Operators for Block Spin Transformations
Tadeusz Balaban
Department of Mathematics
Rutgers, The State University of New Jersey
110 Frelinghuysen Rd
Piscataway, NJ 08854-8019
tbalaban@math.rutgers.edu
Joel Feldman∗
Department of Mathematics
University of British Columbia
Vancouver, B.C.
CANADA V6T 1Z2
feldman@math.ubc.ca
http://www.math.ubc.ca/∼feldman/
Horst Knörrer, Eugene Trubowitz
Mathematik
ETH-Zentrum
CH-8092 Zürich
SWITZERLAND
horst.knoerrer@math.ethz.ch, eugene.trubowitz@math.ethz.ch
http://www.math.ethz.ch/∼knoerrer/
March 15, 2015
∗
Research supported in part by the Natural Sciences and Engineering Research Council of Canada and
the Forschungsinstitut für Mathematik, ETH Zürich.
Table of Contents
§I Introduction
p 2
§II Block Spin Operators
p 6
§III Differential Operators
p 17
§IV The Covariance
p 22
§V The Green’s Functions
p 32
§VI The Degree One Part of the Critical Field
p 43
Appendices
§A Bloch Theory for Periodic Operators
p 52
Periodic Operators in “Position Space”
and “Momentum Space” Environments
p 52
Periodized Operators
p 55
Averaging Operators
p 59
Analyticity of the Fourier Transform and L1 –L∞ Norms
p 62
Functions of Periodic Operators
p 64
Scaling of Periodized Operators
p 66
§B Trigonometric Inequalities
p 68
§C Lattice and Operator Summary
p 71
References
p 73
1
I. Introduction
In [parabolic-all.tex], we exhibit, for a many particle system of weakly interacting
Bosons in three space dimensions, the formation of a potential well of the type that typically
leads to symmetry breaking in the thermodynamic limit. To do so, we use the block spin
renormalization group approach. In previous papers [BFKT4,BFKT1,BFKT2,BFKT3]
(followed by a simple change of variables) we have written the partition function of such a
system on a discrete torus* in terms of a functional integral on a 1 + 3 dimensional space
X0 = ZZ/Ltp ZZ × ZZ3 /Lsp ZZ3
with positive integers Ltp , Lsp . Up to corrections which are exponentially small in the
coupling constant, and up to a multiplicative normalization factor, this representation is
of the form
Z h
Q dψ(x)∗ ∧dψ(x) i A0 (ψ ∗ ,ψ)
e
χ0 (ψ)
(I.1)
2πı
x∈X0
with an action A0 of the form
A0 (ψ∗ , ψ) = − hψ∗ , D0 ψi0 − V0 (ψ∗ , ψ) + µ0 hψ∗ , ψi0 + E0′ (ψ∗ , ψ)
(I.2)
Here
◦ D0 = 1l − e−h0 − e−h0 ∂0 , where ∂0 the forward time derivative (see (II.8) below) and
h0 is – up to a scaling – the single particle Hamiltonian.
◦ V0 (ψ∗ , ψ) is a quartic monomial that describes the coupling between the particles
◦ µ0 is related to the chemical potential of the system
◦ E0′ (ψ∗ , ψ) is perturbatively small
◦ χ0 (ψ) is a “small field cut off function”.
See [parabolic-all.tex, (I.3), (I.4)].
For the block spin renormalization group action, we pick a “block rectangle” of length
2
L in the “time direction” and L in “space directions”, where L is a sufficiently large odd
positive integer, and a corresponding nonnegative, compactly supported function q(x) on
ZZ × ZZ3 (the averaging profile). The choice of this kind of rectangle is characteristic of
the “parabolic scaling”, see [parabolic-all.tex, Definition I.3, Remark I.4, Definition
I.13.d]. For simplicity we assume that Lsp and Ltp are powers of L.
The block spin averaging operator, which we denote Q, maps functions on the lattice
(1)
X0 to functions on the “coarse” lattice X−1 = L2 ZZ/Ltp ZZ × LZZ3 /Lsp ZZ3 . After each
* All bounds achieved so far are uniform in the volume of this torus.
2
renormalization group step we scale, to again give functions on a unit lattice. After the first
(1)
RG step this unit lattice is X0 = ZZ/ L12 Ltp ZZ × ZZ3 / L1 Lsp ZZ3 . The “scaled” block spin
averaging operator maps functions on the lattice X1 = L12 ZZ/ L12 Ltp ZZ × L1 ZZ3 / L1 Lsp ZZ3
(1)
to functions on the unit lattice X0 .
In the nth renormalization group step, we end up consdering functions on the chain
of lattices
(n+1)
(n)
(n−1)
(1)
X−1
⊂ X0 ⊂ X1
⊂ · · · ⊂ Xn−1 ⊂ Xn(0)
where, for integers j ≥ −1 and n ≥ 0 ,
(n)
Xj
= ε2j ZZ/ε2n+j Ltp ZZ × εj ZZ3 /εn+j Lsp ZZ3
with εj =
1
Lj
(n)
The subscript in Xj determines the “coarseness” of the lattice — nearest neighbour
points are a distance ε2j = L12j apart in the time direction and a distance εj = L1j apart in
(n)
spatial directions. The superscript in Xj determines the number of points in the lattice
(n)
(0)
— |Xj | = |X0 |/L5n for all j. We usually write Xn = Xn . See [ parabolic-all.tex,
Definition I.5.a].
(n)
The (n+1)st block spin transformation involves the passage from X0 to its sublattice
(n+1)
X−1 . The averaging operations determine linear maps
(n)
Q : H0
(n+1)
7→ H−1
(n)
Qn : Hn = Hn(0) 7→ H0
and
(n) denotes the the (finite dimensional) Hilbert space of functions on
= L2 Xj
R
P
with integral X (n) du = ε5j u∈X (n) and the real inner product
(n)
where Hj
(n)
Xj
j
j
hα1 , α2 ij =
Z
(n)
α1 (u) α2 (u) du
Xj
Again see [parabolic-all.tex, Definition I.5.a]. In §II we pick a specific averaging profile
and give bounds on the operators Q, Qn , their Fourier transforms, and related operators.
Scaling is performed by the linear isomorphisms
(n)
L : Xj
(n)
(u0 , u) 7→ (L2 u0 , Lu)
→ Xj−1
(n)
(n)
For a function α ∈ Hj , define the function L∗ (α) ∈ Hj−1 by L∗ (α)(Lu) = α(u) . In
particular, after rescaling and multiplication with the “scaling factor” L2n , the differential
operator D0 in (I.2) becomes the operator
Dn = L2n L−n
1l − e−h0 − e−h0 ∂0 Ln∗
∗
3
on Hn . This operator is discussed in §III.
(n)
As mentioned above, the passage from a functional integral on X0 to a functional
(n+1)
integral on X−1
is an averaging procedure over, roughly speaking, a rectangle of size L2
in the time direction and size L in the spatial directions. This passage is analyzed using
stationary phase techniques that involve
(n)
◦ the determination of critical fields on X0 (that are functions of external fields on
(n+1)
X−1 ) for an appropriate action, and
◦ a functional integral over “fluctuation fields” around the critical field.
The covariance for the integral over the fluctuation fields has been identified in [parabolicall.tex, (I.15)] and is bounded in §IV.
The composition of the critical fields of n renormalization group steps is – after rescaling – a field on Xn , called the “background field”, that is a function of an external field
(n)
on X0 . It is crucial in our representation of the partition function. See [parabolicall.tex, Theorem I.19]. The “leading order” part of the background field is linear in the
external field and has been identified in [parabolic-all.tex, Proposition I.16]. It is the
(n)
composition of an operator, from H0 to Hn determined by the averaging profile q, and an
operator Sn on Hn which can be viewed as a Green’s function for the differential operator
Dn (plus a mass term). This operator, Sn , is discussed in §V.
To get bounds on the critical fields in the fluctuation integral at step n + 1, we use a
well known algebraic relation between these critical fields and the background fields at step
n + 1 given in [blockspinRGalg.tex, Proposition B.8], [parabolic-all.tex, Proposition
IV.4.a]. The operators in the linearization of this relation, and various other linearizations,
are studied in §VI.
By construction, many of the operators discussed in this paper are linear operators
defined on the Hilbert space of functions on a lattice that are invariant under translations
with respect to a sublattice. It is natural to use Bloch decompositions and Fourier transforms for an analysis of such operators. Abstract basic results about Bloch decompositions
and their connections to Fourier transform are summarized in Appendix A, giving special
emphasis to averaging operators.
Most operator estimates we obtain in this paper are with respect to a norm of the
following kind.
(n−j)
Definition I.1 For any operator A : Hj
mass m ≥ 0, we define the norm
kAkm = max
n
sup
(n−k)
u∈Xk
Z
′
du e
(n−j)
Xj
m|u−u′ |
(n−k)
→ Hk
′
|A(u, u )| ,
4
, with kernel A(u, u′ ), and for any
sup
(n−j)
u′ ∈Xj
Z
′
du em|u−u | |A(u, u′ )|
(n−k)
Xk
o
In the special case that m = 0, this is just the usual ℓ1 –ℓ∞ norm of the kernel.
We see in Lemmas A.11 and A.12 that this norm is related to the analyticity properties
of the Fourier transform. In this paper we use the following Fourier transform conventions.
(n)
The dual lattice of Xj
is
(n)
Xˆj =
2π
ZZ/ 2π
ZZ
ε2n+j Ltp
ε2j
(n)
For a function α ∈ Hj
α̂(p) =
where
R
(n)
X̂j
dp
(2π)4
=
Z
(n)
Xj
1
−ip·u
α(u)e
ε5n+j Ltp L3sp
P
du
(n)
p∈X̂j
×
2π
ZZ3 / 2π
ZZ3
εn+j Lsp
εj
α(u) =
Z
(n)
X̂j
α̂(p)eiu·p
dp
(2π)4
. The maps
(n)
(n)
L : Xˆj−1 → Xˆj
(q0 , q) 7→ (L2 q0 , Lq)
(n)
are again linear isomorphisms, and, for a function α ∈ Hj
,
5
\
L
∗ (α)(q) = L α̂(Lq)
(n)
The quotient map dual to the inclusion Xj
(j+k,j)
π̂n+j
(n−k)
⊂ Xj+k
(I.3)
is
(n−k)
(n)
: Xˆj+k → Xˆj
(I.4)
When the indices are clear from the context we suppress them and write π̂.
The estimates of this paper are used in [parabolic-all.tex]. In particular, the construction of the background fields and the critical fields in [parabolic-all.tex, Appendix
G] uses a contraction mapping argument around the linearizations of §VI and §V in this
paper.
For the readers’ convenience we have included, in Appendix C, a list of most of the
operators and lattices that appear in this paper.
Convention I.2 Most estimates in this paper are bounds on norms of operators as in
Definition I.1. The (finite number of ) constants that appear in these bounds are consecutively labelled Γ1 , Γ2 , · · ·, γ1 , γ2 , · · ·, m1 , m2 , · · ·. All of these constants Γj , γj , mj are
independent of L and the scale index n. We define Γop to be the maximum of the Γj ’s,
and, in [parabolic-all.tex], refer to the estimates using this constant Γop .
5
II. Block Spin Operators
In this chapter, we analyze the block spin “averaging operators” Q of [parabolicall.tex, Definitions I.1.a and I.13.d] and Qn of [parabolic-all.tex, Definition I.13.d] as
(n)
well as the operator Qn of [parabolic-all.tex, Definition I.5.b]. Recall that Q : H0 →
(n+1)
H−1
is defined by
P
q(y − x)ψ([x])
(II.1)
(Qψ)(y) =
x∈ZZ×ZZ3
(n)
where [x] denotes the class of x ∈ ZZ × ZZ3 in the quotient space X0 . The averaging
profile q is the q–fold convolution of the characteristic function, 1 (x), of the rectangle
L2 −1 L2 −1 L−1 L−1 3
× − 2 , 2
, normalized to have integral one. That is,
− 2 , 2
q times
q=
1
L5q
z
}|
{
1 ∗1 ∗···∗1
See Example A.8 and Remark A.10. Here q ≥ 4 is a fixed even natural number.(1)
The operator
Qn = Q(1) · · · Q(n) = L−1
∗ Q
(n−j)
(n−j+1)
j
where Q(j) = L−j
∗ Q L∗ : Hj
→ Hj−1
Qn = a 1l +
n
(n)
Ln∗ : Hn = Hn(0) → H0
. The operator
n−1
P
j=1
∗ −1
1
Q
Q
2j
j
j
L
The Fourier transform of the characteristic function 1 is
with
σ(k) = sin
1
2 k0
3
Y
sin
1
2 kν
ν=1
Therefore
q̂(k) = u+ (k)q
with
and, by Lemma A.9.a
[
(Qψ)(k)
=
X
(n)
0
π̂(k)=k
k∈X̂
(n)
for all ψ ∈ H0
(1)
(II.2)
u+ (k) =
q̂(k)ψ̂(k)
σ(Lk)
σ(k)
σ(Lk)
L5 σ(k)
(n)
with k ∈ Xˆ0 and
(II.3)
(II.4)
(II.5)
(n+1)
and k ∈ Xˆ−1 .
See Remark II.7 for a discussion of the condition q > 2. The condition that q be even is imposed
purely for convenience.
6
Remark II.1
(a) Since q is even, σ(k)q is an entire function of k ∈ C × C3 that is periodic with respect
to the lattice 2π(ZZ × ZZ3 ). Also
q
(n−j)
σ(pj )q = σ π̂n(j,0) (pj )
for all pj ∈ Xˆj
(n)
(b) For all φ ∈ Hn and k ∈ Xˆ0 ,
X
\
un (p)q φ̂(p)
(Q
φ)(k)
=
n
with
un (p) = ε5n
p∈X̂n
π̂(p)=k
(n)
(c) For all ψ ∈ H0
σ(p)
σ(L−n p)
[
and k ∈ X̂0 , Q
n ψ(k) = Q̂n (k)ψ̂(k) where
−1
n−1
X X
2q
1
Q̂n (k) = a 1 +
L2j uj (pj )
(n)
j=1
(n−j)
j
π̂(pj )=k
pj ∈X̂
(d) The functions un (p) and u+ (p) are entire in p and are invariant under pν → −pν for
each 0 ≤ ν ≤ 3.
(e) Set, with the notation of (I.4), the “single period” lattices and their duals
(0,−1)
2π
Z/2πZZ × 2π
Z3 /2πZZ3 = ker π̂n−1
B+ = ZZ/L2 ZZ × ZZ3 /LZZ3
B̂+ = L
2Z
L Z
3 2π 3
Z
Z
= ker π̂n(j,0)
Bj = ε2j ZZ/ZZ × εj ZZ3 /ZZ3
B̂j = 2πZZ/ 2π
Z
×
2πZ
Z
/
2 Z
εj
ε
j
for each integer j ≥ 0. In this notation, the representations of Q, Qn and Qn of (II.5) and
parts (b) and (c) are
X
[
(Qψ)(k)
=
u+ (k + ℓ)q ψ̂(k + ℓ)
\
(Q
n φ)(k) =
ℓ∈B̂ +
X
un (k + ℓ)q φ̂(k + ℓ)
ℓ∈B̂n
n−1
X P
Q̂n (k) = a 1 +
j=1 ℓj ∈B̂j
1
L2j uj (k
+ ℓj )
2q
−1
P
(n+1)
[
is represented by
Here in (Qψ)(k)
= ℓ∈B̂+ u+ (k + ℓ)q ψ̂(k + ℓ), for example, k ∈ X̂−1
3
2π
2π
the element of ε2 Ltp ZZ × εn Lsp ZZ having minimal components and ℓ is represented by the
n
element of
2π
ZZ
ε2−1
×
2π
Z3
ε−1 Z
having minimal components. Similarly
∗ θ (k + ℓ) = u (k + ℓ)q θ̂(k)
d
Q
+
q
∗
[
Q
n ψ (k + ℓn ) = un (k + ℓn ) ψ̂(k)
7
(n−j)
(n)
(j,0)
Proof: (a) Any two points of Xˆj
with the same image in Xˆ0 under π̂n
differ by
2π times an integer vector. The formula follows.
(n−j)
(b) By (I.3) and (II.5), we have, for α ∈ Hj
\
(j) α)(p
(Q
j−1 ) =
=
(n−j+1)
and pj−1 ∈ Xˆj−1
\
j
−j
1
pj−1 )
L5j (QL∗ α)(L
X
1
L5j
(n−j)
0
π̂(k)=L−j pj−1
\
j
q̂(k)(L
∗ α)(k)
k∈X̂
1
= 5q
L
=
1
L5q
X
(n−j)
k∈X̂
0
π̂(k)=L−j pj−1
X
(n−j)
pj ∈X̂
j
π̂(pj )=pj−1
σ(Lk)q
j
α̂
L
k
σ(k)q
σ(L−j+1 pj )q
α̂(pj )
σ(L−j pj )q
so that, by part (a),
\
(Q
n φ)(p0 ) =
X
1
L5qn
(n−j)
pj ∈X̂
j
π̂(pj )=pj−1
1≤j≤n
X
= ε5q
n
(n−j)
j
π̂(pj )=pj−1
1≤j≤n
pj ∈X̂
= ε5q
n
X
pn ∈X̂n
π̂(pn )=p0
σ(L−n+1 pn )q
σ(p1 )q σ(L−1 p2 )q
·
·
·
φ̂(pn )
σ(L−1 p1 )q σ(L−2 p2 )q
σ(L−n pn )q
σ(pn )q σ(L−1 pn )q
σ(L−n+1 pn )q
·
·
·
φ̂(pn )
σ(L−1 pn )q σ(L−2 pn )q
σ(L−n pn )q
σ(pn )q
φ̂(pn )
σ(L−n pn )q
(c) follows from part (b) and Lemma A.9.a.
(d) is obvious since
sin z
z
sin m
is even and entire for any nonzero integer m.
In Lemmas II.2 and II.3 we derive a number of bounds on the kernels un and u+
that appear in the representations for Qn and Q of Remark II.1.e. In Proposition II.4 we
analyze the operator Q̂n . Then in Remark II.5 and Lemma II.6 we study how to move
derivatives past Q and Qn .
8
When dealing with the asymmetry between “temporal” and “spatial” scaling we set,
for convenience,
( 2
)
1
1
2
=
for
ν
=
0
ε
=
n
2n
n
L for ν = 0
L0
L
Lν =
εn,ν =
(II.6)
εn = L1n = L1n
L for ν = 1, 2, 3
for ν = 1, 2, 3
ν
Lemma II.2 Assume that |Re kν | ≤ π, |Im kν | ≤ 2 for each 0 ≤ ν ≤ 3.
Q3
(a) un (k + ℓ) ≤ ν=0 |ℓν24
for all ℓ ∈ B̂n . We use |ℓν | to denote the magnitude of
|+π
the smallest representative of ℓν in its equivalence class, as an element of B̂n . There is a
constant Γ1 , depending only on q, such that kQn km=1 ≤ Γ1 .
Q3
if 0 6= ℓ ∈ B̂n .
(b) un (k + ℓ) ≤ |k| ν=0 |ℓν24
|+π
(c) un (k) − 1 ≤ 43 |k|2 .
(d) If ℓ ∈ B̂n and ℓν̃ 6= 0 for some 0 ≤ ν̃ ≤ 3, then un (k + ℓ) = sin 12 kν̃ vn,ν̃ (k + ℓ) with
Q
24
vn,ν̃ (k + ℓ) ≤ 3
ν=0 |ℓν |+π .
(e) For all ℓ ∈ B̂n ,
3
Im un (k + ℓ) ≤ 16 |Im k| Q
ν=0
24
|ℓν |+π
h Q
3
Im un (k + ℓ)q ≤ 16 q |Im k|
ν=0
(f ) If k is real
2 4
π
Proof: Set s(x) =
εn,ν in (II.6),
sin x
x .
un (p) =
≤ un (k) ≤
24
|ℓν |+π
iq
π 4
2
By the definitions of un (p) in Remark II.1.b, σ(k) in (II.3) and
1
ε2n
3
Y
sin 21 p0
sin 21 ε2n p0 ν=1
3
Y
sin 12 pν
s(pν /2)
=
1
1
s(εn,ν pν /2)
εn sin 2 εn pν
ν=0
(II.7)
(a) We may assume without loss of generality that ℓν is bounded, as a real number, by
1
π
Re kν + ℓν is always bounded
−
π.
(Recall
that
is
an
odd
natural
number.)
So
εn,ν
εn,ν
π
by εn,ν
. Consequently, the hypotheses of Lemma B.1.c, with x + iy = kν + ℓν and ε = εn,ν ,
are satisfied and
4
if ℓν = 0
sin 12 (kν + ℓν )
24
≤
≤
8
1 sin 1 ε (k + ℓ ) if
|ℓ
|
≥
2π
ν
|ℓν | + π
|Re kν +ℓν |
ν
εn,ν
2 n,ν ν
9
since |Re kν | ≤ π and ℓν ∈ 2πZZ. As q > 1, |un (k + ℓ)|q is summable in ℓ and the bound
on kQn km=1 follows from Lemma A.11.c.
(b) If ℓν 6= 0,
1
k
2 ν
+ ℓν ≥ 21 Re kν + ℓν ≥
1
6
π + |ℓν |
so that, by Lemma B.1.a, the denominator
sin 1 εn,ν (kν + ℓν ) ≥
εn,ν
2
1
1
εn,ν
π + |ℓν | =
√
21
π 6 εn,ν
√1
3 2π
π + |ℓν |
On the other hand, the numerator, by Lemma B.1.b,
sin 1 (kν + ℓν ) = sin
2
1
2 kν
≤ |kν |
As ℓ 6= 0, there is at least one ν with ℓν 6= 0. Pick one and call it ν̃. Bound the factor
√
3 2π|kν̃ |
sin 21 (kν̃ + ℓν̃ )
24|k|
≤
≤
1
1
|ℓν̃ | + π
|ℓν̃ | + π
εn,ν̃ sin 2 εn,ν̃ (kν̃ + ℓν̃ )
Bound the remaining factors, with ν 6= ν̃, by
3 Y
ν=0
24
|ℓν |+π
3
Y
sin 12 (kν + ℓν )
24
≤ |k|
1
1
|ℓ | + π
sin 2 εn,ν (kν + ℓν )
εn,ν
ν=0 ν
(c) For both z = 21 εn,ν kν and z = 21 kν , |z|2 ≤
Using
a
b
−1=
(a−1)−(b−1)
b
as in part (a). All together
π2
4
+ 1 < 4 so that, by Lemma B.1.e,
sin z
1 2
≤ |z|
−
1
z
2
we have, by Lemma B.1.a,
1 1
1 1
2
2
sin( 21 kν )
≤ 2 | 2 kν | + 2 | 2 εn,ν kν | ≤
−
1
1
1
| sin( 1 ε k )|/| 1 ε k |
εn,ν sin( 2 εn,ν kν )
2 n,ν ν
2 n,ν ν
π 1
√
(1
28
+ ε2n,ν )|kν |2 ≤ |kν |2
Finally, using
3
Q
ν=0
Aν − 1 = (A0 − 1)
3
Q
ν=1
Aν + (A1 − 1)
3
Q
ν=2
Aν + · · · + (A3 − 1)
we have, by Lemma B.1.c,
un (k) − 1 ≤ 43 |k0 |2 + 43−1 |k1 |2 + · · · + |k3 |2 ≤ 43 |k|2
10
(d) The proof is the same as that for part (b), except that the factor sin
numerator
ℓν̃
sin 12 (kν̃ + ℓν̃ ) = (−1) 2π sin 12 kν̃
1
2 kν̃
in the
is pulled out of un , leaving vn,ν̃ , rather than being bounded by |kν̃ |.
(e) By Lemma B.1.c
Im
sin 21 (kν + ℓν )
≤ 8|Im kν | 24
1
1
|ℓν | + π
sin 2 εn,ν (kν + ℓν ) εn,ν
In general, for any complex numbers zj = rj eiθj , 1 ≤ j ≤ J,
P
Q
J
J
Q
J
zj = sin
θj rj
Im
j=1
Repeatedly using
we have
J
J
X
J
J
X
Q
Q
Im zj Q |zj ′ |
zj ≤
sin(θj )
rj =
Im
j=1
j=1
Q
3
3
P
Im un (k + ℓ) ≤
8|Im kν |
ν=0
and
j=1
sin(θ + θ ′ ) = sin(θ) cos(θ ′ ) + cos(θ) sin(θ ′ ) ≤ | sin(θ)| + | sin(θ ′ )|
j=1
So
j=1
ν=0
j ′ 6=j
j=1
24
|ℓν |+π
≤ 16 |Im k|
h Q
3
Im un (k + ℓ)q ≤ 16 q |Im k|
ν=0
24
|ℓν |+π
iq
3
Q
ν=0
24
|ℓν |+π
(f) Just apply Lemma B.1.d separately to all of the numerators and denominators in the
right hand side of (II.7).
Lemma II.3 Assume that |Re kν | ≤
3
Q
(a) u+ (k + ℓ) ≤
ν=0
24
Lν |ℓν |+π
π
Lν
and |Im kν | ≤
2
Lν
for each 0 ≤ ν ≤ 3.
for all ℓ ∈ B̂+ . We use |ℓν | to denote the magnitude of the
smallest representative of ℓν in its equivalence class, as an element of B̂+ .
ih Q
i
3
h Q
24
+
(b) u+ (k + ℓ) ≤
Lν |kν |
with ℓ 6= 0. Here I+ is any
Lν |ℓν |+π for all ℓ ∈ B̂
ν=0
ν∈I+
subset of ν 0 ≤ ν ≤ 3, ℓν 6= 0 .
11
3
P
(c) u+ (k) − 1 ≤ 43
L2ν |kν |2 .
ν=0
(d) If ℓ ∈ B̂+ and ℓν̃ 6= 0 for some 0 ≤ ν̃ ≤ 3, then u+ (k + ℓ) = sin
3
Q
24
v+,ν̃ (k + ℓ) ≤
Lν |ℓν |+π .
Lν̃
k
2 ν̃
v+,ν̃ (k + ℓ) with
ν=0
(e) For all ℓ ∈ B̂+ ,
3
Q
Im u+ (k + ℓ) ≤ 16 |Im Lk|
24
Lν |ℓν |+π
ν=0
h Q
3
Im u+ (k + ℓ)q ≤ 16 q |Im Lk|
ν=0
24
Lν |ℓν |+π
iq
Proof: (a) We may assume without loss of generality that ℓν is bounded, as a real
number, by π − Lπν . (Recall that L is an odd natural number.) So we will always have
Re kν + ℓν ≤ π. So, by Lemma B.1.c with ε = 1 and x + iy = Lν (kν + ℓν ),
Lν
sin L2ν (kν + ℓν ) 4
≤
8
L sin 1 (k + ℓ ) Lν |Re kν +ℓν |
ν
ν
2 ν
since |Re kν | ≤
π
Lν ,
|Im kν | ≤
(b) If ℓν 6= 0,
and 21 Re kν + ℓν ≤
π
2
1
2 kν
2
Lν
and ℓν ∈
if ℓν = 0
if |Lν ℓν | ≥ 2π
≤
24
Lν |ℓν | + π
2π
Z.
Lν Z
+ ℓν ≥ 12 Re kν + ℓν ≥
1
6
π
Lν
+ |ℓν |
so that, by Lemma B.1.a, the denominator
Lν sin 21 (kν + ℓν ) ≥
√
2
6π
π + Lν |ℓν |
On the other hand, the numerator, by Lemma B.1.b,
sin Lν (kν + ℓν ) = sin
2
Lν
2 kν
≤ 2 Lν |kν | = Lν |kν |
2
As ℓ 6= 0, there is at least one ν with ℓν 6= 0, so that I+ is not empty. Bound each factor
with ν ∈ I+ by
sin L2ν (kν + ℓν ) 6π Lν |kν |
L sin 1 (k + ℓ ) ≤ √2 Lν |ℓν | + π
ν
ν
2 ν
Bound the remaining factors by
24
Lν |ℓν |+π .
(c) has the same proof as that of Lemma II.2.c, with kν replaced by Lν kν and
by Lν .
12
1
εn,ν
replaced
(d) The proof is similar to that for part (b), except that the factor sin
numerator
Lν̃ ℓν̃
sin L2ν̃ (kν̃ + ℓν̃ ) = (−1) 2π sin L2ν̃ kν̃
out of u+ , leaving v+,ν̃ , rather than being bounded by Lν̃ |kν̃ |.
is pulled
sin L2ν (kν +ℓν ) .
Lν sin 1 (kν +ℓν ) with ν 6= ν̃ is bounded by Lν |ℓ24
ν |+π
Lν̃
2 kν̃
in the
Also, each
2
(e) By Lemma B.1.c
Lν
sin
(k
+
ℓ
)
24
ν
ν
2
Im
L sin 1 (k + ℓ ) ≤ 8|Im Lν kν | Lν |ℓν | + π
ν
ν
2 ν
The proof now continues as in Lemma II.2.e, just by substituting kν → Lν kν , ℓν → Lν ℓν
and εn,ν = L1ν .
Proposition II.4 There are constants Γ2 , depending only on q, and Γ3 , depending only
on a, such that the following hold for all L > Γ2 .
(a) On the domain k ∈ C × C3 |Im kν | < 2 for each 0 ≤ ν ≤ 3 Q̂n (k) is analytic,
and invariant under kν → −kν for each 0 ≤ ν ≤ 3, and obeys
5
a
6
≤ |Q̂n (k)| ≤ 45 a
Re Q̂n (k) ≥
a
2
If k is real 65 a ≤ Q̂n (k) ≤ a.
(b) If |Re kν | ≤ π and |Im kν | ≤ 2 for each 0 ≤ ν ≤ 3, then
Q̂n (k) − an ≤
a
|k|2
3003
where
−2
1−L
an = a 1−L
−2n
(c) kQn km=1 ≤ Γ3
n−1
P P
Proof: (a) Recall, from Remark II.1.e, that Q̂n (k) = a 1 +
j=1 ℓj ∈B̂j
By Lemma II.2.a,
n−1
X
P
j=1 ℓj ∈B̂j
where cq =
|z| ≤
1
5
hP
j∈ZZ
1
|uj (k
L2j
+ ℓj )|
24/π 2q
[2|j|+1
i4
2q
≤
∞
X
1
L2j
j=1
3
Q
P
ℓ∈2πZZ×2πZZ3 ν=0
2q
24
|ℓν |+π
. Just pick L large enough that
1
=
Re 1+z
Re (1+z)
|1+z|2
13
≥
4/5
(6/5)2
=
20
36
1
u (k + ℓj )2q
L2j j
cq
L2 −1
<
=
1
5
−1
.
cq
L2 −1
and use that, if
(b) Using O(|k|2 ) to denote any function that is bounded by a constant, depending only
on q,
1+
n−1
X
P
1
u (k
L2j j
+ ℓj )2q =
j=1 ℓj ∈B̂j
n−1
X
1
L2j
+
j=0
≤
≤
So
Q̂n (k) = a
1−L−2n
1−L−2
n−1
X
1
[uj (k)2q
L2j
n−1
X
1
O
L2j
j=1
1−L−2n
1−L−2
+
j=1
−2n
1
L2 O
|k|2
−1
= an 1 +
1−L
1−L−2
+
1
L2 O
|k|2
+
|k|2
and it suffices to choose Γ2 large enough that
1 1−L−2
2 9
1−L−2
11
L2 1−L−2n O |k| ≤
10 ≤ 1−L−2n ≤ 10
− 1] +
n−1
X
j=1
P
ℓj ∈B̂j
ℓ6=0
1
u (k
L2j j
+ ℓj )2q
by Lemma II.2.b,c
1 1−L−2
L2 1−L−2n O
|k|2
−1
1 1−L−2
2 L2 1−L−2n O |k| ≤
2
10 1
22 3003 |k|
for all allowed k’s and L’s.
1
2
(c) follows immediately from part (a) and Lemma A.11.b.
(n)
For a function α ∈ Hj
and ν = 0, 1, 2, 3 , we define the forward derivative by
1
(∂ν α)(x) = εj,ν
α(x + εj,ν eν ) − α(x)
(II.8)
where eν is a unit vector in the ν th direction. The Fourier transforms
iεn,ν pν /2 sin(εn,ν pν /2)
φ̂(p) for all φ ∈ Hn
and p ∈ Xˆn
∂d
ν φ (p) = 2ie
εn,ν
(n)
(n)
ikν /2
∂d
sin(kν /2)ψ̂(k)
for all ψ ∈ H0
and k ∈ X̂0
ν ψ (k) = 2ie
(n+1)
(n+1)
iLν kν /2 sin(Lν kν /2)
∂d
θ̂(k)
for all θ ∈ H
and k ∈ Xˆ
ν θ (k) = 2ie
−1
Lν
(II.9)
−1
(−)
(−)
We now define operators like, for example, Qn,ν chosen so that ∂ν Qn = Qn,ν ∂ν . Set
u(+)
n,ν (p) =
Y
0≤ν ′ ≤3
ν ′ 6=ν
(−)
un,ν
(p)
1
εn,ν
sin 21 pν ′
sin 12 εn,ν ′ pν ′
′
3
Y
sin 12 pν
= 1
sin 21 εn,ν pν ν ′ =0
εn,ν
1
εn,ν
(+)
u+,ν (k) =
Y
0≤ν ′ ≤3
ν ′ 6=ν
sin 12 pν ′
sin 12 εn,ν ′ pν ′
′
(−)
u+,ν (k)
sin 21 Lν ′ kν ′
Lν ′ sin 21 kν ′
3
sin 21 Lν kν Y sin 12 Lν ′ kν ′
=
Lν sin 12 kν ν ′ =0 Lν ′ sin 21 kν ′
and
(+)
(+)
ζn,ν
(k, ℓn ) = eiεn,ν (k+ℓn )ν /2 e−ikν /2 cos 12 ℓn,ν
ζ+,ν (k, ℓ) = ei(k+ℓ)ν /2 e−iLν kν /2 cos 21 Lν ℓν
(−)
ζn,ν
(k, ℓn ) = eikν /2 e−iεn,ν (k+ℓn )ν /2 cos 12 ℓn,ν
ζ+,ν (k, ℓ) = eiLν kν /2 e−i(k+ℓ)ν /2 cos 21 Lν ℓν
14
(−)
(+)
(n)
Define the operators Qn,ν : H0
(−)
(n)
→ Hn and Qn,ν : Hn → H0
by
\
(+) (+)
q−1
Qn,ν ψ (k + ℓn ) = ζn,ν
(k, ℓn)u(+)
ψ̂(k)
n,ν (k + ℓn )un (k + ℓn )
X
\
(−) (−)
(−)
ζn,ν
(k, ℓn)un,ν
(k + ℓn )un (k + ℓn )q−1 φ̂(k + ℓn )
Qn,ν φ (k) =
(II.10)
ℓn ∈B̂n
(+)
(n+1)
and the operators Q+,ν : H−1
(n)
→ H0
(−)
(n)
and Q+,ν : H0
(n+1)
→ H−1
by
\
(+)
(+)
(+) Q+,ν θ (k + ℓ) = ζ+,ν (k, ℓ)u+,ν (k + ℓ)u+ (k + ℓ)q−1 θ̂(k)
X (−)
\
(−)
(−) ζ+,ν (k, ℓ)u+,ν (k + ℓ)u+ (k + ℓ)q−1 ψ̂(k + ℓ)
Q+,ν ψ (k) =
(II.11)
ℓ∈B̂ +
Remark II.5 Let 0 ≤ ν ≤ 3. We have
∂ν Q∗n = Q(+)
n,ν ∂ν
(+)
(−)
∂ν Qn = Qn,ν
∂ν
(n)
∂ν Q∗ = Q+,ν ∂ν
(−)
∂ν Q = Q+,ν ∂ν
(n)
If S : Hn → Hn and T : H0 → H0 are linear operators that are translation invariant
(n)
with respect to Xn and X0 , respectively, then
(−)
(−)
∗
Qn,ν
SQ(+)
n,ν = Qn SQn
Proof:
(+)
Q+,ν T Q+,ν = QT Q∗
For the “Q∗n ” and “Qn ” cases, it suffices to observe that
2ieiεn,ν (k+ℓ)ν /2
sin(εn,ν (k+ℓ)ν /2) un (k
εn,ν
(+)
ikν /2
+ ℓ) = ζn,ν
(k, ℓ)u(+)
sin(kν /2)
n,ν (k + ℓ) 2ie
and
(−)
(−)
2ieikν /2 sin(kν /2) un (k + ℓ) = ζn,ν
(k, ℓ)un,ν
(k + ℓ) 2ieiεn,ν (k+ℓ)ν /2
sin(εn,ν (k+ℓ)ν /2) εn,ν
and
(−)
(−)
(+)
2
ζn,ν
(k, ℓ)un,ν
(k + ℓ)ζn,ν
(k, ℓ)u(+)
n,ν (k + ℓ) = un (k + ℓ)
(−)
(+)
for all k, ℓ, ν. We remark that “Qn,ν SQn,ν = Qn SQ∗n ” should not be surprising since
(−)
(+)
∂ν Qn SQ∗n = Qn,ν SQn,ν ∂ν and Qn SQ∗n is translation invariant on the unit scale and so
commutes with ∂ν . The proof for the “Q∗ ” and “Q” cases are virtually identical.
15
Lemma II.6 Let 0 ≤ ν ≤ 3, ℓ ∈ B̂+ and ℓn ∈ B̂n .
(+)
(+)
(−)
(−)
(+)
(+)
(a) ζn,ν (k, ℓn )un,ν (k+ℓn ) and ζn,ν (k, ℓn )un,ν (k+ℓn ) are entire in k and ζ+,ν (k, ℓ)u+,ν (k+ℓ)
(−)
(−)
and ζ+,ν (k, ℓ)u+,ν (k + ℓ) are entire in k
(b) Assume that |Re kν ′ | ≤ π, |Im kν ′ | ≤ 2, |Re kν ′ | ≤
0 ≤ ν ′ ≤ 3. Then
Y
(+)
ζ (k, ℓn)u(+) (k + ℓn ) ≤ e2
n,ν
n,ν
0≤ν ′ ≤3
ν ′ 6=ν
Y
(+)
ζ (k, ℓ)u(+) (k + ℓ) ≤ e2
+,ν
+,ν
0≤ν ′ ≤3
ν ′ 6=ν
π
Lν ′
and |Im kν ′ | ≤
2
Lν ′
for each
24
|ℓn,ν ′ |+π
24
Lν |ℓν ′ |+π
3
(−)
Q
ζ (k, ℓn)u(−) (k + ℓn ) ≤ e2 24
n,ν
n,ν
|ℓn,ν |+π
ν ′ =0
24
|ℓn,ν ′ |+π
3
(−)
Q
ζ (k, ℓ)u(−) (k + ℓ) ≤ e2 24
+,ν
+,ν
Lν |ℓν |+π
ν ′ =0
24
Lν ′ |ℓν ′ |+π
(±) (c) There is a constant Γ4 , depending only on q, such that Qn,ν m=1 ≤ Γ4 .
Proof:
(a) The proof is virtually identical to that of Remark II.1.d.
(b) The proof is virtually identical to that of Lemmas II.2.a and II.3.a.
(±)
(±)
(±)
(c) By (II.10), the Fourier transform of Qn,ν is ζn,ν (k, ℓn )un,ν (k + ℓn )un (k + ℓn )q−1 , which
by part (b) and Lemma II.2.a, is bounded in magnitude by
e2
Y
0≤ν ′ ≤3
ν ′ 6=ν
24
|ℓn,ν ′ |+π
3
Q
ν=0
q−1
24
|ℓν |+π
As q > 2, the claim now follows by Lemma A.11.c.
Remark II.7 The principle obstruction to allowing q = 1 arises when a differential
operator ∂ν is intertwined with the block spin averaging operator Qn , as happens in Remark
II.5. See, for example, the proof of Lemma II.6.c. We use the condition q > 1 in Lemma
II.2.a and then starting at Lemma IV.3 in §IV and in §V, VI. (See Lemma V.5.) We use
the condition q > 2 in Proposition VI.1 and Lemma II.6.c.
16
III. Differential Operators
In [parabolic-all.tex, Definition I.5.a] we associated to an operator h0 on
L2 ZZ3 /Lsp ZZ3 the operators
Dn = L2n L−n
1l − e−h0 − e−h0 ∂0 Ln∗
∗
(III.1)
Here ∂0 is the forward time derivative of (II.8). In this chapter we assume that h0 is the
periodization of a translation invariant operator h0 on L2 ZZ3 whose Fourier transform
ĥ0 (p)
◦ is entire in p and invariant under pν → −pν for each 1 ≤ ν ≤ 3
◦ is nonnegative when p is real and is strictly positive when p ∈ IR3 \ 2πZZ3
ĥ0
(0) = 0 for 1 ≤ ν ≤ 3 and has strictly positive Jacobian matrix
◦ obeys ĥ0 (0) = ∂∂p
h 2
i ν
∂ ĥ0
H = ∂p
(0)
.
µ ∂pν
1≤µ,ν≤3
Remark III.1
(a) The operator Dn is the periodization of a translation invariant operator Dn , acting on
L2 εn ZZ × ε2n ZZ3 , whose Fourier transform is
D̂n (p) =
1 2 2 −ĥ0 (εn p)
2 εn p0 e
sin 21 ε2n p0
1 2
2 εn p0
2
+ p2
2
1 − e−ĥ0 (εn p)
−ĥ0 (εn p) sin εn p0
−
ip
e
0
ε2n p2
ε2n p0
with p = (p0 , p) ∈ C × C3 .
(b) D̂n (p) is entire in p and invariant under pν → −pν for each 1 ≤ ν ≤ 3.
(c) D̂n (p) has nonnegative real part when p is real.
Proof: (a) follows from (I.3) and the observation, by (II.9), that the Fourier transform
of ∂0 , on ZZ, is
2ieik0 /2 sin(k0 /2) = −2 sin2 (k0 /2) + i sin(k0 )
(b) and (c) are obvious.
17
Lemma III.2 There are constants γ1 , Γ5 and a function m̄(c) > 0 that depend only on
ĥ0 and in particular are independent of n and L, such that the following hold.
(a) For all p ∈ IR × IR3 ,
3
D̂n (p) ≥ γ1 |p0 | + P |pν |2
ν=1
We use |p0 |, |pν | and |p| to refer to the magnitudes of the smallest representatives of
p0 ∈ C, pν ∈ C and p ∈ C3 in C/ ε2π2 ZZ, C/ 2π
Z and C3 / 2π
Z3 , respectively.
εn Z
εn Z
n
(b) For all p ∈ C × C3 with ε2n p0 , εn p having modulus less than one,
D̂n (p) = −ip0 + 12 ε2n p20 +
1
2
3
P
Hν,ν ′ pν pν ′ + O εn |p|3 + ε4n |p0 |3
ν,ν ′ =1
The higher order part O( · ) is uniform in n and L.
(c) We have, for all p ∈ C × C3 with ε2n |Im p0 | ≤ 1 and εn |Im p| ≤ 1,
3
P
D̂n (p) ≤ Γ5 |p0 | +
|pν |2
ν=1
and
∂ ℓi
ℓ D̂n (p) ≤ Γ5
i
∂pν
(
1
[1+|p0 |+|p|2 ]ℓi −1
if ν = 0, ℓi = 1, 2
1
[1+|p0 |+|p|2 ]ℓi /2−1
if 1 ≤ ν ≤ 3, 1 ≤ ℓi ≤ 4
(d) For all c > 0 and p ∈ C × C3 , with |p0 | + |p| ≥ c and |Im p| ≤ m̄(c),
3
P
D̂n (p) ≥ γ1 |p0 | +
|pν |2
ν=1
(e) For all c > 0 and all p in the set
p ∈ C × C3 |Im p| ≤ m̄(c), |p| ≥ c ∪ p ∈ C × C3 |Im p| ≤ m̄(c), |εn p0 | ≥ c
we have
Re D̂n (p) ≥ γ1 ε2n |p0 |2 +
Proof:
3
P
ν=1
|pν |2
(a) By the hypotheses on ĥ,
1−e−ĥ0 (εn p)
ε2n p2
=
ĥ0 (εn p)+O(ĥ0 (εn p)2 )
ε2n p2
18
=
3
1 2
2 εn p·Hp+O(|εn p| )
ε2n p2
for εn p is a real neighbourhood of 0. This is strictly positive and bounded away from 0 on
some real neighbourhood of 0, uniformly in εn . Since ĥ is continuous and strictly positive
on IR3 \ 2πZZ3 , there is a constant γ1′ > 0, independent of εn , such that
1−e−ĥ0 (εn p)
ε2n |p|2
≥ γ1′
e−ĥ0 (εn p) ≥ γ1′
and
for all p ∈ IR3 . The claim now follows from Lemma B.1.a.
(b) Expanding
sin z
z
e−ĥ(εn p)
= 1 + O(|z|2 ), for |z| ≤ 1, and
3
3 P
Hν,ν ′ pν pν ′ + O εn |p|
= 1 − 21 ε2n
ν,ν ′ =1
gives, using Remark III.1.a,
h
ih
2
2 i
D̂n (p) = − ip0 + 21 ε2n p20 1 + O εn |p| + ε2n p0
+
= −ip0 + 21 ε2n p20 +
1
2
The claim now follows from
3
P
for
εn |p| ≤ 1
3
P
1
Hν,ν ′ pν pν ′ +
2
ν,ν ′ =1
O εn |p|3
Hν,ν ′ pν pν ′ + O εn |p|3 + ε2n |p0 ||p|2 + ε4n |p0 |3
ν,ν ′ =1
ε2n |p0 ||p|2 ≤ ε4n |p0 |3
1/3
εn |p|3
2/3
ZZ × 2π
ZZ3 , we may assume that ε2n Re p0 and
(c) Since D̂n (p) is periodic with respect to 2π
ε2n
εn
each εn Re pν , 1 ≤ ν ≤ 3 is bounded in magnitude by π. By Lemma B.1.b, sinz z ≤ 2 for
all z ∈ C with |Im z| ≤ 1. Since εn p runs over a compact set (independently of n and L),
e−ĥ(εn p) is bounded. So
2
1 2 2 −ĥ (ε p) sin 12 ε2n p0 2 ε p e 0 n
, ip0 e−ĥ0 (εn p) sin εn p0 ≤ const |p0 |
1 2
2 n 0
2
εn p0 ε p
2 n 0
and
1 − e−ĥ0 (εn p) 2
≤ const |εn2p| ≤ const |p|2
ε
2
n
εn
This gives the bound on D̂n (p).
The bounds
i2
h
h
i
1 2
sin(ε2n p0 )
2
2 sin( 2 εn p0 )
∂
∂
= 2 sin(εn p0 ) = O(1)
ε
= cos(ε2n p0 ) = O(1)
1 2
∂p0 n
∂p0
ε2n
2 εn
i
i2
h
h
1 2
sin(ε2n p0 )
2
2
2
∂2
∂ 2 2 sin( 2 εn p0 )
= −ε2n sin(ε2n p0 ) = O(ε2n )
=
2ε
cos(ε
p
)
=
O(ε
)
ε
1
2
2
2
0
2
n
n
n
εn
∂p0 n
∂p0
2 εn
h
h
i
i
1 −ĥ0 (εn p)
1 −ĥ0 (εn p)
∂2
∂
= O(|p|)
= O(1 + ε2n |p|2 ) = O(1)
∂pν ε2n e
∂p2ν ε2n e
h
i
h
i
2
4
3
∂3
1 −ĥ0 (εn p)
∂4
1 −ĥ0 (εn p)
e
=
O(ε
|p|+ε
|p|
)
e
= O(ε2n +ε4n |p|2 +ε6n |p|4 )
n
n
∂p3 ε2
∂p4 ε2
ν
n
ν
n
= O(ε2n )
= O(εn )
19
together with
ε2n ≤
ε2n |p0 | ≤ const
const
1+|p0 |+|p|2
yield the bounds on the derivatives.
(d) Write p = P + iQ with P = (P 0 , P), Q = (Q0 , Q) ∈ IR × IR3 . We may choose m̄(c)
sufficiently small that, if |P + iQ| ≥ c and |Q| ≤ m̄(c), then
|P 0 + iQ0 | +
3
P
ν=1
|Pν
3
3
3
P
P
P
2
2
2
+ iQν | ≤ |P 0 | +
|Pν | + |Q0 | +
|Qν | ≤ 2 |P 0 | +
|Pν |
2
ν=1
ν=1
ν=1
If γ1 > 0 is chosen small enough and Γ5 is chosen large enough, then, for all such P , Q, we
have, by parts (a) and (c),
3
P
D̂n (P ) ≥ 4γ1 |P 0 | +
|Pν |2
ν=1
≥ 2γ1 |P 0 + iQ0 | +
3
P
ν=1
|Pν + iQν |2
3
P
|Pν + iQν |2 |Q|
D̂n (P + iQ) − D̂n (P ) ≤ 2 Γ5 1 + |P 0 + iQ0 | +
ν=1
Recalling that |P 0 + iQ0 | +
small enough that
P3
ν=1 |Pν
2 Γ5 m̄(c) ≤ 21 γ1 min
2
+ iQν | ≥ min
n
c c2
,
2 4
o
n
and
c c2
,
2 4
o
, it now suffices to choose m̄(c)
2 Γ5 m̄(c) ≤ 12 γ1
(e) Again write p = P + iQ with P = (P 0 , P), Q = (Q0 , Q) ∈ IR × IR3 . Since D̂n (P + iQ) is
periodic with respect to P ∈ 2π
Z × 2π
Z3 , we may assume that ε2n P 0 ≤ π and εn Pν ≤ π,
ε2n Z
εn Z
for each 1 ≤ ν ≤ 3. If the constant γ1 was chosen small enough, then, as in part (a),
Re D̂n (P ) =
1 2 2 −ĥ0 (εn P)
2 εn P 0 e
sin 12 ε2n P 0
1 2
2 εn P 0
2
+ P2
1 − e−ĥ0 (εn P)
2 2
2
≥
2γ
ε
P 0 + |P|
1
n
ε2n P2
Hence it suffices to prove that it is possible to choose m̄ = m̄(c) so that
Re D̂n (P + iQ) − Re D̂n (P ) ≤ γ1 |P|2
when |P| ≥ c and |Q| ≤ m̄
(III.2)
and that
Re D̂n (P + iQ) − Re D̂n (P ) ≤ γ1 c2
when |P| ≤ c, |εn P 0 | ≥ c and |Q| ≤ m̄
20
(III.3)
This is a consequence of the following bounds on the derivatives of the real parts of the
three terms making up D̂n (P + iQ) in Remark III.1.a. For the first term,
2 1 2
d 2
ε (P 0 + itQ0 )2 e−ĥ0 (εn P) sin 2 εn (P 0 + itQ0 ) 1 2
dt n
ε (P 0 + itQ0 )
2 n
2
≤ const εn |P 0 + iQ0 | |Q0 | + ε2n |P 0 + iQ0 |2 ε2n |Q0 |
≤ const m̄
h
i sin 1 ε2 (P + itQ ) 2 0
0
−ĥ0 (εn P+itεn Q)
−ĥ0 (εn P)
2 n
=0
+ itQ0 ) e
−e
1 2
ε
(
P
+
it
Q
)
0
0
n
t=0
2
i sin 1 ε2 (P + itQ ) 2 h
d2 2
0
0
2 n
2 εn (P 0 + itQ0 )2 e−ĥ0 (εn P+itεn Q) − e−ĥ0 (εn P)
dt
1 2
ε
(
P
+
it
Q
)
0
0
2 n
2 ≤ const ε2n |Q0 |2 + ε2n |P 0 +iQ0 | |Q0 | εn |Q| + ε2n |Q0 | + ε2n |P 0 +iQ0 |2 εn |Q| + ε2n |Q0 |
d
Re ε2n (P 0
dt
2
≤ const m̄2
For the second term,
1−e−ĥ0 (εn P+itεn Q) d
dt Re
ε2n
=0
t=0
d2 1−e−ĥ0 (εn P+itεn Q) ≤ const 12 εn |Q| 2 ≤ const m̄2
2
2
εn
εn
dt
For the third term,
2
d
sin
ε
(
P 0 + itQ0 ) n
−
ĥ
(ε
P
)
Re i(P 0 + itQ0 ) e 0 n
dt
ε2 (P 0 + itQ ) n
≤ const |Q0 | + const |P 0 +
0
2
iQ0 | εn |Q0 |
≤ const m̄
i sin ε2 (P + itQ ) h
d
0
0 n
−
ĥ
(ε
P
+itε
Q)
−
ĥ
(ε
P
)
n
Re i(P 0 + itQ0 ) e 0 n
−e 0 n
dt
2
εn (P 0 + itQ0 ) t=0 h
i
2
−ĥ0 (εn P) sin εn (P 0 ) d
= Re iP 0 dt ĥ0 (εn P + itεn Q)
e
2
εn P 0 t=0
−ĥ (ε P) sin ε2n (P 0 ) ′
= εn P 0 Q · ĥ0 (εn P) e 0 n
ε2n P 0 ≤ const ε2n |P 0 | |P| m̄
≤ const |P| m̄
i sin ε2 (P + itQ ) h
d2
0
0 n
2 i(P 0 + itQ0 ) e−ĥ0 (εn P+itεn Q) − e−ĥ0 (εn P)
dt
2
εn (P 0 + itQ0 ) 2 ≤ const |Q0 | εn |Q| + ε2n |Q0 | + |P 0 +iQ0 | εn |Q| + ε2n |Q0 |
≤ const m̄2
Now choose m̄ = m̄(c) small enough that (III.2) and (III.3) are satisfied.
21
IV. The Covariance
The covariance for the fluctuation integral in [parabolic-all.tex] is
C (n) = ( La2 Q∗ Q + ∆(n) )−1
where
∆(n) =
(
1l + Qn Qn Dn−1 Q∗n
−1
if n ≥ 1
Qn
)
(n)
: H0
(n)
→ H0
D0
if n = 0
See [parabolic-all.tex, (I.15) and (I.14)]. In Lemma IV.2 we study the properties of
∆(n) and in Corollary IV.5 we study the properties of C (n) and its square root.
Remark IV.1 Let n ≥ 1.
(a) The operator ∆(n) is the periodization of a translation invariant operator ∆(n) , acting
on L2 ZZ × ZZ3 , whose Fourier transform is
−1
X
(n)
2q
−1
ˆ
∆ (k) = Q̂n (k) 1 + Q̂n (k)
un (k + ℓ) D̂n (k + ℓ)
ℓ∈B̂n
=
Q̂n (k) D̂n (k)
P
D̂n (k) + Q̂n (k) ℓ un (k + ℓ)2q D̂−1
n (k + ℓ)D̂n (k)
with k ∈ C × C3 , where un (p) and B̂n were defined in parts (b) and (e) of Remark II.1,
respectively.
ˆ (n) (k) is invariant under kν → −kν for each 1 ≤ ν ≤ 3.
(b) ∆
ˆ (n) (k) has nonnegative real part when k is real.
(c) ∆
Lemma IV.2 There are constants(1) m1 > 0, γ2 , Γ6 and Γ7 , such that, for L > Γ2 , the
following hold.
ˆ (n) (k) is analytic on |Im k| < 3m1 .
(a) ∆
(b) For all k ∈ C × C3 with |Im k| ≤ 3m1 .
ˆ (n) (k) = −ik0 +
∆
ˆ (n) (k) D̂−1
∆
n (k) = 1 +
ik0
an
1
an
+
ε2n 2
k0
2
+ O |k|2
+
1
2
The higher order part O( · ) is uniform in n and L.
(1)
Recall Convention I.2.
22
3
P
Hν,ν ′ kν kν ′ + O |k|3
ν,ν ′ =1
(n) ˆ (n) (k), ∂ 2
ˆ (n) (k) ≤ Γ7 for all 0 ≤ ν, ν ′ ≤ 3 and
ˆ (k) ≤ 2a and ∂ ∆
(c) ∆
∂kν
∂kν ∂kν ′ ∆
k ∈ C × C3 with |Im k| < 3m1 .
(d) There is a function ρ(c) > 0, which is defined for all c > 0 and which depends only on
m1 , q, ĥ0 and a and, in particular, is independent of n and L, such that
(n)
ˆ
Re ∆
(k) ≥ ρ(c)
for all k ∈ C × C3 with |k| ≥ c and |Im k| ≤ 3m1 .
(e) For all k ∈ C × C3 with |Im k| ≤ 3m1 and |Re kν | ≤ π for all 0 ≤ ν ≤ 3,
(n) ∆
ˆ (k) ≥ γ2 D̂n (k)
ˆ (n) (p) is analytic on |Im p| ≤ 3m1 . Furthermore, for all p ∈ C × C3 with
(f ) D̂−1
n (p) ∆
|Im p| ≤ 3m1 ,
−1
Γ6
D̂ (p) ∆
ˆ (n) (p) ≤
P3
n
1 + |p0 | + ν=1 |pν |2
Here, as usual, |p0 | and |pν | refer to the magnitudes of the smallest representatives of
Z and C/ 2π
Z respectively.
p0 ∈ C and pν ∈ C in C/ 2π
ε2 Z
εn Z
n
ˆ (n) (k) ≤ Γ7 |kν | for all 1 ≤ ν ≤ 3 and k ∈ C × C3 with |Im k| < 3m1 .
(g) ∂∂kν ∆
Proof: We first prove part (b). Using that
◦ un (k)2q = 1 + O |k|2 by Lemma II.2.b
◦ Q̂n (k) = an + O |k|2 by Proposition II.4.b
◦ |D̂n (k)| ≤ const |k0 | + |k|2 by Lemma III.2.c
and that, for ℓ 6= 0,
Q3
1
◦ |un (k + ℓ)|2q ≤ const |k|2q ν=0 (|ℓ|ν |+π)
2q by Lemma II.2.a
◦ |D̂−1
n (k + ℓ)| ≤ const by Lemma III.2.d
we obtain, by Remark IV.1.a and Lemma III.2.b,
ˆ (n) (k) =
∆
Q̂n (k)D̂n (k)
Q̂n (k)un (k)2q + D̂n (k) + O |k|3
an + O |k|2
= D̂n (k)
an + D̂n (k) + O |k|2
n
o
= D̂n (k) 1 − a1n D̂n (k) + O |k|2
= −ik0 + 21 ε2n k02 +
= −ik0 +
1
an
+
1
2
3
P
Hν,ν ′ kν kν ′ −
ν,ν ′ =1
ε2n 2
k0
2
+
1
2
23
3
P
1
D̂n (k)2
an
+ O |k|3
Hν,ν ′ kν kν ′ + O |k|3
ν,ν ′ =1
ˆ (n) (k) is analytic and bounded
This also shows that, in a neighbourhood of the origin, ∆
in magnitude by 2a. For the second expansion,
2
a
+
O
|k|
(n)
n
ˆ (k)D̂−1 (k) =
∆
n
an + D̂n (k) + O |k|2
= 1 − a1n D̂n (k) + O |k|2
2
0
= 1 + ik
+
O
|k|
an
(n) ˆ (k) ≤ 2a of part (c). We have
(a), (c) We first prove the analyticity and the bound ∆
already done so for a neighbourhood of the origin. So it suffices to consider |k| > c0 , for
ˆ (n) is periodic with respect to 2πZZ × 2πZZ3 , it suffices to
some suitably small c0 . Since ∆
consider k in the set
M (m1 ) =
k ∈ C × C3 |Re kν | ≤ π for all 0 ≤ ν ≤ 3, |Im k| ≤ 3m1 , |k| > c0
Recall, from Remarks II.1.d and III.1.b, that un (p) and D̂n (p) are entire. If m1 is small
enough, then, by Lemma III.2.d, the functions k 7→ D̂n (k+ℓ), ℓ ∈ B̂n , n ≥ 1, have no zeroes
P
in M (m1 ). Hence each term in the infinite sum 1 + Q̂n (k) ℓ∈B̂n un (k + ℓ)2q D̂−1
n (k + ℓ)
is analytic and it suffices to prove that, for k ∈ M (m1 ), the sum converges uniformly and
P
6
2q
1 + Q̂n (k)
D̂−1
n (k + ℓ) ≥ 10 .
ℓ∈B̂n un (k + ℓ)
By Proposition II.4.a, Remark II.1.d and Lemma III.2.d, there is an l > 0 such that
|Q̂n (k)|
X X
un (k + ℓ)2q D̂−1 (k + ℓ) ≤ 6 a
n
5
ℓ∈B̂n
|ℓ|≥l
ℓ∈B̂n
|ℓ|≥l
1
γ1 π
3
Q
ν=0
2q
24
|ℓν |+π
≤
const
l
≤
2
10
(IV.1)
Hence we have uniform convergence and
X
X
2
−1
2
−1
1
+
Q̂
(k)
u
(k
+
ℓ)
D̂
(k
+
ℓ)
≥
1
+
Q̂
(k)
u
(k
+
ℓ)
D̂
(k
+
ℓ)
−
n
n
n
n
n
n
2
10
ℓ∈B̂n
|ℓ|<l
ℓ∈B̂n
For real k and every ℓ ∈ B̂n , the real part of Q̂n (k)un (k + ℓ)2q D̂−1
n (k + ℓ) is nonnegative.
Consequently, if m1 is small enough and |Im k| ≤ 3m1
X
2
(IV.2)
Re Q̂n (k)
un (k + ℓ)2q D̂−1
n (k + ℓ) ≥ − 10
ℓ∈B̂n
|ℓ|<l
since, for all k ∈ M (m1 ) and ℓ ∈ B̂n with |ℓ| < l,
◦ D̂n (k + ℓ) is bounded away from zero (by Lemma III.2.d) and has bounded first
derivative (by Lemma III.2.c) and
24
◦ un (k + ℓ) and Q̂n (k) are bounded with bounded first derivatives (by analyticity, Remark II.1.d and Proposition II.4.a).
So, by (IV.2)
X
2q
−1
8
u
(k
+
ℓ)
D̂
(k
+
ℓ)
1
+
Q̂
(k)
≥ 10
n
n
n
ℓ∈B̂n
|ℓ|<l
All of this is uniform in n and L.
ˆ (n) (k), ∂ 2
ˆ (n) (k) ≤ Γ7 , with
Shrinking m1 by a factor of 2, the bounds ∂∂kν ∆
∆
∂kν ∂kν ′
Γ7 being the maximum of 4a divided by the original 3m1 (for first order derivatives) and
16a divided by the square of the original 3m1 (for second order derivatives), follow by the
Cauchy integral formula.
(d) Denote the real and imaginary parts of the numerator and denominator by
i
h
X
2q
−1
Rn = Re Q̂n (k) Rd = Re 1 + Q̂n (k)
un (k + ℓ) D̂n (k + ℓ)
ℓ∈B̂n
In = Im Q̂n (k)
h
i
X
Id = Im 1 + Q̂n (k)
un (k + ℓ)2q D̂−1
(k
+
ℓ)
n
ℓ∈B̂n
By Proposition II.4.a and (IV.1), (IV.2),
Rn ≥
a
2
Rd ≥
6
10
By Proposition II.4.a, Remark II.1.d and Lemma III.2.d,
X
un (k + ℓ)2q D̂−1
(k
+
ℓ)
Id ≤ Q̂n (k)
≤ aI˜d (c)
n
P
where I˜d (c) = 65 a ℓ∈B̂n
ℓ∈B̂n
1
γ1 min{c/2,c2 /4}
Q3
242q
ν=0 (|ℓν |+π)2q .
When k is real, Q̂n (k) is real. By
analyticity and Proposition II.4.a, the first order derivatives of Q̂n (k) are bounded. So if
m1 is small enough, |In | ≤ 10I˜2 (c) . So
d
(n)
ˆ
Re ∆
(k) =
Rn Rd +In Id
R2d +Id2
≥
a(3/10)−a(2/10)
(1+aI˜d (c))2
=
a
10(1+aI˜d (c))2
(e) By Remark II.1.d, Proposition II.4.a and Lemma III.2.d,
X
2q
−1
un (k + ℓ) D̂n (k + ℓ)
1 + Q̂n (k)
ℓ∈B̂n
X Q̂n (k) un (k + ℓ)2q D̂−1
≤ Q̂n (k) un (k)2q D̂−1
n (k + ℓ)
n (k) + 1 +
≤ 56 a
24 8q
π
06=ℓ∈B̂n
X
−1 D̂n (k) + 1 + 6 a
5
06=ℓ∈B̂n
25
1
γ1 π
3
Q
ν=0
2q
24
|ℓν |+π
and so, by Lemma III.2.c,
X
D̂n (k)1 + Q̂n (k)
un (k + ℓ)2q D̂−1
(k
+
ℓ)
n
ℓ∈B̂n
≤
6
24 8q
a
5
π
h
π + 1 + 3(π + 1) 1 + 65 a
+ Γ5
2
P
06=ℓ∈B̂n
1
γ1 π
3
Q
ν=0
2q
24
|ℓν |+π
i
(f) By part (c) of this lemma, Remark III.1.b and Lemma III.2.d, it suffices to consider
p = k with |Im k| ≤ 3m1 and |k| < c0 < π where c0 is sufficiently small. Now
ˆ (n) (k) =
D̂−1
n (k)∆
Q̂n (k)
P
Q̂n (k)un (k)2q + D̂n (k) 1 + Q̂n (k) 06=ℓ∈B̂n un (k + ℓ)2q D̂−1
n (k + ℓ)
By Proposition II.4.a, Remark II.1.d, Lemma II.2.b and parts (c) and (d) of Lemma III.2,
we may choose c0 > 0 so that
X
a
2q
−1
un (k + ℓ) D̂n (k + ℓ) < 5
D̂n (k) 1 + Q̂n (k)
06=ℓ∈B̂n
and
Q̂n (k) un (k)2q − 1| <
a
5
for all |k| < c0 with |Im k| ≤ 3m1 . That does it.
(g) Fix any 1 ≤ ν ≤ 3. Observe that
(n)
ˆ
∂∆
∂kν
(k) = kν
Z
0
1
(n)
ˆ
∂ 2∆
∂kν2
k(t) dt
with k(t)ν ′ =
kν ′ if ν ′ 6= ν
tkν , if ν = ν ′
ˆ (n)
∆
k(0)
= 0, by Remark IV.1.b. Now apply the bound on the second derivative
since ∂∂k
ν
from part (c).
We next consider the “resolvents”
(n)
Rζ
= ζ1l −
∗
a
L2 Q Q
− ∆(n)
−1
√
in preparation for studying Cn and Cn . We wish to apply Lemma A.13, with A → Dn =
a
Q∗ Q + ∆(n) (scaled). By Lemmas A.9.b and A.5.b, for each ℓ, ℓ′ ∈ B̂ = B̂+ ,
L2
ak (ℓ, ℓ′ ) → dn,k (ℓ, ℓ′ ) =
a
L2 u+ (k
ˆ (n) (k + ℓ)
+ ℓ)q u+ (k + ℓ′ )q + δℓ,ℓ′ ∆
26
(IV.3)
Here we are denoting
3
2π
◦ momenta dual to the L–lattice by k ∈ IR/ L
Z × IR3 / 2π
Z
Z
and
2Z
L
3
◦ momenta dual to the unit lattice ZZ × ZZ by k ∈ IR/2πZZ × IR3 /2πZZ3 and decom
2π
Z × IR3 / 2π
ZZ3
pose k = k + ℓ or k = k + ℓ′ with k in a fundamental cell for IR/ L
2Z
L
3
3
2π
Z/2πZZ × 2π
Z
Z
/2πZ
Z
= B̂+ .
and ℓ, ℓ′ ∈ L
2Z
L
We also use dn,k to denote the B̂+ × B̂+ matrix dn,k (ℓ, ℓ′ ) ℓ,ℓ′ ∈B̂+ . Observe, by Remark
II.1.d and Lemma IV.2.a, that dn,k ℓ, ℓ′ is analytic in the strip |Im k| < 3m1 .
Let vℓ ℓ∈B̂+ and wℓ ℓ∈B̂+ be any vectors in L2 (B̂+ ). Then, if k = L−1 (k),
v̄, dn,k w
h P
i X
ih P
ˆ (n) (L−1 (k)+ℓ)vℓ wℓ
u+ (L−1 (k)+ℓ)qvℓ
= La2
u+ (L−1 (k)+ℓ)qwℓ +
∆
ℓ∈B̂ +
=
X
ℓ∈B̂ +
ℓ∈B̂ +
(s)
vL(ℓ) dn,k (ℓ, ℓ′ ) wL(ℓ′ )
ℓ,ℓ′ ∈B̂1
where the “scaled” matrix
(s)
dn,k (ℓ, ℓ′ ) =
a
L 2 u+
q
q
ˆ (n) L−1 (k+ℓ)
L−1 (k+ℓ) u+ L−1 (k+ℓ′ ) + δℓ,ℓ′ ∆
Lemma IV.3 There are constants m2 , λ0 , Γ8 > 0, such that, for all L > Γ2 and k ∈ C×C3
with |Im k| < 3m2 , the following hold.
(a) Write k = L−1 (k). For both the operator and (matrix) ℓ1 –ℓ∞ norms
(s)
kdn,k k ≤ Γ8
(b) Let λ ∈ C be within a distance
λ0
L2
kdn,k k ≤ Γ8
of the negative real axis. Then the resolvent
−1 λ1l − d(s)
≤ Γ8 L2
n,k
This is true for both the operator and (matrix) ℓ1 –ℓ∞ norms.
Proof: Since dn,k+p (ℓ, ℓ′ ) = dn,k (ℓ + p, ℓ′ + p), for all p, ℓ, ℓ′ ∈ B̂+ , we may always assume
π
for 1 ≤ ν ≤ 3.
that |Re k0 | ≤ Lπ2 and |Re kν | ≤ L
(a) By Lemmas II.3.a and IV.2.c, the matrices
a
a
q
q
′ q
−1
−1
′ q
u
(k
+
ℓ)
u
(k
+
ℓ
)
and
u
L
(k+ℓ)
u
L
(k+ℓ
)
2
2
+
+
+
+
′
+
L
L
ℓ,ℓ ∈B̂
ℓ,ℓ′ ∈B̂1
have finite L1 –L∞ norms and the matrix elements of
ˆ (n) (k + ℓ) ′ + and δℓ,ℓ′ ∆
ˆ (n) L−1 (k+ℓ)
δℓ,ℓ′ ∆
ℓ,ℓ ∈B̂
ℓ,ℓ′ ∈B̂ 1
27
are all bounded.
(b) Case |k| ≤ c0 , with c0 being chosen later in this case: We first consider those diagonal
(s)
matrix elements of dn,k (ℓ, ℓ′ ) having k, ℓ such that |L−1 (k + ℓ)| < c̃0 , where c̃0 > 0 is a
small number to be chosen shortly. This, and all other constants chosen in the course of
this argument are to be independent of L. Then, by Lemma IV.2.b, we have the following.
ˆ (n) L−1 (k +ℓ) ≥ c12 , provided
◦ If at least one of ℓν , 1 ≤ ν ≤ 3 is nonzero, then Re ∆
L
m2 and c̃0 are chosen small enough. Here c1 and the constraints on m2 and c̃0 depend
only on the largest and smallest eigenvalues of Hν,ν ′ , an and the O(|k|3 ). To see
this, denote by k and ℓ the spatial parts of k and ℓ and observe that
∗ Re L−1 (k + ℓ) ≥ L1 max π, 21 |ℓℓ| ,
2
∗ Im L−1 (k + ℓ) = L1 Im k ≤ 3m
L and
2
∗ Im L−1 (k + ℓ)0 = L12 Im k0 ≤ 3m
2 .
L
3 In controlling the contribution from O L−1 (k+ℓ) when |ℓL02| is larger than |ℓLℓ| , we
2
ε2 have to use that, in this case, the real part of a1n + 2n L−1 (k + ℓ) 0 is at least a
strictly positive constant times
ℓ20
.
L4
ˆ (n) L−1 (k + ℓ) ≥
◦ If ℓν = 0 for all 1 ≤ ν ≤ 3 but ℓ0 6= 0, then −sgn ℓ0 Im ∆
provided m2 and c̃0 is chosen small enough. To see this observe that
√ π
∗ Re L−1 (k + ℓ) = L1 |Re k| ≤ 3 L
,
3m
1
−1
2
∗ Im L (k + ℓ) = L Im k ≤ L ,
∗ sgn ℓ0 Re L−1 (k + ℓ)0 ≥ L12 max π, 12 |ℓ0 | ,
∗ Re L−1 (k + ℓ)0 ≤ 2L3 2 |ℓ0 | and
∗ Im L−1 (k + ℓ)0 = 12 Im k0 ≤ 3m22 .
L
π
2L2 ,
L
◦ If ℓ 6= 0, then, by parts (a) and (b) of Lemma II.3,
a L 2 u+
L−1 (k+ℓ)
2q ≤
a
L2 |k|
h Q
3
ν=0
24
|ℓν |+π
i2q
2q
ˆ (n) L−1 (k) ≥ a 2 , provided m2 and c̃0 are
◦ If ℓ = 0 then Re La2 u+ L−1 (k)
+∆
2L
chosen small enough. To see this observe that
4
by Lemma
∗ u+ L−1 (k) −1 ≤ 43 |k|2 , by Lemma II.3.c and u+ L−1 (k) ≤ 24
π
II.3.a.
ˆ (n) L−1 (k) ≥ −c2 m22 + c̃0 |k| 2 where c2 depends only on an , the largest
∗ Re ∆
L
L
eigenvalue of Hν,ν ′ , and the O(|k|3 ).
Note that we have now fixed c̃0 . Now we consider the remaining matrix elements.
◦ For the remaining diagonal matrix elements we have |L−1 (k + ℓ)| ≥ c̃0 and then, by
ˆ (n) L−1 (k+ℓ) ≥ ρ(c̃0 )
Lemma IV.2.d, Re ∆
28
(s)
◦ Finally, the off–diagonal matrix elements of dn,k obey, by parts (a) and (b) of Lemma
II.3,
a u+
L2
−1
L
q
−1
(k+ℓ) u+ L
(s)
q (k+ℓ ) ≤
′
a
|k|
L2
h Q
3
ν=0
24
|ℓν |+π
iq h Q
3
ν=0
24
|ℓ′ν |+π
iq
Hence the off–diagonal part of dn,k has Hilbert-Schmidt, matrix and, as q > 1, L1 –L∞
norms all bounded by a universal constant times La2 |k|.
(s)
Thus λ1l − dn,k has diagonal matrix elements of magnitude at least
min
c
1
L2
,
π
a
2L2 , 2L2 , ρ(c̃0 )
−
a
L2 |k|
h Q
3
ν=0
24
|ℓν |+π
i2q
−
λ0
L2
and off diagonal part with L1 –L∞ norm bounded by a universal constant times La2 |k|. It
now suffices to choose c0 and λ0 small enough that every diagonal matrix element has
magnitude at least 2L1 2 min c1 , π2 , a2 , ρ(c̃0 ) and the off diagonal part has L1 –L∞ norm
bounded by 4L1 2 min c1 , π2 , a2 , ρ(c̃0 ) and then do a Neumann expansion.
(b) Case |k| ≥ c0 , with the c0 just chosen: We may assume that |Re kν | ≤ π for each
0 ≤ ν ≤ 3.
◦ If |L−1 (k + ℓ)| < c̃0 and ℓν 6= 0 for at least one 1 ≤ ν ≤ 3 or if |L−1 (k + ℓ)| ≥ c̃0 , then
ˆ (n) L−1 (k + ℓ) ≥ min ρ(c̃0 ), c12 . The proof of this given in the case |k| ≤ c0
Re ∆
L
applies now too.
◦ If |L−1 (k + ℓ)| < c̃0 , ℓν = 0 for all ν ≥ 1 then
(n)
∆
ˆ
ˆ (n) L−1 (Re k + ℓ) ≥ 0
Re ∆
ˆ (n) L−1 (Re k + ℓ) ≤ 4πΓ7
L−1 (k + ℓ) − ∆
1
L2 |Im k|
The first bound follows immediately from Remark IV.1.c. The second bound follows
from Lemma IV.2.c (for the L12 Im k0 = Im (L−1 (k + ℓ)0 contribution) and Lemma
IV.2.g (for the L1 Im kν contribution, with 1 ≤ ν ≤ 3, — note that on the line segment
∆
ˆ (n) is bounded by Γ7 |L−1 (k + ℓ)ν | = ΓL7 |kν | ≤
from L−1 (Re k + ℓ) to L−1 (k + ℓ), ∂∂k
ν
2
Γ7 π+3m
). Furthermore
L
ˆ (n) L−1 (Re k + ℓ) ≥ π 2
−sgn ℓ0 Im ∆
2L
(n) −1
c0
∆
ˆ
,
L (Re k) ≥ γ1 γ2 min √
2
c20
2
if ℓ0 6= 0
1
L2
if ℓ0 = 0
In the case ℓ0 6= 0, the proof of the bound given in the case |k| ≤ c0 applies now too.
(Just apply it to Re k.) The bound for the case ℓ0 = 0 follows from Lemma IV.2.e
and Lemma III.2.a.
29
◦ For all ℓ, by parts (a) and (e) of Lemma II.3,
3
h Q
u+ L−1 (k+ℓ) q ≤
ν=0
24
|ℓν |+π
iq
h Q
3
Im u+ L−1 (k+ℓ) q ≤ 16q |Im k|
ν=0
We split λ1l −
(s)
dn,k
where
24
|ℓν |+π
into three pieces
(s) λ1l − dn,k (ℓ, ℓ′ ) = D(ℓ, ℓ′ ) − P (ℓ, ℓ′) + I(ℓ, ℓ′)

ˆ (n) L−1 (k+ℓ)

if ℓν 6= 0 for some ν ≥ 1
∆
(n)
−1
ˆ
with dℓ = λ− − ∆
L (k+ℓ)
if |L−1 (k + ℓ)| ≥ c̃0

∆
ˆ (n) L−1 (Re k+ℓ) otherwise
D(ℓ, ℓ′ ) = δℓ,ℓ′ dℓ
λ− = min{Re λ, 0} + iIm λ
q
P (ℓ, ℓ′ ) = La2 v(ℓ)v(ℓ′ ) with v(ℓ) = Re u+ L−1 (k+ℓ)
We have chosen λ− so that Re λ− ≤ 0 and |λ − λ− | ≤
1
′
dℓ δℓ,ℓ
+
X
a 1
L2 dℓ′′
v(ℓ′′ )2
κ=
ℓ′′
λ0
.
L2
As P is a rank one operator
′
1 a v(ℓ) v(ℓ )
1−κ L2 dℓ dℓ′
(D − P )−1 (ℓ, ℓ′ ) =
with
We have shown above that
λ1
|dℓ | ≥ L
2
0
if |L−1 (k + ℓ)| < c̃0 , ℓν = 0 for all ν ≥ 1
Re dℓ ≤
λ1
− L2 otherwise
h Q
iq
3
24
|v(ℓ)| ≤
|ℓν |+π
ν=0
|I(ℓ, ℓ′)| ≤
iq
λ0
L2
+ 4πΓ7 L12 |Im k| δℓ,ℓ′ + 32q |Im k|
a
L2
h Q
3
ν=0
24
|ℓν |+π
iq h Q
3
ν=0
24
|ℓ′ν |+π
iq
c20 c0
,
γ
γ
provided we choose 0 < λ1 ≤ min c1 , ρ(c̃0 ), π2 , γ1 γ2 √
1
2
2 .
2
1
Since Re z ≤ 0 ⇒ Re z ≤ 0, and v(ℓ) ∈ IR for all ℓ, we have that Re κ ≤ 0 so that
1 1−κ ≤ 1 and
2
Γ′
kIkℓ1 −ℓ∞ ≤ L52 λ0 + |Im k|
k(D − P )−1 kℓ1 −ℓ∞ ≤ L
λ2
with the constant λ2 > 0 depending only on a and λ1 and the constant Γ′5 depending only
λ2
on Γ7 , a and q. It now suffices to choose λ0 and m2 smaller than 12Γ
′ and use a Neumann
5
expansion to give
λ1l − d(s) −1 1 ∞ ≤ 2 L2
n,k
λ2
ℓ −ℓ
30
Proposition IV.4 Let m2 , λ0 , Γ8 be as in Lemma IV.3 and use IR− to denote the negative
real axis in C. Set
OC = z ∈ C dist(z, IR− ) >
O = z ∈ C dist(z, IR− ) >
λ0
,
2L2
λ0
3L2 ,
|z| < Γ8 + 1
|z| < Γ8 + 2
and let
◦ C = ∂OC , oriented counterclockwise
◦ f : O → C be analytic.
Γ
λ0
L2
ΓD
Then f ( La2 Q∗ Q + ∆(n) )(s) , defined by (A.11) and Lemma A.14.a, exists and there is a
constant(2) Γ9 such that
kf ( La2 Q∗ Q + ∆(n) )(s) km2 ≤ Γ9 L7 sup |f (ζ)|
ζ∈C
Proof:
(s)
Apply Lemma A.13 with âk (ℓ, ℓ′ ) = dn,k (ℓ, ℓ′ ) and
(n)
Xfin = X1
Zfin = ε21 ZZ × ε1 ZZ3
(n+1)
Zcrs = ZZ × ZZ3
Xcrs = X0
B = B1
and m = 3m2 , m′ = 2m2 and m′′ = m2 . Then volc = 1, B̂1 = L5 and the lemma gives
X f ( a2 Q∗ Q + ∆(n) )(s) ′′ ≤ Cm′ −m′′ |C| sup |f (ζ)| sup
(ζ1l − dˆ(s) )−1 (ℓ, ℓ′ )|
n,k
L
2π volc
m
|Im k|=m′
ζ∈C
ζ∈C
≤
≤
Cm′ −m′′
2π
Cm′ −m′′
2π
(Γ8 + 1)(3π + 2) sup |f (ζ)| L5
ζ∈C
sup
|Im k|=m′
ζ∈C
(Γ8 + 1)(3π + 2) sup |f (ζ)|L5 Γ8 L2
ℓ,ℓ′ ∈B̂1
X (ζ1l − dˆ(s) )−1 (ℓ, ℓ′ )|
sup
n,k
ℓ∈B̂1
ℓ′ ∈B̂1
ζ∈C
by Lemma IV.3.b.
1
z,
Applying Proposition IV.4 with f (z) =
the principal value of the square root gives
f (z) =
√1
z
and f (z) =
√
z, where
√
z is
√
√
−1
Corollary IV.5 The operators C (n) , C (n) and
C (n)
all exist. There is a constant
Γ10 such that
p −1
−1 (n) p −1
L∗ C (n) L∗ −1 ≤ Γ10 L9
L∗ C L∗ L∗ C (n) L∗ ,
,
m
m
m
2
(2)
2
Recall Convention I.2.
31
2
V. The Green’s Functions
In this chapter, we discuss the inverses of the operators
Dn + Q∗n Qn Qn
These inverses, and variations thereof, are constituents of the leading part of the power series expansion of the background fields of [parabolic-all.tex], see [parabolic-all.tex,
Proposition I.16 and Proposition G.3]. In Proposition V.1, below, we show that for sufficiently small µ, the operators Dn + Q∗n Qn Qn − µ are invertible, and we estimate the
decay of the kernels Sn (µ)(x, y) of their inverses
−1
Sn (µ) = Dn + Q∗n Qn Qn − µ
By Remark II.5
∂ν Sn (µ)∗
where
−1
(+)
= Sn,ν
(µ)
−1
−1
(−)
(+)
(µ) = Dn∗ + Q(+)
Sn,ν
n,ν Qn Qn,ν − µ
(+)
(−)
(−)
∂ν Sn (µ)−1 = Sn,ν
(µ)
∂ν
−1
∂ν
(V.1)
−1
(−)
(−)
(V.2)
(µ) = Dn + Q(+)
Sn,ν
n,ν Qn Qn,ν − µ
and Qn,ν , Qn,ν were defined in (II.10). We shall write
(±)
(±)
Sn,ν
= Sn,ν
(0)
Sn = Sn (0)
(V.3)
The main result extends the statement of [parabolic-all.tex, Theorem I.15]. It is
Proposition V.1 There are constants µup , m3 > 0 and Γ11 , depending only on q, h0 and
a, and in particular independent of n and L > Γ2 , such that, for |µ| ≤ µup ,
(+)
(−)
(+)
(−)
the operators Dn + Q∗n Qn Qn − µ and Dn∗ + Qn,ν Qn Qn,ν − µ , Dn + Qn,ν Qn Qn,ν − µ
(+)
(−)
are invertible, and their inverses Sn (µ) and Sn,ν (µ) , Sn,ν (µ) , respectively, fulfill
(±)
k Sn (µ)km3 , k Sn,ν
(µ)km3 ≤ Γ11
(±)
(±)
k Sn (µ) − Sn km3 , k Sn,ν
(µ) − Sn,ν
km3 ≤ |µ| Γ11
This Proposition is proven following the proof of Lemma V.5.
32
Example V.2 As a model computation, we evaluate the inverse transform
s(x) =
Z
IR×IR3
dp0 d3 p
eip·x
−ip0 +m2 +p21 +p22 +p23 (2π)4
1
of ŝ(p) = −ip0 +m
2 +p2 . It is designed to mimic the behaviour of Sn in the limit n → ∞.
Write x = (t, x) ∈ IR × IR3 . We first compute the p0 integral. Observe that the integrand
has exactly one pole, which is at p0 = −i(m2 + p2 ), and that the eip0 t in the integrand
forces us to close the contour in the upper half plane when t > 0 and in the lower half
plane when t < 0. Thus
Z
∞
−∞
dp0
eip0 t+ip·x
−ip0 +m2 +p2 (2π)4
=
0
2
1
e(m+p )t+ip·x (2π)
3
if t > 0
if t < 0
Hence s(x) = 0 for t > 0 and, for t < 0,
s(x) =
Z
IR3
=
2
2
d3 p
e−(m +p )|t| eip·x (2π)
3
−m2 |t|
=e
3 Z
Y
j=1
1
23 (π|t|)3/2
e−m
2
2
x
|t| − 4|t|
∞
2
e−|t|pj eipj xj
−∞
dpj
2π
e
Here are some observations about s(x).
x2
≥ m|x| (the minimum is at |t| =
◦ Since m2 |t| + 4|t|
large |x| in all directions.
|x|
),
2m
s(x) decays exponentially for
x2
◦ For x 6= 0, limtր0 e− 4|t| = 0, so s(x) is continuous everywhere except at x = 0.
◦ s(x) has an integrable singularity at x = 0. There are a number of ways to see this.
x2
|t| κ/2
For example, the inequality e− 8|t| ≤ const |x|
implies that |s(x)| is bounded
2
1
1
near x = 0 by a constant times |t|(3−κ)/2 |x|κ . This is integrable if 1 < κ < 3.
1
◦ If we send x → 0 along a curve with x2 = −4γ|t| ln |t|, s(x) ≈ const |t|3/2−γ
.
x2
P3
≤ const |t|2 , that t2 + j=1 x4j |s(x)| is bounded and expo◦ We see, using
1
nentially decaying. Note that t2 +Σ
4 has an integrable singularity at the origin, since
j xj
R∞
1
const
.
dt = p
−∞ t2 +Σ x4
4
j
x4j e− 8|t|
j
j
Σj x j
As preparation for and in addition to the position space estimates of Proposition V.1,
we also derive bounds on the Fourier transforms of these and related operators. To convert
bounds in momentum space into bounds in position space, we shall use Lemma A.11, with
Xfin = Xn
Zfin = ε2n ZZ × εn ZZ3
(n)
Xcrs = X0
33
Zcrs = ZZ × ZZ3
B = Bn (V.4)
We shall routinely use |p0 |, |pν | and |p| to refer to the magnitudes of the smallest representatives of p0 ∈ C, pν ∈ C and p ∈ C3 in C/ 2π
Z, C/ 2π
Z and C3 / 2π
Z3 , respectively.
ε2n Z
εn Z
εn Z
The operators Sn (µ) act on functions on the lattice Xn , but they are only translation
(n)
invariant with respect to the sublattice X0 . An exponentially decaying operator which
is fully translation invariant, and has the same local singularity as Sn , is the operator
−1
Sn′ = Dn + an exp{−∆n }
where
(V.5)
∆n = ∂0∗ ∂0 + ∂1∗ ∂1 + ∂2∗ ∂2 + ∂3∗ ∂3
−2
1−L
and an = a 1−L
−2n as in Proposition II.4.b, and the forward derivatives ∂ν are defined in
(II.8). Obviously Sn′ has Fourier transform
−1
Sbn′ (p) = D̂n (p) + an exp{−∆n (p)}
where ∆n (p) =
sin 1 ε2n p0 2
2
1 2
2 εn
+
3 P
sin 21 εn pν 2
ν=1
1
2 εn
Before we discuss the properties of Sn′ and of the difference δS = Sn − Sn′ we note
Remark V.3 The Fourier transform ∆n (p) of the four dimensional Laplacian ∆n is
entire. For p ∈ IR × IR3 ,
∆n (p) ≥ π22 |p0 |2 + |p|2
For p ∈ C × C3 with ε2n |Im p0 | ≤ 1 and εn |Im p| ≤ 1
∂
∆n (p)| ≤ 4|pν |
|∆n (p)| ≤ 4 |p0 |2 + |p|2
| ∂p
ν
ℓ
ℓ−2
∂
| ∂p
ℓ ∆n (p)| ≤ 4εn,ν if ℓ ≥ 2
ν
with the εn,ν of (II.6). For p ∈ C × C3 with |Im p| ≤ 1,
Re ∆n (p) ≥ −5π 2 + π12 |p0 |2 + |p|2
Proof: For the first two claims, just apply parts (a) and (b) of Lemma B.1. For the
derivatives, use
sin(ηθ) 2 sin(2ηθ)
sin(ηθ) 2
sin(2ηθ) for ℓ odd
ℓ−1 1
dℓ
d
=
⇒
= ±(2η)
dθ
η
η
η
η
dθℓ
cos(2ηθ) for ℓ even
For the final claim, write p = P + iQ with P , Q ∈ IR × IR3 . Then
Re ∆n (P + iQ) ≥ ∆n (P ) − Re ∆n (P + iQ) − ∆n (P )
≥ π22 |P 0 |2 + |P|2 − 4|Q||P + iQ|
≥ π22 |P + iQ|2 − 2 2 + π22 |Q| |P + iQ| − π22 |Q|2
2
≥ π12 |P + iQ|2 − π 2 2 + π22 + π22 |Q|2
34
By the resolvent identity
δS = Sn − Sn′ = −Sn′ Q∗n Qn Qn − an exp{−∆n } Sn
(n)
Sn and δS are translation invariant with respect to the sublattice X0 of Xn . By “Floquet
c p′ ) , p, p′ ∈ Xˆn vanish
theory” (see Lemma A.1), their Fourier transforms Sbn (p, p′ ) , δS(p,
(n,0)
(n,0) ′
(n)
unless πn (p) = πn (p ), i.e. unless there are k ∈ Xˆ0 and ℓ, ℓ′ ∈ B̂n such that
−1
c k (ℓ, ℓ′ ) =
p = k + ℓ , p′ = k + ℓ′ . The blocks Sbn,k
(ℓ, ℓ′ ) = Sbn−1 (k + ℓ, k + ℓ′ ) and δS
c + ℓ, k + ℓ′ ) are given by
δS(k
−1
Sbn,k
(ℓ, ℓ′ ) = D̂n (k + ℓ) δℓ,ℓ′ + un (k + ℓ)q Q̂n (k)un (k + ℓ′ )q
X
c k (ℓ, ℓ′ ) = −
Sbn′ (k+ℓ) un (k+ℓ)q Q̂n (k)un (k+ℓ′′ )q − an e−∆n (k+ℓ) δℓ,ℓ′′ Sbn,k (ℓ′′ , ℓ′ )
δS
ℓ′′ ∈B̂n
(V.6)
where un and Q̂n are given in parts (b) and (e) of Remark II.1.
Lemma V.4 There are constants(1) m4 > 0 and Γ12 , such that the following hold for all
L > Γ2 .
(a) Sbn′ (p) is analytic in |Im p| < 3m4 and obeys
∂2 ′
′
Γ12
2 Sb (p), ∂ 4 4 Sb′ (p) ≤
Sb (p) ≤
and
2
n
n
n
1+|p0 |+|p|
∂p
∂p
ν
0
Γ12
(1+|p0 |+|p|2 )3
for 1 ≤ ν ≤ 3 there.
(b) For all ℓ, ℓ′ ∈ B̂n , Sbn,k (ℓ, ℓ′ ) is analytic in |Im k| < 3m4 and obeys
n
3
3
Q
Q
Γ12
1
1
Sbn,k (ℓ, ℓ′ ) ≤
′ +
δ
3
′
3
q
ℓ,ℓ
2
′
2
(|ℓν |+1)
1+|ℓ0 |+Σ
|ℓν |
1+|ℓ |+Σ
|ℓ |
ν=1
0
ν=1
ν
ν=0
ν=0
there.
c k (ℓ, ℓ′ ) is analytic in |Im k| < 3m4 and obeys
(c) For all ℓ, ℓ′ ∈ B̂n , δS
1 3
c
δS k (ℓ, ℓ′ ) ≤ Γ12 exp − 40
Σν=0 |ℓν |2 δℓ,ℓ′
n Q
o
3
3
Q
Γ12
1
1
+ 1+|ℓ0 |+Σ
3
(|ℓν |+1)q
(|ℓ′ |+1)q
|ℓν |2
ν=1
ν=0
ν=0
ν
1
(|ℓ′ν |+1)q
o
1
1+|ℓ′0 |+Σ3ν=1 |ℓ′ν |2
there.
(d) For all u, u′ ∈ Xn ,
Sn (u, u′ ) − Sn′ (u, u′ ) ≤ Γ12 e−2m4 |u−u′ |
(1)
n
o
′
′
5n
e−2m4 |u−u |
Sn (u, u′ ) ≤ Γ12 min
|u0 −u′ |2 +|u−u′ |4 , L
0
Recall Convention I.2.
35
Proof:
(a) Obviously Sbn′ (p)−1 is entire. For real p
′
Sb (p)−1 ≥ an exp{−∆n (p)} + const |p0 | + |p|2 ≥ const 1 + |p0 | + |p|2
n
(V.7)
by Remark V.3, Lemma III.2.a, and the fact that ∆n (p), Re D̂n (p) ≥ 0 for real p. The
bound on ∂∂pν D̂n (p) of Lemma III.2.c and Remark V.3 shows that (V.7) is valid for all
|Im p| < 3m4 , if m4 is chosen sufficiently small.
∂ℓ c
′
We now bound the derivatives. For any 0 ≤ ν ≤ 3 and ℓ ∈ IN, ∂p
ℓ Sn (p) is a finite
ν
linear combination of terms of the form
c′ (p)1+j
S
n
j
Y
i=1
∂ ℓi
ℓ
∂pνi
D̂n (p) + an exp{−∆n (p)}
Pj
with each ℓi ≥ 1 and i=1 ℓi = ℓ. By Remark V.3, all derivatives of exp{−∆n (p)} of order
constℓ
i
ℓi ∈ IN are bounded by [1+|p |+|p|
2 ]ℓi . Hence, by Lemma III.2.c,
0
∂ ℓi ℓ D̂n(p) + an exp{−∆n (p)} ≤ const
i
∂pν
As Sbn′ (p) ≤
(
1
[1+|p0 |+|p|2 ]ℓi −1
1
[1+|p0 |+|p|2 ]ℓi /2−1
if ν = 0, ℓi = 1, 2
if ν ≥ 1, 1 ≤ ℓi ≤ 4
const
1+|p0 |+|p|2 ,
j
Y
c′
1+j
∂ ℓi
Sn (p)
ℓi D̂n (p) + an exp{−∆n (p)} i=1
∂pν
≤
const
1+|p0 |+|p|2
j
Y
i=1
(
1
[1+|p0 |+|p|2 ]ℓi
1
[1+|p0 |+|p|2 ]ℓi /2
if ν = 0, ℓi = 1, 2
if ν ≥ 1, 1 ≤ ℓi ≤ 4
and the claim follows.
(b) For any c′′0 > 0 we have, for |k| ≥ c′′0 and |Im k0 | < 3m4 , analyticity and the bound
Sbn,k (ℓ, ℓ′ ) ≤
Γ′10
Γ′
δ ′ + 1+|ℓ0 |+Σ103 |ℓν |2
1+|ℓ0 |+Σ3ν=1 |ℓν |2 ℓ,ℓ
ν=1
3
Q
ν=0
1
(|ℓν |+1)q
3
Q
ν=0
1
(|ℓ′ν |+1)q
1
1+|ℓ′0 |+Σ3ν=1 |ℓ′ν |2
(with Γ′10 depending on c′′0 ) which follows from the representation
Sn = Dn−1 − Dn−1 Q∗n ∆(n) Qn Dn−1
(see [parabolic-all.tex, Remark B.9.b]) and Lemmas III.2.d, II.2.a and IV.2.c.
For |k| < c′0 , with c′0 to be shortly chosen sufficiently small, and |Im k| < 3m4 we use
the representation
−1
Ŝn,k
(ℓ, ℓ′ ) = D̂n (k + ℓ) δℓ,ℓ′ + un (k + ℓ)q Q̂n (k)un (k + ℓ′ )q = Dℓ,ℓ′ + Bℓ,ℓ′
36
(V.8)
with
Dℓ,ℓ′ = D̂n (k + ℓ) δℓ,ℓ′ +
Bℓ,ℓ′ =
if ℓ, ℓ′ = 0
otherwise
an
0
Q̂n (k)un (k)2q − an
un (k + ℓ)q Q̂n (k)un (k + ℓ′ )q
if ℓ = ℓ′ = 0
otherwise
By parts (c) and (d) of Lemma III.2, assuming that |k| < c′0 with c′0 small enough, D is
invertible and the inverse is a diagonal matrix with every diagonal matrix element obeying
−1 Γ′′
10
D ≤
ℓ,ℓ
1+|ℓ0 |+Σ3 |ℓν |2
ν=1
for some Γ′′10 which is independent of c′0 . By parts (b) and (c) of Lemma II.2 and parts (a)
and (b) of Proposition II.4,
3
Q
Bℓ,ℓ′ ≤ consta,q |k|
ν=0
q
24
|ℓν |+π
q
24
|ℓ′ν |+π
3
Q
ν=0
So if |k| < c′0 with c′0 small enough, D + B is invertible with the inverse given by the
p
P∞
Neumann expansion D−1 + p=1 (−1)p D−1 BD−1 . Since D and B are both analytic
on |Im k| < 2 and
X
ℓ∈2πZZ4
consta,q |k|
Q
3
ν=0
24
|ℓν |+π
q
Γ′′
10
1+|ℓ0 |+Σ3ν=1 |ℓν |2
Q
3
ν=0
24
|ℓν |+π
q
<
1
2
if c′0 is small enough, we again get the desired analyticity and bound on Sbn,k (ℓ, ℓ′ ).
(c) Just apply Remark V.3 and parts (a) and (b) of this lemma, Lemma II.2.a and Proposition II.4.a and the fact that
q
q
Q
3
X Q
3
ℓ∈2πZZ4
ν=0
1
|ℓν |+π
1
1+|ℓ0 |+Σ3ν=1 |ℓν |2
ν=0
1
|ℓν |+π
is bounded uniformly in n and L to (V.6).
(d) The bound on Sn (u, u′ ) − Sn′ (u, u′ ) follows from part (c) by Lemma A.11.b with the
replacements (V.4). The bound on Sn′ (u, u′ ) follows from part (a), noting in particular
1
12
that (1+|p Γ|+|p|
2 )3 ∈ L (Ẑfin ), and
0
Z
bn′
∂j S
−ip·(u−u′ ) d4 p
′ j ′
′
|uν − uν | Sn (u, u ) =
j (p) e
(2π)4
∂p
Ẑfin
and
Z
′
Sn (u, u′ ) ≤ Ẑfin
ν
Z
′
−ip·(u−u′ ) d4 p b
Sn (p) e
(2π)4 ≤ Γ12
Ẑfin
37
d4 p
(2π)4
= Γ12 L5n
(+)
(−)
We now prove the analog of Lemma V.4 for the operators Sn,ν and Sn,ν of (V.3). As
in (V.5)–(V.6), we decompose
∗
(+)
Sn,ν
= Sn′ + δSν(+)
with
(−)
Sn,ν
= Sn′ + δSν(−)
(+)
∗
∗
(+)
(−)
δSν(+) = Sn,ν
− Sn′ = −Sn′ Q(+)
n,ν Qn Qn,ν − an exp{−∆n } Sn,ν
(−)
(−)
(−)
δSν(−) = Sn,ν
− Sn′ = −Sn′ Q(+)
n,ν Qn Qn,ν − an exp{−∆n } Sn,ν
They have Fourier representations
X
(+)
(+)
\
(+)
(−)
δSν,k (ℓ, ℓ′ ) = −
Sbn′ (k+ℓ) Un,ν
(k, ℓ)Q̂n (k)Un,ν
(k, ℓ′′ ) − an e−∆n (k+ℓ) δℓ,ℓ′′ Sbn,ν,k (ℓ′′ , ℓ′ )
ℓ′′ ∈B̂n
X
(+)
(−)
\
(−)
(−)
Sbn′ (k+ℓ) Un,ν
(k, ℓ)Q̂n (k)Un,ν
(k, ℓ′′ ) − an e−∆n (k+ℓ) δℓ,ℓ′′ Sbn,ν,k (ℓ′′ , ℓ′ )
δSν,k (ℓ, ℓ′ ) = −
ℓ′′ ∈B̂n
(V.9)
where, by (II.10),
(+)
(+)
q−1
Un,ν
(k, ℓ) = ζn,ν
(k, ℓ)u(+)
n,ν (k + ℓ)un (k + ℓ)
(−)
(−)
(−)
Un,ν
(k, ℓ′′ ) = ζn,ν
(k, ℓ′′ )un,ν
(k + ℓ′′ )un (k + ℓ′′ )q−1
Lemma V.5 There are constants m5 > 0 and Γ13 such that the following hold for all
L > Γ2 .
(±)
(a) For all ℓ, ℓ′ ∈ B̂n , Sbn,ν,k (ℓ, ℓ′ ) is analytic in |Im k| < 3m5 and obeys
(±)
′ Sb
(ℓ,
ℓ
)
≤
n,ν,k
Γ13
1+|ℓ0 |+Σ3ν=1 |ℓν |2
n
δℓ,ℓ′ +
1
1+|ℓ′0 |+Σ3ν=1 |ℓ′ν |2
3
Q
ν=0
1
(|ℓν |+1)q−1
3
Q
ν=0
1
(|ℓ′ν |+1)q
o
there.
c k (ℓ, ℓ′ ) is analytic in |Im k| < 3m5 and obeys
(b) For all ℓ, ℓ′ ∈ B̂n , δS
\
δS (±) (ℓ, ℓ′ ) ≤ Γ13 exp −
ν,k
2
1 3
40 Σν=0 |ℓν |
+
δℓ,ℓ′
Γ13
1+|ℓ0 |+Σ3ν=1 |ℓν |2
3
Q
ν=0
1
(|ℓν |+1)q−1
3
Q
ν=0
1
(|ℓ′ν |+1)q
there.
(±)
′
(c) For all u, u′ ∈ Xn , Sn,ν (u, u′ ) − Sn′ (u, u′ ) ≤ Γ13 e−2m5 |u−u | .
38
1
1+|ℓ′0 |+Σ3ν=1 |ℓ′ν |2
Proof: (a) For any c′′0 > 0 we have, for |k| ≥ c′′0 and |Im k0 | < 3m5 , analyticity and
the desired bound follows from the representations (apply [parabolic-all.tex, Remark
(−)
(+)
B.9.b] with R = Qn,ν , R∗ = Qn,ν and use Remark II.5 to give R D−1 R∗ = Q− D−1 Q∗− )
(+)
(n) (−) ∗ −1
Sn,ν
= Dn∗ −1 − Dn∗ −1 Q(+)
Qn,ν Dn
n,ν ∆
(−)
(n) (−) −1
Sn,ν
= Dn−1 − Dn−1 Q(+)
Qn,ν Dn
n,ν ∆
and Lemmas III.2.d, II.2.a, II.6.b and IV.2.c.
For |k| < c′0 , with c′0 to be shortly chosen sufficiently small, and |Im k| < 3m5 we use
the representation
(−)
Ŝn,ν,k
−1
(+)
(−)
(ℓ, ℓ′ ) = D̂n (k + ℓ) δℓ,ℓ′ + Un,ν
(k, ℓ)Q̂n (k)Un,ν
(k, ℓ′ ) = Dℓ,ℓ′ + Bℓ,ℓ′
with
Dℓ,ℓ′ = D̂n (k + ℓ) δℓ,ℓ′ +
Bℓ,ℓ′ =
(
an
0
if ℓ, ℓ′ = 0
otherwise
Q̂n (k)un (k)2q − an
(+)
(V.10)
(−)
Un,ν (k, ℓ)Q̂n (k)Un,ν (k, ℓ′ )
if ℓ = ℓ′ = 0
otherwise
(+) −1
and the obvious analog for Ŝn,ν,k
— just replace D̂n (k +ℓ) with its complex conjugate.
As in the proof of Lemma V.4.b, D is invertible and the inverse is a diagonal matrix with
every diagonal matrix element obeying
−1 D ≤
ℓ,ℓ
Γ′11
1+|ℓ0 |+Σ3ν=1 |ℓν |2
for some Γ′11 which is independent of c′0 . By parts (b) and (c) of Lemma II.2, parts (a)
and (b) of Proposition II.4, and part (b) of Lemma II.6,
Bℓ,ℓ′ ≤ const
a,q
|k|
3
Q
ν=0
q−1
24
|ℓν |+π
3
Q
ν=0
24
|ℓ′ν |+π
q
So if |k| < c′0 with c′0 small enough, D + B is invertible with the inverse given by the
p
P∞
Neumann expansion D−1 + p=1 (−1)p D−1 BD−1 . As q > 1, the desired analyticity
(±)
and the desired bound on Sb
(ℓ, ℓ′ ) follow as in the proof of Lemma V.4.b
n,ν,k
(b) is proven just as Lemma V.4.c.
(c) follows from part (b) by Lemma A.11.b.
39
Proof of Proposition V.1: Set m3 = min{m4 , m5 }. As |u0 |21+|u|4 is locally integrable
in IR4 , the pointwise bounds on Sn (u, u′ ) − Sn′ (u, u′ ) and Sn′ (u, u′ ), given in Lemma
(±)
V.4.d, and on Sn,ν (u, u′ ) − Sn′ (u, u′ ), given in Lemma V.5.c, imply
(±)
k Sn km3 , k Sn,ν
km3 ≤ Γ̃11
with Γ̃11 a constant, depending only on m3 , times max{Γ12 , Γ13 }. Setting µup =
(±)
1
2Γ̃11
and Γ11 to be the maximum of 2Γ̃11 (for k Sn (µ)km3 andk Sn,ν (µ)km3 ) and 2Γ̃211 (for
(±)
(±)
k Sn (µ)−Sn km3 andk Sn,ν (µ)−Sn,ν km3 ) a Neumann expansion gives the specified bounds.
We now formulate and prove two more technical lemmas that will be used elsewhere.
Lemma V.6 There are constants m6 > 0 and Γ14 such that, for all L > Γ2 ,
Sn Q∗n (y, x) ≤ Γ14 e−2m6 |x−y|
kSn Q∗n km6 ≤ Γ14
Proof: From the definitions of Sn , in (V.3), and ∆(n) , at the beginning of §IV, one sees
directly that
−1
2
2
(V.11)
Sn(∗) Q∗n = Dn(∗) Q∗n ∆(n)(∗) Q−1
n : L (Xcrs ) → L (Xfin )
The Fourier transform of the kernel, b(y, x), of the operator Sn Q∗n is
q ˆ (n)
b̂k (ℓ) = D̂−1
(k) Q̂ 1(k)
n (k + ℓ)un (k + ℓ) ∆
n
By Lemma IV.2.a,c,f, Remark III.1.b and Lemma III.2.d, Proposition II.4.a, Remark
II.1.d and Lemma II.2.a, b̂k (ℓ) is analytic in |Im k| < 3m6 and
b̂k (ℓ) ≤
1 + |k0 +
5
4a Γ6
P3
ℓ0 | + ν=1
|kν + ℓν |2
3 Y
ν=0
24 q
|ℓν | + π
The bound is uniform in n and L and is summable in ℓ, so the claims follow from Lemma
A.11.c.
Lemma V.7 There are constants m7 > 0 and Γ15 such that, for all L > Γ2 , the operators
∗
∗
Sn (µ)Q∗n Qn Sn (µ)Q∗n Qn , Sn (µ)∗ Q∗n Qn Sn (µ)∗ Q∗n Qn
∗ (±)
(±)
Sn,ν
(µ)Q(+)
Sn,ν (µ)Q(+)
n,ν Qn
n,ν Qn
all have bounded inverses. The k · km7 norms of the inverses are all bounded by Γ15 .
40
Proof:
We first consider the case that µ = 0. By (V.11), the operator
Sn Q∗n Qn = Dn−1 Q∗n ∆(n) : L2 (Xcrs ) → L2 (Xfin )
has Fourier transform
q ˆ (n)
b̃k (ℓ) = D̂−1
(k)
n (k + ℓ)un (k + ℓ) ∆
The operator (Sn Q∗n Qn )∗ (Sn Q∗n Qn ) maps L2 (Xcrs ) to L2 (Xcrs ) and has Fourier transform
X
b̃−k (−ℓ)b̃k (ℓ) =
ℓ∈B̂
ℓ∈B̂
For k real,
X
X
−1
2q ˆ (n)
ˆ (n) (k)
D̂−1
(−k)∆
n (−k − ℓ)D̂n (k + ℓ)un (k + ℓ) ∆
b̃−k (−ℓ)b̃k (ℓ) =
X
D̂−1 (k + ℓ)un (k + ℓ)q ∆
ˆ (n) (k)2
n
ℓ∈B̂
ℓ∈B̂
2
q ˆ (n)
(k)
≥ D̂−1
n (k)un (k) ∆
≥ γ22 inf |un (k)|2q
|kν |≤π
≥ γ22
2 8q
π
by Lemmas IV.2.e and II.2.f. To show that half this the lower bound extends into a strip
along the real axis that has width independent of n and L, we observe that
Q3
◦ all first order derivatives of un (k + ℓ) are uniformly bounded by 2 ν=0 |ℓν24
|+π on such
a strip by the Cauchy integral formula and Remark II.1.d and Lemma II.2.a and
Q3
◦ un (k + ℓ) itself is uniformly bounded by ν=0 |ℓν24
|+π on such a strip by Lemma II.2.a
and
ˆ (n) (k) are uniformly bounded on such a strip by Lemma
◦ all first order derivatives of ∆
IV.2.c and
◦ for ℓ 6= 0, all first order derivatives of D̂−1
n (k + ℓ) are uniformly bounded on such a
strip by parts (c) and (d) of Lemma III.2 and
ˆ (n) (k) are uniformly bounded on such
◦ for ℓ = 0, all first order derivatives of D̂−1
n (k)∆
a strip by Lemma IV.2.f.
∗
The operator Sn∗ Q∗n Qn = Dn−1 Q∗n ∆(n)∗ has Fourier transform b̃−k (−ℓ). So
(Sn Q∗n Qn )∗ (Sn Q∗n Qn ) = (Sn∗ Q∗n Qn )∗ (Sn∗ Q∗n Qn )
(−)
(+)
(+)
(+)
(+)
∗
(+)
The operators Sn,ν Qn,ν Qn = Dn−1 Qn,ν ∆(n) and Sn,ν Qn,ν Qn = Dn−1 Qn,ν ∆(n)∗ map
L2 (Xcrs ) to L2 (Xfin ) and have Fourier transforms
(+)
(+)
q−1 ˆ (n)
c̃k (ℓ) = D̂−1
∆ (±k)
n (±k ± ℓ)ζn,ν (k, ℓ)un,ν (k + ℓ)un (k + ℓ)
41
(±)
(+)
(±)
(+)
So the operators (Sn,ν Qn,ν Qn )∗ (Sn,ν Qn,ν Qn ) both map L2 (Xcrs ) to L2 (Xcrs ) and have
Fourier transform
X
c̃−k (−ℓ)c̃k (ℓ)
ℓ∈B̂
=
X
ℓ∈B̂
−1
(+)
2
2(q−1) ˆ (n)
ˆ (n) (k)
D̂−1
∆ (−k)∆
n (−k − ℓ)D̂n (k + ℓ)un,ν (k + ℓ) un (k + ℓ)
P
This is bounded just as ℓ∈B̂ b̃−k (−ℓ)b̃k (ℓ) was. The specified bounds, in the special case
that µ = 0 follow.
By Proposition V.1, a Neumann expansion gives the deisred bounds when µ is nonzero.
42
VI. The Degree One Part of the Critical Field
In [parabolic-all.tex, Proposition G.13 and (G.29)] we derive an expansion for the
critical fields of the form
ψ(∗)n (θ∗ , θ, µ, V) =
a
C (n) (µ)(∗) Q∗ θ(∗)
L2
(≥3)
+ ψ(∗)n (θ∗ , θ, µ, V)
(≥3)
with the C (n) (µ) of [parabolic-all.tex, Proposition I.17] and with ψ(∗)n being of degree
at least 3 in θ(∗) . In this section we derive bounds on a scaled version of C (n) (µ)(∗) Q∗ θ(∗)
and some related operators. To do so we use the representation
a
C (n) (µ)(∗) Q∗
L2
=
a
Q∗ Q
L2
+ Qn
−1 a
Q∗
L2
+ Qn Qn Šn+1 (µ)(∗) Q̌∗n+1 Q̌n+1
of (G.28). Here, as in [parabolic-all.tex, Lemma III.4 and (G.27)],
Q̌n+1 = L∗ Qn+1 L−1
∗
=
1
1l
aL−2
(VI.1)
(n+1)
= QQn
−1
2
−1
Q̌−1
n+1 = L L∗ Qn+1 L∗
: Hn → H−1
(n+1)
∗
+ QQ−1
n Q
: H−1
−1
∗
Šn+1 (µ) = L2 L∗ Sn+1 (L2 µ)L−1
: Hn → Hn
∗ = Dn − µ + Q̌n+1 Q̌n+1 Q̌n+1
These operators are all translation invariant with respect to
Šn+1 (µ) = Šn+1 + µŠn+1 Šn+1 (µ)
(n+1)
X−1 .
(n+1)
→ H−1
(VI.2)
As
with Šn+1 = Šn+1 (0)
we have
a
C (n) (µ)(∗) Q∗
L2
=
(n) (∗) ∗
a
C
Q
2
L
(∗)
+ µAψ,φ Šn+1 Šn+1 (µ)(∗) Q̌∗n+1 Q̌n+1
(VI.3)
where
(n)
Aψ,φ = (aL−2 Q∗ Q + Qn )−1 Qn Qn : Hn → H0
(VI.4)
(≥3)
The operator Aψ,φ is also used in the course of bounding ψ(∗)n in [parabolic-all.tex,
Proposition G.13].
The main results of this section are
Proposition VI.1 There are constants(1) m8 > 0 and Γ16 , Γ17 such that the following
holds, for each L > Γ17 and each µ obeying |L2 µ| ≤ µup .
(1)
Recall Convention I.2.
43
(a)
−1
L∗ Aψ,φ L∗ ≤ Γ16
m=1
−1
L∗
and
a
(n)
(µ)(∗) Q∗
L2 C
L∗ m8 ≤ Γ16
(b) Let 0 ≤ ν ≤ 3. There are operators Aψ,φ,ν and Aψ(∗) θ(∗) ν (µ) such that
∂ν La2 C (n) (µ)(∗) Q∗ = Aψ(∗) θ(∗) ν (µ) ∂ν
∂ν Aψ,φ = Aψ,φ,ν ∂ν
and
−1
L∗ Aψ θ ν (µ)L∗ ≤ Γ16
(∗) (∗)
m
kL−1
∗ Aψ,φ,ν L∗ km=1 ≤ Γ16
8
This proposition is proven at the end of this section, after Lemma VI.6. In this proof we
(∗)
write La2 C (n) Q∗ = Aψ(∗) ,θ(∗) so that
(∗)
(n+1)
(n)
Aψ(∗) ,θ(∗) = (aL−2 Q∗ Q + Qn )−1 aL−2 Q∗ + Qn Qn Šn+1 Q̌∗n+1 Q̌n+1 : H−1 → H0
Remark VI.2
∗
−1
(a) Aψ(∗) θ(∗) = Q−1
n Q Q̌n+1 + Aψ,φ Dn
ˇ (n+1)(∗) =
∆
(∗)
(n+1)(∗) −1
1
L∗
L2 L∗ ∆
ˇ (n+1)(∗) with
Q̌∗n+1 ∆
−1
(∗)
= 1l + Q̌n+1 Q̌n+1 Dn−1 Q̌∗n+1
Q̌n+1
(n+1)
being a fully translation invariant operator on H−1
.
(b) Let 0 ≤ ν ≤ 3. We have ∂ν Aψ,φ = Aψ,φ,ν ∂ν and ∂ν Aψ(∗) θ(∗) (µ) = Aψ(∗) θ(∗) ν (µ)∂ν
where
(+)
(−) −
Aψ,φ,ν = 1l − Q−1
n Q+,ν Q̌n+1 Q+,ν Qn,ν
(+)
−1(∗)
(+)
(n+1)(∗) −1
Aψ(∗) θ(∗) ν = Q−1
L∗
n Q+,ν Q̌n+1 + Aψ,φ,ν L∗ Dn+1 Qn+1,ν ∆
(+)
(+)
(+)
Aψ∗ θ∗ ν (µ) = Aψ∗ θ∗ ν + L2 µAψ,φ,ν L∗ Sn+1,ν Sn+1,ν (L2 µ)Qn+1,ν Qn+1 L−1
∗
(−)
(−)
(+)
Aψθν (µ) = Aψθν + L2 µAψ,φ,ν L∗ Sn+1,ν Sn+1,ν (L2 µ)Qn+1,ν Qn+1 L−1
∗
Proof:
(a) First observe that, by (V.11),
(∗) −1
(∗)
Qn Šn+1 Q̌∗n+1 Q̌n+1 = Qn L∗ Dn+1
−1
Q∗n+1 ∆(n+1)(∗) L−1
∗ = (Qn Dn
(∗)
ˇ (n+1)(∗)
Q∗n ) Q∗ ∆
(VI.5)
Using (VI.5), the operator
(∗)
ˇ (n+1)(∗)
Aψ(∗) θ(∗) = (aL−2 Q∗ Q + Qn )−1 aL−2 Q∗ + Qn Qn Dn−1 Q̌∗n+1 ∆
∗
−1
= Q−1
n (Q QQn +
−1 ∗
1
Q
aL−2 1l)
(∗)
ˇ (n+1)(∗)
Q̌∗n+1 ∆
(∗)
−1
1
+ Aψ,φ Dn−1 Q̌∗n+1
aL−2 1l)
(∗)
ˇ (n+1)(∗)
Aψ,φ Dn−1 Q̌∗n+1 ∆
∗
−1 ∗
= Q−1
n Q (QQn Q +
∗
= Q−1
n Q Q̌n+1 +
+ Aψ,φ Dn−1
44
ˇ (n+1)(∗)
∆
(b) By Remark II.5,
∗
−1
∂ν Aψ,φ = ∂ν (1l + aL−2 Q−1
Qn
n Q Q)
∗
−2
∗ −1
= ∂ν Qn − ∂ν aL−2 Q−1
QQ−1
QQn
n Q (1l + aL
n Q )
(+)
(−)
−2 −1
∗ −1
(−)
= Q−
Qn Q+,ν (1l + aL−2 QQ−1
Q+,ν Qn,ν
∂ν
n,ν ∂ν − aL
n Q )
(+)
(−)
−
= 1l − Q−1
n Q+,ν Q̌n+1 Q+,ν Qn,ν ∂ν
Therefore by part (a), (III.1), (VI.2) and Remark II.5,
−1(∗) ∗
∗
(n+1)(∗) −1
∂ν Aψ(∗) θ(∗) = ∂ν Q−1
L∗ = Aψ(∗) θ(∗)ν ∂ν
n Q Q̌n+1 + Aψ,φ L∗ Dn+1 Qn+1 ∆
since ∂ν L∗ = L1ν L∗ ∂ν by [parabolic-all.tex, Remark III.2.a,b]. To get ∂ν Aψ(∗) θ(∗) (µ) =
Aψ(∗) θ(∗) ν (µ)∂ν when µ 6= 0, write, using (VI.2),
(∗)
Aψ(∗) θ(∗) (µ) = Aψ(∗) θ(∗) + µAψ,φ Šn+1 Šn+1 (µ)(∗) Q̌∗n+1 Q̌n+1
(∗)
= Aψ(∗) θ(∗) + L2 µAψ,φ L∗ Sn+1 Sn+1 (L2 µ)(∗) Q∗n+1 Qn+1 L−1
∗
and use (V.1), Remark II.5 and the fact that Qn+1 is fully translation invariant.
(n+1)
The operators of principal interest, Aψ∗ ,θ∗ and Aψ,θ , act from L2 (Xcrs ) = H−1
(n)
L2 (Xfin ) = H0 with
(n)
Xfin = X0
(n+1)
Xcrs = X−1
to
B = B+
We now give a bunch of Fourier transforms (in the sense of (A.6) and (A.7) – but we shall
suppress the ˆ from the notation). All of the operators above are periodized in the sense
of Definition A.2. As before we denote
2π
Z × IR3 / 2π
ZZ3 ,
◦ momenta dual to the L–lattice L2 ZZ × LZZ3 by k ∈ IR/ L
2Z
L
◦ momenta dual to the unit lattice ZZ × ZZ3 by k ∈ IR/2πZZ × IR3 /2πZZ3 and decom
3
2π
pose k = k + ℓ or k = k + ℓ′ with k in a fundamental cell for IR/ L
Z × IR3 / 2π
Z
Z
2Z
L
3
2π 3
2π
+
Z
Z/2πZ
Z
×
Z
Z
/2πZ
Z
=
B̂
and
and ℓ, ℓ′ ∈ L
2
L
3 2π 3
Z
Z
and
◦ momenta dual to the εj –lattice ε2j ZZ × εj ZZ3 by pj ∈ IR/ 2π
Z
Z
×
IR
/
2
εj
εj
decompose pj = k + ℓj with k in a fundamental cell for IR/2πZZ × IR3 /2πZZ3 and
ℓj ∈ 2πZZ/ 2π
Z3 = B̂j . Here 1 ≤ j ≤ n.
ZZ × 2πZZ3 / 2π
εj Z
ε2j
The Fourier transform of Aψ(∗) θ(∗) is
X
−1
a
Q∗ Q + Qn k (ℓ, ℓ′ )
Aψ(∗) θ(∗) k (ℓ) =
L2
o
ℓ′ ∈B̂ + n
∗ ′
′
−1 (∗) ∗
′
∗ ′ ˇ (n+1)(∗)
a
Q
(ℓ
)
+
Q
(k+ℓ
)
Q
D
Q
(k+ℓ
)Q
(ℓ
)
∆
(k)
n
n n
n
k
L2 k
(VI.6)
45
where, by Remark II.1, Remark VI.2.a and (VI.2)
aL−2 Q∗ Q + Qn k (ℓ, ℓ′ ) = Qn (k + ℓ)δℓ,ℓ′ + aL−2 u+ (k + ℓ)q u+ (k + ℓ′ )q
Q∗k (ℓ) = u+ (k + ℓ)q
n−1
h
X P
Qn (k) = a 1 +
(Qn Dn−1(∗) Q∗n )(k) =
X
1
u (k
L2j j
+ ℓj )2q
j=1 ℓj ∈B̂j
i−1
un (k + ℓn )2q Dn−1(∗) (k + ℓn )
ℓn ∈B̂n
ˇ (n+1)(∗) (k) =
∆
Q̌n+1 (k) =
1 + Q̌n+1 (k)
a
L2
h
P
1+
P
Q̌n+1 (k)
−1(∗)
un (k+ℓ+ℓn)2q u+ (k+ℓ)2q Dn
ℓn ∈B̂n
ℓ∈B̂+
ℓ∈B̂ +
a
u (k
L2 +
2q
+ ℓ) Qn (k + ℓ)
−1
(k+ℓ+ℓn)
i−1
Lemma VI.3 Let |Im kν | ≤ L2ν for each 0 ≤ ν ≤ 3. There is a constant Γ17 , depending
only on q, such that the following hold for all L > Γ17 .
(a) We have
6 a
7 L2
≤ |Q̌n+1 (k)| ≤
6 a
5 L2
and Re Q̌n+1 (k) ≥
a
.
2L2
(b) We have
−1
−2 ∗
′
−1
aL Q Q + Qn k (ℓ, ℓ ) − Qn (k + ℓ) δℓ,ℓ′ ≤
2
aL2
3
Q
ν=0
q
24
Lν |ℓν |+π
3
Q
ν=0

1
q  Q
24
L |k |
Lν |ℓ′ν |+π
 0≤ν≤3 ν ν
for all ℓ
if ℓ 6= 0
ℓν 6=0
(c) Let 0 ≤ ν ≤ 3. Then
−1 (+)
Q Q Q̌n+1 Q(−) (ℓ, ℓ′ ) ≤
n
+,ν
+,ν k
Proof:
3e4
2L2
24 4
π
3
Q
ν=0
q−1
24
Lν |ℓν |+π
(a) is proven much as Proposition II.4.a was.
46
3
Q
ν=0
q
24
′
Lν |ℓν |+π
(b) The straight forward Neumann expansion gives
∗
a
L2 Q Q
+ Qn
−1
k
(ℓ, ℓ′ ) − Qn (k + ℓ)−1 δℓ,ℓ′ ∞ X
X
−1
q
′ q
′ −1 a ≤ L2 Qn (k + ℓ) u+ (k + ℓ) u+ (k + ℓ ) Qn (k + ℓ ) j=0
≤
≤
a
L2
2
aL2
5 2
u+ (k
4a
3
Q
ν=0
3
Q
+ ℓ)q q
24
Lν |ℓν |+π
ν=0
3
Q
ν=0
q
24
Lν |ℓ′ν |+π
∞
X
5Γ′ j
l∈B̂ +
15
a u+ (k+l)
L2 Qn (k+l)
2q
j
4L2
j=0

1
for all ℓ
q  Q
24
L |k | if ℓ =
6 0
Lν |ℓ′ν |+π
 0≤ν≤3 ν ν
ℓν 6=0
by part (a), Lemma II.3.a,b and Proposition II.4.a, with L satisfying the conditions of part
(a).
(c) The specified bound follows from (II.11), part (a), Proposition II.4.a and Lemmas
II.6.b, II.3.a.
Lemma VI.4 For all
3
2π
k = L−1 (k) ∈ IR/ L
Z × IR3 / 2π
Z
Z
2Z
L
3 2π 3
−1
2π
p = L (p) ∈ IR/ ε2 ZZ × IR / εn ZZ
n
k ∈ IR/2πZZ × IR3 /2πZZ3
2π
p ∈ IR/ ε22π ZZ × IR3 / εn+1
ZZ3
n+1
2π
ℓn+1 ∈ 2πZZ/ ε22π ZZ × 2πZZ3 / εn+1
ZZ3
n+1
we have
(a) u+ L−1 (k) = u1 (k) and un L−1 (k) u+ L−1 (k) = un+1 (k) for all n ∈ IN.
(b) Q̌n+1 L−1 (k) = L12 Qn+1 (k)
−1
(c) D−1
(p) = L2 D−1
n L
n+1 (p)
ˇ (n+1)(∗) L−1 (k) = 12 ∆(n+1)(∗) (k)
(d) ∆
L
(e) Šn+1,L−1 (k) L−1 (ℓn+1 ), L−1 (ℓ′n+1 ) = L2 Ŝn+1,k (ℓn+1 , ℓ′n+1 )
Proof: (a) These all follow from (VI.2), Remark VI.2.a and
◦ Q1 = L−1
∗ QL∗ by (II.2)
◦ Dn+1 = L2 L−1
∗ Dn L∗ by (III.1)
and Lemmas A.14.b and A.15.b.
47
Corollary VI.5 There are constants m9 > 0 and Γ18 such that, for all L > Γ17 ,
2
kL−1
∗ Šn+1 L∗ km9 ≤ L Γ18
∗
2
kL−1
∗ Šn+1 Q̌n+1 L∗ km9 ≤ L Γ18
Proof: Set m9 = min{m3 , m6 }. Just combine Lemmas A.14.b and A.15.b and parts (a)
and (e) of Lemma VI.4 to yield
2
kL−1
∗ Šn+1 L∗ km9 = L kSn+1 km9
∗
2
kL−1
∗ Šn+1 Q̌n+1 L∗ km9 = L kSn+1 Qn+1 km9
and then apply Proposition V.1 and Lemma V.6.
Lemma VI.6 Assume that L > Γ17 , the constant of Lemma VI.3. There are constants
m10 > 0 and Γ19 such that the following hold for all k ∈ C4 with |Im k| ≤ m10 .
3
q
Q
24
−1
≤
Γ
(L
(ℓ
))
=
A
(a) L−1
A
L
(ℓ
)
19
1
ψ(∗) θ(∗) L−1 (k)
ψ(∗) θ(∗) ∗ k 1
∗
|ℓ1,ν |+π
ν=0
3
Q
(b) If ℓ1 6= 0, then L−1
A
L
(ℓ
)
≤
Γ
|k|
ψ
θ
∗
1
19
∗
(∗) (∗)
k
ν=0
Proof:
q
24
|ℓ1,ν |+π
By (VI.6), Lemma VI.4 and Remark II.1.e,
Aψ(∗) θ(∗) L−1 (k) (L−1 (ℓ1 ))
n
X
−1
∗
−1
−1 ′
a
(ℓ1 ), L (ℓ1 )) La2 Q∗L−1 (k) (L−1 (ℓ′1 ))
=
L2 Q Q + Qn L−1 (k) (L
+ Qn (L−1 (k + ℓ′1 )) Qn Dn−1(∗) Q∗n (L−1 (k + ℓ′1 ))
o
ˇ (n+1)(∗) (L−1 (k))
Q∗L−1 (k) (L−1 (ℓ′1 ))∆
X
−1
∗
−1
a
(ℓ1 ), L−1 (ℓ′1 )) B1 (k, ℓ′1 )
=
L2 Q Q + Qn L−1 (k) (L
ℓ′1 ∈B̂1
ℓ′1 ∈B̂1
= Qn L−1 (k + ℓ1 )
−1
B1 (k, ℓ1 ) +
X
C(k, ℓ1 , ℓ′1 )B1 (k, ℓ′1 )
ℓ′1 ∈B̂1
where
B1 (k, ℓ′1 ) = u1 (k + ℓ′1 )q
n
a
L2
+ Qn L−1 (k + ℓ′1 )
X
ℓn ∈B̂n
B2 (k, ℓ′1 , ℓn )
o
k + ℓ′1 + L(ℓn ) ∆(n+1)(∗) (k)
B2 (k, ℓ′1 , ℓn ) = un (L−1 (k + ℓ′1 ) +
−1
−1
C(k, ℓ1 , ℓ′1 ) = La2 Q∗ Q + Qn L−1 (k) (L−1 (ℓ1 ), L−1 (ℓ′1 )) − Qn L−1 (k + ℓ1 )
δℓ1 ,ℓ′1
−1(∗)
ℓn )2q Dn+1
48
(a) Choose m10 = min{2, m1 , m̄(π)}. Then
∗
a
L2 Q Q
+ Qn
−1
L−1 (k)
−1
(L
−1
(ℓ1 ), L
(ℓ′1 ))
2
aL2
≤ 65 a δℓ1 ,ℓ′1 +
3
Q
ν=0
q
24
|ℓ1,ν |+π
3
Q
24
ν=0
|ℓ′1,ν |+π
q
(by Lemma VI.3.b, Proposition II.4.a)
3
q
Q
24
u1 (k + ℓ′1 )q ≤
(by Lemma II.2.a)
′
|ℓ1,ν |+π
ν=0
Qn (L−1 (k + ℓ′ )) ≤ 6 a
(by Proposition II.4.a)
1
5
3
2q
Q
24
un (L−1 (k + ℓ′1 ) + ℓn )2q ≤
(by Lemma II.2.a)
|ℓn,ν |+π
ν=0
−1(∗)
D
k + ℓ′1 + L(ℓn ) ≤ γ11π if (ℓ′1 , ℓn ) 6= (0, 0) (by Lemma III.2.d)
n+1
(n+1)(∗) ∆
(k) ≤ 2a
(by Lemma IV.2.c)
−1(∗)
(n+1)(∗)
D
(by Lemma IV.2.f)
(k) ≤ Γ6
n+1 (k) ∆
So
−1
Aψ(∗) θ(∗) L−1 (k) (L (ℓ1 ))
X n
6
′
≤
5 a δℓ1 ,ℓ1 +
ℓ′1 ∈B̂1
3
Q
ν=0
≤ const
≤ Γ19
X n
2
aL2
ν=0
24
|ℓ′1,ν |+π
6
′
5 a δℓ1 ,ℓ1
3
Q
q n
+
ν=0
a
L2
2
aL2
ℓ′1 ∈B̂1
3
Q
q
24
|ℓ1,ν |+π
+
6
5a
3
Q
ν=0
q
24
|ℓ1,ν |+π
3
Q
ν=0
24
|ℓ′1,ν |+π
X Q
3
ℓn ∈B̂n
ν=0
q
24
|ℓ1,ν |+π
3
Q
ν=0
q o
o
2q
2a
24
max
Γ
,
6
|ℓn |+π
γ1 π
q
24
|ℓ′1,ν |+π
o Q
3
ν=0
q
24
|ℓ′1,ν |+π
(b) Using the bounds of part (a) together with
3
Q
u1 (k + ℓ′ )q ≤ |k|
1
ν=0
q
24
|ℓ′1,ν |+π
3
Q
un (L−1 (k + ℓ′ ) + ℓn )2q ≤ |L−1 (k + ℓ′ )|
1
1
ν=0
we have, if ℓ′1 6= 0,
3
Q
B1 (k, ℓ′1 ) ≤ |k|
ν=0
q
24
′
|ℓ1,ν |+π
≤ const |k|
3
Q
ν=0
n
a
L2
+
6
5a
if ℓ′1 6= 0
2q
24
|ℓn |+π
if ℓn 6= 0
X Q
3
2q
24
|ℓn |+π
ℓn ∈B̂n
q
24
|ℓ′1,ν |+π
49
(by Lemma II.2.b)
ν=0
(by Lemma II.2.b)
o
2a
max Γ6 , γ1 π
and
3
Q
B1 (k, 0) ≤
24 q
π
ν=0
n
a
L2
X Q
3
+ 65 a
ℓn ∈B̂n
ν=0
2q
24
max Γ6 , γ2a
|ℓn |+π
1π
o
≤ const
Using these bounds, the first bound of part (a) and Lemma VI.3.b, and assuming that
ℓ1 6= 0,
X
−1
a
Q∗ Q + Qn L−1 (k) (L−1 (ℓ1 ), L−1 (ℓ′1 )) B1 (k, ℓ′1 )
Aψ(∗) θ(∗) L−1 (k) (L−1 (ℓ1 )) =
L2
ℓ′1 ∈B̂1
=
=
∗
a
L2 Q Q
2
aL2
3
Q
+ Qn
3
Q
−1
(L
(ℓ
),
0))
B
(k,
0)
+
O
|k|
1
1
L−1 (k)
−1
24
|ℓ1,ν |+π
ν=0
3
Q
= O |k|
ν=0
ν=0
3
q Q
ν=0
q
24
|ℓ1,ν |+π
Q
24 q
π
0≤ν≤3
ℓ1,ν 6=0
|kν | B1 (k, 0) + O |k|
q
24
|ℓ1,ν |+π
3
Q
ν=0
q
24
|ℓ1,ν |+π
Proof of Proposition VI.1:
Bound on L−1
By Lemmas A.14.b and VI.3.b, if |Im kν ′ | ≤ 2 for each
∗ Aψ,φ L∗ m=1 :
0 ≤ ν ′ ≤ 3 then
−1 −1
L∗
aL−2 Q∗ Q + Qn
− Q−1
L∗ k (ℓ, ℓ′ )
n
−1
−1
− Q−1
ℓ, L−1 ℓ′ )
= aL−2 Q∗ Q + Qn
n L−1 k (L
3
3
q Q
q
Q
24
24
≤ aL2 2
|ℓν |+π
|ℓ′ |+π
ν=0
ν=0
ν
So, by Lemma A.11.b,
−1 −1
L
aL−2 Q∗ Q + Qn
− Q−1
L∗ km=1 ≤ const q
∗
n
By Proposition II.4.a and Lemmas II.2.a and A.11.b,c,
kQn km=1 , kQ−1
n km=1 , kQn km=1 ≤ const q
too. Now just apply Lemmas A.14.c and A.15.c.
By Lemmas A.14.b and VI.3.c, if |Im kν ′ | ≤ 2 for each
Bound on L−1
∗ Aψ,φ,ν L∗ m=1 :
′
0 ≤ ν ≤ 3 then
−1 −1 (+)
(−) (+)
(−) −1
L
Qn Q+,ν Q̌n+1 Q+,ν L∗ k (ℓ, ℓ′ ) = Q−1
ℓ, L−1 ℓ′ )
∗
n Q+,ν Q̌n+1 Q+,ν L−1 k (L
3
3
Q
q−1 Q
q
24
24
3e4 24 4
≤ 2L
2
π
|ℓν |+π
|ℓ′ |+π
ν=0
50
ν=0
ν
As q > 2, Lemma A.11.b yields
−1 −1 (+)
(−)
L∗ Qn Q+,ν Q̌n+1 Q+,ν
L∗ km=1 ≤ const q
(−)
By Lemmas II.6.b and A.11.c, kQn,ν km=1 ≤ const q too, since q > 2. Now just apply
Lemma A.15.c.
: This follows from Lemma VI.6.a by Lemma A.11.c.
Bound on L−1
A
L
ψ
θ
∗
∗
(∗) (∗)
m8
−1 a (n) (∗) ∗
Bound on L∗ L2 C (µ) Q L∗ m : By (VI.3)
8
(n)
a
(µ)(∗) Q∗
L−1
∗
L2 C
L∗ =
L−1
∗ Aψ(∗) θ(∗) L∗
(∗)
2 (∗) ∗
+ L2 µL−1
Qn+1 Qn+1
∗ Aψ,φ L∗ Sn+1 Sn+1 (L µ)
Now just apply Proposition V.1, Lemma II.2 and Proposition II.4.c.
: It suffices to bound
Bound on L−1
A
L
ψ
θ
ν
∗
∗
(∗) (∗)
m8
◦ L−1
A
L
as
above,
ψ,φ,ν ∗
∗
−1 (+)
◦ bound L−1
∗ Qn Q+,ν Q̌n+1 L∗ using
−1 −1 (+)
(+)
−1 L
Qn Q+,ν Q̌n+1 L∗ k (ℓ) = Q−1
ℓ)
∗
n Q+,ν Q̌n+1 L−1 k (L
3
Q
q−1
24
3e2 24 3
≤ 2L
2
π
|ℓν |+π
ν=0
(by Lemma A.15.b, (II.11), Proposition II.4.a and Lemmas II.6.b, II.3.a, VI.3.a) and
Lemma A.11.c, and
−1(∗) (+)
◦ bound Dn+1 Qn+1,ν ∆(n+1)(∗) using
−1(∗) (+)
(n+1)(∗)
D̂
Q
∆
(ℓ
)
n+1
n+1
n+1,ν
k
−1(∗)
(+)
(+)
≤ D̂n+1 k + ℓn+1 ) ζn+1,ν (k, ℓn+1 )un+1,ν (k + ℓn+1 )un+1 (k + ℓn+1 )q−1
ˆ (n+1)(∗) (k)
∆
(by (II.10)) and
(+)
(+)
ζ
(k, ℓn+1 )u
n+1,ν
3
+ ℓn+1 ) ≤ e2 24
π
3
q−1
Q
24
un+1 (k + ℓn+1 )q−1 ≤
|ℓn+1 |+π
n+1,ν (k
ν=0
(by Lemma II.6.b)
(by Lemma II.2.a)
−1(∗)
D̂
k + ℓn+1 ≤ γ11π if ℓn+1 6= 0
(by Lemma III.2.d)
n+1
(n+1)(∗) ˆ
∆
(k) ≤ 2a
(by Lemma IV.2.c)
−1(∗)
D̂
ˆ (n+1)(∗) (k) ≤ Γ6
(by Lemma IV.2.f)
n+1 (k) ∆
and Lemma A.11.c.
Bound on L−1
∗ Aψ(∗) θ(∗) ν (µ)L∗ m8 : This follows from the previous bounds of this Proposition, Remark VI.2.b, Proposition V.1, Lemma II.6.c and Proposition II.4.c.
51
Appendix A: Bloch Theory for Periodic Operators
In the construction of [parabolic-all.tex] there are a good number of linear operators that act on functions defined a finite lattice and that are translation invariant with
(n+1)
(n)
respect to a sub–lattice. For example the lattice X−1
is a sublattice of X0 and the
(n)
(n+1)
block spin averaging operator Q : H0 → H−1
of (II.1) is translation invariant with
(n+1)
(n)
(0)
respect to X−1 . Similarly, X0 is a sublattice of Xn = Xn and the block spin averag(n)
(n)
ing operator Qn : Hn → H0 of (II.2) is translation invariant with respect to X0 . As
(n)
(n)
another example, the fluctuation integral covariance C (n) : H0 → H0 of §IV is transla(n+1)
tion invariant with respect to X−1 . In this appendix, we use the Bloch theory approach
to develop some general machinery for bounding such linear operators. In [parabolicall.tex] the operators of interest tend to be periodizations of operators acting on L2 of
an infinite lattice. An important example is the “differential” operator Dn . See Remark
III.1.a. We also develop general machinery for bounding such periodizations. We use the
results of this appendix in the main body of this paper to bound many of the operators
appearing in [parabolic-all.tex].
Periodic Operators in “Position Space” and “Momentum Space” Environments
We start by setting up a general environment consisting of a “fine” lattice and a
“coarse” sub–lattice. We shall consider operators that act on functions defined on the
former and that are translation invariant with respect to the latter. Let εT , εX > 0,
LT , LX ∈ IN and LT ∈ LT IN, LX ∈ LX IN and define the (finite) (D + 1)–dimensional
lattices
Xfin = εT ZZ/εT LT ZZ × εX ZZD /εX LX ZZD )
Xcrs = LT εT ZZ/εT LT ZZ × LX εX ZZD /εX LX ZZD )
and the corresponding Hilbert spaces
Hf = L2 (Xfin )
hφ∗1 , φ2 if = volf
Hc = L2 (Xcrs )
hψ1∗ , ψ2 ic = volc
X
φ1 (u)∗ φ2 (u)
u∈Xfin
X
ψ1 (x)∗ ψ2 (x)
x∈Xcrs
where we use
volf = εT εD
X
volc = (εT LT )(εX LX )D
to denote the volume of a single cell in Xfin , and Xcrs , respectively. For the Bloch construction, it will also be useful to define the “single period” lattice
B = εT ZZ/LT εT ZZ × εX ZZD /LX εX ZZD ) ∼
= Xfin /Xcrs
52
The lattices dual to Xfin , Xcrs and B are
Xˆfin =
Xˆcrs =
B̂ =
2π
εT LT
2π
εT LT
2π
LT εT
We denote by
D 2π
D
2π
ZZ/ ε2π
Z
Z
×
Z
Z
/
Z
Z
ε
L
ε
T
X
X X2π
D
2π
ZZ/ LT εT ZZ × εX LX ZZ / LX2πεX ZZD
D 2π
D
2π
∼
ZZ/ ε2π
Z
Z
×
Z
Z
/
Z
Z
= Xˆcrs /Xˆfin
LX εX
εX
T
π̂ : X̂fin → X̂crs
the canonical projection from Xˆfin to Xˆcrs . It has kernel B̂. Observe that
for all x ∈ Xcrs , p ∈ Xˆfin
p · x = π̂(p) · x mod 2π
The Fourier and inverse Fourier transforms are, for φ ∈ Hf , ψ ∈ Hc , ζ ∈ L2 (B), p ∈ Xˆfin ,
k ∈ Xˆcrs , ℓ ∈ B̂, u ∈ Xfin , x ∈ Xcrs and w ∈ B,
φ̂(p) = volf
X
φ(u)e−ip·u
φ(u) =
u∈Xfin
ψ̂(k) = volc
X
ψ(x)e−ik·x
ψ(x) =
x∈Xcrs
ζ̂(ℓ) = volf
X
ζ(w)e−iℓ·w
ζ(w) =
w∈B
c
v
olf
(2π)1+D
c
v
olc
(2π)1+D
c
v
olb
(2π)1+D
X
φ̂(p)eiu·p
p∈X̂fin
X
ψ̂(k)eik·x
k∈X̂crs
X
ζ̂(ℓ)eiw·ℓ
ℓ∈B̂
where
cf =
vol
(2π)1+D
(εT LT )(εX LX )D
cc =
vol
(2π)1+D
(εT LT )(εX LX )D
cb =
vol
(2π)1+D
(εT LT )(εX LX )D
denote the volume of a single cell in Xˆfin , Xˆcrs and B̂, respectively. Observe that
c
volc v
olf
(2π)1+D
=
c
volf v
olf
(2π)1+D
c
volc v
olc
(2π)1+D
=
1
LT LD
X
=
LT LD
X
LT LD
X
=
1
|Xfin |
=
1
|Xcrs |
=
1
|X̂fin |
=
1
|X̂crs |
(A.1)
where |Xcrs | denotes the number of points in Xcrs . By (A.1) and the fact that δu,u′ =
P
1
ip·u −ip·u′
e
,
p∈X̂fin e
|X̂ |
fin
hφ1 , φ2 if = volf
hψ1 , ψ2 ic = volc
X
φ1 (u)φ2 (u) =
u∈Xfin
X
ψ1 (x)ψ2 (x) =
x∈Xcrs
53
c
v
olf
(2π)1+D
c
v
olc
(2π)1+D
X
φ̂1 (−p)φ̂2 (p)
p∈X̂fin
X
k∈X̂crs
ψ̂1 (−k)φ̂2 (k)
Let A be any operator on Hf that is translation invariant with respect to Xcrs . We
call such an operator a “periodic operator”. Denote by A(u, u′ ) its kernel, defined so that
(Aφ)(u) = volf
X
A(u, u′ )φ(u′ )
u′ ∈Xfin
By “translation invariant with respect to Xcrs ”, we mean that A(u + x, u′ + x) = A(u, u′ )
for all u, u′ ∈ Xfin and x ∈ Xcrs . Set(1) , for p, p′ ∈ Xˆfin ,
Â(p, p′ ) =
and, for u, u′ ∈ Xfin and k ∈
2π
εT LT
volf
|Xfin |
ZZ ×
X
′
′
e−ip·u A(u, u′ )eip ·u
(A.2)
u,u′ ∈Xfin
2π
ZD ,
εX LX Z
Ak (u, u′ ) = volc
X
the “universal cover” of Xˆcrs ,
′′
e−ik·u A(u, u′′ )eik·u
(A.3)
u′′ ∈Xfin
u′′ −u′ ∈Xcrs
For each fixed u, u′ ∈ Xfin , k 7→ Ak (u, u′ ) is not a function on the torus Xˆcrs since, for
p ∈ LT2πεT ZZ × LX2πεX ZZD ,
′
Ak+p (u, u′ ) = e−ip(u−u ) Ak (u, u′ )
This is why we defined Ak for k ∈ εT2π
Z × εX2π
ZD , rather than k ∈ Xˆcrs . On the other
LT Z
LX Z
′
hand, k 7→ eik(u−u ) Ak (u, u′ ) is a well–defined function on Xˆcrs .
The following lemma is standard.
Lemma A.1 Let A be an operator on Hf that is translation invariant with respect to
Xcrs .
X
′
′
c
v
olf
eip·u Â(p, p′ )e−ip ·u
(a) A(u, u′ ) = (2π)1+D
p,p′ ∈X̂fin
(b) A(u, u′ ) =
Here
P
c
v
olc
(2π)1+D
X
′
′
′
eiℓ·u Â(k + ℓ, k + ℓ′ )e−iℓ ·u eik·(u−u )
[k]∈X̂crs
ℓ,ℓ′ ∈B̂
f (k) means that one sums k over a subset of
[k]∈X̂crs
2π
εT LT
ZZ × εX2π
ZZD that contains
LX
exactly one (arbitrary) representative from each equivalence class of X̂crs . Note that if
k ∈ εT2π
Z × εX2π
ZD and ℓ ∈ B̂, then k + ℓ ∈ Xˆfin .
LT Z
LX Z
(1)
The “normal prefactor” for  would be vol2f . We have chosen
approximate Dirac
(2π)1+D δ(p − p′ )’s
volf
|Xfin |
=
b
volf
vol2f
(2π)1+D
so as to replace
with simple Kronecker δp,p′ ’s in the translation invariant case.
54
[
(c) (Aφ)(p)
=
X
p′ ∈X̂fin
(d) For each k ∈
and u′ and
Â(p, p′ )φ̂(p′ ) for all φ ∈ Hf .
2π
εT LT
ZZ ×
2π
ZD ,
εX LX Z
A(u, u′ ) =
(e) Ak (u, u′ ) =
Ak (u, u′ ) is periodic with respect to Xcrs in both u
c
v
olc
(2π)1+D
X
′
eik·u Ak (u, u′ )e−ik·u
k∈X̂crs
X
′
′
eiℓ·u Â(k + ℓ, k + ℓ′ )e−iℓ ·u
ℓ,ℓ′ ∈B̂
(f ) Define the transpose of A by A∗ (u, u′ ) = A(u′ , u). Then
A∗ (u, u′ ) =
c
v
olc
(2π)1+D
X
[k]∈X̂crs
ℓ,ℓ′ ∈B̂
′
′
′
eiℓ·u Â(−k − ℓ′ , −k − ℓ)e−iℓ ·u eik·(u−u )
Periodized Operators
Define the (infinite) lattices
Zfin = εT ZZ × εX ZZD
Zcrs = LT εT ZZ × LX εX ZZD
Definition A.2 (Periodization) Suppose that a(u, u′ ) is a function on Zfin × Zfin that
◦ is translation invariant with respect to Zcrs in the sense that a(u + x, u′ + x) = a(u, u′ )
for all x ∈ Zcrs and u, u′ ∈ Zfin and
P P a(u, u′ ) and sup
a(u, u′ ) are both
◦ has finite L1 –L∞ norm (i.e. sup
u∈Zfin u′ ∈Zfin
finite)
u′ ∈Zfin u∈Zfin
and that the operator A (on Hf ) acts by
(Aφ)([u]) = volf
X
a(u, u′ )φ([u′ ])
(A.4)
u′ ∈Zfin
Here, for each u ∈ Zfin , the notation [u] means the equivalence class in Xfin that contains
u. Then we say that A is the periodization of a. It is “a with periodic boundary conditions
D
factors
}|
{
z
on a box of size εT LT × εX LX × · · · × εX LX ”.
55
Remark A.3
(a) The right hand side of (A.4) is independent of the representative u chosen from [u] (by
translation invariance with respect to εT LT ZZ × εX LX ZZD ⊂ Zcrs ).
(b) The kernel of A is given by
A [u], [u′ ] =
X
X
a(u, u′′ ) =
u′′ ∈Zfin
[u′′ ]=[u′ ]
a(u, u′ + z)
z∈εT LT ZZ×εX LX ZZD
The sum converges because a has finite L1 –L∞ norm. This is the motivation for the name
the “periodization of a”.
(c) If A is the periodization of a and B is the periodization of b, then C = AB is the
periodization of
X
c(u, u′ ) = volf
a(u, u′′ )b(u′′ , u′ )
u′′ ∈Zfin
D
D
2π
Z
Z
×
IR
/
Z
Z
be the dual space of Zcrs . Its universal cover is
Let Ẑcrs = IR/ εT2π
LT
εX LX
D
D
IR × IR . For each k ∈ IR × IR , set, for u, u′ ∈ Zfin ,
X
ak (u, u′ ) = volc
e−ik·u a(u, u′′ )eik·u
′′
(A.5)
u′′ ∈Zfin
u′′ −u′ ∈Zcrs
and, for ℓ, ℓ′ ∈ B̂,
âk (ℓ, ℓ′ ) =
=
1
|B|2
volf
|B|
X
′
e−iℓ·u ak (u, u′ )eiℓ ·u
′
[u],[u′ ]∈B
X
′
′
′
e−iℓ·u a(u, u′ )eiℓ ·u e−ik·(u−u )
(A.6)
[u]∈B
u′ ∈Zfin
vol
f
1
′
(Recall that |B|
2 = vol |B| . We shall show in Lemma A.5.a, below, that ak (u, u ) is periodic
c
with respect to Zcrs in both u and u′ .) By the L1 –L∞ hypothesis and the Lebesgue
dominated convergence theorem, both ak (u, u′ ) and ak (ℓ, ℓ′ ) are continuous in k.
Remark A.4 As was the case for Ak (u, u′ ), for each fized u, u′ ∈ Zfin , the map k 7→
Z × εX2π
ZD ,
ak (u, u′ ) is not a function on the torus Ẑcrs since, for p ∈ εT2π
LT Z
LX Z
′
ak+p (u, u′ ) = e−ip·(u−u ) ak (u, u′ )
56
However
′
k ∈ Ẑcrs 7→ eik·(u−u ) ak (u, u′ ) = volc
X
a(u, u′ + x)eik·x
x∈Zcrs
is a legitimate function on the torus Ẑcrs and is in fact the Fourier transform of the function
x ∈ Zcrs 7→ a(u, u′ + x)
Correspondingly, for p ∈
2π
εT LT
ZZ ×
2π
ZZD
εX LX
and ℓ, ℓ′ ∈ B̂
âk+p (ℓ, ℓ′ ) = âk (ℓ + p, ℓ′ + p)
The following two lemmas are again standard.
Lemma A.5 Let a(u, u′ ) : Zfin × Zfin → C obey the conditions of Definition A.2, and, in
particular, be translation invariant with respect to Zcrs .
(a) For each k ∈ IR × IRD , ak (u, u′ ) is periodic with respect to Zcrs in both u and u′ and
′
Z
′
1+D
d
k
ak (u, u′ )eik·(u−u ) (2π)
1+D
Ẑcrs
X Z
′ ′
′
d1+D k
eiℓ·u âk (ℓ, ℓ′ )e−iℓ ·u eik·(u−u ) (2π)
=
1+D
a(u, u ) =
ℓ,ℓ′ ∈B̂
Ẑcrs
(b) If, in addition, a(u, u′ ) = α(u − u′ ) is translation invariant with respect to Zfin , then
âk (ℓ, ℓ′ ) = δℓ′ ,ℓ α̂(k + ℓ)
where α̂(p) = volf
P
α(u)e−ip·u .
u∈Zfin
(c) Let A be the periodization of a. Then
Ak ([u], [u′ ]) = ak (u, u′ )
Â(k + ℓ, k + ℓ′ ) = âk (ℓ, ℓ′ )
for all k ∈
2π
εT LT
ZZ ×
2π
ZD ,
εX LX Z
[u], [u′ ] ∈ Xfin and ℓ, ℓ′ ∈ B̂.
Lemma A.6
(a) If a(u, u′ ) =
1
δ ′
volf u,u
is the kernel of the identity operator, then âk (ℓ, ℓ′ ) = δℓ,ℓ′ .
57
(b) Let a(u, u′ ), b(u, u′ ) : Zfin × Zfin → C both obey the conditions of Definition A.2, and
set
X
c(u, u′ ) = volf
a(u, u′′ )b(u′′ , u′ )
u′′ ∈Zfin
Then, for all k ∈
2π
εT LT
ZZ ×
2π
ZD
εX LX Z
and ℓ, ℓ′ ∈ B̂,
ck (ℓ, ℓ′ ) =
X
ak (ℓ, ℓ′′ )bk (ℓ′′ , ℓ′ )
ℓ′′ ∈B̂
We now generalize the above discussion to include periodized operators from L2 (Xcrs )
to L2 (Xfin ) and vice versa. If b(u, x) and c(x, u) are translation invariant with respect
to Zcrs (with x running over Zcrs and with u running over Zfin as usual) and have finite
L1 –L∞ norm, we define, for k ∈ IR × IRD and ℓ, ℓ′ ∈ B̂,
b̂k (ℓ) = volf
X
e−i(k+ℓ)·u b(u, x)eik·x
ĉk (ℓ′ ) = volf
[u]∈B
x∈Zcrs
For p ∈
2π
εT LT
ZZ ×
X
′
e−ik·x c(x, u)ei(k+ℓ )·u
[u]∈B
x∈Zcrs
2π
ZZD
εX LX
(A.7)
and ℓ, ℓ′ ∈ B̂
ĉk+p (ℓ′ ) = ĉk (ℓ′ + p)
b̂k+p (ℓ) = b̂k (ℓ + p)
The inverse transforms are
b(u, x) =
XZ
ℓ∈B̂
c(x, u) =
1+D
d
k
eiℓ·u b̂k (ℓ)eik·(u−x) (2π)
1+D
Ẑcrs
XZ
ℓ′ ∈B̂
′
1+D
d
k
âk (ℓ, ℓ′ )e−iℓ ·u eik·(x−u) (2π)
1+D
Ẑcrs
For ψ ∈ L2 (Xcrs ) and φ ∈ L2 (Xfin )
c + ℓ) = b̂k (ℓ)ψ̂(k)
bψ(k
X
c
cφ(k)
=
ĉk (ℓ′ )φ̂(k + ℓ′ )
(A.8)
ℓ′ ∈B̂
If b∗ (x, u) = b(u, x) and c∗ (u, x) = c(x, u) are the transposes of b and c, respectively, then
b̂∗k (ℓ′ ) = b̂−k (−ℓ′ )
ĉ∗k (ℓ) = ĉ−k (−ℓ)
58
(A.9)
Averaging Operators
In this subsection, we analyze “averaging operators” as examples of periodic operators.
Fix a function q : Xfin → IR and define the “averaging operator” Q : Hf → Hc by
X
(Qφ)(x) = volf
q(x − u)φ(u)
(A.10)
u∈Xfin
Lemma A.7
(a) The adjoint Q∗ is given by
Q∗ ψ)(u) = volc
X
x∈Xcrs
ψ(x)q(x − u)
(b) The composite operators QQ∗ and Q∗ Q are given by
X
(QQ∗ ψ)(x) = volf volc
q(x − u)q(x′ − u)ψ(x′ )
u∈Xfin
x′ ∈Xcrs
(Q∗ Qφ)(u) = volf volc
X
u′ ∈Xfin
x∈Xcrs
Proof:
q(x − u)q(x − u′ )φ(u′ )
trivial.
1
times the
Example A.8 Assume that LT and LX are odd and choose q to be vol
c LT −1 LT −1 LX −1 LX −1
D
characteristic function of the rectangle εT − 2 , 2
× εX − 2 , 2
in Xfin .
Observe that the number of points in this rectangle is exactly LT LD
. For x ∈ Xcrs , denote
LT −1 LT −1 LX −1 LX −1 D X
in Xfin . Also, for u ∈ Xfin ,
by x the rectangle x+εT − 2 , 2 ×εX − 2 , 2
let ξ(u) be the point of Xcrs closest to u. Then
X
(Qφ)(x) = LT1LD
φ(u)
Q∗ ψ)(u) = ψ ξ(u)
X
u∈
x
The composite operators are
(QQ∗ ψ)(x) =
1
LT LD
X
X
u∈
(Q∗ ψ)(u) =
x
(Q∗ Qφ)(u) = (Qφ) ξ(u) =
1
LT LD
X
59
1
LT LD
X
X
u′ ∈
ξ(u)
X
u∈
x
φ(u′ )
ψ(x) = ψ(x)
Lemma A.9 Let Q : Hf → Hc be the averaging operator of (A.10), but with q : Zfin → IR
and q(u) vanishing unless |u0 | < 21 εT LT and |uν | < 21 εX LX for ν = 1, 2, 3.
(a) For all φ ∈ Hf and ψ ∈ Hc ,
X
[
(Qφ)(k)
=
\
∗ ψ)(p) = q̂(p) ψ̂ π̂(p)
(Q
q̂(p)φ̂(p)
p∈X̂fin
π̂(p)=k
\
∗ ψ)(k) =
(QQ
P q̂(p)2 ψ̂(k)
\
∗ Qφ)(p) = q̂(p)
(Q
p∈X̂fin
π̂(p)=k
X
q̂(p′ ) φ̂(p′ )
p′ ∈X̂fin
π̂(p′ )=π̂(p)
(b) For A = Q∗ Q,
âk (ℓ, ℓ′ ) = q(k + ℓ) q(k + ℓ′ )
Proof:
(a) Using the definitions and (A.1),
[
(Qφ)(k)
= volc
X
(Qφ)(x)e−ik·x
= volf volc
x∈Xcrs
=
volf
|Xcrs |
X
x∈Xcrs
u∈Xfin
volf
|Xcrs |
p∈X̂fin
=
=
1
|Xcrs |
X
e−ik·x q(x − u)φ(u)
x∈Xcrs
u∈Xfin
e−ik·x eiu·p q(x − u)φ̂(p) =
X
X
X
e−ik·x ei(x−u)·p q(u)φ̂(p)
x∈Xcrs
u∈Xfin
p∈X̂fin
e−i(k−p)·x q̂(p)φ̂(p)
=
x∈Xcrs
p∈X̂fin
1
|Xcrs |
X
e−i(k−π̂(p))·x q̂(p)φ̂(p)
x∈Xcrs
p∈X̂fin
q̂(p)φ̂(p)
p∈X̂fin
π̂(p)=k
\
∗ ψ)(p) is similar. For the composite operators
The computation for (Q
∗ ψ)(k) =
\
(QQ
X
∗ ψ)(p) =
\
q̂(p)(Q
p∈X̂fin
π̂(p)=k
X
p∈X̂fin
π̂(p)=k
X q̂(p)2 ψ̂(k)
q̂(p)q̂(p) ψ̂ π̂(p) =
\
∗ Qφ)(p).
and similarly for (Q
(b) Since
a(u, u′ ) = volc
X
x∈Zcrs
60
q(x − u)q(x − u′ )
p∈X̂fin
π̂(p)=k
we have
âk (ℓ, ℓ′ ) =
volf volc
|B|
X
′
[u]∈B
u′ ∈Zfin
x∈Zcrs
X
= vol2f
′
e−i(k+ℓ)·(u−x) q(x − u)q(x − u′ )ei(k+ℓ )·(u −x)
′
e−i(k+ℓ)·(u−x) q(x − u)q(u′ )e−i(k+ℓ )·u
′
[u]∈B
u′ ∈Zfin
x∈Zcrs
= vol2f
X
′
ei(k+ℓ)·u q(u)q(u′ )e−i(k+ℓ )·u
′
u,u′ ∈Zfin
= q(k + ℓ) q(k + ℓ′ )
Example A.8 (continued) In the notation of Example A.8, the Fourier transform of q
is
D
X
Y
vol
q̂(p) = volfc
e−ip·u = uLT (εT p0 )
uLX (εX pℓ )
u∈Xfin
u∈ 0
where
ℓ=1
L−1
2
uL (ω) =
1
L
X
k=− L−1
2
e−iωk

Lω
 1 sin 2
= L sin ω2

1
if ω ∈
/ 2πZZ
otherwise
y = uL (ω)
π
L
−π
π
ω
Remark A.10 For the q of Example A.8, which is, up to a multiplicative constant, the
characteristic function of a rectangle, the Fourier transform q̂(p) decays relatively slowly
for large p. Choosing a smoother q increases the rate of decay of q̂(p). A convenient way to
“smooth off” Q is to select an even q ∈ IN and choose q to be the inverse Fourier transform
of
D
Y
q̂(p) = uLT (εT p0 )q
uLX (εX pℓ )q
ℓ=1
For example, when q = 2, q is the convolution of (a constant times) the characteristic
function of a rectangle with itself and so is a “tent” function. In the main text, we use
q ≥ 4.
61
Analyticity of the Fourier Transform and L1 –L∞ Norms
Define, for any m ≥ 0 and a : X × X′ → C, with X and X′ being any of our lattices,
n
o
X
X
′
′
kakm = max sup volX ′
em|y−y | |a(y, y ′ )| , sup volX
em|y−y | |a(y, y ′ )|
y∈X
y ′ ∈X′
y ′ ∈X′
y∈X
Here volX and volX ′ is the volume of a single cell in X and X ′ , respectively.
Lemma A.11 Let a(u, u′ ) : Zfin × Zfin → C, b(u, x) : Zfin × Zcrs → C and c(x, u) :
Zcrs × Zfin → C be translation invariant with respect to Zcrs and have finite L1 –L∞ norms.
Let 0 < m′′ < m′ < m.
(a) If kakm < ∞, then, for each ℓ, ℓ′ ∈ B̂, âk (ℓ, ℓ′ ) is analytic in |Im k| < m and
sup
|Im k|<m
âk (ℓ, ℓ′ ) ≤ kakm
(b) If, for each ℓ, ℓ′ ∈ B̂, âk (ℓ, ℓ′ ) is analytic in |Im k| < m, then,
sup
u,u′ ∈Zfin
a(u, u′ )em′ |u−u′ | ≤
kAkm′′ ≤ kakm′′ ≤
1
volc
sup
|Im k|=m′
Cm′ −m′′
volc
X âk (ℓ, ℓ′ ) ≤
|B|
volf
ℓ,ℓ′ ∈B̂
sup
|Im k|=m′
X
âk (ℓ, ℓ′ ) ≤
sup
|Im k|=m′
ℓ,ℓ′ ∈B̂
Cm′ −m′′ |B|
volf
P
′
u∈Zfin
e−(m −m
′′
)|u|
(c) If, for each ℓ ∈ B̂, b̂k (ℓ) is analytic in |Im k| < m, then,
′
sup b(u, x)em |u−x| ≤
u∈Zfin
x∈Zcrs
1
volc
sup
|Im k|=m′
X
b̂k (ℓ) ≤
1
volf
ℓ∈B̂
sup
|Im k|=m′
ℓ∈B̂
If, for each ℓ′ ∈ B̂, ĉk (ℓ′ ) is analytic in |Im k| < m, then,
′
sup c(x, u)em |x−u| ≤
u∈Zfin
x∈Zcrs
Proof:
1
volc
sup
|Im k|=m′
X
ĉk (ℓ′ ) ≤
âk (ℓ, ℓ′ ) ≤
volf
|B|
X
a(u, u′ )em|u−u′ | ≤
[u]∈B
u′ ∈Zfin
62
1
volf
ℓ′ ∈B̂
(a) If |Im k| < m, then
1
|B|
X
sup âk (ℓ, ℓ′ )
|Im k|=m′
ℓ,ℓ′ ∈B̂
ℓ,ℓ′ ∈B̂
where A is the periodization of a and Cm′ −m′′ = volf
âk (ℓ, ℓ′ )
[u]∈B
sup
|Im k|=m′
ℓ′ ∈B̂
.
b̂k (ℓ)
ĉk (ℓ′ )
kakm ≤ kakm
Analyticity in k follows from the uniform convergence of the series on |Im k| < m.
′
u−u
(b) Fix any u, u′ ∈ Zfin . Set q = m′ |u−u
′ | . Then
m′ |u−u′ |
′
a(u, u )e
=
X Z
ℓ,ℓ′ ∈B̂
=
′
Ẑcrs
X Z
ℓ,ℓ′ ∈B̂
′
âk (ℓ, ℓ′ )ei(k−iq)·(u−u ) eiℓ·u e−iℓ ·u
′
′
âk+iq (ℓ, ℓ′ )eik·(u−u ) eiℓ·u e−iℓ ·u
Ẑcrs
′
′
d1+D k
(2π)1+D
d1+D k
(2π)1+D
where we have applied Stokes’ theorem, using analyticity in k and the fact that
′
eik·(u−u )
X
′
′
′
âk (ℓ, ℓ′ )eiℓ·u e−iℓ ·u = eik·(u−u ) ak (u, u′ )
ℓ,ℓ′ ∈B̂
is periodic in the real part of k with respect to
a(u, u′ )em′ |u−u′ | ≤
Z
Ẑcrs
≤
1
volc
≤
|B|
volf
2π
εT LT
ZZ ×
2π
ZZD .
εX LX
X âk+iq (ℓ, ℓ′ )
ℓ,ℓ′ ∈B̂
sup
k∈Ẑcrs
Hence
d1+D k
(2π)1+D
X âk+iq (ℓ, ℓ′ )
ℓ,ℓ′ ∈B̂
sup âk+iq (ℓ, ℓ′ )
k∈Ẑcrs
ℓ,ℓ′ ∈B̂
The second bound is obvious from
volf
X X P
′′
′
′ m′′ |[u]−y ′ |
A [u], y ′ em |[u]−y | = volf
a(u,
u
)
u′ ∈Z
e
y ′ ∈Xfin
y ′ ∈Xfin
≤ volf
fin
[u′ ]=y ′
X a(u, u′ )em′′ |u−u′ |
u′ ∈Zfin
(with the distance |[u] − y ′ | measured in Xfin and the distance |u − u′ | measure in Zfin ) and
the similar bound with the roles of u and u′ interchanged.
(c) The proof is much the same as that of part (b).
Lemma A.12 Let m > 0. Let, for each ℓ, ℓ′ ∈ B̂, b̂k (ℓ, ℓ′ ) be analytic in |Im k| < m.
Assume that
b̂k+p (ℓ, ℓ′ ) = b̂k (ℓ + p, ℓ′ + p)
63
for all p ∈
2π
εT LT
2π
ZD
εX LX Z
and ℓ, ℓ′ ∈ B̂. Set
X Z
′ ′
′
′
d1+D k
eiℓ·u b̂k (ℓ, ℓ′ )e−iℓ ·u eik·(u−u ) (2π)
a(u, u ) =
1+D
ZZ ×
ℓ,ℓ′ ∈B̂
Ẑcrs
Then a(u, u′ ) : Zfin × Zfin → C obeys the conditions of Definition A.2 and
âk (ℓ, ℓ′ ) = b̂k (ℓ, ℓ′ )
for all k ∈ IR × IRD and ℓ, ℓ′ ∈ B̂.
Proof:
The proof is straightforward.
Functions of Periodic Operators
Let C be a simple, closed, positively oriented, piecewise smooth curve in the complex
plane and denote by OC its interior. Denote by σ(A) the spectrum of the bounded operator
A and assume σ(A) ⊂ OC . Let f (z) be analytic on the closure of OC . Then, by the Cauchy
integral formula,
I
f (A) =
and, for any m ≥ 0,
kf (A)km ≤
1
2πi
C
f (ζ)
ζ1l−A
dζ
(A.11)
1
|C| sup |f (ζ)| sup (ζ1l
2π
ζ∈C
ζ∈C
− A)−1 m
(A.12)
Lemma A.13 Let
◦ a(u, u′ ) : Zfin × Zfin → C obey the conditions of Definition A.2,
◦ C be a simple, closed, positively oriented, piecewise smooth curve in the complex plane
with interior OC ,
◦ O contain the closure of OC and f : O → C be analytic, and
◦ 0 < m′′ < m′ < m.
Suppose that
◦ for each ℓ, ℓ′ ∈ B̂, âk (ℓ, ℓ′ ) is analytic in |Im k| < m.
◦ for each ζ ∈ C \ OC and each k with |Im k| < m, the matrix ζδℓ,ℓ′ − âk (ℓ, ℓ′ ) ℓ,ℓ′ ∈B̂ is
invertible.
Denote by A the periodization of a. Then f (A), defined by (A.11), exists and
X Cm′ −m′′
(ζ1l − âk )−1 (ℓ, ℓ′ )
kf (A)km′′ ≤ 2π
|C|
sup
|f
(ζ)|
sup
volc
|Im k|=m′
ζ∈C
ζ∈C
≤
Cm′ −m′′ |B|
2π volf |C| sup |f (ζ)|
ζ∈C
64
ℓ,ℓ′ ∈B̂
sup
|Im k|=m′
ℓ,ℓ′ ∈B̂
ζ∈C
(ζ1l − âk )−1 (ℓ, ℓ′ )
Here (ζ1l − âk )−1 refers to the inverse of the |B̂| × |B̂| matrix ζδℓ,ℓ′ − âk (ℓ, ℓ′ ) ℓ,ℓ′ ∈B̂ .
Proof:
Each matrix element of ζ1l − âk is continuous on
D=
(ζ, k) ∈ C2 ζ ∈ C \ OC , |Im k| < m
−1
Furthermore det(ζ1l−âk ) does not vanish on D. Hence every matrix element of ζ1l−âk
is also continuous on D and in particular is bounded on compact subsets of D. Set, for
each for each ζ ∈ C \ OC , and u, u′ ∈ Zfin ,
′
rζ (u, u ) =
X Z
ℓ,ℓ′ ∈B̂
′
′
′
1+D
d
k
eiℓ·u (ζ1l − âk )−1 (ℓ, ℓ′ )e−iℓ ·u eik·(u−u ) (2π)
1+D
Ẑcrs
By Lemma A.12, rζ (u, u′ ) obeys the conditions of Definition A.2 and
r̂ζ,k (ℓ, ℓ′ ) = (ζ1l − âk )−1 (ℓ, ℓ′ )
By Lemma A.6, rζ = (ζ1l − a)−1 , as operators on L2 (Zfin ). By Remark A.3.c, for each
−1
. In particular, σ(A) ⊂ OC . By
ζ ∈ C \ OC , the periodization of rζ (u, u′ ) is ζ1l − A
Lemma A.11.b,
k(ζ1l − A)−1 km′′ ≤
≤
Cm′ −m′′
volc
sup
|Im k|=m′
Cm′ −m′′ |B|
volf
X (ζ1l − âk )−1 (ℓ, ℓ′ )
ℓ,ℓ′ ∈B̂
sup
|Im k|=m′
ℓ,ℓ′ ∈B̂
(ζ1l − âk )−1 (ℓ, ℓ′ )
Then, by (A.12),
kf (A)km′′ ≤
≤
≤
1
2π |C| sup |f (ζ)| sup (ζ1l
ζ∈C
ζ∈C
Cm′ −m′′
2π volc
|C| sup |f (ζ)|
ζ∈C
sup
|Im k|=m′
ζ∈C
Cm′ −m′′ |B|
2π volf |C| sup |f (ζ)|
ζ∈C
65
− A)−1 m′′
sup
X (ζ1l − âk )−1 (ℓ, ℓ′ )
ℓ,ℓ′ ∈B̂
|Im k|=m′
ℓ,ℓ′ ∈B̂
ζ∈C
(ζ1l − âk )−1 (ℓ, ℓ′ )
Scaling of Periodized Operators
Scaling plays an important in the construction of [parabolic-all.tex]. See, for
example, [parabolic-all.tex, Definition I.3 and §III] and (I.3). For the abstract setting
of this appendix, select scaling factors σT and σX and define the scaled lattices
(s)
Zfin =
εT
σT
ZZ ×
εX
σX
volf =
εT εD
X
D
σT σX
vol(s)
c =
(LT εT )(LX εX )D
D
σT σX
(s)
ZZD
(s)
ZZD
Zcrs
= LT σεTT ZZ × LX σεX
X
D
D
(s)
2πσX
T
Z
Z
×
IR
/
Z
Z
Ẑcrs
= IR/ ε2πσ
L
ε
L
T T
X X
(s)
(s)
εT
εT
εX
(s)
B = σT ZZ/LT σT ZZ × σX ZZD /LX σεX
ZZD ) ∼
= Xfin /Xcrs
X
T
X
X
T
ZZ/ 2πσ
ZZ × L2πσ
ZZD / 2πσ
ZZD
B̂(s) = L2πσ
εT
εX
T εT
X εX
The map L(τ, x) = (σT τ, σX x) gives bijections
(s)
(s)
L : Zcrs
→ Zcrs
L : B(s) → B
(s) (s) L induces linear bijections L∗ : L2 Zfin → L2 Zfin and L∗ : L2 Zcrs → L2 Zcrs by
L∗ (α)(Lu) = α(u). Observe that
L : Zfin → Zfin
(s)
(s)
D
hL∗ α, L∗ βif = σT σX
hα, βif
D
hL∗ α, L∗ βic = σT σX
hα, βic
Lemma A.14 Let a : L2 Zfin → L2 Zfin have kernel a(u, u′ ).
(a) The kernel of L−1
∗ a L∗ is
D
a(s) (v, v′ ) = σT σX
a Lv, Lv′
(b) The Fourier transform of the kernel of L−1
∗ aL∗ is
(s)
âk (ℓ, ℓ′ ) = âL−1 k (L−1 ℓ, L−1 ℓ′ )
(c) If m ≥ max
Proof:
1
σT
for k ∈ IR × IRD ,
ℓ, ℓ′ ∈ B̂(s)
, σ1X ms , then ka(s) kms ≤ kakm .
(s) (s)
(a) For α ∈ L2 Zfin
and v ∈ Zfin ,
X
(L−1
a(Lv, u′ ) L∗ α (u′ )
∗ aL∗ α)(v) = volf
u′ ∈Zfin
= volf
X
u′ ∈Zfin
(s)
= volf
X
′
a(Lv, u′ ) α(L−1
∗ u )
(s)
v′ ∈Zfin
66
D
σT σX
a(Lv, Lv′ ) α(v′ )
(b) By (A.6) and part (a),
(s)
âk (ℓ, ℓ′ ) =
=
volf
|B̂|
volf
|B̂|
X
′ ′
′
e−iℓ·v a Lv, Lv′ eiℓ ·v e−ik·(v−v )
X
e−i(L
[v]∈B(s)
(s)
v′ ∈Z
fin
−1
ℓ)·u
−1 ′
ℓ )·u′ −i(L−1 k)·(u−u′ )
a(u, u′ )ei(L
e
[u]∈B
u′ ∈Zfin
= âL−1 k (L−1 ℓ, L−1 ℓ′ )
(c) This part follows from the inequality
(s)
sup volf
(s)
v∈Zfin
X
(s)
v′ ∈Zfin
′
(s)
ems |v−v | |a(s) (v, v′ )| = sup volf
(s)
v∈Zfin
= sup volf
u∈Zfin
′
(s)
v′ ∈Zfin
X
D
ems |v−v | σT σX
|a(Lv, Lv′ )|
−1
ems |L
u′ ∈Zfin
≤ sup volf
u∈Zfin
X
X
u′ ∈Zfin
u−L−1 u′ |
|a(u, u′ )|
′
em|u−u | |a(u, u′ )|
and the corresponding inequality with v summed over and v′ supped over.
More generally,
Lemma A.15 Let b : L2 Zcrs → L2 Zfin and c : L2 Zfin → L2 Zcrs have kernels
b(u, x) and c(x, u) respectively.
−1
(a) The kernels of L−1
∗ b L∗ and L∗ c L∗ are
D
b(s) (v, x) = σT σX
b Lv, Lx
D
c(s) (x, v) = σT σX
c Lx, Lv
−1
(b) The Fourier transform of the kernels of L−1
∗ b L∗ and L∗ c L∗ are
(s)
b̂k (ℓ) = b̂L−1 k (L−1 ℓ)
(c) If m ≥ max
1
σT
(s)
ĉk (ℓ′ ) = ĉL−1 k (L−1 ℓ′ )
, σ1X ms , then
kb(s) kms ≤ kbkm
for k ∈ IR × IRD ,
kc(s) kms ≤ kckm
67
ℓ, ℓ′ ∈ B̂(s)
Appendix B: Trigonometric Inequalities
Lemma B.1
(a) For x, y real with |x| ≤
π
2,
(b) For x, y real with |y| ≤ 1,
| sin(x+iy)|
|x+iy|
≤ 2 min 1 ,
sin(x + iy) ≥
1
|x+iy|
√
2
|x
π
+ iy|
sin(x+iy) Im x+iy ≤ 2|y| min |x| ,
2
|x+iy|
(c) For 0 < ε ≤ 1 and x, y real with |εx| ≤ π, |y| ≤ 2
1
sin 12 (x + iy) sin
(x
+
iy)
2
2
≤ 4 min 1,
≤ 6|y| min |x| , 8
Im
1
1
1 sin 1 ε(x + iy) |x|
|x|
ε
2
ε sin 2 ε(x + iy)
(d) For x real with |x| ≤
π
2,
2
π
≤
| sin x|
|x|
≤1
(e) For any complex number z obeying |z| ≤ 2,
sin z
1 2
z − 1 ≤ 2 |z|
Proof:
By the standard trig identity
sin(x + iy) = sin(x) cos(iy) + cos(x) sin(iy) = sin(x) cosh(y) + i cos(x) sinh(y)
(a) For x, y real with |x| ≤ π2
Re sin(x + iy) = sin(x) cosh(y) ≥ | sin(x)| ≥ 2 |x|
π
Re sin(x + iy) = sin(x) cosh(y) ≥ sin(x) sinh(y) ≥ | sin(x)| |y|
Im sin(x + iy) = cos(x) sinh(y) ≥ | cos(x)| |y|
so that
sin(x + iy) ≥ max 2 |x| , |y| ≥
π
(b) For x, y real with |y| ≤ 1,
√
2
|x
π
+ iy|
| sin(x + iy)| = sin(x) cosh(y) + i cos(x) sinh(y) ≤ cosh(1) sin(x) + i cos(x) = cosh(1)
68
P∞
and, since sinh(y) = n=0
y 2n+1 (2n+1)! ≤ |y|
P∞
|y|2n
n=0 (2n)!
= |y| cosh(y) ≤ cosh(1)|y|,
| sin(x + iy)| = sin(x) cosh(y) + i cos(x) sinh(y) ≤ cosh(1)|x| + i|y| = cosh(1)|x + iy|
Thus
| sin(x+iy)|
|x+iy|
giving the first bound.
For the second bound
Using
≤ cosh(1) min 1 ,
1
|x+iy|
≤ 2 min 1 ,
1
|x+iy|
sin(x+iy) x Im sin(x+iy)−y Re sin(x+iy) x cos(x) sinh(y)−y sin(x) cosh(y) =
Im x+iy = x2 +y 2
x2 +y 2
Z
Z
x
Z
x
t
ds cos s
dt
dt sin t
=−
cos x − 1 = −
0
0
0
Z x
Z x Z t Z s
sin x − x =
dt [cos t − 1] = −
dt
ds
du cos u
0
0
0
0
Z y Z t
Z y
ds cosh s
dt
dt sinh t
=
cosh y − 1 =
0
0
0
Z y
Z y Z t Z s
sinh y − y =
dt [cosh t − 1] =
dt
ds
du cosh u
0
0
0
0
and cosh(1) < 2, we have
2
cos x = 1 + α(x) x2
3
cosh y = 1 + γ(y)y 2
sin x = x + β(x) x6
3
sinh y = y + δ(y) y3
with, for |y| ≤ 1, |α(x)| , |β(x)| , |γ(y)| , |δ(y)| ≤ 1. Consequently,
sin(x+iy) ≤ 2|xy|
Im
x+iy Alternatively, using | sin(x)| ≤ |x|, | sinh(y)| ≤ 2|y| and | cosh(y)| ≤ 2,
sin(x+iy) Im x+iy ≤
4|xy|
x2 +y 2
≤ √ 4|y|
2
x +y 2
(c) For 0 < ε ≤ 1, x, y real and |εx| ≤ π, |y| ≤ 2,
sin 21 (x+iy) sin 12 (x + iy) 12 (x+iy)
= 1
1 sin 1 ε(x + iy) sin 2 ε(x+iy) ≤
1 ε(x+iy) ε
2
2
π
√
2
2
cosh(1) min 1,
|x + iy|
69
2
≤ 4 min 1,
|x|
and
Im sin 12 (x+iy) Re sin 21 ε(x+iy) − Re sin 21 (x+iy) Im sin 21 ε(x+iy) 1
1
1
1
1
2 (x+iy)
2 ε(x+iy)
2 (x+iy)
2 ε(x+iy)
Im sin 2 (x + iy) =
sin 1 ε(x+iy) 2
1
1
12
ε sin 2 ε(x + iy)
2 ε(x+iy)
2 sin 1 ε(x+iy) Im 1 2
+ 4 min 1, |x|
2 ε(x+iy)
≤
sin 1 ε(x+iy) 12
2 ε(x+iy)
2
4
+ 4 min 1, |x|
ε|y| min 21 ε|x| ,
|y| min 12 |x| , |x+iy|
√
≤
1
2 |x| ,
|y| min
≤
π
√
|y| min
2
1
2
≤ 6|y| min |x| ,
4
|x+iy|
+ 2ε2 |x| ,
8
2
π
4+16
|x+iy|
|x+iy|
(d) This follows from the observation that, for 0 ≤ x ≤ π2 ,
To see this, observe that
d sin x
= x1 cos x − sinx x
dx x
and
cos x −
sin x
x
=
∞
X
(−1)
n
1
(2n)!
n=1
1
(2n+1)!
−
4
ε|x+iy|
sin x
x
is monotone decreasing.
∞
2n X
2n
x =
(−1)n (2n+1)!
x2n
n=1
This is an alternating series with the first term being negative and with the ratio between
term n + 1 and term n having magnitude
(2n+1)! 2n+2
2
2n
(2n+3)! x
which is less than 1 for all |x| ≤
√
=
x2
2n(2n+3)
≤
x2
10
10 = 3.162.
(e) We have
∞
∞
∞
X
X
X
sin z
n z 2(n+1) 2
n z 2n n z 2n =
(−1)
=
|z|
=
(−1)
(−1)
−
1
z
(2n+1)!
(2n+3)!
(2n+3)! n=0
n=1
As
we have
When |z| ≤ 2
n=0
n+3 factors
}|
{
z
(2n + 3)! = n! (n + 1) · · · (2n + 3) ≥ n! 3! 5n
e|z|
2
/5
sin z
≤
−
1
z
≤e
⇒
|z|2 |z|2 /5
3! e
sin z
1 2
≤ |z|
−
1
z
2
70
Appendix C: Lattice and Operator Summary
The following table gives, for most of the operators considered in this paper,
◦ the definition of the operator
◦ a reference to where in [parabolic-all.tex], the operator is introduced and
◦ the translation invariance properties of the operator.
A later table will specify where, in this paper, bounds on the operators are proven.
Operator
D0 = −e−h0 ∂0 + 1l − e−h0
n
Dn = L2n L−n
∗ D0 L∗
−1
Pn−1
Qn = a 1l + j=1 L12j Qj Q∗j
∆(0) = D0
∗ −1
∆(n) = 1l + Qn Qn Dn−1 Qn
C (n) =
a
Q∗ Q
L2
+∆
Qn , n ≥ 1
(n) −1
Sn−1 = Dn + Q∗n Qn Qn
: H0 → H0
: Hn → Hn
(n)
: H0
(n)
(n)
§I.5
X0
Definition I.5.b
(I.14)
Xn
(n)
X0
X0
(n)
: H0
→ H0
(n)
(I.14)
: H0
(n)
X0
→ H0
(I.15)
X−1
Theorem I.15
X0
Theorem I.15
X0
Lemma III.4.b
X−1
(G.27)
X−1
(n)
: Hn → Hn
Sn (µ)−1 = Dn + Q∗n Qn Qn − µ
: Hn → Hn
2
(n+1)
(n+1)
∗ −1
Q̌n+1 = La 1l + QQ−1
: H−1 → H−1
n Q
−1
Šn+1 (µ) = Dn + Q̌∗n+1 Q̌n+1 Q̌n+1 − µ
: Hn → Hn
(n)
Aψ,φ
Tiwrt
Definition I.5.a
→ H0
: H0 → H0
Definition
: Hn → H0
(n+1)
(n)
(n)
(n+1)
(n+1)
(n+1)
Proposition G.13 X−1
The references in the above table are to [parabolic-all.tex] and “Tiwrt” stands for
“translation invariant with respect to”.
Here is a table giving where some other operators were defined in this paper.
Operator
Definition
∗ −1
Sn Q∗n = Dn−1 Q∗n 1l + Qn Qn Dn−1 Qn
(n)
(n)
Lemma V.6
X0
: Hn → Hn
after (VI.2)
Aψ,φ
(n)
X−1
: Hn → H0
(VI.4)
Aψ,φ,ν
(n)
X−1
: Hn → H0
Remark VI.2
X−1
Šn+1 = Šn+1 (0)
a
(n) (∗) ∗
Q
L2 C
: H0
Tiwrt
→ Hn
(n+1)
(n)
: H−1
→ H0
Aψ∗ ,θ∗ ,ν , Aψ,θ,ν , Aψ∗ ,θ∗ ,ν (µ), Aψ,θ,ν (µ) : H−1
→ H0
Aψ(∗) ,θ(∗) =
(n+1)
71
(n)
(n+1)
(n+1)
(n+1)
(n+1)
before Remark VI.2 X−1
Remark VI.2
(n+1)
X−1
The lattices involved are
Xn = ε2n ZZ/ε2n Ltp ZZ × εn ZZ3 /εn Lsp ZZ3
Xˆn =
(n)
(n)
X0 = ZZ/ε2n Ltp ZZ × ZZ3 /εn Lsp ZZ3
X̂0 =
(n+1)
(n+1)
X−1 = L2 ZZ/ε2n Ltp ZZ × LZZ3 /εn Lsp ZZ3
Xˆ−1 =
where εn =
1
Ln .
The “single period” lattices are
×
2π
Z3 / 2π
Z3
εn Lsp Z
εn Z
×
2π
2π
Z/ L
Z ×
2Z
ε2 Ltp Z
2π
Z3 /2πZZ3
εn Lsp Z
2π
Z/ 2π
Z
ε2n Ltp Z
ε2n Z
2π
Z/2πZZ
ε2n Ltp Z
n
2π
Z3 / 2π
Z3
εn Lsp Z
L Z
3 2π 3
Z
Z
×
2πZ
Z
/
Z
Z
Bn = ε2n ZZ/ZZ × εn ZZ3 /ZZ3
B̂n = 2πZZ/ 2π
2
εn
εn
2π 3
3
3
+
2
+
2π
B = ZZ/L ZZ × ZZ /LZZ
B̂ = L2 ZZ/2πZZ × L ZZ /2πZZ3
The following table specifies where, in this paper, bounds on the various operators are
proven.
Operator
Bound
Qn
Lemma II.2.a
Qn
Proposition II.4
(±)
Qn,ν
Lemma II.6
Dn
Lemma III.2
∆(n)
Lemma IV.2
C (n)
Corollary IV.5
D(n)
Corollary IV.5
Sn (µ), Sn
Proposition V.1
Sn,ν (µ), Sn,ν
(±)
Proposition V.1
Aψ,φ
Proposition VI.1
Aψ,φ,ν
Proposition VI.1
Aψ(∗) θ(∗)
Proposition VI.1
Aψ(∗) θ(∗) ν (µ)
Proposition VI.1
Šn+1
Corollary VI.5
(±)
72
References
[BFKT1] T. Balaban, J. Feldman, H. Knörrer and E. Trubowitz, A Functional Integral Representation for Many Boson Systems I: The Partition Function, Annales Institut
Poincaré 9 (2008), 1229–1273.
[BFKT2] T. Balaban, J. Feldman, H. Knörrer and E. Trubowitz, A Functional Integral Representation for Many Boson Systems II: Correlation Functions, Annales Institut
Poincaré 9 (2008), 1275–1307.
[BFKT3] T. Balaban, J. Feldman, H. Knörrer and E. Trubowitz, Power Series Representations for Complex Bosonic Effective Actions I: A Small Field Renormalization
Group Step, Journal of Mathematical Physics 51 (2010), 053305.
[BFKT4] T. Balaban, J. Feldman, H. Knörrer and E. Trubowitz, The Temporal Ultraviolet Limit for Complex Bosonic Many-body Models, Annales Institut Poincaré 11
(2010), 151–350. DOI: 10.1007/s00023-010-0028-5.
lockspinRGalg.tex]
parabolic-all.tex] T. Balaban, J. Feldman, H. Knörrer and E. Trubowitz, Complex Bosonic Many–
body Models: The Small Field Parabolic Flow, preprint.
73