The Heat Kernel on Hyperbolic Space

Alexander Grigor’yan^{∗} Department of Mathematics Imperial College, London SW7 2BZ United Kingdom email: a.grigoryan@ic.ac.uk

Masakazu Noguchi Department of Mathematics Imperial College, London SW7 2BZ United Kingdom email: m.noguchi@ic.ac.uk

September 1997

1 Introduction and the main result

The purpose of this note is to provide a new proof for the explicit formulas of the heat kernel on hyperbolic space. By definition, the hyperbolic space ℍ^{n} is a (unique) simply connected complete n-dimensional Riemannian manifold with a constant negative sectional curvature −1.

Let Δ denote the Laplacian on a Riemannian manifold X. The heat kernel on X is a function p(x, y, t) on X × X × (0, ∞) which is the minimal positive fundamental solution to the heat equation

\frac{∂v}{∂t} = Δv.

In other words, the Cauchy problem

\{ \begin{matrix}\frac{∂v}{∂t} = Δv, \\ v|_{t=0} = v_{0}(x), \end{matrix}  (1.1)

has a solution

v(x, t) = ∫_{X} p(x, y, t) v_{0}(y) dy,  (1.2)

∗Supported by EPSRC Fellowship B/94/AF/1782

provided that v_{0} is a bounded continuous function. If, in addition, v_{0} ≥ 0, then (1.2) defines the minimal positive solution to (1.1) (see [4] for details).

If X is the Euclidean space ℝ^{n} then the heat kernel is given by the classical formula

p(x, y, t) = \frac{1}{(4πt)^{n/2}} exp (−\frac{ρ^{2}}{4t}) ,  (1.3)

where ρ = |x − y| . Due to homogeneity of the hyperbolic space, the heat kernel on ℍ^{n} also depends only on t and ρ (where ρ = dist(x, y) is now the geodesic distance on ℍ^{n}). Let us denote the heat kernel on ℍ^{n} by p_{n}(ρ, t). Then we have the following.

Theorem 1.1 The heat kernel p_{n}(ρ, t) on the hyperbolic space ℍ^{n} is given by the following formulas.

If n = 2m + 1, then

p_{n}(ρ, t) = \frac{(−1)^{m}}{2^{m}π^{m}} \frac{1}{(4πt)^{\frac{1}{2}}} (\frac{1}{sinh ρ} \frac{∂}{∂ρ})^{m} e^{−m^{2}t− \frac{ρ^{2}}{4t}} .  (1.4)

If n = 2m + 2, then

p_{n}(ρ, t) = \frac{(−1)^{m}}{2^{m+ \frac{5}{2}} π^{m+ \frac{3}{2}}} t^{− \frac{3}{2}} e^{− \frac{(2m+1)^{2}}{4} t} (\frac{1}{sinh ρ} \frac{∂}{∂ρ})^{m} ∫_{ρ}^{∞} \frac{se^{− \frac{s^{2}}{4t}}}{(cosh s − cosh ρ)^{\frac{1}{2}}} ds .

(1.5)

In particular, if n = 1, then (1.4) coincides with the one-dimensional Euclidean heat kernel (1.3). If n = 3, then (1.4) becomes

p_{3}(ρ, t) = \frac{1}{(4πt)^{\frac{3}{2}}} \frac{ρ}{sinh ρ}e^{−t− \frac{ρ^{2}}{4t}} ,  (1.6)

whereas (1.5) yields, for n = 2,

p_{2}(ρ, t) = \begin{matrix}√ & \overline{2} \\ (4πt) & \frac{3}{2} \end{matrix} e^{− \frac{1}{4} t} ∫_{ρ}^{∞} \frac{se^{− \frac{s^{2}}{4t}}}{(cosh s − cosh ρ)^{\frac{1}{2}}} ds .  (1.7)

The formulas (1.4) and (1.5) are not new. The heat kernel in dimension two (formula (1.7)) was found by McKean [6] (see also [1, pp. 242-246]). The three-dimensional case (formula (1.6)) was proved in [3, p. 396]. For n > 3, the formulas (1.4) and (1.5) can be obtained inductively by using the recurrence relation

p_{n+2}(ρ, t) = −\frac{exp (−nt)}{2π sinh ρ} \frac{∂}{∂ρ} p_{n}(ρ, t).  (1.8)

The identity (1.8) is attributed in [3, p. 396] to Millson (unpublished). Its proof can be found in [2, Theorem 2.1].

In this note, we provide an independent proof of (1.4) and (1.5) by using a completely different approach, based on the following two ingredients:

(1) the relation between the heat kernel and the wave kernel which follows from the spectral theory;

(2) the explicit formula for the wave kernel on symmetric spaces which is found in [5].

These are enough to derive (1.4) and (1.5) directly, without using (1.8).

2 Proof of the main theorem

Crucial for the proof is the following relation between the heat equation and the wave equation. Let L denote an elliptic operator on a manifold X, and let us consider the Cauchy problem for the wave equation in X × (−∞, ∞),

\{ \begin{matrix}\frac{∂^{2}u}{∂t^{2}} = Lu, \\ u|_{t=0} = u_{0}(x), \\ \frac{∂u}{∂t} |_{t=0} = 0, \end{matrix}  (2.1)

for a given u_{0} ∈ C(X). Assuming that the operator L is initially defined on C_{0}^{∞}(X) and has a self-adjoint non-positive definite extension in L^{2}(X) (which will also be denoted by L), the solution to (2.1) can be represented as

u(x, t) = cos (t\sqrt{−L}) u_{0}(x).

Similarly, for the Cauchy problem for the heat equation in X × (0, ∞)

\{ \begin{matrix}\frac{∂v}{∂t} = Lv, \\ v|_{t=0} = v_{0}(x), \end{matrix}  (2.2)

one has

v(x, t) = exp (tL) v_{o}(x), t > 0.

On the other hand, we have the following Fourier transform identity (where t > 0):

e^{−tλ^{2}} = \frac{1}{(4πt)^{\frac{1}{2}}} ∫_{−∞}^{∞} e^{− \frac{s^{2}}{4t}} e^{isλ} ds = \frac{1}{(4πt)^{\frac{1}{2}}} ∫_{−∞}^{∞} e^{− \frac{s^{2}}{4t}} cos (sλ) ds,

where λ is either a real number or a self-adjoint operator. Put λ = \sqrt{−L}; then

e^{tL} = \frac{1}{(4πt)^{\frac{1}{2}}} ∫_{−∞}^{∞} e^{− \frac{s^{2}}{4t}} cos (s\sqrt{−L}) ds,

whence we have the following proposition.

Proposition 2.1 Let L be as above, and let v_{0} = u_{0} be a bounded continuous function on X. Then the solutions u(x, t) of (2.1) and v(x, t) of (2.2) are related as follows:

v(x, t) = \frac{1}{(4πt)^{\frac{1}{2}}} ∫_{−∞}^{∞} e^{− \frac{s^{2}}{4t}} u(x, s) ds.  (2.3)

Let us describe the next ingredient of the proof: the explicit formula for u(x, s) on ℍ^{n}. Let us denote by S_{r}(x) the geodesic sphere on ℍ^{n} with centre x ∈ ℍ^{n} and radius r. It is known that the area of S_{r}(x) is equal to A(r) = Ω_{n} sinh^{n−1} r, where

Ω_{n} := \frac{2π^{\frac{n}{2}}}{Γ (\frac{n}{2})}  (2.4)

is the area of the unit sphere in ℝ^{n}. For any number r > 0, denote by M^{r} the averaging operator on ℍ^{n} :

(M^{r}f)(x) := \frac{1}{A(r)} ∫_{S_{r}(x)} f(y) dω(y),  (2.5)

where dω is the area element of S_{r}(x). For convenience, let us denote

(N_{m,k}^{r}f) (x) := (\frac{∂}{∂ cosh r})^{m} (M^{r}f(x) sinh^{k} s) ,  (2.6)

where k and m are non-negative integers.

The next assertion follows from the general mean value theorem for symmetric spaces and can be found in [5, Chapter 2].

Proposition 2.2 Let u(x, s) be the solution of the Cauchy problem on ℍ^{n} × (−∞, ∞)

\{ \begin{matrix}Lu = 0, \\ u|_{t=0} = u_{0}, \\ \frac{∂u}{∂t} |_{t=0} = u_{1}, \end{matrix}  (2.7)

where

L = Δ + (\frac{n − 1}{2})^{2}  (2.8)

and u_{0} and u_{1} are continuous initial functions. Then for any x ∈ ℍ^{n} and s > 0, we have the following.

If n ≥ 3 is odd, then

u(x, s) = c_{n} (\frac{∂}{∂s}N_{\frac{n−3}{2} ,n−2}^{s} u_{0} + N_{\frac{n−3}{2} ,n−2}^{s} u_{1})  (2.9)

where

c_{n} = \frac{Ω_{n}}{2(n − 3)!!Ω_{n−1}} = \frac{1}{(n − 2)!!} .

(We use the notation k!! := 1 · 3 · 5 · ... · k if k is odd, k!! = 2 · 4 · 6 · ... · k if k is even and k!! = 1 if k ≤ 0.)

If n ≥ 2 is even, then the following equation holds:

\frac{1}{2} ∫_{0}^{s} \frac{u(x, r) + u(x, −r)}{(cosh s − cosh r)^{\frac{1}{2}}}dr = c_{n}N_{\frac{n−2}{2} ,n−2}^{s}u_{0}  (2.10)

where c_{n} = \frac{Ω_{n}}{\sqrt{2}(n−3)!!Ω_{n−1}} =\frac{π}{2(n−2)!!} .

Proposition 2.2 simplifies for the case u_{1} = 0, as follows.

Corollary 2.3 Assume, under the hypotheses of Proposition 2.2, that u_{1} = 0. Then, for any x ∈ ℍ^{n} and s > 0, we have the following.

If n ≥ 3 is odd, then

u(x, s) = c_{n} \frac{∂}{∂s}N_{\frac{n−3}{2} ,n−2}^{s}u_{0}(x).  (2.11)

If n ≥ 2 is even, then

u(x, s) = \frac{c_{n}}{π} \frac{∂}{∂s} ∫_{0}^{s} sinh ρ ; (cosh s − cosh ρ)^{\frac{1}{2}} N_{\frac{n−2}{2} ,n−2}^{ρ}u_{0}(x) dρ  (2.12)

Proof. Since (2.11) is obvious from (2.9), let us concentrate on (2.12). Observe that u_{1} = 0 implies that the solution u(x, t) is an even function in t. Therefore (2.10) acquires the form

∫_{0}^{s} K(s, r)u(x, r) dr = c_{n}N_{\frac{n−2}{2} ,n−2}^{s}u_{0}(x),  (2.13)

where we denote

K(s, r) := \frac{1}{(cosh s − cosh r)^{\frac{1}{2}}} , s ≠ r.

By multiplying (2.13) by K(t, s) sinh s and integrating in s, we have, for any t > 0,

∫_{0}^{t} K(t, s) sinh s ∫_{0}^{s} u(x, r)K(s, r) dr ds = c_{n} ∫_{0}^{t} K(t, s) sinh sN_{\frac{n−2}{2} ,n−2}^{s}u_{0}(x) ds .

(2.14)

The left-hand side is computed by changing the order of integration:

∫_{0}^{t} dr ∫_{r}^{t} u(x, r)K(t, s)K(s, r) sinh s ds = π ∫_{0}^{t} u(x, r) dr,  (2.15)

where we have used the identity

∫_{r}^{t} K(t, s)K(s, r) sinh s ds = ∫_{a}^{b} \frac{dξ}{\sqrt{(b − ξ)(ξ − a)}} = π .

By (2.14) and (2.15),

π ∫_{0}^{t} u(x, r) dr = c_{n} ∫_{0}^{t} K(t, s) sinh sN_{\frac{n−2}{2} ,n−2}^{s}u_{0}(x) ds

whence we obtain (2.12) by differentiating in t.

Our next goal will be to extend (2.11) and (2.12) to the case s < 0. Fix x, and denote, for any ρ > 0,

U(ρ) := ∫_{S_{ρ}(x)} u_{0}(y) dω(y) = A(ρ) M^{ρ}u_{0}(x),  (2.16)

and extend U(ρ) to non-positive ρ by U(0) = 0 and U(ρ) = U(|ρ|).

Lemma 2.4 We have U(r) ∈ C^{n−1}(−∞, ∞). Moreover, U^{(l)}(0) = 0 for any l = 0, 1, ..., n − 1.

Proof. The function u_{0} is infinitely smooth, and so is U(ρ) for ρ ≠ 0. To handle the case ρ = 0, let us rewrite (2.16) in the polar coordinates y = (ρ, θ) centred at x:

U(ρ) = ∫_{𝕊^{n}} u_{0}(ρ, θ) A(ρ) dθ.

The trouble is that the distance function ρ(y) := dist(x, y) is not smooth at y = x. However, its square ρ^{2} is infinitely smooth in y, which implies that the function u_{0} is C^{∞} in ρ^{2}. Since

\frac{∂}{∂ρ} = 2ρ \frac{∂}{∂(ρ^{2})},

it is not difficult to see by induction that (\frac{∂}{∂ρ})^{k} is a sum of the terms proportional to

ρ^{2j−k} (\frac{∂}{∂ (ρ^{2})})^{j} ,

where j = 1, 2, ..., k.

Obviously, the function (\frac{∂}{∂ρ})^{m} A has, at ρ = 0, a zero of order ρ^{n−1−m}, that is, it can be represented as

(\frac{∂}{∂ρ})^{m} A = ρ^{n−1−m}A_{m}(ρ),

where A_{m}(ρ) is a continuous function of ρ. When differentiating l times the product u_{0}(ρ, θ)A(ρ) in ρ, we have the sum of the terms proportional to

(\frac{∂}{∂ρ})^{k} u_{0} · (\frac{∂}{∂ρ})^{l−k} A,

which splits further, to the sum of terms as

ρ^{2j−k} (\frac{∂}{∂ (ρ^{2})})^{j} u_{0} · ρ^{n−1−l+k}A_{l−k}(ρ) = ρ^{2j+(n−1−l)} (\frac{∂}{∂ (ρ^{2})})^{j} u_{0} · A_{l−k}(ρ).

If l ≤ n − 1, then the latter function is continuous up to ρ = 0 and vanishes at ρ = 0 whence we obtain U ∈ C^{n−1} and U(0) = U′(0) = U′′(0) = ... = U^{(n−1)}(0) = 0.

Corollary 2.5 We have the following, under the hypotheses of Corollary 2.3, for all x ∈ ℍ^{n} and s ∈ (−∞, ∞).

If n ≥ 3 is odd, then

u(x, s) = \frac{c_{n}}{Ω_{n}} \frac{∂}{∂s} (\frac{1}{sinh s} \frac{∂}{∂s})^{\frac{n−3}{2}} [\frac{1}{sinhs} U(s)] .  (2.17)

If n ≥ 2 is even, then

u(x, s) = \frac{c_{n}}{πΩ_{n}} \frac{∂}{∂s} ∫_{0}^{s} K(s, ρ) sinh ρ (\frac{1}{sinh ρ} \frac{∂}{∂ρ})^{\frac{n−2}{2}} (\frac{1}{sinh ρ}U(ρ)) dρ .

(2.18)

Proof. Indeed, (2.17) coincides with (2.11) if s > 0. Since both sides of (2.17) are even functions in s, (2.17) holds also for s < 0. The case s = 0 follows by continuity. Each derivation \frac{∂}{∂s} of U(s), or division by sinh s, reduces smoothness at 0 by at most 1. After n − 1 such operations as in (2.17), we still have a continuous function on the right-hand side of (2.17). In particular, this argument shows that the right-hand side of (2.17) has meaning for s = 0.

The equation (2.18) follows in the same way from (2.12).

Proof of Theorem 1.1. It is clear from (2.8) that

e^{tΔ} = exp (− (\frac{n − 1}{2})^{2} t) e^{tL},

so the kernels of the semigroups e^{tΔ} and e^{tL} are related by the same equation. Therefore it will suffice to find the kernel of the semigroup e^{tL}. Note that the operator L is non-positive definite, for the top of the spectrum of Δ in L^{2}(ℍ^{n}) is equal to −(\frac{n−1}{2})^{2}.

We shall use the notation u(x, t) and v(x, t) for the solutions to the initial problems (2.1) and (2.2), respectively, for the operator (2.8). The initial function u_{0} = v_{0} is supposed to be in C_{0}^{∞}(ℍ^{n}).

Case I: n = 2m + 1.

We have, by (2.3) and (2.17),

v(x, t) = \frac{c_{n}}{(4πt)^{\frac{1}{2}}} ∫_{−∞}^{∞} e^{− \frac{s^{2}}{4t}} \frac{∂}{∂s} (\frac{1}{sinh s} \frac{∂}{∂s})^{\frac{n−3}{2}} [\frac{1}{sinhs}U(s)] ds ,  (2.19)

whence we obtain, by integration by parts,

v(x, t) = \frac{c_{n}Ω_{n}^{−1}}{(4πt)^{\frac{1}{2}}} ∫_{−∞}^{∞} e^{− \frac{s^{2}}{4t}} \frac{∂}{∂s} (\frac{1}{sinh s} \frac{∂}{∂s})^{m−1} [\frac{1}{sinhs}U(s)] ds
= \frac{(−1)^{m}c_{n}Ω_{n}^{−1}}{(4πt)^{\frac{1}{2}}} ∫_{−∞}^{∞} \{(\frac{1}{sinh s} \frac{∂}{∂s})^{m} e^{− \frac{s^{2}}{4t}} \} U(s) ds
= \frac{2(−1)^{m}c_{n}Ω_{n}^{−1}}{(4πt)^{\frac{1}{2}}} ∫_{0}^{∞} \{(\frac{1}{sinh s} \frac{∂}{∂s})^{m} e^{− \frac{s^{2}}{4t}} \} U(s) ds
= \frac{2(−1)^{m}c_{n}Ω_{n}^{−1}}{(4πt)^{\frac{1}{2}}} ∫_{ℍ^{n}} (\frac{1}{sinh ρ} \frac{∂}{∂ρ})^{m} e^{− \frac{ρ^{2}}{4t}} u_{0}(y) dy,

where ρ = dist(x, y).

Observing that

2c_{n}Ω_{n}^{−1} = \frac{1}{(n − 3)!!Ω_{n−1}} = \frac{1}{2^{m}π^{m}} ,

we conclude that the operator e^{tL} is an integral operator with kernel

\frac{(−1)^{m}}{2^{m}π^{m}} \frac{1}{(4πt)^{\frac{1}{2}}} (\frac{1}{sinh ρ} \frac{∂}{∂ρ})^{m} e^{− \frac{ρ^{2}}{4t}} .

Therefore the heat kernel p_{n}(ρ, t) of e^{tΔ} is

p_{n}(ρ, t) = \frac{(−1)^{m}}{2^{m}π^{m}} \frac{1}{(4πt)^{\frac{1}{2}}} (\frac{1}{sinh ρ} \frac{∂}{∂ρ})^{m} e^{−m^{2}t− \frac{ρ^{2}}{4t}} ,

which was to be proved.

Case II: n = 2m + 2.

We have, by (2.3) and (2.18), by changing the order of the integrals and by integration by parts,

v(x, t)
= \frac{c′_{n}}{\sqrt{t}} ∫_{−∞}^{∞} e^{− \frac{s^{2}}{4t}} [\frac{∂}{∂s} ∫_{0}^{s} K(s, ρ) sinh ρ (\frac{1}{sinh ρ} \frac{∂}{∂ρ})^{m} (\frac{1}{sinh ρ}U(ρ)) dρ] ds
= \frac{c′_{n}}{\sqrt{t}} ∫_{−∞}^{∞} \frac{2s}{4t}e^{− \frac{s^{2}}{4t}} [∫_{0}^{s} K(s, ρ) sinh ρ (\frac{1}{sinh ρ} \frac{∂}{∂ρ})^{m} (\frac{1}{sinh ρ}U(ρ)) dρ] ds
= \frac{c′_{n}}{t^{\frac{3}{2}}} ∫_{0}^{∞} (∫_{ρ}^{∞} se^{− \frac{s^{2}}{4t}} K(s, ρ)ds) sinh ρ (\frac{1}{sinh ρ} \frac{∂}{∂ρ})^{m} (\frac{1}{sinh ρ}U(ρ)) dρ
= \frac{(−1)^{m}c′_{n}}{t^{\frac{3}{2}}} ∫_{0}^{∞} (\frac{1}{sinh ρ} \frac{∂}{∂ρ})^{m} (∫_{ρ}^{∞} se^{− \frac{s^{2}}{4t}} K(s, ρ)ds) U(ρ) dρ
= \frac{(−1)^{m}c′_{n}}{t^{\frac{3}{2}}} ∫_{ℍ^{n}} ((\frac{1}{sinh ρ} \frac{∂}{∂ρ})^{m} ∫_{ρ}^{∞} se^{− \frac{s^{2}}{4t}} K(s, ρ)ds) u_{0}(y) dy,

where c′_{n} = _{\frac{c_{n}}{2π^{\frac{3}{2}} Ω_{n}}} = \frac{1}{2^{m+ \frac{5}{2}} π^{m+ \frac{3}{2}}} .

This shows that the kernel of the integral operator e^{tL} is

\frac{(−1)^{m}}{2^{m+ \frac{5}{2}} π^{m+ \frac{3}{2}}} t^{− \frac{3}{2}} (\frac{1}{sinh ρ} \frac{∂}{∂ρ})^{m} ∫_{ρ}^{∞} \frac{se^{− \frac{s^{2}}{4t}}}{(cosh s − cosh ρ)^{\frac{1}{2}}} ds,

whence the heat kernel is

p_{2m+2}(x, y, t) = \frac{(−1)^{m}}{2^{m+ \frac{5}{2}} π^{m+ \frac{3}{2}}} e^{−t \frac{(2m+1)^{2}}{4}} t^{− \frac{3}{2}} (\frac{1}{sinh ρ} \frac{∂}{∂ρ})^{m} ∫_{ρ}^{∞} \frac{se^{− \frac{s^{2}}{4t}}}{(cosh s − cosh ρ)^{\frac{1}{2}}} ds.

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