Spectral Theory

Kim Klinger-Logan

November 25, 2016

Abstract: The following is a collection of notes (one of many) that compiled in prepa-ration for my Oral Exam which related to spectral theory for automorphic forms. Noneof the information is new and it is rephrased from Rudin’s Functional Analysis, Evans’Theory of PDE, Paul Garrett’s online notes and various other sources.

Background on Spectra

To begin, here is some basic terminology related to spectral theory.

For a continuous linear operator T ∈ EndX, the λ-eigenspace of T is

Xλ := {x ∈ X | Tx = λx}.

If Xλ 6= 0 then λ is an eigenvalue. The resolvent of an operator T is (T − λ)−1. The resolventset

ρ(T ) := {λ ∈ C | λ is a regular value of T}.

The value λ is said to be a regular value if (T − λ)−1 exists, is a bounded linear operator and isdefined on a dense subset of the range of T .

The spectrum of is the collection of λ such that there is no continuous linear resolvent (inverse). Inoter works σ(T ) = C− ρ(T ).

spectrum σ(T ) := {λ | T − λ does not have a continuous linear inverse}discrete spectrum σdisc(T ) := {λ | T − λ is not injective i.e. not invertible}continuous spectrum σcont(T ) := {λ | T−λ is not surjective but does have a dense image}residual spectrum σres(T ) := {λ ∈ σ(T )− (σdisc(T ) ∪ σcont(T ))}

Bounded Operators

Usually we want to talk about bounded operators on Hilbert spaces. Recall that a Hilbert spaceis a vector space with an inner product space that is complete with respect to the distance functioninduced by the metric. We can sometimes deduce nice properties for operators on Banach spaces(a complete normed metric space) when the operator satisfies more specific conditions but most ofthe time we will stick with Hilbert spaces.

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A linear map T : X → Y is bounded if for all ε > 0 there is a δ > 0 such that if |x|X < δ then|Tx|Y < ε. For a linear map of Hilbert spaces, the following are equivalent:

(i) continuous

(ii) continuous at 0

(iii) bounded.

The adjoint of a map T : X → Y is the map T ∗ : Y → X so that 〈Tx, y〉Y = 〈x, T ∗y〉X .Any continuous linear map T have a unique adjoint T ∗ and T ∗∗ = T .

For a self-adjoint maps, all eigenvalues are real.

If T ∈ EndX and T ∗T = TT ∗ then T is normal. For a normal operator there is no residualspectrum–it is all discrete or continuous.

Any normal operator has σres(T ) = ∅.A normal operator is unitary if T ∗T = TT ∗ = I.

Compact Operators

All compact operators are bounded operators. Thus everything that has previously been said stillapplies. There are many ways to define a compact operator T : X → Y :

• maps the unit ball in X to a precompact (has compact closure) set in Y

• maps bounded subsequences in X to sequences in Y with convergent subsequences.

Let {Tn} be a sequence of compact operators on a normed linear space X. Suppose thatTn → T (in the space of bounded operators). Then T is also compact.

For compact operators we do not need to restrict the domain to Hilbert spaces in order to havesome spectral theorem. We have a spectral theorem for compact operators on Banach spaces. Onemethod of proof involves the following two Lemmas:

Reisz’s Lemma: Let X be a Banach space and Y ⊂ X a proper closed subspace. For all ε > 0there is an x ∈ X with ||x|| = 1 and

1 ≥ d(x, Y ) ≥ 1− ε

(We can think of this as the analogue to orthogonality in Hilbert spaces but for Banach spaceswhere there is no notion of orthogonality.)

proof: Let x1 /∈ Y and set R := infy∈Y |y − x1|. Note that R > 0.

For ε > 0 let y1 ∈ Y so that |x1 − y1| < R + ε.

Set x :=x1 − y1

y1 − x1

and note that ||x|| = 1 and

infy∈Y|y−x| = inf

y∈Y

∣∣∣∣y − x1 − y1

y1 − x1

∣∣∣∣ = infy∈Y

∣∣∣∣y +y1

y1 − x1

− x1

y1 − x1

∣∣∣∣ =infy∈Y |y − x1||y1 − x1|

=R

R + ε

We can make this quotient arbitrarily close to 1 by taking smaller ε.

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We can think of this as the analogue to orthogonality in Hilbert spaces but for Banach spaces wherethere is no notion of orthogonality.

Lemma 2: If T is compact then Im(T − I) is closed.

We can also use the Fredholm Alternative to prove the first part of the Spectral Theorem forCompact Operators.

Fredholm Alternative: Let T : X → X be a compact operator on Banach spacesand λ 6= 0, either

(A) T − λ is a bijection or

(B) Im(T − λ) is closed and dim (X/Im(T − λ)) = dim (ker(T − λ)) .

We can also use Reisz’ Lemma to prove the Fredholm Alternative.

Spectral Theorem for Compact Operators on Banach Spaces:Let T be a compact on an infinite-dimensional Banach space then

(1) the nonzero spectrum is discrete (for both T and T ∗)

(2) λi → 0 as i→∞ (and this is the only accumulation point)

(3) the spectrum is countable

(4) the number of eigenvalues outside a disk |λ| ≤ r is finite for r ≥ 0

proof:

(1) This follows nicely from the Fredholm Alternative:

Suppose that λ 6= 0 is in σ(T ) but is not an eigenvalue for T . Then T − λ is injective. Thusdim (ker(T − λ)) = 0 and by the Fredholm Alternative, dim (X/Im(T − λ)) = 0. Thus T − λis surjective and so λ must be in the resolvent.

We can also prove this result using Reisz’ Lemma:

The idea of the proof is by contradiction. Also WLOG, assume λ = 1.

Assume λ ∈ σ(T ) is not an eigenvalue. Thus T − λ = T − I is injective but not surjective.Since T is compact, by Lemma 2, Im(T − I) is a closed proper subspace of X. Call thisY1 := Im(T − I).

Since T − I is injective, Y2 := (T − I)Y1 is a closed proper subspace of Y1.

Create a strictly decreasing sequence of Yn := Im(T − I)n so

Y1 ⊃ Y2 ⊃ · · · ⊃ Yn ⊃ · · · ⊃ Ym ⊃ . . .

Now apply Reisz’s Lemma for Y := Yn+1 and ε = 1/2: Choose yn ∈ Yn so that ||yn|| = 1 andd(yn, Yn+1) > 1/2.

Also note that by compactness of T , {Tyn} must contain a norm convergent subsequence.

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But for n < m,||Tyn − Tym|| = ||(T − I)yn + yn − (T − I)ym − ym||

and (T −I)yn−(T −I)ym−ym ∈ Yn+1 and thus ||Tyn−Tym|| > 1/2 which is our contradiction.

Thus λ must be an eigenvalue.

(2) This will also proceed by contradiction using Reisz’ Lemma.

Assume for the sake of contradiction that there are infinitely many distinct eigenvalues λnoutside a ball centered at the origin of radius ε and each λn has associated eigenvector xn.

Define Yn := span{x1, x2, . . . , xn}.This gives a strictly increasing sequence

Y1 ⊂ Y2 ⊂ · · · ⊂ Yn ⊂ · · · ⊂ Ym ⊂ . . .

Now apply Reisz’ Lemma for Y = Yn−1 and ε = 1/2: Choose yn ∈ Yn so that ||yn|| = 1 andd(yn, Yn−1) > 1/2.

Again by compactness of T , {Tyn} must contain a norm convergent subsequence.

But for n < m,

||Tyn − Tym|| = ||(T − λn)yn + λnyn − (T − λm)ym + λmym||

where (T − λn)yn + λnyn − (T − λm)ym ∈ Ym−1 since ym =∑m

i=1 cixi where∑m−1

i=1 cixi ∈ Ym−1

and (T − λm)xm = 0. Thus ||Tyn − Tym|| > ε/2. This is a contradiction. Thus there areonly finitely many eigenvalues outside a ball centered at 0. This gives us that zero is the onlyaccumulation point.

(3) From (2),

{λn} =⋃n

{|λ| > 1/n}

where each of these sets is finite. Thus σ(T ) is countable.

Let T : X → Y be a compact operator where X is a Banach space and Y is a Hilbertspace, Then T is the limit (in operator norm) of a sequence of finite-dimensional oper-ators.

More can be said about the operators on Hilbert spaces–especially when they are self-adjoint. Forself-adjoint operators (or even symmetric) on Hilbert spaces all of the eigenvalues are real:

Let T be a self-adjoint operator on a Hilbert space with eigenvalue λ so that Tx = λx.Then

λ〈x, x〉 = 〈λx, x〉 = 〈Tx, x〉 = 〈x, Tx〉

〈Tx, x〉 = 〈λx, x〉 = λ〈x, x〉

Thus λ = λ and so λ ∈ R.

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Spectral Theorem for Self-Adjoint Compact Operators on a Hilbert Space:

Let T be a compact self-adjoint operator on a Hilbert space H, then

• H has an orthonormal basis vi of eigenvectors of T andH = the completion of⊕

λHλ.

• λi → 0 as i→∞ (and this is the only accumulation point)

• every eigenspace Hλ is finite-dimensional

• (Ralleigh-Ritz) either ±|T |op is an eigenvalue.

Recall that |T |op := inf{c ≥ 0 | |Tx| ≤ c|x| for all x ∈ X}.

Example: Finite-dimensional

Let T : X → Y be a continuous linear mapping between normed spaces. If Ran(T ) has finitedimension then T is called a finite-dimensional operator.

Finite dimensional operators are compact.

For a bounded set B ⊆ X, T (B) is closed and bounded in the finite-dimensional subspaces Ran(T ) ⊆Y . Heine-Borel says that T (B) is compact in Ran(T ).

Example: Hilbert-Schmidt

Hilbert Schmidt Theorem: Let X be a locally compact space endowed with a positive Borelmeasure and assume that L2(X) is a separable Hilbert space. Let K ∈ L2(X × X)

(i.e.

∫X

∫X

|K(x, y)|2 dxdy <∞). Then the operator

(Tf)(x) =

∫X

K(x, y)f(y) dy

is a compact operator on L2(X).

The basic idea for this proof is to write T as a limit of finite dimensional operators. This operatorT is called the Hilbert-Schmidt operator and K is called the Hilbert-Schmidt kernel.

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Spectral Theorem applied to Elliptic Operators

Consider the eigenvalue problem for the Laplacian with Dirichlet Boundary Conditions on a smoothbounded domain U : {

(∆− λ)u = 0 on Uu = 0 on ∂U

(1)

First we must establish the existence of an inverse to ∆. We can do this by showing the existenceof a solution operator for the following system{

∆u = f on Uu = 0 on ∂U

(2)

where f is a continuous linear function on H10 so f ∈ H−1. The associated bilinear form for ∆ is

B[u, v] =

∫U

∇u · ∇v dx

(the inner product in H10 ). Recall that weak solutions of (2) are u ∈ H1

0 such that for all v ∈ H10 ,

B[u, v] = 〈f, v〉.

For a general elliptic operator we would using Energy Estimates, we see that the hypothesis forLax-Milgram are met and thus by Lax-Milgram we can say that there is a unique weak solutionto (2). However, since ∆ is symmetric this just follows directly from the Reisz RepresentationTheorem by making (u, v); = B(u, v) the new inner product on H.

Lax-Milgram: Assume that B : H × H → R is a bilinear form for which there areconstants α, β > 0 so that

|B[u, v]| ≤ α||u|| · ||v||

for u, v ∈ H and || · || the Hilbert space norm and

β||u||2 ≤ B[u, u]

for u ∈ H. Let f : H → R be a bounded linear functional on H.

Then there is a unique u so that B[u, v] = 〈f, v〉 for all v ∈ H.

The proof of Lax-Milgram essentially boils down to showing that f can be represented by u inB[·, ·] using the Reisz Representation Theorem. This is very simple in the case of ∆ since

Reisz Representation Theorem: H∗ can be cannonically identified with H; moreprecisely for all u∗ ∈ H∗ there is a unique u ∈ H so that 〈u∗, v〉 = (u, v) for all v ∈ H(where 〈·, ·〉 is the pairing of H∗ and H and (·, ·) is the H-inner product).

The linear map u∗ 7→ u is a linear isomorphism.

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Thus there is an inverse operator to ∆, call it K : H−1 → H10 that maps f 7→ u (which is bounded).

Since L2 ⊂ H−1 and H10 compactly embeds into L2 (Rellich compactness), for f ∈ L2 we can rename

the solution operator K to be the composition of a compact map with a continuous map

K : L2 → H10 → L2

so we have that K is compact. Furthermore, K is a self-adjoint operator on Hilbert spaces. Thus weare able to apply the Spectral Theorem for Self-Adjoint Compact Operators on Hilbert Spaces to K.

We can then say of ∆ that its eigenfunctions are equal to those of K and its eigenvalues λn = 1/µnfor µn and eigenvalue of K. Thus for ∆ (and the same can be shown more generally for any sym-metric elliptic operator using roughly the same method):

Spectral Theorem for ∆:

• all eigenvalues are real

• If we repeat each eigenvalue according to its finite multiplicity, 0 < λ1 ≤ λ2 ≤ λ3 ≤. . . and λk →∞ as k →∞.

• There exists an orthonormal basis {uk} of L2(U) where Uk ∈ H10 is an eigenfunction

corresponding to λk.

It is important to note that the theorem above only applies to symmetric elliptic operators. Whenan elliptoc operator is not symmetric, there is no guarantee that all of the eigenvalues will be real.

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Unbounded Operators

A linear map T : D → Y is unbounded if it is not bounded and may not be defined on all of Xfor D ⊂ X. A notion related to self-adjointness exists in this case. Not much can be said aboutunbounded operators that are not densely-defined or at least closed. A closed operator can benaturally defined by having a closed graph in X⊕Y . Explicitly we can also say that every sequence{xn} in D converging to x ∈ X such that Txn → y ∈ Y as n → ∞ we have x ∈ D and Tx = y.Closed operators are more general than bounded operators – they are not necessarily continuous.Note that not every closed operator is densely defined. However, every densely-defined operatorhas a closed adjoint.

An operator T is symmetric when T ⊂ T ∗. Note that all self-adjoint operators are symmetric butnot vice versa.

• If T is symmetric then T ⊂ T ∗∗ ⊂ T ∗.

• If T is closed and symmetric then T = T ∗∗ ⊂ T ∗.

• If T is self-sdjoint then T = T ∗∗ = T ∗.

• If T is essentially self-sdjoint then T ⊂ T ∗∗ = T ∗.

The class of self-adjoint operators is especially important in mathematical physics. Every self-adjointoperator is densely-defined, closed and symmetric. The converse holds for bounded operators butfails in general. An operator is self-adjoint if it is densely-defined, closed symmetric and both T + iand T − i are surjective.

The spectral theorem applies to self-adjoint operators and moreover, to normal operators, but notto densely defined, closed operators in general, since in this case the spectrum can be empty:

Spectral Theorem for Unbounded Operators: Any multiplication operator isa (densely-defined) self-adjoint operator. Any self-adjoint operator is unitarily equiva-lent to a multiplication operator.

Let H be a separable Hilbert space and let (D(T ), T ) ∈ DD∗(H) be a self-adjointoperator on H. Then there exists a finite measure space (X,µ), a measurable real-valuedfunction g : X → R, and a unitary map U : L2(X,µ)→ H such that U−1TU = Mg, theoperator of multiplicative by g, i.e., such that

U−1(D(T )) = D(Mg) = {ϕ ∈ L2(X,µ) | gϕ ∈ L2(X,µ)},

andU(gϕ) = T (U(ϕ))

for ϕ ∈ D(Mg).

Moreover, we have σ(T ) = σ(Mg) = supp(g∗(µ)), which is the essential range of g. Theonly difference with the bounded case is therefore that the spectrum, and the functiong, are not necessarily bounded anymore.

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The only difference with the bounded case is therefore that the spectrum, and the function g, arenot necessarily bounded anymore.

Eigenvalues of symmetric operators are real.

Compact Resolvent

For an unbounded operator T on a Hilbert space, if there is a λ ∈ ρ(T ) so that Rλ = (T − λ)−1 isa compact operator then we say that T has compact resolvent.

If an unbounded operator T on a Hilbert space has compact resolvent then the spectrumof T is discrete.

proof: By the Spectral Theorem for Compact Operators, all nonzero numbers in the spectrum ofa compact operator Rλ0 are eigenvalues of finite multiplicity which have no nonzero accumulationpoint. Also it can be shown that

Lemma: For λ0 ∈ ρ(T ) and λ 6= λ0,

λ is an eigenvalue of T ⇐⇒ (λ − λ0)−1 is an eigenvalue for Rλ0(T ) (with the samemultiplicities).

By this lemma, T has purely discrete spectrum.

If T is also self-adjoint the spectrum is real (in R) and there exists and orthonormalbasis of eigenvectors. (The eigenvalues have no accumulation point though.)

Let Rλ = (T − λ)−1 be the resolvent for λ ∈ C when this inverse exists as a linear operator definedat least on a dense subset of V .

Theorem: Let T be self-adjoint and densely defined. For λ ∈ C−R the operator Rλ is everywheredefined on V , and the operator norm is estimated by

||Rλ|| ≤1

|Imλ|

For T positive, λ /∈ [0,∞), Rλ is everywhere defined on V and the operator norm is estimated by

||Rλ|| ≤

{1|Imλ| Re(λ) ≤ 01|λ| Re(λ) ≥ 0

Theorem: (Hlibert) For points λ, µ off the real line, or, for T positive, for λ, µ off [0,∞)

Rλ −Rµ = (λ− µ)RλRµ

For the operator-norm topology, λ→ Rλ is holomorphic at such points.

Something stronger can be said if the operator is defined on a dense subset of X and is positive(i.e. 〈Tv, v〉 ≥ 0 for all v ∈ D).

Friedrichs Extension: A densely defined, positive, symmetric operator has a self-adjoint extension.

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Friedrichs Extension

A semibounded symmetric operator is one which is satisfies 〈Sv, v〉 ≥ c · 〈v, v〉 or 〈Sv, v〉 ≤ c · 〈v, v〉for some constant c > 0. The operator 1 − ∆ is an canonical example of such an operator since〈∆f, g〉 ≥ 0. We can construct the Friedrichs’ extension of a densely-defined, symmetric operator Sas follows:

Without loss of generality, consider a densely-defined, symmetric operator S with domain DS with

〈Sv, v〉 ≥ 〈v, v〉

for all v ∈ DS. (Note that any semibounded operator can be exhibited this way by multiplying bya constant and adding or subtracting a constant.)

Define the inner product 〈, 〉1 on DS by

〈v, w〉1 := 〈Sv, w〉

for v, w ∈ DS and let V 1 be the completion of DS with respect to the metric induced by 〈, 〉1. Since〈v, v〉1 ≥ 〈v, v〉, the inclusion map DS ↪→ V extends to a continuous map V 1 ↪→ V . Furthermore,since DS is dense in V , we have that V 1 is also dense in V .

For w ∈ V , the functional v 7→ 〈v, w〉 is a continuous linear functional on V 1 with norm

sup|v|1≤1

|〈v, w〉| ≤ sup|v|1≤1

|v| · |w| ≤ sup|v|1≤1

|v|1 · |w| = |w|.

By the Riesz-Frechet Theorem on V 1, there is a w′ ∈ V 1 so that

〈v, w′〉1 = 〈v, w〉

for all v ∈ V 1 and w ∈ V with norm bounded by the norm of v 7→ 〈v, w〉; explicitly, |w′|1 ≤ |w|. The

map A : V → V 1 defined by w 7→ w′ is linear. The inverse of A will be a self-adjoint extension S̃ ofS. This is the Friedrichs extension. We now show that S̃ is in fact self-adjoint and an extension of S.

First note that since |Aw|1 = |w′|1 ≤ |w| from above, the operator norm is sup|w|≤1 |Aw|1 ≤ 1 andso A is continuous.

Also observe that for w′ ∈ DS and all v ∈ V 1,

〈v, w′〉1 = 〈v, Sw′〉 = 〈v, A(Sw′)〉1

so A(Sw′) = w′ for each w′ ∈ DS. Hence AV ⊂ V 1 contains the domain DS of S.

We also see that A is injective since kerA = 0: since V 1 is dense in V , if 0 = 〈v,Aw〉1 = 〈v, w〉 for

all v ∈ V 1 then w = 0. Thus the inverse S̃ of A is defined on DS̃ = AV ⊂ V 1. Hence S̃ is injectiveand is surjective for DS̃ → V .

Now to show that S̃ is an extension of S, it remains to show that A(S̃w) = A(Sw) for w ∈ DS. Forv, w ∈ DS ⊂ DS̃,

〈v, S̃〉 = 〈v, A(S̃w)〉1 = 〈v, w〉1 = 〈Sv, w〉 = 〈v, Sw〉.

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Since Ds is dense in V we have that S̃w = Sw.

We also must show that S̃ is symmetric. First note that A is symmetric: for w′ = Aw ∈ AV since〈v, Aw〉1 = 〈v, w〉 we have 〈v, w′〉1 = 〈v, S̃w′〉 and

〈Av,w〉 = 〈Av, S̃Aw〉 = 〈Av,Aw〉1

which is symmetric in v and w. Thus since DS̃ = AV and

〈S̃Av,Aw〉 = 〈v,Aw〉 = 〈Av,w〉 = 〈Av,Aw〉1 = 〈Av, S̃Aw〉

and so S̃ is symmetric.

Furthermore, this extension S̃ remains semibounded i.e. 〈S̃v, v〉 ≥ 〈v, v〉 for all v = Aw ∈ DS̃ = AVsince

〈S̃v, v〉 = 〈S̃Aw, v〉 = 〈w, v〉 = 〈Aw, v〉1 = 〈v, v〉1 ≥ 〈v, v〉.

It remains to show that S̃ is self-adjoint. Note that any proper extension T ⊃ S̃ is not injectivesince S̃ surjects to V . So if S∗ were a proper extension of S̃ there would be v ∈ DS̃ so that for allw ∈ DS̃,

0 = 〈S̃∗v, w〉 = 〈v, S̃w〉 = 〈v, S̃w〉.

Since S̃ surjects to V , there is a w ∈ DS̃ such that S̃w = v. Hence v = 0 and S̃∗ cannot be a proper

extension of S̃. Thus S̃ is self-adjoint.

This construction serves as a proof for the following theorem of Friedrichs (?, ?).

Theorem 1. A positive, densely-defined, symmetric operator S with domain DS has a positiveself-adjoint extension with the same lower bound.

This extension has useful properties of particular interest to our project. An alternative character-ization of the extension makes this more clear.

Assume that V has a C-conjugate-linear complex conjugation v → vc with the properties: (vc)c = vand 〈vc, wc〉 = 〈v, w〉. Further let S commute with conjugation so that (Sv)c = S(vc). Let V −1 bethe dual of V 1 so that V 1 ⊂ V ⊂ V −1.

Note that given this small specification, there is an alternate characterization of the Friedrichsextension. To specify it, define a continuous, complex-linear map S# : V 1 → V −1 by

(S#v)(w) = 〈v, wc〉1

for v, w ∈ V 1.

Theorem 2. Let X = {v ∈ V 1 | S#v ∈ V }. Then the Friedrichs extension of S is S̃ = S#|X withdomain DS̃ = X.

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Proof. Let T = S#|X . Let : V → V 1 be the inverse of S̃ defined by 〈Av,w〉1 = 〈v, w〉 for all w ∈ V 1

and v ∈ V from the Riesz-Fischer Theorem. Then

〈TAv,w〉 = 〈Av,w〉1 = 〈v, w〉

for v ∈ V and w ∈ V 1. Also,〈ATv,w〉1 = 〈Tv, w〉 = 〈v, w〉1

for v ∈ X and w ∈ V 1. This T = A−1 = S̃.

Extensions of Restrictions

Using the latter characterization of the Friedrichs extension we will now explain how the constructionof the extension works with the case of restricted operators. Assume that S and the related termsare defined as above. Let Θ ⊂ DS be a S-stable subspace. Let the orthogonal complement to Θ inV be

ker Θ = {v ∈ V | 〈v, θ〉 = 0 for all θ ∈ Θ}.

For our purposes, given an operator S as above, we will want to define T = S|DS∩ker Θ so thatDT = DS ∩ ker Θ.

Note that for v ∈ DT and θ ∈ Θ,

〈Tv, θ〉 = 〈Sv, θ〉 = 〈v, Sθ〉 ∈ {〈v, θ′〉 | θ′ ∈ Θ} = {0}

and so T (DT ) ⊂ ker Θ. Furthermore since T is a restriction of S, symmetry and〈Tv, v〉 ≥ 〈v, v〉 are inherited from S.

In contrast with these inherited properties, it is nontrivial to give a simple condition to ensure thatDT is dense in ker Θ and that the V 1-closure of DT is V 1 ∩ ker Θ. This delicacy is demonstratedin Lax and Phillips (?, ?). For cut-off above height a > 1, an argument using the geometry ofthe fundamental domain for Γ shows that Θ ∩ V 1 is dense in V .1 For this reason we will assumeDT = DS ∩ ker Θ is V -dense in ker Θ and V 1-dense in V 1 ∩ ker Θ =: W 1.

Let W−1 be the dual of W so we have W 1 ⊂ ker Θ ⊂ W−1. Define S# : V 1 → V −1 by

(S#v)(w) := 〈v, wc〉1,

for v, w ∈ V 1.

Theorem 3. Let Θ# be the V −1-completion of Θ. The Friedrichs extension T̃ of T has domainDT̃ = {v ∈ W 1 | S#v ∈ V + Θ#} and is characterized by

T̃ v = w ⇐⇒ S#v ∈ w + Θ#

for v ∈ DT̃ and w ∈ ker Θ.

1For a < 1, there are serious complications so we do not address this case.

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Proof. Define T# : W 1 → W−1 by(T#v)(w) := 〈v, wc〉1

we can then define the the domain of the Friedrichs extension T̃ as

DT̃ = {w ∈ W 1 | T#w ∈ ker Θ}

so that T̃ = T#|DT̃. With the inclusion j : W 1 → V 1 for all x, y ∈ W 1

(T#x)(y) = 〈jx, (jy)c〉1 = (S#jx)(jy) =((j∗ ◦ S# ◦ j)x

)(y)

and so T# = j∗ ◦ S# ◦ j and

DT̃ = {w ∈ W 1 | j∗(S#(jw)) = 0}.

The orthogonal compliment ker Θ to Θ in V is a closed subspace of V 1 and the dual W−1 of W 1 is

W−1 = (V 1 ∩ ker Θ)∗ ∼= V −1/Θ#.

The Friedrichs extension makes the following diagram commute

V 1 V V −1

W 1 ker Θ W−1

S#

j−1 j∗

T#

T̃

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Spectral Theory of Automorphic Forms

There are several examples of automorphic forms on Γ\H:(a) Holomorphic modular forms

(b) Maass forms

(c) Constant functions

Representation theory helps us with the question:

Question 1: Why are precisely these the types of automorphic forms on Γ\H and noothers?

Question 2: Where do differential operators come from and why do they work?

The answer to question 2 also comes from representation theory–namely Lie theory and we willaddress it later.

However question 1 is also a question related to spectral theory in that Maass forms are eigen-functions for the Maass operators on C∞(H)

Rk = iy∂

∂x+ y

∂

∂y+k

2= (z − z)

∂

∂z+k

2

Lk = −iy ∂∂x

+ y∂

∂y− k

2= (z − z)

∂

∂z− k

2.

Note that it is easily verified that

∆k = −Lk+2Rk −k

2

(1 +

k

2

)= −Rk−2Lk +

k

2

(1− k

2

).

The Laplacian ∆k is a symmetric operator on L2(H) with domain C∞o (H). Recall that the spec-tral theorem only applies to self-adjoint operators; however, even for symmetric operators wecan conclude that if f is an L2 eigenfunction so that ∆kf = λf then λ is real and eigenvectorscorresponding to distinct eigenvalues are orthogonal.

Also since H has infinite volume, the Laplacian has too many eigenfunctions to be of real interest.A more interesting and tangible situations is to consider the decomposition when Γ\H is compactor finite volume.

Γ\H compact

There are two versions of the spectral problem that are closely related:

Spectral Problem (version 1): Determine the spectrum of the unbounded sym-metric operator ∆k on L2(Γ\H, χ, k).

14

Here L2(Γ\H, χ, k) is the Hilbert space completion of the space of smooth functions on H such that

χ(γ)f(z) =

(cz + d

|cz + d|

)kf

(az + b

cz + d

)for γ ∈ Γ, C∞(Γ\H, χ, k).

Spectral Problem (version 2): Determine the decomposition of the Hilbert spaceL2(Γ\G,χ) into irreducible subspaces.

Where L2(Γ\G,χ) is the square-integrable functions satisfying

f(γgu) = χ(γ)f(g)

for γ ∈ Γ, u ∈ Z+ and g ∈ G that are square integrable with respect to the Haar measure on G/Z+.

Relating the Spectral Problems

To relate these two versions of the spectral problems, we need to describe how these two spacesL2(Γ\H, χ, k) and L2(Γ\G,χ) related to one another. Define the right regular representation

ρ : G→ End(L2(Γ\G,χ)

)meaning that (ρ(g)f) (x) = f(xg) for g, x ∈ G. This is a unitary representation.

It is common to restrict the action of ρ on to the maximal compact K of G. In doing so, we havethat

L2(Γ\G,χ) =⊕k∈Z

L2(Γ\G,χ, k)

where L2(Γ\G,χ, k) is the subspace consisting of functions such that ρ(κθ)f = eikθf (or f(gκθ) =eikθf(g)). There is a Hilbert space isomorphism

σk : L2(Γ\H, χ, k)→ L2(Γ\G,χ, k)

given by (σkf)(g) = (f |kg)(i) for g ∈ G.

In order to understand the relationship between these Hilbert Spaces, we need to define operatorson L2(Γ\G,χ) to play the role of ∆k, Rk and Lk. We define the following differential operators onG

R = e2iθ

(iy∂

∂x+ y

∂

∂y+

1

2i

∂

∂θ

),

L = e−2iθ

(−iy ∂

∂x+ y

∂

∂y− 1

2i

∂

∂θ

)and the Laplace-Beltrami operator

∆ = −y2

(∂2

∂x2+

∂2

∂y2

)+ y

∂2

∂x∂θ.

The direct relationship between these two versions of the spectral problem can best be seen inthe following theorem:

15

Theorem 4. Let Γ be a discontinuous subgroup of G such that −I ∈ Γ and Γ\G/K is compact.Let χ be a character of G. Let χ(−1) = (−1)ε where ε = 0 or 1.

(a) The space H = L2(Γ\G,χ) decomposes into a Hilbert space direct sum of irreducible represen-tations.

The spaces L2(Γ\H, χ, k) each decompose into a Hilbert space direct sum of eigenspaces for ∆k.

(b) Let Hk be any subspace of H that is invariant under the action of G. Then Hk is also invariantunder the action of ∆.

Conversely, let λ ∈ R and let Hλ be the λ-eigenspace of ∆ on H, then Hλ is G-invariant.

(c) If Hk is an irreducible subspace of H then ∆ acts as a scalar on Hk.

The eigenvector λ of ∆ on Hk depends only on the isomorphism class of Hk. If λ is such aneigenvalue than λ is a real number. Either λ ≥ 0 is ε = 0 and λ ≥ 1/4 if ε = 1, otherwiseλ = k

2(1− k

2) where 1 ≤ k ∈ Z and k ≡ εmod 2.

(d) There is only one finite-dimensional representation of G that can occur in H–the trivial repre-sentation:

If χ = 1 then the constant function spans a one-dimensional irreducible subspace of H on whichspace the eigenvalue λ of ∆ equals 0.

All other irreducible constituents of H are infinite dimensional.

(e) Assume λ is not of the form k2(1− k

2) where 1 ≤ k ∈ Z and ≡ εmod 2:

There exists a unique irreducible representation P(λ, ε) of G depending only on λ (and not Γ)such that if Hk is an infinite-dimensional irreducible subrepresentation of H with eigenvalue λthen Hk

∼= P(λ, ε).

Let k ≡ εmod 2 be an integer. The multiplicity of the representation P(λ, ε) is equal to themultiplicity of the eigenvalue in L2(Γ\H, χ, k).

(f) Assume that λ is of the form k2(1− k

2) where 1 ≤ k ∈ Z and k ≡ εmod 2.

There exists two irreducible representations, D+(k) and D−(k) of G depending only on k suchthat if Hk is an infinite-dimensional irreducible representation of H with λ then either Hk

∼=D+(k) or HK

∼= D+(k). The representations D±(k) have the same multiplicity in H; thismultiplicity equals the dimension of the space of holomorphic modular forms of weight k thatsatisfy

f(γz) = χ(γ)(cz + d)kf(z)

for γ ∈ Γ.

Part (a) above unifies these two versions. However understanding the link between the spectral andrepresentation theory can best be seen in the proof of this result.

‘Proof’ of (a):Let χ be a unitary character of Γ and H = L2(Γ\G,χ) Let φ ∈ C∞c (G) and define

ρ(φ) : H → H

16

by ρ(φ)f(g) =

∫G

f(gh)φ(h) dh. Thus we have ρ(φ)f =

∫G

φ(h)ρ(h)f dh where ρ is a right-regular

representation.

the following are some properties of such representations:

Proposition 5. Let φ ∈ C∞c (G).

(a) The operator ρ(φ) is a Hilbert-Schmidt operator. In particular, the operator is compact.

If f ∈ L2(Γ\G,χ) then ρ(φ)f ∈ C∞(Γ\G,χ).

(b) If φ(g) = φ(g−1) then the operator ρ(φ) is self-adjoint.

More generally, if π : G→ End(H) is a unitary representation of G on a Hilbert space H andif φ(g) = φ(g−1) then π(φ) is self-adjoint.

(c) For κθ ∈ K, if φ(κθg) = e−ikθφ(g) then ρ(φ) maps the Hilbert space L2(Γ\G,χ) into C∞(Γ\G,χ, k).

Define π(φ) ∈ End(V ) by

π(φ)v =

∫G

φ(g)π(g)v dg.

We can use the above properties to establish the following lemma.

Lemma 6. Let π : G → End(H) be a unitary representation of G on a Hilbert space H and let0 6= f ∈ H. Let ε > 0 be given. Then there exists φ ∈ C∞c (G) such that π(φ) is self-adjoint and|π(φ)f − f | < ε.In particular if ε < |f | this implies that π(φ)f 6= 0.Moreover if π(κθ) = eikθf for all κθ ∈ K, then we may choose φ so that φ(κθg) = θ(gκθ) = eikθφ(g).

If ρ is unitary and it is reducible (it has a a proper nonzero closed subspace V ) then the orthogonalcomplement U of V is also a nonzero invariant closed subspace and H = U⊕V . On the other hand, ifit is not unitary, it is possible for there to be a closed invariant subspace V that is not complimented.

From this lemma we have

Proposition 7. Let H be a nonzero closed subspace of L2(Γ\G,χ) which is closed under the actionof G. Then H has decomposition as a Hilbert space direct sum

H =⊕k

Hk

where ρ(κθ)f = e2πikθf for f ∈ Hk.Let k be such that Hk 6= 0 then ∆ has a nonzero eigenvector in Hk ∩ C∞(Γ\G,χ).

To prove this result we use the fact that ρ(φ) is a compact operator from Proposition 5 and theSpectral Theorem for compact operatiors.From this we get the first part of (a):

Theorem 8. The space L2(Γ\G,χ) decomposes into Hilbert space direct sum of subspaces that areinvariant and irreducible under the right regular representation ρ.

Similarly we get

17

Theorem 9. Let ξ be a character of C∞c (K\G/K, σ) and let H(ξ) be the space of f ∈ L2(Γ\G,χ, k)that satisfys π(φ)f = ξ(φ)f for all φ ∈ C∞c K\G/K, σ). The space H(ξ) is a finite dimensionalsubspace of C∞(Γ\G,χ, k).If ξ and η are two distinct characters of C∞c (K\G/K, σ) then H(ξ) and H(η) are orthogonal sub-spaces. Furthermore,

L2(Γ\G,χ, k) =⊕ξ

H(ξ)

where H(ξ) 6= 0.

Using this theorem and the fact that because ∆ commutes with the operators in C∞c (K\G/K, σ)the H(ξ) are ∆-invariant and so ∆ induces a self-adjoint operator on each of the finite dimenstionalvector spaces H(ξ) so each of these decomposes into a direct sum of ∆-eigenspaces.

Corollary 10. The space L2(Γ\H, χ, k) decomposes into a Hilbert space direct sum of eigenspacesof ∆k.

Both these results that make up part (a) follow from Proposition 5.

Spectral Decomposition for Γ = SL2(Z)

In this case, Γ\H is non-compact. Thus there is no reason to suspect that the spectrum is purelydiscrete.We will now exhibit the computation of this spectral decomposition for L2(Γ\H) with re-spect to ∆ = y2

(∂2x + ∂2

y

).

Let N be the upper-triangular unipotent matrices in G = SL2(R), A the diagonal matrices, A+

the diagonal matrices with positive diagonal entries, P = NA the parabolic subgroup of upper-triangular matrices, P+ = NA+, Γ = SL2(Z) and Γ∞ = P+ ∩Γ = N ∩Γ. For simplicity, normalizethe total measure of K to 1 rather than 2π.

Pseudo-Eisenstein Series

Pseudo-Eisenstein series are solutions to the adjunction problem: given ϕ ∈ C∞c (N\G), we want tofind Ψϕ ∈ C∞c such that

〈cPf, ϕ〉N\G = 〈f,Ψϕ〉Γ\Gfor f on Γ\G and 〈f, F 〉Γ\G =

∫Γ\G f · F .

We can compute the canonical expression for Ψϕ from this desired equality using the left N -invariance of ϕ and the left Γ-invariance of f as follows: (Note that P ∩ Γ differs from N ∩ Γonly by ±12 which act trivially on H ∼= G/K.)

〈cPf, ϕ〉N\H =

∫N\H

cPf(z)ϕ(Im(z))dx dy

y2=

∫N\H

(∫N∩Γ\N

f(nz) dn

)ϕ(Im(z))

dx dy

y2

=

∫Γ∞\H

f(z)ϕ(Im(z))dx dy

y2=

∫Γ\H

∑γ∈P∩Γ\Γ

f(γz)ϕ(Im(γz))dx dy

y2

18

=

∫Γ\H

f(z)

∑γ∈P∩Γ\Γ

ϕ(Im(γz))

dx dy

y2= 〈f,Ψϕ〉Γ\H

Thus we define the pseudo-Eisenstein series as Ψϕ(z) =∑

γ∈P∩Γ\Γ

ϕ(Im(γz)). Note that the pseudo-

Eisenstein series is absolutely and uniformly convergent for z ∈ C where C is a compact subset ofG. Furthermore, Ψϕ ∈ C∞c (Γ\G).

Now, it is a corollary of the above characterization of psuedo-Eisenstein series that the squareintegrable cuspforms are the orthogonal complement of the (closed) subspace of L2(Γ\H) spannedby pseudo-Eisenstein series with ϕ ∈ C∞0 (N\H) ∼= C∞0 (0,∞). Thus we have

L2(Γ\H) = L2cusp(Γ\H)⊕ L2

p-Eis(Γ\H).

Decomposition of Pseudo-Eisenstein Series

We further decompose L2(Γ\H) by examining the pseudo-Eisenstein series Ψϕ. The spectral decom-position of the data ϕ induces a spectral decomposition for Ψϕ. Identifying N\H ∼= N\G/K ∼= A+,Mellin inversion gives

ϕ(Im z) =1

2πi

∫ σ+i∞

σ−i∞Mϕ(s)(Im z)s ds

for any real σ. This decomposition of ϕ is achieved as follows.

Replacing ξ by ξ/2π in Fourier inversion we get

f(x) =1

2π

∫ ∞−∞

(∫ ∞−∞

f(t)e−itξ dt

)eiξx dξ.

Fourier transforms on R put into multiplicative coordinates are Mellin transforms:For ϕ ∈ C∞c (0,∞), take f(x) = ϕ(ex). Let y = ex and r = et (the exponential in the implied innerintegral) and rewrite Fourier inversion as

f(x) = ϕ(y) =1

2π

∫ ∞−∞

(∫ ∞0

ϕ(r)r−iξdr

r

)yiξ dξ

since dt = drr

. Note that this integral converges as a C∞-function-valued function.

The Fourier transform (inner integral) in these coordinates is Mellin transform. For compactlysupported ϕ, the integral definition extends to all s ∈ C as Mϕ(s) =

∫∞0ϕ(r)r−s dr

r. Mellin

inversion is

ϕ(y) =1

2π

∫ ∞−∞Mϕ(iξ)yiξ dξ.

With ξ the imaginary part of a complex variable s, we can rewrite this as a complex path integral

ϕ(y) =1

2πi

∫ 0+i∞

0−i∞Mϕ(s)ys ds

19

since dξ = −i ds. For f ∈ C∞c (R), f̂(ξ) converges nicely for all complex values of ξ so it extends toan entire function in ξ of rapid decay on horizontal lines (Payley-Wiener Theorem). This extensionproperty applies to ϕ allowing us to move the contour as above.

Thus the pseudo-Eisenstein series is

Ψϕ(z) =∑

γ∈Γ∞\Γ

ϕ(Im(γz)) =1

2πi

∑γ∈Γ∞\Γ

∫ σ+i∞

σ−i∞Mϕ(s) · (Im(γz))s ds.

It would be natural to take σ = 0 however, at σ = 0 the double integral would not be absolutelyconvergent and the two integrals cannot be interchanged. For σ > 1 the double integral is absolutelyconvergent and (using Fubini) we have,

Ψϕ(z) =1

2πi

∫ σ+i∞

σ−i∞Mϕ(s) · Es(z) ds

for Es(z) =∑

γ∈Γ∞\Γ(Im(γz))s the Eisenstein series.

Eisenstein Series

We have a decomposition of Ψϕ in terms of ϕ. We now want to rewrite this piece of the decompo-sition so as to refer only to Ψϕ not ϕ. Note that Es has meromorphic continuation on the entirecomplex plane. Thus we can move the line of integration to the left and choose σ = 1/2 to achieve

Ψϕ =1

2πi

∫ 1/2+i∞

1/2−i∞Mϕ(s)Es ds+

∑s0

ress=s0(Mϕ(s)Es).

As with the pseudo-Eisenstein series, the Eisenstein series Es fits into an adjunction relation

〈Es, f〉Γ\H = 〈ys, cPf〉Γ∞\H

for f on Γ\H. Notice that

〈ys, cPf〉A+ =

∫N\H

cPf(z) · ys dx dyy2

=

∫N\H

(∫Γ∞\N

f(nz) dn

)· ys dx dy

y2

=

∫Γ∞\H

f(z) · ys dx dyy2

=

∫P∩Γ\H

f(z) · ys dx dyy2

=

∫Γ\H

∑γ∈P∩Γ\Γ

f(γz) · Im(γz)sdx dy

y2

=

∫Γ\H

f(z) ·∑

γ∈P∩Γ\Γ

Im(γz)sdx dy

y2= 〈Es, f〉Γ\H.

Thus

〈Es, f〉Γ\H =

∫ ∞0

cPf(iy)ysdy

y2=

∫ ∞0

cPf(iy)y−(1−s) dy

y=M(cPf)(1− s).

20

On the other hand, since cPEs = ys + csy1−s for cs =

ξ(2s− 1)

ξ(2s), we have

〈Es,Ψϕ〉Γ\H = 〈cPEs, ϕ〉Γ∞\H = 〈ys + csy1−s, ϕ〉Γ∞\H =

∫ ∞0

(ys + csy1−s) · ϕ(y)

dy

y2

=

∫ ∞0

(y−(1−s) + csy−s) · ϕ(y)

dy

y=Mϕ(1− s) + csMϕ(s).

So we have the identity,

M(cPΨϕ)(s) = 〈E1−s,Ψϕ〉Γ\H =Mϕ(s) + c1−sMϕ(1− s).

Using this and returning to our equation for Ψϕ above, we get

Ψϕ −∑s0

ress=s0(Mϕ(s)Es) =1

2πi

∫ 1/2+i∞

1/2−i∞Mϕ(s)Es ds

=1

2πi

∫ 1/2+i∞

1/2−i0Mϕ(s)Es +Mϕ(1− s)E1−s ds

=1

2πi

∫ 1/2+i∞

1/2−i0Mϕ(s)Es + c1−sMϕ(1− s)Es ds

=1

2πi

∫ 1/2+i∞

1/2−i0McpΨϕ(s)Es ds

from the functional equation E1−s = c1−sEs and our identity above

=1

4πi

∫ 1/2+i∞

1/2−i∞〈Ψϕ, Es〉Γ\H · Es ds.

Residue

Finally, let us examine the residue. For Γ = SL2(Z), the only pole of Es in the half plane Re(s) ≥ 1/2is at s0 = 1. This pole is simple and the residue is a constant function. Thus we can compute theresidue as follows: ∑

s0

ress=s0(Mϕ(s)Es) =Mϕ(1) · ress=1Es

where

Mϕ(1) =

∫ ∞0

ϕ(y)y−1dy

y=

∫ ∞0

ϕ(y)dy

y2=

∫N\H

ϕ(Imz)dx dy

y2

=

∫N\H

∫Γ∞\N

ϕ(Im(nz)) dndx dy

y2=

∫N\H

ϕ(Im(nz))

(∫Γ∞\N

1 dn

)dx dy

y2

=

∫Γ∞\H

ϕ(Imz)dx dy

y2

21

since the volume of Γ∞\N is 1 and ϕ is left N -invariant

=

∫Γ\H

∑γ∈Γ∞\Γ

ϕ(Imz)dx dy

y2=

∫Γ\H

Ψϕ(z)dx dy

y2= 〈Ψϕ, 1〉Γ\H.

This we have thatL2(Γ\H) = L2

cusp(Γ\H)⊕ C⊕ L2Eis(Γ\H).

22