Hardy Spaces and Boundary Conditions from the Ising Model

0.24 MB·English

by  Clément Hongler

HARDY SPACES AND BOUNDARY CONDITIONS FROM THE
ISING MODEL
2
CLÉMENTHONGLERANDDUONGH.PHONG
1
0
2 Abstract. FunctionsinHardyspacesonmultiply-connecteddomainsinthe
plane are given an explicitcharacterization interms of aboundary condition
n
inspired bythe two-dimensional Isingmodel. The keyunderlying property is
a
J the positivity of a certain operator constructed inductively on the number of
components oftheboundary.
7

]
V
C 1. Introduction
.
h A remarkable property of critical phenomena in two dimensions is their local
t conformal invariance. This has resulted in a rich interaction between statistical
a
m physicsandmanybranchesofmathematics,includingprobability,complexanalysis,
Riemann surfaces,andinfinite-dimensional Lie algebras(see e.g. [1, 3, 4, 2, 13, 14,
[
16] and references therein).

1
The goalof the present paper is to show how ideas from two-dimensionalstatis-
v
ticalphysicscanhelpansweringanimportantquestionincomplexanalysis,namely
6
6 how to explictly characterize the boundary values of holomorphic functions on a
5 smooth multiply-connected domain Ω. See [7] for an elementary introduction to
1 analytic function spaces on planar domains.
.
1 The simply-connected case is well-known. In this case, Ω can be assumed to be
0 theunitdiskDbythe Riemannmappingtheorem. The spaceofholomorphicfunc-
2 tionsadmitting L2 boundaryvaluesisthe HardyspaceH2(D), andtheirboundary
1
: values can be characterized by the condition that all their Fourier coefficients of
v negative index vanish (see e.g. [17]). The projection operator on the space of L2
i
X boundaryvaluesofholomorphicfunctionsisgivenintermsoftheHilberttransform,
r which is the primary example of a singular integral operator. Underlyingl this is
a thebasicfactthat,foranyreal-valuedL2 functionf on∂D,thereisasingle-valued
holomorphic function F on D with f as its real part on ∂D.
In general,there is no suchfunction F if Ω is multiply-connected(see [5], for in-
stance). However,anaturalremedytothisisthefollowingboundaryvalueproblem,
which (as we will show) always possesses a unique solution:
(1.1) F holomorphic on Ω, m((F f)ν12)=0 on ∂Ω
ℑ −
where ν is the outer normal to the boundary, viewed as a complex number. This
boundaryvalueproblemissuggestedbyrecentstudiesoftheIsingmodel[15,8,10].
Ann-connecteddomainΩintheplaneadmits2n−1 differentspinstructures. Since
Ω admits a global frame, one of the spin structures, which we refer to as the
trivial one, can be identified with scalar single-valued functions. We shall show
that the boundary value problem (1.1) can always be solved for the trivial spin
structure. This will give a simple explicit characterization of L2 boundary values
1
HARDY SPACES AND BOUNDARY CONDITIONS FROM THE ISING MODEL 2
of holomorphic functions. In the process, we shall also find an analogue, for each
of the spin structures, of the Hilbert transform for multiply-connected domains.
Inthispaper,weprovideafunctional-analyticproofoftheexistenceandunique-
ness of the solution to the problem 1.1.
OurresultcaninprinciplebeappliedtothestudyoftheSchramm’s-SLEcurves
SLE(3) and SLE(16/3) in finitely-connected geometries, for which solutions of the
boundary value problem (1.1) are crucial [11].
2. Statement of the main results
Let Ω be a bounded domain in R with smooth boundary ∂Ω. Let N and N
in out
be the unit inward and outward pointing normals to the boundary ∂Ω. They can
be identified with complex numbers ν and ν by writing,
in out
∂ ∂
(2.1) N =2 e(ν ), N =2 e(ν ),
in in out out
ℜ ∂z ℜ ∂z
with ν = ν , ν = ν =1.
in out in out
− | | | |
Let L2(∂Ω) denote the space of L2 complex-valued functions defined on ∂Ω.
Even though this is clearly a (complex) Hilbert space, we will most of the time
view it as a real Hilbert space. Define the (real-linear) projection operators P :
in
L2(∂Ω) L2(∂Ω) and P :L2(∂Ω) L2(∂Ω) by
out
→ →
1
P [f](z) := f(z)+ν (z)f(z) ,
in in
2
P [f](z) := 1(cid:0)f(z)+ν (z)f(z(cid:1)) ,
out out
2
Observe that the orthogonal projection(cid:0)ProjeiθR of C on th(cid:1)e real line eiθR in C
acanndbPe ex[fpr](ezs)seadreasjuPsrtojtehiθeRp(ζro)j=ect21ionζs+ofe2tihθζe c,om∀pζle∈xCn,u∀mθb∈erRf.(Tz)huosnPtihne[ft](wzo)
out
(cid:0) (cid:1)
perpendicular lines in C defined by ν−21(z)R and ν−12(z)R. As such, they are
in out
just twisted versions of the projections of complex numbers onto their real and
imaginaryparts. They provideasimple wayofformulatingboundaryconditionsof
the form (1.1), e.g.,
(2.2) m(f(z)ν12(z))=0 P [f](z)=0, z ∂Ω.
ℑ in ⇔ in ∈
Clearly P2 = P , P2 = P , and P +P = Id. We can now define the real
in in out out in out
Hilbert subspaces L2 (∂Ω) and L2 (∂Ω) by
in out
(2.3) L2 (∂Ω)=Ker(P )=Range(P ),
in out in
(2.4) L2 (∂Ω)=Ker(P )=Range(P ),
out in out
and we have the direct-sum decomposition,
(2.5) L2(∂Ω)=L2 (∂Ω) L2 (∂Ω).
in ⊕ out
Let H2(Ω)be the Hardy spaceofholomorphicfunctions on Ω. It canbe defined
inseveralways,andone wayis asthe Banachspaceofholomorphicfunctions F(z)
on Ω satisfying
(2.6) sup F(z)2dσ(z)< ,
0<δ<<1ˆ | | ∞
∂Ωδ
HARDY SPACES AND BOUNDARY CONDITIONS FROM THE ISING MODEL 3
where Ω is the subset of Ω consisting of points at a distance > δ from ∂Ω. For
δ
δ sufficiently small, the orthogonal projection of ∂Ω on ∂Ω is a diffeomorphism,
δ
and functions on ∂Ω can be identified with functions on ∂Ω. What is important
δ
for our purposesis the fact thatfor eachfunction F H2(Ω), the restrictionsofF
∈
to ∂Ω , viewed in this way as functions on ∂Ω, converge in L2(∂Ω) and pointwise
δ
a.e. to a function R F. The “restriction operator” R is a bounded, injective,
∂Ω Ω
operator from H2(Ω) to L2(Ω).
Then our main result can be formulated as follows:
Theorem 1. Let Ω be a bounded domain with smooth boundary. Then for any
function f L2(∂Ω), there exists a unique function F H2(Ω) satisfying the
∈ ∈
boundary condition
(2.7) P (R (F) f)=0.
in Ω
−
This theorem is essentially equivalent to another theorem, which is actually
the one that we shall prove first. Let the operators T : H2(Ω) L2 (∂Ω) and
Ω → in
U :H2(Ω) L2 (∂Ω) be defined by
Ω → out
(2.8) T :=P R , U :=P R .
Ω in ∂Ω Ω out ∂Ω
◦ ◦
It is useful to depict this graphically as
H2(Ω)
R
ւTΩ ↓ ∂Ω ցUΩ
L2 (∂Ω) L2(∂Ω) L2 (∂Ω)
in ← → out
Theorem 2. Let Ω be a bounded domain with smooth boundary. Then
(a) the operators T : H2(Ω) L2 (∂Ω) and U : H2(Ω) L2 (∂Ω) are
Ω → in Ω → out
real-linear isomorphisms.
(b) Define the operator W :L2 (∂Ω) L2 (∂Ω) by
Ω in → out
(2.9) W =U T−1.
Ω Ω◦ Ω
Then W is a one-to-one and onto operator satisfying
W (j) W = Id
Ω Ω
◦ ◦ −
( j) W (j) = T U−1,
− ◦ Ω◦ Ω◦ Ω
where j denotes i Id (multiplication by i operator).
·
ToseehowTheorem1followsfromTheorem2,itsufficestoobservethat,ifT−1
Ω
exists, then for any function f L2(∂Ω), the function F = T−1(P (f)) satisfies
∈ Ω in
the desired property.
Part (b) of Theorem 2 can be easily seen by tracing back the definitions of the
operators T and U , once Part (a) has been proved. Thus we shall henceforth
Ω Ω
concentrate on the proof of Part (a) of Theorem 2.
The spaces L2 (∂Ω) and L2 (∂Ω) can be viewedas the analogues,for multiply-
in out
connected domains, of the spaces of L2 functions with only non-vanishing Fourier
coefficientsofrespectivelypositiveandnegativeindices inthe caseofthe unitdisk.
The operator W is a variant of the Hilbert transform, for the twisted line bundle
Ω
ν−21(z)R on the boundary of Ω.
out
HARDY SPACES AND BOUNDARY CONDITIONS FROM THE ISING MODEL 4
As we had stressed in the introduction, the analogue of the preceding theorem
wouldfailif the projectionoperatorsP andP werereplacedby the projections
in out
onthe trivialline bundles R ∂Ω and iR ∂Ω onthe boundaryof Ω. In this case,
× ×
there would have been an obstruction of a non-trivial finite-dimensional subspace.
That this difficulty could be eliminated by considering boundary conditions of the
form (1.1) is a key insight provided by recent advances in the study of the critical
Ising model. There is a discrete variant of the above theorems can be established
explicitly for discrete fermions (see [9], Section 13).
3. Proof of Theorem 2
We begin by proving the injectivity of the operators T and U .
Ω Ω
Let F H2(Ω), and assume that T (F) = 0. If we set f = R (F), this means
Ω Ω
∈ −1
that f = ρ(z)ν 2(z) for some positive scalar function ρ(z) on the boundary. We
in
claim that ρ(z) = 0 identically, and hence F = 0 identically in Ω. Indeed, the
holomorphicity of F implies
(3.1) F2(z)dz =0
˛
∂Ωδ
for all 0 < δ sufficiently small. But the convergence of the restrictions of F(z) to
∂Ω tof,viewedasL2 functionson∂Ωasexplainedintheprevioussection,implies
δ
in turn that
(3.2) f2(z)dz =0.
˛
∂Ω
On the other hand, a key motivation for the boundary condition (1.1) is the fol-
lowing identity,
dz
(3.3) e(f2(z)dz)=ρ2(z) e( )=ρ2(z)ds
ℜ ℜ ν (z)
in
whereds isthe elementofarc-lengthalong∂Ω. This canbe seenby pickingalocal
defining function r(z) for the boundary ∂Ω. Then ν(z)= 1 (∂ r+i∂ r), and
|∇r| x y
dz
(3.4) =ds+i(∂ rdy ∂ rdx),
x y
ν −
which implies (3.3). It follows that ρ(z) = 0 identically, as was to be shown. The
argument for the injectivity of U is exactly the same.
Ω
The proof of Theorem 2 reduces then to the proof of the surjectivity of the op-
eratorsT andU . Thiswillbe donebyinductiononthenumbernofcomponents
Ω Ω
of the boundary ∂Ω of the domain Ω. The precise statements that we shall prove
are the following. Let
(3.5) ∂Ω=∂ Ω ... ∂ Ω,
1 n
∪ ∪
where the ∂ Ω’s are the connected components of ∂Ω.
j
Foreachj 1,...,n ,wedenotebyH2(Ω,∂ Ω)thesubspaceofH2(Ω)defined
j
∈{ }
by
H2(Ω,∂ Ω):= f H2(Ω):T (f) =0 .
j ∈ Ω |∂Ω\∂jΩ
Let TΩ∂jΩ :H2(Ω)→L2in(∂Ωj) b(cid:8)e the projection onto L2in(∂Ωj)(cid:9)of TΩ,
(3.6) f T∂jΩ(f)=(T (f)) .
7→ Ω Ω |∂Ωj
HARDY SPACES AND BOUNDARY CONDITIONS FROM THE ISING MODEL 5
The restriction of T∂jΩ to H2(Ω,∂ Ω) will be denoted by Tj.
Ω j Ω
Then Part (a) of Theorem 2 is an immediate consequence of the following two
lemmas, the first being the case n = 1, and the second the induction step from n
to n+1:
Lemma3(Simply-connectedcase). LetΩbeasimply-connecteddomain. Thenthe
mappings T :H2(Ω) L2 (∂Ω) and U :H2(Ω) L2 (∂Ω) are isomorphisms.
Ω → in Ω → out
Lemma 4 (Induction step). Let n 1, and assume that for any n-connected
≥
smooth domain Ω, the mapping T :H2(Ω) L2 (∂Ω) is an isomorphism. Let Λ
Ω → in
beany(n+1)-connectedsmoothdomain. ThentheoperatorT :H2(Λ) L2 (∂Λ)
Λ → in
is an isomorphism.
We give now the proofs of Lemma 3 and Lemma 4. An essential ingredient is
the following conformal invariance property. Its proof is straightforward, since a
conformal equivalence between two smooth domains extends to a diffeomorphism
of the boundaries:
Lemma 5 (Conformal equivalence). Let Λ and Ξ be two conformally equivalent
smooth domains and let ψ : Ξ Λ be a conformal map. Let ∂ Ξ be a connected
j
→
component of ∂Ξ and ∂ Λ:=ψ(∂ Ξ). Then the following diagram commutes
j j
H2(Λ,∂ Λ) TΛ∂jΛ L2 (∂Λ )
j −→ in j
Ψ Ψ
↓ ↓ |L2in(∂Ξj)
H2(Ξ,∂ Ξ) TΞ∂jΞ L2 (∂ Ξ),
j −→ in j
where the isomorphism Ψ:H2(Λ) H2(Ξ) is defined by
→
(z f(z)) w f(ψ (w)) ψ′ (w) .
7→ 7→ 7→ j j
(cid:16) q (cid:17)
Weobservethatthesquareroot ψ′ (w)iswell-defined(uptoaglobalharmless
j
sign) even when Ω is multiply-conqnected (see [8], Chapter 4). Thus the trivial
spin structure is mapped into the trivial spin structure under global conformal
transformations.
Proof of Lemma 3: If Ω is simply-connected, then there exists a conformal equiv-
alence between Ω and the unit disk D in C, which extends to a diffeomorphism
between the boundary ∂Ω and the unit circle S. By Lemma 5, it suffices to prove
the desired statement when Ω=D and ∂Ω=S.
Let ψ :D Ω be a conformal mapping. By Lemma 5 (defining the operator Ψ
→
as in that lemma), we have the following commuting diagram
H2(Ω,∂Ω) TΩ∂Ω L2 (∂Ω)
−→ in
Ψ Ψ
↓ ↓ |L2in(∂Ω)
H2(D,S) TDS L2 (∂S),
−→ in
S
We should now solve the problem on the unit disk: we should show that TD = TD
is invertible.
HARDY SPACES AND BOUNDARY CONDITIONS FROM THE ISING MODEL 6
So,letusconstructtheinverseSD. Letf beafunctioninL2in(S). Let(ck)k∈Z be
the Fouriercoefficients of f, so that the Fourier seriesof f reads c e , where
k∈Z k k
e (θ) := eikθ. By definition of L2 (S), we have that P [f] = 1(f +e f) = 0.
k in out P2 −1
It is hence easy to see that we have c +c = 0 for all k Z. We define
k −k−1
SD(f):=g, where g H2(D) is defined by ∈
∈
∞
g(z):=2 c zk.
k
k=0
X
Thisclearlydefinesaboundedoperator. LetuscheckthatTD(g)=f. TheFourier
series of TD(g) reads
∞ −1
c e c e .
k k −k−1 k
−
k=0 k=−∞
X X
The nonnegative Fourier coefficients TD(g) are clearly the same as the ones of f,
and using ck+c−k−1 = 0 for all k Z, we get that TD(g)= f. Using exactly the
∈
same arguments, it is easy to check that SD TD is the identity.
◦
Remark.
ItisalsopossibletoconstructSD bywritingitexplicitlyasaconvolutionkernel.
ThisisinspiritclosertoIsingmodeltechniques: theconvolutionkernelcorresponds
then to a fermionic correlator.
We turn next to the proof of Lemma 4. We need two simple observations. The
firstisasuperpositionprinciple,whichallowstoreducetheinversionoftheoperator
T to the inversion of operators T∂jΩ associated to the components ∂ Ω,...,∂ Ω
Ω Ω 1 n
of ∂Ω.
Lemma 6 (Superposition). Let Ξ be an (n+1)-connected domain with ∂Ξ =
∂ Ξ ... ∂ Ξ. Suppose that for each j 1,...,n+1 , the restriction Tj :
1 ∪ ∪ n+1 ∈ { } Ξ
H2(Ξ,∂ Ξ) L2 (∂ Ξ) of the operator T∂jΞ (originally defined on H2(Ξ)) is an
j → in j Ξ
isomorphism. For each j 1,...,n+1 , denote by S∂Ξj : L2 (∂ Ξ) H2(Ξ)
∈ { } Ξ in j →
−1
the inverse of T∂j, injected into H2(Ξ) (the range of T∂j is contained in
Ξ Ξ
H2(Ξ,∂ Ξ)). Then we have (cid:16) (cid:17)
j
T∂jΞ S∂jΞ = Id j
Ξ ◦ Ξ ∀
T∂jΞ S∂kΞ = 0 j =k,
Ξ ◦ Ξ ∀ 6
and T is invertible, with inverse S := S∂1Ξ ... S∂j+1Ξ in the decomposition
Ξ Ξ Ξ ⊕ ⊕ Ξ
L2(∂Ξ)=L2(∂ Ξ) ... L2(∂ Ξ).
1 n+1
⊕ ⊕
The proof of this lemma is again straightforward.
ThesecondobservationisthefollowingversionoftheRiemannmappingtheorem
for multi-connected domains:
Lemma 7 (Riemannmappingtheorem). Let Ξ bean n+1-connecteddomain, with
n 1. Then for any component ∂ Ξ of the boundary ∂Ξ, there exists a conformal
j
≥
equivalence between Ξ and Ω D, where Ω is an n-connected domain containing the
closure D of the unit disk C,\and ∂ Ξ is mapped onto ∂D.
j
HARDY SPACES AND BOUNDARY CONDITIONS FROM THE ISING MODEL 7
Proof of Lemma 7: Assume first that ∂ Ξ is an inner component of the boundary
j
of Σ. Let D be the connected component of C Σ enclosed by ∂ Ξ. Let I be
1 j p
the inversion map z (z p)−1, for any point p\ C. Choose a point p D
1 1
and a point p C →Ω. If−we apply I , then the d∈omain Σ will be mapped∈to a
0 ∈ \ p1
domain I (Σ) lying within C I (D ). By the Riemann mapping theorem, there
p1 \ p1 1
existsaconformalequivalenceΨbetweenthesimply-connecteddomainC I (D )
and the unit disk D, which extends to a diffeomorphism between I (∂ Σ)\apn1d th1e
p1 1
unit circle S. Then Σ is conformally equivalent to the domain I ΨI (Σ), one of
Ip0 p1
whoseboundarycomponentsisthecircleI ΨI (∂ Σ)=I (S). Translatingand
Ip0 p1 1 Ip0
dilating so that this last circle is the unit circle, we obtain the desired domain Ω.
When∂ Ξis theoutercomponentof∂Ξ,wecanapplyaninversionI withrespect
j p
to a point outside ξ to transformξ into another domainwith ∂ Ξ transformedinto
j
aninnercomponentofthe boundary. This reducesthe problemtothe casealready
treated, and the proof of Lemma 7 is complete.
ThepointofthetwoobservationsLemma6andLemma7isthat,inconjunction
with the conformal equivalence property, it suffices to prove that each individual
operator T∂jΩ for each fixed j, 1 j n, is invertible, when ∂ Ω is an inner
Ω ≤ ≤ j
boundary and a unit circle. This is the content of the next lemma, which is the
hardest part of our argument, and which will be proved in the next section:
Lemma 8 (Key Lemma). Let n 2. Let Ω be an n-connected domain containing
the unit disk D. Let Ξ be the (n+≥1)-connected domain defined by Ω D. Assume
that T : H2(Ω) L2 (∂Ω) is an isomorphism. Then TS : H2(Ξ,S)\ L2 (S) is
Ω → in Ξ → in
an isomorphism.
4. Proof of Lemma 8
Recall that Ξ = Ω D. Our goal is to construct an inverse to the operator TS,
using the operator (T\)−1 and function theory on S. Ξ
Ω
4.1. Function theory on S. It is convenient to identify L2(S) with ℓ2(Z) by the
Fourier transform, :L2(S) ℓ2(Z), defined by
F →
1 2π
f L2(S) c (f):= e−ikxf(x)dx
∈ 7→ k 2π ˆ
(cid:18) 0 (cid:19)
(cid:0) (cid:1)
We denote by −1 the inverse of .
Within ℓ2(ZF), let us introduce tFhe following real-linear subspaces,
ℓ2 (Z) := (c ) :c =0 k 0 ,
− k k∈Z k ∀ ≥
ℓ2+(Z) := (cid:8)(ck)k∈Z :ck =0 ∀k <0(cid:9),
ℓ2in(Z) := (cid:8)(ck)k∈Z :ck−c−1−k =(cid:9)0 ∀k ,
ℓ2out(Z) := (cid:8)(ck)k∈Z :ck+c−1−k =0 ∀k(cid:9).
(cid:8) (cid:9)
HARDY SPACES AND BOUNDARY CONDITIONS FROM THE ISING MODEL 8
We denote by the orthogonalprojectiononℓ2 (Z) andby :ℓ2(Z) ℓ2 (Z)
P± ± Pin → in
the orthogonalprojection on ℓ2 . In coordinates:
in
:(c ) 1 c
P+ k k∈Z 7→ {k≥0} k k∈Z
P− :(ck)k∈Z 7→ (cid:0)1{k<0}ck(cid:1)k∈Z
1
:(c ) (cid:0) (c +c (cid:1) )
Pin k k∈Z 7→ 2 k −k−1
1
:(c ) (c c )
Pout k k∈Z 7→ 2 k− −k−1
Clearly, we have
(4.1) P−◦2Pin =Idℓ2−(Z), 2Pin◦P− =Idℓ2in(Z)
as well as the commuting diagram,
L2(S) L2 (S)
−P→ in
in
↓F ↓F
ℓ2(Z) ℓ2 (Z)
−P→in in
where
L2 (S): = f L2(S): m f eiθ eiθ/2 =0 for almost every θ .
in ∈ ℑ
(this choice of notantion is made as i(cid:16)n o(cid:0)ur (cid:1)case th(cid:17)e inner normal on S is aoctually
pointing towards the exterior of the unit disk D, as D is in the complement of our
domain Ξ).
We also need the operator :ℓ2(Z) ℓ2(Z) defined by
J →
(4.2) :(c ) (c ) .
J k k∈Z 7→ −1−k k∈Z
which exchanges ℓ2 (Z) and ℓ2 (Z), i.e., ℓ2 (Z) =ℓ2 (Z). Finally, we set
+ − J ± ∓
+ := + (cid:0) − := (cid:1)− .
F P ◦F F P ◦F
4.2. The operator Φ : ℓ2(Z) H2(Z,S). We come now to the main building
block of the proof, which i−s the→operator Φ : ℓ2(Z) H2(Z,S) constructed from
− →
the operatorS (which exists by induction hypothesis)and Fourier serieson ℓ2(Z)
Ω
as follows.
For each negative integer k Z , set
−
∈
(4.3)ϕℜe(z):=zk S P ζk , ϕℑ(z):=izk S P iζk (z),
k − Ω in k − Ω in
wNhoteerethSaΩt,:wLh2inil(e∂Ωζk) h→as(cid:0)Ha2p((cid:2)Ωol)e(cid:3)ia(cid:1)stth0,ePinv[eζrks]eiosfaTΩwe(lwl-dheicfihn(cid:0)eedxisL(cid:2)t2s bf(cid:3)uy(cid:1)ncatsisounmopntio∂nΩ).,
in
and hence S (P [ζk]) is well-defined as a holomorphic function on Ω. Thus the
Ω in
functions ϕℜ and ϕℑ are the unique holomorphic functions on Ω 0 such that
k k \{ }
ϕℜ(z) zk and ϕℑ(z) izk are holomorphic in Ω
k − k −
P ϕℜ(z) =P ϕℑ(z) =0 on ∂Ω.
in k in k
Lemma 9. Defi(cid:2)ne the(cid:3)real-line(cid:2)ar oper(cid:3)ator Φ:ℓ2(Z) H2(Ξ,S) by
− →
−1
(4.4) Φ : (c ) e(c )ϕℜ+ m(c )ϕℑ .
k k∈Z 7→ ℜ k k ℑ k k!
k=−∞
X
HARDY SPACES AND BOUNDARY CONDITIONS FROM THE ISING MODEL 9
(a)ThentheoperatorΦiswell-definedandboundedasaboundedoperatorfrom
ℓ2(Z) to H2(Ξ,S).
−
(b) The operatorΦ is invertible, and Φ−1 :H2(Ξ,S) ℓ2(S) is equalto RS .
→ − F ∂Ξ
Proof of Lemma 9. To prove Part (a), we have to show that the series defining Φ
converges and is bounded in H2(Ξ,S) for (c ) ℓ2(Z). Since the boundary ∂Ω
k k∈Z ∈
of Ω lies entirely within the region ζ >ρ for some fixed ρ>1, the functions ζk
{| | }
decay exponentially fast for k negative,
(4.5) ζk ρ−k.
C0(∂Ω)
k k ≤
BytheassumptionofLemma8,theoperatorS isboundedfromL2(∂Ω)toH2(Ω).
Ω
Thus we have
ϕℜ zk Cρ−k
k k − k<0kH2(Ω) ≤
k (cid:0)ϕℑk −izk(cid:1)k<0kH2(Ω) ≤ Cρ−k
for a constant C independ(cid:0)ent of k.(cid:1)It follows that the series
−1
(4.6) e(c ) ϕℜ zk + m(c ) ϕℑ izk
ℜ k k − ℑ k k −
k=−∞
X (cid:0) (cid:1) (cid:0) (cid:1)
converges in H2(Ω) and defines a function in H2(Ω) H2(Ξ). Furthermore, by
⊂
the Cauchy-Schwarzinequality, its H2(Ω) norm is bounded by
−1 −1 −1 −1
(4.7)C ck ρ−k C( ck 2)12( ρ−2k)12 C′( ck 2)12.
| | ≤ | | ≤ | |
k=−∞ k=−∞ k=−∞ k=−∞
X X X X
Next, we consider the series e(c )zk +i m(c )zk. It converges uniformly on
k k
ℜ ℑ
compactsubsets anddefines a holomorphicfunction onΞ. As wehaveseen,onthe
components of ∂Ξ which are also in ∂Ω, it converges exponentially fast. On the
componentSof∂Ξ,itisanL2function,sincethefunctionszk formanorthonormal
system in L2(S),
−1 −1
e(c )zk+i m(c )zk = c 2 < .
k k k
(cid:13) ℜ ℑ (cid:13) | | ∞
(cid:13)k=X−∞ (cid:13)L2(S) k=X−∞
(cid:13) (cid:13)
(cid:13) (cid:13) 1
Itfollows tha(cid:13)t its H2(Ξ) normis also bound(cid:13)edby C −1 c 2 2. Altogether,
k=−∞| k|
the expression (cid:16) (cid:17)
P
−1
Φ (c ) = e(c ) ϕℜ zk + m(c ) ϕℑ izk + e(c )zk+i m(c )zk,
k k∈Z ℜ k k − ℑ k k − ℜ k ℑ k
k=−∞
(cid:0) (cid:1) X (cid:0) (cid:1) (cid:0) (cid:1)
1
defines a function in H2(Ξ), whose H2(Ξ) norm is bounded byC −1 c 2 2.
k=−∞| k|
This shows that Φ is a well-defined and bounded operator, provi(cid:16)ng (a). (cid:17)
P
Next, we prove (b). Checking that F− ◦ Φ = Idℓ2−(Z) is easy. To get that
Φ − = IdH2(Ω,S), it just suffices to check that −is injective. Suppose that
◦ F F
f H2(Ξ,S)issuchthat (f)iszero. SinceitdoesnothaveanynegativeFourier
−
∈ F
coefficients, f can be extended to a holomorphic function on Ω. But T (f) = 0,
Ω
and hence by the injectivity of the operator T established earlier in Section 3, it
Ω
follows that f =0. (cid:3)
HARDY SPACES AND BOUNDARY CONDITIONS FROM THE ISING MODEL 10
4.3. The operators , : ℓ2(Z) ℓ2(Z). Let : ℓ2(Z) ℓ2(Z) be the
O Q − → − O − → −
bounded operator defined by the composition
:ℓ2 (Z) H2(Ξ,S) L2 (S) ℓ2 (Z),
O − −Φ→ −T→S in −F→− −
Ξ
Then we have the following formula:
Lemma 10. (a) The operator TS is invertible with bounded inverse if and only if
Ξ
the operator is invertible with bounded inverse.
O
(b) The operator can be expressed as
O
(4.8) O=Idℓ2−(Z)+Q
where is the operator on ℓ2(Z) defined as the composition
Q −
:ℓ2 (Z) H2(Ξ,S) ℓ2 (Z) ℓ2 (Z),
Q − −Φ→ −F→+ + −J→ −
where is the operator of exchanging Fourier coefficients of positive and negative
J
indices defined in Section 4.1.
Proof of Lemma 10. Part (a) follows immediately from the fact that the other op-
S
erators besides T in the composition defining the operator are invertible with
Ξ O
bounded inverses.
To prove Part (b), let (c ) ℓ2 (Z). Set f :=Φ (c ) H2(Ω,S), i.e.,
k k∈Z ∈ − k k∈Z ∈
−1 (cid:0) (cid:1)
f(z)= e(c )ϕℜ(z)+ m(c )ϕℑ(z).
ℜ k k ℑ k k
k=−∞
X
Set g :=TS(f) L2 (S). Thus
Ξ ∈ in
1
g eiθ = f eiθ +e−iθf(eiθ)
2
We compute the Fourier(cid:0) co(cid:1)efficients(cid:16)of (cid:0)g wi(cid:1)th negative in(cid:17)dices. Set e (z) := zk
k
(defined on S) and f,g := 2πf(eiθ)g eiθ dθ. We have
h i 0
´
e ,g = e ,f + e ,e f (cid:0) (cid:1)
k k k −1
h i h i
= ek,f +(cid:10)ek+1,f (cid:11)
h i
= ek,f +(cid:10)e−k−1,f(cid:11)
h i h i
−1
= e , e(c )ϕℜ+ m(c )ϕℑ + e ,f
* k ℜ j j ℑ k j+ h −k−1 i
j=−∞
X
−1
= e(c ) e ,ϕℜ + m(c ) e ,ϕℑ + e ,f ,
ℜ j k j ℑ k k k h −k−1 i
j=−∞
X (cid:0) (cid:10) (cid:11) (cid:10) (cid:11)(cid:1)
= c + e ,f
k −k−1
h i
where in the before last equation, we use that ϕℜ zj and ϕℑ izj are regular in
j − j −
the unit disk (and hence have no negative Fourier coefficients). From there we get
TS Φ(c )=c + c , which is the desired result. (cid:3)
F ◦ Ξ◦ k k Q k
Wearrivenowatthekeypropertyoftheoperator ,whichisperhapssurprising
Q
in itself: