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Asymptotics for Fermi curves of electric and magnetic periodic fields de Oliveira, Gustavo 2009

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Asymptotics for Fermi curves of electric and magnetic periodic fields  by Gustavo de Oliveira  A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY in The Faculty of Graduate Studies (Mathematics) THE UNIVERSITY OF BRITISH COLUMBIA (Vancouver) July 2009 c Gustavo de Oliveira, 2009  Abstract This work is concerned with some geometrical properties of (complex) Fermi curves of electric and magnetic periodic fields. These are analytic curves in C2 that arise from the study of the eigenvalue problem for periodic Schrödinger operators. More specifically, we characterize a certain class of these curves in the region of C2 where at least one of the coordinates has “large” imaginary part. The new results obtained in this thesis extend previous results in the absence of magnetic field to the case of “small” magnetic field. Our theorems can be used to show that generically these Fermi curves belong to a class of Riemann surfaces of infinite genus.  ii  Table of contents Abstract  . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  ii  Table of contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  iii  List of figures  . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  v  Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  vi  1 Introduction and summary . . . . . . . . . . . . . . . . . . . . . . . . . . . .  1  2 Periodic Schrödinger operators . . . . . . . . . . . . . . . . . . . . . . . . .  5  2.1  Bloch theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  5  2.2  Fermi surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  10  2.3  Electrons in a crystal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  11  2.4  Basic properties of Hk (A, V ) . . . . . . . . . . . . . . . . . . . . . . . . . .  14  2.5  The complex analytic structure of the spectrum . . . . . . . . . . . . . . . .  26  2.6  Gauge invariance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  28  3 Asymptotics for Fermi curves . . . . . . . . . . . . . . . . . . . . . . . . . .  31  3.1  Fermi curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  31  3.2  The free Fermi curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  35  3.3  The ε-tubes about the free Fermi curve . . . . . . . . . . . . . . . . . . . .  41  3.4  Motivation and main results . . . . . . . . . . . . . . . . . . . . . . . . . . .  46  3.5  Strategy outline . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  50  3.6  Notation and remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  53  4 Weak magnetic potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  55  4.1  Invertibility of RG0 G0  . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  iii  55  4.2  Local defining equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  58  4.3  Change of coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  63  4.4  Asymptotics for the coefficients . . . . . . . . . . . . . . . . . . . . . . . . .  67  4.5  Bounds on the derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . .  88  4.6  The regular piece . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  113  4.7  The handles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  120  4.8  Quantitative Morse lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . .  131  4.9  Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  138  5 Exploiting gauge invariance  . . . . . . . . . . . . . . . . . . . . . . . . . . . 141  5.1  A gauge transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  141  5.2  The regular piece revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . .  143  iv  List of figures 2.1  Sketch of band functions En (k). . . . . . . . . . . . . . . . . . . . . . . . . .  10  2.2  Construction of D(m) ⊂ R3 : gluing together an m × m × m set of cells D ⊂ R3 .  13  2.3  Energy bands in conductors and insulators. . . . . . . . . . . . . . . . . . .  14  3.1  b 0) and F(0, 0) when both ik1 and k2 are real. . . . . . . . . . Sketch of F(0,  39  3.2  The pair of helices as a “manifold”. . . . . . . . . . . . . . . . . . . . . . . .  40  3.3  b 0): two copies of C with θ1 (b) and θ2 (b) identified for b2 6= 0. . The full F(0,  40  3.4  On the left: definition of Λ. On the right: possible configuration. . . . . . .  42  3.5  The ε-tubes about the free “lifted” Fermi curve.  . . . . . . . . . . . . . . .  44  3.6  Sketch of (∪d0 ∈G Td0 ) ∩ (C2 \ ∪b∈G0 Tb ) for G = {0} and G = {0, d}. . . . . . .  52  4.1   Sketch of T0 \ ∪b∈Γ# \{0} Tb \ KR (left) and (T0 ∩ Td ) \ KR (right). . . . . .  60  v  Acknowledgements I would like to thank Professor Joel Feldman for suggesting this research topic, and for his help and support during the process of writing this thesis. I am also thankful to UBC and all its staff. This thesis was typeset using LATEX and the many packages that accompany the LATEX distribution. In particular, the graphics were created using the TikZ and pgf packages. All the work was done using a free and open source operating system. My thanks go to the members of the “free software community”, who have made available such powerful tools.  vi  Chapter 1  Introduction and summary This work is concerned with some geometrical properties of (complex) Fermi curves. These are analytic curves in C2 that arise from the study of the eigenvalue problem for periodic Schrödinger operators. More specifically, we characterize a certain class of these curves in the region of C2 where at least one of the coordinates has “large” imaginary part. In order to describe the contents of this thesis we need to introduce the definition of Fermi curve. We stress that our purpose here is only to grasp the main ideas. We will return to the definitions (and results) later in much more detail. Let A1 , A2 and V be functions on R2 that are periodic with respect to Z2 and consider the operator H = (i∇ + A)2 + V acting on L2 (R2 ), where ∇ is the gradient operator in R2 and A = (A1 , A2 ). For k ∈ C2 let En (k) be the n-th eigenvalue of the boundary value problem Hψ = λψ, ψ(x + γ) = eik·γ ψ(x) for all x ∈ R2 and all γ ∈ Z2 . By definition, the n-th band Bn ⊂ C is the range of the function k 7→ En (k). It is known that the spectrum of H is the union of the bands Bn (restricted to real k) for n ≥ 1. The (complex) Fermi curve of A and V with energy λ ∈ C is defined as Fλ,A,V = {k ∈ C2 | the above boundary value problem has a nonzero solution ψ}. Similarly one can define Fermi “surfaces” in any dimension greater than two. 1  The above operator H (and its three-dimensional counterpart) is important in solid state physics. It is the Hamiltonian of a single electron under the influence of the magnetic field with vector potential A, and the electric field with scalar potential V , in the independent electron model of a two-dimensional solid [12]. The classical framework for studying the spectrum of a differential operator with periodic coefficients is the Floquet (or Bloch) theory [12, 8]. Roughly speaking, the main idea of this theory is to “decompose” the original eigenvalue problem, which usually has continuous spectrum, into a family of boundary value problems, each one having discrete spectrum. In our context this leads to decomposing the problem Hψ = λψ (without boundary conditions) into the above k-family of boundary value problems. The Fermi surface—and in particular the Fermi curve—has the following remarkable property. There exists a function F (k, λ, A, V ) analytic on Cd × C × A × V, where A and V are suitable spaces of functions, such that Fλ,A,V = {k ∈ Cd | F (k, λ, A, V ) = 0}. In other words, the Fermi surface is a complex analytic variety. In Chapter 2 we provide a detailed proof of this (well-known) theorem following the proof in [3]. It is believed that for “generic” sufficiently regular potentials A and V the union of the surfaces z = En (k) for n ≥ 1 is a complex analytic manifold. In fact, this statement was made precise in [4] for Fermi curves without magnetic potential (A = 0), and for heat curves. The later are the spectral curves (defined similarly as the Fermi curves) associated to the “heat” equation ψx1 − ψx2 x2 + V ψ = 0 with V periodic, while the former are a particular case of the curves studied in this thesis (where A 6= 0). This picture is believed to hold for other differential operators with sufficiently regular periodic coefficients as well. In Chapters 3 and 4 we prove a theorem, that holds only for “small” A, which is the main step for showing that the above picture is also true for Fermi curves with “small” magnetic potential. This is the main contribution of this thesis. We have followed the same strategy that Feldman, Knörrer and Trubowitz implemented in [4, §16] for proving a similar result for A = 0. Below we provide more details about our results. As we have already mentioned in the last paragraph, there is a relationship between Riemann surfaces and differential operators with periodic coefficients. We briefly mention two examples here. In one dimension, the solution of the KdV equation ut = 3uux − 21 uxxx with initial data u0 is related to the Schrödinger curve S(V ) by S(u( · , t)) = S(u0 ), where  2  S(V ) is the spectral curve associated to the one-dimensional analogue of the operator H with A = 0 [10]. In two dimensions, there is a relationship between Riemann surfaces of finite genus and solutions of the KP equation ux1 x1 + 32 (ut + 3uux2 + 12 ux2 x2 x2 )x2 = 0 [8]. If the initial data for this problem is a function on R2 that is periodic with respect to a certain lattice, this relationship is even more explicit [4]. In this case the solution of the KP equation is related to the heat curve mentioned above. This turns out to be, for “generic” potentials V , a Riemann surface of infinite genus according to the theory proposed in [4]. The class of surfaces introduced in that work yields an extension of the classical theory of finite genus that has analogues of many theorems of finite genus theory. When A and V are zero the (free) Fermi curve can be found explicitly. It consists of two copies of C with the points −b2 + ib1 (in the first copy) and b2 + ib1 (in the second copy) identified for all (b1 , b2 ) ∈ πZ2 with b2 6= 0. The purpose of this thesis is to show that in the region of C2 where k ∈ C2 has “large” imaginary part the Fermi curve (for nonzero A and V ) is “close” to the free Fermi curve. When A is zero this was proved by Feldman, Knor̈rer and Trubowitz in [4, §16]. Very briefly, the main result of this thesis is essentially the following. “Theorem”. Suppose that A and V have some regularity and assume that A is sufficiently small in a suitable norm. Write k in C2 as k = u + iv with u and v in R2 and suppose that |v| is sufficiently large. (Recall that the free Fermi curve is two copies of C with certain points in one copy identified with points in the other one.) Then, in this region of C2 , the Fermi curve of A and V is very close to the free Fermi curve, except that instead of two planes we may have two deformed planes, and identifications between points can open up to handles that look like {(z1 , z2 ) ∈ C2 | z1 z2 = constant} in suitable local coordinates. To prove this theorem we follow the same strategy as [4]. The proof has basically three steps. We first derive very detailed information about the free Fermi curve (which is explicitly known). This is done in Chapter 3. Then, to compute the interacting Fermi curve we have to find the kernel of H − λI in L2 (R2 ) with the above boundary condition. In the second step of the proof we derive a number of estimates for showing that this kernel has finite dimension for small A and k ∈ C2 with large imaginary part. Furthermore, in the complement of this kernel in L2 (R2 ), after a suitable (invertible) change of variables in L2 (R2 ), the operator H − λI multiplied by the inverse of the operator that implements this 3  change of variables is a (compact) perturbation of the identity. This reduces the problem of finding the kernel to finite dimension and thus we can write (local) defining equations for the Fermi curve. In the third step of the proof we use these equations to study the Fermi curve. A few more estimates and the implicit function theorem gives us the deformed planes. The handles are obtained using a quantitative Morse lemma available in [4] that we prove at end of Chapter 4. Steps two and three are contained in Chapter 4, which is the core of this thesis. We have not yet mentioned an important property of H—namely, gauge invariance. Briefly, gauge invariance implies that the spectrum of (i∇ + A)2 + V is the same as the spectrum of (i∇ + A + ∇Ψ)2 + V , where Ψ is function on R2 (under suitable hypotheses), and ∇Ψ is periodic with respect to Z2 . Consequently, the Fermi curve of A and V is equal to the Fermi curve of A + ∇Ψ and V . In Chapter 5, by choosing a convenient gauge Ψ, we are able to indicate how to simplify the proof of the above theorem and improve some constants in it. After performing this gauge transformation “some terms vanish” and the analysis becomes simpler. The critical part of the proof is the second step. The main difficulty arises due to the presence of the term A · i∇ in H. When A is large, taking the imaginary part of k ∈ C2 arbitrarily large is not enough to control this term—it is not enough to make its contribution small and hence have the interacting Fermi curve as a perturbation of the free Fermi curve. (The term V in H is easily controlled by this method.) However, the proof can be implemented by assuming that A is small. We finally make some remarks about the interdependence of chapters. Since we have attempted to write this thesis in order to minimize referencing, the reader may notice some minor repetitions along the text. Chapter 2 is completely independent of the others; Chapter 3 is independent and only needs Chapter 4 to be fulfilled; Chapter 4 depends on Chapter 3, and Chapter 5 depends on Chapters 3 and 4, and can be seen as a bonus chapter. The most demanding (but still elementary) part to read in this thesis is the proof of the lemmas and propositions in Chapter 4. These proofs may be (safely) skipped in a first reading.  4  Chapter 2  Periodic Schrödinger operators 2.1  Bloch theory  Below we briefly describe the main ideas of Bloch (or Floquet) theory without providing proofs. Our discussion is based on [2, 12]. Let Γ be a lattice of static ions in Rd and suppose that the ions generate an electric potential V (x) and a magnetic potential A(x) = (A1 (x), . . . , Ad (x)) that are periodic with respect to Γ. Then the Hamiltonian of a single electron moving in Rd under the influence of this lattice is H = (i∇ + A)2 + V, where ∇ is the gradient operator in Rd . This operator acts on L2 (Rd ). For simplicity we suppress the physical constants in the Hamiltonian. For γ ∈ Γ consider the translation operator Tγ acting on L2 (Rd ) as Tγ : ϕ(x) 7−→ ϕ(x + γ). This operator is unitary on L2 (Rd ) and Tγ Tγ 0 = Tγ+γ 0 for all γ, γ 0 ∈ Γ. Furthermore, it is not difficult to verify that the Hamiltonian H (formally) commutes with all the translation operators, HTγ = Tγ H for all γ ∈ Γ. Under suitable hypotheses on the potentials A and V one can in fact prove that this property holds on an appropriate domain. This translational symmetry is the main ingredient of Bloch theory. 5  Since our goal here is only describing the main ideas of Bloch theory without providing proofs, for the rest of this section we pretend that H and Tγ are matrices. A rigorous version of the procedure below can be implemented for the actual operators. The matrices {H and Tγ for all γ ∈ Γ} form a family of commuting normal matrices. Thus, there exists an orthonormal basis of simultaneous eigenvectors {ϕα } for this family. For all γ ∈ Γ these eigenvectors obey Hϕα = Eα ϕα , Tγ ϕα = λα,γ ϕα , where Eα and λα,γ are numbers. Let {v1 , . . . , vd } be a basis for Γ. Then for any γ ∈ Γ there are integers n1 , . . . , nd such that γ = n1 v1 + · · · + nd vd . If we set γi = ni vi for 1 ≤ i ≤ d, then we can write any element γ ∈ Γ as γ = γ1 + · · · + γd . As the operator Tγ is unitary, all its eigenvalues are complex numbers of modulus one. Hence, there exist real numbers βα,γ such that λα,γ = eiβα,γ . Using the properties of Tγ we find that eiβα,γ+γ 0 ϕα = Tγ+γ 0 ϕα = Tγ Tγ 0 ϕα = Tγ eiβα,γ 0 ϕα = eiβα,γ eiβα,γ 0 ϕα = ei(βα,γ +βα,γ 0 ) ϕα , so that βα,γ+γ 0 = βα,γ + βα,γ 0  mod 2π  for all γ, γ 0 ∈ Γ. Consequently, βα,γ = βα,γ1 + ··· +γd = βα,γ1 + · · · + βα,γd  mod 2π  for all γ ∈ Γ. That is, the number βα,γ is determined (mod 2π) by βα,γ1 , . . . , βα,γd . Now observe that, given any d numbers β1 , . . . , βd , the system of linear equations γi · k = βi  for  1≤i≤d  for the unknowns k1 , . . . , kd has a unique solution. This follows from the fact that γ1 , . . . , γd are linear independent. Hence, for each α there exists a kα ∈ Rd such that γi · kα = βα,γi for 1 ≤ i ≤ d. Therefore, βα,γ = γ · kα 6  mod 2π  for all γ ∈ Γ. Note that, for each α the vector kα is not uniquely determined. Indeed, βα,γ = γ · kα  mod 2π  and  βα,γ = γ · kα0  mod 2π  for all γ ∈ Γ, if and only if (kα − kα0 ) · γ ∈ 2πZ for all γ ∈ Γ. If we define the dual lattice of Γ as Γ# = {b ∈ Rd | b · γ ∈ 2πZ for all γ ∈ Γ}, where b · γ is the usual scalar product on Rd , the last expression can be rewritten as kα − kα0 ∈ Γ# . Summarizing, the numbers {βα,γ for γ ∈ Γ} determine kα up to a vector in Γ# . Hence, the vector kα is unique in Rd /Γ# . This establishes a correspondence between α and kα . Now, relabel the eigenvalues and eigenvectors by replacing the index α by the corresponding vector k ∈ Rd /Γ# and another index n. The index n is needed because many kα with different values of α can be equal. Under the new labelling the eigenvalue-eigenvector equations become Hϕn,k = En (k)ϕn,k , Tγ ϕn,k = eik·γ ϕn,k for all γ ∈ Γ. The eigenvalue of H is denoted En (k) rather than En,k because, while k runs over the continuous set Rd /Γ# , it turns out that n runs over a countable set. Observing the definition of Tγ , for each k ∈ Rd the above equations can be rewritten as Hϕn,k = En (k)ϕn,k , ϕn,k (x + γ) = eik·γ ϕn,k (x) for all x ∈ Rd and all γ ∈ Γ. Under suitable hypotheses on the potentials A and V this boundary value problem is self-adjoint. As we have already mentioned, its spectrum is discrete, it consists of a sequence of real eigenvalues E1 (k) ≤ E2 (k) ≤ · · · ≤ En (k) ≤ · · ·  7  For each integer n ≥ 1 the eigenvalue En (k) defines a continuous function of k that is periodic with respect to the dual lattice Γ# . It is customary to refer to k as the crystal momentum and to En (k) as the n-th band function. The corresponding normalized eigenfunctions ϕn,k are called Bloch eigenfunctions. Let Uk be the unitary transformation on L2 (Rd ) that acts as Uk : ϕ(x) 7−→ eik·x ϕ(x). By applying this transformation we can rewrite the above problem and put the boundary conditions into the operator. Indeed, if we define Hk = Uk−1 H Uk  and  ψn,k = Uk−1 ϕn,k ,  then the above problem is unitary equivalent to Hk ψn,k = En (k)ψn,k , ψn,k (x + γ) = ψn,k (x) for all x ∈ Rd and all γ ∈ Γ, or, using a more compact notation, Hk ψn,k = En (k)ψn,k  for  ψn,k ∈ L2 (Rd /Γ).  To see that these problems are equivalent we proceed (formally) as follows. On the one hand, from the original problem and using the above transformation we find that 0 = Uk−1 0 = Uk−1 (H − En (k))ϕn,k = Uk−1 (H − En (k))Uk Uk−1 ϕn,k = (Uk−1 HUk − En (k))Uk−1 ϕn,k = (Hk − En (k))ψn,k and ψn,k (x + γ) = (Uk−1 ϕn,k )(x + γ) = e−ik·(x+γ) ϕn,k (x + γ) = e−ik·(x+γ) eik·γ ϕn,k (x) = e−ik·x ϕn,k (x) = (Uk−1 ϕn,k )(x) = ψn,k (x). On the other hand, by a similar computation (in the reverse order), using the last two equalities and the above transformation we derive the original problem. This (formally) implies unitary equivalence. Furthermore, a simple (formal) calculation shows that Hk = (i∇ + A − k)2 + V. 8  In fact, Hk ψ = Uk−1 H Uk ψ   = e−ik·x (i∇ + A)2 + V eik·x ψ   = e−ik·x (i∇ + A) · eik·x (−kψ + i∇ψ + Aψ) + V ψ = (−k + A) · (−k + i∇ + A)ψ + i∇ · (−k + i∇ + A)ψ + V ψ   = (i∇ + A − k)2 + V ψ. Of course, the unitary transformation Uk preserves self-adjointness and does not change the spectrum {En (k)}∞ n=1 . Finally, denote by Nk the set of values of n that appear in the pairs α = (k, n) and define Hk = span{ϕn,k | n ∈ Nk }. Then, formally, and in particular ignoring that k runs over an uncountable set, L2 (Rd ) = span{ϕn,k | k ∈ Rd /Γ# and n ∈ Nk } M = Hk . k∈Rd /Γ#  Set ek = span{ψn,k | n ∈ Nk }. H Observe that, as the operator Uk−1 is unitary on L2 (Rd ), the space Hk is unitary equivalent ek . The restriction of U −1 HUk to H ek ek , and L2 (Rd ) is unitary equivalent to ⊕k∈Rd /Γ# H to H k is Hk applied to functions that are periodic with respect to Γ. Therefore, at least formally, to find the spectrum of H on L2 (Rd ) it suffices to find the spectrum of Hk on L2 (Rd /Γ) for all k ∈ Rd /Γ# . Unlike H, the operator Hk has compact resolvent. Thus, the spectrum of Hk is discrete, unlike the spectrum of H that is continuous and is given by σ(H) = {En (k) | n ∈ N and k ∈ Rd /Γ# }. (See Figure 2.1 below.) All these statements can be made precise and rigorously proved. We prove a few of them in §2.4; the proof of some others can be found in [2, 12].  9  E4 E3 σ(H)  E2 E1 k ∈ Rd  Figure 2.1: Sketch of band functions En (k).  2.2  Fermi surfaces  In this section we define the Fermi surfaces in Rd . We first state more precisely some definitions already given above. Let Γ be a lattice of maximal rank in Rd with d ≥ 2. Consider a real number r > d and set A := {(A1 , . . . , Ad ) | Aj ∈ LrR (Rd /Γ) for 1 ≤ j ≤ d}  and  r/2 V := LR (Rd /Γ),  where LrR (Rd /Γ) is the space of real-valued functions f such that |f |r is integrable on the torus Rd /Γ. For real-valued potentials (A, V ) ∈ A × V and for k ∈ Rd define the operator Hk (A, V ) := (i∇ + A − k)2 + V acting on L2 (Rd /Γ), where ∇ is the gradient operator in Rd /Γ. Recall the discussion in the last section and consider the eigenvalue-eigenvector problem Hk (A, V )ψn,k = En (k, A, V )ψn,k  for  ψn,k ∈ L2 (Rd /Γ).  The real “lifted” Fermi surface of (A, V ) with energy λ ∈ R is defined as Fbλ,R (A, V ) := {k ∈ Rd | En (k, A, V ) = λ for some n ≥ 1}. 10  Equivalently, Fbλ,R (A, V ) = {k ∈ Rd | (Hk (A, V ) − λ)ψ = 0 for some ψ ∈ DHk (A,V ) \ {0}}, where DHk (A,V ) ⊂ L2 (Rd /Γ) denotes the (dense) domain of Hk (A, V ). The adjective “lifted” indicates that Fbλ,R (A, V ) is a subset of Rd rather than Rd /Γ# . When d = 2 the Fermi surfaces are called Fermi curves, in particular. This is the main case of interest in this work. We next describe the physical context in which the concept of Fermi surface arises.  2.3  Electrons in a crystal  The purpose of this section is to motivate the definition of Fermi surfaces (for d ≤ 3) and briefly describe its physical meaning. For simplicity we consider the case d = 3. Our discussion is based on [9, 12]. A full rigorous theoretical description of crystalline solids is unavailable. In fact, no one has given an explanation from first principles of why crystals form. That is, no one has proven that a large number of heavy nuclei with enough electrons to produce neutrality, interacting via Coulomb potential, have a ground state that is approximately a crystal. Nevertheless, we may consider a simplified model for solids in attempt to describe some of the observed phenomena. This is what we discuss next. It is observed experimentally that the nuclei in a solid lie more or less in a regular array. For example, the common crystalline phase of iron has crystal structure given approximately by αe1 Z⊕αe2 Z⊕ α2 (e1 +e2 +e2 )Z ⊂ R3 , where α ≈ 2.87Å and {ej }3j=1 is the canonical base of R3 . Thus, we postulate in our model that at each site of a lattice Γ in R3 there is a fixed nucleus with a number of core electrons. Furthermore, we assume that the solid is filling all of R3 . Hence, if we ignore electron-electron interactions, and suppose that the fixed nuclei generate an electric potential V (x) and a magnetic potential A(x) = (A1 (x), . . . , A3 (x)) that are periodic with respect to Γ, we have a cloud of valence electrons moving in R3 subjected only to the influence of this lattice. Each one of these electrons has Hamiltonian H =  1 2me (i~∇  + eA)2 + V , where me and e are the mass and the charge of the electron,  respectively, and ~ is the Planck constant. This model is known as the independent electron model of solids. We shall outline how to use this model to describe the notion of density of states and give a qualitative explanation of the difference between metal and insulators. 11  Let B be a fundamental cell in R3 for the dual lattice Γ# and let {En (k)}∞ n=1 be the eigenvalues of Hk . Given a set X ⊂ R3 , denote by |X| its Lebesgue measure. The (integrated) density of states measure ρ is the measure on R defined by ∞  ρ(−∞, E] :=  2 X {k ∈ B | En (k) ≤ E} . |B| n=1  Since En (k) → ∞ uniformly in k as n → ∞, the number ρ(−∞, E] is finite. Furthermore, one can show that ρ is absolutely continuous with respect to dE, the Lebesgue measure on R. The Radon-Nikodym derivative dρ/dE is usually called the density of states. The (integrated) density of states is a concept of fundamental importance in condensed matter physics. It measures the “number of quantum states per unit volume” below a given energy. In fact, to understand the definition of ρ(−∞, E] given above, suppose for simplicity that for a given energy E there are exactly N bands En (k) such that En (k) < E for all k ∈ B. Then we have ρ(−∞, E] =  ∞  N  n=1  n=1  X 2|B| 2 X {k ∈ B | En (k) ≤ E} = = 2N. |B| |B|  That is, the “number of quantum states per volume |B|” below the energy E is 2N . Here, the factor of 2 comes from the Pauli principle. Since we are considering electrons (fermions), this principle asserts that each quantum state corresponding to the energy level En (k) can have at most two electrons. Thus, in the definition of ρ(−∞, E] above, we include a factor of 2 in the numerator. (We shall explain below how to implement the Pauli principle in our model). To explain the importance of ρ for the study of solids we introduce one more concept. Let D be a fundamental cell in R3 for the lattice Γ and, given an integer m, let D(m) be the set of volume m3 |D| obtained by gluing together an m × m × m set of D’s (see figure 2.2). Consider the operator Hm =  1 (i~∇p + eA)2 + V 2me  on L2 (D(m) ), where ∇p is the gradient operator with periodic boundary conditions. Let Ω be a subset of R and let Pm (Ω) be the spectral projection for Hm . Define ρm (−∞, E] :=  2 dim Pm (−∞, E]. m3  It is possible to show that ρm → ρ as m → ∞ in the sense that ρm (−∞, E] → ρ(−∞, E] for all E ∈ R [12]. We can now return to our model of solids. 12  m D (in R3 )  m  Figure 2.2: Construction of D(m) ⊂ R3 : gluing together an m × m × m set of cells D ⊂ R3 . Suppose that each nucleus in free space is surrounded by l electrons. Then in our model we wish to have l valence electrons per unit cell. While we ignore the interaction between electrons, we cannot ignore the Pauli principle. Again, this principle asserts that the quantum state corresponding to each eigenvalue of H can have at most two electrons. How do we take this into account when H does not have eigenfunctions and when there are infinitely many electrons (in our infinite crystal lattice)? We claim that a reasonable way of taking the Pauli principle into account is to say that in the ground state the electrons fill up the continuum eigenstates up to that energy E where ρ(−∞, E] = l. If we have a large but finite m × m × m crystal with periodic boundary conditions, there are m3 l electrons, and in the ground state these fill up the eigenstates of Hm up to an energy Em determined by ρm (−∞, Em ] = l. The smallest number E with ρ(−∞, E] = l is called the Fermi energy EF . The set of k ∈ B with En (k) = EF for some n is called the Fermi surface. Observe that this agrees with the definition of Fbλ,R (A, V ) given above. This picture is similar to the elementary discussion of the periodic table based on the hydrogen atom but with the complication of continuum states. We are now in a position to explain why electron conduction is hard in some solids (insulators) and easy in others (metals). In the ground state one can use complex conjugation symmetry to prove that there is no net movement of electrons—the expected value of the total momentum is zero. To get flow of electrons one must excite some of the electrons. Usually, a periodic Schrödinger operator has gaps in its spectrum. There is a qualitative difference if EF occurs at the bottom of a gap or not. If EF is at the bottom of a gap, then H has no spectrum in (EF , EF + ε), and there is a discrete amount of energy needed to set up a current. In this case one has an insulator. If EF is not at the bottom of a gap, one 13  has a metal. Of course, if EF is at the bottom of a small gap (ε small), or if EF is not at the bottom of a gap but is fairly close to the bottom of a gap, then one has an intermediate case where the metal/insulator distinction is not sharp (semiconductors). We should also note that in dealing with real solids one must take into account the fact that the solid is not in the ground state but rather in a finite temperature state determined by statistical mechanics. Notice that the gaps in the spectrum are crucial for this theory of insulators versus metals. The Fermi surface plays a fundamental role in this context. We refer the reader to [9] for more details.  Lowest energy for conduction Available for conduction  EF Filled band  Filled band Typical insulator  Typical metal  Figure 2.3: Energy bands in conductors and insulators.  2.4  Basic properties of Hk (A, V )  Recall that r > d ≥ 2 and let AC and VC be the “complexifications” of A and V, respectively. That is, AC := {(A1 , . . . , Ad ) | Aj ∈ Lr (Rd /Γ) for 1 ≤ j ≤ d}  and  VC := Lr/2 (Rd /Γ).  In this section we prove the following properties of the operator Hk (A, V ). Theorem 2.4.1 (Properties of Hk (A, V )). (a) Let k ∈ Cd and let (A, V ) ∈ AC × VC . Then if λ is not in the spectrum of Hk (A, V ) the resolvent (Hk (A, V ) − λ)−1 is a compact operator on L2 (Rd /Γ). (b) For k ∈ Rd and (A, V ) ∈ A × V, the operator Hk (A, V ) is self-adjoint in a dense domain DHk (A,V ) ⊂ L2 (Rd /Γ). 14  In order to prove this theorem we first introduce some notation and prove a number of propositions. Write H := L2 (Rd /Γ) and denote by L (H) the set of all bounded linear operators on H with respect to the operator norm k · k on H. Recall that H is a separable Hilbert space and that the trace class I1 is the set of linear operators A on H such that tr(|A|) < ∞. Here, the operator |A| is the positive square root of A∗ A. The trace ideals Ir and its associated norms are defined as Ir := {A ∈ L (H) | tr(|A|r ) < ∞} I∞ := {A ∈ L (H) | A is compact}  and kAkr := (tr(|A|r ))1/r  for 1 ≤ r < ∞,  and kAk∞ := kAk  for r = ∞.  Let µ(A) = {µn (A)}∞ n=1 be the singular values of A, that is, the eigenvalues of |A|. One can show that Ir is the set of compact operators on H whose singular values are in lr (N) and kAkr = kµ(A)klr (N) . This is true for any 1 ≤ r ≤ ∞ [15]. The properties of k · kr that we need are collected below. Proposition 2.4.2 (Properties of k · kr [15]). Let 1 ≤ r < s ≤ ∞ and suppose that A ∈ L (H) and B ∈ Ir . Then: (a) AB ∈ Ir and kABkr ≤ kAk kBkr ; (b) I1 ⊂ Ir ⊂ Is ⊂ I∞ with k · k ≤ k · ks ≤ k · kr ≤ k · k1 ; (c) B ∗ ∈ Ir and kB ∗ kr = kBkr . For any ϕ ∈ L2 (Rd /Γ) define ϕ̂ : Γ# → C as Z 1 ϕ̂(b) := (Fϕ)(b) := ϕ(x) e−ib·x dx, |Γ| Rd /Γ R where |Γ| := Rd /Γ dx. Then, ϕ(x) = (F −1 ϕ̂)(x) =  X  ϕ̂(b) eib·x  b∈Γ#  and kϕkL2 (Rd /Γ) = |Γ|1/2 kϕ̂kl2 (Γ# ) . To simplify the notation we sometimes write Lp and lp in place of Lp (Rd /Γ) and lp (Γ# ), respectively. 15  For f ∈ Lr and g ∈ lr define the operators g(−i∇) and f on a suitable domain of L2 as g(−i∇) : ϕ 7→  X  g(b)ϕ̂(b) eib·x  f : ψ 7→ f ψ.  and  b∈Γ#  It turns out that f (x)g(−i∇) and g(−i∇)f (x) are in the trace ideal Ir . More precisely, as in [15, Theorem 4.1], using complex interpolation we prove the following result. Proposition 2.4.3. Let 2 ≤ r ≤ ∞ and suppose that f ∈ Lr (Rd /Γ) and g ∈ lr (Γ# ). Then f (x)g(−i∇) ∈ Ir and g(−i∇)f (x) ∈ Ir . Furthermore, kf (x)g(−i∇)kr ≤ |Γ|−1/r kf kLr kgklr  (2.4.1)  kg(−i∇)f (x)kr ≤ |Γ|−1/r kf kLr kgklr .  (2.4.2)  and  Proof. We first prove (2.4.1) and (2.4.2) for r = 2 and r = ∞; then we interpolate using [15, Theorem 2.9]. For r = 2, kf (x)g(−i∇)k22 = tr [(f (x)g(−i∇))∗ (f (x)g(−i∇))] =  X  2  f (x)g(−i∇) |Γ|−1/2 eib·x  L2  b∈Γ#  =  X  f (x) |Γ|−1/2 g(b)eib·x  b∈Γ#  2 L2  X  = |Γ|−1  |g(b)|2 kf k2L2  b∈Γ#  = |Γ|−1 kf k2L2 kgk2l2 (2.4.3) and kg(−i∇)f (x)k22 = tr [(g(−i∇)f (x))∗ (g(−i∇)f (x))] =  X  g(−i∇)f (x) |Γ|−1/2 eib·x  b∈Γ#  2 L2  2  =  X  g(−i∇)  b∈Γ#  =  X c∈Γ#  X X  2  fˆ(c)|Γ|−1/2 ei(b+c)·x  = L2  X  X  b∈Γ#  c∈Γ#  g(b + c)fˆ(c)|Γ|−1/2 ei(b+c)·x L2  |g(b + c)|2 |fˆ(c)|2 = kfˆk2l2 kgk2l2 = |Γ|−1 kf k2L2 kgk2l2 ,  b∈Γ# c∈Γ#  (2.4.4) while for r = ∞, kf (x)g(−i∇)ϕk2L2 ≤ kf k2L∞ kg(−i∇)ϕk2L2 = |Γ| kf k2L∞ k(g(−i∇)ϕ)∧ k2l2 = |Γ| kf k2L∞ kg ϕ̂k2l2 ≤ |Γ| kf k2L∞ kgk2l∞ kϕ̂k2l2 = kf k2L∞ kgk2l∞ kϕk2L2 16  and kg(−i∇)f (x)ϕk2L2 = |Γ| k(g(−i∇)f ϕ)∧ k2l2 = |Γ| kg(f ϕ)∧ k2l2 ≤ |Γ| kgk2l∞ k(f ϕ)∧ k2l2 = kgk2l∞ kf ϕk2L2 ≤ kgk2l∞ kf k2L∞ kϕk2L2 . Hence, kf (x)g(−i∇)k ≤ kf kL∞ kgkl∞  (2.4.5)  kg(−i∇)f (x)k ≤ kf kL∞ kgkl∞ .  (2.4.6)  and  The general 2 ≤ r < ∞ case now follows by interpolation. Consider the family of norms   (tr(|A|1/t ))t for 0 < t ≤ 1 , 2 Φt (A) :=  kAk for t = 0. One can show that lim t↓0 Φt (A) = Φ0 (A) by applying that k · kl∞ = lim p→∞ k · klp to the singular values of A. Thus, we have a continuous family of norms. Let z 7→ F1 (z) := |Γ|z ei arg f (x) |f |zr (x) ei arg g(−i∇) |g|zr (−i∇) and z 7→ F2 (z) := |Γ|z ei arg g(−i∇) |g|zr (−i∇) ei arg f (x) |f |zr (x) be maps from the strip S := {z ∈ C | 0 ≤ Re z ≤ 1/2} to the set of linear operators on L2 . Suppose for the moment that f and g are simple functions. We may assume, without loss of generality, that kf kLr = kgklr = 1. We shall shortly prove the following property. Proposition 2.4.4. For any ϕ, ψ ∈ L2 , and for j ∈ {1, 2}, the function z 7→ hϕ, Fj (z)ψiL2 is continuous in S, analytic in its interior, and bounded. Furthermore, in view of (2.4.3) to (2.4.6), for j ∈ {1, 2} we have Φ0 (Fj (iy)) ≤ k |Γ|iy ei arg f |f |iyr kL∞ kei arg g |g|iyr kl∞ = 1  17  and Φ1/2 Fj  1 2  + iy    1  1  r  r  ≤ |Γ|− 2 k |Γ| 2 +iy ei arg f |f | 2 +iyr kL2 kei arg g |g| 2 +iyr kl2 r  r  r/2  r/2  = k|f | 2 kL2 k|g| 2 kl2 = kf kLr kgklr = 1. Hence, Fj (iy) ∈ I∞ and Fj  1 2   + iy ∈ I2 with kFj (iy)k ≤ 1 and  Fj  1 2  + iy   2  ≤ 1  for all y ∈ R and j ∈ {1, 2}. By complex interpolation [15, Theorem 2.9], it follows that Fj (z) ∈ I(Re z) with Φ(Re z) (Fj (z)) ≤ 1 for all z ∈ S and j ∈ {1, 2}. In particular, for z = 1/r, |Γ|1/r kf (x)g(−i∇)kr = Φ1/r (F1 (1/r)) ≤ 1 and |Γ|1/r kg(−i∇)f (x)kr = Φ1/r (F2 (1/r)) ≤ 1. This proves (2.4.1) and (2.4.2) for f and g as above. Since the set of simple functions is dense in Lr (Rd /Γ) and lr (Γ# ) for 2 ≤ r ≤ ∞, an approximation argument (using the triangle inequality) completes the proof. We now prove Proposition 2.4.4. Proof of Proposition 2.4.4. Recall that f and g are simple functions and suppose for the moment that ψ is a C ∞ -function on Rd /Γ with compact support. We first prove that, for all x ∈ Rd /Γ, the function z 7→ (ei arg g(−i∇) |g|zr (−i∇)ψ)(x) is continuous in S, analytic in its interior and bounded. Indeed, let hx,b (z) := ei arg g(b) |g(b)|zr ψ̂(b)eib·x . Observe that z 7→ hx,b (z) has the desired properties. Then, since |g(b)|r ≤ kgkrlr ≤ 1 for all P P b ∈ Γ# , it follows that b∈Γ# |hx,b (z)| ≤ b∈Γ# |ψ̂(b)| < ∞ (because ψ̂ ∈ l1 since we have assume that ψ is a C ∞ -function). Thus, the sum (ei arg g(−i∇) |g|zr (−i∇)ψ)(x) =  X  hx,b (z)  b∈Γ#  converges uniformly in S (in the uniform norm) by the Weierstrass M-test. Hence it has the same properties as hx,b (z) and the claim follows. 18  Now observe that, for all x ∈ Rd /Γ, the maps z 7→ F1 (z)ψ(x) = |Γ|z ei arg f (x) |f (x)|zr (ei arg g(−i∇) |g|zr (−i∇)ψ)(x) and z 7→ F2 (z)ψ(x) = |Γ|z ei arg g(−i∇) |g|zr (−i∇)ei arg f (x) |f (x)|zr ψ(x) have the same properties as the above functions. (Note that, in the second map, we can think of ei arg f (x) |f (x)|zr ψ(x) as a new ψ̃(x) with the same properties of ψ(x).) Thus, for j ∈ {1, 2}, the function z 7→ hϕ, Fj (z)ψiL2 is continuous and bounded in S. This remains true for any ψ ∈ L2 because the set of C ∞ -functions with compact support is dense in L2 . Furthermore, for j ∈ {1, 2} the function z 7→ hϕ, Fj (z)ψiL2 is also analytic in the interior of S because, by Cauchy’s Formula and Fubini’s Theorem,   Z Z Z Fj (ξ)ψ(x) hϕ, Fj (ξ)ψiL2 1 1 hϕ, Fj (z)ψiL2 = ϕ(x) dξ dx = dξ, 2πi γ ξ − z 2πi γ ξ−z Rd /Γ where γ ⊂ S is a closed path around z. This completes the proof of the proposition. The following notation will be used whenever we consider vector-valued quantities. Let X be a Banach space and let A = (A1 , . . . , Ad ) ∈ X d and B = (B1 , . . . , Bd ) ∈ X d . Then, kAkX := (kA1 k2X + · · · + kAd k2X )1/2  and  A · B := A1 B1 + · · · + Ad Bd .  The next three propositions are simple straightforward calculations. Proposition 2.4.5. Let A ∈ (L (H))d and B ∈ Ird . Then A · B ∈ Ir and B · A ∈ Ir with kA · Bkr ≤ kAk kBkr  and  kB · Akr ≤ kAk kBkr .  Proof. Using the properties of k·kr (see Proposition 2.4.2) and the Cauchy-Schwarz inequality we have kA · Bkr = kA1 B1 + · · · + Ad Bd kr ≤ kA1 B1 kr + · · · + kAd Bd kr ≤ kA1 k kB1 kr + · · · + kAd k kBd kr ≤ kAk kBkr and kB · Akr = kB1 A1 + · · · + Bd Ad kr ≤ kB1 A1 kr + · · · + kBd Ad kr = kA1 B1 kr + · · · + kAd Bd kr ≤ kA1 k kB1 kr + · · · + kAd k kBd kr ≤ kAk kBkr , as desired. 19  Proposition 2.4.6. The following inequalities hold: ≤ d1/2  √ 1 ∇ I−∆  √ 1 k I−∆  and  ≤ |k|.  Proof. Let ϕ ∈ L2 . Then, for 1 ≤ j ≤ d, 2 ∂ √ 1 ϕ ∂x j I−∆ L2  = |Γ|  √  2 bj ϕ̂(b) 2 1+b l2  ≤ |Γ| kϕ̂k2l2 = kϕk2L2  and √ 1 kj I−∆  2  ϕ  L2  = |Γ|  √  2 kj ϕ̂(b) 2 1+b l2  ≤ |Γ| |kj |2 kϕ̂k2l2 = |kj |2 kϕk2L2 .  Now observe that, if T is a bounded linear operator and kT ϕk2L2 ≤ Ckϕk2L2 for all ϕ ∈ L2 then kT k2 ≤ C. Hence, √ 1 ∇ I−∆  d X  =  ∂ √ 1 I−∆ ∂xj  2  !1/2  d X  ≤  ϕ  j=1  =  d X  = d1/2  1  j=1  and √ 1 k I−∆  !1/2  √ 1 kj I−∆  2  ϕ  !1/2  d X  ≤  j=1  !1/2 2  |kj |  = |k|,  j=1  as was to be shown. Proposition 2.4.7. The function g : Γ# → R given by g(b) = (1+b2 )−1/2 is in lr for r > d. Proof. There are constants CΓ# and CΓ# ,d such that kgkrlr =  X  |g(b)|r ≤ CΓ#  b∈Γ#  Z = CΓ# ,d  ∞  Z Rd  (1 + b2 )−r/2 db = CΓ#  Z  ∞  !  Z dS  0  ∂B(0,ρ)  ρd−1 (1 + ρ2 )−r/2 dρ < ∞,  0  since the above integral converges for d − 1 − r < −1. Now write Hk (A, V ) − λ = I − ∆ + u(k, λ) + w(k, A, V ), where ∆ is the Laplace operator in Rd , u(k, λ) := −2ik · ∇ + k 2 − λ − I, and w(k, A, V ) := i∇ · A + iA · ∇ − 2k · A + A2 + V. Applying Propositions 2.4.2 to 2.4.7 we prove the following estimates. 20  (1 + ρ2 )−r/2 dρ  Lemma 2.4.8. Let r > d. There is a constant C = CΓ,r,d such that √ 1 w(k, A, V ) √ 1 I−∆ I−∆ r  ≤ C (1 + |k|)kAkLr + kAk2Lr + kV kLr/2    (a)  and √ 1 u(k, λ) √ 1 I−∆ I−∆ r  Furthermore, let 0 ≤ ε ≤  r−d 2r .  ≤ C (1 + |k|2 + |λ|).  (b)  There is a constant C = CΓ,r,d,k,λ,A,V such that  |h(u(k, λ) + w(k, A, V ))ϕ, ψiL2 | ≤ C k(I − ∆)(1−ε)/2 ϕkL2 k(I − ∆)1/2 ψkL2 + k(I − ∆)1/2 ϕkL2 k(I − ∆)(1−ε)/2 ψkL2    ≤ 2C k(I − ∆)1/2 ϕkL2 k(I − ∆)1/2 ψkL2  (c)  for all ϕ, ψ ∈ L2 (Rd /Γ). Proof. (a) Write √  1 = g(−i∇) I −∆  with  g(b) = √  1 . 1 + b2  Then, using the properties of k · kr , in particular that kgAkr = k(gA)∗ kr = kA∗ g ∗ kr = kA∗ gkr , and Proposition 2.4.6, it follows that kgwgkr = kg(i∇ · A + iA · ∇ − 2k · A + A2 + |V |ei arg V )gkr (by the triangle inequality) ≤ kg(∇ · A)gkr + 2kgA · ∇gkr + 2kgk · Agkr + kgA · Agkr + kg|V |1/2 ei arg V |V |1/2 gkr (by Proposition 2.4.5) ≤ kg∇k kAgkr + k∇gk kgAkr + 2kgkk kAgkr + kgAk kAgkr + kg|V |1/2 k k|V |1/2 gkr = kg∇k kAgkr + k∇gk kA∗ gkr + 2kgkk kAgkr + kA∗ gk kAgkr + k|V |1/2 gk k|V |1/2 gkr ≤ (d1/2 + 2|k|)kAgkr + d1/2 kA∗ gkr + kA∗ gkr kAgkr + k|V |1/2 gk2r . Now, since g ∈ lr for r > d by Proposition 2.4.7, applying Proposition 2.4.3 we obtain kgwgkr ≤ 2(d1/2 + |k|)|Γ|−1/r kAkLr kgklr + |Γ|−2/r kAk2Lr kgk2lr + |Γ|−2/r k|V |1/2 k2Lr kgk2lr  ≤ CΓ,r,d (1 + |k|)kAkLr + kAk2Lr + kV kLr/2 . This proves part (a). 21  (b) The spectrum of g(−i∇)u(k, λ)g(−i∇) is  2k · b + k 2 − λ − 1 1 + b2  #  b∈Γ   .  Hence, 2k · b + k 2 − λ − 1 1 + b2 lr |b| 1 ≤ 2|k| + 2 1 + b lr 1 + b2  kgugkr =  (1 + |k|2 + |λ|) lr  ≤ CΓ,r (1 + |k|2 + |λ|), which proves part (b). √ (c) Write D = I − ∆. The condition on ε implies that r(1 − ε) ≥  r+d 2  Proposition 2.4.7, (1 + b2 )−r(1−ε)/2 ∈ l1 (Γ# ), and hence D−(1−ε) ∈ Ir . Thus, as in part (a), kD−1 ∇ · AD−(1−ε) kr ≤ CΓ,r,d kAkLr , kk · AD−(1−ε) kr ≤ CΓ,r,d |k| kAkLr , kD−(1−ε) A · AD−(1−ε) kr ≤ CΓ,r,d kAk2Lr , kD−(1−ε) V D−(1−ε) kr ≤ CΓ,r,d kV kLr/2 . Furthermore, kD−1 uk ≤ sup b∈Γ#  2k · b + k 2 − λ − 1 √ ≤ 2 (1 + |k|2 + |λ|). 1 + b2  Consequently, |hi∇ · Aϕ, ψi| = |hD−1 i∇ · AD−(1−ε) D1−ε ϕ, Dψi| ≤ kD−1 ∇ · AD−(1−ε) D1−ε ϕkL2 kDψkL2 ≤ kD−1 ∇ · AD−(1−ε) k kD1−ε ϕkL2 kDψkL2 ≤ kD−1 ∇ · AD−(1−ε) kr kD1−ε ϕkL2 kDψkL2 ≤ CΓ,r,d kAkLr kD1−ε ϕkL2 kDψkL2 .  22  > d so that, by  Similarly, |hiA · ∇ϕ, ψi| ≤ CΓ,r,d kAkLr kDϕkL2 kD1−ε ψkL2 , |h2k · A ϕ, ψi| ≤ CΓ,r,d |k| kAkLr kD1−ε ϕkL2 kψkL2 , |hA · A ϕ, ψi| ≤ CΓ,r,d kAk2Lr kD1−ε ϕkL2 kD1−ε ψkL2 , |hV ϕ, ψi| ≤ CΓ,r,d kV kLr/2 kD1−ε ϕkL2 kD1−ε ψkL2 , |huϕ, ψi| ≤ CΓ,r,d (1 + |k|2 + |λ|)kϕkL2 kDψkL2 . Finally, since kDq ϕkL2 ≤ kDs ϕkL2 for all 0 ≤ q ≤ s and all ϕ ∈ L2 , the above estimates imply inequality (c). The proof of the lemma is complete. Lemma 2.4.8 says that the operators given in (a) and (b) belong to Ir and that the quadratic form in (c) is well-defined on the domain H 1 (Rd /Γ) × H 1 (Rd /Γ). Here, the space H 1 (Rd /Γ) is the usual Sobolev space with norm k · kH 1 (Rd /Γ) = k(I − ∆)1/2 · kL2 (Rd /Γ) . We now prove the following general property of Sobolev norms. Proposition 2.4.9. Let 0 < r < s < t. Then, given δ > 0 there is a positive constant C = Cδ,r,s,t such that kϕkH s (Rd /Γ) ≤ δ kϕkH t (Rd /Γ) + C kϕkH r (Rd /Γ) for all ϕ ∈ H t (Rd /Γ). Proof. We first find a constant C > 0 such that (1 + b2 )s ≤ δ 2 (1 + b2 )t + C 2 (1 + b2 )r for all b ∈ Γ# or, equivalently, setting y := 1 + b2 , such that δ 2 y t−s + C 2  1 y s−r  ≥1  1 . Then, for all y ≥ 1. Let φ(y) := δ 2 y t−s + C 2 ys−r  φ0 (y) =  1 y s+1  (δ 2 (t − s)y t + C 2 (r − s)y r ),  so that φ(y) has only one critical point for y > 0, namely, C 2 (s − r) y := δ 2 (t − s) ∗    23  1  t−r  .  Since limy→0 φ(y) = limy→∞ φ(y) = ∞, the point y ∗ is a point of minimum and, for all y ≥ 1, φ(y) ≥ φ(y ∗ ) = C  2(t−s) t−r  " δ2    s−r 2 δ (t − s)   t−s t−r  s−r + 2 δ (t − s)  if we choose " C = δ2    s−r 2 δ (t − s)   t−s t−r      s−r + 2 δ (t − s)   r−s # t−r  =1  r−t  r−s # 2(t−s) t−r  .  Consequently, kϕk2H s = k(I − ∆)s/2 ϕk2L2 = |Γ| k((I − ∆)s/2 ϕ)∧ k2l2 = |Γ|  X  (1 + b2 )s |ϕ̂(b)|2  b∈Γ#  ≤ δ 2 |Γ|  X  (1 + b2 )t |ϕ̂(b)|2 + C 2 |Γ|  X  (1 + b2 )r |ϕ̂(b)|2  b∈Γ#  b∈Γ#  = δ 2 kϕk2H t + C 2 kϕk2H r ≤ (δkϕkH t + CkϕkH r )2 . Taking the square root of both sides of the last expression we obtain the desired inequality.  We now show that Hk (A, V ) is a self-adjoint semibounded operator (on a suitable domain) when k, A and V are real. To prove this property we consider the quadratic form associated to this operator (see [11, §VIII.6]). Proof of Theorem 2.4.1(b). Let D = H 1 (Rd /Γ) be the domain of  √  I − ∆. By Lemma 2.4.8,  the operator (i∇ + A − k)2 + V − λ = I − ∆ + u(k, λ) + w(k, A, V ) gives a well-defined quadratic form q : D × D −→ C, (ϕ, ψ) 7−→ h(I − ∆ + u(k, λ) + w(k, A, V ))ϕ, ψiL2 . Furthermore, for ϕ ∈ D, when k, A and V are real, √ q(ϕ, ϕ) = k I − ∆ ϕk2 + h(u(k, λ) + w(k, A, V ))ϕ, ϕiL2 √ √ ≥ k I − ∆ ϕk2 − Ck(I − ∆)(1−ε)/2 ϕk k I − ∆ ϕk, where C > 0 is a constant and k · k = k · kL2 (to simplify the notation). By Proposition 2.4.9, for any δ > 0 there is a constant Cδ > 0 such that √ k(I − ∆)(1−ε)/2 ϕk ≤ δ k I − ∆ ϕk + Cδ kϕk 24  for all ϕ ∈ D. Choosing δ = (2C)−1 we obtain √ √ q(ϕ, ϕ) ≥ (1 − Cδ)k I − ∆ ϕk2 − CCδ kϕk k I − ∆ ϕk √ √ = 21 k I − ∆ ϕk2 − CCδ kϕk k I − ∆ ϕk =  1 4k  √  2  2  2  I − ∆ ϕk − (CCδ ) kϕk + Ckϕk  !2 √ p k I − ∆ ϕk Cδ Ckϕk − p 2 Ckϕk  √ ≥ 14 k I − ∆ ϕk2 − (CCδ )2 kϕk2 ≥ −(CCδ )2 kϕk2 . Hence, the form q is semibounded and 1 4k  √ √ I − ∆ ϕk2 − (CCδ )2 kϕk2 ≤ q(ϕ, ϕ) ≤ Ck I − ∆ ϕk2 .  (2.4.7)  √ Since D = H 1 (Rd /Γ) is complete and kϕkH 1 = k I − ∆ ϕk, whenever {ϕn } ⊂ D with ϕn → ϕ in L2 and q(ϕn − ϕm , ϕn − ϕm ) → 0 as n, m → ∞, it follows from (2.4.7) that kϕn − ϕm kH 1 → 0 as n, m → ∞ and consequently that ϕ ∈ D with q(ϕn − ϕ, ϕn − ϕ) → 0. This implies that the form q is closed. By [11, Theorem VIII.15], there is a unique self-adjoint semibounded operator Hk (A, V ) densely defined on D ⊂ L2 (Rd /Γ) associated to q. Finally, we outline the proof of Theorem 2.4.1(a). Outline of the proof of Theorem 2.4.1(a). To simplify the notation write Hk = Hk (A, V ), and denote by σ(Hk ) the spectrum of Hk . If λ 6∈ σ(Hk ), the resolvent (Hk − λ)−1 exists and is bounded. This is just the definition of spectrum. We first claim that, if (Hk − λ)−1 is compact for some λ 6∈ σ(Hk ), then it is compact for all λ 6∈ σ(Hk ). In fact, suppose that λ, λ0 6∈ σ(Hk ). Then, by the resolvent identity we have (Hk − λ0 )−1 = (Hk − λ)−1 + (Hk − λ)−1 (λ − λ0 )(Hk − λ0 )−1   = (Hk − λ)−1 I + (λ − λ0 )(Hk − λ0 )−1 .   The factor I + (λ − λ0 )(Hk − λ0 )−1 is a bounded operator. Thus, if (Hk − λ)−1 is compact for some λ ∈ / σ(Hk ), the above product is compact and consequently (Hk − λ0 )−1 is compact for all λ0 ∈ / σ(Hk ). This shows that is enough to prove part (a) of theorem for a single λ∈ / σ(Hk ). This is what we do next. 25  Write Hk = −∆ + Q with Q := i∇ · (A − k) + (A − k) · i∇ + (A − k)2 + V. Then, if λ ∈ / σ(Hk ), and if |λ| is sufficiently large, the operator  −1 1 1 I+√ Q√ −∆ − λ −∆ − λ exists and is bounded. Furthermore, we can choose the λ so that  √ 1 −∆−λ  exists, and similarly  as in the proof of Lemma 2.4.8(c), we can prove that this operator is compact. Thus we can write (Hk − λ)−1 = √  1  1 −∆ − λ I +  and similarly as above conclude that (Hk −  √  1 √ 1 Q √−∆−λ −∆−λ λ)−1 is compact for  1 , −∆ − λ  a suitable λ ∈ / σ(Hk ). In  view of the above remark, this implies the statement for all λ ∈ / σ(Hk ) and completes the outline of the proof.  2.5  The complex analytic structure of the spectrum  For complex-valued potentials (A, V ) ∈ AC × VC and for k ∈ C2 the problem Hϕn,k = En (k)ϕn,k , ϕn,k (x + γ) = eik·γ ϕn,k (x) for all x ∈ Rd and all γ ∈ Γ is no longer self-adjoint. Its spectrum, however, remains discrete. It is a sequence of eigenvalues in the complex plane. From the above boundary condition it is easy to see that for each n ≥ 1 the function k 7→ En (k) remains periodic with respect to Γ# . Furthermore, the transformation Uk is no longer unitary but it is still bounded and invertible and it still preserves the spectrum. That is, we can still rewrite this problem in the form Hk ψn,k = En (k)ψn,k  for  ψn,k ∈ L2 (Rd /Γ)  without modifying the eigenvalues. Thus, as above we define the (complex) “lifted” Fermi surface of (A, V ) with energy λ ∈ C to be Fbλ (A, V ) := {k ∈ Cd | (Hk (A, V ) − λ)ψ = 0 for some ψ ∈ DHk (A,V ) \ {0}}. 26  We shall prove below that this surface has the following (well-known) property. Theorem 2.5.1 ([3]). There exists an analytic function F on Cd × C × AC × VC such that Fbλ (A, V ) = {k ∈ Cd | F (k, λ, A, V ) = 0}. In particular, for k, A and V real, λ ∈ Spec (Hk (A, V ))  if and only if  F (k, λ, A, V ) = 0.  Furthermore, such an analytic function F is given by (2.5.1). To prove this theorem we follow [3]. Another proof of a similar statement can be found in [8, Theorem 4.4.2]. We shall use the definition and some properties of the regularized determinant on Ir (see [15, Chapter 9]). Proof. Since Lp (Rd /Γ) ⊃ Lq (Rd /Γ) for all 1 ≤ p < q, we may assume, without loss of generality, that r ≤ d + 1. Then, since k · kd+1 ≤ k · kr , Lemma 2.4.8 implies that  F (k, λ, A, V ) := detd+1 I +  √ 1 u(k, λ) √ 1 I−∆ I−∆  +  √ 1 w(k, A, V ) √ 1 I−∆ I−∆    (2.5.1)  is a well-defined function on Cd × C × AC × VC . Here detd+1 (I + B) is the regularized determinant for B ∈ Id+1 . This function is Fréchet differentiable [15] and hence analytic on its domain [1, Theorem 14.13]. Analyticity may also be proved as in [6] by approximating B by a sequence of finite rank operators. Now, let µ be a complex number that is not in the spectrum of Hk (A, V ). Then, by Theorem 2.4.1(a), the resolvent (Hk (A, V )−µ)−1 is a compact operator on L2 (Rd /Γ). Thus, the spectrum of (Hk (A, V )−µ)−1 is discrete, and so is the spectrum of Hk (A, V ). Therefore, λ ∈ Spec(Hk (A, V )) if and only if there exists ψ ∈ DHk (A,V ) ⊂ D such that (Hk (A, V ) − λ) √ This is the case if and only if  √ 1 (Hk (A, V I−∆  √ 1 I − ∆ ψ = 0. I −∆  1 )−λ) √I−∆ is not invertible. Since this operator  is the sum of the identity and a compact operator, it fails to be invertible if and only if it has a nontrivial kernel. By [15, Theorem 9.2], this is the case if and only if F (k, λ, A, V ) = 0.  27  2.6  Gauge invariance  In this section we outline the proof of the following (well-known) properties. Theorem 2.6.1 (Gauge invariance). For 1 ≤ j ≤ d, let Aj ∈ C 1 (Rd /Γ), V ∈ C 0 (Rd /Γ), and Ψ ∈ C 2 (Rd /Γ). Then DHk (A,V ) = D−∆+I  eiΨ : DHk (A,V ) → DHk (A,V ) .  and  Furthermore: (a) Ker(Hk (A, V ) − λ) 6= {0} if and only if Ker(e−iΨ Hk (A, V )eiΨ − λ) 6= {0}; (b) e−iΨ Hk (A, V )eiΨ = Hk (A − ∇Ψ, V ); (c) Fbλ (A, V ) = Fbλ (A − ∇Ψ, V ). Outline of the Proof. We first prove parts (a) to (c). (a) Consider the linear transformation eiΨ : ϕ(x) 7−→ eiΨ(x) ϕ(x) acting on DHk (A,V ) . Since the function Ψ is bounded, it is clear that eiΨ is a bounded operator with bounded inverse on L2 (Rd /Γ) given by e−iΨ . Define HkΨ (A, V ) := e−iΨ Hk (A, V )eiΨ  and  DH Ψ (A,V ) := e−iΨ DHk (A,V ) . k  Observe that, by hypothesis, DH Ψ (A,V ) = DHk (A,V ) . Furthermore, we claim that k  Hk (A, V )ψ = λψ for ψ ∈ DHk (A,V ) with ψ 6= 0, if and only if HkΨ (A, V )ϕ = λϕ, where ϕ = e−iΨ ψ with ϕ ∈ DHk (A,V ) and ϕ 6= 0. Indeed, observe that the former equation implies the last one, 0 = e−iΨ 0 = e−iΨ (Hk (A, V ) − λ)ψ = e−iΨ (Hk (A, V ) − λ)eiΨ e−iΨ ψ = (e−iΨ Hk (A, V )eiΨ − λ)e−iΨ ψ = (HkΨ (A, V ) − λ)ϕ, where ϕ 6= 0 since ψ 6= 0, and that a similar calculation (in the reverse order) implies the converse statement. This proves part (a). 28  (b) To simplify the notation write A0 = A − ∇Ψ. Observe that, formally (without worry about domains and weak derivatives), e−iΨ Hk (A, V )eiΨ ϕ = e−iΨ ((i∇ + A − k)2 + V )eiΨ ϕ = e−iΨ (i∇ + A − k) · eiΨ (−(∇Ψ) + i∇ + A − k))ϕ + V ϕ = −(∇Ψ) + A − k) · (−(∇Ψ) + i∇ + A − k)ϕ + i∇ · (−(∇Ψ) + i∇ + A − k)ϕ + V ϕ = (k 2 − 2ik · ∇ − ∆ + A0 · (i∇ − k) + (i∇ − k) · A0 + (A0 )2 + V )ϕ = ((i∇ − k)2 + (i∇ − k) · A0 + A0 · (i∇ − k) + (A0 )2 + V )ϕ = ((i∇ + A0 − k)2 + V )ϕ. However, since e−iΨ DHk (A,V ) = DHk (A,V ) , we claim that this calculation can be rigorously justified indeed. Assuming this we have e−iΨ Hk (A, V )eiΨ = Hk (A − ∇Ψ, V ), which proves part (b). (c) As a consequence of parts (a) and (b) we obtain Fbλ (A, V ) = {k ∈ Cd | (Hk (A, V ) − λ)ψ = 0 for some ψ ∈ DHk (A,V ) \ {0}} = {k ∈ Cd | (e−iΨ Hk (A, V )eiΨ − λ)ϕ = 0 for some ϕ ∈ DHk (A,V ) \ {0}} = {k ∈ Cd | (Hk (A − ∇Ψ, V ) − λ)ϕ = 0 for some ϕ ∈ DHk (A−∇Ψ,V ) \ {0}} = Fbλ (A − ∇Ψ, V ), as desired. We now outline the proof of the first part of the theorem. First observe that, since D−∆+I = H 2 (Rd /Γ), and H 2 (Rd /Γ) is dense in L2 (Rd /Γ), the set D−∆+I is dense in DHk (A,V ) . Furthermore, it is easy to verify that D−∆+I ⊂ DHk (A,V ) . Thus, if we can prove that Hk (A, V ) with domain D−∆+I is closed, then it follows that DHk (A,V ) = D−∆+I . That is, we can choose D−∆+I as a domain for Hk (A, V ). Furthermore, since Ψ ∈ C 2 (Rd /Γ), it is clear then that eiΨ maps DHk (A,V ) to DHk (A,V ) . Therefore, to conclude the proof of the theorem it suffices to show that Hk (A, V ) is closed in D−∆+I . 29  The operator Hk (A, V ) is closed in D−∆+I if its graph is closed in D−∆+I × D−∆+I . Equivalently, its graph is closed if for any {ϕn } ⊂ D−∆+I such that limn→∞ ϕn =: ϕ exists and limn→∞ Hk (A, V )ϕn =: ψ exists, then ϕ ∈ D−∆+I and Hk (A, V )ϕ = ψ with ψ ∈ D−∆+I . To prove that Hk (A, V ) is closed write Hk (A, V ) = −∆ + Q with Q := i∇ · (A − k) + (A − k) · i∇ + (A − k)2 + V, and assume that limn→∞ ϕn =: ϕ and limn→∞ Hk (A, V )ϕn =: ψ exist. Then, for a suitable constant λ,   (Hk (A, V ) + λ)ϕn = (−∆ + λ + Q)ϕn = I + Q(−∆ + λ)−1 (−∆ + λ)ϕn , where the operator Q(−∆ + λ)−1 is bounded. Furthermore, we can choose λ sufficiently large so that the operator norm of Q(−∆ + λ)−1 is strictly less than 1. Consequently, the   operator I + Q(−∆ + λ)−1 is bounded and has a bounded inverse. Since by hypothesis {(Hk (A, V ) + λ)ϕn } converges, we conclude that limn→∞ (−∆ + λ)ϕn exists. It follows then that ϕ ∈ D−∆+I and (−∆ + λ)ϕ ∈ D−∆+I because −∆ + λ is closed in D−∆+I (we are using   this fact without proof). Finally, since I + Q(−∆ + λ)−1 is bounded and has a bounded inverse, it follows that limn→∞ (Hk (A, V ) + λ)ϕn is in D−∆+I . All this together shows that Hk (A, V ) is closed in D−∆+I and completes the proof of the theorem.  30  Chapter 3  Asymptotics for Fermi curves 3.1  Fermi curves  Below we define the Fermi curves and briefly describe some of its properties. Let Γ be a lattice in R2 and let A1 , A2 and V be real-valued functions in L2 (R2 /Γ). Set A := (A1 , A2 ) and define the operator H(A, V ) := (i∇ + A)2 + V acting on L2 (R2 ), where ∇ is the gradient operator in R2 . For k ∈ R2 consider the following self-adjoint eigenvalue-eigenvector problem with boundary conditions, H(A, V )ϕ = λϕ, ϕ(x + γ) = eik·γ ϕ(x) for all x ∈ R2 and all γ ∈ Γ. The spectrum of this problem is discrete. It consists of a sequence of real eigenvalues E1 (k, A, V ) ≤ E2 (k, A, V ) ≤ · · · ≤ En (k, A, V ) ≤ · · · For each integer n ≥ 1 the eigenvalue En (k, A, V ) defines a continuous function of k. From the above boundary condition it is easy to see that this function is periodic with respect to the dual lattice Γ# := {b ∈ R2 | b · γ ∈ 2πZ for all γ ∈ Γ},  31  where b · γ is the usual scalar product on R2 . It is customary to refer to k as the crystal momentum and to En (k, A, V ) as the n-th band function. The corresponding normalized eigenfunctions ϕn,k are called Bloch eigenfunctions. Let Uk be the unitary transformation on L2 (R2 ) that acts as Uk : ϕ(x) 7−→ eik·x ϕ(x). By applying this transformation we can rewrite the above problem and put the boundary conditions into the operator. Indeed, if we define Hk (A, V ) := Uk−1 H(A, V ) Uk  and  ψ := Uk−1 ϕ,  then the above problem is unitary equivalent to Hk (A, V )ψ = λψ, ψ(x + γ) = ψ(x) for all x ∈ R2 and all γ ∈ Γ, or, using a more compact notation, Hk (A, V )ψ = λψ  for  ψ ∈ L2 (R2 /Γ).  To see that these problems are equivalent we proceed (formally) as follows. On the one hand, from the original problem and using the above transformation we find that 0 = Uk−1 0 = Uk−1 (H(A, V ) − λ)ϕ = Uk−1 (H(A, V ) − λ)Uk Uk−1 ϕ = (Uk−1 H(A, V )Uk − λ)Uk−1 ϕ = (Hk (A, V ) − λ)ψ and ψ(x + γ) = (Uk−1 ϕ)(x + γ) = e−ik·(x+γ) ϕ(x + γ) = e−ik·(x+γ) eik·γ ϕ(x) = e−ik·x ϕ(x) = (Uk−1 ϕ)(x) = ψ(x). On the other hand, by a similar computation (in the reverse order), using the last two equalities and the above transformation we derive the original problem. This (formally) implies unitary equivalence. Furthermore, a simple (formal) calculation shows that Hk (A, V ) = (i∇ + A − k)2 + V.  32  In fact, Hk (A, V )ψ = Uk−1 H(A, V )Uk ψ   = e−ik·x (i∇ + A)2 + V eik·x ψ   = e−ik·x (i∇ + A) · eik·x (−kψ + i∇ψ + Aψ) + V ψ = (−k + A) · (−k + i∇ + A)ψ + i∇ · (−k + i∇ + A)ψ + V ψ   = (i∇ + A − k)2 + V ψ. Of course, the unitary transformation Uk preserves self-adjointness and does not change the spectrum {En (k, A, V )}∞ n=1 . The real “lifted” Fermi curve of (A, V ) with energy λ ∈ R is defined as Fbλ,R (A, V ) := {k ∈ R2 | En (k, A, V ) = λ for some n ≥ 1}. Equivalently, Fbλ,R (A, V ) = {k ∈ R2 | (Hk (A, V ) − λ)ϕ = 0 for some ϕ ∈ DHk (A,V ) \ {0}}, where DHk (A,V ) ⊂ L2 (R2 /Γ) denotes the (dense) domain of Hk (A, V ). The adjective “lifted” indicates that Fbλ,R (A, V ) is a subset of R2 rather than R2 /Γ# . As we may replace V by V − λ, we only discuss the case λ = 0 and write FbR (A, V ) in place of Fb0,R (A, V ) to simplify the notation. Furthermore, since Hk (A, V ) = Hk−Â(0) (A − Â(0), V ), if we perform the change of coordinates k → k + Â(0) and redefine A − Â(0) → A we may assume, without loss of generality, that Â(0) =  1 |Γ|  Z A(x) dx = 0. R2 /Γ  The dual lattice Γ# acts on R2 by translating k 7→ k + b for b ∈ Γ# . This action maps FbR (A, V ) to itself because for each n ≥ 1 the function k 7→ En (k, A, V ) is periodic with respect to Γ# . In other words, the real lifted Fermi curve “is periodic” with respect to Γ# . Define FR (A, V ) := FbR (A, V )/Γ# . We call FR (A, V ) the real Fermi curve of (A, V ). It is a curve in the torus R2 /Γ# . The above definitions and the real Fermi curve have physical meaning. It is useful and interesting, however, to study the “complexification” of these curves. Knowledge about the complexified curves may provide information about the real counterparts. 33  For complex-valued functions A1 , A2 and V in L2 (R2 /Γ) and for k ∈ C2 the above problem is no longer self-adjoint. Its spectrum, however, remains discrete. It is a sequence of eigenvalues in the complex plane. From the boundary condition in the original problem it is easy to see that the family of functions k 7→ En (k, A, V ) remains periodic with respect to Γ# . Furthermore, the transformation Uk is no longer unitary but it is still bounded and invertible and it still preserves the spectrum, that is, we can still rewrite the original problem in the form Hk (A, V )ψ = λψ  for  ψ ∈ L2 (R2 /Γ)  without modifying the eigenvalues. Thus, it makes sense to define b F(A, V ) := {k ∈ C2 | Hk (A, V )ϕ = 0 for some ϕ ∈ DHk (A,V ) \ {0}} and b F(A, V ) := F(A, V )/Γ# . b We call F(A, V ) and F(A, V ) the (complex) “lifted” Fermi curve and the (complex) Fermi curve, respectively. When there is no risk of confusion we shall refer to either simply as Fermi curve. Let {γ1 , γ2 } be a basis of Γ and set C∗ := C \ {0}. Define the exponential map E as E : C2 −→ C∗ × C∗ , k 7−→ (eik·γ1 , eik·γ2 ). This map is holomorphic and the pair (C2 , E) is a covering space of C∗ × C∗ . In fact, every point of C∗ × C∗ has an open neighbourhood W ⊂ C∗ × C∗ such that the inverse image of W under E is a disjoint union of open sets Uj ⊂ C2 , with the map E sending each Uj b homeomorphically onto W . If we recall that F(A, V ) is invariant under the action of Γ# and observe that b · γ1 ∈ 2πZ and b · γ2 ∈ 2πZ for all b ∈ Γ# , it is not difficult to see that  b E F(A, V) ∼ = F(A, V ). That is, up to the isomorphism  b J : E F(A, V ) −→ F(A, V ), (eik·γ1 , eik·γ2 ) 7−→ [k], 34  where [k] denotes a point (or equivalence class) in R2 /Γ# , the curve F(A, V ) is the image b of F(A, V ) under the exponential map. We sometimes assume that the isomorphism J is understood and simply write  b V ) = F(A, V ). E F(A, Alternatively, we could have used this expression to define F(A, V ) (and avoid talking about the isomorphism J).  3.2  The free Fermi curve  b When the potentials A and V are zero the curve F(A, V ) can be found explicitly. In this section we collect some properties of this curve. For ν ∈ {1, 2} and b ∈ Γ# set Nb,ν (k) := (k1 + b1 ) + i(−1)ν (k2 + b2 ), Nν (b) := {k ∈ C2 | Nb,ν (k) = 0}, Nb (k) := Nb,1 (k)Nb,2 (k), Nb := N1 (b) ∪ N2 (b), θν (b) := 12 ((−1)ν b2 + ib1 ). Observe that Nν (b) is a line in C2 . The free lifted Fermi curve is an union of these lines. Here is the precise statement. b 0) is the locally finite union Theorem 3.2.1 (The free Fermi curve). The curve F(0, [  Nν (b).  b∈Γ# ν∈{1,2}  In particular, the curve F(0, 0) is a complex analytic curve in C2 /Γ# and F(0, 0) ∼ = E(N0 ). Before we prove this theorem we shall prove some simple properties of the lines Nν (b). Proposition 3.2.2 (Properties of Nν (b)). Let ν ∈ {1, 2} and let b, c, d ∈ Γ# . Then: (a) Nν (b) ∩ Nν (c) = ∅  if  b 6= c ;  (b) dist(Nν (b), Nν (c)) =  √1 |b 2  − c|; 35  (c) N1 (b) ∩ N2 (c) = {(iθ1 (c) + iθ2 (b), θ1 (c) − θ2 (b))}; (d) the map k 7→ k + d maps Nν (b) to Nν (b − d); (e) the map k 7→ k + d maps N1 (b) ∩ N2 (c) to N1 (b − d) ∩ N2 (c − d). Proof. (a) By contradiction. Suppose that Nν (b) ∩ Nν (c) is not empty. Then there is at least one k ∈ C2 such that k1 + b1 + i(−1)ν (k2 + b2 ) = 0, k1 + c1 + i(−1)ν (k2 + c2 ) = 0. This is true if and only if b1 − c1 + i(−1)ν (b2 − c2 ) = 0. But this is impossible because b 6= c and b1 , b2 , c1 and c2 are real numbers. Thus, the intersection Nν (b) ∩ Nν (c) is empty. This proves part (a). (b) Let k ∈ Nν (b) and k 0 ∈ Nν (c). Then, k1 + b1 + i(−1)ν (k2 + b2 ) = 0, k10 + c1 + i(−1)ν (k20 + c2 ) = 0. Write d := c − b. Hence, using the above equations we obtain dist(Nν (b), Nν (c)) = inf{|k − k 0 | | k ∈ Nν (b) and k 0 ∈ Nν (c)} = inf{|(k1 − k10 , k2 − k20 )| | k ∈ Nν (b) and k 0 ∈ Nν (c)} = inf{|(c1 − b1 + i(−1)ν (c2 − b2 + k20 − k2 ), k2 − k20 )| | k2 , k20 ∈ C} = inf{|(d1 + i(−1)ν (d2 + z), −z)| | z ∈ C} = inf{(d2 − 2 Re(i(−1)ν d1 z̄ − d2 z̄) + 2|z|2 )1/2 | z ∈ C} = inf{(d2 + 2(d2 x − (−1)ν d1 y) + 2(x2 + y 2 ))1/2 | x, y ∈ R} √  2 1/2 = inf 12 d2 + √12 (d2 , −(−1)ν d1 ) + 2 (x, y) (x, y) ∈ R2 =  √1 |d| 2  =  √1 |b 2  − c|,  as claimed. (c) Let k ∈ N1 (b) ∩ N2 (c). Then, k1 + b1 − i(k2 + b2 ) = 0, k1 + c1 + i(k2 + c2 ) = 0, 36  which hold if and only if k1 =  1 2  (−b1 − c1 + i(b2 − c2 )) = iθ1 (c) + iθ2 (b),  k2 =  1 2  (−b2 − c2 + i(c1 − b1 )) = θ1 (c) − θ2 (b).  This proves part (c). (d) Observe that Nν (b) + d = {k + d ∈ C2 | k1 + b1 + i(−1)ν (k2 + b2 ) = 0} = {k 0 ∈ C2 | k10 + b1 − d1 + i(−1)ν (k20 + b2 − d2 ) = 0} = {k 0 ∈ C2 | Nb−d,ν (k 0 ) = 0} = Nν (b − d). This shows that the map k 7→ k + d maps Nν (b) to Nν (b − d), as desired. (e) Similarly as in part (d), the statement of part (e) follows from the equality N1 (b) ∩ N2 (c) + d =    1 2  (−b1 − c1 + i(b2 − c2 )) + d1 ,  =    1 2  (−(b1 − d1 ) − (c1 − d1 ) + i((b2 − d2 ) − (c2 − d2 ))) , 1 2  1 2  (−b2 − c2 + i(c1 − b1 )) + d2  (−(b2 − d2 ) − (c2 − d2 ) + i((c1 − d1 ) − (b1 − d1 )))      = N1 (b − d) ∩ N2 (c − d). The proof of the proposition is complete. We now prove Theorem 3.2.1. Proof of Theorem 3.2.1. For all k ∈ C2 the functions {eib·x | b ∈ Γ# } form a complete set of eigenfunctions for Hk (0, 0) in L2 (R2 /Γ) satisfying Hk (0, 0)eib·x = (i∇ − k)2 eib·x = (b + k)2 eib·x = Nb (k)eib·x . Hence, b 0) = {k ∈ C2 | Nb (k) = 0 for some b ∈ Γ# } = F(0,  [ b∈Γ#  Nb =  [  Nν (b).  b∈Γ# ν∈{1,2}  b 0). We next prove that this union is locally finite. This is the desired expression for F(0, First observe that, since the lattice Γ# is discrete, there is a constant C > 0 such that |b − c| ≥ C for all distinct elements b, c ∈ Γ# . Thus, by Proposition 3.2.2(b), the distance 37  between any two distinct lines Nν (b) and Nν (c) is bounded below by 1 C dist(Nν (b), Nν (c)) = √ |b − c| ≥ √ > 0. 2 2 Now, let U be an open bounded subset of C2 . We claim that only a finite number of lines Nν (b) can intersect U . Indeed, suppose this is not the case. Then for at least one ν ∈ {1, 2} there is an infinite number of lines Nν (b) crossing U . In particular, there is at least one point of each of these lines inside U . By the above inequality, all these points are apart from √ each other by at least C/ 2. Since we have an infinite number of such points inside U this implies that U is unbounded. But this is a contradiction. Therefore, only a finite number of lines Nν (b) can intersect U . Consequently, given U there exits a finite set B ⊂ Γ# such that [  b 0) ∩ U = U ∩ F(0,  Nb =  [ b∈Γ#  b∈Γ#  Nb ∩ U =  [  Nb ∩ U  b∈B  = {k ∈ U | Nb (k) = 0 for b ∈ B}. b 0) is locally finite. Furthermore, the curve F(0, b 0) is locally the zero That is, the union F(0, set of a finite number of polynomials Nb (k) and hence it is a complex analytic curve in C2 . Clearly the same conclusion holds for F(0, 0) in C2 /Γ# . Indeed, consider F(0, 0) ∩ W for b 0) ⊂ C2 “around some open bounded subset W in C2 /Γ# . Then if we embed this set in F(0, some b ∈ Γ# \ {0}”, we obtain a finite number of defining equations for F(0, 0) ∩ W . This can be properly done by exploiting the covering property F(0, 0) ∼ = E(N0 ). To prove this relation we proceed as follows. Let N0 − b be the translation of N0 by −b ∈ Γ# . Then, by Proposition 3.2.2(d) we have Nb = N0 − b for all b ∈ Γ# . Hence,  [  [ [  ∼ b 0) = E F(0, 0) = E F(0, Nb = E(Nb ) = E(N0 − d) b∈Γ#  b∈Γ#  b∈Γ#  = E(N0 ) = E(N1 (0) ∪ N2 (0)) = E(N1 (0)) ∪ E(N2 (0)) = E({(ik2 , k2 ) | k2 ∈ C}) ∪ E({(−ik2 , k2 ) | k2 ∈ C})   = (eik2 (i,1)·γ1 , eik2 (i,1)·γ2 ) k2 ∈ C ∪ (eik2 (−i,1)·γ1 , eik2 (−i,1)·γ2 ) k2 ∈ C . In particular this shows that F(0, 0) ∼ = E(N0 ) and completes the proof of the theorem. Let us briefly describe what the free Fermi curve looks like. In the Figure 3.1 there is a b 0) for which both ik1 and k2 are real, for the case where sketch of the set of (k1 , k2 ) ∈ F(0, 38  the lattice Γ# has points over the coordinate axes, that is, it has points of the form (b1 , 0) and (0, b2 ). Observe that, in particular, Proposition 3.2.2 yields N1 (0) ∩ N2 (b) = {(iθ1 (b), θ1 (b))}, N1 (−b) ∩ N2 (0) = {(iθ2 (−b), θ2 (b))}, the map k 7→ k + b maps N1 (0) ∩ N2 (b) to N1 (−b) ∩ N2 (0). b 0) that differ by elements of Γ# correspond to the same point Recall that points in F(0, in F(0, 0). Thus, in the sketch on the left, we should identify the lines k2 = −b2 /2 and k2 = b2 /2 for all b ∈ Γ# with b2 6= 0, to get a pair of helices climbing up the outside of a cylinder, as illustrated by the figure on the right. The helices intersect each other twice on each cycle of the cylinder—once on the front half of the cylinder and once on the back half. Hence, viewed as a “manifold” (with singularities), the pair of helices are just two copies of R with points that corresponds to intersections identified. We can use k2 as a coordinate in each copy of R and then the pairs of identified points are k2 = b2 /2 and k2 = −b2 /2 for all b ∈ Γ# with b2 6= 0 (see Figure 3.2).  ik1 N2 (0)  N2 (−b)  ik1 N1 (b)  N1 (0)  N2 (b)  N1 (−b)  k2  k2  b 0) and F(0, 0) when both ik1 and k2 are real. Figure 3.1: Sketch of F(0, b 0) is just two copies of C with k2 So far we have only considered k2 real. The full F(0, as a coordinate in each copy, provided we identify the points θ1 (b) = 12 (−b2 + ib1 ) (in the first copy) and θ2 (b) =  1 2 (b2  + ib1 ) (in the second copy) for all b ∈ Γ# with b2 6= 0 (see  Figure 3.3). 39  R  R  Figure 3.2: The pair of helices as a “manifold”.  b 0): two copies of C with θ1 (b) and θ2 (b) identified for b2 6= 0. Figure 3.3: The full F(0, b 0). For b ∈ Γ# set To conclude this section we give a (global) defining equation for F(0, "  2 # 2k · b + k 2 − 1 1 2k · b + k 2 − 1 Rb (k) := exp − + . 1 + b2 2 1 + b2 We have the following proposition. b 0)). The following equality holds: Proposition 3.2.3 (Global defining equation for F(0, ) ( Y (k + b)2 b 0) = k ∈ C2 Rb (k) = 0 , F(0, 1 + b2 # b∈Γ  where the infinite product inside brackets converges to an entire function on C2 . Before we prove this proposition we remark that the above product is analogous to the canonical product associated to a sequence of complex numbers {zn }. Such product defines an entire function on C which has a zero at each point zn (see [13, p 302]). 40  Proof. We give only a sketch of the proof. By Theorem 2.5.1 we have  b 0) = k ∈ C2 | F (k, 0, 0, 0) = 0 F(0, with  F (k, 0, 0, 0) = det3 I +  √ 1 u(k, 0) √ 1 I−∆ I−∆    .  Write T :=  √ 1 u(k, 0) √ 1 I−∆ I−∆  and set  =  √ 1 (−2ik I−∆  1 · ∇ + k 2 − I) √I−∆      2 j X (−1) R3 (T ) := (I + T ) exp  T j  − I. j j=1  The eigenvalues of R3 (T ), which we denote by {λb (R3 (T ))}b∈Γ# , are given by λb (R3 (T )) =  (k + b)2 Rb (k) − 1. 1 + b2  Hence (see [15, Chapter 9] for details), F (k, 0, 0, 0) = det3 (I + T ) = det(I + R3 (T )) Y Y (k + b)2 = (1 + λb (R3 (T ))) = Rb (k). 1 + b2 # # b∈Γ  b∈Γ  Furthermore, according to Theorem 2.5.1 the function F (k, 0, 0, 0) is analytic on C2 . This proves the proposition.  3.3  The ε-tubes about the free Fermi curve  We now introduce real and imaginary coordinates in C2 and define ε-tubes about the free Fermi curve. We derive some properties of the ε-tubes as well. For k ∈ C2 write k1 = u1 + iv1  and  k2 = u2 + iv2 ,  where u1 , u2 , v1 and v2 are real numbers. Then, Nb,ν (k) = (k1 + b1 ) + i(−1)ν (k2 + b2 ) = i(v1 + (−1)ν (u2 + b2 )) − (−1)ν (v2 − (−1)ν (u1 + b1 )),  41  so that |Nb,ν (k)| = |v + (−1)ν (u + b)⊥ |, where (y1 , y2 )⊥ := (y2 , −y1 ). Observe that (y ⊥ )⊥ = −y and |y ⊥ | = |y|. Furthermore, for any real number λ we have (λy)⊥ = λy ⊥ . Since Nb (k) = Nb,1 (k)Nb,2 (k), it follows that Nb (k) = 0 if and only if v − (u + b)⊥ = 0  or  v + (u + b)⊥ = 0.  Let 2Λ be the length of the shortest nonzero “vector” in Γ# . Then there is at most one b ∈ Γ# with |v + (u + b)⊥ | < Λ and at most one b ∈ Γ# with |v − (u + b)⊥ | < Λ. Indeed, suppose there is another b0 6= b such that |v + (u + b0 )⊥ | < Λ or |v − (u + b0 )⊥ | < Λ. Then, |b − b0 | = |(b − b0 )⊥ | = |v ± (u + b)⊥ − (v ± (u + b0 )⊥ )| < Λ + Λ = 2Λ, which contradicts the definition of Λ. Thus, there is no such b0 (see Figure 3.4).  v⊥ + u Λ  v⊥ − u  Figure 3.4: On the left: definition of Λ. On the right: possible configuration. Let ε be a constant satisfying 0<ε<  Λ . 6  For ν ∈ {1, 2} and b ∈ Γ# define the ε-tube about Nν (b) as Tν (b) := {k ∈ C2 | |Nb,ν (k)| = |v + (−1)ν (u + b)⊥ | < ε}, 42  and the ε-tube about Nb = N1 (b) ∪ N2 (b) as Tb := T1 (b) ∪ T2 (b). Since (v + (u + b)⊥ ) + (v − (u + b)⊥ ) = 2v, at least one of the factors |v + (u + b)⊥ | or |v − (u + b)⊥ | in |Nb (k)| must always be greater or equal to |v|. If k 6∈ Tb both factors are also greater or equal to ε. If k ∈ Tb one factor is bounded by ε and the other must lie within ε of |2v|. Thus, k 6∈ Tb  =⇒  |Nb (k)| ≥ ε|v|,  (3.3.1)  k ∈ Tb  =⇒  |Nb (k)| ≤ ε(2|v| + ε).  (3.3.2)  The pairwise intersection T b ∩ T b0 is compact whenever b 6= b0 . Here T b denotes the closure of Tb . Indeed, to prove this first observe that, the intersection T ν (b) ∩ T ν (b0 ) is empty if b 6= b0 because, if it were not, we would have |v + (−1)ν (u + b)⊥ − v − (−1)ν (u + b0 )⊥ | ≤ 2ε <  Λ , 3  which contradicts |v + (−1)ν (u + b)⊥ − v − (−1)ν (u + b0 )⊥ | = |b − b0 | ≥ 2Λ, which is certainly true according to the definition of Λ. Furthermore, if k ∈ T 1 (b) ∩ T 2 (b0 ) then |u + 12 (b + b0 )| = 12 |v − (u + b)⊥ − v − (u + b0 )⊥ | ≤  ε 2  +  ε 2  = ε,  |v − 12 (b − b0 )⊥ | = 12 |v − (u + b)⊥ + v + (u + b0 )⊥ | ≤  ε 2  +  ε 2  = ε,  which defines a compact set. Thus, the intersection T b ∩ T b0 = (T 1 (b) ∩ T 2 (b0 )) ∪ (T 2 (b) ∩ T 1 (b0 ))  (3.3.3)  is compact. Finally, it is easy to see that T b ∩ T b0 ∩ T b00 is empty for all distinct elements b, b0 , b00 ∈ Γ# . In fact, in view of (3.3.3),     T b ∩ T b0 ∩ T b00 = T 1 (b) ∩ T 2 (b0 ) ∩ T 1 (b00 ) ∪ T 1 (b) ∩ T 2 (b0 ) ∩ T 2 (b00 )     ∪ T 2 (b) ∩ T 1 (b0 ) ∩ T1 (b00 ) ∪ T 2 (b) ∩ T 1 (b0 ) ∩ T2 (b00 ) = ∅ ∪ ∅ ∪ ∅ ∪ ∅ = ∅. 43  ik1 T2 (−b)  T1 (b)  T2 (0)  T1 (0)  T2 (b)  T1 (−b)  k2  Figure 3.5: The ε-tubes about the free “lifted” Fermi curve. If a point k belongs to the free Fermi curve the function Nb (k) vanishes for some b ∈ Γ# . To conclude this section we give a lower bound for this function when (b, k) is away from the zero set. Proposition 3.3.1 (Lower bound for |Nb (k)|). (a) If |b + u + v ⊥ | ≥ Λ and |b + u − v ⊥ | ≥ Λ, then |Nb (k)| ≥ (b) If |v| > 2Λ and k ∈ T0 , then |Nb (k)| ≥  Λ 2 (|v|  Λ 2 (|v|  + |u + b|).  + |u + b|) for all b 6= 0 but at most one  b 6= 0. This exceptional b̃ obeys |b̃| > |v| and | |u + b̃| − |v| | < Λ. (c) If |v| > 2Λ and k ∈ T0 ∩ Td with d 6= 0, then |Nb (k)| ≥  Λ 2 (|v| + |u + b|)  for all b 6∈ {0, d}.  Furthermore we have |d| > |v| and | |u + d| − |v| | < Λ. Proof. (a) By hypothesis, both factors in |Nb (k)| = |v + (u + b)⊥ | |v − (u + b)⊥ | are greater or equal to Λ. We now prove that at least one of the factors must also be greater or equal to 12 (|v| + |u + b|). Suppose that |v| ≥ |u + b|. Then, since (v + (u + b)⊥ ) + (v − (u + b)⊥ ) = 2v, 44  at least one of the factors must also be greater or equal to 1 1 |v| = (|v| + |v|) ≥ (|v| + |u + b|). 2 2 Now suppose that |v| < |u + b|. Then, since (v + (u + b)⊥ ) − (v − (u + b)⊥ ) = 2(u + b)⊥ , at least one of the factors must also be greater or equal to 1 1 |u + b| = (|u + b| + |u + b|) > (|v| + |u + b|). 2 2 All this together implies that |Nb (k)| ≥  Λ (|v| + |u + b|), 2  which proves part (a). (b) By hypothesis ε < Λ/6 < 2Λ < |v|. Let k ∈ T0 . Then, by (3.3.2), |N0 (k)| ≤ ε(2|v| + ε) < 3ε|v| <  Λ |v|. 2  (3.3.4)  Thus we have either |u + v ⊥ | < Λ or |u − v ⊥ | < Λ (otherwise apply part (a) to get a contradiction). Suppose that |u+v ⊥ | < Λ. Then there is no b ∈ Γ# \{0} with |b+u+v ⊥ | < Λ and there is at most one b̃ ∈ Γ# \ {0} satisfying |b̃ + u − v ⊥ | < Λ. This inequality implies | |u + b̃| − |v| | < Λ. Furthermore, for this b̃, |b̃| = |2v ⊥ − (u + v ⊥ ) + (b̃ + u − v ⊥ )| > 2|v| − 2Λ > |v|, since −2Λ > −|v|. Now suppose that |u − v ⊥ | < Λ. Then there is no b ∈ Γ# \ {0} obeying |b + u − v ⊥ | < Λ and there is at most one b̃ ∈ Γ# \ {0} satisfying |b̃ + u + v ⊥ | < Λ. This inequality implies | |u + b̃| − |v| | < Λ. Consequently, |b̃| = |2v ⊥ − (v ⊥ − u) − (v ⊥ + b̃ + u)| > 2|v| − 2Λ > |v|. Finally observe that, if b 6∈ {0, b̃} then |b + u + v ⊥ | ≥ Λ and |b + u − v ⊥ | ≥ Λ. Hence, applying part (a) it follows that |Nb (k)| ≥  Λ 2 (|v|  + |u + b|). This proves part (b).  (c) As in the proof of part (b), if k ∈ T0 ∩ Td then in addition to (3.3.4) we have |Nd (k)| ≤ ε(2|v| + ε) < 3ε|v| <  Λ |v|. 2  Thus, applying part (b) we conclude that d must be the exceptional b̃ of part (b). The statement of part (c) follows then from part (b). This completes the proof of the proposition.  45  3.4  Motivation and main results  Below we state our main results. The proofs come later divided in many steps. In [4], the authors introduced a class of Riemann surfaces of infinite genus that are “asymptotic to” a finite number of complex lines joined by infinite many handles. These surfaces are constructed by pasting together a compact submanifold of finite genus, plane domains, and handles. More precisely, these surfaces can be decomposed into X com ∪ X reg ∪ X han , where X com is a compact submanifold with smooth boundary and finite genus, X reg is a finite union of open “regular pieces”, and X han is an infinite union of closed “handles”. All these components satisfy a number of geometric/analytic hypotheses stated in [4] that specify the asymptotic holomorphic structure of the surface. The class of surfaces obtained in this way yields an extension of the classical theory of compact Riemann surfaces that has analogues of many theorems of the classical theory. The choice of geometric/analytic hypotheses was guided by two requirements. First, that the classical theory of compact Riemann surfaces could be developed in the new context. Secondly, that a number of interesting examples satisfy the hypotheses. In fact, it was proven in [4] that this new class of surfaces includes quite general hyperelliptic surfaces, heat curves (which are related to the Kadomcev-Petviashvilli equation), and Fermi curves with zero magnetic potential F(A = 0, V ). In order to verify the geometric/analytic hypotheses for F(0, V ) the authors proved two “asymptotic” theorems similar to the ones we prove below. This is the main step needed to verify the geometric/analytic hypotheses. In this thesis we extend their results to Fermi curves F(A, V ) with “small” magnetic potential A. We have followed their strategy of analysis. The main idea is to consider the eigenvalueeigenvector problem for Hk (A, V ) for k ∈ C2 with large imaginary part as a perturbation of the problem for Hk (0, 0). When A is zero, they are able to prove asymptotic theorems for F(0, V ) for arbitrary large V . When A is not zero, however, new difficulties arise due to the presence of the term A · (i∇ − k) in the Hamiltonian Hk . When A is large, taking the imaginary part of k ∈ C2 arbitrarily large is not enough to control this term—it is not enough to make its contribution small and hence have the interacting Fermi curve as a perturbation of the free Fermi curve. (The term V in Hk is easily controlled by this 46  method.) However, as we shall see below, the proof can be implemented by assuming that A is small. Before we proceed we need to introduce some notation. For any ϕ ∈ L2 (R2 /Γ) define ϕ̂ : Γ# → C as ϕ̂(b) := (Fϕ)(b) := where |Γ| :=  R  R2 /Γ dx.  1 |Γ|  Z  ϕ(x) e−ib·x dx,  R2 /Γ  Then,  ϕ(x) = (F −1 ϕ̂)(x) =  X  ϕ̂(b) eib·x  kϕkL2 (R2 /Γ) = |Γ|1/2 kϕ̂kl2 (Γ# ) .  and  b∈Γ#  Let ρ be a positive constant and set Kρ := {k ∈ C2 | |v| ≤ ρ}. Consider the projection pr : C2 −→ C, (k1 , k2 ) 7−→ k2 , and define q := (i∇ · A) + A2 + V. Finally, recall that Tν (b) = {k ∈ C2 | |Nb,ν (k)| = |v + (−1)ν (u + b)⊥ | < ε}  and  Tb := T1 (b) ∪ T2 (b).  Clearly, the set Kρ is invariant under the action of Γ# and Kρ /Γ# is compact. Hence, b b the image of F(A, V ) ∩ Kρ under the exponential map E : F(A, V ) → F(A, V ) is compact in F(A, V ). It will essentially play the role of X com in the decomposition of F(A, V ). Our first theorem characterizes the regular piece X reg . Theorem 3.4.1 (The regular piece). Let 0 < ε < Λ/6 and suppose that A1 , A2 and V are functions in L2 (R2 /Γ) obeying kb2 q̂(b)kl1 (Γ# ) < ∞ and k(1 + b2 )Â(b)kl1 (Γ# \{0}) < 2ε/63. Then there is a constant ρ = ρΛ,ε,q,A such that, for ν ∈ {1, 2}, the projection pr induces a biholomorphic map between      b F(A, V ) ∩ Tν (0) \ Kρ ∪   [ b∈Γ# \{0}  47  Tb   and its image in C. This image component contains n o z ∈ C 8|z| > ρ and |z + (−1)ν θν (b)| > ε for all b ∈ Γ# \ {0} and is contained in ( z∈C  1 |z + (−1)ν θν (b)| > 2    ε2 ε− 40Λ    ) for all b ∈ Γ# \ {0} ,  where θν (b) = 12 ((−1)ν b2 + ib1 ). Furthermore, pr−1 : Image(pr) −→ Tν (0), (1,0)  y 7−→ (−β2 (1,0)  where β2  − i(−1)ν y − r(y), y),  is a constant given by (4.6.8) that depends only on ρ and A, (1,0)  |β2  |<  ε2 100Λ  and  |r(y)| ≤  ε3 C + , 2 50Λ ρ  where C = CΛ,ε,q,A is a constant.  b Since Tb + c = Tb+c for all b, c ∈ Γ# , the complement of E F(A, V ) ∩ Kρ in F(A, V ) is the disjoint union of !! E    b F(A, V ) ∩ T0 \  Kρ ∪  [  Tb  b∈Γ# b2 6=0  and [   b E F(A, V ) ∩ T0 ∩ Tb .  b∈Γ# b2 6=0  Basically, the first of the two sets will be the regular piece of F(A, V ), while the second set will be the handles. The map Φ parametrizing the regular part will be the composition of the exponential map E with the inverse of the map discussed in the above theorem. The detailed information about the handles comes from our second main theorem. Theorem 3.4.2 (The handles). Let 0 < ε < Λ/6 and suppose that A1 , A2 and V are functions in L2 (R2 /Γ) with kb2 q̂(b)kl1 (Γ# ) < ∞ and k(1 + b2 )Â(b)kl1 (Γ# \{0}) < 2ε/63. Then, for every sufficiently large constant ρ and for every d ∈ Γ# \ {0} with 2|d| > ρ, there are maps n (z1 , z2 ) ∈ C2 n : (z1 , z2 ) ∈ C2  φd,1 : φd,2  εo ε and |z2 | ≤ −→ T1 (0) ∩ T2 (d), 2 2 o ε |z1 | ≤ and |z2 | ≤ ε −→ T1 (−d) ∩ T2 (0), 2  |z1 | ≤  and a complex number td with |td | ≤  C |d|4  such that: 48  (i) For ν ∈ {1, 2} the domain of the map φd,ν is biholomorphic to its image, and the image contains n k ∈ C2  |k1 + i(−1)ν k2 | ≤  ε εo and |k1 + (−1)ν+1 d1 − i(−1)ν (k2 + (−1)ν+1 d2 )| ≤ . 8 8  Furthermore, Dφ̂d,ν   1 1 = 2 −i(−1)ν  1       1  I +O |d|2 i(−1)ν  and φd,ν (0) = (iθν (d), (−1)ν+1 θν (d)) + O     ε  1 . +O 900 ρ  (ii) n b (T (0) ∩ T (d) ∩ F(A, V )) = (z1 , z2 ) ∈ C2 φ−1 2 d,1 1 n 2 b φ−1 d,2 (T1 (−d) ∩ T2 (0) ∩ F(A, V )) = (z1 , z2 ) ∈ C  ε and |z2 | ≤ 2 ε z1 z2 = td , |z1 | ≤ and |z2 | ≤ 2  z1 z2 = td , |z1 | ≤  εo , 2 εo . 2  (iii) φd,1 (z1 , z2 ) = φd,2 (z2 , z1 ) − d. These are the main new results of this thesis. Using these theorems one should be able to verify that F(A, V ) satisfies the geometric/analytic hypotheses as was done in [4] for F(0, V ). In the next section we outline the strategy for proving these results. The proofs are given in the next chapter divided in many steps. We finally mention a small simplification (or modification) that we were able to make in the Theorem 3.4.1 (the regular piece). We shall not go into details here. We refer the reader b to Chapter 5. First, recall an important property of F(A, V )—namely, gauge invariance. b b + ∇Ψ, V ), where Ψ is function on R2 Briefly, gauge invariance implies that F(A, V ) = F(A (under suitable hypotheses), and ∇Ψ is periodic with respect to Γ. In Chapter 5, by choosing a convenient gauge Ψ, we are able to indicate how to simplify the proof of Theorem 3.4.1 and improve some constants in it. After performing this gauge transformation “some terms vanish” and the analysis becomes simpler. We do not provide this “new” proof because it is essentially the same as the proof given below, up to some minor modifications. In fact, we believe that the simplifications introduced by the gauge will become clear after the reader gets familiar with the proof of the above results. 49  3.5  Strategy outline  Below we briefly describe the general strategy of analysis used to prove our results. It was applied by Feldman, Knörrer and Trubowitz in [4, §16] for studying the case where A = 0. We implement it in detail in the subsequent sections for A 6= 0. Let us first introduce some notation and definitions. Observe that Hk (A, V )ϕ = ((i∇ + A − k)2 + V )ϕ = ((i∇ − k)2 + A · (i∇ − k) + (i∇ − k) · A + A2 + V )ϕ = ((i∇ − k)2 + A · (i∇ − 2k) + (i∇ · A) + A · i∇ + A2 + V )ϕ = ((i∇ − k)2 + 2A · (i∇ − k) + (i∇ · A) + A2 + V )ϕ, and write Hk (A, V ) = ∆k + h(k, A) + q(A, V ) with ∆k := (i∇ − k)2 ,  h(k, A) := 2A · (i∇ − k)  q(A, V ) := (i∇ · A) + A2 + V.  and  For each finite subset G of Γ# set G0 := Γ# \ G  and  C2G := C2 \  [  Nb ,  b∈G0  L2G := span{eib·x | b ∈ G}  and  L2G0 := span{eib·x | b ∈ G0 }.  To simplify the notation write L2 in place of L2 (R2 /Γ), let I be the identity operator on L2 , and let πG and πG0 be the orthogonal projections from L2 onto L2G and L2G0 , respectively. Then, L2 = L2G ⊕ L2G0  and  I = πG + πG0 .  2 For k ∈ C2G define the partial inverse (∆k )−1 G on L as −1 (∆k )−1 G := πG + ∆k πG0 .  Its matrix elements are (∆k )−1 G     b,c  :=  eib·x |Γ|1/2  , (∆k )−1 G  eic·x |Γ|1/2  where b, c ∈ Γ# . 50   = L2    δb,c  if c ∈ G,   1 δ b,c Nc (k)  if c 6∈ G,  b Here is the main idea: By definition, a point k is in F(A, V ) if Hk (A, V ) has a nontrivial kernel in L2 . Hence, to study the part of the curve in the intersection of ∪d0 ∈G Td0 with C2 \ ∪b∈G0 Tb for some finite subset G of Γ# (see Figure 3.6), it is natural to look for a nontrivial solution of (∆k + h + q)(ψG + ψG0 ) = 0, where ψG ∈ L2G and ψG0 ∈ L2G0 . Equivalently, if we make the following (invertible) change of variables in L2 , (ψG + ψG0 ) = (∆k )−1 G (ϕG + ϕG0 ), where ϕG ∈ L2G and ϕG0 ∈ L2G0 , we may consider the equation (∆k + h + q)ϕG + (I + (h + q)∆−1 k )ϕG0 = 0.  (3.5.1)  The projections of this equation onto L2G0 and L2G are, respectively, πG0 (h + q)ϕG + πG0 (I + (h + q)∆−1 k )ϕG0 = 0,  (3.5.2)  πG (∆k + h + q)ϕG + πG (h + q)∆−1 k ϕG0 = 0.  (3.5.3)  Now define RG0 G0 on L2 as RG0 G0 := πG0 (I + (h + q)∆−1 k )πG0 . Observe that RG0 G0 is the zero operator on L2G . Then, if RG0 G0 has a bounded inverse on L2G0 , the equation (3.5.2) is equivalent to −1 ϕG0 = −RG 0 G0 πG0 (h + q)ϕG .  Substituting this into (3.5.3) yields −1 πG (∆k + h + q − (h + q)∆−1 k RG0 G0 πG0 (h + q))ϕG = 0.  This equation has a nontrivial solution if and only if the (finite) |G| × |G| determinant −1 det [ πG (∆k + h + q − (h + q)∆−1 k RG0 G0 πG0 (h + q))πG ] = 0  or, equivalently, expressing all operators as matrices in the basis {|Γ|−1/2 eib·x | b ∈ Γ# },   X wd0 ,b −1  (RG = 0, (3.5.4) det Nd0 (k)δd0 ,d00 + wd0 ,d00 − 0 G0 )b,c wc,d00 N (k) b 0 b,c∈G  d0 ,d00 ∈G  51  where wb,c := hb,c + q̂(b − c) = −2(c + k) · Â(b − c) + q̂(b − c). Therefore, if RG0 G0 has a bounded inverse on L2G0 —which is in fact the case under suitable conditions—in the region under consideration we can study the Fermi curve in detail using the (local) defining equation (3.5.4). In order to implement this strategy we shall first derive a number of analytic estimates.  ik1 T2 (0)  T1 (0)  k2  T0 ∩ (C2 \ ∪b∈Γ# \{0} Tb )  T2 (0)  T1 (d)  ik1  T2 (d)  T1 (0)  k2  (T0 ∪ Td ) ∩ (C2 \ ∪b∈Γ# \{0,d} Tb )  Figure 3.6: Sketch of (∪d0 ∈G Td0 ) ∩ (C2 \ ∪b∈G0 Tb ) for G = {0} and G = {0, d}.  52  3.6  Notation and remarks  We summarize here some notation and remarks that will be used henceforth. (i) For ν ∈ {1, 2} define the (complementary) index ν 0 as ν 0 := ν − (−1)ν . Observe that ν0 =    2  if ν = 1   1  if ν = 2  and  0  (−1)ν = −(−1)ν .  (ii) Let f (x) and g(x) be multivariable functions and let p be a real number. The notation f (x) = O(|x|p ) means that there is a constant C > 0 such that |f (x)| ≤ C|x|p for all x in a suitable domain. Similarly, the statement O(f (x)) = O(g(x)) is equivalent to say that there are constants C1 > 0 and C2 > 0 such that C1 |f (x)| ≤ |g(x)| ≤ C2 |f (x)|. (iii) Let z and w be vectors in C2 . We denote the length of z by |z| := (z1 z̄1 + z2 z̄2 )1/2 = (|z1 |2 + |z2 |2 )1/2 . Note that z · w := z1 w1 + z2 w2 is not an inner product on C2 . However, we still have the property |z · w| = |z1 w1 + z2 w2 | ≤ |z1 | |w1 | + |z2 | |w2 | = (|z1 |, |z2 |) · (|w1 |, |w2 |) ≤ |z| |w|. Here we have used the Schwarz inequality (for the inner product on R2 ).  53  (iv) Let T be a linear operator from L2C to L2B with B, C ⊂ Γ# . We denote its matrix elements in the basis {|Γ|−1/2 eib·x | b ∈ Γ# } by   ib·x eic·x e ,T , Tb,c := |Γ|1/2 |Γ|1/2 L2 where b ∈ B and c ∈ C. The operator T represented as a matrix [Tb,c ] acts on l2 (Γ# ). (v) Consider a linear operator T : X → Y . The operator norm of T is defined as kT k := sup x∈X  kT xkY . kxkX  (vi) In general, we denote by CX1 ,...,Xn a constant that depends only on X1 , . . . , Xn . We may use the same symbol CX1 ,...,Xn to denote different constants that change from line to line in our calculations without further notice. (vii) The following notation will be used whenever we consider vector-valued quantities. Let X be a Banach space and let A = (A1 , . . . , Ad ) ∈ X d and B = (B1 , . . . , Bd ) ∈ X d . Then, kAkX := (kA1 k2X + · · · + kAd k2X )1/2  A · B := A1 B1 + · · · + Ad Bd .  and  (viii) Finally, recall the Neumann series for bounded linear operators, (I − T )  −1  =  ∞ X  T j,  j=0  and the the geometric series for complex numbers, ∞  X 1 = zj , 1−z j=0  which are convergent if kT k < 1 and |z| < 1.  54  Chapter 4  Weak magnetic potential 4.1  Invertibility of RG0 G0  A simple strategy for inverting RG0 G0 is the following. Since this operator has the form I + T , it is easily invertible if kT k < 1. In this section we manage to get this bound and prove that for a suitable choice of G ⊂ Γ# , large |v|, and weak magnetic potential, the operator RG0 G0 is invertible on L2G0 . In general, for any B, C ⊂ Γ# (C such that ∆−1 k πC exists) define the operator RBC as RBC := πB (I + (h + q)∆−1 k )πC −1 −1 = πB πC + πB q ∆−1 k πC + πB (2A · i∇)∆k πC − πB (2k · A)∆k πC .  (4.1.1)  Its matrix elements are (RBC )b,c = δb,c +  q̂(b − c) 2c · Â(b − c) 2k · Â(b − c) − − , Nc (k) Nc (k) Nc (k)  (4.1.2)  where b ∈ B and c ∈ C. We shall first estimate the norm of the last three terms on the right hand side of (4.1.1). We begin with the following proposition. Proposition 4.1.1. Let k ∈ C2 and let B, C ⊂ Γ# with C ⊂ {b ∈ Γ# | Nb (k) 6= 0}. Then, 1 , |N c (k)| c∈C |c| kπB (A · i∇)∆−1 k πC k ≤ kÂkl1 sup |N (k)| , c c∈C 1 kπB (k · A)∆−1 k πC k ≤ kÂkl1 |k| sup |N (k)| . c c∈C kπB q ∆−1 k πC k ≤ kq̂kl1 sup  55  To prove this proposition we apply the following inequality, which we prove later in §4.9. Proposition 4.1.2. Consider a linear operator T : L2C → L2B with matrix elements Tb,c . Then, (  )  kT k ≤ max sup  X  |Tb,c |, sup  c∈C b∈B  X  |Tb,c | .  b∈B c∈C  Proof of Proposition 4.1.1. Write T1 := πB q ∆−1 k πC . Then, in view of (4.1.1) and (4.1.2), sup  X  |(T1 )b,c | ≤ sup  c∈C b∈B  sup  X  X |q̂(b − c)|  c∈C b∈B  |(T1 )b,c | ≤ sup  b∈B c∈C  |Nc (k)|  X |q̂(b − c)|  b∈B c∈C  |Nc (k)|  ≤ sup c∈C  ≤ sup c∈C  1 kq̂kl1 , |Nc (k)| 1 kq̂kl1 . |Nc (k)|  By Proposition 4.1.2, these estimates imply the first inequality. −1 Now, let T2 := πB (A · i∇)∆−1 k πC and T3 := πB (k · A)∆k πC . Similarly as above, the  second and third inequalities follow from the estimates sup  X  |(T2 )b,c | ≤ sup  c∈C b∈B  sup  X  X |c| |Â(b − c)| |Nc (k)|  c∈C b∈B  |(T2 )b,c | ≤ sup  c∈C  X |c| |Â(b − c)| |Nc (k)|  b∈B c∈C  b∈B c∈C  ≤ sup ≤ sup c∈C  |c| kÂkl1 , |Nc (k)| |c| kÂkl1 , |Nc (k)|  and sup  X  |(T3 )b,c | ≤ sup  c∈C b∈B  c∈C b∈B  sup  X  X |k| |Â(b − c)|  |(T3 )b,c | ≤ sup  b∈B c∈C  |Nc (k)|  X |k| |Â(b − c)|  b∈B c∈C  |Nc (k)|  ≤ sup c∈C  ≤ sup c∈C  1 |k| kÂkl1 , |Nc (k)| 1 |k| kÂkl1 . |Nc (k)|  This proves the proposition. −1 The key estimate for the existence of RG 0 G0 is given below.  Proposition 4.1.3 (Estimate of kRSS − πS k). Let k ∈ C2 with |u| ≤ 2|v| and |v| > 2Λ. Suppose that S ⊂ {b ∈ Γ# | |Nb (k)| ≥ ε|v|}. Then, kRSS − πS k ≤ kq̂kl1  1 14 + kÂkl1 . ε|v| ε  (4.1.3)  If A = 0 the right hand side of (4.1.3) can be made arbitrarily small for any q(0, V ) = V by taking |v| sufficiently large. If A 6= 0, however, we need to take kÂkl1 small to make that quantity less than 1. The term  14 ε kÂkl1  in (4.1.3) comes from the estimate we have for  kπG0 h ∆−1 k πG0 k. 56  Proof. By hypothesis, for all b ∈ S, 1 1 ≤ . |Nb (k)| ε|v|  (4.1.4)  |b| 4 ≤ . |Nb (k)| ε  (4.1.5)  We now show that, for all b ∈ S,  First suppose that |b| ≤ 4|v|. Then, |b| 4|v| 4 ≤ = . |Nb (k)| ε|v| ε Now suppose that |b| ≥ 4|v|. Again, by hypothesis we have |u| ≤ 2|v|  and  |v| > 2Λ > Λ/6 > ε.  Hence, |b| 3 |v ± (u + b)⊥ | ≥ |b| − |u| − |v| ≥ |b| − 3|v| ≥ |b| − |b| = . 4 4 Consequently, |b| |b| 4 4 16 4 4 = ≤ |b| = ≤ ≤ . |Nb (k)| |b| |b| |b| |v| ε |v + (u + b)⊥ | |v − (u + b)⊥ | This proves (4.1.5). The expression for RSS − πS is given by (4.1.1). Observe that |k| = |u + iv| ≤ |u| + |v| ≤ 3|v|. Then, applying Proposition 4.1.1 and using (4.1.4) and (4.1.5) we obtain |c| 1 + 2kÂkl1 sup |N (k)| |N c c (k)| b∈S b∈S 1 8 1 14 ≤ (6|v| kÂkl1 + kq̂kl1 ) + kÂkl1 = kq̂kl1 + kÂkl1 . ε|v| ε ε|v| ε  kRSS − πS k ≤ (6|v| kÂkl1 + kq̂kl1 ) sup  This is the desired inequality. From the last proposition it follows easily that RSS has a bounded inverse for large |v| and weak magnetic potential. Lemma 4.1.4 (Invertibility of RSS ). Let k ∈ C2 ,   2 |u| ≤ 2|v|, |v| > max 2Λ, kq̂kl1 , kq̂kl1 < ∞ ε 57  and  kÂkl1 <  2 ε. 63  Suppose that S ⊂ {b ∈ Γ# | |Nb (k)| ≥ ε|v|}. Then the operator RSS has a bounded inverse with kRSS − πS k < kq̂kl1  1 14 17 + kÂkl1 < ε|v| ε 18  and −1 kRSS − πS k < 18kRSS − πS k.  Proof. Write RSS = πS + T with T = RSS − πS . Then, by Proposition 4.1.3, kT k = kRSS − πS k ≤ kq̂kl1  1 14 1 4 17 + kÂkl1 < + = < 1. ε|v| ε 2 9 18  −1 Hence, the Neumann series for RSS = (πS + T )−1 converges (and is a bounded operator).  Furthermore, −1 − πS k = k(πS + T )−1 − πS k = k(πS + T )−1 − (πS + T )−1 (πS + T )k kRSS  = k(πS + T )−1 T k ≤ (1 − kT k)−1 kT k < 18kRSS − πS k, as was to be shown. Lemma 4.1.4 says that if G is such that G0 ⊂ {b ∈ Γ# | |Nb (k)| ≥ ε|v|}, the operator RG0 G0 has a bounded inverse on L2G0 for |u| ≤ 2|v|, large |v|, and weak magnetic potential. b We are now able to write (local) defining equations for F(A, V ) under such conditions.  4.2  Local defining equations  In this section we derive (local) defining equations for the Fermi curve. We begin with a simple proposition. Proposition 4.2.1. Suppose either (i) or (ii) or (iii) where: (i) G = {0} and k ∈ T0 \ ∪b∈Γ# \{0} Tb ; (ii) G = {0, d} and k ∈ T0 ∩ Td ; (iii) G = ∅ and k ∈ C2 \ ∪b∈Γ# Tb . Then G0 = Γ# \ G = {b ∈ Γ# | |Nb (k)| ≥ ε|v|}.  58  Proof. The proposition follows easily if we observe that G0 = Γ# \ G and recall from (3.3.1) that k 6∈ Tb  |Nb (k)| ≥ ε|v|.  =⇒  We now need some notation. Let B be a fundamental cell for Γ# ⊂ R2 (see [12, p 310]). Then any vector u ∈ R2 can be written as u = ξ + u, for some ξ ∈ Γ# and u ∈ B. Define   α := sup{|u| | u ∈ B},  2 R := max α, 2Λ, kq̂kl1 ε   ,  KR := {k ∈ C2 | |v| ≤ R}.  We first show that in C2 \ KR the Fermi curve is contained in the union of ε-tubes about the free Fermi curve. b Proposition 4.2.2 (F(A, V ) \ KR is contained in the union of ε-tubes). b F(A, V ) \ KR ⊂  [  Tb .  b∈Γ#  b b Proof. Recall that F(A, V ) is invariant under the action of Γ# , that is, that k ∈ F(A, V ) if b b and only if k + ξ ∈ F(A, V ) for all ξ ∈ Γ# . Hence, given any k ∈ F(A, V ), we can always b perform a change of coordinates k → k − ξ for some suitable ξ ∈ Γ# so that k − ξ ∈ F(A, V) and Re(k − ξ) ∈ B. Here (Re k) ∈ R2 denotes the real part of k ∈ C2 . Thus, without loss of generality we may assume that (Re k) ∈ B. We now prove that any point outside the region KR and outside the union of ε-tubes S b does not belong to F(A, V ). Suppose that k ∈ C2 \ (KR ∪ b∈Γ# Tb ) and recall that k is in b F(A, V ) if and only if (3.5.1) has a nontrivial solution. If we choose G = ∅ then G0 = Γ# and this equation reads RG0 G0 ϕG0 = 0. By Proposition 4.2.1(iii) we have G0 = Γ# = {b ∈ Γ# | |Nb (k)| ≥ ε|v|}. Furthermore, since u ∈ B and |v| > R ≥ α, it follows that |u| ≤ α < |v| < 2|v|. Consequently, the operator RG0 G0 has a bounded inverse by Lemma 4.1.4. Thus, the only solution of the above equation is ϕG0 = 0. That is, there is no nontrivial solution of this equation and therefore b k 6∈ F(A, V ). 59  We are left to study the Fermi curve inside the ε-tubes. There are two types of regions to consider: intersections and non-intersections of tubes. To study non-intersections we choose G = {0} and consider the region (T0 \ ∪b∈Γ# \{0} Tb ) \ KR . For intersections we take G = {0, d} for some d ∈ Γ# \ {0} and consider (T0 ∩ Td ) \ KR (see Figure 4.1). Observe that, since the tubes Tb have the following translational property, Tb + c = Tb+c for all b, c ∈ Γ# , b and the curve F(A, V ) is invariant under the action of Γ# , there is no loss of generality in considering only the two regions above. Any other part of the curve can be reached by translation.  KR  KR   Figure 4.1: Sketch of T0 \ ∪b∈Γ# \{0} Tb \ KR (left) and (T0 ∩ Td ) \ KR (right). Recall that G0 = Γ# \ G, and for d0 , d00 ∈ G and i, j ∈ {1, 2} set 0 00  dd Bij (k; G) := −4  X Âi (d0 − b) −1 00 (RG 0 G0 )b,c Âj (c − d ), N (k) b 0  b,c∈G 0 00  Cid d (k; G) := −2Âi (d0 − d00 ) + 2  X q̂(d0 − b) − 2b · Â(d0 − b) −1 00 (RG 0 G0 )b,c Âi (c − d ) N (k) b 0  b,c∈G  +2  X Âi (d0 − b) −1 00 00 00 (RG 0 G0 )b,c (q̂(c − d ) − 2d · Â(c − d )), N (k) b 0  b,c∈G d0 d00  C0  (k; G) := q̂(d0 − d00 ) − 2d00 · Â(d0 − d00 ) −  X q̂(d0 − b) − 2b · Â(d0 − b) −1 00 00 00 (RG 0 G0 )b,c (q̂(c − d ) − 2d · Â(c − d )). N (k) b 0  b,c∈G  (4.2.1) 60  Then, Dd0 ,d00 (k; G) := wd0 ,d00 −  X wd0 ,b −1 (RG 0 G0 )b,c wc,d00 N (k) b 0  b,c∈G  = q̂(d0 − d00 ) − 2(d00 + k) · Â(d0 − d00 ) −  X  (q̂(d0 − b) − 2(b + k) · Â(d0 − b))  b,c∈G0 0 00  0 00  0 00  −1 (RG 0 G0 )b,c (q̂(c − d00 ) − 2(d00 + k) · Â(c − d00 )) Nb (k)  0 00  0 00  0 00  0 00  dd 2 dd 2 dd dd = B11 k1 + B22 k2 + (B12 + B21 )k1 k2 + C1d d k1 + C2d d k2 + C0d d .  Furthermore, we shall shortly prove the following property. 0 00  0 00  0 00  d d , C d d , C d d (and Proposition 4.2.3. For d0 , d00 ∈ G and i, j ∈ {1, 2}, the functions Bij 0 i  consequently Dd0 ,d00 ) are analytic on (T0 \ ∪b∈Γ# \{0} Tb ) \ KR and (T0 ∩ Td ) \ KR for G = {0} and G = {0, d}, respectively. Using the above functions we can write (local) defining equations for the Fermi curve. b Lemma 4.2.4 (Local defining equations for F(A, V )). b (i) Let G = {0} and k ∈ (T0 \ ∪b∈Γ# \{0} Tb ) \ KR . Then k ∈ F(A, V ) if and only if N0 (k) + D0,0 (k) = 0. b (ii) Let G = {0, d} and k ∈ (T0 ∩ Td ) \ KR . Then k ∈ F(A, V ) if and only if (N0 (k) + D0,0 (k))(Nd (k) + Dd,d (k)) − D0,d (k)Dd,0 (k) = 0. We now prove this lemma and then Proposition 4.2.3. The proof of this lemma is easy once we have that RG0 G0 is invertible. Proof. (i) First, by Proposition 4.2.1(i) we have G0 = Γ# \ {0} = {b ∈ Γ# | |Nb (k)| ≥ ε|v|}. Furthermore, since k ∈ T0 , we have either |v − u⊥ | < ε or |v + u⊥ | < ε. In either case this implies |u| < ε + |v| < 2Λ + |v| < 2|v|.  (4.2.2)  Hence, the operator RG0 G0 has a bounded inverse by Lemma 4.1.4. Thus, in the region b under consideration F(A, V ) is given by (3.5.4): 0 = N0 (k) + w0,0 −  X b,c∈G0  w0,b (R−10 0 )b,c wc,0 = N0 (k) + D0,0 (k). Nb (k) G G  This proves part (i). 61  (ii) Similarly, by Proposition 4.2.1(ii), G0 = Γ# \ {0, d} = {b ∈ Γ# | |Nb (k)| ≥ ε|v|}. Furthermore, since k ∈ T0 ∩ Td ⊂ T0 , similarly as above we obtain (4.2.2). Thus, by Lemma 4.1.4 the operator RG0 G0 has a bounded inverse. Hence, in the region under consideration b F(A, V ) is given by (3.5.4): (N0 (k) + D0,0 (k))(Nd (k) + Dd,d (k)) − D0,d (k)Dd,0 (k) = 0. This proves part (ii) and completes the proof of the lemma. As promised, here is the (sketchy) proof of Proposition 4.2.3. 0 00  0 00  0 00  d d , C d d and C d d are analytic funcProof of Proposition 4.2.3. It suffices to show that Bij 0 i  tions. This property follows from the fact that all the series involved in the definition of these functions are uniformly convergent sums of analytic functions (see [16, Theorem 4.1]). The argument is similar for all cases. We give only a sketch of the proof. First observe that, in view of Lemma 4.1.4, for each b, c ∈ G0 , −1 |(RG 0 G0 )b,c | =  (πG0 − (πG0 − RG0 G0 ))−1 ∞ X  =  ≤  (πG0 − RG0 G0 )j  j=0 ∞ X  b,c   b,c  kπG0 − RG0 G0 kj <  j=0     ∞  X 17 j j=0  18  = 18.  Hence, the above sum converges uniformly by the Weierstrass M-test. Since (πG0 − RG0 G0 )b,c −1 is an analytic function of k, so is (RG 0 G0 )b,c . 0 00  0 00  0 00  d d , C d d and C d d are given by sums of the form Now observe that Bij 0 i  const +  X f (b, d0 ) −1 00 (RG 0 G0 )b,c g(c, d ), N (k) b 0  b,c∈G  where f and g are known functions. Furthermore, all the terms in these series are analytic functions, and the sum converges uniformly because of the uniform bounds |b| 4 ≤ , |Nb (k)| ε  1 1 1 ≤ < |Nb (k)| ε|v| εR  and  −1 −1 |(RG 0 G0 )b,c | ≤ kRG0 G0 k ≤ 18  for all b, c ∈ G0 , and because f (·, d0 ) and g(·, d00 ) are in l1 (Γ# ) in all cases by hypothesis. 0 00  0 00  0 00  d d , C d d and C d d are analytic functions. Consequently, all the limits Bij 0 i  62  To study in detail the defining equations above we shall estimate the asymptotic be0 00  0 00  0 00  d d , C d d , C d d and D 0 00 for large |v|. (We sometimes refer to haviour of the functions Bij d ,d 0 i  these functions as coefficients.) Since all these functions have a similar form it is convenient to prove these estimates in a general setting and specialize them later. This is the contents of §4.4 and §4.5. We next introduce a change of variables in C2 that will be useful for proving these bounds.  4.3  Change of coordinates  The following change of coordinates in C2 will be useful for our analysis. For ν ∈ {1, 2} and d0 , d00 ∈ G define the functions wν,d0 , zν,d0 : C2 → C as wν,d0 (k) := k1 + d01 + i(−1)ν (k2 + d02 ),  (4.3.1)  zν,d0 (k) := k1 + d01 − i(−1)ν (k2 + d02 ). Observe that, the transformation (k1 , k2 ) 7→ (wν,d0 , zν,d0 ) is just a translation composed with a rotation. Furthermore, if k ∈ Tν (d0 ) \ KR then |wν,d0 (k)| is “small” and |zν,d0 (k)| is “large”. Indeed, |wν,d0 (k)| = |Nd0 ,ν (k)| < ε  |zν,d0 (k)| = |Nd0 ,ν 0 (k)| ≥ |v| > R.  and  Define also 0 00  0 00  0 00  0 00  0 00  0 00  0 00  0 00  dd dd dd dd − B22 + i(−1)ν (B12 + B21 )), Jνd d := 14 (B11 dd dd K d d := 21 (B11 + B22 ), 0 00  0 00  0 00  0 00  0 00  dd dd dd dd Ldν d := −d01 B11 − i(−1)ν d02 B22 − 21 (d02 + i(−1)ν d01 )(B12 + B21 ) 0 00  0 00  + 21 (C1d d + i(−1)ν C2d d ), 0 00  0 00  0 00  0 00  0 00  0 00  0 00  0 00  dd 02 d d 0 0 dd dd 0 dd M d d := d02 − d02 C2d d + C0d d , 1 B11 + d2 B22 + d1 d2 (B12 + B21 ) − d1 C1 0 00  0 00  0 00  0 00  where Jνd d , K d d , Ldν d and M d d are functions of k ∈ C2 that also depend on the choice of G ⊂ Γ# . Using these functions we can express Nd0 (k) + Dd0 ,d0 (k) and Dd0 ,d00 (k) as follows. Proposition 4.3.1. Let ν ∈ {1, 2} and let d0 , d00 ∈ G. Then, 0 0  0 0  0 0  0 0  0 0  0 0  2 dd 2 Nd0 + Dd0 ,d0 = Jνd0 d wν,d zν,d0 + (1 + K d d )wν,d0 zν,d0 + Ldν 0d wν,d0 + Ldν d zν,d0 + M d d 0 + Jν  63  and 0 00  0 00  0 00  0 00  0 00  0 00  2 dd 2 Dd0 ,d00 = Jνd0 d wν,d zν,d0 + K d d wν,d0 zν,d0 + Ldν 0d wν,d0 + Ldν d zν,d0 + M d d . 0 + Jν  Furthermore, X (1, −i(−1)ν ) · Â(d0 − b) −1 ν 00 (RG 0 G0 )b,c (1, −i(−1) ) · Â(c − d ), N (k) b 0  0 00  Jνd d (k) = −  b,c∈G 0 00  K d d (k) = −2  X Â(d0 − b) · Â(c − d00 ) −1 (RG 0 G0 )b,c , N (k) b 0  b,c∈G 0 00  Ldν d (k) =  X q̂(d0 − b) + 2(d0 − b) · Â(d0 − b) −1 ν 00 (RG 0 G0 )b,c (1, i(−1) ) · Â(c − d ) N (k) b 0  b,c∈G  +  X (1, i(−1)ν ) · Â(d0 − b) −1 00 0 00 00 (RG 0 G0 )b,c (q̂(c − d ) + 2(d − d ) · Â(c − d )) N (k) b 0  b,c∈G  − (1, i(−1)ν ) · Â(d0 − d00 ), and 0 00  M d d (k) = −  X q̂(d0 − b) + 2(d0 − b) · Â(d0 − b) −1 00 (RG 0 G0 )b,c q̂(c − d ) N (k) b 0  b,c∈G  + q̂(d0 − d00 ) + 2(d0 − d00 ) · Â(d0 − d00 ). Proof. To simplify the notation write w = wν,d0 ,  z = zν,d0 ,  0 00  dd Bij = Bij  and  0 00  Ci = Cid d .  First observe that, in view of (4.3.1), Nd0 = (k1 + d01 + i(−1)ν (k2 + d02 ))(k1 + d01 − i(−1)ν (k2 + d02 )) = wz. Furthermore, k1 = 12 (w + z) − d01 , k2 =  (−1)ν 2i (w  − z) − d02 ,  k12 = 14 (w2 + z 2 ) + 12 wz − d01 (w + z) + d02 1, k22 = − 41 (w2 + z 2 ) + 12 wz + i(−1)ν d02 (w − z) + d02 2, k1 k2 =  i(−1)ν 2 4 (z  − w2 ) − 21 (d02 − i(−1)ν d01 )w − 21 (d02 + i(−1)ν d01 ) + d01 d02 .  64  Hence, Dd0 ,d00 = B11 k12 + B22 k22 + (B12 + B21 )k1 k2 + C1 k1 + C2 k2 + C0 = 41 (B11 − B22 − i(−1)ν (B12 + B21 ))w2 + 14 (B11 − B22 + i(−1)ν (B12 + B21 ))z 2  + − d01 B11 + i(−1)ν d02 B22 − 12 (d02 − i(−1)ν d01 )(B12 + B21 ) + 12 (C1 − i(−1)ν C2 ) w  + − d01 B11 + i(−1)ν d02 B22 − 21 (d02 + i(−1)ν d01 )(B12 + B21 ) + 12 (C1 + i(−1)ν C2 ) z 02 0 0 0 0 1 + d02 1 B11 + d2 B22 + d1 d2 (B12 + B21 ) − d1 C1 − d2 C2 + C0 + 2 (B11 + B22 )wz 0 00  0 00  0 00  0 00  0 00  0 00  = Jνd0 d w2 + Jνd d z 2 + K d d wz + Ldν 0d w + Ldν d z + M d d . This proves the first claim. Consequently, 0 0  0 0  0 0  0 0  0 0  0 0  Nd0 + Dd0 ,d0 = Jνd0 d w2 + Jνd d z 2 + (1 + K d d )wz + Ldν 0d w + Ldν d z + M d d , which proves the second claim. Now, again to simplify the notation write fg =  X fˆ(b, d0 ) −1 00 (RG 0 G0 )b,c ĝ(c, d ), N (k) b 0  b,c∈G  that is, to represent sums of this form suppress the summation and the other factors. Note that f g 6= gf according to this notation. Then, substituting (4.2.1) into the definitions of 0 00  0 00  0 00  0 00  Jνd d , K d d , Ldν d and M d d we obtain 0  00  Jνd ,d = 14 (B11 − B22 + i(−1)ν (B12 + B21 )) = −A1 A1 + A2 A2 − i(−1)ν (A1 A2 + A2 A1 ) = (A1 − i(−1)ν A2 )(−A1 + i(−1)ν A2 ) = −((1, −i(−1)ν ) · A) ((1, −i(−1)ν ) · A) =−  X (1, −i(−1)ν ) · Â(d0 − b) −1 ν 00 (RG 0 G0 )b,c (1, −i(−1) ) · Â(c − d ) N (k) b 0  b,c∈G  and 0  00  K d ,d = 21 (B11 + B22 ) = −2(A1 A1 + A2 A2 ) = −2  X Â(d0 − b) · Â(c − d00 ) −1 (RG 0 G0 )b,c N (k) b 0  b,c∈G  and 65  0 00  Ldν d = −d01 B11 − i(−1)ν d02 B22 − 21 (d02 + i(−1)ν d01 )(B12 + B21 ) + 12 (C1 + i(−1)ν C2 ) = 4d01 A1 A1 + 4i(−1)ν d02 A2 A2 + 2d02 (A1 A2 + A2 A1 ) + 2i(−1)ν d01 (A1 A2 + A2 A1 ) + (q − 2b · A)A1 + A1 (q − 2d00 · A) + i(−1)ν (q − 2b · A)A2 + i(−1)ν A2 (q − 2d00 · A) − (Â1 (d0 − d00 ) + i(−1)ν Â2 (d0 − d00 )) = 2(2d01 − d001 )A1 A1 + 2i(−1)ν (2d02 − d002 )A2 A2 + 2(d02 − d002 + i(−1)ν d01 )A1 A2 + 2(d02 + i(−1)ν d01 − i(−1)ν d001 )A2 A1 + (q − 2b · A)A1 + A1 q + i(−1)ν (q − 2b · A)A2 + i(−1)ν A2 q − (1, i(−1)ν ) · Â(d0 − d00 ) = 2(d01 A1 + d02 A2 )(A1 + i(−1)ν A2 ) + 2(1, i(−1)ν ) · A ((d0 − d00 ) · A) + (q − 2b · A)(A1 + i(−1)ν A2 ) + (A1 + i(−1)ν A2 )q − (1, i(−1)ν ) · Â(d0 − d00 ) = (q + 2(d − b) · A)(1, i(−1)ν ) · A + (1, i(−1)ν ) · A (q + 2(d0 − d00 ) · A) − (1, i(−1)ν ) · Â(d0 − d00 ) so that 0 00  Lνd d =  X q̂(d0 − b) + 2(d0 − b) · Â(d0 − b) −1 ν 00 (RG 0 G0 )b,c (1, i(−1) ) · Â(c − d ) N (k) b 0  b,c∈G  +  X (1, i(−1)ν ) · Â(d0 − b) −1 00 0 00 00 (RG 0 G0 )b,c (q̂(c − d ) + 2(d − d ) · Â(c − d )) N (k) b 0  b,c∈G  − (1, i(−1)ν ) · Â(d0 − d00 ), 0 00  02 0 0 0 0 M d d = d02 1 B11 + d2 B22 + d1 d2 (B12 + B21 ) − d1 C1 − d2 C2 + C0 02 0 0 0 0 0 0 00 = −4(d02 1 A1 A1 + d2 A2 A2 + d1 d2 A1 A2 + d1 d2 A2 A1 ) + 2d · Â(d − d )  − 2d01 (q − 2b · A)A1 − 2d01 A1 (q − 2d0 · A) − 2d02 (q − 2b · A)A2 − 2d02 A2 (q − 2d0 · A) − (q − 2b · A)(q − 2d0 · A) + q̂(d0 − d00 ) − 2d00 · Â(d0 − d00 ) = −4(d0 · A)(d0 · A) − 2(q − 2b · A)(d0 · A) − 2(d0 · A)(q − 2d0 · A) − (q − 2b · A)(q − 2d0 · A) + q̂(d0 − d00 ) + 2(d0 − d00 ) · Â(d0 − d00 ) = −qq − 2(d − b) · A q + q̂(d0 − d00 ) + 2(d0 − d00 ) · Â(d0 − d00 ) =−  X q̂(d0 − b) + 2(d0 − b) · Â(d0 − b) −1 00 (RG 0 G0 )b,c q̂(c − d ) N (k) b 0  b,c∈G  + q̂(d0 − d00 ) + 2(d0 − d00 ) · Â(d0 − d00 ). We next use this change of variables for deriving asymptotics for certain functions. 66  4.4  Asymptotics for the coefficients  Let f and g be functions on Γ# and for k ∈ C2 and d0 , d00 ∈ G set Φd0 ,d00 (k; G) :=  X f (d0 − b) −1 00 (RG 0 G0 )b,c g(c − d ). N (k) b 0  (4.4.1)  b,c∈G  In this section we study the asymptotic behaviour of the function Φd0 ,d00 (k) for k in the union of ε-tubes with large |v|. We first give all the statements and then the proofs. Reset the constant R as   4 2 := , R max 1, α, 2Λ, 140kÂkl1 , k(1 + b )q̂(b)kl1 ε  (4.4.2)  and make the following hypothesis. Hypothesis 4.4.1. kb2 q̂(b)kl1 < ∞  k(1 + b2 )Â(b)kl1 <  and  2 ε. 63  Our first lemma provides and expansion for Φd0 ,d0 (k) “in powers of 1/|zν,d0 (k)|”. Lemma 4.4.1 (Asymptotics for Φd0 ,d0 (k)). Under Hypothesis 4.4.1, let ν ∈ {1, 2} and let f and g be functions on Γ# with kb2 f (b)kl1 < ∞ and kb2 g(b)kl1 < ∞. Suppose either (i) or (ii) where: (i) G = {0} and k ∈ (Tν (0) \ ∪b∈G0 Tb ) \ KR ; (ii) G = {0, d} and k ∈ (Tν (0) ∩ Tν 0 (d)) \ KR . Then, for (µ, d0 ) = (ν, 0) if (i) or (µ, d0 ) ∈ {(ν, 0), (ν 0 , d)} if (ii), (1)  (2)  (3)  Φd0 ,d0 (k) = αµ,d0 (k) + αµ,d0 (k) + αµ,d0 (k), where for 1 ≤ j ≤ 2, (j)  |αµ,d0 (k)| ≤  Cj (2|zµ,d0 (k)| − R)j  and  (3)  |αµ,d0 (k)| ≤  C3 , |zµ,d0 (k)|R2  where Cj = Cj;Λ,A,q,f,g and C3 = C3;ε,Λ,A,q,f,g are constants. Furthermore, the functions (j)  αµ,d0 (k) are given by (4.4.25) and (4.4.28) and are analytic in the region under consideration.  67  (1)  Below we have more information about the function αµ,d0 (k). (1)  Lemma 4.4.2 (Asymptotics for αµ,d0 (k)). Consider the same hypotheses of Lemma 4.4.1. Then, for (µ, d0 ) = (ν, 0) if (i) or (µ, d0 ) ∈ {(ν, 0), (ν 0 , d)} if (ii), (1)  (1,0)  (1,1)  (1,2)  (1,3)  zµ,d0 (k) αµ,d0 (k) = αµ,d0 + αµ,d0 (w(k)) + αµ,d0 (k) + αµ,d0 (k), (1,0)  (1,j)  where αµ,d0 is a constant given by (4.4.39), and the remaining functions αµ,d0 are given by (4.4.38). Furthermore, for 0 ≤ j ≤ 2, (1,j)  |αµ,d0 | ≤ Cj  (1,3)  |αµ,d0 | ≤  and  C3 , 2|zµ,d0 (k)| − R  where Cj = Cj;Λ,A,f,g and C3 = C3;Λ,A,f,g are constants given by (4.4.40). The next lemma estimates the decay of Φd0 ,d00 (k) with respect to zν 0 ,d (k) for d0 6= d00 . Lemma 4.4.3 (Decay of Φd0 ,d00 (k) for d0 6= d00 ). Under Hypothesis 4.4.1, let ν ∈ {1, 2} and let f and g be functions on Γ# with kb2 f (b)kl1 < ∞ and kb2 g(b)kl1 < ∞. Suppose further that G = {0, d} and k ∈ (Tν (0) ∩ Tν 0 (d)) \ KR . Then, for d0 , d00 ∈ G with d0 6= d00 , |Φd0 ,d00 (k)| ≤  CΓ# ,ε,f,g |zν 0 ,d (k)|3−10−1  ,  where CΓ# ,ε,f,g is a constant. The above lemmas are the main statements of this section. Before we move to the proofs we give one more proposition that provides relations between the quantities |v|, |k2 |, |zν,d0 (k)| and |d| for k in the ε-tubes with large |v|. Proposition 4.4.4. For ν ∈ {1, 2} we have: (i) Let k ∈ Tν (0) \ KR . Then, 1 1 3 ≤ ≤ |zν,0 (k)| |v| |zν,0 (k)|  and  1 1 8 ≤ ≤ . 4|v| |k2 | |v|  (ii) Let k ∈ (Tν (0) ∩ Tν 0 (d)) \ KR . Then, 1 1 3 ≤ ≤ , |zν,0 (k)| |v| |zν,0 (k)| and  1 2|zν 0 ,d (k)|  1 1 3 ≤ ≤ |zν 0 ,d (k)| |v| |zν 0 ,d (k)| ≤  1 2 ≤ . |d| |zν 0 ,d (k)|  This proposition will be used several times henceforth. We next prove all the above statements. 68  Proof of Proposition 4.4.4 Proof of Proposition 4.4.4. We first derive a more general inequality and then we prove parts (i) and (ii). First observe that, if k ∈ Tµ (d0 ) \ KR then |v + (−1)µ (u + d0 )⊥ | = |Nd0 ,µ (k)| < ε < |v|. Hence, |v| ≤ |2v − (v + (−1)µ (u + d0 )⊥ )| ≤ 3|v|. But |2v − (v + (−1)µ (u + d0 )⊥ )| = |v − (−1)µ (u + d0 )⊥ | = |k1 + d01 − i(−1)µ (k2 + d02 )| = |zµ,d0 (k)|. Thus, |v| ≤ |zµ,d0 (k)| ≤ 3|v|. Therefore, 1 |z  µ,d0  (k)|  ≤  3 1 ≤ . |v| |zµ,d0 (k)|  (4.4.3)  (i) The first inequality of part (i) follows from the above estimate setting (µ, d0 ) = (ν, 0). To prove the second inequality observe that, since |v| > R ≥ 2Λ > 12ε by hypothesis and |v| ≤ |zν,0 (k)| by (4.4.3), on the one hand we have 1 4 |v|  ≤  11 12 |v|  = |v| −  1 12 |v|  ≤ |v| − 61 Λ ≤ |v| − ε ≤ |zν,0 (k)| − |k1 + i(−1)ν k2 | ≤ |zν,0 (k) − k1 − i(−1)ν k2 | = 2|k2 |.  On the other hand, since |zν,0 (k)| < 3|v| by (4.4.3), |k2 | = |2i(−1)ν k2 | = |k1 + i(−1)ν k2 − (k1 − i(−1)ν k2 )| = |k1 + i(−1)ν k2 − zν,0 (k)| ≤ ε + 3|v| ≤ 4|v|. Combining these estimates we obtain the second inequality of part (i). (ii) Similarly, in view of (4.4.3), if k ∈ Tµ (d0 ) \ KR for (µ, d0 ) ∈ {(ν, 0), (ν 0 , d)} then 1 1 3 ≤ ≤ |zν,0 (k)| |v| |zν,0 (k)|  1 1 3 ≤ ≤ . |zν 0 ,d (k)| |v| |zν 0 ,d (k)|  and  69  (4.4.4)  These are the first two inequalities of part (ii). Now, since 0  0  zν 0 ,d (k) = k1 − i(−1)ν k2 + d1 − i(−1)ν d2 0  0  = zν 0 ,0 (k) + d1 − i(−1)ν d2 = wν,0 (k) + d1 − i(−1)ν d2 , 0  |wν,0 (k)| < ε, and |d1 − i(−1)ν d2 | = |d|, it follows that |zν 0 ,d (k)| − ε ≤ |d| ≤ |zν 0 ,d (k)| + ε. Furthermore, by (4.4.4), ε<  |zν 0 ,d (k)| Λ |v| ≤ ≤ . 6 12 12  Thus, 1 |zν 0 ,d (k)| ≤ |d| ≤ 2|zν 0 ,d (k)|. 2 This yields the third inequality of part (ii) and completes the proof.  Proof of Lemma 4.4.1 Proof of Lemma 4.4.1. We consider all cases at the same time. Therefore, we have either hypothesis (i) with (µ, d0 ) = (ν, 0) or hypothesis (ii) with (µ, d0 ) ∈ {(ν, 0), (ν 0 , d)}. Note that either (ν, ν 0 ) = (1, 2) or (ν, ν 0 ) = (2, 1). Step 1 Recall the change of variables wµ,d0 (k) = k1 + d01 + i(−1)µ (k2 + d02 ), zµ,d0 (k) = k1 + d01 − i(−1)µ (k2 + d02 ), and set  G01 := b ∈ G0 | |b − d0 | < 14 R ,  G02 := b ∈ G0 | |b − d0 | ≥ 14 R . Observe that G0 = G01 ∪ G02 . By Proposition 4.2.1, G01 , G02 ⊂ G0 = Γ# \ G = {b ∈ Γ# | |Nb (k)| ≥ ε|v|}.  70  Furthermore, by Proposition 4.4.4, for (µ, d0 ) = (ν, 0) if (i) or (µ, d0 ) ∈ {(ν, 0), (ν 0 , d)} if (ii), 1 3 ≤ . |v| |zµ,d0 | Thus, using the above remarks and observing the definition of G02 we have X X f (d0 − b) −1 0 (RG 0 G0 )b,c g(c − d ) N (k) b 0 0  |R1 (k)| :=  (4.4.5)  b∈G1 c∈G2  ≤  X X |c − d0 |2 1 −1 |g(c − d0 )| kRG |f (d0 − b)| 0 G0 k ε|v| |c − d0 |2 0 0 b∈G1  ≤  c∈G2  Cε,f,g 1 16 2 −1 kc g(c)kl1 ≤ kRG 0 G0 k kf kl1 2 ε|v| R |zµ,d0 |R2  and |R2 (k)| :=  X X f (d0 − b) −1 0 (RG 0 G0 )b,c g(c − d ) N (k) b 0 0  (4.4.6)  b∈G2 c∈G  ≤  X |d0 − b|2 X 1 −1 0 kRG |f (d − b)| |g(c − d0 )| 0 G0 k 0 − b|2 ε|v| |d 0 0 c∈G  b∈G2  ≤  Cε,f,g 16 1 −1 2 kRG kgkl1 ≤ . 0 G0 k k|b| f (b)kl1 2 ε|v| R |zµ,d0 |R2  Hence, #  " Φd0 ,d0 (k) =  X b,c∈G01  +  X  +  b∈G01 c∈G02  X b∈G02 c∈G0  f (d0 − b) −1 (RG0 G0 )b,c g(c − d0 ) Nb (k) (4.4.7)  X f (d0 − b) −1 0 = (RG 0 G0 )b,c g(c − d ) + R1 (k) + R2 (k) N (k) b 0 b,c∈G1  with |R1 (k) + R2 (k)| ≤  Cε,f,g . |zµ,d0 |R2  (4.4.8)  Now, if we set TG0 G0 := πG0 − RG0 G0 and recall the convergent series expansion −1 −1 RG = 0 G0 = (πG0 − TG0 G0 )  ∞ X  TGj 0 G0 ,  j=0  we can write ∞ X X f (d0 − b) X f (d0 − b) j −1 0 (RG ) g(c − d ) = (TG0 G0 )b,c g(c − d0 ). 0 G0 b,c N (k) N (k) b b 0 0  b,c∈G1  j=0 b,c∈G1  71  (4.4.9)  Note, the above equality is fine because G01 is finite set. Let G03 := {b ∈ G0 | |b − d0 | < 12 R}, G04 := {b ∈ G0 | |b − d0 | ≥ 12 R}. Again, observe that G0 = G03 ∪ G04 . Thus, we can break TG0 G0 into TG0 G0 = πG0 T πG0 = (πG03 + πG04 )T (πG03 + πG04 ) = T33 + T43 + T34 + T44 , where Tij := πGi T πGj for i, j ∈ {3, 4}. Using this decomposition we are able to prove the following. Proposition 4.4.5. Under the hypotheses of Lemma 4.4.1 we have ∞ X ∞ X X X f (d0 − b) j f (d0 − b) j (TG0 G0 )b,c g(c − d0 ) = (T33 )b,c g(c − d0 ) + R3 (k) N (k) N (k) b b 0 0 j=0 b,c∈G1  j=0 b,c∈G1  with R3 (k) given by (4.4.34) and |R3 (k)| ≤  CΛ,f,g . |zµ,d0 |R2  (4.4.10)  This proposition will be proved below. Combining this with (4.4.7) and (4.4.9) we obtain Φ  d0 ,d0  ∞ X 3 X X f (d0 − b) j 0 (k) = (T33 )b,c g(c − d ) + Rj (k). Nb (k) 0 j=0 b,c∈G1  (4.4.11)  j=1  j . Recall that Step 2 We now look in detail to the operator T33 and its powers T33  1 θµ (b) = ((−1)µ b2 + ib1 ) 2 0  and set µ0 := µ − (−1)µ so that (−1)µ = −(−1)µ . Then, 0  Nb (k) = Nb,µ (k)Nb,µ0 (k) = (k1 + b1 + i(−1)µ (k2 + b2 ))(k1 + b1 + i(−1)µ (k2 + b2 )) = (k1 + d01 + i(−1)µ (k2 + d02 ) + b1 − d01 + i(−1)µ (b2 − d02 )) 0  0  × (k1 + d01 + i(−1)µ (k2 + d02 ) + b1 − d01 + i(−1)µ (b2 − d02 )) 0  = (k1 + d01 + i(−1)µ (k2 + d02 ) + b1 − d01 − i(−1)µ (b2 − d02 )) × (k1 + d01 − i(−1)µ (k2 + d02 ) + b1 − d01 − i(−1)µ (b2 − d02 )) = (wµ,d0 − 2iθµ0 (b − d0 ))(zµ,d0 − 2iθµ (b − d0 )). 72  Extend the definition of θµ (y) to any y ∈ C2 . Thus, 2(k + d0 ) · Â(b − c) = 2(k1 + d01 )Â1 (b − c) + 2(k2 + d02 )Â2 (b − c) = (wµ,d0 + zµ,d0 )Â1 (b − c) + (i(−1)µ (zµ,d0 − wµ,d0 ))Â2 (b − c) = (Â1 (b − c) − i(−1)µ Â2 (b − c))wµ,d0 + (Â1 (b − c) + i(−1)µ Â2 (b − c))zµ,d0 = −2iθµ (Â(b − c)) wµ,d0 − 2iθµ0 (Â(b − c)) zµ,d0 . Hence, Tb,c =  1 (2(c + k) · Â(b − c) − q̂(b − c)) Nc (k)  2(c − d0 ) · Â(b − c) − q̂(b − c) + 2(k + d0 ) · Â(b − c) = = Xb,c + Yb,c , (wµ,d0 − 2iθµ0 (c − d0 ))(zµ,d0 − 2iθµ (c − d0 ))  (4.4.12)  where Xb,c := Yb,c :=  2(c − d0 ) · Â(b − c) − q̂(b − c) − 2iθµ (Â(b − c)) wµ,d0 , (wµ,d0 − 2iθµ0 (c − d0 ))(zµ,d0 − 2iθµ (c − d0 )) (wµ,d0  −2iθµ0 (Â(b − c)) zµ,d0 . − 2iθµ0 (c − d0 ))(zµ,d0 − 2iθµ (c − d0 ))  (4.4.13) (4.4.14)  Let X and Y be the operators whose matrix elements are, respectively, Xb,c and Yb,c . Set X33 := πG03 XπG03  and  Y33 := πG03 Y πG03 .  We next prove the following estimates,   4 1 1 kX33 k ≤ 20kÂkl1 + kq̂kl1 < , Λ |zµ,d0 |R 3 8 1 kY33 k ≤ kθµ0 (Â)kl1 < , Λ 14  (4.4.15)  where |zµ,d0 |R := 2|zµ,d0 | − R. First observe that the “vector” b ∈ Γ# has the same length as the complex number 2iθµ (b): |b| = |(b1 , b2 )| = |b1 + i(−1)µ b2 | = |2iθµ (b)|. Thus, for b ∈ G03 ,  (4.4.16)  |2iθµ (b − d0 )| |b − d0 | 1 = < . R R 2  Consequently, |z  µ,d0  1 1 1 2 ≤ < = . 1 0 0 0 − 2iθµ (b − d )| |zµ,d | − |2iθµ (b − d )| |zµ,d0 |R |zµ,d0 | − 2 R 73  (4.4.17)  Furthermore, for b ∈ G0 , 1 1 1 ≤ ≤ 0 0 − 2iθµ0 (b − d )| |b − d | − |wµ,d0 | |b − d0 | − ε 1 1 ≤ = . 2Λ − Λ Λ  |wµ,d0  (4.4.18) (4.4.19)  Here we have used that |wµ,d0 | < ε < Λ and |b − d0 | ≥ 2Λ for all b ∈ G0 . Using again that ε < Λ ≤ |c − d0 |/2 for all c ∈ G0 we have |c − d0 | < 2. |c − d0 | − ε  (4.4.20)  Finally recall that ε 1 < Λ 6  and  1 |z  µ,d0  |  ≤  1 1 < , |v| R  (4.4.21)  where the last inequality follows from Proposition 4.4.4 since |v| > R by hypothesis. Then, using the above inequalities and Proposition 4.1.2, the bounds (4.4.15) for kX33 k and kY33 k follow from the estimates   X X  sup  |Xb,c | + sup c∈G03 b∈G0  b∈G03 c∈G0  3  3     ≤  sup  X  c∈G03 b∈G0  3  + sup  0   2|c − d | |Â(b − c)| + |q̂(b − c)| + |2iθµ (Â(b − c))| |wµ,d | |wµ,d0 − 2iθµ0 (c − d0 )| |zµ,d0 − 2iθµ (c − d0 )| 0  X  0  b∈G03 c∈G  3  (now apply (4.4.16), (4.4.17) and |wµ,d0 | < ε) "  # √ 0 | |Â(b − c)| X X 2|c − d |q̂(b − c)| + ε 2 | Â(b − c)| 2   sup + sup + ≤ |zµ,d0 |R c∈G03 |wµ,d0 − 2iθµ0 (c − d0 )| |wµ,d0 − 2iθµ0 (c − d0 )| b∈G03 0 0 b∈G3  c∈G3  (now apply (4.4.18) and (4.4.19))  " # √ 0 | |Â(b − c)| X X 2 2|c − d |q̂(b − c)| + ε 2 | Â(b − c)|  sup  ≤ + sup + |zµ,d0 |R c∈G03 |c − d0 | − ε Λ b∈G03 0 0 b∈G3  c∈G3  (now apply (4.4.20), (4.4.21) and (4.4.2)) "" #  √ #  2 ε 2 kq̂kl1 4kq̂kl1 1 ≤ 2 4+ kÂkl1 + ≤ 20kÂkl1 + |zµ,d0 |R Λ Λ Λ |zµ,d0 |R   4kq̂kl1 1 ≤ 20kÂkl1 + Λ R 1 1 1 < + = 7 4 3  74  and     sup  X  + sup  c∈G03 b∈G0  X  b∈G03 c∈G0  3   |Yb,c |  3      ≤  sup  X  X  + sup  c∈G03 b∈G0  b∈G03 c∈G0  3    |wµ,d0  |2iθµ0 (Â(b − c))| |zµ,d0 | − 2iθµ0 (c − d0 )| |zµ,d0 − 2iθµ (c − d0 )|  3  (now apply (4.4.16), (4.4.17) and (4.4.19)) "  # X X 0 (Â(b − c))| |zµ,d0 | |2θ 2 µ  + sup ≤ sup 0 Λ c∈G03 |zµ,d0 |R b∈G 0 0 3 c∈G b∈G3 3 √ √ 8 8 2 16 2 ε ≤ kθµ0 (Â)kl1 ≤ kÂkl1 ≤ Λ√ Λ 63 Λ 16 2 1 1 < < . 63 6 14 j . For each integer j ≥ 1 write Step 3 We now look in detail to T33 j j , = (X33 + Y33 )j = Zj + Wj + Y33 T33  (4.4.22)  where Wj is the sum of the j terms containing only one factor X33 and j − 1 factors Y33 , Wj := X33 Y33 · · · Y33 + Y33 X33 Y33 · · · Y33 + · · · + Y33 · · · Y33 X33 =  j X  (Y33 )m−1 X33 (Y33 )j−m ,  m=1  and j Zj := (X33 + Y33 )j − Wj − Y33 .  In view of (4.4.15) we have j  kY33 k ≤    1 14  j ,   j−1 CΛ,A,q 1 , kWj k ≤ jkX33 k kY33 k ≤ j |zµ,d0 |R 14  j−2   CΛ,A,q 2 j j 2 1 kZj k ≤ (2 − j − 1) kX33 k ≤ . 3 |zµ,d0 |2R 3 j−1  Hence, the series S :=  ∞ X j=0  j Y33  = (I − Y33 )  −1  ,  W :=  ∞ X j=1  75  Wj  and  Z :=  ∞ X j=2  Zj  (4.4.23)  converge, and the operator norm of W and Z decay with respect to |zµ,d0 |. Indeed, kSk ≤  ∞ X j=0   ∞  X 1 j kY33 k ≤ < C, 14 j  j=0  ∞ X   j−1 ∞ X CΛ,A,q 1 kW k ≤ , kWj k ≤ j < 2|zµ,d0 | − R 14 |zµ,d0 |R j=1 j=1 ∞ ∞  j 0 X X CΛ,A,q CΛ,A,q 2 kZk ≤ kZj k ≤ ≤ 2 2 . 0| 0| 3 |z |z µ,d µ,d R R j=2 j=2 Thus, we have the expansion  0 CΛ,A,q  ∞ X  j T33 = S + W + Z.  j=0  Step 4 Consequently, ∞ X X X f (d0 − b) (S + W + Z)b,c g(c − d0 ) f (d0 − b) j (T33 )b,c g(c − d0 ) = Nb (k) (wµ,d0 − 2iθµ0 (b − d0 ))(zµ,d0 − 2iθµ (b − d0 )) 0 0 j=0 b,c∈G1  b,c∈G1  (2)  (1)  = αµ,d0 + αµ,d0 + R4 , (4.4.24) where (1)  αµ,d0 (k) :=  X b,c∈G01  (2) αµ,d0 (k)  :=  X b,c∈G01  f (d0 − b) Sb,c (k) g(c − d0 ) , (wµ,d0 (k) − 2iθµ0 (b − d0 ))(zµ,d0 (k) − 2iθµ (b − d0 )) f (d0 − b) Wb,c (k) g(c − d0 ) (wµ,d0 (k) − 2iθµ0 (b − d0 ))(zµ,d0 (k) − 2iθµ (b − d0 ))  (4.4.25)  and R4 (k) :=  X b,c∈G01  f (d0 − b) Zb,c (k) g(c − d0 ) . (wµ,d0 (k) − 2iθµ0 (b − d0 ))(zµ,d0 (k) − 2iθµ (b − d0 ))  (4.4.26)  By a short calculation as in (4.4.33), using (4.4.17) and (4.4.19) we find that CΛ,f,g 1 2 kf kl1 kgkl1 kSk ≤ , 0 Λ 2|zµ,d | − R |zµ,d0 |R CΛ,A,q,f,g 1 2 (2) |αµ,d0 (k)| ≤ kf kl1 kgkl1 kW k ≤ , Λ 2|zµ,d0 | − R |zµ,d0 |2R CΛ,A,q,f,g 1 2 |R4 (k)| ≤ kf kl1 kgkl1 kZk ≤ . 0 Λ 2|zµ,d | − R |zµ,d0 |3R (1)  |αµ,d0 (k)| ≤  Hence, recalling (4.4.11) we conclude that (1)  (2)  (3)  Φd0 ,d0 = αµ,d0 + αµ,d0 + αµ,d0 , 76  (4.4.27)  where (3)  αµ,d0 (k) :=  4 X  Rj (k).  (4.4.28)  j=1  Furthermore, in view of (4.4.8), (4.4.10) and (4.4.27), since 1 |zµ,d0 |3R  =  (2|z  1 1 < , 3 | − R) |zµ,d0 |R2  µ,d0  for 1 ≤ j ≤ 2 we have (j)  |αµ,d0 (k)| ≤  Cj  C3 , |zµ,d0 (k)|R2  (3)  |αµ,d0 (k)| ≤  and  |zµ,d0 (k)|jR  where Cj = Cj;Λ,A,q,f,g and C3 = C3;ε,Λ,A,q,f,g are constants. This proves the main statement of the lemma. Finally observe that, since G03 is a finite set, the matrices X33 and Y33 are analytic in k because their matrix elements are analytic functions of k. (Note, the functions wµ,d0 (k) and zµ,d0 (k) are analytic.) Consequently, the matrices Wj and Zj are also analytic and so are Sb,c , Wb,c and Zb,c because the series (4.4.23) converge uniformly with respect (j)  to k. Thus, all the functions αµ,d0 (k) are analytic in the region under consideration. This completes the proof of the lemma. We now prove Proposition 4.4.5, which was used in the above proof. Proof of Proposition 4.4.5. Step 1 Recall that TG0 G0 = T33 + T34 + T43 + T44  Tij = πG0i T πG0j ,  with  and set (0) X33 := 0,  (0) Y34 := T34 ,  (0) W43 := T43  (0) Z44 := T44 .  and  First observe that (1)  (1)  (1)  (1)  2 TG2 0 G0 = T33 + X33 + Y34 + W43 + Z44 ,  where (1) (0) (0) X33 := T33 X33 + T34 W43 (1) Y34  :=  (0) T33 Y34  +  (0) T34 Z44  (1) (0) (0) W43 := T43 T33 + T43 X33 + T44 W43 (1) Z44  :=  (0) T43 Y34  +  (0) T44 Z44  77  : L2G0 → L2G0 , 3  :  L2G0 3  3  →  L2G0 , 4  : L2G0 → L2G0 , 4  :  L2G0 4  3  →  L2G0 . 4  Now suppose that (j−1)  + Y34  (j−1)  : L2G0 → L2G0 ,  j TGj 0 G0 = T33 + X33  (j−1)  (j−1)  + W43  (j−1)  + Z44  with X33  (j−1) Y34 (j−1)  W43  (j−1) Z44  3  :  L2G0 3  3  →  L2G0 , 4  : L2G0 → L2G0 , 4  :  L2G0 4  3  →  L2G0 . 4  It is straightforward to verify that (j)  (j)  (j)  (j)  j+1 + X33 + Y34 + W43 + Z44 , TGj+1 0 G0 = T33  (4.4.29)  where (j) (j−1) (j−1) X33 := T33 X33 + T34 W43 (j) Y34  :=  (j−1) T33 Y34  +  : L2G0 → L2G0 , 3  3  (j−1) T34 Z44  :  L2G0 3  →  L2G0 , 4  (j) (j−1) (j−1) j W43 := T43 T33 + T43 X33 + T44 W43 : L2G0 → L2G0 , 4  (j) Z44  :=  (j−1) T43 Y34  +  (j−1) T44 Z44  (4.4.30)  3  : L2G0 → L2G0 . 4  4  Thus, it follows by induction that (4.4.29) and (4.4.30) hold for any j ≥ 0. Step 2 Since πG01 πG04 = πG04 πG01 = 0 and πG01 πG03 = πG03 πG01 = πG01 , substituting (4.4.29) (0)  into the sum below for the terms where j ≥ 1 we have, recalling that X33 = 0,   0 ∞ ∞ X 0 X X X f (d − b) j  (· · · ) + (TG0 G0 )b,c g(c − d0 ) =  N (k) b 0 j=0  j=0 b,c∈G1  =  ∞ X  X  j=0 b,c∈G01  f (d0  − b) j (T33 )b,c g(c − d0 ) + Nb (k)  j=1  ∞ X  X f (d0 − b) (j−1) (X33 )b,c g(c − d0 ) . (4.4.31) N (k) b j=1 b,c∈G01 | {z } ∞ X 0 X f (d − b) (j) (X33 )b,c g(c − d0 ) N (k) b 0 j=1 b,c∈G1  Now recall from (4.4.17) and (4.4.19) that, for all b ∈ G03 , 1 2 1 ≤ , |Nb (k)| Λ |zµ,d0 |R  78  (4.4.32)  and observe that G01 ⊂ G03 . Let M be either TG0 G0 or T33 . Then, the estimate X f (d0 − b) (Mj )b,c g(c − d0 ) Nb (k) 0  b,c∈G1   ic·x X f (d0 − b) X  eib·x j e = ,M g(c − d0 ) 1/2 1/2 N (k) |Γ| |Γ| b 0 0 b∈G1  (4.4.33)  c∈G1  1 X ≤ |Γ| 0  |f (d0  b∈G1  − b)| X ib·x ke kL2 kMj k keic·x kL2 |g(c − d0 )| |Nb (k)| 0 c∈G1  1 2 kf kl1 kgkl1 kMkj ≤ Λ |zµ,d0 |R implies that the left hand side and the first term on the right hand side of (4.4.31) converge because kMk < 17/18. Thus, the last term in (4.4.31) also converges. Hence, we are left to show that R3 (k) :=  ∞ X X f (d0 − b) (j) (X33 )b,c g(c − d0 ) N (k) b 0  (4.4.34)  j=1 b,c∈G1  obeys |R3 (k)| ≤  CΛ,f,g . |zµ,d0 | R2  In order to do this we need the following inequality, which we prove later. Proposition 4.4.6. Consider a constant β ≥ 0 and suppose that k(1 + |b|β )q̂(b)kl1 < ∞ and k(1 + |b|β )Â(b)kl1 < 2ε/63. Suppose further that |v| > 2ε k(1 + |b|β )Â(b)kl1 . Then, for any B, C ⊂ G0 and m ≥ 1, kπB TGm0 G0 πC k  ≤ (1 + (2Λ)  β−dβe  dβe−1  dβem   )  17 18  m sup b∈B c∈C  1 , 1 + |b − c|β  where dβe is the smallest integer greater or equal than β. Step 3 Now observe that, if b ∈ G01 and c ∈ G04 then |b − c| = |b − d0 − (c − d0 )| ≥ |c − d0 | − |b − d0 | ≥  R R R − = . 2 4 4  Thus, applying the last proposition with β = 2 and recalling that G03 ⊂ G0 , for m ≥ 0 we have m kπG01 T33 T34 k  ≤  kπG01 TGm0 G0 TG0 G04 k  =  kπG01 TGm+1 0 G0 π G0 k 4  3(m + 1) ≤ 1 2 1 + 16 R    17 18  m+1 .  Furthermore, since πG04 πG03 = πG04 πG01 = 0 and πG03 πG01 = πG01 , from (4.4.29) we obtain (j)  j+1 W43 πG01 = πG04 TGj+1 0 G0 πG0 πG0 = πG0 TG0 G0 πG0 . 3 1 4 1  79  Hence, (j) kW43 πG01 k  =  kπG04 TGj+1 0 G0 π G0 k 1  ≤ kTG0 G0 k  j+1   <  17 18  j+1 .  Therefore, for 0 ≤ m < j, kπ  G01  m T33 T34 W (j−m−1) πG01 k  ≤ kπ  G01  (j−m−1) m T33 T34 k kW43 πG01 k  3(m + 1) ≤ 1 2 1 + 16 R    17 18  j+1 .  Iterating the first expression in (4.4.30) we find that (j)  (j−1)  + T33 X33  (j−1)  + T33 T34 W43  X33 = T34 W43  = T34 W43  (j−1) (j−2)  (j−2)  2 + T33 X33  .. . =  (4.4.35) (j−1) T34 W43 j−1 X  =  +  (j−2) T33 T34 W43 (j−m−1)  m T33 T34 W43  + ··· +  (1) j−2 T34 W43 T33  +  (0) j−1 T34 W43 T33  .  m=0  Thus, using the above inequality, (j)  kπG01 X33 πG01 k =  j−1 X  (j−m−1)  m T34 W43 πG01 T33  πG01 ≤  m=0  ≤  3 1+  1 2 16 R  j−1 X  (j−m−1)  m T34 W43 kπG01 T33  πG01 k  m=0    17 18  j+1 X j−1  (m + 1) =  m=0  3 2 + 18 R2  2  (j + j)    17 18  j+1 .  Consequently, πG01    ∞ X (j)  X33  πG0  1  ≤  j=1  ∞ X  (j)  kπG01 X33 πG01 k  j=1   j+1 ∞ X 17 C 2 ≤ (j + j) ≤ 2, 1 2 18 R 2 + 8 R j=1 3  where C is an universal constant. Finally, using this and (4.4.32), since |zµ,d0 | ≤ 3|v| we have   ∞ X f (d0 − b) X (j)  |R3 (k)| = X33  g(c − d0 ) Nb (k) j=1 b,c∈G01 b,c   ∞ X 2 6C 1 (j) ≤ kf kl1 πG01  X33  πG01 kgkl1 ≤ kf kl1 kgkl1 . Λ|v| Λ |zµ,d0 | R2 j=1  In view of (4.4.31) and (4.4.34) this completes the proof. 80  We now prove Proposition 4.4.6, which was used above and left behind without proof. This is the last step we need to finish the proof of Lemma 4.4.1 indeed. Proof of Proposition 4.4.6. For any b, c ∈ Γ# set Qb,c := (1 + |b − c|β )Tb,c . We first claim that, for any B, C ⊂ G0 , X 17 sup |Qb,c | < 18 b∈B  and  sup  X  |Qb,c | <  c∈C b∈B  c∈C  17 . 18  (4.4.36)  In fact, using the bounds (4.1.4), (4.1.5) and (4.2.2), namely, for all b ∈ G0 , |b| 4 ≤ |Nb (k)| ε  1 1 ≤ , |Nb (k)| ε|v|  |k| ≤ |u| + |v| ≤ 3|v|,  and  it follows that X X q̂(b − c) 2c · Â(b − c) 2k · Â(b − c) sup |Qb,c | = sup (1 + |b − c|β ) − − Nc (k) Nc (k) Nc (k) b∈B b∈B c∈C  c∈C  ≤ k(1 + |b|β )q̂(b)kl1  1 4 1 14 17 + k(1 + |b|β )Â(b)kl1 < + = ε|v| ε 2 9 18  and sup  X  |Qb,c | = sup  c∈C b∈B  X  (1 + |b − c|β )  c∈C b∈B  ≤ k(1 + |b|β )q̂(b)kl1  q̂(b − c) 2c · Â(b − c) 2k · Â(b − c) − − Nc (k) Nc (k) Nc (k)  1 14 1 4 17 + k(1 + |b|β )Â(b)kl1 < + = . ε|v| ε 2 9 18  Furthermore, since |Tb,c | ≤ |Qb,c | for all b, c ∈ Γ# , for any integer m ≥ 1 we have  m  m X X 17 17 m m |(TBC )b,c | < |(TBC )b,c | < and sup . sup 18 18 c∈C b∈B c∈C  b∈B  Now, let p be the smallest integer greater or equal than β, and for any integer m ≥ 1 and any ξ0 , ξ1 , . . . , ξm ∈ Γ# , let b = ξ0 and c = ξm . Then,  β  p m (2Λ)β X β β |b − c| β |b − c| |b − c| = (2Λ) ≤ (2Λ) = 2Λ 2Λ (2Λ)p  |ξi1 −1 − ξi1 | · · · |ξip −1 − ξip |  i1 ,...,ip =1  ≤ (2Λ)  β−p  ≤ (2Λ)β−p  m X   max |ξi1 −1 − ξi |p , . . . , |ξip −1 − ξip |p  i1 ,...,ip =1 m X  (|ξi1 −1 − ξi1 |p + · · · + |ξip −1 − ξip |p )  i1 ,...,ip =1 m X β−p p−1  = (2Λ)  pm  p  |ξi−1 − ξi | ≤ (2Λ)  i=1  β−p  p−1  pm  m Y  (1 + |ξi−1 − ξi |p ).  i=1  (4.4.37) 81  To simplify the notation write s := sup b∈B c∈C  1 . 1 + |b − c|β  Hence, sup  X  |(TGm0 G0 )b,c | ≤ sup  b∈B c∈C  b∈B c∈C  X 1 sup (1 + |b − c|β )|(TGm0 G0 )b,c | β 1 + |b − c| b∈B c∈C   ≤ s sup  X  b∈B c∈C  X  |(TGm0 G0 )b,c | + (2Λ)β−p p mp−1 sup  b∈B ξ ∈G0 1  (1 + |b − ξ1 |β )|Tb,ξ1 |   X  ×  (1 + |ξ1 − ξ2 |2 )|Tξ1 ,ξ2 | · · ·  ξ2 ∈G0  X  (1 + |ξm−1 − c|2 )|Tξm−1 ,c |  c∈C    m X 17  + (2Λ)β−p p mp−1 sup (1 + |b − ξ1 |2 )|Tb,ξ1 | ≤s 18 b∈B 0 ξ1 ∈G   × sup  X  ξ1 ∈G0 ξ ∈G0 2  (1 + |ξ1 − ξ2 |2 )|Tξ1 ,ξ2 | · · ·  sup  X  ξm−1 ∈G0 c∈C  (1 + |ξm−1 − c|2 )|Tξm−1 ,c |     m X X 17 = s |Qξm−1 ,c | |Qb,ξ1 | · · · sup + (2Λ)β−p p mp−1 sup 18 b∈B ξ ∈G0 ξm−1 ∈G0 c∈C 1  m 17 ≤ s (1 + (2Λ)β−p p mp−1 ) . 18 Similarly, sup  X  c∈C b∈B  |(TGm0 G0 )b,c | ≤ sup b∈B c∈C  X 1 sup (1 + |b − c|2 )|(TGm0 G0 )b,c | 2 1 + |b − c| c∈C b∈B   ≤ s sup  X  c∈C b∈B  |(TGm0 G0 )b,c | + (2Λ)β−p p mp−1 sup c∈C  X  (1 + |ξm−1 − c|2 )|Tξm−1,c |  ξm−1 ∈G0   ×  X  (1 + |ξm−2 − ξm−1 |2 )|Tξm−2 ,ξm−1 | · · ·  ξm−2 ∈G0  X  (1 + |b − ξ1 |2 )|Tb,ξ1 |  b∈B    m 17 ≤ s + (2Λ)β−p p mp−1 sup 18 c∈C  X  (1 + |ξm−1 − c|2 )|Tξm−1,c |  ξm−1 ∈G0   ×  X  sup ξm−1 ∈G0 ξ  ≤ s (1 + (2Λ)  (1 + |ξm−2 − ξm−1 |2 )|Tξm−2 ,ξm−1 | · · · sup  m−2 ∈G  β−p  pm  ξ1 ∈G0 b∈B  0  p−1  X   )  17 18  m .  82  (1 + |b − ξ1 |2 )|Tb,ξ1 |  Therefore, by Proposition 4.1.2, kπB TGm0 G0 πC k  ≤ (1 + (2Λ)  β−dβe  dβe−1  dβem   )  17 18  m sup b∈B c∈C  1 , 1 + |b − c|β  where dβe is the smallest integer greater or equal than β. This is the desired inequality.  Proof of Lemma 4.4.2 Proof of Lemma 4.4.2. To simplify the notation write w = wµ,d0 ,  z = zµ,d0  |z|R = 2|z| − R.  and  First observe that 1 w − 2iθ (c − µ0  d0 )  =  −1 w + , 0 0 0 2iθ (c − d ) 2iθµ (c − d )(w − 2iθµ0 (c − d0 )) µ0  so that z z = 0 Nc (k) (w − 2iθµ0 (c − d ))(z − 2iθµ (c − d0 ))   2iθµ (c − d0 ) 1 1+ = w − 2iθµ0 (c − d0 ) z − 2iθµ (c − d0 ) 2iθµ (c − d0 ) 1 1 = + 0 0 w − 2iθµ0 (c − d ) w − 2iθµ0 (c − d ) z − 2iθµ (c − d0 ) 2iθµ (c − d0 ) −1 w 1 = + + 0 0 0 0 2iθµ0 (c − d ) 2iθµ0 (c − d )(w − 2iθµ0 (c − d )) w − 2iθµ0 (c − d ) z − 2iθµ (c − d0 ) =: ηc(0) + ηc(w) + ηc(z) , where, in view of (4.4.17) to (4.4.20), since |w| < ε, |ηc(0) | ≤  1 , 2Λ  |ηc(w) | ≤  ε 2Λ2  and  |ηc(z) | ≤  4 . |z|R  Hence, Yb,c =  −2iθµ0 (Â(b − c))z Nc (k)  = −2iθµ0 (Â(b − c))ηc(0) − 2iθµ0 (Â(b − c))ηc(w) − 2iθµ0 (Â(b − c))ηc(z) (0)  (w)  (z)  =: Yb,c + Yb,c + Yb,c . (·) (·) Let Y ( · ) be the operator whose matrix elements are Yb,c and set Y33 := πG03 Y ( · ) πG03 .  Then, similarly as we estimated kY33 k, using (4.4.17) to (4.4.20) and Proposition 4.1.2, it 83  follows easily that (0)  kY33 k ≤  1 kθµ0 (Â)kl1 , 2Λ  (w)  kY33 k ≤  ε kθµ0 (Â)kl1 , 2Λ2  (z)  kY33 k ≤  4 kθµ0 (Â)kl1 . |z|R  Furthermore, S = (I − Y33 )−1 = 1 + (1 − Y33 )−1 Y33 = 1 + SY33 2 = 1 + (1 + SY33 )Y33 = 1 + Y33 + SY33 (0)  (w)  (z)  2 = 1 + Y33 + Y33 + Y33 + SY33 ,  where, recalling (4.4.15), 2 kSY33 k ≤ k(1 − Y33 )−1 k kY33 k2 ≤  14 < 13  kY33 k2 1 − kY33 k   2 8 kθµ0 (Â)k2l1 . Λ  Combining all this we have z Sb,c (0) (w) (z) = (ηb + ηb + ηb )Sb,c Nb (k) (0)  (w)  (0)  (w)  (z)  (z)  2 = (ηb + ηb )(δb,c + Yb,c + Yb,c + Yb,c + (SY33 )b,c ) + ηb Sb,c h i h i (0) (0) (0) (w) (w) (0) (w) = ηb (δb,c + Yb,c ) + ηb Yb,c + ηb (δb,c + Yb,c + Yb,c ) i i h h (0) (w) (z) (z) (0) (w) 2 + (ηb + ηb )(SY33 )b,c + (ηb + ηb )Yb,c + ηb Sb,c (0)  (1)  (2)  (3)  =: Kb,c + Kb,c + Kb,c + Kb,c with (0) |Kb,c |  ≤  (1)  |Kb,c | ≤ < (2)  |Kb,c | ≤ (3)  |Kb,c | ≤    1 1 0 1+ kθµ (Â)kl1 , 2Λ 2Λ   ε ε 1 ε kθµ0 (Â)kl1 + 2 kθµ0 (Â)kl1 kθµ0 (Â)kl1 + 1+ 4Λ3 2Λ2 2Λ Λ   ε 7 1+ kθµ0 (Â)kl1 , 2 2Λ 6Λ   1 8 2 64 kθµ0 (Â)k2l1 < 3 kθµ0 (Â)k2l1 , Λ Λ Λ CΛ,A 3 4 14 4 kθµ0 (Â)kl1 + < 2Λ |z|R 13 |z|R |z|R (1)  for all b, c ∈ G03 . Here, to estimate |Kb,c | we have used that ε < Λ/6.  84  Finally, recalling (4.4.25) and using the above estimates we find that z Sb,c g(c − d0 ) N (k) b b,c∈G01 " 3 # X X (j) = f (d0 − b) Kb,c g(c − d0 )  (1)  zµ,d0 (k)αµ,d0 (k) =  X  f (d0 − b)  b,c∈G01 (1,0)  (4.4.38)  j=0 (1,1)  (1,2)  (1,3)  =: αµ,d0 + αµ,d0 (w(k)) + αµ,d0 (k) + αµ,d0 (k), where, in particular, (1,0) αµ,d0  =−  X b,c∈G01  " # θµ0 (Â(b − c)) f (d0 − b) δb,c + g(c − d0 ). 2iθµ0 (b − d0 ) θµ0 (c − d0 )  (4.4.39)  Furthermore, it follows easily from (4.4.38) that, for 0 ≤ j ≤ 2, (1,j)  |αµ,d0 | ≤ Cj with    1 1 C0 := 1+ kθµ0 (Â)kl1 kf kl1 kgkl1 , 2Λ 2Λ   ε 7 C1 := 1+ kθµ0 (Â)kl1 kf kl1 kgkl1 , 2Λ2 6Λ 64 C2 := 3 kθµ0 (Â)k2l1 kf kl1 kgkl1 , Λ  (4.4.40)  while for j = 3, (1,3)  |αµ,d0 | ≤ CΛ,A,f,g  1 . |z|R  This completes the proof of the lemma.  Proof of Lemma 4.4.3 We first derive the following inequality. Proposition 4.4.7. Let α and δ be constants with 1 < α ≤ 2 and 1 < δ ≤ 2. Suppose that f is a function on Γ# obeying k|b|α f (b)kl1 < ∞. Then, for any ξ1 , ξ2 ∈ Γ# with ξ1 6= ξ2 ,   1 X if α, δ < 2, |f (b − ξ1 )| C ≤ × |b − ξ2 |δ |ξ1 − ξ2 |α+δ−2  ln |ξ − ξ | if α = 2 or δ = 2, b∈Γ# \{ξ1 ,ξ2 } 1 2 where C = CΓ# ,α,δ,f is a constant.  85  Proof. We can obtain the desired estimate as follows: X b∈Γ# \{ξ1 ,ξ2 }   |f (b − ξ1 )| ≤ sup |b − ξ1 |α |f (b − ξ1 )| δ |b − ξ2 | b∈Γ#  X b∈Γ# \{ξ1 ,ξ2 }  1 |b − ξ1 |α  X  ≤ k|b|α f (b)kl1  b∈Γ# \{ξ1 ,ξ2 }  |b − ξ2  X  = k|b|α f (b)kl1  b∈Γ# \{0,ξ1 −ξ2 } α  |b|δ  |δ  1 |b − (ξ1 − ξ2 )|α  Z  ≤ k|b| f (b)kl1 CΓ#  |b − ξ2  1 |b − ξ1 |α  |δ  |x|≥Λ |x−(ξ1 −ξ2 )|≥Λ  d2 x |x|δ |x − (ξ1 − ξ2 )|α  (by a rotation in R2 such that ξ1 − ξ2 → |ξ1 − ξ2 |(1, 0)) Z d2 y α = k|b| f (b)kl1 CΓ# |y|≥Λ |y|δ |y − |ξ1 − ξ2 |(1, 0)|α |y−|ξ1 −ξ2 |(1,0)|≥Λ Z d2 z k|b|α f (b)kl1 CΓ# = |z|≥Λ/|ξ1 −ξ2 | |ξ1 − ξ2 |δ+α−2 |z|δ |z − (1, 0)|α |z−(1,0)|≥Λ/|ξ1 −ξ2 |   1 if α, δ < 2, CΓ# ,α,δ,f ≤ × δ+α−2  |ξ1 − ξ2 | ln |ξ − ξ | if α = 2 or δ = 2. 1 2  We now apply this proposition for proving Lemma 4.4.3. Proof of Lemma 4.4.3. First observe that kπ{b} TGm0 G0 π{c} k =  sup kπ{b} TGm0 G0 π{c} ϕkL2 = π{b} TGm0 G0  ϕ∈L2 kϕkL2 =1   =  eib·x eic·x m , T 0 G0 G |Γ|1/2 |Γ|1/2   L2  eib·x |Γ|1/2  eic·x |Γ|1/2   = L2  L2  eib·x eic·x m , T 0 G0 G |Γ|1/2 |Γ|1/2  = |(TGm0 G0 )b,c |. Hence, by Proposition 4.4.6 with β = 2, for all b, c ∈ G0 and m ≥ 1,  m 1 17 m m . |(TG0 G0 )b,c | = kπ{b} TG0 G0 π{c} k ≤ (1 + 2m) 18 1 + |b − c|2 This inequality is also valid for m = 0 because    1 if b = c   1 |(TG0 0 G0 )b,c | = |δb,c | = ≤ .   1 + |b − c|2 0 if b =  6 c 86   L2  Thus, ∞ X X f (d0 − b) m (TG0 G0 )b,c g(c − d00 ) N (k) b m=0 b,c∈G0 " ∞  m # X X |g(c − d00 )| 1 X 17 ≤ (1 + 2m) |f (d0 − b)| ε|v| 18 1 + |b − c|2 m=0 b∈G0 c∈G0   00 X X C |g(c − d )|  ≤ |f (d0 − b)| |g(b − d00 )| + , ε|v| |b − c|2 0 0  |Φd0 ,d00 (k)| =  (4.4.41)  c∈G \{b}  b∈G  where C is an universal constant. Now, by the triangle inequality, Hölder’s inequality, and since k · kl2 ≤ k · kl1 , X  X |d0 − d00 |2 |f (d0 − b)| |g(b − d00 )| |d0 − d00 |2 0 b∈G X 4 ≤ 0 (|d0 − b|2 + |b − d00 |2 ) |f (d0 − b)| |g(b − d00 )| |d − d00 |2 0  |f (d0 − b)| |g(b − d00 )| =  b∈G0  b∈G  4 (kb2 f (b)kl2 kgkl2 + kf kl2 kb2 g(b)kl2 ) |d0 − d00 |2 Cf,g 4 ≤ 0 (kb2 f (b)kl1 kgkl1 + kf kl1 kb2 g(b)kl1 ) ≤ 0 . 00 2 |d − d | |d − d00 |2  ≤  (4.4.42) Furthermore, by Proposition 4.4.7 with α = δ = 2, for any 0 < 1 < 2, X c∈G0 \{b}  CΓ# ,g,1 |g(c − d00 )| ln |b − d00 | ≤ C ≤ . # Γ ,g 2 00 2 |b − c| |b − d | |b − d00 |2−1  Applying this inequality and (4.4.42) to (4.4.41) we obtain " # X |f (d0 − b)| Cf,g C + CΓ# ,g,1 . |Φd0 ,d00 (k)| ≤ ε|v| |d0 − d00 |2 |b − d00 |2−1 0 b∈G  Applying again Proposition 4.4.7 with α = 2 and δ = 2 − 1 we conclude that, for any 0 < 2 < 2 − 1 ,   Cε,Γ# ,f,g,1 ,2 Cf,g C ln |d0 − d00 | |Φd0 ,d00 (k)| ≤ + CΓ# ,f,g,1 0 ≤ . 0 00 2 00 2− 1 ε|v| |d − d | |d − d | |v| |d0 − d00 |2−1 −2 Finally, recall from Proposition 4.4.4(ii) that |zν 0 ,d | < 3|d| and |zν 0 ,d | < 3|v|, observe that |d0 − d00 | = |d|, and set  = 1 + 2 . Then, for any 0 <  < 2, |Φd0 ,d00 (k)| ≤  Cε,Γ# ,f,g,1 ,2 Cε,Γ# ,f,g, ≤ . |d| |d|2−1 −2 |zν 0 ,d |3−  Choosing  = 10−1 we obtain the desired inequality. 87  4.5  Bounds on the derivatives (j)  In the last section we expressed Φd0 ,d00 (k) as a sum of certain functions αµ,d0 (k) for k in the ε-tubes with large |v|. In this section we provide bounds for the derivatives of all these functions. We first give all the statements and then the proofs. Our first lemma concerns the derivatives of Φd0 ,d00 (k). Lemma 4.5.1 (Derivatives of Φd0 ,d00 (k)). Under Hypothesis 4.4.1, let f and g be functions in l1 (Γ# ) and suppose either (i) or (ii) where: (i) G = {0} and k ∈ (T0 \ ∪b∈G0 Tb ) \ KR ; (ii) G = {0, d} and k ∈ (T0 ∩ Td ) \ KR . Then, for any integers n and m with n + m ≥ 1 and for any d0 , d00 ∈ G, ∂ n+m C Φd0 ,d00 (k) ≤ , m n ∂k1 ∂k2 |v| where C is a constant with C = Cε,Λ,A,f,g,m,n if (i) or C = CΛ,A,f,g,m,n if (ii). We now improve the estimate of Lemma 4.5.1(ii) for d0 6= d00 . Lemma 4.5.2 (Derivatives of Φd0 ,d00 (k) for d0 6= d00 ). Consider a constant β ≥ 2 and suppose that k|b|β q̂(b)kl1 < ∞ and k(1 + |b|β )Â(b)kl1 < 2ε/63. Let ν ∈ {1, 2} and let f and g be functions on Γ# obeying k|b|β f (b)kl1 < ∞ and k|b|β g(b)kl1 < ∞. Suppose further that G = {0, d} and k ∈ T0 ∩ Td with |v| > 2ε k|b|β q̂(b)kl1 . Then, for any integers n and m with n + m ≥ 0 and for any d0 , d00 ∈ G with d0 6= d00 , ∂ n+m C Φd0 ,d00 (k) ≤ 1+β , ∂k1n ∂k2m |d| where C = Cε,Λ,A,f,g,m,n is a constant. Observe that, in particular, this Lemma with m = n = 0 generalizes Lemma 4.4.3. (j)  We next have bounds for the derivatives of αµ,d0 (k). (j)  Lemma 4.5.3 (Derivatives of αµ,d0 (k)). Under Hypothesis 4.4.1, let ν ∈ {1, 2} and let f and g be functions in l1 (Γ# ). Suppose either (i) or (ii) where: (i) G = {0} and k ∈ (Tν (0) \ ∪b∈G0 Tb ) \ KR ; (ii) G = {0, d} and k ∈ (Tν (0) ∩ Tν 0 (d)) \ KR . 88  Then, there is a constant ρ = ρε,A,q,m,n with ρ ≥ R such that, for |v| ≥ ρ and for (µ, d0 ) = (ν, 0) if (i) or (µ, d0 ) ∈ {(ν, 0), (ν 0 , d)} if (ii), for any integers n and m with n + m ≥ 1 and for 1 ≤ j ≤ 2, Cj ∂ n+m (j) αµ,d0 (k) ≤ n m ∂k1 ∂k2 (2|zµ,d0 (k)| − R)j  ∂ n+m (3) C3 αµ,d0 (k) ≤ , n m ∂k1 ∂k2 |zµ,d0 (k)|R2  and  where Cl = Cl;f,g,Λ,A,q,n,m for 1 ≤ l ≤ 3 are constants. Furthermore, C1;f,g,Λ,A,1,0 , C1;f,g,Λ,A,0,1 ≤ 13Λ−2 kf kl1 kgkl1  and  C1;f,g,Λ,A,1,1 ≤ 65Λ−3 kf kl1 kgkl1 .  Proof of Lemma 4.5.1 Proof of Lemma 4.5.1. Step 0 When there is no risk of confusion we shall use the same notation to denote an operator or its matrix. Define FBC := [f (b − c)]b∈B,c∈C ,  GBC := [g(b − c)]b∈B,c∈C ,    ΦG (k) := Φd0 ,d00 (k; G) d0 ,d00 ∈G .  Here FBC and GBC are |B| × |C| matrices and ΦG (k) is a |G| × |G| matrix. First observe that   X f (d0 − b) −1 00  (RG ΦG (k) =  0 G0 )b,c g(c − d ) N (k) b 0   b,c∈G  d0 ,d00 ∈G  can be written as the product of matrices −1 FGG0 ∆−1 k RG0 G0 GG0 G . −1 −1 = H −1 , we can write Φ (k) Furthermore, since on L2G0 we have ∆−1 G k k RG0 G0 = (RG0 G0 ∆k )  as FGG0 Hk−1 GG0 G . Hence,  ∂ n+m Hk−1 ∂ n+m 0 Φ (k) = F GG0 G . G GG ∂k1n ∂k2m ∂k1n ∂k2m  (4.5.1)  This is the quantity we want to estimate. Step 1 Let T = T (k) be an invertible matrix. Then applying T T −1 = I and using the Leibniz rule for  ∂ m0 m ∂ki 0  (T T −1 ) we find that   m 0 −1  X m0 ∂ m0 −m1 T ∂ m1 T −1 ∂ m0 T −1 −1 = −T . ∂kim0 m1 ∂kim0 −m1 ∂kim1 m1 =0  89  ∂ m0 m ∂ki 0  to the identity  Iterating this formula m0 − 1 times we obtain    −1  m0 mj−1 m −m m −1 Y X 0 j−1 j ∂ mj−1 T ∂ mm0 T −1 ∂ T  (−T −1 ) mj−1 −mj  mm m0 = mj ∂ki ∂ki 0 ∂ki j=1 mj =0   mY 0 −1 mj−1 m −m X−1 mj−1  j−1 j ∂ T (−T −1 ) mj−1 −mj  = mj ∂k mj =0  j=1  i  (4.5.2)  mm0 −1 −1    X ∂ mm0 −1 −mm0 T ∂ mm0 T −1 mm0 −1 (−T −1 ) mm −1 −mm × mm 0 mm0 ∂ki 0 ∂ki 0 mm0 =0   mY 0 −1 mj−1 m −m X−1 mj−1  j−1 j ∂ T ∂ mm0 −1 T T −1 mj−1 −mj  T −1 mm0 −1 T −1 . = (−1)m0  mj ∂ki ∂k j=1  mj =0  i  Step 2 In view of (4.5.2), it is not difficult to see that  ∂ m Hk−1 ∂k2m  is given by a finite linear  combination of terms of the form   m Y   j=1  n  Pm  ∂ m H −1  (4.5.3) n  ∂ ∂ k = m. Thus, when we compute ∂k n ∂k2m , the derivative ∂k1n acts either on 1  nj n ∂ n ∂ j Hk k or ∂ nHjk . However, since ∂H =0 nj ∂k2 b,c = 2(k2 + b2 )δb,c − 2Â2 (b − c), we have ∂kn  where Hk−1   nj H ∂ k  −1 Hk , Hk−1 n ∂k2 j  j=1 nj  ∂k2  if nj ≥ 1 and ∂ n Hk−1 ∂k1n  1  n ∂ n ∂ j Hk n ∂k1 ∂knj 2  =  ∂ n Hk ∂k1n  ∂k2  if nj = 0. Similarly, using again (4.5.2) one can see that  is given by a finite linear combination of terms of the form (4.5.3), with m and k2 P replaced by n and k1 , respectively, and nj=1 nj = n. Therefore, combining all this we conclude that  where  Pn+m j=1  ∂ n+m Hk−1 ∂k1n ∂k2m  nj δ2,ij  is given by a finite linear combination of terms of the form   n+m nj H Y ∂ k  −1 −1 −1  (4.5.4) ∆k RG0 G0 , ∆−1 nj k RG0 G0 ∂k i j j=1 Pn+m = m and j=1 nj δ1,ij = n, that is, where the sum of nj for which  ij = 2 is equal to m, and the sum of nj for which ij = 1 is equal to n. Step 3 The first step in bounding (4.5.4) is to estimate ∂ nj Hk −1 n ∆ πG0 . ∂kijj k A simple calculation shows that  ∂ nj Hk −1 n ∆ ∂kijj k  ! b,c     2(kij + bij )δb,c + 2Âij (b − c) if nj = 1,    1 = × 2δb,c if nj = 2, Nc (k)      0 if nj ≥ 3. 90  Furthermore, by Proposition 4.2.1, 1 1 ≤ |Nb (k)| ε|v| for all b ∈ G0 , while by Proposition 3.3.1 we have 1 2 ≤ |Nb (k)| Λ|v|  (4.5.5)  and |ki + bi | ≤ |ui + bi | + |vi | ≤ |v| + |u + b| ≤  2 |Nb (k)| Λ  for all b ∈ G0 if G = {0, d}, and for all b ∈ G0 \ {b̃} if G = {0}. Furthermore, |b̃| ≤ Λ + |u| + |v| (4.5.6) < Λ + 3|v|, since |u| < 2|v| because k ∈ T0 (see (4.2.2)). Now, let 1B (x) be the characteristic function of the set B. Then, using the above estimates,  X  ∂ nj Hk −1 n j ∆k π G0 b,c ∂kij c∈G0 b∈G0 " # X 2|kij + bij |δnj ,1 + 2δnj ,2 2|Âij (b − c)| ≤ sup δb,c + δnj ,1 |Nb (k)| |Nb (k)| c∈G0 b∈G0 " # 2|kij + b̃ij | + 2 2|Âij (b̃ − c)| ≤ sup δb̃,c + 1G0 (b̃) |Nb̃ (k)| |Nb̃ (k)| c∈G0 " # X 2|kij + bij | + 2 2|Âij (b − c)| + sup δb,c + |Nb (k)| |Nb (k)| c∈G0 0 b∈G \{b̃} " #  X 2|kij + b̃ij | + 2 + 2kÂkl1 2|Âij (b − c)| 4 2 ≤ 1G0 (b̃) + sup + δb,c + ε|v| Λ |Nb (k)| |Nb (k)| c∈G0 sup  b∈G0 \{b̃}  2 4 4 4 (2(|u| + |v| + |b̃|) + 2 + 2kÂkl1 )1G0 (b̃) + + + kÂkl1 ε|v| Λ Λ|v| Λ|v| 2 4 4 4 ≤ (12|v| + 2Λ + 2 + 2kÂkl1 )1G0 (b̃) + + + kÂkl1 ε|v| Λ Λ|v| Λ|v| ≤  (recall that |v| > 1) ≤ 1G0 (b̃) ε−1 CΛ,A + CΛ,A . 91  Similarly, " #  X  ∂ nj Hk X 2|kij + bij |δnj ,1 + 2δnj ,2 2|Âij (b − c)| −1 sup ≤ sup δb,c + δnj ,1 n ∆ πG0 |Nb (k)| |Nb (k)| b,c ∂kijj k b∈G0 c∈G0 b∈G0 c∈G0 " # 2|kij + bij | + 2 2|Âij (b − b̃)| ≤ sup 1G0 (b̃) δb,b̃ + |Nb (k)| |Nb (k)| b∈G0 " # X 2|kij + bij | + 2 2|Âij (b − c)| + sup δb,c + |Nb (k)| |Nb (k)| b∈G0 c∈G0 \{b̃} " #  X 2|Âij (b − c)| 2|kij + b̃ij | + 2 + 2kÂkl1 4 2 δb,c + ≤ 1G0 (b̃) + sup + ε|v| Λ |Nb (k)| |Nb (k)| b∈G0 c∈G0 \{b̃}  ≤  2 4 4 4 (2(|u| + |v| + |b̃|) + 2 + 2kÂkl1 )1G0 (b̃) + + + kÂkl1 ε|v| Λ Λ|v| Λ|v|  ≤ 1G0 (b̃) ε−1 CΛ,A + CΛ,A . Hence, by Proposition 4.1.2, ∂ nj Hk −1 −1 n ∆ πG0 ≤ 1G0 (b̃) ε CΛ,A + CΛ,A . ∂kijj k Step 4 By a similar (and much simpler) calculation (using Proposition 4.1.2) we get kFGG0 k ≤ kf kl1 , kGGG0 k ≤ kgkl1 , k∆−1 k πG0 k ≤ 1G0 (b̃)  (4.5.7) 1 2 + (1 − 1G0 (b̃)) . ε|v| Λ|v|  From Lemma 4.1.4 we have k(RG0 G0 )−1 k ≤ 18. Thus, the operator norm of (4.5.4) is bounded by     n+m n+m n n Y Y j j ∂ Hk −1 −1 −1 ∂ Hk  −1 −1 −1    ∆−1 ∆k RG0 G0 ≤ k∆−1 nj nj ∆k πG0 kRG0 G0 k , k RG0 G0 k k kRG0 G0 k ∂k ∂k i i j j j=1 j=1 which is bounded either by   n+m Y 1 1 18  (ε−1 CΛ,A + CΛ,A ) 18 ≤ ε−(n+m+1) CΛ,A,n,m ε|v| |v| j=1  if G = {0}, or by   n+m Y  1 18  Λ|v|   CΛ,A 18 kgkl1 ≤ CΛ,A,n,m  j=1  92  1 |v|  if G = {0, d}. Therefore, ∂ n+m Hk−1 ≤ ∂k1n ∂k2m  X finite sum where # of terms depend on n and m  C0 C0 C ≤ Cn,m ≤ , |v| |v| |v|  (4.5.8)  with C = Cε,Λ,A,n,m if G = {0} or C = CΛ,A,n,m if G = {0, d}. Finally, recalling (4.5.1) and (4.5.7) we have ∂ n+m Hk−1 ∂ n+m 0 Φ (k) = F GG0 G G GG ∂k1n ∂k2m ∂k1n ∂k2m ≤ kFGG0 k  ∂ n+m Hk−1 C0 C 0 G k ≤ kf kl1 kG kgkl1 ≤ , G m n ∂k1 ∂k2 |v| |v|  where C = Cε,Λ,A,n,m,f,g if G = {0} or C = CΛ,A,n,m,f,g if G = {0, d}. This is the desired inequality. The proof of the lemma is complete.  Proof of Lemma 4.5.2 Let R+ be the set of non-negative real numbers and let σ be a real-valued function on R+ such that: (i) σ(t) ≥ 1 for all t ∈ R+ with σ(0) = 1; (ii) σ(s)σ(t) ≥ σ(s + t) for all s, t ∈ R+ ; (iii) σ increases monotonically. For example, for any β ≥ 0 the functions t 7→ eβt and t 7→ (1 + t)β satisfy these properties. Now, let T be a linear operator from L2C to L2B with B, C ⊂ Γ# , (or a matrix T = [Tb,c ] with b ∈ B and c ∈ C), and consider the σ-norm ( ) X X kT kσ := max sup |Tb,c |σ(|b − c|), sup |Tb,c |σ(|b − c|) . c∈C b∈B  b∈B c∈C  In §4.9 we prove that this norm has the following properties. Proposition 4.5.4 (Properties of k · kσ ). Let S and T be linear operators from L2C to L2B with B, C ⊂ Γ# . Then: (a) kT k ≤ kT kσ≡1 ≤ kT kσ ; (b) If B = C, then kS T kσ ≤ kSkσ kT kσ ; 93  (c) If B = C, then k(I + T )−1 kσ ≤ (1 − kT kσ )−1 if kT kσ < 1; (d) |Tb,c | ≤  1 σ(|b−c|) kT kσ  for all b ∈ B and all c ∈ C.  Using these properties we prove Lemma 4.5.2. Proof of Lemma 4.5.2. We follow the same notation as above. First observe that, similarly as in the last proof we can write −1 −1 Φd0 ,d00 (k) = F{d0 }G0 ∆−1 k RG0 G0 GG0 {d00 } = F{d0 }G0 Hk GG0 {d00 } .  Now, let σ(|b|) = (1 + |b|)β , and observe that there is a positive constant Cβ such that σ(|b|) ≤ Cβ (1 + |b|β ) for all b ∈ Γ# . Then, it is easy to see that kF{d0 }G0 kσ = kf kσ ≤ Cβ k(1 + |b|β )f (b)kl1 and kGG0 {d00 } kσ = kgkσ ≤ Cβ k(1 + |b|β )g(b)kl1 . Furthermore, by (4.4.36) and Proposition 4.1.2, −1 −1 kσ ≤ kRG 0 G0 kσ = k(I + TG0 G0 )  ∞ X  kTG0 G kjσ < 18,  (4.5.9)  j=0  and since for diagonal operators the σ-norm and the operator norm agree, from (4.5.7) we have k∆−1 k π G0 kσ ≤  2 . Λ|v|  Hence, in view of Proposition 4.5.4(b) and Proposition 4.4.4(ii), −1 |Φd0 ,d00 (k)| ≤ kF{d0 }G0 ∆−1 k RG0 G0 GG0 {d00 } k ≤ Cβ,f,g,Λ,A,m,n  1 , |d|  and by repeating the proof of Lemma 4.5.1 with the operator norm replaced by the σ-norm we obtain ∂ n+m Φd0 ,d00 (k) ∂k1n ∂k2m  ≤ Cβ,f,g,Λ,A,m,n σ  1 . |d|  Therefore, by Proposition 4.5.4(d), for any integers n and m with n + m ≥ 0, ∂ n+m 1 Φd0 ,d00 (k) ≤ n m 0 ∂k1 ∂k2 1 + |d − d00 |β  ∂ n+m Φd0 ,d00 (k) ∂k1n ∂k2m  This is the desired inequality. 94  ≤ Cβ,f,g,Λ,A,m,n σ  1 . |d|1+β  Proof of Lemma 4.5.3 Define the operator M (j) : L2G0 → L2G0 as 3 3    S    (j) M := W      Z  if j = 1, if j = 2, if j = 3,  where S, W and Z are given by (4.4.23). In order to prove Lemma 4.5.3 we first prove the following proposition. Proposition 4.5.5. Assume the same hypotheses of Lemma 4.5.3. Then, for any integers n and m with n + m ≥ 1 and for 1 ≤ j ≤ 3, Cj ∂ n+m ∆−1 M (j) ≤ , k n m ∂k1 ∂k2 (2|zµ,d0 (k)| − R)j where C1 = C1;Λ,A,n,m and Cj = Cj;Λ,A,q,n,m for 2 ≤ j ≤ 3 are constants. Furthermore, C1;Λ,A,1,0 ≤  13 , Λ2  C1;Λ,A,0,1 ≤  13 Λ2  and  C1;Λ,A,1,1 ≤  65 . Λ3  Proof. Step 0 To simplify the notation write w = wµ,d0 ,  z = zµ,d0  and  |z|R = 2|z| − R.  First observe that, for any analytic function of the form h(k) = h̃(w(k), z(k)), we have     ∂w ∂ ∂ ∂ ∂z ∂ ∂ h= + h̃ = + h̃, ∂k1 ∂k1 ∂w ∂k1 ∂z ∂w ∂z     ∂ ∂z ∂ ∂ ∂ ∂w ∂ ν + h̃ = i(−1) − h̃. h= ∂k2 ∂k2 ∂w ∂k2 ∂z ∂w ∂z Thus, m X n    n−r+m−p ∂ r+p X ∂ n+m m n −1 ν m m−p ∂ (j) ∆ = (i(−1) ) (−1) M ∆−1 M (j) p r ∂k1n ∂k2m k ∂z n−r+m−p ∂wr+p k p=0 r=0  ≤ 2n+m sup sup p≤r r≤n  ∂ n−r+m−p ∂ r+p −1 (j) ∆ M . ∂z n−r+m−p ∂wr+p k  Now, by the Leibniz rule, m X n    n−r+m−p −1 r+p X ∆k ∂ M (j) ∂ n ∂ m −1 (j) m n ∂ ∆ = M ∂z n ∂wm k p r ∂z n−r ∂wm−p ∂z r ∂wp p=0 r=0  ≤ 2n+m sup sup p≤m r≤n  95  ∂ n−r+m−p ∆−1 k ∂z n−r ∂wm−p  ∂ r+p M (j) . ∂z r ∂wp  Furthermore, we shall prove below that sup sup p≤m r≤n  ∂ n−r+m−p ∆−1 k ∂z n−r ∂wm−p  Cj,n,m ∂ r+p M (j) , ≤ ∂z r ∂wp |z|n+j R  (4.5.10)  with constants C1,n,m = C1,n,m;Λ,A and Cj,n,m = Cj,n,m;Λ,A,q for 2 ≤ j ≤ 3. Hence, Cj,n,m ∂ n ∂ m −1 (j) ∆ M ≤ 2n+m n+j . ∂z n ∂wm k |z|R Therefore, being careful with the indices, Cj,n−r+m−p,r+p Cj ∂ n+m ∆−1 M (j) ≤ 2n+m sup sup 2n−r+m−p+r+p ≤ j , k m n n−r+m−p+j ∂k1 ∂k2 p≤m r≤n |z|R |z|R where C1 = C1;Λ,A,n,m and Cj = Cj;Λ,A,q,n,m for 2 ≤ j ≤ 3. This is the desired inequality. We are left to prove (4.5.10) and estimate the constants C1;Λ,A,i,j for i, j ∈ {0, 1} to finish the proof of the proposition. Step 1 The first step for obtaining (4.5.10) is to estimate ∂ r+p ∆−1 k π 0 . ∂z r ∂wp G3 Observe that ∂ r+p ∆−1 k ∂z r ∂wp  ! = b,c  ∂ r+p (∆−1 δb,c 1 ∂r ∂p k )b,c = r p p 0 r ∂z ∂w ∂w w − 2iθµ0 (b − d ) ∂z z − 2iθµ (b − d0 )  (−1)r r! δb,c (−1)p p! (w − 2iθµ0 (b − d0 ))p+1 (z − 2iθµ (b − d0 ))r+1 p! r! δb,c , ≤ 0 0 |w − 2iθµ (b − d )|p+1 |z − 2iθµ (b − d0 )|r+1 =  and recall from (4.4.17) and (4.4.18) that, for all b ∈ G03 , 1 2 ≤ 0 |z − 2iθµ (b − d )| |z|R  1  and  |w − 2iθµ0 (b −  d0 )|  ≤  1 . Λ  (4.5.11)  Then, ∂ r+p ∆−1 k ∂z r ∂wp and consequently,   X X  sup  + sup b∈G03 c∈G0  3  c∈G03 b∈G0  3  ∂ r+p ∆−1 k ∂z r ∂wp  ! ≤ b,c  ! ≤ b,c  p! r! 2r+1 δb,c , Λp+1 |z|r+1 R  p! r! 2r+1 Λp+1 |z|r+1 R   sup  p! r! 2r+2 = p+1 r+1 . Λ |z|R 96     X  b∈G03 c∈G0  3  + sup  X  c∈G03 b∈G0  3   δb,c  Therefore, by Proposition 4.1.2, ∂ r+p ∆−1 p! r! 2r+2 1 k 0 π ≤ . ∂z r ∂wp G3 Λp+1 |z|r+1 R  (4.5.12)  Step 2 We now estimate the second factor in (4.5.10). Let us first consider the case j = 1, that is, M (1) = S. Since S = (I − Y33 )−1 , the operator S is clearly invertible. Thus, by applying (4.5.2) with T = S −1 , one can see that  ∂pS ∂wp  is given by a finite linear  combination of terms of the form   p Y   j=1  where or  Pp  j=1 nj  ∂ nj S −1 nj ∂w  S  ∂ nj S −1 ∂wnj  = p. Hence, when we compute    S,  ∂r ∂pS ∂z r ∂wp ,  (4.5.13) the derivative  ∂r ∂z r  . Similarly, using again (4.5.2) with T = S −1 , one can see that  acts either on S ∂r S ∂z r  is given by a  finite linear combination of terms of the form (4.5.13), with p and w replaced by r and z, P ∂ r+p S respectively, and rj=1 mj = r. Thus, we conclude that ∂z r ∂w p is given by a finite linear combination of terms of the form   r+p Y   j=1  where  Pr+p j=1  mj = r and  Pr+p j=1  S  ∂ mj +nj S −1 ∂z mj ∂wnj    S,  (4.5.14)  nj = p. Indeed, observe that the general form of the terms  (4.5.14) follows directly from (4.5.2) because that identity is also valid for mixed derivatives. Since S = (I − Y33 )−1 with kY33 k < 1/14 and Yb,c =  −2iθµ0 (Â(b − c)) z , (w − 2iθµ0 (c − d0 ))(z − 2iθµ (c − d0 ))  (4.5.15)  we have kSk = k(I − Y33 )−1 k ≤  1 14 ≤ 1 − kY33 k 13  (4.5.16)  and   ∂ j+l −1 S ∂z j ∂wl     = b,c  =   ∂ j+l ∂ j+l (I − Y ) = Yb,c 33 ∂z j ∂wl ∂z j ∂wl b,c  ∂ j −2iθµ0 (Â(b − c)) z ∂ l 1 . j 0 l ∂z z − 2iθµ (c − d ) ∂w w − 2iθµ0 (c − d0 )  Furthermore, (−1)j−1 j! 2iθµ0 (Â(b − c)) 2iθν (c − d0 ) ∂ j −2iθµ0 (Â(b − c)) z = ∂z j z − 2iθµ (c − d0 ) (z − 2iθν (c − d0 ))j+1 (−1)l l! ∂l 1 = ∂wl w − 2iθµ0 (c − d0 ) (w − 2iθµ0 (c − d0 ))l+1 97  for j ≥ 1, for l ≥ 0.  Recall from (4.4.18) and (4.4.20) that, for all c ∈ G0 , |c − d0 | |c − d0 | ≤ ≤ 2. |w − 2iθµ0 (c − d0 )| |c − d0 | − ε  (4.5.17)  Then, using this and (4.5.11), for j ≥ 1 and l ≥ 0,   ∂ j+l −1 S ∂z j ∂wl   ≤ b,c  ≤  |c − d0 | j! l! |Â(b − c)| |z − 2iθµ (c − d0 )|j+1 |w − 2iθµ0 (c − d0 )|l |w − 2iθµ0 (c − d0 )| 2j+2 j! l! |Â(b − c)| Λl |z|j+1 R  (4.5.18)  ,  while for j = 0 and l ≥ 0,   ∂ j+l −1 S ∂z j ∂wl   ≤ b,c  l! |Â(b − c)| |z| 2 l! |Â(b − c)| ≤ . 0 0 l+1 |z − 2iθµ (c − d )| |w − 2iθµ0 (c − d )| Λl+1  Consequently,   sup   X  b∈G03 c∈G0  3  + sup  X  c∈G03 b∈G0      ∂ j+l −1 S ∂z j ∂wl  (4.5.19)    3  b,c     j+2  X X 2 j! l!  |z|R  |Â(b − c)| sup + sup δ0,j ≤ 1 − δ0,j + 2Λ Λl |z|j+1 b∈G03 c∈G0 c∈G03 b∈G0 R 3 3   j+3 |z|R 2 j! l! ≤ 1 − δ0,j + kÂkl1 . δ0,j 2Λ Λl |z|j+1 R Therefore, by Proposition (4.1.2), ∂ j+l −1 S ≤ ∂z j ∂wl   1 − δ0,j  Thus, for r ≥ 1, in view of (4.5.14) where   r+p Y  ∂ r+p  |z|R + δ0,j 2Λ  Pr+p j=1  ∂ mj +nj    2j+3 j! l! Λl |z|j+1 R  kÂkl1 .  (4.5.20)  mj = r,   S ≤ Cr,p  kSk S −1  kSk ∂z r ∂wp ∂z mj ∂wnj j=1    r+p r+p m +3 Y Y j 2 mj ! nj ! |z|R 1   ≤ Cr,p CΛ,A kÂkl1 CΛ,A 1 − δ0,mj + δ0,mj n m +1 j Λ 2Λ |z| j j=1  ≤ CΛ,A,r,p  j=1  1 , |z|r+1 R  since mj ≥ 1 for at least one 1 ≤ j ≤ r + p. Similarly, if r = 0 then ∂ r+p S ≤ CΛ,A,r,p . ∂z r ∂wp 98  R  Hence, in view of (4.5.12), sup sup p≤m r≤n  ∂ n−r+m−p ∆−1 k ∂z n−r ∂wm−p  ∂ r+p M (1) ∂z r ∂wp    (m − p)! (n − r)! 2n−r+2 1 |z|R C k Âk 1 − δ + δ 1 0,r 0,r Λ,A,r,p l n−r+1 m−p+1 2Λ Λ |z|R |z|r+1 p≤m r≤n R 1 ≤ CΛ,A,n,m n+1 . |z|R  ≤ sup sup  This proves (4.5.10) for j = 1. Step 3 We now estimate the constant C1;Λ,A,i,j for i, j ∈ {0, 1}. First observe that ∂w = |δ1,j + i(−1)ν δ2,j | = 1 ∂kj  and  ∂z = |δ1,j − i(−1)ν δ2,j | = 1. ∂kj  Thus, in view of (4.5.16) and (4.5.20), since |z| ≥ |v| > R ≥ 2Λ,   ∂S ∂z ∂S −1 ∂S −1 ∂w ∂S −1 S = −S + S = −S ∂kj ∂kj ∂kj ∂w ∂kj ∂z !   2  −1 2 kÂk −1 4 kÂk ∂S 3 2 ∂S 2 1 1 l l ≤ kSk2 + + ≤ ∂w ∂z 2 Λ2 |z|2R  2 3 8kÂkl1 18kÂkl1 ≤ = . 2 2 Λ Λ2 Similarly,     ∂2S ∂z ∂S −1 ∂w ∂S −1 ∂z ∂S −1 ∂S ∂S ∂w ∂S −1 + S−S + =− ∂ki ∂kj ∂ki ∂kj ∂w ∂kj ∂z ∂kj ∂w ∂kj ∂z ∂ki      2 −1 2 −1 2 −1 ∂w ∂w ∂ S ∂z ∂ S ∂z ∂w ∂ S ∂z ∂ 2 S −1 −S + + + S, ∂kj ∂ki ∂w2 ∂ki ∂z∂w ∂kj ∂ki ∂w∂z ∂ki ∂z 2 so that, using the above inequality as well,   ∂2S ∂S ∂S −1 ∂S −1 ≤ 2kSk + ∂ki ∂kj ∂ki ∂w ∂z  2 −1  2 ∂ S ∂ S −1 ∂ 2 S −1 2 + kSk +2 + ∂w2 ∂z∂w ∂z 2 !  2 3 18kÂkl1 8kÂkl1 3 23 kÂkl1 25 kÂkl1 26 kÂkl1 ≤2 + + + 2 Λ2 Λ2 2 Λ3 Λ|z|2R |z|3R ! 432 54 55kÂkl1 8kÂkl1 ≤ 4 kÂk2l1 + 3 kÂkl1 ≤ +1 . Λ Λ Λ3 Λ Furthermore, by (4.5.12), ∂∆−1 ∂∆−1 ∂∆−1 22 23 8 k k k ≤ + ≤ 2 + ≤ 2 2 ∂kj ∂w ∂z Λ |z|R Λ|z|R Λ |z|R 99  and ∂ 2 ∆−1 k ∂ki ∂kj  ≤ ≤  ∂ 2 ∆−1 ∂ 2 ∆−1 ∂ 2 ∆−1 k k k + 2 + ∂w2 ∂z∂w ∂z 2 23 24 26 5 · 23 1 + . + < Λ3 |z|R Λ2 |z|2R Λ|z|3R Λ3 |z|R  Hence, since kÂkl1 < 2ε/63 and ε < Λ/6, ∂∆−1 ∂ −1 ∂S k ∆k S ≤ kSk + k∆−1 k k ∂kj ∂kj ∂kj  ≤  8 Λ2 |z|R  3 2 18kÂkl1 13 1 + ≤ 2 2 2 Λ|z|R Λ Λ |z|R  and ∂∆−1 ∂S k + ∂ki ∂ki  ∂ 2 ∆−1 ∂∆−1 ∂2 k k ∆−1 S ≤ kSk + ∂ki ∂kj k ∂ki ∂kj ∂kj ≤  1 |z|R  ∂2S ∂S + k∆−1 k k ∂kj ∂ki ∂kj !  5 · 23 3 8 18kÂkl1 2 55kÂkl1  8kÂkl1 65 1 + 2 + + 1 . < 3 3 2 2 3 Λ 2 Λ Λ Λ Λ Λ Λ |z|R  Therefore, C1;Λ,A,1,0 ≤  13 , Λ2  13 Λ2  C1;Λ,A,0,1 ≤  and  C1;Λ,A,1,1 ≤  65 , Λ3  as was to be shown. Step 4 To prove (4.5.10) for j = 2 we need to bound ∂ r+p M (2) ∂ r+p W = . ∂z r ∂wp ∂z r ∂wp Recall from (4.4.23) that W =  ∞ X j=1  Wj =  j ∞ X X  (Y33 )m−1 X33 (Y33 )j−m ,  j=1 m=1  where Yb,c is given above by (4.5.15) and kX33 k ≤ C/|z| < 1/3 with Xb,c =  (c − d0 ) · Â(b − c) − q̂(b − c) − 2iθµ (Â(b − c)) w . (w − 2iθµ0 (c − d0 ))(z − 2iθµ (c − d0 ))  First observe that ∂ r+p (Y33 )m−1 X33 (Y33 )j−m ∂z r wp is given by a sum of j r+p terms of the form ∂ l1 +n1 Y33 ∂ lm−1 +nm−1 Y33 ∂ lm +nm X33 ∂ lm+1 +nm+1 Y33 ∂ lj +nj Y33 · · · · · · , ∂z l1 ∂wn1 ∂z lm−1 ∂wnm−1 ∂z lm ∂wnm ∂z lm+1 ∂wnm+1 ∂z lj ∂wnj 100  where there are j factors ordered as in the product (Y33 )m−1 X33 (Y33 )j−m . Furthermore, for P P each term in the sum we have ji=1 li = r and ji=1 ni = p. Thus, j ∞ X ∞ X X ∂ r+p ∂ r+p ∂ r+p = W W (Y33 )m−1 X33 (Y33 )j−m = j ∂z r wp ∂z r wp ∂z r wp j=1 m=1  j=1  ≤  ∞ X  j X  j=1 m=1  ≤  ∞ X  j  r+p  ≤  j=1  ∂ r+p (Y33 )m−1 X33 (Y33 )j−m ∂z r wp j X  sup  m=1 I  j=1 ∞ X  j  (4.5.21)  r+p  j X  sup  m=1 I  ∂ lm +nm X33 ∂ lj +nj Y33 ∂ l1 +n1 Y33 · · · · · · ∂z l1 ∂wn1 ∂z lm ∂wnm ∂z lj ∂wnj ∂ lm +nm X33 ∂ lj +nj Y33 ∂ l1 +n1 Y33 · · · · · · , ∂z l1 ∂wn1 ∂z lm ∂wnm ∂z lj ∂wnj  (4.5.22)  where ( I :=  (li , ni ) li ≤ r and ni ≤ p for 1 ≤ i ≤ j with  j X  li = r and  i=1  j X  ) ni = p .  i=1  (4.5.23) P∞  Note, we can differentiate the series (4.5.21) term-by-term because the sum j=1 Wj conP verges uniformly and the sum jm=1 is finite. We next estimate the factors in (4.5.22). Combining (4.5.18) and (4.5.19) we have   li +2 |z|R 2 l i ! ni ! ∂ li +ni Yb,c ≤ 1 − δ0,li + δ0,li |Â(b − c)|. n l li +1 i i n 2Λ ∂z ∂w Λ i |z|R  (4.5.24)  Furthermore, using (4.5.11) and (4.5.17), ∂ li +ni Xb,c ∂z li ∂wni =  ∂ li 1 ∂ ni (c − d0 ) · Â(b − c) − q̂(b − c) − 2iθµ (Â(b − c)) w w − 2iθµ0 (c − d0 ) ∂z li z − 2iθµ (c − d0 ) ∂wni  =  (−1)li li ! (−1)ni ni ! (2θµ (Â(b − c)) 2θµ0 (c − d0 ) − (c − d0 ) · Â(b − c) − q̂(b − c)) (z − 2iθµ (c − d0 ))li +1 (w − 2iθµ0 (c − d0 ))ni +1  ≤  li ! ni ! (2|Â(b − c)| |c − d0 | + |q̂(b − c)|) |z − 2iθµ (c − d0 )|li +1 |w − 2iθµ0 (c − d0 )|ni +1  2li +1 li ! ni ! 2|Â(b − c)| |c − d0 | + |q̂(b − c)| |w − 2iθµ0 (c − d0 )| Λni |z|lRi +1  2li +1 li ! ni !  1 ≤ 4| Â(b − c)| + |q̂(b − c)| . Λ Λni |z|lRi +1  ≤  (4.5.25) 101  Hence,      sup  X  + sup  b∈G03 c∈G0  c∈G03 b∈G0  3  ∂ li +ni Yb,c ∂z li ∂wni  X   3      li +2 X X 2 li ! ni !  |z|R  |Â(b − c)| ≤ 1 − δ0,li + δ0,li sup + sup li +1 2Λ Λni |z|R b∈G03 c∈G0 c∈G03 b∈G0 3 3  li +3  2 li ! ni ! |z|R δ0,li kÂkl1 ≤ 1 − δ0,li + li +1 n 2Λ Λ i |z|R and      sup  X  b∈G03 c∈G0  3  + sup  X  c∈G03 b∈G0    ∂ li +ni Xb,c ∂z li ∂wni  3      ≤  ≤  2li +1 li ! ni !  Λni |z|lRi +1  2li +2 li ! ni ! Λni |z|lRi +1  sup  X  b∈G01 c∈G0  1  + sup    |q̂(b − c)|  4|Â(b − c)| + Λ 0  X  c∈G01 b∈G  1    kq̂kl1 4kÂkl1 + . Λ  Thus, by Proposition (4.1.2), since |z| ≥ |v| > R ≥ 2Λ,   li +3 |z|R 2 li ! ni ! ∂ li +ni Y33 ≤ 1 − δ0,li + δ0,li kÂkl1 l n li +1 i i n 2Λ ∂z ∂w Λ i |z|R   1 1 2li +3 li ! ni ! 2li +3 li ! ni ! ≤ + k Âk ≤ kÂkl1 1 l |z|R 2Λ Λni |z|lRi Λni +1 |z|lRi  (4.5.26)  and !  1 2li +3 li ! ni ! kÂkl1 . 2kÂkl1 |z|R Λni +1 |z|lRi (4.5.27) Pj Pj Applying these estimates to (4.5.22) and recalling that i=1 li = r and i=1 ni = p we ∂ li +ni 2li +2 li ! ni ! X ≤ 33 ∂z li ∂wni Λni |z|lRi +1    kq̂kl1 4kÂkl1 + = Λ  2Λ +  kq̂kl1  have j ∞ X X ∂ r+p ∂ l1 +n1 Y33 ∂ lm +nm X33 ∂ lj +nj Y33 r+p W ≤ j sup · · · · · · r p l n l n ∂z w ∂z 1 ∂w 1 ∂z m ∂w m ∂z lj ∂wnj m=1 I j=1 ! ( ) j j ∞ X X kq̂kl1 1 Y 2li +3 li ! ni ! r+p ≤ j sup 2Λ + kÂkl1 ni +1 |z|li |z|R 2k Âk 1 Λ I l R m=1 j=1 i=1 ! !j ( j ) j j ∞ Y Y X 2r X r+p 8kÂkl1 kq̂kl1 1 = 2Λ + j sup l ! n ! 1 i m Λ 2kÂkl1 |z|R Λp |z|rR j=1 I m=1 m=1 i=1 !   ∞ 2r r!p! X r+p+1 1 j 1 kq̂kl1 0 ≤ 2Λ + j ≤ C . Λ,A,q,r,p r+1 21 |z|r+1 2kÂkl1 Λp |z|R j=1 R  102  This is the inequality we needed to prove (4.5.10) for j = 2. In fact, using (4.5.12) we obtain ∂ n−r+m−p ∆−1 k sup sup n−r ∂w m−p ∂z p≤m r≤n  0 ∂ r+p M (2) (m − p)! (n − r)! 2n−r+2 CΛ,A,q,r,p ≤ sup sup ∂z r ∂wp Λm−p+1 |z|n−r+1 |z|r+1 p≤m r≤n R R 1 ≤ CΛ,A,q,m,n n+2 . |z|R  Step 5 To prove (4.5.10) for j = 3 we need to estimate ∂ r+p M (3) ∂ r+p Z = , ∂z r ∂wp ∂z r ∂wp where Z=  ∞ X  Zj =  j=2  ∞ X  j (X33 + Y33 )j − Wj − Y33 .  j=2  First observe that ∂ r+p ∂ r+p j ) Z = ((X33 + Y33 )j − Wj − Y33 j ∂z r ∂wp ∂z r ∂wp is given by a sum of (2j − j − 1) · j r+p terms of the form ∂ l1 +n1 Y33 ∂ lj +nj Y33 ∂ lm +nm X33 · · · , · · · l1 n1 ∂z lm{z ∂wnm ∂z lj ∂wnj} |∂z ∂w  (4.5.28)  j factors  where there are j − 2 factors involving X33 or Y33 and two factors containing X33 . FurtherP P more, for each term in the sum we have ji=1 li = r and ji=1 ni = p. Thus, ∂ lj +nj Y33 ∂ l1 +n1 Y33 ∂ lm +nm X33 ∂ r+p j r+p · · · Z , ≤ (2 − j − 1) j sup · · · j ∂z r ∂wp ∂z l1 ∂wn1 ∂z lm ∂wnm ∂z lj ∂wnj I where the set I is given above by (4.5.23). Now observe that, the estimate for the derivatives of X33 in (4.5.27) is better then the estimate for the derivatives of Y33 in (4.5.26) because the former has an extra factor CΛ,A,q /|z|R < 1. Since the product (4.5.28) has at least two factors containing X33 , we can estimate any of these products by considering the worst case. This happens when there are exactly two factors involving X33 . Hence, by proceeding in this way, for each j ≥ 2 we have ∂ r+p ∂z r ∂wp  Zj ≤ (2j − j − 1) j r+p sup I  ≤ 2j j r+p  2Λ +  kq̂kl1     2Λ +   !2  kq̂kl1 2kÂkl1  1 2r r!p! |z|2R Λp |z|rR  2kÂkl1  j 2 1 0 ≤ CΛ,A,q,r,p j r+p , 21 |z|r+2 R 103   j  l +3 Y i 1 2 li !ni ! k Âk 1 l  |z|2R i=1 Λni +1 |z|lRi !j 8kÂkl1 Λ  !2  since kAkl1 ≤ 2ε/63 and ε < Λ/6. Thus,  j ∞ ∞ 0 X X CΛ,A,q,r,p CΛ,A,q,r,p ∂ r+p 2 ∂ r+p r+p Z ≤ Zj ≤ ≤ j r+2 r+2 . r p r p ∂z ∂w ∂z ∂w 21 |z| |z| R R j=2 j=2 Therefore, recalling (4.5.12), sup sup p≤m r≤n  0 ∂ r+p M (3) (m − p)! (n − r)! 2n−r+2 CΛ,A,q,r,p ≤ sup sup ∂z r ∂wp Λm−p+1 |z|n−r+1 |z|r+2 p≤m r≤n R R 1 ≤ CΛ,A,q,m,n n+3 . |z|R  ∂ n−r+m−p ∆−1 k ∂z n−r ∂wm−p  This is the desired inequality for j = 3. The proof of the proposition is complete. We can now prove Lemma 4.5.3. We first prove it for 1 ≤ j ≤ 2 and then for j = 3 separately. Proof of Lemma 4.5.3 for 1 ≤ j ≤ 2. Define the |B| × |C| matrices FBC := [f (b − c)]b∈B,c∈C  GBC := [g(b − c)]b∈b,c∈C ,  and  and write w = wµ,d0 ,  z = zµ,d0  and  |z|R = 2|z| − R.  First observe that, for 1 ≤ j ≤ 2, the functions   (j) X f (d0 − b)Mb,c g(c − d0 )   (j)  αµ,d0 (k) d0 ∈G =  (w − 2iθµ0 (b − d0 ))(z − 2iθµ (b − d0 )) 0 b,c∈G1  d0 ∈G  are the diagonal entries of the matrix (j) GG01 G . FGG01 ∆−1 k M  Thus, similarly as in the proof of Lemma 4.5.1, by Proposition 4.5.5, for 1 ≤ j ≤ 2, ∂ n+m (j) ∂ n ∂ m −1 (j) 0 ≤ kF k kGG01 G k α (k) ∆ M 0 GG1 ∂k1n ∂k2m µ,d ∂k1n ∂k2m k ≤ kf kl1 kgkl1  Cj ∂ n ∂ m −1 (j) ∆ M ≤ j , ∂k1n ∂k2m k |z|R  where C1 = C1;Λ,A,n,m,f,g and C2 = C2;Λ,A,q,n,m,f,g are constants. Furthermore, C1;Λ,A,1,0,f,g ≤  13 kf kl1 kgkl1 , Λ2  and  C1;Λ,A,0,1,f,g ≤  C1;Λ,A,1,1,f,g ≤  This proves the lemma for 1 ≤ j ≤ 2. 104  13 kf kl1 kgkl1 Λ2  65 kf kl1 kgkl1 . Λ3  Proof of Lemma 4.5.3 for j = 3. We need to estimate 4  X ∂ n+m ∂ n+m (3) αµ,d0 (k) = Rj (k), n m ∂k1 ∂k2 ∂k1n ∂k2m j=1  where R1 , . . . , R4 are given by (4.4.5), (4.4.6), (4.4.34) and (4.4.26), respectively. Step 1 We begin with the terms involving R1 and R2 , which are easier. We follow the same notation as above. First observe that, similarly as in the proof of Lemma 4.5.1, since −1 −1 2 ∆−1 k RG0 G0 = Hk on LG0 , we have  ∂ n+m Hk−1 ∂ n+m Hk−1 ∂ n+m 0 0 0 R (k) G ≤ kF k kGG02 {d0 } k, = F 0 0 0 1 {d }G1 {d }G1 ∂k1n ∂k2m ∂k1n ∂k2m G2 {d } ∂k1n ∂k2m ∂ n+m Hk−1 ∂ n+m Hk−1 ∂ n+m 0 0 0 0 R (k) = F G ≤ kF k kGG0 {d0 } k. 0 0 2 {d }G2 {d }G2 ∂k1n ∂k2m ∂k1n ∂k2m G {d } ∂k1n ∂k2m Furthermore, we have already proved that (see (4.5.7) and (4.5.8)) kGG0 {d0 } k ≤ kgkl1  kF{d0 }G01 k ≤ kf kl1 , and, since |z| ≤ 3|v| by Proposition 4.4.4,  ∂ n+m Hk−1 1 ≤ ε−(n+m+1) CΛ,A,n,m . n m ∂k1 ∂k2 |z| Now recall that G02 = {b ∈ G0 | |b − d0 | > 41 R}. Then, sup  X  b∈{d0 } c∈G0  2  sup  X  c∈G02 b∈{d0 }  |f (b − c)| ≤  X |d0 − c|2 1 16 |f (d0 − c)| ≤ kb2 f (b)kl1 sup 0 ≤ 2 kb2 f (b)kl1 , 0 − c|2 2 0 |d |d − c| R c∈G2 0  c∈G2  |f (b − c)| ≤ sup c∈G02  |d0 − c|2 1 16 |f (d0 − c)| ≤ kb2 f (b)kl1 sup 0 ≤ 2 kb2 f (b)kl1 . 0 2 2 |d − c| R c∈G02 |d − c|  Hence, by Proposition (4.1.2), kF{d0 }G02 k ≤ 16kb2 f (b)kl1  1 . R2  kGG02 {d0 } k ≤ 16kb2 f (b)kl1  1 . R2  Similarly,  Therefore, combining all this, for 1 ≤ j ≤ 2 we obtain ∂ n+m 1 Rj (k) ≤ ε−(n+m+1) CΛ,A,n,m,f,g . ∂k1n ∂k2m |z|R2 105  Step 2 Recall from (4.4.26) the expression for R4 . Then, similarly as above, by applying Proposition 4.5.5 for j = 3 we find that ∂ n+m 1 ∂ n+m R (k) ∆−1 Z kGG01 {d0 } k ≤ kf kl1 kgkl1 CΛ,A,q,n,m 3 . ≤ kF 0 }G0 k 4 {d k n m n m 1 ∂k1 ∂k2 ∂k1 ∂k2 |z|R Step 3 To bound the derivatives of R3 (which is given by (4.4.34)) we need a few more (j)  estimates. Recall from (4.4.29) that W43 = πG04 TGj+1 0 G0 πG0 . First observe that 3 ∂ r+p ∂ r+p (j−m−1) −1 m 0 ∆ T T W = π ∆−1 π 0 T m T34 TGj−m 0 G0 πG0 34 43 3 ∂k1r ∂k2p G1 k 33 ∂k1r ∂k2p k G1 33 is given by a sum of (j + 2)r+p terms of the form ∂ l1 +n1 ∆−1 k ∂k1l1 ∂k2n1  πG01  ∂ l2 +n2 T33 ∂k1l2 ∂k2n2  ···  ∂ lm+2 +nm+2 T34 ∂ lm+3 +nm+3 TG0 G0 l  n  l  ∂k1m+2 ∂k2 m+2  Moreover, for each term in the sum we have  n  ∂k2m+3 ∂k2 m+3  Pj+2  i=1 li  = r and  ∂ lj+2 +nj+2 TG0 G0 l  n  ∂k1j+2 ∂k2 j+2  πG03 .  Pj+2  ni = p. Thus, ! j+2 Y ∂ li +ni T(i) πG03 , (4.5.29) ni li i=1 ∂k1 ∂k2  ∂ r+p (j−m−1) −1 m ≤ (j + 2)r+p sup p πG01 ∆k T33 T34 W43 r ∂k1 ∂k2 I0 where the set I 0 is given by (4.5.23) with    ∆−1  k πG01      T33 := T(i)    T34       TG0 G0  ···  i=1  j replaced by j + 2 and for i = 1, 2 ≤ i ≤ m + 1,  for  (4.5.30) for i = m + 2, for m + 3 ≤ i ≤ j + 2.  Step 3a The first step in bounding (4.5.29) is to estimate ∂ r+p ∆−1 k π 0 . ∂k1r ∂k2p G1 We follow the same argument that we have used in the proof of Lemma 4.5.1 to bound ∂ n+m Hk−1 . ∂k1n ∂k2m In fact, in view of (4.5.2) one can see that ∂ p ∆−1 k ∂k2p   =  X  p Y   finite sum where # of terms depend on p  106  j=1  ∆−1 k  ∂ nj ∆ n  ∂k2 j   k  ∆−1 k ,  (4.5.31)  ∆−1 k and  r  Pp  ∂ p ∆−1  r  ∂ ∂ k , the derivative ∂k = p. Hence, when we compute ∂k r r acts either on ∂k2p 1 1   nj nj r ∂ ∂ k = 2(k2 + c2 )δb,c , we have ∂k or ∂ n∆jk . However, since ∂∆ nj ∆k = 0 if nj ≥ 1 r ∂k2  where  j=1 nj  ∂k2  ∂r ∂k1r  ∂ nj n ∂k2 j  b,c  ∆k =  ∂r ∂k1r  1  ∂k2  ∆k if nj = 0. Similarly, using again (4.5.2) one can see that  ∂ r ∆−1 k ∂k1r  is  given by a finite sum as in (4.5.31), with p and k2 replaced by r and k1 , respectively, and Pr j=1 nj = r. Thus, combining all this we conclude that   r+p −1 r+p n Y X j ∂ ∆k ∂ ∆k  −1  ∆−1 ∆k , (4.5.32) p = n k r ∂k1 ∂k2 ∂kijj j=1 finite sum where # of terms depend on r and p  where  Pr+p j=1  nj δ2,ij = p and  Pr+p  nj δ1,ij = r. If we observe     2(kij + cij )δb,c if  !   n j ∂ ∆k = 2δb,c if n  ∂kijj   b,c   0 if j=1  that nj = 1, nj = 2, nj ≥ 3,  and extract the “leading term” from the summation in (4.5.32), in a sense that will be clear below, we can rewrite (4.5.32) in terms of matrix elements as     1 (−1)r+p (r + p)! 2(k1 + c1 ) r 2(k2 + c2 ) p ∂ r+p = Nc (k) Nc (k) Nc (k) ∂k1r ∂k2p Nc (k) X (2(k1 + c1 ))αj (2(k2 + c2 ))βj + , Nc (k)r+p+1 finite sum where # of terms depend on r and p  where αj + βj < r + p for every j in the summation. Recall from (4.5.5) and (4.5.6) that, for all c ∈ G0 \ {c̃}, |ki + ci | 2 1 7 ≤ < < |Nc (k)| Λ 3ε 2ε  |ki + c̃i | Λ + 3|v| 7 ≤ ≤ . |Nc̃ (k)| ε|v| 2ε  and  (4.5.33)  Hence, ∂ r+p 1 (r + p)! ≤ |Nc (k)| ∂k1r ∂k2p Nc (k)   r+p 7 + ε  X finite sum where # of terms depend on r and p   αj +βj 7 1 ε |Nc (k)|2 (4.5.34)    (r + p)! 7 r+p 1 + Cε,r,p . ≤ |Nc (k)| ε |Nc (k)|2 Thus, by Proposition 4.1.2, since |Nc (k)| ≥ ε|v| ≥ ε|z|/3 for all c ∈ G0 , we have ∂ r+p ∆−1 Cε,r,p 7r+p (r + p)! 3 k 0 π ≤ + . p G1 r r+p+1 ε |z| |z|2 ∂k1 ∂k2 107  (4.5.35)  Now, let ρ1 = ρ1;ε,r,p be the constant ρ1;ε,r,p := max l1 ≤r n1 ≤p  εl1 +n1 +1 Cε,l1 ,n1 , 4(l1 + n1 )! 7l1 +n1  where Cε,l1 ,n1 is the constant in (4.5.35). Then, for |z| > ρ1 and for any l1 ≤ r and any n1 ≤ p, ∂ l1 +n1 ∆−1 k ∂k1l1 ∂k2n1  7l1 +n1 (l1 + n1 )! 3 7l1 +n1 (l1 + n1 )! 4 + |z| |z| εl1 +n1 +1 εl1 +n1 +1  l1 +n1 +1 7 1 = (l1 + n1 )! . ε |z|  πG01 ≤  (4.5.36)  This is the first inequality we need to bound (4.5.29). We next estimate the other factors in that expression. Step 3b Recall from (4.4.12) that Tb,c =  1 (2(c + k) · Â(b − c) − q̂(b − c)). Nc (k)  By direct calculation we have  r+p  ∂ r+p Tb,c ∂ 1 = (2(c + k) · Â(b − c) − q̂(b − c)) ∂k1r ∂k2p ∂k1r ∂k2p Nc (k) !  r−1+p  1 ∂ ∂ r+p−1 1 +r 2Âj (b − c) + p 2Âj (b − c). ∂k1r−1 ∂k2p Nc (k) ∂k1r ∂k2p−1 Nc (k) Hence, using (4.5.33) and (4.5.34), since |Nc (k)| ≥ ε|v| ≥ ε|z|/3 for all c ∈ G0 and |v| > 1, !  r+p  ∂ r+p Tb,c Cε,r,p Cε,r,p 7 7 |q̂(b − c)| + |Â(b − c)| + + |Â(b − c)| p ≤ (r + p)! r ε ε|v| ε ε|v| |v| ∂k1 ∂k2  r+p+1 Cε,r,p 7 |Â(b − c)| + (|Â(b − c)| + |q̂(b − c)|). ≤ (r + p)! ε |z| (4.5.37) Therefore, by Proposition 4.1.2, ∂ r+p TG0 G0 ∂k1r ∂k2p  ≤ Θr,p ,  (4.5.38)   r+p+1 7 1 kÂkl1 + Cε,A,q,r,p . ε |z|  (4.5.39)  where Θr,p  := (r + p)!  This is the second estimate we need to bound (4.5.29). We next derive one more inequality.  108  Step 3c Set 2 Qr,p b,c := (1 + |b − c| )  ∂ r+p Tb,c . ∂k1r ∂k2p  We first prove that, for any B, C ⊂ G0 , sup  X  b∈B c∈C  |Qr,p b,c | ≤ Ωr,p  and sup  X  c∈C b∈B  |Qr,p b,c | ≤ Ωr,p ,  where Ωr,p := (r + p)!   r+p+1 7 1 k(1 + b2 )Â(b)kl1 + Cε,A,q,r,p . ε |z|  (4.5.40)  In fact, in view of (4.5.37) we have sup  X  b∈B c∈C  |Qr,p b,c | = sup  X  (1 + |b − c|2 )  b∈B c∈C  ≤ sup  X  ∂ r+p Tb,c ∂k1r ∂k2p  (1 + |b − c|2 )  b∈B c∈C  "  #  r+p+1 Cε,r,p 7 × (r + p)! |Â(b − c)| + (|Â(b − c)| + |q̂(b − c)|) ε |z|  r+p+1 1 7 k(1 + b2 )Â(b)kl1 + Cε,A,q,r,p , ≤ (r + p)! ε |z| and similarly we estimate supc∈C  P  b∈B  |Qr,p b,c |. Now observe that, as in (4.4.37), for any  integer m ≥ 0 and for any ξ0 , ξ1 , . . . , ξm+2 ∈ Γ# , let b = ξ0 and c = ξm+2 . Then, |b − c|2 ≤ 2(m + 2)  m+2 X  |ξi−1 − ξi |2 .  i=1  To simplify the notation write ∂ li ,ni =  ∂ li +ni ∂k1li ∂k2ni  ,  and recall from (4.5.30) and (4.5.40) the definition of T(i) and Ωr,p . Hence, similarly as in the proof of Proposition 4.4.6, since |b − c| ≥ R/4 for all b ∈ G01 and c ∈ G04 ,  109  X  m+2 Y  b∈G01 c∈G0  i=2  sup  4  ! ∂ li ,ni T(i) b,c m+2 Y  X 1 sup (1 + |b − c|2 ) ≤ sup 2 b∈G01 1 + |b − c| b∈G01 0 ≤  ∂ li ,ni T(i)  i=2  c∈G4  c∈G04  ! b,c  X 2(m + 2) (1 + |b − ξ1 |2 ) ∂ l2 ,n2 Tb,ξ1 1 2 sup0 1 + 16 R b∈G1 0 ξ1 ∈G3  X  ×  (1 + |ξ1 − ξ2 |2 ) ∂ l3 ,n3 Tξ1 ,ξ2 · · ·  ξ2 ∈G03  ≤  X  (1 + |ξm+1 − c|2 ) ∂ lm+2 ,nm+2 Tξm+1 ,c  c∈G04  X X 2(m + 2) 2 l2 ,n2 sup (1 + |b − ξ | ) ∂ T sup (1 + |ξ1 − ξ2 |2 ) ∂ l3 ,n3 Tξ1 ,ξ2 1 b,ξ 1 1 2 1 + 16 R b∈G01 ξ1 ∈G03 0 0 ξ1 ∈G3  sup  X  ξm+1 ∈G03  c∈G04  ×  =  ξ2 ∈G3  (1 + |ξm+1 − c|2 ) ∂ lm+2 ,nm+2 Tξm+1 ,c  m+2 X X l ,n 2(m + 2) Y 2(m + 2) l2 ,n2 m+2 m+2 |Q |Q sup | · · · sup | ≤ Ωl ,n b,ξ1 ξm+1 ,c 1 2 1 2 R b∈G01 R i=2 i i 1 + 16 1 + 16 ξm+1 ∈G03 0 0 ξ1 ∈G3  c∈G4  and X  m+2 Y  c∈G04 b∈G0  i=2  sup  1  ! ∂ li ,ni T(i) b,c m+2 Y  X 1 ≤ sup sup (1 + |b − c|2 ) 2 b∈G01 1 + |b − c| c∈G04 0 ≤  2(m + 2) 1 2 sup0 1 + 16 R c∈G4 ×  X  X  ∂ li ,ni T(i)  i=2  b∈G1  c∈G04  ! b,c  (1 + |ξm+1 |2 ) ∂ lm+2 ,nm+2 Tξm+1 ,c  ξm+1 ∈G03  (1 + |ξm − ξm+1 |2 ) ∂ lm+1 ,nm+1 Tξm ,ξm+1 · · ·  ξm ∈G03  ≤  (1 + |b − ξ1 |2 ) ∂ l2 ,n2 Tb,ξ1  b∈G01  2(m + 2) 1 2 sup0 1 + 16 R c∈G4 ×  X  X  sup  X  ξm+1 ∈G03  ξm ∈G03  ξ1 ∈G03  (1 + |ξm − ξm+1 |2 ) ∂ lm+1 ,nm+1 Tξm ,ξm+1  X  × sup  (1 + |ξm+1 |2 ) ∂ lm+2 ,nm+2 Tξm+1 ,c  ξm+1 ∈G03  (1 + |b − ξ1 |2 ) ∂ l2 ,n2 Tb,ξ1  b∈G01  2(m + 2) = 1 2 sup0 1 + 16 R c∈G4  X ξm+1 ∈G03  l ,nm+2 |Qξm+2 |··· m+1 ,c  sup  X  ξ1 ∈G03 b∈G0 1  110  2 ,n2 |Qlb,ξ | 1  m+2 2(m + 2) Y ≤ Ωl ,n . 1 2 1 + 16 R i=2 i i  Therefore, by Proposition 4.1.2, πG01  m+2 Y  ∂ li +ni T(i)  i=2  ∂k1li ∂k2ni  ≤  m+2 2(m + 2) Y Ωl ,n . 1 2 1 + 16 R i=2 i i  We have all we need to bound (4.5.29). Step 3d From (4.5.38) and (4.5.36) it follows that j+2 Y  ∂ li +ni T(i)  i=m+3  ∂k1li ∂k2ni  ≤  j+2 Y  Θli ,ni  i=m+3  and ∂ l1 +n1 T(1) ∂k1l1 ∂k2n1   r+p+1 7 1 ≤ (r + p)! . ε |z|  Thus, recalling (4.5.29) we get ! j+2 li +ni T Y ∂ ∂ r+p (i) (j−m−1) ∆−1 π 0 T m T34 W43 ≤ (j + 2)r+p sup πG03 ni li ∂k1r ∂k2p k G1 33 ∂k ∂k I0 2 1 i=1 ( # j+2 )  r+p+1 "m+2 Y Y 1 2(m + 2) 7 ≤ (j + 2)r+p sup Ωli ,ni Θli ,ni 1 2 (r + p)! ε |z| 1 + 16 R I0 i=2 i=m+3 ( # j+2 )  l1 +n1 +1 "m+2 Y Y 7 C ≤ (j + 2)r+p (m + 2) sup (l1 + n1 )! Ωli ,ni Θli ,ni , |z|R2 I 0 ε i=2  i=m+3  where C is an universal constant. Now, recall the definition of Θr,p and Ωr,p in (4.5.39) and (4.5.40), observe that kÂkl1 < k(1 + b2 )Âkl1 , and let ρ2 = ρ2;ε,A,q,r,p be a sufficiently large constant such that, for |z| > ρ2 and for any li ≤ r and any ni ≤ p,  li +ni +1 7 Θli ,ni , Ωli ,ni ≤ 2(li + ni )! k(1 + b2 )Â(b)kl1 . ε Then, ∂ r+p (j−m−1) −1 m p ∆k πG01 T33 T34 W43 r ∂k1 ∂k2 # j+2 ) (  l1 +n1 +1 "m+2 Y Y 7 C r+p Ωli ,ni Θli ,ni ≤ (j + 2) (m + 2) sup (l1 + n1 )! |z|R2 I 0 ε i=2 i=m+3 ( Pj+2 )  j+2 j+2 i=1 (li +ni ) Y 7 r+p (m + 2)C 2 j+1 7 ≤ (j + 2) (2k(1 + b )Â(b)kl1 ) sup (li + ni )! |z|R2 ε ε I0 i=1  (since  Pj+2  i=1 li  Pj+2  Qj+2  = r, i=1 ni = p and i=1 (li + ni )! < (r + p)!)  r+p+1  j+1 7 1 r+p 14 2 ≤ C(r + p)! (m + 2)(j + 2) k(1 + b )Â(b)kl1 ε ε |z|R2  j+1 Cε,r,p r+p 4 ≤ (m + 2)(j + 2) , |z|R2 9 111  since k(1 + b2 )Â(b)kl1 < 2ε/63. This establishes a bound for (4.5.29). Step 4 We now apply the last inequality for deriving an estimate for the derivatives of R3 and complete the proof of the lemma for j = 3. Recall from (4.4.35) that (j) X33  j−1 X  =  (j−m−1)  m T33 T34 W43  .  m=0  Then, j−1 X ∂ r+p ∂ r+p (j−m−1) −1 (j) −1 m 0 π ∆ X ≤ p p ∆k πG01 T33 T34 W43 G1 k 33 r r ∂k1 ∂k2 ∂k1 ∂k2 m=0  j+1  j+1 X j−1 j−1 X Cε,r,p Cε,r,p r+p 4 r+p 4 (m + 2) (m + 2)(j + 2) ≤ (j + 2) ≤ |z|R2 9 |z|R2 9 m=0 m=0  j+1 Cε,r,p 1 2 r+p 4 = (j + 2) (j + 3j). |z|R2 9 2  Thus, since G01 ⊂ G03 , πG01  ∂ r+p ∂k1r ∂k2p   ∆−1 k  ∞ X   (j) X33  πG01  ≤  j=1  ≤  Cε,r,p |z|R2  ∞ X j=1  ∞ X j=1  ∂ r+p −1 (j) p πG01 ∆k X33 r ∂k1 ∂k2   j+1 4 1 2 1 (j + 2)r+p (j + 3j) ≤ CCε,r,p , 9 2 |z|R2  where C is an universal constant. Therefore,   ∞ ∂ r+p ∂ r+p  −1 X (j)  R3 (k) = F{d0 }G01 r p ∆k X33 GG01 {d0 } ∂k1r ∂k2p ∂k1 ∂k2 j=1   ∞ ∂ r+p  −1 X (j)  ≤ kF{d0 }G01 k πG01 r p ∆k X33 πG01 kGG01 {d0 } k ∂k1 ∂k2 j=1  ≤ CCε,r,p kf kl1 kgkl1  1 . |z|R2  Finally, combining all the estimates we have 4  X ∂ n+m ∂ n+m (3) α (k) ≤ Rj (k) 0 ∂k1n ∂k2m µ,d ∂k1n ∂k2m j=1  ≤3  C C 4C + 3 ≤ , 2 |z|R |z|R2 |z|R  where C = Cε,Λ,A,q,f,g,m,n is a constant. Set ρε,A,q,m,n := max{ρ1;ε,m,n , ρ2;ε,A,q,m,n }. The proof of the lemma for j = 3 is complete. 112  4.6  The regular piece  Proof of Theorem 3.4.1. Step 1 (defining equation) We first derive a defining equation for the Fermi curve. Without loss of generality we can assume that Â(0) = 0 (see the discussion in §3.1). Let G = {0}, recall that G0 = Γ# \{0}, and consider the region (Tν (0)\∪b∈G0 Tb )\Kρ , where ρ is a constant to be chosen sufficiently large obeying ρ ≥ R. By Proposition 4.2.1(i) we have G0 = {b ∈ Γ# | |Nb (k)| ≥ ε|v|}. To simplify the notation write     [ b Mν := F(A, V ) ∩ Tν (0) \ Kρ ∪ Tb  . b∈Γ# \{0}  By Lemma 4.2.4(i), a point k is in Mν if and only if N0 (k) + D0,0 (k) = 0. By Proposition 4.3.1, if we set w(k) := wν,0 (k) = k1 + i(−1)ν k2 and z(k) := zν,0 (k) = k1 − i(−1)ν k2 , this equation becomes β1 w2 + β2 z 2 + (1 + β3 )wz + β4 w + β5 z + β6 + q̂(0) = 0,  (4.6.1)  where β1 := Jν000 ,  β2 := Jν00 ,  β3 := K 00 ,  β4 := L00 ν0 ,  β5 := L00 ν ,  β6 := M 00 − q̂(0),  00 given by Proposition 4.3.1. Observe that all the coefficients with Jν00 , K 00 , L00 ν and M  β1 , . . . , β6 have exactly the same form as the function Φ0,0 (k) of Lemma 4.4.1(i) (see (4.4.1)). Thus, by this lemma, for 1 ≤ i ≤ 6 we have (1)  βi = βi (j)  where the functions βi (j)  |βi (k)| ≤  (2)  + βi  (3)  + βi ,  (4.6.2)  are analytic in the region under consideration with  C C ≤ j (2|z(k)| − ρ) |z(k)|j  for  1≤j≤2 113  and  (3)  |βi (k)| ≤  C , |z(k)|ρ2  (j)  where C = Cε,Λ,q,A is a constant. The exact expressions for βi  can be easily obtained from  the definitions and from Lemma 4.4.1(i). Substituting (4.6.2) into (4.6.1) and dividing both sides of the equation by z yields (1)  w + β2 z + g = 0,  (4.6.3)  where g :=  β1 w2 β4 w β6 q̂(0) (2) (3) + (β2 + β2 )z + β3 w + + β5 + + z z z z  (4.6.4)  obeys |g(k)| ≤  C ρ  (4.6.5)  with a constant C = Cε,Λ,q,A . Therefore, a point k is in Mν if and only if F (k) = 0, where (1)  F (k) := w(k) + β2 (k) z(k) + g(k) is an analytic function (in the region under consideration). Step 2 (candidates for a solution) Let us now identify which points are candidates to solve the equation F (k) = 0. First observe that, by Proposition 3.2.2(c) we have N1 (0) ∩ N2 (d) = {(iθ1 (d), θ1 (d))} and N1 (d) ∩ N2 (0) = {(iθ2 (d), −θ2 (d))}. Thus, the lines Nν (0) and Nν 0 (d) intersect at 0  Nν (0) ∩ Nν 0 (d) = {(iθν (d), (−1)ν θν (d))}. Hence, the second coordinate of this point and the second coordinate of a point k differ by 0  pr(k) − pr(Nν (0) ∩ Nν 0 (d)) = k2 − (−1)ν θν (d) = k2 + (−1)ν θν (d). Now observe that, if k ∈ Tν (0) ∩ Tν 0 (d) then |k1 + i(−1)ν k2 | < ε  114  and |k2 + (−1)ν θν (d)| = |i(−1)ν k2 + iθν (d)| = i(−1)ν k2 + 12 (i(−1)ν d2 − d1 ) = ≤  1 2 (k1 1 2  + i(−1)ν k2 ) − 21 (k1 + d1 − i(−1)ν (k2 + d2 )  N0,ν (k) − Nd,ν 0 (k) <  ε 2  +  ε 2  = ε.  That is, the second coordinate of k and the second coordinate of Nν (0) ∩ Nν 0 (d) must be apart from each other by at most ε. This gives a necessary condition on the second coordinate of a point k for being in Mν . Conversely, if a point k is in the (ε/4)-tube inside Tν (0), that is, ε |k1 + i(−1)ν k2 | < , 4 and its second coordinate differ from the second coordinate of Nν (0) ∩ Nν 0 (d) by at most ε/4, that is, ε |k2 + (−1)ν θν (d)| < , 4 then |Nd,ν 0 (k)| = N0,ν (k) − 2(k2 + (−1)ν θν (d))| ≤  ε ε + 2 < ε, 4 4  that is, the point k is also in Tν 0 (d) and hence lie in the intersection Tν (0) ∩ Tν 0 (d). This gives a sufficient condition on the first and second coordinates of a point k for being in Tν (0) ∩ Tν 0 (d). For y ∈ C define the set of candidates for a solution of F (k) = 0 as    [ [ Mν (y) := pr−1 (y) ∩ Tν (0) \ Tb  = pr−1 (y) ∩ Tν (0) \ b∈Γ# \{0}   Tν 0 (b) .  b∈Γ# \{0}  Observe that, if |y + (−1)ν θν (b)| ≥ ε for all b ∈ Γ# \ {0} then Mν (y) = pr−1 (y) ∩ Tν (0) = {(k1 , y) ∈ C2 | |k1 + i(−1)ν y| < ε}.  (4.6.6)  On the other hand, if |y + (−1)ν θν (d)| < ε for some d ∈ Γ# \ {0}, then there is at most one such d and consequently Mν (y) = pr−1 (y) ∩ (Tν (0) \ Tν 0 (d)) 2  ν0  ν  = {(k1 , y) ∈ C | |k1 + i(−1) y| < ε and |k1 + d1 + i(−1) (y + d2 )| ≥ ε}.  115  (4.6.7)  Indeed, suppose there is another d0 6= 0 such that |y + (−1)ν θν (d0 )| < ε. Then, |d − d0 | = |2(−1)ν θν (d − d0 )| = |y + (−1)ν θν (d) − (y + (−1)ν θν (d0 ))| ≤ 2ε < 2Λ, which contradicts the definition of Λ. Thus, there is no such d0 6= 0. Step 3 (uniqueness) We now prove that, given k2 , if there exists a solution k1 (k2 ) of F (k1 , k2 ) = 0, then this solution is unique and it depends analytically on k2 . This follows easily using the implicit function theorem and the estimates below, which we prove later. Proposition 4.6.1. Under the hypotheses of Theorem 3.4.1 we have ε C1 + 900 ρ  |F (k) − w(k)| ≤  (a)  and 1 C2 ∂F (k) − 1 ≤ + , 4 ∂k1 7·3 ρ  (b)  where the constants C1 and C2 depend only on ε, Λ, q and A. Now, suppose that (k1 , y) ∈ Mν (y). Then, C2 ∂F 1 (k1 , y) − 1 ≤ + . 4 ∂k1 7·3 ρ Hence, by the implicit function theorem [16, Theorem 3.7.1], by choosing the constant ρ ≥ R sufficiently large, if F (k1∗ , y) = 0 for some (k1∗ , y) ∈ Mν (y), then there is a neighbourhood U × V  ⊂ C2 which contains (k1∗ , y), and an analytic function η : V → U such that  F (k1 , k2 ) = 0 for all (k1 , k2 ) ∈ U × V if and only if k1 = η(k2 ). In particular, this implies that the equation F (k1 , k2 ) = 0 has at most one solution (η(y), y) in Mν (y) for each y ∈ C. We next look for conditions on y to have a solution or have no solution in Mν (y). Step 4 (existence) We first state an improved version of Proposition 4.6.1(a). Proposition 4.6.2. Under the hypotheses of Theorem 3.4.1 we have (1,0)  F (k) − w(k) = β2 where (1,0) β2  (1,1)  + β2  (1,2)  (w(k)) + β2  (k) + h(k),  " # X θν 0 (Â(−b)) θν 0 (Â(b − c)) δb,c + θν (Â(c)) = −2i θν 0 (b) θν 0 (c) 0 b,c∈G1  is a constant that depends only on ρ and A and (1,3)  h := β2  116  + g.  (4.6.8)  Furthermore, 1 ε2 , 100Λ 1 (1,2) |β2 (k)| < 4 3 ε4 , 7 Λ (1,0)  |β2  1 ε3 , 40Λ2 1 |h(k)| ≤ Cε,Λ,q,A . ρ (1,1)  |<  |β2  (k)| <  We now derive conditions for the existence of solutions. Suppose that F (η(y), y) = 0. Then, since η(y) + i(−1)ν y = w(η(y), y) and ε < Λ/6, using the above proposition we obtain |η(y) + i(−1)ν y| = |w(η(y), y)| = |F (η(y), y) − w(η(y), y)| ≤  ε4 C ε3 ε2 C ε2 + + + ≤ + . 2 4 3 100Λ 40Λ 7 Λ ρ 50Λ ρ  Hence, by choosing the constant ρ sufficiently large we find that |η(y) + i(−1)ν y| <  ε2 . 40Λ  In view of (4.6.7), there is no solution in Mν (y) if for some d ∈ Γ# \ {0} we have |y + (−1)ν θν (d)| < ε  and  0  |η(y) + d1 + i(−1)ν (y + d2 )| < ε.  This happens if 1 |y + (−1) θν (d)| ≤ 2 ν    ε2 ε− 40Λ    because in this case 0  |η(y) + d1 + i(−1)ν (y + d2 )| = |η(y) + i(−1)ν y − 2i(−1)ν y + d1 − i(−1)ν d2 | ≤ |η(y) + i(−1)ν y| + 2|y + (−1)ν θν (d)| < ε. Therefore, the image set of pr is contained in ( )   2 ε 1 ε− for all b ∈ Γ# \ {0} . Ω1 := z ∈ C |z + (−1)ν θν (b)| > 2 40Λ On the other hand, in view of (4.6.6), there is a solution in Mν (y) if |y + (−1)ν θν (b)| > ε for all b ∈ Γ# \ {0}. Recall from Proposition 4.4.4(a) that ρ < |v| < 8|k2 |. Thus, the image set of pr contains the set n o Ω2 := z ∈ C 8|z| > ρ and |z + (−1)ν θν (b)| > ε for all b ∈ Γ# \ {0} . Step 5 (conclusion) Summarizing, we have the following biholomorphic correspondence: pr  Mν 3 k −−−−−−−−→ k2 ∈ Ω, pr−1  Mν 3 (η(y), y) ←−−−−−−−−−− y ∈ Ω, 117  where Ω2 ⊂ Ω ⊂ Ω1 and (1,0)  η(y) = −β2 (1,0)  with the constant β2  − i(−1)ν y − r(y),  given by (4.6.8), (1,0)  |β2  |<  ε2 100Λ  and  |r(y)| ≤  ε3 C + . 2 50Λ ρ  This completes the proof of the theorem.  Proof of Propositions 4.6.1 and 4.6.2 We follow the same notation as above. Proof of Proposition 4.6.1. (a) Recall that β2 = Jν00 . First observe that, by Proposition 4.3.1, Lemma 4.4.1, and (4.4.25), we have (1)  β2 (k) = (Jν00 )(1) (k) =  X (1, i(−1)ν ) · Â(−b) Sb,c (1, −i(−1)ν ) · Â(c). N (k) b 0  (4.6.9)  b,c∈G1  Thus, by (4.5.11) and (4.5.16), (1) |β2 (k)|  ≤  √  2 45 √ 4 44 2kÂkl1 2kÂkl1 ≤ Λ(2|z(k)| − R) 44 Λ|z(k)| 45    2ε 63  2 ≤  ε 1 . 900 |z(k)| (4.6.10)  Now recall that 1 |g(k)| ≤ Cε,Λ,q,A . ρ Hence, (1)  |F (k) − w(k)| = |β2 (k)z(k) + g(k)| ≤  ε 1 + Cε,Λ,q,A . 900 ρ  This proves part (a). (b) We first compute   ∂g ∂ β1 w2 β4 w β6 q̂(0) (2) (3) = + (β2 + β3 )z + β3 w + + β5 + + ∂k1 ∂k1 z z z z ! (3) (2) 2 2 ∂β1 w 2wz − w ∂β ∂β3 ∂β2 (2) (3) = + β1 + + 2 z + β2 + β2 + w + β3 (4.6.11) 2 ∂k1 z z ∂k1 ∂k1 ∂k1 +  ∂β4 w z − w ∂β5 ∂β6 1 β6 q̂(0) + β4 + + − 2 − 2 . ∂k1 z z2 ∂k1 ∂k1 z z z 118  Now observe that, since k ∈ Tν (0) \ Kρ we have |w(k)| < ε, 3|v| ≥ |z| and ρ < |v| ≤ |z|. Furthermore, by Lemmas 4.4.1(i), 4.5.1(i) and 4.5.3(i), for 1 ≤ i ≤ 6 and 1 ≤ j ≤ 2, |βi (k)| ≤  C , |z(k)|  (j)  |βi (k)| ≤  C , |z(k)|j  (3)  |βi (k)| ≤ (3)  (j)  ∂βi (k) C ≤ , ∂k1 |z(k)|ρ2  ∂βi (k) C ≤ , ∂k1 |z(k)|j  C ∂βi (k) ≤ , ∂k1 |z(k)|  C , |z(k)|ρ2 (4.6.12)  where C = Cε,Λ,q,A in all cases. Hence, ∂g(k) 1 ≤ Cε,Λ,q,A . ∂k1 ρ  (4.6.13)  By Lemma 4.5.3(i) with f = g = (1, −i(−1)ν ) · Â, we obtain (1)  13 ∂β (k) ≤ |z(k)| 2 k(1, −i(−1)ν ) · Âk2l1 z(k) 2 ∂k1 Λ |z(k)| 26 4ε2 26 · 4 1 1 26 ≤ < . ≤ 2 kÂk2l1 ≤ 2 2 2 2 Λ Λ (63) (63) 6 7 · 34  (4.6.14)  Therefore, ∂F ∂ ∂ (1) (k) − 1 = (F (k) − w(k)) = (β (k)z(k) + g(k)) ∂k1 ∂k1 ∂k1 2 (1)  1 ∂g ∂β2 1 (1) (k)z(k) + β2 (k) + (k) ≤ + Cε,Λ,q,A . = 4 ∂k1 ∂k1 7·3 ρ This proves part (b) and completes the proof of the proposition. Here is the proof of Proposition 4.6.2. Proof of Proposition 4.6.2. First observe that 1 0 0 (1, i(−1)ν ) · A = A1 + i(−1)ν A2 = A1 − i(−1)ν A2 = −2i (iA1 + (−1)ν A2 ) = −2iθν 0 (A). 2 Thus, recalling (4.6.9), (1)  β2 (k) = (Jν00 )(1) (k) =  X 2iθν 0 (Â(−b)) Sb,c 2iθν (Â(c)). Nb (k) 0  b,c∈G1  Now, by Lemma 4.4.2 we have (1)  (1,0)  z(k)β2 (k) = β2 where (1,0) β2  (1,1)  + β2  (1,2)  (w(k)) + β2  (1,3)  (k) + β3  (k),  " # X θν 0 (Â(−b)) θν 0 (Â(b − c)) = −2i δb,c + θν (Â(c)) θν 0 (b) θν 0 (c) 0 b,c∈G1  119  and (1,3)  |β3  (k)| ≤ CΛ,A  1 1 < CΛ,A . |z(k)| ρ  Hence, (1)  (1,0)  F (k) − w(k) = z(k)β2 (k) + g(k) = β2 (1,3)  with h := β3  (1,1)  + β2  (1,2)  (w(k)) + β2  (k) + h(k)  + g. Furthermore, in view of (4.6.5), (1,3)  |h(k)| ≤ |β3  1 (k)| + |g(k)| < Cε,Λ,q,A . ρ  This proves the first part of the proposition. Finally, by (4.4.40), since kÂkl1 < 2ε/63 and ε < Λ/6, we find that (1,0) |β2 |   1 ≤ 1+ 2Λ  1 1+ ≤ 2Λ   1 kθν 0 (Â)kl1 k2iθν 0 (Â)kl1 k2iθν (Â)kl1 2Λ  4 1 1 2ε 4kÂk2l1 ≤ kÂk2l1 < ε2 2Λ 63 Λ 100Λ   ε ≤ 2 1+ Λ  ε ≤ 2 1+ Λ   7 kθν 0 (Â)kl1 k2iθν 0 (Â)kl1 k2iθν (Â)kl1 6Λ  7 2ε 8 1 4kÂk2l1 ≤ 2 εkÂk2l1 < ε3 6Λ 63 Λ 40Λ2  and (1,1) |β2 |  and (1,2)  |β2  |≤  256 1 64 kθν 0 (Â)k2l1 k2iθν 0 (Â)kl1 k2iθν (Â)kl1 ≤ 3 kÂk4l1 < 4 3 ε4 . Λ3 Λ 7 Λ  This completes the proof.  4.7  The handles  Proof of Theorem 3.4.2. Step 1 (defining equation) We first derive a defining equation for the Fermi curve. Without loss of generality we can assume that Â(0) = 0. Let G = {0, d}, recall that G0 = Γ# \ {0, d}, and consider the region (Tν (0) ∩ Tν 0 (d)) \ Kρ , where ρ is a constant to be chosen sufficiently large obeying ρ ≥ R. Observe that, this requires d being sufficiently large for (Tν (0) ∩ Tν 0 (d)) \ Kρ being not empty. In fact, by Proposition 4.4.4(ii), for k in this region we have ρ < |v| ≤ 2|d|.  120  Now, recall from Proposition 4.2.1(ii) that G0 = {b ∈ Γ# | |Nb (k)| ≥ ε|v|}, and to simplify the notation write b Hν := F(A, V ) ∩ (Tν (0) ∩ Tν 0 (d)) \ Kρ . By Lemma 4.2.4(ii), a point k is in Hν if and only if (N0 (k) + D0,0 (k))(Nd (k) + Dd,d (k)) − D0,d (k)Dd,0 (k) = 0.  (4.7.1)  Define w1 (k) := wν,0 = k1 + i(−1)ν k2 , z1 (k) := zν,0 = k1 − i(−1)ν k2 ,  (4.7.2)  0  w2 (k) := wν 0 ,d = k1 + d1 + i(−1)ν (k2 + d2 ), 0  z2 (k) := zν 0 ,d = k1 + d1 − i(−1)ν (k2 + d2 ). Note that, by Proposition 4.4.4(ii), |v| ≤ |z1 | ≤ 3|v|,  |v| ≤ |z2 | ≤ 3|v|  and  |d| ≤ |z2 | ≤ 2|d|.  By Proposition 4.3.1, N0 + D0,0 = β1 w12 + β2 z12 + (1 + β3 )w1 z1 + β4 w1 + β5 z1 + β6 + q̂(0),  (4.7.3)  Nd + Dd,d = η1 w22 + η2 z22 + (1 + η3 )w2 z2 + η4 w2 + η5 z2 + η6 + q̂(0), where β1 := Jν000 ,  β2 := Jν00 ,  β3 := K 00 ,  β4 := L00 ν0 ,  β5 := L00 ν ,  β6 := M 00 − q̂(0),  η1 := Jνdd ,  η2 := Jνdd0 ,  η3 := K dd ,  η4 := Ldd ν ,  η5 := Ldd ν0 ,  η6 := M dd − q̂(0),  and  0 0  0 0  0 0  0 0  with Jνd d , K d d , Ldν d and M d d given by Proposition 4.3.1. Observe that all the coefficients β1 , . . . , β6 and η1 , . . . , η6 have exactly the same form as the function Φd0 ,d0 (k) of Lemma 4.4.1(ii) (see (4.4.1)). Thus, by this lemma, for 1 ≤ i ≤ 6 we have (1)  + βi  (1)  + ηi  βi = βi ηi = ηi  (2)  + βi ,  (2)  + ηi ,  121  (3)  (3)  (4.7.4)  (j)  where the functions βi  (j)  and ηi  are analytic in the region under consideration with  C C ≤ j (2|z1 (k)| − ρ) |z1 (k)|j C C (j) |ηi (k)| ≤ ≤ j (2|z2 (k)| − ρ) |z2 (k)|j (j)  |βi (k)| ≤  for  1≤j≤2  and  for  1≤j≤2  and  C , |z1 (k)|ρ2 C (3) |ηi (k)| ≤ , |z2 (k)|ρ2 (3)  |βi (k)| ≤  (j)  where C = Cε,Λ,q,A is a constant. The exact expressions for βi  (j)  and ηi  can be easily  obtained from the definitions and from Lemma 4.4.1(ii). Substituting (4.7.4) into (4.7.3) yields  where  1 (1) (N0 + D0,0 ) = w1 + β2 z1 + g1 , z1 1 (1) (Nd + Dd,d ) = w2 + η2 z2 + g2 , z2  (4.7.5)  β1 w12 β 4 w1 β6 q̂(0) (2) (3) + (β2 + β2 )z1 + β3 w1 + + β5 + + , z1 z1 z1 z1 η4 w2 η6 q̂(0) η1 w22 (2) (3) + (η2 + η2 )z2 + η3 w2 + + η5 + + g2 := z2 z2 z2 z2  (4.7.6)  g1 :=  obey |g1 (k)| ≤  C ρ  |g2 (k)| ≤  and  C ρ  (4.7.7)  with a constant C = Cε,Λ,q,A . This gives us more information about the first term in (4.7.1). We next consider the second term in that equation. Write D0,d = c1 (d) + p1  and  Dd,0 = c2 (d) + p2  (4.7.8)  with c1 (d) := q̂(−d) − 2d · Â(−d),  p1 := D0,d − q̂(−d) + 2d · Â(−d),  c2 (d) := q̂(d) + 2d · Â(d),  p2 := Dd,0 − q̂(d) − 2d · Â(d).  We shall shortly prove the following estimates. Proposition 4.7.1. Under the hypotheses of Theorem 3.4.2 we have, for any integers n and m with n + m ≥ 0 and for 1 ≤ j ≤ 2, ∂ n+m C1 pj (k) ≤ n m ∂k1 ∂k2 |d|  and  |cj (d)| ≤  where the constants C1 and C2 depend only on ε, Λ, q and A.  122  C2 , |d|  Thus, by dividing both sides of (4.7.1) by z1 z2 and substituting (4.7.5) and (4.7.8) we find that  1  (N0 + D0,0 )(Nd + Dd,d ) − D0,d Dd,0 z1 z 2 1 (1) (1) = (w1 + β2 z1 + g1 )(w2 + η2 z2 + g2 ) − (c1 (d) + p1 )(c2 (d) + p2 ). z1 z2  0=  (4.7.9)  We now introduce a (nonlinear) change of variables in C2 . Set (1)  x1 (k) := w1 (k) + β2 (k) z1 (k) + g1 (k), x2 (k) := w2 (k) +  (1) η2 (k) z2 (k)  (4.7.10)  + g2 (k).  We shall prove below that this transformation satisfies the following estimates. Proposition 4.7.2. Under the hypotheses of Theorem 3.4.2 we have: (i) For 1 ≤ j ≤ 2 and for ρ sufficiently large, |xj (k) − wj (k)| ≤  ε C ε + < . 900 ρ 8  (ii)     ∂x1  ∂k1  ∂x1 ∂k2     ∂k1 ∂x2   ∂x2 ∂k1  ∂x2 ∂k2   =  1  i(−1)ν  1  0 i(−1)ν    (I + M )  and ∂k1  ∂x1 ∂k2 ∂x1      ∂k2 ∂x2  =  1  1 2 i(−1)ν 0    1 i(−1)ν   (I + N )  with kM k ≤  4 C 1 + < 4 7·3 ρ 2  and  kN k ≤ 4kM k.  Furthermore, for all m, i, j ∈ {1, 2}, ∂ 2 km 3 C ≤ 3 ε2 + . ∂xi ∂xj Λ ρ Here, all the constants C depend only on ε, Λ, q and A. By the inverse function theorem, these estimates imply that the above transformation is invertible. Therefore, by rewriting the equation (4.7.9) in terms of these new variables, we conclude that a point k is in Hν if and only if x1 (k) and x2 (k) satisfy the equation x1 x2 + r(x1 , x2 ) = 0, 123  (4.7.11)  where r(x1 , x2 ) := −  1 (c1 (d) + p1 )(c2 (d) + p2 ). z1 z 2  This is the defining equation we shall study in detail below. In order to do this we need some estimates. Step 2 (estimates) Using the above inequalities we have, for i, j, l ∈ {1, 2}, 2 X ∂pj ∂km C ∂ pj (k(x)) ≤ ≤ ∂xi ∂km ∂xi |d| m=1  and  2 X ∂2 pj (k(x)) ≤ ∂xi ∂xl  m,n=1  2 X ∂ 2 pj ∂km ∂kn ∂pj ∂ 2 km C , + ≤ ∂km ∂kn ∂xi ∂xl ∂km ∂xi ∂xl |d| m=1  so that |r(x)| ≤ C  1 1 1 C ≤ 4, |d|2 |d| |d| |d|  1 1 1 C ∂ 1 1 1 +C 2 ≤ 4 r(x) ≤ C 3 ∂xi |d| |d| |d| |d| |d| |d| |d| and ∂2 C r(x) ≤ 4 . ∂xi ∂xj |d| Here, all the constants depend only on ε, Λ, q and A. Step 3 (Morse lemma) We now apply the quantitative Morse lemma in §4.8 for studying the equation (4.7.11). We consider this lemma with C , |d|4  a=b=  and d sufficiently large so that b < max{ 23  δ = ε,  1 ε 55 , 4 }.  Observe that, under this condition we  have (δ − a)(1 − 19b) >  ε 2  ε (δ − a)(1 − 55b) > . 4  and  According to this lemma, there is a biholomorphism Φν defined on n Ω1 := (z1 , z2 ) ∈ C2  |z1 | <  ε εo and |z2 | < 2 2  with range containing n (x1 , x2 ) ∈ C2  |x1 | <  124  ε εo and |x2 | < 4 4  (4.7.12)  such that kDΦν − Ik ≤  C , |d|2  ((x1 x2 + r) ◦ Φν )(z1 , z2 ) = z1 z2 + td , (4.7.13)  C , |d|4 C |Φν (0)| ≤ 4 , |d| |td | ≤  where DΦν is the derivative of Φν and td is a constant that depends on d. Hence, if for ν = 1 we define φd,1 : Ω1 −→ T1 (0) ∩ T2 (d) as φd,1 (z1 , z2 ) := (k1 (Φ1 (z1 , z2 )), k2 (Φ1 (z1 , z2 ))), where k(x) is the inverse of the transformation (4.7.10), we obtain the desired map. Note that the conclusion (ii) of the theorem is immediate. We next prove (i) and (iii). Step 4 (proof of (i)) By Proposition 4.7.2(i), for 1 ≤ j ≤ 2, ε |xj (k) − wj (k)| ≤ . 8 Now, recall from (4.7.2) the definition of w1 (k) and w2 (k). Then, since |xj (k)| ≤ |xj (k) − wj (k)| + |wj (k)| <  ε + |wj (k)|, 8  the set n (k1 , k2 ) ∈ C2  |w1 (k)| <  ε εo and |w2 (k)| < 8 8  is contained in the set (4.7.12). This proves the first part of (i). To prove the second part we use Proposition 4.7.2 and (4.7.13). First observe that   ∂k 1 1 1  Dφd,1 = DΦ1 = (I + N )(I + DΦ1 − I) ∂x 2 i −i   1 1 1  (I + N + R), =  2 i −i where kN k ≤  1 C + 3 3 ρ  and  125  kRk ≤  C . |d|2  Furthermore, from (4.7.2) and (4.7.10) we have 1 1 (1) (1) k1 = iθν (d) + (w1 + w2 ) = iθν (d) + (x1 + x2 + β2 z1 + η2 z2 + g1 + g2 ), 2 2 (−1)ν k2 = −(−1)ν θν (d) + (w1 − w2 ) 2i (−1)ν (1) (1) = −(−1)ν θν (d) + (x1 − x2 − β2 z1 + η2 z2 − g1 + g2 ), 2i so that       ε  1 1 ν = (iθν (d), −(−1) θν (d)) + O . +O φd,1 (0) = k(Φ1 (0)) = k O 4 |d| 900 ρ b Step 5 (proof of (iii)) To prove part (iii) it suffices to note that T1 (0) ∩ T2 (d) ∩ F(A, V) b is mapped to T1 (−d) ∩ T2 (0) ∩ F(A, V ) by translation by d and define φd,2 by φd,2 (z1 , z2 ) := φd,1 (z2 , z1 ) + d. This completes the proof of the theorem.  Proof of Propositions 4.7.1 and 4.7.2 We follow the same notation as above. Proof of Proposition 4.7.1. It suffices to estimate cd0 ,d00 := q̂(d0 − d00 ) − 2(d0 − d00 ) · Â(d0 − d00 )  pd0 ,d00 := Dd0 ,d00 − cd0 ,d00  and  for d0 , d00 ∈ {0, d} with d0 6= d00 . Define 0 00  lνd d := (1, i(−1)ν ) · Â(d0 − d00 ). Observe that, since |q̂(d0 − d00 )| =  |d0  X 1 1 1 |d0 − d00 |2 |q̂(d0 − d00 )| ≤ 0 |b|2 |q̂(b)| ≤ kb2 q̂(b)kl1 2 , 00 2 00 2 −d | |d − d | |d| # b∈Γ  and similarly |Â(d0 − d00 )| ≤ kb2 Â(b)kl1  1 , |d|2  it follows that |cd0 ,d00 | ≤  CA,q |d|  and  This gives the desired bounds for c1 and c2 . 126  0 00  |lνd d | ≤  CA . |d|2  Now, by Proposition 4.3.1 we have 0 00  0 00  0 00  0 00  0 00  0 00  0 00  0 00  2 dd 2 p = Jνd0 d wν,d zν,d0 + K d d wν,d zν,d0 + (L̃dν 0d − lνd0 d )wν,d0 + (L̃dν d − lνd d )zν,d0 + M̃ d d 0 + Jν  with 0 00  0 00  0 00  L̃dν d := Ldν d + lνd d 0 00  0 00  0 00  0 00  M̃ d d := M d d − c.  and 0 00  0 00  Observe that all the coefficients Jνd d , K d d , L̃dν d and M̃ d d have exactly the same form as the function Φd0 ,d00 (k) of Lemma 4.5.2 (see Proposition 4.3.1 and (4.4.1)). Thus, by this lemma with β = 2, for any integers n and m with n + m ≥ 0, the absolute value of the ∂ n+m ∂k1n ∂k2m -derivative  of each of these functions is bounded above by Cε,Λ,A,q,m,n  1 . |d|3  Hence, if we recall from Proposition 4.4.4(ii) that |z1 (k)| ≤ 6|d|  and  |z2 (k)| ≤ 2|d|,  and apply the Leibniz rule we find that ∂ n+m C . pd0 ,d00 (k) ≤ Cm,n ∂k1n ∂k2m |d| This yields the desired bounds for p1 and p2 and completes the proof. We now prove Proposition 4.7.2. Proof of Proposition 4.7.2. (i) Similarly as in (4.6.10) we have (1)  |β2 (k)| ≤  1 ε 900 |z1 (k)|  and  (1)  |η2 (k)| ≤  ε 1 . 900 |z2 (k)|  Thus, in view of (4.7.7), and by choosing ρ sufficiently large, ε C ε + < , 900 ρ 8 ε C ε (1) |x2 (k) − w2 (k)| ≤ |η2 (k) z2 (k) + g2 (k)| ≤ + < . 900 ρ 8 (1)  |x1 (k) − w1 (k)| ≤ |β2 (k) z1 (k) + g1 (k)| ≤  This proves part (i). (ii) Recall (4.7.2) and (4.7.10). Then, for 1 ≤ j ≤ 2, (1)  ∂x1 ∂ ∂w1 ∂β ∂z1 (1) ∂g1 (1) = (w1 + z1 β2 + g1 ) = + z1 2 + β + , ∂kj ∂kj ∂kj ∂kj ∂kj 2 ∂kj (1)  ∂x2 ∂ ∂w2 ∂η ∂z2 (1) ∂g2 (1) = (w2 + z2 η2 + g2 ) = + z2 2 + η + . ∂kj ∂kj ∂kj ∂kj ∂kj 2 ∂kj 127  To estimate the terms above on the right hand side we use a calculation that we have already done in the proof of Proposition 4.6.1. First observe that the functions g1 and g2 are similar to the function g (see (4.7.6) and (4.6.4)). Thus, it is easy to see that ∂g2 ∂kj  ∂g1 ∂kj  and  are given by expressions similar to (4.6.11). Since k ∈ Tν (0) ∩ Tν 0 (d) we have |w1 (k)| < ε  and |w2 (k)| < ε. Recall also the inequalities in Proposition 4.4.4(ii). Hence, by Lemmas 4.4.1(ii), 4.5.1(ii) and 4.5.3(ii), we obtain (4.6.12) with k1 and z(k) replaced by kj and z1 (k), respectively, and for k1 , z(k) and β replaced by kj , z2 (k) and η, respectively. Consequently, similarly as in (4.6.13) and using again Lemma 4.4.1(ii), for 1 ≤ j ≤ 2 we have ∂z1 (1) ∂g1 1 β2 + ≤ Cε,Λ,q,A ∂kj ∂kj ρ  and  1 ∂z2 (1) ∂g2 η2 + ≤ Cε,Λ,q,A . ∂kj ∂kj ρ  Now recall that β2 = Jν00 and η2 = Jνdd0 . Then, by Proposition 4.3.1, Lemma 4.4.1(ii), and (4.4.25), it follows that (1)  β2 (k) = (Jν00 )(1) (k) =  X (1, i(−1)ν ) · Â(−b) Sb,c (1, −i(−1)ν ) · Â(c), N (k) b 0  b,c∈G1 (1)  η2 (k) = (Jνdd0 )(1) (k) =  X (1, i(−1)ν 0 ) · Â(d − b) 0 Sb,c (1, −i(−1)ν ) · Â(c − d). N (k) b 0  b,c∈G1  Hence, by Lemma 4.5.3(ii), similarly as in (4.6.14), for 1 ≤ j ≤ 2, (1)  z1 (k)  ∂β2 (k) 1 13 , ≤ 2 k(1, −i(−1)ν ) · Âk2l1 < ∂kj Λ 7 · 34  z2 (k)  13 1 ∂η2 (k) 0 ≤ 2 k(1, −i(−1)ν ) · Âk2l1 < . ∂kj Λ 7 · 34  (1)  Therefore,   ∂x1  ∂k1 ∂x2 ∂k1    ∂x1 ∂k2  ∂x2 ∂k2      (1) (1) ∂β2 (k) ∂β2 (k) 1 i(−1)ν z1 (k) ∂k z (k) 1 ∂k2  1 + = (1) (1) 0 ∂η (k) ∂η 2 2 (k) 1 i(−1)ν z2 (k) ∂k z (k) 2 ∂k2 1     (1) (1) ∂g1 ∂g1 β2 −i(−1)ν β2  +  ∂k1 ∂k2  + 0 (1) (1) ∂g2 ∂g2 η2 −i(−1)ν η2 ∂k1 ∂k2   1 i(−1)ν  (I + M1 + M2 + M3 ), =:  0 1 i(−1)ν  where kM1 k ≤ 2  2 7 · 34  1 kM2 + M3 k ≤ Cε,Λ,q,A . ρ  and  Set M := M1 + M2 + M3 . This proves the first claim. 128  Now, by choosing ρ sufficiently large we can make 1 kM k < . 2 Write    1 i(−1)ν . P :=  0 1 i(−1)ν  Then, by the inverse function theorem and using the Neumann series,   ∂k1  ∂x1 ∂k2 ∂x1    ∂k1 ∂x2  ∂k2 ∂x2   =  ∂x1  ∂k1 ∂x2 ∂k1  −1  ∂x1 ∂k2  ∂x2 ∂k2  = (I + M )−1 P −1 = (I + M̃ )P −1   =: P −1 (I + P M̃ P −1 ) =  1  1 2 i(−1)ν 0  1 i(−1)ν    (I + P M̃ P −1 ),  with kP M̃ P −1 k ≤ 2kM̃ k1 ≤  2kM k ≤ 4kM k. 1 − kM k  Set N := P M̃ P −1 . This proves the second claim. Differentiating the matrix identity T T −1 = I and applying the chain rule we find that   2 2 X X ∂km ∂ ∂km ∂ 2 xl ∂kr ∂kp ∂ 2 km ∂xl ∂kp =− =− . ∂xi ∂xj ∂xl ∂xi ∂kp ∂xj ∂xl ∂kr ∂xp ∂xi ∂xj l,p=1  l,p=1  Furthermore, in view of the above calculations we have ∂ki 1 1 ≤ (1 + kN k) ≤ (1 + 4kM k) ∂xj 2 2   1 1 3 ≤ 1+4 < . 2 2 2 Thus, ∂ 2 km ≤4 ∂xi ∂xj   3 ∂ 2 xl 3 sup . 2 l,r,p ∂kr ∂xp  We now estimate (1)  (1)  (1)  (1)  (1)  (1)  ∂ 2 x1 ∂z1 ∂β2 ∂ 2 β2 ∂z1 ∂β2 ∂ 2 g1 = + z1 + + ∂ki ∂kj ∂ki ∂kj ∂ki ∂kj ∂kj ∂ki ∂ki ∂kj and ∂ 2 x2 ∂z2 ∂η2 ∂ 2 η2 ∂z2 ∂η2 ∂ 2 g2 = + z2 + + . ∂ki ∂kj ∂ki ∂kj ∂ki ∂kj ∂kj ∂ki ∂ki ∂kj  129  From (4.6.11) with g, w and z replaced by g1 , w1 and z1 , respectively, we obtain (2)  (3)  ∂ 2 g1 ∂ 2 β1 w12 2z12 − 6w1 z1 + 4w12 ∂ 2 β2 ∂ 2 β2 ∂β1 2w1 z1 − w12 = + β + + + 2 1 ∂k1 ∂k12 ∂k12 z1 z12 z13 ∂k12 ∂k12 ! (2) (3) ∂β2 ∂β ∂ 2 β3 ∂β4 z1 − w1 ∂β3 ∂ 2 β4 w1 +2 + 2 + + +2 w + 2 1 2 2 ∂k1 ∂k1 ∂k1 ∂k1 ∂k1 ∂k1 z1 z12 + β4  ! z1  β6 2q̂(0) 2(w1 − z1 ) ∂ 2 β5 ∂ 2 β6 1 ∂β6 1 + + +2 3 + 3 . −2 3 2 2 2 ∂k1 z1 z1 ∂k1 ∂k1 z1 z1 z1  Hence, by Lemmas 4.4.1(ii), 4.5.1(ii) and 4.5.3(ii), ∂ 2 g1 1 ≤ Cε,Λ,q,A . 2 ρ ∂k1 Similarly we prove that ∂ 2 gl 1 ≤ Cε,Λ,q,A ∂ki ∂kj ρ for all l, i, j ∈ {1, 2} because all the derivatives acting on gl are essentially the same (up to constant factors). Indeed, ∂ gl = ∂ki    ∂wl ∂ ∂zl ∂ + ∂ki ∂wl ∂ki ∂zl   gl  with   ∂w1  ∂k1 ∂w2 ∂k1    ∂w1 ∂k2  ∂w2 ∂k2   =  i(−1)ν  1  ν0  1 i(−1)       ∂z1  ∂k1  and  ∂z2 ∂k1    ∂z1 ∂k2  ∂z2 ∂k2   =  1  −i(−1)ν  1  0 −i(−1)ν  In particular this implies ∂zi = 1. ∂kj Furthermore, again by Lemma 4.5.3(ii), (1)  ∂β2 ∂kj  ≤ Cε,Λ,q,A  (1)  1 , ρ  ∂η2 ∂kj  1 ≤ Cε,Λ,q,A , ρ  and (1)  ∂ 2 β2 (k) 65 1 2 z1 (k) ≤ 3 k(1, −i(−1)ν ) · Âk2l1 < ε , ∂k1 ∂kj Λ 5Λ3 (1)  z2 (k)  ∂ 2 η2 (k) 65 1 2 0 ≤ 3 k(1, −i(−1)ν ) · Âk2l1 < ε . ∂ki ∂kj Λ 5Λ3  Hence, ∂ 2 xl 1 2 1 ≤ ε + Cε,Λ,q,A . 3 ∂ki ∂kj 5Λ ρ Therefore,  3 3 1 ∂ 2 km ∂ 2 xl 3 ≤4 sup ≤ 3 ε2 + Cε,Λ,q,A . ∂xi ∂xj 2 Λ ρ l,r,p ∂kr ∂xp This completes the proof of the proposition. 130   .  4.8  Quantitative Morse lemma  In this section we prove a quantitative Morse lemma from [4] that is used above for proving Theorem 3.4.2. Lemma 4.8.1 (Quantitative Morse lemma [4]). Let δ be a constant with 0 < δ < 1 and assume that f (x1 , x2 ) = x1 x2 + r(x1 , x2 ) is an holomorphic function on Dδ = {(x1 , x2 ) ∈ C2 | |x1 | ≤ δ and |x2 | ≤ δ}. Suppose further that, for all x ∈ Dδ and 1 ≤ i ≤ 2 the function r satisfies   ∂r (x) ≤ a < δ ∂xi  and   ∂2r 1 (x) ≤b< , ∂xi ∂xj 55 i,j∈{1,2}  where a and b are constants. Then f has a unique critical point ξ = (ξ1 , ξ2 ) ∈ Dδ with |ξ1 | ≤ a  and  |ξ2 | ≤ a.  Furthermore, let s = max{|ξ1 |, |ξ2 |}. Then there is a biholomorphic map Φ from the domain D(δ−s)(1−19b) to a neighbourhood of ξ ∈ Dδ that contains {(z1 , z2 ) ∈ C2 | |zi − ξi | < (δ − s)(1 − 55b) for 1 ≤ i ≤ 2} such that (f ◦ Φ)(z1 , z2 ) = z1 z2 + c, where c ∈ C is a constant fulfilling |c − r(0, 0)| ≤ a2 . The differential DΦ obeys kDΦ − Ik ≤ 18b. If  ∂r ∂x1 (0, 0)  = 0 and  ∂r ∂x2 (0, 0)  = 0, then ξ = 0 and s = 0.  Proof. Step 1 For 1 ≤ i ≤ 2 set  Ci :=  (x1 , x2 ) ∈ Dδ  131   ∂f (x) = 0 . ∂xi  To prove the first claim we show that C1 and C2 have a unique point of intersection. For each x1 with |x1 | ≤ δ consider the functions u(x2 ) := x2  v(x2 ) :=  and  ∂r (x1 , x2 ) ∂x1  on the domain Ω := {x2 ∈ C | |x2 | ≤ δ} with boundary ∂Ω := {x2 ∈ C | |x2 | = δ}. By hypothesis, for all x2 ∈ ∂Ω we have |u(x2 )| = δ  |v(x2 )| ≤ a < δ,  and  so that |u(x2 )| > |v(x2 )|  and  u(x2 ) 6= 0.  Thus, by Rouche’s theorem [13, Theorem 10.43], the functions u(x2 ) and u(x2 ) + v(x2 ) have the same number of zeros in Ω. Since u(x2 ) has only one zero, the equation 0 = u(x2 ) + v(x2 ) = x2 +  ∂r ∂f (x1 , x2 ) = (x1 , x2 ) ∂x1 ∂x1  has a unique solution x̃2 (x1 ). By the implicit function theorem, this solution is an holomorphic function of x1 . Furthermore, differentiating with respect to x1 the identity x̃2 (x1 ) = −  ∂r (x1 , x̃2 (x1 )) ∂x1  we find that  −1 ∂ x̃2 (x1 ) ∂2r ∂2r = − 2 (x1 , x̃2 (x1 )) 1 + (x1 , x̃2 (x1 )) . ∂x1 ∂x2 ∂x1 ∂x1  (4.8.1)  Hence, using the hypotheses, |x̃2 (x1 )| ≤ a  ∂ x̃2 (x1 ) b ≤ . ∂x1 1−b  and  Similarly we can parametrize the curve C2 by a map x2 7→ (x̃1 (x2 ), x2 ) satisfying |x̃1 (x2 )| ≤ a  ∂ x̃1 (x2 ) b ≤ . ∂x2 1−b  and  Therefore, the curves C1 and C2 intersect in a unique point (ξ1 , ξ2 ) with x̃1 (ξ2 ) = ξ1 and x̃2 (ξ1 ) = ξ2 . Thus we have |ξ1 | ≤ a and |ξ2 | ≤ a. This proves the first claim. Step 2 Without loss of generality we may assume that r(0, 0) = 0. (If r(0, 0) 6= 0 then instead of f consider f0 (x1 , x2 ) = x1 x2 +r0 (x1 , x2 ) with f0 := f −r(0, 0) and r0 := r−r(0, 0).) Define f˜(x1 , x2 ) := f (x1 + ξ1 , x2 + ξ2 ) 132  and r̃(x1 , x2 ) := x1 ξ2 + ξ1 x2 + ξ1 ξ2 + r(x1 + ξ1 , x2 + ξ2 ). Then, f˜(x1 , x2 ) = x1 x2 + r̃(x1 , x2 ), where r̃ is an holomorphic function on Dδ−s with s := max{|ξ1 |, |ξ2 |}. Since (ξ1 , ξ2 ) is a critical point of f , we have r̃(0, 0) = ξ1 ξ2 + r(ξ1 , ξ2 ), ∂ r̃ (0, 0) = ξ2 + ∂x1 ∂ r̃ (0, 0) = ξ1 + ∂x2  ∂r (ξ1 , ξ2 ) = 0, ∂x1 ∂r (ξ1 , ξ2 ) = 0. ∂x2  Thus, if we translate the system of coordinates in C2 so that (ξ1 , ξ2 ) is mapped to (0, 0), we obtain ∂ r̃ (0, 0) = 0 ∂x1  r̃(0, 0) = r(0, 0) = 0,  and  ∂ r̃ (0, 0) = 0. ∂x2  Furthermore, since ∂ 2 r̃ ∂2r = , ∂xi ∂xj ∂xi ∂xj we still have the bound   ∂ 2 r̃ (x) ∂xi ∂xj   ≤ b. i,j∈{1,2}  This shows that it suffices to prove the lemma in the special case that r(0, 0) =  ∂r ∂r (0, 0) = (0, 0) = 0 ∂x1 ∂x2  and then replace δ by δ − s. Step 3 For 0 ≤ t ≤ 1 set ft (x1 , x2 ) := x1 x2 + t r(x1 , x2 ). Below we construct a t-dependent vector field X t (x) that is holomorphic on Dδ(1−4b) and satisfies r(x) + (∇ft )(x) · X t (x) = 0,  (4.8.2)  kX t (x)k ≤ 5b(|x1 | + |x2 |) ≤ 10bδ,  (4.8.3)  ∂X t (x) ≤ 8b ∂xi 133  for 1 ≤ i ≤ 2.  (4.8.4)  Now, for 0 ≤ τ ≤ 1 consider the map Φτ : Dδ(1−10b) −→ C2 , where Φτ (x) is the solution of the initial value problem d Φτ (x) = X τ (Φτ (x)) dτ  for  0 ≤ τ ≤ 1,  Φ0 (x) = x. This solution exists because X τ (x) is holomorphic. Consequently, in view of (4.8.2), Φ0 (x) = x  and  d d fτ (Φτ (x)) = r(Φτ (x)) + (∇fτ )(Φτ (x)) · Φτ (x) = 0. dτ dτ  Furthermore, by (4.8.3) we have Φτ (x) d Φτ (x) d kΦτ (x)k = · Φτ (x) = · X τ (Φτ (x)) ≤ kX τ (Φτ (x))k ≤ 10bkΦτ (x)k, dτ kΦτ (x)k dτ kΦτ (x)k so that kΦτ (x)k ≤ e10bτ kΦ0 (x)k = e10bτ kxk. This implies d Φτ (x) = kX τ (Φτ (x))k ≤ 10bkΦτ (x)k ≤ 10be10b kxk. dτ Hence, after integrating with respect to τ , kΦτ (x) − xk ≤ 10be10b kxk ≤ 15bδ. This shows that Φt (x) remains in the domain of X t (x) for all 0 ≤ t ≤ 1 and x ∈ Dδ(1−19b) . Now observe that d (Dx Φτ )(x) = Dx (X τ (Φτ (x))) = (Dx X τ )(Φτ (x))(Dx Φτ )(x), dτ so that, by (4.8.4), d d k(Dx Φτ )(x)k ≤ (Dx Φτ )(x) dτ dτ ≤ k(Dx X τ )(Φτ (x))k k(Dx Φτ )(x)k ≤  √  2 8bk(Dx Φτ )(x)k.  Consequently, k(Dx Φτ )(x)k ≤ e12b k(Dx Φ0 )(x)k = e12b , d (Dx Φτ )(x) ≤ 18b dτ 134  and Z kDx Φt − Ik ≤ 0  t  d (Dx Φτ )(x) dτ ≤ 18b < 1. dτ  Therefore, by the inverse function theorem, the map Φt is biholomorphic into its image. If we set Φ := Φ1 , then Φ has the desired properties because Z f1 ◦ Φ1 = f1 ◦ Φ1 − f0 + f0 = 0  1  d fτ (Φτ ) dτ + f0 = f0 dτ  and, by the contraction mapping theorem, the image of Dδ(1−19b) under Φ contains Bδ0 with 1 − 18b (1 − 19b)δ ≥ (1 − 55b)δ. 1 + 18b  δ0 =  Step 4 To construct X t (x) observe that the equation (4.8.2) is * r(x1 , x2 ) +    + ∂r t (x) x2 + t ∂x X (x) 1   ,  1  = 0. ∂r X2t (x) x1 + t ∂x2 (x)  (4.8.5)  By the assumptions on r, for 1 ≤ i ≤ 2 we have ∂r (x) ≤ b(|x1 | + |x2 |) ≤ 2bδ. ∂xi  (4.8.6)  Hence, |r(x)| ≤ b(|x1 | + |x2 |)2 < 4bδ 2 < 4bδ. Thus, since b < 1/55, by the inverse function theorem, for 0 ≤ t ≤ 1 the map Pt : Dδ −→ C2   ∂r ∂r (x), x1 + t (x) (x1 , x2 ) 7−→ x2 + t ∂x1 ∂x2 is biholomorphic into its image, and the image contains Dδ(1−2b) . Furthermore,   2 0 1  = t ∂ r Dx Pt −  ∂xi ∂xj 1 0 and  ≤ tb    0 1  ≤ tb . Dx Pt−1 −  1 − tb 1 0  (4.8.7)  (4.8.8)  The last inequality follows by inverse function theorem and a estimate similar to (4.8.1).  135  Set g(y) := −(r ◦ Pt−1 )(y). To solve (4.8.5) we first solve the equation + *   t Y1 (y) y1  g(y1 , y2 ) =   ,  t Y2 (y) y2 on Dδ(1−2b) . This is done by the functions Y1t (y) =  1 g(y1 , 0) y1  and  Y2t (y) =  1 (g(y1 , y2 ) − g(y1 , 0)). y2  We now show that these functions are holomorphic and derive some estimates for them. Observe that, in view of (4.8.6), the change of variables   ∂r ∂r y = Pt (x) = x2 + t (x), x1 + t (x) ∂x1 ∂x2 obeys (1 − 2b)(|x1 | + |x2 |) ≤ |y1 | + |y2 | ≤ (1 + 2b)(|x1 | + |x2 |).  (4.8.9)  By the chain rule, (4.8.6), (4.8.8), and the last inequality, for 1 ≤ i ≤ 2, √ ! ∂g 2b 1 + 2b (|y1 | + |y2 |), (y) ≤ b(|x1 | + |x2 |) 1 + ≤b ∂yi 1−b 1 − 2b since  √  2b/(1 − b) < 2b because b < 1/55. Therefore, |g(y1 , y2 )| ≤  b 1 + 2b (|y1 | + |y2 |)2 . 2 1 − 2b  This shows that Y1t and Y2t are not singular at y = 0 and thus are holomorphic. Furthermore, it is easy to see that |Y1t (y)| ≤  b 1 + 2b |y1 |, 2 1 − 2b  ∂Y1t 3b 1 + 2b (y) ≤ ∂y1 2 1 − 2b  and  ∂Y2t (y) = 0. ∂y1  These are the estimates we need for Y1t . We next consider Y2t . To derive bounds for Y2t we consider the regions |y2 | ≥ |y1 | and |y2 | ≤ |y1 | separately. First, if |y2 | ≥ |y1 | then |Y2t (y)| ≤  1 1 + 2b (|y1 | + |y2 |)2 1 + 2b (|g(y1 , y2 )| + |g(y1 , 0)|) ≤ b ≤ 2b (|y1 | + |y2 |), |y2 | 1 − 2b |y2 | 1 − 2b  and similarly   ∂Y2t 1 ∂g ∂g 1 + 2b (y) ≤ (y1 , y2 ) + (y1 , 0) ≤ 4b , ∂y1 |y2 | ∂y1 ∂y1 1 − 2b ∂Y2t 1 1 ∂g 1 + 2b (y) ≤ |Y2t (y)| + (y1 , y2 ) ≤ 6b . ∂y2 |y2 | |y2 | ∂y2 1 − 2b 136  Observe that, in particular, these estimates hold for |y1 | = |y2 |. Now, for fixed y1 we can apply the maximum modulus principle to derive bounds for the functions Y2t (y1 , ·) and ∂ ∂yi Y2 (y1 , ·)  in the disk |z| ≤ |y1 |. (Note, this is the case |y2 | ≤ |y1 |.) By this principle,  the modulus of these functions inside the disk is bounded by the maximum modulus at the boundary |z| = |y1 |. Thus, using the above estimates, which are valid for |y1 | = |y2 |, we obtain |Y2t (y1 , z)| ≤ max |Y2t (y1 , z)| ≤ max 2b |z|=|y1 |  |z|=|y1 |  1 + 2b (|y1 | + |z|) 1 − 2b  1 + 2b 1 + 2b |y1 | ≤ 4b (|y1 | + |z|). = 4b 1 − 2b 1 − 2b Similarly, ∂Y2t 1 + 2b (y1 , z) ≤ 4b ∂y1 1 − 2b  and  ∂Y2t 1 + 2b (y1 , z) ≤ 6b . ∂y2 1 − 2b  We have all the bounds we need for Y2t . Let Y t (y) = (Y1t (y), Y2t (y)). Then, combining all the above estimates we find that, for 1 ≤ i ≤ 2 and for all y ∈ Dδ(1−2b) , q 1 + 2b kY (y)k ≤ 16 41 b (|y1 | + |y2 |), 1 − 2b ∂Y t 1 + 2b . (y) ≤ 6b ∂yi 1 − 2b t  (4.8.10) (4.8.11)  Furthermore, by construction the vector field Y t (y) satisfies the equation hy, Y t (y)i = −h(Pt−1 (y)). If we recall the change of variables y = Pt (x), this equation is equivalent to hPt (x), Y t (Pt (x))i = −r(x). Finally, if we set X t := Yt ◦ Pt , then X t (x) satisfies the desired equation hPt (x), X t (x)i = −r(x) on Pt−1 (Dδ(1−2b) ). Note, this is in fact equation (4.8.5). That is, we have proved (4.8.2). Furthermore, in view of (4.8.6), the region Pt−1 (Dδ(1−2b) ) contains Dδ(1−4b) . From (4.8.10), (4.8.11), (4.8.7) and (4.8.9), we obtain the estimates (4.8.3) and (4.8.4), namely, q (1 + 2b)2 (|x1 | + |x2 |) < 5b(|x1 | + |x2 |) < 10bδ, kX t (x)k ≤ 16 14 b 1 − 2b ∂X t 1 + 2b (x) ≤ 6b < 8b. ∂xi 1 − 2b This completes the proof of the theorem. 137  4.9  Appendix  We now prove Propositions 4.1.2 and 4.5.4. Proposition 4.1.2 follows from part (a) of Proposition 4.5.4, which we reproduce below. Let T be a linear operator from L2C to L2B with B, C ⊂ Γ# , and recall the definition ( ) X X kT kσ := max sup |Tb,c |σ(|b − c|), sup |Tb,c |σ(|b − c|) , c∈C b∈B  b∈B c∈C  where σ satisfies the hypotheses stated in p. 93. We next prove that this norm has the following properties. Proposition 4.5.4 (Properties of k · kσ ). Let S and T be linear operators from L2C to L2B with B, C ⊂ Γ# . Then: (a) kT k ≤ kT kσ≡1 ≤ kT kσ ; (b) If B = C, then kS T kσ ≤ kSkσ kT kσ ; (c) If B = C, then k(I + T )−1 kσ ≤ (1 − kT kσ )−1 if kT kσ < 1; (d) |Tb,c | ≤  1 σ(|b−c|) kT kσ  for all b ∈ B and all c ∈ C.  Proof. (a) By the Cauchy-Schwarz inequality we have |(T ϕ)∧ (b)| =  X c∈C  X  c∈C  #1/2 X  |Tb,c | |ϕ̂(c)|  |Tb,c0 |  c0 ∈C  c∈C  #1/2 "  " ≤  X   |Tb,c |1/2 |ϕ̂(c)| |Tb,c |1/2  |Tb,c ϕ̂(c)| =  c∈C #1/2 " 2  " ≤  X  Tb,c ϕ̂(c) ≤  sup  X  #1/2 X  |Tb,c0 |  b∈B c0 ∈C  2  |Tb,c | |ϕ̂(c)|  .  c∈C  Hence, " k(T ϕ)∧ k2l2 =  X  |(T ϕ)∧ (b)|2 ≤  sup  |Tb,c0 |  b∈B c0 ∈C  b∈B  " ≤  # X  #" sup  X  |Tb,c0 |  c∈C b∈B  sup  X  |Tb,c |  X  |ϕ̂(c)|2  c∈C  )2  ( ≤ max  |Tb,c | |ϕ̂(c)|2  # c∈C b∈B  b∈B c0 ∈C  XX  sup  X  b∈B c0 ∈C  |Tb,c0 |, sup  X  c∈C b∈B  138  |Tb,c |  kϕ̂kl2 .  Thus, kT ϕkL2  B  kϕkL2  C  k(T ϕ)∧ kl2 ≤ max = kϕ̂k2l2  (  ) sup  X  |Tb,c0 |, sup  X  |Tb,c |  c∈C b∈B  b∈B c0 ∈C  for all ϕ ∈ L2C . This implies that (  )  kT k ≤ max  sup  X  |Tb,c0 |, sup  X  |Tb,c | .  c∈C b∈B  b∈B c0 ∈C  That is, kT k ≤ kT kσ≡1 . This is the first inequality of part (a). The second inequality, namely, kT kσ≡1 ≤ kT kσ , follows immediately if we observe that σ(t) ≥ 1 for all t ≥ 0 by hypothesis (see p. 93). This proves part (a). (b) By hypothesis B = C. First observe that, since σ increases monotonically, and σ(s + t) ≤ σ(s)σ(t) for all s, t ∈ R+ , we have σ(|b − c|) = σ(|b − d + d − c|) ≤ σ(|b − d| + |d − c|) ≤ σ(|b − d|)σ(|d − c|) for all b, c, d ∈ B. Hence, sup  X  |(ST )b,c |σ(|b − c|) = sup  b∈B c∈B  X X  Sb,d Td,c σ(|b − c|)  b∈B c∈B d∈B  ≤ sup  X  |Sb,d |σ(|b − d|)  b∈B d∈B  |Td,c |σ(|d − c|)  c∈B  " ≤  X  #  #" sup  X  |Td,c |σ(|d − c|)  sup  d∈B c∈B  X  |Sb,d |σ(|b − d|) ,  b∈B d∈B  and similarly we prove that " sup  X  c∈B b∈B  |(ST )b,c |σ(|b − c|) ≤  #" sup  X  |Td,c |σ(|d − c|)  c∈B d∈B  # sup  X  d∈B b∈B  Thus, kST kσ ≤ kSkσ kT kσ , as claimed.  139  |Sb,d |σ(|b − d|) .  (c) By hypothesis B = C. Since kT kσ < 1, the Neumann series for (I + T )−1 converges (and its σ-norm is bounded). Hence, using part (b), k(I + T )−1 kσ =  ∞ X j=0  Tj  ≤ σ  ∞ X  kT kjσ ≤ (1 − kT kσ )−1 ,  j=0  as was to be shown. (d) Since σ(t) ≥ 1 for all t ≥ 0, observing the definition of k · kσ it is easy to see that |Tb,c | =  1 Tb,c σ(|b − c|) ≤ kT kσ σ(|b − c|)  for all b ∈ B and all c ∈ C. This completes the proof. We finally prove Proposition 4.1.2. Proof of Proposition 4.1.2. By applying Proposition 4.5.4(a) we find that ( ) X X kT k ≤ kT kσ≡1 = max sup |Tb,c |, sup |Tb,c | . b∈B c∈C  This is the desired inequality.  140  c∈C b∈B  Chapter 5  Exploiting gauge invariance 5.1  A gauge transformation  In this section we introduce a gauge transformation A → A(ν) that simplify (or modify) the main estimates and results in Chapter 4. For ν ∈ {1, 2} define Ψ̂ν : Γ# → C as  ν   (1, i(−1) ) · Â(b)  i(1, i(−1)ν ) · b Ψ̂ν (b) :=   0  if b 6= 0, if b = 0.  Observe that, for all b ∈ Γ# \ {0} and ν ∈ {1, 2}, |(1, i(−1)ν ) · b| = |b1 + i(−1)ν b2 | = |b| ≥ 2Λ and |Ψ̂ν (b)| =  |(1, i(−1)ν ) · Â(b)| |(1, i(−1)ν ) · Â(b)| √ |Â(b)| = ≤ 2 . ν |b1 + i(−1) b2 | |b| |b|  Thus, the function Ψ̂ν is well-defined and for any β ≥ 0 we have k|b|1+β Ψ̂ν (b)kl1 ≤  √  2 k|b|β Â(b)kl1 .  Now set A(ν) := A − ∇Ψν where Ψν := (Ψ̂ν )∨ ,  141  and to simplify the notation write ζ := (1, i(−1)ν ). Since without loss of generality we have assumed that Â(0) = 0, it follows that Â(ν) (0) = 0. Furthermore, for all b ∈ Γ# \ {0}, Â(ν) (b) = Â(b) − ibΨ̂ν (b) = Â(b) − b =  ζ · Â(b) (ζ · b)Â(b) − (ζ · Â(b))b = ζ ·b ζ ·b  1 (ζ1 b1 + ζ2 b2 )Â1 (b) − (ζ1 Â1 (b) + ζ2 Â2 (b))b1 , ζ ·b (ζ1 b1 + ζ2 b2 )Â2 (b) − (ζ1 Â1 (b) + ζ2 Â2 (b))b2  (b2 Â1 (b) − b1 Â2 (b))ζ2 , (b2 Â1 (b) − b1 Â2 (b))(−ζ1 )  =  1 ζ ·b  =  1 b⊥ · Â(b) ⊥ (b2 , −b1 ) · (Â1 (b), Â2 (b)) (ζ2 , −ζ1 ) = ζ ζ ·b ζ ·b  =  b⊥ · Â(b) (1, i(−1)ν )⊥ . (1, i(−1)ν ) · b   (5.1.1)  Hence, for all b ∈ Γ# , |Â(ν) (b)| ≤  √  2 |Â(b)|,  and for any β ≥ 0, k|b|β Â(ν) (b)kl1 ≤  √  2 k|b|β Â(b)kl1 .  The transformation A → A(ν) is useful because of the following proposition, which is a particular case of Theorem 2.6.1 (gauge invariance). Proposition 5.1.1 (Gauge invariance). Let ν ∈ {1, 2} and assume that A1 , A2 ∈ C 1 (R2 /Γ) and V ∈ C 0 (R2 /Γ) so that Ψν ∈ C 2 (R2 /Γ). Then, Ker(Hk (A, V )) 6= {0}  if and only if  Ker(Hk (A(ν) , V )) 6= {0},  and consequently b b (ν) , V ). F(A, V ) = F(A Therefore, to study the Fermi curve of (A, V ) we may replace A by A(1) or A(2) . We shall exploit this property below.  142  5.2  The regular piece revisited  From now on we shall make the following hypothesis. Hypothesis 5.2.1. A1 , A2 ∈ C 1 (R2 /Γ)  and  V ∈ C 0 (R2 /Γ).  We now describe the main simplifications (or modifications) introduced by the gauge transformation A → A(ν) . Our first observation is that the expressions for Nd0 + Dd0 ,d0 and Dd0 ,d00 (in Proposition 4.3.1) become simpler. 0  Proposition 5.2.1. Under hypothesis 5.2.1, let ν ∈ {1, 2} and replace A by A(ν ) . Then 0 00  0 00  Jνd d = 0 and Ldν 0d = 0 for any d0 , d00 ∈ G and consequently 0  0  0  2 d d d Nd0 + Dd0 ,d0 = Jνd0 wν,d 0 + (1 + K )wν,d0 zν,d0 + Lν zν,d0 + M  0  and 0 00  0 00  0 00  0 00  2 dd Dd0 ,d00 = Jνd0 d wν,d wν,d0 zν,d0 + Ldν d zν,d0 + M d d . 0 + K  Proof. First recall from (5.1.1) that 0  A(ν ) (b) =  b⊥ · Â(b) 0 (i(−1)ν , −1) (1, i(−1)ν 0 ) · b  0  for all b ∈ Γ# \{0}, and that A(ν ) (0) = 0. Now, substitute this expression into the definitions 0  0  of Jνd and Ldν 0 in Proposition 4.3.1. Then, if we observe that 0  (−1)ν = −(−1)ν , and compute 0  0  (1, −i(−1)ν ) · (i(−1)ν , −1) = i((−1)ν + (−1)ν ) = 0, 0  0  it follows easily that Jνd = 0 and Ldν 0 = 0. Thus, by Proposition 4.3.1, 0  0  0  0  2 d d d Nd0 + Dd0 ,d0 = Jνd0 wν,d 0 + (1 + K )wν,d0 zν,d0 + Lν zν,d0 + M  and 0 00  0 00  0 00  0 00  2 dd Dd0 ,d00 = Jνd0 d wν,d wν,d0 zν,d0 + Ldν d zν,d0 + M d d , 0 + K  as was to be shown. 143  We now inspect the proof of Theorem 3.4.1 (the regular piece) to see what we have gained by performing this transformation. The last proposition immediately implies that the defining equation (4.6.1), namely, β1 w2 + β2 z 2 + (1 + β3 )wz + β4 w + β5 z + β6 + q̂(0) = 0, is reduced to β1 w2 + (1 + β3 )wz + β5 z + β6 + q̂(0) = 0, because β2 = Jν00 = 0 and β4 = L00 ν 0 = 0. Thus, instead of equation (4.6.3), namely, (1)  w + β2 z + g = 0, we have w + g = 0, where |g(k)| ≤  C . ρ (1)  In the above equation, observe the absence of term β2 z which does not decay with respect to ρ (it is only O(1)). This yields better bounds and makes the analysis simpler. In fact, Proposition 4.6.1 becomes Proposition 5.2.2. Under the hypotheses of Theorem 3.4.1 and Hypothesis 5.2.1, after the 0  gauge transformation A → A(ν ) we have |F (k) − w(k)| ≤  C1 ρ  (a)  and ∂F C2 (k) − 1 ≤ , ∂k1 ρ  (b)  where the constants C1 and C2 depend only on ε, Λ, q and A. Consequently, after replacing Proposition 4.6.1 by the above proposition, we no longer need Proposition 4.6.2, which is an improvement of Proposition 4.6.1 necessary to take care (1)  of the term β2 z (which vanishes after the gauge transformation). Thus, Lemma 4.4.2 that was used to prove Proposition 4.6.2 is not necessary anymore. This simplifies the analysis. Then, using the new bounds, the proof of Theorem 3.4.1 can be carried out in exactly the same way yielding the following improved version of Theorem 3.4.1. 144  0  Theorem 5.2.3 (The regular piece after the transformation A → A(ν ) ). Let 0 < ε < Λ/6 and assume that A1 , A2 and V satisfy Hypothesis 5.2.1 with k(1+b2 )Â(b)kl1 (Γ# \{0}) < 2ε/63 0  and kb2 q̂(b)kl1 (Γ# ) < ∞. Then, after performing the transformation A → A(ν ) , there is a constant ρ = ρΛ,ε,q,A such that, for ν ∈ {1, 2}, the projection pr induces a biholomorphic map between     b F(A, V ) ∩ Tν (0) \ Kρ ∪   [  Tb   b∈Γ# \{0}  and its image in C. This image component contains n z∈C  o 8|z| > ρ and |z + (−1)ν θν (b)| > ε for all b ∈ Γ# \ {0}  and is contained in ( z∈C  ) ε # |z + (−1) θν (b)| > for all b ∈ Γ \ {0} , 2 ν  where θν (b) = 12 ((−1)ν b2 + ib1 ). Furthermore, pr−1 : Image(pr) −→ Tν (0), y 7−→ (−i(−1)ν y − r(y), y), where |r(y)| ≤  C ρ  and C = CΛ,ε,q,A is a constant. This theorem provides a simpler picture than Theorem 3.4.1 because here |r(y)| decays with respect to ρ.  145  Bibliography [1] S. B. Chae, Holomorphy and calculus in normed spaces, Marcel Dekker, 1985. [2] J. Feldman, The spectrum of periodic Schrödinger operators, UBC class notes. [3] J. Feldman, H. Knörrer and E. Trubowitz, Asymmetric Fermi surfaces for magnetic Schrödinger operators, Commun. Part. Diff. Eq. 26, 319-336 (2000). [4] J. Feldman, H. Knörrer and E. Trubowitz, Riemann surfaces of infinite genus, Amer. Math. Soc., 2003. [5] D. Gieseker, H. Knörrer and E. Trubowitz, The geometry of algebraic Fermi curves, Perspectives in mathematics 14, 1992. [6] H. Knörrer and E. Trubowitz, A directional compactification of the complex Bloch variety, Comment. Math. Helvetici 65, 114-149 (1990). [7] I. Krichever, Spectral theory of two-dimensional periodic operators and its applications, Russian Math. Surveys 44:2, 145-225 (1989). [8] P. Kuchment, Floquet theory for partial differential equations, Birkhäuser Verlag, 1993. [9] O. Madelung, Introduction to solid state theory, Springer, 1995. [10] H. McKean, Integrable systems and algebraic curves, in Global Analysis, Proceedings, 1978, LNM 755, Springer Verlag, 1979. [11] M. Reed and B. Simon, Methods of modern mathematical physics I: functional analysis, Academic Press, 1972. [12] M. Reed and B. Simon, Methods of modern mathematical physics IV: analysis of operators, Academic Press, 1978. 146  [13] W. Rudin, Real and complex analysis, McGraw-Hill, 1987. [14] M. U. Schmidt, A proof of the Willmore conjecture, preprint, arXiv:math/0203224 (2002). [15] B. Simon, Trace ideals and their applications, 2nd edition, Amer. Math. Soc., 2005. [16] J. L. Taylor, Several complex variables with connections to algebraic geometry and Lie groups, Amer. Math. Soc., 2002.  147  

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