Spectral Theory. Kim Klinger-Logan. November 25, 2016

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1 Spectral Theory Kim Klinger-Logan November 25, 2016 Abstract: The following is a collection of notes (one of many) that compiled in preparation for my Oral Exam which related to spectral theory for automorphic forms. None of 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 X λ := {x X T x = λx}. If X λ 0 then λ is an eigenvalue. The resolvent of an operator T is (T λ) 1. The resolvent set 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 is defined 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). In oter works σ(t ) = C ρ(t ). spectrum σ(t ) := {λ T λ does not have a continuous linear inverse} discrete spectrum σ disc (T ) := {λ T λ is not injective i.e. not invertible} continuous spectrum σ cont (T ) := {λ T λ is not surjective but does have a dense image} residual spectrum σ res (T ) := {λ σ(t ) (σ disc (T ) σ cont (T ))} Bounded Operators Usually we want to talk about bounded operators on Hilbert spaces. Recall that a Hilbert space is a vector space with an inner product space that is complete with respect to the distance function induced 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 of the time we will stick with Hilbert spaces. 1

2 A linear map T : X Y is bounded if for all ɛ > 0 there is a δ > 0 such that if x X < δ then T x Y < ɛ. 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 T x, 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 = T T then T is normal. spectrum it is all discrete or continuous. Any normal operator has σ res (T ) =. A normal operator is unitary if T T = T T = I. For a normal operator there is no residual Compact Operators All compact operators are bounded operators. Thus everything that has previously been said still applies. 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 {T n } be a sequence of compact operators on a normed linear space X. Suppose that T n 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 have some spectral theorem. We have a spectral theorem for compact operators on Banach spaces. One method 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 ɛ > 0 there is an x X with x = 1 and 1 d(x, Y ) 1 ɛ (We can think of this as the analogue to orthogonality in Hilbert spaces but for Banach spaces where there is no notion of orthogonality.) proof: Let x 1 / Y and set R := inf y Y y x 1. Note that R > 0. For ɛ > 0 let y 1 Y so that x 1 y 1 < R + ɛ. Set x := x 1 y 1 y 1 x 1 and note that x = 1 and inf y x = inf y Y y Y y x 1 y 1 y 1 x 1 = inf y Y y + y 1 x 1 y 1 x 1 y 1 x 1 = inf y Y y x 1 y 1 x 1 We can make this quotient arbitrarily close to 1 by taking smaller ɛ. = R R + ɛ 2

3 We can think of this as the analogue to orthogonality in Hilbert spaces but for Banach spaces where there 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 for Compact Operators. Fredholm Alternative: Let T : X X be a compact operator on Banach spaces and λ 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 proof: (1) the nonzero spectrum is discrete (for both T and T ) (2) λ i 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 (1) This follows nicely from the Fredholm Alternative: Suppose that λ 0 is in σ(t ) but is not an eigenvalue for T. Then T λ is injective. Thus dim (ker(t λ)) = 0 and by the Fredholm Alternative, dim (X/Im(T λ)) = 0. Thus T λ is 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 this Y 1 := Im(T I). Since T I is injective, Y 2 := (T I)Y 1 is a closed proper subspace of Y 1. Create a strictly decreasing sequence of Y n := Im(T I) n so Y 1 Y 2 Y n Y m... Now apply Reisz s Lemma for Y := Y n+1 and ɛ = 1/2: Choose y n Y n so that y n = 1 and d(y n, Y n+1 ) > 1/2. Also note that by compactness of T, {T y n } must contain a norm convergent subsequence. 3

4 But for n < m, T y n T y m = (T I)y n + y n (T I)y m y m and (T I)y n (T I)y m y m Y n+1 and thus T y n T y m > 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 λ n outside a ball centered at the origin of radius ɛ and each λ n has associated eigenvector x n. Define Y n := span{x 1, x 2,..., x n }. This gives a strictly increasing sequence Y 1 Y 2 Y n Y m... Now apply Reisz Lemma for Y = Y n 1 and ɛ = 1/2: Choose y n Y n so that y n = 1 and d(y n, Y n 1 ) > 1/2. Again by compactness of T, {T y n } must contain a norm convergent subsequence. But for n < m, T y n T y m = (T λ n )y n + λ n y n (T λ m )y m + λ m y m where (T λ n )y n + λ n y n (T λ m )y m Y m 1 since y m = m i=1 c ix i where m 1 i=1 c ix i Y m 1 and (T λ m )x m = 0. Thus T y n T y m > ɛ/2. This is a contradiction. Thus there are only finitely many eigenvalues outside a ball centered at 0. This gives us that zero is the only accumulation point. (3) From (2), {λ n } = 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 Hilbert space, Then T is the limit (in operator norm) of a sequence of finite-dimensional operators. More can be said about the operators on Hilbert spaces especially when they are self-adjoint. For self-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 T x = λx. Then λ x, x = λx, x = T x, x = x, T x Thus λ = λ and so λ R. T x, x = λx, x = λ x, x 4

5 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 v i of eigenvectors of T and H = the completion of λ H λ. λ i 0 as i (and this is the only accumulation point) every eigenspace H λ is finite-dimensional (Ralleigh-Ritz) either ± T op is an eigenvalue. Recall that T op := inf{c 0 T x c x for all x X}. Example: Finite-dimensional Let T : X Y be a continuous linear mapping between normed spaces. dimension then T is called a finite-dimensional operator. If Ran(T ) has finite 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 Borel measure and assume that L 2 (X) is a separable Hilbert space. Let K L 2 (X X) (i.e. K(x, y) 2 dxdy < ). Then the operator X X is a compact operator on L 2 (X). (T f)(x) = X K(x, y)f(y) dy The basic idea for this proof is to write T as a limit of finite dimensional operators. This operator T is called the Hilbert-Schmidt operator and K is called the Hilbert-Schmidt kernel. 5

6 Spectral Theorem applied to Elliptic Operators Consider the eigenvalue problem for the Laplacian with Dirichlet Boundary Conditions on a smooth bounded domain U: { ( λ)u = 0 on U (1) u = 0 on U First we must establish the existence of an inverse to. We can do this by showing the existence of a solution operator for the following system { u = f on U (2) u = 0 on U where f is a continuous linear function on H0 1 so f H 1. The associated bilinear form for is B[u, v] = u v dx U (the inner product in H 1 0). Recall that weak solutions of (2) are u H 1 0 such that for all v H 1 0, B[u, v] = f, v. For a general elliptic operator we would using Energy Estimates, we see that the hypothesis for Lax-Milgram are met and thus by Lax-Milgram we can say that there is a unique weak solution to (2). However, since is symmetric this just follows directly from the Reisz Representation Theorem 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 are constants 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 in B[, ] using the Reisz Representation Theorem. This is very simple in the case of since Reisz Representation Theorem: H can be cannonically identified with H; more precisely 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 u is a linear isomorphism. 6

7 Thus there is an inverse operator to, call it K : H 1 H 1 0 that maps f u (which is bounded). Since L 2 H 1 and H 1 0 compactly embeds into L 2 (Rellich compactness), for f L 2 we can rename the solution operator K to be the composition of a compact map with a continuous map K : L 2 H 1 0 L 2 so we have that K is compact. Furthermore, K is a self-adjoint operator on Hilbert spaces. Thus we are 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 λ n = 1/µ n for µ n and eigenvalue of K. Thus for (and the same can be shown more generally for any symmetric 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 < λ 1 λ 2 λ 3... and λ k as k. There exists an orthonormal basis {u k } of L 2 (U) where U k H 1 0 is an eigenfunction corresponding to λ k. It is important to note that the theorem above only applies to symmetric elliptic operators. When an elliptoc operator is not symmetric, there is no guarantee that all of the eigenvalues will be real. 7

8 Unbounded Operators A linear map T : D Y is unbounded if it is not bounded and may not be defined on all of X for D X. A notion related to self-adjointness exists in this case. Not much can be said about unbounded operators that are not densely-defined or at least closed. A closed operator can be naturally defined by having a closed graph in X Y. Explicitly we can also say that every sequence {x n } in D converging to x X such that T x n y Y as n we have x D and T x = 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 operator has a closed adjoint. An operator T is symmetric when T T. Note that all self-adjoint operators are symmetric but not 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-adjoint operator is densely-defined, closed and symmetric. The converse holds for bounded operators but fails in general. An operator is self-adjoint if it is densely-defined, closed symmetric and both T + i and T i are surjective. The spectral theorem applies to self-adjoint operators and moreover, to normal operators, but not to densely defined, closed operators in general, since in this case the spectrum can be empty: Spectral Theorem for Unbounded Operators: Any multiplication operator is a (densely-defined) self-adjoint operator. Any self-adjoint operator is unitarily equivalent to a multiplication operator. Let H be a separable Hilbert space and let (D(T ), T ) DD (H) be a self-adjoint operator on H. Then there exists a finite measure space (X, µ), a measurable real-valued function g : X R, and a unitary map U : L 2 (X, µ) H such that U 1 T U = M g, the operator of multiplicative by g, i.e., such that U 1 (D(T )) = D(M g ) = {ϕ L 2 (X, µ) gϕ L 2 (X, µ)}, and for ϕ D(M g ). U(gϕ) = T (U(ϕ)) Moreover, we have σ(t ) = σ(mg) = supp(g (µ)), which is the essential range of g. The only difference with the bounded case is therefore that the spectrum, and the function g, are not necessarily bounded anymore. 8

9 The only difference with the bounded case is therefore that the spectrum, and the function g, are not 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 R λ = (T λ) 1 is a compact operator then we say that T has compact resolvent. If an unbounded operator T on a Hilbert space has compact resolvent then the spectrum of T is discrete. proof: By the Spectral Theorem for Compact Operators, all nonzero numbers in the spectrum of a compact operator R λ0 are eigenvalues of finite multiplicity which have no nonzero accumulation point. Also it can be shown that Lemma: For λ 0 ρ(t ) and λ λ 0, λ is an eigenvalue of T (λ λ 0 ) 1 is an eigenvalue for R λ0 (T ) (with the same multiplicities). By this lemma, T has purely discrete spectrum. If T is also self-adjoint the spectrum is real (in R) and there exists and orthonormal basis of eigenvectors. (The eigenvalues have no accumulation point though.) Let R λ = (T λ) 1 be the resolvent for λ C when this inverse exists as a linear operator defined at least on a dense subset of V. Theorem: Let T be self-adjoint and densely defined. For λ C R the operator R λ is everywhere defined on V, and the operator norm is estimated by R λ 1 Imλ For T positive, λ / [0, ), R λ is everywhere defined on V and the operator norm is estimated by R λ { 1 Imλ Re(λ) 0 1 λ Re(λ) 0 Theorem: (Hlibert) For points λ, µ off the real line, or, for T positive, for λ, µ off [0, ) R λ R µ = (λ µ)r λ R µ For the operator-norm topology, λ R λ 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. T v, v 0 for all v D). Friedrichs Extension: A densely defined, positive, symmetric operator has a selfadjoint extension. 9

10 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 S as follows: Without loss of generality, consider a densely-defined, symmetric operator S with domain D S with Sv, v v, v for all v D S. (Note that any semibounded operator can be exhibited this way by multiplying by a constant and adding or subtracting a constant.) Define the inner product, 1 on D S by v, w 1 := Sv, w for v, w D S and let V 1 be the completion of D S with respect to the metric induced by, 1. Since v, v 1 v, v, the inclusion map D S V extends to a continuous map V 1 V. Furthermore, since D S is dense in V, we have that V 1 is also dense in V. For w V, the functional v v, w is a continuous linear functional on V 1 with norm sup v, w sup v w sup v 1 w = w. v 1 1 v 1 1 v 1 1 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 v, w ; explicitly, w 1 w. The map A : V V 1 defined by w w is linear. The inverse of A will be a self-adjoint extension S of S. This is the Friedrichs extension. We now show that S 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 and so A is continuous. Also observe that for w D S and all v V 1, v, w 1 = v, Sw = v, A(Sw ) 1 so A(Sw ) = w for each w D S. Hence AV V 1 contains the domain D S of S. We also see that A is injective since ker A = 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 S of A is defined on D S = AV V 1. Hence S is injective and is surjective for D S V. Now to show that S is an extension of S, it remains to show that A( Sw) = A(Sw) for w D S. For v, w D S D S, v, S = v, A( Sw) 1 = v, w 1 = Sv, w = v, Sw. 10

11 Since D s is dense in V we have that Sw = Sw. We also must show that S is symmetric. First note that A is symmetric: for w = Aw AV since v, Aw 1 = v, w we have v, w 1 = v, Sw and Av, w = Av, SAw = Av, Aw 1 which is symmetric in v and w. Thus since D S = AV and and so S is symmetric. SAv, Aw = v, Aw = Av, w = Av, Aw 1 = Av, SAw Furthermore, this extension S remains semibounded i.e. Sv, v v, v for all v = Aw D S = AV since Sv, v = SAw, v = w, v = Aw, v 1 = v, v 1 v, v. It remains to show that S is self-adjoint. Note that any proper extension T S is not injective since S surjects to V. So if S were a proper extension of S there would be v D S so that for all w D S, 0 = S v, w = v, Sw = v, Sw. Since S surjects to V, there is a w D S such that Sw = v. Hence v = 0 and S cannot be a proper extension of S. Thus S 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 D S self-adjoint extension with the same lower bound. has a positive This extension has useful properties of particular interest to our project. An alternative characterization of the extension makes this more clear. Assume that V has a C-conjugate-linear complex conjugation v v c with the properties: (v c ) c = v and v c, w c = v, w. Further let S commute with conjugation so that (Sv) c = S(v c ). Let V 1 be the dual of V 1 so that V 1 V V 1. Note that given this small specification, there is an alternate characterization of the Friedrichs extension. To specify it, define a continuous, complex-linear map S # : V 1 V 1 by for v, w V 1. (S # v)(w) = v, w c 1 Theorem 2. Let X = {v V 1 S # v V }. Then the Friedrichs extension of S is S = S # X with domain D S = X. 11

12 Proof. Let T = S # X. Let : V V 1 be the inverse of S defined by Av, w 1 = v, w for all w V 1 and v V from the Riesz-Fischer Theorem. Then for v V and w V 1. Also, for v X and w V 1. This T = A 1 = S. T Av, w = Av, w 1 = v, w AT v, w 1 = T v, w = v, w 1 Extensions of Restrictions Using the latter characterization of the Friedrichs extension we will now explain how the construction of the extension works with the case of restricted operators. Assume that S and the related terms are defined as above. Let Θ D S be a S-stable subspace. Let the orthogonal complement to Θ in V 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 that D T = D S ker Θ. Note that for v D T and θ Θ, T v, θ = Sv, θ = v, Sθ { v, θ θ Θ} = {0} and so T (D T ) ker Θ. Furthermore since T is a restriction of S, symmetry and T v, v v, v are inherited from S. In contrast with these inherited properties, it is nontrivial to give a simple condition to ensure that D T is dense in ker Θ and that the V 1 -closure of D T is V 1 ker Θ. This delicacy is demonstrated in Lax and Phillips (?,?). For cut-off above height a > 1, an argument using the geometry of the fundamental domain for Γ shows that Θ V 1 is dense in V. 1 For this reason we will assume D T = D S 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 for v, w V 1. (S # v)(w) := v, w c 1, Theorem 3. Let Θ # be the V 1 -completion of Θ. The Friedrichs extension T of T has domain D T = {v W 1 S # v V + Θ # } and is characterized by for v D T and w ker Θ. T v = w S # v w + Θ # 1 For a < 1, there are serious complications so we do not address this case. 12

13 Proof. Define T # : W 1 W 1 by (T # v)(w) := v, w c 1 we can then define the the domain of the Friedrichs extension T as D T = {w W 1 T # w ker Θ} so that T = T # D T. 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 D T = {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 S # V 1 V V 1 j 1 j T W 1 ker Θ W 1 T # 13

14 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 no others? 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 will address it later. However question 1 is also a question related to spectral theory in that Maass forms are eigenfunctions for the Maass operators on C (H) Note that it is easily verified that R k = iy x + y y + k 2 = (z z) z + k 2 L k = iy x + y y k 2 = (z z) z k 2. k = L k+2 R k k 2 ( 1 + k ) = R k 2 L k + k ( 1 k ) The Laplacian k is a symmetric operator on L 2 (H) with domain C o (H). Recall that the spectral theorem only applies to self-adjoint operators; however, even for symmetric operators we can conclude that if f is an L 2 eigenfunction so that k f = λf then λ is real and eigenvectors corresponding 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 compact or 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 symmetric operator k on L 2 (Γ\H, χ, k). 14

15 Here L 2 (Γ\H, χ, k) is the Hilbert space completion of the space of smooth functions on H such that for γ Γ, C (Γ\H, χ, k). χ(γ)f(z) = ( ) k cz + d f cz + d ( ) az + b cz + d Spectral Problem (version 2): Determine the decomposition of the Hilbert space L 2 (Γ\G, χ) into irreducible subspaces. Where L 2 (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 spaces L 2 (Γ\H, χ, k) and L 2 (Γ\G, χ) related to one another. Define the right regular representation ρ : G End ( L 2 (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 have that L 2 (Γ\G, χ) = k Z L 2 (Γ\G, χ, k) where L 2 (Γ\G, χ, k) is the subspace consisting of functions such that ρ(κ θ )f = e ikθ f (or f(gκ θ ) = e ikθ f(g)). There is a Hilbert space isomorphism given by (σ k f)(g) = (f k g)(i) for g G. σ k : L 2 (Γ\H, χ, k) L 2 (Γ\G, χ, k) In order to understand the relationship between these Hilbert Spaces, we need to define operators on L 2 (Γ\G, χ) to play the role of k, R k and L k. We define the following differential operators on G ( R = e 2iθ iy x + y y + 1 ), 2i θ ( L = e 2iθ iy x + y y 1 ) 2i θ and the Laplace-Beltrami operator ( ) = y 2 2 x y 2 2 y 2 x θ. The direct relationship between these two versions of the spectral problem can best be seen in the following theorem: 15

16 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) ɛ where ɛ = 0 or 1. (a) The space H = L 2 (Γ\G, χ) decomposes into a Hilbert space direct sum of irreducible representations. The spaces L 2 (Γ\H, χ, k) each decompose into a Hilbert space direct sum of eigenspaces for k. (b) Let H k be any subspace of H that is invariant under the action of G. Then H k is also invariant under the action of. Conversely, let λ R and let H λ be the λ-eigenspace of on H, then H λ is G-invariant. (c) If H k is an irreducible subspace of H then acts as a scalar on H k. The eigenvector λ of on H k depends only on the isomorphism class of H k. If λ is such an eigenvalue than λ is a real number. Either λ 0 is ɛ = 0 and λ 1/4 if ɛ = 1, otherwise λ = k(1 k ) where 1 k Z and k ɛmod (d) There is only one finite-dimensional representation of G that can occur in H the trivial representation: If χ = 1 then the constant function spans a one-dimensional irreducible subspace of H on which space the eigenvalue λ of equals 0. All other irreducible constituents of H are infinite dimensional. (e) Assume λ is not of the form k(1 k ) where 1 k Z and ɛmod 2: 2 2 There exists a unique irreducible representation P(λ, ɛ) of G depending only on λ (and not Γ) such that if H k is an infinite-dimensional irreducible subrepresentation of H with eigenvalue λ then H k = P(λ, ɛ). Let k ɛmod 2 be an integer. The multiplicity of the representation P(λ, ɛ) is equal to the multiplicity of the eigenvalue in L 2 (Γ\H, χ, k). (f) Assume that λ is of the form k(1 k ) where 1 k Z and k ɛmod There exists two irreducible representations, D + (k) and D (k) of G depending only on k such that if H k is an infinite-dimensional irreducible representation of H with λ then either H k = D + (k) or H K = D + (k). The representations D ± (k) have the same multiplicity in H; this multiplicity equals the dimension of the space of holomorphic modular forms of weight k that satisfy f(γz) = χ(γ)(cz + d) k f(z) for γ Γ. Part (a) above unifies these two versions. However understanding the link between the spectral and representation theory can best be seen in the proof of this result. Proof of (a): Let χ be a unitary character of Γ and H = L 2 (Γ\G, χ) Let φ C c (G) and define ρ(φ) : H H 16

17 by ρ(φ)f(g) = representation. G f(gh)φ(h) dh. Thus we have ρ(φ)f = the following are some properties of such representations: G φ(h)ρ(h)f dh where ρ is a right-regular 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 L 2 (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 and if φ(g) = φ(g 1 ) then π(φ) is self-adjoint. (c) For κ θ K, if φ(κ θ g) = e ikθ φ(g) then ρ(φ) maps the Hilbert space L 2 (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 let 0 f H. Let ɛ > 0 be given. Then there exists φ C c (G) such that π(φ) is self-adjoint and π(φ)f f < ɛ. In particular if ɛ < f this implies that π(φ)f 0. Moreover if π(κ θ ) = e ikθ f for all κ θ K, then we may choose φ so that φ(κ θ g) = θ(gκ θ ) = e ikθ φ(g). If ρ is unitary and it is reducible (it has a a proper nonzero closed subspace V ) then the orthogonal complement U of V is also a nonzero invariant closed subspace and H = U V. On the other hand, if it 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 L 2 (Γ\G, χ) which is closed under the action of G. Then H has decomposition as a Hilbert space direct sum H = k H k where ρ(κ θ )f = e 2πikθ f for f H k. Let k be such that H k 0 then has a nonzero eigenvector in H k C (Γ\G, χ). To prove this result we use the fact that ρ(φ) is a compact operator from Proposition 5 and the Spectral Theorem for compact operatiors. From this we get the first part of (a): Theorem 8. The space L 2 (Γ\G, χ) decomposes into Hilbert space direct sum of subspaces that are invariant and irreducible under the right regular representation ρ. Similarly we get 17

18 Theorem 9. Let ξ be a character of Cc (K\G/K, σ) and let H(ξ) be the space of f L 2 (Γ\G, χ, k) that satisfys π(φ)f = ξ(φ)f for all φ Cc K\G/K, σ). The space H(ξ) is a finite dimensional subspace of C (Γ\G, χ, k). If ξ and η are two distinct characters of Cc (K\G/K, σ) then H(ξ) and H(η) are orthogonal subspaces. Furthermore, L 2 (Γ\G, χ, k) = H(ξ) ξ where H(ξ) 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 dimenstional vector spaces H(ξ) so each of these decomposes into a direct sum of -eigenspaces. Corollary 10. The space L 2 (Γ\H, χ, k) decomposes into a Hilbert space direct sum of eigenspaces of k. Both these results that make up part (a) follow from Proposition 5. Spectral Decomposition for Γ = SL 2 (Z) In this case, Γ\H is non-compact. Thus there is no reason to suspect that the spectrum is purely discrete.we will now exhibit the computation of this spectral decomposition for L 2 (Γ\H) with respect to = y 2 ( 2 x + 2 y). Let N be the upper-triangular unipotent matrices in G = SL 2 (R), A the diagonal matrices, A + the diagonal matrices with positive diagonal entries, P = N A the parabolic subgroup of uppertriangular matrices, P + = NA +, Γ = SL 2 (Z) and Γ = P + Γ = N Γ. For simplicity, normalize the total measure of K to 1 rather than 2π. Pseudo-Eisenstein Series Pseudo-Eisenstein series are solutions to the adjunction problem: given ϕ Cc find Ψ ϕ Cc such that c P f, ϕ N\G = f, Ψ ϕ Γ\G for f on Γ\G and f, F Γ\G = f F. Γ\G (N\G), we want to We can compute the canonical expression for Ψ ϕ 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 Γ only by ±1 2 which act trivially on H = G/K.) c P f, ϕ N\H = = N\H Γ \H ( dx dy c P f(z)ϕ(im(z)) = y 2 N\H dx dy f(z)ϕ(im(z)) = y 2 Γ\H γ P Γ\Γ 18 N Γ\N ) f(nz) dn ϕ(im(z)) f(γz)ϕ(im(γz)) dx dy y 2 dx dy y 2

19 = Γ\H f(z) γ P Γ\Γ ϕ(im(γz)) Thus we define the pseudo-eisenstein series as Ψ ϕ (z) = dx dy y 2 γ P Γ\Γ = f, Ψ ϕ Γ\H ϕ(im(γz)). Note that the pseudo- Eisenstein series is absolutely and uniformly convergent for z C where C is a compact subset of G. Furthermore, Ψ ϕ C c (Γ\G). Now, it is a corollary of the above characterization of psuedo-eisenstein series that the square integrable cuspforms are the orthogonal complement of the (closed) subspace of L 2 (Γ\H) spanned by pseudo-eisenstein series with ϕ C 0 (N\H) = C 0 (0, ). Thus we have L 2 (Γ\H) = L 2 cusp(γ\h) L 2 p-eis(γ\h). Decomposition of Pseudo-Eisenstein Series We further decompose L 2 (Γ\H) by examining the pseudo-eisenstein series Ψ ϕ. The spectral decomposition of the data ϕ induces a spectral decomposition for Ψ ϕ. Identifying N\H = N\G/K = A +, Mellin inversion gives ϕ(im z) = 1 σ+i Mϕ(s)(Im z) s ds 2πi σ i 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 ( ) f(t)e itξ dt e iξx dξ. 2π Fourier transforms on R put into multiplicative coordinates are Mellin transforms: For ϕ Cc (0, ), take f(x) = ϕ(e x ). Let y = e x and r = e t (the exponential in the implied inner integral) and rewrite Fourier inversion as f(x) = ϕ(y) = 1 ( ) iξ dr ϕ(r)r y iξ dξ 2π 0 r since dt = dr r. Note that this integral converges as a C -function-valued function. The Fourier transform (inner integral) in these coordinates is Mellin transform. For compactly supported ϕ, the integral definition extends to all s C as Mϕ(s) = s dr ϕ(r)r. Mellin 0 r inversion is ϕ(y) = 1 Mϕ(iξ)y iξ dξ. 2π 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 19 Mϕ(s)y s ds

20 since dξ = i ds. For f Cc (R), ˆf(ξ) converges nicely for all complex values of ξ so it extends to an entire function in ξ of rapid decay on horizontal lines (Payley-Wiener Theorem). This extension property applies to ϕ allowing us to move the contour as above. Thus the pseudo-eisenstein series is Ψ ϕ (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 absolutely convergent and the two integrals cannot be interchanged. For σ > 1 the double integral is absolutely convergent and (using Fubini) we have, Ψ ϕ (z) = 1 2πi σ+i σ i for E s (z) = γ Γ \Γ (Im(γz))s the Eisenstein series. Mϕ(s) E s (z) ds Eisenstein Series We have a decomposition of Ψ ϕ in terms of ϕ. We now want to rewrite this piece of the decomposition so as to refer only to Ψ ϕ not ϕ. Note that E s has meromorphic continuation on the entire complex plane. Thus we can move the line of integration to the left and choose σ = 1/2 to achieve Ψ ϕ = 1 2πi 1/2+i 1/2 i Mϕ(s)E s ds + s 0 res s=s0 (Mϕ(s)E s ). As with the pseudo-eisenstein series, the Eisenstein series E s fits into an adjunction relation for f on Γ\H. Notice that y s, c P f A + = Thus E s, f Γ\H = = = 0 N\H Γ \H Γ\H E s, f Γ\H = y s, c P f Γ \H ( ) s dx dy s dx dy c P f(z) y = f(nz) dn y y 2 N\H Γ \N y 2 = s dx dy f(z) y Γ\H γ P Γ\Γ f(z) γ P Γ\Γ c P f(iy)y s dy y 2 = y 2 = P Γ\H s dx dy f(z) y y 2 s dx dy f(γz) Im(γz) y 2 s dx dy Im(γz) = E y 2 s, f Γ\H. 0 (1 s) dy c P f(iy)y y = M(c P f)(1 s). 20

21 On the other hand, since c P E s = y s + c s y 1 s ξ(2s 1) for c s =, we have ξ(2s) E s, Ψ ϕ Γ\H = c P E s, ϕ Γ \H = y s + c s y 1 s, ϕ Γ \H = = So we have the identity, 0 0 (y s + c s y 1 s ) ϕ(y) dy y 2 (y (1 s) + c s y s ) ϕ(y) dy y = Mϕ(1 s) + c smϕ(s). M(c P Ψ ϕ )(s) = E 1 s, Ψ ϕ Γ\H = Mϕ(s) + c 1 s Mϕ(1 s). Using this and returning to our equation for Ψ ϕ above, we get Ψ ϕ s 0 res s=s0 (Mϕ(s)E s ) = 1 2πi = 1 2πi = 1 2πi 1/2+i 1/2 i0 1/2+i 1/2 i0 = 1 2πi 1/2+i 1/2 i Mϕ(s)E s ds Mϕ(s)E s + Mϕ(1 s)e 1 s ds Mϕ(s)E s + c 1 s Mϕ(1 s)e s ds 1/2+i 1/2 i0 Mc p Ψ ϕ (s)e s ds from the functional equation E 1 s = c 1 s E s and our identity above = 1 4πi 1/2+i 1/2 i Ψ ϕ, E s Γ\H E s ds. Residue Finally, let us examine the residue. For Γ = SL 2 (Z), the only pole of E s in the half plane Re(s) 1/2 is at s 0 = 1. This pole is simple and the residue is a constant function. Thus we can compute the residue as follows: res s=s0 (Mϕ(s)E s ) = Mϕ(1) res s=1 E s where = N\H Mϕ(1) = Γ \N s dy ϕ(y)y y = 0 ϕ(y) dy y = 2 N\H ( dx dy ϕ(im(nz)) dn = ϕ(im(nz)) y 2 N\H dx dy = ϕ(imz) y 2 Γ \H dx dy ϕ(imz) Γ \N y 2 ) dx dy 1 dn y 2 21

22 = ϕ(imz) Γ\H γ Γ \Γ since the volume of Γ \N is 1 and ϕ is left N-invariant dx dy dx dy = Ψ y 2 ϕ (z) = Ψ y 2 ϕ, 1 Γ\H. Γ\H This we have that L 2 (Γ\H) = L 2 cusp(γ\h) C L 2 Eis(Γ\H). 22