Absolutely Continuous Spectrum for Quantum Trees
Nalini Anantharaman, Maxime Ingremeau, Mostafa Sabri, Brian Winn
aa r X i v : . [ m a t h . SP ] M a r ABSOLUTELY CONTINUOUS SPECTRUM FOR QUANTUM TREES
NALINI ANANTHARAMAN, MAXIME INGREMEAU, MOSTAFA SABRI, BRIAN WINN
Abstract.
We study the spectra of quantum trees of finite cone type. These are quan-tum graphs whose geometry has a certain homogeneity, and which carry a finite set ofedge lengths, coupling constants and potentials on the edges. We show the spectrumconsists of bands of purely absolutely continuous spectrum, along with a discrete setof eigenvalues. Afterwards, we study random perturbations of such trees, at the levelof edge length and coupling, and prove the stability of pure AC spectrum, along withresolvent estimates. Introduction
Our aim in this paper is to establish the existence of bands of purely absolutely con-tinuous (AC) spectrum for a large family of quantum trees. One of our motivations is toprovide a collection of examples relevant for the Quantum Ergodicity result proven in [7].For discrete trees, the problem is quite well understood when the tree is somehowhomogeneous. The adjacency matrix of the ( q + 1)-regular tree T q has pure AC spectrum[ − √ q, √ q ] as is well-known [23]. If we fix a root o ∈ T q and regard the tree as descendingfrom o , then the subtree descending from any offspring is the same (each is a q -ary tree),except for the subtree at the origin (which has ( q + 1) children). We say that T q hastwo “cone types”. It was shown in [21] that if T is a general tree with finitely manycone types, such that each vertex has a child of its own type, and all types arise in eachprogeny subtree, then the spectrum consists of bands of pure AC spectrum. This problemwas revisited in [6], where these assumptions were relaxed to allow T to be any universalcover of a finite graph of minimal degree at least 2. In this case however, besides the bandsof AC spectrum, a finite number of eigenvalues may appear.A natural question is whether AC spectrum survives if we add a potential. This ismotivated by the famous Anderson model [9] where random independent, identically-distributed potentials are attached at lattice sites. It remains a major open problem toprove such stability for the Anderson model on the euclidean lattice Z d , d ≥ L e andwe study differential operators acting on the edges with appropriate boundary conditionsat the vertices specified by certain coupling constants. The presence of AC spectrumfor quantum trees appears to have been studied less systematically than in the case ofdiscrete Schr¨odinger operators. In case of regular trees T q , it was shown in [12] that thequantum tree obtained by endowing each edge with the same length L , the same symmetricpotential W on the edges and the same coupling constant α at the vertices, has a spectrumconsisting of bands of pure AC spectrum, along with eigenvalues between the bands. The Mathematics Subject Classification.
Primary 81Q10, 34B45. Secondary 47B80.
Key words and phrases.
Absolutely continuous spectrum, quantum graphs, random operators, trees. setting was a bit generalized quite recently in [13], where each vertex in a 2 q -regular tree issurrounded by the same set of lengths ( L , . . . , L q ), each length repeated twice, similarlythe same set of symmetric potentials ( W , . . . , W q ), and the boundary conditions are takento be Kirchhoff. The nature of the spectrum is partly addressed, but the possibility thatit consists of a discrete set of points is not excluded. Finally, it was shown in [1] thatthe AC spectrum of the equilateral quantum tree [12] remains stable under weak randomperturbation of the edge lengths. The theorem however does not yield purity of the ACspectrum in some interval; one can only infer that the Lebesgue measure of the perturbedAC spectrum is close to the unperturbed one. We also mention the papers [17, 30] whichconsider radial quantum trees, for which a reduction to a half-line model can be performed.Our aim here is twofold. First, go beyond regular graphs. We are mainly interested inthe case where the tree is the universal cover of some compact quantum graph. This impliesthe set of different lengths, potentials and coupling constants is finite, but the situationcan be much more general than the special Cayley graph setting considered in [13]. Weshow in this framework that the spectrum will consist of (nontrivial) bands of pure ACspectrum, plus some discrete set of eigenvalues. Next, we consider random perturbationsof these trees. We can perturb both the edge lengths and coupling constants. This settingis more general than [1], where the tree was regular and the coupling constants werezero. But our main motivation here is especially to derive the purity of the perturbed ACspectrum, along with a strong control on the resolvent, which is an important ingredientto prove quantum ergodicity for large quantum graphs. We do this in a companion paper[7].1.1. Some definitions.
Quantum graphs.
Let G = ( V, E ) be a graph with vertex set V and edge set E . Wewill assume that there are no self-loops and that there is at most one edge between anytwo vertices, so that we can see E as a subset of V × V . For each vertex v ∈ V , we denoteby d ( v ) the degree of v . We let B = B ( G ) be the set of oriented edges (or bonds), so that | B | = 2 | E | . If b ∈ B , we shall denote by ˆ b the reverse bond. We write o b for the origin of b and t b for the terminus of b . We define the map e : B −→ E by e (( v, v ′ )) = { v, v ′ } . An orientation of G is a map or : E −→ B such that e ◦ or = Id E .A length graph ( V, E, L ) is a connected combinatorial graph (
V, E ) endowed with a map L : E → (0 , ∞ ). If b ∈ B , we denote L b := L ( e ( b )).A quantum graph Q = ( V, E, L, W, α ) is the data of: • A length graph (
V, E, L ), • A potential W = ( W b ) b ∈ B ∈ L b ∈ B C ([0 , L b ]; R ) satisfying for x ∈ [0 , L b ],(1.1) W b b ( L b − x ) = W b ( x ) . • Coupling constants α = ( α v ) v ∈ V ∈ R V .The underlying metric graph is the quotient G := { x = ( b, x b ); b ∈ B, x b ∈ [0 , L b ] } / ∼ , where ( b, x b ) ∼ ( b ′ , x ′ b ′ ) if b ′ = ˆ b and x ′ b ′ = L b − x b .A function on the graph will be a map f : G −→ R . It can also be identified with acollection of maps ( f b ) b ∈ B such that f b ( L b − · ) = f ˆ b ( · ). We say that f is supported on e for some e ∈ E if f b ≡ e ( b ) = e .If each f b is positive and measurable, we define R G f ( x )d x := P b ∈ B R L b f b ( x b )d x b . Wemay then define the spaces L p ( G ) for p ∈ [1 , ∞ ] in the usual way.Condition (1.1) simply requires W to be well-defined on G (no symmetry assumption). BSOLUTELY CONTINUOUS SPECTRUM FOR QUANTUM TREES 3
When G = ( V, E ) is a tree, i.e., contains no cycles (which will be the case in most ofthe paper), we say that Q is a quantum tree , and we denote it by the letter T rather than Q , while the set G is called a metric tree and is denoted by T .1.1.2. Orienting quantum trees.
Let T be a combinatorial tree, that is, a graph containingno cycles. We denote its vertex set by V ( T ) or just V , its edge set by E ( T ), and its setof oriented edges by B ( T ). In all the paper, we will often write v ∈ T instead of v ∈ V ( T )to lighten the notations.In this paragraph, we explain how we can present the tree T in a coherent view , that isto say, fix an oriented edge b o ∈ B ( T ), and give an orientation to all the other edges of T ,by asking that they “point in the same direction as b o ”.More precisely, let us fix once and for all an oriented edge b o ∈ B ( T ), corresponding toan edge e o ∈ E ( T ). If we remove the edge e o from T , we obtain two connected compo-nents which are still combinatorial trees. We will write T + b o for the connected componentcontaining t b o , and T − b o for the component containing o b o .Let v ∈ T + b o be at a distance n from t b o . Amongst the neighbours of v , one of them is atdistance n − t b o : we denote it by v − , and say that v − is the parent of v . The otherneighbours of v are at a distance n + 1 from t b o , and are called the children of v . The setof children of v is denoted by N + v . On the contrary, if v ∈ T − b o is at distance n from o b o , itsunique neighbour at a distance n − o b o is called the child of v , and denoted by v + ,and its other neighbours are its parents, whose set we denote by N − v . These definitionsare natural if we see the tree at the left of Figure 1 as a genealogical tree.Let V ∗ = (cid:0) V ( T ) \ { o b o , t b o } (cid:1) ∪ { o } . We define a map b : V ∗ −→ B ( T ) as follows: we set b ( o ) = b o , and, if v ∈ T + b o , then b ( v ) = ( v − , v ), while if v ∈ T − b o , then b ( v ) = ( v, v + ). Oneeasily sees that e ◦ b : V ∗ −→ E ( T ) is a bijection, so that b ◦ ( e ◦ b ) − is an orientation of T . The map b serves to index all oriented edges: those in T + b o by their terminus, those in T − b o by their origin, and b o by its “midpoint” o . The latter makes sense once we turn T into a quantum tree T . We denote L v := L b ( v ) and W v := W b ( v ) . The metric tree T canbe identified with the set T ≡ G v ∈ V ∗ [0 , L v ] = { x = ( v, x v ) | v ∈ V ∗ , x v ∈ [0 , L v ] } . A function on T will then be the data of ψ = ( ψ v ) v ∈ V ∗ , where each ψ v is a function ofthe variable x v ∈ [0 , L v ].On a quantum tree, we consider the Schr¨odinger operator(1.2) ( H T ψ v )( x v ) = − ψ ′′ v ( x v ) + W v ( x v ) ψ v ( x v )with domain D ( H T ), the set of functions ( ψ v ) ∈ L v ∈ V ∗ W , (0 , L v ) satisfying the so-called δ -conditions. Namely, for all v ∈ T + b o ,(1.3) ψ v ( L v ) = ψ v + (0) ∀ v + ∈ N + v and X v + ∈N + v ψ ′ v + (0) = ψ ′ v ( L v ) + α v ψ v ( L v ) , while for all v ∈ T − b o ,(1.4) ψ v − ( L v − ) = ψ v (0) ∀ v − ∈ N − v and X v − ∈N − v ψ ′ v − ( L v − ) + α v ψ v (0) = ψ ′ v (0) . Finally, ψ o ( L o ) = ψ o + (0) ∀ o + ∈ N + o , P o + ψ ′ o + (0) = ψ ′ o ( L o ) + α t bo ψ o ( L o ), and ψ o − ( L o − ) = ψ o (0), P o − ψ ′ o − ( L o − ) + α o bo ψ o (0) = ψ ′ o (0). In a common convention we will refer to the α v = 0 case as the Kirchhoff-Neumann condition . NALINI ANANTHARAMAN, MAXIME INGREMEAU, MOSTAFA SABRI, BRIAN WINN ot b o o b o FlowCoherent view ot b o o b o Twisted viewFlow Flow
Figure 1.
The two views of a tree.
Remark 1.1.
The above conventions mean that we see T as a doubly infinite genealogicaltree. This is what we called the coherent view; it can also be pictured by saying that weimagine an electric flow moving from T − b o to T + b o .There is another way of orienting the graph which we call the twisted view . This isdone by turning b o into a V-shape and viewing V ( T ) as offspring of o . See Figure 1 for anillustration; here one should think that o is a source from which the electric flows movesoutwards. When necessary to highlight this genealogical structure, we will write T + o forthe set of offsprings of o . Each vertex v has a single parent v − and several children; allthe edges take the form { v, v − } for a unique v .The link between the two views is immediate: functions on T + b o in both views coincide,while on T − b o , one replaces b by ˆ b and derivatives take a sign. Here ˆ b = ( t b , o b ) is the edgereversal of b . Hence, in the twisted view , all functions in the domain of H satisfy (1.3).Given v ∈ V ∗ , z ∈ C , let C z ( x ) and S z ( x ) be a basis of solutions of the problem − ψ ′′ v + W v ψ v = zψ v satisfying(1.5) (cid:18) C z (0) S z (0) C ′ z (0) S ′ z (0) (cid:19) = (cid:18) (cid:19) . Then any solution ψ v of the problem satisfies(1.6) (cid:18) ψ v ( L v , z ) ψ ′ v ( L v , z ) (cid:19) = M z ( v ) (cid:18) ψ v (0 , z ) ψ ′ v (0 , z ) (cid:19) where M z ( v ) = (cid:18) C z ( L v ) S z ( L v ) C ′ z ( L v ) S ′ z ( L v ) (cid:19) . If W v ≡ S z ( x ) = sin √ zx √ z , C z ( x ) = cos √ zx ;if W v ( x ) = c + c cos(2 πx/L v ) then S z , C z would be Mathieu functions. It is a standardfact that S z ( x ) , C z ( x ) are analytic functions of z ∈ C (see for instance [28, Chapter 1]). These views of the tree have the advantage of avoiding the assumption that T has a special “root”vertex of degree one [1, 30]. Such assumption simplifies the orientation a bit, but is not satisfied in manynatural situations. As will be clear later, we will only need to study functions supported in T ± b o , which iswhy we did not specify what happens on b o . But one could specify that ψ o from the coherent view becomes( ψ (1) o , ψ (2) o ) in the twisted view, with ψ (1) o ( x ) = ψ o ( x + L o ) and ψ (2) o ( x ) = ψ o ( L o − x ), for x ∈ [0 , L o ]. BSOLUTELY CONTINUOUS SPECTRUM FOR QUANTUM TREES 5
Trees of finite cone type.
We define a cone in T to be a subtree of the form T + b or T − b , for some b ∈ B ( T ). Each cone T + b has an origin t b , and each cone T − b has an end o b .We say that two quantum cones T + b and T + b ′ are isomorphic if there is an isomorphismof combinatorial graphs ϕ : T + b → T + b ′ such that L ϕ ( v ) = L v , W ϕ ( v ) = W v and α ϕ ( v ) = α v for all v ∈ T + b . Isomorphic T − b and T − b ′ are defined the same way.We say that T is a tree of finite cone type if there exists b o ∈ B ( T ) such that:(i) There are finitely many non-isomorphic quantum cones T +( v − ,v ) as v ∈ T + b o .(ii) There are finitely many non-isomorphic quantum cones T − ( w,w + ) as w ∈ T − b o .Here ( t b o ) − = o b o and ( o b o ) + = t b o . Note that in a regular tree, all cones T ± b are isomorphic,but a necessary condition for it to be a quantum tree of finite cone type, is that its edgesand vertices be endowed with finitely many lengths, potentials and coupling constants. If T is a tree of finite cone type, with b o ∈ B ( T ) fixed, we may introduce a type function ℓ : T + b o → N = N ∪ { } , taking values in a finite set, such that ℓ ( v ) = ℓ ( w )iff T +( v − ,v ) ≡ T +( w − ,w ) as quantum trees. Similarly, ℓ : T − b o → N satisfies ℓ ( v ) = ℓ ( w ) iff T − ( v,v + ) ≡ T − ( w,w + ) . Note that if ℓ ( v ) = ℓ ( w ), then W v = W w , L v = L w and α v = α w , sincethe corresponding isomorphism respects this information. Hence, any coherent quantumtree T of finite cone type comes with the following structure:(a) A fixed b o ∈ B ( T ).(b) Two finite sets of labels A + = { i , . . . , i m } , A − = { j , . . . , j n } and two matrices M = ( M i,j ) i,j ∈ A + , N = ( N i,j ) i,j ∈ A − . If v ∈ T + b o has type j , it has M j,k children of type k . If w ∈ T − b o has type j , it has N j,k parents of type k .(c) Finite sets { L i } i ∈ A ± , { W i } i ∈ A ± and { α i } i ∈ A ± encoding the lengths, potentials andcoupling constants, respectively. More precisely, b o is endowed a special length L o andpotential W o . If ( v − , v ) ∈ T + b o with ℓ ( v ) = i , then L v = L i , W v = W i and α v = α i .The same attribution is made if ( v, v + ) ∈ T − b o with ℓ ( v ) = i .If we take the twisted view instead, we only need one alphabet A = A + ∪ A − and onecorresponding matrix M = ( M i,j ) i,j ∈ A .A trivial example is the equilateral, ( q + 1)-regular quantum tree, with identical poten-tials W on each edge and identical coupling constant α on each vertex [12]. In this case,all vertices in T ± b o have the same type, and we get two 1 × M = N = (cid:0) q (cid:1) .An important class of examples comes from universal covers of finite undirected graphs.More precisely, if G is a finite undirected graph and T is its universal cover, then T is acombinatorial tree satisfying condition (i). If we endow G with a quantum structure G and lift it to T in the natural way, then the corresponding T will be a quantum tree offinite cone type.Quantum trees of finite cone type satisfying (a)–(c) will be our basic, “unperturbed”trees. We denote the Schr¨odinger operator (1.2) acting in this setting as H . Later on, weshall study random perturbations of these trees, and denote the corresponding operatorby H ωǫ , where ǫ is the strength of the disorder.We make the following assumption on T : (C1*) For any k, l ∈ A + , there is n = n ( k, l ) such that ( M n ) k,l ≥
1. Similarly, for i, j ∈ A − , there is n = n ( i, j ) with ( N n ) i,j ≥ Also note that it is not required that there are finitely many non-isomorphic quantum trees T − ( v − ,v ) as v ∈ T + b o . To illustrate this point, consider the binary tree (so each vertex has 3 neighbors except forthe special root ⋆ with 2 neighbors), let b o = ( ⋆, v ), with v either neighbor. Then all cones T + b ⊂ T + b o lookthe same; they are binary trees. However, the backward cones T − b are distinct in each generation, becausethey “see” the special root at distinct distances. Despite this, T has finite cone type. NALINI ANANTHARAMAN, MAXIME INGREMEAU, MOSTAFA SABRI, BRIAN WINN
See Remark 1.3 below for a discussion of this condition. We may now state a firsttheorem, which describes the structure of the spectrum of H = H T on a tree T of finitecone type. We denote by G z ( x, y ) = ( H − z ) − ( x, y ) the Green’s function of H . Theorem 1.2.
Let
M, N satisfy (C1*) . Then the spectrum of H consists of a disjointunion of closed intervals and of isolated points: σ ( H ) = ( F r I r ) ∪ P , where the I r areclosed intervals, and P is a discrete set. The spectrum is purely absolutely continuous inthe interior of each band ˚ I r . For λ ∈ ˚ I r , and for any v ∈ T , the limit G λ +i00 ( v, v ) existsand satisfies Im G λ +i00 ( v, v ) > , where G z is the Green’s function of H . Let R ± z, be the Weyl-Titchmarsh functions of H as defined in [1], see (2.2). Let R ± λ, = R ± λ +i0 , when the limit exists. Theorem 1.2 implies that Im R + λ, ( v )+Im R − λ, ( v ) > I r . We will need the stronger property that Im R + λ, ( v ) > v . For this, weintroduce the following strengthening of (C1*) . (C1) The quantum tree T is the universal cover of a finite quantum graph G of minimaldegree ≥ Remark 1.3.
Condition (C1*) means that on T + b o , any cone type l ∈ A + appears asoffspring of any k ∈ A + after a finite number of generations, and similarly for T − b o . It isnot required that cone types in A − appear in T + b o - we only need the matrices M and N tobe separately irreducible. We can also allow for “rooted” trees where the root o has degreeone. In this case the situation is a bit simpler actually; we only have to deal with onematrix M . Condition (C1*) applies in particular to trees with a “radial periodic” data,i.e. data that are periodic functions of the distance to the origin (such as some examplesappearing in [30]).Assumption (C1) implies (C1*) (see Remark 3.7), and is in fact more restrictive. Inparticular, T is “unimodular”, that is, all data is somehow homogeneous as we move alongthe tree. This excludes for example the binary tree and more generally radial periodictrees, where the root plays a special role. However, such unimodular trees are still verygeneral, they are actually the most interesting for us, and many techniques (such as areduction to a half-line model) fail to tackle them. Even in the very simple case where thebase graph G is regular but the edge lengths are not equal, the lifted structure in generalwill be neither radial periodic, nor identical around each vertex (in contrast to [13]).Note that the case where G is a cycle is already known when the couplings are zero.In this case H T is just a periodic Schr¨odinger operator on R (of period ≤ | G | ), it iswell-known that the spectrum is purely AC in this case [29, Section XIII.16]. Theorem 1.4. If T satisfies (C1) , then the spectrum of H consists of a disjoint unionof closed intervals and of isolated points: σ ( H ) = ( F r I r ) ∪ P , where the I r are closedintervals, and P is a discrete set. The spectrum is purely absolutely continuous in theinterior of each band ˚ I r . For λ ∈ ˚ I r , the limit R + λ +i0 , ( v ) exists for any v ∈ V and satisfies Im R + λ, ( v ) > . Remark 1.5.
In Theorems 1.2 and 1.4, it is not excluded in principle that S r I r = ∅ ,i.e. the spectrum consists of isolated points. We think this never happens for infinitequantum trees of finite cone type with the δ -conditions we consider, i.e. we believe theseshould always have some continuous spectrum. We did not find such a result in theliterature however. This is why we dedicate Section 4 to prove the following: if T satisfieseither:(1) assumption (C1) and has a single data ( L, α, W ) (all edges carry the same length,coupling and symmetric potential),(2) or has a general data ( L e , α o e , W e ) e ∈ E ( G ) , but the finite graph G is moreover Hamil-tonian , BSOLUTELY CONTINUOUS SPECTRUM FOR QUANTUM TREES 7 then H T always has some continuous spectrum, i.e. S r ˚ I r = ∅ . Recall that a finite graphis Hamiltonian if it has a cycle that visits each vertex exactly once. Note that as a discretetree, T may cover many different graphs. We only need one of these finite graphs to beHamiltonian. For example, we can consider any regular tree, despite the fact that someregular graphs (like the Petersen graph) are not Hamiltonian.In particular, the Cayley tree considered in [13] can be realized as the universal coverof the complete bipartite graph K q, q , which is Hamiltonian. For this, use the fact that K q, q has a proper 2 q -edge-colouring and put the same length/potential on edges of thesame colour. The lift of this is then a tree which has the same data around each vertex,and we may take L q + j = L j , W q + j = W j to be in the setting [13]. Then our theoremsimply this tree has nontrivial bands of pure AC spectrum, thus enriching the results of[13]. Again, this is just one very special application of our framework.1.3. Random perturbations of trees of finite cone type.
Fix a quantum tree T satisfying (C1) . As explained in Remark 3.7, any such tree is a tree of finite cone type.We fix an edge e ∈ E ( T ), and see our quantum tree in the twisted view (in which all verticesare descendent of a vertex o ), so as to deal with a single alphabet A and a correspondingmatrix M . We denote the lengths and coupling constants of the unperturbed tree T by( L v ) v ∈ V ∗ and ( α v ) v ∈ V . These can also be denoted ( L i ) i ∈ A ∪{ o } and ( α i ) i ∈ A . We assumethere are no potentials on the edges and the couplings are nonnegative: W v ≡ α v ≥ . We now want to analyze random perturbations of T . For this purpose, we introducea probability space (Ω , F , P ), a family of random variables ω ∈ Ω ( L ωv ) v ∈ V ∗ repre-senting random lengths, and a family of random variables ω ∈ Ω ( α ωv ) v ∈ V represent-ing random coupling constants. In principle we could also consider random potentials ω ∈ Ω ( W ωv ) v ∈ V ∗ , however here we assume there are no potentials on the edges evenafter perturbation. We also assume the perturbed couplings are nonnegative: W ωv ≡ α ωv ≥ . We make the following assumptions on the random perturbation (see Remark 1.1 for thenotation T + o ): (P0) The operator H ωǫ is the Laplacian on the edges acting on L W , (0 , L ωv ), satisfying δ -conditions with coupling constants ( α ωv ) v ∈ T , which are assumed to satisfy L ωv ∈ (cid:2) L v − ǫ, L v + ǫ (cid:3) and α ωv ∈ (cid:2) α v − ǫ, α v + ǫ (cid:3) . (P1) For all v, w ∈ T + o , the random variables ( α ωv , L ωv ) and ( α ωw , L ωw ) are independent ifthe forward trees of v and w do not intersect, i.e. if T +( v − ,v ) ∩ T +( w − ,w ) = ∅ . (P2) For all v, w ∈ T + o that share the same label, the restrictions of the random variables( α ω , L ω ) to the isomorphic forward trees of v and w are identically distributed. Remark 1.6.
Assumptions (P1) and (P2) hold, in particular, for independent identicallydistributed random variables (which is the main case we have in mind).We shall consider intervals I lying in the interior of the unperturbed AC spectrum:(1.7) Σ = S r ˚ I r , where I r are given in Theorem 1.4.We will also need to ensure that the various sin (cid:0) √ λL v (cid:1) do not vanish. More precisely,by (P0) , the perturbed lengths all lie in S j ∈ A ∪{ } [ L j, min ( ǫ ) , L j, max ( ǫ )], where L j, min ( ǫ ) = L j − ǫ and L j, max ( ǫ ) = L j + ǫ . We then assume(1.8) I ∩ D = ∅ , NALINI ANANTHARAMAN, MAXIME INGREMEAU, MOSTAFA SABRI, BRIAN WINN where the set D = D ǫ is a “thickening” of the Dirichlet spectrum, given by D = [ j ∈ A ∪{ o } [ n ≥ " π n L j, max ( ǫ ) , π n L j, min ( ǫ ) . This ensures that sin (cid:0) √ λL ωv (cid:1) , sin (cid:0) √ λL v (cid:1) = 0 for any λ ∈ I , v ∈ T and ω .Recall that the Weyl-Titchmarsh functions R + z ( v ) will be introduced in (2.2). Introducethe following condition: (Green -s) There is a non-empty open set I and some s > b ∈ T ,sup λ ∈ I ,η ∈ (0 , E (cid:18)(cid:12)(cid:12)(cid:12) Im R + λ +i η ( o b ) (cid:12)(cid:12)(cid:12) − s (cid:19) < ∞ . Condition (Green -s) implies in particular that the spectrum in I is purely AC, as longas it stays away from the Dirichlet spectrum, see Appendix A.2. Here (Green -s) refersto “Green’s function” and the moment value s . In fact, such inverse bounds on the WTfunction imply moments bounds on the Green’s function; see Corollary 2.5.Introduce the following assumptions: (C0) The minimal degree of T is at least 3. (C2) For each k ∈ A , there is k ′ with M k,k ′ ≥ l ∈ A : M k,l ≥ M k ′ ,l ≥ v ∈ T has at least one child v ′ such thateach label found in N + v can also be found in N + v ′ . See [6] for examples of such trees. Theorem 1.7.
Let T satisfy (C0), (C1) , (C2) and ( α, L ) satisfy (P0) , (P1) and (P2) ,and be without edge potentials. Let I ⊂ Σ be compact with I ∩ D = ∅ . Then for any s > ,we may find ǫ ( I, s ) such that (Green -s) holds on I for any ǫ ≤ ǫ . In particular, σ ( H ωǫ ) has purely absolutely continuous spectrum almost-surely in I . The “in particular” part is due to Theorem A.6.In the above theorem, the disorder window ǫ ( I, s ) depends on the value of the moment s . We can actually obtain a disorder window valid uniformly for all s , but at the price ofassuming some regularity on the δ -potential : (P3) For any v ∈ T , ℓ ( v ) = j , j ∈ A , the distribution ν j of α ωv is H¨older continuous :there exist C ν > β ∈ (0 ,
1] such that for any bounded I ⊂ R ,max j ∈ A ν j ( I ) ≤ C ν · | I | β . This holds e.g. if the ν j are absolutely continuous with a bounded density (then β = 1). Theorem 1.8.
Suppose in addition to the assumptions of Theorem 1.7 that (P3) issatisfied. Then there exists ǫ ( I ) such that for any ǫ ≤ ǫ and any s ≥ , (Green -s) holdson I . Green’s function on quantum trees
The aim of this section is to derive quantum analogs for the well-known recursive for-mulas of the Green’s function on combinatorial trees. These identities will play a key rolein the spectral analysis of the quantum tree, and may be of independent interest. In fact,we shall also need them when studying quantum ergodicity in [7]. Some of these identitiesappeared before in [1].In all this section, we fix a quantum tree T , and denote by W , ( T ) the set of ψ = ( ψ v )such that ψ v ∈ W , (0 , L v ), P v k ψ v k W , < ∞ .If b ∈ B ( T ), recall the notation T ± b of § x = ( b, x b ) ∈ T , we define a quantum tree T + x by T + x = [ x b , t b ] ∪ T + b . More precisely, add a vertex v x at x , let V ( T + x ) = V ( T + b ) ∪{ v x } , BSOLUTELY CONTINUOUS SPECTRUM FOR QUANTUM TREES 9 E ( T + x ) = E ( T + b ) ∪ { v x , t b } , L { v x ,t b } = L b − x b , W ( v x ,t b ) = ( W b ) | [ L b − x b ,L b ] , α v x = 0, and thelengths, potentials and coupling constants be the same as those of T + b on the rest of theedges. In a similar fashion, we define T − x = T − b ∪ [ o b , x b ].Let u = ( b, u b ) ∈ T . By [1, Theorem 2.1], which remains true in our context, if wedefine H max T ± u on T ± u to be the Schr¨odinger operator − ∆ + W with domain D ( H max T ± u ),the set of ψ ∈ W , ( T ± u ) satisfying δ -conditions on inner vertices of T ± u , then for any z ∈ C + := H := { z ∈ C : Im z > } , there are unique z -eigenfunctions V + z ; u ∈ D ( H max T + u ), U − z ; u ∈ D ( H max T − u ) satisfying U − z ; u ( u ) = V + z ; u ( u ) = 1. Complex eigenvalues exist because H max T ± u is not self-adjoint, as there are no domain conditions at u . Lemma 2.1.
Let z ∈ C + . The resolvent G z of H T is an integral operator with kernel G z ( x, y ) defined as follows. Given x, y ∈ T , fix o, v such that x, y ∈ T + o ∩ T − v . Then (2.1) G z ( x, y ) = U − z ; v ( x ) V + z ; o ( y ) W zv,o ( y ) if y ∈ T + x , U − z ; v ( y ) V + z ; o ( x ) W zv,o ( y ) if y ∈ T − x , where W zv,o ( x ) is the Wronskian W zv,o ( x ) = V + z ; o ( x )( U − z ; v ) ′ ( x ) − ( V + z ; o ) ′ ( x ) U − z ; v ( x ) . Versions of this lemma previously appeared in [1, Lemma A.2] and [17, Lemma D.15].We give the proof in Appendix A for completeness.Since for each z ∈ C + , G z satisfies the δ -boundary conditions in each of its argu-ments, we deduce that, whenever o b = o b ′ = v , we have G z (( b, , · ) = G z (( b ′ , , · ) and G z ( · , ( b, G z ( · , ( b ′ , G z ( v, · ) and G z ( · , v ) respectively.As in [1], we define the Weyl-Titchmarch (WT) functions for x ∈ T by(2.2) R + z ( x ) = ( V + z ; o ) ′ ( x ) V z ; o ( x ) and R − z ( x ) = − ( U − z ; v ) ′ ( x ) U − z ; v ( x ) . Note that we take here the coherent point of view, which is why there is a negative signin the definition of R − z ( x ).Given an oriented edge b = ( o b , t b ), we define(2.3) ζ z ( b ) = G z ( o b , t b ) G z ( o b , o b ) . Remark 2.2. If a < inf σ ( H T ), then ζ z ( b ) is well-defined on C \ ( a, ∞ ), i.e. the denom-inator does not vanish, as follows from (2.1) and the proof of [1, Theorem 2.1(ii)]. Theproof also shows that z ζ z ( b ) is holomorphic on C \ ( a, ∞ ) and real-valued on ( −∞ , a ].In the case of combinatorial trees, ζ z ( b ) coincides with what was denoted ζ zo b ( t b ) in[4]. In fact, by the multiplicative property of the Green function, we have G z ( o b ,t b ) G z ( o b ,o b ) = G z ( o b ,o b ) ζ zob ( t b ) G z ( o b ,o b ) = ζ zo b ( t b ).In the case that T is the ( q + 1)-regular tree with equilateral edges, with identical cou-pling constants and potentials, then ζ z ( b ) is the quantity µ − ( z ) in [12], and is independentof b . Moreover, the limit µ − ( λ ) = lim η ↓ µ − ( λ + i η ) exists in this case, provided that λ isnot in the Dirichlet spectrum, i.e., that sin( λL ) = 0.Finally, for the quantum Cayley graphs of [13], the ζ z ( b ) coincide with the multipliers µ m ( z ). Hence, there are finitely many distinct ζ z ( b ). Moreover, in this setting, ζ z (ˆ b ) = ζ z ( b ), and ζ z ( gb ) = ζ z ( b ), where g is an element of the group acting on the graph.Given an oriented edge b , we will denote by N + b the set of outgoing edges from b , i.e.the set of b ′ with o b ′ = t b and b ′ = ˆ b . Lemma 2.3.
Let z ∈ C + . We have the following relations between ζ z and the WTfunctions R ± z : (2.4) ζ z ( b ) = C z ( L b ) + R + z ( o b ) S z ( L b ) , ζ z (ˆ b ) = S ′ z ( L b ) + R − z ( t b ) S z ( L b ) , (2.5) R + z ( t b ) = S ′ z ( L b ) S z ( L b ) − S z ( L b ) ζ z ( b ) , R − z ( o b ) = C z ( L b ) S z ( L b ) − S z ( L b ) ζ z (ˆ b ) . Moreover, (2.6) 1 ζ z ( b ) S z ( L b ) + X b + ∈N + b ζ z ( b + ) S z ( L b + ) = X b + ∈N + b C z ( L b + ) S z ( L b + ) + S ′ z ( L b ) S z ( L b ) + α t b , (2.7) 1 ζ z ( b ) − ζ z (ˆ b ) = S z ( L b ) G z ( t b , t b ) , ζ z (ˆ b ) ζ z ( b ) = G z ( o b , o b ) G z ( t b , t b ) , and (2.8) X b + ∈N + b C z ( L b + ) S z ( L b + ) + S ′ z ( L b ) S z ( L b ) + α t b = X t b ′ ∼ t b ζ z ( b ′ ) S z ( L b ′ ) + 1 G z ( t b , t b ) , where b ′ = ( t b , t b ′ ) . Given a non-backtracking path b , . . . , b k (that is to say, o b i +1 = t b i and t b i +1 = o b i for all i ∈ { , . . . , k − } ), we have the multiplicative property (2.9) G z ( o b , t b k ) = G z ( o b , o b ) ζ z ( b ) · · · ζ z ( b k ) = G z ( t b k , t b k ) ζ z (ˆ b ) · · · ζ z (ˆ b k ) . Finally, for any path b , . . . , b k , we have (2.10) G z ( o b , t b k ) = G z ( t b k , o b ) . Proof.
By (2.1), ζ z ( b ) = V + z ; o ( t b ) V + z ; o ( o b ) · W zv,o ( o b ) W zv,o ( t b ) . But the Wronskian is constant on b , as checkedby differentiating it. Moreover, since V + z ; o ( x b ) is a z -eigenfunction on b , we have V + z ; o ( t b ) = C z ( L b ) V + z ; o ( o b ) + S z ( L b )( V + z ; o ) ′ ( o b ). Hence, ζ z ( b ) = C z ( L b ) + R + z ( o b ) S z ( L b ) as claimed.Next, ζ z (ˆ b ) = G z ( t b ,o b ) G z ( t b ,t b ) = U − z ; v ( o b ) U − z ; v ( t b ) again by constancy of W zv,o on b . Writing (1.6) in theform(2.11) (cid:18) ψ ( o b , z ) ψ ′ ( o b , z ) (cid:19) = M z ( b ) − (cid:18) ψ ( t b ) ψ ′ ( t b ) (cid:19) = (cid:18) S ′ z ( L b ) − S z ( L b ) − C ′ z ( L b ) C z ( L b ) (cid:19) (cid:18) ψ ( t b ) ψ ′ ( t b ) (cid:19) , we get U − z ; v ( o b ) = S ′ z ( L b ) U − z ; v ( t b ) − S z ( L b )( U − z ; v ) ′ ( t b ), so ζ z (ˆ b ) = S ′ z ( L b ) + R − z ( t b ) S z ( L b ) asclaimed.Next, V + z ; o ( y b ) = V + z ; o ( o b ) C z ( y b ) + ( V + z ; o ) ′ ( o b ) S z ( y b ), so R + z ( t b ) = R + z ( o b ) S ′ z ( L b ) + C ′ z ( L b ) R + z ( o b ) S z ( L b ) + C z ( L b ) . Using R + z ( o b ) = ζ z ( b ) − C z ( L b ) S z ( L b ) , we get R + z ( t b ) = ζ z ( b ) S ′ z ( L b ) − C z ( L b ) S ′ z ( L b ) + C ′ z ( L b ) S z ( L b ) ζ z ( b ) S z ( L b ) . The first part of (2.5) follows by the Wronskian identity C z ( x ) S ′ z ( x ) − C ′ z ( x ) S z ( x ) = 1.For the second part, by (2.11), R − z ( o b ) = − ( U − z ; v ) ′ ( o b ) U − z ; v ( o b ) = C ′ z ( L b ) U − z ; v ( t b ) − C z ( L b )( U − z ; v ) ′ ( t b ) S ′ z ( L b ) U − z ; v ( t b ) − S z ( L b )( U − z ; v ) ′ ( t b )= C z ( L b ) R − z ( t b ) + C ′ z ( L b ) S z ( L b ) R − z ( t b ) + S ′ z ( L b ) . BSOLUTELY CONTINUOUS SPECTRUM FOR QUANTUM TREES 11
The claim now follows as before using (2.4).Since V + z ; o satisfies the δ -conditions, we have(2.12) X b + ∈N + b R + z ( o b + ) = R + z ( t b ) + α t b = S ′ z ( L b ) S z ( L b ) − S z ( L b ) ζ z ( b ) + α t b . Recalling (2.4), this proves (2.6).By (2.1), 1 G z ( t b , t b ) = W zv,o ( t b ) U − z ; v ( t b ) V + z ; o ( t b ) = ( U − z ; v ) ′ ( t b ) U − z ; v ( t b ) − ( V + z ; o ) ′ ( t b ) V + z ; o ( t b ) , so(2.13) 1 G z ( t b , t b ) = − ( R + z ( t b ) + R − z ( t b )) . We have R + z ( t b ) = S ′ z ( L b ) S z ( L b ) − S z ( L b ) ζ z ( b ) and R − z ( t b ) = ζ z (ˆ b ) − S ′ z ( L b ) S z ( L b ) by (2.4) and (2.5). Hence,(2.14) R + z ( t b ) + R − z ( t b ) = − S z ( L b ) ζ z ( b ) + ζ z (ˆ b ) S z ( L b ) , proving the first part of (2.7).For the second part, we showed that S z ( L b ) G z ( t b ,t b ) = − ζ z ( b ) ζ z (ˆ b ) ζ z ( b ) , so replacing b by ˆ b wededuce that S z ( L b ) G z ( o b ,o b ) = − ζ z (ˆ b ) ζ z ( b ) ζ z (ˆ b ) , so ζ z (ˆ b ) ζ z ( b ) = G z ( o b ,o b ) G z ( t b ,t b ) .It follows from (2.13) and (2.14) that1 ζ z ( b ) S z ( L b ) = ζ z (ˆ b ) S z ( L b ) + 1 G z ( t b , t b ) . Inserting this expression in (2.6), we obtain (2.8).As previously observed, the Wronskian is constant on each b , so ζ z (ˆ b ) = U − z ; v ( o b ) U − z ; v ( t b ) . Hence, G z ( t b k , t b k ) ζ z (ˆ b ) · · · ζ z (ˆ b k ) = U − z ; v ( v k ) V + z ; o ( v k ) W zv,o ( v k ) U − z ; v ( v ) U − z ; v ( v ) · · · U − z ; v ( v k − ) U − z ; v ( v k )= U − z ; v ( v ) V + z ; o ( v k ) W zv,o ( v k ) = G z ( o b , t b k ) . By (2.7), ζ z ( b ) · · · ζ z ( b k ) = ζ z (ˆ b ) · · · ζ z (ˆ b k ) · G z ( t bk ,t bk ) G z ( o b ,o b ) , proving the other equality.Finally, by the first part of (2.9) we have G z ( t b k , o b ) = G z ( t b k , t b k ) ζ z (ˆ b k ) · · · ζ z (ˆ b ) = G z ( o b , t b k ) by the second part. (cid:3) For the following lemma, fix o, v ∈ V and consider the WT functions (2.2). Assume that o b , t b ∈ V ( T + o ) ∩ V ( T − v ), that is to say, that b ∈ B ( T + o ) and ˆ b ∈ B ( T − v ). Let T + o b ⊆ T + o and T − t b ⊆ T − v be the subtrees starting at o b and t b , respectively. Let G T + ob ( x, y ) be theGreen kernel of the δ -problem on T + o b . This means the usual δ -conditions at v ∈ V ( T + o b ),with α o b = 0. Similarly, G T − tb ( x, y ) is the Green kernel of the δ -problem on T − t b .We will need the notion of Herglotz functions [14] throughout the paper. A Herglotzfunction (a.k.a.
Nevanlinna function or
Pick function) is an analytic function from C + to C + . Herglotz functions form a positive cone: if f , f are Herglotz and a , a are positiveconstants, then a f + a f is Herglotz. Composition of two Herglotz functions is again aHerglotz function. The functions z
7→ √ z and z
7→ − /z for example are Herglotz. Every Herglotz function f has a canonical representation [14, Theorem II.I] of the form(2.15) f ( z ) = Az + B + Z R (cid:18) t − z − t t (cid:19) d m ( t ) , where A and B are constants and m is a Borel measure satisfying R R (1 + t ) − d m < ∞ . Lemma 2.4.
Let b ∈ T and z ∈ C + . Let o, v ∈ V be such that b ∈ T + o and ˆ b ∈ T − v . Thenwe may express (2.16) R + z ( o b ) = − G z T + ob ( o b , o b ) and R − z ( t b ) = − G z T − tb ( t b , t b ) , where G z T ± v ( v, v ) are defined with the Neumann condition at v .The functions F ( z ) = R + z ( o b ) , R − z ( t b ) and G z ( v, v ) are Herglotz functions. If all W v ≥ and α v ≥ , then e F ( z ) = R + z ( o b ) √ z , R − z ( t b ) √ z are also Herglotz.Moreover, we have the following “current” relations: (2.17) X b + ∈N + b Im R + z ( o b + ) ≤ Im R + z ( o b ) | ζ z ( b ) | and X b − ∈N − b Im R − z ( t b − ) ≤ Im R − z ( t b ) | ζ z (ˆ b ) | . Equality holds in both cases if Im z = 0 , whenever defined. Most statements of this lemma appear in [1]. We give the proof in Appendix A forcompleteness. We also deduce that − S ′ z ( L b ) S z ( L b ) and S z ( L b ) ζ z ( b ) ∈ C + , see Remark A.3.The following corollary says that the inverse moments of the imaginary part of the WTfunctions essentially control all relevant spectral quantities on the tree : Corollary 2.5.
Let I ⊂ R be compact, I ∩ D = ∅ , and z ∈ C + . Fix c , c , c > suchthat for all z ∈ I + i[0 , , L b ∈ [ L min , L max ] , (2.18) c ≤ | S z ( L b ) | ≤ c and | C z ( L b ) | ≤ c . Then for any p ≥ , and b ∈ T , | G z ( o b , o b ) | p ≤ | Im R + z ( o b ) | − p , | ζ z ( b ) | p ≤ c − p X b + ∈N + b | Im R + z ( o b + ) | − p and | R + z ( o b ) | p ≤ c − p p − c − p X b + ∈N + b | Im R + z ( o b + ) | − p + c p . Also note that R − z ( t b ) = R + z ( o b b ) + C z ( L b ) − S ′ z ( L b ) S z ( L b ) using (2.4), so up to choosing c > | S ′ z ( L b ) | ≤ c , a control over all R + z ( o b ), b ∈ T , implies a control over all R − z ( t b ). Proof.
We have by (2.13), | G z ( o b , o b ) | p = | R + z ( o b )+ R − z ( o b ) | − p ≤ | Im R + z ( o b ) | − p . By (2.12), | ζ z ( b ) | p ≤ c − p | S z ( L b ) ζ z ( b ) | p ≤ c − p (cid:12)(cid:12)(cid:12) α t b + S ′ z ( L b ) S z ( L b ) − X b + ∈N + b R + z ( o b + ) (cid:12)(cid:12)(cid:12) − p ≤ c − p (cid:12)(cid:12)(cid:12) Im (cid:16) α t b + S ′ z ( L b ) S z ( L b ) − X b + ∈N + b R + z ( o b + ) (cid:17)(cid:12)(cid:12)(cid:12) − p ≤ c − p (cid:16) X b + ∈N + b Im R + z ( o b + ) (cid:17) − p ≤ c − p | Im R + z ( o b + ) | − p , If W b is symmetric, i.e. W b ( L b − x b ) = W b ( x b ), then S ′ z ( L b ) = C z ( L b ), so R − z ( t b ) = R + z ( o b b ). BSOLUTELY CONTINUOUS SPECTRUM FOR QUANTUM TREES 13 where we used that − S ′ z ( L b ) S z ( L b ) and R + z ( o e ) are Herglotz in the last line, with b + ∈ N + b arbitrary. Hence, | R + z ( o b ) | p = (cid:12)(cid:12)(cid:12) ζ z ( b ) − C z ( L b ) S z ( L b ) (cid:12)(cid:12)(cid:12) p ≤ c − p p − ( | ζ z ( b ) | p + c p ) ≤ c − p p − (cid:16) c − p | Im R + z ( o b + ) | − p + c p (cid:17) . (cid:3) AC spectrum for the unperturbed tree
The aim of this section is to prove Theorems 1.2 and 1.4.Let T be a quantum tree of finite cone type, with the structure described in § v − , v ) ∈ B ( T + b o ), we denote ζ z ( v ) = G z ( v − , v ) G z ( v − , v − ) . This notation is simply analogous to the one introduced in Section 1.1.2, and does notmean that ζ z is a function of the terminus alone. It simply means that each discrete edgein T + b o can be specified by indicating the terminus alone. We also let ζ z ( t b o ) = ζ z ( b o ).Denote ζ zj = ζ z ( v ) if ℓ ( v ) = j . Then (2.6) says that for each j ∈ A + ,(3.1) 1 ζ zj S z ( L j ) + m X k =1 M j,k ζ zk S z ( L k ) = m X k =1 M j,k C z ( L k ) S z ( L k ) + S ′ z ( L j ) S z ( L j ) + α j . The matrix elements M j,k were defined in § ζ zj = ζ zv − ( v ).In order to put it in a nicer form, we denote h j = S z ( L j ) ζ zj . Then we get the followingsystem of polynomial equations:(3.2) m X k =1 M j,k S z ( L k ) h k h j − F j ( z ) h j + 1 = 0 , j = 1 , . . . , m where F j ( z ) = α j + P mk =1 M j,k C z ( L k ) S z ( L k ) + S ′ z ( L j ) S z ( L j ) .An analogous system of equations involving the matrix N = ( N i,j ) arises when consid-ering cones in T − b o . We restrict ourselves to the above system; the other one is analyzedsimilarly.We mention that a similar system of equations in a more special framework appearedrecently in [13, eq. (4.8)]. In this case, one has M j,j = 1 for each j and M j,k = 2 for k = j .Our aim in the following is to control the values of ζ λ +i ηj as η ↓
0. For the models[12, 13], the ζ zj are uniformly bounded. The following simple criterion gives a sufficientcondition for this to happen. Note the condition M j,j > j has at least one offspring of its own type. Later we will relax that restriction. Lemma 3.1.
Suppose M j,j > for some j . Then | ζ zj | < for any z ∈ C \ R . In fact, | ζ zj | < M j,j . This lemma parallels the combinatorial case [21, Lemma 3], see [13, Lemma 3.9] for aspecial case.
Proof.
Let z ∈ C + and b with ℓ ( t b ) = j . Then (2.17) becomes P mk =1 M j,k Im R + z ( k ) ≤ Im R + z ( j ) | ζ zj | , where R + z ( k ) := R + z ( o e ) if ℓ ( t e ) = k . The inequality is actually strict if Im z > | ζ zj | < Im R + z ( j ) M j,j Im R + z ( j ) = M j,j .The case Im z < R + z ( o e ) should bereplaced by | Im R + z ( o e ) | in (2.17). (cid:3) The lemma implies in particular that | ζ λ +i0 j | ≤ M j,j for any λ ∈ R .There are many models of interest for which the condition of Lemma 3.1 is not satistfied,so we next consider the general case. Now the limit ζ λ +i0 j may no longer exist, but we aimto show this problem can only occur on a discrete subset of R . Proposition 3.2.
There is a discrete set D ⊂ R such that, for all j = 1 , . . . , m , thesolutions h j ( λ + i η ) = S λ +i η ( L j ) ζ λ +i ηj of (3.2) have a finite limit as η ↓ for all λ ∈ R \ D .The map λ S λ ( L j ) ζ λ +i0 j is continuous on R \ D , and there is a discrete set D ′ such thatit is analytic on R \ ( D ∪ D ′ ) .Proof. We follow the strategy in [6, § h j satisfies an algebraic equation Q j ( h j ) = 0. For this, we willuse an algebraic tool from [25].Let λ ∈ R , and let P j ( h , . . . , h m ) = P mk =1 M j,k S z ( L k ) h k h j − F j ( z ) h j + 1. Clearly, P j ∈ K [ h , . . . , h m ], where K = K λ is the field of functions f ( z ) possessing a convergentLaurent series f ( z ) = P ∞ j = − n a j ( λ )( z − λ ) j in some neighbourhood N λ ⊂ C of λ .Let K ′ = J λ be the field of functions f which are meromorphic on N λ ∩ C + for someneighbourhood N λ of λ . Then K ′ is an extension of K , and we know that S z ( L j ) ζ zj belongs to K ′ (see Remark 2.2) and satisfy P j ( S z ζ z , . . . , S z ζ zm ) = 0. Calculating theJacobian (cid:16) ∂P j ∂h k ( h ) (cid:17) , we find ∂P j ∂h k = ( M j,k S z ( L j ) h j , k = j, P mℓ =1 M ℓ,j S z ( L ℓ ) h ℓ + M j,j S z ( L j ) h j − F j ( z ) , k = j. Let J z = det (cid:18) ∂P j ∂h k (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) h = S z ( L k ) ζ zk . We will show that z J z is not the zero element of K ′ . For this, we first study theasymptotics of J z as z → −∞ .Take z = − r with r > r →∞ C − r ( L k ) rS − r ( L k ) = 1 and lim r →∞ S ′− r ( L k ) rS − r ( L k ) = 1 . This follows from classical estimates [28, Chapter 1]. In fact, S − r ( L ) ≈ sin i rL i r = sinh rLr , C − r ( L ) ≈ cos i rL = cosh rL ≈ S ′− r ( L ) . More precisely, we write C − r ( L ) rS − r ( L ) = cosh rL + R ( r, L )sinh rL + rR ′ ( r, L ) , where R ( r, L ) = C − r ( L ) − cosh rL and R ′ ( r, L ) = S − r ( L ) − sinh rLr . By [28, p. 13], rR ′ ( r,L )sinh rL → R ( r,L )sinh rL → r → ∞ . Since cosh rL sinh rL →
1, (3.3) follows. Hence,(3.4) F j ( − r ) ∼ α j + r + r P k M j,k ∼ C j r as r → ∞ . On the other hand, since h j is Herglotz (see Remark A.3), it has a representation of theform (2.15). If t = inf σ ( H T ), we also know from Remark 2.2 that h j ( λ ) is well-definedand real-valued for λ < t . By [34, Theorem 3.23], the measure m is thus supported on[ t , ∞ ). Hence, for large r (say r > − t + 1), h j ( − r ) = − Ar + B + Z ∞ t − r tt + r d m ( t )1 + t , BSOLUTELY CONTINUOUS SPECTRUM FOR QUANTUM TREES 15 where we used that h j ( − r ) = lim η ↓ h j ( − r + i η ) and dominated convergence (recall that d m ( t )1+ t is a finite measure). Thus, h j ( − r ) − r = A − Br + Z ∞ t t − r t + r d m ( t )1 + t . Using dominated convergence again, we see that h j ( − r ) / ( − r ) → A as r → ∞ . Thisimplies that h j ( − r ) S − r ( L k ) = O ( r e − rL k ) = O (e − rL k ) . Therefore, recalling that the C j were defined in (3.4), we find that as r → ∞ , J − r = det − C r + O (e − rL ) O (e − rL ) · · · O (e − rL m ) O (e − rL ) − C r + O (e − rL ) · · · O (e − rL m )... ... . . . ... O (e − rL ) O (e − rL ) · · · − C m r + O (e − rL m ) Hence, J − r ∼ ( − m C · · · C m r m = Cr m = 0 for r large enough. Since z J z is holo-morphic on C \ [ a , ∞ ), it follows that it cannot vanish identically on any neighbourhood N λ ∩ C + . Hence, J z is not the zero element of K ′ .It follows by [25, Proposition VIII.5.3] that each S z ζ zj is algebraic over K . By theNewton-Puiseux theorem (see e.g. [31, Theorem 3.5.2]), each h j thus has an expansion ofthe form(3.5) h j ( z ) = X n ≥ m a n ( z − λ ) n/d in some neighbourhood N λ of λ . Here m ∈ Z , d ∈ N , and the entire series P n ≥ a n z n has a positive radius of convergence. In particular, z S z ζ zj is analytic near any λ ∈ N λ \ { λ } . The set D corresponds to those λ for which m < λ , and the set D ′ corresponds to those λ for which d > (cid:3) Our next aim is to show that all WT functions have a positive imaginary part on mostof the spectrum.Let σ D be the union of the Dirichlet spectra: σ D = m [ j =1 { λ ∈ R : S λ ( L j ) = 0 } . We would like to index the WT functions by vertices, but the notation R + z ( v ) is a bitambiguous since R + z ( t b ) = R + z ( o b + ) even if t b = o b + = v . So we take the convention that(3.6) R ± z ( v ) := ( R ± z ( o b ) if b = ( v − , v ) ∈ T + b o ,R ± z ( t b ) if b = ( v, v + ) ∈ T − b o . Here ( t b o ) − = o b o and ( o b o ) + = t b o . This keeps with the convention of § ψ ( b ) by their terminus on T + b o and their origin on T − b o .As there are finitely many types of ζ z ( b ) for b = ( v − , v ) ∈ B ( T + b o ), we see by (2.4)there are finitely many types of R + z ( o b ) (this may not be true for R − z ( o b ) for such b ). Wedenote R + z ( j ) := R + z ( o b ) if b = ( v − , v ) ∈ T + b o and ℓ ( v ) = j ∈ A + . Similarly, we denote R − z ( k ) = R − z ( t b ) if b = ( v, v + ) ∈ T − b o and ℓ ( v ) = k ∈ A − . The notation is probably a bit awkward since for v ∈ T + b o , v is the terminus of b yet R ± z ( v ) := R ± z ( o b ).We stress however that R ± z is not a function of the vertex o b alone but depends on the whole directed edge b , so it should really be read as a function ψ ( b ), which we index by the terminus. By some abuse of notation, we assume the discrete sets D , D ′ of Proposition 3.2 arethe same for the system analogous to (3.2) which involves the matrix ( N i,j ). Remark 3.3.
Denote R ± λ := R ± λ +i0 . Then the limits R + λ ( j ) exist for λ ∈ R \ ( D ∪ σ D )and j ∈ A + . This follows from Proposition 3.2 and (2.4), which implies that R + λ ( j ) = ζ λj − C λ ( L j ) S λ ( L j ) . Similarly, the limits R − λ ( k ) exist for k ∈ A − .It follows that R ± λ ( v ) exist for v ∈ { o b o , t b o } and λ / ∈ D ∪ σ D . In fact, ζ λ ( b o ) = h jo S λ ( L o ) for some j o ∈ A + , which exists by Proposition 3.2, so R + λ ( o b o ) exists by (2.4). Similarlythe result for the ( N ij ) system implies the existence of ζ λ (ˆ b o ) and R − λ ( t b o ). Finally if t b o has type j o ∈ A + , then R + λ ( t b o ) = P mk =1 M j o ,k R + λ ( k ) − α t bo by (2.12), which exists by theprevious paragraph. Similarly R − λ ( o b o ) = P nk =1 N j ′ o ,k R − λ ( k ) + α o bo exists.Proposition 3.2 tells us moreover that R ± λ +i0 ( v ) are analytic on R \ ( D ∪ D ′ ∪ σ D ). Inparticular, their zeroes do not accumulate. Hence, there is a discrete set D ′′ such that R + λ ( v ) + R − λ ( v ) = 0 for λ ∈ R \ ( D ∪ D ′ ∪ D ′′ ∪ σ D ) and v ∈ { o b o , t b o } . We therefore define D := D ∪ D ′ ∪ D ′′ ∪ σ D . We may actually generalize the result of Remark 3.3 as follows.
Lemma 3.4. (a) If λ / ∈ D ∪ σ D , then R ± λ ( o b ) exists for all b = ( v − , v ) ∈ T + b o and R ± λ ( t b ) exists for all b = ( w, w + ) ∈ T − b o . (b) If moreover λ / ∈ D , then G λ ( v, v ) exists for any v ∈ T , and G λ ( v, v ) = 0 .Proof. By symmetry it suffices to prove (a) for T + b o . Consider b = ( v − , v ) ∈ T + b o . Wealready know that R + λ ( o b ) = R + λ ( j ) exists from Remark 3.3. Next, we show by inductionthat ζ λ ( b b ) S λ ( L b ) is finite. Note that we already know ζ λ ( b b o ) S λ ( L o ) is finite by (2.5). Soconsider any oriented edge b = ( t b o , v ). Applying (2.6) to b b instead of b , we may express ζ z ( b b ) S z ( L b ) in terms of some C z , S z functions, plus ζ z ( b ′ ) for b ′ ∈ N + b b . One of them is ζ z ( b b o ), whose limit on the real axis exists from Remark 3.3. The rest are precisely thosewith b ′ ∈ N + b o \ { b } , which also exist by Proposition 3.2. Thus, ζ λ ( b b ) S λ ( L b ) exists for any b = ( t b o , v ). By induction we get existence for any ( v − , v ) ∈ T + b o . It follows from (2.5) that R − λ ( o b ) exists for all such b = ( v − , v ) ∈ T + b o .(b) By Remark 3.3, R + λ ( v ) + R − λ ( v ) = 0 for v = o b o , t b o . Using (2.13), this implies G λ ( v, v ) exists. We now observe that(3.7) G z ( v + , v + ) = S z ( L v ) ζ z ( v, v + ) + ζ z ( v, v + ) G z ( v, v ) . In fact, by (2.7), G z ( v + , v + ) = ζ z ( v, v + ) ζ z ( v + , v ) G z ( v, v ) = (cid:16) ζ z ( v, v + ) + S z ( L v ) G z ( v, v ) (cid:17) ζ z ( v, v + ) G z ( v, v )as claimed. Using Proposition 3.2, we thus deduce the existence of G λ ( w, w ) for all w ∈ T + b o .Similarly the existence of w ∈ T − b o follows from the analog of Proposition 3.2 with the ( N i,j )system. Finally using (2.13) we see that (a) implies G λ ( v, v ) = 0. (cid:3) We now observe that under (C1*) , all the WT functions are related as follows:
Lemma 3.5.
Suppose T satisfies (C1*) and let λ ∈ R \ ( D ∪ σ D ) . (i) If Im R ± λ +i0 ( j ) = 0 for some j ∈ A ± , then Im R ± λ +i0 ( j ) = 0 for all j ∈ A ± . (ii) Assume λ ∈ R \ D .If Im R + λ +i0 ( j ) = 0 for some j ∈ A + and Im R − λ +i0 ( k ) = 0 for some k ∈ A − , then Im G λ +i0 ( v, v ) = 0 and Im R ± λ +i0 ( v ) = 0 for all v ∈ T . BSOLUTELY CONTINUOUS SPECTRUM FOR QUANTUM TREES 17
The same conclusion holds if Im G λ +i0 ( w, w ) = 0 for some w ∈ T . (iii) For any v ∈ T , we have σ ( H ) \ D = { λ ∈ R \ D : Im G λ +i0 ( v, v ) > } \ D . Proof.
We first note that ζ λj = 0, due to the relation ζ λj = h j S λ ( L j ) = − /S λ ( L j ) P mk =1 M j,k S λ ( L k ) h k − F j ( λ )and the fact that the h k are finite.Denote R λk = R + λ +i0 ( k ) . Suppose that Im R λk > k ∈ A + and let l ∈ A + . Thenby (C1*) , ( M n ) l,k ≥
1, so if v has label l and w has label k , there is a path ( u , . . . , u r )with u = v and u r = w . Denote b j = ( u j − , u j ). Then applying (2.17) repeatedly,Im R λl = Im R λ ( o b ) ≥ X ( e ; e r ) | ζ λ ( e ) · · · ζ λ ( e r − ) | Im R λ ( o e r ) ≥ | ζ λ ( b ) · · · ζ λ ( b r − ) | Im R λ ( o b r ) > , where the sum runs over all ( r − e ; e r ) outgoing from b , and the last inequalityholds because Im R λ ( o b r ) = Im R λk > ζ λj = 0.So under (C1*) , if Im R λj > j ∈ A + , then Im R λk > k ∈ A + . So ifIm R λj = 0 for some j ∈ A + , then it must be zero for all j ∈ A + .The proof for R − λ +i0 is the same.For (ii), say t b o has type j o . Since R + λ ( t b o ) = P mk =1 M j o ,k R + λ ( k ) − α t bo by (2.12), we getIm R + λ ( t b o ) = 0 by (i). Using (2.5), this implies Im ζ λ ( b o ) = 0, which by (2.4) implies thatIm R + λ ( o b o ) = 0. Similarly, if o b o has type j , then R − λ ( o b o ) = P nk =1 N j,k R − λ ( k ) + α o bo , soIm R − λ ( o b o ) = 0 by (i), also implying Im R − λ ( t b o ) = 0 via (2.5), (2.4). Now R + λ ( v )+ R − λ ( v ) =0 for v = o b o , t b o . Using (2.13), this implies G λ ( v, v ) exists, and Im G λ ( v, v ) = 0 for v = o b o , t b o . Since Im G λ ( t b o , t b o ) = 0 and Im ζ λ ( t b o , v + ) = Im ζ λj = 0, then using (3.7), weget Im G λ ( v + , v + ) = 0 for any v + ∈ N + t bo , so Im G λ ( w, w ) = 0 for all w ∈ T + b o by induction.Similarly, we may use Im G λ ( o b o , o b o ) = 0 along with (2.7) to deduce that Im G λ ( v, v ) = 0for all v ∈ T − b o . This proves claim (ii) for the Green function.Next, if v ∈ T + b o , we know that Im R + λ ( v ) = Im R + λ ( j ) = 0. By Lemma 3.4 G λ ( v, v )exists, and we showed Im G λ ( v, v ) = 0. Using (2.13), it follows that Im R − λ ( v ) = 0. Hence,Im R ± λ ( v ) = 0 for v ∈ T + b o . The claim for T − b o follows similarly.Now suppose that Im G λ ( w, w ) = 0 for some w ∈ T . By symmetry we may assume w ∈ T + b o . Recall that G λ ( v, v ) , R ± λ ( v ) exist by Lemma 3.4. By (2.13), we get Im R + λ ( w ) =Im R − λ ( w ) = 0. Consider w + ∈ N + w . By Proposition 3.2, ζ λ ( w, w + ) exists, so using (2.4),we get Im ζ λ ( w, w + ) = 0, hence Im G λ ( w + , w + ) = 0. On the other hand Im R + λ ( w ) = 0implies Im R + λ ( w − ) = 0 by (i), so we similarly get Im ζ λ ( w − , w ) = 0 and Im G λ ( w − , w − ) =0. This shows that Im G λ ( v, v ) = 0 for all v ∈ T + b o and also for v = o b o , since by definition(3.6), Im R + λ ( o b o ) =: Im R + λ ( j o ) = 0 for v = t b o , if ℓ ( t b o ) = j o . Now if u ∈ N − o bo , then asin (3.7) we have G λ ( u, u ) = S λ ( L u ) ζ λ ( o b o , u ) + ζ λ ( o b o , u ) G λ ( o b o , o b o ), but Im R − λ ( o b o ) =0 implies P b − ∈N − bo Im R − λ ( t b − ) = Im R − λ ( o b o ) = 0, so Im R − λ ( t b − ) = 0 for each b − , soIm R − λ ( k ) = 0 for all k ∈ A − by (i) and we deduce again that Im G λ ( u, u ) = 0 for u ∈ N − o bo , hence for all u ∈ T − b o by induction.Finally, to prove (iii), recall that if E H is the projection-valued measure E H ( J ) = χ J ( H ), then σ ( H ) = supp E H . Moreover, E H ( J ) = 0 if and only if µ f ( J ) = 0 for all f ∈ L ( T ), where µ f ( J ) = h f, χ J ( H ) f i . By [34, Lemma 3.13], we know that supp µ f = { λ ∈ R : Im h f, G λ f i > } . Since D is a discrete set, we deduce that supp µ f \ D = { λ ∈ R \ D : Im h f, G λ f i > } \ D .Let λ ∈ R \ D and suppose there is ǫ such that Im G λ ( v, v ) = 0 for all I := ( λ − ǫ, λ + ǫ ). Then Im R ± λ ( w ) = 0 for all w by (ii), so Im h f, G λ f i = 0 for any f by Lemma A.2.Thus, µ f ( I ) = 0 for all f ∈ L ( T ), so E H ( I ) = 0 and thus λ / ∈ σ ( H ).Conversely, fix λ ∈ R \ D and v ∈ T . If Im G λ ( v, v ) >
0, then Im R + λ ( v ) > R − λ ( v ) >
0, say the former holds and let v = o e . Then by Lemma A.2, Im h f, G λ f i > f = φ − λ ; e , since g − φ − λ ; e ( λ ) ≥ k φ − λ ; e k >
0. In fact, k φ − λ ; e k = 0 would imply C λ ( x ) = R − λ ( o e ) S λ ( x ) for all x ∈ e , contradicting the fact that C λ and S λ are linearly independent.We thus showed that { λ ∈ R \ D : Im G λ ( v, v ) > }\ D ⊆ supp µ f \ D ⊆ σ ( H ) \ D . (cid:3) Lemma 3.6. If T satisfies (C1*) , then: (i) For any j ∈ A + , k ∈ A − , the map σ ( H ) \ D ∋ λ Im R + λ +i0 ( j ) + Im R − λ +i0 ( k ) has adiscrete set of zeroes. The same holds for σ ( H ) \ D ∋ λ Im G λ +i0 ( v, v ) , for any v ∈ T . (ii) σ ( H ) is a union of closed intervals and isolated points, S r I r ∪ P . The limits G λ +i0 ( v, v ) exist in the interior ˚ I r and satisfy Im G λ +i0 ( v, v ) > , for any v ∈ T . (iii) The spectrum of H T is purely absolutely continuous in any compact subset K ⊂ ˚ I r .Proof. We know that f ( λ ) = R + λ ( j ) + R − λ ( k ) is analytic on R \ D , so if Im f ( λ ) = 0for some λ / ∈ D , we may expand Im f ( λ ) = P n ≥ b n ( λ − λ ) n for λ ∈ ( λ − ǫ, λ + ǫ ),where b n = Im a n and ( a n ) are the coefficients for f ( λ ). Suppose λ ∈ σ ( H ) \ D . If all b n = 0 then Im f is identically zero on ( λ − ǫ, λ + ǫ ). In view of Lemma 3.5 (ii)-(iii),this contradicts that λ ∈ σ ( H ). Hence let k be the smallest index with b k = 0. ThenIm f ( λ ) = ( λ − λ ) k g ( λ ), where g ( λ ) = P n ≥ b n + k ( λ − λ ) n . Clearly g ( λ ) = b k = 0 and g is continuous, so we may find ǫ ′ ≤ ǫ such that both ( λ − λ ) k and g ( λ ) are nonzero on( λ − ǫ ′ , λ + ǫ ′ ) \ { λ } . This shows that λ is an isolated zero of Im f , as required.This proves the first part of (i). For the second part, suppose Im G λ ( w, w ) = 0 for some w ∈ T . By Lemma 3.5, this implies Im R + λ ( j ) + Im R − λ ( k ) = 0. Hence, λ must lie in thepreceding discrete set of zeroes.For (ii), recall that σ ( H ) \ D = { λ ∈ R \ D : Im G λ ( v, v ) > } \ D by Lemma 3.5.By Proposition 3.2 and (2.4), we know λ R ± λ ( v ) is continuous, so R \ D ∋ λ G λ ( v, v )is continuous by (2.13). Hence, { λ ∈ R \ D : Im G λ ( v, v ) > } is a union of intervals S r J r which is independent of v by Lemma 3.5. We take I r as the closure of J r and P = σ ( H ) ∩ D .Finally, if K is a compact subset of ˚ I r , we know that G λ ( v, v ) is uniformly bounded, andthe same holds for R ± λ ( v ). In particular, if v = o e and ψ is supported in e , we get usingrespresentation (A.2) along with (2.13) that sup λ ∈ K |h ψ, G λ ψ i| < ∞ . The claim followsby the density of the linear span of such ψ . (cid:3) This completes the proof of Theorem 1.2. We next move to Theorem 1.4.
Remark 3.7.
Condition (C1) implies (C1*) . In fact, as remarked in [6], all cone typesare indexed by the directed edges of the finite graph G . If we consider the universal cover T rooted at the midpoint o of some b o ∈ B ( G ) (here o is not viewed as an added vertex,just a reference point), this means that the type of each vertex v ∈ T is determined bya directed edge, so there are at most | B ( G ) | types. By [27, Lemma 3.1], we know thenon-backtracking matrix of B ( G ) is irreducible. This implies that if T is considered in the twisted view, and if M is the single matrix over some alphabet A encoding all cone types,then M satisfies: for any k, l ∈ A , there is n ( k, l ) such that ( M n ) k,l ≥
1. In particular, (C1*) holds if we take the matrices ˜
M , N encoding the types in T + b o and T − b o , respectively. BSOLUTELY CONTINUOUS SPECTRUM FOR QUANTUM TREES 19
Proof of Theorem 1.4.
Since (C1) implies (C1*) , we already know that σ ( H ) has thestructure given in Theorem 1.2. Let λ ∈ ˚ I r be in the interior of an AC band.Within the twisted view, all vertices are offspring of o and we deal with the single,combined alphabet A . Under the stronger assumption (C1) , we know the larger matrix M is irreducible. Consequently, if we suppose that Im R + λ +i0 ( j ) = 0 for some j ∈ A , thenthe statement in Lemma 3.5 (i) now implies that Im R + λ +i0 ( j ) = 0 for all j ∈ A .Now let v ∈ T . We know Im G λ ( v, v ) >
0, so by (2.13), Im R + λ ( v ) + Im R − λ ( v ) > R + λ ( v ) > R − λ ( v ) > R + λ ( v ) = 0. Say v = t b for some b ∈ B ( T ) and ℓ ( v ) = j .Then 0 = Im R + λ ( t b ) = P mk =1 M j,k Im R + λ ( k ) implies that Im R + λ ( k ) = 0 for some, henceall, k ∈ A . But by (2.4), R − λ ( t b ) = R + λ ( o b b ) + C λ ( L b ) − S ′ λ ( L b ) S λ ( L b ) , so Im R − λ ( t b ) = Im R + λ ( o b b ).As mentioned in Remark 3.7, under (C1) , all cone types are indexed by the directededges of G , in particular T + b b is one of the finitely many nonisomorphic cones . In otherwords, Im R + λ ( o b b ) = Im R + λ ( r ) for some r ∈ A . Hence, Im R − λ ( t b ) = 0. We thus getIm R + λ ( v ) + Im R − λ ( v ) = 0, a contradiction. Thus, Im R + λ ( v ) > (cid:3) Examples of nontrivial spectrum
For Theorem 1.4 to be interesting, we’ll need to know that σ ( H ) is not reduced to theisolated points P . Our aim in this section is to give some examples in which this can beproved. We believe the phenomenon to be true for a wider class of examples.4.1. Equilateral trees.
Let G be a discrete graph of minimal degree ≥ T = e G itsuniversal cover. We know from [10, Section 1.6] that the spectrum of the adjacency matrix σ ( A T ) has a continuous part. Actually their argument remains valid for the normalizedadjacency matrix P f ( x ) = d ( x ) ( A f )( x ) (and also if we add potentials). Consequently,using [11, Theorem 3.18], the induced quantum tree with equilateral edge length, identicalsymmetric potentials, and identical coupling constants, will also have some continuousspectrum. Using Theorem 1.4, we can now conclude: If G is a graph of degree ≥ , if T = e G is its universal cover, and we endow each edge of T with the same length L and potential W , and each vertex with the same coupling constant α , then σ ( H T ) consists of non-empty bands of purely absolutely continuous spectrum, andpossibly some isolated eigenvalues. This generalizes the case of regular trees previously considered in [12, 33].We may easily extend this to graphs with several lengths which are rationally dependent.More precisely, if in G , we have L j = n j L for some n j ∈ N ∗ , add n j vertices of degree 2 tothe edge e j , with Kirchhoff-Neumann conditions. Then using the previous claim, we seethat T also has nontrivial AC spectrum in this case.4.2. An argument of Bordenave-Sen-Vir´ag.
We now consider the non-equilateralcase. For this, we start by adapting an argument from [10] to quantum graphs.We begin with some definitions, which appear in a more general framework in [10].Let G be a discrete graph and T = e G its universal cover.A labeling (or colouring ) of the vertices of T is a map η : V ( T ) → Z . With respect to agiven labeling, we call a vertex v :(a) prodigy if it has a neighbour w with η ( w ) < η ( v ) and such that all other neighboursof w also have label less than η ( v ),(b) level if it is not a prodigy and if all of its neighbours have the same or lower labels,(c) bad if it is neither prodigy nor level. This property is why we need (C1) ; it is not necessarily true under (C1*) . cf. footnote in § The tree T = e G is equipped with a natural unimodular measure on the space of rootedgraphs, namely P = | G | P x ∈ G δ [ e G, ˜ x ] .We say the labeling η on T is invariant if there exists a unimodular probability measureon the set of coloured rooted graphs, which is concentrated on { [ T , v, η ] } v ∈ T . See e.g. [4,Appendix A] for some background on coloured rooted graphs.Let S ⊂ ℓ ( e G ) be a subspace and let P S be the orthogonal projection onto S . We saythat S is invariant if P S ( gv, gw ) = P S ( v, w ) for any g ∈ Γ, where Γ is the group of coveringtransformations with e G/ Γ ≡ G .Given an invariant subspace S ⊂ ℓ ( e G ), we define its von-Neumann dimension bydim VN S = E P [ h δ o , P S δ o i ] = 1 | G | X x ∈ G P S (˜ x, ˜ x ) . A line ensemble in T is a disjoint union of bi-infinite lines ( l i ). More precisely, L : V ( T ) × V ( T ) → { , } is a line ensemble if: • L ( u, v ) = 0 if { u, v } / ∈ E ( T ), • L ( u, v ) = L ( v, u ), • for any v ∈ V ( T ), we have P u L ( u, v ) ∈ { , } .Abusing notation, we then let L = { e : L ( e ) = 1 } , which gives a subgraph consisting ofdisjoint lines.We say a line ensemble L is invariant if there exists a unimodular probability measureon the space of weighted rooted graphs, which is concentrated on { [ T , v, L ] } v ∈ T .We say that T is Hamiltonian if there exists an invariant line ensemble L that containsthe root with probability 1. Remark 4.1.
Recall that a finite graph G is Hamiltonian if there is a cycle in G which vis-its each vertex exactly once. If G is Hamiltonian, then e G is Hamiltonian in the above sense.In fact, if C = ( x , . . . , x m ) is a cycle in G , then its lift to e G is a line ensemble L which gen-erally consists of a disjoint union of countable lines ( l i ), where l i = ( . . . , ˜ x , . . . , ˜ x m , ˜ x , . . . )(see Figure 2). Since it is a lift, this line ensemble is invariant. More precisely, if [ H, v, R ]denotes an equivalence class of graph H with root v and edge weight R ( e ) for e ∈ E ( H ),then [ e G, v, L ] = [ e G, gv, L ] for any covering transformation g . This by definition of the uni-versal cover and L . So the measure ˜ P = | G | P x ∈ G δ [ e G, ˜ x, L ] is well-defined and unimodularityfollows from P ( x,y ) ∈ B ( G ) f ( x, y ) = P ( x,y ) ∈ B ( G ) f ( y, x ).Moreover, ˜ P ( o ∈ L ) = | G | P x ∈ G ˜ x ∈L = | G | P x ∈ G x ∈ C = | C || G | . If C covers G , we thusget ˜ P ( o ∈ L ) = 1.In particular, the ( q + 1)-regular tree T q is Hamiltonian, since it covers the completebipartite ( q + 1)-regular graph on 2( q + 1) vertices, which is Hamiltonian.We may now state our adaptation of [10, Theorem 1.5] to quantum trees. Here, if G isa finite graph, we denote by G = G ( α, L , W ) the quantum graph obtained by endowingeach edge with a length L e , a potential W e and each vertex a coupling constant α v , so theSchr¨odinger operator H = − ∆ + W acts with δ -conditions. We say that T = e G if T = e G is endowed with the lifted structure α v = α πv , L ( u,v ) = L ( πu,πv ) and W ( u,v ) = W ( πu,πv ) .Recall definition (1.5) of S z ( x b ). Proposition 4.2.
Suppose G is a finite Hamiltonian graph. Endow G with a quantumstructure G with δ -conditions and let T = e G . If λ is an eigenvalue of H T , then λ mustbe a Dirichlet value, i.e. S λ ( L b ) = 0 for some b . A measure is unimodular if it invariant under the moving of the root. More precisely, it should satisfy
R P o ′ ∼ o f ([ G, o, o ′ ]) d P ([ G, o ]) =
R P o ′ ∼ o f ([ G, o ′ , o ]) d P ([ G, o ]). See [10, § BSOLUTELY CONTINUOUS SPECTRUM FOR QUANTUM TREES 21 K , . 1 286 4 3578 6 4 4684 86 157157753 642642642 731753 842842842 7 5 3862 8 6 2 8 6 2 Figure 2.
The lift of a Hamiltonian cycle is a line ensemble covering allvertices of T . (Each coloured bold line is infinite.) Proof.
Suppose H T has an eigenvalue, say H T ϕ λ = λϕ λ for some ϕ λ ∈ L ( T ).Suppose on the contrary that S λ ( L b ) = 0 for all b .Let ˚ ϕ = ϕ | V . We claim that ( A λ ˚ ϕ λ )( v ) = W λ ( v )˚ ϕ λ ( v ) , where ( A λ ψ )( v ) = X u ∼ v ψ ( u ) S λ ( L uv ) and W λ ( v ) = α v + X u ∼ v C λ ( L uv ) S λ ( L uv )and uv := ( u, v ). In fact, X u ∼ v ϕ λ ( u ) S λ ( L vu ) = X u ∼ v ϕ λ ( t vu ) S λ ( L vu ) = X u ∼ v (cid:20) ϕ λ ( o vu ) C λ ( L vu ) S λ ( L vu ) + ϕ ′ λ ( o vu ) (cid:21) = ϕ λ ( v ) X u ∼ v C λ ( L vu ) S λ ( L vu ) + α v ϕ λ ( v ) . Let E λ ⊂ ℓ ( T ) be the set of functions satisfying this eigenvalue equation, i.e. ψ ∈ E λ ifand only if(4.1) W λ ( w ) ψ ( w ) = X u ∼ w ψ ( u ) S λ ( L uw ) . Note that E λ is invariant. In fact, let M λ = A λ − W λ . Then M λ is self-adjoint. Thisis because all weights are real-valued and symmetric. Moreover, E λ = ker M λ . Now, if g ∈ Γ, let ( U g f )( v ) = f ( g − v ). Using that α gv = α v L ( gu,gv ) = L ( u,v ) , W ( gu,gv ) = W ( u,v ) ,it easily follows that U − g M λ U g = M λ . Standard arguments imply that U − g P E λ U g = P E λ ,so E λ is indeed invariant.Let C be a Hamiltonian cycle in G , so its lift L is a line ensemble as in Remark 4.1.Using the line ensemble, we may use the construction of [10, Theorem 1.5] to define forany k ∈ N ∗ an invariant labeling η k : V ( T ) → Z k of the vertices of T by integers whichsatisfies: • b := P ( o is bad ) ≤ /k , • vertices in L with η k ( v ) = 0 are prodigy. Vertices in L with η k ( v ) = 0 are bad, • vertices outside L are level. In our case, all vertices are in L , so there are no level vertices.We now argue as in [10, Theorem 2.3]. Here the situation is simpler as there are nolevel vertices. Let B be the space of vectors which vanish on the set of bad vertices. Thendim VN B = P ( o is not bad ) = 1 − b . Let E ′ = E λ ∩ B . Then using dim VN ( R ∩ Q ) ≥ dim VN R + dim VN Q −
1, we havedim VN E λ ≤ b + dim VN E ′ . We show E ′ is the trivial subspace by induction on the label j , showing that from lowto high, any f ∈ E ′ vanishes on vertices with label j . Remember vertices v can only beprodigy or bad.Recall that we have finitely many labels. Let j be the smallest label and let v be oflabel j . If v is a bad vertex, then f ( v ) = 0 since f ∈ B . Note that v cannot be a prodigy.Hence f ( v ) = 0 on vertices of smallest label.Now assume f ∈ E ′ vanishes on all vertices with label strictly below j . Since f ∈ B , weknow f vanishes on bad vertices. If v is a prodigy vertex of label j , then v has a neighbour w such that f vanishes on w and all neighbours of w , except perhaps v . But (4.1) gives f ( v ) S λ ( L vw ) = W λ ( w ) f ( w ) − P u ∼ w,u = v f ( u ) S λ ( L uv ) , so if the RHS is zero, then f ( v ) = 0.Recalling that b ≤ /k , we have showed that dim E λ ≤ /k . As k is arbitrary, weget dim E λ = 0. It follows that E λ = { } . Indeed, we have P E λ ( v, v ) = 0 for all v , sotr P E λ = 0, so k P E λ k op ≤ k P E λ k = 0, implying E λ = { } . It follows that there is no ℓ function on T such that A λ ψ = W λ ψ . By [15], it follows that there is no L function on T such that H T ϕ = λϕ . In other words, λ is not an eigenvalue of H T (contradiction). (cid:3) More examples.
Let T = e G . We now show the spectral bottom a = inf σ ( H T ) isstrictly below the smallest Dirichlet value. By virtue of Proposition 4.2, if G is Hamilton-ian, this implies a is not an eigenvalue. In particular, a is not an isolated spectral value,so in view of Theorem 1.4, there is some pure AC spectrum near a .Recall that if Q j are the quadratic forms associated to operators H j , then H ≥ H if D ( Q ) ⊆ D ( Q ) and Q ( f, f ) ≥ Q ( f, f ) for f ∈ D ( Q ). If H ≥ H then inf σ ( H ) ≥ inf σ ( H ).In T there are finitely many different kinds of edges (lengths and potentials). To eachoriented edge b , we associate the smallest Dirichlet: the smallest E such that S E ( L b ) = 0.Denote the least of those values by E D and let ( v, w ) be the edge on which it is attained(choose one of them if there is more than one edge with the same lowest Dirichlet value).Consider the quantum star graph around v , with the usual δ -condition at v , and Dirich-let conditions at the extremities w ′ ∼ v . Denote this (compact) graph by ⋆ and let E be its smallest eigenvalue. We claim that a ≤ E < E D . For the first inequality, let H ⋆ f = E f . Then E = Q ⋆ ( f,f ) k f k , where Q ⋆ is the quadraticform associated to H ⋆ . Let e f be the extension of f by zero to T . Then e f ∈ D ( Q T ), where Q T corresponds to H T . So a = inf g =0 Q T ( g, g ) k g k ≤ Q T ( e f , e f ) k e f k = Q ⋆ ( f, f ) k f k = E . For the second inequality, note that if f ∈ D ( H ⋆ ) and H ⋆ f = Ef , then on the edge b , f ( x b ) = AC E ( x b ) + BS E ( x b ) . Due to the Dirichlet conditions at extremities of the star, f ( L b ) = 0. If E < E D then S E ( L b ) = 0 and B = − AC E ( L b ) /S E ( L b ). Evaluating at x b = 0, the centre of the star, BSOLUTELY CONTINUOUS SPECTRUM FOR QUANTUM TREES 23 reveals that A = f ( v ). Thus, f ( x b ) = f ( v ) C E ( x b ) S E ( L b ) − C E ( L b ) S E ( x b ) S E ( L b ) . The condition at v , P b ∈ ⋆ f ′ ( o b ) = α v f ( v ), leads to(4.2) X w ′ ∼ v − C E ( L vw ′ ) S E ( L vw ′ ) = α v . Denote the left hand side of (4.2) as a function of E as Z ( E ). A solution Z ( E ) = α v willbe an eigenvalue of the star graph.Let us consider the behaviour of Z ( E ) as E increases to E D . We first show that as E approaches E D from below, S E ( L vw ) → • by [28, Theorem 6(a)], S E D ( x ) has exactly two zeros on [0 , L vw ]. These are thus { , L vw } . If E < E D , since S E (0) = 0 and S ′ E (0) = 1, we know that S E is positivenear 0. If we show that its first zero on (0 , ∞ ) occurs after L vw , this will implythat S E ( L vw ) >
0, which is what we seek. • We thus check that if
E < E D , then the first zero of S E on (0 , ∞ ) occurs afterthe first zero of S E D . For this, let f ( x ) = S E ( x ) S ′E D ( x ) − S ′ E ( x ) S E D ( x ). Then f ′ ( x ) = S E ( x ) S E D ( x )( E − E D ), which is negative until the first zero of S E or S E D .Suppose for contradiction that the first zero of S E (call it L E ) is before the firstzero of S E D . Then we get that f (0) = 0, f ( L E ) >
0, but f ′ ( x ) < , L E ),which is absurd. This proves the claim.Next, S ′E D ( L vw ) < L vw is the first zero of S E D after x = 0. By the Wronskianrelation C E D ( L vw ) S ′E D ( L vw ) = 1, so C E D ( L vw ) < E D is the smallestDirichlet eigenvalue, all the terms in Z ( E ) are either finite or diverge to + ∞ as E ր E D ,so that Z ( E ) diverges to + ∞ as E ր E D .On the other hand, from the proof of Proposition 3.2, we know that C E ( L b ) /S E ( L b ) → + ∞ as E → −∞ , so Z ( E ) → −∞ in the same limit.Together we have Z ( E ) → −∞ as E → −∞ and Z ( E ) → + ∞ as E ր E D . Since Z is continuous, there is a solution to Z ( E ) = α v strictly below E D , as claimed.5. AC spectrum under perturbations
We now aim to prove Theorem 1.7. For this, we adapt the approach of [22], see also[16] for some earlier ideas.A very sketchy outline of the argument is as follows. Our results in the previous sectionstell us that we have a good control over the unperturbed operator: it has pure AC spectrumin Σ, and all relevant spectral quantities such as the Green’s functions and WT functionshave a limit on Σ, which has a strictly positive imaginary part. Let H zv be such a spectralquantity to be chosen later, where z ∈ C + and v ∈ V ( T ). The aim is now to prove an L p -continuity estimate in mean with respect to the disorder ǫ . More precisely, in somesemi-metric γ on H , see (5.1), we aim to show that lim ǫ ↓ sup z ∈ I +i(0 , E ( γ ( h zv , H zv ) p ) = 0,where h zv is the analogous spectral quantity for the perturbed operator (equal to H zv when ǫ = 0). Such a uniform stability result directly implies almost-sure pure AC spectrumin the combinatorial case, by classical results. In our case we will have to work further(Section 5.3 and Appendix A.2).The important question now is which Herglotz function to choose for H zv . In the com-binatorial case it is natural to take ζ z ( b ), with b = ( v − , v ). We considered somethingclose in Proposition 3.2, namely S z ( L b ) ζ z ( b ). For the present continuity considerations, H zv = R + z ( o b ) √ z seems to behave better. Still, the function √ zS z ( L b ) ζ z ( b ) will also play an important role in fixing the disorder window later on (Appendix B). Of course it can beargued that all such quantities are related in Section 2, but one needs to be careful be-cause the aim is roughly to prove strict contraction estimates on E ( γ ( h zv , H zv ) p ) in terms of P w ∈N + v E ( γ ( h zw , H zw ) p ), so adding/multiplying terms to h zw is not a very good operation,although there are partial answers (Lemma 5.1 and Lemma 5.2).We have not discussed how the L p -continuity actually proceeds; we outline the proof in § § The two step expansion estimate.
Consider the hyperbolic disc D = { z ∈ C : | z | < } equipped with the usual hyperbolic distance metric d ( z, z ′ ) = cosh − (1 + δ ( z, z ′ )) , where δ ( z, z ′ ) = 2 | z − z ′ | (1 − | z | )(1 − | z ′ | ) , | z | , | z ′ | < . We will use the M¨obius transformation C ( z ) = z − i z +i that sends the upper half planemodel H isometrically to the disk model. Its inverse is C − ( u ) = i u − u . Note that if, for g, h ∈ H , we set(5.1) γ ( g, h ) = | g − h | Im g Im h , then(5.2) γ ( g, h ) = 2 δ ( C ( g ) , C ( h )) . In fact, δ ( C g, C h ) = 2 | ( g − i)( h + i) − ( g + i)( h − i) | ( | g + i | − | g − i | )( | h + i | − | h − i | ) = 12 γ ( g, h ) . The following is a more adequate replacement of [22, Lemma 1] to our framework.
Lemma 5.1.
Let K be a compact subset of the hyperbolic disc D . Then there exists acontinuous function C K : R + −→ R + , such that C K (0) = 0 , and δ ( λ z, λ z ′ ) ≤ (cid:0) | λ | + C K ( | λ − λ | ) (cid:1) δ ( z, z ′ ) + C K ( | λ − λ | ) for all z ∈ K and for all z ′ ∈ D , for all λ i ∈ C such that | λ i | ≤ .More explicitly, if r K = max z ∈ K | z | < , we can take C K ( t ) = − r K ) · t .Proof. First assume λ = 0. Suppose z = z ′ . We have δ ( λ z, λ z ′ ) = | λ | | z − λ λ − z ′ | | z − z ′ | · (1 − | z | )(1 − | z ′ | )(1 − | λ z | )(1 − | λ z ′ | ) · δ ( z, z ′ ) ≤ | λ | | z − λ λ − z ′ | | z − z ′ | · δ ( z, z ′ ) ≤ | λ | (cid:18) | − λ λ − | | z ′ || z − z ′ | (cid:19) δ ( z, z ′ ) , where, to obtain the first inequality, we used that | λ i | ≤
1. Let δ K = 1 − r K . If | z − z ′ | ≥ δ K ,we have δ ( λ z, λ z ′ ) ≤ (cid:18) | λ | + | λ − λ | h | z || z − z ′ | i(cid:19) δ ( z, z ′ ) ≤ (cid:0) | λ | + (1 + 2 r K δ − K ) | λ − λ | (cid:1) δ ( z, z ′ ) . (5.3) BSOLUTELY CONTINUOUS SPECTRUM FOR QUANTUM TREES 25
Now assume | z − z ′ | < δ K . Then | z ′ | < δ K + r K = r K , so δ ( λ z, λ z ′ ) ≤ | λ | ( | z − z ′ | + | − λ λ − | · | z ′ | ) (1 − | z | )(1 − | z ′ | )= | λ | δ ( z, z ′ ) + 2 2 | z − z ′ | · | λ || λ − λ | · | z ′ | + | λ − λ | · | z ′ | (1 − | z | )(1 − | z ′ | ) ≤ | λ | δ ( z, z ′ ) + 2 δ K (1+ r K )2 · | λ − λ | + ( r K ) | λ − λ | (1 − r K )(1 − ( r K ) ) . (5.4)The first assertion follows. For the explicit formula, we check the relevant terms in (5.3),(5.4) are bounded by 4 r K (1 − r K ) | λ − λ | . As r K <
1, this will complete the proof.Note that | λ − λ | ≤
2, so2(1 + 2 r K δ − K ) | λ || λ − λ | + (1 + 2 r K δ − K ) | λ − λ | ≤ n r K − r K + (cid:16) r K − r K (cid:17) o | λ − λ | = 4 1 + r K (1 − r K ) | λ − λ | . For the other term, using − x ≤ − x for x ∈ [0 , δ K (1+ r K )2 · | λ − λ | + ( r K ) | λ − λ | (1 − r K )(1 − ( r K ) ) ≤ − r K + (1 + r K ) (1 − r K )(1 − ( r K )) · | λ − λ | = 4 1 + r K (1 − r K ) | λ − λ | . Finally, if λ = 0, δ ( λ z, λ z ′ ) = 2 | λ z ′ | −| λ z ′ | ≤ | λ | −| z ′ | . If | z − z ′ | ≥ δ K then 1 ≤ | z − z ′ | (1 − r K ) and the claim follows. Otherwise | z ′ | < r K , so −| z ′ | ≤ − r K , proving the claim. (cid:3) We will also need [20, Lemma 2.16], which we recall below:
Lemma 5.2.
For any g, h, z ∈ H , max { γ ( g, h + z ) , γ ( g + z, h ) } ≤ (1 + c g ( z )) γ ( g, h ) + c g ( z ) , where c g ( z ) = | z | Im g + | z | (Im g ) . We now consider T in the twisted view. If T is a tree with parameters ( { α v } , { L v } ) beforeperturbation and ( { α ωv } , { L ωv } ) after perturbation, and if b = ( v − , v ) ∈ B ( T ) = B ( T + o ),we set h zv = R + z,ω ( o b ) √ z and H zv = R + z, ( o b ) √ z , where R + z,ω ( o b ) is the WT function of ( T , { L ωv } , { α ωv } ) and R + z, ( o b ) the WT function of( T , { L v } , { α v } ). The notation may seem a bit confusing since the WT function is evaluatedat v − instead of v . However, this is in accordance with the notations of § φ v instead of φ b if v = t b .We also define for b = ( v − , v ),g zv = R + z,ω ( L ωb ) √ z and Γ zv = R + z, ( L b ) √ z . Then the δ -conditions (1.3) applied to V + z ; o ( x ) give(5.5) X v + ∈N + v h zv + = g zv + α ωv √ z and X v + ∈N + v H zv + = Γ zv + α v √ z . We shall assume the coupling constants α v ≥ W ≥
0. In this case wecan ensure that h zv and H zv are Herglotz functions (Lemma 2.4), so their Cayley transformlies in D . In this case, g zv and Γ zv are also Herglotz by (5.5). Remark 5.3.
Assume there is no potential on the edges: W v ≡
0. Then the functions h zv and g zv are also connected by the following relations: if b = ( v − , v ) and we expand V + z ; o ( x v )in the basis C z ( x v ) = cos √ zx v and S z ( x v ) = sin √ zx v √ z , we get R + z ( t b ) = R + z ( o b ) S ′ z ( L b ) + C ′ z ( L b ) R + z ( o b ) S z ( L b ) + C z ( L b ) = R + z ( o b ) cos √ zL b − √ z sin √ zL b R + z ( o b ) sin √ zL b √ z + cos √ zL b , so g zv = h zv cos √ zL v − sin √ zL v h zv sin √ zL v +cos √ zL v . From this, we find C (g zv ) = g zv − ig zv + i = C ( h zv ) · cos √ zL v − i sin √ zL v cos √ zL v + i sin √ zL v = e − √ zL v C ( h zv )as previously observed in [1]. We also remark that we can invert the M¨obius identitydefining g zv in terms of h zv to get h zv = g zv cos √ zL v +sin √ zL v − g zv sin √ zL v +cos √ zL v .We finally define γ v ( h ) = γ ( h zv , H zv ) . The following plays the analog of [22, Lemma 2].
Lemma 5.4.
Let K be a compact subset of the hyperbolic disc, and assume that z variesin a compact subset such that H zv ∈ H and C (Γ zv ) ∈ K for all v . Then (see equations (5.6) , (5.7) and (5.8) below for the definition of the quantities q , α and Q ) γ v ( h ) ≤ (1+ C K ( ǫ ′ ))(1+ c H ( ǫ )) X v ′ ∈N + v Im H zv ′ · γ v ′ ( h ) P u ∈N + v Im H zu (cid:16) X w ∈N + v q w ( h ) Q v ′ ,w ( h ) cos α v ′ ,w ( h ) (cid:17) + 2 C K ( ǫ ′ ) + (1 + C K ( ǫ ′ )) c H ( ǫ ) , if sup z,v | α ωv − α v √ z | ≤ ǫ and sup z,v | e √ zL ωv − e √ zL v | ≤ ǫ ′ , where C K ( t ) = − r K ) · t and c H ( t ) = sup z,v t P v + ∈N + v Im H zv + t P v + ∈N + v Im H zv + ! . We will apply this when K = S v {C (Γ zv ) : z ∈ I +i[0 , } , where I is a compact interval onwhich Γ λ +i0 v exists and Im Γ λ +i0 v >
0. Note that the union here is finite as the unperturbedmodel is of finite cone type. For the same reason, the supremum in c H is a maximum.The quantities q, Q, α are defined by formulas similar to those in [22]: for x, y ∈ N + v ,(5.6) q y ( h ) = Im h y P u ∈N + v Im h u (5.7) cos α x,y ( h ) = cos arg( h x − H zx )( h y − H zy )(5.8) Q x,y ( h ) = q Im h x Im h y Im H zx Im H zy γ x ( h ) γ y ( h ) (Im h x Im H zy γ y ( h ) + Im h y Im H zx γ x ( h )) . assuming h x = H zx and h y = H zy , otherwise we let Q x,y ( h ) = cos α x,y ( h ) = 0. BSOLUTELY CONTINUOUS SPECTRUM FOR QUANTUM TREES 27
Proof of Lemma 5.4.
Using (5.2) and Remark 5.3, we have γ v ( h ) = 2 δ ( C ( H zv ) , C ( h zv )) = 2 δ (cid:16) e √ zL v C (Γ zv ) , e √ zL ωv C (g zv ) (cid:17) . By hypothesis, C (Γ zv ) stays in K . Applying Lemma 5.1 and (5.2), we deduce that(5.9) γ v ( h ) ≤ (1 + C K ( ǫ ′ )) γ (Γ zv , g zv ) + 2 C K ( ǫ ′ ) . Suppose α ωv ≥ α v . Using (5.5), we need to estimate(5.10) γ X v + ∈N + v H zv + − α v √ z , X v + ∈N + v h zv + − α ωv √ z ≤ γ X v + ∈N + v H zv + , X v + ∈N + v h zv + − α ωv − α v √ z , where the inequality holds by γ ( ξ + z, ξ ′ + z ) ≤ γ ( ξ, ξ ′ ) if Im z ≥
0, as easily checked. Sincewe are assuming α ωv ≥ α v ≥
0, we have − α ωv − α v √ z ∈ H . Using Lemma 5.2, we may drop theterm α ωv − α v √ z . If α v ≥ α ωv , we simply replace (5.10) by γ X v + ∈N + v H zv + − α v √ z , X v + ∈N + v h zv + − α ωv √ z ≤ γ X v + ∈N + v H zv + − α v − α ωv √ z , X v + ∈N + v h zv + and remove α v − α ωv √ z using Lemma 5.2. Finally, as calculated in [22, Lemma 2], (cid:12)(cid:12)(cid:12) X v + ∈N + v H zv + − X v + ∈N + v h zv + (cid:12)(cid:12)(cid:12) = X v ′ ∈N + v Im H zv ′ · γ v ′ ( h ) (cid:16) X w ∈N + v Im h zw Q v ′ ,w ( h ) cos α v ′ ,w ( h ) (cid:17) . Dividing by (Im P v + H v + )(Im P v + h v + ) completes the proof. (cid:3) Remark 5.5.
Note that P v + q v + = 1. On the other hand, Q v ′ ,w is a quotient of ageometric and arithmetic mean, so 0 ≤ Q v ′ ,w ≤
1. Since − ≤ cos α v ′ ,w ≤
1, this showsthat − ≤ P w ∈N + v q w Q v ′ ,w cos α v ′ ,w ≤ (C2) . Inother words, for each vertex v , there is a vertex v ′ ∈ N + v such that every label found in N + v can also be found in N + v ′ . We then say that v ′ is chosen w.r.t. (C2) .From now on, we denote S v = N + v . If ∗ = t b o and ∗ ′ is the vertex chosen w.r.t. (C2) corresponding to ∗ , we let S ∗ , ∗ ′ = (cid:0) N + ∗ \ {∗ ′ } (cid:1) ∪ N + ∗ ′ . Given x ∈ S ∗ , let(5.11) c x ( h ) = X y ∈ S ∗ q y ( h ) Q x,y ( h ) cos α x,y ( h )and for x ∈ S ∗ ′ , let c x ( h ) = (cid:16) X y ∈ S ∗ q y ( h ) Q ∗ ′ ,y ( h ) cos α ∗ ′ ,y ( h ) (cid:17)(cid:16) X y ∈ S ∗′ q y ( h ) Q x,y ( h ) cos α x,y ( h ) (cid:17) (5.12) = c ∗ ′ ( h ) X y ∈ S ∗′ q y ( h ) Q x,y ( h ) cos α x,y ( h ) . It follows from Remark 5.5 that c x ( h ) ∈ [ − , x ∈ S ∗ ,(5.13) p x = Im H zx P y ∈ S ∗ Im H zy , and for x ∈ S ∗ ′ ,(5.14) p x = Im H z ∗ ′ Im H zx ( P y ∈ S ∗ Im H zy )( P u ∈ S ∗′ Im H zu ) = p ∗ ′ · Im H zx P u ∈ S ∗′ Im H zu . Then P x ∈ S ∗ , ∗′ p x = 1. Note that c x ( h ) is a quantity that depends on the random param-eters of the perturbed graph, whereas p x is non-random.Recall definition (1.7) of the set Σ. Using (5.5), we also have Im Γ λ +i0 v > Proposition 5.6.
Let I ⊂ Σ be a compact interval. There exists a continuous function C I,H : [0 , ∞ ) → [0 , ∞ ) with C I,H (0 ,
0) = 0 such that if sup z ∈ I +i[0 , ,v | α ωv − α v √ z | ≤ ǫ and sup z ∈ I +i[0 , ,v | e √ zL ωv − e √ zL v | ≤ ǫ ′ , then γ ∗ ( h ) ≤ (1 + C I,H ( ǫ, ǫ ′ )) X x ∈ S ∗ , ∗′ p x c x ( h ) γ x ( h ) + C I,H ( ǫ, ǫ ′ ) . Proof.
Let K = S e {C (Γ ze ) : z ∈ I + i[0 , } . We apply Lemma 5.4 to v = ∗ , then to v = ∗ ′ . If c I,H ( ǫ, ǫ ′ ) = 2 C K ( ǫ ′ ) + c H ( ǫ ) + C K ( ǫ ′ ) c H ( ǫ ), the statement follows by taking C I,H ( ǫ, ǫ ′ ) = 2 c I,H ( ǫ, ǫ ′ ) + c I,H ( ǫ, ǫ ′ ) . (cid:3) To use this result under assumption (P0) , note that (cid:12)(cid:12)(cid:12) e √ zL ωv − e √ zL v (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) e i √ z ( L ωv + L v ) (cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12) e i √ z ( L ωv − L v ) − e i √ z ( L v − L ωv ) (cid:12)(cid:12)(cid:12) ≤ √ z ( L ωv − L v ) ≤ c I ǫ. Thus, | e √ zL ωv − e √ zL v | ≤ ǫ ′ , with ǫ ′ = 2 c I ǫ .5.2. L p continuity of the WT function. The aim of this subsection is to establish thefollowing uniform continuity result in L p -norm: Theorem 5.7.
Let T satisfy (C0), (C1), (C2) and ( α, L ) satisfy (P0) , (P1) and (P2) .For all compact I ⊂ Σ , I ∩ D = ∅ , and p > , there is ǫ ( I, p ) > , η ( I, ε D ) > and C p : [0 , ǫ ) → [0 , ∞ ) with lim ǫ → C p ( ǫ ) = 0 such that for any v ∈ T , ǫ ≤ ǫ , (5.15) sup z ∈ I +i(0 ,η ] E ( γ ( h zv , H zv ) p ) ≤ C p ( ǫ ) , sup z ∈ I +i(0 ,η ] E ( | h zv − H zv | p ) ≤ C p ( ǫ ) . Given the key results of § T = T + o is defined by a cone matrix M on a set of labels A satisfying (C0) , (C1) and (C2) . We previously denoted ∗ = t b o . More generally, given ( v − , v ) ∈ B ( T + o )with ℓ ( v ) = j , if we consider the subtree T +( v − ,v ) , we shall denote v = ∗ j . The set S ∗ j , ∗ ′ j isthen constructed analogously, and all results of § ∗ j = o b o ).Let us now discuss the proof of Theorem 5.7 in several steps: Step 1:
The Euclidean bound follows from the hyperbolic one.In fact, if the γ -bound is proved, then using the Cauchy-Schwarz inequality,(5.16) E ( | h zv − H zv | p ) ≤ E ( γ ( h zv , H zv ) p ) E ((Im h zv Im H zv ) p ) ≤ C ( ǫ ) E ( | h zv | p ) | H zv | p . To bound the moment E ( | h zv | p ), one uses the simple inequality(5.17) | ξ | ≤ γ ( ξ, ζ ) Im ζ + 2 | ζ | , ξ, ζ ∈ H , applied to ξ = h zv and ζ = H zv . BSOLUTELY CONTINUOUS SPECTRUM FOR QUANTUM TREES 29
Step 2:
To prove the γ -bound, it suffices to show that for each j ∈ A ,(5.18) E (cid:16) γ ( h z ∗ j , H z ∗ j ) p (cid:17) ≤ (1 + c ( ǫ ))(1 − δ ) X k ∈ A P j,k E ( γ ∗ k ( h z ) p ) + C ( ǫ ) , as long as P = ( P j,k ) forms a nonnegative irreducible matrix (and c ( ǫ ) , C ( ǫ ) → ǫ → u ∈ R A such that P ⊺ u = u . If we consider the vector E p γ := ( E ( γ [ h z ∗ j , H z ∗ j ] p )) j ∈ A , then (5.18)implies that h u, E p γ i C A ≤ (1 − δ ) h u, P E p γ i C A + C ( ǫ ) = (1 − δ ) h u, E p γ i C A + C ( ǫ ) , so h u, E p γ i C A ≤ C ( ǫ ) δ , and the γ -bound easily follows. Step 3:
To prove (5.18), we apply the two-step expansion (Proposition 5.6) to get(5.19) E (cid:16) γ ( h z ∗ j , H z ∗ j ) p (cid:17) ≤ (1 + C I,H ) p − E (cid:16)(cid:16) X x ∈ S ∗ j, ∗′ j p x c x ( h z ) γ x ( h z ) (cid:17) p (cid:17) + (1 + C I,H ) p − C I,H
The idea now is that for p ≥ (cid:16) X x ∈ S ∗ j, ∗′ j p x c x ( h ) γ x ( h ) (cid:17) p ≤ X x ∈ S ∗ j, ∗′ j p x γ x ( h ) p ≤ max x ∈ S ∗ j, ∗′ j γ ( H zx , h zx ) p as follows from Jensen’s inequality and the facts P p x = 1 and | c x ( h ) | ≤
1. However thefirst inequality is generally strict for p >
1, and this is what provides the (1 − δ ) in (5.18)if we choose P = P ( z ) to be the matrix P j,k := X x ∈ S ∗ j, ∗′ j ,ℓ ( x )= k p x for j, k ∈ A , which satisfies the requirements of Step 2.To derive the strict contraction for (5.20) more precisely, the authors in [22] introducean additional averaging over the permutations of S ∗ j , ∗ ′ j preserving the labels. LetΠ j := { π : S ∗ j , ∗ ′ j → S ∗ j , ∗ ′ j | π is bijective and ℓ ( π ( x )) = ℓ ( x ) for all x ∈ S ∗ j , ∗ ′ j } . Fix ∗ = ∗ j . Recall we denote functions φ b on [0 , L b ] by φ v if b = ( v − , v ). If v ∈ S ∗ , ∗ ′ and π ∈ Π, then φ πv := φ ( ∗ ,πv ) if πv ∈ S ∗ , and φ πv := φ ( ∗ ′ ,πv ) if πv ∈ S ∗ ′ .Denote H z = ( H zx ) x ∈ S ∗ , ∗′ ∈ C S ∗ , ∗′ . Given g ∈ H S ∗ , ∗′ , π ∈ Π, denote g ◦ π = ( g π ( x ) ) x ∈ S ∗ , ∗′ .By (2.16) and the symmetry of the tree, we get for the unperturbed WT function(5.21) H z = H z ◦ π for all π ∈ Π. By (P1) , (P2) and (2.16), h zx and h zy are independent and identicallydistributed for x, y ∈ S ∗ , ∗ ′ that carry the same label. Hence,(5.22) E (cid:0) f ( h z ) (cid:1) = E (cid:0) f ( h z ◦ π ) (cid:1) for any integrable function f and all π ∈ Π.Recall from Remark 5.3 that h zv = g zv cos √ zL v + sin √ zL v − g zv sin √ zL v + cos √ zL v . On the other hand, g zv = P v + h zv + − α v √ z . Given g ∈ H S ∗ , ∗′ , it is therefore natural to define g ∗ ′ = g ∗ ′ ( z, α, L ) by(5.23) g ∗ ′ = φ ∗ ′ ( g ) cos √ zL + sin √ zL − φ ∗ ′ ( g ) sin √ zL + cos √ zL , φ ∗ ′ ( g ) = − α √ z + X x ∈ S ∗′ g x . In particular, for g = H z = ( H zx ), we get ( H z ) ∗ ′ ( z, α , L ) = H z ∗ ′ .Introduce the notation g ( π ) x = g π ( x ) for x ∈ S ∗ , ∗ ′ . We also define(5.24) g ( π ) ∗ ′ = ( g π ) ∗ ′ = φ ∗ ′ ( g ( π ) ) cos √ zL + sin √ zL − φ ∗ ′ ( g ( π ) ) sin √ zL + cos √ zL . Note that g ( π ) ∗ ′ = g π ∗ ′ , in fact π ∗ ′ is not even defined. On the other hand, H ( π ) ∗ ′ ( z, α ∗ ′ , L ∗ ′ ) = H ∗ ′ by (5.21).Given g ∈ H S ∗ , ∗′ , π ∈ Π and g ∗ ′ ( z, α, L ) as in (5.23)–(5.24), we denote γ ( π ) x ( g ) = γ x ( g ( π ) ) = (cid:26) γ ( g π ( x ) , H zx ) : x ∈ S ∗ , ∗ ′ ,γ ( g ( π ) ∗ ′ , H z ∗ ′ ) : x = ∗ ′ . Then in view of (5.22), the mean in the RHS of (5.19) becomes(5.25) E (cid:16)(cid:16) X x ∈ S ∗ j, ∗′ j p x c x ( h ) γ x ( h ) (cid:17) p (cid:17) = 1 | Π j | E (cid:16) X π ∈ Π j (cid:16) X x ∈ S ∗ j, ∗′ j p x c ( π ) x ( h ) γ ( π ) x ( h ) (cid:17) p (cid:17) . To implement the strict inequality in (5.20), given p ≥ z ∈ H , α, L >
0, one defines κ ( p ) ∗ := κ ( p ) ∗ ( z, α, L, · ) : H S ∗ , ∗′ → R , by(5.26) κ ( p ) ∗ ( g ) := P π ∈ Π ( P x ∈ S ∗ , ∗′ p x c ( π ) x ( g ) γ ( π ) x ( g )) p P π ∈ Π P x ∈ S ∗ , ∗′ p x γ ( π ) x ( g ) p . Here, κ ( p ) ∗ depends on α, L via c ( π ) x ( g ) := c x ( g ( π ) ). Indeed , recall that c y ( g ) involves c ∗ ′ ( g )for y ∈ S ∗ ′ , and c ∗ ′ ( g ) involves g ∗ ′ , which in turns depends on α, L .The RHS of (5.25) now becomes(5.27) 1 | Π j | E (cid:16) κ ( p ) ∗ j (cid:0) z, α ω ∗ ′ j , L ω ∗ ′ j , h (cid:1) X π ∈ Π j X x ∈ S ∗ j, ∗′ j p x γ ( π ) x ( h ) p (cid:17) . If all γ ( π ) x ( h ) ≈ h / ∈ B r ( H ), where B r ( H ) := { g ∈ H S ∗ j, ∗′ j : γ ( g x , H zx ) ≤ r ∀ x ∈ S ∗ j , ∗ ′ j } . This case is controlled by Proposition 5.8 below. To state it, recall that we assume I isaway from the Dirichlet spectrum in (1.8). This implies, for the L in κ ( p ) ∗ ,(5.28) min z ∈ I +i[0 , | sin (cid:0) √ zL (cid:1) sin (cid:0) √ zL ∗ ′ (cid:1) | ≥ ε D > . Proposition 5.8.
Let I ⊂ Σ be compact and satisfy (1.8) and let p > . For any ∗ = ∗ j ,there exist δ ∗ ( I, p ) > , ǫ ∗ ( I, ε D ) > , η ∗ ( I, ε D ) > and R ∗ : [0 , ǫ ∗ ] → [0 , ∞ ) with lim ǫ → R ∗ ( ǫ ) = 0 such that for all ǫ ∈ [0 , ǫ ∗ ] , sup z ∈ I +i[0 ,η ∗ ] sup | L − L ∗′ |≤ ǫ sup | α − α ∗′ |≤ ǫ sup g ∈ H S ∗ , ∗′ \ B R ∗ ( ǫ ) ( H ) κ ( p ) ∗ ( z, α, L, g ) ≤ − δ ∗ . Proposition 5.8 is remarkable as it gives a uniform contraction estimate on the randomvariable. Using it in (5.27), we get the (1 − δ ) we were seeking in (5.18) (take δ =min j ∈ A δ ∗ j ), thus completing the proof of Theorem 5.7.The proof of Proposition 5.8 is given in Appendix B. Note that c ( π ) x ( g ) = c πx ( g ) in general. For example, if x ∈ S ∗ and πx ∈ S ∗ ′ , then c ( π ) x ( g ) is defined bya single sum while c πx ( g ) is a product of two sums, as in (5.11)–(5.12). BSOLUTELY CONTINUOUS SPECTRUM FOR QUANTUM TREES 31
Proof of pure AC spectrum.
We may finally conclude with the proof of Theo-rem 1.7, which we recall:
Theorem 5.9.
Let T satisfy (C0), (C1), (C2) . For all compact I ⊂ Σ , I ∩ D = ∅ , and p > , there is ǫ ( I, p ) > , η ( I, ε D ) > , C ′ p ( I, ε D ) > such that for any v ∈ T , (5.29) sup z ∈ I +i(0 ,η ] E ( | Im R ± z ( v ) | − p ) ≤ C ′ p and sup z ∈ I +i(0 ,η ] E ( | R ± z ( v ) | ± p ) ≤ C ′ p for all ǫ ∈ [0 , ǫ ) and all ( α ω , L ω ) satisfying (P0) , (P1) and (P2) .In particular, H ωǫ has pure AC spectrum almost surely in I .Proof. We showed after (5.16) that sup z ∈ I +i(0 ,η ] E ( | h zx | p ) < ∞ . So for b = ( v − , v ), E (cid:0) | R + z,ω ( v ) | p (cid:1) = | z | p/ E ( | h zv | p ) ≤ C p for any z ∈ I + i(0 , η ]. On the other hand, observe that γ (cid:18) −
1i Im h zv , −
1i Im H zv (cid:19) = γ (i Im h zv , i Im H zv ) ≤ γ ( h zv , H zv ) . In particular, using (5.17) with ξ = i / Im h zv and ζ = i / Im H zv , we get (cid:12)(cid:12)(cid:12)(cid:12) h zv (cid:12)(cid:12)(cid:12)(cid:12) p ≤ p − (cid:26) p γ ( h zv , H zv ) p (Im H zv ) p + 2 p (cid:12)(cid:12)(cid:12)(cid:12) H zv (cid:12)(cid:12)(cid:12)(cid:12) p (cid:27) . Hence, E (cid:0) | Im h zv | − p (cid:1) ≤ C ′ p for any z ∈ I + i(0 , η ], by (5.15). In particular, E (cid:0) | R + z,ω ( v ) | − p (cid:1) = | z | − p/ E (cid:0) | h zv | − p (cid:1) ≤ C ′′ p . On the other hand, for any b , R − z ( t b ) = R + z ( o b b ), and we know all results hold true for thetree T − t b = T + o b b (recall we are assuming (C1) ). So we also have E (cid:0) | R − z,ω ( v ) | p (cid:1) < ∞ forany p ∈ Z .Finally, for (Im R + z ( v )) − p , denote √ z = a + i b . ThenIm h zv = a Im R + z ( v ) − b Re R + z ( v ) | z | =: F z | z | . Hence, E (cid:16) R + z ( v ) p (cid:17) = a p E (cid:16) | F z + b Re R + z ( v ) | p (cid:17) = a p E (cid:16) | F z + b Re R + z ( v ) | p {| F z | > | b Re R + z ( v ) |} ( ω ) (cid:17) + a p E (cid:16) | F z + b Re R + z ( v ) | p {| F z |≤| b Re R + z ( v ) |} ( ω ) (cid:17) . For the first expectation, we have | F z + b Re R + z ( v ) | ≥ | F z | − b | Re R + z ( v ) | ≥ | F z | . Forthe second expectation, we make use of the lower-bound for Im R + z ( v ) / Im z proved in theAppendix (Lemma A.4) to conclude that there is some C > | F z + b Re R + z ( v ) | = a Im R + z ( v ) ≥ aC Im z for any z, v, ω . Summarizing, we get E (cid:0) Im R + z ( v ) − p (cid:1) ≤ (2 a ) p E (cid:0) | F z | − p (cid:1) + C − p (Im z ) − p P (cid:0) | F z | ≤ | b Re R + z ( v ) | (cid:1) ≤ (2 a ) p | z | − p E (cid:0) (Im h zv ) − p (cid:1) + C − p | z | − p (Im z ) − p (2 b ) p E (cid:18)(cid:12)(cid:12)(cid:12)(cid:12) Re R + z ( v )Im h zv (cid:12)(cid:12)(cid:12)(cid:12) p (cid:19) , using Markov’s inequality. The right-hand side is bounded uniformly in z ∈ I + i(0 , η ],since b Im z = √ z Im z = √ z . This completes the proof of (5.29).The consequence on pure AC spectrum follows from Theorem A.6. (cid:3) Uniform inverse moments.
We now aim to prove Theorem 1.8. The arguments ofthis section are inspired from their analogs [2, 5] in case of regular combinatorial graphs.Note that (P0) implies that the distributions ν j of α ωv have compact support supp ν j ⊆ [ α min − ǫ, α max + ǫ ]. In particular, all moments M ς = max v E [ | α ωv | ς ] < ∞ exist.As in Section 3, we denote R + z ( k ) := R + z ( o b ) if ℓ ( t b ) = k . Note that here R + z ( o b )is random; even if ℓ ( t b ) = ℓ ( t b ′ ) this doesn’t imply that R + z ( o b ) and R + z ( o b ′ ) are equal.However their distribution are the same, which justifies the notation for the purposes ofthis section. We also let Q = max v ∈ T deg( v ) − q = min v ∈ T deg( v ) − I ⊂ Σ be compact with I ∩ D = ∅ . Fix η ( I, ε D ) as in Theorem 5.9. Given δ ∈ (0 , ǫ >
0, we introduce(5.30) σ ǫac ( δ ) = { λ ∈ R : P (Im R + λ +i η ( j ) > δ ) > δ ∀ j ∈ A , ∀ η ∈ (0 , η ) } . Lemma 5.10.
There exist ǫ , δ > such that I ⊆ σ ǫac ( δ ) for ǫ ≤ ǫ .Proof. By Theorem 5.9, fixing p = 2 we may find ǫ ( I ) and C such that E ( | Im R ± z ( j ) | − ) ≤ C so by Jensen E (Im R ± z ( j )) ≥ E ( | Im R ± z ( j ) | − ) − / ≥ C − / > P (Im R + λ +i η ( j ) > δ ) ≤ δ for some λ ∈ I , j ∈ A and η ∈ (0 , η ). Then E h Im R + λ +i η ( j ) i = E h Im R + λ +i η ( j )1 Im R + λ +i η ( j ) >δ i + E h Im R + λ +i η ( j )1 Im R + λ +i η ( j ) ≤ δ i ≤ ( C δ ) / + δ < C − / if δ > E [ | Im R + λ +i η ( j ) | ] ≤ C . Hence, I ⊆ σ ǫac ( δ ) for some δ > (cid:3) Given z ∈ C + and j ∈ A , let F jz ( x ) = P (cid:0) Im R + z ( j ) ≤ x (cid:1) , F z ( x ) = max j ∈ A F jz ( x ) and H jz ( x ) = P (cid:0) | ζ zj | ≤ x (cid:1) . Lemma 5.11.
Let I ⊂ R be compact, I ∩ D = ∅ , ς ≥ , λ ∈ I , η > and z = λ + i η .Then F z ( x ) ≤ F z ( xy − ) q + C ν,Q,ς (cid:18) y β F z (cid:16) Qc c y (cid:17) q + y ς (cid:19) for all x > and y ∈ (0 , c I ] , where β is from (P3) , c I = c Qc c and c , c , c > arechosen as in (2.18) .Proof. By (2.17) we have(5.31) Im R + z ( o b ) ≥ | ζ z ( b ) | X b + ∈N + b Im R + z ( o b + ) . So given y >
0, if Im R + z ( o b ) ≤ x , then either | ζ z ( b ) | ≤ y , or Im R + z ( o b + ) ≤ xy − for all b + ∈ N + b . Recalling the matrix M = ( M i,j ) we thus have by independence, F jz ( x ) ≤ F z ( xy − ) M j, · · · F mz ( xy − ) M j,m + H jz ( y ) ≤ F z ( xy − ) q + H jz ( y ) . BSOLUTELY CONTINUOUS SPECTRUM FOR QUANTUM TREES 33
Now, assuming c ≤ | S z ( L b ) | ≤ c and | C z ( L b ) | ≤ c Qc y ≤ y we have by (2.6) H jz ( y ) = P (cid:16)(cid:12)(cid:12)(cid:12) S z ( L b ) (cid:16) α t b + S ′ z ( L b ) S z ( L b ) − X b + ∈N + b R + z ( o b + ) (cid:17)(cid:12)(cid:12)(cid:12) ≥ y − (cid:17) ≤ P (cid:16) | α t b S z ( L b ) | + | C z ( L b ) | + | S z ( L b ) | X b + ∈N + b | R + z ( o b + ) | ≥ y − (cid:17) ≤ P (cid:16) c | α t b | + (4 y ) − + c X b + ∈N + b | R + z ( o b + ) | ≥ y − (cid:17) ≤ P (cid:0) | α t b | ≥ (4 c y ) − (cid:1) + P (cid:16) X b + ∈N + b | R + z ( o b + ) | ≥ (2 c y ) − (cid:17) ≤ (4 c y ) ς M ς + m X k =1 M j,k P (cid:0) | R + z ( k ) | ≥ (2 Qc y ) − (cid:1) , where M ς = max v E [ | α ωv | ς ] < ∞ . But for t = Qc y ,(5.32) P (cid:0) | R + z ( o b ) | ≥ t (cid:1) = P (cid:18) | ζ z ( b ) − C z ( L b ) || S z ( L b ) | ≥ t (cid:19) ≤ P (cid:18) | ζ z ( b ) | + ( c / Qc y ) c ≥ t (cid:19) ≤ P (cid:16) | ζ z ( b ) | ≥ c t (cid:17) ≤ P (cid:16) | ζ z ( b ) S z ( L b ) | ≥ c t (cid:17) ≤ P (cid:16)(cid:12)(cid:12) α t b + Re S ′ z ( L b ) S z ( L b ) − X b + ∈N + b Re R + z ( o b + ) (cid:12)(cid:12) ≤ c t and X b + ∈N + b Im R + z ( o b + ) ≤ c t (cid:17) ≤ P (cid:16) α t b lies in an interval of length 4 c t and X b + ∈N + b Im R + z ( o b + ) ≤ c t (cid:17) ≤ C ν (cid:16) c t (cid:17) β P (cid:16) X b + ∈N + b Im R + z ( o b + ) ≤ c t (cid:17) ≤ C ν (cid:16) c t (cid:17) β F z (cid:18) Qc yc (cid:19) q . Here we used that − S ′ z ( L ) S z ( L ) is Herglotz and that { R + z ( o b + ) } are independent of α t b , as followsfrom (2.16), and bounded the probability by first conditioning over the random variablesdifferent from α ωt b , so that the “interval” above is fixed.Thus, if c I = c qc c , where | C z ( L b ) | ≤ c , then for any 0 < y ≤ c I ≤ c qc | C z ( L b ) | , H z ( y ) ≤ (4 c ) ς M ς · y ς + C ν (cid:16) Qc c (cid:17) β Q · y β · F z (cid:16) Qc yc (cid:17) q . (cid:3) Theorem 5.12.
Let I ⊂ σ ǫac ( δ ) be bounded, λ ∈ I , ς ≥ , η ∈ (0 , η ) and z = λ + i η .Then F z ( x ) ≤ C ν,Q,δ,ς x βς/ for all x ∈ (0 , x δ ] , for some C ν,Q,δ < ∞ and x δ = x ( δ, ν, Q, ς, I ) > .Proof. The theorem is proved by gradually improving on the decay. First, take y = x / in Lemma 5.11 to get for x ∈ (0 , c I ], F z ( x ) ≤ F z ( x / ) q + C ν,Q (cid:16) x β/ F z (cid:16) Qc c x / (cid:17) q + x ς/ (cid:17) , so using the bounds x β/ [ F z ( Qc c x / )] q ≤ x β/ and x ς/ ≤ x β/ for small x , we get(5.33) F z ( x ) ≤ F z ( x ) q + C ′ ν,Q x β/ . Since λ ∈ σ ǫac ( δ ), we have F z ( x ) = max j P (Im R + z ( j ) ≤ δ ) ≤ − δ for any x ∈ [0 , δ ]. Choose x ≤ min( δ, c I , ( δ C ′ ν,Q ) /β ). For α small enough, we have (1 − δ ) ≤ (1 − δ ) x α . So there issome α ∈ (0 , β/
16] such that F z ( x ) ≤ (cid:16) − δ (cid:17) x α . Now define recursively x n = x n − . We show by induction that F ( x n ) ≤ (1 − δ ) x α n . Weknow this for n = 0. Next, for n ≥
1, using (5.33), induction and q ≥
2, we have F z ( x n ) ≤ F z ( x n − ) + Cx β/ n ≤ (cid:16) − δ (cid:17) x α n + x α n · Cx β/ − α ≤ h(cid:16) − δ (cid:17) + Cx β/ i x α n ≤ (cid:16) − δ (cid:17) x α n , where we used x n ≤ x , 2 α ∈ (0 , β/
8] and x ≤ ( δ C ) /β . Hence, if x ∈ (0 , x ], so that x ∈ ( x n +1 , x n ] for some n , using that F z is monotone increasing, we get F z ( x ) ≤ F z ( x n ) ≤ (cid:16) − δ (cid:17) x α n +1 ≤ (cid:16) − δ (cid:17) x α ≤ x α . This proves a first power decay which we now improve. Suppose we have F z ( x ) ≤ cx α for some c > x ∈ (0 , x ]. Taking y = x ς ς in Lemma 5.11, we get F z ( x ) ≤ F z ( x ς ς ) ) q + Cx β ς ς F z (cid:16) Qc c x ς ς (cid:17) q + Cx ς ς ≤ cx q ς ς α + ˜ Cx ςς +1 ( β + qα ) + Cx ς ς . If q ≥
3, then q ς ς ≥ ς ς >
1. Also, if α ≤ βς , then β + qα ς ς ≥ β +3 α ς ς > α , since βς + 3 ας > α + 3 ας . Finally, ς ς ) ≥ ≥ β , since ς ≥
3, so ς ς ) ≥ βς . We thus showedthat if q ≥
3, then the decay power is strictly increased as long as α ≤ βς . Iteration thusproves the theorem when q ≥ x βς/ ≤ x βς/ ).The rest of the proof is devoted to the case q = 2, where we need to improve again.Using (2.17) twice, we haveIm R + z ( o b ) ≥ | ζ z ( b ) | X b + | ζ z ( b + ) | X b ++ Im R + z ( o b ++ ) ≥ X b + c − | ζ z ( b + ) | ( | α t b | + | S ′ z ( L b ) S z ( L b ) | + P b + | ζ z ( b + ) || S z ( L b + ) | + P b + | C z ( L b + ) S z ( L b + ) | ) X b ++ Im R + z ( o b ++ ) ≥ X b + c c − | ζ z ( b + ) | ( c | α t b | + | C z ( L b ) | + P b + | ζ z ( b + ) | + P b + | C z ( L b + ) | ) X b ++ Im R + z ( o b ++ ) . Define the events E = {| α t b | ≤ y − } , E = {| ζ z ( b + ) | > y − for at least two b + } ,E = {| ζ z ( b +0 ) | > y − for exactly one b +0 } ,E = {| ζ z ( b + ) | ≤ y − for all b + } . Using an estimate from (5.32), we have P ( E ) ≤ Q ( Q + 1)2 (cid:2) P ( | ζ z ( b + ) | > y − ) (cid:3) ≤ c ν,Q (cid:16) c y − (cid:17) β F z (cid:16) c y − (cid:17) . The exponent of the first term is increased by ς at each step, the second one by at least α ς ) , asfor the last one, it already has the required decay. After finitely many steps, the exponent thus reaches βς . BSOLUTELY CONTINUOUS SPECTRUM FOR QUANTUM TREES 35
For E ∩ E , | ζ z ( b + ) | ≤ y − for all b + = b +0 , so we have | ζ z ( b +0 ) | c | α t b | + P t b ′ ∼ t b | C z ( L b ′ ) | + P b + | ζ z ( b + ) | = 11 + c | α tb | + P tb ′ ∼ tb | C z ( L b ′ ) | + P b + = b +0 | ζ z ( b + ) || ζ z ( b +0 ) | ≥ Q + c + y ( Q + 1) c , since c | α tb | + P tb ′ ∼ tb | C z ( L b ′ ) | + P b + = b +0 | ζ z ( b + ) || ζ z ( b +0 ) | ≤ c y − +( Q +1) c + Qy − y − . Assuming y ≤
1, theRHS is ≥ c . Hence, for ˜ c = c c − c we have P (cid:0) { Im R + z ( o b ) ≤ x } ∩ E ∩ E (cid:1) ≤ P (cid:16) ˜ c X b ++0 ∈N b +0 Im R + z ( o b ++0 ) ≤ x (cid:17) ≤ F z (˜ c − x ) . Finally, for E ∩ E , we use (5.31) along with | ζ z ( b ) | ≥ c c − ( Q + c ) y − +( Q +1) c . If y is small,this is ≥ Cy , so we get P (cid:0) { Im R + z ( o b ) ≤ x } ∩ E ∩ E (cid:1) ≤ P (cid:16) C y X b + Im R + z ( o b + ) ≤ x (cid:17) ≤ F z ( C − xy − ) . Estimating P ( E c ) ≤ y ς M ς by Chebyshev, we thus showed that F z ( x ) ≤ Cy β F z ( c − y ) + F z (˜ c − x ) + F z ( C − xy − ) + c ′ y ς . Assuming we showed that F z ( x ) ≤ c x α , then choosing y = x ς ς , we get F z ( x ) ≤ ˜ Cx βς ς x ας ς + ˜ cx α + ˜ C ′ x ς ς (2 α ) + c ′ x ς ς . To get βς +4 ας ς > α we must have 2 βς + 4 ας > α + 4 ας , i.e. α < βς , so α ≤ βς suffices.Next, ς ς >
1. Finally, ς ς ≥ ≥ β since ς ≥
3. We thus showed the decay exponentcan be strictly improved up to βς . (cid:3) Proof of Theorem 1.8.
Given p ≥
1, choose ς such that p < βς . By Lemma 5.10, we have I ⊆ σ ǫac ( δ ) for some δ > ǫ ≤ ǫ . As p ≥
1, given λ ∈ I , we have by the layer-cakerepresentation, E (cid:0) | Im R + z ( o b ) | − p (cid:1) = p Z ∞ t p − P (cid:0) | Im R + z ( o b ) | − ≥ t (cid:1) d t = p Z ∞ t p − F z ( t − ) d t . By Theorem 5.12, denoting t δ = x − δ , we know F z ( t − ) ≤ C ν,Q t − βς/ for all t ≥ t δ . Hence E (cid:0) | Im R + z ( o b ) | − p (cid:1) ≤ p Z t δ t p − d t + pC Z ∞ t δ t p − − βς/ d t = t pδ + 5 pβς − p Ct βς/ − pδ . We may assume t δ ≥ x δ if necessary. Since this holds for any λ ∈ I and η ∈ (0 , η ), we getsup λ ∈ I sup η ∈ (0 ,η ) E (cid:0) | Im R + z ( o b ) | − p (cid:1) ≤ t pδ (cid:16) pβς − p C ν,Q (cid:17) . (cid:3) Appendix A. Proofs of some technical facts
A.1.
General results.
Proof of Lemma 2.1.
Given f ∈ L ( T ), define G z f := R T G z ( x, y ) f ( y ) d y . We shouldshow that G z f ∈ D ( H T ) and ( H T − z ) G z f = f .Assume f is continuous on T and supported in a ball Λ ⊂ T . Let x = ( b, x b ) ∈ T , with x b ∈ (0 , L b ). Fix o, v ∈ T such that b ∈ B ( T + o ) ∩ B ( T − v ) and Λ ⊂ T + o ∩ T − v . By definition,( G z f )( x ) = V + z ; o ( x ) Z T − x U − z ; v ( y ) W zv,o ( y ) f ( y ) d y + U − z ; v ( x ) Z T + x V + z ; o ( y ) W zv,o ( y ) f ( y ) d y . Hence,(A.1) ( G z f ) ′ ( x ) = ( V + z ; o ) ′ ( x ) Z T − x U − z ; v ( y ) W zv,o ( y ) f ( y ) d y + ( U − z ; v ) ′ ( x ) Z T + x V + z ; o ( y ) W zv,o ( y ) f ( y ) d y , where the term V + z ; o ( x ) U − z ; v ( x − ) W zv,o ( x − ) f ( x − ) − U − z ; v ( x ) V + z ; o ( x + ) W zv,o ( x + ) f ( x + ) from Leibniz’s rule canceled, allfunctions being continuous. Next,( G z f ) ′′ ( x ) = ( V + z ; o ) ′′ ( x ) Z T − x U − z ; v ( y ) W zv,o ( y ) f ( y ) d y + ( U − z ; v ) ′′ ( x ) Z T + x V + z ; o ( y ) W zv,o ( y ) f ( y ) d y − f ( x ) , where we used that ( V + z ; o ) ′ ( x ) U − z ; v ( x − ) W zv,o ( x − ) f ( x − ) − ( U − z ; v ) ′ ( x ) V + z ; o ( x + ) W zv,o ( x + ) f ( x + ) = − f ( x ). Recalling that ψ ′′ = ( W − z ) ψ for ψ = V + z ; o , U − z ; v , we get ( G z f ) ′′ = ( W − z ) G z f − f , so ( H T − z ) G z f = f .For the boundary conditions, note that G z f ( x ) = R T G z ( x, y ) f ( y ) d y and ( G z f ) ′ ( x ) = R T ∂ x G z ( x, y ) f ( y ) d y by (A.1). So it suffices to check that x G z ( x, y ) satisfies the δ -conditions. But this follows immediately since V + z ; o ∈ D ( H max T + o ) and U − z ; v ∈ D ( H max T − v ).Finally, as H T is self-adjoint, we have by the spectral theorem k f k = k ( H T − z ) G z f k = R T | λ − z | d µ G z f ≥ | Im z | R T d µ G z f = | Im z | k G z f k , i.e. k G z f k ≤ z k f k . This holdson the subspace of continuous f of compact support. By the density of such functions, G z extends to a bounded operator on L ( T ) satisfying k G z k ≤ z .We proved that G z f ∈ D ( H T ) and ( H T − z ) G z f = f assuming f is continuous ofcompact support. For general f ∈ L ( T ), take a sequence ( f j ) of such functions with f j → L f . We showed G z f j ∈ W , ( T ) for each j ; this space being complete, we obtaina subsequence ( G z f j k ) converging in W , ( T ). The limit must be G z f since G z f isthe L limit of G z f j . It follows that ( H T − z ) G z f = f a.e., and for each x , we have G z f ( x ) = lim k G z f j k ( x ) and ( G z f ) ′ ( x ) = lim k ( G z f j k ) ′ ( x ), so we deduce that G z f satisfiesthe boundary conditions. (cid:3) Remark A.1.
Fix an edge b ∈ B ( T ) and choose o, v with b ∈ B ( T + o ) ∩ B ( T − v ). Then for x = ( b, x b ) , y = ( b, y b ) ∈ T , we can also express(A.2) G z T ( x, y ) = − φ − z ; b ( x ) φ + z ; b ( y ) R + z ( o b )+ R − z ( o b ) if y ∈ T + x , − φ − z ; b ( y ) φ + z ; b ( x ) R + z ( o b )+ R − z ( o b ) if y ∈ T − x , where φ − z ; b ( x ) = U − z ; v ( x ) U − z ; v ( o b ) and φ + z ; b ( x ) = V + z ; o ( x ) V + z ; o ( o b ) . This follows immediately from (2.1), (2.2),since the Wronskian W zv,o ( y ) = W zv,o ( b, y b ) is constant for y b ∈ [0 , L b ]. Note that(A.3) φ ± z ; b ( x ) = C z ( x b ) ± R ± z ( o b ) S z ( x b )for x = ( b, x b ) ∈ T . On the other hand, G z T ( y, x ) = G z T ( x, y ) BSOLUTELY CONTINUOUS SPECTRUM FOR QUANTUM TREES 37 for x = ( b, x b ) , y = ( b, y b ) ∈ T , since x ∈ T ± y ⇐⇒ y ∈ T ∓ x . This implies h f , G z f i = h f , G z f i for any real-valued f j supported on e ( b ). Hence, for any f = f + i f supported in e ,(A.4) h f, G z f i = h f , G z f i + h f , G z f i . Lemma A.2.
Fix b ∈ B ( T ) and suppose R ± λ ( o b ) := R ± λ +i0 ( o b ) exist and are not both zero.Then for any f supported in e ( b ) , (A.5) Im h f, G λ f i = Im R + λ ( o b ) g − f ( λ ) + Im R − λ ( o b ) g + f ( λ ) | R + λ ( o b ) + R − λ ( o b ) | , where g ± f ( λ ) = (cid:12)(cid:12)(cid:12) h f, Re φ ± λ ; b i L [0 ,L b ] (cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12) h f, Im φ ± λ ; b i L [0 ,L b ] (cid:12)(cid:12)(cid:12) . In particular, if Im R ± λ ( o b ) = 0 for all b ∈ B , then Im h f, G λ f i = 0 for all f ∈ L ( T ) . This lemma was stated without proof in [1, eq (A.15)].
Proof.
First note that it suffices to prove this for real-valued f . In fact, for f = f + i f ,we then use (A.4) and deduce the result, since g + f ( λ ) + g + f ( λ ) = g + f ( λ ) as easily checkedand similarly for g − f ( λ ). So assume f is real.Since b and λ are fixed, we shall denote R ± := R ± λ ( o b ) and φ ± := φ ± λ ; b , and drop b indices in x b and f b for simplicity.We have by (A.2), h f, G λ f i = − R + + R − Z L b f ( x ) (cid:16) Z x φ + ( x ) φ − ( y ) f ( y ) d y + Z L b x φ − ( x ) φ + ( y ) f ( y ) d y (cid:17) d x . But by (A.3), φ + ( x ) φ − ( y ) = C ( x ) C ( y ) − R − C ( x ) S ( y ) + R + S ( x ) C ( y ) − R + R − S ( x ) S ( y ) ,φ − ( x ) φ + ( y ) = C ( x ) C ( y ) + R + C ( x ) S ( y ) − R − S ( x ) C ( y ) − R + R − S ( x ) S ( y ) . Thus, φ − ( x ) φ + ( y ) = φ + ( x ) φ − ( y ) + ( R + + R − )( C ( x ) S ( y ) − S ( x ) C ( y )). Hence, h f, G λ f i = − R + + R − Z [0 ,L b ] f ( x ) f ( y ) φ + ( x ) φ − ( y ) d y d x − Z L b Z L b x f ( x ) f ( y ) ( C ( x ) S ( y ) − S ( x ) C ( y )) d y d x . Since f is real-valued, we getIm h f, G λ f i = Im( R + + R − ) | R + + R − | Z [0 ,L b ] f ( x ) f ( y ) Re[ φ + ( x ) φ − ( y )] d y d x − Re( R + + R − ) | R + + R − | Z [0 ,L b ] f ( x ) f ( y ) Im[ φ + ( x ) φ − ( y )] d y d x = 1 | R + + R − | (cid:8) Im( R + + R − )[ h f, Re φ + ih f, Re φ − i − h f, Im φ + ih f, Im φ − i ](A.6) − Re( R + + R − )[ h f, Im φ + ih f, Re φ − i + h f, Re φ + ih f, Im φ − i ] (cid:9) . Now Re φ ± = C ( x ) ± (Re R ± ) S ( x ) and Im φ ± = ( ± Im R ± ) S ( x ). Hence, the term incurly brackets isIm( R + + R − ) (cid:2) h f, C i + Re R + h f, S i (cid:3) (cid:2) h f, C i − Re R − h f, S i (cid:3) (A.7) − Im( R + + R − ) (cid:2) Im R + h f, S i (cid:3) (cid:2) − Im R − h f, S i (cid:3) − Re( R + + R − ) (cid:2) Im R + h f, S i (cid:3) (cid:2) h f, C i − Re R − h f, S i (cid:3) − Re( R + + R − ) (cid:2) h f, C i + Re R + h f, S i (cid:3) (cid:2) − Im R − h f, S i (cid:3) = h f, C i Im( R + + R − ) + h f, C ih f, S i (cid:2) (Im R + + Im R − )(Re R + − Re R − ) − (Re R + + Re R − )(Im R + − Im R − ) (cid:3) + h f, S i (cid:2) (Im R + + Im R − )(Im R + Im R − − Re R + Re R − )+ (Re R + + Re R − )(Im R + Re R − + Im R − Re R + ) (cid:3) = h f, C i Im( R + + R − ) + 2 h f, C ih f, S i (cid:2) Im R − Re R + − Im R + Re R − (cid:3) + h f, S i (cid:2) (Im R + ) Im R − + Im R + (Im R − ) + Im R − (Re R + ) + Im R + (Re R − ) (cid:3) . On the other hand,Im R + g − f + Im R − g + f = Im R + h(cid:0) h f, C i − Re R − h f, S i (cid:1) + (Im R − ) h f, S i i + Im R − h(cid:0) h f, C i + Re R + h f, S i (cid:1) + (Im R + ) h f, S i i , which is exactly the expression at the end of (A.7). This completes the proof of (A.5).Finally, if Im R ± λ ( o b ) = 0 for all b ∈ B , then Im h f b , G λ f b i = 0 for all f b supported in b , by(A.5). The same ideas show that Im h f b , G λ f b ′ i = 0 in this case. In fact, we don’t need to gothrough all the above calculations, just note that in (A.6), we get Im φ ± = ( ± Im R ± ) S = 0.It follows that Im h f, G λ f i = 0 for all f ∈ L ( T ). (cid:3) Proof of Lemma 2.4.
We may find an L solution e U − z ; v on T − v satisfying the δ -conditionsat vertices w / ∈ { o b , v } , the Neumann condition at o b and e U − z ; v ( v ) = 1. As in (2.1), we get G z T + ob ( x, y ) = e U − z ; v ( x ) V + z ; o ( y ) W zv,o,ob ( x ) if y ∈ T + x , e U − z ; v ( y ) V + z ; o ( x ) W zv,o,ob ( x ) if y ∈ T − x , where W zv,o,o b ( x ) = V + z ; o ( x )( e U − z ; v ) ′ ( x ) − ( V + z ; o ) ′ ( x ) e U − z ; v ( x ).In particular, G z T + ob ( o b ,o b ) = W zv,o,ob ( o b ) e U − z ; v ( o b ) V + z ; o ( o b ) = 0 − ( V z ; o ) ′ ( o b ) V z ; o ( o b ) = − R + z ( o b ).Similarly, we find e V + z ; o on T + o satisfying Neumann’s condition at t b , yielding G z T − tb ( x, y ) = U − z ; v ( x ) e V + z ; o ( y ) W zv,o,tb ( x ) if y ∈ T + x , U − z ; v ( y ) e V + z ; o ( x ) W zv,o,tb ( x ) if y ∈ T − x , so we get similarly R − z ( t b ) = − G z T − tb ( t b ,t b ) .To see the Herglotz property [17], note that if H max T ± u is as in §
2, then h V + z ; u , H max T + u V + z ; u i L ( T + u ) − hH max T + u V + z ; u , V + z ; u i L ( T + u ) = 2i Im z · k V + z ; u k L ( T + u ) . On the other hand, the left-hand side can also be computed by integration by parts onevery edge. All the boundary terms except the one at u cancel thanks to the self-adjoint BSOLUTELY CONTINUOUS SPECTRUM FOR QUANTUM TREES 39 conditions. We thus obtain h V + z ; u , H max T + u V + z ; u i L ( T + u ) − hH max T + u V + z ; u , V + z ; u i L ( T + u ) = V + z ; u ( u )( V + z ; u ) ′ ( u ) − ( V + z ; u ) ′ ( u ) V + z ; u ( u ) . Since V + z ; u ( u ) = 1, this reduces to 2i Im( V + z ; u ) ′ ( u ) = 2i Im ( V + z ; o ) ′ ( u ) V + z ; o ( u ) . We thus showedIm R + z ( u ) = Im z k V + z ; u k T + u , implying the result by taking u = o b . The claim for R − z issimilar: the preceding proof shows that in the twisted view, ( U + z ; o ) ′ ( u ) U + z ; o ( u ) is Herglotz, and thenegative sign is there to pass to the coherent view. These claims in turn show by (2.13)that z G z ( v, v ) is Herglotz To show that R + z ( o b ) √ z is Herglotz, we use the same approach. First, h V + z ; u , H max T + u V + z ; u i L ( T + u ) + hH max T + u V + z ; u , V + z ; u i L ( T + u ) = 2 Re z · k V + z ; u k L ( T + u ) . On the other hand, the left-hand side is 2 Re h V + z ; u , H max T + u V + z ; u i . Integrating by parts yields2 (cid:16) k ( V + z ; u ) ′ k L ( T + u ) + h V + z ; u , W V + z ; u i L ( T + u ) + X v ∈ V ( T + u ) \{ u } α v | V + z ; u ( v ) | + Re V + z ; u ( u )( V + z ; u ) ′ ( u ) (cid:17) . As before, V + z ; u ( u ) = 1 and ( V + z ; u ) ′ ( u ) = R + z ( u ). Hence, Re R + z ( u ) = Re z k V + z ; u k L ( T + u ) −k ( V + z ; u ) ′ k L ( T + u ) − h V + z ; u , W V + z ; u i L ( T + u ) − P v ∈ V ( T + u ) \{ u } α v | V + z ; u ( v ) | .Let √ z = r + i s . It follows thatIm (cid:16) R + z ( u ) √ z (cid:17) = Im R z Re 1 √ z + Im 1 √ z Re R z = Im z k V + z ; u k r − s Re z k V + z ; u k + s ( k ( V + z ; u ) ′ k + h V + z ; u , W V + z ; u i + P α v | V + z ; u ( v ) | ) r + s = Im( z √ z ) k V + z ; u k + s ( k ( V + z ; u ) ′ k + h V + z ; u , W V + z ; u i + P α v | V + z ; u ( v ) | ) | z | . Since Im( z √ z ) = | z | Im √ z , the RHS is clearly positive if W ≥ α v ≥ v .We next prove the current relations. Since V + z ; o satisfies the δ -conditions, we have P b + ∈N + b R + z ( o b + ) = R + z ( t b ) + α t b , so P b + ∈N + b Im R + z ( o b + ) = Im R + z ( t b ). Similarly, as P w − ∈N − w U ′ w − ( L w − ) + α w U w (0) = U ′ w (0), we get P b − ∈N − b Im R − z ( t b − ) = Im R − z ( o b ).Suppose Hf = zf and let J z ( x b ) = Im[ f ( x b ) f ′ ( x b )]. Then J ′ z ( x b ) = Im[ | f ′ ( x b ) | + f ( x b )[ W ( x b ) f ( x b ) − zf ( x b )] = − Im z | f ( x b ) | . Hence, J ( x b ) decreases on b , in particular, J ( t b ) ≤ J ( o b ). Thus, | V + z ; v ( t b ) | Im R + z ( t b ) = Im[ V + z ; v ( t b )( V + z ; v ) ′ ( t b )] ≤ | V + z ; v ( o b ) | Im R + z ( o b ).Thus, Im R + z ( t b ) ≤ | V + z ; v ( o b ) | | V + z ; v ( t b ) | Im R + z ( o b ) = Im R + z ( o b ) | ζ z ( b ) | . This proves (2.17) for R + z . Since | U − z ; v ( o b ) | Im R − z ( o b ) = − Im[ U − z ; v ( o b )( U − z ; v ) ′ ( o b )], the claim for R − z follows similarly.Finally, if the terms are defined for Im z = 0, then equality follows from J ′ z ( x b ) = 0. (cid:3) Remark A.3.
The method above also shows that z
7→ − S ′ z ( L ) S z ( L ) is Herglotz.In fact, Im( S ′ z ( L ) S z ( L ) ) = | S z ( L ) | Im[ S z ( L ) S ′ z ( L )] ≤ | S z ( L ) | Im[ S z (0) S ′ z (0)] = 0. If the poten-tials are symmetric, so that S ′ z ( L ) = C z ( L ), we also get Im( C z ( L ) S z ( L ) ) ≤ Though it is known that z
7→ h ψ, G z ψ i is Herglotz for any ψ ∈ L ( T ) by the spectral theorem, wefollowed this somehow roundabout argument to deduce the same holds for z G z ( v, v ). See the appendixof [8] for a more general result. Recall that Im R + z ( t b ) = P b + ∈N + b Im R + z ( o b + ) ≥
0. Using (2.5), Im( − S z ( L b ) ζ z ( b ) ) =Im R + z ( t b ) − Im S ′ z ( L b ) S z ( L b ) ≥
0. Thus, we also get Im( S z ( L b ) ζ z ( b )) ≥ b .Above we have shown that Im R + z ( v ) = Im z k V + z ; v k L ( T + v ) . We would like to replace the L norm by a lower bound that does not depend on z, v or any of the potentials. This isproved in the following lemma: Lemma A.4.
Let the potential W = W b be bounded, k W k ∞ := sup b ∈ B k W b k L ∞ ([0 ,L b ]) < ∞ , and let K ⊂ C + be a compact set. Then there exists C = C ( K, k W k ∞ ) > such thatfor all z ∈ K and all v ∈ V ( T ) , we have Im R + z ( v ) ≥ C Im z. Proof.
We begin with the following general fact: let
M, ℓ > . There exists c > suchthat for any potential Q on [0 , ℓ ] with k Q k ∞ ≤ M , and any solution of − f ′′ + Qf = 0 on [0 , ℓ ] with f (0) = 1 , we have k f k L (0 ,ℓ ) ≥ c . Indeed, suppose to the contrary there are Q n , f n with k Q n k ∞ ≤ M and f n solutions with f n (0) = 1 such that k f n k L (0 ,ℓ ) →
0. Then k f ′′ n k L →
0. This implies k f n k W , → k · k W , is equivalentto k u k L + k u ′′ k L ). But W , (0 , ℓ ) ⊂ C [0 , ℓ ], in particular this implies k f n k C [0 ,ℓ ] → f n (0) →
0, a contradiction.Taking Q = W − z and [0 , ℓ ] the edge in T + v outgoing from v we find for f = V + z ; v , k V + z ; v k T + v ≥ k f k L (0 ,ℓ ) ≥ c. (cid:3) We showed in Remark A.3 that z − S ′ z ( L ) S z ( L ) is Herglotz. We also have: Lemma A.5.
If the potentials W are all non-negative, the function z
7→ − S ′ z ( L ) √ zS z ( L ) isHerglotz.Proof. Considerdd x S ′ z ( x ) S z ( x ) √ ¯ z − S ′ z ( x ) S z ( x ) √ z ! = S ′′ z ( x ) S z ( x ) √ ¯ z + | S ′ z ( x ) | √ ¯ z − | S ′ z ( x ) | √ z − S ′′ z ( x ) S z ( x ) √ z = ( W ( x ) − ¯ z ) | S z ( x ) | √ ¯ z − | S ′ z ( x ) | Im (cid:26) √ z (cid:27) − ( W ( x ) − z ) | S z ( x ) | √ z = − W ( x ) | S z ( x ) | Im (cid:26) √ z (cid:27) − | S ′ z ( x ) | Im (cid:26) √ z (cid:27) − ( √ ¯ z − √ z ) | S z ( x ) | . Integrating this from 0 to L we get2i Im √ z Z L | S z ( x ) | d x −
2i Im (cid:26) √ z (cid:27) Z L (cid:0) W ( x ) | S z ( x ) | + | S ′ z ( x ) | (cid:1) d x = S ′ z ( L ) S z ( L ) √ ¯ z − S ′ z ( L ) S z ( L ) √ z − S ′ z (0) S z (0) √ ¯ z + S ′ z (0) S z (0) √ z = −
2i Im ( S ′ z ( L ) S z ( L ) √ z ) = 2i | S z ( L ) | Im (cid:26) − S ′ z ( L ) √ zS z ( L ) (cid:27) . As W ≥
0, we deduce that − S ′ z ( L ) √ zS z ( L ) ∈ C + . (cid:3) BSOLUTELY CONTINUOUS SPECTRUM FOR QUANTUM TREES 41
A.2.
A criterion for AC spectrum.
We recall the following fact [29, Theorem XIII.20].
Suppose H is a self-adjoint operator on a Hilbert space H , with resolvent G z . Supposethere exists p > such that for any ϕ in a dense subset of H , we have (A.8) lim inf η ↓ Z ba | Im h ϕ, G λ +i η ϕ i| p d λ < ∞ . Then H has purely absolutely continuous spectrum in ( a, b ) . This criterion also appeared later in [24] for p = 2. In [29], lim inf η ↓ is replaced bysup <η< , but one sees from the proof that the above statement holds. Theorem A.6.
Suppose a Schr¨odinger operator H T on a quantum tree satisfies the fol-lowing: there exists p > such that for any b ∈ T , (A.9) lim inf η ↓ Z I (cid:16) Im R + λ +i η ( o b ) | S λ +i η ( L b ) | (cid:17) p d λ < ∞ . Then H T has purely absolutely continuous spectrum in I .In particular, if conditions (1.8) and (Green -p) hold, then P -a.e. operator H T haspure AC spectrum in I .Proof. Let ϕ ∈ L ( T ) be continuous, compactly supported in a ball Λ ⊂ T . Z E E | Im h ϕ, G λ +i η ϕ i| p d λ = Z E E (cid:12)(cid:12)(cid:12)(cid:12) Im Z Λ × Λ ϕ ( x ) G λ +i η ( x, y ) ϕ ( y ) d x d y (cid:12)(cid:12)(cid:12)(cid:12) p d λ ≤ k ϕ k p ∞ · | Λ | p − Z E E Z Λ × Λ | G λ +i η ( x, y ) | p d x d y d λ . (A.10)Now R Λ × Λ | G z ( x, y ) | p d x d y = P b,b ′ ∈ B (Λ) R L b R L b ′ | G z ( x b , y b ′ ) | p d x b d y b ′ .Fix b, b ′ ∈ B (Λ). Since for any v ∈ V , y b ′ G z ( v, y b ′ ) is an eigenfunction witheigenvalue z , we have G z ( v, y b ′ ) = S z ( L b ′ − y b ′ ) G z ( v, o b ′ ) + S z ( y b ′ ) G z ( v, t b ′ ) S z ( L b ′ ) , similarly for the first argument, so we deduce that for ( b, x b ) , ( b ′ , y b ′ ) ∈ T , b = b ′ , G z ( x b , y b ′ ) = S z ( L b − x b ) G z ( o b , y b ′ ) + S z ( x b ) G z ( t b , y b ′ ) S z ( L b )(A.11) = S z ( L b − x b ) S z ( L b ′ − y b ′ ) G z ( o b , o b ′ ) + S z ( L b − x b ) S z ( y b ′ ) G z ( o b , t b ′ ) S z ( L b ) S z ( L b ′ )+ S z ( x b ) S z ( L b ′ − y b ′ ) G z ( t b , o b ′ ) + S z ( x b ) S z ( y b ′ ) G z ( t b , t b ′ ) S z ( L b ) S z ( L b ′ ) . Let b , . . . , b k be a path with b = b and b k = b ′ . We observe that for any b k +1 ∈ N + b k , | ζ z ( b ) · · · ζ z ( b k ) | ≤ X ( b ′ ; b ′ k +1 ) | ζ z ( b ) · · · ζ z ( b ′ k ) | Im R + z ( o b ′ k +1 ) / | Im R + z ( o b k +1 ) | / ≤ | Im R + z ( o b ) || Im R + z ( o b k +1 ) | ! / , This holds if the potentials W are symmetric, otherwise one should replace S z ( L b − x b ) by S z ( L b ) C z ( x b ) − C z ( L b ) S z ( x b ). This does not affect the argument. where we used (2.17) repeatedly in the last step. On the other hand, by (2.13), | G z ( o b , o b ) | = 1 | R + z ( o b ) + R − z ( o b ) | ≤ R + z ( o b ) , where we used the fact that R ± z ( v ) is Herglotz. Hence, | G z ( o b ; t b k ) | ≤ | Im R + z ( o b ) Im R + z ( o b k +1 ) | / . The other G z ( v ; v r ) appearing in (A.11) are bounded similarly.For b = b ′ , the first equality in (A.11) should be modified as we don’t have an eigen-function at the point x b = y b . In fact, assuming without loss that x b ≤ y b , we have(A.12) G z ( x b , y b ) = 1 S z ( L b ) (cid:0) [ S z ( L b ) C z ( x b ) − C z ( L b ) S z ( x b )] G z ( o b , y b )+ S z ( x b )[ G z ( t b , y b ) + S z ( L b ) C z ( y b ) − C z ( L b ) S z ( y b )] (cid:1) . This can be checked by explicit calculation from (A.2) and (2.13): these tell us that G z ( o b , y b ) = G z ( o b , o b ) φ − ( o b ) φ + ( y b ) = G z ( o b , o b ) φ + ( y b ) ,G z ( t b , y b ) = G z ( o b , o b ) φ − ( y b ) φ + ( L b )= G z ( o b , o b )[ φ + ( y b ) φ − ( L b ) − ( R + + R − )( C z ( L b ) S z ( y b ) − S z ( L b ) C z ( y b ))]= G z ( o b , o b )[ C z ( L b ) − R − S z ( L b )] φ + ( y b ) + C z ( L b ) S z ( y b ) − S z ( L b ) C z ( y b ) . Here we passed from φ − ( y b ) φ + ( L b ) to φ + ( y b ) φ − ( L b ) as in Lemma A.2. Inserting this intothe RHS of (A.12) gives G z ( o b , o b ) φ − ( x b ) φ + ( y b ) = G z ( x b , y b ) as asserted.In any case, for any b ∈ B , R L b | S z ( x b ) | p d x b ≤ c , uniformly in λ + i η ∈ ( E , E ) + i(0 , λ S λ ( x ) is analytic). Recalling (A.10), (A.11), we have proved that Z E E | Im h ϕ, G λ +i η ϕ i| p d λ ≤ ˜ c k ϕ k p ∞ · | Λ | p − X b,b ′ ∈ B (Λ) Z E E d λ | Im R + z ( o b ) Im R + z ( o b ′′ ) | p/ | S z ( L b ) S z ( L b ′ ) | p , where b ′′ is any edge with o b ′′ = t b ′ . Applying Cauchy-Schwarz and (A.9), we see that (A.8)is satisfied for any continuous ϕ of compact support. Hence, H T has pure AC spectrumin ( E , E ).In particular, if (Green -p) holds for a random tree, then by Fatou’s lemma and Fubini,we have E lim inf η ↓ Z I X t b ∼ o b (cid:16) Im R + λ +i η ( o b ) | S λ +i η ( L b ) | (cid:17) p d λ ≤ lim inf η ↓ Z I E X t b ∼ o b (cid:16) Im R + λ +i η ( o b ) | S λ +i η ( L b ) | (cid:17) p d λ < ∞ , where we applied (Green -p) in the last step. It follows that the lim inf on the left-handside is finite for P -a.e. operator H T . Hence, P -a.e. H T has pure AC spectrum in I . (cid:3) BSOLUTELY CONTINUOUS SPECTRUM FOR QUANTUM TREES 43
Appendix B. The uniform contraction estimate
The proof of Proposition 5.8 goes by analyzing different cases. Following [22] we intro-duce “visible sets of vertices” Vis γ ( g, ε ) and Vis i Im ( g, ε ). The idea is that we seek a strictcontraction for κ ( π ) ∗ by estimating some terms in the weighted sum (5.26). By controllingthe visibility we ensure that estimates on Q ( π ) x,y ( g ) cos α ( π ) x,y ( g ) do not become “invisible”,for example by the weightings q y ( g ) or γ x ( g ) becoming too small (which would jeopardizean implied control over c ( π ) x and κ ( p ) ∗ , respectively (recall (5.11)-(5.12)).One sees that Lemmas 4 and 5 in [22] hold verbatim for quantum graphs.We then study three cases to prove Proposition 5.8. In the first case, certain γ ( π ) x ( g ) isvery small. This implies the terms in the sum defining κ ( p ) ∗ ( g ) have different magnitudes,and the result follows from (an improved) Jensen inequality. In the second case it isassumed that all γ ( π ) x ( x ) have essentially the same size, but certain Im g ( π ) y , y ∈ S ∗ ′ is verysmall. In this case one proves that certain c ( π ) x ( g ) must be small, so the result followsfrom [22, Lemma 5]. Finally in the third case, it is proved that if g / ∈ B r ( H ), then thereare always π, x, y such that Q ( π ) x,y ( g ) cos α ( π ) x,y ( g ) is uniformly smaller than one. Assumingwe are neither in the first nor the second case, one deduces that some c ( π ) x ( g ) is small,concluding the proof again using [22, Lemma 5]. See [22, Section 4.7] for more details asto how these cases are put together to prove Proposition 5.8.For quantum graphs, the first and second cases are the same as in [22]. Up to thispoint we are only using calculus and the definition of κ ( p ) ∗ ( g ) and c x ( g ). So more precisely,Propositions 4 and 5 from [22] hold without change to quantum graphs.For the third case however, we need some effort to carry out the adaptation.We start with the following lemma. Lemma B.1.
Given z ∈ H , we have − cot ( √ zL ) ∈ H . In particular, assuming W ≡ ,we have ζ z ( b ) sin ( √ zL b ) ∈ H .Proof. In Lemma A.5 the more general fact that − S ′ z ( L ) √ zS z ( L ) is Herglotz is proved. Special-ising to W ≡ − cot √ zL is Herglotz.For the second claim, from (2.5) it follows that − √ zS z ( L b ) ζ z ( b ) = R + z ( t b ) √ z − S ′ z ( L b ) √ zS z ( L b ) , so by the previous part and the fact that R + z ( t b ) √ z ∈ H , we get that − √ zS z ( L b ) ζ z ( b ) ∈ H . Again,specialising to W ≡ (cid:3) In analogy to (2.4), saying that ζ z ( b ) = C z ( L b ) + S z ( L b ) R + z ( o b ), given g ∈ H S o,o ′ , wedefine ζ zg ( ∗ ′ ) = ζ zg ( ∗ ′ , α, L ) by(B.1) ζ zg ( ∗ ′ ) = cos √ zL + g ∗ ′ sin √ zL . Then ζ zH ( ∗ ′ , α ∗ ′ , L ∗ ′ ) = ζ z ( ∗ , ∗ ′ ) as expected. Moreover, from (5.23), one easily checks that ζ zg ( ∗ ′ ) = − φ ∗′ ( g ) sin √ zL +cos √ zL , so that − ( √ zL ) ζ zg ( ∗ ′ ) = φ ∗ ′ ( g ) − cot √ zL , which implies thatsin( √ zL ) ζ zg ( ∗ ′ ) ∈ H by Lemma B.1.We consider the argument of vectors related to the operator’s Green function. The con-vention is that arg z ∈ ( − π, π ]. Moreover, we denote by d S ( · , · ) the translation invariantmetric in S which is normalized by d S (0 , π ) = π .Denote Z zv = ζ z ( v − , v ) sin √ zL v , the unperturbed ζ z sin √ zL of the edge ( v − , v ). Wedefine a quantity, related to the ‘minimal angle’ of this with the real axis, by(B.2) θ := 110 min { d S (arg Z zk , β ) | β ∈ { , π } , z ∈ I + i[0 , , k ∈ A } , where we denoted Z zk = Z zv if v has label k . In view of Lemma B.1, since I ⊂ Σ is chosencompact, the minimum exists and we have θ >
0. We also let ς = min { Im H zk : z ∈ I + i[0 , , k ∈ A } , (B.3) ς = max {| Γ zk | : z ∈ I + i[0 , , k ∈ A } , where we denoted H zk = H zv for v of label k and similarly for Γ. Again ς > ε = ε ( r ) = min x ∈ S ∗ , ∗′ inf z ∈ I +i[0 , inf g x ∈ H \ B r ( H zx ) | g x − H zx | . We may now state the main result of this appendix (corresponding to case 3 in theabove discussion).
Proposition B.2.
Let I ⊂ Σ be compact and satisfy (5.28) . There is c = c ( θ ) < , ǫ ∗ ( θ , ς , I, ε D ) > and R : [0 , ǫ ∗ ) → [0 , ∞ ) with lim ǫ → R ( ǫ ) = 0 such that for all ǫ ∈ [0 , ǫ ∗ ) , if • g ∈ H S ∗ , ∗′ \ B R ( ǫ ) ( H ) and g ∗ ′ = g ∗ ′ ( z, α, L ) is defined by (5.23) – (5.24) , • | α − α ∗ ′ | ≤ ǫ and | L − L ∗ ′ | ≤ ǫ , • Im z ∈ [0 , η ∗ ] for some η ∗ ( θ , ε D ) ,then there is π ∈ Π with Q ( π ) x,y ( g ) ≤ c or cos α ( π ) x,y ( g ) ≤ c, either for some x, y ∈ S ∗ ′ or for x = ∗ ′ and all y ∈ S ∗ \ {∗ ′ } .Proof. First recall that for ξ i ∈ C (B.5) d S (arg( ξ ) , arg( ξ )) = d S (cid:0) arg( ξ ¯ ξ ) , (cid:1) = d S (cid:18) arg (cid:0) ξ ξ (cid:1) , (cid:19) . Also, for α, β, γ ∈ S , we have by the triangle inequality and translation invariance,(B.6) d S ( α + β, γ ) ≥ d S ( α, − d S ( β, γ ) . Moreover, as α x,y ( g ) = arg(( g x − H zx )( g y − H zy )), we have by (B.5),(B.7) d S ( α x,y ( g ) ,
0) = d S (arg( g x − H zx ) , arg( g y − H zy )) . By Lipschitz continuity, we may find c I such that for all z ∈ I +i[0 ,
1] and L ∈ S i ∈ A [ L i − , L i + 1], we have(B.8) (cid:12)(cid:12) sin √ z ( L − L i ) (cid:12)(cid:12) ≤ c I · | L − L i | . We also take c ′ I such that(B.9) (cid:12)(cid:12)(cid:12)(cid:12) α − α i √ z (cid:12)(cid:12)(cid:12)(cid:12) ≤ c ′ I · | α − α i | . We then choose(B.10) ǫ ∗ := min (cid:18) θ c ′ I (1 + θ ) ς , θ ε D c I ς (1 + θ ) , θ ε D c I (1 + ς )(1 + θ ) ς (cid:19) , (B.11) M I,D = max (cid:18) c ′ I , c I (1 + ς ) ε D (cid:19) and define R : [0 , ǫ ∗ ) → [0 , ∞ ) by R ( ǫ ) = ε − ( θ ǫ ) , with θ ǫ = (1 + θ ) θ M I,D ǫ ,
BSOLUTELY CONTINUOUS SPECTRUM FOR QUANTUM TREES 45 where ε ( r ) is defined in (B.4). By [22, Lemma 10] or [20, Lemma 2.18], R ( ǫ ) = θ ǫ ς ( ς − θ ǫ ) is well-defined, since ς − θ ǫ > ǫ ∗ .Take g ∈ H S ∗ , ∗′ \ B R ( ǫ ) and ǫ ∈ [0 , ǫ ∗ ). If there is x ∈ S ∗ , ∗ ′ such that g x = H zx , then Q x,y ( g ) = 0 by definition for all y ∈ S ∗ , ∗ ′ and we are done. So assume that g x = H zx forall x ∈ S ∗ , ∗ ′ . We also assume(B.12) d S ( α ( π ) x,y ( g ) , ≤ θ for all π ∈ Π and x, y ∈ S ∗ ′ since otherwise we are done. Our aim is to show d S ( α ( π ) ∗ ′ ,y ( g ) , > θ for some π ∈ Π andall y ∈ S ∗ \ {∗ ′ } .Recall that H zπ ( x ) = H zx for π ∈ Π and x ∈ S ∗ , ∗ ′ . We set τ ( π ) ∗ ′ := X y ∈ S ∗′ (cid:16) g ( π ) y − H zy (cid:17) . Recalling (B.7), our assumption (B.12) implies that(B.13) (cid:12)(cid:12) τ ( π ) ∗ ′ (cid:12)(cid:12) ≥ (cid:12)(cid:12) g ( π ) x − H zx (cid:12)(cid:12) for all π ∈ Π and x ∈ S ∗ ′ , and(B.14) d S (cid:0) arg( τ ( π ) ∗ ′ ) , arg( g ( π ) x − H zx ) (cid:1) ≤ θ for all π ∈ Π and x ∈ S ∗ , ∗ ′ . See [22, Lemma 9] or [20, Lemma 4.19] for a proof.We now proceed with two claims. Claim 1: There exists π ∈ Π such that | τ ( π ) ∗ ′ | ≥ (1+ θ ) θ M I,D ǫ. Proof of Claim 1. We assumed that g / ∈ B R ( ǫ ) , so there is ˆ x ∈ S ∗ , ∗ ′ such that γ ˆ x ( g ) > R ( ǫ ).By the choice of ∗ ′ w.r.t. (C2) there is π such that π (ˆ x ) ∈ S ∗ ′ . By (B.13) and thedefinitions (B.4) of ε and R above, we obtain | τ ( π − ) ∗ ′ | ≥ | g ( π − ) π (ˆ x ) − H zπ (ˆ x ) | = | g ˆ x − H z ˆ x | ≥ ε ( R ( ǫ )) = θ ǫ = (1 + θ ) θ M I,D ǫ .
Claim 2: Denote Z z ( v ) = ζ z ( v − , v ) sin √ zL v and w = α − α ∗′ √ z . Then we have g ( π ) ∗ ′ − H z ∗ ′ = Z zg ( π ) ( ∗ ′ ) Z z ( ∗ ′ ) (cid:20) (cid:16) τ ( π ) ∗ ′ − w (cid:17) n √ zL cot √ zL ∗ ′ + Γ z ∗ ′ · sin √ z ( L − L ∗ ′ )sin √ zL sin √ zL ∗ ′ o + ((Γ z ∗ ′ ) + 1) sin √ z ( L − L ∗ ′ )sin √ zL sin √ zL ∗ ′ (cid:21) . Proof of Claim 2. We have by (5.23), g ( π ) ∗ ′ − H z ∗ ′ = φ ∗ ′ ( g ( π ) ) cos √ zL + sin √ zL − φ ∗ ′ ( g ( π ) ) sin √ zL + cos √ zL − Γ z ∗ ′ cos √ zL ∗ ′ + sin √ zL ∗ ′ − Γ z ∗ ′ sin √ zL ∗ ′ + cos √ zL ∗ ′ = 1( − φ ∗ ′ ( g ( π ) ) sin √ zL + cos √ zL )( − Γ z ∗ ′ sin √ zL ∗ ′ + cos √ zL ∗ ′ ) × h φ ∗ ′ ( g ( π ) ) { cos √ zL cos √ zL ∗ ′ + sin √ zL sin √ zL ∗ ′ + Γ z ∗ ′ (sin √ zL cos √ zL ∗ ′ − cos √ zL sin √ zL ∗ ′ ) }− Γ z ∗ ′ (cos √ zL cos √ zL ∗ ′ + sin √ zL sin √ zL ∗ ′ )+ sin √ zL cos √ zL ∗ ′ − cos √ zL sin √ zL ∗ ′ i = ( φ ∗ ′ ( g ( π ) ) − Γ z ∗ ′ ) cos √ z ( L − L ∗ ′ ) + ( φ ∗ ′ ( g ( π ) )Γ z ∗ ′ + 1) sin √ z ( L − L ∗ ′ )( − φ ∗ ′ ( g ( π ) ) sin √ zL + cos √ zL )( − Γ z ∗ ′ sin √ zL ∗ ′ + cos √ zL ∗ ′ ) . But by reverse M¨obius transformation, φ ∗ ′ ( g ( π ) ) = g ( π ) ∗′ cos √ zL − sin √ zLg ( π ) ∗′ sin √ zL +cos √ zL . Hence, − φ ∗ ′ ( g ( π ) ) sin √ zL + cos √ zL = sin √ zL − g ( π ) ∗ ′ sin √ zL cos √ zLg ( π ) ∗ ′ sin √ zL + cos √ zL + cos √ zL = 1 g ( π ) ∗ ′ sin √ zL + cos √ zL = 1 ζ zg ( π ) ( ∗ ′ )by definition (B.1). Thus, g ( π ) ∗ ′ − H z ∗ ′ = ζ zg ( π ) ( ∗ ′ ) ζ z ( ∗ , ∗ ′ ) (cid:2) ( φ ∗ ′ ( g ( π ) ) − Γ z ∗ ′ ) { cos √ z ( L − L ∗ ′ ) + Γ z ∗ ′ sin √ z ( L − L ∗ ′ ) } + ((Γ z ∗ ′ ) + 1) sin √ z ( L − L ∗ ′ ) (cid:3) . Next, φ ∗ ′ ( g ( π ) ) = P x ∈ S ∗′ g ( π ) x − α √ z , so ( φ ∗ ′ ( g ( π ) ) − Γ z ∗ ′ ) = τ ( π ) ∗ ′ − w . Claim 2 follows. Conclusion:
By Claim 2, using arg( ξζ ) = arg( ξ ) + arg( ζ ) and arg( ξ + ζ ) = arg( ξ ) +arg(1 + ζ/ξ ), we getarg( g ( π ) ∗ ′ − H z ∗ ′ ) = arg (cid:0) Z zg ( π ) ( ∗ ′ ) Z z ( ∗ ′ ) (cid:1) + arg (cid:20)(cid:16) − wτ ( π ) ∗ ′ (cid:17)n √ zL cot √ zL ∗ ′ + Γ z ∗ ′ · sin √ z ( L − L ∗ ′ )sin √ zL sin √ zL ∗ ′ o + (Γ z ∗ ′ ) + 1 τ ( π ) ∗ ′ · sin √ z ( L − L ∗ ′ )sin √ zL sin √ zL ∗ ′ (cid:21) + arg( τ ( π ) ∗ ′ )We thus get for the permutation π ∈ Π taken from Claim 1 and all y ∈ S ∗ \ {∗ ′ } d S (cid:0) α ( π ) ∗ ′ ,y ( g ) , (cid:1) = d S (cid:16) arg( g ( π ) ∗ ′ − H z ∗ ′ ) , arg( g ( π ) y − H zy ) (cid:17) ≥ d S (cid:16) arg( Z zg ( π ) ( ∗ ′ ) Z z ( ∗ ′ )) , (cid:17) − d S (cid:18) arg (cid:20) − wτ ( π ) ∗ ′ ++ − wτ ( π ) ∗ ′ ! Γ z ∗ ′ · sin √ z ( L − L ∗ ′ )sin √ zL sin √ zL ∗ ′ + (Γ z ∗ ′ ) + 1 τ ( π ) ∗ ′ · sin √ z ( L − L ∗ ′ )sin √ zL sin √ zL ∗ ′ + − wτ ( π ) ∗ ′ ! cot √ zL cot √ zL ∗ ′ (cid:21) , (cid:19) − d S (cid:16) arg( τ ( π ) ∗ ′ ) , arg ( g ( π ) y − H zy ) (cid:17) where we used (B.6) twice. To bound the first distance, note that we have Z z ( ∗ ′ ) ∈ H and d S (arg Z z ( ∗ ′ ) , β ) ≥ θ for β ∈ { , π } , by the definition (B.2) of θ . Moreover,since Z zg ( π ) ( ∗ ′ ) ∈ H by the argument after (B.1), we have d S (arg( Z zg ( π ) ( ∗ ′ )) , β ) > β ∈ { , π } . Hence, d S (arg( Z z ( ∗ ′ ) Z zg ( π ) ( ∗ ′ )) , > θ . The last distance can be estimatedby (B.14).For the second distance we begin by noticing | wτ ( π ) ∗′ | ≤ θ θ . In fact, | w | ≤ c ′ I ǫ by (B.9),so this follows from Claim 1 and (B.11). Also, using (B.8), (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (Γ z ∗ ′ ) + 1 τ ( π ) ∗ ′ · sin (cid:0) √ z ( L − L ∗ ′ ) (cid:1) sin ( √ zL ) sin (cid:0) √ zL ∗ ′ (cid:1) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ( ς + 1) θ M I,D ǫ (1 + θ ) c I ǫε D and (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:16) − wτ ( π ) ∗ ′ (cid:17) Γ z ∗ ′ · sin (cid:0) √ z ( L − L ∗ ′ ) (cid:1) sin ( √ zL ) sin (cid:0) √ zL ∗ ′ (cid:1) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ς c I ǫε D . BSOLUTELY CONTINUOUS SPECTRUM FOR QUANTUM TREES 47
Now as noted in [22, Lemma 8], for any ζ, ξ ∈ C , | ζ | <
1, we have(B.15) d S (arg (1 + ξ + ζ ) , ≤ ( | ζ | −| ζ | if ξ = 0 ,d S (arg ξ,
0) + | ζ | −| ζ | if ξ = 0 . We use this with ξ = (cid:0) − wτ ( π ) ∗′ (cid:1) cot ( √ zL ) cot (cid:0) √ zL ∗ ′ (cid:1) and ζ = − wτ ( π ) ∗ ′ + (cid:16) − wτ ( π ) ∗ ′ (cid:17) Γ z ∗ ′ · sin (cid:0) √ z ( L − L ∗ ′ ) (cid:1) sin ( √ zL ) sin (cid:0) √ zL ∗ ′ (cid:1) + (Γ z ∗ ′ ) + 1 τ ∗ ′ · sin (cid:0) √ z ( L − L ∗ ′ ) (cid:1) sin ( √ zL ) sin (cid:0) √ zL ∗ ′ (cid:1) . From the estimates above, | ζ | ≤ θ θ + c I ε D ( ς + 1) θ M I,D (1 + θ ) + 2 c I ǫς ε D ≤ θ θ + θ θ ) + θ θ ) ≤ θ θ , using (B.10), (B.11). Here, we used θ ≤ (which holds since θ ≤ π ).Applying (B.15) thus yields d S (cid:0) α ( π ) ∗ ′ ,y ( h ) , (cid:1) > θ − θ − θ − d S (cid:18) arg (cid:20) − wτ ( π ) ∗ ′ ! cot √ zL cot √ zL ∗ ′ (cid:21) , (cid:19) ≥ θ − d S (cid:18) arg − wτ ( π ) ∗ ′ ! , (cid:19) − d S (cid:18) arg (cid:20) cot √ zL cot √ zL ∗ ′ (cid:21) , (cid:19) The first distance is controlled again by (B.15) with ξ = 0. For the second distance, weobserve that Re z is not a Dirichlet value for L ∗ ′ , L by assumption (5.28), so the argumenttends to 0 as Im z ↓
0. So for Im z small enough, the last distance is ≤ θ . This yields d S (cid:0) α ( π ) ∗ ′ ,y ( g ) , (cid:1) > θ − θ − θ > θ . The assertion follows by letting c := cos θ . (cid:3) Acknowledgments:
M.I. was funded by the Labex IRMIA during part of the writing ofthis article.M.S. was supported by a public grant as part of the
Investissement d’avenir project,reference ANR-11-LABX-0056-LMH, LabEx LMH. He thanks the Universit´e Paris Saclayfor excellent working conditions, where part of this work was done.We thank Konstantin Pankrashkin and Simone Warzel for useful discussions.
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Universit´e de Strasbourg, CNRS, IRMA UMR 7501, F-67000 Strasbourg, France.
E-mail address : [email protected] Laboratoire J.A.Dieudonn´e, UMR CNRS-UNS 7351, Universit´e Cˆote d’Azur, 06108 Nice,France
E-mail address : [email protected] Department of Mathematics, Faculty of Science, Cairo University, Cairo 12613, Egypt.
E-mail address : [email protected] Department of Mathematical Sciences, Loughborough University, Leicestershire, LE113TU, United Kingdom.
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