A H^{3}(G,{\mathbb T})-valued index of symmetry protected topological phases with on-site finite group symmetry for two-dimensional quantum spin systems
aa r X i v : . [ m a t h - ph ] J a n A H ( G, T )-valued index of symmetry protected topologicalphases with on-site finite group symmetry fortwo-dimensional quantum spin systems Yoshiko Ogata ∗ January 8, 2021
Abstract
We consider SPT-phases with on-site finite group G symmetry β for two-dimensional quan-tum spin systems. We show that they have H ( G, T )-valued invariant. The notion of symmetry protected topological (SPT) phases was introduced by Gu and Wen [GW].It is defined as follows: we consider the set of all Hamiltonians with some symmetry, which havea unique gapped ground state in the bulk, and can be smoothly deformed into a common trivialgapped Hamiltonian without closing the gap. We say two such Hamiltonians are equivalent, if theycan be smoothly deformed into each other, without breaking the symmetry. We call an equivalenceclass of this classification, a symmetry protected topological (SPT) phase. Based on tensor networkor TQFT analysis, [CGLW], [MGSC] it is conjectured that SPT phases with on-site finite group G symmetry for ν -dimensional quantum spin systems have a H ν +1 ( G, T )-valued invariant. Weproved their conjecture affirmatively in [O1] for ν = 1. In this paper, We show that the conjectureis also true for ν = 2.We start by summarizing standard setup of 2-dimensional quantum spin systems on the twodimensional lattice Z [BR1, BR2]. We will use freely the basic notation in section A. Throughoutthis paper, we fix some 2 ≤ d ∈ N . We denote the algebra of d × d matrices by M d .For each subset Γ of Z , we denote the set of all finite subsets in Γ by S Γ . We introducethe Euclidean metric on Z , inherited from R . We denote by d( S , S ) the distance between S , S ⊂ Z . For a subset Γ of Z and r ∈ R ≥ , ˆΓ( r ) denotes the all points in R whose distancefrom Γ is less than or equal to r . We also set Γ( r ) := ˆΓ( r ) ∩ Z . When we take a complement ofΓ ⊂ Z , it means Γ c := Z \ Γ. For each n ∈ N , we denote [ − n, n ] ∩ Z by Λ n .For each z ∈ Z , let A { z } be an isomorphic copy of M d , and for any finite subset Λ ⊂ Z ,we set A Λ = N z ∈ Λ A { z } . For finite Λ, the algebra A Λ can be regarded as the set of all boundedoperators acting on the Hilbert space N z ∈ Λ C d . We use this identification freely. If Λ ⊂ Λ ,the algebra A Λ is naturally embedded in A Λ by tensoring its elements with the identity. For aninfinite subset Γ ⊂ Z , A Γ is given as the inductive limit of the algebras A Λ with Λ ∈ S Γ . We call A Γ the quantum spin system on Γ. For simplicity we denote the two dimensional quantum spinsystem A Z by A . We also set A loc := S Λ ∈ S Z A Λ . For a subset Γ of Γ ⊂ Z , the algebra A Γ canbe regarded as a subalgebra of A Γ . With this identification, for A ∈ A Γ , we occasionally use thesame symbol A to denote A ⊗ I A Γ \ Γ1 ∈ A Γ . Similarly, an automorphism γ on A Γ can be naturally ∗ Graduate School of Mathematical Sciences The University of Tokyo, Komaba, Tokyo, 153-8914, Japan Sup-ported in part by the Grants-in-Aid for Scientific Research, JSPS. γ ⊗ id A Γ \ Γ1 on A Γ . We use this identification freely and with a slightabuse of notation we occasionally denote γ ⊗ id A Γ \ Γ1 by γ . Similarly, for disjoint Γ − , Γ + ⊂ Z and α ± ∈ Aut A Γ ± , we occasionally write α − ⊗ α + to denote (cid:16) α − ⊗ id Γ c − (cid:17) (cid:16) α + ⊗ id Γ c + (cid:17) , under theabove identification.Throughout this paper we fix a finite group G and its unitary representation U on C d . LetΓ ⊂ Z be a non-empty subset. For each g ∈ G , there exists a unique automorphism β Γ on A Γ such that β Γ g ( A ) = Ad O I U ( g ) ! ( A ) , A ∈ A I , g ∈ G, (1.1)for any finite subset I of Γ. We call the group homomorphism β Γ : G → Aut A Γ , the on-site actionof G on A Γ given by U . For simplicity, we denote β Z g by β g .A mathematical model of a quantum spin system is fully specified by its interaction Φ. Auniformly bounded interaction on A is a map Φ : S Z → A loc such thatΦ( X ) = Φ( X ) ∗ ∈ A X , X ∈ S Z , (1.2)and sup X ∈ S Z k Φ( X ) k < ∞ . (1.3)It is of finite range with interaction length less than or equal to R ∈ N if Φ( X ) = 0 for any X ∈ S Z whose diameter is larger than R . An on-site interaction, i.e., an interaction with Φ( X ) = 0 unless X consists of a single point, is said to be trivial. An interaction Φ is β -invariant if β g (Φ( X )) = Φ( X )for any X ∈ S Z . For a uniformly bounded and finite range interaction Φ and Λ ∈ S Z define thelocal Hamiltonian ( H Φ ) Λ := X X ⊂ Λ Φ( X ) , (1.4)and denote the dynamics τ (Λ)Φ t ( A ) := e it ( H Φ ) Λ Ae − it ( H Φ ) Λ , t ∈ R , A ∈ A . (1.5)By the uniform boundedness and finite rangeness of Φ, for each A ∈ A , the following limit existslim Λ → Z ν τ (Λ) , Φ t ( A ) =: τ Φ t ( A ) , t ∈ R , (1.6)and defines the dynamics τ Φ on A . (See [BR2].) For a uniformly bounded and finite rangeinteraction Φ, a state ϕ on A is called a τ Φ -ground state if the inequality − i ϕ ( A ∗ δ Φ ( A )) ≥ A in the domain D ( δ Φ ) of the generator δ Φ . Let ϕ be a τ Φ -ground state, witha GNS triple ( H ϕ , π ϕ , Ω ϕ ). Then there exists a unique positive operator H ϕ, Φ on H ϕ such that e itH ϕ, Φ π ϕ ( A )Ω ϕ = π ϕ ( τ t Φ ( A ))Ω ϕ , for all A ∈ A and t ∈ R . We call this H ϕ, Φ the bulk Hamiltonianassociated with ϕ . Definition 1.1.
We say that an interaction Φ has a unique gapped ground state if (i) the τ Φ -groundstate, which we denote as ω Φ , is unique, and (ii) there exists a γ > σ ( H ω Φ , Φ ) \ { } ⊂ [ γ, ∞ ), where σ ( H ω Φ , Φ ) is the spectrum of H ω Φ , Φ . We denote by P UG the set of all uniformlybounded finite range interactions, with unique gapped ground state. We denote by P UGβ the setof all uniformly bounded finite range β -invariant interactions, with unique gapped ground state.2n this paper we consider a classification problem of a subset of P UGβ , models with short rangeentanglement. To describe the models with short range entanglement, we need to explain theclassification problem of unique gapped ground state phases, without symmetry. For Γ ⊂ Z ,we denote by Π Γ : A → A Γ the conditional expectation with respect to the trace state. Let f : (0 , ∞ ) → (0 , ∞ ) be a continuous decreasing function with lim t →∞ f ( t ) = 0. For each A ∈ A ,let k A k f := k A k + sup N ∈ N (cid:18) k A − Π Λ N ( A ) k f ( N ) (cid:19) . (1.7)We denote by D f the set of all A ∈ A such that k A k f < ∞ .The classification of unique gapped ground state phases P UG without symmetry is the following. Definition 1.2.
Two interactions Φ , Φ ∈ P UG are equivalent if there is a path of interactionsΦ : [0 , → P UG satisfying the following:1. Φ(0) = Φ and Φ(1) = Φ .2. For each X ∈ S Z , the map [0 , ∋ s → Φ( X ; s ) ∈ A X is C . We denote by ˙Φ( X ; s ) thecorresponding derivatives. The interaction obtained by differentiation is denoted by ˙Φ( s ),for each s ∈ [0 , R ∈ N such that X ∈ S Z and diam X ≥ R imply Φ( X ; s ) = 0, for all s ∈ [0 , C Φ b := sup s ∈ [0 , sup X ∈ S Z (cid:16) k Φ ( X ; s ) k + | X | (cid:13)(cid:13)(cid:13) ˙Φ ( X ; s ) (cid:13)(cid:13)(cid:13)(cid:17) < ∞ . (1.8)5. Setting b ( ε ) := sup Z ∈ S Z sup s,s ∈ [0 , , < | s − s | <ε (cid:13)(cid:13)(cid:13)(cid:13) Φ( Z ; s ) − Φ( Z ; s ) s − s − ˙Φ( Z ; s ) (cid:13)(cid:13)(cid:13)(cid:13) (1.9)for each ε >
0, we have lim ε → b ( ε ) = 0.6. There exists a γ > σ ( H ω Φ( s ) , Φ( s ) ) \ { } ⊂ [ γ, ∞ ) for all s ∈ [0 , σ ( H ω Φ( s ) , Φ( s ) ) is the spectrum of H ω Φ( s ) , Φ( s ) .7. There exists an 0 < η < ζ ( t ) := e − t η . Then for each A ∈ D ζ , ω Φ( s ) ( A ) is differentiable with respect to s , and there is a constant C ζ such that: (cid:12)(cid:12)(cid:12)(cid:12) dds ω Φ( s ) ( A ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ C ζ k A k ζ , (1.10)for any A ∈ D ζ .(Recall (1.7)).We write Φ ∼ Φ if Φ and Φ are equivalent. If Φ , Φ ∈ P UGβ and if we can take the path in P UGβ , i.e., so that β g (Φ( X ; s )) = Φ( X ; s ), g ∈ G for all s ∈ [0 , and Φ are β -equivalent and write Φ ∼ β Φ .The object we classify in this paper is the following: Definition 1.3.
Fix a trivial interaction Φ ∈ P UG . We denote by P SLβ the set of all Φ ∈ P
UGβ such that Φ ∼ Φ . Connected components of P SLβ with respect to ∼ β are the symmetry protectedtopological (SPT)-phases. 3ecause we have Φ ∼ ˜Φ for any trivial Φ , ˜Φ ∈ P UG , the set P SLβ does not depend on thechoice of Φ .The main result of this paper is as follows. Theorem 1.4.
There is a H ( G, T ) -valued index on P SLβ , which is an invariant of the classifi-cation ∼ β of P SLβ . This paper is organized as follows. In section 2, we define the H ( G, T )-valued index for aclass of states which are created from a fixed product state via “factorizable” automorphisms,and satisfying some additional condition. This additional condition is the existence of the setof automorphisms 1.which do not move the state, and 2. are almost like β -action restricted tothe upper half-plane except for some one-dimensional perturbation. In section 3, we show thatthe existence of such set of automorphisms are guaranteed in suitable situation. Furthermore, insection 4, we show the stability of the index, i.e., suitably β -invariant automorphism does notchange this index. Finally in section 5, we show our main Theorem Theorem 1.4, showing that inour setting of Theorem 1.4, all the conditions required in section 2, 3, 4 are satisfied. Althoughthe index is defined in terms of GNS representations, in some good situation, we can calculate itwithout going through GNS representation. This is shown in section 6.The present result and the main idea of the proof were announced publicly on 15 December2020 at IAMP One World Mathematical Physics Seminar (see you-tube video)[O2]. It was alsopresented in the international meeting Theoretical studies of topological phases of matter on 17December 2020, and Current Developments in Mathematics 4th January 2021 via zoom with alecture note. [O2]. H ( G, T ) -valued index in -dimensional systems In this section, we associate an H ( G, T )-index for some class of states. It will turn out later thatthis class includes SPT phases. For 0 < θ < π , a cone C θ is defined by C θ := (cid:8) ( x, y ) ∈ Z | | y | ≤ tan θ · | x | (cid:9) . (2.1)For 0 < θ < θ ≤ π , we use a notation C ( θ ,θ ] := C θ \ C θ and C [0 ,θ ] := C θ . Left, right, upper,lower half planes are denoted by H L , H R , H U , H D , i.e., H L := (cid:8) ( x, y ) ∈ Z | x ≤ − (cid:9) , H R := (cid:8) ( x, y ) ∈ Z | ≤ x (cid:9) , (2.2) H U := (cid:8) ( x, y ) ∈ Z | ≤ y (cid:9) , H D := (cid:8) ( x, y ) ∈ Z | y ≤ − (cid:9) . (2.3)We use a notation β g := β Z g , β Ug := β H U g , β RUg := β H R ∩ H U g , β LUg := β H L ∩ H U g .For each subset S of Z , we set S σ := S ∩ H σ , S ζ := S ∩ H ζ , S σζ := S ∩ H σ ∩ H ζ σ = L, R, ζ = U, D. (2.4)We ocationally write A S,σ , A S,ζ , A S,σ,ζ to denote A S σ , A S ζ , A S σζ . For an automorphism α on A and 0 < θ < π , we denote by D θα a set of all triples ( α L , α R , Θ) with α L ∈ Aut ( A H L ) , α R ∈ Aut ( A H R ) , Θ ∈ Aut (cid:0) A ( C θ ) c (cid:1) (2.5)decomposing α as α = (inner) ◦ ( α L ⊗ α R ) ◦ Θ . (2.6)4or ( α L , α R , Θ) ∈ D ( θ ) α , we set α := α L ⊗ α R . (2.7)The class of automorphisms which allow such decompositions for any directions are denoted byQAut ( A ) := n α ∈ Aut( A ) | D θα = ∅ for all 0 < θ < π o . (2.8)Furthermore, for each0 < θ . < θ < θ . < θ . < θ < θ . < θ . < θ < θ . < π , (2.9)we consider decompositions of α ∈ Aut( A ) such that α = (inner) ◦ (cid:16) α [0 ,θ ] ⊗ α ( θ ,θ ] ⊗ α ( θ ,θ ] ⊗ α ( θ , π ] (cid:17) ◦ (cid:0) α ( θ . ,θ . ] ⊗ α ( θ . ,θ . ] ⊗ α ( θ . ,θ . ] (cid:1) (2.10)with α X := O σ = L,R,ζ = D,U α X,σ,ζ , α [0 ,θ ] := O σ = L,R α [0 ,θ ] ,σ , α ( θ , π ] := O ζ = D,U α ( θ , π ] ,ζ α X,σ,ζ ∈ Aut (cid:0) A C X,σ,ζ (cid:1) , α
X,σ := O ζ = U,D α X,σ,ζ , α
X,ζ := O σ = L,R α X,σ,ζ α [0 ,θ ] ,σ ∈ Aut (cid:16) A C [0 ,θ ,σ (cid:17) , α ( θ , π ] ,ζ ∈ Aut (cid:16) A C ( θ , π ,ζ (cid:17) , (2.11)for X = ( θ , θ ] , ( θ , θ ] , ( θ . , θ . ] , ( θ . , θ . ] , ( θ . , θ . ] , σ = L, R, ζ = D, U. (2.12)The class of automorphisms on A which allow such decompositions for any directions θ . , θ , θ . , θ . , θ , θ . , θ . , θ , θ . (satisfying (2.9)) is denoted by SQAut( A ). Note that SQAut( A ) ⊂ QAut( A ).The set of all α ∈ SQAut( A ) with each of α I in the decompositions required to commute with β Ug , g ∈ G , is denoted by GSQAut( A )GSQAut( A ) := α ∈ SQAut( A ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) for any θ . , θ , θ . , θ . , θ , θ . , θ . , θ , θ . satisfying (2.9)there is a decomposition (2 . , (2 . , (2 .
12) satisfying α I ◦ β Ug = β Ug ◦ α I , g ∈ G, for all I = [0 , θ ] , ( θ , θ ] , ( θ , θ ] , (cid:16) θ , π i , ( θ . , θ . ] , ( θ . , θ . ] , ( θ . , θ . ] . (2.13)We also defineHAut ( A ) := α ∈ Aut( A ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) for any 0 < θ < π , there exist α σ ∈ Aut (cid:0) A ( C θ ) σ (cid:1) , σ = L, R such that α = (inner) ◦ ( α L ⊗ α R ) . (2.14)In section 5, we will see that quasi-local automorphisms corresponding to paths in symmetricgapped phases belong to the following set:GUQAut ( A ) := ( γ ∈ Aut ( A ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) there are γ H ∈ HAut( A ) , γ C ∈ GSQAut( A )such that γ = γ C ◦ γ H ) . (2.15)5e fix a reference state ω as follows. We fix a unit vector ξ x ∈ C d and let ρ ξ x be the vectorstate on M d given by ξ x , for each x ∈ Z . Then our reference state ω is given by ω := O Z ρ ξ x . (2.16)Throughout this section this ω is fixed. Let ( H , π , Ω ) be a GNS triple of ω . Because of theproduct structure of ω , it is decomposed as H = H L ⊗ H R , π = π L ⊗ π R , Ω = Ω L ⊗ Ω R , (2.17)where ( H σ , π σ , Ω σ ) is a GNS triple of ω σ := ω | A Hσ for σ = L, R . As ω | A Hσ is pure, π σ isirreducible. What we consider in this section is the set of states created via elements in QAut( A )from our reference state ω : SL := { ω ◦ α | α ∈ QAut( A ) } . (2.18)Because any pure product states can be transformed to each other via an automorphism of productform ˜ α = N x ∈ Z ˜ α x and ˜ αα belongs to QAut( A ) for any α ∈ QAut( A ), SL does not depend onthe choice of ω . For each ω ∈ SL , we setEAut( ω ) := { α ∈ QAut( A ) | ω = ω ◦ α } . (2.19)By the definition of SL , EAut( ω ) is not empty.For 0 < θ < π and a set of automorphisms ( ˜ β g ) g ∈ G ⊂ Aut( A ), we introduce a set T ( θ, ( ˜ β g )) := ( η σg ) g ∈ G,σ = L,R (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) η σg ∈ Aut (cid:0) A ( C θ ) σ (cid:1) , ˜ β g = (inner) ◦ (cid:0) η Lg ⊗ η Rg (cid:1) ◦ β Ug , for all g ∈ G, σ = L, R . (2.20)In a word, it is a set of decompositions of ˜ β g ◦ (cid:0) β Ug (cid:1) − into tensor of Aut (cid:0) A ( C θ ) L (cid:1) , Aut (cid:0) A ( C θ ) R (cid:1) modulo inner automorphisms. For ( η σg ) g ∈ G,σ = L,R ∈ T ( θ, ( ˜ β g )), we set η g := η Lg ⊗ η Rg , g ∈ G. (2.21)The following set of automorphisms is the key ingredient for the definition of our index. For ω ∈ SL and 0 < θ < π , we setIG ( ω, θ ) := ( ( ˜ β g ) g ∈ G ∈ Aut ( A ) × G (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ω ◦ ˜ β g = ω for all g ∈ G, and T ( θ, ( ˜ β g )) = ∅ ) . (2.22)We also set IG ( ω ) := ∪ <θ< π IG ( ω, θ ) . (2.23)In this section we associate some third cohomology h ( ω ) for each ω ∈ SL with IG( ω ) = ∅ . Z ( G, T ) In this subsection, we derive 3-cocycles out of ω , α , θ , ( ˜ β g ), ( η σg ) ( α L , α R , Θ).
Lemma 2.1.
Let ω ∈ SL , α ∈ EAut( ω ) , < θ < π , ( ˜ β g ) ∈ IG ( ω, θ ) , ( η σg ) ∈ T ( θ, ( ˜ β g )) , ( α L , α R , Θ) ∈ D θα . Then i) There are unitaries W g , g ∈ G on H such that Ad ( W g ) ◦ π = π ◦ α ◦ Θ ◦ η g β Ug ◦ Θ − ◦ α − , g ∈ G (2.24) with notation (2.7), (2.21). (ii) There exist a unitary u σ ( g, h ) on H σ , for each σ = L, R , g, h ∈ G , such that Ad ( u σ ( g, h )) ◦ π σ = π σ ◦ α σ ◦ η σg β σUg η σh (cid:0) β σUg (cid:1) − (cid:0) η σgh (cid:1) − ◦ α − σ , (2.25) and Ad ( u L ( g, h ) ⊗ u R ( g, h )) π = π ◦ α ◦ η g β Ug η h (cid:0) β Ug (cid:1) − ( η gh ) − ◦ α − . (2.26) Furthermore, u σ ( g, h ) commutes with any element of π σ ◦ α σ (cid:0) A (( C θ ) c ) σ (cid:1) . Definition 2.2.
For ω ∈ SL , α ∈ EAut( ω ), 0 < θ < π , ( ˜ β g ) ∈ IG ( ω, θ ), ( η σg ) g ∈ G,σ = L,R ∈T ( θ, ( ˜ β g )), ( α L , α R , Θ) ∈ D θα , we denote byIP (cid:16) ω, α, θ, ( ˜ β g ) , ( η σg ) , ( α L , α R , Θ) (cid:17) (2.27)the set of (( W g ) g ∈ G , ( u σ ( g, h )) g,h ∈ G,σ = L,R ) with W g ∈ U ( H ) and u σ ( g, h ) ∈ U ( H σ ) satisfyingAd ( W g ) ◦ π = π ◦ α ◦ Θ ◦ η g β Ug ◦ Θ − ◦ α − , g ∈ G, and (2.28)Ad ( u σ ( g, h )) ◦ π σ = π σ ◦ α σ ◦ η σg β σUg η σh (cid:0) β σUg (cid:1) − (cid:0) η σgh (cid:1) − ◦ α − σ , g, h ∈ G, σ = L, R. (2.29)(Here we used notation (2.7) and (2.21).) By Lemma 2.1, it is non-empty.
Proof.
For a GNS triple ( H , π ◦ α, Ω ) of ω = ω ◦ α there are unitaries ˜ W g on H such thatAd (cid:16) ˜ W g (cid:17) ◦ π ◦ α = π ◦ α ◦ ˜ β g , g ∈ G (2.30)because ω ◦ ˜ β g = ω .Because ( η σg ) g ∈ G,σ = L,R ∈ T ( θ, ( ˜ β g )), and ( α L , α R , Θ) ∈ D θα , there are unitaries v g , V ∈ U ( A )such that ˜ β g = Ad ( v g ) ◦ (cid:0) η Lg ⊗ η Rg (cid:1) ◦ β Ug , α = Ad V ◦ α ◦ Θ . (2.31)Substituting these, we haveAd (cid:16) ˜ W g π ( V ) (cid:17) π ◦ α ◦ Θ = π ◦ α ˜ β g = π ◦ α ◦ Ad( v g ) ◦ η g β Ug = Ad (( π ◦ α ( v g )) π ( V )) π ◦ α ◦ Θ ◦ η g β Ug . (2.32)Therefore, setting W g := π ( V ) ∗ (cid:0) π ◦ α ( v ∗ g ) (cid:1) ˜ W g π ( V ) ∈ U ( H ), we obtain (2.24).Using this (2.24), we haveAd (cid:0) W g W h W ∗ gh (cid:1) π = π ◦ α ◦ Θ ◦ η g β Ug η h (cid:0) β Ug (cid:1) − η − gh Θ − α − . (2.33)Note that because conjugation by β Ug does not change the support of automorphisms, η g β Ug η h (cid:0) β Ug (cid:1) − η − gh belongs to Aut ( A C θ ). On the other hand, Θ belongs to Aut (cid:0) A ( C θ ) c (cid:1) . Therefore, they commuteand we obtainAd (cid:0) W g W h W ∗ gh (cid:1) π = (2 .
33) = π ◦ α ◦ η g β Ug η h (cid:0) β Ug (cid:1) − η − gh α − = O σ = L,R π σ ◦ α σ ◦ η σg β σUg η σh (cid:0) β σUg (cid:1) − (cid:0) η σgh (cid:1) − ◦ α − σ (2.34)7rom this and the irreducibility of π R , we see that Ad (cid:16) W g W h W ∗ gh (cid:17) gives rise to a ∗ -isomorphism τ on B ( H R ).It is implemented by some unitary u R ( g, h ) on H R by the Wigner theorem and weobtain I H L ⊗ (Ad ( u R ( g, h )) ◦ π R ( A )) = I H L ⊗ τ ( π R ( A )) = Ad (cid:0) W g W h W ∗ gh (cid:1) ( I H L ⊗ π R ( A ))= I H L ⊗ π R ◦ α R ◦ η Rg β RUg η Rh (cid:0) β RUg (cid:1) − (cid:0) η Rgh (cid:1) − ◦ α − R ( A ) , (2.35)for any A ∈ A H R . Hence we obtain (2.25) for σ = R .To see that u R ( g, h ) belongs to (cid:0) π R ◦ α R (cid:0) A (( C θ ) c ) R (cid:1)(cid:1) ′ , let A ∈ A (( C θ ) c ) R . Then because η Rg β RUg η Rh (cid:0) β RUg (cid:1) − (cid:16) η Rgh (cid:17) − belongs to Aut (cid:0) A ( C θ ) R (cid:1) , we haveAd ( u R ( g, h )) π R ( α R ( A )) = π R α R η Rg β RUg η Rh (cid:0) β RUg (cid:1) − (cid:0) η Rgh (cid:1) − α − R α R ( A ) = π R α R ( A ) . (2.36)This proves that u R ( g, h ) belongs to (cid:0) π R ◦ α R (cid:0) A (( C θ ) c ) R (cid:1)(cid:1) ′ . Analogous statement for u L ( g, h ) canbe shown exactly the same way. The last statement (2.26) of (ii) is trivial from (2.25). (cid:3) Lemma 2.3.
Let ω ∈ SL , α ∈ EAut( ω ) , < θ < π , ( ˜ β g ) ∈ IG ( ω, θ ) , ( η σg ) ∈ T ( θ, ( ˜ β g )) , ( α L , α R , Θ) ∈ D θα . Let (( W g ) , ( u R ( g, h ))) be an element of IP (cid:16) ω, α, θ, ( ˜ β g ) , ( η σg ) , ( α L , α R , Θ) (cid:17) .Then the followings hold. (i) For any g, h, k ∈ G , Ad (cid:0) W g ( I H L ⊗ u R ( h, k )) W ∗ g (cid:1) ◦ π = π ◦ (cid:16) id A HL ⊗ α R η Rg β RUg (cid:16) η Rh β RUh η Rk (cid:0) β RUh (cid:1) − (cid:0) η Rhk (cid:1) − (cid:17) (cid:0) η Rg β RUg (cid:1) − α − R (cid:17) . (2.37) (ii) For any g, h ∈ G , Ad (( u L ( g, h ) ⊗ u R ( g, h )) W gh ) = Ad ( W g W h ) , (2.38) on B ( H ) . (iii) For any g, h, k ∈ G , Ad ( W g ) ( I H L ⊗ u R ( h, k )) ∈ CI H L ⊗ B ( H R ) . (2.39) (iv) For any g, h, k, f ∈ G , Ad ( W g W h ) ( I H L ⊗ u R ( k, f )) = (Ad (( I H L ⊗ u R ( g, h )) W gh )) ( I H L ⊗ u R ( k, f )) . (2.40) Proof.
We use the notation (2.7), (2.21).(i) Substituting (2.28) (2.29), we haveAd (cid:0) W g ( I H L ⊗ u R ( h, k )) W ∗ g (cid:1) ◦ π = π ◦ α ◦ Θ ◦ η g β Ug ◦ Θ − ◦ α − ◦ (cid:16) id A HL ⊗ α R ◦ η Rh β RUh η Rk (cid:0) β RUh (cid:1) − (cid:0) η Rhk (cid:1) − ◦ α − R (cid:17) ◦ α ◦ Θ ◦ (cid:0) η g β Ug (cid:1) − ◦ Θ − ◦ α − = π ◦ α ◦ Θ ◦ η g β Ug ◦ Θ − ◦ (cid:16) id A HL ⊗ η Rh β RUh η Rk (cid:0) β RUh (cid:1) − (cid:0) η Rhk (cid:1) − (cid:17) ◦ Θ ◦ (cid:0) η g β Ug (cid:1) − ◦ Θ − ◦ α − . (2.41)8ecause η Rh β RUh η Rk (cid:0) β RUh (cid:1) − (cid:0) η Rhk (cid:1) − belongs to Aut (cid:0) A ( C θ ) R (cid:1) , it commutes with Θ ∈ Aut (cid:0) A ( C θ ) c (cid:1) .Hence we obtain(2 .
41) = π ◦ α ◦ Θ ◦ η g β Ug ◦ (cid:16) I A HL ⊗ η Rh β RUh η Rk (cid:0) β RUh (cid:1) − (cid:0) η Rhk (cid:1) − (cid:17) ◦ (cid:0) η g β Ug (cid:1) − ◦ Θ − ◦ α − = π ◦ α ◦ Θ ◦ (cid:16) I A HL ⊗ η Rg β RUg ◦ η Rh β RUh η Rk (cid:0) β RUh (cid:1) − (cid:0) η Rhk (cid:1) − ◦ (cid:0) η Rg β RUg (cid:1) − (cid:17) ◦ Θ − ◦ α − . (2.42)Again, the term in the round braket in the last line is localized at ( C θ ) R , and it commutes withΘ. Therefore, we haveAd (cid:0) W g ( I H L ⊗ u R ( h, k )) W ∗ g (cid:1) ◦ π = π ◦ (cid:16) id A HL ⊗ α R ◦ η Rg β RUg ◦ η Rh β RUh η Rk (cid:0) β RUh (cid:1) − (cid:0) η Rhk (cid:1) − ◦ (cid:0) η Rg β RUg (cid:1) − ◦ α − R (cid:17) (2.43)(ii) Again by (2.28) and (2.29), we haveAd (( u L ( g, h ) ⊗ u R ( g, h )) W gh ) ◦ π = π ◦ α ◦ η g β Ug η h (cid:0) β Ug (cid:1) − ( η gh ) − ◦ Θ ◦ η gh β Ugh ◦ Θ − ◦ α − = π ◦ α ◦ Θ ◦ η g β Ug η h (cid:0) β Ug (cid:1) − ( η gh ) − ◦ η gh β Ugh ◦ Θ − ◦ α − = π ◦ α ◦ Θ ◦ η g β Ug η h β Uh ◦ Θ − ◦ α − = Ad ( W g W h ) ◦ π . (2.44)Here, for the second equality, we again used the commutativity of η s and Θ, due to their disjointsupport. Because π is irreducible, we obtain (2.38).(iii) For any A ∈ A H L , we haveΘ − ◦ α − (cid:16) A ⊗ I A HR (cid:17) = Θ − ◦ (cid:16) α − L ( A ) ⊗ I A HR (cid:17) ∈ Θ − (cid:16) A H L ⊗ CI A HR (cid:17) ⊂ A H L ∪ ( C cθ ) R , (2.45)because Θ ∈ Aut (cid:0) A ( C θ ) c (cid:1) . Therefore, η Rg ∈ Aut( A ( C θ ) R ) acts trivially on it and we have (cid:0) β Ug (cid:1) − ( η g ) − ◦ Θ − ◦ α − (cid:16) A ⊗ I A HR (cid:17) ∈ A H L ∪ ( C cθ ) R . (2.46)As Θ preserves A H L ∪ ( C cθ ) R ,Θ ◦ (cid:0) β Ug (cid:1) − ( η g ) − ◦ Θ − ◦ α − (cid:16) A ⊗ I A HR (cid:17) (2.47)also belongs to A H L ∪ ( C cθ ) R . As a result,Ad (cid:0) W ∗ g (cid:1) ( π L ( A ) ⊗ I H R ) = π ◦ α ◦ Θ ◦ (cid:0) β Ug (cid:1) − ( η g ) − ◦ Θ − ◦ α − (cid:16) A ⊗ I A HR (cid:17) (2.48)belongs to π L ( A H L ) ⊗ π R ◦ α R (cid:16) A ( C cθ ) R (cid:17) , hence commutes with I H L ⊗ u R ( h, k ). Hence Ad( W g ) ( I H L ⊗ u R ( h, k ))commutes with any elements in π L ( A L ) ⊗ CI H R . Because π L is irreducible, Ad( W g ) ( I H L ⊗ u R ( h, k ))belongs to CI H L ⊗ B ( H R ).(iv) By (iii), Ad ( W gh ) ( I H L ⊗ u R ( k, f )) belongs to CI H L ⊗ B ( H R ). Therefore, from (ii), we haveAd ( W g W h ) ( I H L ⊗ u R ( k, f )) = Ad (( u L ( g, h ) ⊗ u R ( g, h )) W gh ) ( I H L ⊗ u R ( k, f ))= Ad (( I H L ⊗ u R ( g, h )) W gh ) ( I H L ⊗ u R ( k, f )) , (2.49)obtaining (iv). (cid:3) With this preparation we may obtain some element of Z ( G, T ) from (( W g ) , ( u σ ( g, h ))).9 emma 2.4. Let ω ∈ SL , α ∈ EAut( ω ) , < θ < π , ( ˜ β g ) ∈ IG ( ω, θ ) , ( η σg ) ∈ T ( θ, ( ˜ β g )) , ( α L , α R , Θ) ∈ D θα . Let (( W g ) , ( u σ ( g, h ))) be an element of IP (cid:16) ω, α, θ, ( ˜ β g ) , ( η σg ) , ( α L , α R , Θ) (cid:17) .Then there is a c R ∈ Z ( G, T ) such that I H L ⊗ u R ( g, h ) u R ( gh, k ) = c R ( g, h, k ) (cid:0) W g ( I H L ⊗ u R ( h, k )) W ∗ g (cid:1) ( I H L ⊗ u R ( g, hk )) , (2.50) for all g, h, k ∈ G . Definition 2.5.
We denote this 3-cocycle c R in the Lemma by c R (cid:16) ω, α, θ, ( ˜ β g ) , ( η σg ) , ( α L , α R , Θ) , (( W g ) , ( u σ ( g, h ))) (cid:17) (2.51)and its cohomology class by h (1) (cid:16) ω, α, θ, ( ˜ β g ) , ( η σg ) , ( α L , α R , Θ) , (( W g ) , ( u σ ( g, h ))) (cid:17) := h c R (cid:16) ω, α, θ, ( ˜ β g ) , ( η σg ) , ( α L , α R , Θ) , (( W g ) , ( u σ ( g, h ))) (cid:17)i H ( G, T ) . (2.52) Proof.
First we prove that there is a number c R ( g, h, k ) ∈ T satisfying (2.50). From (2.29),wehaveAd ( I H L ⊗ u R ( g, h ) u R ( gh, k )) π = π L ⊗ π R ◦ α R ◦ (cid:0) η Rg β RUg (cid:1) (cid:0) η Rh β RUh (cid:1) (cid:0) η Rk β RUk (cid:1) (cid:0) η Rghk β RUghk (cid:1) − α − R . (2.53)On the other hand, using (i) of Lemma 2.3, we haveAd (cid:0)(cid:0) W g ( I H L ⊗ u R ( h, k )) W ∗ g (cid:1) ( I H L ⊗ u R ( g, hk )) (cid:1) π (2.54)is also equal to the right hand side of (2.53). Because π is irreducible, this means that there is anumber c R ( g, h, k ) ∈ T satisfying (2.50).Now let us check that this c R is a 3-cocycle. For any g, h, k, f ∈ G , by repeated use of (2.50),we get I H L ⊗ u R ( g, h ) u R ( gh, k ) u R ( ghk, f ) = [ I H L ⊗ u R ( g, h ) u R ( gh, k )] · ( I H L ⊗ u R ( ghk, f )) (2.55)= (cid:0) c R ( g, h, k ) (cid:0) W g ( I H L ⊗ u R ( h, k )) W ∗ g (cid:1) ( I H L ⊗ u R ( g, hk )) (cid:1) · ( I H L ⊗ u R ( ghk, f ))= (cid:0) c R ( g, h, k ) (cid:0) W g ( I H L ⊗ u R ( h, k )) W ∗ g (cid:1)(cid:1) · [ I H L ⊗ u R ( g, hk ) u R ( ghk, f )]= (cid:0) c R ( g, h, k ) (cid:0) W g ( I H L ⊗ u R ( h, k )) W ∗ g (cid:1)(cid:1) · (cid:0) c R ( g, hk, f ) (cid:0) W g ( I H L ⊗ u R ( hk, f )) W ∗ g (cid:1) ( I H L ⊗ u R ( g, hkf )) (cid:1) = c R ( g, h, k ) c R ( g, hk, f ) (cid:0) W g [ I H L ⊗ u R ( h, k ) u R ( hk, f )] W ∗ g (cid:1) · ( I H L ⊗ u R ( g, hkf ))= c R ( g, h, k ) c R ( g, hk, f ) c R ( h, k, f ) W g ( W h ( I H L ⊗ u R ( k, f )) W ∗ h ( I H L ⊗ u R ( h, kf ))) W ∗ g · ( I H L ⊗ u R ( g, hkf ))= c R ( g, h, k ) c R ( g, hk, f ) c R ( h, k, f ) · W g W h ( I H L ⊗ u R ( k, f )) W ∗ h W ∗ g · (cid:2)(cid:0) W g ( I H L ⊗ u R ( h, kf )) W ∗ g (cid:1) I H L ⊗ u R ( g, hkf ) (cid:3) = c R ( g, h, k ) c R ( g, hk, f ) c R ( h, k, f ) c ( g, h, kf ) · (cid:8) W g W h ( I H L ⊗ u R ( k, f )) W ∗ h W ∗ g (cid:1) }· ( I H L ⊗ u R ( g, h ) u R ( gh, kf )) . (2.56)Here, (and below) we apply (2.50) for terms in [ · ] to get the succeeding equality. Applying (iv) ofLemma 2.3 to the {·} part of (2.56), we obtain(2 .
56) = c R ( g, h, k ) c R ( g, hk, f ) c R ( h, k, f ) c ( g, h, kf ) (Ad (( I H L ⊗ u R ( g, h )) W gh )) ( I H L ⊗ u R ( k, f )) ( I H L ⊗ u R ( g, h ) u R ( gh, kf ))= c R ( g, h, k ) c R ( g, hk, f ) c R ( h, k, f ) c ( g, h, kf ) ( I H L ⊗ u R ( g, h )) (cid:2) W gh ( I H L ⊗ u R ( k, f )) W ∗ gh ( I H L ⊗ u R ( gh, kf )) (cid:3) = c R ( g, h, k ) c R ( g, hk, f ) c R ( h, k, f ) c ( g, h, kf ) c R ( gh, k, f ) ( I H L ⊗ u R ( g, h ) u R ( gh, k ) u R ( ghk, f )) . (2.57)10ence, we obtain c R ( g, h, k ) c R ( g, hk, f ) c R ( h, k, f ) c ( g, h, kf ) c R ( gh, k, f ) = 1 , for all g, h, k, f ∈ G. (2.58)This means c R ∈ Z ( G, T ). (cid:3) H ( G, T ) -valued index From the previous subsection, we remark the following fact.
Lemma 2.6.
For any ω ∈ SL with IG( ω ) = ∅ , there are α ∈ EAut( ω ) , < θ < π , ( ˜ β g ) ∈ IG ( ω, θ ) , ( η σg ) ∈ T ( θ, ( ˜ β g )) , ( α L , α R , Θ) ∈ D θα , (( W g ) , ( u R ( g, h ))) ∈ IP (cid:16) ω, α, θ, ( ˜ β g ) , ( η σg ) , ( α L , α R , Θ) (cid:17) . (2.59) Proof.
Because IG( ω ) = ∅ , there is some 0 < θ < π such that IG( ω, θ ) = ∅ , and hence ( ˜ β g ) ∈ IG( ω, θ ) and ( η σg ) ∈ T ( θ, ( ˜ β g )) exist. Because ω ∈ SL , by definition there exists some α ∈ EAut( ω ) and by the definition of EAut( ω ), there is some ( α L , α R , Θ) ∈ D θα . The existence of(( W g ) , ( u R ( g, h ))) ∈ IP (cid:16) ω, α, θ, ( ˜ β g ) , ( η σg ) , ( α L , α R , Θ) (cid:17) is given by Lemma 2.1. (cid:3) By Lemma 2.4, for ω ∈ SL with IG( ω ) = ∅ , for each choice of (2.59), we can associate someelement of H ( G, T ): h (1) (cid:16) ω, α, θ, ( ˜ β g ) , ( η σg ) , ( α L , α R , Θ) , (( W g ) , ( u σ ( g, h ))) (cid:17) . (2.60)In this subsection, we show that the third cohomology class does not depend on the choice of(2.59): Theorem 2.7.
For any ω ∈ SL with IG( ω ) = ∅ , h (1) (cid:16) ω, α, θ, ( ˜ β g ) , ( η σg ) , ( α L , α R , Θ) , (( W g ) , ( u σ ( g, h ))) (cid:17) is independent of the choice of α, θ, ( ˜ β g ) , ( η σg ) , ( α L , α R , Θ) , (( W g ) , ( u σ ( g, h ))) . Definition 2.8.
Let ω ∈ SL with IG( ω ) = ∅ . We denote the third cohomology given in Theorem2.7 by h ( ω ) := h (1) (cid:16) ω, α, θ, ( ˜ β g ) , ( η σg ) , ( α L , α R , Θ) , (( W g ) , ( u σ ( g, h ))) (cid:17) . First we show the independence from (( W g ) , ( u σ ( g, h ))). Lemma 2.9.
Let ω ∈ SL , α ∈ EAut( ω ) , < θ < π , ( ˜ β g ) ∈ IG( ω, θ ) , ( η σg ) ∈ T (cid:16) θ, ( ˜ β g ) (cid:17) , ( α L , α R , Θ) ∈ D θα , (2.61)(( W g ) , ( u σ ( g, h ))) , (cid:16) ( ˜ W g ) , (˜ u σ ( g, h )) (cid:17) ∈ IP (cid:16) ω, α, θ, ( ˜ β g ) , ( η σg ) , ( α L , α R , Θ) (cid:17) . (2.62) Then we have h (1) (cid:16) ω, α, θ, ( ˜ β g ) , ( η σg ) , ( α L , α R , Θ) , (( W g ) , ( u σ ( g, h ))) (cid:17) = h (1) (cid:16) ω, α, θ, ( ˜ β g ) , ( η σg ) , ( α L , α R , Θ) , (cid:16) ( ˜ W g ) , (˜ u σ ( g, h )) (cid:17)(cid:17) . (2.63)11 efinition 2.10. From this lemma and because there is always (( W g ) , ( u R ( g, h ))) in IP (cid:16) ω, α, θ, ( ˜ β g ) , ( η σg ) , ( α L , α R , Θ) (cid:17) by Lemma 2.1, we may define h (2) (cid:16) ω, α, θ, ( ˜ β g ) , ( η σg ) , ( α L , α R , Θ) (cid:17) := h (1) (cid:16) ω, α, θ, ( ˜ β g ) , ( η σg ) , ( α L , α R , Θ) , (( W g ) , ( u σ ( g, h ))) (cid:17) (2.64)for any ω ∈ SL , α ∈ EAut( ω ) , < θ < π , ( ˜ β g ) ∈ IG( ω, θ ) , ( η σg ) ∈ T (cid:16) θ, ( ˜ β g ) (cid:17) , ( α L , α R , Θ) ∈ D θα , (2.65)independent of the choice of (( W g ) , ( u σ ( g, h ))). Proof.
BecauseAd ( W g ) ◦ π = π ◦ α ◦ Θ ◦ η g β Ug ◦ Θ − ◦ α − = Ad (cid:16) ˜ W g (cid:17) ◦ π , (2.66)Ad ( u R ( g, h )) ◦ π R = π R ◦ α R ◦ η Rg β RUg η Rh (cid:0) β RUg (cid:1) − (cid:0) η Rgh (cid:1) − ◦ α − R = Ad (˜ u R ( g, h )) ◦ π R (2.67)and π , π R are irreducible, there are b ( g ) , a ( g, h ) ∈ T , g, h ∈ G such that W g = b ( g ) ˜ W g , ˜ u R ( g, h ) = a ( g, h ) u R ( g, h ) . (2.68)Set c R := c R (cid:16) ω, α, θ, ( ˜ β g ) , ( η σg ) , ( α L , α R , Θ) , (( W g ) , ( u σ ( g, h ))) (cid:17) , ˜ c R := c R (cid:16) ω, α, θ, ( ˜ β g ) , ( η σg ) , ( α L , α R , Θ) , (cid:16) ( ˜ W g ) , (˜ u σ ( g, h )) (cid:17)(cid:17) . (2.69)Then from the definition of these values and (2.68), we have a ( g, h ) a ( gh, k ) ( I H L ⊗ u R ( g, h ) u R ( gh, k )) = I H L ⊗ ˜ u R ( g, h )˜ u R ( gh, k )= ˜ c R ( g, h, k ) (cid:16) ˜ W g ( I H L ⊗ ˜ u R ( h, k )) ˜ W ∗ g (cid:17) ( I H L ⊗ ˜ u R ( g, hk ))= ˜ c R ( g, h, k ) a ( h, k ) a ( g, hk ) (cid:0) W g ( I H L ⊗ u R ( h, k )) W ∗ g (cid:1) ( I H L ⊗ u R ( g, hk ))= ˜ c R ( g, h, k ) a ( h, k ) a ( g, hk ) c R ( g, h, k ) ( I H L ⊗ u R ( g, h ) u R ( gh, k )) . (2.70)Hence we have ˜ c R ( g, h, k ) = c R ( g, h, k ) a ( h, k ) a ( g, hk ) a ( g, h ) a ( gh, k ), and we get [ c R ] H ( G, T ) =[˜ c R ] H ( G, T ) , proving the claim. (cid:3) Next we show the independence from α, ( α L , α R , Θ).
Lemma 2.11.
Let ω ∈ SL , α , α ∈ EAut( ω ) , < θ < π , ( ˜ β g ) ∈ IG( ω, θ ) , ( η σg ) ∈ T (cid:16) θ, ( ˜ β g ) (cid:17) , (2.71)( α L, , α R, , Θ ) ∈ D θα , ( α L, , α R, , Θ ) ∈ D θα . (2.72) Then we have h (2) (cid:16) ω, α , θ, ( ˜ β g ) , ( η σg ) , ( α L, , α R, , Θ ) (cid:17) = h (2) (cid:16) ω, α , θ, ( ˜ β g ) , ( η σg ) , ( α L, , α R, , Θ ) (cid:17) . (2.73)12 efinition 2.12. From this lemma and because there are always α ∈ EAut( ω ) and ( α L , α R , Θ) ∈D θα for ω ∈ SL and 0 < θ < π by the definition, we may define h (3) (cid:16) ω, θ, ( ˜ β g ) , ( η σg ) (cid:17) := h (2) (cid:16) ω, α, θ, ( ˜ β g ) , ( η σg ) , ( α L , α R , Θ) (cid:17) (2.74)for any ω ∈ SL , , < θ < π , ( ˜ β g ) ∈ IG( ω, θ ) , ( η σg ) ∈ T (cid:16) θ, ( ˜ β g ) (cid:17) , (2.75)independent of the choice of α, ( α L , α R , Θ).
Proof.
By Lemma 2.1, there are(( W g, ) , ( u σ, ( g, h ))) ∈ IP (cid:16) ω, α , θ, ( ˜ β g ) , ( η σg ) , ( α L, , α R, , Θ ) (cid:17) . (2.76)For each i = 1 ,
2, we have Θ i ∈ Aut A C cθ and α i = (inner) ◦ α ,i ◦ Θ i , (2.77)setting α ,i := α L,i ⊗ α R,i . (2.78)Because ω ◦ α = ω = ω ◦ α , we have ω ◦ α ◦ α − = ω . Therefore, there is a unitary ˜ V on H such that π ◦ α ◦ α − = Ad (cid:16) ˜ V (cid:17) ◦ π . Substituting (2.77) to this, we see that there is a unitary V on H satisfying π ◦ α , ◦ Θ = Ad ( V ) ◦ π ◦ α , ◦ Θ . (2.79)From this, we obtainAd ( V W g, V ∗ ) ◦ π = Ad ( V W g, ) π ◦ α , ◦ Θ ◦ Θ − ◦ α − , = Ad ( V ) ◦ π ◦ α , ◦ Θ ◦ η g β Ug ◦ Θ − ◦ α − , ◦ α , ◦ Θ ◦ Θ − ◦ α − , = π ◦ α , ◦ Θ ◦ Θ − ◦ α − , ◦ α , ◦ Θ ◦ η g β Ug ◦ Θ − ◦ α − , ◦ α , ◦ Θ ◦ Θ − ◦ α − , = π ◦ α , ◦ Θ ◦ η g β Ug ◦ Θ − ◦ α − , , (2.80)for all g ∈ G . Furthermore, we haveAd ( V ( I H L ⊗ u R, ( g, h )) V ∗ ) ◦ π = Ad ( V ( I H L ⊗ u R, ( g, h ))) ◦ π ◦ α , ◦ Θ ◦ Θ − ◦ α − , = Ad ( V ) ◦ π ◦ (cid:16) id A HL ⊗ α R, η Rg β RUg η Rh (cid:0) β RUg (cid:1) − (cid:0) η Rgh (cid:1) − α − R, (cid:17) α , ◦ Θ ◦ Θ − ◦ α − , = π ◦ α , ◦ Θ ◦ Θ − ◦ α − , (cid:16) id A HL ⊗ α R, η Rg β RUg η Rh (cid:0) β RUg (cid:1) − (cid:0) η Rgh (cid:1) − α − R, (cid:17) ◦ α , ◦ Θ ◦ Θ − ◦ α − , = π ◦ α , ◦ Θ ◦ Θ − ◦ (cid:16) id A HL ⊗ η Rg β RUg η Rh (cid:0) β RUg (cid:1) − (cid:0) η Rgh (cid:1) − (cid:17) ◦ Θ ◦ Θ − ◦ α − , (2.81)Now, because η Rg β RUg η Rh (cid:0) β RUg (cid:1) − (cid:16) η Rgh (cid:17) − is an automorphism on A C θ and Θ ◦ Θ − is an auto-morphism on A C cθ , they commute. Therefore, we haveAd ( V ( I H L ⊗ u R, ( g, h )) V ∗ ) ◦ π = (2 .
81) = π ◦ α , ◦ (cid:16) id A HL ⊗ η Rg β RUg η Rh (cid:0) β RUg (cid:1) − (cid:0) η Rgh (cid:1) − (cid:17) ◦ α − , = π L ⊗ (cid:16) π R ◦ α R, η Rg β RUg η Rh (cid:0) β RUg (cid:1) − (cid:0) η Rgh (cid:1) − ( α R, ) − (cid:17) . (2.82)13rom this equality and the fact that π L is irreducible, we see that V ( I H L ⊗ u R, ( g, h )) V ∗ is of theform I H L ⊗ u R, ( g, h ) with some unitary u R, ( g, h ) on H R . This u R, ( g, h ) satisfiesAd ( u R, ( g, h )) ◦ π R = π R ◦ α R, η Rg β RUg η Rh (cid:0) β RUg (cid:1) − (cid:0) η Rgh (cid:1) − ( α R, ) − . (2.83)Analogously, we obtain a unitary u L, ( g, h ) on H L such that V ( u L, ( g, h ) ⊗ I H R ) V ∗ = u L, ( g, h ) ⊗ I H R , (2.84)Ad ( u L, ( g, h )) ◦ π L = π L ◦ α L, η Lg β LUg η Lh (cid:0) β LUg (cid:1) − (cid:0) η Lgh (cid:1) − ( α L, ) − . (2.85)From (2.80), (2.83), (2.84), we see that(( V W g, V ∗ ) , ( u σ, ( g, h ))) ∈ IP (cid:16) ω, α , θ, ( ˜ β g ) , ( η σg ) , ( α L, , α R, , Θ) (cid:17) . (2.86)Set c R, := c R (cid:16) ω, α , θ, ( ˜ β g ) , ( η σg ) , ( α L, , α R, , Θ ) , (( W g, ) , ( u σ, ( g, h ))) (cid:17) ,c R, := c R (cid:16) ω, α , θ, ( ˜ β g ) , ( η σg ) , ( α L, , α R, , Θ ) , ( V W g, V ∗ ) , ( u σ, ( g, h )) (cid:17) . (2.87)It suffices to show that c R, = c R, . This can be checked directly as follows: V ( I H L ⊗ u R, ( g, h ) u R, ( gh, k )) V ∗ = I H L ⊗ u R, ( g, h ) u R, ( gh, k )= c R, ( g, h, k ) (cid:0) V W g, V ∗ ( I H L ⊗ u R, ( h, k )) V W ∗ g, V ∗ (cid:1) ( I H L ⊗ u R, ( g, hk ))= c R, ( g, h, k ) V (cid:0) W g, ( I H L ⊗ u R, ( h, k )) W ∗ g, (cid:1) ( I H L ⊗ u R, ( g, hk )) V ∗ = c R, ( g, h, k ) c R, ( g, h, k ) V ( I H L ⊗ u R, ( g, h ) u R, ( gh, k )) V ∗ . (2.88) (cid:3) Lemma 2.13.
Let ω ∈ SL , < θ < π , ( ˜ β g ) ∈ IG( ω, θ ) , ( η σg ) , (˜ η σg ) ∈ T (cid:16) θ, ( ˜ β g ) (cid:17) . (2.89) Then we have h (3) (cid:16) ω, θ, ( ˜ β g ) , ( η σg ) (cid:17) = h (3) (cid:16) ω, θ, ( ˜ β g ) , (˜ η σg ) (cid:17) . (2.90) Definition 2.14.
From this lemma and the definition of IG( ω, θ ), we may define h (4) (cid:16) ω, θ, ( ˜ β g ) (cid:17) := h (3) (cid:16) ω, θ, ( ˜ β g ) , ( η σg ) (cid:17) (2.91)for any ω ∈ SL , < θ < π , ( ˜ β g ) ∈ IG( ω, θ ) , ( η σg ) ∈ T (cid:16) θ, ( ˜ β g ) (cid:17) , (2.92)independent of the choice of ( η σg ). Proof.
There are α ∈ EAut( ω ) and ( α L , α R , Θ) ∈ D θα for ω ∈ SL by the definition. We set α := α L ⊗ α R and η g := η Lg ⊗ η Rg , ˜ η g := ˜ η Lg ⊗ ˜ η Rg . By Lemma 2.1, there is some(( W g ) , ( u σ ( g, h ))) ∈ IP (cid:16) ω, α, θ, ( ˜ β g ) , ( η σg ) , ( α L , α R , Θ) (cid:17) . (2.93)14ecause ( η σg ) , (˜ η σg ) ∈ T (cid:16) θ, ( ˜ β g ) (cid:17) , we have˜ β g = (inner) ◦ (cid:0) η Lg ⊗ η Rg (cid:1) ◦ β Ug = (inner) ◦ (cid:0) ˜ η Lg ⊗ ˜ η Rg (cid:1) ◦ β Ug . (2.94)From this, we obtain ˜ η Lg ◦ (cid:0) η Lg (cid:1) − ⊗ ˜ η Rg ◦ (cid:0) η Rg (cid:1) − = (inner) , (2.95)hence there are unitaries v σg ∈ A H σ , σ = L, R such that˜ η σg ◦ (cid:0) η σg (cid:1) − = Ad (cid:0) v σg (cid:1) . (2.96)Because ˜ η σg , η σg are automorphisms on A C θ ,σ , v σg belongs to A C θ ,σ . (See Lemma B.1.) Setting v g := v Lg ⊗ v Rg , we obtain ˜ η g = Ad ( v g ) ◦ η g .Set ˜ W g := (cid:0)(cid:0) π L α L (cid:0) v Lg (cid:1)(cid:1) ⊗ (cid:0) π R α R (cid:0) v Rg (cid:1)(cid:1)(cid:1) W g , (2.97)˜ u σ ( g, h ) := π σ (cid:0) α σ (cid:0) v σg · (cid:0) η σg β σUg (cid:1) ( v σh ) (cid:1)(cid:1) · u σ ( g, h ) · π σ (cid:16) α σ (cid:16)(cid:0) v σgh (cid:1) ∗ (cid:17)(cid:17) (2.98)for each g, h ∈ G and σ = L, R . We claim that (cid:16)(cid:16) ˜ W g (cid:17) , (˜ u σ ( g, h )) (cid:17) ∈ IP (cid:16) ω, α, θ, ( ˜ β g ) , (˜ η σg ) , ( α L , α R , Θ) (cid:17) . (2.99)First, we have π ◦ α ◦ Θ ◦ ˜ η g β Ug ◦ Θ − ◦ α − = π ◦ α ◦ Θ ◦ Ad ( v g ) ◦ η g β Ug ◦ Θ − ◦ α − = π ◦ α ◦ Ad ( v g ) ◦ Θ ◦ η g β Ug ◦ Θ − ◦ α − = Ad (cid:0)(cid:0) π L α L (cid:0) v Lg (cid:1)(cid:1) ⊗ (cid:0) π R α R (cid:0) v Rg (cid:1)(cid:1)(cid:1) π ◦ α ◦ Θ ◦ η g β Ug ◦ Θ − ◦ α − = Ad (cid:16) ˜ W g (cid:17) ◦ π . (2.100)For the first equality, we substituted ˜ η g = Ad ( v g ) ◦ η g , and for the second equality, we used thefact that v σg belongs to A C θ ,σ , while Θ is an automorphism on A ( C θ ) c ,σ . The last equality followsfrom the definition of W g . On the other hand, we have π σ ◦ α σ ◦ ˜ η σg β σUg ˜ η σh (cid:0) β σUg (cid:1) − (cid:0) ˜ η σgh (cid:1) − ◦ α − σ = π σ ◦ α σ ◦ Ad (cid:0) v σg (cid:1) ◦ η σg β σUg Ad ( v σh ) ◦ η σh (cid:0) β σUg (cid:1) − (cid:0) η σgh (cid:1) − Ad (cid:0) v σgh ∗ (cid:1) ◦ α − σ = Ad (cid:0) π σ ◦ α σ (cid:0)(cid:0) v σg (cid:1) η σg β σUg ( v σh ) (cid:1)(cid:1) π σ ◦ α σ η σg β σUg η σh (cid:0) β σUg (cid:1) − (cid:0) η σgh (cid:1) − ◦ α − σ ◦ Ad (cid:0) α σ (cid:0) v σgh ∗ (cid:1)(cid:1) = Ad (cid:0) π σ ◦ α σ (cid:0)(cid:0) v σg (cid:1) η σg β σUg ( v σh ) (cid:1)(cid:1) ◦ Ad ( u σ ( g, h )) π σ ◦ Ad (cid:0) α σ (cid:0) v σgh ∗ (cid:1)(cid:1) = Ad (˜ u σ ( g, h )) ◦ π σ , (2.101)for all g, h ∈ G . For the first equality, we substituted ˜ η g = Ad ( v g ) ◦ η g . The third equality is thedefinition of u ( g, h ). Hence we have proven (2.99).Set c R := c R (cid:16) ω, α, θ, ( ˜ β g ) , ( η σg ) , ( α L , α R , Θ) , (( W g ) , ( u σ ( g, h ))) (cid:17) , ˜ c R := c R (cid:16) ω, α, θ, ( ˜ β g ) , (˜ η σg ) , ( α L , α R , Θ) , (cid:16) ( ˜ W g ) , (˜ u σ ( g, h )) (cid:17)(cid:17) . (2.102)15n order to show the statement of the Lemma, it suffices to show that c R = ˜ c R . Substituting thedefinition of ˜ u R , we obtain˜ u R ( g, h )˜ u R ( gh, k )= π R (cid:0) α R (cid:0) v Rg (cid:0) η Rg β RUg (cid:1) (cid:0) v Rh (cid:1)(cid:1)(cid:1) · u R ( g, h ) · π R (cid:16) α R (cid:16)(cid:0) v Rgh (cid:1) ∗ (cid:17)(cid:17) π R (cid:0) α R (cid:0) v Rgh (cid:0) η Rgh β RUgh (cid:1) (cid:0) v Rk (cid:1)(cid:1)(cid:1) · u R ( gh, k ) · π R (cid:16) α R (cid:16)(cid:0) v Rghk (cid:1) ∗ (cid:17)(cid:17) = π R (cid:0) α R (cid:0) v Rg (cid:0) η Rg β RUg (cid:1) (cid:0) v Rh (cid:1)(cid:1)(cid:1) · (cid:2) u R ( g, h ) · π R (cid:0) α R (cid:0)(cid:0) η Rgh β RUgh (cid:1) (cid:0) v Rk (cid:1)(cid:1)(cid:1)(cid:3) u R ( gh, k ) π R (cid:16) α R (cid:16)(cid:0) v Rghk (cid:1) ∗ (cid:17)(cid:17) = π R (cid:0) α R (cid:0) v Rg (cid:0) η Rg β RUg (cid:1) (cid:0) v Rh (cid:1)(cid:1)(cid:1) · (cid:2) Ad ( u R ( g, h )) (cid:0) π R (cid:0) α R (cid:0)(cid:0) η Rgh β RUgh (cid:1) (cid:0) v Rk (cid:1)(cid:1)(cid:1)(cid:1) · u R ( g, h ) (cid:3) u R ( gh, k ) · π R (cid:16) α R (cid:16)(cid:0) v Rghk (cid:1) ∗ (cid:17)(cid:17) = π R (cid:0) α R (cid:0) v Rg (cid:0) η Rg β RUg (cid:1) (cid:0) v Rh (cid:1)(cid:1)(cid:1) · (cid:16) π R (cid:16) α R ◦ η Rg β RUg η Rh (cid:0) β RUg (cid:1) − (cid:0) η Rgh (cid:1) − ◦ α − R α R (cid:0)(cid:0) η Rgh β RUgh (cid:1) (cid:0) v Rk (cid:1)(cid:1)(cid:17)(cid:17) · u R ( g, h ) u R ( gh, k ) · π R (cid:16) α R (cid:16)(cid:0) v Rghk (cid:1) ∗ (cid:17)(cid:17) = π R (cid:0) α R (cid:0) v Rg · (cid:0) η Rg β RUg (cid:1) (cid:0) v Rh (cid:1) · η Rg β RUg η Rh β RUh (cid:0) v Rk (cid:1)(cid:1)(cid:1) · u R ( g, h ) u R ( gh, k ) · π R (cid:16) α R (cid:16)(cid:0) v Rghk (cid:1) ∗ (cid:17)(cid:17) (2.103)For the fourth equality, we used the definition of u R . From the above equation, applying (2.50) tothe [ · ] part below, we have I H L ⊗ ˜ u R ( g, h )˜ u R ( gh, k )= I H L ⊗ π R (cid:0) α R (cid:0) v Rg · (cid:0) η Rg β RUg (cid:1) (cid:0) v Rh (cid:1) · η Rg β RUg η Rh β RUh (cid:0) v Rk (cid:1)(cid:1)(cid:1) · [ u R ( g, h ) u R ( gh, k )] · π R (cid:16) α R (cid:16)(cid:0) v Rghk (cid:1) ∗ (cid:17)(cid:17) = c R ( g, h, k ) (cid:0) I H L ⊗ π R (cid:0) α R (cid:0) v Rg · (cid:0) η Rg β RUg (cid:1) (cid:0) v Rh (cid:1) · η Rg β RUg η Rh β RUh (cid:0) v Rk (cid:1)(cid:1)(cid:1)(cid:1)(cid:8) W g ( I H L ⊗ u R ( h, k )) W ∗ g (cid:9) ( I H L ⊗ u R ( g, hk )) · π R (cid:16) α R (cid:16)(cid:0) v Rghk (cid:1) ∗ (cid:17)(cid:17) . (2.104)Now from the definition of ˜ u R , the {·} part above becomes W g ( I H L ⊗ u R ( h, k )) W ∗ g = Ad ( W g ) ◦ π ◦ (cid:16) id L ⊗ α R (cid:16)(cid:0) v Rh · η Rh β RUh (cid:0) v Rk (cid:1)(cid:1) ∗ (cid:17)(cid:17) · Ad ( W g ) ( I H L ⊗ ˜ u R ( h, k )) · (cid:0) Ad ( W g ) π (cid:0) id L ⊗ α R (cid:0) v Rhk (cid:1)(cid:1)(cid:1) . (2.105)Because v Rg belongs to A C θ ,R and η Rg is an automorphism on A C θ ,R while Θ is an automorphismon A ( C θ ) c and β Ug ( A C θ ,R ) = A C θ ,R , we haveAd ( W g ) ◦ π ◦ (cid:16) id L ⊗ α R (cid:16)(cid:0) v Rh · η Rh β RUh (cid:0) v Rk (cid:1)(cid:1) ∗ (cid:17)(cid:17) = π ◦ α ◦ Θ ◦ η g β Ug ◦ Θ − ◦ α − ◦ (cid:16) id L ⊗ α R (cid:16)(cid:0) v Rh · η Rh β RUh (cid:0) v Rk (cid:1)(cid:1) ∗ (cid:17)(cid:17) = π (cid:16) id L ⊗ α R ◦ η Rg β RUg (cid:16)(cid:0) v Rh · η Rh β RUh (cid:0) v Rk (cid:1)(cid:1) ∗ (cid:17)(cid:17) , andAd ( W g ) ◦ π (cid:0) id L ⊗ α R (cid:0) v Rhk (cid:1)(cid:1) = π (cid:0) id L ⊗ α R ◦ η Rg β RUg (cid:0) v Rhk (cid:1)(cid:1) . (2.106)Substituting this to (2.105), we obtain W g ( I H L ⊗ u R ( h, k )) W ∗ g = π (cid:16) id L ⊗ α R ◦ η Rg β RUg (cid:16)(cid:0) v Rh · η Rh β RUh (cid:0) v Rk (cid:1)(cid:1) ∗ (cid:17)(cid:17) · Ad ( W g ) ( I H L ⊗ ˜ u R ( h, k )) · π (cid:0) id L ⊗ α R ◦ η Rg β RUg (cid:0) v Rhk (cid:1)(cid:1) . (2.107)16ubstituting this, {·} part of (2.104), we obtain I H L ⊗ ˜ u R ( g, h )˜ u R ( gh, k )= c R ( g, h, k ) (cid:0) I H L ⊗ π R (cid:0) α R (cid:0) v Rg · (cid:0) η Rg β RUg (cid:1) (cid:0) v Rh (cid:1) · η Rg β RUg η Rh β RUh (cid:0) v Rk (cid:1)(cid:1)(cid:1)(cid:1) π (cid:16) id L ⊗ α R ◦ η Rg β RUg (cid:16)(cid:0) v Rh · η Rh β RUh (cid:0) v Rk (cid:1)(cid:1) ∗ (cid:17)(cid:17) · Ad ( W g ) ( I H L ⊗ ˜ u R ( h, k )) · π (cid:0) id L ⊗ α R ◦ η Rg β RUg (cid:0) v Rhk (cid:1)(cid:1)(cid:16) I H L ⊗ u R ( g, hk ) · π R (cid:16) α R (cid:16)(cid:0) v Rghk (cid:1) ∗ (cid:17)(cid:17)(cid:17) = c R ( g, h, k ) (cid:0) I H L ⊗ π R (cid:0) α R (cid:0) v Rg (cid:1)(cid:1)(cid:1) Ad ( W g ) ( I H L ⊗ ˜ u R ( h, k )) · π (cid:0) id L ⊗ α R ◦ η Rg β RUg (cid:0) v Rhk (cid:1)(cid:1) π ◦ (cid:16) id L ⊗ α R (cid:16)(cid:0) v Rg · η Rg β RUg (cid:0) v Rhk (cid:1)(cid:1) ∗ (cid:17)(cid:17) · ( I H L ⊗ ˜ u R ( g, hk ) · ) π (cid:16) id L ⊗ α R (cid:0) v Rghk (cid:1) α R (cid:16)(cid:0) v Rghk (cid:1) ∗ (cid:17)(cid:17) = c R ( g, h, k ) Ad (cid:0)(cid:0) I H L ⊗ π R (cid:0) α R (cid:0) v Rg (cid:1)(cid:1)(cid:1) W g (cid:1) ( I H L ⊗ ˜ u R ( h, k )) · ( I H L ⊗ ˜ u R ( g, hk ))= c R ( g, h, k ) n Ad (cid:16) π L (cid:16) α L (cid:16) v Lg ∗ (cid:17) ⊗ I H R (cid:17)(cid:17) Ad ˜ W g ( I H L ⊗ ˜ u R ( h, k )) o · ( I H L ⊗ ˜ u R ( g, hk )) (2.108)Because of (iii) of Lemma 2.3, the {·} part of the last equation is equal to Ad ˜ W g ( I H L ⊗ ˜ u R ( h, k )).Hence we obtain I H L ⊗ ˜ u R ( g, h )˜ u R ( gh, k ) = c R ( g, h, k ) Ad ˜ W g ( I H L ⊗ ˜ u R ( h, k )) · ( I H L ⊗ ˜ u R ( g, hk )) . (2.109)This proves c R = ˜ c R , completing the proof. (cid:3) Lemma 2.15.
Let ω ∈ SL , < θ < π , ( ˜ β (1) g ) , ( ˜ β (2) g ) ∈ IG( ω, θ ) . (2.110) Then we have h (4) (cid:16) ω, θ, ( ˜ β (1) g ) (cid:17) = h (4) (cid:16) ω, θ, ( ˜ β (2) g ) (cid:17) . (2.111) Definition 2.16.
From this lemma we may define h (5) ( ω, θ ) := h (4) (cid:16) ω, θ, ( ˜ β g ) (cid:17) (2.112)for any ω ∈ SL , < θ < π , with IG( ω, θ ) = ∅ (2.113)independent of the choice of ( ˜ β g ). Proof.
By the definition of IG( ω, θ ), there are( η σg,i ) g ∈ G,σ = L,R ∈ T ( θ, ( ˜ β ( i ) g )) , for i = 1 , . (2.114)We set η g,i := η Lg,i ⊗ η Rg,i , for i = 1 ,
2. There are α ∈ EAut( ω ) and ( α L , α R , Θ) ∈ D θα for ω ∈ SL bythe definition. Setting α := α L ⊗ α R , we have α = (inner) ◦ α ◦ Θ. By Lemma 2.1, there is some (cid:16) ( W g, ) , ( u (1) σ ( g, h )) (cid:17) ∈ IP (cid:16) ω, α, θ, ( ˜ β (1) g ) , ( η σg, ) , ( α L , α R , Θ) (cid:17) . (2.115)17et K σg := η σg, ◦ (cid:0) η σg, (cid:1) − ∈ Aut ( A C θ ,σ ) , for σ = L, R, g ∈ G, K g := K Lg ⊗ K Rg ∈ Aut ( A C θ ) . (2.116)We claim that there are unitaries V σg , g ∈ G, σ = L, R on H σ such thatAd (cid:0) V σg (cid:1) ◦ π σ = π σ ◦ α σ ◦ K σg ◦ ( α σ ) − . (2.117)To see this, note that ω = ω ◦ ˜ β ( i ) g = ω ◦ α ◦ ˜ β ( i ) g ∼ q.e. ω ◦ α ◦ Θ ◦ (cid:0) η Lg,i ⊗ η Rg,i (cid:1) ◦ β Ug , i = 1 , . (2.118)Therefore, we have ω ◦ α ◦ Θ ◦ (cid:0) η Lg, ⊗ η Rg, (cid:1) ∼ q.e. ω ◦ (cid:0) β Ug (cid:1) − ∼ q.e. ω ◦ α ◦ Θ ◦ (cid:0) η Lg, ⊗ η Rg, (cid:1) , (2.119)and then using the fact that Θ ∈ Aut( A C cθ ) and K g ∈ Aut( A C θ ), ω ∼ q.e. ω ◦ α ◦ Θ ◦ K g ◦ Θ − ◦ α − = ω ◦ α ◦ K g ◦ ( α ) − = O σ = L,R ω σ ◦ α σ K σg ( α σ ) − . (2.120)This implies that ω σ and ω σ ◦ α σ K σg ( α σ ) − are quasi-equivalent. Because π σ is irreducible, thisimplies the existence of a unitary V σg on H σ satisfying (2.117), proving the claim.Next we claim that there are unitaries v σg,h on H σ , for g, h ∈ G and σ = L, R such thatAd W g, (cid:0) I H L ⊗ V Rh (cid:1) = I H L ⊗ v Rg,h , Ad W g, (cid:0) V Lh ⊗ I H R (cid:1) = v Lg,h ⊗ I H R , (2.121)and Ad (cid:16) V σg v σg,h u (1) σ ( g, h ) (cid:0) V σgh (cid:1) ∗ (cid:17) π σ = π σ ◦ α σ η σg, β σUg η σh, (cid:0) β σUg (cid:1) − (cid:0) η σgh, (cid:1) − α − σ , (2.122)for any g, h ∈ G and σ = L, R . To see this, first we calculateAd (cid:0) W g, (cid:0) I H L ⊗ V Rh (cid:1) ( W g, ) ∗ (cid:1) ◦ π = Ad (cid:0) W g, (cid:0) I H L ⊗ V Rh (cid:1)(cid:1) π ◦ α ◦ Θ ◦ (cid:0) η g, β Ug (cid:1) − ◦ Θ − ◦ α − = π ◦ α ◦ Θ ◦ η g, β Ug ◦ Θ − ◦ α − ◦ (cid:16) id L ⊗ α R ◦ K Rh ◦ ( α R ) − (cid:17) ◦ α ◦ Θ ◦ (cid:0) η g, β Ug (cid:1) − ◦ Θ − ◦ α − = π ◦ α ◦ Θ ◦ η g, β Ug ◦ Θ − ◦ (cid:0) id L ⊗ K Rh (cid:1) ◦ Θ ◦ (cid:0) η g, β Ug (cid:1) − ◦ Θ − ◦ α − = π ◦ α ◦ Θ ◦ η g, β Ug ◦ (cid:0) id L ⊗ K Rh (cid:1) ◦ (cid:0) η g, β Ug (cid:1) − ◦ Θ − ◦ α − = π ◦ α ◦ Θ ◦ (cid:16) id L ⊗ η Rg, β RUg K Rh (cid:0) η Rg, β RUg (cid:1) − (cid:17) ◦ Θ − ◦ α − = π ◦ (cid:16) id L ⊗ α R ◦ η Rg, β RUg K Rh (cid:0) η Rg, β RUg (cid:1) − α − R (cid:17) . (2.123)Here, in the fourth and sixth equality, we used the fact that K Rh , η Rg, β RUg K Rh (cid:0) η Rg, β RUg (cid:1) − ∈ Aut ( A C θ ) and Θ ∈ Aut (cid:0) A C cθ (cid:1) commute, in order to remove Θ. Equation (2.123) and the fact that π L is irreducible imply that there is a unitary v Rg,h satisfying (2.121). The same argument impliesthe existence of v Lg,h satisfying (2.121).For this v Rg,h , we would like to show (2.122). Rewriting η σg, β σUg η σh, (cid:0) β σUg (cid:1) − (cid:0) η σgh, (cid:1) − = K σg · (cid:16) η σg, β σUg K σh (cid:0) η σg, β σUg (cid:1) − (cid:17) · η σg, β σUg η σh, (cid:0) β σUg (cid:1) − (cid:0) η σgh, (cid:1) − · (cid:0) K σgh (cid:1) − , (2.124)18e obtain π L ⊗ π R ◦ α R η Rg, β RUg η Rh, (cid:0) β RUg (cid:1) − (cid:0) η Rgh, (cid:1) − α − R = π ◦ (cid:16) id L ⊗ α R ◦ K Rg · (cid:16) η Rg, β RUg K Rh (cid:0) η Rg, β RUg (cid:1) − (cid:17) · η Rg, β RUg η Rh, (cid:0) β RUg (cid:1) − (cid:0) η Rgh, (cid:1) − · (cid:0) K Rgh (cid:1) − α − R (cid:17) = π L ⊗ Ad (cid:16) V Rg v Rg,h u (1) R ( g, h ) (cid:0) V Rgh (cid:1) ∗ (cid:17) π R , (2.125)substituting (2.117), (2.123), (2.121). This proves (2.122) for σ = R . Analogous result for σ = L can be proven by the same argument. Hence we have proven the claim (2.123) and (2.122).Setting V g := V Lg ⊗ V Rg ∈ U ( H ) , (2.126)we have Ad ( V g W g, ) ◦ π = π ◦ α ◦ K g ◦ α − ◦ α ◦ Θ ◦ η g, ◦ β Ug ◦ Θ − ◦ α − = π ◦ α ◦ Θ ◦ η g, ◦ β Ug ◦ Θ − ◦ α − . (2.127)In the last equality, we used the definition of K g and the commutativity of Θ and K g again. From(2.127) and (2.122), setting u (2) σ ( g, h ) := V σg v σg,h u (1) σ ( g, h ) (cid:0) V σgh (cid:1) ∗ , (2.128)we see that (cid:16) ( V g W g, ) , (cid:16) u (2) R ( g, h ) (cid:17)(cid:17) ∈ IP (cid:16) ω, α, θ, ( ˜ β (2) g ) , ( η σg, ) , ( α L , α R , Θ) (cid:17) , (2.129)and I H L ⊗ u (2) R ( g, h ) = (cid:0) I H L ⊗ V Rg (cid:1) W g, (cid:0) I H L ⊗ V Rh (cid:1) ( W g, ) ∗ (cid:16) I H L ⊗ u (1) R ( g, h ) (cid:0) V Rgh (cid:1) ∗ (cid:17) . (2.130)Now we set c R, := c R (cid:16) ω, α, θ, ( ˜ β (1) g ) , ( η σg, ) , ( α L , α R , Θ) , (cid:16) ( W g, ) , ( u (1) σ ( g, h )) (cid:17)(cid:17) ,c R, := c R (cid:16) ω, α, θ, ( ˜ β (2) g ) , ( η σg, ) , ( α L , α R , Θ) , (cid:16) ( V g W g, ) , (cid:16) u (2) R ( g, h ) (cid:17)(cid:17)(cid:17) . (2.131)To prove the Lemma, it suffices to show c R, = c R, . By (2.130), we have I H L ⊗ u (2) R ( g, h ) u (2) R ( gh, k )= (cid:0) I H L ⊗ V Rg (cid:1) W g, (cid:0) I H L ⊗ V Rh (cid:1) ( W g, ) ∗ (cid:16) I H L ⊗ u (1) R ( g, h ) (cid:0) V Rgh (cid:1) ∗ (cid:17) · (cid:0) I H L ⊗ V Rgh (cid:1) W gh, (cid:0) I H L ⊗ V Rk (cid:1) ( W gh, ) ∗ (cid:16) I H L ⊗ u (1) R ( gh, k ) (cid:0) V Rghk (cid:1) ∗ (cid:17) = (cid:0) I H L ⊗ V Rg (cid:1) W g, (cid:0) I H L ⊗ V Rh (cid:1) ( W g, ) ∗ (cid:16) I H L ⊗ u (1) R ( g, h ) (cid:17) · W gh, (cid:0) I H L ⊗ V Rk (cid:1) ( W gh, ) ∗ (cid:16) I H L ⊗ u (1) R ( gh, k ) (cid:0) V Rghk (cid:1) ∗ (cid:17) = (cid:0) I H L ⊗ V Rg (cid:1) W g, (cid:0) I H L ⊗ V Rh (cid:1) ( W g, ) ∗ n Ad (cid:16)(cid:16) I H L ⊗ u (1) R ( g, h ) (cid:17) · W gh, (cid:17) (cid:0) I H L ⊗ V Rk (cid:1)o · (2.132) (cid:16) I H L ⊗ h u (1) R ( g, h ) u (1) R ( gh, k ) i (cid:0) V Rghk (cid:1) ∗ (cid:17) = c R, ( g, h, k ) (cid:0) I H L ⊗ V Rg (cid:1) W g, (cid:0) I H L ⊗ V Rh (cid:1) ( W g, ) ∗ (cid:8) Ad ( W g, W h, ) (cid:0) I H L ⊗ V Rk (cid:1)(cid:9) · (cid:16) W g, (cid:16) I H L ⊗ u (1) R ( h, k ) (cid:17) W ∗ g, (cid:17) (cid:16) I H L ⊗ u (1) R ( g, hk ) (cid:0) V Rghk (cid:1) ∗ (cid:17) . (2.133)19e used (2.50) for [ · ] part and Lemma 2.3 (ii) and (2.121) for {·} part for the fourth equality.Again using (2.130), we have I H L ⊗ u (2) R ( g, h ) u (2) R ( gh, k ) = (2 . c R, ( g, h, k ) (cid:0) I H L ⊗ V Rg (cid:1) W g, (cid:0) I H L ⊗ V Rh (cid:1) (cid:8) Ad ( W h, ) (cid:0) I H L ⊗ V Rk (cid:1)(cid:9) · (cid:16) W h, (cid:16) I H L ⊗ (cid:0) V Rk (cid:1) ∗ (cid:17) ( W h, ) ∗ (cid:0) I H L ⊗ V Rh (cid:1) ∗ (cid:16) I H L ⊗ u (2) R ( h, k ) (cid:17) (cid:0) I H L ⊗ (cid:0) V Rhk (cid:1)(cid:1) ( W g, ) ∗ (cid:17) W g, (cid:16) I H L ⊗ (cid:0) V Rhk (cid:1) ∗ (cid:17) ( W g, ) ∗ (cid:0) I H L ⊗ V Rg (cid:1) ∗ (cid:16) I H L ⊗ u (2) R ( g, hk ) (cid:17) (cid:0) I H L ⊗ (cid:0) V Rghk (cid:1)(cid:1) (cid:16) I H L ⊗ (cid:0) V Rghk (cid:1) ∗ (cid:17) = c R, ( g, h, k ) (cid:0) I H L ⊗ V Rg (cid:1) W g, · (cid:16)(cid:16) I H L ⊗ u (2) R ( h, k ) (cid:17)(cid:17) · ( W g, ) ∗ (cid:0) I H L ⊗ V Rg (cid:1) ∗ (cid:16) I H L ⊗ u (2) R ( g, hk ) (cid:17) = c R, ( g, h, k ) · Ad (cid:0)(cid:0) I H L ⊗ V Rg (cid:1) W g, (cid:1) (cid:16)(cid:16) I H L ⊗ u (2) R ( h, k ) (cid:17)(cid:17) · (cid:16) I H L ⊗ u (2) R ( g, hk ) (cid:17) = c R, ( g, h, k ) · Ad (cid:16)(cid:16) V Lg ∗ ⊗ I H R (cid:17) V g W g, (cid:17) (cid:16)(cid:16) I H L ⊗ u (2) R ( h, k ) (cid:17)(cid:17) · (cid:16) I H L ⊗ u (2) R ( g, hk ) (cid:17) = c R, ( g, h, k ) · Ad ( V g W g, ) (cid:16)(cid:16) I H L ⊗ u (2) R ( h, k ) (cid:17)(cid:17) · (cid:16) I H L ⊗ u (2) R ( g, hk ) (cid:17) (2.135)In the last line we used (2.129) and Lemma 2.3 (iii) to remove V Lg ∗ . From this, we see that c R, = c R, , completing the proof. (cid:3) Lemma 2.17.
Let ω ∈ SL , < θ < θ < π , with IG( ω, θ ) , IG( ω, θ ) = ∅ . (2.136) Then we have h (5) ( ω, θ ) = h (5) ( ω, θ ) . (2.137) Definition 2.18.
From this lemma, for any ω ∈ SL with IG( ω ) = ∅ , we may define h ( ω ) := h (5) ( ω, θ ) (2.138)independent of the choice of θ . This is the index we associate to ω ∈ SL with IG( ω ) = ∅ . Proof.
By the assumption, there are some ( ˜ β g ) ∈ IG( ω, θ ) and ( η σg ) ∈ T (cid:16) ( θ , ˜ β g ) (cid:17) . Because ω ∈ SL , there are α ∈ EAut( ω ) and ( α L , α R , Θ) ∈ D θ α by the definition. Setting α := α L ⊗ α R ,we have α = (inner) ◦ α ◦ Θ. Because 0 < θ < θ < π , we also have ( η σg ) ∈ T (cid:16) ( θ , ˜ β g ) (cid:17) , and( ˜ β g ) ∈ IG( ω, θ ). For the same reason, we also have ( α L , α R , Θ) ∈ D θ α .By Lemma 2.1, there is some(( W g ) , ( u σ ( g, h ))) ∈ IP (cid:16) ω, α, θ , ( ˜ β g ) , ( η σg ) , ( α L , α R , Θ) (cid:17) . (2.139)However, we also have(( W g ) , ( u σ ( g, h ))) ∈ IP (cid:16) ω, α, θ , ( ˜ β g ) , ( η σg ) , ( α L , α R , Θ) (cid:17) . (2.140)Therefore, we obtain h (5) ( ω, θ ) = h (5) ( ω, θ ). (cid:3) This completes the proof of Theorem 2.7. 20
The existence of ˜ β for SPT phases In this section, we give a sufficient condition for IG( ω ) to be non-empty. We consider the samesetting as in subsection 2.1. Theorem 3.1.
For any < θ < π and α ∈ SQAut( A ) satisfying ω ◦ α ◦ β g = ω ◦ α for all g ∈ G , IG( ω ◦ α, θ ) is not empty. In order to prove this theorem, we first show several general lemmas.
Lemma 3.2.
Let A , B be UHF-algebras. Let ω be a pure state on A ⊗ B and ϕ A , ϕ B states on A , B respectively. Assume that ω is quasi-equivalent to ϕ A ⊗ ϕ B . Then for any pure states ψ A , ψ B on A , B , there are automorphisms γ A ∈ Aut ( A ) , γ B ∈ Aut ( B ) and a unitary u ∈ U ( A ⊗ B ) such that ω = (( ψ A ◦ γ A ) ⊗ ( ψ B ◦ γ B )) ◦ Ad( u ) . (3.1) If ψ A and ϕ A are quasi-equivalent, then we may set γ A = id A . Proof.
Let ( H ω , π ω , Ω ω ) ( H ϕ A , π ϕ A , Ω ϕ A ) ( H ϕ B , π ϕ B , Ω ϕ B ) be GNS triples of ω , ϕ A , ϕ B re-spectively. Then ( H ϕ A ⊗ H ϕ B , π ϕ A ⊗ π ϕ B , Ω ϕ A ⊗ Ω ϕ B ) is a GNS triple of ϕ A ⊗ ϕ B . As ω isquasi-equivalent to ϕ A ⊗ ϕ B , there is a ∗ -isomorphism τ : π ω ( A ⊗ B ) ′′ → π ϕ A ( A ) ′′ ⊗ π ϕ B ( B ) ′′ such that τ ◦ π ω = π ϕ A ⊗ π ϕ B . Because ω is pure, we have π ω ( A ⊗ B ) ′′ = B ( H ω ) and from theisomorphism τ , π ϕ A ( A ) ′′ ⊗ π ϕ B ( B ) ′′ is also a type I factor. Then from Theorem 2.30 V [T], bothof π ϕ A ( A ) ′′ and π ϕ B ( B ) ′′ are type I factors. The restriction of τ to π ω ( A ⊗ CI B ) ′′ implies a ∗ -isomorphism from π ω ( A ⊗ CI B ) ′′ onto the type I factor π ϕ A ( A ) ′′ . Hence we see that π ω ( A ⊗ CI B ) ′′ is a type I -factor. Therefore, from Theorem 1.31 V of [T], there are Hilbert spaces K A , K B anda unitary W : H ω → K A ⊗ K B such that Ad ( W ) (cid:0) π ω ( A ⊗ CI B ) ′′ (cid:1) = B ( K A ) ⊗ CI K B . Because ω is pure, we also have Ad ( W ) (cid:0) π ω ( CI A ⊗ B ) ′′ (cid:1) = CI K A ⊗ B ( K B ). From this, we see that thereare irreducible representations ρ A , ρ B of A and B on K A , K B such that Ad( W ) ◦ π ω = ρ A ⊗ ρ B .Fix some unit vectors ξ A ∈ K A , ξ B ∈ K B . Then because of the irreducibility of ρ B and ρ B , ω A := h ξ A , ρ A ( · ) ξ A i and ω B := h ξ B , ρ B ( · ) ξ B i are pure states on A , B . By Theorem 1.1 of [KOS](originally proved by Powers [P] for UHF-algebras) for any pure states ψ A , ψ B on A , B , thereexist automorphisms γ A ∈ Aut( A ) γ B ∈ Aut( B ) such that ω A = ψ A ◦ γ A ω B = ψ B ◦ γ B . Now forunit vectors W ∗ ( ξ A ⊗ ξ B ) , Ω ω ∈ H ω , by Kadison’s transitivity theorem and the irreducibility of π ω , there exists a unitary u ∈ U ( A ⊗ B ) such that π ω ( u )Ω ω = W ∗ ( ξ A ⊗ ξ B ). Substituting this,we obtain ω = h Ω ω , π ω ( · ) Ω ω i = h π ω ( u ∗ ) W ∗ ( ξ A ⊗ ξ B ) , π ω ( · ) π ω ( u ∗ ) W ∗ ( ξ A ⊗ ξ B ) i = h W ∗ ( ξ A ⊗ ξ B ) , π ω ◦ Ad( u ) ( · ) W ∗ ( ξ A ⊗ ξ B ) i = h ( ξ A ⊗ ξ B ) , ( ρ A ⊗ ρ B ) ◦ Ad( u ) ( · ) ( ξ A ⊗ ξ B ) i = ( ω A ⊗ ω B ) ◦ Ad( u ) = ( ψ A ◦ γ A ⊗ ψ B ◦ γ B ) ◦ Ad( u ) . (3.2)Now assume that ψ A and ϕ A are quasi-equivalent, i.e., the GNS representations of ψ A , ϕ A , denotedby π ψ A and π ϕ A are quasi-equivalent. From the above argument, π ω | A and π ϕ A are quasi-equivalent.At the same time, π ω | A and ρ A are quasi-equivalent. Therefore, π ψ A and ρ A are quasi-equivalent.Because both of them are irreducible, we see that a pure state ψ A can be represented by a unitvector ζ ∈ K A as ψ A = h ζ, ρ A ( · ) ζ i . Because ρ A is irreducible, by Kadison’s transitivity theorem,there exists a unitary w ∈ U ( A ) such that ρ A ( w ∗ ) ζ = ξ A . Hence we obtain ψ A ◦ Ad( w ) = ω A .Substituting this instead of ω A = ψ A ◦ γ A in (3.2), we obtain ω = ( ψ A ⊗ ψ B ◦ γ B ) ◦ Ad (( w ⊗ id B ) u ) , (3.3)proving the last claim. (cid:3) emma 3.3. Let B , A ,L , A ,L , A ,R , A ,R be UHF-algebras. Set A := A ,L ⊗ A ,R , A := A ,L ⊗ A ,R , A L := A ,L ⊗ A ,L , and A R := A ,R ⊗ A ,R . Let ω , ϕ (1 , L , ϕ (1 , R , ψ be pure states on B ⊗ A , A L , A R , B , respectively. Suppose that ω is quasi-equivalent to (cid:16) ψ ⊗ ϕ (1 , L ⊗ ϕ (1 , R (cid:17)(cid:12)(cid:12)(cid:12) B ⊗ A . Thenfor any pure states ϕ (1) L , ϕ (1) R on A ,L , A ,R respectively, there are automorphisms γ (1) L ∈ Aut ( A ,L ) , γ (1) R ∈ Aut ( A ,R ) , and a unitary u ∈ U ( B ⊗ A ) such that ω = (cid:16) ψ ⊗ (cid:16) ϕ (1) L ◦ γ (1) L (cid:17) ⊗ (cid:16) ϕ (1) R ◦ γ (1) R (cid:17)(cid:17) ◦ Ad u. (3.4) Proof.
Because the pure state ω is quasi-equivalent to (cid:16) ψ ⊗ ϕ (1 , L ⊗ ϕ (1 , R (cid:17)(cid:12)(cid:12)(cid:12) B ⊗ A = ψ ⊗ (cid:16) ϕ (1 , L ⊗ ϕ (1 , R (cid:17)(cid:12)(cid:12)(cid:12) A ,applying Lemma 3.2, for any pure states ϕ (1) L , ϕ (1) R on A ,L , A ,R , there exist an automorphism S ∈ Aut A and a unitary v ∈ U ( B ⊗ A ) such that ω = (cid:16) ψ ⊗ (cid:16)(cid:16) ϕ (1) L ⊗ ϕ (1) R (cid:17) ◦ S (cid:17)(cid:17) ◦ Ad v. (3.5)From (3.5) and ω ∼ q.e. (cid:16) ψ ⊗ ϕ (1 , L ⊗ ϕ (1 , R (cid:17)(cid:12)(cid:12)(cid:12) B ⊗ A , we get (cid:16) ψ ⊗ (cid:16)(cid:16) ϕ (1) L ⊗ ϕ (1) R (cid:17) ◦ S (cid:17)(cid:17) ∼ q.e. (cid:16) ψ ⊗ ϕ (1 , L ⊗ ϕ (1 , R (cid:17)(cid:12)(cid:12)(cid:12) B ⊗ A ,which implies (cid:16) ϕ (1) L ⊗ ϕ (1) R (cid:17) ◦ S ∼ q.e. (cid:16) ϕ (1 , L ⊗ ϕ (1 , R (cid:17)(cid:12)(cid:12)(cid:12) A . (3.6)Applying Lemma 3.2 to (3.6), there are automorphisms γ (1) L ∈ Aut ( A ,L ), γ (1) R ∈ Aut ( A ,R ), anda unitary w ∈ U ( A ) such that (cid:16) ϕ (1) L ⊗ ϕ (1) R (cid:17) ◦ S = (cid:16)(cid:16) ϕ (1) L ◦ γ (1) L (cid:17) ⊗ (cid:16) ϕ (1) R ◦ γ (1) R (cid:17)(cid:17) ◦ Ad w. (3.7)Substituting this to (3.5), we obtain (3.4). (cid:3) Lemma 3.4.
Let A L , A R , B LU , B LD , B RU , B RD , C U , C D be UHF-algebras, and set B U := B LU ⊗ B RU , B D := B LD ⊗ B RD , B L := B LD ⊗ B LU , B R := B RD ⊗ B RU , A := A L ⊗ A R , B := B D ⊗ B U = B L ⊗ B R , C := C D ⊗ C U , D := A ⊗ B ⊗ C . (3.8) Let ω X be a pure state on each X = A L , A R , B LU , B LD , B RU , B RD , C U , C D , and set ω U BC := ω B LU ⊗ ω B RU ⊗ ω C U , on B U ⊗ C U ω D BC := ω B LD ⊗ ω B RD ⊗ ω C D , on B D ⊗ C D ,ω A := ω A L ⊗ ω A R on A ω L AB := ω A L ⊗ ω B LU ⊗ ω B LD on A L ⊗ B L ω R AB := ω A R ⊗ ω B RU ⊗ ω B RD on A R ⊗ B R ω := O X = A L , A R , B LU , B LD , B RU , B RD , C U , C D ω X , on D . (3.9) Let α, ˆ α be automorphisms on D which allow the following decompositions ˆ α = (cid:0) ρ U BC ⊗ id A ⊗ ρ D BC (cid:1) ◦ (cid:0) id C U ⊗ ˆ γ L AB ⊗ ˆ γ R AB ⊗ id C D (cid:1) ◦ (inner) (3.10) α = (cid:0) ρ U BC ⊗ id A ⊗ id B D ⊗ C D (cid:1) ◦ (cid:0) id C U ⊗ γ L AB ⊗ γ R AB ⊗ id C D (cid:1) ◦ (inner) . (3.11)22 ere, ρ U BC / ρ D BC are automorphisms on B U ⊗ C U / B D ⊗ C D respectively. For each σ = L, R , γ σ AB , ˆ γ σ AB are automorphisms on A σ ⊗ B σD ⊗ B σU . Suppose that ω ◦ ˆ α = ω . Then there areautomorphisms η L , η R on A L ⊗ B LD ⊗ B LU , A R ⊗ B RD ⊗ B RU such that ω ◦ α is quasi-equivalentto ω ◦ (id C U ⊗ η L ⊗ η R ⊗ id C D ) . Proof.
First we claim that there are automorphisms θ LU B ∈ Aut B LU , θ RU B ∈ Aut B RU and aunitary u ∈ U (cid:0) B U ⊗ C U (cid:1) such that ω U BC ◦ ρ U BC = ω U BC ◦ (cid:0) θ LU B ⊗ θ RU B ⊗ id C U (cid:1) ◦ Ad ( u ) . (3.12)To prove this, we first note that from ω ◦ ˆ α = ω and the decomposition (3.10), we have ω U BC ◦ ρ U BC ⊗ ω A ⊗ ω D BC ◦ ρ D BC ∼ q.e. ω C U ⊗ ω L AB ◦ (cid:16) d γ L AB (cid:17) − ⊗ ω R AB ◦ (cid:16) d γ R AB (cid:17) − ⊗ ω C D . (3.13)From this, because both of the states above are pure, (hence the restrictions of their GNS repre-sentations onto C U ⊗ B U are factors) we have ω U BC ◦ ρ U BC = (cid:0) ω U BC ◦ ρ U BC ⊗ ω A ⊗ ω D BC ◦ ρ D BC (cid:1)(cid:12)(cid:12) C U ⊗ B U ∼ q.e. ω C U ⊗ (cid:18) ω L AB ◦ (cid:16) d γ L AB (cid:17) − ⊗ ω R AB ◦ (cid:16) d γ R AB (cid:17) − (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) B U . (3.14)We apply Lemma 3.3 for B , A L , A R , A L , A R , ω , ϕ (1 , L , ϕ (1 , R , ψ replaced by C U , B LU , B RU , A L ⊗ B LD , A R ⊗ B RD , ω U BC ◦ ρ U BC , ω L AB ◦ (cid:16) d γ L AB (cid:17) − , ω R AB ◦ (cid:16) d γ R AB (cid:17) − , ω C U respectively. From (3.14),they satisfy the conditions in Lemma 3.3. Applying Lemma 3.3, (for pure states ϕ (1) L = ω B LU and ϕ (1) R = ω B RU ) we obtain automorphisms θ LU B ∈ Aut ( B LU ), θ RU B ∈ Aut ( B RU ), and a unitary u ∈ U ( B U ⊗ C U ) satisfying (3.12).We set η L := (cid:0) θ LU B ⊗ id A L ⊗ id B LD (cid:1) ◦ γ L AB ∈ Aut ( B LU ⊗ A L ⊗ B LD ) η R := (cid:0) θ RU B ⊗ id A R ⊗ id B RD (cid:1) ◦ γ R AB ∈ Aut ( B RU ⊗ A R ⊗ B RD ) . (3.15)Then we have ω ◦ α = (cid:0) ω A L ⊗ ω A R ⊗ ω U BC ⊗ ω D BC (cid:1) ◦ α ∼ q.e. (cid:0) ω A L ⊗ ω A R ⊗ ω U BC ◦ ρ U BC ⊗ ω D BC (cid:1) ◦ (cid:0) id C U ⊗ γ L AB ⊗ γ R AB ⊗ id C D (cid:1) ∼ q.e. (cid:0) ω A L ⊗ ω A R ⊗ ω U BC ⊗ ω D BC (cid:1) ◦ (cid:0) id C U ⊗ (cid:0)(cid:0) θ LU B ⊗ id A L ⊗ id B LD (cid:1) ◦ γ L AB (cid:1) ⊗ (cid:0)(cid:0) θ RU B ⊗ id A R ⊗ id B RD (cid:1) ◦ γ R AB (cid:1) ⊗ id C D (cid:1) = ω ◦ (id C U ⊗ η L ⊗ η R ⊗ id C D ) . (3.16)This completes the proof. (cid:3) Now we are ready to prove Theorem 3.1.
Proof of Theorem 3.1.
Let 0 < θ < π and α ∈ SQAut( A ) satisfying ω ◦ α ◦ β g = ω ◦ α forall g ∈ G . We would like to show that IG( ω ◦ α, θ ) is not empty.Let us set θ . := θ and consider θ . , θ , θ . , θ . , θ , θ . , θ , θ . satisfying (2.9) for this θ . . Because α ∈ SQAut( A ), there is a decompotision given by (2.10), (2.11), (2.12). Using thisdecomposition, set α := α D ⊗ α U α ζ := (cid:16) α ( θ ,θ ] ,ζ ⊗ α ( θ ,θ ] ,ζ ⊗ α ( θ , π ] ,ζ (cid:17) ◦ (cid:0) α ( θ . ,θ . ] ,ζ ⊗ α ( θ . ,θ . ] ,ζ ⊗ α ( θ . ,θ . ] ,ζ (cid:1) ∈ Aut (cid:18) A (( C θ . ) c ) ζ (cid:19) , ζ = U, D,α := α [0 ,θ ] ∈ Aut (cid:0) A C θ (cid:1) . (3.17)23e have α = (inner) ◦ α ◦ α .We would like to show that (cid:0) α ◦ β Ug ◦ α − , α ◦ β g ◦ α − (cid:1) satisfy the conditions of ( α, ˆ α ) inLemma 3.4. We first show that they satisfy a decomposition corresponding to (3.10) and (3.11).For Γ = Z , H U , we have (cid:0) β Γ g (cid:1) − α ◦ β Γ g ◦ α − = (inner) ◦ (cid:0) β Γ g (cid:1) − ◦ (cid:0) α β Γ g α − (cid:1) (cid:0) α β Γ g α − (cid:1) − α α β Γ g α − α − . (3.18)The latter part (cid:0) α β Γ g α − (cid:1) − α α β Γ g α − α − decomposes to left and right. To see this, first notethat α − α α = α − θ . ,θ . ] α [0 ,θ ] α ( θ . ,θ . ] ∈ Aut (cid:0) A C θ . (cid:1) . (3.19)Because the conjugation (cid:0) β Γ g (cid:1) − · β Γ g does not change the support of an automorphism, (cid:0) β Γ g (cid:1) − (cid:0) α − α α (cid:1) β Γ g is also supported on C θ . . Therefore, we have α (cid:16)(cid:0) β Γ g (cid:1) − (cid:0) α − α α (cid:1) β Γ g (cid:17) α − = α ( θ ,θ ] α ( θ . ,θ . ] (cid:0) β Γ g (cid:1) − α − θ . ,θ . ] α [0 ,θ ] α ( θ . ,θ . ] β Γ g α − θ . ,θ . ] α − θ ,θ ] (3.20)Hence we get the left-right decomposition: (cid:0) α β Γ g α − (cid:1) − α α β Γ g α − α − = α (cid:16)(cid:0) β Γ g (cid:1) − (cid:0) α − α α (cid:1) β Γ g (cid:17) α − α − = α ( θ ,θ ] α ( θ . ,θ . ] (cid:0) β Γ g (cid:1) − α − θ . ,θ . ] α [0 ,θ ] α ( θ . ,θ . ] β Γ g α − θ . ,θ . ] α − θ ,θ ] ◦ α − ,θ ] = O σ = L,R (cid:16) α ( θ ,θ ] ,σ α ( θ . ,θ . ] ,σ (cid:0) β Γ σ g (cid:1) − α − θ . ,θ . ] ,σ α [0 ,θ ] ,σ α ( θ . ,θ . ] ,σ β Γ σ g α − θ . ,θ . ] ,σ α − θ ,θ ] ,σ ◦ α − ,θ ] ,σ (cid:17) =: O σ = L,R Ξ Γ ,g,σ . (3.21)Here we setΞ Γ ,g,σ = (cid:16) α ( θ ,θ ] ,σ α ( θ . ,θ . ] ,σ (cid:0) β Γ σ g (cid:1) − α − θ . ,θ . ] ,σ α [0 ,θ ] ,σ α ( θ . ,θ . ] ,σ β Γ σ g α − θ . ,θ . ] ,σ α − θ ,θ ] ,σ ◦ α − ,θ ] ,σ (cid:17) ∈ Aut (cid:16) A ( C θ ) σ (cid:17) . (3.22)On the other hand, the first part of (3.18) with Γ = Z , H U satisfies β − g α β g α − = ξ D ⊗ ξ U , (cid:0) β Ug (cid:1) − α β Ug α − = id A HD ⊗ ξ U (3.23)where ξ ζ := (cid:0) β ζg (cid:1) − α ,ζ β ζg α − ,ζ ∈ Aut (cid:18) A (( C θ . ) c ) ζ (cid:19) , ζ = U, D (3.24)Hence we obtain decompositions (cid:0) β Ug (cid:1) − ◦ α ◦ β Ug ◦ α − = (inner) ◦ (cid:16) id A HD ⊗ ξ U (cid:17) ◦ (Ξ H U ,g,L ⊗ Ξ H U ,g,R ) , ( β g ) − ◦ α ◦ β g ◦ α − = (inner) ◦ ( ξ D ⊗ ξ U ) ◦ (cid:0) Ξ Z ,g,L ⊗ Ξ Z ,g,R (cid:1) . (3.25)Because ξ ζ ∈ Aut (cid:18) A (( C θ . ) c ) ζ (cid:19) commutes with β C [0 ,θ . g and β C [0 ,θ . ,U g , we get α ◦ β Ug ◦ α − = (inner) ◦ (cid:18) id A HD ⊗ β C ( θ . , π ,U g ξ U (cid:19) ◦ (cid:16) β C [0 ,θ . ,L,U g Ξ H U ,g,L ⊗ β C [0 ,θ . ,R,U g Ξ H U ,g,R (cid:17) ,α ◦ β g ◦ α − = (inner) ◦ (cid:18) β C ( θ . , π ,D g ξ D ⊗ β C ( θ . , π ,U g ξ U (cid:19) ◦ (cid:16) β C [0 ,θ . ,L g Ξ Z ,g,L ⊗ β C [0 ,θ . ,R g Ξ Z ,g,R (cid:17) . (3.26)24urthermore, from β g -invariance of ω ◦ α , we have ω ◦ α ◦ β g ◦ α − = ω . (3.27)Now we apply Lemma 3.4 for A σ , B σζ , C ζ replaced by A ( C [0 ,θ . ) σ A ( C ( θ . ,θ ) σ,ζ A (cid:16) C ( θ , π (cid:17) ζ ,for σ = L, R , ζ = D, U . By (3.27) and (3.26), (cid:0) α ◦ β Ug ◦ α − , α ◦ β g ◦ α − (cid:1) satisfy the con-ditions of ( α, ˆ α ) in Lemma 3.4, for ω and its restrictions. Applying Lemma 3.4, there are˜ η σ,g ∈ Aut (cid:16) A ( C θ ) σ (cid:17) , g ∈ G , σ = L, R such that ω ◦ α ◦ β Ug ◦ α − ∼ q.e. ω ◦ (˜ η Lg ⊗ ˜ η Rg ) , g ∈ G. (3.28)Because both of ω ◦ α ◦ β Ug ◦ α − and ω ◦ (˜ η Lg ⊗ ˜ η Rg ) are pure, by Kadison’s transitibity theorem,there exists a unitary ˜ v g ∈ U ( A ) such that ω ◦ α ◦ β Ug ◦ α − = ω ◦ Ad ˜ v g ◦ (˜ η Lg ⊗ ˜ η Rg ) , g ∈ G. (3.29)We define ˜ β g := Ad (cid:0) α − (cid:0) ˜ v g − (cid:1)(cid:1) ◦ α − ◦ (cid:0) ˜ η Lg − ⊗ ˜ η Rg − (cid:1) ◦ α ◦ β Ug , g ∈ G. (3.30)It suffices to show that ( ˜ β g ) ∈ IG( ω ◦ α, θ ) = IG( ω ◦ α, θ . ). By (3.29), we have ω ◦ α ◦ ˜ β g = ω ◦ α .Therefore, what is left to be proven is that there are η σg ∈ Aut (( C θ ) σ ), g ∈ G , σ = L, R such that˜ β g = (inner) ◦ (cid:0) η Lg ⊗ η Rg (cid:1) ◦ β Ug , for all g ∈ G (3.31)By the decomposition (2.10) and the fact that ˜ η Lg − ⊗ ˜ η Rg − has support in C θ , we have α − ◦ (cid:0) ˜ η Lg − ⊗ ˜ η Rg − (cid:1) ◦ α = (inner) ◦ (cid:0) α ( θ . ,θ . ] ⊗ α ( θ . ,θ . ] (cid:1) − (cid:0) α [0 ,θ ] ⊗ α ( θ ,θ ] (cid:1) − (cid:0) ˜ η Lg − ⊗ ˜ η Rg − (cid:1) (cid:0) α [0 ,θ ] ⊗ α ( θ ,θ ] (cid:1) ◦ (cid:0) α ( θ . ,θ . ] ⊗ α ( θ . ,θ . ] (cid:1) = (inner) ◦ (cid:0) η Lg ⊗ η Rg (cid:1) , (3.32)where η σg = (cid:0) α ( θ . ,θ . ] ,σ ⊗ α ( θ . ,θ . ] ,σ (cid:1) − (cid:0) α [0 ,θ ] ,σ ⊗ α ( θ ,θ ] ,σ (cid:1) − (cid:0) ˜ η σg − (cid:1) (cid:0) α [0 ,θ ] ,σ ⊗ α ( θ ,θ ] ,σ (cid:1) ◦ (cid:0) α ( θ . ,θ . ] ,σ ⊗ α ( θ . ,θ . ] ,σ (cid:1) ∈ Aut (( C θ . ) σ ) , σ = L, R (3.33)Substituting this to (3.30), we obtain (3.31). This completes the proof. (cid:3) h ( ω ) In this section we prove the stability of the index h ( ω ) with respect to γ ∈ GUQAut( A ). Theorem 4.1.
Let ω ∈ SL with IG( ω ) = ∅ . Let γ ∈ GUQAut ( A ) . Then we have ω ◦ γ ∈ SL with IG( ω ◦ γ ) = ∅ and h ( ω ◦ γ ) = h ( ω ) . (4.1)25 roof. Step 1.
From ω ∈ SL , there is an α ∈ EAut( ω ). For any 0 < θ < π fixed, we show that D θα ◦ γ = ∅ , hence α ◦ γ ∈ QAut( A ) and ω ◦ γ = ω ◦ αγ ∈ SL . Set θ . := θ and choose0 < θ < θ . < θ < θ . := θ < θ . < θ < θ . < θ . < θ < θ . < π . (4.2)Because α ∈ QAut( A ), there exists some ( α L , α R , Θ) ∈ D θ α . Setting α := α L ⊗ α R , we have α = (inner) ◦ α ◦ Θ. Because γ ∈ GUQAut ( A ), there are γ H ∈ HAut( A ) and γ C ∈ GSQAut( A )such that γ = γ C ◦ γ H . (4.3)Because γ H ∈ HAut( A ), we may decompose γ H as γ H = (inner) ◦ ( γ H,L ⊗ γ H,R ) = (inner) ◦ γ (4.4)with some γ H,σ ∈ Aut (cid:16) A ( C θ ) σ (cid:17) , σ = L, R . We set γ := γ H,L ⊗ γ H,R ∈ Aut (cid:0) A C θ (cid:1) . Bydefinition, γ C ∈ GSQAut( A ), allows a decomposition γ C = (inner) ◦ γ CS γ CS = (cid:16) γ [0 ,θ ] ⊗ γ ( θ ,θ ] ⊗ γ ( θ ,θ ] ⊗ γ ( θ , π ] (cid:17) ◦ (cid:0) γ ( θ . ,θ . ] ⊗ γ ( θ . ,θ . ] ⊗ γ ( θ . ,θ . ] (cid:1) (4.5)with γ X := O σ = L,R,ζ = D,U γ X,σ,ζ , γ [0 ,θ ] := O σ = L,R γ [0 ,θ ] ,σ , γ ( θ , π ] := O ζ = D,U γ ( θ , π ] ,ζ γ X,σ,ζ ∈ Aut (cid:0) A C X,σ,ζ (cid:1) , γ
X,σ := O ζ = U,D γ X,σ,ζ , γ
X,ζ := O σ = L,R γ X,σ,ζ γ [0 ,θ ] ,σ ∈ Aut (cid:16) A C [0 ,θ ,σ (cid:17) , γ ( θ , π ] ,ζ ∈ Aut (cid:16) A C ( θ , π ,ζ (cid:17) , (4.6)for X = ( θ , θ ] , ( θ , θ ] , ( θ . , θ . ] , ( θ . , θ . ] , ( θ . , θ . ] , σ = L, R, ζ = D, U. (4.7)Here we have γ I ◦ β Ug = β Ug ◦ γ I for all g ∈ G, (4.8)for any I = [0 , θ ] , ( θ , θ ] , ( θ , θ ] , (cid:16) θ , π i , ( θ . , θ . ] , ( θ . , θ . ] , ( θ . , θ . ] . (4.9)SetˆΘ := Θ ◦ (cid:16) γ ( θ ,θ ] ⊗ γ ( θ , π ] (cid:17) ◦ (cid:0) γ ( θ . ,θ . ] ⊗ γ ( θ . ,θ . ] (cid:1) ∈ Aut (cid:16) A C cθ . (cid:17) ⊂ Aut (cid:16) A C cθ . (cid:17) , (4.10)and ˆ α σ := α σ ◦ (cid:0) γ [0 ,θ ] ,σ ⊗ γ ( θ ,θ ] ,σ (cid:1) ◦ γ ( θ . ,θ . ] ,σ ◦ γ H,σ ∈ Aut( A H σ ) , σ = L, R. (4.11)We claim α ◦ γ = (inner) ◦ ( ˆ α L ⊗ ˆ α R ) ◦ ˆΘ . (4.12)26his means (ˆ α L , ˆ α R , ˆΘ) ∈ D θ . αγ , hence D θαγ = D θ . αγ = ∅ . The claim (4.12) can be checked asfollows. Note that γ ( θ ,θ ] ⊗ γ ( θ , π ] and γ ( θ . ,θ . ] commute because of their disjoint supports.Because Θ ∈ Aut (cid:16) A C cθ (cid:17) , it commutes with γ [0 ,θ ] ⊗ γ ( θ ,θ ] and γ ( θ . ,θ . ] . Therefore, we have α ◦ γ = (inner) ◦ α ◦ Θ ◦ (cid:16) γ [0 ,θ ] ⊗ γ ( θ ,θ ] ⊗ γ ( θ ,θ ] ⊗ γ ( θ , π ] (cid:17) ◦ (cid:0) γ ( θ . ,θ . ] ⊗ γ ( θ . ,θ . ] ⊗ γ ( θ . ,θ . ] (cid:1) ◦ γ = (inner) ◦ α ◦ (cid:0) γ [0 ,θ ] ⊗ γ ( θ ,θ ] (cid:1) ◦ γ ( θ . ,θ . ] ◦ Θ ◦ (cid:16) γ ( θ ,θ ] ⊗ γ ( θ , π ] (cid:17) ◦ (cid:0) γ ( θ . ,θ . ] ⊗ γ ( θ . ,θ . ] (cid:1) ◦ γ = (inner) ◦ α ◦ (cid:0) γ [0 ,θ ] ⊗ γ ( θ ,θ ] (cid:1) ◦ γ ( θ . ,θ . ] ◦ ˆΘ ◦ γ (4.13)Because γ ∈ Aut (cid:0) A C θ (cid:1) and ˆΘ ∈ Aut (cid:16) A C cθ . (cid:17) commute, we have α ◦ γ = (4 .
13) = (inner) ◦ α ◦ (cid:0) γ [0 ,θ ] ⊗ γ ( θ ,θ ] (cid:1) ◦ γ ( θ . ,θ . ] ◦ γ ◦ ˆΘ = (inner) ◦ (ˆ α L ⊗ ˆ α R ) ◦ ˆΘ , (4.14)proving (4.12). Step 2.
From IG( ω ) = ∅ , we fix a 0 < θ < π such that IG( ω, θ ) = ∅ . We choose θ . , θ , θ . , θ . , θ , θ . , θ . , θ , θ . such that 0 < θ < θ . < θ < θ . < θ . < θ < θ . < θ . < θ < θ . < π . (4.15)For these θ s, we associate the decomposition of γ in Step 1 . Fix (cid:16) ˜ β g (cid:17) ∈ IG( ω, θ ) and ( η σg ) ∈T ( θ , ( ˜ β g )). Set η g := η Lg ⊗ η Rg . Note that ( η σg ) also belongs to T ( θ , ( ˜ β g )). Setˆ η σg := (cid:0) γ [0 ,θ ] ,σ γ ( θ . ,θ . ] ,σ γ H,σ (cid:1) − η σg (cid:16) β σUg γ [0 ,θ ] ,σ γ ( θ . ,θ . ] ,σ γ H,σ (cid:0) β σUg (cid:1) − (cid:17) ∈ Aut (cid:16) A ( C θ . ) σ (cid:17) , (4.16)for σ = L, R . We also set ˆ η g := ˆ η Lg ⊗ ˆ η Rg . We claim that (cid:16) γ − ˜ β g γ (cid:17) ∈ IG( ω ◦ γ, θ . ) with(ˆ η σg ) ∈ T (cid:16) θ . , (cid:16) γ − ˜ β g γ (cid:17)(cid:17) . Clearly we have ω ◦ γ ◦ (cid:16) γ − ˜ β g γ (cid:17) = ω ◦ ˜ β g ◦ γ = ω ◦ γ. (4.17)Therefore, what remains to be shown is γ − ˜ β g γ = (inner) ◦ (cid:0) ˆ η Lg ⊗ ˆ η Rg (cid:1) ◦ β Ug (4.18)To see this, we first have γ − ◦ η g ◦ γ = (inner) ◦ γ − ◦ (cid:0) γ ( θ . ,θ . ] ⊗ γ ( θ . ,θ . ] ⊗ γ ( θ . ,θ . ] (cid:1) − ◦ (cid:16) γ [0 ,θ ] ⊗ γ ( θ ,θ ] ⊗ γ ( θ ,θ ] ⊗ γ ( θ , π ] (cid:17) − ◦ η g ◦ (cid:16) γ [0 ,θ ] ⊗ γ ( θ ,θ ] ⊗ γ ( θ ,θ ] ⊗ γ ( θ , π ] (cid:17) ◦ (cid:0) γ ( θ . ,θ . ] ⊗ γ ( θ . ,θ . ] ⊗ γ ( θ . ,θ . ] (cid:1) γ (4.19)from the decomposition (4.3), (4.4) (4.5). Because γ ( θ ,θ ] ⊗ γ ( θ ,θ ] ⊗ γ ( θ , π ] commutes with η g ∈ Aut (cid:0) A C θ (cid:1) and γ ( θ . ,θ . ] ⊗ γ ( θ . ,θ . ] commutes with (cid:0) γ [0 ,θ ] (cid:1) − η g γ [0 ,θ ] ∈ Aut (cid:0) A C θ (cid:1) , we have γ − ◦ η g ◦ γ = (4 .
19) = (inner) ◦ γ − ◦ (cid:0) γ ( θ . ,θ . ] ⊗ γ ( θ . ,θ . ] ⊗ γ ( θ . ,θ . ] (cid:1) − ◦ (cid:0) γ [0 ,θ ] (cid:1) − ◦ η g ◦ (cid:0) γ [0 ,θ ] (cid:1) ◦ (cid:0) γ ( θ . ,θ . ] ⊗ γ ( θ . ,θ . ] ⊗ γ ( θ . ,θ . ] (cid:1) γ = (inner) ◦ γ − ◦ (cid:0) γ ( θ . ,θ . ] (cid:1) − ◦ (cid:0) γ [0 ,θ ] (cid:1) − ◦ η g ◦ (cid:0) γ [0 ,θ ] (cid:1) ◦ (cid:0) γ ( θ . ,θ . ] (cid:1) γ (4.20)27n the other hand, because γ CS and β Ug commute, we have γ − ◦ β Ug ◦ γ = (inner) γ − ◦ γ − CS β Ug γ CS γ = (inner) γ − ◦ β Ug γ . (4.21)Combining (4.20) and (4.21), we obtain γ − ˜ β g γ = (inner) ◦ γ − (cid:0) γ ( θ . ,θ . ] (cid:1) − ◦ (cid:0) γ [0 ,θ ] (cid:1) − ◦ η g ◦ (cid:0) γ [0 ,θ ] (cid:1) ◦ (cid:0) γ ( θ . ,θ . ] (cid:1) γ ◦ γ − ◦ β Ug γ = (inner) ◦ γ − (cid:0) γ ( θ . ,θ . ] (cid:1) − ◦ (cid:0) γ [0 ,θ ] (cid:1) − ◦ η g β Ug ◦ (cid:0) γ [0 ,θ ] (cid:1) ◦ (cid:0) γ ( θ . ,θ . ] (cid:1) ◦ γ = (inner) ◦ (cid:0) ˆ η Lg ⊗ ˆ η Rg (cid:1) ◦ β Ug . (4.22)In the second equality, we used the fact that γ [0 ,θ ] γ ( θ . ,θ . ] and β Ug commute. This completes theproof of the claim. Step 3.
We use the setting and notation of
Step 1. (with θ chosen in Step 2. ) and
Step. 2 . ByLemma 2.1, there exists(( W g ) , ( u σ ( g, h ))) ∈ IP (cid:16) ω, α, θ , ( ˜ β g ) , ( η σg ) , ( α L , α R , Θ) (cid:17) . (4.23)Now we have ω ◦ γ ∈ SL , α ◦ γ ∈ EAut( ω ◦ γ ) , (cid:16) γ − ◦ ˜ β g ◦ γ (cid:17) ∈ IG( ω ◦ γ, θ . ) , (ˆ η σg ) ∈ T (cid:16) θ . , (cid:16) γ − ˜ β g γ (cid:17)(cid:17) , (ˆ α L , ˆ α R , ˆΘ) ∈ D θ . αγ . (4.24)We claim (( W g ) , ( u σ ( g, h ))) ∈ IP (cid:16) ω ◦ γ, α ◦ γ, θ . , ( γ − ˜ β g γ ) , (ˆ η σg ) , (ˆ α L , ˆ α R , ˆΘ) (cid:17) . (4.25)This immediately implies h ( ω ) = h ( ω ◦ γ ). To prove the claim, we first see from (4.10) and (4.11)that(ˆ α L ⊗ ˆ α R ) ◦ ˆΘ ◦ γ − (cid:0) γ ( θ . ,θ . ] (cid:1) − ◦ (cid:0) γ [0 ,θ ] (cid:1) − = α ◦ (cid:0) γ [0 ,θ ] ⊗ γ ( θ ,θ ] (cid:1) ◦ γ ( θ . ,θ . ] ◦ γ ◦ Θ ◦ (cid:16) γ ( θ ,θ ] ⊗ γ ( θ , π ] (cid:17) ◦ (cid:0) γ ( θ . ,θ . ] ⊗ γ ( θ . ,θ . ] (cid:1) ◦ γ − (cid:0) γ ( θ . ,θ . ] (cid:1) − ◦ (cid:0) γ [0 ,θ ] (cid:1) − = α ◦ (cid:0) γ [0 ,θ ] ⊗ γ ( θ ,θ ] (cid:1) ◦ Θ ◦ (cid:16) γ ( θ ,θ ] ⊗ γ ( θ , π ] (cid:17) ◦ (cid:0) γ ( θ . ,θ . ] ⊗ γ ( θ . ,θ . ] (cid:1) ◦ (cid:0) γ [0 ,θ ] (cid:1) − (4.26)because γ ( θ . ,θ . ] ◦ γ ∈ Aut (cid:0) A C θ . (cid:1) and Θ ◦ (cid:16) γ ( θ ,θ ] ⊗ γ ( θ , π ] (cid:17) ◦ (cid:0) γ ( θ . ,θ . ] ⊗ γ ( θ . ,θ . ] (cid:1) ∈ Aut (cid:0) A C θ . c (cid:1) commute. Furthermore, because γ [0 ,θ ] and Θ ◦ (cid:16) γ ( θ ,θ ] ⊗ γ ( θ , π ] (cid:17) ◦ (cid:0) γ ( θ . ,θ . ] ⊗ γ ( θ . ,θ . ] (cid:1) ∈ Aut (cid:0) A C θ . c (cid:1) commute, while γ ( θ ,θ ] and Θ ∈ Aut (cid:0) A C θ c (cid:1) commute, we have(ˆ α L ⊗ ˆ α R ) ◦ ˆΘ ◦ γ − (cid:0) γ ( θ . ,θ . ] (cid:1) − ◦ (cid:0) γ [0 ,θ ] (cid:1) − = (4 .
26) = α ◦ γ ( θ ,θ ] ◦ Θ ◦ (cid:16) γ ( θ ,θ ] ⊗ γ ( θ , π ] (cid:17) ◦ (cid:0) γ ( θ . ,θ . ] ⊗ γ ( θ . ,θ . ] (cid:1) = α ◦ Θ ◦ γ ( θ ,θ ] ◦ (cid:16) γ ( θ ,θ ] ⊗ γ ( θ , π ] (cid:17) ◦ (cid:0) γ ( θ . ,θ . ] ⊗ γ ( θ . ,θ . ] (cid:1) = α ◦ Θ ◦ ˆ γ. (4.27)Here ˆ γ := γ ( θ ,θ ] ◦ (cid:16) γ ( θ ,θ ] ⊗ γ ( θ , π ] (cid:17) ◦ (cid:0) γ ( θ . ,θ . ] ⊗ γ ( θ . ,θ . ] (cid:1) ∈ Aut (cid:0) A C θ c (cid:1) commutes with β Ug .Combining this and ˆ η g β Ug = (cid:0) γ [0 ,θ ] γ ( θ . ,θ . ] γ (cid:1) − η g β Ug γ [0 ,θ ] γ ( θ . ,θ . ] γ , (4.28)we obtain π ◦ (ˆ α L ⊗ ˆ α R ) ◦ ˆΘ ◦ ˆ η g β Ug (cid:16) ˆΘ (cid:17) − ( ˆ α L ⊗ ˆ α R ) − = π ◦ α ◦ Θ ◦ ˆ γ ◦ η g β Ug ◦ ˆ γ − ◦ Θ − ◦ α − . (4.29)28ecause ˆ γ commutes with β Ug and η g ∈ Aut (cid:0) A C θ (cid:1) commutes with ˆ γ ∈ Aut (cid:0) A C θ c (cid:1) , we have π ◦ (ˆ α L ⊗ ˆ α R ) ◦ ˆΘ ◦ ˆ η g β Ug (cid:16) ˆΘ (cid:17) − ( ˆ α L ⊗ ˆ α R ) − = (4 .
29) = π ◦ α ◦ Θ ◦ η g β Ug ◦ Θ − ◦ α − = Ad ( W g ) ◦ π (4.30)Hence the condition for W g in (4.25) is checked. On the other hand, substituting (4.11) and (4.16),we get π R ◦ ˆ α R ◦ ˆ η Rg β RUg ˆ η Rh (cid:0) β RUg (cid:1) − (cid:0) ˆ η Rgh (cid:1) − ˆ α − R = π R ◦ α R ◦ (cid:0) γ [0 ,θ ] ,R ⊗ γ ( θ ,θ ] ,R (cid:1) ◦ γ ( θ . ,θ . ] ,R ◦ γ H,R ◦ (cid:0) γ [0 ,θ ] ,R ◦ γ ( θ . ,θ . ] ,R ◦ γ H,R (cid:1) − η Rg β RUg η Rh (cid:0) β RUg (cid:1) − (cid:0) η Rgh (cid:1) − ◦ γ [0 ,θ ] ,R ◦ γ ( θ . ,θ . ] ,R ◦ γ H,R ◦ (cid:0)(cid:0) γ [0 ,θ ] ,R ⊗ γ ( θ ,θ ] ,R (cid:1) ◦ γ ( θ . ,θ . ] ,R ◦ γ H,R (cid:1) − α − R = π R ◦ α R ◦ γ ( θ ,θ ] ,R ◦ η Rg β RUg η Rh (cid:0) β RUg (cid:1) − (cid:0) η Rgh (cid:1) − ◦ (cid:0) γ ( θ ,θ ] ,R (cid:1) − ◦ α − R . (4.31)Because η Rg β RUg η Rh (cid:0) β RUg (cid:1) − (cid:16) η Rgh (cid:17) − ∈ Aut (cid:0) A C θ (cid:1) commutes with γ ( θ ,θ ] ,R , we obtain π R ◦ ˆ α R ◦ ˆ η Rg β RUg ˆ η Rh (cid:0) β RUg (cid:1) − (cid:0) ˆ η Rgh (cid:1) − ˆ α − R = (4 .
31) = π R ◦ α R ◦ η Rg β RUg η Rh (cid:0) β RUg (cid:1) − (cid:0) η Rgh (cid:1) − α − R = Ad ( u R ( g, h )) ◦ π R . (4.32)An analogous statement for σ = L also holds. This completes the proof of (4.25). Hence thestatement of the Theorem is proven. (cid:3) In this section, we prove Theorem 1.4. The proof relies heavily on the machinery of quasi-localautomorphisms developed in [BMNS] [NSY], [MO]. (Summary is given in Appendix D.) We useterminology and facts from Appendix C, D, freely. We introduce a set of F -functions with fastdecay, F a as Definition C.2. Crucial point for us is the following. Theorem 5.1.
Let Φ , Φ ∈ P UG and ω Φ , ω Φ be their unique gapped ground states. Supposethat Φ ∼ Φ holds, via a path Φ : [0 , → P UG . Then there exists some Ψ ∈ ˆ B F ([0 , with Ψ ∈ ˆ B F ([0 , for some F ∈ F a of the form F ( r ) = exp ( − r θ ) (1+ r ) with a constant < θ < , such that ω Φ = ω Φ ◦ τ Ψ1 , . If Φ , Φ ∈ P UGβ and Φ ∼ β Φ , we may take Ψ to be β -invariant. For the proof, see Appendix D.From this and Theorem 3.1 and Theorem 4.1, in order to show Theorem 1.4, it suffices to showthe following.
Theorem 5.2.
Let F ∈ F a be an F -function of the form F ( r ) = exp ( − r θ ) (1+ r ) with a constant <θ < . Let Ψ ∈ ˆ B F ([0 , be a path of interactions satisfying Ψ ∈ ˆ B F ([0 , . Then we have τ Ψ1 , ∈ SQAut( A ) . Furthermore, if Ψ is β Ug -invariant, i.e., β Ug (Ψ( X ; t )) = Ψ( X ; t ) for any X ∈ S Z , t ∈ [0 , , and g ∈ G , then we have τ Ψ1 , ∈ GSQAut( A ) . Proof.
Fix arbitrary0 < θ . < θ < θ . < θ . < θ < θ . < θ . < θ < θ . < π . (5.1)29e show the existence of the decomposition τ Ψ1 , = Ad( u ) ◦ (cid:16) α (0 ,θ ] ⊗ α ( θ ,θ ] ⊗ α ( θ ,θ ] ⊗ α ( θ , π ] (cid:17) ◦ (cid:0) α ( θ . ,θ . ] α ( θ . ,θ . ] ⊗ α ( θ . ,θ . ] (cid:1) , (5.2)with α s of the form (2.11) and (2.12). We follow the strategy in [NO]. Step 1.
Fix some 0 < θ ′ < θ , and set ˜ F ( r ) := exp (cid:16) − r θ ′ (cid:17) (1 + r ) . (5.3)With suitably chosen constant c >
0, we havemax (cid:26) F (cid:16) r (cid:17) , (cid:16) F (cid:16)h r i(cid:17)(cid:17) (cid:27) ≤ c ˜ F ( r ) , r ≥ . (5.4)Namely, c ˜ F ( r ) satisfy the condition of ˜ F θ in Definition C.2 (ii) for our F = exp ( − cr θ ) (1+ r ) and θ = .Set C := ( C [0 ,θ ] ,σ , C ( θ ,θ ] ,σ,ζ , C ( θ ,θ ] ,σ,ζ , C ( θ , π ] ,ζ ,σ = L, R, ζ = D, U. ) , (5.5) C := ( C ( θ . ,θ . ) ,σ,ζ , C ( θ . ,θ . ) ,σ,ζ , C ( θ . ,θ . ) ,σ,ζ ,σ = L, R, ζ = D, U. ) . (5.6)Define Ψ (0) , Ψ (1) ∈ ˆ B F ([0 , (0) ( X ; t ) := ( Ψ ( X ; t ) , if there exists a C ∈ C such that X ⊂ C , otherwise , Ψ (1) ( X ; t ) := Ψ (0) ( X ; t ) − Ψ ( X ; t ) , (5.7)for each X ∈ S Z , t ∈ [0 , (cid:16) τ Ψ (0) , (cid:17) − ◦ τ Ψ1 , as some quasi-local automorphism. Let t, s ∈ [0 , (1) ,and ˜Ψ by Ψ. Hence we setΞ ( s ) ( Z, t ) := X m ≥ X X ⊂ Z, X ( m )= Z ∆ X ( m ) (cid:16) τ Ψ t,s (cid:16) Ψ (1) ( X ; t ) (cid:17)(cid:17) (5.8)and Ξ ( n )( s ) ( Z, t ) := X m ≥ X X ⊂ Z,X ( m ) ∩ Λ n = Z ∆ X ( m ) (cid:16) τ (Λ n )Ψ t,s (cid:16) Ψ (1) ( X ; t ) (cid:17)(cid:17) . (5.9)Corresponding to (D.31), we obtain τ (Λ n ) , Ψ t,s (cid:0) H Λ n , Ψ (1) ( t ) (cid:1) = H Λ n , Ξ ( n )( s ) ( t ) . (5.10)Applying Proposition D.6. we have Ξ ( n )( s ) , Ξ ( s ) ∈ ˆ B ˜ F ([0 , n →∞ (cid:13)(cid:13)(cid:13) τ Ξ ( n )( s ) t,u ( A ) − τ Ξ ( s ) t,u ( A ) (cid:13)(cid:13)(cid:13) = 0 , A ∈ A , t, u ∈ [0 ,
1] (5.11)30olds. Two functions ˆ τ (Λ n ) , Ξ ( n )( s ) t,s ( A ) and τ (Λ n ) , Ψ t,s ◦ (cid:16) τ (Λ n ) , Ψ (0) t,s (cid:17) − ( A ) satisfy the same differentialequation and the initial condition. Therefore we obtainˆ τ (Λ n ) , Ξ ( n )( s ) t,s ( A ) = τ (Λ n ) , Ψ t,s ◦ (cid:16) τ (Λ n ) , Ψ (0) t,s (cid:17) − ( A ) , t ∈ [0 , , A ∈ A . (5.12)From the fact that ˆ τ (Λ n ) , Ξ ( n )( s ) t,u = τ (Λ n ) , Ξ ( n )( s ) u,t = τ Ξ ( n )( s ) u,t converges strongly to an automorphism τ Ξ ( s ) u,t on A (5.11), we havelim n →∞ (cid:13)(cid:13)(cid:13) ˆ τ (Λ n )Ξ ( n )( s ) t,s ( A ) − τ Ξ ( s ) s,t ( A ) (cid:13)(cid:13)(cid:13) = 0 , A ∈ A . (5.13)On the other hand, by Theorem D.3, we have for t ∈ [0 ,
1] and A ∈ A lim n →∞ (cid:13)(cid:13)(cid:13)(cid:13) τ (Λ n ) , Ψ t,s ◦ (cid:16) τ (Λ n ) , Ψ (0) t,s (cid:17) − ( A ) − τ Ψ t,s ◦ (cid:16) τ Ψ (0) t,s (cid:17) − ( A ) (cid:13)(cid:13)(cid:13)(cid:13) = 0 . (5.14)Therefore, taking n → ∞ limit in (5.12), we obtain τ Ξ ( s ) s,t ( A ) = τ Ψ t,s ◦ (cid:16) τ Ψ (0) t,s (cid:17) − ( A ) , t, s ∈ [0 , , A ∈ A . (5.15)Hence we have τ Ψ s,t = (cid:0) τ Ψ t,s (cid:1) − = (cid:16) τ Ψ (0) t,s (cid:17) − (cid:16) τ Ξ ( s ) s,t (cid:17) − = τ Ψ (0) s,t τ Ξ ( s ) t,s (5.16)In particular, we get τ Ψ1 , = τ Ψ (0) , τ Ξ (1) , . (5.17) Step 2.
We show X Z ∈ S ( Z ) , C ∈C s.t.Z ⊂ C sup t ∈ [0 , (cid:13)(cid:13)(cid:13) Ξ (1) ( Z, t ) (cid:13)(cid:13)(cid:13) < ∞ . (5.18)From this, V ( t ) := X Z ∈ S ( Z ) , C ∈C s.t.Z ⊂ C Ξ (1) ( Z, t ) ∈ A (5.19)converges absolutely in the norm topology and define an element in A . Furthermore, for V n ( t ) := X Z ∈ S ( Z ) , Z ⊂ Λ n C ∈C s.t.Z ⊂ C Ξ (1) ( Z, t ) ∈ A Λ n , n ∈ N , (5.20)we get lim n →∞ sup t ∈ [0 , k V n ( t ) − V ( t ) k = 0 , (5.21)from (5.18). 31o prove (5.18), we first bound X Z ∈ S ( Z ) , C ∈C s.t.Z ⊂ C sup t ∈ [0 , (cid:13)(cid:13)(cid:13) Ξ (1) ( Z, t ) (cid:13)(cid:13)(cid:13) ≤ X Z ∈ S ( Z ) , C ∈C s.t.Z ⊂ C X m ≥ X X : X ⊂ Z,X ( m )= Z " sup t ∈ [0 , (cid:13)(cid:13)(cid:13) ∆ X ( m ) (cid:16) τ Ψ t, (cid:16) Ψ (1) ( X ; t ) (cid:17)(cid:17)(cid:13)(cid:13)(cid:13) ≤ X m ≥ X X : C ∈C s.t.X ( m ) ⊂ C sup t ∈ [0 , (cid:13)(cid:13)(cid:13) ∆ X ( m ) (cid:16) τ Ψ t, (cid:16) Ψ (1) ( X ; t ) (cid:17)(cid:17)(cid:13)(cid:13)(cid:13) ≤ X m ≥ X X : C ∈C s.t.X ( m ) ⊂ C " sup t ∈ [0 , (cid:13)(cid:13) Ψ (1) ( X ; t ) (cid:13)(cid:13) C F (cid:16) e I F (Ψ) − (cid:17) | X | G F ( m ) = 8 C F (cid:16) e I F (Ψ) − (cid:17) X m ≥ X X : C ∈C s.t.X ( m ) ⊂ C " sup t ∈ [0 , (cid:16)(cid:13)(cid:13)(cid:13) Ψ (1) ( X ; t ) (cid:13)(cid:13)(cid:13)(cid:17) | X | G F ( m ) . (5.22)For the third inequality, we used Theorem D.3 3. For any cone C , C of Z with apex at theorigin, we set M ( C , C ) := X m ≥ X X : ∀ C ∈C ,X ∩ (( C c )( m )) = ∅ ,X ∩ C = ∅ , X ∩ C = ∅ " sup t ∈ [0 , (cid:16)(cid:13)(cid:13)(cid:13) Ψ (1) ( X ; t ) (cid:13)(cid:13)(cid:13)(cid:17) | X | G F ( m ) . (5.23)From the definition of Ψ (1) , we have Ψ (1) ( X ; t ) = 0 , unless X has a non-empty intersection withat least two of elements in C . Therefore, if X gives a non-zero contribution in (5.22), then it hasto satisfy X ∩ (( C c ) ( m )) = ∅ , for all C ∈ C , ∃ C , C ∈ C such that , C = C , X ∩ C = ∅ , X ∩ C = ∅ . Hence we have (5 . ≤ C F (cid:16) e I F (Ψ) − (cid:17) X C ,C ∈C C = C M ( C , C ) (5.24)Hence it suffice so show that M ( C , C ) < ∞ for all C , C ∈ C with C = C .In order to proceed, we prepare two estimates. We will freely identify C and R in an obviousmanner. In particular, arg z of z ∈ Z ⊂ R in the following definition is considered with thisidentification: for ϕ < ϕ , we setˇ C [ ϕ ,ϕ ] := (cid:8) z ∈ Z | arg z ∈ [ ϕ , ϕ ] (cid:9) . (5.25)We define ˇ C ( ϕ ,ϕ ) etc. analogously. Set c (0) ζ ,ζ ,ζ ,ζ := p − max { cos( ζ − ζ ) , cos( ζ − ζ ) , } , ζ , ζ , ζ , ζ ∈ R . (5.26)32 emma 5.3. Let ϕ < ϕ < ϕ < ϕ with ϕ − ϕ < π . Then b ( ϕ , ϕ , ϕ , ϕ ):= X m ≥ X X : X ∩ ˇ C [ ϕ ,ϕ = ∅ ,X ∩ ˇ C [ ϕ ,ϕ = ∅ " sup t ∈ [0 , ( k Ψ ( X ; t ) k ) | X | G F ( m ) ≤ (64) κ , ,F (cid:16) c (0) ϕ ,ϕ ,ϕ ,ϕ (cid:17) ( k| Ψ k| F ) X m ≥ G F ( m ) < ∞ . Proof.
Substituting Lemme C.4, we obtain b ( ϕ , ϕ , ϕ , ϕ ) (5.27):= X m ≥ X X : X ∩ ˇ C [ ϕ ,ϕ = ∅ ,X ∩ ˇ C [ ϕ ,ϕ = ∅ " sup t ∈ [0 , ( k Ψ ( X ; t ) k ) | X | G F ( m ) ≤ X m ≥ X x ∈ ˇ C [ ϕ ,ϕ ,y ∈ ˇ C [ ϕ ,ϕ X X ∋ x,y " sup t ∈ [0 , ( k Ψ ( X ; t ) k ) | X | G F ( m ) ≤ ( k| Ψ k| F ) X x ∈ ˇ C [ ϕ ,ϕ ,y ∈ ˇ C [ ϕ ,ϕ F (d( x, y )) X m ≥ G F ( m ) ≤ (64) κ , ,F (cid:16) c (0) ϕ ,ϕ ,ϕ ,ϕ (cid:17) ( k| Ψ k| F ) X m ≥ G F ( m ) < ∞ . (5.28)We used Lemma 5.3 at the last inequality. The last value is finite by (C.14) for our F ∈ F a . (cid:3) Set c (1) ζ ,ζ ,ζ := p − max { cos( ζ − ζ ) , cos( ζ − ζ ) } , ζ , ζ , ζ ∈ [0 , π ) . (5.29) Lemma 5.4.
For ϕ < ϕ < ϕ with ϕ − ϕ < π , we have b ( ϕ , ϕ , ϕ ) := X m ≥ X X : X ⊂ ˇ C [ ϕ ,ϕ X ∩ ˇ C [ ϕ ,ϕ = ∅ X ∩ ˇ C [ ϕ ,ϕ = ∅ X ∩ ((( ˇ C ( ϕ ,ϕ ) c ) ( m ) ) = ∅ " sup t ∈ [0 , ( k Ψ ( X ; t ) k ) | X | G F ( m ) ≤ · · · ( πκ , ,F + F (0)) ( k| Ψ k| F ) X m ≥ ( m + 1) G F ( m ) (cid:18)(cid:16) c (1) ϕ ,ϕ ,ϕ (cid:17) − + (cid:16) c (1) ϕ ,ϕ ,ϕ (cid:17) − (cid:19) < ∞ . (5.30) Proof.
Set L ϕ := (cid:8) z ∈ R | arg z = ϕ (cid:9) , ϕ ∈ [0 , π ) . (5.31)33ote that if X ∈ S Z satisfies X ⊂ ˇ C [ ϕ ,ϕ ] and X ∩ (cid:16)(cid:16)(cid:0) ˇ C ( ϕ ,ϕ ) (cid:1) c (cid:17) ( m ) (cid:17) = ∅ , then we have d ( X, L ϕ ) ≤ m, or d ( X, L ϕ ) ≤ m. (5.32)Therefore, we have X m ≥ X X : X ⊂ ˇ C [ ϕ ,ϕ X ∩ ˇ C [ ϕ ,ϕ = ∅ X ∩ ˇ C [ ϕ ,ϕ = ∅ X ∩ ((( ˇ C ( ϕ ,ϕ ) c ) ( m ) ) = ∅ " sup t ∈ [0 , ( k Ψ ( X ; t ) k ) | X | G F ( m ) ≤ X m ≥ G F ( m ) X X : X ∩ ˇ C [ ϕ ,ϕ = ∅ d ( X,L ϕ ) ≤ m + X X : X ∩ ˇ C [ ϕ ,ϕ = ∅ d ( X,L ϕ ) ≤ m " sup t ∈ [0 , ( k Ψ ( X ; t ) k ) | X | ≤ X m ≥ G F ( m ) X x ∈ ˇ C [ ϕ ,ϕ y ∈ L ϕ ( m ) + X x ∈ ˇ C [ ϕ ,ϕ y ∈ L ϕ ( m ) X X : X ∋ x,y " sup t ∈ [0 , ( k Ψ ( X ; t ) k ) | X | ≤ ( k| Ψ k| F ) X m ≥ G F ( m ) X x ∈ ˇ C [ ϕ ,ϕ y ∈ L ϕ ( m ) + X x ∈ ˇ C [ ϕ ,ϕ y ∈ L ϕ ( m ) F (d( x, y )) ≤ · · · ( πκ , ,F + F (0)) ( k| Ψ k| F ) X m ≥ ( m + 1) G F ( m ) (cid:18)(cid:16) c (1) ϕ ,ϕ ,ϕ (cid:17) − + (cid:16) c (1) ϕ ,ϕ ,ϕ (cid:17) − (cid:19) (5.33)At the last inequality, we used Lemma C.5 with ϕ − ϕ < π . Because of ϕ − ϕ < π and (C.14),the last value is finite. (cid:3) Now let us go back to the estimate of (5.23). If C , C ∈ C are C = ˇ C [ ϕ ,ϕ ] , C = ˇ C [ ϕ ,ϕ ] with ϕ < ϕ < ϕ < ϕ , ϕ − ϕ < π , then from Lemma 5.3, we have M ( C , C ) ≤ b ( ϕ , ϕ , ϕ , ϕ ) < ∞ . (5.34)Now suppose that C , C ∈ C are C = ˇ C [ ϕ ,ϕ ] , C = ˇ C [ ϕ ,ϕ ] with ϕ < ϕ < ϕ , ϕ − ϕ < π .By the definition of C and C , there is some C = C ( ζ ,ζ ) ∈ C such that ϕ < ζ < ϕ < ζ < ϕ and ζ − ζ < π . For X ∈ S Z to give a nonzero contribution in (5.23), it have to satisfy X ( m ) ∩ (cid:0) ˇ C [ ζ ,ζ ] (cid:1) c = ∅ , X ∩ ˇ C [ ϕ ,ϕ ] = ∅ , X ∩ ˇ C [ ϕ ,ϕ ] = ∅ . (5.35)For such an X , one of the following occurs: (i) X ∩ ˇ C [ ζ ,ϕ ] = ∅ and X ∩ ˇ C [ ϕ ,ϕ ] = ∅ (ii) X ∩ ˇ C [ ϕ ,ζ ] = ∅ and X ∩ ˇ C [ ϕ ,ϕ ] = ∅ iii) X ∩ ˇ C [ ϕ ,ζ ] = ∅ (and X ∩ ˇ C [ ζ ,ϕ ] = ∅ ) and X ∩ ˇ C [ ϕ ,ϕ +2 π ] = ∅ , (iv) X ⊂ ˇ C ζ ,ζ , X ∩ (cid:16)(cid:0) ˇ C ζ ,ζ (cid:1) c (cid:17) ( m ) = ∅ , X ∩ ˇ C [ ϕ ,ζ ] = ∅ , and X ∩ ˇ C [ ζ ,ϕ ] = ∅ .Hence we get M ( C , C ) ≤ b ( ϕ , ϕ , ζ , ϕ ) + b ( ϕ , ζ , ϕ , ϕ ) + b ( ϕ , ζ , ϕ , ϕ + 2 π )+ b ( ζ , ϕ , ζ ) < ∞ . (5.36)Hence we have proven the claim of Step 2 . Step 3.
Next we set ˜Ξ(
Z, t ) := ( Ξ (1) ( Z, t ) , if ∃ C ∈ C s.t. Z ⊂ C . (5.37)Clearly, we have ˜Ξ ∈ ˆ B ˜ F ([0 , H Λ n , ˜Ξ ( t ) + V n ( t ) = H Λ n , Ξ (1) ( t ) . (5.38)As a uniform limit of [0 , ∋ t V n ( t ) ∈ A , (5.21), [0 , ∋ t V ( t ) ∈ A is norm-continuous.Because of ˜Ξ ∈ ˆ B ˜ F ([0 , , ∋ t τ ˜Ξ t,s ( V ( t )) ∈ A is also norm-continuous, for each s ∈ [0 , s ∈ [0 , , ∋ t W ( s ) ( t ) ∈ U ( A )such that ddt W ( s ) ( t ) = − iτ ˜Ξ t,s ( V ( t )) W ( s ) ( t ) , W ( s ) ( s ) = I . (5.39)It is given by W ( s ) ( t ) := ∞ X k =0 ( − i ) k Z ts ds Z s s ds · · · Z s k − s ds k τ ˜Ξ s ,s ( V ( s )) · · · τ ˜Ξ s k ,s ( V ( s k )) . (5.40)Analogously, for each s ∈ [0 ,
1] and n ∈ N , we define a unique norm-differentiable map from [0 , U ( A ) such that ddt W ( s ) n ( t ) = − iτ (Λ n )˜Ξ t,s ( V n ( t )) W ( s ) n ( t ) , W ( s ) n ( s ) = I . (5.41)It is given by W ( s ) n ( t ) := ∞ X k =0 ( − i ) k Z ts ds Z s s ds · · · Z s k − s ds k τ (Λ n )˜Ξ s ,s ( V n ( s )) · · · τ (Λ n )˜Ξ s k ,s ( V n ( s k )) . (5.42)By the uniform convergence (5.21) and Lemma D.3, we havelim n →∞ sup t ∈ [0 , (cid:13)(cid:13)(cid:13) τ (Λ n )˜Ξ t,s ( V n ( t )) − τ ˜Ξ t,s ( V ( t )) (cid:13)(cid:13)(cid:13) = 0 . (5.43)From this and (5.40), (5.42), we obtainlim n →∞ sup t ∈ [0 , (cid:13)(cid:13)(cid:13) W ( s ) n ( t ) − W ( s ) ( t ) (cid:13)(cid:13)(cid:13) = 0 . (5.44)35his and Theorem D.3 4 for Ξ (1) , ˜Ξ ∈ B ˜ F ([0 , n →∞ τ (Λ n ) , ˜Ξ s,t ◦ Ad (cid:16) W ( s ) n ( t ) (cid:17) ( A ) = τ ˜Ξ s,t ◦ Ad (cid:16) W ( s ) ( t ) (cid:17) ( A ) , lim n →∞ τ (Λ n ) , Ξ (1) s,t ( A ) = τ Ξ (1) s,t ( A ) , (5.45)for any A ∈ A .Note that for any A ∈ A ddt τ (Λ n ) , ˜Ξ s,t ◦ Ad (cid:16) W ( s ) n ( t ) (cid:17) ( A )= − i h H Λ n , ˜Ξ ( t ) , τ (Λ n ) , ˜Ξ s,t ◦ Ad (cid:16) W ( s ) n ( t ) (cid:17) ( A ) i − iτ (Λ n ) , ˜Ξ s,t (cid:16)h τ (Λ n ) , ˜Ξ t,s ( V n ( t )) , Ad (cid:16) W ( s ) n ( t ) (cid:17) ( A ) i(cid:17) = − i h H Λ n , ˜Ξ ( t ) + V n ( t ) , τ (Λ n ) , ˜Ξ s,t ◦ Ad (cid:16) W ( s ) n ( t ) (cid:17) ( A ) i = − i h H Λ n , Ξ (1) ( t ) , τ (Λ n ) , ˜Ξ s,t ◦ Ad (cid:16) W ( s ) n ( t ) (cid:17) ( A ) i . We used (D.10) for the second equality and (5.38) for the third equality. On the other hand, forany A ∈ A , we have ddt τ (Λ n ) , Ξ (1) s,t ( A ) = − i h H Λ n , Ξ (1) ( t ) , τ (Λ n ) , Ξ (1) s,t ( A ) i . (5.46)Therefore, τ (Λ n ) , ˜Ξ s,t ◦ Ad (cid:16) W ( s ) n ( t ) (cid:17) ( A ) and τ (Λ n ) , Ξ (1) s,t ( A ) satisfy the same differential equation. Alsonote that we have τ (Λ n ) , ˜Ξ s,s ◦ Ad (cid:16) W ( s ) n ( s ) (cid:17) ( A ) = τ (Λ n ) , Ξ (1) s,s ( A ) = A . Therefore, we get τ (Λ n ) , ˜Ξ s,t ◦ Ad (cid:16) W ( s ) n ( t ) (cid:17) ( A ) = τ (Λ n ) , Ξ (1) s,t ( A ) . (5.47)By (5.45), we obtain τ ˜Ξ s,t ◦ Ad (cid:16) W ( s ) ( t ) (cid:17) ( A ) = τ Ξ (1) s,t ( A ) , A ∈ A , t, s ∈ [0 , . (5.48)Taking inverse, we get Ad (cid:16) W ( s ) ∗ ( t ) (cid:17) ◦ τ ˜Ξ t,s = τ Ξ (1) t,s , t, s ∈ [0 , . (5.49) Step 4.
Combining (5.17) and (5.49) we have τ Ψ1 , = τ Ψ (0) , τ Ξ (1) , = τ Ψ (0) , ◦ Ad (cid:16)(cid:16) W (1) (0) (cid:17) ∗ (cid:17) ◦ τ ˜Ξ0 , . (5.50)By the definition of Ψ (0) and ˜Ξ, we obtain decompositions τ Ψ (0) , = α [0 ,θ ] ⊗ α ( θ ,θ ] ⊗ α ( θ ,θ ] ⊗ α ( θ , π ] τ ˜Ξ0 , = α ( θ . ,θ . ] ⊗ α ( θ . ,θ . ] ⊗ α ( θ . ,θ . ] (5.51)with α s of the form (2.11) and (2.12). This completes the proof of the first part. Step 5.
Suppose that β Ug (Ψ( X ; t )) = Ψ( X ; t ) for any X ∈ S Z , t ∈ [0 , g ∈ G . Thenclearly we have β Ug (cid:0) Ψ (0) ( X ; t ) (cid:1) = Ψ (0) ( X ; t ) for any X ∈ S Z , t ∈ [0 , g ∈ G . By TheoremD.3, 5, this implies τ Ψ (0) , β Ug = β Ug τ Ψ (0) , . From the decomposition (5.51), this means all of α [0 ,θ ] ,σ ,36 ( θ ,θ ] ,σ,ζ , α ( θ ,θ ] ,σ,ζ , α ( θ , π ] ,ζ , σ = L, R , ζ = U, D commute with β Ug . Because Π X commuteswith β Ug τ Ψ t,s commutes with β Ug (Theorem D.3, 5) , and Ψ (1) is β Ug -invariant, Ξ ( s ) is β Ug -invariantfrom the definition (5.8). Therefore, from the definition (5.37), ˜Ξ is also β Ug -invariant. Hence byTheorem D.3, 5 τ ˜Ξ0 , commutes with β Ug . The decomposition (5.51) then implies that α ( θ . ,θ . ] ,σ,ζ α ( θ . ,θ . ] ,σ,ζ α ( θ . ,θ . ] ,σ,ζ , σ = L, R , ζ = U, D commute with β Ug . (cid:3) An analogous proof shows the following.
Proposition 5.5.
Let F ∈ F a be an F -function of the form F ( r ) = exp ( − r θ ) (1+ r ) with a constant < θ < . Let Ψ ∈ ˆ B F ([0 , be a path of interactions satisfying Ψ ∈ ˆ B F ([0 , . Define Ψ (0) ∈ ˆ B F ([0 , by Ψ (0) ( X ; t ) := ( Ψ ( X ; t ) , if X ⊂ H U or X ⊂ H D , otherwise , (5.52) for each X ∈ S Z , t ∈ [0 , . Then (cid:16) τ Ψ (0) , (cid:17) − τ Ψ1 , belongs to HAut( A ) . Proof.
Define ˜ F as in (5.3) with some 0 < θ ′ < θ . The same argument as in Theorem 5.2 Step1. implies that there exists Ξ (1) ∈ ˆ B ˜ F [0 ,
1] with ˜ F ∈ F a , such that τ Ψ1 , = τ Ψ (0) , τ Ξ (1) , . (5.53)This Ξ (1) is given by the formula (5.8) for current Ψ and Ψ (1) ( X ; t ) := Ψ (0) ( X ; t ) − Ψ ( X ; t ). Toprove the theorem, it suffices to show that τ Ξ (1) , belongs to HAut( A ). Indeed, for any 0 < θ < π ,as in Theorem 5.2 Step 2 , we have X Z : Z * C [0 ,θ ,L and Z * C [0 ,θ ,R sup t ∈ [0 , (cid:13)(cid:13)(cid:13) Ξ (1) ( Z, t ) (cid:13)(cid:13)(cid:13) ≤ C F (cid:16) e I F (Ψ) − (cid:17) X m ≥ X X : X ( m ) * C [0 ,θ ,L and X ( m ) * C [0 ,θ ,R " sup t ∈ [0 , (cid:16)(cid:13)(cid:13)(cid:13) Ψ (1) ( X ; t ) (cid:13)(cid:13)(cid:13)(cid:17) | X | G F ( m ) < ∞ . (5.54)To see this, note that if X in the last line has a non-zero contribution to the sum, then at leastone of the following occurs. (i) X ∩ C [ θ , π ] ,U = ∅ , and X ∩ H D = ∅ (ii) X ∩ C [ θ , π ] ,D = ∅ , and X ∩ H U = ∅ (iii) X ⊂ C [0 ,θ ] and (1) X ∩ C [0 ,θ ] ,L = ∅ and X ∩ C [0 ,θ ] ,R = ∅ , or (2) X ⊂ C [0 ,θ ] ,R , X ∩ ˇ C [0 ,θ ] = ∅ , X ∩ ˇ C [ − θ , = ∅ and X ( m ) ∩ (cid:0) C [0 ,θ ] ,R (cid:1) c = ∅ , (3) X ⊂ C [0 ,θ ] ,L , X ∩ ˇ C [ π − θ ,π ] = ∅ , X ∩ ˇ C [ π,π + θ ] = ∅ and X ( m ) ∩ (cid:0) C [0 ,θ ] ,L (cid:1) c = ∅ .Therefore, the summation in the second line of (5.54) is bounded by8 C F (cid:16) e I F (Ψ) − (cid:17) b ( θ , π − θ , π, π ) + b (0 , π, π + θ , π − θ ) + b ( − θ , θ , π − θ , π + θ )+ b ( − θ , , θ ) + b ( π − θ , π, π + θ ) ! < ∞ , Step 3. of Theorem 5.2, setting˜Ξ(
Z, t ) := ( Ξ (1) ( Z, t ) , if Z ⊆ C [0 ,θ ] ,L or Z ⊆ C [0 ,θ ] ,R , (5.55)we obtain τ Ξ (1) , = (inner) ◦ τ ˜Ξ0 , . By the definition, τ ˜Ξ0 , decomposes as τ ˜Ξ0 , = ζ L ⊗ ζ R , with some ζ σ ∈ Aut (cid:16) A C [0 ,θ ,σ (cid:17) , σ = L, R . As this holds for any 0 < θ < π , we conclude τ Ξ (1) , ∈ HAut( A ). (cid:3) Theorem 5.6.
Let F ∈ F a be an F -function of the form F ( r ) = exp ( − r θ ) (1+ r ) with a constant <θ < . Let Ψ ∈ ˆ B F ([0 , be a path of interactions satisfying Ψ ∈ ˆ B F ([0 , . If Ψ is β -invariant,then τ Ψ1 , belongs to GUQAut( A ) . Proof.
Define Ψ (0) as in (5.52) for our Ψ. By Proposition 5.5, we have (cid:16) τ Ψ (0) , (cid:17) − τ Ψ1 , ∈ HAut( A ).On the other hand, applying Theorem 5.2 to Ψ (0) ∈ ˆ B F ([0 , τ Ψ (0) , belongs toSQAut( A ) . Note that Ψ (0) ( X ; t ) is non-zero only if X ⊂ H U or X ⊂ H D , and it coincides withΨ( X ; t ) when it is non-zero. Therefore, if Ψ is β -invariant, Ψ (0) is β Ug -invariant. Therefore, byTheorem 5.2, we have τ Ψ (0) , ∈ GSQAut( A ). Hence we have τ Ψ1 , ∈ GUQAut( A ). (cid:3) Proof of Theorem 1.4.
Let Φ ∈ P UG be the fixed trivial interaction with a unique gappedground state. Its ground state ω := ω Φ is of a product form (2.16). For any Φ ∈ P SLβ , wehave Φ ∼ Φ. Then by Theorem 5.1, there exists some Ψ ∈ ˆ B F ([0 , ∈ ˆ B F ([0 , F ∈ F a of the form F ( r ) = exp ( − r θ ) (1+ r ) with a 0 < θ <
1, such that ω Φ = ω Φ ◦ τ Ψ1 , . FromTheorem 5.2, τ Ψ1 , belongs to SQAut( A ). Because Φ ∈ P SLβ , ω Φ = ω Φ ◦ τ Ψ1 , is β -invariant. Then,by Theorem 3.1, IG( ω Φ ) is not empty. Therefore, we may define h Φ := h ( ω Φ ) by Definition 2.18.To see that h Φ is an invariant of ∼ β , let Φ , Φ ∈ P SLβ with Φ ∼ β Φ . Then by Theorem5.1, there exists some β -invariant Ψ ∈ ˆ B F ([0 , ∈ ˆ B F ([0 , F ∈ F a of the form F ( r ) = exp ( − r θ ) (1+ r ) with a constant 0 < θ < ω Φ = ω Φ ◦ τ Ψ1 , . Applying Theorem 5.6, tothis Ψ, τ Ψ1 , belongs to GUQAut( A ). Then Theorem 4.1 implies h Φ = h ( ω Φ ) = h ( ω Φ ◦ τ Ψ1 , ) = h ( ω Φ ) = h Φ , (5.56)proving the stability. (cid:3) d H U α When α ∈ EAut( ω ) has some good factorization property with respect to the action of β Ug , theindex h ( ω ) can be calculated without going through GNS representations.38 efinition 6.1. For α ∈ Aut ( A ), we set (cid:0) d H U α (cid:1) ( g ) := α − β Ug ◦ α ◦ (cid:0) β Ug (cid:1) − , g ∈ G. (6.1)We say that d H U α is factorized into left and right if there are automorphisms γ g,σ ∈ Aut ( A H σ ), g ∈ G , σ = L, R such that (cid:0) d H U α (cid:1) ( g ) = (inner) ◦ ( γ g,L ⊗ γ g,R ) , g ∈ G. (6.2)For known examples of 2-dimensional SPT-phases like [CGLW] and [MM] and [Y] or injectivePEPS [MGSC], this property holds. From such an automorphism, we can derive an outer actionof G . Lemma 6.2.
Let α ∈ Aut ( A ) be an automorphism. Suppose that d H U α is factorized into left andright i.e., there are automorphisms γ g,σ ∈ Aut ( A H σ ) , g ∈ G , σ = L, R such that (cid:0) d H U α (cid:1) ( g ) = (inner) ◦ ( γ g,L ⊗ γ g,R ) , g ∈ G. (6.3) Then there are unitaries v σ ( g, h ) ∈ U ( A H σ ) , g, h ∈ G , σ = L, R such that γ g,σ β σUg γ h,σ β σUh (cid:0) γ gh,σ β σUgh (cid:1) − = Ad ( v σ ( g, h )) . (6.4) Proof.
Because β Ug is a group action, substituting (6.3), we getid A = α − β Ug α ◦ α − β Uh α ◦ (cid:0) α − β Ugh α (cid:1) − = (inner) ◦ (cid:0) γ g,L β LUg ⊗ γ g,R β RUg (cid:1) ◦ (cid:0) γ h,L β LUh ⊗ γ h,R β RUh (cid:1) ◦ (cid:0) γ gh,L β LUgh ⊗ γ gh,R β RUgh (cid:1) − = (inner) ◦ (cid:16) γ g,L β LUg γ h,L β LUh (cid:0) γ gh,L β LUgh (cid:1) − ⊗ γ g,R β RUg γ h,R β RUh (cid:0) γ gh,R β RUgh (cid:1) − (cid:17) (6.5)By Lemma B.1, we then see that there are unitaries v σ ( g, h ) ∈ Aut ( A H σ ), g ∈ G , σ = L, R satisfying (6.4). (cid:3)
It is well known that a third cohomology class can be associated to cocycle actions [C][J].
Lemma 6.3.
Let α ∈ Aut ( A ) be an automorphism such that d H U α is factorized into left and rightas (6.3). Let v σ ( g, h ) ∈ U ( A H σ ) , g, h ∈ G , σ = L, R be unitaries satisfying (6.4), given in Lemma6.2. Then there is some c σ ∈ C ( G, T ) , σ = L, R such that v σ ( g, h ) v σ ( gh, k ) = c σ ( g, h, k ) (cid:0) γ g,σ ◦ β σUg ( v σ ( h, k )) (cid:1) v σ ( g, hk ) , g, h, k ∈ G. (6.6) Proof.
By (6.4), we have ˆ γ g,σ ˆ γ h,σ = Ad ( v σ ( g, h )) ◦ ˆ γ gh,σ (6.7)for ˆ γ g,σ := γ g,σ β σUg . Using this, we haveAd ( v σ ( g, h )) ◦ Ad ( v σ ( gh, k )) ◦ ˆ γ ghk,σ = Ad ( v σ ( g, h )) ◦ ˆ γ gh,σ ◦ ˆ γ k,σ = ˆ γ g,σ ˆ γ h,σ ˆ γ k,σ = ˆ γ g,σ ◦ Ad ( v σ ( h, k )) ◦ ˆ γ hk,σ = Ad (ˆ γ g,σ ( v σ ( h, k ))) ˆ γ g,σ ◦ ˆ γ hk,σ = Ad (ˆ γ g,σ ( v σ ( h, k )) v σ ( g, hk )) ◦ ˆ γ ghk,σ . (6.8)Because A ′ ∩ A = I A , ˆ γ g,σ ( v σ ( h, k )) v σ ( g, hk ) and v σ ( g, h ) v σ ( gh, k ) are proportional to each other,proving the Lemma. (cid:3) By the same argument as Lemma 2.4, we can show that this c R is actually a 3-cocycle. If ω ∈ SL is given by an automorphism α ∈ EAut( ω ) with factorized d H U α , and if ω is invariant under β Ug ,then we have h ( ω ) = [ c R ] H ( G, T ) , for c R given in Lemma 6.3.39 heorem 6.4. Let ω be a reference state of the form (2.16), and assume that ω ◦ β Ug = ω for any g ∈ G . Let α ∈ QAut ( A ) be an automorphism. Suppose that d H U α is factorized into left and rightas in (6.3) with some γ g,σ ∈ Aut (cid:0) A C θ ,σ (cid:1) and < θ < π , for σ = L, R . Let v σ ( g, h ) ∈ U ( A H σ ) , g, h ∈ G , σ = L, R be unitaries satisfying (6.4) (given in Lemma 6.2) and c R ∈ C ( G, T ) satisfying(6.6) for these v R ( g, h ) (given in Lemma 6.3). Then we have ω ◦ α ∈ SL with IG( ω ◦ α ) = ∅ , c R ∈ Z ( G, T ) , and h ( ω ◦ α ) = [ c R ] H ( G, T ) . Proof.
That ω ◦ α ∈ SL is by definition. BecauseAd ( v σ ( g, h )) = γ g,σ β σUg γ h,σ β σUh (cid:0) γ gh,σ β σUgh (cid:1) − ∈ Aut (cid:0) A C θ ,σ (cid:1) , (6.9)our v σ ( g, h ) belongs to U (cid:0) A C θ ,σ (cid:1) . Because ω α ◦ α − β Ug α = ω β Ug α = ω α (6.10)and α − β Ug α = (inner) ◦ ( γ g,L ⊗ γ g,R ) ◦ β Ug , (6.11)with γ g,σ ∈ Aut (cid:0) A C θ ,σ (cid:1) , we have ( α − β Ug α ) ∈ IG( ω α, θ ), and ( γ g,σ ) ∈ T (cid:0) θ , α − β Ug α (cid:1) . Clearly α ∈ EAut( ω ◦ α ) and there is ( α L , α R , Θ) ∈ D θ α because α ∈ QAut( A ). Set γ g := γ g,L ⊗ γ g,R .From Lemma 2.1, there is some W g ∈ U ( H ) g ∈ G satisfyingAd ( W g ) ◦ π = π ◦ ( α L ⊗ α R ) ◦ Θ ◦ γ g β Ug ◦ Θ − ◦ ( α L ⊗ α R ) − , g ∈ G. (6.12)In particular, because v R ( h, k ) belongs to U (cid:16) A ( C θ ) R (cid:17) , Θ ∈ Aut (cid:16) A C cθ (cid:17) , and γ g β Ug preserves A ( C θ ) R , we haveAd ( W g ) ◦ π ◦ ( α L ⊗ α R ) (id A L ⊗ ( v R ( h, k ))) = π ◦ ( α L ⊗ α R ) ◦ Θ ◦ γ g β Ug ◦ Θ − ◦ (id A L ⊗ ( v R ( h, k )))= π ◦ ( α L ⊗ α R ) (cid:0) id A L ⊗ γ g,R β RUg ( v R ( h, k )) (cid:1) = I H L ⊗ π R ◦ α R ◦ γ g,R β RUg ( v R ( h, k )) (6.13)On the other hand, (6.4) meansAd ( π σ ◦ α σ ( v σ ( g, h ))) π σ = π σ ◦ α σ ◦ γ g,σ β σUg γ h,σ (cid:0) β σUg (cid:1) − ( γ gh,σ ) − ◦ α − σ . (6.14)From (6.12) and (6.14) we have(( W g ) , ( π σ ◦ α σ ( v σ ( g, h )))) ∈ IP (cid:0) ω ◦ α, α, θ , ( α − β Ug α ) , ( γ g,σ ) , ( α L , α R , Θ) (cid:1) . (6.15)Now from (6.6) and then (6.13), we obtain I H L ⊗ π R ◦ α R ( v R ( g, h ) v R ( gh, k )) = c R ( g, h, k ) I H L ⊗ π R ◦ α R (cid:0)(cid:0) γ g,R ◦ β RUg ( v R ( h, k )) (cid:1) v R ( g, hk ) (cid:1) = c R ( g, h, k ) (Ad ( W g ) (id H L ⊗ π R α R ( v R ( h, k )))) · ( I H L ⊗ π R ◦ α R ( v R ( g, hk ))) . (6.16)This means c R = c R (cid:0) ω ◦ α, α, θ , ( α − β Ug α ) , ( γ g,σ ) , ( α L , α R , Θ) , (( W g ) , (( π σ ◦ α σ ( v σ ( g, h ))))) (cid:1) (6.17)in the Definition 2.5. Hence we get c R ∈ Z ( G, T ), and h ( ω ◦ α ) = [ c R ] H ( G, T ) . (cid:3) cknowledgment. The author is grateful to Hal Tasaki for a stimulating discussion over two-dimensional Dijkgraaf-Witten model. The author is grateful to Yasuyuki Kawahigashi for introducing the author variouspapers from operator algebra. This work was supported by JSPS KAKENHI Grant Number16K05171 and 19K03534. It was also supported by JST CREST Grant Number JPMJCR19T2.The present result and the main idea of the proof were announced publicly on 15 December 2020 atIAMP One World Mathematical Physics Seminar (see you-tube video)[O2]. It was also presentedin the international meeting
Theoretical studies of topological phases of matter on 17 December2020, and Current Developments in Mathematics 4th January 2021 via zoom with a lecture note.[O2].
A Basic Notations
For a finite set S , S indicates the number of elements in S . For t ∈ R , [ t ] denotes the smallestinteger less than or equal to t .For a Hilbert space H , B ( H ) denotes the set of all bounded operators on H . If V : H → H isa linear map from a Hilbert space H to another Hilbert space H , then Ad( V ) : B ( H ) → B ( H )denotes the map Ad( V )( x ) := V xV ∗ , x ∈ B ( H ). Occasionally we write Ad V instead of Ad( V ).For a C ∗ -algebra B and v ∈ B , we set Ad( v )( x ) := Ad v ( x ) := vxv ∗ , x ∈ B .For a state ω , ϕ on a C ∗ -algebra B , we write ω ∼ q . e . ϕ when they are quasi-equivalent. (See[BR1].) We denote by Aut B the group of automorphisms on a C ∗ -algebra B . The group of innerautomorphisms on a unital C ∗ -algebra B is denoted by Inn B . For γ , γ ∈ Aut( B ), γ = (inner) ◦ γ means there is some unitary u in B such that γ = Ad( u ) ◦ γ . For a unital C ∗ -algebra B , theunit of B is denoted by I B . For a Hilbert space we write I H := I B ( H ) . For a unital C ∗ -algebra B , by U ( B ), we mean the set of all unitary elements in B . For a Hilbert space we write U ( H ) for U ( B ( H )).For a state ϕ on B and a C ∗ -subalgebra C of B , ϕ | C indicates the restriction of ϕ to C . B Automorphisms on UHF-algebras
Lemma B.1.
Let A , B be UHF-algebras. If automorphisms γ A ∈ Aut( A ) , γ B ∈ Aut( B ) and aunitary W ∈ U ( A ⊗ B ) satisfy ( γ A ⊗ γ B ) ( X ) = Ad W ( X ) , X ∈ A ⊗ B , (B.1) then there are unitaries u A ∈ U ( A ) and u B ∈ U ( B ) such that γ A ( X A ) = Ad u A ( X A ) , X A ∈ A ,γ B ( X B ) = Ad u B ( X B ) , X B ∈ B . (B.2) Proof.
Fix some irreducible representations ( H A , π A ), ( H B , π B ), of A , B . We claim that thereare unitaries v A ∈ U ( H A ) and v B ∈ U ( H B ) such thatAd v A ( π A ( X A )) = π A ◦ γ A ( X A ) , X A ∈ A , Ad v B ( π B ( X B )) = π B ◦ γ B ( X B ) , X B ∈ B . (B.3)To see this, note that ( π A ◦ γ A ⊗ π B ◦ γ B ) = Ad ( π A ⊗ π B )( W ) ◦ ( π A ⊗ π B ) . (B.4)From this, π A ◦ γ A (resp. π B ◦ γ B ) is quasi-equivalent to π A (resp. π B ). Because π A and π B areirreducible, by the Wigner theorem, there are unitaries v A ∈ U ( H A ) and v B ∈ U ( H B ) satisfying(B.3). 41e then haveAd ( π A ⊗ π B )( W ) ◦ ( π A ⊗ π B ) = ( π A ◦ γ A ) ⊗ ( π B ◦ γ B ) = (Ad v A ◦ π A ) ⊗ (Ad v B ◦ π B ) = Ad v A ⊗ v B ◦ ( π A ⊗ π B ) . (B.5)Because π A ⊗ π B is irreducible, there is a c ∈ T such that( π A ⊗ π B ) ( W ) = c ( v A ⊗ v B ) . (B.6)We claim there is a unitary u B ∈ U ( B ) such that π B ( u B ) = v B . (B.7)Choose a unit vector ξ ∈ H A with h ξ, v A ξ i 6 = 0. For each x ∈ B ( H A ⊗ H B ), the map H B × H B ∋ ( η , η )
7→ h ( ξ ⊗ η ) , x ( ξ ⊗ η ) i (B.8)is a bounded sesquilinear form. Therefore, there is a unique Φ ξ ( x ) ∈ B ( H B ) such that h η , Φ ξ ( x ) η i = h ( ξ ⊗ η ) , x ( ξ ⊗ η ) i , ( η , η ) ∈ H B × H B . (B.9)The map Φ ξ : B ( H A ⊗ H B ) → B ( H B ) is linear and k Φ ξ ( x ) k ≤ k x k , x ∈ B ( H ) . (B.10)Because W belongs to A ⊗ B , there are sequence z N = n N X i =1 a ( N ) i ⊗ b ( N ) i , with a ( N ) i ∈ A , b ( N ) i ∈ B (B.11)such that k W − z N k < N . (B.12)Because of (B.10), we have k Φ ξ (( π A ⊗ π B ) ( W − z N )) k < N . (B.13)Note that Φ ξ (( π A ⊗ π B ) ( z N )) = n N X i =1 D ξ, π A (cid:16) a ( N ) i (cid:17) ξ E π B (cid:16) b ( N ) i (cid:17) ∈ π B ( B ) . (B.14)Therefore, we have c h ξ, v A ξ i v B = Φ ξ ( c ( v A ⊗ v B )) = Φ ξ (( π A ⊗ π B ) ( W )) ∈ π B ( B ) n , (B.15)where · n denotes the norm closure. Because π B ( B ) is norm-closed, we have π B ( B ) n = π B ( B ).Hence we have v B ∈ π B ( B ), i.e., there is a unitary u B ∈ B such that v B = π B ( u B ).We then have π B ◦ Ad u B ( X ) = Ad π B ( u B ) ◦ π B ( X ) = Ad v B ◦ π B ( X ) = π B ◦ γ B ( X ) , X ∈ B . (B.16)As B is simple, Ad u B ( X ) = γ B ( X ) for all X ∈ B .The proof for A is the same. (cid:3) F -functions In this section, we collect various estimates about F -functions. Let us first start from the definition. Definition C.1. An F -function F on ( Z , d) is a non-increasing function F : [0 , ∞ ) → (0 , ∞ )such that (i) k F k := sup x ∈ Z (cid:16)P y ∈ Z F (d( x, y )) (cid:17) < ∞ , and (ii) C F := sup x,y ∈ Z (cid:16)P z ∈ Z F (d( x,z )) F (d( z,y )) F (d( x,y )) (cid:17) < ∞ .These are called uniform integrability and the convolution identity , respectively.We denote by F a a class of F -functions which decay quickly. Definition C.2.
We say an F -function F belongs to F a if (i) for any k ∈ N ∪ { } and 0 < θ ≤
1, we have κ θ,k,F := ∞ X n =0 ( n + 1) k ( F ( n )) θ < ∞ , (C.1)and (ii) for any 0 < θ <
1, there is a F -function ˜ F θ such thatmax (cid:26) F (cid:16) r (cid:17) , (cid:16) F (cid:16)h r i(cid:17)(cid:17) θ (cid:27) ≤ ˜ F θ ( r ) , r ≥ . (C.2)For example, a function F ( r ) = exp ( − r θ ) (1+ r ) with a constant 0 < θ < F a . (See section8 of [NSY].)In this section, we derive inequalities about F ∈ F a . In order for that the following Lemma isuseful. We will freely identify C and R in an obvious manner. Lemma C.3.
For ≤ θ < θ ≤ π , c > , and r ≥ , set S [ θ ,θ ] r,c := (cid:8) se iθ ∈ R | r ≤ s < r + c, θ ∈ [ θ , θ ] (cid:9) . (C.3) Then we have (cid:16) S [ θ ,θ ] r,c ∩ Z (cid:17) ≤ π (cid:16) √ c (cid:17) ( r + 1) . (C.4) In particular, we have (cid:16) S [ θ ,θ ] r, ∩ Z (cid:17) ≤ r + 1) . (C.5) Proof.
Because the diameter of a 2-dimensional unit square is √
2, any unit square B of Z with B ∩ S [ θ ,θ ] r,c ∩ Z = ∅ satisfies B ⊂ ˆ S [ θ ,θ ] r,c ( √ n B | unit square of Z with B ∩ S [ θ ,θ ] r,c ∩ Z = ∅ o = X B : B ∩ S [ θ ,θ r,c ∩ Z = ∅ ≤ (cid:12)(cid:12)(cid:12) ˆ S [ θ ,θ ] r,c ( √ (cid:12)(cid:12)(cid:12) . (C.6)43ote that the area of ˆ S [ θ ,θ ] r,c ( √ (cid:12)(cid:12)(cid:12) ˆ S [ θ ,θ ] r,c ( √ (cid:12)(cid:12)(cid:12) is less than (cid:12)(cid:12)(cid:12) ˆ S [ θ ,θ ] r,c ( √ (cid:12)(cid:12)(cid:12) ≤ π (cid:16) ( r + c + √ − ( r − √ ) (cid:17) ≤ π (2 r + c )(2 √ c ) ≤ π (cid:16) √ c (cid:17) ( r + 1)(C.7)if r > √
2. For r ≤ √
2, we have (cid:12)(cid:12)(cid:12) ˆ S [ θ ,θ ] r,c ( √ (cid:12)(cid:12)(cid:12) ≤ π (cid:16) ( r + c + √ (cid:17) ≤ π · (2 √ c ) ≤ π (cid:16) √ c (cid:17) ( r + 1) . (C.8)Hence in any case, we have (cid:12)(cid:12)(cid:12) ˆ S [ θ ,θ ] r,c ( √ (cid:12)(cid:12)(cid:12) ≤ π (cid:16) √ c (cid:17) ( r + 1) . (C.9)Substituting this to (C.6), we obtain n B | unit square of Z with B ∩ S [ θ ,θ ] r,c ∩ Z = ∅ o ≤ π (cid:16) √ c (cid:17) ( r + 1) . (C.10)On the other hand, we have n S [ θ ,θ ] r,c ∩ Z o = X z ∈ S [ θ ,θ r,c ∩ Z X z ∈ S [ θ ,θ r,c ∩ Z X B :unit square of Z I z ∈ B = X B :unit square of Z X z ∈ S [ θ ,θ r,c ∩ Z I z ∈ B ≤ X B :unit square of Z B ∩ S [ θ ,θ r,c ∩ Z = ∅
1= n B | unit square of Z with B ∩ S [ θ ,θ ] r,c ∩ Z = ∅ o ≤ π (cid:16) √ c (cid:17) ( r + 1) . (C.11) (cid:3) For an F -function F ∈ F a , define a function G F on t ≥ G F ( t ) := sup x ∈ Z X y ∈ Z , d( x,y ) ≥ t F (d( x, y )) , t ≥ . (C.12)Note that by uniform integrability the supremum is finite for all t . In particular, for any 0 < θ < G F ( t ) ≤ ∞ X r =[ t ] X y ∈ Z : r ≤ d(0 ,y ) 1, 0 < θ, ϕ < 1, and k ∈ N ∪ { } , we have ∞ X n =0 (1 + n ) k ( G F ( n )) α ≤ (64 · κ θ, ,F ) α ∞ X n =0 (1 + n ) k · F ( n ) α (1 − θ ) = (64 · κ θ, ,F ) α κ α (1 − θ ) ,k,F < ∞ , ∞ X n =[ r ] (1 + n ) k ( G F ( n )) α ≤ (64 · κ θ, ,F ) α ∞ X n =[ r ] (1 + n ) k · (cid:16) F ( n ) α (1 − θ ) (cid:17) (1 − ϕ ) (cid:16) F ( n ) α (1 − θ ) (cid:17) ϕ = (64 · κ θ, ,F ) α κ α (1 − θ )(1 − ϕ ) ,k,F F (cid:16)h r i(cid:17) α (1 − θ ) ϕ . (C.14)For any 0 < c ≤ 1, we have ∞ X r =0 F ( cr )( r + 2) = ∞ X l =0 X r ∈ Z ≥ l ≤ cr 1) + 1 (cid:19) ≤ ∞ X l =0 F ( l ) (cid:18) l + 1 c + 2 (cid:19) ≤ c ∞ X l =0 F ( l ) ( l + 3) ≤ κ , ,F c < ∞ . (C.15)We also have for m ∈ Z ≥ and 0 < c ≤ ∞ X r =0 X r ∈ Z ≥ : √ r + r c ≥ ( m +1) ( r + 1) F (cid:18)q r + r c − ( m + 1) (cid:19) ≤ ∞ X l =0 X r ,r ∈ Z ≥ l ≤ √ r + r c − ( m +1) Let x = s e iφ ∈ ˇ C [ ϕ ,ϕ ] and y = s e iφ ∈ ˇ C [ ϕ ,ϕ ] with s , s ≥ 0. If cos ( φ − φ ) ≥ x, y ) = q s + s − s s cos ( φ − φ ) ≥ q s + s p − cos ( φ − φ ) ≥ p − max { cos ( ϕ − ϕ ) , cos ( ϕ − ϕ ) , } q s + s . (C.18)If cos ( φ − φ ) < 0, then we haved( x, y ) = q s + s − s s cos ( φ − φ ) ≥ q s + s . (C.19)Hence for any x = s e iφ ∈ ˇ C [ ϕ ,ϕ ] and y = s e iφ ∈ ˇ C [ ϕ ,ϕ ] with s , s ≥ x, y ) ≥ p − max { cos ( ϕ − ϕ ) , cos ( ϕ − ϕ ) , } q s + s = c (0) ϕ ,ϕ ,ϕ ,ϕ q s + s . (C.20)Substituting this estimate, we obtain X x ∈ ˇ C [ ϕ ,ϕ ,y ∈ ˇ C [ ϕ ,ϕ F (d( x, y )) ≤ ∞ X r =0 ∞ X r =0 X x ∈ S [ ϕ ,ϕ r , ∩ Z y ∈ S [ ϕ ,ϕ r , ∩ Z F (d( x, y )) ≤ ∞ X r =0 ∞ X r =0 F (cid:18) c (0) ϕ ,ϕ ,ϕ ,ϕ q r + r ) (cid:19) (cid:16) S [ ϕ ,ϕ ] r , ∩ Z (cid:17) (cid:16) S [ ϕ ,ϕ ] r , ∩ Z (cid:17) ≤ (64) ∞ X r =0 ∞ X r =0 F (cid:18) c (0) ϕ ,ϕ ,ϕ ,ϕ q r + r ) (cid:19) ( r + 1)( r + 1) ≤ (64) ∞ X r =0 X r ,r ∈ Z ≥ ( r ,r ) ∈ S [0 , π r, ∩ Z F (cid:18) c (0) ϕ ,ϕ ,ϕ ,ϕ q r + r ) (cid:19) ( r + 1)( r + 1) ≤ (64) ∞ X r =0 F (cid:16) c (0) ϕ ,ϕ ,ϕ ,ϕ r (cid:17) ( r + 2) · (cid:16) S [0 , π ] r ∩ Z (cid:17) ≤ (64) ∞ X r =0 F (cid:16) c (0) ϕ ,ϕ ,ϕ ,ϕ r (cid:17) ( r + 2) ≤ (64) κ , ,F (cid:16) c (0) ϕ ,ϕ ,ϕ ,ϕ (cid:17) (C.21)We used Lemma C.3 to bound (cid:16) S [0 , π ] r, ∩ Z (cid:17) etc. At the last inequality we used (C.15) (cid:3) L ϕ := (cid:8) z ∈ R | arg z = ϕ (cid:9) , ϕ ∈ [0 , π ) . (C.22)and c (1) ζ ,ζ ,ζ := p − max { cos( ζ − ζ ) , cos( ζ − ζ ) } , ζ , ζ , ζ ∈ [0 , π ) . (C.23) Lemma C.5. Let ϕ, θ , θ ∈ R with θ < θ and < | ϕ − θ | < π for all θ ∈ [ θ , θ ] . Then wehave X x ∈ ˇ C [ θ ,θ X y ∈ L ϕ ( m ) F (d( x, y )) ≤ · · · (cid:16) c (1) ϕ,θ ,θ (cid:17) − ( πκ , ,F + F (0)) ( m + 1) , (C.24) for any m ∈ N ∪ { } Proof. For each r ∈ Z , set T ϕ,r,m := (cid:8) se iθ ∈ R | r ≤ s cos( θ − ϕ ) ≤ r + 1 , − m ≤ s sin( θ − ϕ ) ≤ m (cid:9) . (C.25)Note that s cos( θ − ϕ ) is a projection of se iθ onto L ϕ and | s sin( θ − ϕ ) | is the distance of se iθ fromthe line including L ϕ . Then we have L ϕ ( m ) ⊂ ∪ ∞ r = − m T ϕ,r,m ∩ Z , and (cid:12)(cid:12)(cid:12) ˆ T ϕ,r,m ( √ (cid:12)(cid:12)(cid:12) ≤ (2 √ m + 2 √ ≤ m + 1) . (C.26)Because the diameter of a 2-dimensional unit square is √ 2, any unit square B of Z with B ∩ T ϕ,r,m ∩ Z = ∅ satisfies B ⊂ ˆ T ϕ,r,m ( √ (cid:8) B | unit square of Z with B ∩ T ϕ,r,m ∩ Z = ∅ (cid:9) = X B : B ∩ T ϕ,r,m ∩ Z = ∅ ≤ (cid:12)(cid:12)(cid:12) ˆ T ϕ,r,m ( √ (cid:12)(cid:12)(cid:12) ≤ m + 1) . (C.27)On the other hand, we have (cid:8) T ϕ,r,m ∩ Z (cid:9) = X z ∈ T ϕ,r,m ∩ Z X z ∈ T ϕ,r,m ∩ Z X B :unit square of Z I z ∈ B = X B :unit square of Z X z ∈ T ϕ,r,m ∩ Z I z ∈ B ≤ X B :unit square of Z B ∩ T ϕ,r,m ∩ Z = ∅ 1= (cid:8) B | unit square of Z with B ∩ T ϕ,r,m ∩ Z = ∅ (cid:9) ≤ m + 1) . (C.28)If x ∈ ˇ C [ θ ,θ ] , we have x = r e iθ for some r ≥ θ ∈ [ θ , θ ]. By the assumption, we have0 < | θ − ϕ | < π hence 0 < cos( ϕ − θ ) < 1. Therefore, for any r ∈ R , we haved( x, re iϕ ) = q r + r − r r cos( θ − ϕ ) ≥ q r + r p − cos( θ − ϕ ) ≥ q r + r p − max { cos( θ − ϕ ) , cos( θ − ϕ ) } . (C.29)Therefore, for any x ∈ ˇ C [ θ ,θ ] and y ∈ T ϕ,r,m , we haved( x, y ) ≥ d( x, re iϕ ) − ( m + 1) = q r + r c (1) ϕ,θ ,θ − ( m + 1) . (C.30)47rom this and (C.26) and (C.28), for any x = r e iθ ∈ C [ θ ,θ ] , r ≥ 0, we have X y ∈ L ϕ ( m ) F (d( x, y )) ≤ ∞ X r = − m X y ∈ ( T ϕ,r,m ∩ Z ) F (d( x, y )) ≤ ∞ X r = −∞ X y ∈ ( T ϕ,r,m ∩ Z ) F (d( x, y )) ≤ X r ∈ Z : √ r + r c (1) ϕ,θ ,θ ≥ ( m +1) X y ∈ ( T ϕ,r,m ∩ Z ) F (cid:18)q r + r c (1) ϕ,θ ,θ − ( m + 1) (cid:19) + X r ∈ Z : √ r + r c (1) ϕ,θ ,θ < ( m +1) X y ∈ ( T ϕ,r,m ∩ Z ) F (0) ≤ X r ∈ Z : √ r + r c (1) ϕ,θ ,θ ≥ ( m +1) m + 1) F (cid:18)q r + r c (1) ϕ,θ ,θ − ( m + 1) (cid:19) + X r ∈ Z : √ r + r c (1) ϕ,θ ,θ < ( m +1) m + 1) F (0) ≤ X r ∈ Z ≥ : √ r + r c (1) ϕ,θ ,θ ≥ ( m +1) m + 1) F (cid:18)q r + r c (1) ϕ,θ ,θ − ( m + 1) (cid:19) + 36 ( m + 1) c (1) ϕ,θ ,θ F (0) I r ≤ m +1 c (1) ϕ,θ ,θ (C.31)48e then get X x ∈ ˇ C [ θ ,θ X y ∈ L ϕ ( m ) F (d( x, y )) ≤ ∞ X r =0 X x ∈ S [ θ ,θ r , ∩ Z X r ∈ Z ≥ : √ r + r c (1) ϕ,θ ,θ ≥ ( m +1) m + 1) F (cid:18)q r + r c (1) ϕ,θ ,θ − ( m + 1) (cid:19) +36 ( m + 1) c (1) ϕ,θ ,θ F (0) I r ≤ m +1 c (1) ϕ,θ ,θ ≤ ∞ X r =0 r + 1) X r ∈ Z ≥ : √ r + r c (1) ϕ,θ ,θ ≥ ( m +1) m + 1) F (cid:18)q r + r c (1) ϕ,θ ,θ − ( m + 1) (cid:19) +36 ( m + 1) c (1) ϕ,θ ,θ F (0) I r ≤ m +1 c (1) ϕ,θ ,θ ≤ · · (cid:18) c (1) ϕ,θ ,θ (cid:19) (cid:18) √ c (1) ϕ,θ ,θ (cid:19) π ( m + 1) κ , ,F + 64 · · ( m + 1) c (1) ϕ,θ ,θ F (0) (cid:18) m + 1 c (1) ϕ,θ ,θ + 1 (cid:19) ≤ · · · (cid:16) c (1) ϕ,θ ,θ (cid:17) − ( πκ , ,F + F (0)) ( m + 1) . (C.32)We used (C.16). (cid:3) D Quasi-local automorphisms In this section we collect some results from [NSY], and prove Theorem 5.1. Definition D.1. A norm-continuous interaction on A defined on an interval [0 , 1] is a map Φ : S Z × [0 , → A loc such that (i) for any t ∈ [0 , · , t ) : S Z → A loc is an interaction, and (ii) for any Z ∈ S Z , the map Φ( Z, · ) : [0 , → A Z is norm-continuous.To ensure that the interactions induce quasi-local automorphisms we need to impose sufficientdecay properties on the interaction strength. Definition D.2. Let F be an F -function on ( Z , d). We denote by B F ([0 , A defined on an interval [0 , 1] such that the function k Φ k F : [0 , → R defined by k Φ k F ( t ) := sup x,y ∈ Z F (d( x, y )) X Z ∈ S Z ,Z ∋ x,y k Φ( Z ; t ) k , t ∈ [0 , , (D.1)49s uniformly bounded, i.e., sup t ∈ [0 , k Φ k ( t ) < ∞ . It follows that t 7→ k Φ k F ( t ) is integrable, andwe set I F (Φ) := I , (Φ) := C F Z dt k Φ k F ( t ) , (D.2)with C F given in Definition C.1. We also set k| Φ k| F := sup x,y ∈ Z F (d( x, y )) X Z ∈ S Z ,Z ∋ x,y sup t ∈ [0 , ( k Φ( Z ; t ) k ) (D.3)and denote by ˆ B F ([0 , ∈ B F ([0 , k| Φ k| < ∞ .We will need some more notation. For Φ ∈ B F ([0 , ≤ m ∈ R , we introduce a path ofinteractions Φ m by Φ m ( X ; t ) := | X | m Φ ( X ; t ) , X ∈ S ( Z ) , t ∈ [0 , . (D.4)An interaction gives rise to local (and here, time-dependent) Hamiltonians, via H Λ , Φ ( t ) := X Z ∈ Λ Φ( Z, t ) , t ∈ [0 , , Λ ∈ S Z . (D.5)We denote by U Λ , Φ ( t ; s ), the solution of ddt U Λ , Φ ( t ; s ) = − iH Λ , Φ ( t ) U Λ , Φ ( t ; s ) , s, t ∈ [0 , 1] (D.6) U Λ , Φ ( s ; s ) = I . (D.7)We define corresponding automorphisms τ (Λ) , Φ t,s , ˆ τ (Λ) , Φ t,s on A by τ (Λ) , Φ t,s ( A ) := U Λ , Φ ( t ; s ) ∗ AU Λ , Φ ( t ; s ) , (D.8)ˆ τ (Λ) , Φ t,s ( A ) := U Λ , Φ ( t ; s ) AU Λ , Φ ( t ; s ) ∗ , (D.9)with A ∈ A . Note that ˆ τ (Λ) , Φ t,s = τ (Λ) , Φ s,t , (D.10)by the uniqueness of the solution of the differential equation. Theorem D.3 ([NSY]) . Let F be an F -function on ( Z , d) . Suppose that Φ ∈ B F ([0 , . Thenthe following holds:1. The limits τ Φ t,s ( A ) := lim Λ ր Z τ (Λ) , Φ t,s ( A ) , ˆ τ Φ t,s ( A ) := lim Λ ր Z ˆ τ (Λ) , Φ t,s ( A ) , A ∈ A , t, s ∈ [0 , 1] (D.11) exist and defines strongly continuous families of automorphisms on A such that ˆ τ Φ t,s = τ Φ s,t = τ Φ t,s − and ˆ τ Φ t,s ◦ ˆ τ Φ s,u = ˆ τ Φ t,u , τ Φ t,t = id A , t, s, u ∈ [0 , . (D.12) 2. For any X, Y ∈ S Z with X ∩ Y = ∅ the bound (cid:13)(cid:13)(cid:2) τ Φ t,s ( A ) , B (cid:3)(cid:13)(cid:13) ≤ k A k k B k C F (cid:16) e I F (Φ) − (cid:17) | X | G F ( d ( X, Y )) (D.13) holds for all A ∈ A X , B ∈ A Y , and t, s ∈ [0 , .If Λ ∈ S Z and X ∪ Y ⊂ Λ , a similar bound holds for τ (Λ) , Φ t,s . . For any X ∈ S Z we have (cid:13)(cid:13) ∆ X ( m ) (cid:0) τ Φ t,s ( A ) (cid:1)(cid:13)(cid:13) ≤ k A k C F (cid:16) e I F (Φ) − (cid:17) | X | G F ( m ) , (D.14) for A ∈ A X . Here we set ∆ X (0) := Π X and ∆ X ( m ) := Π X ( m ) − Π X ( m − for m ∈ N . Asimilar bound holds for τ (Λ) , Φ t,s . (See (C.12) for the definition of G F .)4. For any X, Λ ∈ S ( Z ) with X ⊂ Λ , and A ∈ A X we have (cid:13)(cid:13)(cid:13) τ (Λ) , Φ t,s ( A ) − τ Φ t,s ( A ) (cid:13)(cid:13)(cid:13) ≤ C F k A k e I F (Φ) I F (Φ) | X | G F (cid:0) d (cid:0) X, Z \ Λ (cid:1)(cid:1) . (D.15) 5. If β Ug (Φ( X ; t )) = Φ( X ; t ) for any X ∈ S Z , t ∈ [0 , , and g ∈ G , then we have β Ug ◦ τ Φ t,s = τ Φ t,s ◦ β Ug for any t, s ∈ [0 , and g ∈ G , Proof. Item 1 is Theorem 3.5 of [NSY], while 2 and 4 follow from Corollary 3.6 of the same paperby a straightforward bounding of D ( X, Y ) and the summation in eq. (3.80) of [NSY] respectively.Item 3 can be obtained using 2 and [NSY, Cor. 4.4].Suppose that β Ug (Φ( X ; t )) = Φ( X ; t ) for any X ∈ S Z , t ∈ [0 , g ∈ G . Then we have ddt β Ug ( U Λ , Φ ( t ; s )) = − β Ug ( iH Λ , Φ ( t )) β Ug ( U Λ , Φ ( t ; s )) = − iH Λ , Φ ( t ) β Ug ( U Λ , Φ ( t ; s )) , t ∈ [0 , β Ug ( U Λ , Φ ( s ; s )) = I . Hence β Ug ( U Λ , Φ ( t ; s )) and U Λ , Φ ( t ; s ) satisfy the same differential equa-tion and initial condition. Therefore we get β Ug ( U Λ , Φ ( t ; s )) = U Λ , Φ ( t ; s ). From this, we obtain β Ug τ (Λ) , Φ t,s = τ (Λ) , Φ t,s β Ug , and taking Λ ↑ Z , we obtain β Ug ◦ τ Φ t,s = τ Φ t,s ◦ β Ug . (cid:3) The following is slightly strengthened version of Assumption 5.15. of [NSY]. Assumption D.4. [[NSY]] We assume that the family of linear maps {K t : A loc → A} t ∈ [0 , is norm continuous and satisfy the followings: There is a family of linear maps {K ( n ) t : A Λ n →A Λ n } t ∈ [0 , for each n ≥ (i) For each n ≥ 1, the family {K ( n ) t : A Λ n → A Λ n } t ∈ [0 , satisfies (a) For each t ∈ [0 , (cid:16) K ( n ) t ( A ) (cid:17) ∗ = K ( n ) t ( A ∗ ) for all A Λ n . (b) For each A ∈ A Λ n , the function [0 , ∋ t → K ( n ) t ( A ) is norm continuous. (c) For each t ∈ [0 , 1] the map K ( n ) t : A Λ n → A Λ n is norm continuous and moreover, thiscontinuity is uniform on [0 , (ii) There is some p ≥ B > X ∈ S Z and n ≥ X ⊂ Λ n (cid:13)(cid:13)(cid:13) K ( n ) t ( A ) (cid:13)(cid:13)(cid:13) ≤ B | X | p k A k , for all A ∈ A X and t ∈ [0 , . (iii) There is some q ≥ 0, a non-negative, non-increasing function G with G ( x ) → x → ∞ ,and a constant C > X, Y ∈ S Z and n ≥ X ∪ Y ⊂ Λ n , (cid:13)(cid:13)(cid:13)h K ( n ) t ( A ) , B i(cid:13)(cid:13)(cid:13) ≤ C | X | q k A k k B k G (d( X, Y )) , for all A ∈ A X , B ∈ A Y and t ∈ [0 , . iv) There is some r ≥ 0, a non-negative, non-increasing function H with H ( x ) → x → ∞ ,and a constant D > X ∈ S Z there exists N ≥ n ≥ N (cid:13)(cid:13)(cid:13) K ( n ) t ( A ) − K t ( A ) (cid:13)(cid:13)(cid:13) ≤ D | X | r k A k H (cid:0) d( X, Z \ Λ n ) (cid:1) for all A ∈ A X and t ∈ [0 , Theorem D.5. Let F ∈ F a , with ˜ F θ in (C.2) for each < θ < . Assume that {K t } t ∈ [0 , isa family of linear maps satisfying Assumption D.4, with G = G F in (iii). (Recall Definition C.2and (C.12)). Let Φ ∈ B F ([0 , be an interaction satisfying Φ m ∈ B F ([0 , for m = max { p, q, r } where p, q, r are numbers in Assumption D.4. Then, the right hand side of the following sum Ψ ( Z, t ) := X m ≥ X X ⊂ Z, X ( m )= Z ∆ X ( m ) ( K t (Φ ( X ; t ))) , Z ∈ S Z , t ∈ [0 , 1] (D.17) defines a path of interaction such that Ψ ∈ B ˜ F θ ([0 , , for any < θ < . Furthermore, the formula Ψ ( n ) ( Z, t ) := X m ≥ X X ⊂ Z,X ( m ) ∩ Λ n = Z ∆ X ( m ) (cid:16) K ( n ) t (Φ ( X ; t )) (cid:17) (D.18) defines Ψ ( n ) ∈ B ˜ F θ ([0 , , for any < θ < such that Ψ ( n ) ( Z, t ) = 0 unless Z ⊂ Λ n , and satisfies K ( n ) t ( H Λ n , Φ ( t )) = H Λ n , Ψ ( n ) ( t ) . (D.19) For any t, u ∈ [0 , , we have lim n →∞ (cid:13)(cid:13)(cid:13) τ Ψ ( n ) t,u ( A ) − τ Ψ t,u ( A ) (cid:13)(cid:13)(cid:13) = 0 , A ∈ A . (D.20) Furthermore, if Φ m + k ∈ ˆ B F ([0 , for k ∈ N ∪ { } , then we have Ψ ( n ) k , Ψ k ∈ ˆ B ˜ F θ ([0 , for ant < θ < . Proof. Because of F ∈ F a , we see from (C.14) that for any 0 < α < k ∈ N , G αF has a finite k -moment. We also recall (C.2) and (C.14) to see thatmax F (cid:16) r (cid:17) , ∞ X n =[ r ] (1 + n ) G F ( n ) α ≤ ˜ C ˜ F α (1 − θ ′ ) ϕ ( r ) , r ≥ , (D.21)for any 0 < α, θ ′ , ϕ < 1. As this holds for any 0 < α, θ ′ , ϕ < 1, the condition in (ii) of Theorem 5.17[NSY] holds for any ˜ F θ . Therefore, from (ii) of Theorem 5.17 [NSY], we get Ψ , Ψ ( n ) ∈ B ˜ F θ ([0 , ( n ) converges locally in F -norm to Ψ with respect to ˜ F θ , for any 0 < θ < n Z (cid:13)(cid:13)(cid:13) Ψ ( n ) (cid:13)(cid:13)(cid:13) ˜ F θ ( t ) dt < ∞ , (D.22)see also [NSY, eq. (5.101)]. Therefore, from Theorem 3.8 of [NSY], we obtain (D.20).52y the proof of Theorem 5.17 and Theorem 5.13 (5.87) of [NSY], if Φ k + m ∈ ˆ B F ([0 , k ∈ N , then we have Ψ ( n )( s ) k , Ψ ( s ) k ∈ ˆ B ˜ F ([0 , X Z ∈ S Z Z ∋ x,y | Z | k sup t ∈ [0 , k Ψ( Z ; t ) k≤ B X Z ∈ S Z Z ∋ x,y | Z | k + p sup t ∈ [0 , k Φ( Z ; t ) k + 4 C ∞ X n =0 G F ( n ) X X : X ( n +1) ∋ x,y | X | q | X ( n + 1) | k sup t ∈ [0 , k Φ( X ; t ) k≤ B k| Φ k + p |k F F (d( x, y )) + 4 C ∞ X n =0 G F ( n )(2 n + 3) k X X : X ( n +1) ∋ x,y | X | q + k sup t ∈ [0 , k Φ( X ; t ) k≤ B k| Φ k + p |k F F (d( x, y )) + ˜ C θ ˜ F θ (d( x, y )) k| Φ q + k |k F < ∞ (D.23)with some constant ˜ C θ , for each 0 < θ < 1. In the last line we used (C.14) and Lemma 8.9 of[NSY]. Hence we get Ψ ( n ) k , Ψ k ∈ ˆ B ˜ F θ ([0 , (cid:3) Proof of Theorem 5.1. Suppose Φ ∼ Φ via a path Φ. Our definition of Φ ∼ Φ means theexistence of a path of interactions satisfying Assumption 1.2 of [MO]. Therefore, Theorem 1.3 of[MO] guarantees the existence of a path of quasi-local automorphism α t satisfying ω Φ = ω Φ ◦ α .From the proof in [MO], the automorphism α t is given by a family of interactionsΨ ( Z, t ) := X m ≥ X X ⊂ Z, X ( m )= Z ∆ X ( m ) (cid:16) K t (cid:16) ˙Φ ( X ; t ) (cid:17)(cid:17) , Z ∈ S Z , t ∈ [0 , , (D.24)with K t ( A ) := − Z duW γ ( u ) τ Φ( t ) u ( A ) , (D.25)as α t = τ Ψ t, . (Note that by partial integral of (1.19) of [MO], we obtain (6.103) of [NSY] withfunction W γ in (6.35) of [NSY]).) The interaction Ψ actually belongs to ˆ B F ([0 , F ∈ F a . To see this, note that the path Φ in Definition 1.2 satisfy Assumption 6.12 of [NSY] forany F -function because X X ∈ S Z X ∋ x,y (cid:16) k Φ ( X ; s ) k + | X | (cid:13)(cid:13)(cid:13) ˙Φ ( X ; s ) (cid:13)(cid:13)(cid:13)(cid:17) ≤ (2 R +1) C Φ b F ( R ) F (d( x, y )) , (D.26)with C Φ b , R , given in 3, 4 of Definition 1.2. In particular, it satisfies Assumption 6.12 of [NSY],with respect to the F -function (see section 8 of [NSY]) F ( r ) := e − r (1+ r ) . By section 8 of [NSY], F belongs to F a . Fix any 0 < α < 1. Then by Proposition 6.13 and its proof of [NSY], the familyof maps given by (D.25) ((6.102) of [NSY]) satisfies Assumption D.4, with p = 0, q = 1, r = 1and G = G F , where F ( r ) = (1 + r ) − exp ( − r α ). Furthermore, we have ˙Φ m ∈ ˆ B F ([0 , m ∈ N , because (cid:13)(cid:13)(cid:13)(cid:12)(cid:12)(cid:12) ˙Φ m (cid:13)(cid:13)(cid:13)(cid:12)(cid:12)(cid:12) F := sup x,y ∈ Z F (d( x, y )) X Z ∈ S Z ,Z ∋ x,y sup t ∈ [0 , | Z | m (cid:16)(cid:13)(cid:13)(cid:13) ˙Φ( Z ; t ) (cid:13)(cid:13)(cid:13)(cid:17) ≤ (2 R +1) (2 R + 1) m C Φ b F ( R ) < ∞ . (D.27)53e have F ∈ F a , and fixing any 0 < α ′ < α , ˜ F ( r ) := (1 + r ) − exp (cid:16) − r α ′ (cid:17) satisfymax (cid:26) F (cid:16) r (cid:17) , (cid:16) F (cid:16)h r i(cid:17)(cid:17) θ (cid:27) ≤ C ,θ,α ′ ˜ F ( r ) , r ≥ , (D.28)for a suitable constant C ,θ,α ′ .Therefore, by Theorem D.5, Ψ given by (D.24) for this K t and ˙Φ satisfy Ψ , Ψ ∈ ˆ B ˜ F ([0 , F ∈ F a above.If Φ is β g -invariant, then τ Φ( t ) commutes with β g , hence K t commutes with β g . As Π X com-mutes with β g and ˙Φ is β g -invariant, we see that Ψ is β g -invariant. (cid:3) Proposition D.6. Let F, ˜ F ∈ F a be F -functions of the form F ( r ) = (1 + r ) − exp (cid:0) − r θ (cid:1) , ˜ F ( r ) :=(1 + r ) − exp (cid:16) − r θ ′ (cid:17) with some constants < θ ′ < θ < . Let Ψ , ˜Ψ ∈ B F ([0 , be a path ofinteractions such that Ψ ∈ B F ([0 , . Finally, let τ ˜Ψ t,s and τ (Λ n ) , ˜Ψ t,s be automorphisms given by Ψ , ˜Ψ from Theorem D.3.Then, with s ∈ [0 , , the right hand side of the following sum Ξ ( s ) ( Z, t ) := X m ≥ X X ⊂ Z, X ( m )= Z ∆ X ( m ) (cid:16) τ ˜Ψ t,s (Ψ ( X ; t )) (cid:17) , Z ∈ S Z , t ∈ [0 , 1] (D.29) defines a path of interaction such that Ξ ( s ) ∈ B ˜ F ([0 , . Furthermore, the formula Ξ ( n )( s ) ( Z, t ) := X m ≥ X X ⊂ Z,X ( m ) ∩ Λ n = Z ∆ X ( m ) (cid:16) τ (Λ n ) , ˜Ψ t,s (Ψ ( X ; t )) (cid:17) (D.30) defines Ξ ( n )( s ) ∈ B ˜ F ([0 , such that Ξ ( n ) ( Z, t ) = 0 unless Z ⊂ Λ n , and satisfies τ (Λ n ) , ˜Ψ t,s ( H Λ n , Ψ ( t )) = H Λ n , Ξ ( n )( s ) ( t ) . (D.31) For any t, u ∈ [0 , , we have lim n →∞ (cid:13)(cid:13)(cid:13) τ Ξ ( n )( s ) t,u ( A ) − τ Ξ ( s ) t,u ( A ) (cid:13)(cid:13)(cid:13) = 0 , A ∈ A . (D.32) Furthermore, if Ψ ∈ ˆ B F ([0 , , then we have Ξ ( n )( s ) , Ξ ( s ) ∈ ˆ B ˜ F ([0 , . Proof. 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