On the analytic continuation of the Poisson kernel

Matthew B. Stenzel

Abstract

We give a “heat equation” proof of a theorem which says that for

all ε sufficiently small, the map Sε : f 7→ exp(−ε√

∆)f extends to an

isomorphism fromHs(X) toOs+(n−1)/4(∂Mε). This result was announced

by L. Boutet de Monvel in 1978 but only recently has a proof, due to S.

Zelditch [23], appeared in the literature. The main tools in our proof are

the subordination formula relating the Poisson kernel to the heat kernel,

and an expression for the singularity of the Poisson kernel in the complex

domain in terms of the Laplace transform variable s = d2(z, y) + ε2 where

d2 is the analytic continuation of the distance function squared on X,

z ∈Mε, and y ∈ X.

1 Introduction

Let (X, g) be a compact, connected, n-dimensional real analytic Riemannian

manifold without boundary and with Laplace operator ∆ ≥ 0. The Poisson

operator, exp(−τ√

∆), is the solution operator at time t to the pseudodifferential

initial value problem

(∂τ +√

∆)u = 0

u|τ=0 = f.

L. Boutet de Monvel observed in [3] that this operator has a remarkable analytic

continuation property. Let X ↪→ M be an embedding of X as a totally real

submanifold of a complex manifold M . Then for each sufficiently small τ >

0 there is a complex neighborhood Mτ of X such that the Schwartz kernel

of exp(−τ√

∆) can be extended to a smooth function on Mτ × X which is

holomorphic on Mτ . The restriction to ∂Mτ × X of this kernel is a Fourier

integral distribution of complex type and the corresponding Fourier integral

operator, which we denote by Sτ , is a continuous bijection between Hs(X) and

the Sobolev space Os+n−14 (∂Mτ ) of boundary values of holomorphic functions

1

on Mτ :

Sτ : Hs(X)∼=−→ Os+

n−14 (∂Mτ ).

The Mτ are the sublevel sets of a certain non-negative strictly plurisubharmonic

function, Mτ = {φ < τ}, and X is identified with φ−1(0). One consequence of

this is a Paley-Wiener type theorem on manifolds [9]: a function on X can be

analytically continued to a holomorphic function on Mτ with Sobolev regularity

of order s + n−14 at the boundary if and only if it is in the image of e−τ

√∆

acting onHs(X). Other applications include the distribution of complex zeros of

Laplace eigenfunctions [22], Weyl-type estimates on the growth of eigenfunctions

in the complex domain [23], and microlocal analysis of analytic singularities [18].

The announcement in [3] gave a sketch of the proof of this result, but a proof

could not be found in the literature until the work of S. Zelditch [23]. Since

the goal of this paper is to give another proof we first review Zelditch’s proof.

Let UF be the Hadamard-Feynman fundamental solution to the wave equation,

(∂2t + ∆)u = 0. Hadamard [12] showed that if (X, g) is real analytic then the

Schwartz kernel of UF has the following form. Let Γ(t, x, y) = t2 − d2(x, y). If

n = dim(X) is even, then

UF (t, x, y) = Γ−n−12

∞∑j=0

Uj(t, x, y)Γj (1)

and if n is odd, then

UF (t, x, y) = Γ−n−12

n∑j=0

Uj(t, x, y)Γj+log(Γ)

∞∑j=0

Vj(t, x, y)Γj+

∞∑j=1

Wj(t, x, y)Γj .

(2)

If we choose the complexification M small enough, then all of the infinite series

above converge to functions which are holomorphic in a neighborhood of the

complex zero set of (the analytic continuation of) Γ in C ×M ×M . Zelditch

writes the expressions (1), (2) as a Fourier integrals on X with complex valued

phase function ψ(t+ iτ, x, y, θ) = θ((t+ iτ)2 − d2(x, y)), τ > 0, and shows that

the amplitude is a formal analytic symbol of degree (n− 3)/2. He observes that

the Poisson-wave operator for t > 0 can be obtained by differentiating UF :

eit√

∆ =d

idtUF (t) if t > 0.

This gives an expresion for the Schwartz kernel of ei(t+iτ)√

∆ of the form (1),

(2) (with the exponent −(n− 1)/2 replaced by −(n+ 1)/2) and also as Fourier

integrals on X with the same phase and formal analytic symbol of degree (n−

2

1)/2 (for t > 0 and τ > 0). The parametrix can be modified (by factoring the

phase function t2 − d2) so that it is valid when t ≥ 0, τ > 0 and d 6= 0.

Zelditch then considers the analytic continuation of the Schwartz kernel

ei(·)√

∆(t + iτ, x, y), to complex values of x (with t ≥ 0, τ > 0). Since the

formal series in (1), (2) converge to a holomorphic function in a neighborhood

of the complex zero set of Γ in C×M×M , the amplitude of the Fourier integral

expression for ei(·)√

∆(t + iτ, x, y) is also a formal analytic symbol for complex

values of x. Setting t = 0 Zelditch concludes that the operator Sτ described

above is a complex Fourier integral operator.

In this article we re-prove this result using a slightly different approach and

fill in some details which have not appeared in the literature to date. Instead of

starting with the Hadamard parametrix we start with the Minakshisundaram-

Pleijel parametrix for the heat equation and use the subordination formula to

relate the Poisson kernel to the heat kernel. This method, already suggested

by Zelditch in [23], has much in common with the wave equation approach

and leads to the same results. A possible advantage is it shows how the well-

known Minakshisundaram-Pleijel parametrix is related to the parametrix for

the Poisson kernel. We also prove the bijectivity1 of Sτ (following the outline

in [4]) which does not seem to have appeared in the literature.

Let P ∈ Cω((0,∞) × X × X) be the Schwartz kernel of exp(−t√

∆). For

s ∈ R, let Os(∂Mε) be the closure in the L2-Sobolev space Hs(∂Mε) of the set

of restrictions to ∂Mε of holomorphic functions on Mε which are smooth on the

closure of Mε (or, equivalently, the closure of the kernel of ∂b in Hs(∂Mε)). Our

goal is to give a “heat equation” proof of the following theorem.

Theorem 1 (Boutet de Monvel [3; 4; 5, Appendix 6.4; 11, Theorem 5.1]). There

exists an ε0 > 0 such that for each fixed ε ∈ (0, ε0):

1. For each fixed y ∈ X, the map x→ P (ε, x, y) can be analytically continued

to Mε.

2. The restriction of P (ε, ·, ·) to ∂Mε × X is a Fourier integral distribution

with complex phase, of degree −(n− 1)/4.

3. Let Sε be the operator whose Schwartz kernel is the restriction of P (ε, ·, ·)to ∂Mε×X. For all s ∈ R, Sε is a continuous bijection from Hs(X) onto

Os+(n−1)/4(∂Mε).

1It’s easy to see that the kernel and cokernel of Sτ are finite dimensional by elliptic theory

(we thank S. Zelditch for this observation).

3

To establish the analytic continuation of P (ε, ·, y) to Mε, we use the subor-

dination formula to relate the Poisson kernel to the heat kernel: if E(t, x, y) is

the heat kernel, then for (ε, x, y) ∈ (0,∞)×X ×X

P (ε, x, y) =ε√4π

∫ ∞0

E(t, x, y)e−ε2/4tt−1/2 dt

t(3)

(see [17, Theorem 2.1]). For z complex and close to y the properties of the

analytic continuation of the heat kernel proved in [18] can be used to construct

the analytic continuation of P . If z is not close to y then the relationship

between P and the wave kernel together with the finite propagation speed of

analytic singularities of the wave kernel gives the analytic continuation of P

(Corollary 2).

The restriction of P (ε, ·, y) to ∂Mε will become singular and we must show

that the restriction is a Fourier integral distribution of complex type. To do so

we use the results of [18] (see Theorem 3) to re-write (3) as

P (ε, z, y) ≈ ε(4π)−n+12

∫ ∞0

e−(d2(z,y)+ε2)θ/4a(θ, z, y) dθ (4)

where d2(·, y) is the analytic continuation of the distance function in the first

variable and a(θ, z, y) is an analytic symbol of order (n+1)/2. Here z, y are in a

complex neighborhood X ⊂M ×X of a neighborhood of the diagonal in X×X(see Remark 1, and Remark 4 for a oscillatory integral description), and≈means

modulo a function which is smooth in (ε, z, y) ∈ {|ε| < ε0}×X and holomorphic

in ε and z. In the proof of Theorem 1, part 2, we show that φ(θ, z, y) =

i(d2(z, y) + ε2)θ/4 is a regular complex phase function of positive type and that

the operator Sε associated with P by the Schwartz kernel theorem is in fact

a Fourier integral operator of complex type. In Remark 2 we verify that the

isotropic submanifold of (T ∗∂Mε\0)× (T ∗X\0) generated by φ is the (twisted)

graph of an isomorphism of symplectic cones as in [5, Appendix A.6.4]. To show

that Sε is a continuous map between the appropriate Sobolev spaces we verify

Hormander’s criteria for the L2 continuity of Fourier integral operators with

complex phase [14, Theorem 3.5]. To do so requires considering the complex

Lagrangian submanifold Λφ. We verify it has the appropriate structure in the

proof of Theorem 1, part 3.

To verify that Sε is a bijection from Hs(X) to Os+(n−1)/4(∂Mε) we show,

following the outline in [4], that exp(−ε√

∆) maps Hs(X) onto the space of

restrictions to X of functions in Os+(n−1)/4(∂Mε) by showing that the inverse

operator, exp(+ε√

∆), is well-defined on such restrictions (Lemma 6). The main

idea is to use (4) to write the Poisson kernel as the Laplace transform of the

4

symbol a in the Laplace transform variable s = (d2(z, y) + ε2)/4 (here s plays

the role of Γ in the wave equation approach). This allows us to use the results

of [1] to obtain a detailed description of the branched analytic continuation of

P to complex values of s in a deleted neighborhood of zero in C (Proposition

1). We fill in the details of the outline in [4], using Stoke’s theorem to move the

integral which defines exp(−t√

∆)(F |X) into the complex domain and show that

the resulting integral is well defined for complex values of t in a neighborhood

of zero in C (slit along the nonpositive imaginary axis) containing −ε.

2 Grauert Tubes

Let M be a connected n-dimensional complex manifold containing X as a

totally real embedded submanifold. Such an embedding is always possible,

and any two are locally biholomorphically equivalent (see [7]). Let T εX =

{v ∈ TX : |v| < ε}. For ε > 0 sufficiently small, the Riemannian metric on X

determines a real analytic diffeomorphism of T εX with an open set Mε ⊂M by

analytic continuation of the exponential map:

v ∈ TxX ∩ T εX → Expxiv ∈Mε. (5)

This identification is the “adapted complex structure” on T εX (or T ∗εX; see

[10, 11, 19]). We will refer to the complex manifolds Mε, TεX, and T ∗εX as

“Grauert tubes.” We will always choose ε small enough that ε is less than the

maximal value for which this identification is possible, and that ∂Mε is the

image under (5) of {v ∈ TX : |v| = ε}.The distance squared function, d2(x, y), is real analytic in a neighborhood of

the diagonal ∆X ⊂ X ×X and can be analytically continued to a holomorphic

function, still denoted by d2, on a neighborhood of ∆X in M ×M . Since X is

compact, there is an ε0 > 0 such that if v ∈ TyX and |v| < ε0, then (Expyiv, y)

is in this neighborhood.

Lemma 1. Let ε < ε0. For all z = Expyiv ∈ ∂Mε we have d2(z, y) = −ε2.

Proof. This follows by analytic continuation in t from the identity

d2(Expy(tv/|v|), y) = t2 (v 6= 0).

5

3 Analytic Continuation of the Wave, Poisson

and Heat Kernels

Let 0 = λ0 < λ1 ≤ λ2 ≤ . . . be the spectrum of ∆ (with multiplicity) and let

φk be a real valued orthonormal basis of L2(X) satisfying ∆φk = λkφk. The

φk are real analytic (see [15, Theorem 8.6.1]). We denote the wave kernel by

W , the Poisson kernel by P , and the heat kernel by E. The wave kernel is the

distribution kernel of the operator exp(it√

∆),

exp(it√

∆)f(x) =∑

φk(x)fkeit√λk , f ∈ C∞(X), (6)

where fk are the Fourier coefficients, fk =< f, φk > and t ∈ R. In the sense of

distributions,

W (t, x, y) =∑

φk(x)φk(y)eit√λk . (7)

For any f ∈ C∞(X), (6) has an analytic continuation in t to the upper half plane

{t + is : s > 0}, and so (7) does as well in the sense of distributions. It is well

known that the analytic singular support of W is the set of (t, x, y) ∈ R×X×Xsuch that x and y can be joined by a geodesic of length |t| (see [8, Theorem 1]

and the remarks following it). The Poisson kernel is the distribution kernel of

the operator exp(−s√

∆),

exp(−s√

∆)f(x) =∑

φk(x)fke−s√λk , f ∈ C∞(X), s ≥ 0.

For s > 0, P is given by the convergent sum

P (s, x, y) =∑

φk(x)φk(y)e−s√λk ,

and lims→0+ P (s, x, y) = δx(y). In fact the sum converges for Re(s) > 0, and

(in the sense of distributions)

limu→0+

P (u+ iv, x, y) = W (v, x, y).

The heat kernel, E, has the same description as P if we replace√

∆, resp.√λk,

by ∆, resp. λk.

3.1 The Analytic Continuation of the Poisson Kernel

Let W denote the formal series

W (ζ, z, w)def=∑k

φk(z)φk(w)eiζ√λk , (ζ, z, w) ∈ C×M ×M. (8)

The following is contained in the work of Boutet de Monvel [3; 4; 5, Appendix

6.4] and gives an analytic continuation of the Poisson kernel.

6

Lemma 2.

1. For all s0 > 0 there is a neighborhood U (depending on s0) of X in M such

that the formal series W converges absolutely and uniformly on compacta

to a holomorphic function on {Im ζ > s0} × U × U .

2. There is a neighborhood V of i(0,∞)×X×X in C×M ×M such that W

converges absolutely and uniformly on compacta to a holomorphic function

on V.

3. The restriction of W to i(0,∞)×X ×X is P .

We note there is no neighborhood U of X in M such that W converges on

i(0,∞)× U × U .

Proof. P is real analytic on (0,∞) × X × X because it satisfies the elliptic

equation with real analytic coefficients,

(−2∂2t + ∆x + ∆y)P = 0.

Thus for each fixed s0 > 0 there is a neighborhood Us0 of X in M such that

P (s0, ·, ·) can be analytically continued to a holomorphic function on Us0 ×Us0 .

The representation

e−s0√λkφk(x) =

∫X

P (s0, x, y)φk(y) dy (9)

shows that all φk can be analytically continued to Us0 . Shrinking Us0 we may

assume that P (s0, z, w) is bounded on Us0 × Us0 . We obtain for some Cs0(independent of k) the rough estimate

supz∈Us0

|φk(z)| ≤ es0√λkCs0 . (10)

Then for all ζ = t+ is with s ≥ 3s0,

supUs0×Us0

|φk(z)φk(w)eiζ√λk | ≤ e−s0

√λkC2

s0 .

The estimate λk ∼ (k/C)2/n shows that∑φk(z)φk(w)eiζ

√λk converges ab-

solutely and uniformly on compacta to a holomorphic function on {Im ζ >

3s0} × Us0 × Us0 . We may take V = ∪s0>0{Im ζ > 3s0} × Us0 × Us0 . Clearly

the restriction of W to i(0,∞)×X ×X is P .

7

3.2 The Analytic Continuation of the Wave Kernel

The following lemma shows that, for small t and (x, y) sufficiently far from the

diagonal in X × X, W can be analytically continued in (t, x, y) variables to a

holomorphic function equal to W on an open set where both are defined. For

any positive number α let

(X ×X)≥αdef= {(x, y) ∈ X ×X : d(x, y) ≥ α}.

Lemma 3. Given any α > 0, there is a β > 0 and a neighborhood Y of (X ×X)≥α in M ×M such that W can be analytically continued from (−α/2, α/2)×(X × X)≥α to a holomorphic function, W≥α, on (−α/2, α/2) × i(−β, β) × Y.

W≥α is equal to W on a neighborhood of i(0, β)× (X ×X)≥α.

Proof. Since no points x, y, with d(x, y) ≥ α can be joined by a geodesic of

length |t| < α, the distribution W given by (7) is a real analytic function on

(−α, α) × (X × X)≥α. So W can be analytically continued to a holomorphic

function, which we denote by W≥α, on a neighborhood of the form (−α/2, α/2)×i(−β, β) × Y for some β > 0, where Y is a neighborhood of (X × X)≥α in

M ×M . We will show that W≥α is equal to the series W given by (8) on a

neighborhood of i(0, β)× (X×X)≥α (on which W converges by Lemma 2, item

2). Clearly W−W≥α is holomorphic on a neighborhood of i(0, β)×(X×X)≥α in

C×M×M . Note also that if (x, y) ∈ (X×X)≥α is fixed, then (W−W≥α)(·, x, y)

is a holomorphic function on (−α/2, α/2)× i(0, β) (the series in equation (8) is

absolutely convergent for ζ ∈ i(0, β), so it certainly converges for ζ ∈ R×i(0, β)).

From Equations (7) and (8) we have

lims→0+

(W − W≥α)(t+ is, x, y) = 0

for (t, x, y) ∈ (−α/2, α/2)× (X×X)≥α, in the sense of distributions. It follows,

from the one-dimensional version of the “Edge of the Wedge Theorem” or the

distributional version of Painleve’s Theorem, that (W − W≥α)(ζ, x, y) = 0 for

all (ζ, x, y) ∈ (−α/2, α/2)× i(0, β)× (X ×X)≥α. Since i(0, β)× (X ×X)≥α is

a totally real submanifold of the domain of definition of W − W≥α, it follows

that W = W≥α on a neighborhood of i(0, β)× (X ×X)≥α in C×M ×M .

Corollary 2. Given any α > 0, there is a β > 0 and a neighborhood Y in M×Mof (X×X)≥α such that P can be analytically continued from (0, β)×(X×X)≥α

to (−β, β)× i(−α/2, α/2)× Y.

Proof. W is the analytic continuation of P , and W≥α provides the desired an-

alytic continuation of W .

8

3.3 Analytic Continuation of the Heat Kernel

The essential difference between the analytic continuation of the heat and Pois-

son kernels is that there is an ε0 such that for all s, the heat kernel at time s

can be analytically continued to a Grauert tube Mε0 (the Poisson kernel can

only be continued to a tube whose radius depends on s; see Theorem 1).

Lemma 4. There is a positive ε0 such that the heat kernel can be analytically

continued from (0,∞) × X × X to {Re ζ > 0} × Mε0 × Mε0 as E(t, z, w) =∑k e−λktφk(z)φk(w), with uniform convergence on compact subsets of {Re ζ >

0} ×Mε0 ×Mε0 .

Proof. This follows from the estimates (10) and λk ∼ (k/C)2/n (the compact-

ness of X allows us to find a Mε0 ⊂ Us0 where Us0 is as in the proof of Lemma

2).

4 Proof of Theorem 1

We will need some results from the proof of [18], Theorem 0.1, on the analytic

continuation of the Minakshisundaram-Pleijel parametrix. Although Boutet de

Monvel’s theorem was cited in [18], the following result is independent of it. Let

(X ×X)≤αdef= {(x, y) ∈ X ×X : d(x, y) ≤ α}.

Theorem 3 ([18]). For all sufficiently small α > 0, we can find a neighborhood

X in M ×X, containing (X ×X)≤α with the property that:

1. For (x, y) ∈ (X ×X)≤α, x is in the domain of a geodesic coordinate chart

centered at y.

2. The distance function squared and the coefficients in the Minakshisundaram-

Pleijel parametrix for the heat equation, uk, can be analytically continued

to a neighborhood of X in M ×M .

3. There is an L > 0 such that for all k and all (z, y) ∈ X , the estimate

|uk(z, y)| ≤ Lk+1k! holds ([18, Proposition 3.1]).

4. Let C > 1 satisfy2 L/(Ce) < 1/4. Then there is an η > 0 such that for

all (t, z, y) ∈ (0, 1)×X ,

E(t, z, y) = (4πt)−n/2e−d2(z,y)/4t

∞∑k=0

tkuk(z, y)χ(t−1 − kC) +O(e−η/t)

2Our choice of C is larger than in [18, Definition 4.2], where L/(Ce) < 1/2.

9

where the O(·) is uniform as t→ 0+ on X . Here χ ∈ C∞(R, [0, 1]), χ ≡ 0

on (−∞, 0], χ ≡ 1 on [1/2,∞) so that χ(t−1 − kC) truncates the series

after a finite, t-dependent number of terms.

5. (4πt)−n/2∑∞k=0 t

kuk(z, y)χ(t−1 − kC) is an analytic symbol of order n/2

in the parameter 1/t.

6. If (z, y) ∈ X and z = Expxiv, v ∈ TxX, then

Re d2(Expxiv, y) ≥ −|v|2. (11)

Proof. For items 1–5 see the proof of Theorem 0.1 in [18, Section 4], in which

the result of [3, 4] is used only to show that items 4 and 5 hold for both z

and y complex (see the Lemma following [18, Proposition 4.11]); we do not

need this part of the result. For item 6, note from [18], Proposition 4.12 and

Remark 4.13, that for z = Expxiv, with v ∈ TxX sufficiently small, the map

X 3 y 7→ Re d2(Expxiv, y) has a non-degenerate local minimum at y = x with

minimum value −|v|2. After possibly shrinking X and α, we may assume that

(11) holds for all (Expxiv, y) ∈ X .

Proof of Theorem 1, part 1. Choose α > 0 small enough that we can find a

neighborhood X as in Theorem 3. From Corollary 2 there is a β > 0 and

an open subset Y in M × X containing (X × X)≥α such that for all y ∈ X,

the map (ε, x) → P (ε, x, y) can be analytically continued from (0, β) × {x ∈X : d(x, y) ≥ α} to (−β, β) × i(−α/2, α/2) × {z ∈ M : (z, y) ∈ Y}. Let us

further assume that β < α/2, so that the disk of radius β, D(β), is contained

in (−β, β) × i(−α/2, α/2) (this will be used in the proof of Lemma 6). We

will show that there is an ε1 ∈ (0, β) such that for all positive ε less than ε1

and all y ∈ X, the map x → P (ε, x, y) can be analytically continued from

{x ∈ X : d(x, y) ≤ α} to {z ∈ Mε : (z, y) ∈ X}. Then, since X ∪ Y is an open

subset of M ×X containing the compact set X ×X, we can find an ε2 so that

Mε2×X is contained in X ∪Y. If ε0 is the smaller of ε1, ε2 (and smaller than β),

then for all y ∈ X and all ε ∈ (0, ε0) the map x→ P (ε, ·, y) can be analytically

continued from X to Mε. This will complete the proof of Theorem 1, part 1.

To show the existence of ε1 we use the subordination formula to relate the

Poisson and heat kernels. The subordination formula says that for (ε, x, y) ∈(0,∞)×X ×X,

P (ε, x, y) =ε√4π

∫ ∞0

E(t, x, y)e−ε2/4tt−1/2 dt

t(12)

10

(see [17, Theorem 2.1]). We first consider the integral over [1/2,∞). On this

interval the factor χ(t−1 − kC) is zero for all k > 2/C, so the sum in item 4

of Theorem 3 has no more than b2/Cc terms. Then the integrand is uniformly

bounded for (ε, z, w) in compact subsets of C×Mε1 ×Mε1 and so the integral

ε√4π

∫ ∞1/2

E(t, z, w)e−ε2/4tt−1/2 dt

t(13)

is a holomorphic function of (ε, z, w) ∈ C × Mε1 × Mε1 . Thus it suffices to

consider the (formal) integral

ε√4π

∫ 1/2

0

E(t, z, y)e−ε2/4tt−1/2 dt

t(14)

Let us now shrink ε1 so that ε1 < 2√η. Setting θ = 1/t we can write (14) as,

for (ε, z, y) ∈ (0, ε1)×X ∩ (X ×X) (z “real”),

ε√4π

∫ 1/2

0

E(t, z, y)e−ε2/4tt−1/2 dt

t

= ε(4π)−n+12

∫ ∞2

e−(d2(z,y)+ε2)θ/4∞∑k=0

θn−12 −kuk(z, y)χ(θ−kC) dθ+R(ε, z, y)

(15)

where R(ε, z, y) =∫∞

2O(e−(η+ε2/4)θ)θ−1/2 dθ extends to a smooth function on

(ε, z, y) ∈ {|ε| < ε1} × X , holomorphic in both z and ε. The integrand of

(15) extends to a holomorphic function of z for (ε, z, y) ∈ (0, ε1) × X , and

the sum in (15) is a symbol in θ of order (n − 1)/2. In particular it can be

estimated by a constant times θn−12 , locally uniformly in (z, y). To show that

the integral on the right hand side of (15) converges and is holomorphic in z

for fixed y, (z, y) ∈ X ∩ (Mε × X), it suffices to show that for all ε ∈ (0, ε1),

Re d2(z, y)+ε2 > 0 on X ∩(Mε×X). Since Mε = {Expxiv : |v| < ε} (see Section

2), this follows from Theorem 3, item 6. This completes the proof of Theorem

1, part 1.

Remark 1. We can extend the interval of integration in (15) to (0,∞) modulo a

function which is smooth on (ε, z, y) ∈ {|ε| < ε1} × X and holomorphic in both

z and ε, because χ vanishes to infinite order at θ = 0.

For (θ, z, y) ∈ (0,∞) × X , let a(θ, z, y) be the analytic symbol of order

(n− 1)/2,

a(θ, z, y) = θn−12

∞∑k=0

θ−kuk(z, y)χ(θ − kC).

11

By (15) and Remark 1, the Poisson kernel restricted to the set of all (ε, z, y)

such that (z, y) ∈ X and Re d2(z, y) + ε2 > 0 is

P (ε, z, y) ≈ ε(4π)−n+12 L[a(·, z, y)]

((d2(z, y) + ε2)/4

)(16)

where L is the Laplace transform and ≈ means modulo a function which is

smooth in (ε, z, y) ∈ {|ε| < ε0} × X , and holomorphic in ε and z. We will use a

result of of Beyer and Heller [1] (see also [13]) to give an explicit expression for

the singular part of the Poisson kernel in terms of the Laplace transform variable

s = (d2(z, y) + ε2)/4 (here s corresponds to Γ in the Hadamard parametrix

approach in [23]). Although we could not find a convenient reference to the

behavior near the origin of the Laplace transform of an analytic symbol in

the analytic microlocal analysis literature, we note the Laplace transform of a

symbol has been used in the C∞ setting to obtain the asymptotic expansion

of the Bergman kernel at the boundary of a strictly pseudoconvex domain [2;

6, Corollaire 1.7]. Presumably the singularity of the Poisson kernel could also

be analyzed by showing that it satisfies a holonomic system of microdifferential

equations as in [2, 16]. Instead we will give a classical proof using the result of

[1] which we now recall. Let D(η) be the disk of radius η centered at the origin

in C.

Theorem 4 (W. A. Beyer and L. Heller [1, Theorem 1]). Let F ∈ C0((0,∞))).

Suppose there exists K, σ > 0 such that F is integrable on [0,K] and for all

θ > K,

F (θ) = θ−β

[N∑k=0

ukθ−k +RN (θ)

]where RN (θ) = O(N !(σ/θ)N+1), uniformly in N and θ > K. Then L[F ](s) is

analytic for |arg s| < π/2, s 6= 0, and there exists η > 0 (proportional to σ−1)

such that L[F ] has a branched analytic continuation to D(η)\{0} of the form:

1. If β 6= Z, then L[F ](s) = sβ−1g(s) + h(s) where g and h are analytic at

s = 0 and g(s) =∑∞i=0 uiΓ(1− i− β)si.

2. If β ∈ Z, then L[F ](s) = sβ−1∑−βi=0 uiΓ(1 − i − β)si + (log s)g(s) + h(s)

where g and h are analytic at s = 0 and g(s) =∑∞i=1

(−1)i

(i−1)!ui−βsi−1.

The idea of the proof is the following. Let u(v) =∑∞k=0 v

kuk/k! be the Borel

transform of the formal sum∑∞k=0 θ

−kuk. This sum converges for |v| < 1/σ.

For τ = 1/(4σ) let F (t) = t1−β∫ τ

0e−tvu(v) dv. Then it is shown using classical

analysis that L[F ] satisfies the conclusion of the Theorem and differs from L[F ]

by a function which is analytic at s = 0.

12

Using this result we show that P can be analytically continued in s to a

deleted neighborhood of zero in C. This will be used in the proof of Theorem

1, part 3.

Proposition 1. There is an ε0 such that for all fixed (z, y) ∈ X , the map

s 7→ L[a(·, z, y)](s) has a (branched) analytic continuation from Re(s) > 0 to

D(ε20)\{0} ∪ Re(s) > 0 of the following form:

1. If n is even, then

L[a(·, z, y)](s) = s−n+12 g(s, z, y) + h(s, z, y)

where g and h are smooth on D(ε20)×X and analytic in s and z.

2. If n is odd, then

L[a(·, z, y)](s) = s−n+12

n−12∑

k=0

uk(z, y)Γ((n+ 1)/2− k)sk

+ (log s)g(s, z, y) + h(s, z, y)

where g and h are smooth on D(ε20)×X and analytic in s and z.

Moreover

g(s, z, y) =

∑∞k=0 uk(z, y)Γ(n+1

2 − k)sk if n is even∑∞k=1 uk+n−1

2(z, y) (−1)k

(k−1)!sk−1 if n is odd.

Proof. We verify the hypotheses of Theorem 4 with β = −(n+ 1)/2. We must

show that there are positive numbers K and σ such that for all (z, y) ∈ X ,

a(·, z, y) is defined and continuous on (0,∞), integrable on [0,K], and such that

for θ > K,

a(θ, z, y) = θn−12

[N∑k=0

uk(z, y)θ−k +RN (θ, z, y)

]whereRN (θ, z, y) = O(N !(σ/θ)N+1), uniformly inN and θ > K, and (z, y) ∈ X .

The continuity of a(·, z, y) on (0,∞) and integrability over [0,K] for any positive

K are clear (since χ(t) = 0 for t ≤ 0 and vanishes to infinite order at t = 0).

For simplicity we will take K = 1. Write

RN (x, y, θ) =

N∑k=0

uk(x, y)θ−k(χ(θ − kC)− 1) +

∞∑k=N+1

uk(x, y)θ−kχ(θ − kC).

(17)

13

We first estimate the first term on the right hand side of (17). Using item

3 in Theorem 3 we have |uk(x, y)| ≤ Lk+1k! for all (x, y) ∈ X . Note that

χ(θ − kC) − 1 is zero if θ > kC + 1/2, so that we may assume θ ≤ kC + 1/2.

Furthermore we have 1/(2C) < 1. This gives∣∣∣∣∣N∑k=0

uk(x, y)θ−k(χ(θ − kC)− 1)

∣∣∣∣∣ ≤ L(L/θ)N+1N∑k=0

(C/L)N+1−k (k + 1)N+1−k

k!.

Note (k+1)N+1−kk! ≤ (N +1)! for k = 0, 1, . . . , N . Then if α1 = max(1, C/L)

we have ∣∣∣∣∣N∑k=0

|uk(x, y)(χ(θ − kC)− 1)θ−k

∣∣∣∣∣ ≤ L(α1L/θ)N+1N !(N + 1)2.

We can choose α2 so that (N + 1)2 ≤ αN+12 for all N ≥ 0. Setting σ1 = Lα1α2

gives the desired estimate for the first term on the right hand side of (17).

To estimate the second term on the right hand side of (17), we note that

χ(θ − kC) is zero if k ≥ θ/C and write∣∣∣∣∣∞∑

k=N+1

uk(x, y)θ−kχ(θ − kC)

∣∣∣∣∣ ≤ L(L/θ)N+1N !

bθ/Cc∑k=N+1

k!

N !(L/θ)

k−N−1.

Since k ≤ θ/C, we can estimate k!/N ! ≤ (N + 1)(θ/C)k−N−1. Then∣∣∣∣∣∞∑

k=N+1

uk(x, y)θ−kχ(θ − kC)

∣∣∣∣∣ ≤ L(L/θ)N+1N !(N + 1)

bθ/Cc∑k=N+1

(L/C)k−N−1

.

Since L/C < e/4, the series converges and is bounded independent of θ and N .

We can choose α3 so that N + 1 ≤ αN+13 for all N ≥ 0. Setting σ2 = Lα3 gives

the desired bound for the second term on the right hand side of (17). We can

take σ to be the larger of σ1 and σ2. The conclusion of [1, Theorem 1] then

gives the existence of ε0. Since the estimate on RN holds uniformly, it follows

from the proof of [1, Theorem 1] that g and h are smooth on D×X and analytic

in z.

Proof of Theorem 1, part 2. Since P (ε, ·, ·) is smooth on Y, we need only con-

sider X∩(Mε×X). As in (15) we can write the restriction of P (ε, ·, ·) to ∂Mε×Xformally as

P (ε, z, y) = ε(4π)−n+12

∫ ∞0

e−(d2(z,y)+ε2)θ/4∞∑k=0

θn−12 −kuk(z, y)χ(θ − kC) dθ,

(18)

14

plus a smooth function on ∂Mε×X. We will show that (18) is a Fourier integral

distribution with complex phase. The proof of [18, Theorem 0.1] shows that the

amplitude appearing in (18) is a symbol of order (n−1)/2 on X∩(∂Mε×X)×R+.

We need to show that the phase function3

φ(z, y, θ) =iθ

4(d2(z, y) + ε2) (19)

defined on the cone

Vdef= X ∩ (∂Mε ×X)× R+

is a regular phase function of positive type on V (see [20, Definition 3.5]),

and that real locus of the complex Lagrangian submanifold generated by φ is

contained in (T ∗∂Mε\0)× (T ∗X\0). Here d2 is the analytic continuation of the

distance squared function from a neighborhood of the diagonal in X × X to

X ∩ (Mε0 ×X).

Let Cφ denote the set of real θ-critical points of φ,

Cφ = {(z, y, θ) ∈ V : d2(z, y) + ε2 = 0}.

From (11) and the remarks preceding it,

Cφ = {(Expyiv, y, θ) ∈ V : v ∈ TyX, |v| = ε, θ > 0.}.

The following lemma shows that dφ 6= 0 on V and d(φθ) 6= 0 on Cφ. Let ρ be

the function on Mε0 defined by ρ(Expyiv) = |v|g, so that Mε = {ρ(z) < ε}.

Lemma 5. Let ıX : X ∩ (∂Mε ×X) → X be the inclusion map. Then there is

an ε1 > 0 such that for all 0 < ε < ε1 and all v ∈ TyX with |v| = ε,

ı∗Xd(id2)(Expyiv,y) = 2(−ı∗∂Mε

(Im ∂ρ2

), v[)

(Expyiv,y)

where v[ is the covector obtained from the metric identification of TyX and

T ∗yX.

Proof. Let v =∑viei where e1, . . . , en is an orthonormal basis for TyX, and let

p(s1, . . . , sn) = Expy(∑skek) be a normal real analytic geodesic chart centered

at y. Then

ds(d2(p(iv1, . . . , ivn), p(s1, . . . , sn))

) ∣∣s=0

= −2i

n∑k=1

vk dsk∣∣s=0

(20)

3We will suppress the dependence on ε in the notation.

15

(see [18], proof of Proposition 4.12 (c)). This can be interpreted as−2i(∑vkek)[,

and so

dy(id2)(Expyiv,y) = 2v[.

We now fix y ∈ X and compute d(d2(·, y)). Let Dε,y = {z ∈ ∂Mε : (z, y) ∈ X}.By (11),

Re d2(z, y) ≥ −ε2 for all z ∈ Dε,y

and equality holds if z = Expyiv. Thus d(Re d2(·, y)) is zero on vectors tangent

to Dε,y at points of the form z = Expyiv, and so d(Re d2(·, y)) must be a

multiple of dρ2 at these points. To determine the multiple we evaluate both

on the vector ddt

∣∣t=0

Expyietv and find the multiple is −1. Since d2(z, y) is a

holomorphic function of z, we have d(Im d2(·, y)) = Jdρ2 at these points and

the Lemma follows.

Clearly φ is homogeneous of degree one in θ, and Imφ ≥ 0 by (11). This

together with Lemma 5 shows that φ is a regular phase function of positive type

on V . The isotropic submanifold generated by φ,

Λφdef= {((z, φz), (y, φy)) : (z, y, θ) ∈ Cφ} ,

can be computed using Lemma 5 as

Λφ =

{((Expyiη

],−θ2ı∗∂Mε

Im ∂ρ2), (y,θ

2η)) : η ∈ T ∗yX, |η| = ε, θ > 0

},

or, replacing η by ε|ξ|−1ξ ∈ T ∗yX\0,

Λφ ={

((Expyiε|ξ|−1ξ],−ε−1|ξ|ı∗∂MεIm ∂ρ2), (y, ξ)) : y ∈ X, ξ ∈ T ∗yX\0,

},

(21)

which is clearly contained in (T ∗∂Mε\0) × (T ∗X\0). Thus (15) is a Fourier

integral distribution with complex phase, and the operator Sε associated with it

by the Schwartz kernel theorem is a continuous map from C∞(X) to C∞(∂Mε).

Its degree of is easily calculated to be −(n − 1)/4 from the definition [20, p.

177–178]. This concludes the proof of Theorem 1, part 2.

Remark 2. Λφ can be interpreted as the twisted graph of an isomorphism of

symplectic cones in the following way (see [5, Appendix A.6.4]). Let α0 be the

pullback of the canonical one-form on T ∗X to S∗εX and let α = ı∗∂Mε

(Im ∂ρ2

).

Then (S∗εX,α0) and (∂Mε, α) are contact manifolds. The “adapted complex

structure” identification

Φ: S∗εX 3 η → Expyiη] ∈ ∂Mε,

16

where y is the base point of η, is a contact isomorphism. The half-line bundle

Σ0 ⊂ T ∗(S∗εX)\0 consisting of positive multiples of α0 is a symplectic cone

canonically isomorphic with T ∗X\0.4 Similarly the half-line bundle Σ ⊂ T ∗∂Mε

consisting of positive multiples of α is a symplectic cone. Composing with Φ∗

gives an isomorphism of symplectic cones,

χ : T ∗X\0→ Σ (22)

ξ → ε−1|ξ|Φ∗(

(α0)ε ξ|ξ|

)= ε−1|ξ|αExpy(iε|ξ|−1ξ]) (23)

where y is the base point of ξ. Comparing (23) with (21) shows that Λφ is the

graph of χ, twisted by multiplication by −1 in the fibers of ∂Mε.

Remark 3. If X is n-dimensional Euclidean space, then the sum in (18) is

identically one and d2(x, y) = (x− y)2. For real x, y, and ε > 0, the integral in

(18) gives the classical Poisson kernel,

P (x, y, ε) = ε(4π)−n+12

∫ ∞0

e−((x−y)2+ε2)θ/4θn−12 dθ

= επ−n+12 Γ ((n+ 1)/2) ((x− y)2 + ε2)−

n+12

with the correct constant factor.

Remark 4. We can express the Poisson kernel near the diagonal as an oscillatory

integral (see (27)) in the following way. We can write (c.f. (15) and Remark 1)

for (real) (ε, x, y) ∈ (0, ε0)× (X ×X)≤α,

P (ε, x, y) ≈ ε(4π)−n+12

∫ ∞0

e−(d2(x,y)+ε2)θ/4θn−12 a′(x, y, θ) dθ.

where a′(x, y, θ) =∑∞k=0 θ

−kuk(x, y)χ(θ − kC) is a symbol of order zero. Let

ej(x) be an orthonormal basis of TxX and write y = Expx(∑nj=1 t

jej(x)), t =

(t1, . . . , tn) ∈ Rn. Then d2(x, y) =< g(x)t, t > with g(x) the matrix of the

metric tensor at x and < ·, · > is the Euclidean inner product. Then

P (ε, x, y) ≈ ε(4π)−n+12

∫ ∞0

e−<g(x)t,t>θ/4e−ε2θ/4θ

n−12 a′(x, y, θ) dθ. (24)

Using standard results about the Fourier transform of a Gaussian ([15, Theorem

7.6.1]) we have, for t ∈ Rn and g(x) a symmetric non-singular matrix with

4The isomorphism consists of writing ξ ∈ T ∗X\0 in “polar form” as ξ = θη, with η ∈ S∗εX

and θ > 0, and identifying ξ ∈ T ∗X\0 with θ(α0)η ∈ T ∗(S∗εX).

17

Re g(x) ≥ 0,

e−<g(x)t,t>θ/4 = (2π)−n(4π)n2 θ−

n2 (det g(x))−

12

∫ξ∈Rn

eit·ξ−|ξ|2g/θ dξ

where |ξ|2g =< g−1(x)ξ, ξ >. Then

P (ε, x, y) ≈ (2π)−n(det g(x))−1/2∫ξ∈Rn

eit·ξ(

ε√4π

∫ ∞0

e−|ξ|2g/θ−ε

2θ/4θ−12 a′(x, y, θ) dθ

)dξ.

Let

b(x, y, ξ, ε)def= eε|ξ|g

ε√4π

∫ ∞0

e−|ξ|2g/θ−ε

2θ/4θ−12 a′(x, y, θ) dθ.

Then b can be expressed as a Laplace transform,

b(x, y, ξ, ε) = eε|ξ|gε√4πL [F (x, y, ξ, ·)]

(ε2/4

)(25)

where

F (x, y, ξ, θ)def= e−|ξ|

2g/θθ−1/2a′(x, y, θ), (26)

so that

P (ε, x, y) ≈ (2π)−n(det g(x))−1/2

∫ξ∈Rn

eit·ξ−ε|ξ|gb(x, y, ξ, ε) dξ (27)

where y = Expx(∑nj=1 t

jej(x)). Note t depends on both y and x but the integral

is independent of the choice of orthonormal basis ej(x) because b is invariant

under the orthogonal group action on ξ ∈ Rn. It can be shown (using the same

technique as in the proof of Proposition 1) that b can be extended to a branched

analytic function of d2(x, y) + ε2 in a deleted neighborhood of zero in C.

Proof of Theorem 1, part 3. To prove Hs-continuity of Sε we must consider the

complex Lagrangian submanifold Λφ associated with Sε. We think of V ⊂∂Mε × X × R+ as a real manifold and φ as a complex valued function of the

(real) variables in V . Since V and φ are real analytic, we may consider their

analytic extensions, which we denote by tildes.5 We may assume V has the

form

V = (X ∩ (∂Mε ×X)) × C\0.

Let Cφ denote the set of θ-critical points for φ, i.e.,

Cφ = {(z, y, θ) ∈ V : d2(z, y) + ε2 = 0}.

5I.e., V is a complex manifold containing V as a totally real submanifold and φ is the

analytic continuation of φ to V (which we may assume exists after possibly shrinking V ).

18

Note Cφ is Cφ ∩ V . The associated (complex) Lagrangian submanifold is

Λφdef= {((z, φz), (y, φy)) ∈ (T ∗∂Mε\0)˜× (T ∗X\0)˜ : (z, y, θ) ∈ Cφ}.

Λφ is a positive conic immersed complex Lagrangian submanifold. Note Λφ is

equal to ΛφRdef= Λφ ∩ T

∗(∂Mε ×X) (see (21) and [20, Theorem 3.6])).

To verify that Sε is a continuous map from Hs(X) to Hs+n−14 (∂Mε), it

suffices to show that Sε ◦ (1 + ∆)n−18 is continuous from L2(X) into L2(∂Mε).

Since (1 + ∆)n−18 is a pseudodifferential operator of degree (n − 1)/4, Sε ◦

(1 + ∆)n−18 is a complex Fourier integral operator of degree zero associated

with Λφ. According to [14, Theorem 3.5] we must show that for for every

γ0 = ((z0, ζ0), (y0,−ξ0)) ∈ Λ′φR

, the projections from

(Tγ0Λ′φ)R

def= (Tγ0Λ′

φ) ∩ Tγ0(T ∗(∂Mε ×X))

to T(z0,ζ0)(T∗∂Mε) and T(y0,ξ0)(T

∗X) are injective.6 It suffices to prove this for

Λφ instead of Λ′φ. Let c0 = (z0, y0, θ0) ∈ Cφ, let

F : (z, y, θ) ∈ Cφ → ((z, φz), (y, φy)) ∈ Λφ = ΛφR,

and let γ0 = F (c0). Then Tγ0Λφ = Tc0F (Tc0Cφ), i.e.,

Tγ0Λφ ={

(($,φzz$ + φzyϑ+ θ−10 φzd), (ϑ, φyz$ + φyyϑ+ θ−1

0 φyd)) :

($,ϑ, d) ∈ TCc0V , (dφθ)c0($,ϑ, d) = 0

}(see [21, p. 547-8]; we have used that φ is homogeneous of degree one in θ).

Suppose ($,ϑ, d) ∈ Tc0Cφ are complex tangent vectors, Tc0F ($,ϑ, d) is in

(Tγ0Λφ)R, and the projection of Tc0F ($,ϑ, d) onto T(z0,ζ0)(T∗∂Mε) is zero. We

will show ($,ϑ, d) = 0. By Lemma 5, φy is real and non-zero on Cφ. Thus

Im (φyy)ϑ+ θ−10 φy Im d = 0. (28)

The tangency condition d(φθ)c0($,ϑ, d) = 0 implies, since $ = 0 and ∂φθ/∂θ =

0, that φyϑ = 0. Since Im (φyy) is positive definite by [18, Proposition 4.12 (a)],

taking the inner product of (28) with ϑ gives ϑ = 0 and hence d = 0, so

($,ϑ, d) = 0.

Now suppose Tc0F ($,ϑ, d) is real and its projection onto T(y0,ξ0)(T∗X) is

zero. Then

Im (φzz)$ + θ−10 φz Im d = 0. (29)

6Note (Tγ0Λφ

)R is not a priori equal to Tγ0 (ΛφR). It is obvious that the projections onto

each factor of Tγ0 (ΛφR) are injective because Λ

φR = Λφ is the graph of (22).

19

Note Im (φzz) is not positive definite. Let (x1, . . . , xn) → Expy0(∑xkek) be

a normal real analytic geodesic chart centered at y0 as in Lemma 5. Then

(x + iτ) → Expy0(∑

(xk + iτk)ek) are local coordinates on M near y0. By

compactness, after shrinking ε we may assume that the analytic continua-

tion exists and gives local coordinates near z0 = Expy0(iv0), |v0| = ε. Let

(s1, . . . , sn−1) → (τ1(s), . . . , τn(s)) be local coordinates on the sphere |v| = ε

centered at v0. Then (x, s) are local coordinates on ∂Mε centered at z0 and

Im (φzz) = Im

(φxx φxs

φsx φss

)

Since Im(φxx) is positive definite when τ = 0, by compactness and shrinking ε we

may assume that Im(φxx) is positive definite at z0. Since (Expy0(i∑τk(s)ek), y0, θ0)

is in Cφ for all s in a neighborhood of zero and dzφ is real on Cφ, we conclude

Im(φxs) = Im(φxs) = Im(φss) = 0 at z0. Write $ in these coordinates as (a, b)

and take the inner product of (29) with $. The condition (dφθ)c0($,ϑ, d) = 0

means that φz$ = 0. We obtain at Im(φxx)a = 0 and so a = 0. From this it

follows that Im d = 0 and so TcF ($,ϑ, d) ∈ Tγ0Λφ. Since Λφ is the graph of a

diffeomorphism, the projection onto T(y0,ξ0)(T∗X) is injective and ($,ϑ, d) = 0.

This shows that Sε◦(1+∆)n−18 is a continuous map from L2(X) to L2(∂Mε)

and Sε is continuous from Hs(X) to Hs+n−14 (∂Mε). It remains to show that Sε

is a bijection from Hs(X) onto Os+n−14 (∂Mε).

If Sεf = 0, then e−ε√

∆f is a holomorphic function on Mε whose restriction

to ∂Mε (in the sense of distributions) is zero. Since a holomorphic function is

harmonic for the Kahler Laplacian and there is a unique solution to the Dirichlet

problem on Mε, we must have e−ε√

∆f = 0 on Mε and so on X. Since e−ε√

∆ is

injective on Hs(X) (as can be seen from the eigenfunction expansion), f = 0.

This shows that Sε is injective.

We now show, following the outline in [4], that for ε sufficiently small, Sε

is onto Os+n+14 (∂Mε). Since Sε is continuous, it suffices to show Sε is onto

the dense subspace of restrictions to ∂Mε of functions holomorphic on some

neighborhood of Mε.7 So let F be in this dense subspace of Os+n+1

4 (∂Mε).

Since ∂Mε is compact we may suppose F ∈ O(M ε1) for some ε1 with ε < ε1 < ε0

(i.e., F is holomorphic in some neighborhood of Mε1). It suffices to show that

there is a g ∈ C∞(X) such that F |X = e−ε√

∆g. The following lemma says that

the operator e+ε√

∆ is well-defined on F |X .

7Real analytic functions are dense in ker ∂b ∩Hs(∂Mε), and each real analytic function in

ker ∂b extends to a two-sided neighborhood of ∂Mε, and to the interior also by pseudoconvexity.

20

Lemma 6. For each x ∈ X, the map t 7→ exp(−t√

∆) (F |X) (x) can be an-

alytically continued from t > 0 to a single valued analytic function of t on

D(ε1)\i(−ε1, 0], smooth in (t, x) ∈ D(ε1)\i(−ε1, 0]×X.

Proof. Fix x ∈ X. Let Cut(x) be the cut locus of x and let X ′x = X\(Cut(x) ∪{x}). Since the cut locus has measure zero and P is symmetric in x and y, we

have for t > 0 that

exp(−t√

∆)(F |X)(x) =

∫X′x

P (t, y, x)F |X(y) dy. (30)

If x and y are both real then it is not clear that P (t, x, y) can be continued

to values of t with Re t < 0. To get around this difficulty we take advantage

of the fact that F extends to Mε1 and deform the integration in y into Mε1 .

We may assume (after possibly shrinking ε0) that dy is the pullback to X of a

holomorphic volume form, ω, on Mε0 . Let

Cx,ε1 = {Expy(−isv(y, x)) : y ∈ X ′x, 0 ≤ s ≤ ε1}

where v(y, x) is the vector in TyX pointing from y to x and v(y, x) is the

corresponding unit vector. For t > ε1, P (t, ·, x)F (·) is holomorphic on Mε1 ⊂Mt. Thus the integral of dz (P (t, z, x)F (z)ω(z)) over the (n + 1)-chain Cx,ε1 is

zero. By Stoke’s Theorem, for t > ε1 we have∫∂Cx,ε1

P (t, ·, x)F (·)ι∗∂Cx,ε1ω = 0. (31)

The boundary ∂Cx,ε1 consists of four pieces:

Cx,ε1 = ∂Cut(x)Cx,ε1 ∪X ′x ∪ Γx,ε1 ∪ Expx(iB(ε1))

where ∂Cut(x)Cx,ε1 consists of points of the form Expyiv, y ∈ Cut(x) and |v| ≤ ε1,

B(ε1) is the closed ball of radius ε1 in TxX, and

Γx,ε1 ={

Expy(−iε1v(y, x)) : y ∈ X ′x}.

Using (30) and (31) with the proper orientations, we have for t > ε1

exp(−t√

∆)(F |X)(x) =

∫∂Cut(x)Cx,ε1+Γx,ε1+Expx(iB(ε1))

P (t, z, x)F (z)ι∗ω(z).

(32)

We consider each of the three boundary integrals separately.

I. Integral over ∂Cut(x)Cx,ε1 . We will show that this does not contribute

any analytic singularities for t ∈ D(ε1). Recall that after having fixed α in

21

the proof of Theorem 1, part 1, we choose ε0 less than the β of Corollary 2

(and of course ε1 < ε0) . After possibly shrinking ε0 we may assume that

{(z, x) : z ∈ ∂Cut(x)Cx,ε1 , x ∈ X} ⊂ Y (since {(y, x) : y ∈ Cut(x), x ∈ X} ⊂ Y).

So modulo a function which is smooth on D(ε0)×X and (for fixed x) analytic

in t ∈ D(ε0), we can write

exp(−t√

∆)(F |X)(x) ≈∫

Γx,ε1+Expx(iB(ε1))

P (t, z, x)F (z)ι∗ω(z).

II. Integral over Γx,ε1 . If (z, x) ∈ Y, then P (t, z, x) is analytic (for fixed x) in

(t, z) ∈ D(ε0)×{z ∈M : (z, x) ∈ Y}. So we need only consider points z ∈ Γx,ε1such that (z, x) ∈ X .

Lemma 7. Fix z, x such that z ∈ Γx,ε1 and (z, x) ∈ X . Then P (t, z, x)

can be analytically continued to a holomorphic function of t in the disk D(ε1),

depending smoothly on (z, x).

Proof. We have from (16)

P (t, z, x) ≈ t(4π)−n+12 L[a(·, z, x)]

((d2(z, x) + t2)/4

).

On Γx,ε1 we have z = Expy(−iε1v(y, x)). By analytically continuing the identity

d2(Expy(lv(y, x)), x) = (|v|g− l)2 we have d2(z, x)+ t2 = |v|2g−ε21 +2i|v|gε1 + t2.

Here ε1 is fixed and 0 < |v|g < α. Let us choose the branch of the square root

and logarithm that is holomorphic on C\(−∞, 0]i and agrees with the principal

branch on (0,∞). By Proposition 1, L[a(·, z, x)](w) is a holomorphic, single

valued function of w on the region(C+ ∪D(ε20)

)\i(−∞, 0] where C+ is the

open right half-plane (see Figure 1). We must show that for all |v| ∈ (0, α)

and all u ∈ D(ε21), the quantity 14 (|v| − ε21 + 2i|v|ε1 + u) lies in the region

shown in Figure 1 (we’ll write simply |v| instead of |v|g). We will check that

if |v|2 − ε21 + Re(u) = 0, then 2|v|ε1 + Im(u) > 0; and if |v|2 − ε21 + Re(u) ≤ 0

then | 14 (|v|2 − ε21 + 2i|v|ε1 + u)| < ε20. First suppose |v|2 − ε21 + Re(u) = 0. Then

Im(u)2 < ε41−Re(u)2, and so Im(u)2 < 2ε21|v|2 < 4ε21|v|2, hence 2|v|ε1 +Im(u) >

0. Next suppose |v|2 − ε21 + Re(u) ≤ 0. Then |v|2 − ε21 ≤ ε21 and∣∣∣∣14(|v|2 − ε21 + 2i|v|ε1 + u)

∣∣∣∣ < 1

4

√(|v|2 − ε21)2 + 4|v|2ε21 +

1

4ε21

≤ 1

43ε21 +

1

4ε21.

The smooth dependence on (z, x) is clear.

22

�Ε02Re�w�

Im�w�

Figure 1: Region where

L[a(·, z, x)](w) is analytic and

single valued.

Ε02Re�s�

4 Ε02

Im�s�

Figure 2: Region where

L[a(·, z, x)]((|v|2g−ε20+s+2i|v|gε0)/4)

is analytic for all |v|g ≥ 0 (together

with the disk D(ε20)).

Remark 5. The set of s such that L[a(·, z, x)]((|v|2g − ε21 + 2i|v|gε1 + s)/4) is

analytic in s for all |v|g ≥ 0, together with the disk D(ε20), is illustrated in

Figure 2 in the case where ε1 = ε0.

So modulo a function which is smooth on D(ε0)×X and (for fixed x) analytic

in t ∈ D(ε0), we can write

exp(−t√

∆)(F |X)(x) ≈∫

Expx(iB(ε1))

P (t, z, x)F (z)ι∗ω(z). (33)

III. Integral over Expx(iB(ε1)). Let I(t, x) be the integral on the right hand

side of (33). Identifying TxX with Rn we have

I(t, x) =

∫|u|≤ε1

P (t,Expx(iu), x)F (Expx(iu))J(u) du

where J(u) is the Jacobian of the map (u1, . . . , un) ∈ Rn 7→ Expx(iu) ∈ M

(which is a real analytic diffeomorphism for |u| < ε0). Writing this in polar

coordinates gives

I(t, x) =

∫ ε1

0

Q(t, r, x)rn−1 dr (34)

where

Q(t, r, x) =

∫η∈Sn−1

P (t, zr, x)F (zr)J(rη) dη

23

and zr = Expx(irη). Note d2(z, x) = −r2. From Equation (16) and Proposition

1 we can write I = I1 + I2 + I3 where

I1(t, x) = t

∫ ε1

0

(−r2 + t2)−(n+1)/2Q1( 14 (−r2 + t2), zr, x)rn−1 dr

I2(t, x) = t

∫ ε1

0

log(−r2 + t2)Q2( 14 (−r2 + t2), zr, x)rn−1 dr

I3(t, x) = t

∫ ε1

0

Q3( 14 (−r2 + t2), zr, x)rn−1 dr

Qi(14 (−r2 + t2), zr, x) =

∫η∈Sn−1

gi(14 (−r2 + t2), zr, x)F (zr)J(rη) dη

and gi(s, z, x) is smooth in D(ε20)×X and analytic in s and z, and F ∈ O(Mε1).

If t ∈ D(ε1), then 14 (−r2 + t2) lies in D(ε20) for all r ∈ D(ε1). In particular for

each fixed real r with |r| ≤ 1, Qi(14 (−r2 + t2), z, x) is a holomorphic function

of t ∈ D(ε1), smooth in (t, x). It follows I3(t, x) is a holomorphic function of

t ∈ D(ε1), smooth in (t, x).

We claim that for each η ∈ Sn−1, zr, F (zr) and J(rη) can be analytically

continued in r from [0, ε1] to D(ε1). After possibly shrinking ε0 it is clear this is

true for zr and J(rη). To see that it is true for F (zr), it suffices to show that the

analytic continuation of zr to r ∈ D(ε1) takes values in Mε1 . We note that the

analytic continuation of the map r 7→ γη(r), where γη is the geodesic associated

with η ∈ TxX, into the tangent bundle with the adapted complex structure is

r+is 7→ sγη(r). Since |sγη(r)|g = s, the analytic continuation of zr = Expx(irη)

in r to D(ε1) takes values in Mε1 . Then the path of integration in I1 and I2

can be changed to the quarter circle {−iα : 0 ≤ α ≤ ε1} ∪ {ε1eiα : 3π/2 ≤ α ≤2π}. This shows that there is an analytic continuation of I1 and I2 in t to

D(ε1)\i(−ε1, 0], and that (32) is analytic in t on D(ε1)\i(−ε1, 0].

We now conclude the proof of Theorem 1, part 3, by showing Sε is onto. As

noted above it suffices to show that F |X is in the range of e−ε√

∆. Fix ε > 0 and

x ∈ X. For Re(s) > 0 we have e−ε√

∆(e−s√

∆(F |X))(x) = e−(ε+s)√

∆(F |X)(x).

Both sides are analytic functions of s ∈ D(ε1)\i(−ε1, 0], so equality continues to

hold for such s. Letting s → −ε+ gives e−ε√

∆(eε√

∆(F |X))(x) = F |X(x), i.e.,

F |X is in the range of e−ε√

∆.

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