploiting again the insulator hypothesis guaranteeing \(E_n(k)-E_m(k)\neq 0\) for all \(k\in \mathbb{T}^2_*\). ◻

Proof. Observe that \([X_j,P]\in \mathcal{P}_{0}(\mathcal{H})\) by Lemma 3 and Lemma 2. Note that the operator \([X_j,P]\) is off-diagonal with respect to the decomposition induced by the projection \(P\), see 23 . Therefore, Lemma 4[it:IA] implies that \(I([X_j,P])\equiv P_1\) exists in \(\mathcal{P}(\mathcal{H}).\) ◻

Proof. Since \(\sigma^{\mathrm{s}}_{ij}= \lim _{\eta \to 0^{+}} \tau(J_{i}^{\mathrm{conv}} L_{\eta,j})\), it suffices to prove that the quantity \(\tau(J_{i}^{\mathrm{conv}} L_{\eta,j})\) is antisymmetric. First of all, we show \(\tau(J_{i}^{\mathrm{conv}} L_{\eta,i}) = 0\). This is a simple consequence of the symmetry under inversion. Indeed, since \[\label{eqn:rvhops} \Pi_a \,H \,\Pi_a^{-1} = H, \qquad \Pi_a\, X_i\, \Pi_a^{-1} =(-)^{i+a} X_i, \qquad\Pi_a\, S_{z}\, \Pi_a^{-1} = - S_{z}\tag{30}\] we have⁵ \[\Pi_a \,J_{i}^{\mathrm{conv}} \,\Pi_a ^{-1} = (-)^{i+a+1}J_{i}^{\mathrm{conv}}\;, \qquad \Pi_a\, L_{\eta,j}\, \Pi_a^{-1} = (-)^{j+a}L_{\eta,j} \;.\] Therefore, since \(J_{i}^{\mathrm{conv}} L_{\eta,j} \in \mathcal{P}(\mathcal{H})\) by using Lemma 1[it:tau4] and the identity \(\Pi_a \,\chi_1\, \Pi_a^{-1}=\chi_1\), we conclude that \[\begin{align} \tau(J_{i}^{\mathrm{conv}} L_{\eta,i}) = \tau(\Pi_a\, J_{i}^{\mathrm{conv}} L_{\eta,i}\,\Pi_a^{-1}) = -\tau(J_{i}^{\mathrm{conv}} L_{\eta,i}) \;. \end{align}\] Now, we proceed by proving that \(\tau(J_{1}^{\mathrm{conv}} L_{\eta,2})=-\tau(J_{2}^{\mathrm{conv}} L_{\eta,1})\). Observe that \[R_{2\pi/3} H R_{2\pi/3}^{-1} = H, \qquad R_{2\pi/3} S_{z} R_{2\pi/3}^{-1} = S_{z},\] \[R_{2\pi/3} \begin{pmatrix} X_1 \\ X_2 \end{pmatrix} R_{2\pi/3}^{-1}= \begin{pmatrix} -\frac{1}{2} & +\frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & -\frac{1}{2} \end{pmatrix} \begin{pmatrix} X_1 \\ X_2 \end{pmatrix}.\] Therefore, we get that \[\begin{align} \tau(J_{1}^{\mathrm{conv}} L_{\eta,2})&=\frac{\sqrt{3}}{4}\tau(J_{1}^{\mathrm{conv}} L_{\eta,1})+\frac{1}{4}\tau(J_{1}^{\mathrm{conv}} L_{\eta,2})-\frac{3}{4}\tau(J_{2}^{\mathrm{conv}} L_{\eta,1})-\frac{\sqrt{3}}{4}\tau(J_{2}^{\mathrm{conv}} L_{\eta,2})\\ &=\frac{1}{4}\tau(J_{1}^{\mathrm{conv}} L_{\eta,2})-\frac{3}{4}\tau(J_{2}^{\mathrm{conv}} L_{\eta,1}), \end{align}\] in the first equality we have used that \(R_{2\pi/3} \chi_1 R_{2\pi/3}^{-1}=\widetilde{\chi}_1\) being the characteristic function of \(\widetilde{C}_1=\mathrm{R}_{2\pi/3} \, C_1\) with \(\mathrm{R}_{2\pi/3}\) defined in 13 (in other words, \(\widetilde{C}_1\) is the fundamental cell with reference to the basis \(\{a_2, a_3\}\)) and Lemma 1[it:tau2], and in the second equality we have exploited that \(\tau(J_{i}^{\mathrm{conv}} L_{\eta,i}) = 0\). ◻

Proof. [it:vanstorqueonP] First of all, note that \(\mathscr{T}_z,L_{\eta,j}\in\mathcal{P}(\mathcal{H})\) by Lemmas 2 and 3, thus the operator \(\mathscr{T}_z\, L_{\eta,j} \,\chi_{\mathcal{P}_{\gamma_i}}\) is trace class. Consider the operator \(\chi_{\{T_\nu \mathcal{P}_{\gamma_i}:\, \nu\in\Gamma \}}\) being in \(\mathcal{P}(\mathcal{H})\). By Lemma 1[it:tau1] we have that \[\left\lvert C_1\right\rvert\tau(\chi_{\{T_\nu \mathcal{P}_{\gamma_i}:\, \nu\in\Gamma \}}\mathscr{T}_z\, L_{\eta,j})=\mathop{\mathrm{Tr}}(\chi_1\,\chi_{\{T_\nu \mathcal{P}_{\gamma_i}:\, \nu\in\Gamma \}}\,\mathscr{T}_z\, L_{\eta,j}\,\chi_1)=\mathop{\mathrm{Tr}}(\mathscr{T}_z\, L_{\eta,j}\,\chi_{\mathcal{P}_{\gamma_i}}).\] So we conclude by showing that \(\tau(\chi_{\{T_\nu \mathcal{P}_{\gamma_i}:\, \nu\in\Gamma \}}\mathscr{T}_z\, L_{\eta,j})=0\). Introducing \(\widetilde{\chi}_1\) as the characteristic function of \(\widetilde{C}_1=\mathrm{R}_{2\pi/3} \, C_1\) with \(\mathrm{R}_{2\pi/3}\) defined in 13 , we observe that \[\begin{align} \left\lvert C_1\right\rvert\tau(\chi_{\{T_\nu \mathcal{P}_{\gamma_i}:\, \nu\in\Gamma \}}\mathscr{T}_z\, L_{\eta,j})&=\mathop{\mathrm{Tr}}(\chi_1\,\chi_{\{T_\nu \mathcal{P}_{\gamma_i}:\, \nu\in\Gamma \}}\,\mathscr{T}_z\, L_{\eta,j}\,\chi_1)\\ &=\mathop{\mathrm{Tr}}(R_{2\pi/3}\,\chi_1\,\chi_{\{T_\nu \mathcal{P}_{\gamma_i}:\, \nu\in\Gamma \}}\,\mathscr{T}_z\, L_{\eta,j}\,\chi_1\, R_{2\pi/3}^{-1})\\ &=\mathop{\mathrm{Tr}}(\widetilde{\chi}_1\,\chi_{\{T_\nu \mathcal{P}_{\gamma_i}:\, \nu\in\Gamma \}}\,\mathscr{T}_z\, R_{2\pi/3}\, L_{\eta,j}\,R_{2\pi/3}^{-1}\,\widetilde{\chi}_1)\\ &=|\widetilde{C}_1|\tilde{\tau}(\chi_{\{T_\nu \mathcal{P}_{\gamma_i}:\, \nu\in\Gamma \}}\mathscr{T}_z\, R_{2\pi/3}\, L_{\eta,j}\,R_{2\pi/3}^{-1})\\ &=\left\lvert{C}_1\right\rvert\tau(\chi_{\{T_\nu \mathcal{P}_{\gamma_i}:\, \nu\in\Gamma \}}\mathscr{T}_z\, R_{2\pi/3}\, L_{\eta,j}\,R_{2\pi/3}^{-1}) \;, \end{align}\] where we used the invariance of the trace under unitary conjugation, the identities \(R_{2\pi/3}\,\mathscr{T}_z\, R_{2\pi/3}^{-1}= \mathscr{T}_z\), \(R_{2\pi/3}\,\chi_{\{T_\nu \mathcal{P}_{\gamma_i}:\, \nu\in\Gamma \}}\, R_{2\pi/3}^{-1}=\chi_{\{T_\nu \mathcal{P}_{\gamma_i}:\, \nu\in\Gamma \}}\), and Lemma 1[it:tau2]. We proceed by noting that \[\begin{align} R_{2\pi/3}\, L_{\eta,1}\,R_{2\pi/3}^{-1}&=\mathrm{i}\int_{-\infty}^{0}\mathrm{d}s\, \mathrm{e}^{\eta s} \mathrm{e}^{\mathrm{i}s H}[R_{2\pi/3}\, X_1\,R_{2\pi/3}^{-1},P]\mathrm{e}^{-\mathrm{i}s H}\\ &=\mathrm{i}\int_{-\infty}^{0}\mathrm{d}s\, \mathrm{e}^{\eta s} \mathrm{e}^{\mathrm{i}s H} \left[\frac{\sqrt{3}}{2}X_2-\frac{1}{2}X_1,P\right]\mathrm{e}^{-\mathrm{i}s H}, \end{align}\] and similarly \[R_{2\pi/3}\, L_{\eta,2}\,R_{2\pi/3}^{-1}=-\mathrm{i}\int_{-\infty}^{0}\mathrm{d}s\, \mathrm{e}^{\eta s} \mathrm{e}^{\mathrm{i}s H} \left[\frac{\sqrt{3}}{2}X_1+\frac{1}{2}X_2,P\right]\mathrm{e}^{-\mathrm{i}s H}.\] Therefore, defining \(\alpha_j:=\tau(\chi_{\{T_\nu \mathcal{P}_{\gamma_i}:\, \nu\in\Gamma \}}\mathscr{T}_z\, L_{\eta,j}),\) we obtain the following system in \(\alpha_j\)’s: \[\begin{cases} \alpha_1&=\frac{\sqrt{3}}{2}\alpha_2-\frac{1}{2}\alpha_1\\ \alpha_2&=-\frac{\sqrt{3}}{2}\alpha_1-\frac{1}{2}\alpha_2, \end{cases}\] from which it follows that \(\alpha_1=0=\alpha_2\).
[it:vansrotoncell] Lemma 1[it:tau6] together with the previous point [it:vanstorqueonP] implies the statement. ◻

Proof of Proposition 7. [it:main1] & [it:main2] Note that the operator \(J_{i}^{\mathrm{conv}} \, L_{\eta,j}\) is periodic and bounded since both \(J_{i}^{\mathrm{conv}}\) and \([P,X_j]\) are short-range by Lemmas 2 and 3. Thus, Lemma 1[it:tau1] ensures that \(\sigma_{ij}^{\mathrm{s},\mathrm{conv}}\) is well-defined. On the other hand, for the proper spin conductivity \(\sigma_{ij}^{\mathrm{s},\mathrm{prop}}\) we have that \(\tau(X_i \mathscr{T}_z\, L_{\eta,j} )=0\) in view of Proposition 10[it:vansrotoncell], thus only \(\mathrm{i}[H,X_i]S_z L_{\eta,j}\in\mathcal{P}(\mathcal{H})\) contributes to its value. Specifically, we get that \[\label{eqn:sigma1} \begin{align} \mathop{\mathrm{Re}}\tau(J_{i}^{\mathrm{prop}}\, L_{\eta,j})&=\mathop{\mathrm{Re}}\tau(\mathrm{i}[H,X_i]S_z \,L_{\eta,j})\\ &=\frac{1}{2}\left( \tau(\mathrm{i}[H,X_i]S_z L_{\eta,j})+ \overline{\tau(\mathrm{i}[H,X_i]S_z L_{\eta,j}) } \right)\\ &=\frac{1}{2}\left( \tau(\mathrm{i}[H,X_i]S_z L_{\eta,j})+ \tau(S_z \mathrm{i}[H,X_i]L_{\eta,j} ) \right)\\ &=\tau(J_{i}^{\mathrm{conv}}\, L_{\eta,j}) \end{align}\tag{31}\] where we have used that \(\overline{\mathop{\mathrm{Tr}}(A)}=\mathop{\mathrm{Tr}}(A^*)\) for every trace class operator \(A\) and the cyclicity of the trace per unit volume, see Lemma 1[it:tau5]. Therefore, we shall compute the adiabatic limit \(\eta\to 0^+\) of \(\tau(J_{i}^{\mathrm{conv}}\, L_{\eta,j})\).

To this end, it is convenient to rewrite the argument of the real part in 31 by using that the operator \(L_{\eta,j}\) is off-diagonal due to the off-diagonality of \([X_j,P]\). By exploiting the cyclicity of the trace per unit volume, we have that \[\begin{align} \tau(\mathrm{i}[H,X_i]S_z \,L_{\eta,j})&=\tau\left(\mathrm{i}{([H,X_i]S_z)}^{\mathrm{OD}} \,{(L_{\eta,j})}^{\mathrm{OD}}\right)\\ &=\tau\left(\mathrm{i}[H,X_i^{\mathrm{OD}}]S_z^{\mathrm{D}} \,{(L_{\eta,j})}^{\mathrm{OD}}\right)+\tau\left(\mathrm{i}[H,X_i^{\mathrm{D}}]S_z^{\mathrm{OD}} \,{(L_{\eta,j})}^{\mathrm{OD}}\right)\\ &=:T_1(\eta)+T_2(\eta), \end{align}\] where we have defined the two terms \(T_1(\eta)\) and \(T_2(\eta)\) according to the order of appearance. We further split \(T_1(\eta)\) as sum of two terms: \[\begin{align} T_1(\eta)&=-\int_{-\infty}^{0}\mathrm{d}s\, \mathrm{e}^{\eta s}\tau\left( [H,X_i^{\mathrm{OD}}]S_z \mathrm{e}^{\mathrm{i}s H} [X_j,P]\mathrm{e}^{-\mathrm{i}s H}\right)\\ &=-\int_{-\infty}^{0}\mathrm{d}s\, \mathrm{e}^{\eta s}\left\{\tau\left(X_i^{\mathrm{OD}}S_z \mathrm{e}^{\mathrm{i}s H} [X_j,P]\mathrm{e}^{-\mathrm{i}s H}H\right)\right.\\ &\qquad\left.+\tau\left( X_i^{\mathrm{OD}} [H,S_z] \mathrm{e}^{\mathrm{i}s H}[X_j,P]\mathrm{e}^{-\mathrm{i}s H}\right)+\tau\left( X_i^{\mathrm{OD}} S_zH \mathrm{e}^{\mathrm{i}s H}[X_j,P]\mathrm{e}^{-\mathrm{i}s H}\right)\right\}\\ &=\int_{-\infty}^{0}\mathrm{d}s\, \mathrm{e}^{\eta s}\left\{\tau\left( X_i^{\mathrm{OD}}S_z [H,\mathrm{e}^{\mathrm{i}s H}[X_j,P]\mathrm{e}^{-\mathrm{i}s H}]\right)+\tau\left( X_i^{\mathrm{OD}} [H,S_z] \mathrm{e}^{\mathrm{i}s H}[X_j,P]\mathrm{e}^{-\mathrm{i}s H}\right)\right\}\\ &=:T_{11}(\eta)+T_{12}(\eta), \end{align}\] by using that \(X_i^{\mathrm{OD}} =[[X_i,P],P]\in \mathcal{P}(\mathcal{H})\) in view of 23 , Lemma 2 and 3, and repeatedly the cyclicity of the trace per unit volume. For the term \(T_{11}(\eta)\) we perform an integration by parts: \[\label{eqn:eqn:sigma3} \begin{align} T_{11}(\eta) &=-\mathrm{i}\int_{-\infty}^{0}\mathrm{d}s\, \mathrm{e}^{\eta s}\tau\left( X_i^{\mathrm{OD}}S_z \frac{\mathrm{d}}{\mathrm{d}s}\mathrm{e}^{\mathrm{i}s H}[X_j,P]\mathrm{e}^{-\mathrm{i}s H}\right)\\ &=-\mathrm{i}\tau\left( X_i^{\mathrm{OD}}S_z [X_j,P]\right)+\mathrm{i}\int_{-\infty}^{0}\mathrm{d}s\, \eta\mathrm{e}^{\eta s}\tau\left( X_i^{\mathrm{OD}}S_z \mathrm{e}^{\mathrm{i}s H}[X_j,P]\mathrm{e}^{-\mathrm{i}s H}\right). \end{align}\tag{32}\] The second term on the right-hand side of 32 does not contribute to the spin conductivity by Lemma 4[it:etaI]. On the other hand, in view of 23 , the first term on the right-hand side of 32 can be rewritten as: \[\label{eqn:eqn:sigma4} \begin{align} -\mathrm{i}\tau\left( X_i^{\mathrm{OD}}S_z^{\mathrm{D}} [X_j,P]\right)&=\mathrm{i}\tau\left( P[X_i,P]S_z [X_j,P]\right)-\mathrm{i}\tau\left( [X_i,P]P S_z [X_j,P]\right)\\ &=\mathrm{i}\tau\left( P[X_i,P][S_z, [X_j,P]]\right)+\mathrm{i}\tau\left( S_z P[[X_i,P], [X_j,P]]\right)\\ &=\mathrm{i}\tau\left( P[X_i,P][[P,S_z],X_j]\right)+\mathrm{i}\tau\left( S_z P[[X_i,P], [X_j,P]]\right) \end{align}\tag{33}\] where we have used repeatedly the cyclicity of the trace per unit volume, and in the last equality we have exploited that \([S_z, [X_j,P]]=-[P, [S_z,X_j]]-[X_j,[P,S_z]]=[[P,S_z],X_j]\). By taking the real part of the second term on the last line of 33 , we obtain the term in 28 ; while in the last line of 33 the real part of the first summand gives the third term in the expression for \(\Sigma^{\mathrm{snc}}_{ij}\), see 29 . Next, we collect the remaining terms contributing to the spin conductivity: \[\label{eqn:firsttwoinsnc} \lim_{\eta\to 0^+}\mathop{\mathrm{Re}}\left\{T_2(\eta)+T_{12}(\eta)\right\}=\mathop{\mathrm{Re}}\tau\left(\mathrm{i}[H,X_i] [[S_z,P],P] P_{1}\right)+ \mathop{\mathrm{Re}}\tau\left( \mathrm{i}[[X_i,P],P][S_z,H]P_{1} \right)\tag{34}\] where we have used that we can omit the diagonal part of \(X_i\) due to the cyclicity of the trace per unit volume and definition of the operator \(P_{1}\) in ?? . The right-hand side of 34 gives the remaining first two terms in the expression for \(\Sigma^{\mathrm{snc}}_{ij}\).

[it:main3] follows immediately from [it:main2] by using that \(\Sigma^{\mathrm{snc}}_{ij}=0\) due to \([H,S_z]=0=[P,S_z]\). ◻

Proof. For brevity, we write \(P^{\upuparrows}:= p^\uparrow P p^\uparrow\) and \(P^{\downdownarrows}:= p^\downarrow P p^\downarrow\), so that \([H,S_{z}] = 0\) implies \(P= P^{\upuparrows}+ P^{\downdownarrows}\). Thus, by using Proposition 7[it:main3], Lemma 1[it:tau1] and [it:tau3] one obtains that \[\label{eqn:sigmaspinchern} \begin{align} \sigma_{ij}^{\mathrm{s}}&=\mathop{\mathrm{Re}}\tau\big( \mathrm{i}S_z P[[X_i,P], [X_j,P]]\big)\\ &=\frac{1}{2}\tau\big( \mathrm{i}P^{\upuparrows}[[X_i,P^{\upuparrows}], [X_j,P^{\upuparrows}]]\big)-\frac{1}{2}\tau\big( \mathrm{i}P^{\downdownarrows}[[X_i,P^{\downdownarrows}], [X_j,P^{\downdownarrows}]]\big)\\ &=\frac{1}{2\pi}\left(\frac{1}{2} Chern(P^{\upuparrows})_{ij} -\frac{1}{2} Chern(P^{\downdownarrows})_{ij}\right)=\frac{1}{2\pi} S-Chern(P)_{ij} \in \frac{1}{4\pi}\mathbb{Z}\;, \end{align}\tag{36}\] where we used that the fibre of \([X_i,P]\) is \(\mathrm{i}\partial_{k_{i}}P(k)\). Moreover, if \(H\) is time-reversal symmetric then we note the identity \(Chern(P^{\downdownarrows})=-Chern(P^{\upuparrows})\) and conclude that \(\sigma_{ij}^{\mathrm{s}}=\frac{1}{2\pi} Chern(P^{\upuparrows})_{ij}\in \frac{1}{2\pi}\mathbb{Z}\). ◻

Proof. We note that \(\tau (A[S_z, F] B )\) is well-defined by Lemma 1[it:tau1]. By using identity 38 for the operator \(F\), we obtain that \[\begin{align} \frac{1}{4}\tau(A[S_z, F] B)&=-\tau(A S_z[S_z, F] S_z B)\\ &=-\tau([A, S_z][S_z, F] S_z B)-\tau( S_z A [S_z, F] S_z B)\\ &=-\tau([A, S_z][S_z, F] [S_z, B])-\tau([A, S_z][S_z, F] B S_z)\\ &\phantom{=}\quad-\tau( S_z A [S_z, F] [S_z, B])-\tau( S_z A [S_z, F] B S_z ). \end{align}\] Therefore, by using the cyclicity of \(\tau(\cdot)\) and the identity \(S_z^2=\frac{1}{4}\mathbb{1}_{\mathcal{H}}\), we obtain that \[\begin{align} \label{eqn:terms-tauAB} -\frac{1}{2}\tau(A[S_z, F] B)&=\tau([A, S_z][S_z, F] [S_z, B])+\tau([A, S_z][S_z, F] B S_z) \nonumber \\ &\phantom{=}\quad+\tau( S_z A [S_z, F] [S_z, B]) \;. \end{align}\tag{41}\] Thus, by using ?? and the hypothesis that \(A,B,F\in \mathcal{Z}(S_z)\), inequality ?? follows. ◻

Proof. First of all, we note that \(P, {(H-z)}^{-1} ,[H,X_j], [P,X_j], P_1\in \mathcal{P}(\mathcal{H})\) by Lemmas 2, 3, and Corollary 1. Notice that \[\label{eqn:commresxj} [{(H-z)}^{-1},X_j]={(H-z)}^{-1}[X_j,H]{(H-z)}^{-1}\tag{42}\] which is in \(\mathcal{Z}(S_z)\) as well. Observe that obviously \(H\in \mathcal{Z}(S_z)\) by the very definition of the space \(\mathcal{Z}(S_z)\). Note that \[\label{eqn:resolventinZ} [(H - z )^{-1},S_z]=(H - z )^{-1}[S_z,H] (H - z )^{-1},\tag{43}\] thus by the boundedness of the resolvent operators we get that \((H - z )^{-1}\in \mathcal{Z}(S_z)\) uniformly in \(z\) which varies in the compact set \(\mathcal{C}\). Therefore, by using 19 , we have that \[\label{eqn:[P,Sz]} [P,S_z]=\frac{\mathrm{i}}{2\pi} \oint_{\mathcal{C}} \mathrm{d}z\, [(H - z )^{-1},S_z] \in \mathcal{Z}(S_z).\tag{44}\] By Jacobi’s identity we note that \[[[H,X_j],S_z]=[[H,S_z],X_j],\] so we obtain that \([H,X_j]\in\mathcal{Z}(S_z)\) as well. By Jacobi’s identity, 43 and 42 , we have that \[\begin{align} [[(H - z )^{-1},X_j],S_z]&=[[(H - z )^{-1},S_z],X_j]= [(H - z )^{-1}[S_z,H] (H - z )^{-1},X_j]\\ &=[(H - z )^{-1},X_j][S_z,H] (H - z )^{-1} +(H - z )^{-1}[[S_z,H],X_j] (H - z )^{-1}\\ &\qquad+(H - z )^{-1}[S_z,H] [(H - z )^{-1},X_j]\\ &= (H - z )^{-1} [X_j,H](H - z )^{-1}[S_z,H] (H - z )^{-1}\\ &\qquad+(H - z )^{-1}[[S_z,H],X_j] (H - z )^{-1}\\ &\qquad+(H - z )^{-1}[S_z,H](H - z )^{-1} [X_j,H](H - z )^{-1}, \end{align}\] thus one estimates the right-hand side terms by \({\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert [H,S_z] \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}\) up to a constant, being uniform in \(z\in\mathcal{C}\). Thus, again in view of 19 , we get that \[[[P,X_j],S_z]=\frac{\mathrm{i}}{2\pi} \oint_{\mathcal{C}} \mathrm{d}z\, [[(H - z )^{-1},X_j],S_z]\in\mathcal{Z}(S_z).\] We are left to show that \(P_1\in \mathcal{Z}(s_z)\). By employing identity ?? , we get that \[[P_1,S_z]=\frac{\mathrm{i}}{2\pi} \oint_{\mathcal{C}} \mathrm{d}z\, [(H - z )^{-1}[[X_j,P],P] (H - z )^{-1},S_z],\] thus Leibniz’s rule and \({(H-z)}^{-1},P,[P,X_j]\in \mathcal{Z}(S_z)\) imply that \(P_1\in \mathcal{Z}(s_z)\). ◻

Proof of Theorem 14. Let us first prove ?? . By Proposition 7, we have \[\label{eqn:spuriousterms} \sigma^{\mathrm{s}}_{ij}-\Sigma^{\mathrm{sc}}_{ij}=\mathop{\mathrm{Re}}\tau\left(\mathrm{i}[H,X_i] [[S_z,P],P] P_{1} +\mathrm{i}[[X_i,P],P][S_z,H]P_{1} + \mathrm{i}P[X_i,P][[P,S_z],X_j] \right).\tag{45}\] The conclusion follows by applying Lemma 6 on each of the terms on the r.h.s.  of 45 . Indeed, considering the first term, we write \[\mathop{\mathrm{Re}}\tau\left(\mathrm{i}[H,X_i] [[S_z,P],P] P_{1}\right) =\mathop{\mathrm{Re}}\tau\left(\mathrm{i}[H,X_i] [S_z,P]PP_{1}\right)-\mathop{\mathrm{Re}}\tau\left(\mathrm{i}[H,X_i] P[S_z,P] P_{1}\right) \;.\] By Lemma 7 and Remark 15 \([H,X_i], P, P P_1, [H,X_i]P, P_1 \in \mathcal{Z}(S_z)\), so Lemma 6 implies that \[\big| \tau\left(\mathrm{i}[H,X_i] [[S_z,P],P] P_{1}\right) \big| \leq C {\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert [H,S_z] \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}^{2} \;.\] The bound for the second term in 45 follows similarly. For the third term we notice that \[\begin{align} \tau\left( P[X_i,P][[P,S_z],X_j]\right)&= \tau\left( P[X_i,P][S_z,[X_j,P]]P\right), \end{align}\] where we have used the cyclicity of the trace per unit volume and the Jacobi’s identity; thus, by noticing that \(P[X_i,P], P, [X_j,P]\in \mathcal{Z}(S_z)\) thanks to Lemma 7 and Remark 15, and employing again Lemma 6 estimate ?? follows.

We are left with proving ?? . By Proposition 7 and by Lemma 5, we have that \[\label{eqn:diffsigmaschern} \begin{align} \Sigma^{\mathrm{sc}}_{ij} -\frac{1}{2\pi}Chern(P^{\mathrm{sc}})_{ij} & = \mathop{\mathrm{Re}}\tau\left( \mathrm{i}S_z P[[X_i,P], [X_j,P]]\right) - \mathop{\mathrm{Re}}\tau\left( \mathrm{i}S_z P^{\mathrm{sc}}[[X_i,P^{\mathrm{sc}}], [X_j,P^{\mathrm{sc}}]]\right) \\ & = \mathop{\mathrm{Re}}\tau\left( \mathrm{i}S_z (P-P^{\mathrm{sc}})[[X_i,P], [X_j,P]]\right)\\ &\phantom{=}+\mathop{\mathrm{Re}}\tau\left( \mathrm{i}S_z P^{\mathrm{sc}}[[X_i,(P-P^{\mathrm{sc}})], [X_j,P]]\right) \\ &\phantom{=}+\mathop{\mathrm{Re}}\tau\left( \mathrm{i}S_z P^{\mathrm{sc}}[[X_i,P^{\mathrm{sc}}], [X_j,(P-P^{\mathrm{sc}})]]\right). \end{align}\tag{46}\] By Remark 13[it:diffPPsc] and 37 it holds true that \[\label{eqn:P-Psc} P-P^{\mathrm{sc}}=\frac{\mathrm{i}}{\pi} \oint_{\mathcal{C}} \mathrm{d}z\,(H - z )^{-1}[S_z,H]S_z (H^{\mathrm{sc}} - z )^{-1}.\tag{47}\] Therefore, to estimate the first summand on the r.h.s. of 46 we notice that \[\begin{align} \tau\left( S_z (P-P^{\mathrm{sc}})[[X_i,P], [X_j,P]]\right)&= \frac{\mathrm{i}}{\pi}\oint_{\mathcal{C}} \mathrm{d}z\,\tau\left( S_z (H - z )^{-1}[S_z,H]S_z (H^{\mathrm{sc}} - z )^{-1}[[X_i,P], [X_j,P]]\right). \end{align}\] Observe that the operator \(S_z (H^{\mathrm{sc}} - z )^{-1}[[X_i,P], [X_j,P]]\) is in \(\mathcal{Z}(S_z)\) in view of Lemma 7 and Remark 15. Also the operators \(H, S_z (H - z )^{-1}\in \mathcal{Z}(S_z)\) again by Lemma 7, thus Lemma 6 implies that the first summand on the r.h.s. of 46 is bounded, up to a constant, by \({{\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert [H,S_z] \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}}^2\). For the analysis of the second term on the r.h.s. of 46 , we note that by Leibniz’s rule and 47 , we get that \[\begin{align} [X_i,P-P^{\mathrm{sc}}]&= \frac{\mathrm{i}}{\pi} \oint_{\mathcal{C}} \mathrm{d}z\Bigl([X_i,(H - z )^{-1}][S_z,H]S_z (H^{\mathrm{sc}} - z )^{-1} \Bigr.\\ &\phantom{=}\Bigl.+(H - z )^{-1}][[H,X_i],S_z]S_z (H^{\mathrm{sc}} - z )^{-1} +(H - z )^{-1}][S_z,H]S_z[X_i, (H^{\mathrm{sc}} - z )^{-1}]\Bigr)\\ &=:T_1+T_2+T_3, \end{align}\] where the definitions of the three terms \(T_1,T_2,T_3\) is understood with respect to the order of the appearance. Let us proceed by analyzing the first contribution for the second term on the r.h.s. of 46 : \[\label{eqn:T1} \begin{align} \tau\left( \mathrm{i}S_z P^{\mathrm{sc}}[T_1, [X_j,P]]\right)&=\frac{\mathrm{i}}{\pi} \oint_{\mathcal{C}} \mathrm{d}z\,\Bigl(\tau\left( \mathrm{i}S_z P^{\mathrm{sc}} [X_i,(H - z )^{-1}][S_z,H]S_z (H^{\mathrm{sc}} - z )^{-1} [X_j,P]\right)\Bigr.\\ &\phantom{=}\Bigl.-\tau\left( \mathrm{i}S_z P^{\mathrm{sc}}[X_j,P] [X_i,(H - z )^{-1}][S_z,H]S_z (H^{\mathrm{sc}} - z )^{-1} \right)\Bigr). \end{align}\tag{48}\] Since on the r.h.s of the last equality the operators \(S_z P^{\mathrm{sc}} [X_i,(H - z )^{-1}],\,H,\,S_z (H^{\mathrm{sc}} - z )^{-1} [X_j,P]\in\mathcal{Z}(S_z)\) and also the operators \(S_z P^{\mathrm{sc}}[X_j,P] [X_i,(H - z )^{-1}],\,S_z (H^{\mathrm{sc}} - z )^{-1}\in\mathcal{Z}(S_z)\) by using again Lemma 7 and Remark 15, then Lemma 6 implies that the l.h.s. of 48 is bounded up to a constant by \({\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert [H,S_z] \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}^2\). The analysis of the contributions for the second term on the r.h.s. of 46 coming from \(T_2\) and \(T_3\) is analogous. The last summand on the r.h.s. of 46 can be estimated similarly. ◻

Proof. Note that the operators \(\chi_{A}\), \(\chi_{B}\), and \(T_{v}\) for any fixed \(v \in \mathbb{R}^{2}\), are short-range, so is \(H_{\mathrm{KM}}\). Furthermore, \(H_{\mathrm{KM}}\) is periodic since \([T_{ v},T_{u}] =0\) for any vectors \(v,u\in \mathbb{R}^2\) and \([\chi_\sharp, T_\gamma]=0\) for every \(\sharp\in\{A,B\}\).

To verify that \(H_{\mathrm{KM}}\) is \(2\pi/3\)-rotationally symmetric, one notes that \[\label{eqn:actionexp1} \big[S_{j},S_{k}\big] = \mathrm{i}\, \sum_{\ell} \varepsilon_{j k \ell} S_{\ell} \;,\tag{50}\] \(\varepsilon_{j k \ell}\) being the totally antisymmetric tensor, so that \[\mathrm{e}^{-\mathrm{i}\frac{2\pi}{3}S_z}S_x =\left(\frac{\sqrt{3}}{2}S_y-\frac{1}{2}S_x\right)\mathrm{e}^{-\mathrm{i}\frac{2\pi}{3}S_z} \;, \qquad \mathrm{e}^{-\mathrm{i}\frac{2\pi}{3}S_z}S_y= -\left(\frac{\sqrt{3}}{2}S_x+\frac{1}{2}S_y \right)\mathrm{e}^{-\mathrm{i}\frac{2\pi}{3}S_z} \;.\] We also observe that a translation operator \(T_v\) and a rotation \(R_{2 \pi /3}\) are intertwined by the following relation: \[\label{eqn:transrot} R_{2 \pi /3}\,T_v=T_{\mathrm{R}_{2 \pi /3} v}\,R_{2 \pi /3}.\tag{51}\] Accordingly, setting for brevity \(\sigma_{\pm} := (\sqrt{3} \sigma_{x} \pm \sigma_{y})/2\) we have that \[\begin{align} 3R_{2\pi/3}H_{\mathrm{R1}} &= \mathrm{i}R_{2\pi/3}\Big[- \big( T_{ d_1} - T_{-d_1} \big) \otimes \sigma_{+} +\big( T_{ d_2} - T_{-d_2} \big) \otimes \sigma_{-} + \big( T_{ d_3} - T_{-d_3} \big) \otimes \sigma_y \Big]\\ &=\mathrm{i}\Big[\big( T_{ d_2} - T_{-d_2} \big) \otimes \sigma_{-} + \big( T_{ d_3} - T_{-d_3} \big) \otimes \sigma_y - \big( T_{ d_1} - T_{-d_1} \big) \otimes \sigma_{+} \Big] R_{2\pi/3}\\ &=3H_{\mathrm{R1}}R_{2\pi/3}. \end{align}\] Analogously, one checks that \(R_{2\pi/3}H_{\mathrm{R2}}=H_{\mathrm{R2}}R_{2\pi/3}\). Using again 51 , we obtain \([R_{2\pi/3},H_{\mathrm{NN}}]=[R_{2\pi/3},H_{\mathrm{SO}}]=0\), and obviously \([R_{2\pi/3}, H_v]=0\) as well. Hence, \(H_{\mathrm{KM}}\) is \(2\pi/3\)-rotation symmetric.

Let us proceed to check that \(H_{\mathrm{KM}}\) is inversion symmetric. The relevant operator in this case is \[(\Pi_{1}\psi)(x):= \mathrm{e}^{-\mathrm{i}\pi S_{y}}\psi(\mathrm{R}_{v}^{-1} x) \;, \qquad \forall \psi \in \mathcal{H} \;,\] where \(\mathrm{R}_{v}\) is the restriction to \(\mathcal{C}\) of the vertical reflection on \(\mathbb{R}^{2}\), namely \(\mathrm{R}_{v}(x_{1},x_{x})= (x_{1},-x_{2})\). It is straightforward to check that \(\Pi_{1}\) satisfies 15 with \(a=1\). By 50 , we have \[\label{eqn:acthref} \mathrm{e}^{-\mathrm{i}\pi S_y}S_x=-S_x\mathrm{e}^{-\mathrm{i}\pi S_y},\quad\mathrm{e}^{-\mathrm{i}\pi S_y}S_y=S_y\mathrm{e}^{-\mathrm{i}\pi S_y},\quad \mathrm{e}^{-\mathrm{i}\pi S_y}S_z=-S_z\mathrm{e}^{-\mathrm{i}\pi S_y}.\tag{52}\] Furthermore, we observe that the \(\Pi_{1}\) and translation operators are intertwined by the following identity: \[\label{eqn:transref} \Pi_{1}\,T_v=T_{\mathrm{R}_{\mathrm{v}} v}\,\Pi_{1}.\tag{53}\] By identity 53 , one easily verifies that \([H_{\mathrm{NN}}, \Pi_{1}]=[H_v,\Pi_{1}]=0\). By using the third equality in 52 and 53 , one has that \([H_{\mathrm{SO}},\Pi_{1}]=0\). By employing the first two equalities in 52 and again 53 , one also has that \[\begin{align} 3\Pi_{1}H_{\mathrm{R1}} &= \mathrm{i}\Pi_{1} \Big[ -\big( T_{ d_1} - T_{-d_1} \big) \otimes \sigma_{+} + \big( T_{ d_2} - T_{-d_2} \big) \otimes \sigma_{-} +\big( T_{ d_3} - T_{-d_3} \big) \otimes \sigma_y \Big]\\ &=\mathrm{i}\Big [ \big( T_{ d_2} - T_{-d_2} \big) \otimes \sigma_{-} +\big( T_{ d_1} - T_{-d_1} \big) \otimes\left(-\sigma_{+}\right) + \big( T_{ d_3} - T_{-d_3} \big) \otimes \sigma_y \Big] \Pi_{1}\\ &=3H_{\mathrm{R1}}\Pi_{1} . \end{align}\] Similarly, one checks that \([H_{\mathrm{R2}},\Pi_{1}]=0\). Finally, analogous computations show that \([H_{\mathrm{KM}},\Theta]=0\). ◻

Proof. First of all, we provide the dimerisation \([H_{\mathrm{KM}}]_{{\mathcal{D}}_i}\). With abuse of notation, we let \(T_{v}\) denote the translation operator on \(\ell^{2}(\Gamma)\), whose definition is as in 9 with \(\Gamma\) in place of \(\mathcal{C}\). By straightforward computations, we obtain: \[\begin{align} [H_{\mathrm{NN}}]_{\mathcal{D}_i} & = \sum_{j=1}^{3} \begin{pmatrix} 0 & T_{d_{i}-d_{j}}\\ T_{d_{j}-d_{i}} & 0\end{pmatrix} \otimes \mathbb{1}_{\mathbb{C}^{2}} \;, \\ [H_{\mathrm{SO}}]_{\mathcal{D}_i}& = - \frac{\mathrm{i}}{3 \sqrt{3}} \sum_{j=1}^{3} \begin{pmatrix} T_{a_{j}} -T_{-a_{j}} & 0 \\ 0 & T_{-a_{j}} -T_{a_{j}} \end{pmatrix} \otimes \sigma_{z} \;, \\ [H_{\mathrm{W}}]_{\mathcal{D}_i} & = \begin{pmatrix} \mathbb{1}_{\ell^{2}(\mathcal{C})} & 0\\ 0 & -\mathbb{1}_{\ell^{2}(\mathcal{C})} \end{pmatrix} \otimes \mathbb{1}_{\mathbb{C}^{2}} \;, \end{align}\] together with \[\begin{align} [H_{\mathrm{R1}}]_{\mathcal{D}_i} & = \frac{\mathrm{i}}{3} \sum_{j=1}^{3} \begin{pmatrix} 0& -T_{d_{i}-d_{j}} \\ T_{d_{j}-d_{i}} & 0\end{pmatrix} \otimes ( d_{j} \wedge \sigma)_{z} \;, \\ [H_{\mathrm{R2}}]_{\mathcal{D}_i} & = \frac{\mathrm{i}}{3} \sum_{j=1}^{3} \begin{pmatrix} T_{a_{j}} -T_{-a_{j}} & 0 \\ 0 & T_{a_{j}} - T_{-a_{j}} \end{pmatrix} \otimes ( a_{j} \wedge \sigma)_{z} \;. \end{align}\] \(\sigma= (\sigma_{x},\sigma_{y},\sigma_{z})\) being the vector of Pauli matrices, and where \(\{ d_{j}\}_{j}\) and \(\{ a_{j}\}_{j}\) are the vectors introduced in 1 and 2 . Let \(H_{\mathrm{\sharp}}(k)\) denote the fibration of \(H_{\mathrm{\sharp}}\) via the Bloch–Floquet transform at the momentum \(k\in\mathbb{T}^2_*\), with \(\sharp \in \{\text{NN}, \text{SO},\text{W},\text{R1}, \text{R2} \}\), having fixed the dimerisation in with \(i=3\). Because \(\big(\mathcal{F}\,T_{v}\psi\big)(k) = \mathrm{e}^{-\mathrm{i}k \cdot v}\, \mathcal{F}\psi(k)\) for any \(v\in\Gamma\), it follows that \[tH_{\mathrm{NN}}(k) = \begin{pmatrix} 0 & \Omega(k) \\ \overline{\Omega(k)} & 0 \end{pmatrix} \otimes \mathbb{1}_{\mathbb{C}^{2}} \;, \quad \lambda_{\mathrm{SO}}H_{\mathrm{SO}}(k) = \begin{pmatrix} \alpha_{\mathrm{SO}}(k) & 0 \\ 0 & - \alpha_{\mathrm{SO}}(k) \end{pmatrix} \otimes \sigma_{z} \;.\] The term \(H_{\mathrm{W}}(k)\) is trivial, so that we are left with determining \(H_{\mathrm{R2}}(k)\) and \(H_{\mathrm{R1}}(k)\). By simple computations, we obtain the identity \[\mathrm{i}(\sigma \wedge d_{j})_{z} = \begin{pmatrix} 0 & \mathrm{e}^{\mathrm{i}\frac{2 \pi}{3}(3-j)} \\ -\mathrm{e}^{-\mathrm{i}\frac{2 \pi}{3}(3-j)} & 0 \end{pmatrix} \qquad \quad j =1,2,3 \;,\] which allows us to check the correctness of the formula for \(\Omega_{\mathrm{R}}(k)\). Similarly, we obtain \[{(\sigma \wedge a_{j})}_z = -\sqrt{3}\begin{pmatrix} 0 & \mathrm{e}^{\mathrm{i}\frac{2 \pi}{3}(3-j)} \\ \mathrm{e}^{-\mathrm{i}\frac{2 \pi}{3}(3-j)} & 0 \end{pmatrix} \qquad \quad j =1,2,3 \;,\] and the formula for \(\alpha_{\mathrm{R}}\) ensues by simple manipulations. ◻

Proof. By time-reversal symmetry, we have \[H_{\mathrm{KM}}(k) = \begin{pmatrix} \mathbb{0}_2 &\mathbb{1}_{2} \\ -\mathbb{1}_{2} & \mathbb{0}_2 \end{pmatrix} \overline{H_{\mathrm{KM}}(-k)} \begin{pmatrix} \mathbb{0}_2 & \mathbb{1}_{2} \\ -\mathbb{1}_{2} & \mathbb{0}_2 \end{pmatrix}^{-1} \;.\] Since \(H_{\mathrm{KM}}(k) = H_{\mathrm{KM}}(k)^{*}\) and since \(\begin{pmatrix} \mathbb{0}_2 &\mathbb{1}_{2} \\ -\mathbb{1}_{2} & \mathbb{0}_2 \end{pmatrix}\) is unitary, \(H_{\mathrm{KM}}(-k)^{\mathrm{T}}\) is unitarily equivalent to \(H_{\mathrm{KM}}(k)\), hence the claim. ◻

Proof. In view of Lemma 9, by using 59 and 60 we obtain \[H_{\mathrm{KM}}(k_F^+)= \begin{pmatrix} (w+ \lambda_{\mathrm{SO}})\,\sigma_{z} & \lambda_{\mathrm{R}} \,\sigma_+ \\ \, \lambda_{\mathrm{R}} \,\sigma_- & (w - \lambda_{\mathrm{SO}})\,\sigma_{z} \end{pmatrix}\] where \(\sigma_\pm = (\sigma_{x} \pm \mathrm{i}\sigma_{y})/2\). By computing the characteristic polynomial, we obtain the following expressions for the energy bands, denoted by \(\big(\mathcal{E}_{i,l}\big)_{i=1,2, l = \pm}\), at \(k_{F}^{+}\): \[\begin{align} \mathcal{E}_{1,\pm}(k_{F}^{+})&= \lambda_{\mathrm{SO}}\pm \sqrt{w^2+\lambda_{\mathrm{R}}^{2}} \;, \qquad \mathcal{E}_{2,\pm}(k_{F}^{+})= -\lambda_{\mathrm{SO}} \pm |w|\;. \end{align}\] Note that the spin is not conserved and thus is not a good label for the energy bands. To compute the local internal gap \(\Delta(k_{F}^{+})\), it is actually easier to first determine which bands are the external ones. Without loss of generality, we assume \(\lambda_{\mathrm{SO}} >0\): otherwise, we could flip the sign of \(w\) and \(\lambda_{\mathrm{R}}\) to flip the sign of \(H_{\mathrm{KM}}(k)\), the latter operation leaving the set of internal/external bands invariant. Note that \(\mathcal{E}_{1,+}(k_{F}^{+})\) is always the positive external band. When \(\mathcal{E}_{1,-}(k_{F}^{+})\geq\mathcal{E}_{2,-}(k_{F}^{+})\), the negative external band is \(\mathcal{E}_{2,-}(k_{F}^{+})\), so that the local internal spectral gap is \[\Delta(k_{F}^{+})= |\mathcal{E}_{1,-}(k_{F}^{+})-\mathcal{E}_{2,+}(k_{F}^{+}) | \;.\] The condition \(\mathcal{E}_{1,-}(k_{F}^{+})\geq\mathcal{E}_{2,-}(k_{F}^{+})\) is equivalent to \(|\lambda_{\mathrm{R}}|\geq 2\sqrt{\lambda_{\mathrm{SO}}^{2} + |w |\lambda_{\mathrm{SO}}}\). In contrast, if \(\mathcal{E}_{1,-}(k_{F}^{+})<\mathcal{E}_{2,-}(k_{F}^{+})\), \(\mathcal{E}_{2,\pm}(k_{F}^{+})\) become the internal bands, and therefore \(\Delta(k_{F}^{+}) =2|w|\). The expression for \(w_{c}\) and \(\mu _{c}\) is a straightforward computation from \(\Delta(k_{F}^{+})\), noticing that \(\mathcal{E}_{2,+}(k_{F}^{+})\) is always an internal band. ◻

Proof. Equivalently, we prove that the operator \(H_{\mathrm{KM}}(k) - \mu _{c} \mathbb{1}_{4}\) is invertible for any \(k\neq k_{F}^{\pm}\), and for \(k = k_{F}^{\pm}\) but \(w \neq w_{c}^{\pm}\). Because of the time-reversal invariance, namely Lemma 10, we can restrict ourselves to the upper-half Brillouin torus, \(\mathbb{T}^{2}_{+}:=\{(k_{1},k_{2}) \in \mathbb{T}^2_*\, | \, k_{2}\geq 0\}\). We let \(B_{\rho}(k):=\{q \in \mathbb{R}^{2} \,| \,|q-k|<\rho\}\) and \(\widetilde{B}_{\rho}(k):=B_{\rho}(k) \setminus \{ k\}\). By using the explicit expression for the energy bands at \(\lambda_{\mathrm{R}} =0\), see Remark 18, we know that for \(\rho\) small \[\sup _{q \in \mathbb{T}^{2}_{+} \setminus B_{\rho}(k_{F}^{+})} \Big\| \big(H_{\mathrm{KM}}(k)\big|_{\lambda_{\mathrm{R}} = 0} - \mu _{c} \mathbb{1}_{4}\big)^{-1}\Big\| \lesssim \rho^{-1} \;.\] We use the identity \[\label{eqn:vonneumann} {(A+B)}^{-1}=A^{-1}{(\mathbb{1}+BA^{-1})}^{-1}\tag{61}\] with \(A = H_{\mathrm{KM}}(k)\big|_{\lambda_{\mathrm{R}} = 0} - \mu _{c} \mathbb{1}_{4}\), \(B = \lambda_{\mathrm{R}} H_{\mathrm{R}}\) and the smallness of \(BA^{-1}\), to obtain \[\begin{align} \sup _{k \in \mathbb{T}^{2}_{+} \setminus B_{\rho}(k_{F}^{+})} \Big\| \big(H_{\mathrm{KM}}(k) - \mu _{c} \mathbb{1}_{4}\big)^{-1}\Big\| & \leq \sup _{k \in \mathbb{T}^{2}_{+}\setminus B_{\rho}(k_{F}^{+})}\frac{\Big\| \big(H_{\mathrm{KM}}(k)\big|_{\lambda_{\mathrm{R}} = 0} - \mu _{c} \mathbb{1}_{4}\big)^{-1}\Big\|}{1 -|\lambda_{\mathrm{R}}|\| H_{\mathrm{R}}\| \Big\| \big(H(k)\big|_{\lambda_{\mathrm{R}} = 0} - \mu _{c} \mathbb{1}_{4}\big)^{-1} \Big\|} \\ &\lesssim \rho^{-1} \end{align}\] provided that \(|\lambda_{\mathrm{R}}|\) is small enough depending on \(\rho\), where we used that \(H_{R}\) is a bounded operator. Now, we proceed by investigating the invertibility of \(H_{\mathrm{KM}}(k)-\mu_c\mathbb{1}_{4}\) inside the pierced ball \(\widetilde{B}_{\rho}(k_F^+)\). We note that if \(|w|\) is large enough the Hamiltonian \(H_{\mathrm{KM}}(k) - \mu _{c}\mathbb{1}_{4}\) is invertible for any \(k \in \mathbb{T}^2_*\). This follows by using equality 61 again, where \(A=w H_{\mathrm{W}}\) and \(B=H_{\mathrm{KM}}-w H_{\mathrm{W}}\). Therefore, the claim of the proposition follows if we can prove that \(H_{\mathrm{KM}}(k_{F}^{+}+q) - \mu _{c}\mathbb{1}_{4}\) is invertible for \(q \in \widetilde{B}_{\rho}(0)\) with \(\rho>0\) small enough, for \(w\) in a compact set and for \(|\lambda_{\mathrm{R}}|\) small enough, depending only on \(t\), \(\lambda_{\mathrm{SO}}\) and \(r\). To this end, we compute the determinant and obtain: \[\label{eqn:detHKM} \begin{align} &\det\Big( H_{\mathrm{KM}}(k_{F}^{+}+q) - \mu _{c} \Big) \\ &\,= \Big(w^2 - \Big(\lambda_{\mathrm{SO}} - \frac{\lambda_{\mathrm{R}}^{2}}{4\lambda_{\mathrm{SO}}} \Big)^{2}\Big)^{2}+2 |q|^{2}\Big(w^2 + \lambda_{\mathrm{SO}}^{2}+O_{w,\lambda_{\mathrm{SO}},r}(\lambda_{\mathrm{R}}^{2})\Big) + o_{w,\lambda_{\mathrm{SO}},\lambda_{\mathrm{R}},r}(|q|^2), \end{align}\tag{62}\] where, without loss of generality we set \(t = 2/3\) (compare with 57 ). This expression was obtained by expanding the matrix elements of \(H_{\mathrm{KM}}\) see ?? as in 59 and 60 . If \(w\neq w_c^{\pm}\) then the determinant is obviously non-zero since the first term in 62 does not vanish. Otherwise, we need to analyze the second-order term in \(\left\lvert q\right\rvert\). By inspection of \(H_{\mathrm{KM}}(k^{+}_{F}+q)\), we have that \(O_{w,\lambda_{\mathrm{SO}},r}(\,\cdot \,)\) and \(o_{w,\lambda_{\mathrm{SO}},\lambda_{\mathrm{R}},r}(\,\cdot \,)\) depend continuously on the parameters provided that \(\lambda_{\mathrm{SO}} \neq 0\). Therefore, since \(\lambda_{\mathrm{SO}}\) is fixed and \(w\) is in a compact set, we can choose \(|\lambda_{\mathrm{R}}|\) small enough so that the term proportional to \(|q|^{2}\) is non-zero and, if \(\rho\) is small enough (recall \(q \in \widetilde{B}_{\rho}(0)\)), dominates the term \(o_{w,\lambda_{\mathrm{SO}},\lambda_{\mathrm{R}},r}(|q|^2)\). This implies that \(\det\Big( H_{\mathrm{KM}}(k_{F}^{+}+q) - \mu _{c}\mathbb{1}_{4} \Big) \neq 0\) unless \(q=0\) and the claim is proven. ◻

Proof of Theorem 21. Assume the contrary, that is, \(\sigma^{\mathrm{s}}_{12}(H,\mu) \in \frac{1}{2 \pi} \mathbb{Z}\) for any insulator \((H,\mu)\) with \(H\) time-reversal symmetric and belonging to \(\mathcal{P}_{0}^{\hexagon}(\mathcal{H})\). Then, \(\delta \sigma_{12}^{\mathrm{s}}(t,\lambda_{\mathrm{SO}},\lambda_{\mathrm{R}},r) \in \frac{1}{2 \pi} \mathbb{Z}\) for any value of \(t,\lambda_{\mathrm{SO}},\lambda_{\mathrm{R}},r\), which is in contradiction with ?? .

We are left with proving the lower bound in ?? , the upper bound being stated in Theorem 14. Once more, we do this by considering the pair \((H_{\mathrm{KM}},\mu_{c})\) at fixed \(t,\lambda_{\mathrm{SO}},r\) non-zero, at \(\lambda_{\mathrm{R}} > 0\) sufficiently small and at \(w\) sufficiently close to \(w_{c}^{+}(\lambda_{\mathrm{SO}},\lambda_{\mathrm{R}})\), that is, for \(m \neq 0\), \(m\) sufficiently small. Since all parameters but \(m\) and \(\lambda_{\mathrm{R}}\) are fixed, let us abridge 64 to \(\sigma_{12}^{s}(m,\lambda_{\mathrm{R}})\), and likewise \(P^{\mathrm{sc}}= P^{\mathrm{sc}}(m)\), \(w^{+}_{c} = w^{+}_{c}(\lambda_{\mathrm{R}})\), and \(\delta\sigma_{12}^{s}=\delta\sigma_{12}^{s}(\lambda_{\mathrm{R}})\) (notice that \(H^{\mathrm{sc}}=H_{\mathrm{KM}}|_{\lambda_{\mathrm{R}}=0}\)).

We introduce the function \[f(m,\lambda_{\mathrm{R}}):= \sigma_{12}^{s}(m,\lambda_{\mathrm{R}}) - \frac{1}{2\pi}S-Chern\big(P^{\mathrm{sc}}(m)\big)_{12} \;\] for \(m\neq 0\) and \(\lambda_R>0\), both sufficiently small. We observe that \({\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert H^{\mathrm{snc}} \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert} = O(\lambda_{\mathrm{R}})\) (see Remark 17 and 37 ), so that the claim we need to prove is that there exist a \(m\neq 0\), a constant \(C>0\) and an open interval \(J\ni 0\), such that \(|f(m,\lambda_{\mathrm{R}})| \geq C \lambda_{\mathrm{R}}^{2}\) holds true for all \(\lambda_{\mathrm{R}} \in J\).

We now assume the negation of the claim and show that this leads to a contradiction. We first of all note that the negation of the claim implies that \[f(m,\lambda_{\mathrm{R}}) = o(\lambda_{\mathrm{R}}^{2}) .\] In fact, the negation of the claim is that for all \(m\neq 0\), constants \(C>0\), and open intervals \(J \ni 0\), there exists at least one point \(\lambda_0 \in J \setminus \{ 0\}\), such that \(|f(m,\lambda_0)|< C \lambda_0^{2}\). Since this holds for any open interval \(J \ni 0\), we can iterate the reasoning and find a sequence \((\lambda_n)_n\in \mathbb{N}\) such that \(\lambda_n \to 0\) and \(|f(m,\lambda_n)|< C \lambda_n^{2}\). By Remark 28, the limit \(\lim_{\lambda_{\mathrm{R}}\to 0}f(m,\lambda_{\mathrm{R}})\) exists for any \(m\neq0\). As the constant \(C>0\) can be chosen arbitrarily small, \[\lim_{\lambda_{\mathrm{R}} \to 0}f(m,\lambda_{\mathrm{R}})/\lambda_{\mathrm{R}}^2 = 0,\] that is, \(f(m,\lambda_{\mathrm{R}}) = o(\lambda_{\mathrm{R}}^{2})\).

It remains to show that the condition \(f(m,\lambda_{\mathrm{R}}) = o(\lambda_{\mathrm{R}}^{2})\) leads to a contradiction. To this end, we define \[g(m,\lambda_{\mathrm{R}}):=f(m,\lambda_{\mathrm{R}}) - f(-m,\lambda_{\mathrm{R}}) \;, \qquad m>0 \;.\] Since \(f=o(\lambda_{\mathrm{R}}^{2})\) for all \(m\neq 0\) and \(\lambda_{\mathrm{R}}>0\) small enough, we also have \(g = o(\lambda_{\mathrm{R}}^{2})\). However, we can compute \(g\) more precisely by means of Proposition 24 and prove a contradiction. First of all, we compute the spin Chern number of \(P^{\mathrm{sc}}\). The model at \(\lambda_{\mathrm{R}} = 0\) consists of two copies of the Haldane model, see 55 and text around it. Then, as noted at the end of Section 4.2 it is straightforward to see that at \(\lambda_{\mathrm{R}} = 0\), for \(m\) small \[\label{eq:free-spin-chern} S-Chern(P^{\mathrm{sc}}(m))_{12} = \begin{cases} 0 & \qquad m >0 \\ 1 & \qquad m< 0 \;. \end{cases}\tag{66}\] Accordingly, by using this and Proposition 24 we have \[\begin{align} g(m,\lambda_{\mathrm{R}}) & = \frac{1}{2\pi} + \sigma_{12}^{s}(m,\lambda_{\mathrm{R}}) -\sigma_{12}^{s}(-m,\lambda_{\mathrm{R}}) \\ &= \frac{1}{2\pi} + \delta\sigma_{12}^{s}(\lambda_{\mathrm{R}}) + \Big(\sigma_{12}^{s}(m,\lambda_{\mathrm{R}}) - \sigma_{12}^{s}(0^{+},\lambda_{\mathrm{R}}) \Big) - \Big(\sigma_{12}^{s}(-m,\lambda_{\mathrm{R}}) - \sigma_{12}^{s}(0^{-},\lambda_{\mathrm{R}}) \Big) \\ & = -\frac{1}{2\pi}\frac{ \lambda_{\mathrm{R}}^{2} r}{2t\lambda_{\mathrm{SO}} - \lambda_{\mathrm{R}}^{2}r} + \Big(\sigma_{12}^{s}(m,\lambda_{\mathrm{R}}) - \sigma_{12}^{s}(0^{+},\lambda_{\mathrm{R}}) \Big) - \Big(\sigma_{12}^{s}(-m,\lambda_{\mathrm{R}}) - \sigma_{12}^{s}(0^{-},\lambda_{\mathrm{R}}) \Big) \;. \end{align}\] We now show that \(\sigma_{12}^{s}( \pm m,\lambda_{\mathrm{R}}) - \sigma_{12}^{s}(0^{\pm},\lambda_{\mathrm{R}}) =o(\lambda_{\mathrm{R}}^{2})\). To see this, for any \(m'=m'(\lambda_{\mathrm{R}}) >0\) small enough, we write, using 66 and the definition of \(f\) \[\begin{align} \sigma_{12}^{s}( m,\lambda_{\mathrm{R}}) - \sigma_{12}^{s}(0^{+},\lambda_{\mathrm{R}}) & = \sigma_{12}^{s}( m,\lambda_{\mathrm{R}}) - \sigma_{12}^{s}( m',\lambda_{\mathrm{R}}) +\sigma_{12}^{s}( m',\lambda_{\mathrm{R}}) - \sigma_{12}^{s}(0^{+},\lambda_{\mathrm{R}}) \\ & = f(m,\lambda_{\mathrm{R}}) - f(m',\lambda_{\mathrm{R}}) + \sigma_{12}^{s}( m',\lambda_{\mathrm{R}}) - \sigma_{12}^{s}(0^{+},\lambda_{\mathrm{R}}) \\ & = o(\lambda_{\mathrm{R}}^{2}) \;, \end{align}\] where we used that \(\sigma_{12}^{s}( m',\lambda_{\mathrm{R}}) - \sigma_{12}^{s}(0^{+},\lambda_{\mathrm{R}})\) can be made \(o(\lambda_{\mathrm{R}}^2)\) for \(m'\) small enough depending on \(\lambda_{\mathrm{R}}\). Then, by inspection, for all \(m>0\) we get that \(|g(m,\lambda_{\mathrm{R}})| \geq C \lambda_{\mathrm{R}}^{2}\), for some constant \(C>0\) and \(\lambda_{\mathrm{R}}\) small enough, which is a contradiction. ◻

Proof of Proposition 24. For the sake of brevity, we drop the explicit dependence on the fixed parameters \(t,\lambda_{\mathrm{SO}}\), \(\lambda_{\mathrm{R}}\), \(r\). Because of time-reversal symmetry (see Lemma 10), we restrict to the upper-half Brillouin torus, \(\mathbb{T}^{2}_{+}:=\{(k_{1},k_{2}) \in \mathbb{T}^2_*\, | \, k_{2}\geq 0\}\), the contribution coming from the negative one being the same, that is, we have by Theorem 26 \[\label{eqn:AAA-int} \begin{align} \sigma^{\mathrm{s}}_{12} = -2 \int_{\mathbb{R}\times \mathbb{T}_{+}^{2}} \frac{\mathrm{d}\mathbf{k}}{(2 \pi)^{3}} \mathop{\mathrm{Tr}}_{\mathbb{C}^{4}} \Big(A^{\mathrm{s}}_{1}(\mu_{c};\mathbf{k})A_{0}(\mu_{c};\mathbf{k})A_{2}(\mu_{c};\mathbf{k})\Big) \;. \end{align}\tag{67}\] Note that, if the integrand in 67 were well-behaved, by the dominated convergence theorem we could take the limit inside the integral, thus proving that the limits \(m\to 0^{\pm}\) exist and that \(\delta \sigma_{12}^{\mathrm{s}}= 0\). However, the integrand has an non-integrable divergence, which makes the existence of the limit non-trivial and \(\delta \sigma_{12}^{\mathrm{s}}\neq 0\). To capture the singularity of the integrand, we introduce additional notation. By Proposition 20 we know that \((H_{\mathrm{KM}}(k)-\mu)^{-1}\) can be singular only at the Fermi points \(k_F^\pm\), defined in 56 ; thus the only singular point in \(\mathbb{R} \times \mathbb{T}^{2}_{+}\) for \(\widehat{G}(\mu;\mathbf{k})\) is \(\mathbf{k}_{F}^{+} := (0,k_{F}^{+}) \in \mathbb{R} \times \mathbb{T}^{2}_{+}\). We extract the most singular contribution from the Green function around \(\mathbf{k}^{+}_{F}\). We use the notation \(\mathbf{q}=(q_{0},q) \in \mathbb{R} \times \mathbb{R}^{2}\), and denote \(|\, \cdot \, |\) the Euclidean norm. Recalling that \(m = w - w_{c}^{+}\), see 63 , and using the notation \(O_{m,\mathbf{q}}(1)\) to denote terms that remain bounded as \(|m|, |\mathbf{q}| \to 0\), we can write \[\widehat{G}(\mu _{c};\mathbf{k}_{F}^{+} +\mathbf{q}) = S(m,\mathbf{q}) + O_{m,\mathbf{q} }(1) \;, \qquad \big |S(m,\mathbf{q}) \big| \lesssim \big(|\mathbf{q}|+ |m|\big)^{-1} \;,\] which, by direct computations with ?? , 59 and 60 , is given by \[S(m,\mathbf{q}) = \chi(m,\mathbf{q})^{-1} \begin{pmatrix} z_{1}(\mathrm{i}q_0+m) & 0 & v_{1}(\mathrm{i}q_1 + q_2) & z_{2} (m+\mathrm{i}q_0) \\ 0 & 0 & 0 & 0\\ v_{1} (-\mathrm{i}q_1 + q_2) & 0 & \mathrm{i}z_{3} q_0 + z_{4} m & v_{2}(\mathrm{i}q_1 + q_2) \\ z_{2} (m+\mathrm{i}q_0) & 0 & v_{2}(-\mathrm{i}q_1 + q_2) & z_{5} (\mathrm{i}q_{0}+m) \end{pmatrix} ,\] where: \[\alpha_{\pm}:= 4 \lambda_{\mathrm{SO}}^{2} \pm \lambda_{\mathrm{R}}^{2} \;, \qquad \widetilde{\alpha}:= 3 t \lambda_{\mathrm{SO}} -\frac{3}{2} \lambda_{\mathrm{R}}^{2}r \;,\] \[\begin{align} z_{1} &= - \frac{\lambda_{\mathrm{R}}^{2}}{4 \lambda_{\mathrm{SO}}^{2} }\alpha_{-} \;,& z_{2} & := \frac{\lambda_{\mathrm{R}}}{2 \lambda_{\mathrm{SO}}} \alpha_{-} \;,& z_{3}&:= -\frac{1}{4 \lambda_{\mathrm{SO}}^{2}} \alpha_{+} \alpha_{-} \;,& z_{4}&:= \frac{1}{4 \lambda_{\mathrm{SO}}^{2}} \alpha_{-}^{2} \;,\\ z_{5}&:= \alpha_{-} \;, & v_{1}&:= - \frac{\lambda_{\mathrm{R}}}{4 \lambda_{\mathrm{SO}}^{2}} \widetilde{\alpha}\alpha_{-} \;, & v_{2}& := \frac{1}{2\lambda_{\mathrm{SO}}} \widetilde{\alpha}\alpha_{-} , \end{align}\] and \[\chi(m,\mathbf{q}) =\frac{\alpha_{-}}{4 \lambda_{\mathrm{SO}}^{2}} \big(m^{2}\alpha_{-} -2\mathrm{i}q_{0} m \lambda_{\mathrm{R}}^{2}+ q_{0}^{2}\alpha_{+} +|q|^{2} \widetilde{\alpha}^{2}\big).\] Then, we extract the most singular contribution from the connection associated with \(\widehat{G}\) \[A_{\nu}(\mu _{c};\mathbf{k}_{F}^{+} + \mathbf{q}) = \widetilde{A}_{\nu}(m,\mathbf{q})+ O_{m,\mathbf{q}}(1) \;, \qquad \widetilde{A}_{\nu}(m,\mathbf{q}):=\partial_{k_{\nu}} \big[ \widehat{G}(\mu _{c}; \mathbf{k}_{F}^{+})^{-1} \big] S(m,\mathbf{q}) \;\] and similarly for \(A^{\mathrm{s}}_{\nu}\). We have \[\partial_{k_{\nu}}\big[ \widehat{G}(\mu _{c}; \mathbf{k}_{F}^{+})^{-1}\big] = \partial_{k_{\nu}}H(k_{F}^{+}) + \mathrm{i}\delta_{\nu,0} \mathbb{1}_{\mathbb{C}^{4}}\] By inspection, \(H_{\mathrm{KM}}(k)\) and \(S(m,\mathbf{q})\) are rational functions in \(m\); therefore, \(\widehat{G}(\mu_c;\mathbf{k})\) and the terms \(O_{m,\mathbf{q}}(1)\) are also rational functions in \(m\). Since the latter terms are, in particular, bounded, they are also continuous in \(m\). Since the singular part of \(A_{\nu}\) and \(A^{s}_{\nu}\) diverges as \((|\mathbf{q}| + |m|)^{-1}\), all the sub-leading contributions to \(\mathop{\mathrm{Tr}}_{\mathbb{C}^{4}} \Big(A^{\mathrm{s}}_{i}(\mu_{c};\mathbf{k})A_{0}(\mu_{c};\mathbf{k})A_{j}(\mu_{c};\mathbf{k})\Big)\), namely those with at least one \(O_{\mathbf{q},m}(1)\), are uniformly integrable over \(\mathbb{T}^{2}_{+}\) as \(m \to 0\). Therefore, by the dominated convergence theorem, these terms provide the same finite contribution to the limits \(m \to 0^{\pm}\); in particular, they do not contribute to \(\delta \sigma_{12}^{\mathrm{s}}\). Accordingly, we can write \[\label{diff-singular-contr} \delta \sigma_{12}^{\mathrm{s}}= \Big( \lim _{m \to 0^{+}} - \lim _{m \to 0^{-}}\Big)\widetilde{\sigma}^{\mathrm{s}}_{12}(m) \;,\tag{68}\] where, denoting \(\tilde{\mathbb{T}}_{+}^{2}:=\{q \in \mathbb{R}^{2} \, | \, k_{F}^{+}+q \in \mathbb{T}_{+}^{2}\}\), \[\label{singular-Pontryagin-formula} \begin{align} \widetilde{\sigma}^{\mathrm{s}}_{12}(m) = -2 \int_{\mathbb{R}\times \tilde{\mathbb{T}}_{+}^{2}} \frac{\mathrm{d}\mathbf{q}}{(2 \pi)^{3}} \mathop{\mathrm{Tr}}_{\mathbb{C}^{4}} \Big(\widetilde{A}^{\mathrm{s}}_{1}(m;\mathbf{q})\widetilde{A}_{0}(m;\mathbf{q})\widetilde{A}_{2}(m;\mathbf{q})\Big) \;. \end{align}\tag{69}\] We now show that both limits in 68 exist (and, as a matter of fact, also proving that the limits \(\lim _{m \to 0^{\pm}} \sigma^{\mathrm{s}}_{12}(t,\lambda_{\mathrm{SO}},m,\lambda_{\mathrm{R}},r)\) exist) and compute their exact values. With the aid of a computer software [60], we compute the trace in 69 and obtain \[\mathop{\mathrm{Tr}}_{\mathbb{C}^{4}} \Big(\widetilde{A}^{\mathrm{s}}_{1}(m;\mathbf{q})\widetilde{A}_{0}(m;\mathbf{q})\widetilde{A}_{2}(m;\mathbf{q})\Big) = -\frac{ 3 m t \widetilde{\alpha}\alpha_{-}^{2}}{4 \lambda_{\mathrm{SO}} \chi(m,\mathbf{q})^{2}} - \frac{3 t \widetilde{\alpha}^{3} \alpha_{-}^{3}}{4 \lambda_{\mathrm{SO}}^{5} \chi(m,\mathbf{q})^{3}}\big(\alpha_{+}q_{0} -\mathrm{i}m \lambda_{\mathrm{R}}^{2}\big) q_{1}q_{2} \;.\] The second term does not contribute to 69 , because it is odd in \(q_{1}\), whereas the integration domain \(\tilde{\mathbb{T}}_{+}^{2}\) is even. Therefore, \[\begin{align} \delta \sigma_{12}^{\mathrm{s}} & = \Big( \lim _{m \to 0^{+}} - \lim _{m \to 0^{-}}\Big) \frac{3 m t \widetilde{\alpha} \alpha_{-}^{2}}{16 \lambda_{\mathrm{SO}} \pi^{3}} \int_{\mathbb{R}\times \tilde{\mathbb{T}}_{+}^{2}} \frac{\mathrm{d}\mathbf{q}}{\chi(m,\mathbf{q})^{2}} \\ & = \Big( \lim _{m \to 0^{+}} - \lim _{m \to 0^{-}}\Big) \frac{3 m t \widetilde{\alpha}\lambda_{\mathrm{SO}}^{3}}{4 \pi^{3}} \int_{\mathbb{R}^{3} } \mathrm{d}\mathbf{q} \big( q_{0}^{2}\alpha_{+} -2\mathrm{i}q_{0} m \lambda_{\mathrm{R}}^{2} +m^{2}\alpha_{-}+|q|^{2} \widetilde{\alpha}^{2}\big)^{-2} \;. \end{align}\] where in the second line we performed some algebraic manipulations and extended the integral to the whole \(\mathbb{R}^{2}\) by dominated convergence, using that the integrand is uniformly integrable on \(\mathbb{R}^{2} \setminus \tilde{\mathbb{T}}_{+}^{2}\) and continuous in \(m\).

We compute the integral in \(\mathrm{d}q_{0}\) using that, for any \(a,b,c\in \mathbb{R}\) such that \(4ac + b^{2} >0\), we have \(\int_{\mathbb{R}} \mathrm{d}z \big(a z^{2} + \mathrm{i}b z + c\big)^{-2} = -4 \pi a (4 a c +b^{2})^{-\frac{3}{2}}\). Then, we note that for \(a',b'>0\) we have \(\int_{\mathbb{R}^{2}} \mathrm{d}x (a' |x|^{2} +b')^{-\frac{3}{2}} = 2 \pi \big(a' b'^{1/2}\big)^{-1}\), hence we have \[\begin{align} & \int_{\mathbb{R}^{2}} \mathrm{d}q \int_{\mathbb{R}} \mathrm{d}q_{0} \big( q_{0}^{2}\alpha_{+} -2\mathrm{i}q_{0} m \lambda_{\mathrm{R}}^{2} +m^{2}\alpha_{-}+|q|^{2} \widetilde{\alpha}^{2}\big)^{-2} \\ & \qquad = - 4 \pi \alpha_{+} \int_{\mathbb{R}^{2}} \mathrm{d}q \big( 4 |q|^{2} \alpha_{+} \widetilde{\alpha}^{2} + (2 m \lambda_{\mathrm{SO}}^{2})^{2} \big)^{-\frac{3}{2}} = - \frac{\pi^{2}}{\lambda_{\mathrm{SO}}^{2} \widetilde{\alpha}^{2}|m|} \;, \end{align}\] implying the existence of the limits and the sought result ?? . ◻

Proof. Letting \(C^{c}_{L}:={\mathcal{C}}\setminus C_{L}\) and \(\chi_{L}^{c} :=\mathbb{1}_{ C_{L}^{c}}\), we have the identity \[\begin{align} \mathop{\mathrm{Tr}}_{\mathcal{H}} \big( \chi_{L} A B \chi_{L} \big) = \mathop{\mathrm{Tr}}_{\mathcal{H}} \big( \chi_{L} A \chi_{L} B \chi_{L} \big) + \mathop{\mathrm{Tr}}_{\mathcal{H}} \big( \chi_{L} A \chi^{c}_{L} B \chi_{L} \big) \;. \end{align}\] We need to prove that the second piece gives a vanishing contribution when divided by \(|C_{L}|\) as \(L \to \infty\). To this end, we let \(\widetilde{C}_{L}:= \{ x \in C_{L} \, | \, \mathrm{dist}(x, C_{L}^{c}) \leq \log L \}\), where \(\mathrm{dist}\) is in the sense of \(|\, \cdot \, |_1\), be the internal annulus of thickness \(L \log L\). Then, \[\begin{align} \big|\mathop{\mathrm{Tr}}_{\mathcal{H}} \big( \chi_{L} A \chi^{c}_{L} B \chi_{L} \big) \big| & \leq \| A\| \Big( \sum_{y \in C_{L}^{c}, x \in \widetilde{C}_{L}} |B_{y,x} | + \sum_{y \in C_{L}^{c}, x \in C_{L} \setminus \widetilde{C}_{L}} |B_{y,x} |\Big) \;. \end{align}\] In the first term, we are summing over \(x\) belonging to a region of area \(L \log L\), thus \[\sum_{y \in C_{L}^{c}, x \in \widetilde{C}_{L}} |B_{y,x} | \lesssim (L \log L) \sup_{x \in \mathbb{Z}^{2}} \sum_{y \in \mathbb{Z}^{2}}|B_{y,x}| \lesssim L \log L \;,\] where in the last step we used that \(B\) is short-range, see 6 . In the other region, we note that \(B_{y,x}\) is small because the points \(y\) and \(x\) are at distance at least \(\log L\), thus for \(\xi '>\xi\), with \(\xi\) being the constant in 6 for the operator \(B\), we have \[\begin{align} \sum_{y \in C_{L}^{c}, x \in C_{L} \setminus \widetilde{C}_{L}} |B_{y,x} | & \leq \left( \sup _{y \in C_{L}^{c}, x \in C_{L} \setminus \widetilde{C}_{L}} \mathrm{e}^{- |x-y|_{1}/\xi'} \right) \sum_{y \in C_{L}^{c}, x \in C_{L} \setminus \widetilde{C}_{L}} \mathrm{e}^{|x-y|_{1}/\xi'} |B_{y,x} | \\ & \leq \left( \sup _{y \in C_{L}^{c}, x \in C_{L} \setminus \widetilde{C}_{L}} \mathrm{e}^{- |x-y|_{1}/\xi'} \right) L^{2} \sup_{x\in\mathbb{Z}^{2}}\sum_{y \in\mathbb{Z}^{2}} \mathrm{e}^{|x-y|_{1}/\xi'} |B_{y,x} | \\ & \lesssim L^{2- 1/\xi'} \;, \end{align}\] where we used that \(\sup _{y \in C_{L}^{c}, x \in C_{L} \setminus \widetilde{C}_{L}} \mathrm{e}^{- |x-y|_{1}/\xi'} \leq L^{-1/\xi'}\). Since \(|C_{L}| \sim L^{2}\), this concludes the proof. ◻

Proof. To prove ?? , we can open the commutator \([A,B]\) and consider the terms \(ABC\) and \(BAC\) separately. Since \(BC \in \mathcal{P}_{0}(\mathcal{H})\), by Lemma 14 we have \[\tau (ABC) = \lim_{L\to \infty} |C_{L}|^{-1} \mathop{\mathrm{Tr}}_{\mathcal{H}}(A_{L} BC) \;.\] For the other term, we obtain instead \[\tau (BAC) = \lim_{L\to \infty} |C_{L}|^{-1} \mathop{\mathrm{Tr}}_{\mathcal{H}}(\chi_{L}BA \chi_{L} C) \;.\] We write \[\mathop{\mathrm{Tr}}_{\mathcal{H}}(\chi_{L}BA \chi_{L} C) = \mathop{\mathrm{Tr}}_{\mathcal{H}}(B \chi_{L}A \chi_{L} C) +\mathop{\mathrm{Tr}}_{\mathcal{H}}(\chi_{L}B \chi^{c}_{L}A \chi_{L} C) -\mathop{\mathrm{Tr}}_{\mathcal{H}}(\chi_{L}^{c}B \chi_{L}A \chi_{L} C) \;.\] Following the proof of Lemma 14, one can see that the second and third terms are \(O(L^{2-1/\xi})\), \(\xi\) being the constant associated with \(B\) according to 6 , so that the first claim is proven. Concerning ?? , since by Lemma 3 \([B,X] \in \mathcal{P}(\mathcal{H})\) we can apply Lemma 14 and use that \(\chi_{L}\) commutes with \(X\). ◻

Proof of Lemma 13. By 72 we have \[\label{eq:one-body-expectation} \begin{align} \omega_{\beta}^{H,\mu} \big( \mathrm{d}\Gamma(A_{L}) \big) & =\omega_{\beta}^{H,\mu} \left( \sum_{i,j} \langle f_{i}, A_{L} f_{j} \rangle a^{*}(f_{i}) a(f_{j}) \right) \\ & = \sum_{i,j} \langle f_{i}, A_{L} f_{j} \rangle \omega_{\beta}^{H,\mu} \left( a^{*}(f_{i}) a(f_{j}) \right) \\ & = \mathop{\mathrm{Tr}}_{\mathcal{H}} \big( A_{L} \gamma_{\beta}^{H,\mu} \big) \end{align}\tag{74}\] where \((f_{i})_{i}\) is any basis of localised functions, so that the sums above are all finite, and where \(\gamma_{\beta}^{H,\mu}\) was introduced in 70 . Since \((H,\mu)\) is an insulator, we have \(\gamma^{H,\mu}_{\beta} \to P\) in the operator topology as \(\beta \to \infty\), and that \(P \in \mathcal{P}_{0}(\mathcal{H})\) (see Lemma 2). Therefore, since \(A_{L}\) is trace-class we have \[\label{eq:lim-beta} \lim _{\beta \to \infty} \omega_{\beta}^{H,\mu} \big( \mathrm{d}\Gamma(A_{L}) \big) = \mathop{\mathrm{Tr}}_{\mathcal{H}} \big( A_{L} P\big)\tag{75}\] and we obtain ?? by Lemma 14.

Regarding the commutator, one uses 73 and note that since \(B \in \mathcal{P}_{0}(\mathcal{H})\) and since \(a(f_{i})\) are uniformly bounded in \(i\) \[[\mathrm{d}\Gamma(A_{L}), \mathrm{d}\Gamma (B)] = \mathrm{d}\Gamma([A_{L}, B]) = \lim _{R \to \infty} \mathrm{d}\Gamma([A_{L}, B_{R}])\] where the limit is in the operator topology. Therefore, proceeding as in 74 , we get \[\begin{align} \omega^{H,\mu}_{\beta} ([\mathrm{d}\Gamma(A_{L}), \mathrm{d}\Gamma (B)]) & = \lim_{R \to \infty} \omega^{H,\mu}_{\beta} (\mathrm{d}\Gamma([A_{L}, B_{R}]) \\ & = \lim _{R \to \infty}\mathop{\mathrm{Tr}}_{\mathcal{H}} \big([A_{L},B_{R}] \gamma^{H,\mu}_{\beta} \big) . \end{align}\] To take the limit in the last line, we note that both \(A_{L}B_{R}\) and \(B_{R}A_{L}\) converge in the trace-class topology to \(A_{L}B\) and \(BA_{L}\) respectively. To see this, note that for \(R\) large enough \((B- B_{R})A_{L} = \chi_{R}^{c}B \chi_{L} A_{L}\), so that by Hölder’s inequality for Schatten norms \(\|(B-B_{R})A_{L} \|_{\mathrm{tr}} \leq \|\chi_{R}^{c}B \chi_{L} \| \|A_{L} \|_{\mathrm{tr}}\) with \(\|\chi_{R}^{c}B \chi_{L} \| \to 0\) as \(R \to \infty\) since \(B \in \mathcal{P}_{0}(\mathcal{H})\), where \(\|\, \cdot \, \|_{\mathrm{tr}}\) denotes the trace norm. Therefore, \[\lim _{R \to \infty}\mathop{\mathrm{Tr}}_{\mathcal{H}} \big([A_{L},B_{R}] \gamma^{H,\mu}_{\beta} \big) = \mathop{\mathrm{Tr}}_{\mathcal{H}} \big([A_{L},B] \gamma^{H,\mu}_{\beta} \big)\] since \(\gamma^{H,\mu}_{\beta}\) is bounded. Then, we can take the \(\beta \to \infty\) as in 75 since \([A_{L},B]\) is trace class, and conclude by ?? .

Finally, to prove the last claim we use that \(\chi_{L}\) is a projection and that it commutes with \(X\), so that \([B_{L}, X] = [B_{L}, X_{L}]\). But \(X_{L}\) is bounded hence by 73 \[\mathrm{d}\Gamma([B_{L},X]) = [\mathrm{d}\Gamma(B_{L}), \mathrm{d}\Gamma(X_{L})] = [\mathrm{d}\Gamma(B_{L}), \mathrm{d}\Gamma(X)] \;,\] where we also used that \(X = X_{L} + \chi^{c}_{L} X\) and that \([\mathrm{d}\Gamma(B_{L}), \mathrm{d}\Gamma (\chi^{c}_{L} X)]= \mathrm{d}\Gamma\big( [B_{L},\chi^{c}_{L} X] \big) = 0\). The conclusion follows by arguing as before, and by using ?? . ◻

Proof. We start from Definition 4 with \(\sharp = \mathrm{conv}\). We write \(X(t):= \mathrm{e}^{\mathrm{i}t H} X \mathrm{e}^{-\mathrm{i}t H}\). We note that \(J_{i}^{\mathrm{conv}} [X_{j}(s),P] =J_{i}^{\mathrm{conv}} \mathrm{e}^{\mathrm{i}s H} [X_{j},P] \mathrm{e}^{-\mathrm{i}s H} \in \mathcal{P}(\mathcal{H})\) by Lemma 2 and 3, uniformly in \(s\), thus by Lemma 1[it:tau4] \[\begin{align} \sigma^{\mathrm{s}}_{ij} =\mathop{\mathrm{Re}}\lim_{\eta\to 0^+}\tau( J_{i}^{\mathrm{conv}} \, L_{\eta,j}) = \mathrm{i}\lim _{\eta \to 0^{+}} \int_{-\infty}^{0} \mathrm{e}^{\eta s} \tau (J_{i}^{\mathrm{conv}}[X_{j}(s),P]) \mathrm{d}s \;. \end{align}\] By the Leibniz rule, we have \[\begin{align} \tau \big(J_{i}^{\mathrm{conv}} [X_{j}(s),P ] \big) & + \tau \big( [J_{i}^{\mathrm{conv}},X_{j}(s)]P \big) = \tau([X_{j}(s),J_{i}^{\mathrm{conv}} P]) \;. \end{align}\] We now prove that the r.h.s. of the latter equation is identically null. Since \(J(s)= \mathrm{i}[H,X(s)] = \frac{\mathrm{d}}{\mathrm{d}t} X(s)\), we can write \[\label{eq:integration-X-null-term} \tau([X_{j}(s),J_{i}^{\mathrm{conv}} P]) = \tau \big( [X_{j},J_{i}^{\mathrm{conv}} P] \big) - \tau \Big( \int_{s}^{0} [ J_{j}(s_{1}), J_{i}^{\mathrm{conv}} P ]\mathrm{d}s_{1} \Big) \;.\tag{76}\] Because \(\chi_{L}\) commutes with \(X\) and \(\chi_{L} X_{j} \chi_{L}\) is bounded we have \[\mathop{\mathrm{Tr}}( \chi_{L} [X,J_{i}^{\mathrm{conv}}P]\chi_{L}) = \mathop{\mathrm{Tr}}( \chi_{L}X\chi_{L}J_{i}^{\mathrm{conv}}P\chi_{L} - \chi_{L}J_{i}^{\mathrm{conv}}P\chi_{L} X \chi_{L}) = 0 \;.\] Concerning the second term in 76 , note that \(J_{j}(s_{1}), J_{i}^{\mathrm{conv}}, P \in \mathcal{P}(\mathcal{H})\) and that \(J_{j}(s_{1})\) is uniformly bounded in \(s_{1}\). Thus, we exchange \(\tau(\cdot)\) with integration by Lemma 1[it:tau4] and since by Lemma 1[it:tau5] \(\tau([ J_{j}(s_{1}), J_{i}^{\mathrm{conv}} P ]) = 0\) we conclude that 76 is in fact identically null. Therefore, \[\sigma^{\mathrm{s}}_{ij} = -\mathrm{i}\lim _{\eta \to 0^{+}} \int_{-\infty}^{0} \mathrm{e}^{\eta s} \tau \big( [J_{i}^{\mathrm{conv}},X_{j}(s)]P \big) \mathrm{d}s \;.\] To obtain the two terms in ?? , we write \(X_{j}(s)\) in terms of the charge current once more. The term with \(X_{j}(0)\) can be integrated in \(\mathrm{d}s\) directly whereas we write \(\int_{-\infty}^{0} \mathrm{d}s \int_{s}^{0} \mathrm{d}s_{1} = \int_{-\infty}^{0} \mathrm{d}s_{1} \int_{-\infty}^{s_{1}} \mathrm{d}s\) to integrate the other term. ◻

Proof. Since \(\omega^{H,\mu}_{\beta}\) is quasi-free \(K^{\beta,L}_{ij}\) can be expressed as a sum of products of two-point functions by the Wick rule, compare with Definition 6. Because \(H\) is bounded so is \(\exp (\mathrm{i}t H)\) for any \(t \in \mathbb{C}\) and thus the two-point function in ?? has analytic extension. Thus, also \(K^{\beta,L}_{ij}\) has an analytic extension and ?? follows by the KMS property ?? . ◻

Proof. Consider the first term in the Kubo formula ?? . By Lemma 2 and Lemma 3, we note that \(J_{i}^{\mathrm{conv}}\), \(P\) and \(J_{j}(t)\) are in \(\mathcal{P}_{0}(\mathcal{H})\) for any \(t\in \mathbb{R}\). It is then easy to check that \([\mathcal{J}^{\mathrm{spin}}_{L;i} ,\mathcal{J}_{j}(t)]\) is uniformly (w.r.t.\(t\in\mathbb{R}\)) bounded as an operator on \(\mathcal{F}(\mathcal{H})\), as long as \(L\) is finite. Thus, by using the first identity in ?? we have \[\label{eq:Kubo-1-I} \begin{align} \int_{-\infty}^{0}\mathrm{e}^{\eta t}\tau \big( [J_{i}^{\mathrm{conv}} ,J_{j}(t)] P \big) \mathrm{d}t & = \int_{-\infty}^{0}\mathrm{e}^{\eta t}\lim_{L \to \infty} \lim _{\beta \to \infty} |C_{L}|^{-1} \omega^{H,\mu}_{\beta}\big( [\mathcal{J}^{\mathrm{spin}}_{L;i} ,\mathcal{J}_{j}(t)] \big) \mathrm{d}t \\ & = \lim_{L \to \infty} \lim _{\beta \to \infty} |C_{L}|^{-1}\int_{-\infty}^{0}\mathrm{e}^{\eta t} \omega^{H,\mu}_{\beta}\big( [\mathcal{J}^{\mathrm{spin}}_{L;i} ,\mathcal{J}_{j}(t)] \big) \mathrm{d}t \;, \end{align}\tag{77}\] where we swapped the limit with the integral by the dominated convergence theorem. Using once more that \([\mathcal{J}^{\mathrm{spin}}_{L;i} ,\mathcal{J}_{j}(t)]\) is uniformly bounded in \(t\), we have \[\label{eq:Kubo-1-II} \begin{align} \Big| \int_{-\infty}^{0} &\big(\mathrm{e}^{\eta t} - \mathrm{e}^{\eta_{\beta} t} \big) \omega^{H,\mu}_{\beta}\big( [\mathcal{J}^{\mathrm{spin}}_{L;i} ,\mathcal{J}_{j}(t)] \big) \mathrm{d}t \Big| \\ & \lesssim \int_{-\infty}^{0} \big(\mathrm{e}^{\eta t} -\mathrm{e}^{\eta_{\beta} t}\big) \mathrm{d}t =\eta^{-1} - \eta_{\beta}^{-1} \leq \frac{2\pi}{\eta^{2} \beta} \;, \end{align}\tag{78}\] which is vanishing as \(\beta \to \infty\), having fixed \(\eta >0\). On the other hand, by Lemma 15, noting that \(\mathrm{e}^{\eta_{\beta} t} = \mathrm{e}^{\eta_{\beta}( t + \mathrm{i}\beta)}\) we write \[\label{eq:contour-deformation} \begin{align} \int_{-\infty}^{0}\mathrm{e}^{\eta_{\beta} t}& \omega^{H,\mu}_{\beta}\Big( [\mathcal{J}^{\mathrm{spin}}_{L;i} ,\mathcal{J}_{j}(t)] \Big) \mathrm{d}t \\ & = \int_{-\infty}^{0}\mathrm{e}^{\eta_{\beta} t} \Big( K^{\beta,L}_{ij}(t) - K^{\beta,L}_{ij}(t+\mathrm{i}\beta)\Big) \mathrm{d}t \\ & = \lim _{T \to \infty} \int_{-T}^{0} \Big(\mathrm{e}^{\eta_{\beta} t} K^{\beta,L}_{ij}(t) -\mathrm{e}^{\eta_{\beta}(t+\mathrm{i}\beta)} K^{\beta,L}_{ij}(t+\mathrm{i}\beta) \Big)\mathrm{d}t \\ & = \mathrm{i}\int_{0}^{\beta} \mathrm{e}^{ \mathrm{i}\eta _{\beta} t} K_{ij}^{\beta,L}(\mathrm{i}t) \mathrm{d}t - \mathrm{i}\lim _{T \to \infty} \mathrm{e}^{-\eta_{\beta} T} \int_{0}^{\beta} \mathrm{e}^{\mathrm{i}\eta _{\beta} t} K_{ij}^{\beta,L}(-T+\mathrm{i}t) \mathrm{d}t \end{align}\tag{79}\] where we used the Cauchy integral formula for \(K^{\beta,L}_{ij}\) being analytic. Because \([\mathcal{J}^{\mathrm{spin}}_{L;i} ,\mathcal{J}_{j}(-T+\mathrm{i}t)]\) is uniformly bounded in \(T \in \mathbb{R}\) and locally bounded in \(t\), so is \(K_{ij}^{\beta,L}\). Therefore, since \(\eta_{\beta}>0\) by definition, the last term in 79 is identically zero. We also note that \(\mathcal{J}^{\mathrm{spin}}_{L;i}\mathcal{J}_{j}(\mathrm{i}t) = \mathbf{T}\mathcal{J}^{\mathrm{spin}}_{L;i}\mathcal{J}_{j}(\mathrm{i}t)\) for \(t>0\) and thus \[\label{eq:T-ordering} K^{\beta,L}_{ij}(\mathrm{i}t) = \mathcal{K}^{\beta,L}_{ij}(t) \;, \qquad t \in (0,\beta] \;.\tag{80}\] Putting together 77 , 78 , 79 and 80 we obtain \[\label{eq:first-term-final} \int_{-\infty}^{0}\mathrm{e}^{\eta t}\tau \big( [J_{i}^{\mathrm{conv}} ,J_{j}(t)]P \big) \mathrm{d}t = \mathrm{i}\lim_{L \to \infty} \lim _{\beta \to \infty} \frac{1}{|C_{L}|} \int_{0}^{\beta} \mathrm{e}^{ \mathrm{i}\eta _{\beta} t} \mathcal{K}_{ij}^{\beta,L}(t)\mathrm{d}t \;.\tag{81}\] Let us now analyse the second term in the Kubo formula ?? . Let \(\mathcal{X}_{j}:= \mathrm{d}\Gamma(X_{j})\). By employing the second identity in ?? , we have \[\tau\big( [J_{i}^{\mathrm{conv}},X_{j} ] P \big) = \lim _{L \to \infty} \lim_{\beta \to \infty} |C_{L}|^{-1} \omega^{H,\mu}_{\beta} \big( \big[ \mathcal{J}^{\mathrm{spin}}_{L;i}, \mathcal{X}_{j} \big]\big).\] We want to open the commutator on the right-hand side term in the last equality and this can be done provided that we localize the operator \(\mathcal{X}_{j}\) so that it is bounded. By the argument at the end of the proof of Lemma 13 we note that for any \(R \geq L\) we have \(\omega^{H,\mu}_{\beta} \big( \big[ \mathcal{J}^{\mathrm{spin}}_{L;i}, \mathcal{X}_{j} \big]\big) = \omega^{H,\mu}_{\beta} \big( \big[ \mathcal{J}^{\mathrm{spin}}_{L;i}, \mathcal{X}_{R;j} \big]\big)\), where \(\mathcal{X}_{R;j}:= \mathrm{d}\Gamma( \chi_{R} X_{j})\) is bounded when \(R\) is finite. Then, by the KMS property we can write \[\begin{align} \omega^{H,\mu}_{\beta} \big( \big[ \mathcal{J}^{\mathrm{spin}}_{L;i}, \mathcal{X}_{j} \big]\big) = \omega^{H,\mu}_{\beta} \Big( \mathcal{J}^{\mathrm{spin}}_{L;i} \big( \mathcal{X}_{R;j} - \mathcal{X}_{R;j} (\mathrm{i}\beta) \big) \Big) \;, \end{align}\] and, by slight abuse of notation, letting \(\mathcal{J}_{R;j}(\mathrm{i}t) := - \mathrm{i}\frac{\mathrm{d}}{\mathrm{d}t}\mathcal{X}_{R;j}(\mathrm{i}t) = \alpha_{\mathrm{i}t}\big( \mathrm{d}\Gamma(\mathrm{i}[H,\chi_{R}X_{j}])\big)\), we can write \[\omega^{H,\mu}_{\beta} \big( \big[ \mathcal{J}^{\mathrm{spin}}_{L;i}, \mathcal{X}_{j} \big]\big) = -\mathrm{i}\, \omega^{H,\mu}_{\beta} \Big( \int_{0} ^{\beta}\mathcal{J}^{\mathrm{spin}}_{L;i} \mathcal{J}_{R;j}(\mathrm{i}t) \mathrm{d}t\Big) = -\mathrm{i}\,\int_{0} ^{\beta} \omega^{H,\mu}_{\beta} \big(\mathcal{J}^{\mathrm{spin}}_{L;i} \mathcal{J}_{R;j}(\mathrm{i}t) \big) \mathrm{d}t \;,\] where in the last step we swapped integration with expectation over the grand canonical state because the operator \(\mathcal{J}^{\mathrm{spin}}_{L;i} \mathcal{J}_{R;j}( \mathrm{i}t)\) is bounded, locally in \(t\), and thus Bochner integrable over \([0,\beta]\). We shall now remove the localisation \(R\) by taking the limit \(R \to \infty\). This can be done by noting that \(\mathcal{J}_{L;i} \mathcal{J}_{R;j} ( \mathrm{i}t)\) converges in the operator topology to \(\mathcal{J}_{L;i} \mathcal{J}_{j}( \mathrm{i}t)\), which is Bochner integrable over \([0,\beta]\). Thus, we can take the limit inside the integral by the Lebesgue dominated convergence theorem as well as inside \(\omega^{H,\mu}_{\beta}\). Therefore, using 80 we can write \[\tau\big( [J_{i}^{\mathrm{conv}},X_{j} ] P \big) = - \mathrm{i}\lim _{L \to \infty} \lim_{\beta \to \infty} |C_{L}|^{-1} \int_{0}^{\beta} \mathcal{K}^{\beta,L}_{ij}(t) \mathrm{d}t \;,\] which, together with 81 , implies the claim. ◻

Proof of Theorem 26. We compute \(\mathcal{K}^{\beta,L}_{ij}(t) = \omega^{H,\mu}_{\beta} \big( \mathbf{T} \mathcal{J}^{\mathrm{spin}}_{L;i} \mathcal{J}_{j}(\mathrm{i}t) \big) = \omega^{H,\mu}_{\beta} \big( \mathcal{J}^{\mathrm{spin}}_{L;i} \mathcal{J}_{j}(\mathrm{i}t) \big)\), \(t \in (0,\beta]\), by using that \(\omega^{H,\mu}_{\beta}\) is quasi-free. First of all, we note that \[\label{eq:truncation-K-function} \omega^{H,\mu}_{\beta}(\mathcal{J}_{j}(\mathrm{i}t)) = 0 \;,\qquad j=1,2 \;.\tag{82}\] This follows by observing that \(H\) is invariant under \(2\pi/3\)-rotations, whereas \(J_{i}\) transform like vectors. By the Wick rule and simple algebraic manipulations⁷, for \(t \in (0,\beta]\) we find \[\begin{align} &\omega^{H,\mu}_{\beta} \big(\mathbf{T}\mathcal{J}^{\mathrm{spin}}_{L;i} \mathcal{J}_{j}(\mathrm{i}t) \big) = \mathop{\mathrm{Tr}}_{\mathcal{H}} \Big( \big(J_{i}^{\mathrm{conv}} \big)_{L} G^{H,\mu}_{\beta}(t) J_{j}G^{H,\mu}_{\beta}(-t)\Big) \;. \end{align}\] where the operator \(G^{H,\mu}_{\beta}(t)\) is the finite-temperature imaginary-time-ordered Green function, that is, for any \(f,g \in \mathcal{H}\) \[\label{eq:T-ordered-GF} \langle f, G^{H,\mu}_{\beta}(t) g \rangle _{\mathcal{H}} :=\omega^{H,\mu}_{\beta} \big(\mathbf{T} a(f) \alpha_{\mathrm{i}t}(a^{*}(g)) \big) \;.\tag{83}\] By ?? and by noting that \(\alpha_{\mathrm{i}t}(a^{*}(f)) = a^{*}(\mathrm{e}^{- t(H - \mu \mathbb{1})}f)\), we find that \(G^{H,\mu}_{\beta}(t)\) is given by \[G^{H,\mu}_{\beta}(t) : = \frac{ \mathrm{e}^{-t (H-\mu \mathbb{1})}}{\mathbb{1} + \mathrm{e}^{-\beta (H - \mu \mathbb{1})}} ,\qquad t\in [0,\beta) \;,\] while in the rest of the interval \([-\beta,\beta]\) it is determined by the anti-periodicity condition \(G^{H,\mu}_{\beta}(t-\beta) = - G^{H,\mu}_{\beta}(t)\). Note that, by Lemma 3, \(G^{H,\mu}_{\beta}(t) \in \mathcal{P}_{0}(\mathcal{H})\) for any \(t \in \mathbb{R}\) and any \(\beta >0\). We also introduce the Fourier components of \(G^{H,\mu}_{\beta} (t)\); because \(G^{H,\mu}_{\beta}(t-\beta) = - G^{H,\mu}_{\beta}(t)\), only odd frequencies are non-zero; thus, for \(\omega \in \frac{2 \pi}{\beta} (\mathbb{Z} + \frac{1}{2})\) we set⁸ \[\begin{align} \widehat{G}^{H,\mu}(\omega) &: = \int_{0}^{\beta} \mathrm{e}^{-\mathrm{i}\omega t} G^{H,\mu}_{\beta} (t)\mathrm{d}t = \big(H - (\mu-\mathrm{i}\omega) \mathbb{1}\big)^{-1} \;. \end{align}\] In particular, \(\widehat{G}^{H,\mu}(\omega) \in \mathcal{P}_{0}(\mathcal{H})\), and its matrix elements decay as \(|\omega|^{-1}\). Also, note that \(\widehat{G}^{H,\mu}\) does not depend explicitly on \(\beta\), which on the other hand only enters the frequency domain \(\frac{2 \pi}{\beta} (\mathbb{Z} + \frac{1}{2})\).

Since \(J_{i}^{\mathrm{conv}}\) is localised and since \(G^{H,\mu}_{\beta},J_{j} \in \mathcal{P}_{0}(\mathcal{H})\), the operator inside the trace is trace-class, uniformly in \(t\) we exchange the trace with integration and obtain \[\begin{align} \int_{0}^{\beta} \mathrm{e}^{\mathrm{i}\eta _{\beta} t} & \mathop{\mathrm{Tr}}_{\mathcal{H}} \Big( \big(J_{i}^{\mathrm{conv}} \big)_{L} G^{H,\mu}_{\beta}(t) J_{j}G^{H,\mu}_{\beta}(-t) \Big) \mathrm{d}t \\ & = \frac{1}{\beta} \sum_{\omega \in (2 \pi / \beta) \mathbb{Z}} \mathop{\mathrm{Tr}}_{\mathcal{H}} \Big( \big(J_{i}^{\mathrm{conv}} \big)_{L} \widehat{G}^{H,\mu}(\omega -\eta_{\beta} ) J_{j}\widehat{G}^{H,\mu}(\omega) \Big) \;, \end{align}\] where we also swapped trace and summation over \(\omega\) because the matrix elements of \(\widehat{G}^{H,\mu}\) decay as \(|\omega|^{-1}\). Taking the \(\beta \to \infty\) limit gives \[\widehat{\mathcal{K}}_{ij}(\eta) = (2 \pi)^{-1} \lim _{L \to \infty} |C_{L}|^{-1} \int_{\mathbb{R}} \mathrm{d}\omega\mathop{\mathrm{Tr}}_{\mathcal{H}} \Big( \big(J_{i}^{\mathrm{conv}} \big)_{L} \widehat{G}^{H,\mu}(\omega - \eta ) J_{j} \widehat{G}^{H,\mu}(\omega) \Big) \;.\] By similar arguments we take the limit \(L \to \infty\) inside the integral and, because \(\widehat{G}^{H,\mu}\), \(J_{i}^{\mathrm{conv}}\) and \(J_{j}\) are short-range, by Lemma 14 and Lemma 1[it:tau3] we obtain \[\widehat{\mathcal{K}}_{ij}(\eta) = \int_{\mathbb{R} \times \mathbb{T}^{*}_{2}} \frac{\mathrm{d}\mathbf{k}}{(2 \pi)^{3}} \mathop{\mathrm{Tr}}_{\mathbb{C}^{4}} \Big( \widehat{J}^{\mathrm{conv}}_{i}(k) \widehat{G}^{H,\mu}(\omega - \eta;k) \widehat{J}_{j}(k) \widehat{G}^{H,\mu}(\omega;k) \Big) \;,\] where \(\mathbf{k} \equiv (\omega, k)\) and where \(\widehat{G}^{H,\mu}(\omega;k):= \big(H(k) - (\mu - \mathrm{i}\omega) \mathbb{1} \big)^{-1}\). To conclude, since \(\partial_{k_{i}} H(k) = \partial_{k_{i}}\left( \widehat{G}^{H,\mu}(\omega;k) ^{-1}\right)\), it is easy to check that \[\begin{align} \widehat{J}^{\mathrm{conv}}_{i}(k) = \frac{1}{2} \big\{ \partial_{k_{i}} \widehat{G}^{H,\mu}(\omega;k) ^{-1}, S_{z} \big\} \;, \qquad \widehat{J}_{j}(k) = \partial_{k_{j}} \widehat{G}^{H,\mu}(\omega;k) ^{-1} \;, \\ -\partial_{\eta} \widehat{G}^{H,\mu}(\omega - \eta;k)\big|_{\eta =0} = - \widehat{G}^{H,\mu}(\omega ;k) \big[ \partial_{\omega} \widehat{G}^{H,\mu}(\omega;k) ^{-1} \big] \widehat{G}^{H,\mu}(\omega ;k) \;. \end{align}\] The claim then follows because \((H,\mu)\) is an insulator so that we can take the limit \(\eta \to 0^{+}\) inside the integral by the Lebesgue dominated convergence theorem. ◻