CIMPA School 2018
"Elliptic problems and applications in geometry"
Singular Spectral Perturbations
Colette Anné, Laboratoire de Mathématiques Jean Leray, Nantes February 26 - March 07 – Beyrouth (March 12, 2018)
Contents
1 Introduction 1
2 Spectral theory of bounded self-adjoint operators 2 3 Self-Adjoint unbounded operators and quadratic forms 8 4 Sobolev embedding theorem and Rellich theorem 14
5 Excision 19
6 Partial collapsing 20
7 Adding handles 22
8 Sierpinski gasket 24
1 Introduction
I will concider spectral perturbations of the operator Laplacian such as mak- ing small holes or adding thin handles. These perturbations are not smooth and can be tackled with the general theory of self-adjoint operators and pos- itive quadratic forms (see Fig. 1 below for an example).
General and good references are
• Reed M. & Simon B. Functional Analysis I, Academic Press 1980.
• Kato T. Perturbation Theory for Linear Operators, Springer 1976.
• Simon B. A comprehensive Course in Analysis IV, AMS 2015.
Figure 1: Adding handles (the vertical dotlines realize identifications).
2 Spectral theory of bounded self-adjoint operators
2.1 Requirements
Hilbert spaces, bounded (or continuous) operators, open map theorem.
2.1.1 Open Map Theorem:
Let X and Y be two Banach spaces andT :X→Y a bounded linear operator.
If T is surjective then T is open.
2.1.2 Corollary: Inverse Theorem:
Let X and Y be two Banach spaces andT :X→Y a bounded linear operator.
If T is bijective then T−1 is continuous.
2.1.3 Corollary: Closed Graph Theorem:
Let X andY be two Banach spaces and T :X→Y a linear operator. Then T is continuous (or bounded) if and only if
Gr(T) ={(x, T x), x∈X} ⊂X×Y is closed.
2.1.4 Banach-Steinhaus Theorem:
Let X be a Banach space, Y a normed space and F a family of bounded operators from X toY. If for anyx∈X the set {A(x), A∈ F } is bounded (inY) then F is bounded (in norm of operator).
2.2 Adjoint of an operator
His a separable Hilbert space onCandAan operator defined on it. By the Riesz Lemma, for allx∈ Hthere is a unique vector, denotedA∗x, such that
∀y∈ H< Ay, x >=< y, A∗x >
A∗ is the adjoint of A.
2.2.1 Properties
• A∗ is a bounded operator
• kA∗k=kAk
• if B is an other operator, (AB)∗ =B∗A∗
• if Ais invertible, then A∗ is invertible and (A−1)∗= (A∗)−1.
• kAA∗k=kAk2. 2.2.2 Exercise
• calculate the adjoint of the operator An:l2(N)→l2(N) defined by
An(x1, x2, . . .) = (
n
z }| {
0,0, . . ., x1, x2, . . .)
• Do An andA∗n have strong limit ?
• conclusion ?
2.2.3 Self-Adjointness
We say thatA is self-adjoint if A=A∗ or
∀x, y∈ H< Ax, y >=< x, Ay >
Example. Let P be a projector : P2 = P then P is self-adjoint if and only ifP is an orthogonal projector : KerP andImP are orthogonal spaces.
2.2.4 Theorem of Hellinger-Toeplitz:
A self-adjoint operator defined on all the Hilbert space is bounded.
Proof: it is sufficient to prove that its graph is closed.
2.3 Spectral theory
LetH=Rn. We know that any symmetric matrix can be diagonized in an orthonormal basis. In the infinite dimensional case it is not always the case.
2.3.1 Example: Combinatorial Laplacian LetH=l2(Z) and concider the following operator
f ∈l2(Z), A(f)(n) = 2f(n)−f(n+ 1)−f(n−1)
We can consider this operator through the Fourier transform, if fˆ(θ) = P
n∈Zf(n)einθ then
Afc(θ) = (2−e−iθ−eiθ) ˆf(θ) = 2(1−cos(θ)) ˆf(θ)
For anyλ∈Rthe equation2(1−cos(θ))F =λF has no non zero solution for F ∈L2([0,2π])A has no eigenvalue.
2.3.2 Spectrum
LetA be a bounded operator defined onH, theresolvent setofA is ρ(A) ={λ∈C,(A−λId) invertible}
The spectrum of A is σ(A) = C \ ρ(A). In the preceding example
$σ(A)=[0,4].$
λ∈σ(A) is aneigenvalueofA if (A−λId) is not injective.
Forλ∈ρ(A) the operator (A−λId)−1 is theresolvent ofA at λ.
1. Theorem: The resolvent set of a bounded operator is an open subset of Cand the resolvent is an analytic function on it.
2. Corollary: The spectrum of a bounded operator is never empty.
3. Exercise: the resolvent set of a bounded operator is never empty.
4. Definition: Thespectral radiusof Ais r(A) = sup
λ∈σ(A)
|λ|
5. Proposition:
• r(A) = limn→∞kAnk1/n
• If A is self-adjoint, then the spectrum of A is real.
indication: show that(Im(A))⊥= KerA∗ and that if A is self-adjoint k(A−λId)(x)k ≥ |=(λ)| kxk.
6. Example: LetH=l2(N) and look at the operator A
(xn)n∈N
= ( 1
n+ 1xn)n∈N
A is injective but0 ∈σ(A). Indeed, the image of A is dense but A is not surjective.
2.3.3 Compact operator
1. Definition: LetX be a Banach space,Y a normed space andA:X→ Y a bounded operator. The operator is compact if for any bounded subset B of X,A(B) is relatively compact in Y.
2. Example: LetKbe a continuous function on the product space[0,1]× [0,1]and define the operatorA acting inC([0,1],R)by:
A(f)(x) = Z
K(x, y)f(y)dy Ais compact by the theorem of Ascoli.
3. Theorem: If the operatorA:X→ Y is compact and if (xn)n∈N is a sequence, weakly convergent tox, then (A(xn))n∈N converges in norm toA(x).
4. Properties:
• a limit in norm of a sequence of compact operators is compact (diagonal principle)
• if Aor B are compact, then A◦B is compact
• if X=Y Hilbert, the operatorA is compact if and only ifA∗ is compact (admitted: use the polar decomposition)
2.3.4 Compact operators in a Hilbert space LetHbe a separable Hilbert space.
1. Theorem: If A is a compact operator in H, then A is the limit (in norm of operators) of operators with finite rank.
Proof: Let ϕn, n ∈N be an orthonormal basis of H and consider the operatorsAn(f) =P
j≤n< f, φj > A(φj). As A is compact, we have limn→∞kA−Ank= 0.
2. Fredholm alternative: If A is a compact operator acting in H and (Id−A) is injective then (Id−A) is invertible (and continuous).
Proof: DefineΦ = (Id−A) and show that ifΦis injective there exists a constantC >0 such that for allx∈ H,kxk ≤CkΦ(x)k. It means, in particular, that Im(Φ) is closed, as well as Im(Φk) for all integer k show now that the decreasing sequence of spaces Im(Φk), k ∈ N is stationnary, and conclude. (a proof due to Terence Tao).
indication: IfIm(Φk), k∈Nis not stationnary there exists a sequence yk∈Im(Φk) withkykk= 1 and d(yk,Im(Φk+1))≥1/2.
3. Theorem [Riesz-Schauder] Let A be a compact operator acting in H, then σ(A) is a discrete set, with only 0 as possible accumulation point, the non zero elements of σ(A) are all eigenvalues with finite multiplicity.
Proof: Indeed for any >0 the set {λ∈σ(A),|λ| ≥} is finite: if it is infinite take λn, n ∈N all distincts in this space. By the Fredholm alternative there existsxn,kxnk= 1 and A(xn) =λnxn.
(a) Thexn, n∈Nare linearly independant: if
PN
j=0αjxj = 0 withαN 6= 0, then
N−1
Y
j=0
(A−λj)(
N
X
j=0
αjxj) =αN
N−1
Y
j=0
(λN −λj)xN = 0.
(a) As a consequence, ifEnis the vector space generated by
xj,0 ≤ j ≤ n, it defines a strctly increasing sequence. Thus, there existyn ∈En withkynk= 1 and d(yn, En−1)≥1/2. We remark that En is stable byAand (A−λn)(En)⊂En−1 and if m > n
kA(ym)−A(yn)k=kλmym+ (A−λm)ym−A(yn)k ≥ 2 because
(A−λm)ym −A(yn)
/λm is in Em−1. But A is compact, this inequality cannot hold.
4. Question: Is the operator of the example 2.3.1 compact?
5. Theorem [Hilbert-Schmidt] Let A be a self-adjoint compact oper- ator acting in H of infinite dimension, then there is an orthonormal basis (φn)n∈N of eigenvectors:
A(φn) =λnφn lim
n→∞λn= 0
Moreover, the non zero eigenvalues have finite multiplicity.
Proof: as A is self-adjoint its spectrum is real, as it is compact it is a set of eigenvalues with finite multiplicity (why?), plus 0. As H is separable, the set of eigenvalues is at most countable. Let H0 =
⊕λ∈σ(A)Ker(A−λ)
(a) This direct sum is orthonormal (b) A(H0)⊂ H0
(c) A(H⊥0)⊂ H⊥0 and A|H⊥
0 is self-adjoint
and compact, its spectrum consists on eigenvalues ofA, so H⊥0 ={0}. 6. Exercise: In l2(N) let consider the Hilbert spaces
hs ={(xn)∈l2; ((1 +n2)s/2 ∈l2} for which scalar product ? (takes≥0)
Show that Id:hs → hs0 is compact if s > s0 (as a limit of finite rank operators).
2.3.5 Fondamental theorem: Spectral representation
(admitted, see Reed Simon I)Let Abe a bounded self-adjoint operator in the Hilbert spaceH there exists a unique continuous morphism of ∗-algebra
Φ :ˆ B(R)→ L(H)
if B(R) denote the set of bounded borelian real functions and L(H) is the space of bounded operator onH, such that
• if the support of f is included in the resolvent set ρ(A)thenΦ(f) = 0ˆ
• Φ(xˆ 7→x) =A
• Aψ=λψ⇒Φ(fˆ )ψ=f(λ)ψ
• If B commute with A then it commutes with all the Φ(fˆ )
• If f ≥0 then the operator Φ(fˆ ) is positive We denote Φ(fˆ ) =f(A).
Iff =χI the characteristic function of the intervalI ⊂R thenχI(A) is an orthogonal projector, theorthogonal projector on the total eigenspace corresponding toσ(A)∩I and denotedEI(A).
2.3.6 Example: Let us turn back to the example H=L2([0,2π]) A(ϕ) = 2(1−cos(θ))ϕ. We fix F(θ) = 2(1−cos(θ)) then for any interval I, χI(A)(ϕ) = (χI ◦F).ϕ = χF−1I.ϕ. As a consequence, C0(F−1I) is a subspace ofEI(A) which is always of infinite dimension.
Ahas only an essential spectrum.
3 Self-Adjoint unbounded operators and quadratic forms
Look at the operator Laplacian: for f ∈C2(Rn)
∆(f) =−
n
X
1
∂2
∂x2j(f)
C2 is not an Hilbert space, we can look at the Sobolev space H2(Rn) ={f ∈L2(Rn),(1 +|ξ|2) ˆf ∈L2(Rn)}
where∆is well defined, but ∆(f) is not inH2,
That is why we introduce the concept of unbounded operators: H is a separable Hilbert space andS an operator defined on its domain D(S) which isdense inH.
example: H=L2(Rn),∆with domain D(∆) =H2(Rn). 3.1 Definitions
• Let (S, D(S))be an operator on the Hilbert space H, thegraphof S is the subset ofH × H
Γ(S) ={(f, S(f)), f ∈D(S)}
• the operator (S, D(S))is closedif Γ(S) is closed inH × H
• Let (S, D(S)) and (T, D(T)) be two operators on the same Hilbert space,T is anextensionof S (denoted S⊂T) ifD(S)⊂D(T) and
∀f ∈D(S), T(f) =S(f)
• Proposition: If the operator (S, D(S)) admits a closed extension, we say that S is closable, (S, D(S)) is closable if and only if Γ(S) is a graph, the graph of S¯, the closure of S and the smallest closed extension ofS.
3.2 Examples
• H=L2([0,1]). Let ϕ0∈ H,ϕ0 6= 0, and
define D(T) = C0(]0,1[) and T(f) = f(1/2)ϕ0. This operator is not clos- able: there exists a sequence fn ∈ C0(]0,1[) such that fn(1/2) = 1 and limn→∞R1
0 |fn|2 = 0.
• H=L2(R), functions with complex values,
D(T) =C0∞(R)andT(f) =i∂x∂ f. This operator is closable and the domain of its closure isH1(R).
This operator is also symmetric:
∀f1, f2 ∈D(T), hT(f1), f2i=hf1, T(f2)i
3.3 Self-Adjointness
Let (S, D(S)) be an operator on the Hilbert space H, the adjoint of S, denoted(S∗, D(S∗))is defined by
D(S∗) = n
f ∈ H;∃h∈ H such that∀g∈D(S),hS(g), fi=hg, hio Iff ∈D(S∗) we defineS∗(f) =h.
3.3.1 Proposition:
• The operator S is closable if and only ifD(S∗)is dense in H.
• If S is closable, then S= (S∗)∗.
• If S is symmetric,S∗ is a closed extension ofS, and so,S is closable.
3.3.2 Examples
1. H = L2(R), D(T) = L1(R)∩L2(R), T(f) = (R
f)ψ0 for some ψ0 ∈ L2(R), ψ0 6= 0. Then D(T∗) =ψ⊥0 is not dense! So T is not closable.
2. H = L2(]0,1[), D(T) = C0∞(]0,1[), T(f) = i∂x∂ f T is symmetric, D(T∗) =H1([0,1]) and D(T) =H01([0,1]).
3.3.3 Definition
The operator(S, D(S))isself-adjointifSis symmetric andD(S∗) =D(S). (S, D(S))isessentially self-adjointif S¯ is self-adjoint.
3.3.4 Fondamental property
Let (S, D(S)) be a symmetric operator on the Hilbert space H, then the following properties are equivalent
• (S, D(S))is self-adjoint
• (S, D(S))is closed andKer(S∗±i) ={0}
• Im(S±i) =H.
3.3.5 Exercise:
The operator of the previous example (3.3.2-2) has infinitly many self-adjoint extensions:
Let D(Tα) ={f ∈ H1([0,1]), f(1) = αf(0)} for which α is (Tα, D(Tα)) withTα(f) =i∂x∂f self-adjoint ?
3.4 Spectral theory of self-adjoint operators
Let(S, D(S))be a closed operator on the Hilbert space H. ThenD(S) is a Banach space for the norm of operator
∀f ∈D(S) kfk2S=kfk2+kS(f)k2 indeed a Hilbert space. Theresolvent setof S is
ρ(S) ={λ∈C,(S−λId) :D(S)→ H bijective} and thespectrum set is σ(S) =C\ρ(S).
3.4.1 Exercise
If the self-adjoint operator (S, D(S)) is positive (any f ∈ D(S) satisfies hS(f), fi ≥0) then −1∈ρ(S) (or any strictly negative number).
3.4.2 Remark
As D(S) is a Banach space if S is closed and for any λ ∈ ρ(S) the map (S−λId) :D(S)→ His continuous for this norm, the resolvent(S−λId)−1 : H →D(S) is continuous, and it is also continuous if we regard it to arrive inH.
Forλ∈ρ(S), we define Rλ(S) = (S−λId)−1. 3.4.3 Proposition
ρ(S) is open, (Rλ(S))λ∈ρ(S) is a commutative family of bounded operators satisfying
Rλ(S)−Rµ(S) = (λ−µ)Rµ(S)Rλ(S).
3.4.4 Proposition
If(S, D(S))is self-adjoint, then σ(S)⊂R Letλ∈C such that=λ6= 0then
kRλ(S)kH→H≤ 1
|=λ|
3.4.5 Fondamental theorem: Spectral representation
Let (S, D(S))be a self-adjoint operator in the Hilbert space Hthere exists a unique continuous morphism of ∗-algebra
Φ :ˆ B(R)→ L(H)
(B(R) /the set of bounded borelian real functions,/ L(H) is the space of bounded operator on H),/ /such that
• Sψ=λψ⇒Φ(fˆ )ψ=f(λ)ψ
• If B commute with S then it commutes with all the Φ(fˆ )
• If f ≥0 then the operator Φ(fˆ ) is positive We denote Φ(fˆ ) =f(S).
It is an application of the theorem on bounded operators to Rλ(S), but Rλ(S)is not self-adjoint if=λ6= 0it is justnormal: the operator commutes with its adjoint, so we have first to extend the previous theorem to normal operators.
3.5 Quadratic forms
Any self-adjoint operator (S, D(S)) defines a quadratic form in H with do- main (containing) D(S) : q(f, f) =hS(f), fi. But for the Laplacian acting onRn for instance we have, forf ∈C0∞(Rn)
h∆(f), fi= Z
Rn
|df|2
which can be defined in a bigger space than H2, namelyH1(Rn). So there are correspondance between self-adjoint operators and quadratic forms if we take care of the domains.
3.5.1 Definitions
1. A quadratic form (q, D(q)) on its domain D(q), a dense subspace in the Hilbert space H, is q : D(q)× D(q) → C sesquilinear and symmetric:
q(f1, f2) =q(f2, f1)
The quadratic form issemibounded if there exists M ≥0 such that
∀f ∈D(q), q(f, f)≥ −Mkfk2 and if we can take M = 0we say that q ispositive.
2. Remark. A sesquilinear semibounded form is symmetric (exercise) 3. The semibounded (by−M) quadratic form isclosedif the spaceD(q)
is complete as a normed space with normk.kq
f ∈D(q),kfk2q =q(f, f) + (1 +M)kfk2
Thuse D(q) is an Hilbert space for the scalar product hf1, f2iq = q(f1, f2) + (1 +M)hf1, f2i.
4. Example(E, H1(Rn))withE(f) =R
|df|2is closed. It is the quadratic form of energy.
3.5.2 Fondamental Theorem
If (q, D(q)) is a closed semibounded quadratic form, then q is the quadratic form of a unique self-adjoint operator.
There exists (S, D(S)) unique such that
D(S) ={f ∈D(q), ∃ϕ∈ H,∀ψ∈D(q)q(f, ψ) =hϕ, ψi}
and S(f) =ϕ.
Indeed, let us suppose for simplicity that q is positive (ie. M = 0), for anyψ∈ H, the linear formlψ(f) =hf, ψiis continuous on the Hilbert space (D(q),h., .iq). Let us denoteA(ψ) the element of D(q) satisfying
∀f ∈D(q), hf, ψi=hf, A(ψ)iq
The operator A is bounded: kA(ψ)kq ≤ kψk, injective, andIm(A) is dense inD(q). The solution isD(S) = Im(A) and iff =A(ψ)thenS(f) =ψ−f. 3.5.3 Examples
The Laplacian is the operator associated to the energy form.
1. Let H = L2(]0,1[), and q(f1, f2) = R
df1df2 on D(q) = H01([0,1]) The self-adjoint operator associated is the Laplacian with Dirichlet boundary conditions.
2. Let H = L2(]0,1[), and q(f1, f2) = R
df1df2 on D(q) = H1([0,1]) The self-adjoint operator associated is the Laplacian with Neumann boundary conditions.
3. These examples can be generalized to all the Riemannian manifolds with boundary.
3.5.4 Friedrich’s extension
Let(S, D(S))be a positive symmetric operator, it defines a quadratic form D(q) =D(S) and q(f, f) =hS(f), fi.
1. This quadratic form is closable: AsS is symmetric and positive, hf1, f2i1=hf1, f2i+q(f1, f2)
defines a scalar product, let denote k.k1 the norm associated and H1 the completion of D(S) for this norm. All we have to do is to show thatH1is a subspace ofH. Letι:D(S)→ Hbe the natural injection, andˆιits prolongation to H1.
Lemma: ˆιis injective.
Indeed, let f = lim(H1)fn and fn ∈ D(S). If ˆι(f) = 0 then limn→∞kfnk= 0. But
kfk21 = lim
m→∞( lim
n→∞hfn, fmi1) = lim
m→∞( lim
n→∞hfn,(1 +S)fmi) = 0 2. The self-adjoint operator associated to this closure isthe Friedrich’s
extensionof(S, D(S)).
3. Example: The Friedrich’s extension of(∆, C0∞(Rn))is (∆, H2(Rn)).
4 Sobolev embedding theorem and Rellich theorem
These theorems permit the Laplacian to have acompact resolvent in certain cases and then to have a discret spectrum. References:
4.1 References for this part
• Roe J. Elliptic operators, topology and asymptotic methods Pitman 1998.
• Grigor’yan A.Heat Kernel and Analysis on Manifolds AMS 2009.
• Hebey E. Nonlinear Analysis on Manifolds: Sobolev Spaces and In- equalities AMS 1999.
• Hörmander L. The Analysis of Linear Partial Differential Operators Springer 1985.
Requirement: Manifolds, Riemannian metrics, Fourier transform.
4.2 The case of Tn =Rn/Zn 4.2.1 Plancherel formula
Letf ∈L2(Tn)then theFourier coefficients off are f(ν) =ˆ
Z
Tn
f(x) exp−i2πν.xdx, ν∈Zn
The functions eν(x) = expi2πν.x form an orthonormal basis of the Hilbert spaceL2(Tn) and we can write inL2(Tn): f =P
ν∈Znfˆ(ν)eν. 4.2.2 Proposition
If f ∈ C∞(Tn) then its Fourier coefficients are rapidly decreasing: for any positive integerN there exists C(N)>0 such that for anyν ∈Zn,|fˆ(ν)| ≤ CN(1 +|ν|2)−N.
There are theorems in the other direction.
4.2.3 Definition k’th Sobolev space Wk
Letkbe a positive integer,Wkis the closure ofC∞(Tn)for the norm coming from the scalar product
hf, fik = X
ν∈Zn
|fˆ(ν)|2(1 +|ν|2)k
PropositionCk(Tn) is continously included inWk. (exercise)
4.2.4 The Sobolev embedding theorem
For any integer p > n/2 the (k+p)’th Sobolev space is continously included in Ck(Tn).
Proof: there existsck>0such that for any f ∈C∞(Tn), kfkCk ≤ck X
ν∈Zn
|fˆ(ν)|(1 +|ν|2)k/2
but by assumption on p the serie of general term(1 +|ν|2)−p is convergent, the Cauchy-Schwarz inequality gives then
kfkCk ≤ckkfkWk+p
s X
ν∈Zn
(1 +|ν|2)−p
4.2.5 The Rellich’s theorem
If k1 < k2 the inclusion Wk2 ,→Wk1 is compact.
Proof: see that this inclusion is the limit of operators of finit rank. (compare with the exercise 2.3.4.6)
4.3 On a compact Riemannian manifold M (general case) The manifold M of dimension n can be covered by a finite atlas where it works as before, so the same theorem holds because we have uniform comparison estimates.
Be careful that this is in general not true whenM is not compact.
More precisely, let(Ui, φi)1≤i≤N be a finite atlas of M andψ1 a partition of 1 subordinated to this cover. We define the Sobolevk-norm off ∈C∞(M) by
kfkk= X
1≤i≤N
k(ψi.f)◦φ−1i kk.
Indeed we can suppose that the open set Vi = φi(Ui) satisfies Vi ⊂ [0,1]n, thuse the function(ψi.f)◦φ−1i a priori defined onVican be prolongated by 0 and define a function onTn. The k’th Sobolev space of M,Wk(M), is then the completion ofC∞(M)for this norm. Remark that the norm depends on the atlas but given two atlas give two equivalent norms. The same theorems hold forM as for Tn:
4.3.1 Ck(M) is continously included in Wk(M).
4.3.2 Sobolev embedding theorem
For any integer p > n/2, W(k+p)(M) is continously included in Ck(M).
4.3.3 Rellich’s theorem
If k1 < k2 the inclusion Wk2(M),→Wk1(M) is compact.
4.4 The spectrum of the Laplacian of a compact Riemannian manifold
Let (M, g) be a compact Riemannian manifold. In local coordinates (x1, . . . , xn)the metric defines a symmetric, positive defined matrixgij(x) = gx(∂xi, ∂xj). It defines also a scalar product on the cotangent space by [gij(x)], the inverse matrix of [gij(x)], if ξ ∈Tx∗(M) we denote it (ξ, ξ)g(x).
It defines also a volume form dvolg = (det[gij)])1/2dx1. . . dxn. Let ρ(x) = (det[gij)])1/2(x).
We can now define the energy form q(f, f) =
Z
M
|df|2gdvolg
The local expression of the Laplacian is then
∆(f)(x) = −1 ρ(x)
X
1≤i,j≤n
∂xj(ρgij∂xi(f))(x).
The domain ofq is H1(M, g) ={f ∈L2(M, g),|df|g ∈L2(M, g)}. We have clearlyH1(M, g) =W1(M)and q, D(q) =H1(M, g) is closed.
4.4.1 Proposition:
The self-adjoint operator associated to the closed quadratic form(q, H1(M, g)) is∆ with domainW2(M).
Proof (sketch): By the theorem about closed quadratic form the operator (∆, D(∆))is self-adjoint on the domain
D(∆) ={f ∈H1(M, g),∆(f)∈L2(M, g)}
The inclusion W2(M)⊂D(∆) is clear but the inverse inclusion is far from trivial and not systematic. It uses theellipticity of the Laplacian onM: the principal symbol of∆is invertible.
The point is to obtain this formula: there exists a constant c >0 such that
∀f ∈C∞(M),kfkW2(M)≤c(kfk+k∆(f)k)
Remark. This control is clear on Tn: we have to control the L2-norm of any derivative of f of order less than 2 by the right hand term. It is a consequence (via the Fourier transform) of the fact that for any homogeneous polynomialP onRnof degree less than 2 there exists a constant c >0such thatP(x)≤c(x21+· · ·+x2n).
Few methodscan be used, one is to construct a parametrix (going back to Hadamard) in the framework of pseudodifferential operators, see Hörmander.
One is to make local elliptic estimates, see Grigor’yan Thm. 6.9. One is to do Riemannian geometry, see Hebey:
Bochner-Weitzenböck formula. Let∇denote the Levi-Civita connection of (M, g), we define the rough Laplacian ∆¯ acting on 1-forms (and more
generally on p-forms) by the formula ∆ =¯ ∇∗∇. On the other hand the Laplacian on p-forms is defined by the formula∆ = (d+δ)2ifδis the adjoint ofdcalled the co-differential. Letα a 1-form onM these two operators are related by the Weitzenböck formula
∆(α) = ¯∆(α) +Ricci(α) whereRicciis a curvature term, without derivative.
applying it to the 1-formdf and making the scalar product withdf gives
∀f ∈C∞(M) Z
|∇(df)|2 = Z
M
|∆(f)|2− Z
M
Ric(∇f,∇f)
where the gradiant∇f off is defined by: g(x)(∇f, X) =dfx(X)and Ric is the Ricci curvature ofM.
The left handle term is equivalent to the term concerning the second derivatives in the norm ofW2(M). So we are done if the Ricci curvature is bounded from below.
By this kind of calculus you see how to obtain such results in the non compact case : you need curvature hypothesis!
RemarkThe same control holds also inWk(M)
∀f ∈C∞(M),kfkW
k+2(M)≤c(kfkW
k(M)+k∆(f)kW
k(M)) 4.4.2 Theorem
Let (M, g) be a compact connected Riemannian manifold, then there exists an orthonormal basis (ϕj)j∈N of L2(M) consisting of eigenfunctions of its Laplacian ∆: there exists
0 =λ0 < λ1≤. . . with limj→∞λj = +∞ and ∀j,∆(ϕj) =λjϕj
Indeed, the resolvent (∆ + 1)−1 :L2(M, g) → L2(M, g) is compact self- adjoint.
Appying the previous remark and the Sobolev embedding theorem, we see that the eigenfunctions are smooth.
4.4.3 Minimax formula
With the same notation than in the theorem one has
∀j ∈N, λj = infn sup
f∈E\{0}
R
M|df|2 R
M|f|2 ;E subspaceH1(M, g) dimE= (j+ 1)o
TheRayleigh-Ritz quotient off ∈H1(M, g)\ {0}is R(f) =
R
M|df|2 R
M|f|2. Proof: Forj∈Nlet Ej be the space generated byϕi,0≤i≤j then
λj = sup
f∈Ej\{0}
R
M|df|2 R
M|f|2
On the other hand, ifE ⊂H1(M, g)satisfiesdimE = (j+ 1)then consider πj :E →Ej the orthogonal projector onEj restricted toE.
Ifπj is not a bijection, then there existsf ∈Ej⊥∩Enon zero,R(f)≥λj andsupf∈E\{0}R(f)≥supf∈Ej\{0}R(f).
Ifπj is a bijection, then there existsfj ∈Ej⊥such thatf =fj+ϕj ∈E. But againR(f)≥λj.
(Remark that ϕj, j ∈Nis also orthonormal for<, >q.)
4.4.4 Relatively compact manifolds or domains with boundary The same result holds if the Laplacian on study is selfadjoint (boundary conditions as Dirichlet or Neumann).
5 Excision
5.1 M=M \B.
Let(Mn, g) be a compact Riemannian manifold, x0 ∈M We construct the following perturbation: For >0we denoteB ={x∈M, d(x0, x)< }and defineM=M \B a manifold with boundary.
5.2 Proposition
If n ≥ 2 the Laplacian with Dirichlet or Neumann boundary conditions converges to the Laplacian onM. For which convergence ?
5.2.1 convergence of the eigenvalues and the eigenfunctions.
5.2.2 generalization to M =M\T U B(Y)
for Y ⊂M a submanifold of codimension bigger than 2.
5.2.3 convergence in norm of the resolvent.
We suppose that two closed positive quadratic formsq and eq, living in the Hilbert spacesHresp. He areδ- quasi unitarily equivalent (of orderk≥1):
there exist linear bounded operators J : H → He and J0 :H → He and also J1 : H1 → He1 and J01 :He1 → H1 on the energy form domains which satisfy:
1. for allf ∈ H, u∈He
kJ fk ≤(1 +δ)kf,k |hJ f , ui − hf, J0ui| ≤δkfkkuk (1a) 2. for allf ∈ H1, u∈He1
kf −J0J fk ≤δkfk1, ku−J0J uk ≤δkuk (1b) 3. for allf ∈ H1, u∈He1
kJ1f−J fk ≤δkfk1, kJ01u−J0uk ≤δkuk1 (1c) 4. for allf ∈ Hk, u∈He1
|q(Je 1f, u)−q(f, J01u)| ≤δkfkkkuk1 (1d) Then the following holds true:
k J R−RJe
R(k−2)/2k ≤7δ and (2a)
k(∆ + 1)e 1/2 J1R−R(Je 01)∗
R(k−2)/2k ≤4δ. (2b)
whereR:= (∆ + 1)−1 resp. Re:= (∆ + 1)e −1 denotes the resolvent of∆resp.
∆e. Via functionnal calculus, these kind of estimates give the convergence of the spectrum and of the eigenfunctions. It applies for excision, see
• Anné C. & Post O. Wildly perturbed manifolds: norm resolvent and spectral convergence arXiv:1802.01124
6 Partial collapsing
I present here a result of Takahashi:
• J. Takahashi Collapsing of connected sums and the eigenvalues of the Laplacian, J. Geom. Phys. 40no 3-4(2002), 201–208.
He conciders the following perturbation: Let M1, M2 be two compact con- nected Riemannian manifolds (metrics g1, g2) of dimension n ≥ 2 and two points x1 ∈ M1 and x2 ∈ M2. Let M1() = M1 \ B(x1, ) and M2() =.(M2 \B(x2,1). It means that we concider on M2\B(x2,1) the metric2g2. To make this construction we need that the injectivity radius of M2 is bigger than 1, what is always possible by a scaling. But we ask more:
the ballsB(x1, 0) andB(x2,1)must be euclidean.
Then we can construct M =M1()∪M2() and we ask What happens at the limit?
6.1 the quadratic form
We fix the Hilbert spaces as follows. Let
H=L2(M1)×L2(M2\B(x2,1)) H =L2(M1())×L2(M2\B(x2,1)) using the isometry between L2(M2())and L2(M2\B(x2,1))given byh→ n/2h. We fix normal coordinates around the two points x1, x2. Then the quadratic form of the energy is
D(q) =n
(f, h)∈H1(M1())×H1(M2\B(x2,1)),
f(x1+θ) =−n/2h(x2+θ),∀θ∈Sn−1 o
q((f, h),(f, h)) = Z
M1()
|df|2dvolg1 + 1 2
Z
M2\B(x2,1)
|dh|2dvolg2
6.2 the result
The limit spectrum is the spectrum of the Laplacian ∆1 on (M1, g1).
Indeed, it is a consequence of the Minimax formula. We denote by 0 =λ0()< λ1()≤λ2(). . .
the spectrum of the perturbed manifold and by 0 = λ0 < λ1 ≤λ2. . . the spectrum of∆1.
1. Any function f ∈ H01(M1()) defines (f,0) ∈ D(q), so the Minimax formula tells us that for anyk∈Nλk()is bounded by the kth eigen- value of the Dirichlet problem onM1()and finally, using the previous result
∀k∈N lim sup
→0
λk()≤λk.
2. Now, for k ∈ N fixed, let (f, h) be a normed eigenfunction of the operator defined byqrelative to the eigenvalueλk(). By the previous point we haveR
M2\B(x2,1)|dh|2dvolg2 =O(2)thusehis approximately constant, and because of the boundary condition this constant is 0. More precisely, by the Rellich theorem we have lim→0khk = 0 in L2(M2\B(x2,1)).
On the other handf ∈H1(M1())satisfiesR
M1()|df|2dvolg1 ≤λk(). To conclude, by the Minimax formula, that
λk≤lim inf
→0 λk()
we have to proceed by recurrence on k and to do these estimates for an orthonormal family
(fj, hj)
0≤j≤k of eigenfunctions relative to
λj()
0≤j≤k. We restrict to a subfamily such thatlimm→∞λk(m) = lim inf→0λk()and for this subfamily apply the previous estimates on hjm. Then the family(fjm)0≤j≤kis almost orthonormal and the space Em they generate is a good candidate to estimate the kth eigenvalue of the Neumann problem onM1(). The result of the previous section gives then the conclusion.
6.3 applications
6.3.1 we cannot hear the topology of a manifold.
First proof by Chavel & Feldmann (Spectra of manifolds with small handles Commen. Math. Helv. 56, 83–102 (1981).
6.3.2 minoration of the conformal spectrum
On a compact Riemannian manifold by the one of a Sphere, volume fixed. See Colbois & El Soufi : Extremal Eigenvalues of the Laplacian in a Conformal Class of Metrics: The ‘Conformal Spectrum’ Annals Global An. and Geo.
24, 337–349, 2003.
7 Adding handles
7.1 The construction
Let(Mn, g) be a compact Riemannian manifold, x0, x1 ∈M. We construct the following perturbation: For >0we denoteB(xj) ={x∈M, d(xj, x)<
} and define M0 = M\(B(x0)∪B(x1)) a manifold with boundary, and an (abstract) handle of lenghL: C = [0, L]×Sn−1.
The perturbation M results from the glueing of C on M0, we remark that they have thesame boundary: we fix normal charts aroundx0 and x1
the boundary are asymptotically isometric. We define
L2(M) =L2(M0, g)×L2([0, L]×Sn−1, ds2+2gcan) (3) H1(M) ={(f, h)∈H1(M0)×H1(C)f =h on ∂C} (4) q
(f, h),(f, h)
= Z
M0
|df|2+ Z
[0,L]×Sn−1
(|∂sh|2+ 1
2|∂θh|2) (5) 7.2 Proposition
The spectrum, and the eigenfunctions ofMconverge, as→0, to the union of the spectrum ofM and the Dirichlet spectrum of the interval[0, L].
7.2.1 Proof:
We define the limit spaces H = L2(M, g) × L2([0, L], ds2) and H1 = H1(M, g)×H01([0, L], ds2). We concider the union of the two limit spec- tra repeted as the total multiplicity and denote it
0 =µ0 < µ1 ≤µ2. . . In a same way we denote the spectrum onM
0 =λ0()< λ1()≤λ2(). . .
1. If ϕ ∈ H1(M, g) is an eigenfunction ∆M(ϕ) = λϕ, it corresponds to the element (ϕ,0) ∈ H1 and we transplant it on M as follows: we know that there exists an eigenfunctionϕ∈H02(M0)
∆(ϕ) =λϕ onM0
for the Dirichlet boundary problem. Indeed(ϕ,0)∈H1(M). In the same way if ψ ∈ H01([0, L], ds2) is an eigenfunction of the Dirichlet problem∆(ψ) = λψ on the interval [0, L], we can see it as a function on the handle constant on the fibers (Sn−1) and finally as an element (0, ψ)∈H1(M). The minimax formula then tells us
∀j∈N lim sup
→0
λj()≤µj.
2. Concider now, for j ∈ N, µ = lim inf→0λj(). If f = (ϕ, ψ) ∈ H1(M) is a normed eigenfunction ∆(f) = λj()f, we define fϕ the harmonic prolongation ofϕ on (B(x0)∪B(x1))
We know that the family (fϕ, ψ) is bounded in the fixed space H1(M, g)×H1([0, L]×Sn−1, ds2+gcan)and that
k∂θψk=O()
by the Rellich theorem we can extract a sequence (m) such that ϕgm, m ∈ N converges in L2 and wealky in H1 to an eigenfunction φon M with eigenvalueµ and in the same way ψm, m∈N converge inL2 and wealky inH1to a functionψon[0, L]satisfying∆(ψ) =µψ weakly.
To obtain the Dirichlet boundary condition we apply the following estimate given in polar coordinates: let f ∈H1(B1(0)\B(0))with 0 value on the external boundary∂(B1(0))
n−1 Z
Sn−1
|f(, θ)|2 ≤kfk21
if the dimensionn≥3, for n= 2we have to multiply the write hand side by |log|. Any way the boundary term goes to 0 with and the limit functionψ is inH01([0, L]).
Finally as the eigenfunctions were taken with L2-norm equal to 1 we cannot have in the same timeφand ψ equal to 0.
Doing this for thejfirst eigenvalues, we obtain by the minimax formula that
lim inf
→0 λj()≥µj. This conclude the proof.
7.3 Convergence in norm of the resolvent
The same general result apply here, as for the excision problem.
8 Sierpinski gasket
The Sierpinski gasket, denoted SG, is an example of a self similar fractal set which can be viewed as a limit of more regular set. We introduce the approach of
Figure 2: Sierpinski gasket
• R. Strichartz Differential Equations on Fractals Princeton university Press (2006).
We considerSG as the limit of finite graphs:
8.1 Description
Let Γ0 be the triangle graph: 3 vertices qi,0 ≤ i ≤ 2, and 3 edges and constructΓ1 fromΓ0 by taking 3 new vertices: the middle of the edges and 3 new edges: the edges between them. We introduce the three mappings of R2 defined by the three vertices of Γ0
Fi(x) = 1
2(x−qi) +qi
We remark thatΓ1 =∪iFi(Γ0) and we define recursivelyΓm+1=∪iFi(Γm).
Hence, every vertexvofΓm can be defined by a word in the letters0,1,2 of lenghm+ 1: ij,0≤j≤m by
v=Fim◦Fim−1· · · ◦Fi1(qi0) but the writing is not unique in general.
We denoteVmthe set of vertices ofΓmandv=F¯ı(qi0)for¯ı= (i1, . . . , im) and|¯ı|=m.
Each word ij,1 ≤ j ≤ m of lengh m defines a cell F¯ı(Γ0) = Fim ◦ Fim−1· · · ◦Fi1(Γ0) andx∼y inΓm if and only if they are in the same cell.
The graphΓm has boundeddegree: each vertex has 2 or 4 neighbors. We concider on it a combinatoric Laplacian: so we need a weight on the vertices and on the edges. What is the good choice ? Indeed we begin by the energy form.
8.2 Functions on SG
LetC(SG)denote the set of continuous functions defined onSG =∪m∈NΓm. A continuous function onSGis caracterized by its values on the dense subset V∞=∪m∈NVm.
8.2.1 Self-similar measure µ
We choose µ as a regular probability measure on SG such that any cell of depthm has measure3−m. Thus,
• any point has measure 0
• µis self-similar in the sense that for anyA⊂SG µ(A) =1
3 X
0≤i≤2
µ(Fi−1A) or equivalently
∀f ∈C(SG) Z
SG
f dµ= 1 3
X
0≤i≤2
Z
SG
f ◦Fidµ
• ∀f ∈C(SG),R
SGf dµ= limm→∞ 1 3m+1
P
j=0,1,2
P
|¯ı|=mf(F¯ıqj) Indeed, by the Riemann rule we haveR
SGf dµ= limm→∞ 1 3m
P
|¯ı|=mf(x¯ı) where we have for all¯ı : x¯ı ∈ F¯ı(SG) and we can take x¯ı = F¯ı(qj) which gives the formula by taking the threeqj.
By this last formula, we can define a weight µm on Vm which satisfy R
SGf dµ= limm→∞P
v∈Vmf(v)µm(v) this is obtained withµm(v) = 3m+11 if v∈V0 and µm(v) = 3m+12 for the other points which are indeed always in 2 cells of levelm.
• The points of V0 are concidered as the boundary points ofSG: indeed we have alsoR
SGf dµ= limm→∞P
v∈Vm\V0
2 3m+1f(v).
8.3 Quadratic form of energy
The Hilbert space for Γm is l2(Vm, µm)and the energy form:
qm(f, f) = cm 2
X
x∼y
|f(x)−f(y)|2 wherecm is a constant ofnormalization to be explained.
To construct a limit we need a map Hm :l2(Vm)→l2(Vm+1)such that qm+1(Hmf, Hmf) =qm(f, f)
We choose theharmonic prolongation which indeed minimizes the energy (definition).
• exercise: calculate the harmonic prolongation on one triangle.
This imposes the value ofcm:
cm= 5
3 m
• Property: let f a function on Vm+1 and udefined on Vm. Then qm+1(f, Hm(u)) =qm(f|Vm, u).
Indeed, qm(f|Vm, u) = qm+1(Hm(f|Vm), Hm(u)) thus concider φ = f − Hm(f|Vm), it is a function on Vm+1 which is 0 on Vm. It is easy to see thatqm+1(φ, Hm(u)) = 0: letu˜=Hm(u)
2
cmqm+1(φ,u) =˜ X
x∼m+1y
(φ(x)−φ(y))(˜u(x)−u(y))˜
= 2 X
x∈Vm+1\Vm
φ(x) X
y∼m+1x
(˜u(x)−u(y))˜
becauseφis 0 onVm. Butu˜ is harmonic atx∈Vm+1\Vm, this conclude the proof.
8.3.1 Laplacian on Γm
The polarization formula gives then, for all functionf defined onVm
∆m(f)(x) = cm
µm(x) X
y∼x
(f(x)−f(y)).
This gives two different formula
x∈V0 ⇒∆m(f)(x) = 3.5mX
y∼x
(f(x)−f(y))
x∈Vm\V0 ⇒∆m(f)(x) = 3.5m 2
X
y∼x
(f(x)−f(y))
• exercise: verify the formula of∆m
(We have already concidered this kind of formula, see 2.3.1) 8.3.2 Limit
1. Property ∀f ∈l2(Vm+1)
qm+1(f, f)≥qm+1(Hm(f|Vm), Hm(f|Vm)) =qm(f|Vm, f|Vm).
Forf ∈C(SG)the family(qm(f|Vm, f|Vm))m∈Nis increasing. We define the limit energyq by this limit, when it exists:
D(q) = n
f ∈C(SG), lim
m→∞qm(f|Vm, f|Vm)<∞o
2. Lemma. (q, D(q)) is closed. We see that iff is a function on SGwith finite energyE then f is continuous. Denote fm = f|Vm, we have for x∼y∈Vm
(5
3)m|f(x)−f(y)|2≤2qm(fm, fm)≤2E Thus|f(x)−f(y)| ≤rm√
2E withr= q3
5 <1.
Now ifxm+i ∈Vm+i,0≤i≤k is a path (xm+i ∼xm+i+1 in Γm+i+1) then
|f(xm)−f(xm+k)| ≤
k
X
i=1
|f(xm+i−1)−f(xm+i)| ≤
k
X
i=1
rm+i√ 2E
It follows that ifx, y∈V∞andd(x, y)≤2−m they are in the same cell of level m and|f(x)−f(y)| ≤Crm for a constant C ≥√
2E/(1−r). (Indeedf is Hölder)
Finally, ifq(f, f) = 0then for allm, qm(fm, fm) = 0andf is constant.
So we conciderq onD0(q) ={f ∈D(q), f|V0 = 0.
• Property (D0(q), q)is an Hilbert space.
3. Laplacian∆µonSGWe define onΓm,∆em(f)(x) = 3.52mP
y∼mx(f(x)−
f(y)) and put
∆µ(f)(x) = lim
m→∞∆em(f)(x)
with domain the functions f ∈ D(q) such that this limit exists in L2(SG, dµ).
4. Green’s formula Letf ∈D(q),for all m ∈ N, considering the formula of∆m in section 8.3.1, we have
qm(f, f) = X
x∈Vm
µm(x)f(x)3.5m 2
X
y∼mx
(f(x)−f(y))+
X
x∈V0
µm(x)f(x)3.5m 2
X
y∼mx
(f(x)−f(y)) Whenm→ ∞ we find
q(f, f) = Z
SG
f∆µ(f)dµ+ X
x∈V0
f(x)∂ν(f)(x)
where ∂ν(f)(x) is the normal derivative of f at the boundary point x∈V0 given by
∂ν(f)(x) = lim
m→∞
5m 2.3m
X
y∼mx
(f(x)−f(y)) If we writeV0 ={qi, i∈Z/3Z} then
∂ν(f)(qi) = lim
m→∞
5m 3m
f(qi)−f(Fim(qi+1)) +f(Fim(qi−1)) 2
and if f is harmonic
∂ν(f)(qi) =f(qi)−f(qi+1) +f(qi−1)
2 .
8.4 Spectrum
We consider the Dirichlet Laplacian with domain included inD0(q). Thus0 is not an eigenvalue.
An eigenvalue λm on Γm defines an eigenvalue λm+1 solution of λm = λm+1(5−λm+1) The sequence defines a limit eigenvalue λ = 32lim 5mλm. But the eigenfunctions may be localized. . .
8.5 exercise
Do the same withI = [0,1] concidered as the limit of the graphs Γm with Vm ={2km,0≤k≤2m}and edges(2km ∼ k+12m ).
