Proof. The main failure here is operator norm closure. To see it, consider for instance a translation-invariant operator whose integral kernel has sufficiently fast polynomial decay so as to make it a bounded operator. Its finite-hopping truncations, at hopping distance \(R\in\mathbb{N}\), converge to it in operator norm. But of course finite-hopping operators are exponentially local. ◻

Proof. If \(A\) is spherically-local, i.e., \([A,\widehat{X}_j\otimes \mathbb{1}_N]\in\mathcal{K}\) for all \(j\), then \[\begin{align} [A\otimes \mathbb{1}_{\mathbb{C}^{2^k}},W_d] = [A\otimes \mathbb{1}_{\mathbb{C}^{2^k}},\sum_{j=1}^d \widehat{X}_j\otimes \mathbb{1}_N\otimes \Gamma_j] = \sum_{j=1}^d [A,\widehat{X}_j\otimes \mathbb{1}_N] \otimes \Gamma_j \in \mathcal{K}\,. \end{align}\] Conversely, suppose \(A\) is Dirac-local. Since the matrices \(\Gamma_1,\dots,\Gamma_d\) satisfy the \(\{\Gamma_j,\Gamma_k\}=2\delta_{jk}\mathbb{1}_{2^k}\) for all \(j,k=1,\dots, d\), it follows that \[\begin{align} \left\{\sum_{j=1}^d [A,\widehat{X}_j\otimes \mathbb{1}_N]\otimes \Gamma_j, \mathbb{1}_{\mathcal{H}_d}\otimes \mathbb{1}_N\otimes \Gamma_k \right\} &= \sum_{j=1}^d [A,\widehat{X}_j\otimes \mathbb{1}_N]\otimes \{\Gamma_j,\Gamma_k\}\\ &= 2[A,\widehat{X}_k\otimes \mathbb{1}_N]\otimes \mathbb{1}_{2^k} \\&\in \mathcal{K}\,. \end{align}\] Thus \([A,\widehat{X}_k\otimes \mathbb{1}_N]\in\mathcal{K}\) for all \(k=1,\dots,d\). ◻

Proof. Two identities will be useful. The first: for any \(A\in\mathcal{B}(\mathcal{H}_d\otimes\mathbb{C}^N)\) and the position basis \(\Set{\delta_x}_{x\in\mathbb{Z}^d}\), we have \[\begin{align} \left\lVert A\right\rVert_p \leq \sum_{k\in\mathbb{Z}^d}\left(\sum_{x\in\mathbb{Z}^d}\left\lVert A_{x+k,x}\right\rVert^p\right)^{1/p}\qquad(p\geq 1) \end{align}\] where \(\left\lVert A\right\rVert_p\equiv\left(\operatorname{tr}\left(|A|^p\right)\right)^{1/p}\) is the Schatten-\(p\) norm. We shall use the fact that if \(\left\lVert A\right\rVert_p<\infty\) for some \(p<\infty\) then \(A\) is compact.

The second identity is \[\begin{align} \left\lVert\hat{x}-\hat{y}\right\rVert^2\leq 2\frac{\left\lVert x-y\right\rVert^2}{1+\left\lVert x\right\rVert\left\lVert y\right\rVert}\qquad(x,y\in\mathbb{Z}^d)\,. \end{align}\] Together with the exponential locality of \(A\), 3 , these identities imply that the commutator \([\hat{X}_i,A]\) is Schatten-\((d+1)\) and as such compact indeed.

That the converse is false is clear from the \(d=1\) case, where we could take an operator that fails to be exponentially-local \(A\), then truncate it to \[\begin{align} A_{\mathrm{trun.}} := \Lambda A \Lambda + \Lambda^\perp A \Lambda^\perp \end{align}\] and it would still fail to be exponentially local, but is obviously trivially \(\Lambda\)-local. ◻

Proof. Let \(I,J\) be two closed disjoint subsets of \(\mathbb{S}^{d-1}\) and \(\Lambda_I,\Lambda_J\) the projections to the associated cones in \(\ell^2(\mathbb{Z}^d)\), which are disjoint by hypothesis. We will show \(\Lambda_I A \Lambda_J\) is Hilbert-Schmidt, and hence compact.

Since \(I,J\) are disjoint, there exists some constant \(c>0\) such that \[\begin{align} \left\lVert x-y\right\rVert^2 \geq c \left(\left\lVert x\right\rVert+\left\lVert y\right\rVert\right)^2\,. \end{align}\] Moreover, \(\left(1+\left\lVert x\right\rVert\right)^\nu\leq\left(1+\left\lVert x\right\rVert+\left\lVert y\right\rVert\right)^\nu\). Then pick \(\mu> \nu+d\) on the estimate \[\begin{align} \left\lVert\Lambda_I A \Lambda_J\right\rVert_{\mathrm{HS}}^2 &\leq \sum_{x\in C_I, y\in C_J}\left\lVert A_{xy}\right\rVert^2\\ &\leq \sum_{x,y\in\mathbb{Z}^d}C_\mu^2\left(1+\left\lVert x\right\rVert+\left\lVert y\right\rVert\right)^{2\left(\nu-\mu\right)} \\ &< \infty\,. \end{align}\]

As above, it is clear we could come up with operators which are spherically-local but not weakly-local. ◻

Proof. Choose \(a<E_F\) such that \[\begin{align} [a,E_F]\subset\Delta . \end{align}\] Let \(f\in C_b(\mathbb{R})\) satisfy \[\begin{align} 0\leq f\leq 1, \qquad f(t)=1\quad(t\leq a), \qquad f(t)=0\quad(t\geq E_F). \end{align}\] Define \[\begin{align} g_0:=\chi_{(-\infty,E_F)}-f . \end{align}\] Then \(g_0\) is a bounded Borel function satisfying \[\begin{align} \left\lVert g_0\right\rVert_\infty\leq 1, \qquad \operatorname{supp}(g_0)\subset [a,E_F]\subset\Delta . \end{align}\] Therefore \(g_0\) is one of the functions allowed in ?? . Hence, for each \(j=1,\dots,d\), \[\begin{align} \mathbb{E}\left[ \left\| [\widehat X_j\otimes\mathbb{1}_N,g_0(H_\omega)] \right\|_{\mathfrak S_{d+1}}^{d+1} \right] <\infty . \end{align}\] Since the random variable inside the expectation is non-negative, it follows that \[\begin{align} \left\| [\widehat X_j\otimes\mathbb{1}_N,g_0(H_\omega)] \right\|_{\mathfrak S_{d+1}} <\infty \end{align}\] for almost every \(\omega\). Equivalently, \[\begin{align} [\widehat X_j\otimes\mathbb{1}_N,g_0(H_\omega)] \in\mathfrak S_{d+1} \subset\mathcal{K}(\mathcal{H}_{d,N}) \end{align}\] for almost every \(\omega\). Taking the intersection of these finitely many full-measure events, we obtain \[\begin{align} \label{eq:g032compact32commutators32almost32surely} \mathbb{P}\left[\Set{ [\widehat X_j\otimes\mathbb{1}_N,g_0(H_\omega)] \in\mathcal{K}(\mathcal{H}_{d,N}) \text{ for all }j=1,\dots,d }\right]=1 . \end{align}\tag{10}\]

On the other hand, by ?? , \[\begin{align} \mathbb{P}\left[\Set{H_\omega\in\mathcal{L}_{d,N}}\right]=1 . \end{align}\] Since \(\mathcal{L}_{d,N}\) is a \(C^\ast\)-algebra, it is closed under continuous functional calculus. Thus \[\begin{align} \label{eq:fH32spherically32local32almost32surely} \mathbb{P}\left[\Set{f(H_\omega)\in\mathcal{L}_{d,N}}\right]=1 . \end{align}\tag{11}\] Equivalently, \[\begin{align} [\widehat X_j\otimes\mathbb{1}_N,f(H_\omega)] \in\mathcal{K}(\mathcal{H}_{d,N}), \qquad j=1,\dots,d, \end{align}\] almost surely.

Now \[\begin{align} P_\omega = \chi_{(-\infty,E_F)}(H_\omega) = f(H_\omega)+g_0(H_\omega). \end{align}\] Therefore, on the full-measure event on which both 10 and 11 hold, we have, for every \(j=1,\dots,d\), \[\begin{align} [\widehat X_j\otimes\mathbb{1}_N,P_\omega] &= [\widehat X_j\otimes\mathbb{1}_N,f(H_\omega)] + [\widehat X_j\otimes\mathbb{1}_N,g_0(H_\omega)] \\ &\in \mathcal{K}(\mathcal{H}_{d,N})+\mathcal{K}(\mathcal{H}_{d,N}) \\ &\subset \mathcal{K}(\mathcal{H}_{d,N}) . \end{align}\] Hence \[\begin{align} P_\omega\in\mathcal{L}_{d,N} \end{align}\] almost surely. ◻

Proof. Since \(\mathcal{X}_d\) is an Abelian \(C^\ast\)-algebra, by the Gelfand-Naimark theorem [49], it is \(\ast\)-isomorphic to the space of continuous functions \(C(\Delta_d)\) where \(\Delta_d\) is the maximal ideal space of \(\mathcal{X}_d\), defined as the space all multiplicative linear functionals on \(\mathcal{X}_d\) equipped with weak-\(\ast\) topology. Consider the map \[\begin{align} \label{eq:homeomorphism32from32maximal32ideal32space32to32sphere} \Delta_d \ni \varphi\mapsto (\varphi(\widehat{X}_1),\dots,\varphi(\widehat{X}_d)) \in \mathbb{R}^d\,. \end{align}\tag{13}\] Note that it indeed takes values in \(\mathbb{R}^d\) thanks to the self-adjointness of \(\widehat{X}_j\) and the \(\ast\)-homomorphism property of \(\varphi\). The map is a homeomorphism onto its image. The map is continuous by definition of weak-\(\ast\) topology. If \(\varphi(\hat{X}_j)=\psi(\hat{X}_j)\) for all \(j\), then \(\varphi(Y)=\psi(Y)\) whenever \(Y\) is a polynomial in \(X_1,\dots,X_d\), and since these polynomials are dense in \(\mathcal{X}_d\), it follows that \(\varphi=\psi\) and hence the map 13 is injective. We show that the image is \(\mathbb{S}^{d-1}\), and hence \(\Delta_d\cong \mathbb{S}^{d-1}\). To that end, let \(\varphi\in \Delta_d\). Since \(\widehat{X}^2_1+\dots+\widehat{X}_d^2=\mathbb{1}\), it follows that \[\begin{align} \varphi(\widehat{X}_1)^2 + \dots+\varphi(\widehat{X}_d)^2 = 1\,. \end{align}\] Therefore, the image must lie in \(\mathbb{S}^{d-1}\). To show that each point on \(\mathbb{S}^{d-1}\) corresponds to some functional in \(\Delta_d\), we consider the evaluation functional: for each \(x\in \mathbb{Z}^d\), define \(\varphi_x(\widehat{X}_i)=x_i/\|(x_1,\dots,x_d)\|\). Thus, we have shown that \[\begin{align} \mathcal{X}_d \cong C(\mathbb{S}^{d-1})\,. \end{align}\]

Now we verify ?? . The Gelfand-Naimark theorem gives an isomorphism, the Gelfand transform [49], that maps \(T\in\mathcal{X}_d\) onto \(\varphi\mapsto \varphi(T)\) for \(\varphi\in \Delta_d\). Fix \(k\in \{1,\dots,d\}\). The map \(x\mapsto x_k\) in \(C(\mathbb{S}^{d-1})\) corresponds via 13 to the map \(\varphi\mapsto \varphi(\widehat{X}_k)\) in \(C(\Delta_d)\). On the other hand, the Gelfand transform maps \(\widehat{X}_k\) to \(\varphi\mapsto \varphi(\widehat{X}_k)\). Thus, we have established \(\rho(x\mapsto x_k)=\widehat{X}_k\). Finally, the identity \(\rho(x\mapsto 1)=\mathbb{1}\) follows from the fact that all the isomorphisms constructed are unital.

Suppose \(\rho(f)\) is compact for \(f\in C(\mathbb{S}^{d-1})\), then \(\operatorname{im}f\) only accumulates at zero, and hence by continuity, \(f=0\). Thus the only compact element in \(C(\mathbb{S}^{d-1})\), and hence in \(\mathcal{X}_d\), is the zero element. ◻

Proof. Define the operators \[\begin{align} \widehat{Y}_j := \frac{(M\widehat{X})_j}{\|M\widehat{X}\|} \end{align}\] acting on \(\mathcal{H}_d\). Define the \(C^\ast\)-algebra \[\begin{align} \mathcal{Y}_{d,N}:= \Set{A\in\mathcal{B}(\mathcal{H}_d\otimes \mathbb{C}^N) | [A,\widehat{Y}_j\otimes \mathbb{1}_N]\in\mathcal{K},\;\forall j=1,\dots,d}. \end{align}\] We first show that \(\Set{UAU^\ast | A\in\mathcal{L}_d} = \mathcal{Y}_{d,N}\). The operator \(U^* (\widehat{Y}_j\otimes \mathbb{1}_N) U\) is diagonal \[\begin{align} U^* (\widehat{Y}_j\otimes \mathbb{1}_N) U \delta_x = U^* (\widehat{Y}_j\otimes \mathbb{1}_N) \delta_q\otimes e_r = \frac{(Mq)_j}{\|Mq\|}\delta_x \end{align}\] with eigenvalues \(\lambda_{x}=(Mq)_j/\|Mq\|\). Let us analyze the difference \(D_j = \widehat{X}_j - U^\ast \widehat{Y}_j\otimes\mathbb{1}_N U\in\mathcal{B}(\mathcal{H}_d)\). We verify that the operator \(D_j\) is compact. Since \(D_j\) is diagonal, it is compact if and only if its eigenvalues decay to zero as the index goes to infinity (\(\|x\| \to \infty\)). Using the vector inequality \(\|u/\|u\| - v/\|v\|\|\leq 2\|u-v\|/\|v\|\), we have \[\begin{align} \left\|x - \frac{Mq}{\|Mq\|}\right\| = \left\|\frac{Mq+r}{\|Mq+r\|} - \frac{Mq}{\|Mq\|}\right\| \leq \frac{2\|r\|}{\|Mq\|}. \end{align}\] Since \(R\) is a finite set, we have \(2\|r\|\leq 2\max_{r\in R}\|r\|\), and since \(M\) is non-singular, it has a smallest singular value \(\sigma_{\min} > 0\), and hence \(\|Mq\| \ge \sigma_{\min} \|q\|\). Thus \(2\|r\|/\|Mq\|\to 0\) as \(\|q\|\to \infty\) and hence as \(\|x\|\to\infty\). This shows that \(D_j\in\mathcal{K}(\mathcal{H}_d)\). Suppose \(A\in\mathcal{L}_d\). Then \[\begin{align} [UAU^\ast,\widehat{Y}_j\otimes \mathbb{1}_N] = U[A,U^\ast (\widehat{Y}_j\otimes \mathbb{1}_N)U ]U^\ast = U[A,\widehat{X}_j-D_j]U^\ast \in\mathcal{K}(\mathcal{H}_d\otimes \mathbb{C}^N). \end{align}\] This shows that \(\Set{UAU^\ast | A\in\mathcal{L}_d} \subset \mathcal{Y}_{d,N}\). The other direction is argued in the same way.

Next we show that \[\begin{align} \mathcal{Y}_{d,N} = \mathcal{L}_{d,N}. \end{align}\] Intuitively, the algebra \(\mathcal{Y}_{d,N}\) are those “elliptically”-local operators, i.e., those operators that essentially commute with \(\widehat{Y}_j\otimes \mathbb{1}_N\). Consider the smooth map \(\varphi:\mathbb{S}^{d-1}\to\mathbb{S}^{d-1}\) defined as \(\varphi(u) = Mu/\|Mu\|\) whose inverse is smooth as well. Using continuous functional calculus, we have \(\widehat{Y}_j\otimes\mathbb{1}_N = \varphi(\widehat{X}_j\otimes\mathbb{1}_N)\) and \(\widehat{X}_j\otimes\mathbb{1}_N = \varphi^{-1}(\widehat{Y}_j\otimes\mathbb{1}_N)\). If \(A\in\mathcal{B}(\mathcal{H}_d\otimes \mathbb{C}^N)\) satisfies \([A,\widehat{X}_j\otimes\mathbb{1}_N]\in\mathcal{K}\), then \([A,\widehat{Y}_j\otimes\mathbb{1}_N]=[A,\varphi(\widehat{X}_j\otimes\mathbb{1}_N)]\in\mathcal{K}\) which shows that \(\mathcal{L}_{d,N}\subset \mathcal{Y}_{d,N}\). This other direction is analogous. ◻

Proof. It follows from 23 that we can describe \(\mathcal{L}(\mathcal{Y}_d)\) as \[\begin{align} \mathcal{L}(\mathcal{Y}_d) = \Set{A\in\mathcal{B}(\mathcal{H}) | [A,Y]\in \mathcal{K},\, \forall Y\in \mathcal{Y}_d}. \end{align}\] We can analogously describe the algebra \(\mathcal{L}(\mathcal{Y}'_d)\). Since \(\mathcal{Y}_d\) is generated by some operators \(Y_1,\dots,Y_d\) whose joint spectrum (as defined in 15 ) is \(\mathbb{S}^{d-1}\), it follows by Gelfand-Naimark theorem that \(\mathcal{Y}_d\cong C(\mathbb{S}^{d-1})\). Analogously, we have \(\mathcal{Y}'_d\cong C(\mathbb{S}^{d-1})\). Let \(\rho:C(\mathbb{S}^{d-1})\to\mathcal{Y}_d\) and \(\rho':C(\mathbb{S}^{d-1})\to\mathcal{Y}'_d\) be isomorphic representations given by the Gelfand-Naimark theorem. Then we have \[\begin{align} \mathcal{L}(\mathcal{Y}_d) = \Set{A\in\mathcal{B}(\mathcal{H}) | [A,\rho(f)]\in \mathcal{K},\, \forall f\in C(\mathbb{S}^{d-1})}\,. \end{align}\] We can analogously describe the algebra \(\mathcal{L}(\mathcal{Y}'_d)\). Let \(\mathcal{Y}_d\subset \mathcal{B}(\mathcal{H})\) and \(\mathcal{Y}'_d\subset\mathcal{B}(\mathcal{H}')\) be algebras acting on the Hilbert spaces \(\mathcal{H},\mathcal{H}'\). Crucially, the algebras \(\mathcal{Y}_d,\mathcal{Y}'_d\) of operators do not contain compact ones except zero (similarly shown in 21). In this setting, we can apply [43] (which is itself a consequence of the Voiculescu’s Theorem) to get a unitary isomorphism \(U:\mathcal{H}\to\mathcal{H}'\) such that \[\begin{align} \label{eq:tmp32unitary32isomorphism32in32showing32isomorphic32spherically-local} U\rho(f)U^\ast - \rho'(f)\in\mathcal{K}(\mathcal{H}),\quad \forall f\in C(\mathbb{S}^{d-1})\,. \end{align}\tag{16}\] We claim that \(\mathcal{L}(\mathcal{Y}_d)\ni A\mapsto UAU^\ast \in\mathcal{L}(\mathcal{Y}'_d)\) is well-defined and gives the sought-after isomorphism. Indeed, let \(A\in \mathcal{L}(\mathcal{Y}_d)\) and \(f\in C(\mathbb{S}^{d-1})\) be arbitrary, then \[\begin{align} [UAU^\ast,\rho'(f)] = U[A,U^\ast \rho'(f)U]U^\ast \in \mathcal{K} \end{align}\] where we apply 16 and the assumption that \(A\in\mathcal{L}(\mathcal{Y}_d)\) in the last step. Thus \(UAU^\ast\in\mathcal{L}(\mathcal{Y}'_d)\). ◻

Proof of 26. Let \(\rho:C(\mathbb{S}^{d-1})\to \mathcal{B}(\mathcal{H}_d)\) be the representation induced by the composition of the isomorphism \(C(\mathbb{S}^{d-1})\to \mathcal{X}_d\) in ?? and the inclusion \(\mathcal{X}_d\hookrightarrow \mathcal{B}(\mathcal{H}_d)\). By [52], there exists a spectral measure \(E\) on the Borel sets of \(\mathbb{S}^{d-1}\) such that \[\begin{align} \label{eq:spectral32measure32on32Xd} \rho(f) = \int f dE,\quad \forall f\in C(\mathbb{S}^{d-1}) \end{align}\tag{18}\] and \(\rho\) extends to a representation on \(B(\mathbb{S}^{d-1})\), the space of bounded Borel-measurable functions on \(\mathbb{S}^{d-1}\).

Let \(F\subset \mathbb{S}^{d-1}\) be any Borel set. We argue that \[\begin{align} \label{eq:lambda32is32spectral32measure} \Lambda_F = E(F)\,. \end{align}\tag{19}\] Fix a point \(z\in \mathbb{S}^{d-1}\). Let \(f\in C(\mathbb{S}^{d-1})\) be the continuous function defined by \(f_z(x) = \sum_{i=1}^d (x_i-z_i)^2\). Using ?? , \(f_z\) corresponds to the operator \(\rho(f_z)=\sum_{i=1}^d (\widehat{X}_i-z_i\mathbb{1})^2\) in \(\mathcal{X}_d\) whose kernel is \(\operatorname{im}\Lambda_{\{z\}}\). Using [53], it follows that \(\operatorname{im}\Lambda_{\{z\}} = \ker \rho(f_z) = \operatorname{im}E(\{z\})\). Since there are countably many points in \(\mathbb{Z}^d\), there are only countably many points \(z\) on \(\mathbb{S}^{d-1}\) for which \(\operatorname{im}\Lambda_{\{z\}}\) is non-vanishing. Thus, using the countable additivity of the spectral measure \(E\), it follows that 19 holds for all Borel sets in \(\mathbb{S}^{d-1}\).

Using the representation \(\rho:B(\mathbb{S}^{d-1})\to \mathcal{B}(\mathcal{H}_d)\) and the characterization 19 of the spectral measures, we are ready to prove the result. Suppose \(A\in\mathcal{L}_d\). Let \(F,G\subset\mathbb{S}^{d-1}\) be a disjoint pair of closed subsets. Then, there exists a continuous function \(f\in C(\mathbb{S}^{d-1})\) such that \(F\subset f^{-1}(1)\) and \(G\subset f^{-1}(0)\). By 14 , we have \(\rho(f)A-A\rho(f)\in\mathcal{K}(\mathcal{H}_d)\), and hence \[\begin{align} \Lambda_F (\rho(f)A-A\rho(f))\Lambda_G \in \mathcal{K}(\mathcal{H}_d)\,. \end{align}\] Using 19 , it follows that \[\begin{align} \rho(\chi_F f) A \Lambda_G - \Lambda_F A \rho(f\chi_G) \in \mathcal{K}(\mathcal{H}_d)\,. \end{align}\] By construction of \(f\), we have \(\chi_F f=\chi_F\) and \(f\chi_G = 0\). Thus, we arrived at \(\Lambda_FA\Lambda_G\in\mathcal{K}(\mathcal{H}_d)\).

We now prove the necessity of the statement. Let \(f\in C(\mathbb{S}^{d-1})\) be arbitrary. Fix \(\varepsilon>0\). There exists a partition of \(\mathbb{R}^d\) into cubes fine enough such that if we let \(\{F_k\}_{k=1}^{n}\) be the collection of the sets of intersection of cube and \(\mathbb{S}^{d-1}\), then we have \[\begin{align} \label{eq:function32evaluate32at32cubes32difference32small} \sup_{x,y\in F_k} |f(x)-f(y)| < \varepsilon,\quad \forall k\in\{1,\dots,n\}\,. \end{align}\tag{20}\] Pick any \(x_k\in F_k\). It follows from [52] that \[\begin{align} \label{eq:approximate32spectral32integral32with32cubes} \|\rho(f)-\sum_{k=1}^n f(x_k) E(F_k)\| < \varepsilon\,. \end{align}\tag{21}\] Since \(\mathbb{S}^{d-1} = \cup_{k=1}^n F_k\) is a disjoint union, it follows that \(\mathbb{1}= \sum_{k=1}^n E(F_k)\) and, using 19 , we have \[\begin{align} A\rho(f)-\rho(f) A &= \sum_{j=1}^n\sum_{k=1}^n \Lambda_{F_j} (A\rho(f)-\rho(f)A) \Lambda_{F_k} \\ &= \sum_{j=1}^n \left(\sum_{k\sim j}+\sum_{k\not\sim j}\right) (\Lambda_{F_j} A\Lambda_{F_k}\rho(f) - \rho(f)\Lambda_{F_j} A\Lambda_{F_k}) \end{align}\] where we have used the notation \(k\sim j\) to denote those \(k\) such that the cube \(F_k\) is adjacent to \(F_j\), and \(k\not\sim j\) to those \(k\) where \(F_k\) is not adjacent to \(F_j\). In particular, if \(F_k\) and \(F_j\) are not adjacent, then there are disjoint closed sets that separate them, and hence \(\Lambda_{F_j} A\Lambda_{F_k}\in\mathcal{K}(\mathcal{H}_d)\) by assumption. Therefore, the operator relating to \(\sum_{j=1}^n\sum_{k\not\sim j}\) is compact, and we denote it by \(K_\varepsilon\). Now we break up the terms \[\begin{align} A\rho(f)-\rho(f)A - K_\varepsilon= & \sum_{j=1}^n\sum_{k\sim j} \Lambda_{F_j}A\Lambda_{F_k}\left(\rho(f)-\sum_{l=1}^n f(x_l)\Lambda_{F_l}\right) \\ + & \sum_{j=1}^n\sum_{k\sim j} \Lambda_{F_j}A\Lambda_{F_k}\sum_{l=1}^n f(x_l)\Lambda_{F_l} \\ - & \sum_{j=1}^n\sum_{k\sim j} \left(\rho(f)-\sum_{l=1}^n f(x_l)\Lambda_{F_l}\right) \Lambda_{F_j}A\Lambda_{F_k} \\ - & \sum_{j=1}^n\sum_{k\sim j} \left(\sum_{l=1}^n f(x_l)\Lambda_{F_l}\right) \Lambda_{F_j}A\Lambda_{F_k} \end{align}\] and proceed to bound the operator norms of the first, second, third and the fourth terms individually. For the first term, let \(T:= A(\rho(f)-\sum_{l=1}^n f(x_l)\Lambda_{F_l})\) for convenience, and let \(\xi\in\mathcal{H}_d\) be arbitrary, and we have \[\begin{align} \left\|\sum_{j=1}^n\sum_{k\sim j} \Lambda_{F_j}T\Lambda_{F_k}\xi \right\|^2 = \sum_{j=1}^n\left\| \Lambda_{F_j}T\sum_{k\sim j}\Lambda_{F_k}\xi \right\|^2 \leq \max_j \|\Lambda_{F_j} T\|^2 \sum_{j=1}^n \sum_{k\sim j} \|\Lambda_{F_k}\xi\|^2\,. \end{align}\] Observe that for each cube in \(\mathbb{R}^d\), there are \(3^{d}-1\) cubes that surrounds the center cube. It follows that \[\begin{align} \label{eq:adjacent32cubes32estimate} \sum_{j=1}^n \sum_{k\sim j} \|\Lambda_{F_k}\xi\|^2 \leq 3^d \|\xi\|^2\,. \end{align}\tag{22}\] Thus, using the above 22 and 21 , the operator norm of the first term is bounded above by \(\varepsilon 3^{d/2}\|A\|\), and similarly, so is the operator norm of the third term. Now, consider the second and the fourth terms, let \(\xi\in\mathcal{H}_d\) be arbitrary, we have \[\begin{align} \left\|\sum_{j=1}^n \sum_{k\sim j} (f(x_k)-f(x_j)) \Lambda_{F_j}A\Lambda_{F_k} \xi\right\|^2 = & \sum_{j=1}^n \left\|\Lambda_{F_j} A \sum_{k\sim j}(f(x_k)-f(x_j))\Lambda_{F_k}\xi \right\|^2 \\ \leq & \max_{j}\|\Lambda_{F_j}A\|^2 \sum_{j=1}^n |f(x_k)-f(x_j)|^2\|\Lambda_{F_k}\xi\|^2\,. \end{align}\] Using 20 and 22 , the operator norm of the second and fourth terms combined is bounded by \(\varepsilon 3^{d/2}\|A\|\). Thus \[\begin{align} \|A\rho(f)-\rho(f)A - K_\varepsilon\| \leq \varepsilon 3^{d/2-1} \|A\|\,. \end{align}\] We can construct a sequence of compact operators \(K_\epsilon\) that converges to \(A\rho(f)-\rho(f)A\) with \(\epsilon\to 0\). Thus \(A\rho(f)-\rho(f)A\in\mathcal{K}(\mathcal{H}_d)\). ◻

Proof. The proof is the same as 26. ◻

Proof. From 14 , we have \(\mathcal{L}_d=\mathcal{D}_\rho(C(\mathbb{S}^{d-1}))\). Then \[\begin{align} K_1(\mathcal{D}_\rho(C(\mathbb{S}^{d-1}))) \cong K^0(C_0(\mathbb{R}^{d-1})) \cong K^0(S^{d-1}\mathbb{C}) =\begin{cases} \mathbb{Z}& \text{d is odd} \\ 0 & \text{d is even} \end{cases} \end{align}\] where in the first isomorphism, we use the definition of \(K\)-homology groups ([43]) and the fact that the \(C(\mathbb{S}^{d-1})\) is the unitization of \(C_0(\mathbb{R}^{d-1})\); in the second isomorphism, we use that \(C_0(\mathbb{R}^{d-1})\cong S^{d-1}\mathbb{C}\); and in the third isomorphism, we repeatedly use the Bott periodicity (with \(S\) being the suspension) to reduce the calculations to the cases of \(K^1(\mathbb{C})\) or \(K^0(\mathbb{C})\), which are given by \[\begin{align} K^1(\mathbb{C})=0,\quad K^0(\mathbb{C})=\mathbb{Z}\,. \end{align}\] Similarly, we also have \[\begin{align} K_0(\mathcal{D}(C(\mathbb{S}^{d-1}))) \cong K^1(C_0(\mathbb{R}^{d-1})) = K^1(S^{d-1}\mathbb{C})=\begin{cases} 0 & \text{d is odd} \\ \mathbb{Z}& \text{d is even} \end{cases}\,. \end{align}\]

We proceed to the second part of the claim. The argument follows from [43]. Let \([P]_0-[Q]_0\) be an element in \(K_0(\mathcal{L}_d)\), where \(P\in\mathcal{P}_n(\mathcal{L}_d)\) and \(Q\in\mathcal{P}_m(\mathcal{L}_d)\). Using \([\mathbb{1}_m]_0=0\) from 32, it follows that \(-[Q]_0 = [Q^\perp]_0\). Thus \([P]_0-[Q]_0 = [P\oplus Q^\perp]_0\). Using \(\mathbb{1}_{n+m}\sim_0 \mathbb{1}\) from 32, there exists \(V\in M_{1,n+m}(\mathcal{L}_d)\) such that \(\mathbb{1}_{n+m}=V^\ast V\) and \(\mathbb{1}= VV^\ast\). Let \(T:= V(P\oplus Q^\perp)V^\ast\). Then \(T\in \mathcal{P}(\mathcal{L}_d)\) and \([T]_0=[P\oplus Q^\perp]_0=[P]_0-[Q]_0\).

Suppose an element \([U]_1\in K_1(\mathcal{L}_d)\) is represented by \(U\in\mathcal{U}_n(\mathcal{L}_d)\) for some \(n\in\mathbb{N}\). We need to find \(T\in\mathcal{U}(\mathcal{L}_d)\) such that \([T]_1=[U]_1\). The proof technique follows [43] and [54]. Using 32, we have \(\mathbb{1}_n\sim_0\mathbb{1}\) and hence there exists an element \(V\in M_{1,n}(\mathcal{L}_d)\) such that \(\mathbb{1}_n = V^\ast V\) and \(\mathbb{1}= VV^\ast\). Consider the elements \[\begin{align} W= \begin{bmatrix} 1 \\ 0 \\ \vdots\\ 0 \end{bmatrix} V\in M_n(\mathcal{L}_d),\quad S = \begin{bmatrix} W & \mathbb{1}_n-WW^\ast \\ 0 & W^\ast \end{bmatrix}\in M_{2n}(\mathcal{L}_d)\,. \end{align}\] It is straightforward to check that \(S\in \mathcal{U}_{2n}(\mathcal{L}_d)\) and \[\begin{align} WUW^\ast + \mathbb{1}_n-WW^\ast &= \begin{bmatrix} VUV^\ast & 0 \\ 0 & \mathbb{1}_{n-1} \end{bmatrix}\in \mathcal{U}_n(\mathcal{L}_d) \\ S\begin{bmatrix} U & 0 \\ 0 & \mathbb{1}_n \end{bmatrix} S^\ast &= \begin{bmatrix} WUW^\ast + \mathbb{1}_n-WW^\ast & 0 \\ 0 & \mathbb{1}_n \end{bmatrix}\in \mathcal{U}_{2n}(\mathcal{L}_d)\,. \end{align}\] Thus we have \[\begin{align} [VUV^\ast]_1 = [WUW^\ast + \mathbb{1}_n-WW^\ast]_1 = [S (U\oplus \mathbb{1}_n)S^\ast]_1 = [S]_1 + [U]_1 + [S^\ast]_1 = [U]_1 \end{align}\] and \(VUV^\ast \in \mathcal{U}(\mathcal{L}_d)\) is the sought-after operator. ◻

Proof. Denote \(\mathcal{A}_d:= C_0(\mathbb{R}^{d-1})\) for convenience. The index homomorphisms 27 and 28 originate from the pairing between \(K\)-theory group \(K_j(\mathcal{A}_d)\) and the \(K\)-homology \(K^j(\mathcal{A}_d)\) that lead to the bilinear maps 26 and 25 respectively. Indeed, these follow from noting \(K_1(\mathcal{L}_d)=K_1(\mathcal{D}_\rho(\widetilde{\mathcal{A}}_d))=K^0(\mathcal{A}_d)\) and \(K_0(\mathcal{L}_d)=K_0(\mathcal{D}_\rho(\widetilde{\mathcal{A}}_d))=K^1(\mathcal{A}_d)\); and that \(K_1(C(\mathbb{S}^{d-1}))\cong K_1(\mathcal{A}_d)\), and \(K_0(\mathcal{A}_d)\cong \widetilde{K}_0(C(\mathbb{S}^{d-1}))\). To show that the index homomorphisms are in fact isomorphisms, we invoke the universal coefficient theorem [43]: for each \(j\in\{0,1\}\) there is a natural short exact sequence \[0\longrightarrow \operatorname{Ext}^{1}_{\mathbb{Z}}\bigl(K_{j-1}(\mathcal{A}_d),\mathbb{Z}\bigr) \longrightarrow K^{j}(\mathcal{A}_d) \longrightarrow \mathrm{Hom}_{\mathbb{Z}}\bigl(K_{j}(\mathcal{A}_d),\mathbb{Z}\bigr) \longrightarrow 0,\] where the right-hand map is the index homomorphism. The \(K\)-groups of \(\mathcal{A}_d\) are \[K_{0}(\mathcal{A}_d)= \begin{cases} \mathbb{Z}& d-1\text{ odd}\\ 0 & d-1\text{ even} \end{cases}, \qquad K_{1}(\mathcal{A}_d)= \begin{cases} 0 & d-1\text{ odd}\\ \mathbb{Z}& d-1\text{ even} \end{cases}.\] In particular, the non-zero group \(K_{j}(\mathcal{A}_d)\) is free abelian of rank \(1\), hence \(\operatorname{Ext}^{1}_{\mathbb{Z}}(K_{j-1}(A),\mathbb{Z})=0\) in the relevant parity. Therefore, we have the index isomorphisms \[\begin{align} K^j(\mathcal{A}_d)\cong \mathrm{Hom}_{\mathbb{Z}}(K_{j}(A),\mathbb{Z}). \end{align}\] Evaluating at the generators for \(K_j(\mathcal{A}_d)\), which are exactly the Dirac phase and projection when \(d\) is even and odd respectively (see [55]), give the desired result. ◻

Proof. Let \(P\in\mathcal{P}_n(\mathcal{L}_d)\) be spherically-local and \(P\rho^n(a)P\) be never compact unless \(a=0\). Since \([P,\rho^n(a)]\in\mathcal{K}(\mathcal{H}_d)\) for all \(a\in C(\mathbb{S}^{d-1})\) by construction, we may consider the \(\ast\)-homomorphism \[\begin{align} \varphi_P:C(\mathbb{S}^{d-1})\to \mathcal{Q}(\operatorname{im}P) \end{align}\] defined by \(a\mapsto \pi(P\rho^n(a)P)\), see [43]. Since \(P\rho^n(a)P\) is never compact unless \(a=0\), it follows that \(\varphi_P\) is injective, and hence is an extension of the compact operators by \(C(\mathbb{S}^{d-1})\), see [56]. Since \(\varphi_{\mathbb{1}_m}\) is a trivial extension, it follows from Voiculescu’s Theorem that \(\varphi_P\oplus\varphi_{\mathbb{1}_m}\) is unitarily equivalent to \(\varphi_P\). By [43], the projections \(P\oplus \mathbb{1}_m\) and \(P\) are Murray-von Neumann equivalent. ◻

Proof. Let \(P\in\mathcal{P}(\mathcal{L}_1)\). Suppose \(P\) is bulk-non-trivial in the sense of 34. Recall the notation \(\Lambda=\chi_\mathbb{N}(X_1)=\Lambda_{\{1\}}\), and \(\Lambda^\perp = \Lambda_{\{-1\}}\). Then the operators \[\begin{align} \label{eq:1d32projections32to32half32line} \Lambda P\Lambda,\Lambda^\perp P\Lambda^\perp,\Lambda P^\perp\Lambda,\Lambda^\perp P^\perp \Lambda^\perp \end{align}\tag{30}\] are not compact. In one-dimension, since \([\Lambda,P]\in\mathcal{K}(\mathcal{H}_1)\), it follows that the operators 30 are essential projections, i.e., they are orthogonal projections in the Calkin algebra. Indeed, take the operator \(\Lambda P\Lambda\) for instance, we have \[\begin{align} \pi(\Lambda P\Lambda)^2 = \pi(\Lambda P\Lambda P\Lambda) = \pi(\Lambda [P,\Lambda] P\Lambda + \Lambda P\Lambda) = \pi(\Lambda P\Lambda),\quad \pi(\Lambda P\Lambda)^\ast = \pi(\Lambda P\Lambda) \end{align}\] where \(\pi\) is the quotient map to the Calkin algebra. In particular, \(\sigma_{\mathrm{ess}}(\Lambda P\Lambda:\operatorname{im}\Lambda\to\operatorname{im}\Lambda)\) must be one of \(\{0\},\{1\}\) or \(\{0,1\}\). Since \(\Lambda P\Lambda\) is not compact, we rule out the case \(\{0\}\). Suppose it is \(\{1\}\). Then \(\sigma_{\mathrm{ess}}(\Lambda-\Lambda P\Lambda:\operatorname{im}\Lambda\to\operatorname{im}\Lambda)=\{0\}\). This leads to the contradiction that \(\Lambda P^\perp\Lambda\) is compact. Thus it must be the case that \(\sigma_{\mathrm{ess}}(\Lambda P\Lambda:\operatorname{im}\Lambda\to\operatorname{im}\Lambda)=\{0,1\}\). Analogously, one can verify \(\sigma_{\mathrm{ess}}(\Lambda^\perp P\Lambda^\perp:\operatorname{im}\Lambda^\perp\to\operatorname{im}\Lambda^\perp)=\{0,1\}\).

Conversely, suppose \(P\) is bulk-non-trivial in the sense of 33. It follows from definition that \(\Lambda P\Lambda\) and \(\Lambda^\perp P\Lambda^\perp\) are not compact. Suppose \(\Lambda P^\perp \Lambda\) is compact. Then \(\sigma_{\mathrm{ess}}(\Lambda P^\perp \Lambda:\operatorname{im}\Lambda\to\operatorname{im}\Lambda)=\{0\}\) and hence \[\begin{align} \sigma_{\mathrm{ess}}(\Lambda P\Lambda:\operatorname{im}\Lambda\to\operatorname{im}\Lambda)= \sigma_{\mathrm{ess}}(\Lambda-\Lambda P^\perp \Lambda:\operatorname{im}\Lambda\to\operatorname{im}\Lambda)=\{1\}\,. \end{align}\] This contradicts that we assume \(\sigma_{\mathrm{ess}}(\Lambda P\Lambda:\operatorname{im}\Lambda\to\operatorname{im}\Lambda)=\{0,1\}\). Analogously, one verify that \(\Lambda^\perp P^\perp \Lambda^\perp\) is not compact, or else \(\sigma_{\mathrm{ess}}(\Lambda^\perp P\Lambda^\perp:\operatorname{im}\Lambda^\perp\to\operatorname{im}\Lambda^\perp)=\{1\}\), leading to a contradiction. ◻

Proof. Let \(\mathcal{H}:=\mathcal{H}_d\otimes \mathbb{C}^N\) and view the chiral projection \(P\) in 29 as acting on the graded Hilbert space \(\mathcal{H}\oplus \mathcal{H}\cong \mathcal{H}\otimes \mathbb{C}^2\). We will write \(\Lambda_I\) also for the projection \(\Lambda_I\oplus \Lambda_I\) (on \(\mathcal{H}\oplus\mathcal{H}\)); this is consistent with the convention that \(\Lambda_I\) acts only on the \(\ell^2(\mathbb{Z}^d)\) factor and as the identity on internal degrees.

First we establish the locality of \(P\). Fix \(j\in\{1,\dots,d\}\) and write \(\widehat{X}_j\) for \(\widehat{X}_j\otimes \mathbb{1}_N\) (on \(\mathcal{H}\)) and also for \(\widehat{X}_j\oplus \widehat{X}_j\) (on \(\mathcal{H}\oplus\mathcal{H}\)). Since \(U\) is spherically-local, \([U,\widehat{X}_j]\in\mathcal{K}(\mathcal{H})\) and hence also \([U^\ast,\widehat{X}_j]\in\mathcal{K}(\mathcal{H})\). Using 29 , \[\Bigl[P,\widehat{X}_j\Bigr] =\frac{1}{2} \begin{bmatrix} 0 & [U^\ast,\widehat{X}_j]\\ [U,\widehat{X}_j] & 0 \end{bmatrix}\in \mathcal{K}(\mathcal{H}\oplus\mathcal{H}).\] Thus \(P\in\mathcal{P}(\mathcal{L}_d)\) (and similarly \(P^\perp\in\mathcal{P}(\mathcal{L}_d)\)).

Next, we show that \(\Lambda_I\) has infinite rank for every non-empty open \(I\subset\mathbb{S}^{d-1}\). Indeed, there are infinitely many \(x\in \mathbb{Z}^d\) for which \(\hat{x}:=x/\|x\|\in I\) if \(I\) is non-empty and open subset of \(\mathbb{S}^{d-1}\).

Finally, we show that \(\Lambda_I P\Lambda_I\) and \(\Lambda_I P^\perp\Lambda_I\) are not compact. Using 29 and the fact that \(\Lambda_I\) acts diagonally on \(\mathcal{H}\oplus\mathcal{H}\), \[\Lambda_I P\Lambda_I =\frac{1}{2} \begin{bmatrix} \Lambda_I & \Lambda_I U^\ast \Lambda_I\\ \Lambda_I U \Lambda_I & \Lambda_I \end{bmatrix}, \qquad \Lambda_I P^\perp\Lambda_I =\frac{1}{2} \begin{bmatrix} \Lambda_I & -\Lambda_I U^\ast \Lambda_I\\ -\Lambda_I U \Lambda_I & \Lambda_I \end{bmatrix}.\] Let \(E_{11}:=\mathbb{1}_\mathcal{H}\oplus 0\) be the projection onto the first chiral block. We have \[E_{11}(\Lambda_I P\Lambda_I)E_{11}=\frac{1}{2}\,\Lambda_I,\qquad E_{11}(\Lambda_I P^\perp\Lambda_I)E_{11}=\frac{1}{2}\,\Lambda_I.\] Since \(\Lambda_I\) is not compact by the above, neither \(\Lambda_I P\Lambda_I\) nor \(\Lambda_I P^\perp\Lambda_I\) can be compact. ◻

Proof. Suppose \(P\) is not bulk-non-trivial. There exists an open subset \(I\) of \(\mathbb{S}^{d-1}\) such that \(\Lambda_I P\Lambda_I\in\mathcal{K}(\mathcal{H}_d)\). There exists a non-zero continuous function \(f\in C(\mathbb{S}^{d-1})\) such that \(f(z)=0\) for \(z\in I^c\); e.g., we can apply Urysohn’s lemma to the sets \(\{z_0\}\) and \(I^c\) for some \(z_0\in I\). Consider the decomposition \[\begin{align} P\rho(f)P=(\Lambda_I+\Lambda_{I^c})P\rho(f)P(\Lambda_I+\Lambda_{I^c})\,. \end{align}\] Recall that \([P,\rho(f)]\in\mathcal{K}(\mathcal{H}_d)\) since \(P\) is spherically-local, and hence we may interchange \(P\) and \(\rho(f)\) modulo a compact operator. Also \([\Lambda_I,\rho(f)]=0\) using 19 . Thus \(\Lambda_IP\rho(f)P\Lambda_I = \Lambda_IP\Lambda_I \rho(f)+K\) holds for some \(K\in\mathcal{K}(\mathcal{H}_d)\) and is compact by assumption. And the terms involving \(\Lambda_{I^c}\) vanish since \(\Lambda_{I^c}\rho(f)=0\).

Suppose \(P\rho(f)P\in\mathcal{K}(\mathcal{H}_d)\) for some non-zero \(f\in C(\mathbb{S}^{d-1})\). Then, there exists a point \(z\in \mathbb{S}^{d-1}\) such that \(f(z)\neq 0\); furthermore, there exists an open neighborhood \(I\subset \mathbb{S}^{d-1}\) of \(z\) such that \(|f(z)|\geq c>0\) for some positive number \(c\). We can construct a continuous function \(g\) such that \(g(z)=1/f(z)\) for \(z\in I\). Then \[\begin{align} \Lambda_IP\rho(f)\rho(g)\Lambda_I = \Lambda_IP\rho(fg)\Lambda_I=\Lambda_IP\Lambda_I \end{align}\] is compact since \(P\rho(f)\) is compact by assumption. This leads to a contradiction. ◻

Proof of 38. We establish the set bijection \[\begin{align} \label{eq:pi032unitary32bijective32to32K1} \pi_0(\mathcal{U}(\mathcal{L}_d)) \to K_1(\mathcal{L}_d) \end{align}\tag{31}\] that maps unitary in path-connected component to its \(K\)-class. First we show it is injective. It suffices to show that if \(U\in\mathcal{U}(\mathcal{L}_d)\) satisfies \([U]_1=0\), then \(U\sim_h \mathbb{1}\) in \(\mathcal{U}(\mathcal{L}_d)\). Indeed, for \(U,V\in\mathcal{U}(\mathcal{L}_d)\) with \([U]_1=[V]_1\), we have \([UV^\ast]_1=0\), and the homotopy \(UV^\ast\sim_h\mathbb{1}\) induces the homotopy \(U\sim_h V\) in \(\mathcal{U}(\mathcal{L}_d)\). Suppose \(U\in\mathcal{U}(\mathcal{L}_d)\) and \([U]_1=0\). Using 43, we may assume \(U\) takes the form of \(P_0UP_0+P_0^\perp\) for some spherically-proper projection \(P_0\). Since \([U]_1=0\), it follows that \(U\oplus \mathbb{1}_n\sim_h \mathbb{1}_{n+1}\) in \(\mathcal{U}_{n+1}(\mathcal{L}_d)\) for some \(n\in\mathbb{N}\). Then, with the assumptions in 44 satisfied, we have \(U\sim_h \mathbb{1}\). This completes the proof for 31 .

Let us show that 31 is surjective. Let \([U]_1\in K_1(\mathcal{L}_d)\) be a \(K\)-class with \(U\in\mathcal{U}_n(\mathcal{L}_d)\) for some \(n\in\mathbb{N}\). We’d like to find \(W\in\mathcal{U}(\mathcal{L}_d)\) such that \([W]_1=[U]_1\). Let \(\mathbb{Z}^d=F_1\cup \dots\cup F_n\) be a partition of \(\mathbb{Z}^d\) into spherically-proper subsets. Let \(P_j=\Lambda_{F_j}\). Using 42, there exists partial isometry \(V_j\in\mathcal{L}_d\) such that \(V_j^\ast V_j = \mathbb{1}\) and \(V_j V_j^\ast = P_j\). Let \(V=\begin{bmatrix}V_1 &\dots & V_n\end{bmatrix}\in M_{1,n}(\mathcal{L}_d)\), and consider \(W:=VUV^\ast\in\mathcal{U}(\mathcal{L}_d)\). It is straightforward to show that \[\begin{align} \begin{bmatrix} \cos t \mathbb{1}_n & -\sin t V^\ast \\ \sin t V & \cos t \mathbb{1} \end{bmatrix} \begin{bmatrix} U & 0 \\ 0 & \mathbb{1} \end{bmatrix} \begin{bmatrix} \cos t \mathbb{1}_n & \sin t V^\ast \\ -\sin t V & \cos t \mathbb{1} \end{bmatrix} \end{align}\] for \(t\in[0,\pi/2]\) provides the homotopy \(U\oplus \mathbb{1}\sim_h \mathbb{1}_n \oplus W\). Therefore \([U]_1=[W]_1\).

Using 31 and 29, we establish that \(\pi_0(\mathcal{U}(\mathcal{L}_d))\cong \mathbb{Z}\) if \(d\in 2\mathbb{N}+1\) and \(\pi_0(\mathcal{U}(\mathcal{L}_d))\cong \Set{0}\) if \(d\in 2\mathbb{N}\).

Next, we show that for \(d\in 2\mathbb{N}+1\), the index is given by \(\mathcal{N}:\mathcal{U}(\mathcal{L}_d)\to \mathbb{Z}\), which amounts to showing that \(\mathcal{N}\) is continuous and bijective. It is clear that \(\mathcal{N}\) is continuous. If \(U,V\in\mathcal{U}(\mathcal{L}_d)\) have the same index \(\mathcal{N}(U)=\mathcal{N}(V)\), using 31, we have \([U]_1=[V]_1\), and hence \(U,V\) belong to the same path-connected component by 31 . For any \(m\in\mathbb{Z}\), using 31, there exists \([U]_1\in K_1(\mathcal{L}_d)\) for some \(U\in\mathcal{U}(\mathcal{L}_d)\) such that \(\mathcal{N}(U)=m\). This completes the proof. ◻

Proof of 39. We establish the set bijection \[\begin{align} \label{eq:pi032projection32same32as32K0} \pi_0(\mathcal{P}^\mathrm{nt}(\mathcal{L}_d))\to K_0(\mathcal{L}_d) \end{align}\tag{32}\] that maps projection in each path-connected component to its \(K\)-class. First we show that it is injective. Consider the case when \(d\) is even. Suppose \(P,Q\in\mathcal{P}^\mathrm{nt}(\mathcal{L}_d)\) belong to the same class \([P]_0=[Q]_0\). Then \(P\oplus \mathbb{1}_n \sim Q\oplus \mathbb{1}_n\) for some \(n\in\mathbb{N}\). Using 32, it follows that \(P\sim_0 P\oplus \mathbb{1}_n\sim Q\oplus \mathbb{1}_n\sim_0 Q\) and hence \(P\sim Q\). Since \([\mathbb{1}]_0=0\) by 32, we have \([P^\perp]_0=[\mathbb{1}]_0-[P]_0=-[P]_0=-[Q]_0=[\mathbb{1}]_0-[Q]_0= [Q^\perp]_0\). Thus \(P^\perp \sim Q^\perp\) based on the same argument (in showing \(P\sim Q\)). Using [57], we have \(P\sim_u Q\), i.e., there exists \(U\in\mathcal{U}(\mathcal{L}_d)\) such that \(Q=UPU^\ast\). Since by 38 we have that \(\pi_0(\mathcal{U}(\mathcal{L}_d))=\{0\}\) when \(d\) is even, it follows that \(U\sim_h \mathbb{1}\) in \(\mathcal{U}(\mathcal{L}_d)\). Thus \(P\sim_h Q\) in \(\mathcal{P}(\mathcal{L}_d)\), which is in fact in \(\mathcal{P}^\mathrm{nt}(\mathcal{L}_d)\) by 54.

Now consider the case when \(d\) is odd. Let \(P\) be a spherically-proper projection, e.g., that from 6. Let \(Q\in\mathcal{P}^\mathrm{nt}(\mathcal{L}_d)\) be arbitrary. We show that \(Q\sim_h P\), establishing that \(\pi_0(\mathcal{P}^\mathrm{nt}(\mathcal{L}_d))=\Set{0}\). Since \(K_0(\mathcal{L}_d)=\Set{0}\) from 29 and hence \([Q]_0=[P]_0\), analogous to the injectivity proof in the case when \(d\) is even, it follows that \(Q=UPU^\ast\) for some \(U\in\mathcal{U}(\mathcal{L}_d)\). Since \(P\) is spherically-proper, using 42, we have \(P\sim \mathbb{1}\), and hence there exists spherically-local partial isometry \(V\) such that \(P=VV^\ast\) and \(\mathbb{1}=V^\ast V\). Consider the spherically-local unitary \(W=VUV^\ast + P^\perp\in\mathcal{U}(\mathcal{L}_d)\). We claim that \(U\sim_h W\) in \(\mathcal{U}(\mathcal{L}_d)\). To that end, we consider \[\begin{align} R:=\begin{bmatrix} V & P^\perp \\ 0 & V^\ast \end{bmatrix}\in \mathcal{U}_2(\mathcal{L}_d) \end{align}\] Then we have \(R (U\oplus \mathbb{1}) R^\ast = W\oplus \mathbb{1}\), and hence \([U]_1 = [W]_1\). Using 31 , we have \(U\sim_h W\) in \(\mathcal{U}(\mathcal{L}_d)\). In particular, this provides the desired homotopy \[\begin{align} Q = UPU^\ast \sim_h WPW^\ast = VUV^\ast P VU^\ast V^\ast = P \end{align}\] in \(\mathcal{P}(\mathcal{L}_d)\), which is in fact in \(\mathcal{P}^\mathrm{nt}(\mathcal{L}_d)\) by 54.

We now show that 32 is surjective. Using 29, an element in \(K_0(\mathcal{L}_d)\) is of the form \([Q]_0\) for \(Q\in\mathcal{P}(\mathcal{L}_d)\). However, \(Q\) may not be bulk-non-trivial; therefore, we seek \(P\in\mathcal{P}^\mathrm{nt}(\mathcal{L}_d)\) such that \([P]_0=[Q]_0\). To that end, let \(S\) be a spherically-proper projection, which is also bulk-non-trivial, see 41. Consider \(R:=Q\oplus S\in\mathcal{P}_2(\mathcal{L}_d)\). In fact \(R\) is bulk-non-trivial since \(S\) is. (Although 34 is defined only in \(\mathcal{P}(\mathcal{L}_d)\), it naturally extends to \(\mathcal{P}(M_2(\mathcal{L}_d))\equiv \mathcal{P}_2(\mathcal{L}_d)\).) Moreover, we have \([R]_0=[Q]_0+[S]_0 = [Q]_0\) where we use \([S]_0=[\mathbb{1}]_0=0\) with the help of 42 and 32. Since \(S\) and \(S^\perp\) are spherically-proper, using 42, there exist partial isometry \(V_1,V_2\in\mathcal{L}_d\) such that \(V_1V_1^\ast = S\), \(V_2V_2^\ast = S^\perp\) and \(V_1^\ast V_1=V_2^\ast V_2=\mathbb{1}\). Let \(V:=\begin{bmatrix} V_1 & V_2 \end{bmatrix}\in M_{1,2}(\mathcal{L}_d)\). Then \(VV^\ast=\mathbb{1}\) and \(V^\ast V=\mathbb{1}_2\). Consider \(P:=VRV^\ast\in\mathcal{P}(\mathcal{L}_d)\). We have \([P]_0=[R]_0=[Q]_0\). Moreover, \(P\in\mathcal{P}^\mathrm{nt}(\mathcal{L}_d)\) is bulk-non-trivial. Indeed, we use 37 and suppose that \(P\rho(a)P\) is compact. Consider \(V^\ast P\rho(a)PV = RV^\ast \rho(a) V R\). Now \(\rho(a)V- V\rho_2(a)\) is compact since \(V\) is spherically-local and \(\rho_2(a):=\rho(a)\oplus \rho(a)\). Thus \(R\rho_2(a)R\) is compact. Since \(R\) is bulk-non-trivial, it follows that \(a=0\) and we conclude that \(P\) is bulk-non-trivial.

Next, we show that for \(d\in 2\mathbb{N}\), the index is given by \(\mathcal{N}:\mathcal{P}^\mathrm{nt}(\mathcal{L}_d)\to \mathbb{Z}\), which amounts to showing that \(\mathcal{N}\) is continuous and bijective. It is clear that \(\mathcal{N}\) is continuous. That \(\mathcal{N}\) is bijective follows from 31 and 32 . ◻

Proof. Consider an increasing sequence of finite subsets \(F_1\subset F_2\cdots\) such that \(\cup_{k\in\mathbb{N}}F_k=\mathbb{Z}^d\). Indeed, we can construct \(F_k\) to be \(B_r \cap \mathbb{Z}^d\) for larger and larger \(r>0\). Then \(\Lambda_{F_k}\) converges to \(\mathbb{1}\) in the strong operator topology. Since \(K\) is compact, it follows that \(\Lambda_{F_k}K\) converges in norm to \(K\). Therefore, for \(k\) large enough, we have \(\|\Lambda_{F_k}K-K\|\leq \varepsilon\). ◻

Proof. Let \(J\) be a closed subset of \(\mathbb{S}^{d-1}\). Define \(N_k\) by \[\begin{align} N_k = \Set{x\in \mathbb{S}^{d-1} | \|x-y\|<1/k \text{ for some y in J}} \supset J. \end{align}\] It is clear that \(N_k\) is open in \(\mathbb{S}^{d-1}\), decreasing (\(N_{k}\supset N_{k+1}\) holds for each \(k\)), and \(J = \cap_k N_k\). In particular, \(N_k^c\) is closed and \(J^c = \cup_k N_k^c\). Since \(N_k^c\) and \(J\) are disjoint and closed, by 26, it follows that \(\Lambda_{N^c_k}A\Lambda_J \in \mathcal{K}\). Using 48, we can decompose \(C_{N^c_k}\) into two disjoint subsets \[\begin{align} C_{N^c_k} = E_k \cup F_k \end{align}\] such that \(|F_k|<\infty\) and \(\| \Lambda_{E_k} A \Lambda_J\|\leq \varepsilon/2^k\). Let \[\begin{align} E = \cup_{k\in\mathbb{N}} E_k \,. \end{align}\] We can rewrite \(E\) as \(\cup_{k\in\mathbb{N}} G_k\) where \(G_k = E_k\setminus (\cup_{i=1}^{k-1}E_i)\) are disjoint. Then \(\Lambda_E A\Lambda_J = (\sum_k \Lambda_{G_k}) A \Lambda_J\). In fact the series \(\sum_k \Lambda_{G_k}A\Lambda_J\) converge in operator norm. Indeed, we have \[\begin{align} \sum_{k=1}^\infty \| \Lambda_{G_k} A \Lambda_J\| \leq \sum_{k=1}^\infty \| \Lambda_{E_k} A \Lambda_J\| \leq \varepsilon\,. \end{align}\] Since each \(\Lambda_{G_k}A\Lambda_J\) is compact, it follows that \(\Lambda_EA\Lambda_J\) is compact. Moreover, \(\|\Lambda_EA\Lambda_J\|\leq \varepsilon\). Define \(F\) by \[\begin{align} F = C_{J^c}\setminus E\,. \end{align}\] We argue that \(|F\cap C_{N_k^c}|<\infty\) for all \(k\in\mathbb{N}\). Indeed, we have \[\begin{align} F\cap C_{N_k^c}=F\cap(E_k\cup F_k) = F\cap F_k\subset F_k \end{align}\] and we have \(|F\cap C_{N_k^c}|\leq |F_k|<\infty\). Let \(I\) be any closed subset of \(\mathbb{S}^{d-1}\) disjoint from \(J\). Let \(\delta = \mathrm{dist}(I,J)>0\). If we pick \(k\in\mathbb{N}\) large enough so that \(1/k < \delta\), then we have \(I\subset N_{k}^c\). Thus, \(|F\cap C_I|\leq |F\cap C_{N_k^c}|<\infty\). ◻

Proof. Let \(\{I_i\}_{i\in\mathbb{N}}\) be a countable basis for \(\mathbb{S}^{d-1}\). Note in \(d=1\) we simply alternate between two points of \(\mathbb{S}^0\). We do the construction iteratively. Let us pick an element \(x_1\in C_{I_1}\subset \mathbb{Z}^d\). There exists \(r_1>\|x_1\|\) such that \[\begin{align} \|\Lambda_{B_{r_1}^c} A \Lambda_{\{x_1\}}\|\leq \varepsilon_1 \,. \end{align}\] Indeed, this follows from 48 with the fact that \(A \Lambda_{\{x_1\}}\) is compact and \(\Lambda_{B_{r}}\) converges to \(\mathbb{1}\) strongly as \(r\to\infty\). Since \(|B_{r_1}|\) is finite, which implies \(A^\ast \Lambda_{B_{r_1}}\) is compact, it follows from 48 again that there exists \(t_1>r_1\) such that \(\|\Lambda_{B_{t_1}^c}A^\ast\Lambda_{B_{r_1}}\|\leq \varepsilon_2/2\), or \[\begin{align} \| \Lambda_{B_{r_1}} A \Lambda_{B_{t_1}^c}\|\leq \varepsilon_2/2 \,. \end{align}\] Pick \(x_2\in C_{I_2}\subset \mathbb{Z}^d\) with \(\|x_2\|> t_1\). There exists \(r_2>\|x_2\|\) such that \[\begin{align} \|\Lambda_{B_{r_2}^c} A \Lambda_{\{x_2\}}\| \leq \varepsilon_2/2 \end{align}\] which follows again from 48. Therefore, we have \[\begin{align} \|\Lambda_{B_{r_2}^c\cup B_{r_1}} A \Lambda_{\{x_2\}}\| \leq \|\Lambda_{B_{r_2}^c} A \Lambda_{\{x_2\}}\| + \|\Lambda_{B_{r_1}} A \Lambda_{B_{t_1}^c}\| \leq \varepsilon_2 \end{align}\] where we used \(\Lambda_{\{x_2\}}\leq \Lambda_{B_{t_1}^c}\) in the first inequality.

We iterate the procedure: we pick \(t_2>r_2\) such that \(\|\Lambda_{B_{t_2}^c} A^\ast \Lambda_{B_{r_2}}\|\leq \varepsilon_3/2\); pick \(x_3\in C_{I_3}\subset \mathbb{Z}^d\) with \(\|x_3\|>t_2\); and pick \(r_3>\|x_3\|\) such that \(\|\Lambda_{B_{r_3}^c} A (\delta_{x_3}\otimes \delta_{x_3}^\ast)\| \leq \varepsilon_3/2\); and we get \(\|\Lambda_{B_{r_3}^c\cup B_{r_2}} A \Lambda_{\{x_3\}}\|\leq \varepsilon_3\); and so on. ◻

Proof. Let \(A\) be a spherically-local operator. Consider the collection \(\{G_k\}\) of all closed dyadic cubes in \(\mathbb{S}^{d-1}\). For each cube \(G_k\), we can use 49 to partition \(C_{G_k^c}\) into two disjoint subsets \(E_k\) and \(F_k\) such that \(\Lambda_{E_k}A\Lambda_{G_k}\) has small norm and is compact, and \(|F_k\cap C_I|<\infty\) for all closed set \(I\subset \mathbb{S}^{d-1}\) disjoint from \(F_k\). On the other hand, using 50, there exists a spherically-proper sequence of points \(\{x_k\}\) and a sequence of increasing balls \(\{B_{r_k}\}\) (with \(0=:r_0<r_1<r_2<\dots\)) such that \(x_k\in B_{r_k}\setminus B_{r_{k-1}}\) and \(\Lambda_{\mathbb{Z}^d\setminus(B_{r_k}\setminus B_{r_{k-1}})} A \Lambda_{\{x_k\}}\) has small norm and is compact. Now, apply 56 to get a spherically-local operator \(B\) such that \(\|A-B\|\leq \varepsilon\) and \(A-B\in\mathcal{K}(\mathcal{H}_d)\) and \[\begin{align} \Lambda_{\mathbb{Z}^d\setminus(B_{r_k}\setminus B_{r_{k-1}})} B \Lambda_{\{x_k\}} &= 0 \tag{33}\\ \Lambda_{E_k} B\Lambda_{G_k} & = 0 \tag{34} \end{align}\] for all \(k\in\mathbb{N}\). In other words, \(\mathbb{Z}^d\setminus(B_{r_k}\setminus B_{r_{k-1}})\) is decoupled from \(\{x_k\}\) via \(B\); and the complement of \(C_{G_k}\) is almost decoupled from \(C_{G_k}\) except for a set \(F_k\). Since each \(x_k\) lies in disjoint annulus \(B_{r_k}\setminus B_{r_{k-1}}\) and the support of \(B\delta_{x_k}\) is also contained in the same annulus via 33 , then \[\begin{align} R_k\subset B_{r_k}\setminus B_{r_{k-1}} \end{align}\] and it follows that \(R_k\) as defined in ?? have the properties that \(|R_k|<\infty\) for all \(k\), and the sets \(R_k\) are pairwise disjoint.

Let \(I\) and \(J\) be arbitrary disjoint closed dyadic cubes. By way of contradiction, suppose there are infinitely many \(k\) such that \(R_{k}\cap C_I\neq \varnothing\) and \(R_{k}\cap C_J\neq \varnothing\) for all \(i\in\mathbb{N}\); we denote the subsequence as \(\{k_1\}\). Let \(\mathcal{D}_l\) be the collection of all closed dyadic cubes of the \(l\)-th generation as defined in 27 where \(l\) is large enough such that no cubes in \(\mathcal{D}_l\) intersects both \(I\) and \(J\). Indeed, this is possible using the fact that \(I\) and \(J\) are disjoint closed dyadic cubes, and the construction of closed dyadic cubes in 27. Since \(\mathcal{D}_l\) is a finite covering of \(\mathbb{S}^{d-1}\) but \(\{x_{k_1}\}\) contains infinitely many points of \(\mathbb{Z}^d\), there exists a cube \(Q\in \mathcal{D}_l\) such that \(\hat{x}_{k_1}\in Q\) for infinitely many \(k_1\); we denote the sub-subsequence as \(\{k_2\}\). Without loss of generality, suppose \(Q\) is disjoint from \(I\).

Since \(R_{k_2}\cap C_I\neq \varnothing\) by assumption, we pick a point \(y_{k_2}\) in \(R_k\cap C_I\) for each \(k_2\). Since \(\{x_{k_2}\}\subset C_Q\) and \(C_Q\) is disjoint from \(C_I\supset R_k\cap C_I\), we cannot have \(y_{k_2}\) being the same as \(x_{k_2}\). Therefore \[\begin{align} y_{k_2}\in \operatorname{supp}(B\delta_{x_{k_2}})\cap C_I \subset C_{Q^c} \end{align}\] for each \(k_2\). Denote the particular partition, as performed in the beginning, of \(C_{Q^c}\) as \(E\) and \(F\), where \(E\) is decoupled from \(C_Q\) via \(B\) (with 34 ), and for all closed subsets \(K\subset \mathbb{S}^{d_1}\) disjoint from \(Q\), we have \(|F\cap C_K|<\infty\) (as promised by 49). Thus \(y_{k_2}\subset E\cup F\). However, \(y_{k_2}\) cannot be in \(E\); otherwise, we have \[\begin{align} \braket{\delta_{y_{k_2}},B\delta_{x_{k_2}}} = \braket{\Lambda_{E}\delta_{y_{k_2}},B\Lambda_{Q}\delta_{x_{k_2}}} = \braket{\delta_{y_{k_2}},\Lambda_{E}B\Lambda_{Q}\delta_{x_{k_2}}}=0 \end{align}\] where the last equality follows from 34 , and we would have \(y_{k_2}\notin \operatorname{supp}(B\delta_{x_{k_2}})\), a contradiction. Thus \(y_{k_2}\in F\cap C_I\) for all \(k_2\). This contradicts the fact that \(|F\cap C_I|<\infty\). ◻

Proof of 42. We show that \(P\sim \mathbb{1}\). The proof for \(P^\perp\sim \mathbb{1}\) is similar since \(P\) is spherically-proper iff \(P^\perp\) is. Let \(\Set{y_k}_{k=1}^\infty=\mathbb{Z}^d\) be an enumeration of the lattice. Let \(F\) be the spherically-proper set that correspond to \(P=\Lambda_F\). Define subsets \(N_k\) by \[\begin{align} N_k=\Set{x\in\mathbb{S}^{d-1} | \|x-\hat{y}_k\|<1/k}\,. \end{align}\] Iteratively, for each \(k\in \mathbb{N}\), pick \(x_k\in F\) such that \(x_k\) minimizes \[\begin{align} \Set{ \left\lVert x\right\rVert | x\in C_{N_k}\cap F\setminus \{x_1,\dots,x_{k-1}\}}\,. \end{align}\] This is possible since \(|C_{N_k}\cap F|=\infty\) for all \(k\) by definition of spherical properness. Consider the mapping \(\mathbb{Z}^d \ni y_k\mapsto x_k \in F\). It is clear that the map is bijective. The map is injective since at each step we exclude previously chosen points from \(F\). Assume by way of contradiction that the map is not surjective. Choose any \(z^\ast \in F\setminus \Set{x_k}_{k=1}^\infty\) that is not picked by the algorithm. Let \(K\subset\mathbb{N}\) be all indices where \(\hat{y}_k=\hat{z}^\ast\) for \(k\in K\). The index set \(K\) is infinite as there are infinitely points in \(\mathbb{Z}^d\) in the same direction of \(z^\ast\). Moreover we have \(z^\ast \in C_{N_k} \cap F\setminus \Set{x_1,\dots,x_{k-1}}\) for all \(k\in K\), i.e., \(z^\ast\) is a valid candidate for every step \(k\in K\) but never get picked. The algorithm picks \(x_k\) that minimizes the norm among candidates. Since \(z^\ast\) is a candidate, the chosen \(x_k\) must satisfy \(\|x_k\|\leq \|z^\ast\|\). Thus for every step \(k\in K\), the algorithm selects distinct \(x_k\in F\) with norm \(\|x_k\|\leq \|z^\ast\|\). Since \(K\) is an infinite set, it implies that there are infinitely many distinct elements in \(F\subset \mathbb{Z}^d\) with norm less than or equal to \(\|z^\ast\|\), which is impossible.

Let \(V\) maps \(\delta_{y_k}\) to \(\delta_{x_k}\) and extend linearly to an operator from \(\mathcal{H}_d\) to \(\operatorname{im}P\). Then \(V^\ast V=\mathbb{1}\) and \(VV^\ast = P\). We now show that \(V\) is spherically-local. Let \(I\) and \(J\) be a pair of disjoint closed subsets of \(\mathbb{S}^{d-1}\), and we consider the operator \(\Lambda_J V\Lambda_I\). The distance between the set \(\mathrm{dist}(I,J)=\delta > 0\) is strictly positive. Let \(k_0\) be large enough so that \(1/k_0 < \delta\). Then, for all \(k\geq k_0\) and \(y_k\in C_I\), we have \(C_J\cap C_{N_k}=\varnothing\). Thus \(\Lambda_J V\Lambda_I = \Lambda_J V \Lambda_{\{x_k\}_{k<k_0}}\) is finite-rank. ◻

Proof of 43. Let \(U\in\mathcal{U}(\mathcal{L}_d)\) be a spherically-local unitary operator. Let \(\varepsilon>0\) be some small number to be determined. Using 51, there exists a spherically-local operator \(G\in\mathcal{L}_d\) such that \(\|U-G\|\leq \varepsilon\), and a spherically-proper sequence of points \(\{x_k\}_{k\in \mathbb{N}}\) such that the subsets \(R_k\subset\mathbb{Z}^d\) as defined in ?? has those properties specified in the lemma. In particular, we can choose \(\varepsilon\) small enough so that \(G\in\mathcal{G}(\mathcal{L}_d)\) is invertible. Denote \(P\) the spherically-proper projection \(\Lambda_{\{x_k\}_{k\in\mathbb{N}}}\).

Let us define a spherically-local invertible operator \(V\) such that \(VG\) acts as identity on \(\operatorname{im}P\). To that end, we define \(V\) on \(\operatorname{im}\Lambda_{R_k}\) as any invertible operator that maps \(G\delta_{x_k}\mapsto \delta_{x_k}\). Define \(V\) to be identity on \((\oplus_{k}\operatorname{im}\Lambda_{R_k})^\perp\). The operator \(V\) is well-defined since \(\{R_k\}_{k\in\mathbb{N}}\) consists of mutually disjoint subsets. We argue that \(V\) is spherically-local. To that end, we show that for any pair of disjoint closed dyadic cubes \(I\) and \(J\) in \(\mathbb{S}^{d-1}\), we have that \(\Lambda_J V \Lambda_I\) is finite-rank. We can decompose \(\mathbb{Z}^d\) into four disjoint subsets \[\begin{align} \textstyle C_{I^c},\;C_I\cap X,\;C_I\cap Y,\;C_I\cap Z \end{align}\] where \(X\) is the \(\mathbb{Z}^d \setminus \cup_{k} R_k\); and \(Y\) is the union of all \(R_k\) such that \(R_k\cap C_I\neq \varnothing\) and \(R_k\cap C_J\neq \varnothing\); and \(Z\) is the union of all \(R_k\) such that \(R_k\cap C_I=\varnothing\) or \(R_k\cap C_J=\varnothing\). With the decompositions, we write \[\begin{align} \Lambda_J V\Lambda_I = \Lambda_J V \Lambda_I (\Lambda_{I^c} + \Lambda_{C_I\cap X} + \Lambda_{C_I\cap Y} + \Lambda_{C_I\cap Z}) \end{align}\] and study each term. It is clear that \(\Lambda_JV\Lambda_I\Lambda_{I^c}=0\). Since \(V\) is identity on \((\oplus_{k\in\mathbb{N}} \operatorname{im}\Lambda_{R_k})^\perp\), it follows that \(\Lambda_J V\Lambda_I\Lambda_{C_I\cap X} = \Lambda_J\Lambda_{C_I\cap X} = 0\). Suppose \(y\in C_I\cap Z\), then \(y\in C_I\); and either \(y\in C_J\cap R_k\) for some \(k\in\mathbb{N}\) such that \(R_k\cap C_I=\varnothing\); or \(y\in C_I\cap R_k\) for some \(k\in\mathbb{N}\) such that \(R_k\cap C_J=\varnothing\). The former case is vacuous. In the latter case, we have \(\Lambda_JV\Lambda_I\delta_y=\Lambda_JV\delta_y\); and since \(V\delta_y\in \operatorname{im}\Lambda_{R_k}\) and \(R_k\cap C_J=\varnothing\), it follows that \(\Lambda_JV\delta_y=0\). Thus \(\Lambda_J V\Lambda_{I}\Lambda_{C_I\cap U}=0\). Finally, using 51, the set \(C_I\cap Y\) is finite and hence \(\Lambda_JV\Lambda_I=\Lambda_J V\Lambda_I \Lambda_{C_I\cap Y}\) is finite-rank.

The argument in the previous paragraph works for all operators \(A\in\mathcal{B}(\mathcal{H}_d)\) such that \(V\operatorname{im}\Lambda_{R_k}\subset \operatorname{im}\Lambda_{R_k}\) for all \(k\in \mathbb{N}\), and are identity on \((\oplus_{k\in\mathbb{N}} \operatorname{im}\Lambda_{R_k})^\perp\). Therefore, we can construct \(V\sim_h \mathbb{1}\) by deforming each invertible matrices of \(V\) on each \(\operatorname{im}\Lambda_{Y_k}\) to the identity matrices. So far, we have made the deformations \(U\sim_h G\sim_h VG\) in \(\mathcal{G}(\mathcal{L}_d)\).

By construction \(VG\) take the form \[\begin{align} VG =P + PVGP^\perp + P^\perp VG P^\perp = (P + P^\perp VG P^\perp)(\mathbb{1}+ PVGP^\perp)\,. \end{align}\] Since \(VG\in\mathcal{L}_d\) and \([P,L]=0\), it is clear that \(P + P^\perp VG P^\perp\) and \(\mathbb{1}+ PVGP^\perp\) are both spherically-local. Observe that \((\mathbb{1}+ t PVGP^\perp)\) is always in \(\mathcal{G}(\mathcal{L}_d)\) for \(t\in[0,1]\), with inverse \(\mathbb{1}- t PVGP^\perp\). Therefore, we have \(VG\sim_h P + P^\perp VG P^\perp\) in \(\mathcal{G}(\mathcal{L}_d)\). Let \(S=P+P^\perp SP^\perp\in\mathcal{U}(\mathcal{L}_d)\) be the polar part of \(P + P^\perp VG P^\perp\). It follows from [57] that we have \(P + P^\perp VG P^\perp\sim_h S\) in \(\mathcal{G}(\mathcal{L}_L)\).

Since \(P\) is spherically-proper, using 42, we have \(P\sim P^\perp\). Using 53, we have \(S=P+P^\perp SP^\perp \sim_h W = PWP+P^\perp\) in \(\mathcal{U}(\mathcal{L}_d)\) for some \(W\in\mathcal{U}(\mathcal{L}_d)\).

We have constructed \(U\sim_h W=PWP+P^\perp\) in \(\mathcal{G}(\mathcal{L}_d)\). Using [57], we obtain the homotopy \(U\sim_h W\) in \(\mathcal{U}(\mathcal{L}_L)\) and concludes the proof. ◻

Proof of 44. Let \(W_t\in\mathcal{U}_{n+1}(\mathcal{L}_d)\) be the homotopy implementing \(U\oplus \mathbb{1}_n\sim_h \mathbb{1}_{n+1}\).

Let \(P_0:=P\). Since \(P_0\) is spherically-proper, so is \(P_0^\perp\). Using 55, we may decompose \(P_0^\perp\) into \(n+1\) spherically-proper projections \[\begin{align} P_0^\perp = Q_0+ P_1 + \dots + P_{n-1} + P_n. \end{align}\]

Using 42, we have \(Q_0\sim P_0^\perp\) and \(P_k\sim \mathbb{1}\) for all \(k=1,\dots,n\). Let \(V_k\in\mathcal{L}_d\) be the spherically-local partial isometry such that \(P_k=V_kV_k^\ast\) and \(\mathbb{1}= V_k^\ast V_k\) for \(k=1,\dots,n\). Since \(Q_0\sim P_0^\perp\), there exists a spherically-local partial isometry \(V_0\in\mathcal{L}_d\) such that \(P_0^\perp = V_0^\ast V_0\) and \(Q_0=V_0V_0^\ast\). Then \[\begin{align} \begin{bmatrix} P_0+V_0 & V_1 & \dots & V_n \end{bmatrix} W_t \begin{bmatrix} (P_0+V_0)^\ast \\ V_1^\ast \\ \vdots \\ V_n^\ast \end{bmatrix} \end{align}\] gives a homotopy in \(\mathcal{U}(\mathcal{L}_d)\) that deforms \(U\) to \(\mathbb{1}\). ◻

Proof. Let \(V:\operatorname{im}P\to\operatorname{im}Q\) be the partial isomtery such that \(P=V^\ast V\) and \(Q=VV^\ast\). Let \(R=\mathbb{1}-P-Q\). In the decomposition of \(\mathcal{H}_d = \operatorname{im}P \oplus \operatorname{im}Q \oplus \operatorname{im}R\), consider the rotation \[\begin{align} \label{eq:rotation32between32P32perp32Q} R_t:= \begin{bmatrix} (\cos t)P & -(\sin t)V^\ast & 0 \\ (\sin t) V & (\cos t) Q & 0 \\ 0 & 0 & R \end{bmatrix}. \end{align}\tag{35}\] Then the homotopy \[\begin{align} R_t \begin{bmatrix} PUP & 0 & 0 \\ 0 & Q & 0 \\ 0 & 0 & R \end{bmatrix} R_t^\ast \end{align}\] deforms \(PUP+P^\perp\) to \(QVUV^\ast Q + Q^\perp=:W\) in \(\mathcal{U}(\mathcal{L}_d)\) for \(t\in [0,\pi/2]\). ◻

Proof. Here, we will mainly use the alternative characterization of bulk non-triviality described by 37.

Let \(P,Q\in\mathcal{P}(\mathcal{L}_d)\) and suppose \(P\) is bulk-non-trivial and \(P\sim_h Q\). It follows that \(P\sim_u Q\), i.e., there exists \(U\in\mathcal{U}(\mathcal{L}_d)\) such that \(P=U^\ast QU\). Let \(a\in C(\mathbb{S}^{d-1})\) be such that \(Q\rho(a)Q\in\mathcal{K}(\mathcal{H}_d)\). Then \[\begin{align} P\rho(a)P = U^\ast QU\rho(a)U^\ast QU = U^\ast Q[U,\rho(a)]U^\ast QU + U^\ast Q\rho(A)Q U\in\mathcal{K}(\mathcal{H}_d) \end{align}\] where we use \([U,\rho(a)]\in\mathcal{K}(\mathcal{H}_d)\). Since \(P\in\mathcal{P}^\mathrm{nt}(\mathcal{L}_d)\), it follows that \(a=0\). Now \(P\sim_h Q\) implies \(P^\perp\sim_h Q^\perp\). Follow the same argument as before, we have that \(Q^\perp\rho(a)Q^\perp\) is not compact unless \(a=0\). Thus \(Q\in\mathcal{P}^\mathrm{nt}(\mathcal{L}_d)\). ◻

Proof. Let \(F\subset\mathbb{Z}^d\) be spherically-proper. Let \(\Set{I_n}\) be a countable basis for \(\mathbb{S}^{d-1}\). It suffices to show spherical properness with respect to the collection of (overlapping) cones \(\Set{C_{I_n}}\). Let \(A_n:= F\cap C_{I_n}\). The goal is to distribute points in \(A_n\) to two disjoint sets \(F_1\) and \(F_2\) where each \(A_n\) contributes infinitely many points to both sides. To that end, consider \(\Set{(n_i,k_i)}_{i=1}^\infty\) an enumeration of \(\mathbb{N}\times \mathbb{N}\). For \(i=1\), pick \(x_1\in A_{n_1}\) for \(F_1\), and pick \(y_1\in A_{n_1}\setminus\Set{x_1}\) for \(F_2\). For \(i=2\), pick \(x_2\in A_{n_2}\setminus\Set{x_1,y_1}\) for \(F_1\) and \(y_2\in A_{n_2}\setminus \Set{x_1,y_1,x_2}\) for \(F_2\), and so on. Since \(A_n\) contains infinitely many points by spherical properness of \(F\), at each step \(i\), the set \(A_{n_i}\) is nonempty after excluding finitely many points, and hence the algorithm is well-defined. Since we exclude previously chosen points, the sets \(F_1\) and \(F_2\) are disjoint. Moreover, for each \(n\in\mathbb{N}\), points in \(A_n\) are chosen infinitely many times due to the enumeration of \(\mathbb{N}\times \mathbb{N}\), i.e., the set \(\Set{(n_i,k_i) | i\in \mathbb{N},\;n_i=n}\) is an infinite set. Therefore, \(|F_1\cap A_n|\) and \(|F_2\cap A_n|\) are both infinite for all \(n\in\mathbb{N}\). Finally, we toss all the remaining points in \(F\) that are not selected in any steps to \(F_1\) to make \(F_1\) and \(F_2\) a partition of \(F\).

Since \(|F^c\cap C_I|=\infty\) for all non-empty open subsets \(I\subset\mathbb{S}^{d-1}\) and \(F_1,F_2\) are both subsets of \(F\), it follows that \(|F_i^c\cap C_I|=\infty\) as well for \(i\in\Set{1,2}\). ◻

Proof. We would have liked to define \(B\) as \[\begin{align} A - \sum_{i} P_i A Q_i\,. \end{align}\] However, this formula may fail to represent the operator we want since the range of the projections \(P_i,P_j\) or \(Q_i,Q_j\) may overlap, which would mean we over-delete elements. To remedy this problem, inspired by the inclusion–exclusion formula, we define \[\begin{align} S_n&:=\sum_{i=1}^n P_i AQ_i - \sum_{1\leq i<j\leq n}P_iP_j A Q_iQ_j \notag \\ &\quad + \sum_{1\leq i<j<k\leq n}P_iP_jP_k A Q_iQ_jQ_k - \dots + (-1)^{n-1}P_1\dots P_n A Q_1\dots Q_n \notag \\ &= \sum_{i=1} (-1)^{i+1}\left(\sum_{1\leq k_1<\dots<k_i\leq n} P_{k_1}\dots P_{k_i} A Q_{k_1}\dots Q_{k_i}\right) \label{eq:inclusion32exclusion} \end{align}\tag{36}\] which corrects all the over-counting. More precisely, let \(\varepsilon_k:=\|P_k AQ_k\|\). Then \[\begin{align} \label{eq:norm32bound32on32S95n} \|S_n\|\leq \sum_{k=1}^n 2^{k-1}\varepsilon_k\,. \end{align}\tag{37}\] For example, we have \(\|S_1\|=\|P_1 AQ_1\|=\varepsilon_1\), and \[\begin{align} \|S_2\|=\|P_1AQ_1+P_2 AQ_2-P_1P_2AQ_1Q_2\| \leq \varepsilon_1+2\varepsilon_2 \end{align}\] where we used \(\|P_1P_2AQ_1Q_2\|\leq \|P_2AQ_2\|=\varepsilon_2\). Fix \(l\), we count the number of terms \(P_{k_1}\dots P_{k_l}AQ_{k_1}\dots Q_{k_l}\) in \(S_n\) with \(k_1<\dots <k_l\) having \(k_l=m\). Then use the fact that \[\begin{align} \|P_{k_1}\dots P_{k_l}AQ_{k_1}\dots Q_{k_l}\|\leq \|P_{k_l}AQ_{k_l}\|=\varepsilon_m \,. \end{align}\] There are \(2^{m-1}\) number of terms of that form. Let \(S=\lim_{n\to\infty} S_n\). We need to show that the limit exists. To that end, for \(m>n\), consider \(S_m-S_n\). Using the formula 36 and idea leading to upper bound 37 , all the terms \(P_{k_1}\dots P_{k_l}AQ_{k_1}\dots Q_{k_l}\) in \(S_m-S_n\) will have \(k_l\geq n+1\). Using ?? , we have \[\begin{align} \|S_m-S_n\| &\leq 2^n\varepsilon_{n+1}+2^{n+1}\varepsilon_{n+2}+\dots + 2^{m-1}\varepsilon_m \\ &\leq \frac{\varepsilon}{2^{{n+1}}} +\dots + \frac{\varepsilon}{2^m} \\ &\leq \varepsilon\sum_{k=n+1}^\infty \frac{1}{2^k} \end{align}\] which converges to \(0\) as \(n\to\infty\) (independent of \(m\)).

We have \(\|S\|\leq \varepsilon\). Indeed, from 37 and ?? , we have \[\begin{align} \|S\|\leq \sum_{k=1}^\infty 2^{k-1}\frac{1}{2^{k-1}}\frac{\varepsilon}{2^k} \leq \varepsilon\,. \end{align}\] Define \[\begin{align} B = A-S \,. \end{align}\] Let us now show that \[\begin{align} P_k B Q_k = 0,\;\forall k\in\mathbb{N} \end{align}\] to prove that \(B\) is indeed unaffected by the interactions originally present in \(P_kAQ_k\). We show that \(P_kS_nQ_k=P_kAQ_k\) for all \(n\geq k\) by induction. From 36 , we have the recursion relation \[\begin{align} S_{n+1} = S_n + P_{n+1}AQ_{n+1} - P_{n+1}S_nQ_{n+1} \label{eq:recursion32relation} \end{align}\tag{38}\] where \(P_{n+1}AQ_{n+1}\) is from the first sum in 36 , and \(P_{n+1}S_nQ_{n+1}\) is from the rest of sums. Let \(k\geq 1\) be arbitrary. It holds that \(P_1S_1Q_1=P_1AQ_1\). Take \(n=k\) in 38 and consider \[\begin{align} P_{k+1} S_{k+1} Q_{k+1} &= P_{k+1} S_kQ_{k+1} + P_{k+1}P_{k+1}AQ_{k+1}Q_{k+1} - P_{k+1}P_{k+1}S_kQ_{k+1}Q_{k+1} \\ &= P_{k+1}AQ_{k+1} \,. \end{align}\] Thus \(P_kS_kQ_k=P_kAQ_k\) holds for all \(k\geq 1\). Suppose \(P_kS_{n}Q_k=P_kAQ_k\) holds. Using 36 , we have \[\begin{align} P_kS_{n+1}Q_k &= P_kS_nQ_k + P_kP_{n+1}AQ_{n+1}Q_k - P_kP_{n+1}S_nQ_{n+1}Q_k \\ &=P_kS_nQ_k + P_{n+1}P_kAQ_kQ_{n+1} - P_{n+1}P_kS_nQ_kQ_{n+1} \\ &= P_kAQ_k \end{align}\] where in the last equality we used the induction assumption.

Since \(P_kAQ_k\) is compact, it follows that \(S_n\) in 36 is compact, and that its norm limit \(S\) is compact. Therefore \(A-B=S\) is compact. ◻

Proof. Let us prove the case for class \(\mathrm{DIII}\) and leave the rest for the reader. We claim that there exists a hyper-spherically-local unitary operator \(U\) on \(\mathcal{H}_d\otimes \mathbb{C}^N\) that commutes with \(\Pi\) and satisfies \(U^\ast\Theta U=-i\sigma_y\mathcal{C}\).

Let \(\mathcal{V}_{\pm} = \mathcal{H}_d\otimes \mathbb{C}^W\) be the positive and negative chiral spaces. Since \(\Theta\) anti-commutes with \(\Pi\), it follows that \(\Theta\) maps anti-unitarily \(\mathcal{V}_\pm\) to \(\mathcal{V}_\mp\). Let \(\Lambda^\pm_I\) be as in 1 by on the positive and negative chiral subspaces. For each \(z\in \mathbb{S}^{d-1}_\mathbb{Z}\), pick an orthonormal basis \(\{\varphi^+_{j,z}\}_{j\in\mathbb{N}}\) for \(\operatorname{im}\Lambda_{\{z\}}^+\). Since \(\mathcal{V}_+=\oplus_{z\in\mathbb{S}^{d-1}_\mathbb{Z}} \operatorname{im}\Lambda_{\{z\}}^+\), the set \(\{\varphi_{j,z}\}_{j\in\mathbb{N},z\in\mathbb{S}^{d-1}_\mathbb{Z}}\) forms an orthonormal basis for \(\mathcal{V}_+\). Define \[\begin{align} \varphi^-_{j,z}:=\Theta \varphi^+_{j,z}. \end{align}\] We have \(\Theta \varphi^-_{j,z} = -\varphi^+_{j,z}\). Since \(\Theta\) is hyper-spherically-local, by 59, it follows that \(\varphi^-_{j,z}\in \operatorname{im}\Lambda_{\{z\}}^-\). Let \(\{\eta_{j,z}^{\pm}\}_{j\in\mathbb{N}}\) be an enumeration of the standard basis (those of the form \(\delta_x\otimes e_i\)) in \(\operatorname{im}\Lambda_{\{z\}}^\pm\). Let \(U\) be the unitary operator that sends the standard basis \(\eta_{j,z}^\pm\) in \(\operatorname{im}\Lambda_{\{z\}}^\pm\) to \(\varphi^{\pm}_{j,z}\) for the respective chirality and points \(z\) in \(\mathbb{S}^{d-1}_\mathbb{Z}\). By construction \(U\) is spherically-hyper-local. We have \(U\Pi=\Pi U\). Indeed, on each invariant subspaces \(\operatorname{im}\Lambda_{\{z\}}\) we have \[\begin{align} U\Pi \left(\sum_j\alpha_j\eta_{j,z}^+ + \sum_j\beta_j\eta_{j,z}^-\right) &= U\left(\sum_j\alpha_j\eta_{j,z}^+ - \sum_j\beta_j\eta_{j,z}^-\right) \\ &= \left(\sum_j\alpha_j\varphi_{j,z}^+ - \sum_j\beta_j\varphi_{j,z}^-\right) \\ &= \Pi U \left(\sum_j\alpha_j\eta_{j,z}^+ + \sum_j\beta_j\eta_{j,z}^-\right). \end{align}\] Let us verify \(U^\ast\Theta U=-i\sigma_y\mathcal{C}\). We have \[\begin{align} \Theta U \left(\sum_j\alpha_j\eta_{j,z}^+ + \sum_j\beta_j\eta_{j,z}^-\right) & = \Theta \left(\sum_j\alpha_j\varphi_{j,z}^+ + \sum_j\beta_j\varphi_{j,z}^-\right) \\ &= \left(\sum_j\bar{\alpha}_j\varphi_{j,z}^- - \sum_j\bar{\beta}_j\varphi_{j,z}^+\right) \\ &= U\left(\sum_j-\bar{\beta}_j\eta_{j,z}^+ + \sum_j\bar{\alpha}_j\eta_{j,z}^-\right) \\ &= U (-i\sigma_y\mathcal{C}) \left(\sum_j\alpha_j\eta_{j,z}^+ + \sum_j\beta_j\eta_{j,z}^-\right) . \end{align}\] ◻

Proof. This follows from 18 and 19 . Indeed, if \(A\) is hyper-spherically-local, then \([A,\rho(f)]=0\) for all \(f\in C(\mathbb{S}^{d-1})\) where \(\rho:C(\mathbb{S}^{d-1})\to \mathcal{X}_{d,N}\) is the isomorphism ?? and \(\mathcal{X}_{d,N}\) is the \(C^\ast\)-algebra generated by \(\widehat{X}_1\otimes \mathbb{1}_N,\dots,\widehat{X}_d\otimes \mathbb{1}_N\). In particular, we have \([A,\rho(\lambda\mapsto\lambda)]=[A,\int\lambda dE(\lambda)]=0\) where we use 18 . It follows from [59] that \([A,E(I)]=0\) for all measurable subsets \(I\subset \mathbb{S}^{d-1}\). Using 19 , it follows that \([A,\Lambda_I]=0\). The converse follows from [59]. ◻

Proof of 61. The result follows from using the canonical form 58 and the representations of real Clifford algebra in 6, and the re-dimerization 22. We show for the case of DIII and leave the rest for the reader. Let the internal degrees of freedom \(N=2W\) be even, let \(\Pi\) be the 39 , and let \(\Theta\) be any anti-linear operator such that it is hyper-spherically-local, \(\Theta^2=-\mathbb{1}\) and \([\Theta,\Pi]=0\). We are interested in the space \[\begin{align} \mathcal{AZ}_{d,N}^{\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}} = \Set{H\in\mathcal{SU}(\mathcal{L}_{d,N}) | \{H,\Pi\}=0,\, [H,\Theta]=0}. \end{align}\] Using 58, we can choose \(\Theta\) to be \[\begin{align} \mathcal{C}\otimes \begin{bmatrix} 0 & -\mathcal{C}_W \\ \mathcal{C}_W & 0 \end{bmatrix} =: -i\sigma_y\mathcal{C} \end{align}\] with respect to the chiral grading. Using the choice of \(\Theta= -i\sigma_y \mathcal{C}\) and re-dimerization, we have the homeomorphism \[\begin{align} \label{eq:canonical32space32DIII} \mathcal{AZ}_{d,N}^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}\cong \Set{U\in\mathcal{U}(\mathcal{L}_d) | U^\ast=-\overline{U}}=:\mathcal{U}^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}_d . \end{align}\tag{41}\] Indeed, if \(H\) is self-adjoint, unitary and anti-commutes with \(\Pi\), then \(H\) is of the form \[\begin{align} H=\begin{bmatrix} 0 & U^\ast \\ U & 0 \end{bmatrix} \end{align}\] for \(U\in\mathcal{U}(\mathcal{L}_{d,W})\). Using \([H,\Theta]=0\), it follows that \(U^\ast = -\overline{U}\). Consider the re-dimerization unitary \(R:\mathcal{H}_d\to \mathcal{H}_{d,W}\) in 22. Since it maps between standard bases, we have \(\overline{R}=R\). If \(U\in\mathcal{U}(\mathcal{L}_{d,N})\) satisfies \(U^\ast=-\overline{U}\), then \(R^\ast U R\in\mathcal{U}(\mathcal{L}_d)\) and \[\begin{align} (R^\ast U R)^\ast = R^\ast U^\ast R = R^\ast (-\overline{U})R = -\overline{R^\ast U R}. \end{align}\] where we use \(\overline{R}=R\) in the last equality. This establishes 41 . The homeomorphism 41 also implies that we have the homeomorphism \(\mathcal{AZ}_{d,N}^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}\cong \mathcal{AZ}_{d,M}^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}\) for all \(M,N\in 2\mathbb{N}\). On the other hand, using the representations of Clifford algebra in 6, we have \[\label{eq:diii32SUo} \mathcal{SU}_o(\mathcal{L}_{d}^\mathbb{R}\otimes \mathrm{C\ell}_{0,2}) = \Set{\begin{bmatrix} 0 & B \\ -\overline{B} & 0 \end{bmatrix} \in \mathcal{SU}(\mathcal{L}_{d}^\mathbb{R}\otimes M_2(\mathbb{C}))}\tag{42}\] which is homeomorphic to \(\Set{B\in \mathcal{U}(\mathcal{L}_{d}) | B^\ast=-\overline{B}} = \mathcal{U}_d^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}\), exactly 41 . ◻

Proof. We have \[\begin{align} D\kern-0.1emK_n(\mathcal{L}_d^\mathbb{R}) \cong D\kern-0.1emK_n(\mathcal{D}(C(\mathbb{S}^{d-1},\mathbb{R}))) \cong D\kern-0.1emK_n(\mathcal{D}(\widetilde{C_0(\mathbb{R}^{d-1}))}) \cong K\kern-0.1emK\kern-0.1emO_{n-1}(C_0(\mathbb{R}^{d-1}),\mathbb{R}) \end{align}\] where the first isomorphism uses 14 and the second one identifies the unitization of \(C_0(\mathbb{R}^{d-1},\mathbb{R})\) as \(C(\mathbb{S}^{d-1},\mathbb{R})\), and the third isomorphism uses [44] that identified van Daele’s \(K\)-theory of the dual of an algebra with Kasparov’s \(K\)-homology in the real case. The rest of the calculation is standard \[\begin{align} K\kern-0.1emK\kern-0.1emO_{n-1}(C_0(\mathbb{R}^{d-1}),\mathbb{R}) &\cong K\kern-0.1emK\kern-0.1emO_{n-1-(d-1)}(\mathbb{R},\mathbb{R})\\ &\cong K\kern-0.1emO_{n-d}(\mathbb{R})\cong \begin{cases} \mathbb{Z}& n-d\equiv 0,4 \ (\mathrm{mod}\ 8) \\ \mathbb{Z}_2 & n-d\equiv 1,2\ (\mathrm{mod}\ 8) \\ 0 & n-d\equiv 3,5,6,7 \ (\mathrm{mod}\ 8) \end{cases} \end{align}\] where the first isomorphism uses [63] and the second isomorphism uses [63] that relates \(KK\)-group to the \(K\)-group of real \(C^\ast\)-algebra, and the last isomorphism can be found on [63]. ◻

Proof. We have \[\begin{align} (VA\overline{V}^{-1})^{-1} = \overline{V} A^{-1} V^{-1} = -\overline{V}\mathcal{C}A \mathcal{C}V^{-1} = -\mathcal{C}VA\overline{V}^{-1}\mathcal{C}. \end{align}\] ◻

Proof. Let \(B=-\overline{A}A\). Using holomorphic functional calculus, we have \[\begin{align} \overline{B^{-1/2}} = (\overline{B})^{-1/2} = (-A\overline{A})^{-1/2} = (ABA^{-1})^{-1/2} = AB^{-1/2}A^{-1}. \end{align}\] Then \[\begin{align} -\overline{\Psi(A)}\Psi(A) = \overline{AB^{-1/2}}AB^{-1/2} = -\overline{A} (AB^{-1/2}A^{-1}) AB^{-1/2} = -\overline{A} A B^{-1} = \mathbb{1}. \end{align}\] ◻

Proof. Write \(A\delta_x = \alpha\delta_x + w_x\) for \(w_x\perp \delta_x\) and \(\alpha=\braket{\delta_x,A\delta_x}\). We claim that \[\begin{align} \mathrm{dist}(A\delta_x,\mathbb{C}\delta_x) = \|w_x\|\geq \frac{1}{\|A\|} \end{align}\] for all \(x\in \mathbb{Z}^d\). We have \[\begin{align} A(\mathcal{C}A\delta_x) = A\mathcal{C}(\alpha\delta_x + w_x) = \bar{\alpha}A\delta_x + A\mathcal{C}w_x. \end{align}\] Then \[\begin{align} \braket{\delta_x,A(\mathcal{C}A\delta_x)} = \braket{\delta_x,\bar{\alpha}A\delta_x} + \braket{\delta_x,A\mathcal{C}w_x} = |\alpha|^2 + \braket{\delta_x,A\mathcal{C}w_x}. \end{align}\] On the other hand, using \(-A\overline{A}\delta_x=\delta_x\), we have \[\begin{align} A (\mathcal{C}A\delta_x) = A\overline{A}\delta_x =-\delta_x. \end{align}\] Thus \(\braket{\delta_x,A (\mathcal{C}A\delta_x)}=-1\) as well. This gives \[\begin{align} 1 + |\alpha|^2 = |\braket{\delta_x,A\mathcal{C}w_x}|\leq \|A\|\|w_x\| \end{align}\] and hence the bound ?? . ◻

Proof. Let \(A\in\mathcal{G}^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}_d\) and let \(\varepsilon>0\) be small to be determined. We use 51 to perturb \(A\) by at most \(\varepsilon\) to a spherically-local \(G\) with \(A-G\in\mathcal{K}(\mathcal{H}_d)\) and produce a spherically-proper sequence of points \(\{x_k\}_{k\in\mathbb{N}}\) with disjoint finite subsets \(R_k:=\operatorname{supp}(G\delta_{x_k})\cup \{x_k\}\) satisfying the cone-separation property. In particular, we may enlarge the island \(R_k\) \[\begin{align} R_k = \operatorname{supp}(G\delta_{x_k}) \cup \{x_k\} \cup \{y_k\} \end{align}\] by adding a nearest-neighbor point \(y_k\) to \(x_k\), while keeping spherical properness of \(\cup_k\{x_k,y_k\}\) and the properties \(R_k\) satisfies.

In general, \(G\notin \mathcal{G}_d^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}\) and \(-G\overline{G}=\mathbb{1}\) does not hold. Nonetheless, we can recover it locally along the sequence \(\{x_k\}_{k\in\mathbb{N}}\), while keeping the separation structure of \(R_k\) and the invertibility and locality of operator. Let \(x\in\mathbb{Z}^d\). Consider the defect operator \(D:=G\overline{G}+\mathbb{1}\). Since \(A-G\) is compact and \(A\overline{A}+\mathbb{1}=0\), it follows that \(D\) is compact. Consider the defect vector \(r_x:= D\delta_x\). Since \(\delta_x\) converges to zero weakly and \(D\) is compact, the sequence \(r_x\) converges in norm to zero as \(\|x\|\to \infty\). Consider the defect vector along the spherically-proper sequence \(\{x_k\}_{k\in\mathbb{N}}\). Since \(r_{x_k}\to 0\), pick a subsequence, still spherically-proper, such that \[\begin{align} \|r_{x_k}\| \leq \varepsilon 2^{-k}. \end{align}\] We can do this without losing spherical properness by a diagonal selection argument over a countable basis \(\{I_n\}\) of open sets on \(\mathbb{S}^{d-1}\): enumerate pairs \((n,k)\in \mathbb{N}^2\), and at stage \((n,k)\) choose a point in the cone \(C_{I_n}\) far enough so that \(\|r_{x_k}\| \leq \varepsilon 2^{-k}\), and also not previously used. Re-index to a single sequence. The inherited \(R_{x_k}\) remain disjoint and retain cone-separation because we only pass to a subsequence. So, after relabeling, assume \[\begin{align} \sum_{k=1}^\infty \|r_{x_k}\|^2 \leq \frac{\varepsilon^2}{3} . \end{align}\] For \(x\in\mathbb{Z}^d\), let \(\eta_x :=\Lambda^\perp_{\{x\}} \overline{G}\delta_x\). Using 67, we have \[\begin{align} \|\eta_x\|=\mathrm{dist}(\overline{G}\delta_x,\mathbb{C}\delta_x) = \mathrm{dist}(G\delta_x,\mathbb{C}\delta_x)\geq \mathrm{dist}(A\delta_x,\mathbb{C}\delta_x)-\varepsilon\geq \frac{1}{2\|A\|}. \end{align}\] where we use \(\|A-G\|\leq \varepsilon\) and choose \(\varepsilon\) to be smaller than \(1/(2\|A\|)\). Define the rank-one operator \[\begin{align} K_x = -\frac{1}{\|\eta_x\|^2} r_x\otimes \eta_x^\ast. \end{align}\] Then \(K_x\delta_x =0\) since \(\eta_x\perp \delta_x\), and \[\begin{align} K_x \overline{G}\delta_x = -r_x\left\langle\frac{\eta_x}{\|\eta_x\|^2},\overline{G}\delta_x\right\rangle = -r_x\left\langle\frac{\eta_x}{\|\eta_x\|^2},\Lambda_{\{x\}} \overline{G}\delta_x + \eta_x\right\rangle = -r_x. \end{align}\] Now set \(K:=\sum_{k=1}^\infty K_{x_k}\) along the spherically-proper sequence \(\{x_k\}_{k\in\mathbb{N}}\). The series converges in Hilbert-Schmidt norm. Indeed, we have \[\begin{align} \left\lVert\sum_{k=m}^n K_{x_k}\right\rVert_{\mathrm{HS}} = \sum_{j,k=m}^n \braket{r_{x_j},r_{x_k}} \left\langle\frac{\eta_{x_k}}{\|\eta_{x_k}\|^2},\frac{\eta_{x_j}}{\|\eta_{x_j}\|^2}\right\rangle = \sum_{k=m}^n \frac{\|r_{x_k}\|^2}{\|\eta_{x_k}\|^2}\leq \sum_{k=1}^\infty \frac{\|r_{x_k}\|^2}{\|\eta_{x_k}\|^2} \leq \frac{4}{3}\|A\|^2\varepsilon^2 \end{align}\] where in the first equality, we have crucially used the fact that \(\{\eta_{x_k}\}_{k\in\mathbb{N}}\) are pairwise orthogonal. Indeed, since \(\operatorname{supp}(G\delta_{x_k})\subset R_k\) and hence \(\operatorname{supp}(\eta_{x_k})\subset R_k\), this follows from the disjointness of \(\{R_k\}_{k\in\mathbb{N}}\). Then \[\begin{align} \|K\|\leq \|K\|_\mathrm{HS} \leq \frac{2 \|A\|}{\sqrt{3}}\varepsilon. \end{align}\]

Let \(T:=G+K\). Then \(A-T=(A-G)-K\) is compact, and \(\|A-T\|\leq C_A\varepsilon\) where \(C_A=1+2\|A\|/\sqrt{3}\) if we choose \(\varepsilon\leq 1/(2\|A\|)\). We now verify ?? : \[\begin{align} -T\overline{T}\delta_{x_k} = -T\mathcal{C}(G+K)\delta_{x_k} = -T\overline{G}\delta_{x_k} = -(G+K)\overline{G}\delta_{x_k} = -G\overline{G}\delta_{x_k} + r_{x_k} = \delta_{x_k}. \end{align}\] ◻

Proof. Let \(A\in\mathcal{G}^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}_d\) and let \(\varepsilon>0\) be small to be determined. We use 68 to perturb \(A\) by \(\varepsilon\) to a \(T\in\mathcal{G}(\mathcal{L}_d)\) and produce a spherically-proper sequence of nearest-neighbor dimers \(\{x_k,y_k\}_{k\in\mathbb{N}}\) with the relation ?? and a sequence of disjoint finite subsets \(R_k:=\operatorname{supp}(T\delta_{x_k})\cup \{x_k\}\cup \{y_k\}\) satisfying the cone-separation property. Let \(P\) be \(\Lambda_{\cup_k\{x_k,y_k\}}\).

We construct a pinning operator \(V\in\mathcal{G}(\mathcal{L}_d)\) such that \(VT\overline{V}^{-1}\) agrees with the dimer operator \(E\) on \(\operatorname{im}P\). On the finite-dimensional space \(\operatorname{im}\Lambda_{R_k}\), define \(V_k\) to be any invertible operator that acts as \[\begin{align} V_k\delta_{x_k}= \delta_{x_k} ,\quad V_k (T\delta_{x_k})= \delta_{y_k}. \end{align}\] Define \(V:= \oplus_k V_k \oplus \mathbb{1}_{(\oplus_k\operatorname{im}\Lambda_{R_k})^\perp} \in\mathcal{G}(\mathcal{L}_d)\). We now verify that \(V T\overline{V}^{-1}\) agrees with \(E\) on \(\operatorname{im}P\). Indeed, we have \[\begin{align} VT\overline{V}^{-1} \delta_{x_k} = VT\mathcal{C}V^{-1}\delta_{x_k} = VT\delta_{x_k} = \delta_{y_k} \end{align}\] and \[\begin{align} VT\overline{V}^{-1} \delta_{y_k} =VT\mathcal{C}V^{-1}\delta_{y_k} = VT\mathcal{C}T\delta_{x_k} = VT\overline{T}\delta_{x_k}=-V\delta_{x_k} = -\delta_{x_k} \end{align}\] where in the second to last equality we use ?? . Let \(V\) be identity on \((\oplus \operatorname{im}\Lambda_{R_k})^\perp\). Using argument exactly the same as in 43, the operator \(V\) is spherically-local and we can construct a homotopy \(V\sim_h\mathbb{1}\) in \(\mathcal{G}(\mathcal{L}_d)\).

Since \(\|A-T\|\leq\varepsilon\) can be made arbitrarily small, and \(A\in\mathcal{G}_d^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}\), it follows that \(\sigma(-\overline{T}T)\) lies in a neighborhood of \(1\) and avoids \((-\infty,0]\). Therefore, we can perform symmetrization ?? on the straight-line homotopy \(t\mapsto (1-t)A+tT\) and get \(A\sim_h \Psi(T)\) in \(\mathcal{G}_d^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}\). Using the previously constructed homotopy \(V\sim_h \mathbb{1}\) in \(\mathcal{G}_d^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}\) and 65, we get \(\Psi(T)\sim_h V\Psi(T)\overline{V}^{-1}=\Psi(VT\overline{V}^{-1})\) in \(\mathcal{G}_d^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}\), where the last equality follows from 66. Let \(S:=\Psi(VT\overline{V}^{-1})\). We argue that \(S\) agrees with \(E\) on \(\operatorname{im}P\). Indeed, let \(Z:=VT\overline{V}^{-1}\). Since \(Z\) agrees with \(E\) on \(\operatorname{im}P\), then \(-\overline{Z}Z=\mathbb{1}\) on \(\operatorname{im}P\), and so is \((-\overline{Z}Z)^{-1/2}\). Thus \(S=\Psi(Z)=Z(-\overline{Z}Z)^{-1/2}\) agrees with \(E\) on \(\operatorname{im}P\).

With respect to the decomposition \(\operatorname{im}P\oplus \operatorname{im}P^\perp\), we may then write \[\begin{align} S = \begin{bmatrix} E & X \\ 0 & B \end{bmatrix} \end{align}\] where \(E,X,B\), viewed on \(\mathcal{H}_d\), are \(PSP,PSP^\perp,P^\perp SP^\perp\). Using \(S\overline{S}=-\mathbb{1}\), we have the relations \(E\overline{X}+X\overline{B}=0\) and \(B\overline{B}=-\mathbb{1}_{\operatorname{im}P^\perp}\). Let \[\begin{align} W_t:= \begin{bmatrix} \mathbb{1}& t X\overline{B} \\0 & \mathbb{1} \end{bmatrix}. \end{align}\] Then \(W_t\in\mathcal{G}(\mathcal{L}_d)\) and we have \[\begin{align} W_tS\overline{W}_t^{-1} = \begin{bmatrix} E & -tE\overline{X}B + X + tX\overline{B}B \\ 0 & B \end{bmatrix}. \end{align}\] Setting \(t=1/2\), the off-diagonal part vanishes \[\begin{align} -\frac{1}{2}E\overline{X}B + X + \frac{1}{2}X\overline{B}B = X + X\overline{B}B = 0. \end{align}\] Thus we obtain \(S\sim_h PEP+P^\perp S P^\perp\). At least point we may conclude the proposition by swapping \(P\) and \(P^\perp\). Alternatively, to keep the chosen sequence \(\{x_k\}_{k\in\mathbb{N}}\) as support, we may use the rotation \(R_t\) in 35 to get \(PEP+P^\perp S P^\perp\sim_h PSP+P^\perp EP^\perp\) in \(\mathcal{G}_d^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}\). Indeed, the rotation \(R_t\) is real \(R_t=\overline{R}_t\), and hence the congruence homotopy stays within \(\mathcal{G}_d^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}\) by 65. ◻

Proof. Let \(\{x_k,y_k\}_{k\in\mathbb{N}}\) and \(\{\tilde{x}_k,\tilde{y}_k\}_{k\in\mathbb{N}}\) be the nearest-neighbor pairs for \(P=\Lambda_{\cup_k \{x_k,y_k\}}\) and \(Q=\Lambda_{\cup_k\{\tilde{x}_k,\tilde{y}_k\}}\). Analogously to 42, we can pick a bijection \(f:\mathbb{N}\to\mathbb{N}\) between the sequence of dimerization, such that if we define \[\begin{align} V\delta_{x_k} = \delta_{\tilde{x}_{f(k)}},\quad V\delta_{y_k}=\delta_{\tilde{y}_{f(k)}} \end{align}\] then \(V\) is spherically-local. One readily verifies that \(VE=EV\). ◻

Proof. Let \(P_0:=P\). Since \(P_0\) is spherically-proper, so is \(P_0^\perp\). Using 55, we may decompose \(P_0^\perp\) into \(n+1\) spherically-proper projections \[\begin{align} P_0^\perp = Q_0+ P_1 + \dots + P_{n-1} + P_n. \end{align}\] Using 70, we have \(Q_0\sim P_0^\perp\) and \(P_k\sim \mathbb{1}\) for all \(k=1,\dots,n\), implemented by real, spherically-local partial isometries \(V_0,V_1,\dots,V_n\) such that \(P_0^\perp = V_0^\ast V_0\), \(Q_0=V_0V_0^\ast\), and \(P_k=V_kV_k^\ast\) and \(\mathbb{1}= V_k^\ast V_k\) for \(k=1,\dots,n\). Moreover, \(V_kE=EV_k\) for \(k=0,\dots,n\). Consider \[\begin{align} V:= [P_0+V_0,V_1,\dots,V_n]\in M_{1,n+1}(\mathcal{L}_d^\mathbb{R}). \end{align}\] Direct computation shows that \(V^\ast V=\mathbb{1}_{n+1}\) and \(VV^\ast = \mathbb{1}\) and \(VE_{n+1}=EV\).

Let \(W_t\) be the path for \(A\oplus E_{n}\sim_h B\oplus E_n\) in \(\mathcal{G}_{d,n+1}^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}\). Then \(VW_t\overline{V}^{-1} = VW_tV^\ast\) gives the path \(A\sim_h B\) in \(\mathcal{G}_d^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}\). ◻

Proof of 62 for class \(\mathrm{DIII}\). Using 61 and 63, it suffices to show that the map \[\begin{align} \label{eq:diii32pi032to32K32group32map} \pi_0(\mathcal{SU}_o(\mathcal{L}_d^\mathbb{R}\otimes \mathrm{C\ell}_{0,2})) \to D\kern-0.1emK_e(\mathcal{L}_d^\mathbb{R}\otimes \mathrm{C\ell}_{0,2}) \end{align}\tag{45}\] which sends path-connected component to its van Daele \(K\)-group, is a bijection. With specific representation of \(\mathrm{C\ell}_{0,2}\), the set \(\mathcal{SU}_o(\mathcal{L}_d^\mathbb{R}\otimes \mathrm{C\ell}_{0,2})\) takes the form in 42 . Here, we choose \(e\) to be the element \[\begin{align} e:=\begin{bmatrix} 0 & E \\ -E & 0 \end{bmatrix} \end{align}\] where \(E\) is the nearest-neighbor dimer operator. For \(D\kern-0.1emK_e(\mathcal{L}_d^\mathbb{R}\otimes \mathrm{C\ell}_{0,2})\) to make sense, we need to show that \(e\sim_h-e\) in \(\mathcal{SU}_o(\mathcal{L}_d^\mathbb{R}\otimes \mathrm{C\ell}_{0,2})\). To that end, for each dimer \((x,y)\), with respect to \(\mathbb{C}\delta_x\oplus \mathbb{C}\delta_y\), we consider the rotation \[\begin{align} \begin{bmatrix} \cos t & -\sin t \\ \sin t & \cos t \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} \cos t & -\sin t \\ \sin t & \cos t \end{bmatrix} \end{align}\] for \(t\in [0,\pi/2]\). Apply the rotation to each dimer, the homotopy stays spherically-local. We get \(E\sim_h -E\) in \(\mathcal{U}_d^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}\) and hence, using 42 , we have \(e\sim_h -e\).

We show that if \(U,V\in\mathcal{SU}_o(\mathcal{L}_d^\mathbb{R}\otimes \mathrm{C\ell}_{0,2})\) has the same \([U]_e=[V]_e\) van Daele \(K\)-group, then \(U\sim_h V\) in \(\mathcal{SU}_o(\mathcal{L}_d^\mathbb{R}\otimes \mathrm{C\ell}_{0,2})\). Write \(U=\begin{bmatrix} 0 & X \\ -\overline{X} & 0 \end{bmatrix}\) and \(V = \begin{bmatrix} 0 & Y \\ -\overline{Y} & 0 \end{bmatrix}\) for \(X,Y\in\mathcal{U}_d^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}\). Using 69, there exist spherically-proper projections \(P,Q\) reducing \(E\) such that \(X\sim_h A=PAP+P^\perp EP^\perp\) and \(Y\sim_h \widetilde{B}=Q\widetilde{B}Q+Q^\perp EQ^\perp\) in \(\mathcal{G}_d^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}\). In fact, we may choose \(Q\) to be orthogonal to \(P\). Indeed, after constructing \(P\), we may construct \(Q\) within \(P^\perp\), where the construction of spherically-proper sequence of points traced back to 50. Using 35 , we then have \(Y\sim_h Q\widetilde{B}Q+Q^\perp EQ^\perp \sim_h B=PBP+P^\perp EP^\perp\) in \(\mathcal{G}_d^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}\), supported on the same \(P\) as in \(A\). Using 72 and 46 , there exists \(S,T\in\mathcal{U}_d^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}\) such that \(A\sim_h S=PSP+P^\perp EP^\perp\) and \(B\sim_h T=PTP+P^\perp EP^\perp\) in \(\mathcal{G}_d^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}\). Using 72 again, we get \(X\sim_h S\) and \(Y\sim_h T\) in \(\mathcal{U}_d^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}\).

Since \([U]_e=[V]_e\), it follows that \[\begin{align} \begin{bmatrix} 0 & S \\ -\overline{S} & 0 \end{bmatrix} \oplus \begin{bmatrix} 0 & E \\ -E & 0 \end{bmatrix}_n \sim_h \begin{bmatrix} 0 & T \\ -\overline{T} & 0 \end{bmatrix} \oplus \begin{bmatrix} 0 & E \\ -E & 0 \end{bmatrix}_n \end{align}\] in \(\mathcal{SU}_o(M_{n+1}(\mathcal{L}_d^\mathbb{R}\otimes \mathrm{C\ell}_{0,2}))\). Equivalently, under a conjugation by some elementary matrix, we have \[\begin{align} S\oplus E_n \sim_h T\oplus E_n \end{align}\] in \(\mathcal{U}_{d,n+1}^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}:= \Set{U\in\mathcal{U}_{n+1}(\mathcal{L}_d) | U^\ast = -\overline{U}}\). Now we use 71 and get \(S\sim_h T\) in \(\mathcal{G}_d^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}\), and then we use 72 to get \(S\sim_h T\) in \(\mathcal{U}_d^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}\). This provides the desired homotopy \(U\sim_h V\) in \(\mathcal{SU}_o(\mathcal{L}_d^\mathbb{R}\otimes \mathrm{C\ell}_{0,2})\).

We show that the map 45 is surjective. Let \(\xi\in D\kern-0.1emK_e(\mathcal{L}_d^\mathbb{R}\otimes \mathrm{C\ell}_{0,2})\). Then by construction there exists \(U\in\mathcal{SU}_o(M_n(\mathcal{L}_d^\mathbb{R}\otimes \mathrm{C\ell}_{0,2}))\) for some \(n\) such that \(\xi = [U]_e\). The goal is to find \(W\in \mathcal{SU}_o(\mathcal{L}_d^\mathbb{R}\otimes \mathrm{C\ell}_{0,2})\) such that \([W]_e=[U]_e\). Write \(U=\begin{bmatrix} 0 & Z \\ -\overline{Z} & 0 \end{bmatrix}\) for \(Z\in \mathcal{U}_{d,n}^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}\). Let \(\mathbb{Z}^d=\bigcup_{k\in\mathbb{N}}\{x_k,y_k\}\) be an enumeration of dimers. Let \(\mathbb{N}=I_1\cup I_2\cup \dots\cup I_n\) be a partition of index set in to \(n\) disjoint infinite subsets such that that for each \(j\), the associated set of lattice sites \(F_j:=\bigcup_{k\in I_j}\{x_k,y_k\}\) is spherically-proper. This is possible because every cone contains infinitely many dimers; distribute dimers among \(n\) bins so that each bin receives infinitely many dimers in every cone. Let \(P_j:=\Lambda_{F_j}\). Using 70, there are \(V_1,\dots,V_n\in\mathcal{L}_d^\mathbb{R}\) such that \(V_j^\ast V_j=\mathbb{1}\) and \(V_jV_j^\ast = P_j\) and \(V_jE=EV_j\). Let \(V:=(V_1,\dots,V_n)\in M_{1,n}(\mathcal{L}_d^\mathbb{R})\). Then \(VV^\ast = \mathbb{1}\) and \(V^\ast V=\mathbb{1}_n\) and \(VE_n=EV\). Consider \(VZV^\ast \in\mathcal{U}_d^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}\). We argue that \[\begin{align} Z\oplus E \sim_h E_n\oplus VZV^\ast \end{align}\] in \(\mathcal{U}_{d,n+1}^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}\). Indeed, this is achieved by the rotation \[\begin{align} \begin{bmatrix} \cos t \mathbb{1}_n & -\sin t V^\ast \\ \sin t V & \cos t \mathbb{1} \end{bmatrix} \begin{bmatrix} Z & 0 \\ 0 & E \end{bmatrix} \begin{bmatrix} \cos t \mathbb{1}_n & \sin t V^\ast \\ -\sin t V & \cos t \mathbb{1} \end{bmatrix} \end{align}\] for \(t\in[0,\pi/2]\). Let \(W=\begin{bmatrix}0 & VZV^\ast \\ - \overline{VZV^\ast} & 0 \end{bmatrix} \in\mathcal{SU}_o(\mathcal{L}_d^\mathbb{R}\otimes \mathrm{C\ell}_{0,2})\). Then we have \(U\oplus e\sim_h e_n\oplus W\) in \(\mathcal{SU}_o(M_{n+1}(\mathcal{L}_d^\mathbb{R}\otimes \mathrm{C\ell}_{0,2}))\) and hence \([U]_e=[W]_e\). ◻

Proof. The proof is based on the polar decomposition. We show that \[\begin{align} \label{eq:diii32polar32map} \mathcal{G}_d^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}\ni A\mapsto A|A|^{-1} \in \mathcal{U}_d^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI} \end{align}\tag{46}\] maps into \(\mathcal{U}_d^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}\). Let \(A\in\mathcal{G}_d^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}\) and consider its polar decomposition \(A=U|A|\). We first show the identity \(|A^{-1}|=U|A|^{-1}U^\ast\) which is true regardless of symmetry condition. Indeed, we have \[\begin{align} |A^{-1}|=(AA^\ast)^{-1/2} = (U|A||A|U^\ast)^{-1/2} = U|A|^{-1}U^\ast. \end{align}\] So the polar decomposition of \(A^{-1}\) is \[\begin{align} U^\ast |A^{-1}|=U^\ast( U|A|^{-1}U^\ast) = |A|^{-1}U^\ast = A^{-1}. \end{align}\] Now \[\begin{align} \mathcal{C}A|A|^{-1}\mathcal{C}= \mathcal{C}A \mathcal{C}\mathcal{C}|A|^{-1}\mathcal{C}= -A^{-1}|A^{-1}|^{-1} = -U^\ast = -(A|A|^{-1})^{-1} \end{align}\] which shows that \(A|A|^{-1}\in\mathcal{U}_d^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}\). The conclusion of the lemma follows from continuity of the map 46 .

To prove the other assertion, if \(G\in\mathcal{G}_d^\mathrm{D\kern-0.05emI\kern-0.05emI\kern-0.05emI}\), we consider the path \(t\mapsto G|G|^{-t}\) for \(t\in[0,1]\). ◻