Spectral Theory

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Spectral Theory

Michael E. Taylor

Contents 0. Introduction 1. The spectral theorem 2. Self-adjoint diﬀerential operators 3. Heat asymptotics and eigenvalue asymptotics 4. The Laplace operator on Sn 5. The Laplace operator on hyperbolic space 6. The harmonic oscillator 7. The quantum Coulomb problem 8. The Laplace operator on cones

Introduction

This chapter is devoted to the spectral theory of self-adjoint, differential operators. We cover a number of different topics, beginning in §1 with a proof of the spectral theorem. It was an arbitrary choice to put that material here, rather than in Appendix A, on functional analysis. The main motivation for putting it here is to begin a line of reasoning that will be continued in subsequent sections, using the great power of studying unitary groups as a tool in spectral theory. After we show how easily this study leads to a proof of the spectral theorem in §1, in later sections we use it in various ways: as a tool to establish self-adjointness, as a tool for obtaining specific formulas, including basic identities among special functions, and in other capacities. Sections 2 and 3 deal with some general questions in spectral theory, such as when does a differential operator define a self-adjoint operator, when does it have a compact resolvent, and what asymptotic properties does its spectrum have? We tackle the latter question, for the Laplace operator ∆, by examining the asymptotic behavior of the trace of the solution operator et∆ for the heat equation, showing that (0.1) Tr et∆ = (4πt)−n/2 vol Ω + o(t−n/2), t ↘ 0, when Ω is either a compact Riemannian manifold or a bounded domain in Rn (and has the Dirichlet boundary condition). Using techniques developed in §13 of Chapter 7, we could extend (0.1) to general compact Riemannian 2 manifolds with smooth boundary and to other boundary conditions, such as the Neumann boundary condition. We use instead a different method here in §3, one that works without any regularity hypotheses on ∂Ω. In such generality, (0.1) does not necessarily hold for the Neumann boundary problem. The study of (0.1) and refinements got a big push from [Kac]. As pursued in [MS], it led to developments that we will discuss in Chapter 10. The problem of to what extent a Riemannian manifold is determined by the spectrum of its Laplace operator has led to much work, which we do not include here. Some is discussed in [Ber], [Br], [BGM], and [Cha]. We mention particularly some distinct regions in R2 whose Laplace operators have the same spectra, given in [GWW]. We have not included general results on the spectral behavior of ∆ obtained via geometrical optics and its refinement, the theory of Fourier integral operators. Results of this nature can be found in Volume 3 of [Ho], in [Shu], and in Chapter 12 of [T1]. Sections 4 through 7 are devoted to specific examples. In §4 we study the Laplace operator on the unit spheres Sn. We specify precisely the spectrum of ∆ and discuss explicit formulas for certain functions of ∆, particularly ( ) K 1/2 (0.2) A−1 sin tA, A = −∆ + (n − 1)2 . 4 with K = 1, the sectional curvature of Sn. In §5 we obtain an explicit formula for (0.2), with K = −1, on hyperbolic space. In §6 we study the spectral theory of the harmonic oscillator (0.3) H = −∆ + |x|2. We obtain an explicit formula for e−tH , an analogue of which will be useful in Chapter 10. In §7 we study the operator (0.4) H = −∆ − K|x|−1 on R3, obtaining in particular all the eigenvalues of this operator. This operator arises in the simplest quantum mechanical model of the hydrogen atom. In §8 we study the Laplace operator on a cone. Studies done in these sections bring in a number of special functions, including Legendre functions, Bessel functions, and hypergeometric functions. We have included two auxiliary problem sets, one on confluent hypergeometric functions and one on hypergeometric functions.

1. The spectral theorem

Appendix A contains a proof of the spectral theorem for a compact, self- adjoint operator A on a Hilbert space H. In that case, H has an orthonormal basis {uj} such that Auj = λjuj, λj being real numbers having only 0 as an accumulation point. The vectors uj are eigenvectors. 1. The spectral theorem 3

A general bounded, self-adjoint operator A may not have any eigenvectors, and the statement of the spectral theorem is somewhat more subtle. The following is a useful version.

Theorem 1.1. If A is a bounded, self-adjoint operator on a separable Hilbert space H, then there is a σ-compact space Ω, a Borel measure µ, a unitary map (1.1) W : L2(Ω, dµ) −→ H, and a real-valued function a ∈ L∞(Ω, dµ) such that (1.2) W −1AW f(x) = a(x)f(x), f ∈ L2(Ω, dµ).

Note that when A is compact, the eigenvector decomposition above yields (1.1) and (1.2) with (Ω, µ) a purely atomic measure space. Later in this section we will extend Theorem 1.1 to the case of unbounded, self-adjoint operators. In order to prove Theorem 1.1, we will work with the operators (1.3) U(t) = eitA, deﬁned by the power-series expansion ∞ ∑ (it)n (1.4) eitA = An. n! n=0 This is a special case of a construction made in §4 of Chapter 1. U(t) is uniquely characterized as the solution to the diﬀerential equation d (1.5) U(t) = iAU(t),U(0) = I. dt We have the group property (1.6) U(s + t) = U(s)U(t), which follows since both sides satisfy the ODE (d/ds)Z(s) = iAZ(s),Z(0) = U(t). If A = A∗, then applying the adjoint to (1.4) gives (1.7) U(t)∗ = U(−t), which is the inverse of U(t) in view of (1.6). Thus {U(t): t ∈ R} is a group of unitary operators. For a given v ∈ H, let Hv be the closed linear span of {U(t)v : t ∈ R}; we say Hv is the cyclic space generated by v. We say v is a cyclic vector ⊥ for H if H = Hv. If Hv is not all of H, note that Hv is invariant under ⊥ ⊂ ⊥ U(t), that is, U(t)Hv Hv for all t, since for a linear subspace V of H, generally (1.8) U(t)V ⊂ V =⇒ U(t)∗V ⊥ ⊂ V ⊥. 4

Using this observation, we can prove the next result.

Lemma 1.2. If U(t) is a unitary group on a separable Hilbert space H, then H is an orthogonal direct sum of cyclic subspaces.

Proof. Let {wj} be a countable, dense subset of H. Take v1 = w1 and ̸ ≥ H1 = Hv1 . If H1 = H, let v2 be the ﬁrst nonzero element P1wj, j 2, ⊥ where P1 is the orthogonal projection of H onto H1 , and let H2 = Hv2 . Continue.

In view of this, Theorem 1.1 is a consequence of the following:

Proposition 1.3. If U(t) is a strongly continuous, unitary group on H, having a cyclic vector v, then we can take Ω = R, and there exists a positive Borel measure µ on R and a unitary map W : L2(R, dµ) → H such that

(1.9) W −1U(t)W f(x) = eitxf(x), f ∈ L2(R, dµ).

The measure µ on R will be the Fourier transform

(1.10) µ = ζ,ˆ where

(1.11) ζ(t) = (2π)−1/2 (eitAv, v).

It is not clear a priori that (1.10) defines a measure; since ζ ∈ L∞(R), we see that µ is a tempered distribution. We will show that µ is indeed a positive measure during the course of our argument. As for the map W , we first define

(1.12) W : S(R) −→ H, where S(R) is the Schwartz space of rapidly decreasing functions, by

(1.13) W (f) = f(A)v, where we deﬁne the operator f(A) by the formula

∫ ∞ (1.14) f(A) = (2π)−1/2 fˆ(t)eitA dt. −∞ 1. The spectral theorem 5

The reason for this notation will become apparent shortly; see (1.20). Mak- ing use of (1.10), we have ∫ ∫ ( ) ( ) f(A)v, g(A)v = (2π)−1 fˆ(s)eisAv ds, gˆ(t)eitAv dt ∫∫ ( ) = (2π)−1 fˆ(s)gˆ(t) ei(s−t)Av, v ds dt ∫∫ (1.15) = (2π)−1/2 fˆ(s)gˆ(σ − s)ζ(σ) ds dσ d = ⟨(fg), ζ⟩ = ⟨fg, µ⟩. Now, if g = f, the left side of (1.15) is ∥f(A)v∥2, which is ≥ 0. Hence ⟨ ⟩ (1.16) |f|2, µ ≥ 0, for all f ∈ S(R). With this, we can establish:

Lemma 1.4. The tempered distribution µ, deﬁned by (1.10)–(1.11), is a positive measure on R. √ Proof. Apply (1.16) with f = Fs,σ, where

−1/2 −(τ−σ)2/4s Fs,σ(τ) = (4πs) e , s > 0, σ ∈ R. Note that this is a fundamental solution to the heat equation. For each s > 0,Fs,0 ∗ µ is a positive function. We saw in Chapter 3 that Fs,0 ∗ µ converges to µ in S′(R) as s → 0, so this implies that µ is a positive measure.

Now we can ﬁnish the proof of Proposition 1.3. From (1.15) we see that W has a unique continuous extension (1.17) W : L2(R, dµ) −→ H, and W is an isometry. Since v is assumed to be cyclic, the range of W must be dense in H, so W must be unitary. From (1.14) it follows that if f ∈ S(R), then isA isτ (1.18) e f(A) = fs(A), with fs(τ) = e f(τ). Hence, for f ∈ S(R), −1 isA −1 isτ (1.19) W e W f = W fs(A)v = e f(τ). Since S(R) is dense in L2(R, dµ), this gives (1.9). Thus the spectral theorem for bounded, self-adjoint operators is proved. Given (1.9), we have from (1.14) that (1.20) W −1f(A)W g(x) = f(x)g(x), f ∈ S(R), g ∈ L2(R, dµ), 6 which justiﬁes the notation f(A) in (1.14). Note that (1.9) implies

(1.21) W −1AW f(x) = x f(x), f ∈ L2(R, dµ), since (d/dt)U(t) = iAU(t). The essential supremum of x on (R, µ) is equal to ∥A∥. Thus µ has compact support in R if A is bounded. If a self-adjoint operator A has the representation (1.21), one says A has simple spectrum. It follows from Proposition 1.3 that A has simple spectrum if and only if it has a cyclic vector. One can generalize the results above to a k-tuple of commuting, bounded, self-adjoint operators A = (A1,...,Ak). In that case, for t = (t1, . . . , tk) ∈ Rk, set

it·A (1.22) U(t) = e , t · A = t1A1 + ··· + tkAk.

The hypothesis that the Aj all commute implies U(t) = U1(t1) ··· Uk(tk), isAj where Uj(s) = e .U(t) in (1.22) continues to satisfy the properties (1.6) and (1.7); we have a k-parameter unitary group. As above, for v ∈ H, k we set Hv equal to the closed linear span of {U(t)v : t ∈ R }, and we say v is a cyclic vector provided Hv = H. Lemma 1.2 goes through in this case. Furthermore, for f ∈ S(Rk), we can deﬁne ∫ (1.23) f(A) = (2π)−k/2 fˆ(t)eit·A dt, and if H has a cyclic vector v, the proof of Proposition 1.3 generalizes, giving a unitary map W : L2(Rk, dµ) → H such that

(1.24) W −1U(t)W f(x) = eit·xf(x), f ∈ L2(Rk, dµ), t ∈ Rk.

Therefore, Theorem 1.1 has the following extension.

Proposition 1.5. If A = (A1,...,Ak) is a k-tuple of commuting, bounded, self-adjoint operators on H, there is a measure space (Ω, µ), a unitary map 2 ∞ W : L (Ω, dµ) → H, and real-valued aj ∈ L (Ω, dµ) such that

−1 2 (1.25) W AjW f(x) = aj(x)f(x), f ∈ L (Ω, dµ), 1 ≤ j ≤ k.

A bounded operator B ∈ L(H) is said to be normal provided B and B∗ commute. Equivalently, if we set 1( ) 1 ( ) (1.26) A = B + B∗ ,A = B − B∗ , 1 2 2 2i then B = A1 + iA2, and (A1,A2) is a 2-tuple of commuting, self-adjoint operators. Applying Proposition 1.5 and setting b(x) = a1(x) + ia2(x), we have: 1. The spectral theorem 7

Corollary 1.6. If B ∈ L(H) is a normal operator, there is a unitary map W : L2(Ω, dµ) → H and a (complex-valued) b ∈ L∞(Ω, dµ) such that (1.27) W −1BW f(x) = b(x)f(x), f ∈ L2(Ω, dµ).

In particular, Corollary 1.6 holds when B = U is unitary. We next extend the spectral theorem to an unbounded, self-adjoint operator A on a Hilbert space H, whose domain D(A) is a dense linear subspace of H. This extension, due to von Neumann, uses von Neumann’s unitary trick, described in (8.18)–(8.19) of Appendix A. We recall that, for such A, the following three properties hold: A i : D(A) −→ H bijectively, (1.28) U = (A − i)(A + i)−1 is unitary on H, A = i(I + U)(I − U)−1, where the range of I − U = 2i(A + i)−1 is D(A). Applying Corollary 1.6 to B = U, we have the following theorem:

Theorem 1.7. If A is an unbounded, self-adjoint operator on a separable Hilbert space H, there is a measure space (Ω, µ), a unitary map W : L2(Ω, dµ) → H, and a real-valued measurable function a on Ω such that (1.29) W −1AW f(x) = a(x)f(x), W f ∈ D(A). In this situation, given f ∈ L2(Ω, dµ), W f belongs to D(A) if and only if the right side of (1.29) belongs to L2(Ω, dµ).

The formula (1.29) is called the “spectral representation” of a self-adjoint operator A. Using it, we can extend the functional calculus defined by (1.14) as follows. For a Borel function f : R → C, define f(A) by (1.30) W −1f(A)W g(x) = f(a(x))g(x). If f is a bounded Borel function, this is defined for all g ∈ L2(Ω, dµ) and provides a bounded operator f(A) on H. More generally, ( ) { } (1.31) D f(A) = W g ∈ H : g ∈ L2(Ω, dµ) and f(a(x))g ∈ L2(Ω, dµ) . In particular, we can define eitA, for unbounded, self-adjoint A, by W −1eitAW g = eita(x)g(x). Then eitA is a strongly continuous unitary group, and we have the following result, known as Stone’s theorem (stated as Proposition 9.5 in Appendix A):

Proposition 1.8. If A is self-adjoint, then iA generates a strongly continuous, unitary group, U(t) = eitA. 8

Note that Lemma 1.2 and Proposition 1.3 are proved for a strongly continuous, unitary group U(t) = eitA, without the hypothesis that A be bounded. This yields the following analogue of (1.2):

(1.32) W −1U(t)W f(x) = eita(x)f(x), f ∈ L2(Ω, dµ), for this more general class of unitary groups. Sometimes a direct construction, such as by PDE methods, of U(t) is fairly easy. In such a case, the use of U(t) can be a more convenient tool than the unitary trick involving (1.28). We say a self-adjoint operator A is positive, A ≥ 0, provided (Au, u) ≥ 0, for all u ∈ D(A). In terms of the spectral representation, this says we have (1.29) with a(x) ≥ 0 on Ω. In such a case, e−tA is bounded for t ≥ 0, even for complex t with Re t ≥ 0, and also deﬁnes a strongly continuous semigroup. This proves Proposition 9.4 of Appendix A. Given a self-adjoint operator A and a Borel set S ⊂ R, deﬁne P (S) = χS(A), that is, using (1.29),

−1 2 (1.33) W P (S)W g = χS(a(x))g(x), g ∈ L (Ω, dµ), where χS is the characteristic∪ function of S. Then each P (S) is an orthogonal projection. Also, if S = j≥1 Sj is a countable union of disjoint Borel sets Sj, then, for each u ∈ H, ∑n (1.34) lim P (Sj)u = P (S)u, n→∞ j=1 with convergence in the H-norm. This is equivalent to the statement that ∑n → 2 ∈ 2 χSj (a(x))g χS(a(x))g in L -norm, for each g L (Ω, dµ), j=1 which in turn follows from Lebesgue’s dominated convergence theorem. By (1.34), P (·) is a strongly countably additive, projection-valued measure. Then (1.30) yields ∫ (1.35) f(A) = f(λ) P (dλ).

P (·) is called the spectral measure of A. One useful formula for the spectral measure is given in terms of the jump −1 of the resolvent Rλ = (λ−A) , across the real axis. We have the following.

Proposition 1.9. For bounded, continuous f : R → C, ∫ ∞ [ ] 1 (1.36) f(A)u = lim f(λ) (λ − iε − A)−1 − (λ + iε − A)−1 u dλ. ε↘0 2πi −∞ 1. The spectral theorem 9

Proof. Since W −1f(A)W is multiplication by f(a(x)), (1.36) follows from the fact that ∫ ∞ 1 εf(λ) −→ (1.37) 2 2 dλ f(a(x)), π −∞ (λ − a(x)) + ε pointwise and boundedly, as ε ↘ 0.

An important class of operators f(A) are the fractional powers f(A) = Aα, α ∈ (0, ∞), deﬁned by (1.30)–(1.31), with f(λ) = λα, provided A ≥ 0. Note that if g ∈ C([0, ∞)) satisﬁes g(0) = 1, g(λ) = O(λ−α) as λ → ∞, then, for u ∈ H,

α α (1.38) u ∈ D(A ) ⇐⇒ ∥A g(εA)u∥H is bounded, for ε ∈ (0, 1], as follows easily from the characterization (1.31) and Fatou’s lemma. We note that Proposition 2.2 of Chapter 4 applies to D(Aα), describing it as an interpolation space. We particularly want to identify D(A1/2), when A is a positive, self- adjoint operator on a Hilbert space H constructed by the Friedrichs method, as described in Proposition 8.7 of Appendix A. Recall that this arises as follows. One has a Hilbert space H1, a continuous injection J : H1 → H with dense range, and one deﬁnes A by ( ) (1.39) A(Ju), Jv H = (u, v)H1 , with { D(A) = Ju ∈ JH1 ⊂ H : v 7→ (u, v)H is (1.40) 1 } continuous in Jv, in the H-norm . We establish the following.

Proposition 1.10. If A is obtained by the Friedrichs extension method (1.39)–(1.40), then

1/2 (1.41) D(A ) = J(H1) ⊂ H.

Proof. D(A1/2) consists of elements of H that are limits of sequences in 1/2 D(A), in the norm ∥A u∥H +∥u∥H . As shown in the proof of Proposition 8.7 in Appendix A, D(A) = R(JJ ∗). Now (1.42) ∥A1/2JJ ∗f∥2 = (AJJ ∗f, JJ ∗f) = ∥J ∗f∥2 . H H H1 ∗ 1/2 Thus a sequence (JJ fn) converges in the D(A )-norm (to an element g) ∗ if and only if (J fn) converges in the H1-norm (to an element u), in which ∗ case g = Ju. Since J : H → H1 has dense range, precisely all u ∈ H1 ∗ arise as limits of such (J fn), so the proposition is proved. 10

Exercises

1. The deﬁnition (1.33) of the spectral measure P (·) of a self-adjoint operator A depends a priori on a choice of the spectral representation of A. Show that any two spectral representations of A yield the same spectral measure. (Hint: For f ∈ S(R), f(A) is well deﬁned by (1.14), or alternatively by (1.36).)

2. Self-adjoint diﬀerential operators

In this section we present some examples of diﬀerential operators on a manifold Ω which, with appropriately speciﬁed domains, give unbounded, self-adjoint operators on L2(Ω, dV ), dV typically being the volume element determined by a Riemannian metric on Ω. We begin with self-adjoint operators arising from the Laplacian, making use of material developed in Chapter 5. Let Ω be a smooth, compact Riemannian manifold with boundary, or more generally the closure of an open subset Ω of a compact manifold M without boundary. Then, as shown in Chapter 5, − 1 −→ 1 ∗ (2.1) I ∆ : H0 (Ω) H0 (Ω) is bijective, with inverse we denote T ; if we restrict T to L2(Ω), (2.2) T : L2(Ω) −→ L2(Ω) is compact and self-adjoint. Denote by R(T ) the image of L2(Ω) under T . We can apply Proposition 8.2 of Appendix A to deduce the following.

Proposition 2.1. If Ω is a region in a compact Riemannian manifold M, 2 D R ⊂ 1 then ∆ is self-adjoint on L (Ω), with domain (∆) = (T ) H0 (Ω) described above.

For a further description of D(∆), note that D { ∈ 1 ∈ 2 } (2.3) (∆) = u H0 (Ω) : ∆u L (Ω) . If ∂Ω is smooth, we can apply the regularity theory of Chapter 5 to obtain D 1 ∩ 2 (2.4) (∆) = H0 (Ω) H (Ω). Instead of relying on Proposition 8.2, we could use the Friedrichs construction, given in Proposition 8.7 of Appendix A. This construction can be applied more generally. Let Ω be any Riemannian manifold, with Laplace 1 ∞ operator ∆. We can deﬁne H0 (Ω) to be the closure of C0 (Ω) in the space { ∈ 2 ∈ 2 1 } 1 u L (Ω) : du L (Ω, Λ ) . The inner product on H0 (Ω) is

(2.5) (u, v)1 = (u, v)L2 + (du, dv)L2 . 2. Self-adjoint diﬀerential operators 11

1 → 2 We have a natural inclusion H0 (Ω) , L (Ω), and the Friedrichs method gives a self-adjoint operator A on L2(Ω) such that ∈ D ∈ 1 (2.6) (Au, v)L2 = (u, v)1, for u (A), v H0 (Ω), with { D ∈ 1 7→ 1 → C (A) = u H0 (Ω) : v (u, v)1 extends from H0 (Ω) to a (2.7) } continuous linear functional L2(Ω) → C , that is, { D ∈ 1 ∃ ∈ 2 (A) = u H0 (Ω) : f L (Ω) such that (2.8) } ∀ ∈ 1 (u, v)1 = (f, v)L2 , v H0 (Ω) . ∈ ∞ − D Integrating (2.5) by parts for v C0 (Ω), we see that A = I ∆ on (A), so we have a self-adjoint extension of ∆ in this general setting, with domain again described by (2.3). The process above gives one self-adjoint extension of ∆, initially deﬁned ∞ on C0 (Ω). It is not always the only self-adjoint extension. For example, suppose Ω is compact with smooth boundary; consider H1(Ω), with inner product (2.5), and apply the Friedrichs extension procedure. Again we have a self-adjoint operator A, extending I − ∆, with (2.8) replaced by { D(A) = u ∈ H1(Ω) : ∃f ∈ L2(Ω) such that (2.9) } 1 (u, v)1 = (f, v)L2 , ∀v ∈ H (Ω) . In this case, Proposition 7.2 of Chapter 5 yields the following.

Proposition 2.2. If Ω is a smooth, compact manifold with boundary and ∆ the self-adjoint extension just described, then 2 (2.10) D(∆) = {u ∈ H (Ω) : ∂ν u = 0 on ∂Ω}.

In case (2.10), we say D(∆) is given by the Neumann boundary condition, while in case (2.4) we say D(∆) is given by the Dirichlet boundary condition. In both cases covered by Propositions 2.1 and 2.2, (−∆)1/2 is deﬁned as a self-adjoint operator. We can specify its domain using Proposition 1.10, obtaining the next result:

D − 1/2 1 Proposition 2.3. In case (2.3), (( ∆) ) = H0 (Ω); in case (2.10), D((−∆)1/2) = H1(Ω).

∞ Though ∆ on C0 (Ω) has several self-adjoint extensions when Ω has a boundary, it has only one when Ω is a complete Riemannian manifold. This is a classical result, due to Roelcke; we present an elegant proof due 12 to Chernoﬀ [Chn]. When an unbounded operator A0 on a Hilbert space H, with domain D0, has exactly one self-adjoint extension, namely the closure of A0, we say A0 is essentially self-adjoint on D0.

Proposition 2.4. If Ω is a complete Riemannian manifold, then ∆ is ∞ essentially self-adjoint on C0 (Ω). Thus the self-adjoint extension with ∞ domain given by (2.3) is the closure of ∆ on C0 (Ω).

Proof. We will obtain this as a consequence of Proposition 9.6 of Appendix A, which states the following. Let U(t) = eitA be a unitary group on a Hilbert space H which leaves invariant a dense linear space D; U(t)D ⊂ D. If A is an extension of A0 and A0 : D → D, then A0 and all its powers are essentially self-adjoint on D. In this case, U(t) will be the solution operator for a wave equation, and we will exploit ﬁnite propagation speed. Set ( ) 0 I ∞ ∞ (2.11) iA = , D(A ) = C (Ω) ⊕ C (Ω). 0 ∆ − I 0 0 0 0 The group U(t) will be the solution operator for the wave equation ( ) ( ) f u(t) (2.12) U(t) = , g ut(t) where u(t, x) is determined by ∂2u − (∆ − 1)u = 0; u(0, x) = f, u (0, x) = g. ∂t2 t It was shown in §2 of Chapter 6 that U(t) is a unitary group on H = 1 ⊕ 2 H0 (Ω) L (Ω); its generator is an extension of (2.11), and ﬁnite propaga- ∞ ⊕ ∞ tion speed implies that U(t) preserves C0 (Ω) C0 (Ω) for all t, provided k Ω is complete. Thus each A0 is essentially self-adjoint on this space. Since ( ) ∆ − I 0 (2.13) −A2 = , 0 0 ∆ − I

2k we have the proof of Proposition 2.3. Considering A0 , we deduce further- k ∞ more that each power ∆ is essentially self-adjoint on C0 (Ω), when Ω is complete.

∞ Though ∆ is not essentially self-adjoint on C0 (Ω) when Ω is compact, we do have such results as the following:

Proposition 2.5. If Ω is a smooth, compact manifold with boundary, then ∆ is essentially self-adjoint on (2.14) {u ∈ C∞(Ω) : u = 0 on ∂Ω}, 2. Self-adjoint diﬀerential operators 13 its closure having domain described by (2.3). Also, ∆ is essentially self- adjoint on ∞ (2.15) {u ∈ C (Ω) : ∂ν u = 0 on ∂Ω}, its closure having domain described by (2.10).

Proof. It suﬃces to note the simple facts that the closure of (2.14) in H2(Ω) is (2.3) and the closure of (2.15) in H2(Ω) is (2.10).

We note that when Ω is a smooth, compact Riemannian manifold with boundary, and D(∆) is given by the Dirichlet boundary condition, then ∩∞ (2.16) D(∆j) = {u ∈ C∞(Ω) : ∆ku = 0 on ∂Ω, k = 0, 1, 2,... }, j=1 and when D(∆) is given by the Neumann boundary condition, then ∩∞ j ∞ k (2.17) D(∆ ) = {u ∈ C (Ω) : ∂ν (∆ u) = 0 on ∂Ω, k ≥ 0}. j=1 We now derive a result that to some degree amalgamates Propositions 2.4 and 2.5. Let Ω be a smooth Riemannian manifold with boundary, and set ∞ { ∈ ∞ } (2.18) Cc (Ω) = u C (Ω) : supp u is compact in Ω ; we do not require elements of this space to vanish on ∂Ω. We say that Ω is complete if it is complete as a metric space.

Proposition 2.6. If Ω is a smooth Riemannian manifold with boundary which is complete, then ∆ is essentially self-adjoint on { ∈ ∞ } (2.19) u Cc (Ω) : u = 0 on ∂Ω . In this case, the closure has domain given by (2.3).

Proof. Consider the following linear subspace of (2.19): D { ∈ ∞ j } (2.20) 0 = u Cc (Ω) : ∆ u = 0 on ∂Ω for j = 0, 1, 2,... . 1 ⊕ 2 Let U(t) be the unitary group on H0 (Ω) L (Ω) deﬁned as in (2.12), with u also satisfying the Dirichlet boundary condition, u(t, x) = 0 for x ∈ ∂Ω. Then, by ﬁnite propagaton speed, U(t) preserves D0 ⊕ D0, provided Ω is complete, so as in the proof of Proposition 2.4, we deduce that ∆ is essentially self-adjoint on D0; a fortiori it is essentially self-adjoint on the space (2.19). 14

By similar reasoning, we can show that if Ω is complete, then ∆ is essentially self-adjoint on { ∈ ∞ } (2.21) u Cc (Ω) : ∂ν u = 0 on ∂Ω . The results of this section so far have involved only the Laplace operator ∆. It is also of interest to look at Schr¨odingeroperators, of the form −∆ + V , where the “potential” V (x) is a real-valued function. In this section we will restrict attention to the case V ∈ C∞(Ω) and we will also suppose that V is bounded from below. By adding a constant to −∆ + V , we may as well suppose (2.22) V (x) ≥ 1 on Ω.

1 ∞ We can deﬁne a Hilbert space HV 0(Ω) to be the closure of C0 (Ω) in the space 1 { ∈ 2 ∈ 2 1 1/2 ∈ 2 } (2.23) HV (Ω) = u L (Ω) : du L (Ω, Λ ),V u L (Ω) , with inner product

(2.24) (u, v)1,V = (du, dv)L2 + (V u, v)L2 . 1 → 2 Then there is a natural injection HV 0(Ω) , L (Ω), and the Friedrichs extension method provides a self-adjoint operator A. Integration by parts ∈ ∞ − in (2.24), with v C0 (Ω), shows that such A is an extension of ∆ + V . For this self-adjoint extension, we have D 1/2 1 (2.25) (A ) = HV 0(Ω). In case Ω is a smooth, compact Riemannian manifold with boundary and ∈ ∞ 1 1 V C (Ω), one clearly has HV 0(Ω) = H0 (Ω). In such a case, we have an immediate extension of Proposition 2.1, including the characterization (2.4) of D(−∆+V ). One can also easily extend Proposition 2.2 to −∆+V in this case. It is of substantial interest that Proposition 2.4 also extends, as follows:

Proposition 2.7. If Ω is a complete Riemannian manifold and the function V ∈ C∞(Ω) satisﬁes V ≥ 1, then −∆ + V is essentially self-adjoint on ∞ C0 (Ω).

Proof. We can modify the proof of Proposition 2.4; replace ∆−1 by ∆−V 1 ⊕ 2 in (2.11) and (2.12). Then U(t) gives a unitary group on HV 0(Ω) L (Ω), and the ﬁnite propagation speed argument given there goes through. As − ∞ before, all powers of ∆ + V are essentially self-adjoint on C0 (Ω).

Some important classes of potentials V have singularities and are not bounded below. In §7 we return to this, in a study of the quantum mechanical Coulomb problem. Exercises 15

We record here an important compactness property when V ∈ C∞(Ω) tends to +∞ at inﬁnity in Ω.

Proposition 2.8. If the Friedrichs extension method described above is used to construct the self-adjoint operator −∆ + V for smooth V ≥ 1, as above, and if V → +∞ at inﬁnity (i.e., for each N < ∞, ΩN = {x ∈ Ω: V (x) ≤ N} is compact), then −∆ + V has compact resolvent.

1 → Proof. Given (2.25), it suﬃces to prove that the injection HV 0(Ω) 2 L (Ω) is compact, under the current hypotheses on V . Indeed, if {un} is 1 { } { 1/2 } bounded in HV 0(Ω), with inner product (2.24), then dun and V un are bounded in L2(Ω). By Rellich’s theorem and a diagonal argument, one { } 2 has a subsequence unk whose restriction to each ΩN converges in L (ΩN )- 1/2 2 norm. The boundedness of {V un} in L (Ω) then gives convergence of this subsequence in L2(Ω)-norm, proving the proposition.

The following result extends Proposition 2.4 of Chapter 5.

Proposition 2.9. Assume that Ω is connected and that either Ω is compact or V → +∞ at inﬁnity. Denote by λ0 the ﬁrst eigenvalue of −∆ + V . Then a λ0-eigenfunction of −∆ + V is nowhere vanishing on Ω. Conse- quently, the λ0-eigenspace is one-dimensional.

Proof. Let u be a λ0-eigenfunction of −∆ + V . As in the proof of Propo- sition 2.4 of Chapter 5, we can write u = u+ + u−, where u+(x) = u(x) for u(x) > 0 and u−(x) = u(x) for u(x) ≤ 0, and the variational characterization of the λ0-eigenspace implies that u are eigenfunctions (if nonzero). Hence it suﬃces to prove that if u is a λ0-eigenfunction and u(x) ≥ 0 on Ω, then u(x) > 0 on Ω. To this end, write ∫

t(∆−V +λ0) u(x) = e u(x) = pt(x, y)u(y) dV (y). Ω

We see that this forces pt(x, y) = 0 for all t > 0, when x ∈ Σ = {x : u(x) = 0}, y ∈ O, O = {x : u(x) > 0}, since pt(x, y) is smooth and ≥ 0. The strong maximum principle (see Exercise 3 in §1 of Chapter 6) forces Σ = ∅.

Exercises

1 ∞ 1. Let HV (Ω) be the space (2.23). If V ≥ 1 belongs to C (Ω), show that the Friedrichs extension also deﬁnes a self-adjoint operator A1, equal to −∆ + V ∞ D 1/2 1 on C0 (Ω), such that (A1 ) = HV (Ω). If Ω is complete, show that this 16

operator coincides with the extension A deﬁned in (2.25). Conclude that, in 1 1 this case, HV (Ω) = HV 0(Ω). 2. Let Ω be complete, V ≥ 1 smooth. Show that if A is the self-adjoint extension of −∆ + V described in Proposition 2.7, then (2.26) D(A) = {u ∈ L2(Ω) : −∆u + V u ∈ L2(Ω)}, where a priori we regard −∆u + V u as an element of D′(Ω). 2 2 1 2 1 1/2 3. Deﬁne T : L (Ω) → L (Ω, Λ )⊕L (Ω) by D(T ) = HV 0(Ω), T u = (du, V u). Show that

∗ 2 1 2 2 1/2 2 (2.27) D(T ) = {(v1, v2) ∈ L (Ω, Λ ) ⊕ L (Ω) : δv1 ∈ L (Ω),V v2 ∈ L (Ω)}. Show that T ∗T is equal to the self-adjoint extension A of −∆ + V defined by the Friedrichs extension, as in (2.25). 4. If Ω is complete, show that the self-adjoint extension A of −∆ + V in Propo- sition 2.7 satisfies (2.28) D(A) = {u ∈ L2(Ω) : ∆u ∈ L2(Ω), V u ∈ L2(Ω)}. (Hint: Denote the right side by W. Use Exercise 3 and A = T ∗T to show that D(A) ⊂ W. Use Exercise 2 to show that W ⊂ D(A).) 5. Let D = −i d/dx on C∞(R), and let B(x) ∈ C∞(R) be real-valued. Define the unbounded operator L on L2(R) by (2.29) D(L) = {u ∈ L2(R): Du ∈ L2(R), Bu ∈ L2(R)}, Lu = Du + iB(x)u. Show that L∗ = D − iB, with ∗ D(L ) = {u ∈ L2(R): Du − iBu ∈ L2(R)}.

∗ 2 2 ′ Deduce that A0 = L L is given by A0u = D u + B u + B (x)u on 2 2 2 2 2 ′ 2 D(A0) = {u ∈ L (R): Du ∈ L (R), Bu ∈ L (R),D u+B u+B (x)u ∈ L (R)}. 6. Suppose that |B′(x)| ≤ ϑB(x)2 + C, for some ϑ < 1, C < ∞. Show that

2 2 2 ′ 2 D(A0) = {u ∈ L (R): D u + (B + B )u ∈ L (R)}. (Hint: Apply Exercise 2 to D2 + (B2 + B′) = A, and show that D(A1/2) is given by D(L), deﬁned in (2.29).) 7. In the setting of Exercise 6, show that the operator L of Exercise 5 is closed. (Hint: L∗L = A is a self-adjoint extension of D2 + (B2 + B′). Show that D 1/2 D D D ∗ D (A1 ) = (L) and also = (L).) Also show that (L ) = (L) in this case.

3. Heat asymptotics and eigenvalue asymptotics

In this section we will study the asymptotic behavior of the eigenvalues of the Laplace operator on a compact Riemannian manifold, with or without boundary. We begin with the boundaryless case. Let M be a compact Riemannian manifold without boundary, of dimension n. In §13 of Chapter 7 we have 3. Heat asymptotics and eigenvalue asymptotics 17 constructed a parametrix for the solution operator et∆ of the heat equation

( ) ∂ (3.1) − ∆ u = 0 on R+ × M, u(0, x) = f(x) ∂t and deduced that ( t∆ −n/2 2 (3.2) Tr e ∼ t a0 + a1t + a2t + ··· ), t ↘ 0, for certain constants aj. In particular, −n/2 (3.3) a0 = (4π) vol M. This is related to the behavior of the eigenvalues of ∆ as follows. Let the eigenvalues of −∆ be 0 = λ0 ≤ λ1 ≤ λ2 ≤ · · · ↗ ∞. Then (3.2) is equivalent to ∑∞ ( −tλj −n/2 2 (3.4) e ∼ t a0 + a1t + a2t + ··· ), t ↘ 0. j=0 We will relate this to the counting function

(3.5) N(λ) ∼ #{λj : λj ≤ λ}, establishing the following:

Theorem 3.1. The eigenvalues {λj} of −∆ on the compact Riemannian manifold M have the behavior (3.6) N(λ) ∼ C(M)λn/2, λ → +∞, with a vol M (3.7) C(M) = 0 = . n n n/2 Γ( 2 + 1) Γ( 2 + 1)(4π)

That (3.6) follows from (3.4) is a special case of a result known as Kara- mata’s Tauberian theorem. The following neat proof follows one in [Si3]. Let µ be( a positive) (locally ﬁnite) Borel measure on [0, ∞); in the example above, µ [0, λ] = N(λ).

Proposition 3.2. If µ is a positive measure on [0, ∞), α ∈ (0, ∞), then ∫ ∞ (3.8) e−tλ dµ(λ) ∼ at−α, t ↘ 0, 0 implies ∫ x (3.9) dµ(λ) ∼ bxα, x ↗ ∞, 0 18 with a (3.10) b = . Γ(α + 1)

α −1 Proof. Let dµt be the measure given by µt(A) = t µ(t A), and let α−1 dν(λ) = αλ dλ; then νt = ν. The hypothesis (3.8) becomes ∫ ∫ −λ −λ (3.11) lim e dµt(λ) = b e dν(λ), t→0 with b given by (3.10), and the desired conclusion becomes ∫ ∫

(3.12) lim χ(λ) dµt(λ) = b χ(λ) dν(λ) t→0 when χ is the characteristic function of [0, 1]. It would suffice to show that (3.12) holds for all continuous χ(λ) with compact support in [0, ∞). −λ From (3.11) we deduce that the measures e dµt are uniformly bounded, for t ∈ (0, 1]. Thus (3.12) follows if we can establish ∫ ∫ −λ −λ (3.13) lim g(λ)e dµt(λ) = b g(λ)e dν(λ), t→0 + for g in a dense subspace of C0(R ), the space of continuous functions on [0, ∞) that vanish at infinity. Indeed, the hypothesis implies that (3.13) holds for all g in A, the space of finite, linear combinations of functions of −sλ λ ∈ [0, ∞) of the form φs(λ) = e , s ∈ (0, ∞), as can be seen by dilating the variables in (3.11). By the Stone-Weierstrass theorem, A is dense in + Co(R ), so the proof is complete.

We next want to establish similar results on N(λ) for the Laplace operator ∆ on a compact manifold Ω with boundary, with Dirichlet boundary condition. At the end of §13 in Chapter 7 we sketched a construction of a parametrix for et∆ in this case which, when carried out, would yield an expansion ( ) t∆ −n/2 1/2 (3.14) Tr e ∼ t a0 + a1/2t + a1t + ··· , t ↘ 0, extending (3.2). However, we will be able to verify the hypothesis of Propo- sition 3.2 with less eﬀort than it would take to carry out the details of this construction, and for a much larger class of domains. For simplicity, we will restrict attention to bounded domains in Rn and to the ﬂat Laplacian, though more general cases can be handled similarly. Now, let Ω be an arbitrary bounded, open subset of Rn, with closure Ω. The Laplace operator on Ω, with Dirichlet boundary condition, was studied in §5 of Chapter 5.

Lemma 3.3. For any bounded, open Ω ⊂ Rn, ∆ with Dirichlet boundary condition, et∆ is trace class for all t > 0. 3. Heat asymptotics and eigenvalue asymptotics 19

Proof. Let Ω ⊂ B, a large open ball. Then the variational characterization of eigenvalues shows that the eigenvalues λj(Ω) of −∆ on Ω and λj(B) of L = −∆ on B, both arranged in increasing order, have the relation

(3.15) λj(Ω) ≥ λj(B). But we know that e−tL has integral kernel in C∞(B × B) for each t > 0, − − hence is trace class. Since e tλj (Ω) ≤ e tλj (B), this implies that the positive self-adjoint operator et∆ is also trace class.

Limiting arguments, which we leave to the reader, allow one to show that, even in this generality, if H(t, x, y) ∈ C∞(Ω × Ω) is, for ﬁxed t > 0, the integral kernel of et∆ on L2(Ω), then ∫ (3.16) Tr et∆ = H(t, x, x) dx.

Ω See Exercises 1–5 at the end of this section.

Proposition 3.4. If Ω is a bounded, open subset of Rn and ∆ has the Dirichlet boundary condition, then (3.17) Tr et∆ ∼ (4πt)−n/2 vol Ω, t ↘ 0.

−n/2 |x−y|2/4t Proof. We will compare H(t, x, y) with H0(t, x, y) = (4πt) e , the free-space heat kernel. Let E(t, x, y) = H0(t, x, y) − H(t, x, y). Then, for ﬁxed y ∈ Ω, ∂E (3.18) − ∆ E = 0 on R+ × Ω,E(0, x, y) = 0, ∂t x and

(3.19) E(t, x, y) = H0(t, x, y), for x ∈ ∂Ω. To make simple sense out of (3.19), one might assume that every point of ∂Ω is a regular boundary point, though a further limiting argument can be made to lift such a restriction. The maximum principle for solutions to the heat equation implies −n/2 −δ(y)2/4s (3.20) 0 ≤ E(t, x, y) ≤ sup H0(s, z, y) ≤ sup (4πs) e , 0≤s≤t,z∈Ω 0≤s≤t where δ(y) = dist(y, ∂Ω). Now the function −n/2 −δ2/4s ψδ(s) = (4πs) e on (0, ∞) vanishes at 0 and ∞ and has a unique maximum at s = δ2/2n; 2 −n we have ψδ(δ /2n) = Cnδ . Thus ( ) −n/2 −δ(y)2/4t −n (3.21) 0 ≤ E(t, x, y) ≤ max (4πt) e ,Cnδ(y) . 20

Of course, E(t, x, y) ≤ H0(t, x, y) also. Now, let O ⊂⊂ Ω be such that vol(Ω \O) < ε. For t small enough, ≤ 2 O namely for t δ1/2n where δ1 = dist( , ∂Ω), we have 2 (3.22) 0 ≤ E(t, x, x) ≤ (4πt)−n/2e−δ(x) /4t, x ∈ O, while of course 0 ≤ E(t, x, x) ≤ (4πt)−n/2, for x ∈ Ω \O. Therefore, ∫ (3.23) lim sup (4πt)n/2 E(t, x, x) dx ≤ ε, t→0 Ω so ∫ vol Ω − ε ≤ lim inf (4πt)n/2 H(t, x, x) dx t→0 Ω (3.24) ∫ ≤ lim sup (4πt)n/2 H(t, x, x) dx ≤ vol Ω. t→0 Ω As ε can be taken arbitrarily small, we have a proof of (3.17).

Corollary 3.5. If Ω is a bounded, open subset of Rn,N(λ) the counting function of the eigenvalues of −∆, with Dirichlet boundary condition, then (3.6) holds.

Note that if Oε is the set of points in Ω of distance ≥ ε from ∂Ω and we deﬁne v(ε) = vol(Ω \Oε), then the estimate (3.24) can be given the more precise reformulation √ (3.25) 0 ≤ vol Ω − (4πt)n/2 Tr et∆ ≤ ω( 2nt), where ∫ ∞ 2 2 (3.26) ω(ε) = v(ε) + e−ns /2ε dv(s). ε The fact that such a crude argument works, and works so generally, is a special property of the Dirichlet problem. If one uses the Neumann boundary condition, then for bounded Ω ⊂ Rn with nasty boundary, ∆ need not even have compact resolvent. However, Theorem 3.1 does extend to the Neumann boundary condition provided ∂Ω is smooth. One can do this via the sort of parametrix for boundary problems sketched in §13 of Chapter 7. We now look at the heat kernel H(t, x, y) on the complement of a smooth, bounded region K ⊂ Rn. We impose the Dirichlet boundary condition on ∂K. As before, 0 ≤ H(t, x, y) ≤ H0(t, x, y), where H0(t, x, y) is the free- space heat kernel. We can extend H(t, x, y) to be Lipschitz continuous on (0, ∞)×Rn ×Rn by setting H(t, x, y) = 0 when either x ∈ K or y ∈ K. We now estimate E(t, x, y) = H0(t, x, y) − H(t, x, y). Suppose K is contained in the open ball of radius R centered at the origin. 3. Heat asymptotics and eigenvalue asymptotics 21

Lemma 3.6. For |x − y| ≤ |y| − R, we have 2 (3.27) E(t, x, y) ≤ Ct−1/2e−(|y|−R) /4t.

Proof. With y ∈ Ω = Rn \ K, write ∫ ∞ 2 (3.28) H(t, x, y) = (4πt)−1/2 e−s /4t cos sΛ ds, −∞ √ where Λ = −∆ and ∆ is the Laplace operator on Ω, with the Dirichlet boundary√ condition. We have a similar formula for H0(t, x, y), using instead Λ0 = −∆0, with ∆0 the free-space Laplacian. Now, by ﬁnite propagation speed,

cos sΛ δy(x) = cos sΛ0 δy(x), provided |s| ≤ d = dist(y, ∂K), and |x − y| ≤ d. Thus, as long as |x − y| ≤ d, we have ∫ [ ] −1/2 −s2/4t (3.29) E(t, x, y) = (4πt) e cos sΛ0 δy(x) − cos sΛ δy(x) ds.

|s|≥d Then the estimate (3.27) follows easily, along the same lines as estimates on heat kernels discussed in Chapter 6, §2.

When we combine (3.27) with the obvious inequality −n/2 −|x−y|2/4t (3.30) 0 ≤ E(t, x, y) ≤ H0(t, x, y) = (4πt) e , we see that, for each t > 0,E(t, x, y) is rapidly decreasing as |x|+|y| → ∞. Using this and appropriate estimates on derivatives, we can show that E(t, x, y) is the integral kernel of a trace class operator on L2(Rn). We can write ∫ ( ) (3.31) Tr et∆0 − et∆P = E(t, x, x) dx,

Rn where P is the projection of L2(Rn) onto L2(Ω) deﬁned by restriction to Ω. Now, as t ↘ 0, (4πt)n/2E(t, x, x) approaches 1 on K and 0 on Rn \ K. Together with the estimates (3.27) and (3.30), this implies ∫ (3.32) (4πt)n/2 E(t, x, x) dx −→ vol K,

Rn as t ↘ 0. This establishes the following:

Proposition 3.7. If K is a closed, bounded set in Rn, ∆ is the Laplacian 2 n on L (R \K), with Dirichlet boundary condition, and ∆0 is the Laplacian 22 on L2(Rn), then et∆0 − et∆P is trace class for each t > 0 and ( ) − (3.33) Tr et∆0 − et∆P ∼ (4πt) n/2 vol K, as t ↘ 0.

This result will be of use in the study of scattering by an obstacle K, in Chapter 9. It is also valid for the Neumann boundary condition if ∂K is smooth.

Exercises

n In Exercises 1–4, let Ω ⊂ R be a bounded, open set and let Oj be open with smooth boundary such that

O1 ⊂⊂ O2 ⊂⊂ · · · ⊂⊂ Oj ⊂⊂ · · · ↗ Ω.

Let Lj be −∆ on Lj , with Dirichlet boundary condition; the corresponding operator on Ω is simply denoted −∆. 1. Using material developed in §5 of Chapter 5, show that, for any t > 0, f ∈ L2(Ω),

−tLj t∆ 2 e Pj f −→ e f strongly in L (Ω),

as j → ∞, where Pj is multiplication by the characteristic function of Oj . Don’t peek at Lemma 3.4 in Chapter 11! 2. If λν (Oj ) are the eigenvalues of Lj , arranged in increasing order for each j, show that, for each ν,

λν (Oj ) ↘ λν (Ω), as j → ∞. 3. Show that, for each t > 0, − Tr e tLj ↗ Tr et∆.

+ + 4. Let Hj (t, x, y) be the heat kernel on R ×Oj ×Oj . Extend Hj to R ×Ω×Ω so as to vanish if x or y belongs to Ω\Oj . Show that, for each x ∈ Ω, y ∈ Ω, t > 0,

Hj (t, x, y) ↗ H(t, x, y), as j → ∞. Deduce that, for each t > 0, ∫ ∫

Hj (t, x, x) dx ↗ H(t, x, x) dx.

Oj Ω 5. Using Exercises 1–4, give a detailed proof of (3.16) for general bounded Ω ⊂ Rn. 6. Give an example of a bounded, open, connected set Ω ⊂ R2 (with rough boundary) such that ∆, with Neumann boundary condition, does not have compact resolvent. 4. The Laplace operator on Sn 23

4. The Laplace operator on Sn

n A key tool in the analysis of the Laplace operator ∆S on S is the formula for the Laplace operator on Rn+1 in polar coordinates: ∂2 n ∂ 1 (4.1) ∆ = + + ∆ . ∂r2 r ∂r r2 S

In fact, this formula is simultaneously the main source of interest in ∆S and the best source of information about it. To begin, we consider the Dirichlet problem for the unit ball in Euclidean space, B = {x ∈ Rn+1 : |x| < 1}: (4.2) ∆u = 0 in B, u = f on Sn = ∂B, given f ∈ D′(Sn). In Chapter 5 we obtained the Poisson integral formula for the solution: ∫ 1 − |x|2 f(y) (4.3) u(x) = n+1 dS(y), An |x − y| Sn n where An is the volume of S . Equivalently, if we set x = rω with r = |x|, ω ∈ Sn, ∫ 1 − r2 f(ω′) (4.4) u(rω) = dS(ω′). ′ 2 (n+1)/2 An (1 − 2rω · ω + r ) Sn Now we can derive an alternative formula for the solution of (4.2) if we use (4.1) and regard ∆u = 0 as an operator-valued ODE in r; it is an Euler equation, with solution (4.5) u(rω) = rA−(n−1)/2f(ω), r ≤ 1, where A is an operator on D′(Sn), deﬁned by ( ) (n − 1)2 1/2 (4.6) A = −∆ + . S 4 If we set r = e−t and compare (4.5) and (4.4), we obtain a formula for the semigroup e−tA as follows. Let θ(ω, ω′) denote the geodesic distance on Sn from ω to ω′, so cos θ(ω, ω′) = ω · ω′. We can rewrite (4.4) as 2 u(rω) = sinh(log r−1) r−(n−1)/2 A n ∫ (4.7) f(ω′) · dS(ω′). [ ]−(n+1)/2 2 cosh(log r−1) − 2 cos θ(ω, ω′) Sn In other words, by (4.5), ∫ 2 f(ω′) (4.8) e−tAf(ω) = sinh t dS(ω′). ( )(n+1)/2 An 2 cosh t − 2 cos θ(ω, ω′) Sn 24

Identifying an operator on D′(Sn) with its Schwartz kernel in D′(Sn ×Sn), we write 2 sinh t (4.9) e−tA = , t > 0. (n+1)/2 An (2 cosh t − 2 cos θ) Note that the integration of (4.9) from t to ∞ produces the formula −1 −tA −(n−1)/2 (4.10) A e = 2Cn(2 cosh t − 2 cos θ) , t > 0, provided n ≥ 2, where ( − ) 1 1 −(n+1)/2 n 1 Cn = = π Γ . (n − 1)An 4 2 With the exact formula (4.9) for the semigroup e−tA, we can proceed to give formulas for fundamental solutions to various important PDE, particularly ∂2u (4.11) − Lu = 0 (wave equation) ∂t2 and ∂u (4.12) − Lu = 0 (heat equation), ∂t where (n − 1)2 (4.13) L = ∆ − = −A2. S 4

If we prescribe Cauchy data u(0) = f, ut(0) = g for (4.11), the solution is (4.14) u(t) = (cos tA)f + A−1(sin tA)g. Assume n ≥ 2. We obtain formulas for these terms by analytic continuation of the formulas (4.9) and (4.10) to Re t > 0 and then passing to the limit t ∈ iR. This is parallel to the derivation of the fundamental solution to the wave equation on Euclidean space in §5 of Chapter 3. We have [ ] −1 (it−ε)A −(n−1)/2 A e = −2Cn 2 cosh(it − ε) − 2 cos θ ,

(4.15) 2 [ ]−(n+1)/2 e(it−ε)A = sinh(it − ε) 2 cosh(it − ε) − 2 cos θ . An Letting ε ↘ 0, we have A−1 sin tA = (4.16) −(n−1)/2 lim −2Cn Im (2 cosh ε cos t − 2i sinh ε sin t − 2 cos θ) ε↘0 and cos tA = (4.17) −2 lim Im(sin t)(2 cosh ε cos t − 2i sinh ε sin t − 2 cos θ)−(n+1)/2. ε↘0 An 4. The Laplace operator on Sn 25

For example, on S2 we have, for 0 ≤ t ≤ π,

A−1 sin tA = −2C (2 cos θ − 2 cos t)−1/2, θ < |t|, (4.18) 2 0, θ > |t|, with an analogous expression for general t, determined by the identity (4.19) A−1 sin(t + 2π)A = −A−1 sin tA on D′(S2k), plus the fact that sin tA is odd in t. The last line on the right in (4.18) reﬂects the well-known ﬁnite propagation speed for solutions to the hyperbolic equation (4.11). To understand how the sign is determined in (4.19), note that, in (4.15), with ε > 0, for t = 0 we have a real kernel, produced by taking the −(n − 1)/2 = −k + 1/2 power of a positive quantity. As t runs from 0 to 2π, the quantity 2 cosh(it − ε) = 2 cosh ε cos t − 2i sinh ε sin t moves once clockwise around a circle of radius 2(cosh2 ε + sinh2 ε)1/2, centered at 0, so 2 cosh ε cos t − 2i sinh ε sin t − 2 cos θ describes a curve winding once clockwise about the origin in C. Thus taking a half-integral power of this gives one the negative sign in (4.14). On the other hand, when n is odd, the exponents on the right side of (4.15)–(4.17) are integers. Thus (4.20) A−1 sin(t + 2π)A = A−1 sin tA on D′(S2k+1). Also, in this case, the distributional kernel for A−1 sin tA must vanish for |t| ̸= θ. In other words, the kernel is supported on the shell θ = |t|. This is the generalization to spheres of the strict Huygens principle. In case n = 2k + 1 is odd, we obtain from (4.16) and (4.17) that ( ) 1 1 ∂ k−1( ) (4.21) A−1 sin tA f(x) = sin2k−1 s f(x, s) (2k − 1)!! sin s ∂s s=t and ( ) 1 1 ∂ k( ) (4.22) cos tA f(x) = sin s sin2k−1 s f(x, s) , (2k − 1)!! sin s ∂s s=t where, as in (5.66) of Chapter 3, (2k − 1)!! = 3 · 5 ··· (2k − 1) and

n (4.23) f(x, s) = mean value of f on Σs(x) = {y ∈ S : θ(x, y) = |s|}. We can examine general functions of the operator A by the functional calculus ∫ ∞ ∫ ∞ (4.24) g(A) = (2π)−1/2 gˆ(t)eitA dt = (2π)−1/2 gˆ(t) cos tA dt, −∞ −∞ where the last identity holds provided g is an even function. We can rewrite this, using the fact that cos tA has period 2π in t on D′(Sn) for n odd, period 4π for n even. In concert with (4.22), we have the following formula 26 for the Schwartz kernel of g(A) on D′(S2k+1), for g even: ( ) ∞ 1 1 ∂ k ∑ (4.25) g(A) = (2π)−1/2 − gˆ(θ + 2kπ). 2π sin θ ∂θ k=−∞ As an example, we compute the heat kernel on odd-dimensional spheres. 2 2 Take g(λ) = e−tλ . Theng ˆ(s) = (2t)−1/2e−s /4t and ∑ ∑ 2 (4.26) (2π)−1/2 gˆ(s + 2kπ) = (4πt)−1/2 e−(s+2kπ) /4t = ϑ(s, t), k k

2 where ϑ(s, t) is a “theta function.” Thus the kernel of e−tA on S2k+1 is given by ( ) 2 1 1 ∂ k (4.27) e−tA = − ϑ(θ, t). 2π sin θ ∂θ A similar analysis on S2k gives an integral, with the theta function appear- ing in the integrand. The operator A has a compact resolvent on L2(Sn), and hence a discrete set of eigenvalues, corresponding to an orthonormal basis of eigenfunctions. Indeed, the spectrum of A has the following description.

Proposition 4.1. The spectrum of the self-adjoint operator A on L2(Sn) is { } 1 (4.28) spec A = (n − 1) + k : k = 0, 1, 2,... . 2

Proof. Since 0 is the smallest eigenvalue of −∆S, the deﬁnition (4.6) shows that (n − 1)/2 is the smallest eigenvalue of A. Also, (4.20) shows that all eigenvalues of A are integers if n is odd, while (4.19) implies that all eigenvalues of A are (nonintegral) half-integers if n is even. Thus spec A is certainly contained in the right side of (4.28). Another way to see this containment is to note that since the function u(x) given by (4.5) must be smooth at x = 0, the exponent of r in that formula can take only integer values. Let Vk denote the eigenspace of A with eigenvalue νk = (n−1)/2+k. We want to show that Vk ≠ 0 for k = 0, 1, 2,... . Moreover, we want to iden- A−(n−1)/2 tify Vk. Now if f ∈ Vk, it follows that u(x) = u(rω) = r f(ω) = rkf(ω) is a harmonic function deﬁned on all of Rn+1, which, being homogeneous and smooth at x = 0, must be a harmonic polynomial, homogeneous of degree k in x. If Hk denotes the space of harmonic polynomials, homogeneous of degree k, restriction to Sn ⊂ Rn+1 produces an isomorphism:

≈ (4.29) ρ : Hk −→ Vk.

To show that each Vk ≠ 0, it suﬃces to show that each Hk ≠ 0. 4. The Laplace operator on Sn 27

n+1 Indeed, for c = (c1, . . . , cn+1) ∈ C , consider k pc(x) = (c1x1 + ··· + cn+1xn+1) . A computation gives

k−2 ∆pc(x) = k(k − 1)⟨c, c⟩(c1x1 + ··· + ckxk) , ⟨ ⟩ 2 ··· 2 c, c = c1 + + ck.

Hence ∆pc = 0 whenever ⟨c, c⟩ = 0, so the proposition is proved.

2 n We now want to specify the orthogonal projections Ek of L (S ) on Vk. We can attack this via (4.10), which implies ∑∞ − − − − 1 tνk − (n 1)/2 (4.30) νk e Ek(x, y) = 2Cn(2 cosh t 2 cos θ) , k=0 where θ = θ(x, y) is the geodesic distance from x to y in Sn. If we set −t r = e and use νk = (n − 1)/2 + k, we get the generating function identity ∑∞ k −1 − 2 −(n−1)/2 r νk Ek(x, y) = 2Cn(1 2r cos θ + r ) k=0 (4.31) ∑∞ k = r pk(cos θ); k=0 in particular,

(4.32) Ek(x, y) = νk pk(cos θ). These functions are polynomials in cos θ. To see this, set t = cos θ and write ∑∞ − 2 −α α k (4.33) (1 2tr + r ) = Ck (t) r , k=0 α thus deﬁning coeﬃcients Ck (t). To compute these, use ∞ ( ) ∑ j + α − 1 (1 − z)−α = zj, j j=0 with z = r(2t − r), to write the left side of (4.33) as

∞ ( ) ∞ j ( )( ) ∑ α ∑ ∑ j + α − 1 j rj(2t − r)j = (−1)ℓrj+ℓ(2t)j−ℓ j j ℓ j=0 j=0 ℓ=0 ∞ [k/2] ( )( ) ∑ ∑ k − ℓ + α − 1 k − ℓ = (−1)ℓ (2t)k−2ℓrk. k − ℓ ℓ k=0 ℓ=0 28

Hence [k/2] ( )( ) ∑ k − ℓ + α − 1 k − ℓ (4.34) Cα(t) = (−1)ℓ (2t)k−2ℓ. k k − ℓ ℓ ℓ=0 These are called Gegenbauer polynomials. Therefore, we have the following:

2 n Proposition 4.2. The orthogonal projection of L (S ) onto Vk has kernel 1 (4.35) E (x, y) = 2C ν Cα(cos θ), α = (n − 1), k n k k 2 with Cn as in (4.10).

In the special case n = 2, we have C2 = 1/4π, and νk = k + 1/2; hence 2k + 1 2k + 1 (4.36) E (x, y) = C1/2(cos θ) = P (cos θ), k 4π k 4π k 1/2 where Ck (t) = Pk(t) are the Legendre polynomials. The trace of Ek is easily obtained by integrating (4.35) over the diagonal, to yield

− 2ν − (4.37) Tr E = 2C A ν C(n 1)/2(1) = k C(n 1)/2(1). k n n k k n − 1 k Setting t = 1 in (4.33), so (1 − 2r + r2)−α = (1 − r)−2α, we obtain ( ) k + 2α − 1 (4.38) Cα(1) = , e.g., P (1) = 1. k k k

Thus we have the dimensions of the eigenspaces Vk:

n Corollary 4.3. The eigenspace Vk of −∆S on S , with eigenvalue 1 λ = ν2 − (n − 1)2 = k2 + (n − 1)k, k k 4 satisﬁes ( ) ( ) ( ) 2k + n − 1 k + n − 2 k + n − 2 k + n − 1 (4.39) dim V = = + . k n − 1 k k − 1 k 2 In particular, on S we have dim Vk = 2k + 1.

Another natural approach to Ek is via the wave equation. We have ∫ 1 T −iνkt itA Ek = e e dt 2T −T (4.40) ∫ 1 T = cos t(A − νk) dt, 2T −T Exercises 29 where T = π or 2π depending on whether n is odd or even. (In either case, one can take T = 2π.) In the special case of S2, when (4.18) is used, comparison of (4.36) with the formula produced by this method produces the identity ∫ 1 θ cos(k + 1 )t (4.41) P (cos θ) = 2 dt, k 1/2 π −θ (2 cos t − 2 cos θ) for the Legendre polynomials, known as the Mehler-Dirichlet formula.

Exercises

Exercises 1–5 deal with results that follow from symmetries of the sphere. The group SO(n + 1) acts as a group of isometries of Sn ⊂ Rn+1, hence as 2 n a group of unitary operators on L (S ). Each eigenspace Vk of the Laplace operator is preserved by this action. Fix p = (0,..., 0, 1) ∈ Sn, regarded as the “north pole.” The subgroup of SO(n + 1) ﬁxing p is a copy of SO(n). 1. Show that each eigenspace Vk has an element u such that u(p) ≠ 0. Conclude by forming ∫ u(gx) dg

SO(n)

that each eigenspace Vk of ∆S has an element zk ≠ 0 such that zk(x) = zk(gx), for all g ∈ SO(n). Such a function is called a spherical function. 2. Suppose Vk has a proper subspace W invariant under SO(n + 1). (Hence ⊥ W ⊂ Vk is also invariant.) Show that W must contain a nonzero spherical function. 3. Suppose zk and yk are two nonzero spherical functions in Vk. Show that they must be multiples of each other. Hence the unique spherical functions (up to constant multiples) are given by (4.35), with y = p.(Hint: zk and yk are 2 eigenfunctions of −∆S , with eigenvalue λk = k + (n − 1)k. Pick a sequence of surfaces

n n Σj = {x ∈ S : θ(x, p) = εj } ⊂ S , → ̸ | with εj 0, on which zk = αj = 0. With βj = yk Σj , it follows that βj zk − αj yk is an eigenfunction of −∆S that vanishes on Σj . Show that, for j large, this forces βj zk − αj yk to be identically zero.) 4. Using Exercises 2 and 3, show that the action of SO(n+1) on each eigenspace Vk is irreducible, that is, Vk has no proper invariant subspaces. 5. Show that each Vk is equal to the linear span of the set of polynomials of the k form pc(x) = (c1x1 + ··· + cn+1xn+1) , with ⟨c, c⟩ = 0. (Hint: Show that this linear span is invariant under SO(n + 1).) 6. Using (4.9), show that

− 2 sinh t (4.42) Tr e tA = . (2 cosh t − 2)(n+1)/2 30

Find the asymptotic behavior as t ↘ 0. Use Karamata’s Tauberian theorem to determine the asymptotic behavior of the eigenvalues of A, hence of −∆S . Compare this with the general results of §3 and also with the explicit results of Corollary 4.3. 7. Using (4.27), show that, for A on Sn with n = 2k + 1, ( ) k −tA2 A2k+1 1 1 ∂ −θ2/4t ∞ Tr e = √ − e + O(t ) 4πt 2π sin θ ∂θ (4.43) ( ) θ=0 −n/2 −n/2+1 = (4πt) A2k+1 + O t ,

as t ↘ 0. Compare the general results of §3. 8. Show that − − − (4.44) e πi(A (n 1)/2)f(ω) = f(−ω), f ∈ L2(Sn).

n (Hint: Check it for f ∈ Vk, the restriction to S of a homogeneous harmonic polynomial of degree k.)

Exercises 9–13 deal with analysis on Sn when n = 2. When doing them, look for generalizations to other values of n. 9. If Ξ(A) has integral kernel KΞ(x, y), show that when n = 2, ∞ ( ) 1 ∑ 1 (4.45) K (x, y) = (2ℓ + 1)Ξ ℓ + P (cos θ), Ξ 4π 2 ℓ ℓ=0

where cos θ = x · y and Pℓ(t) are the Legendre polynomials. 10. Demonstrate the Rodrigues formula for the Legendre polynomials: ( ) ( ) 1 d k k (4.46) P (t) = t2 − 1 . k 2kk! dt (Hint: Use Cauchy’s formula to get ∫ 1 2 −1/2 −k−1 Pk(t) = (1 − 2zt + z ) z dz 2πi γ from (4.33); then use the change of variable 1 − uz = (1 − 2tz + z2)1/2. Then appeal to Cauchy’s formula again, to analyze∑ the resulting integral.) 2 2 2 11. If f ∈ L (S ) has the form f(x) = g(x · y) = φℓPℓ(x · y), for some y ∈ S , show that ∫ ( ) ∫ 2ℓ + 1 1 1 (4.47) φℓ = f(z)Pℓ(y · z) dS(z) = ℓ + g(t)Pℓ(t) dt. 4π 2 −1 S2 ∫ · (Hint: Use S2 Ek(x, z)Eℓ(z, y) dS(z) = δkℓ Eℓ(x, y).) Conclude that g(x y) is the integral kernel of ψ(A − 1/2), where ∫ 4π 1 (4.48) ψ(ℓ) = φℓ = 2π g(t)Pℓ(t) dt. 2ℓ + 1 −1 This result is known as the Funk-Hecke theorem. Exercises 31

12. Show that, for x, y ∈ S2, ∑∞ ikx·y ℓ (4.49) e = (2ℓ + 1) i jℓ(k) Pℓ(x · y), ℓ=0 where ( ) ( ) ∫ 1/2 ℓ 1 π 1 1 z 2 ℓ izt (4.50) jℓ(z) = Jℓ+1/2(z) = (1 − t ) e dt. 2z 2 ℓ! 2 −1 (Hint: Take g(t) = eikt in Exercise 11, apply the Rodrigues formula, and integrate by parts.) Thus eikx·y is the integral kernel of the operator (1/2)πi(A−1/2) 4π e jA−1/2(k). For another approach, see Exercises 10 and 11 in §9 of Chapter 9. 13. Demonstrate the identities [ ] 2 d (4.51) (1 − t ) + ℓt P (t) = ℓP − (t) dt ℓ ℓ 1 and [ ] d d (4.52) (1 − t2) P (t) + ℓ(ℓ + 1)P (t) = 0. dt dt ℓ ℓ

2 Relate (4.52) to the statement that, for ﬁxed y ∈ S , φ(x) = Pℓ(x·y) belongs to the ℓ(ℓ + 1)-eigenspace of −∆S .

2 Exercises 14–19 deal with formulas for an orthogonal basis of Vk (for S ). We will make use of the structure of irreducible representations of SO(3), obtained in §9 of Appendix B, Manifolds, Vector Bundles, and Lie Groups. 14. Show that the representation of SO(3) on Vk is equivalent to the representation Dk, for each k = 0, 1, 2,... . 15. Show that if we use coordinates (θ, ψ) on S2, where θ is the geodesic distance 3 from (1, 0, 0) and ψ is the angular coordinate about the x1-axis in R , then [ ] ∂ iψ ∂ ∂ (4.53) L = ,L = i e + i cot θ . 1 ∂ψ ∂θ ∂ψ 16. Set k k ikψ (4.54) wk(x) = (x2 + ix3) = sin θ e .

Show that wk ∈ Vk and that it is the highest-weight vector for the representation, so

L1wk = ik wk.

17. Show that an orthogonal basis of Vk is given by 2k wk,L−wk,...,L− wk. k−j 18. Show that the functions ζkj = L− wk, j ∈ {−k, −k + 1, . . . , k − 1, k}, listed in Exercise 17 coincide, up to nonzero constant factors, with zkj , given by

zk0 = zk, 32

the spherical function considered in Exercises 1–3, and, for 1 ≤ j ≤ k,

j j zk,−j = L−zk, zkj = L+zk.

19. Show that the functions zkj coincide, up to nonzero constant factors, with ijψ j − ≤ ≤ (4.55) e Pk (cos θ), k j k, j where Pk (t), called associated Legendre functions, are deﬁned by ( ) |j| | | d (4.56) P j (t) = (−1)j (1 − t2) j /2 P (t). k dt k

5. The Laplace operator on hyperbolic space

The hyperbolic space Hn shares with the sphere Sn the property of having constant sectional curvature, but for Hn it is −1. One way to describe Hn is as a set of vectors with square length 1 in Rn+1, not for a Euclidean metric, but rather for a Lorentz metric ⟨ ⟩ − 2 − · · · − 2 2 (5.1) v, v = v1 vn + vn+1, namely, n n+1 (5.2) H = {v ∈ R : ⟨v, v⟩ = 1, vn+1 > 0}, with metric tensor induced from (5.1). The connected component G of the identity of the group O(n, 1) of linear transformations preserving the quadratic form (5.1) acts transitively on Hn, as a group of isometries. In fact, SO(n), acting on Rn ⊂ Rn+1, leaves invariant p = (0,..., 0, 1) ∈ Hn and n t acts transitively on the unit sphere in TpH . Also, if A(u1, . . . , un, un+1) t tA = (u1, . . . , un+1, un) , then e is a one-parameter subgroup of SO(n, 1) taking p to the curve { 2 − 2 } γ = (0,..., 0, xn, xn+1): xn+1 xn = 1, xn+1 > 0 . Together these facts imply that Hn is a homogeneous space. There is a map of Hn onto the unit ball in Rn, defined in a fashion similar to the stereographic projection of Sn. The map (5.3) s : Hn −→ Bn = {x ∈ Rn : |x| < 1} is defined by −1 (5.4) s(x, xn+1) = (1 + xn+1) x. The metric on Hn defined above then yields the following metric tensor on Bn: ( ) ∑n 2 − | |2 −2 2 (5.5) ds = 4 1 x dxj . j=1 5. The Laplace operator on hyperbolic space 33

Another useful representation of hyperbolic space is as the upper half Rn { ∈ Rn } space + = x : xn > 0 , with a metric we will specify shortly. In fact, with en = (0,..., 0, 1), 1 (5.6) τ(x) = |x + e |−2(x + e ) − e n n 2 n n Rn deﬁnes a map of the unit ball B onto +, taking the metric (5.5) to ∑n 2 −2 2 (5.7) ds = xn dxj . j=1 The Laplace operator for the metric (5.7) has the form ∑n ( ) n 2−n ∆u = xn ∂j xn ∂ju j=1 (5.8) ∑n 2 2 − = xn ∂j u + (2 n)xn ∂nu. j=1 which is convenient for a number of computations, such as (5.9) in the following:

Proposition 5.1. If ∆ is the Laplace operator on Hn, then ∆ is essen- ∞ Hn tially self-adjoint on C0 ( ), and its natural self-adjoint extension has the property [ ) 1 (5.9) spec(−∆) ⊂ (n − 1)2, ∞ . 4

Proof. Since Hn is a complete Riemannian manifold, the essential self- ∞ Hn adjointness on C0 ( ) follows from Proposition 2.4. To establish (5.9), it suﬃces to show that − 2 (n 1) 2 (−∆u, u) 2 Hn ≥ ∥u∥ 2 n , L ( ) 4 L (H ) ∈ ∞ Hn Hn for all u C0 ( ). Now the volume element on , identiﬁed with the −n ··· upper half-space with the metric (5.7), is xn dx1 dxn, so for such u we have ( ) ( 1 ) −∆ − (n − 1)2 u, u 4 L2 ∫ [ ( ) ] − 2 2 − (n 1)u 2−n ··· = (∂nu) xn dx1 dxn (5.10) 2xn n∑−1 ∫ 2 2−n ··· + (∂ju) xn dx1 dxn. j=1 34

Now, by an integration by parts, the ﬁrst integral on the right is equal to ∫ [ ( )]2 −(n−1)/2 ··· (5.11) ∂n xn u xn dx1 dxn. Rn +

Thus the expression (5.10) is ≥ 0, and (5.9) is proved.

We next describe how to obtain the fundamental solution to the wave equation on Hn. This will be obtained from the formula for Sn, via an analytic continuation in the metric tensor. Let p be a ﬁxed point (e.g., the north pole) in Sn, taken to be the origin in geodesic normal coordinates. Consider the one-parameter family of metrics given by dilating the sphere, which has constant curvature K = 1. Spheres dilated to have radius > 1 have constant curvature K ∈ (0, 1). On such a space, the fundamental −1 kernel A sin tA δp(x), with ( ) K 1/2 (5.12) A = −∆ + (n − 1)2 , 4 can be obtained explicitly from that on the unit sphere by a change of scale. The explicit representation so obtained continues analytically to all real values of K and at K = −1 gives a formula for the wave kernel, ( ) 1 1/2 (5.13) A−1 sin tA δ (x) = R(t, p, x),A = −∆ − (n − 1)2 . p 4 We have

[ ]−(n−1)/2 (5.14) R(t, p, x) = lim −2Cn Im 2 cos(it − ε) − 2 cosh r , ε↘0 where r = r(p, x) is the geodesic distance from p to x. Here, as in (4.10), Cn = 1/(n − 1)An. This exhibits several properties similar to those in the case of Sn discussed in §4. Of course, for r > |t|, the limit vanishes, exhibiting the ﬁnite propagation speed phenomenon. Also, if n is odd, the exponent (n − 1)/2 is an integer, which implies that (5.14) is supported on the shell r = |t|. In analogy with (4.25), we have the following formula for g(A)δp(x), for g ∈ S(R), when acting on L2(Hn), with n = 2k + 1: ( ) 1 1 ∂ k (5.15) g(A) = (2π)−1/2 − gˆ(r). 2π sinh r ∂r If n = 2k, we have

g(A) = ∫ ∞( ) (5.16) 1 1 1 ∂ k ( )−1/2 − gˆ(s) cosh s − cosh r sinh s ds. 1/2 π r 2π sinh s ∂s 6. The harmonic oscillator 35

Exercises ( ) −1 1. If n = 2k + 1, show that the Schwartz kernel of −∆ − (n − 1)2/4 − z2 on Hn, for z ∈ C \ [0, ∞), is ( ) 1 1 1 ∂ k G (x, y) = − − eizr, z 2iz 2π sinh r ∂r

2 where r = r(x, y) is geodesic distance, and the integral kernel of et(∆+(n−1) /4), for t > 0, is ( ) k 1 1 1 ∂ −r2/4t Ht(x, y) = √ − e . 4πt 2π sinh r ∂r

6. The harmonic oscillator

We consider the diﬀerential operator H = −∆+|x|2 on L2(Rn). By Propo- ∞ Rn sition 2.7, H is essentially self-adjoint on C0 ( ). Furthermore, as a special case of Proposition 2.8, we know that H has compact resolvent, so L2(Rn) has an orthonormal basis of eigenfunctions of H. To work out the spectrum, it suﬃces to work with the case n = 1, so we consider H = D2 + x2, where D = −i d/dx. The spectral analysis follows by some simple algebraic relations, involving the operators ( ) 1 d a = D − ix = + x , i dx (6.1) ( ) 1 d a+ = D + ix = − x . i dx Note that on D′(R), (6.2) H = aa+ − I = a+a + I, and (6.3) [H, a] = −2a, [H, a+] = 2a+. ∞ Suppose that uj ∈ C (R) is an eigenfunction of H, that is,

(6.4) uj ∈ D(H), Huj = λjuj. Now, by material developed in §2, D(H1/2) = {u ∈ L2(R): Du ∈ L2(R), xu ∈ L2(R)}, (6.5) D(H) = {u ∈ L2(R): D2u + x2u ∈ L2(R)}.

1/2 + Since certainly each uj belongs to D(H ), it follows that auj and a uj belong to L2(R). By (6.3), we have + + (6.6) H(auj) = (λj − 2)auj,H(a uj) = (λj + 2)a uj. 36

+ It follows that auj and a uj belong to D(H) and are eigenfunctions. Hence, if (6.7) Eigen(λ, H) = {u ∈ D(H): Hu = λu}, we have, for all λ ∈ R,

a+ : Eigen(λ, H) → Eigen(λ + 2,H), (6.8) a : Eigen(λ + 2,H) → Eigen(λ, H). ≥ ∥ ∥2 ∈ ∞ R From (6.2) it follows that (Hu, u) u L2 , for all u C0 ( ); hence, in view of essential self-adjointness, (6.9) spec H ⊂ [1, ∞), for n = 1. Now each space Eigen(λ, H) is a ﬁnite-dimensional subspace of C∞(R), + and, by (6.2), we conclude that, in (6.8), a is an isomorphism of Eigen(λj,H) onto Eigen(λj + 2,H), for each λj ∈ spec H. Also, a is an isomorphism of Eigen(λj,H) onto Eigen(λj − 2,H), for all λj > 1. On the other hand, a must annihilate Eigen(λ0,H) when λ0 is the smallest element of spec H, so ′ u0 ∈ Eigen(λ0,H) =⇒ u (x) = −xu0(x) (6.10) 0 −x2/2 =⇒ u0(x) = K e . Thus ( ) −x2/2 (6.11) λ0 = 1, Eigen(1,H) = span e .

2 Since e−x /2 spans the null space of a, acting on C∞(R), and since each nonzero space Eigen(λj,H) is mapped by some power of a to this null space, it follows that, for n = 1, (6.12) spec H = {2k + 1 : k = 0, 1, 2,... } and (( ) ) ∂ k 2 (6.13) Eigen(2k + 1,H) = span − x e−x /2 . ∂x One also writes ( ) ∂ k 2 2 (6.14) − x e−x /2 = H (x) e−x /2, ∂x k where Hk(x) are the Hermite polynomials, given by ( ) 2 d k 2 H (x) = (−1)kex e−x k dx (6.15) [k/2] ∑ k! = (−1)j (2x)k−2j. j!(k − 2j)! j=0 6. The harmonic oscillator 37

We deﬁne eigenfunctions of H: ( ) ∂ k 2 2 (6.16) h (x) = c − x e−x /2 = c H (x)e−x /2, k k ∂x k k where ck is the unique positive number such that ∥hk∥L2(R) = 1. To evaluate ck, note that ∥ + ∥2 + ∥ ∥2 (6.17) a hk L2 = (aa hk, hk)L2 = 2(k + 1) hk L2 . + Thus, if ∥hk∥L2 = 1, in order for hk+1 = γka hk to have unit norm, we −1/2 need γk = (2k + 2) . Hence [ ] 1/2 k −1/2 (6.18) ck = π 2 (k!) . Of course, given the analysis above of H on L2(R), then for H = −∆ + |x|2 on L2(Rn), we have (6.19) spec H = {2k + n : k = 0, 1, 2,... }. In this case, an orthonormal basis of Eigen(2k + n, H) is given by ··· ··· −|x|2/2 ··· (6.20) ck1 ckn Hk1 (x1) Hkn (xn)e , k1 + + kn = k, ∈ { } where kν 0, . . . , k , the Hkν (xν ) are the Hermite polynomials, and the ckν are given by (6.18). The dimension of this eigenspace is the same as the dimension of the space of homogeneous polynomials of degree k in n variables. We now want to derive a formula for the semigroup e−tH , t > 0, called the Hermite semigroup. Again it suﬃces to treat the case n = 1. To some degree paralleling the analysis of the eigenfunctions above, we can produce this formula via some commutator identities, involving the operators 2 − 2 2 (6.21) X = D = ∂x,Y = x ,Z = x∂x + ∂xx = 2x ∂x + 1. Note that H = X + Y . The commutator identities are (6.22) [X,Y ] = −2Z, [X,Z] = 4X, [Y,Z] = −4Y. Thus, X,Y , and Z span a three-dimensional, real Lie algebra. This is isomorphic to sl(2, R), the Lie algebra consisting of 2 × 2 real matrices of trace zero, spanned by ( ) ( ) ( ) 0 1 0 0 1 0 (6.23) n = , n− = , α = . + 0 0 1 0 0 −1 We have

(6.24) [n+, n−] = α, [n+, α] = −2n+, [n−, α] = 2n−. The isomorphism is implemented by

(6.25) X ↔ 2n+,Y ↔ 2n−,Z ↔ −2α. 38

Now we will be able to write − − − − (6.26) e t(2n++2n−) = e 2σ1(t)n+ e 2σ3(t)α e 2σ2(t)n− , as we will see shortly, and, once this is accomplished, we will be motivated to suspect that also − − − (6.27) e tH = e σ1(t)X eσ3(t)Z e σ2(t)Y . To achieve (6.26), write ( ) ( ) − 1 −2σ 1 x e 2σ1n+ = 1 = , 0 1 0 1 ( ) ( ) −2σ − e 3 0 y 0 (6.28) e 2σ3α = = , 0 e2σ3 0 1/y ( ) ( ) − 1 0 1 0 e 2σ2n− = = , −2σ2 1 z 1 and ( ) ( ) − cosh 2t − sinh 2t u v (6.29) e 2t(n++n−) = = . − sinh 2t cosh 2t v u Then (6.26) holds if and only if 1 1 v (6.30) y = = , x = z = = − tanh 2t, u cosh 2t u so the quantities σj(t) are given by 1 (6.31) σ (t) = σ (t) = tanh 2t, e2σ3(t) = cosh 2t. 1 2 2 Now we can compute the right side of (6.27). Note that ∫ 2 −σ1X −1/2 −(x−y) /4σ1 e u(x) = (4πσ1) e u(y) dy,

(6.32) − − 2 e σ2Y u(x) = e σ2x u(x), eσ3Z u(x) = eσ3 u(e2σ3 x). Upon composing these operators we ﬁnd that, for n = 1, ∫ −tH (6.33) e u(x) = Kt(x, y)u(y) dy, with {[ ]/ } − 1 2 2 exp 2 (cosh 2t)(x + y ) + xy sinh 2t (6.34) K (x, y) = . t ( )1/2 2π sinh 2t This is known as Mehler’s formula for the Hermite semigroup. Clearly, for general n, we have ∫ −tH (6.35) e u(x) = Kn(t, x, y)u(y) dy, 6. The harmonic oscillator 39 with

(6.36) Kn(t, x, y) = Kt(x1, y1) ··· Kt(xn, yn). The idea behind passing from (6.26) to (6.27) is that the Lie algebra homomorphism defined by (6.25) should give rise to a Lie group homomorphism from (perhaps a covering group G of) SL(2, R) into a group of operators. Since this involves an infinite-dimensional representation of G (not necessarily by bounded operators here, since e−tH is bounded only for t ≥ 0), there are analytical problems that must be overcome to justify this reasoning. Rather than take the space to develop such analysis here, we will instead just give a direct justification of (6.33)–(6.34). Indeed, let v(t, x) denote the right side of (6.33), with u ∈ L2(R) given. The rapid decrease of Kt(x, y) as |x| + |y| → ∞, for t > 0, makes it easy to show that ( ) (6.37) u ∈ L2(R) =⇒ v ∈ C∞ (0, ∞), S(R) . Also, it is routine to verify that ∂v (6.38) = −Hv. ∂t Simple estimates yielding uniqueness then imply that, for each s > 0, (6.39) v(t + s, ·) = e−tH v(s, ·). Indeed, if w(t, ·) denotes the difference between the two sides of (6.39), then we have w(0) = 0, w ∈ C(R+, D(H)), ∂w/∂t ∈ C(R+,L2(R)), and

d 2 ∥w(t)∥ 2 = −2(Hw, w) ≤ 0, dt L so w(t) = 0, for all t ≥ 0. Finally, as t ↘ 0, we see from (6.31) that each σj(t) ↘ 0. Since v(t, x) is also given by the right side of (6.27), we conclude that (6.40) v(t, ·) → u in L2(R), as t ↘ 0. Thus we can let s ↘ 0 in (6.39), obtaining a complete proof that e−tH u is given by (6.33) when n = 1. It is useful to write down the formula for e−tH using the Weyl calculus, introduced in §14 of Chapter 7. We recall that it associates to a(x, ξ) the operator ∫ a(X,D)u = (2π)−n aˆ(q, p)ei(q·X+p·D)u(x) dq dp (6.41) ∫ ( ) x + y = (2π)−n a , ξ ei(x−y)·ξu(y) dy dξ. 2 In other words, the operator a(X,D) has integral kernel K (x, y), for which ∫ a

a(X,D)u(x) = Ka(x, y)u(y) dy, 40 given by ∫ ( ) x + y K (x, y) = (2π)−n a , ξ ei(x−y)·ξ dξ. a 2

Recovery of a(x, ξ) from Ka(x, y) is an exercise in Fourier analysis. When it is applied to the formulas (6.33)–(6.36), this exercise involves computing a Gaussian integral, and we obtain the formula

−tH (6.42) e = ht(X,D) on L2(Rn), with

−n −(tanh t)(|x|2+|ξ|2) (6.43) ht(x, ξ) = (cosh t) e . It is interesting that this formula, while equivalent to (6.33)–(6.36), has a simpler and more symmetrical appearance. In fact, the formula (6.43) was derived in §15 of Chapter 7, by a diﬀerent method, which we brieﬂy recall here. For reasons of symmetry, involving the identity (14.19), one can write

2 2 (6.44) ht(x, ξ) = g(t, Q),Q(x, ξ) = |x| + |ξ| .

Note that (6.42) gives ∂t ht(X,D) = −Hht(X,D). Now the composition formula for the Weyl calculus implies that ht(x, ξ) satisﬁes the following evolution equation: ∂ h (x, ξ) = −(Q ◦ h )(x, ξ) ∂t t t 1 = −Q(x, ξ)h (x, ξ) − {Q, h } (x, ξ) (6.45) t 2 t 2 ∑( ) 2 2 1 2 2 = −(|x| + |ξ| )ht(x, ξ) + ∂ + ∂ ht(x, ξ). 4 xk ξk k Given (6.44), we have for g(t, Q) the equation ∂g ∂2g ∂g (6.46) = −Qg + Q + n . ∂t ∂Q2 ∂Q

It is easy to verify that (6.43) solves this evolution equation, with h0(x, ξ) = 1. We can obtain a formula for

−tQ(X,D) Q (6.47) e = ht (X,D), for a general positive-deﬁnite quadratic form Q(x, ξ). First, in the case ∑n 2 2 (6.48) Q(x, ξ) = µj(xj + ξj ), µj > 0, j=1 Exercises 41 it follows easily from (6.43) and multiplicativity, as in (6.36), that ∏n ( ) { ∑n } Q −1 · − 2 2 (6.49) ht (x, ξ) = cosh tµj exp (tanh tµj)(xj + ξj ) . j=1 j=1 Now any positive quadratic form Q(x, ξ) can be put in the form (6.48) via a linear symplectic transformation, so to get the general formula we need only rewrite (6.49) in a symplectically invariant fashion. This is accomplished 2n using the “Hamilton map” FQ, a skew-symmetric transformation on R deﬁned by 2n (6.50) Q(u, v) = σ(u, FQv), u, v ∈ R , where Q(u, v) is the bilinear form polarizing Q, and σ is the symplectic ′ ′ ′ ′ form on R2n; σ(u, v) = x · ξ − x · ξ if u = (x,( ξ), v = (x), ξ ). When Q has 0 µj the form (6.48), FQ is a sum of 2 × 2 blocks , and we have −µj 0

∏n ( )− ( )−1 1/2 (6.51) cosh tµj = det cosh itFQ . j=1

Passing from FQ to ( ) − 2 1/2 (6.52) AQ = FQ , the unique positive-deﬁnite square root, means passing to blocks ( ) µ 0 j , 0 µj and when Q has the form (6.48), then ∑n ( ) 2 2 (6.53) (tanh tµj)(xj + ξj ) = tQ ϑ(tAQ)ζ, ζ , j=1 where ζ = (x, ξ) and tanh t (6.54) ϑ(t) = . t Thus the general formula for (6.47) is

( )−1/2 Q −tQ(ϑ(tAQ)ζ,ζ) (6.55) ht (x, ξ) = cosh tAQ e .

Exercises

1. Deﬁne an unbounded operator A on L2(R) by D(A) = {u ∈ L2(R): Du ∈ L2(R), xu ∈ L2(R)}, Au = Du − ixu. 42

Show that A is closed and that the self-adjoint operator H satisﬁes ∗ ∗ H = A A + I = AA − I. (Hint: Note Exercises 5–7 of §2.) 2. If Hk(x) are the Hermite polynomials, show that there is the generating function identity ∑∞ 1 − 2 H (x)sk = e2xs s . k! k k=0 (Hint: Use the ﬁrst identity in (6.15).) 3. Show that Mehler’s formula (6.34) is equivalent to the identity ∑∞ j hj (x)hj (y)s = j=0 { [ ]} − − − − 2 2 π 1/2(1 − s2) 1/2 exp (1 − s2) 1 2xys − (x2 + y2)s2 · e (x +y )/2,

for 0 ≤ s < 1. Deduce that

∑∞ j s − 2 H (x)2 = (1 − s2) 1/2e2sx /(1+s), |s| < 1. j 2j j! j=0 4. Using ∫ ∞ − 1 − − H s = e tH ts 1 dt, Re s > 0, Γ(s) 0

ﬁnd the integral kernel As(x, y) such that ∫ −s H u(x) = As(x, y)u(y) dy. ∫ −s Writing Tr H = As(x, x) dx, Re s > 1, n = 1, show that ∫ 1 ∞ ys−1 ζ(s) = dy. y − Γ(s) 0 e 1 See [Ing], pp. 41–44, for a derivation of the functional equation for the Rie- mann zeta function, using this formula. 2 2 2 2 −tHω 5. Let Hω = −d /dx + ω x . Show that e has integral kernel

ω −1/2 1/2 −γ(2ωt)[(cosh 2ωt)(x2+y2)−2xy]/4t Kt (x, y) = (4πt) γ(2ωt) e , where z γ(z) = . sinh z 6. Consider the operator ( ) ( ) ∂ 2 ∂ 2 Q(X,D) = − − iωx − + iωx ∂x 2 ∂x 1 1 ( 2 ) 2 2 ∂ ∂ = −∆ + ω |x| + 2iω x2 − x1 . ∂x1 ∂x2 7. The quantum Coulomb problem 43

Note that Q(x, ξ) is nonnegative, but not deﬁnite. Study the integral kernel Q −tQ(X,D) Kt (x, y) of e . Show that

Q −1 −τ(2ωt)|x|2/4t Kt (x, 0) = (4πt) γ(2ωt) e , where τ(z) = z coth z.

7. Let (ωjk) be an invertible, n × n, skew-symmetric matrix of real numbers (so n must be even). Suppose

n ( ) ∑ ∂ ∑ 2 L = − − i ωjkxk . ∂xj j=1 k

L Evaluate the integral kernel Kt (x, y), particularly at y = 0. 8. In terms of the operators a, a+ given by (6.1) and the basis of L2(R) given by (6.16)–(6.18), show that √ √ + a hk = 2k + 2 hk+1, ahk = 2k hk−1.

7. The quantum Coulomb problem

In this section we examine the operator (7.1) Hu = −∆u − K|x|−1u, acting on functions on R3. Here, K is a positive constant. This provides a quantum mechanical description of the Coulomb force between two charged particles. It is the first step toward a quantum mechanical description of the hydrogen atom, and it provides a decent approx- imation to the observed behavior of such an atom, though it leaves out a number of features. The most important omitted feature is the spin of the electron (and of the nucleus). Giving rise to further small corrections are the nonzero size of the proton, and relativistic effects, which confront one with great subtleties since relativity forces one to treat the electromagnetic field quantum mechanically. We refer to texts on quantum physics, such as [Mes], [Ser], [BLP], and [IZ], for work on these more sophisticated models of the hydrogen atom. We want to define a self-adjoint operator via the Friedrichs method. Thus we want to work with a Hilbert space ∫ { } (7.2) H = u ∈ L2(R3): ∇u ∈ L2(R3), |x|−1|u(x)|2 dx < ∞ , with inner product ∫ −1 (7.3) (u, v)H = (∇u, ∇v)L2 + A(u, v)L2 − K |x| u(x)v(x) dx, 44 where A is a sufficiently large, positive constant. We must first show that A can be picked to make this inner product positive-definite. In fact, we have the following:

Lemma 7.1. For all ε ∈ (0, 1], there exists C(ε) < ∞ such that ∫ | |−1| |2 ≤ ∥∇ ∥2 ∥ ∥2 (7.4) x u(x) dx ε u L2 + C(ε) u L2 , for all u ∈ H1(R3).

Proof. Here and below we will use the inclusion [ ) 2n n (7.5) Hs(Rn) ⊂ Lp(Rn), ∀ p ∈ 2, , 0 ≤ s < , n − 2s 2 from (2.42) of Chapter 4. In Chapter 13 we will establish the sharper result that Hs(Rn) ⊂ L2n/(n−2s)(Rn); for example, H1(R3) ⊂ L6(R3). We will also cite this stronger result in some arguments below, though that could be avoided. We also use the fact that (if B = {|x| < 1} and χB(x) is its characteristic function), q 3 χBV ∈ L (R ), for all q < 3. Here and below we will use V (x) = |x|−1. Thus the left side of (7.4) is bounded by 2 2 2 2 ∥ ∥ q · ∥ ∥ ′ ∥ ∥ ≤ ∥ ∥ ∥ ∥ (7.6) χBV L u L2q + u L2 C u Hσ (R3) + u L2(R3), where we can take any q′ > 3/2; take q′ ∈ (3/2, 3). Then (7.6) holds ′ for some σ < 1, for which L2q (R3) ⊃ Hσ(R3). From this, (7.4) follows immediately.

Thus the Hilbert space H in (7.2) is simply H1(R3), and we see that indeed, for some A > 0, (7.3) deﬁnes an inner product equivalent to the standard one on H1(R3). The Friedrichs method then deﬁnes a positive, self-adjoint operator H + AI, for which ( ) (7.7) D (H + AI)1/2 = H1(R3). Then (7.8) D(H) = {u ∈ H1(R3): −∆u − K|x|−1u ∈ L2(R3)}, where −∆u − K|x|−1u is a priori regarded as an element of H−1(R3) if u ∈ H1(R3). Since H2(R3) ⊂ L∞(R3), we have (7.9) u ∈ H2(R3) =⇒ |x|−1u ∈ L2(R3), so (7.10) D(H) ⊃ H2(R3). 7. The quantum Coulomb problem 45

Indeed, we have:

Proposition 7.2. For the self-adjoint extension H of −∆−K|x|−1 deﬁned above, (7.11) D(H) = H2(R3).

Proof. Pick λ in the resolvent set of H; for instance, λ ∈ C \ R. If u ∈ D(H) and (H − λ)u = f ∈ L2(R3), we have

(7.12) u − KRλV u = Rλf = gλ,

−1 −1 where V (x) = |x| and Rλ = (−∆ − λ) . Now the operator of multiplication by V (x) = |x|−1 has the property

1 3 2−ε 3 (7.13) MV : H (R ) −→ L (R ), for all ε > 0, since H1(R3) ⊂ L6(R3) ∩ L2(R3) and V ∈ L3−ε on |x| < 1. Hence

1 3 −ε 3 MV : H (R ) −→ H (R ), for all ε > 0. Let us apply this to (7.12). We know that u ∈ D(H) ⊂ 1/2 1 3 2−ε 3 2−ε 3 D(H ) = H (R ), so KRλV u ∈ H (R ). Thus u ∈ H (R ), for all ε > 0. But, for ε > 0 small enough,

2−ε 3 2 3 (7.14) MV : H (R ) −→ L (R ),

2 3 2 3 so then u = KRλ(V u) + Rλf ∈ H (R ). This proves that D(H) ⊂ H (R ) and gives (7.11).

Since H is self-adjoint, its spectrum is a subset of the real axis, (−∞, ∞). We next show that there is only point spectrum in (−∞, 0).

Proposition 7.3. The part of spec H lying in C \ [0, ∞) is a bounded, discrete subset of (−∞, 0), consisting of eigenvalues of ﬁnite multiplicity and having at most {0} as an accumulation point.

Proof. Consider the equation (H − λ)u = f ∈ L2(R3), that is, (7.15) (−∆ − λ)u − KV u = f,

−1 −1 with V (x) = |x| as before. Applying Rλ = (−∆ − λ) to both sides, we again obtain (7.12):

(7.16) (I − KRλMV )u = gλ = Rλf.

Note that Rλ is a holomorphic function of λ ∈ C \ [0, ∞), with values in L(L2(R3),H2(R3)). A key result in the analysis of (7.16) is the following: 46

Lemma 7.4. For λ ∈ C \ [0, ∞),

2 3 (7.17) RλMV ∈ K(L (R )), where K is the space of compact operators.

We will establish this via the following basic tool. For λ ∈ C\[0, ∞), φ ∈ 3 C0(R ), the space of continuous functions vanishing at inﬁnity, we have 2 2 (7.18) MφRλ ∈ K(L ) and RλMφ ∈ K(L ). ∈ ∞ R3 To see this, note that, for φ C0 ( ), the ﬁrst inclusion in (7.18) follows from Rellich’s theorem. Then this inclusion holds for uniform limits of such 3 φ, hence for φ ∈ C0(R ). Taking adjoints yields the rest of (7.18). Now, to establish (7.17), write

(7.19) V = V1 + V2, ∈ ∞ R3 | | ≤ ∈ R3 where V1 = ψV, ψ C0 ( ), ψ(x) = 1 for x 1. Then V2 C0( ), so ∈ K ∈ q R3 ∈ RλMV2 . We have V1 L ( ), for all q [1, 3), so, taking q = 2, we have 2 R3 −→ 1 R3 ⊂ −3/2−ε R3 (7.20) MV1 : L ( ) L ( ) H ( ), for all ε > 0, hence 2 R3 −→ 1/2−ε R3 ⊂ 2 R3 (7.21) RλMV1 : L ( ) H ( ) L ( ).

Given V1 supported on a ball BR, the operator norm in (7.21) is bounded 2 by a constant times ∥V1∥L2 . You can approximate V1 in L -norm by a ∈ ∞ R3 sequence wj C0 ( ). It follows that RλMV1 is a norm limit of a sequence of compact operators on L2(R3), so it is also compact, and (7.17) is established. The proof of Proposition 7.4 is ﬁnished by the following result, which can be found as Proposition 7.4 in Chapter 9.

Proposition 7.5. Let O be a connected, open set in C. Suppose C(λ) is a compact-operator-valued holomorphic function of λ ∈ O. If I − C(λ) is invertible at one point p ∈ O, then it is invertible except at most on a discrete set in O, and (I − C(λ))−1 is meromorphic on O.

This applies to our situation, with C(λ) = KRλMV ; we know that I − C(λ) is invertible for all λ ∈ C \ R in this case. One approach to analyzing the negative eigenvalues of H is to use polar coordinates. If −K|x|−1 is replaced by any radial potential V(|x|), the eigenvalue equation Hu = −Eu becomes ∂2u 2 ∂u 1 (7.22) + + ∆ u − V(r)u = Eu. ∂r r ∂r r2 S 7. The quantum Coulomb problem 47

We can use separation of variables, writing u(rθ) = v(r)φ(θ), where φ is 2 an eigenfunction of ∆S, the Laplace operator on S ,

( 1)2 1 (7.23) ∆ φ = −λφ, λ = k + − = k2 + k. S 2 4 Then we obtain for v(r) the ODE 2 λ (7.24) v′′(r) + v′(r) + f(r)v(r) = 0, f(r) = −E − − V(r). r r2 One can eliminate the term involving v′ by setting (7.25) w(r) = rv(r). Then (7.26) w′′(r) + f(r)w(r) = 0. For the Coulomb problem, this becomes [ ] K λ (7.27) w′′(r) + −E + − w(r) = 0. r r2 √ If we set W (r) = w(βr), β = 1/2 E, we get a form of Whittaker’s ODE: [ ] 1 κ 1 − µ2 (7.28) W ′′(z) + − + + 4 W (z) = 0, 4 z z2 with

K 1 ( 1)2 (7.29) κ = √ , µ2 = λ + = k + . 2 E 4 2 This in turn can be converted to the confluent hypergeometric equation (7.30) zψ′′(z) + (b − z)ψ′(z) − aψ(z) = 0 upon setting (7.31) W (z) = zµ+1/2 e−z/2 ψ(z), with 1 K a = µ − κ + = k + 1 − √ , (7.32) 2 2 E b = 2µ + 1 = 2k + 2. Note that ψ and v are related by √ √ √ (7.33) v(r) = (2 E)k+1 rke−2 Erψ(2 Er). Looking at (7.28), we see that there are two independent solutions, one behaving roughly like e−z/2 and the other like ez/2, as z → +∞. Equiv- alently, (7.30) has two linearly independent solutions, a “good” one growing more slowly than exponentially and a “bad” one growing like ez, as z → +∞. Of course, for a solution to give rise to an eigenfunction, we 48 need v ∈ L2(R+, r2 dr), that is, w ∈ L2(R+, dr). We need to have simultaneously w(z) ∼ ce−z/2 (roughly) as z → +∞ and w square integrable near z = 0. In view of (7.8), we also need v′ ∈ L2(R+, r2 dr). To examine the behavior near z = 0, note that the Euler equation associated with (7.28) is ( ) 1 (7.34) z2W ′′(z) + − µ2 W (z) = 0, 4 with solutions z1/2+µ and z1/2−µ, i.e., zk+1 and z−k, k = 0, 1, 2,... . If k = 0, both are square integrable near 0, but for k ≥ 1 only one is. Going to the confluent hypergeometric equation (7.30), we see that two linearly independent solutions behave respectively like z0 and z−2µ = z−2k−1 as z → 0. As a further comment on the case k = 0, note that a solution W behaving like z0 at z = 0 gives rise to v(r) ∼ C/r as r → 0, with c ≠ 0, hence v′(r) ∼ −C/r2. This is not square integrable near r = 0, with respect to r2 dr, so also this case does not produce an eigenfunction of H. If b∈ / {0, −1, −2,... }, which certainly holds here, the solution to (7.30) that is “good” near z = 0 is given by the confluent hypergeometric function

∞ ∑ (a) zn (7.35) F (a; b; z) = n , 1 1 (b) n! n=0 n an entire function of z. Here, (a)n = a(a + 1) ··· (a + n − 1); (a)0 = 1. If also a∈ / {0, −1, −2,... }, it can be shown that

Γ(b) (7.36) F (a; b; z) ∼ ez z−(b−a), z → +∞. 1 1 Γ(a) See the exercises below for a proof of this. Thus the “good” solution near z = 0 is “bad” as z → +∞, unless a is a nonpositive integer, say a = −j. In that case, as is clear from (7.35), 1F1(−j; b; z) is a polynomial in z, thus “good” as z → +∞. Thus the negative eigenvalues of H are given by −E, with K (7.37) √ = j + k + 1 = n, 2 E that is, by

K2 (7.38) E = , n = 1, 2, 3 .... 4n2 Note that, for each value of n, one can write n = j +k +1 using n choices 2 of k ∈ {0, 1, 2, . . . , n − 1}. For each such k, the (k + k)-eigenspace of ∆S has dimension 2k + 1, as established in Corollary 4.3. Thus the eigenvalue 7. The quantum Coulomb problem 49

−E = −K2/4n2 of H has multiplicity n∑−1 (7.39) (2k + 1) = n2. k=0 2 Let us denote by Vn the n -dimensional eigenspace of H, associated to the 2 2 eigenvalue λn = −K /4n . The rotation group SO(3) acts on each Vn, via ρ(g)f(x) = f(g−1x), g ∈ SO(3), x ∈ R3.

By the analysis leading to (7.39), this action on Vn is not irreducible, but rather has n irreducible components. This suggests that there is an extra symmetry, and indeed, as W. Pauli discovered early in the history of quantum mechanics, there is one, arising via the Lenz vector (briefly introduced in §16 of Chapter 1), which we proceed to define. The angular momentum vector L = x × p, with p replaced by the vector operator (∂/∂x1, ∂/∂x2, ∂/∂x3), commutes with H as a consequence of the rotational invariance of H. The components of L are ∂ ∂ (7.40) Lℓ = xj − xk , ∂xk ∂xj where (j, k, ℓ) is a cyclic permutation of (1, 2, 3). Then the Lenz vector is defined by ( ) 1 x (7.41) B = L × p − p × L − , K r with components Bj, 1 ≤ j ≤ 3, each of which is a second-order differential operator, given explicitly by 1 x (7.42) B = (L ∂ + ∂ L − L ∂ − ∂ L ) − j , j K k ℓ ℓ k ℓ k k ℓ r where (j, k, ℓ) is a cyclic permutation of (1, 2, 3). A calculation gives

(7.43) [H,Bj] = 0, in the sense that these operators commute on C∞(R3 \ 0). 3 It follows that if u ∈ Vn, then Bju is annihilated by H − λn, on R \ 0. Now, we have just gone through an argument designed to glean from all functions that are so annihilated, those that are actually eigenfunctions of H. In view of that, it is important to establish the next lemma.

Lemma 7.6. We have

(7.44) Bj : Vn −→ Vn.

2 3 Proof. Let u ∈ Vn. We know that u ∈ D(H) = H (R ). Also, from the analysis of the ODE (7.28), we know that u(x) decays as |x| → ∞, 50

1/2 −|λn| |x| 2 3 roughly like e . It follows from (7.42) that Bju ∈ L (R ). It will 2 be useful to obtain a bit more regularity, using Vn ⊂ D(H ) together with the following.

Proposition 7.7. If u ∈ D(H2), then, for all ε > 0, (7.45) u ∈ H5/2−ε(R3). Furthermore, (7.46) g ∈ S(R3), g(0) = 0 =⇒ gu ∈ H7/2−ε(R3).

Proof. We proceed along the lines of the proof of Proposition 7.2, using (7.12), i.e.,

(7.47) u = KRλV u + Rλf, where f = (H − λ)u, with λ chosen in C \ R. We know that f = (H − λ)u 4 3 2 3 belongs to D(H), so Rλf ∈ H (R ). We know that u ∈ H (R ). Parallel to (7.13), we can show that, for all ε > 0, 2 3 1/2−ε 3 (7.48) MV : H (R ) −→ H (R ), 5/2−ε 3 so KRλV u ∈ H (R ). This gives (7.45). Now, multiply (7.47) by g and write

(7.49) gu = KRλgV u + K[Mg,Rλ]V u + gRλf. This time we have 2 3 3/2−ε 3 MgV : H (R ) −→ H (R ), 7/2−ε 3 so RλgV u ∈ H (R ). Furthermore, s 3 s+3 3 (7.50) [Mg,Rλ] = Rλ [∆,Mg] Rλ : H (R ) −→ H (R ), 7/2−ε 3 so [Mg,Rλ]V u ∈ H (R ). This establishes (7.46).

We can now ﬁnish the proof of Lemma 7.6. Note that the second-order derivatives in Bj have a coeﬃcient vanishing at 0. Keep in mind the ∈ 2 R3 → known exponential decay of u Vn. Also note that Mxj /r : H ( ) H3/2−ε(R3). Therefore, 3/2−ε 3 (7.51) u ∈ Vn =⇒ Bju ∈ H (R ). Consequently, −1/2−ε 3 1 3 2 3 (7.52) ∆(Bju) ∈ H (R ), and V (Bju) ∈ L (R ) + L (R ). 3 Thus (H − λn)(Bju), which we know vanishes on R \ 0, must vanish completely, since (7.52) does not allow for a nonzero quantity supported on {0}. Using (7.8), we conclude that Bju ∈ D(H), and the lemma is proved. 7. The quantum Coulomb problem 51

With Lemma 7.6 established, we can proceed to study the action of Bj and Lj on Vn. When (j, k, ℓ) is a cyclic permutation of (1, 2, 3), we have

(7.53) [Lj,Lk] = Lℓ, and, after a computation, 4 (7.54) [L ,B ] = B , [B ,B ] = − HL . j k ℓ j k K ℓ

Of course, (7.52) is the statement that Lj span the Lie algebra so(3) of SO(3). The identities (7.54), when Lj and Bj act on Vn, can be rewritten as K (7.55) [Lj,Ak] = Aℓ, [Aj,Ak] = Aℓ,Aj = √ Bj. 2 −λn If we set 1 1 (7.56) M = (L + A), N = (L − A), 2 2 we get, for cyclic permutations (j, k, ℓ) of (1, 2, 3),

(7.57) [Mj,Mk] = Mℓ, [Nj,Nk] = Nℓ, [Mj,Nj′ ] = 0, which is clearly the set of commutation relations for the Lie algebra so(3)⊕ so(3). We next aim to show that this produces an irreducible representation of SO(4) on Vn, and to identify this representation. A priori, of course, one certainly has a representation of SU(2) × SU(2) on Vn. 2 We now examine the behavior on Vn of the Casimir operators M = 2 2 2 2 · M1 +M2 +M3 and N . A calculation using the deﬁnitions gives B L = 0, hence A · L = 0, so, on Vn, 1 M 2 = N 2 = (A2 + L2) 4 (7.58) ( ) 1 K2 = L2 − B2 . 4 4λn We also have the following key identity: (7.59) K2(B2 − I) = 4H(L2 + I), which follows from the deﬁnitions by a straightforward computation. If we compare (7.58) and (7.59) on Vn, where H = λn, we get ( 2 ) 2 2 K (7.60) 4M = 4N = − 1 + I on Vn. 4λn

Now the representation σn we get of SU(2)×SU(2) on Vn is a direct sum (possibly with only one summand) of representations Dj/2 ⊗ Dj/2, where j+1 Dj/2 is the standard irreducible representation of SU(2) on C , deﬁned in §9 of Appendix B. The computation (7.60) implies that all the copies in 52 this sum are isomorphic, that is, for some j = j(n), ⊕µ (7.61) σn = Dj(n)/2 ⊗ Dj(n)/2. ℓ=1 ( ) 2 2 A dimension count gives µ j(n) + 1 = n . Note that on Dj/2 ⊗ Dj/2, we have M 2 = N 2 = (j/2)(j/2 + 1). Thus (7.60) implies j(j + 2) = 2 −1 + K /4λn, or K2 (7.62) λ = − , j = j(n). n 4(j + 1)2 Comparing (7.38), we have (j + 1)2 = n2, that is, (7.63) j(n) = n − 1.

2 Since we know that dim Vn = n , this implies that there is just one summand in (7.61), so

(7.64) σn = D(n−1)/2 ⊗ D(n−1)/2. This is an irreducible representation of SU(2) × SU(2), which is a double cover of SO(4), κ : SU(2) × SU(2) −→ SO(4).

It is clear that σn is the identity operator on both elements in ker κ, and so σn actually produces an irreducible representation of SO(4). Let ρn denote the restriction to Vn of the representation ρ of SO(3) on L3(R3), described above. If we regard this as a representation of SU(2), it is clear that ρn is the composition of σn with the diagonal map SU(2) → SU(2) × SU(2). Results established in §9 of Appendix B imply that such a tensor-product representation of SU(2) has the decomposition into irreducible representations: n⊕−1 (7.65) ρn ≈ Dk. k=0

This is also precisely the description of ρn given by the analysis leading to (7.39). There are a number of other group-theoretic perspectives on the quantum Coulomb problem, which can be found in [Eng] and [GS2]. See also [Ad] and [Cor], Vol. 2.

Exercises

1. For H = −∆ − K|x|−1 with domain given by (7.8), show that − (7.66) D(H) = {u ∈ L2(R3): −∆u − K|x| 1u ∈ L2(R3)}, Exercises 53

where a priori, if u ∈ L2(R3), then ∆u ∈ H−2(R3) and |x|−1u ∈ L1(R3) + L2(R3) ⊂ H−2(R3). (Hint: Parallel the proof of Proposition 7.2. If u belongs to the right side of (7.66), and if you pick λ ∈ C \ R, then, as in (7.12),

2 3 (7.67) u − KRλV u = Rλf ∈ H (R ).) Complement (7.13) with ∩ 2 3 −3/2−ε 3 MV : L (R ) −→ H (R ), ε>0 (7.68) ∩ ∩ 1/2−ε 3 −3/4−δ 3 MV : H (R ) −→ H (R ). ε>0 δ>0 (Indeed, sharper results can be obtained.) Then deduce from (7.67) ﬁrst that u ∈ H1/2−ε(R3) and then that u ∈ H5/4−δ(R3) ⊂ H1(R3).) 2. As a variant of (7.4), show that, for u ∈ H1(R3), ∫ ∫ − (7.69) |x| 2|u(x)|2 dx ≤ 4 |∇u(x)|2 dx.

Show that 4 is the best possible constant on the right. (Hint: Use the Mellin transform to show that the spectrum of r d/dr−1/2 on L2(R+, r−1dr) (which coincides with the spectrum of r d/dr on L2(R+, dr)) is {is − 1/2 : s ∈ R}, hence ∫ ∞ ∫ ∞ ′ (7.70) |u(r)|2dr ≤ 4 |u (r)|2r2 dr. 0 0 This is sometimes called an “uncertainty principle” estimate. Why might that be? (Cf. [RS], Vol. 2, p. 169.) 3. Show that H = −∆−K/|x| has no non-negative eigenvalues, i.e., only continuous spectrum in [0, ∞). (Hint: Study the behavior as r → +∞ of solutions to the ODE (7.28), when −E is replaced by +E ∈ [0, ∞). Consult [Olv] for techniques. See also [RS], Vol. 4, for general results.) 4. Generalize the propositions of this section, with modiﬁcations as needed, to other classes of potentials V (x), such as ∞ V ∈ L2 + εL ,

the set of functions V such that, for each ε > 0, one can write V = V1 + 2 V2,V1 ∈ L , ∥V2∥L∞ ≤ ε. Consult [RS], Vols. 2-4, for further generalizations.

Exercises on the conﬂuent hypergeometric function

1. Taking (7.35) as the deﬁnition of 1F1(a; b; z), show that (7.71) ∫ 1 Γ(b) zt a−1 − b−a−1 1F1(a; b; z) = − e t (1 t) dt, Re b > Re a > 0. Γ(a)Γ(b a) 0 54

(Hint: Use the beta function identity, (A.23)–(A.24) of Chapter 3.) Show that (7.71) implies the asymptotic behavior (7.36), provided Re b > Re a > 0, but that this is insuﬃcient for making the deduction (7.37).

Exercises 2–5 deal with the analytic continuation of (7.71) in a and b, and a complete justiﬁcation of (7.36). To begin, write Γ(b) Γ(b) (7.72) F (a; b; z) = A (a, −z) + A (b − a, z)ez, 1 1 Γ(b − a) ψ Γ(a) φ ( ) where, for Re c > 0, ψ ∈ C∞ [0, 1/2] , we set ∫ 1/2 1 −zt c−1 (7.73) Aψ(c, z) = e ψ(t)t dt, Γ(c) 0 and, in (7.72), − − − ψ(t) = (1 − t)b a 1, φ(t) = (1 − t)a 1. 2. Given Re c > 0, show that −c (7.74) Aψ(c, z) ∼ ψ(0)z , z → +∞, and 1 ψ( ) − (7.75) A (c, −z) ∼ 2 z 1ez/2, z → +∞. ψ Γ(c) 3. For j = 0, 1, 2,... , set ∫ 1/2 1 −zt j c−1 (7.76) Aj (c, t) = e t t dt, Γ(c) 0 j so Aj (c, z) = Aψ(c, z), with ψ(t) = t . Show that ∫ ∞ Γ(c + j) −c−j 1 −zt c+j−1 Aj (c, z) = z − e t dt, Γ(c) Γ(c) 1/2

for Re z > 0. Deduce that Aj (c, t) is an entire function of c, for Re z > 0, and that

Γ(c + j) − − A (c, z) ∼ z c j , z → +∞, j Γ(c) if c∈ / {0, −1, −2,... }. 4. Given k = 1, 2, 3,... , write ([ ]) k−1 k ∞ 1 ψ(t) = a + a t + ··· + a − t + ψ (t)t , ψ ∈ C 0, . 0 1 k 1 k k 2 Thus ∫ k∑−1 1/2 1 − − (7.77) A (c, z) = a A (c, z) + e ztψ (t)tk+c 1 dt. ψ j j Γ(c) k j=0 0

Deduce that Aψ(c, z) can be analytically continued to Re c > −k when Re z > 0 and that (7.74) continues to hold if c∈ / {0, −1, −2,... }, a0 ≠ 0. Exercises 55

5. Using tc−1 = c−1(d/dt)tc and integrating by parts, show that

1 − (7.78) A (c, z) = zA (c + 1, z) − e z/2, 0 0 2cΓ(c + 1) for Re c > 0, all z ∈ C. Show that this provides an entire analytic continuation of A0(c, z) and that (7.74)–(7.75) hold, for ψ(t) = 1. Using Γ(c + j) A (c, z) = A (c + j, z) j Γ(c) 0 ( ) and (7.77), verify (7.75) for all ψ ∈ C∞ [0, 1/2] . (Also again verify (7.74).) Hence, verify the asymptotic expansion (7.36).

The approach given above to (7.36) is one the author learned from conversa- tions with A. N. Varchenko. In Exercises 6–15 below, we introduce another solution to the conﬂuent hypergeometric equation and follow a path to the expansion (7.36) similar to one described in [Leb] and in [Olv]. 6. Show that a solution to the ODE (7.30) is also given by

1−b z 1F1(1 + a − b; 2 − b; z),

in addition to 1F1(a; b; z), deﬁned by (7.35). Assume b ≠ 0, −1, −2,... . Set

Γ(1 − b) Γ(b − 1) − (7.79) Ψ(a; b; z) = F (a; b; z)+ z1 b F (1+a−b; 2−b; z). Γ(1 + a − b) 1 1 Γ(a) 1 1 Show that the Wronskian is given by ( ) Γ(b) − W F (a; b; z), Ψ(a; b; z) = − z bez. 1 1 Γ(a) 7. Show that

z (7.80) 1F1(a; b; z) = e 1F1(b − a; b; −z), b∈ / {0, −1, −2,... }. (Hint: Use the integral in Exercise 1, and set s = 1 − t, for the case Re b > Re a > 0.) 8. Show that ∫ ∞ 1 − − − − (7.81) Ψ(a; b; z) = e ztta 1(1 + t)b a 1 dt, Re a > 0, Re z > 0. Γ(a) 0 (Hint: First show that the right side solves (7.30). Then check the behavior as z → 0.) 9. Show that (7.82) Ψ(a; b; z) = zΨ(a + 1; b + 1; z) + (1 − a − b)Ψ(a + 1; b; z). (Hint: To get this when Re a > 0, use the integral expression (7.81) for Ψ(a + 1; b + 1; z), write ze−zt = −(d/dt)e−zt, and integrate by parts.) 10. Show that

Γ(b) Γ(b) − (7.83) F (a; b; z) = e πaiΨ(a; b; z) + e π(a b)i ez Ψ(b − a; b; −z), 1 1 Γ(b − a) Γ(a) 56

where −z = e∓πiz, b ≠ 0, −1, −2,... .(Hint: Make use of (7.80) as well as (7.79).) 11. Using the integral representation (7.81), show that under the hypotheses δ > 0, b∈ / {0, −1, −2,... }, and Re a > 0, we have − (7.84) Ψ(a; b; z) ∼ z α, |z| → ∞, in the sector π (7.85) |Arg z| ≤ − δ. 2 12. Extend (7.84) to the sector |Arg z| ≤ π − δ.(Hint: Replace (7.81) by an integral along the ray γ = {eiαs : 0 ≤ s < ∞}, given |α| < π/2.) 13. Further extend (7.84) to the case where no restriction is placed on Re a. (Hint: Use (7.82).) 14. Extend (7.84) still further, to be valid for 3π (7.86) |Arg z| ≤ − δ. 2 (Hint: See Theorem 2.2 on p. 235 of [Olv], and its application to this problem on p. 256 of [Olv].) 15. Use (7.83)–(7.86) to prove (7.36), that is,

Γ(b) − − (7.87) F (a; b; z) ∼ ez z (b a), z → +∞, 1 1 Γ(a) provided a, b∈ / {0, −1, −2,... }. Remark: For the analysis of Ψ(b−a; b; −z) as z → +∞, the result of Exercise 14 suﬃces, but the result of Exercise 13 does not. This point appears to have been neglected in the discussion of (7.87) on p. 271 of [Leb].

8. The Laplace operator on cones

Generally, if N is any compact Riemannian manifold of dimension m, possibly with boundary, the cone over N,C(N), is the space R+ × N together with the Riemannian metric (8.1) dr2 + r2g, where g is the metric tensor on N. In particular, a cone with vertex at the origin in Rm+1 can be described as the cone over a subdomain Ω of the unit sphere Sm in Rm+1. Our purpose is to understand the behavior of the Laplace operator ∆, a negative, self-adjoint operator, on C(N). If ∂N ≠ ∅, we impose Dirichlet boundary conditions on ∂C(N), though many other boundary conditions could be equally easily treated. The analysis here follows [CT]. The initial step is to use the method of separation of variables, writing ∆ on C(N) in the form ∂2 m ∂ 1 (8.2) ∆ = + + ∆ , ∂r2 r ∂r r2 N 8. The Laplace operator on cones 57 where ∆N is the Laplace operator on the base N. Let µj, φj(x) denote the eigenvalues and eigenfunctions of −∆N (with Dirichlet boundary condition on ∂N if ∂N ≠ ∅), and set m − 1 (8.3) ν = (µ + α2)1/2, α = − . j j 2 If ∑ g(r, x) = gj(r)φj(x), j with gj(r) well behaved, and if we deﬁne the second-order operator Lµ by ( ) ∂2 m ∂ µ (8.4) L g(r) = + − g(r), µ ∂r2 r ∂r r2 then we have ∑

(8.5) ∆g(r, x) = Lµj gj(r)φj(x). j In particular, 2 (8.6) ∆(gjφj) = −λ gjφj provided −(m−1)/2 (8.7) gj(r) = r Jνj (λr).

Here Jν (z) is the Bessel function, introduced in §6 of Chapter 3; there in (6.6) it is defined to be ∫ ν 1 (z/2) − 2 ν−1/2 izt (8.8) Jν (z) = 1 1 (1 t ) e dt, Γ( 2 )Γ(ν + 2 ) −1 for Re ν > −1/2; in (6.11) we establish Bessel’s equation [ ( )] d2 1 d ν2 (8.9) + + 1 − J (z) = 0, dz2 z dz z2 ν which justifies (8.6); and in (6.19) we produced the formula ( ) ∞ ( ) z ν ∑ (−1)k z 2k (8.10) J (z) = . ν 2 k!Γ(k + ν + 1) 2 k=0 We also recall, from (6.56) of Chapter 3, the asymptotic behavior ( ) ( ) 2 1/2 πν π (8.11) J (r) ∼ cos r − − + O(r−3/2), r → +∞. ν πr 2 4 This suggests making use of the Hankel transform, defined for ν ∈ R+ by ∫ ∞ (8.12) Hν (g)(λ) = g(r)Jν (λr)r dr. 0 58 ( ) ∞ ∞ → ∞ R+ Clearly, Hν : C0 (0, ) L ( ). We will establish the following: ( ) ≥ ∞ ∞ Proposition 8.1. For ν 0,Hν extends uniquely from C0 (0, ) to 2 + 2 + (8.13) Hν : L (R , r dr) −→ L (R , λ dλ), unitary. Furthermore, for each g ∈ L2(R+, r dr),

(8.14) Hν ◦ Hν g = g.

To prove this, it is convenient to consider ﬁrst ∫ ∞ e Jν (λr) 2ν+1 (8.15) Hν f(λ) = f(r) ν r dr, 0 (λr) −ν since, by (8.10), (λr) Jν (λr) is a smooth function of λr. Set + (8.16) S(R ) = {f|R+ : f ∈ S(R) is even}.

Lemma 8.2. If ν ≥ −1/2, then

e + + (8.17) Hν : S(R ) −→ S(R ).

ν Proof. By (8.10), Jν (λr)/(λr) is a smooth function of λr. The formula (8.8) yields

J (λr) (8.18) ν ≤ C < ∞, (λr)ν ν for λr ∈ [0, ∞), ν > −1/2, a result that, by the identity ( ) 2 1/2 (8.19) J− (z) = cos z, 1/2 πz established in (6.35) of Chapter 3, also holds for ν = −1/2. This readily yields

e + ∞ + (8.20) Hν : S(R ) −→ L (R ), e whenever ν ≥ −1/2. Now consider the diﬀerential operator Lν , given by ( ) ∂ ∂f Le f(r) = −r−2ν−1 r2ν+1 ν ∂r ∂r (8.21) ∂2f 2ν + 1 ∂f = − − . ∂r2 r ∂r Using Bessel’s equation (8.9), we have ( ) J (λr) J (λr) (8.22) Le ν = λ2 ν , ν (λr)ν (λr)ν 8. The Laplace operator on cones 59 and, for f ∈ S(R+),

He (Le f)(λ) = λ2He f(λ), (8.23) ν ν ν e 2 e e Hν (r f)(λ) = Lν Hν f(λ). Since f ∈ L∞(R+) belongs to S(R+) if and only if arbitrary iterated ap- e 2 ∞ + plications of Lν and multiplication by r to f yield elements of L (R ), the result (8.17) follows. We also have that this map is continuous with respect to the natural Frechet space structure on S(R+).

+ Lemma 8.3. Consider the elements Eb ∈ S(R ), given for b > 0 by −br2 (8.24) Eb(r) = e . We have e (8.25) Hν E1/2(λ) = E1/2(λ), and more generally e −ν−1 (8.26) Hν Eb(λ) = (2b) E1/4b(λ).

Proof. To establish (8.25), plug the power series (8.10) for Jν (z) into (8.15) and integrate term by term, to get ∫ ∑∞ k −ν−2k ∞ (−1) 2 2 (8.27) He E (λ) = λ2k r2k+2ν+1e−r /2 dr. ν 1/2 k!Γ(k + ν + 1) k=0 0 This last integral is seen to equal 2k+ν Γ(k + ν + 1), so we have

∑∞ ( 2 ) 1 λ k 2 (8.28) He E (λ) = − = e−λ /2 = E (λ). ν 1/2 k! 2 1/2 k=0 Having (8.25), we get (8.26) by an easy change√ of variable argument.√ In more detail, set r2/2 = bs2, or s = r/ 2b. Then set µ = 2bλ, so λr = µs. Then (8.28), which we can write as ∫ ∞ −r2/2 ν+1 ν −λ2/2 (8.29) e Jν (λr)r dr = λ e , 0 translates to ∫ ∞ −bs2 (ν+1)/2 ν+1 1/2 −ν/2 ν −µ2/4b (8.30) e Jν (µs)(2b) s (2b) ds = (2b) µ e , 0 or, changing notation back, ∫ ∞ −bs2 ν+1 −ν−1 ν −λ2/4b (8.31) e Jν (λs)s ds = (2b) λ e , 0 which gives (8.26). 60

From (8.26) we have, for each b > 0, e e −ν−1 e (8.32) Hν Hν Eb = (2b) Hν E1/4b = Eb, which veriﬁes our stated Hankel inversion formula for f = Eb, b > 0. To get the inversion formula for general f ∈ S(R+), it suﬃces to establish the following.

Lemma 8.4. The space

(8.33) V = Span {Eb : b > 0} is dense in S(R+).

Proof. Let V denote the closure of V in S(R+). From ( ) 1 2 2 2 (8.34) e−br − e−(b+ε)r → r2e−br , ε 2 we deduce that r2e−br ∈ V, and inductively, we get

2 (8.35) r2je−br ∈ V, ∀ j ∈ Z+. From here, one has

2 (8.36) (cos ξr)e−r ∈ V, ∀ ξ ∈ R. Now each even ω ∈ S′(R) annihilating (8.36) for all ξ ∈ R has the property 2 that e−r ω has Fourier transform zero, which implies ω = 0. The assertion (8.33) then follows by the Hahn-Banach theorem.

Putting the results of Lemmas 8.2–8.4 together, we have

Proposition 8.5. Given ν ≥ −1/2, we have e e (8.37) Hν Hν f = f, for all f ∈ S(R+).

We promote this to

e Proposition 8.6. If ν ≥ −1/2, we have a unique extension of Hν from S(R+) to e 2 + 2ν+1 2 + 2ν+1 (8.38) Hν : L (R , r dr) −→ L (R , λ dλ), as a unitary operator, and (8.37) holds for all f ∈ L2(R+, r2ν+1 dr).

Proof. Take f, g ∈ S(R+), and use the inner product ∫ ∞ (8.39) (f, g) = f(r)g(r)r2ν+1 dr. 0 8. The Laplace operator on cones 61

ν Using Fubini’s theorem and the fact that Jν (λr)/(λr) is real valued and symmetric in (λ, r), we get the ﬁrst identity in e e e e (8.40) (Hν f, Hν g) = (Hν Hν f, g) = (f, g), the second identity following by Proposition 8.5. From here, given that the linear space S(R+) ⊂ L2(R+, r2ν+1 dr) is dense, the assertions of Proposi- tion 8.6 are apparent.

We return to the Hankel transform (8.12). Note that

ν ν e (8.41) Hν (r f)(λ) = λ Hν f(λ),

ν and that Mν f(r) = r f(r) has the property that

2 + 2ν+1 2 + (8.42) Mν : L (R , r dr) −→ L (R , r dr) is unitary. Thus Proposition 8.6 yields Proposition 8.1. Another proof is sketched in the exercises. An elaboration of Hankel’s original proof is given on pp. 456–464 of [Wat]. In view of (8.23) and (8.41), we have ∫ ∞ −α α m Hν (r Lµg) = Lµ(r Jν (λr))g r dr 0 ∫ ∞ (8.43) 2 α m = −λ gr Jν (λr)r dr 0 2 −α = −λ Hν (r g). Now from (8.5)–(8.13), it follows that the map H given by ( ) H −α −α (8.44) g = Hν0 (r g0),Hν1 (r g1),... provides an isometry of L2(C(N)) onto L2(R+, λ dλ, ℓ2), such that ∆ is carried into multiplication by −λ2. Thus (8.44) provides a spectral representation of ∆. Consequently, for well-behaved functions f, we have

f(−∆)g(r, x) ∑ ∫ ∞ ∫ ∞ (8.45) α 2 1−α = r f(λ )Jνj (λr)λ s Jνj (λs)gj(s) ds dλ φj(x). j 0 0 Now we can interpret (8.45) in the following fashion. Deﬁne the operator ν on N by ( ) 2 1/2 (8.46) ν = −∆N + α .

Thus νφj = νjφj. Identifying operators with their distributional kernels, we can describe the kernel of f(−∆) as a function on R+ ×R+ taking values 62 in operators on N, by the formula ∫ ∞ α 2 f(−∆) = (r1r2) f(λ )Jν (λr1)Jν (λr2)λ dλ (8.47) 0

= K(r1, r2, ν), since the volume element on C(N) is rm dr dS(x) if the m-dimensional area element of N is dS(x). At this point it is convenient to have in hand some calculations of Han- kel transforms, including some examples of the form (8.47). We establish some here;∫ many more can be found in [Wat]. Generalizing (8.31), we can ∞ −br2 µ+1 compute 0 e Jν (λr)r dr in a similar fashion, replacing the integral in (8.27) by ∫ ∞ ( ) 2 1 µ ν (8.48) r2k+µ+ν+1e−br dr = b−k−µ/2−ν/2−1Γ + + k + 1 . 0 2 2 2 We get ∫ ∞ −br2 µ+1 e Jν (λr)r dr 0 (8.49) ∞ ( ) ∑ Γ( µ + ν + k + 1) λ2 k = λν 2−ν−1b−µ/2−ν/2−1 2 2 − . k!Γ(k + ν + 1) 4b k=0 We can express the inﬁnite series in terms of the conﬂuent hypergeometric function, introduced in §7. A formula equivalent to (7.35) is ∞ Γ(b) ∑ Γ(a + k) zk (8.50) F (a; b; z) = , 1 1 Γ(a) Γ(b + k) k! k=0 since (a)k = a(a + 1) ··· (a + k − 1) = Γ(a + k)/Γ(a). We obtain, for Re b > 0, Re(µ + ν) > −2, (8.51) ∫ ∞ −br2 µ+1 e Jν (λr)r dr 0 ( ) Γ( µ + ν + 1) µ ν λ2 = λν 2−ν−1b−µ/2−ν/2−1 2 2 F + + 1; ν + 1; − . Γ(ν + 1) 1 1 2 2 4b

2 We can apply a similar attack when e−br is replaced by e−br, obtaining ∫ ∞ −br µ−1 e Jν (λr)r dr 0 (8.52) ( ) ∞ ( ) λ ν ∑ Γ(µ + ν + 2k) λ2 k = b−µ−ν − , 2 k!Γ(ν + k + 1) 2b2 k=0 at least provided Re b > |λ|, ν ≥ 0, and µ + ν > 0; here we use ∫ ∞ (8.53) e−brr2k+µ+ν−1 dr = b−2k−µ−ν Γ(µ + ν + 2k). 0 8. The Laplace operator on cones 63

The duplication formula for the gamma function (see (A.22) of Chapter 3) implies ( ) ( ) µ ν µ ν 1 (8.54) Γ(2k + µ + ν) = π−1/222k+µ+ν−1Γ + + k Γ + + k + , 2 2 2 2 2 so the right side of (8.52) can be rewritten as ∞ ( ) ∑ Γ( µ + ν + k)Γ( µ + ν + 1 + k) λ2 k (8.55) π−1/2λν 2µ−1b−µ−ν 2 2 2 2 2 − . k!Γ(ν + 1 + k) b2 k=0 This inﬁnite series can be expressed in terms of the hypergeometric function, deﬁned by

∑∞ k (a1)k(a2)k z 2F1(a1, a2; b; z) = (b)k k! k=0 (8.56) ∞ Γ(b) ∑ Γ(a + k)Γ(a + k) zk = 1 2 , Γ(a1)Γ(a2) Γ(b + k) k! k=0 for a1, a2 ∈/ {0, −1, −2,... }, |z| < 1. If we put the sum in (8.55) into this form, and use the duplication formula, to write ( ) ( ) µ ν µ ν 1 Γ(a )Γ(a ) = Γ + Γ + + = π1/22−µ−ν+1Γ(µ + ν), 1 2 2 2 2 2 2 we obtain ∫ ∞ −br µ−1 e Jν (λr)r dr 0 (8.57) ( ) ( ) λ ν Γ(µ + ν) µ ν µ ν 1 λ2 = b−µ−ν · F + , + + ; ν + 1; − . 2 Γ(ν + 1) 2 1 2 2 2 2 2 b2 This identity, established so far for |λ| < Re b (and ν ≥ 0, µ + ν > 0), continues analytically to λ in a complex neighborhood of (0, ∞). 2 To evaluate the integral (8.47) with f(λ2) = e−tλ , we can use the power series (8.10) for Jν (λr1) and for Jν (λr2) and integrate the resulting double series term by term using (8.48). We get (8.58) ∫ ∞ −tλ2 e Jν (r1λ)Jν (r2λ)λ dλ 0 ( ) ( ) ( ) 1 r r ν ∑ Γ(ν + j + k + 1) 1 r2 j r2 k = 1 2 − 1 − 2 , 2t 4t Γ(ν + j + 1)Γ(ν + k + 1) j!k! 4t 4t j,k≥0 for any t, r1, r2 > 0, ν ≥ 0. This can be written in terms of the modiﬁed Bessel function Iν (z), given by ( ) ∞ ( ) z ν ∑ 1 z 2k (8.59) I (z) = . ν 2 k!Γ(ν + k + 1) 2 k=0 64

One obtains the following, known as the Weber identity.

Proposition 8.7. For t, r1, r2 > 0, ∫ ∞ ( ) −tλ2 1 −(r2+r2)/4t r1r2 (8.60) e Jν (r1λ)Jν (r2λ)λ dλ = e 1 2 Iν . 0 2t 2t

Proof. The left side of (8.60) is given by (8.58). Meanwhile, by (8.59), ν the right side of (8.60) is equal to (1/2t)(r1r2/4t) times ( ) ( ) ∞ ( ) ∑ 1 r2 ℓ r2 m ∑ 1 r r 2n (8.61) − 1 − 2 1 2 . ℓ!m! 4t 4t n!Γ(ν + n + 1) 4t ℓ,m≥0 n=0 − 2 If we set yj = rj /4t, we see that the asserted identity (8.60) is equivalent to the identity ∑ Γ(ν + j + k + 1) 1 yjyk Γ(ν + j + 1)Γ(ν + k + 1) j!k! 1 2 j,k≥0 (8.62) ∑ 1 1 = yℓ+nym+n. ℓ!m! n!Γ(ν + n + 1) 1 2 ℓ,m,n≥0

j k We compare coeﬃcients of y1y2 in (8.62). Since both sides of 8.62) are symmetric in (y1, y2), it suﬃces to treat the case (8.63) j ≤ k, which we assume henceforth. Then we take ℓ + n = j, m + n = k and sum over n ∈ {0, . . . , j}, to see that (8.62) is equivalent to the validity of (8.64) j ∑ 1 Γ(ν + j + k + 1) 1 = , (j − n)!(k − n)!n!Γ(ν + n + 1) Γ(ν + j + 1)Γ(ν + k + 1) j!k! n=0 whenever 0 ≤ j ≤ k. Using the identity Γ(ν + j + 1) = (ν + j) ··· (ν + n + 1)Γ(ν + n + 1) and its analogues for the other Γ-factors in (8.64), we see that (8.64) is equivalent to the validity of

j ∑ j!k! (8.65) (ν+j) ··· (ν+n+1) = (ν+j+k) ··· (ν+k+1), (j − n)!(k − n)!n! n=0 for 0 ≤ j ≤ k. Note that the right side of (8.65) is a polynomial of degree j in ν, and the general term on the left side of (8.65) is a polynomial of degree j − n in ν. In order to establish (8.65), it is convenient to set (8.66) µ = ν + j 8. The Laplace operator on cones 65 and consider the associated polynomial identity in µ. With

p (µ) = 1, p (µ) = µ, p (µ) = µ(µ − 1),... (8.67) 0 1 2 pj(µ) = µ(µ − 1) ··· (µ − j + 1), we see that {p0, p1, . . . , pj} is a basis of the space Pj of polynomials of degree j in µ, and our task is to write

(8.68) pj(µ + k) = (µ + k)(µ + k − 1) ··· (µ + k − j + 1) as a linear combination of p0, . . . , pj. To this end, deﬁne

(8.69) T : Pj −→ Pj, T p(µ) = p(µ + 1).

By explicit calculation,

p (µ + 1) = p (µ) + p (µ), (8.70) 1 1 0 p2(µ + 1) = (µ + 1)µ = µ(µ − 1) + 2µ = p2(µ) + 2p1(µ), and an inductive argument gives

(8.71) T pi = pi + ipi−1.

k By convention we set pi = 0 for i < 0. Our goal is to compute T pj. Note that

(8.72) T = I + N, Npi = ipi−1, and

j ( ) ∑ k (8.73) T k = N n, n n=0 if j ≤ k. By (8.72),

n (8.74) N pi = i(i − 1) ··· (i − n + 1)pi−n, so we have ( ) ∑j k k T p = j(j − 1) ··· (j − n + 1)p − j n j n n=0 (8.75) j ∑ k! j! = p − . (k − n)!n! (j − n)! j n n=0

This veriﬁes (8.65) and completes the proof of (8.60). 66

Similarly we can evaluate (8.47) with f(λ2) = e−tλ/λ, as an inﬁnite series, using (8.53) to integrate each term of the double series. We get (8.76) ∫ ∞ −tλ e Jν (r1λ)Jν (r2λ) dλ 0 ( ) ( ) ( ) 1 r r ν ∑ Γ(2ν + 2j + 2k + 1) 1 r2 j r2 k = 1 2 − 1 − 2 , t t2 Γ(ν + j + 1)Γ(ν + k + 1) j!k! 4t2 4t2 j,k≥0 provided t > rj > 0. It is possible to express this integral in terms of the Legendre function Qν−1/2(z).

Proposition 8.8. One has, for all y, r1, r2 > 0, ν ≥ 0, ∫ ∞ ( 2 2 2 ) −yλ 1 −1/2 r1 + r2 + y (8.77) e Jν (r1λ)Jν (r2λ) dλ = (r1r2) Qν−1/2 . 0 π 2r1r2

The Legendre functions Pν−1/2(z) and Qν−1/2(z) are solutions to [ ] ( ) d d 1 (8.78) (1 − z2) u(z) + ν2 − u(z) = 0; dz dz 4 Compare with (4.52). Extending (4.41), we can set ∫ θ 2 ( )−1/2 (8.79) Pν−1/2(cos θ) = 2 cos s − 2 cos θ cos νs ds, π 0 and Qν−1/2(z) can be deﬁned by the integral formula ∫ ∞( ) −1/2 −sν (8.80) Qν−1/2(cosh η) = 2 cosh s − 2 cosh η e ds. η The identity (8.77) is known as the Lipschitz-Hankel integral formula.

Proof of Proposition 8.8. We derive (8.77) from the Weber identity (8.60). Recall

−πiν/2 (8.81) Iν (y) = e Jν (iy), y > 0. To work with (8.60), we use the subordination identity ∫ ∞ λ 2 2 (8.82) e−yλ = √ e−y /4te−tλ t−1/2 dt; π 0 cf. Chapter 3, (5.31) for a proof. Plugging this into the left side of (8.77), and using (8.60), we see that the left side of (8.77) is equal to ∫ ∞ ( ) 1 −(r2+r2+y2)/4t r1r2 −3/2 (8.83) √ e 1 2 Iν t dt. 2 π 0 2t 8. The Laplace operator on cones 67

The change of variable s = r1r2/2t gives √ ∫ ∞ 1 2 2 2 −1/2 −s(r +r +y )/2r1r2 −1/2 (8.84) (r1r2) e 1 2 Iν (s)s ds. 2π 0 Thus the asserted identity (8.77) follows from the identity √ ∫ ∞ −sz −1/2 2 (8.85) e Iν (s)s ds = Qν−1/2(z), z > 0. 0 π As for the validity of (8.85), we mention two identities. Recall from (8.57) that ∫ ∞ ( )ν −sz µ−1 λ −µ−ν Γ(µ + ν) e Jν (λs)s ds = z 0 2 Γ(ν + 1) (8.86) ( ) µ ν 1 µ ν λ2 · F + + , + ; ν + 1; − . 2 1 2 2 2 2 2 z2

Next, there is the classical representation of the Legendre function Qν−1/2(z) as a hypergeometric function: (8.87) 1 1 ( ) Γ( 2 )Γ(ν + 2 ) −ν−1/2 ν 3 ν 1 1 Q − (z) = (2z) F + , + ; ν + 1; ; ν 1/2 Γ(ν + 1) 2 1 2 4 2 4 z2 cf. [Leb], (7.3.7). If we apply (8.86) with λ = i, µ = 1/2 (keeping (8.81) in mind), then (8.85) follows.

Remark. Formulas (8.77) and (8.60) are proven in the opposite order in [W].

By analytic continuation, we can treat f(λ2) = e−ελλ−1 sin λt for any ε > 0. We apply this to (8.47). Letting ε ↘ 0, we get for the fundamental solution to the wave equation:

(−∆)−1/2 sin t(−∆)1/2 ∫ ∞ α −(ε+it)λ = − lim (r1r2) Im e Jν (λr1)Jν (λr2) dλ ↘ (8.88) ε 0 0 ( 2 2 2 ) 1 α−1/2 r1 + r2 + (ε + it) = − (r1r2) lim Im Qν−1/2 . π ε↘0 2r1r2

Using the integral formula (8.80), where the path of integration is a suit- able path from η to +∞ in the complex plane, one obtains the following alternative integral representation of (−∆)−1/2 sin t(−∆)1/2. The Schwartz 68 kernel is equal to

(8.89) 0, if t < |r1 − r2|, ∫ β1 [ ] 1 α 2 − 2 2 − −1/2 (8.90) (r1r2) t (r1 + r2 2r1r2 cos s) cos νs ds, π 0 if |r1 − r2| < t < r1 + r2, and ∫ ∞[ ] 1 α 2 2 − 2 −1/2 −sν (8.91) (r1r2) cos πν r1 + r2 + 2r1r2 cosh s t e ds, π β2 if t > r1 + r2, where ( 2 2 − 2 ) ( 2 − 2 − 2 ) −1 r1 + r2 t −1 t r1 r2 (8.92) β1 = cos , β2 = cosh . 2r1r2 2r1r2 Recall that α = −(m − 1)/2, where m = dim N. We next show how formulas (8.89)–(8.91) lead to an analysis of the classical problem of diffraction of waves by a slit along the positive x-axis in the plane R2. In fact, if waves propagate in R2 with this ray removed, on which Dirichlet boundary conditions are placed, we can regard the space as the cone over an interval of length 2π, with Dirichlet boundary conditions at the endpoints. By the method of images, it suffices to analyze the case of the cone over a circle of circumference 4π (twice the circumference of the standard unit circle). Thus C(N) is a double cover of R2 \ 0 in this case. We divide up the spacetime into regions I, II, and III, respectively, as described by (8.89), (8.90), and (8.91). Region I contains only points on C(N) too far away from the source point to be influenced by time t; that the fundamental solution is 0 here is consistent with finite propagation speed. Since the circle has dimension m = 1, we see that ( ) d2 1/2 (8.93) ν = (−∆ )1/2 = − N dθ2 in this case if θ ∈ R/(4πZ) is the parameter on the circle of circumference 4π. On the line, we have 1[ ] (8.94) cos sν δ (θ ) = δ(θ − θ + s) + δ(θ − θ − s) . θ1 2 2 1 2 1 2 To get cos sν on R/(4πZ), we simply make (8.94) periodic by the method of images. Consequently, from (8.90), the wave kernel (−∆)−1/2 sin t(−∆)1/2 is equal to [ ] −1 2 2 2 −1/2 (2π) t − r − r + 2r1r2 cos(θ1 − θ2) if |θ1 − θ2| ≤ π, (8.95) 1 2 0 if |θ1 − θ2| > π, in region II. Of course, for |θ1 − θ2| < π this coincides with the free space fundamental solution, so (8.95) also follows by finite propagation speed. 8. The Laplace operator on cones 69

Figure 8.1

We turn now to an analysis of region III. In order to make this analysis, it is convenent to make simultaneous use both of (8.91) and of another formula for the wave kernel in this region, obtained by choosing another path from η to ∞ in the integral representation (8.80). The formula (8.91) is obtained by taking a horizontal line segment; see Fig. 8.1. If instead we take the path indicated in Fig. 8.2, we obtain the following formula for (−∆)−1/2 sin t(−∆)1/2 in region III: {∫ π( ) −1 −(m−1)/2 2 − 2 − 2 −1/2 π (r1r2) t r1 r2 + 2r1r2 cos s cos sν ds 0 (8.96) ∫ } β2 ( ) − 2 − 2 − 2 − −1/2 −sν sin πν t r1 r2 2r1r2 cosh s e ds . 0

Figure 8.2 70

The operator ν on R/(4πZ) given by (8.93) has spectrum consisting of { } 1 3 (8.97) Spec ν = 0, , 1, , 2,... , 2 2 all the eigenvalues except for 0 occurring with multiplicity 2. The formula (8.91) shows the contribution coming from the half-integers in Spec ν van- 1 ishes, since cos 2 πn = 0 if n is an odd integer. Thus we can use formula (8.96) and compose with the projection onto the sum of the eigenspaces of ν with integer spectrum. This projection is given by (8.98) P = cos2 πν on R/(4πZ). Since sin πn = 0, in the case N = R/(4πZ) we can rewrite (8.96) as ∫ π( ) −1 −(m−1)/2 2 − 2 − 2 −1/2 (8.99) π (r1r2) t r1 r2 + 2r1r2 cos s P cos sν ds. 0 In view of the formulas (8.94) and (8.96), we have

P cos sν δθ1 (θ2) 1[ (8.100) = δ(θ1 − θ2 + s) + δ(θ1 − θ2 − s) 4 ] + δ(θ1 − θ2 + 2π + s) + δ(θ1 − θ2 + 2π − s) mod 4π.

Thus, in region III, we have for the wave kernel (−∆)−1/2 sin t(−∆)1/2 the formula ( ) −1 2 − 2 − 2 − −1/2 (8.101) (4π) t r1 r2 + 2r1r2 cos(θ1 θ2) .

Thus, in region III, the value of the wave kernel at points (r1, θ1), (r2, θ2) of the double cover of R2 \ 0 is given by half the value of the wave kernel on R2 at the image points. The jump in behavior from (8.95) to (8.101) gives rise to a diﬀracted wave.

Figure 8.3 8. The Laplace operator on cones 71

We depict the singularities of the fundamental solution to the wave equation for R2 minus a slit in Figs. 8.3 and 8.4. In Fig. 8.3 we have the situation |t| < r1, where no diﬀraction has occurred, and region III is empty. In Fig. 8.4 we have a typical situation for |t| > r1, with the diﬀracted wave labeled by a “D.”

Figure 8.4

This diffraction problem was first treated by Sommerfeld [Som] and was the first diffraction problem to be rigorously analyzed. For other approaches to the diffraction problem presented above, see [BSU] and [Stk]. Generally, the solution (8.89)–(8.91) contains a diffracted wave on the boundary between regions II and III. In Fig. 8.5 we illustrate the diffraction of a wave by a wedge; here N is an interval of length ℓ < 2π. We now want to provide, for general N, a description of the behavior of the distribution − −1/2 − 1/2 v = ( ∆) sin t( ∆) δ(r2,x2) near this diffracted wave, that is, a study of the limiting behavior as r1 ↘ t − r2 and as r1 ↗ t − r2. We begin with region II. From (8.90), we have v equal to

1 α−1/2 (8.102) (r r ) P − (cos β ) δ in region II, 2 1 2 ν 1/2 1 x2 where Pν−1/2 is the Legendre function deﬁned by (8.79) and β1 is given by (8.92). Note that as r1 ↘ t − r2, β1 ↗ π. To analyze (8.102), replace s by π − s in (8.79), and, with δ1 = π − β1, write ∫ π π ( )−1/2 P − (cos β ) = cos πν 2 cos δ − 2 cos s cos sν ds 2 ν 1/2 1 1 δ1∫ (8.103) π ( )−1/2 + sin πν 2 cos δ1 − 2 cos s sin sν ds. δ1 72

Figure 8.5

As δ1 ↘ 0, the second term on the right tends in the limit to ∫ π sin sν (8.104) sin πν 1 ds. 0 sin 2 s Write the ﬁrst term on the right side of (8.103) as ∫ π −1/2 cos πν (2 cos δ1 − 2 cos s) (cos sν − 1) ds δ1 ∫ (8.105) π −1/2 + cos πν (2 cos δ1 − 2 cos s) ds. δ1

As δ1 ↘ 0, the ﬁrst term here tends in the limit to ∫ π cos sν − 1 (8.106) cos πν 1 ds. 0 sin 2 s The second integral in (8.105) is a scalar, independent of ν, and it is easily seen to have a logarithmic singularity. More precisely, ∫ π − 1 (2 cos δ1 − 2 cos s) 2 ds δ1 (8.107) ( ) ∞ ∞ 2 ∑ ∑ ∼ log A δj + B δj,A = 1. δ j 1 j 1 0 1 j=0 j=1 Consequently, one derives the following. 8. The Laplace operator on cones 73

Proposition 8.9. Fix (r2, x2) and t. Then, as r1 ↘ t − r2,

−1/2 1/2 (−∆) sin t(−∆) δ(r ,x ) { 2 2 1 α−1/2 2 = (r1r2) log cos πν δx2 (8.108) π δ1 ∫ } π cos sν − cos πν + 1 ds δx2 + R1 δx2 , 0 2 cos 2 s where, for s > (m + 1)/2,

1 ∥ ∥ −s−1 ≤ ↘ (8.109) R1 δx2 D Cδ1 log , as δ1 0. δ1

The following result analyzes the second term on the right in (8.108).

Proposition 8.10. We have ∫ ( ) π 1 −1 2 cos s (cos sν − cos πν) ds 0 2 (8.110) { K } ∑ π = cos πν − log ν + a ν−2j + sin πν + S (ν), j 2 K j=0

s s+2K where SK (ν): D → D , for all s.

The spaces Ds are spaces of generalized functions on N, introduced in Chapter 5, Appendix A. We turn to the analysis of v in region III. Using (8.91), we can write v as

1 α−1/2 (8.111) (r r ) cos πν Q − (cosh β ) δ , in region III, π 1 2 ν 1/2 2 x2 where Qν−1/2 is the Legendre function given by (8.80) and β2 is given by (8.92). It is more convenient to use (8.96) instead; this yields for v the formula ∫ { π 1 α−1/2 −1/2 (r1r2) (2 cosh β2 + 2 cos s) cos sν ds π 0 (8.112) ∫ β2 } −1/2 −sν − sin πν (2 cosh β2 − 2 cosh s) e ds . 0

Note that as r1 ↗ t − r2, β2 ↘ 0. 74

The ﬁrst integral in (8.112) has an analysis similar to that arising in (8.103); ﬁrst replace s by π − s to rewrite the integral as ∫ π −1/2 cos πν (2 cosh β2 − 2 cos s) cos sν ds 0 ∫ (8.113) π −1/2 + sin πν (2 cosh β2 − 2 cos s) sin sν ds. 0

As β2 ↘ 0, the second term in (8.113) tends to the limit (8.104), and the ﬁrst term also has an analysis similar to (8.105)–(8.107), with (8.107) replaced by ∫ π −1/2 (2 cosh β2 − 2 cos s) ds 0 (8.114) ( ) ∑ ∑ ∼ 2 ′ j ′ j ′ log Ajβ2 + Bjβ2,A0 = 1. β2 j≥0 j≥1

It is the second term in (8.112) that leads to the jump across r1 = t − r2, hence to the diﬀracted wave. We have ∫ ∫ β2 β2 −1/2 −sν ds π (8.115) (2 cosh β2 − 2 cosh s) e ds ∼ √ = . 2 − 2 2 0 0 β2 s Thus we have the following:

Proposition 8.11. For r1 ↗ t − r2,

−1/2 1/2 (−∆) sin t(−∆) δ(r ,x ) { 2 2 1 2 = (r r )α−1/2 log cos πν δ (8.116) π 1 2 β x2 ∫ 2 π − } cos sν cos πν − π e + 1 ds δx2 sin πν δx2 + R1δx2 , 0 2 cos 2 s 2 where, for s > (m + 1)/2, 1 ∥ ∥ −s−1 ≤ ↘ (8.117) R1δx2 D Cβ2 log , as β2 0. β2

−1 α−1/2 Note that (8.116) diﬀers from (8.108) by the term π (r1r2) times π (8.118) − sin πν δ . 2 x2 This contribution represents a jump in the fundamental solution across the α−1/2 diﬀracted wave D. There is also the logarithmic singularity, (r1r2) times 1 2 (8.119) log cos πν δ , π δ x2 Exercises 75 where δ = δ1 in (8.108) and δ = β2 in (8.116). In the special case where

N is an interval [0,L], so dim C(N) = 2, cos πν δx2 is a sum of two delta functions. Thus its manifestation in such a case is subtle. We also remark that if N is a subdomain of the unit sphere S2k (of even \ dimension), then cos πν δx2 vanishes on the set N N0, where

(8.120) N0 = {x1 ∈ N : for some y ∈ ∂N, dist(x2, y) + dist(y, x1) ≤ π}.

Thus the log blow-up disappears on N \N0. This follows from the fact that 2k cos πν0 = 0, where ν0 is the operator (8.46) on S , together with a ﬁnite propagation speed argument. While Propositions 8.9–8.11 contain substantial information about the nature of the diﬀracted wave, this information can be sharpened in a number of respects. A much more detailed analysis is given in [CT].

Exercises

1. Using (7.36) and (7.80), work out the asymptotic behavior of 1F1(a; b; −z) as z → +∞, given b, b−a∈ / {0, −1, −2,... }. Deduce from (8.51) that whenever ν ≥ 0, s ∈ R, ( ) ∫ ∞ Γ 1 (ν + 1 − is) −br2 −is −is 2 (8.121) lim e Jν (r)r dr = 2 ( ) . ↘ b 0 0 1 Γ 2 (ν + 1 + is) 2. Deﬁne operators

−1 (8.122) Mrf(r) = rf(r), J f(r) = f(r ). Show that (8.123) 2 + 2 + −1 2 + −1 2 + −1 Mr : L (R , r dr) −→ L (R , r dr), J : L (R , r dr) −→ L (R , r dr) are unitary. Show that

# −1 (8.124) Hν = J MrHν Mr is given by

# (8.125) Hν f(λ) = (f ⋆ ℓν )(λ), where ⋆ denotes the natural convolution on R+, with Haar measure r−1dr: ∫ ∞ − − (8.126) (f ⋆ g)(λ) = f(r)g(r 1λ)r 1 dr, 0 and

−1 −1 (8.127) ℓν (r) = r Jν (r ). 3. Consider the Mellin transform: ∫ ∞ − (8.128) M#f(s) = f(r)ris 1 dr. 0 76

As shown in (A.17)–(A.20) of Chapter 3, we have − − (8.129) (2π) 1/2M# : L2(R+, r 1 dr) −→ L2(R, ds), unitary. Show that (8.130) M#(f ⋆ g)(s) = M#f(s) ·M#g(s), and deduce that

# # # (8.131) M Hν f(s) = Ψ(s)M f(s), where (8.132) ( ) ∫ ∫ ∞ ∞ Γ 1 (ν + 1 − is) −1 is−2 −is −is 2 Ψ(s) = Jν (r )r dr = Jν (r)r dr = 2 ( ) . 0 0 1 Γ 2 (ν + 1 + is)

4. From (8.126)–(8.132), give another proof of the unitarity (8.13) of Hν . Using symmetry, deduce that spec Hν = {−1, 1}, and hence deduce again the inversion formula (8.14). 5. Verify the asymptotic expansion (8.107). (Hint: Write 2 cos δ−2 cos s = (s2 − δ2)F (s, δ) with F smooth and positive, F (0, 0) = 1. Then, with G(s, δ) = F (s, δ)−1/2, ∫ ∫ π π − ds (8.133) (2 cos δ − 2 cos s) 1/2 ds = G(s, δ) √ . 2 2 δ δ s − δ Write G(s, δ) = g(s) + δH(s, δ), g(0) = 1, and verify that (8.133) is equal to A1 + A2, where ∫ ( ) π ds 1 1 A = G(s, δ) = g(0) log + O δ log , 1 s δ δ ∫δ [ ] π 1 1 A2 = g(s) √ − ds + O(δ) = B2 + O(δ). 2 2 δ s − δ s Show that ∫ [ ] π/δ 1 1 B2 = g(0) √ − dt + O(δ) = C2 + O(δ), 2 1 t − 1 t with ∫ [ ] ∞ 1 1 C2 = √ − dt. 2 1 t − 1 t

Use the substitution t = cosh u to do this integral and get C2 = log 2.) Next, verify the expansion (8.114).

Exercises on the hypergeometric function Exercises 77

1. Show that 2F1(a1, a2; b; z), deﬁned by (8.56), satisﬁes ∫ 1 Γ(b) − − − − a2 1 − b a2 1 − a1 (8.134) 2F1(a1, a2; b; z) = − t (1 t) (1 tz) dt, Γ(a2)Γ(b a2) 0

for Re b > Re a2 > 0, |z| < 1. (Hint: Use the beta function identity, (A.23)–(A.24) of Chapter 3, to write ∫ 1 (a ) Γ(b) − − − 2 k a2 1+k − b a2 1 = − t (1 t) dt, k = 0, 1, 2,..., (b)k Γ(a2)Γ(b a2) 0 and substitute this into (8.39). Then use ∑∞ (a ) − 1 k (zt)k = (1 − tz) a1 , 0 ≤ t ≤ 1, |z| < 1.) k! k=0

2. Show that, given Re b > Re a2 > 0, (8.134) analytically continues in z to z ∈ C \ [1, ∞). 3. Show that the function (8.134) satisﬁes the ODE { } d2u du z(1 − z) + b − (a + a + 1)z − a a u = 0. dz2 1 2 dz 1 2 ′ Note that u(0) = 1, u (0) = a1a2/b, but zero is a singular point for this ODE. Show that another solution is 1−b u(z) = z 2F1(a1 − b + 1, a2 − b + 1; 2 − b; z). 4. Show that ( ) −a1 −1 2F1(a1, a2; b; z) = (1 − z) 2F1 a1, b − a2; b;(z − 1) z .

(Hint: Make a change of variable s = 1 − t in (8.134).) For many other important transformation formulas, see [Leb] or [WW]. 5. Show that −1 1F1(a; b; z) = lim 2F1(a, c; b; c z). c↗∞

We mention the generalized hypergeometric function, deﬁned by

∑∞ k (a)k z pFq(a; b; z) = , (b)k k! k=0

where p ≤ q + 1, a = (a1, . . . , ap), b = (b1, . . . , bq), bj ∈ C \{0, −1, −2,... }, |z| < 1, and

(a)k = (a1)k ··· (ap)k, (b)k = (b1)k ··· (bq)k,

and where, as before, for c ∈ C, (c)k = c(c + 1) ··· (c + k − 1). For more on this class of functions, see [Bai]. 6. The Legendre function Qν−1/2(z) satisﬁes the identity (8.87), for ν ≥ 0, |z| > 2 2 2 1, and |Arg z| < π; cf. (7.3.7) of [Leb]. Take z = (r1 + r2 + t )/2r1r2, and compare the resulting power series for the right side of (8.77) with the power series in (8.76). 78

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