Random Processes in Hyperbolic Spaces Hyperbolic Brownian ...

Dottorato di Ricerca in Statistica Metodologica
Tesi di Dottorato XXII Ciclo – 2006/2009
Dipartimento di Statistica, Probabilità e Statistiche Applicate
Random Processes in Hyperbolic Spaces
Hyperbolic Brownian Motion and Processes with Finite
Velocity in the Hyperbolic Plane
Valentina Cammarota

Contents
Introduction v
1 Tools of Riemannian Geometry 1
1.1 Settings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Laplace operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.3 Sectional curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.4 Hyperbolic space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2 Hyperbolic Geometry 13
2.1 Möbius transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.2 Poincaré extension of Möbius transformations . . . . . . . . . . . . . . . . . . . . . 16
2.3 Hyperbolic metric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.4 Geodesics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
2.5 Hyperbolic trigonometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
3 Hyperbolic Brownian Motion 29
3.1 Settings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
3.2 Hyperbolic Brownian motion in H 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
3.2.1 Transition function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
3.2.2 Hitting distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
3.2.3 Hyperbolic Brownian bridge . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
3.2.4 Fractional hyperbolic Brownian motion . . . . . . . . . . . . . . . . . . . . 43
3.3 Multidimensional hyperbolic Brownian motion . . . . . . . . . . . . . . . . . . . . 44
3.3.1 Transition function in H 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
3.3.2 The Millson recursive formula . . . . . . . . . . . . . . . . . . . . . . . . . . 45
3.3.3 Bounds for the hyperbolic heat kernel . . . . . . . . . . . . . . . . . . . . . 47
iii

CONTENTS CONTENTS
3.3.4 Gruet’s formula for the heat kernel . . . . . . . . . . . . . . . . . . . . . . . 49
3.3.5 Hyperbolic branching Brownian motion . . . . . . . . . . . . . . . . . . . . 52
3.A Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
4 Travelling Randomly on H 2 with a Pythagorean Compass 59
4.1 Description of the planar random motion on H 2 . . . . . . . . . . . . . . . . . . . . 59
4.2 Equations related to the mean hyperbolic distance . . . . . . . . . . . . . . . . . . 62
4.3 About the higher moments of the hyperbolic distance . . . . . . . . . . . . . . . . 69
4.4 Motions with jumps backwards to the starting point . . . . . . . . . . . . . . . . . 74
4.5 Motion at finite velocity on the surface of a three-dimensional sphere . . . . . . . . 80
5 Cascades of Particles Moving at Finite Velocity in H 2 83
5.1 Description of the randomly moving and branching model . . . . . . . . . . . . . . 83
5.2 Mean hyperbolic distance: the Laplace transform approach . . . . . . . . . . . . . 86
5.3 Equation governing the hyperbolic distance . . . . . . . . . . . . . . . . . . . . . . 92
5.4 Mean hyperbolic distance of the k-th splinter . . . . . . . . . . . . . . . . . . . . . 97
iv
101

Introduction
Euclid wrote his famous Elements around 300 B.C. In the first book of the Elements Euclid develops
plane geometry starting with the following five postulates
• A straight line may be drawn from any point to any other point.
• A finite straight line may be extended continuously in a straight line.
• A circle may be drawn with any center and any radius.
• All right angles are equal.
• If a straight line falling on two straight lines makes the interior angles on the same side less
than two right angles, the two straight lines, if extended indefinitely, meet on the side on
which the angles are less than two right angles.
Euclid’s fifth postulate is equivalent to the modern postulate of Euclidean geometry
• Through a point outside a given infinite line there is one and only one infinite straight line
parallel to the given line.
For two thousand years it was believed that the fifth postulate could be derived from the other
four postulates and for centuries the mathematicians attempted to prove the fifth postulate. It
was not until the nineteenth century that the fifth postulate was finally shown to be independent.
The proof of this independence was the result of a completely unexpected discovery: the denial of
the fifth postulate leads to a new consistent geometry.
It was Gauss who first made this discovery. It was known from the eighteenth century that the
fifth postulate is equivalent to the theorem that the sum of the angles of a triangle is 180 o , Gauss,
assuming that the sum of the angles of a triangle is less than 180 o , discovered the non-Euclidean
geometry.
Only few years later Lobachevsky published the first account of non-Euclidean geometry in 1829
in a paper entitled On the principle of geometry. In 1832, Bolay published an independent account
of non-Euclidean geometry in a paper entitled The absolute science of space.
Today the non-Euclidean geometry of Gauss, Lobachevsky and Bolay is called hyperbolic geometry,
and the term non-Euclidean geometry refers to any geometry that is not Euclidean.
After defining spaces of variable curvature and noting that the Euclidean space is a space of
zero curvature, Riemann considered spaces of constant non zero curvature. An example of a ndimensional
Riemannian space of constant positive curvature is the n-dimensional sphere in the

(n + 1)-dimensional Euclidean space. In the setting of Riemannian geometry an example of a
n-dimensional space of constant negative curvature is the n-dimensional Lobachevskian space. For
a comprehensive history of non-Euclidean geometry see Rosenfeld (38).
In Chapter 1 are introduced some fundamental results of Riemannian geometry, the aim of
this chapter is to prove that the hyperbolic space is a space with constant negative curvature equal
to −1 and to obtain an explicit expression, in local coordinate, of the Laplace operator in the
n-dimensional hyperbolic space. On Riemannian geometry see, e.g., Chavel (8), (9) and Spivak
(40).
Chapter 2 is devoted to hyperbolic geometry. In this chapter are introduced the half-space
model and the disk model of the hyperbolic geometry. We give some explicit formulas for the
hyperbolic metric and a characterization of the geodesics (that is curves of shortest hyperbolic
length) in these models. In the last section of the chapter it is proved the famous Lobachevsky’s
formula and the Hyperbolic Pythagorean Theorem. For a complete account on hyperbolic geometry
see, for example, Beardon (2), Kulczychi (23), Ratcliffe (35) and Terras (41).
Brownian motion on hyperbolic spaces has been studied since the end of the Fifties. In 1959
Gertsenshtein and Vasiliev (15) gave the explicit form of the transition function for the hyperbolic
Brownian motion in two and three dimensional hyperbolic spaces. The studies on hyperbolic Brownian
motion have been revitalized by mathematical finance since some financial products (Asian
options) have a strict connection with the stochastic representation of the hyperbolic Brownian
motion (see, for example, Yor (43)). In Chapter 3 we give a description of the principal results
obtained in the literature about two dimensional and multi-dimensional hyperbolic Brownian
motion: stochastic representation, transition function, hyperbolic Brownian bridge, hitting distribution,
fractional hyperbolic Brownian motion, the Millson recursive formula, uniform estimations
for the transition function and branching hyperbolic Brownian motion.
More recently also works concerning two-dimensional random motions at finite velocity on planar
hyperbolic spaces have been introduced and analyzed, see Orsingher and De Gregorio (34). While
in (34) the components of the motion are supposed to be independent, we present in Chapter 4 a
planar random motion with interacting components. We study the motion of particle that runs on
the geodesic lines of the hyperbolic plane changing direction at Poisson-paced times. The motion
considered here is the non-Euclidean counterpart of the planar motion with orthogonal deviations
studied in Orsingher (33). The main object of the investigation is the hyperbolic distance η(t)
at time t of the moving point from the origin. We are able to give explicit expressions for its
mean value, also under the condition that the number of changes of direction is known. In the
case of motion in the hyperbolic plane with independent components an explicit expression for the
distribution of the hyperbolic distance η(t) has been obtained (Orsingher and De Gregorio (34)).
Here, however, the components of motion are dependent and this excludes any possibility of finding
the distribution of the hyperbolic distance.
We obtain the following explicit formula for the mean value of the hyperbolic distance which
reads
λt −
E{cosh η(t)} = e 2 cosh t λ t
λ2 + 4c2 + √ sinh λ2 + 4c2 2
λ2 + 4c2 2
Section 4.4 is devoted to motions on the hyperbolic plane where the return to the starting point
is admitted and occurs at the instants of changes of direction. The mean distance from the origin of
vi

vii CHAPTER 0. INTRODUCTION
these jumping-back motions is obtained explicitly by exploiting their relationship with the motion
without jumps. In the case where the return to the starting point occurs at the first Poisson event
T1, the mean value of the hyperbolic distance η1(t) reads
E{cosh η1(t)|N(t) ≥ 1} =
λ
√ λ 2 + 4c 2
sinh t
√
2 λ2 + 4c2 sinh λt
2
The last section of the chapter considers the motion at finite velocity, with orthogonal deviations
at Poisson times, on the unit-radius sphere. The main results concern the mean value
E{cos d(P0Pt)}, where d(P0Pt) is the distance of the current point Pt from the starting position
P0. We take profit of the analogy of the spherical motion with its counterpart on the hyperbolic
plane to discuss the different situations due to the finiteness of the space where the random motion
develops. See also Cammarota and Orsingher (5) and (7).
In Chapter 5 we study a random motion of a cloud of particles moving at finite velocity on
geodesic lines of the hyperbolic plane. Such particles are generated by successive splitting out of
a unit-mass particle initially placed at the origin O. The disintegration process of the unit-mass
particle is governed by an underlying Poisson process of rate λ as follows. The original particle
keeps moving on the main geodesic line with constant hyperbolic velocity c; at the first Poisson
event it breaks into two parts of equal mass 1/2. One particle continues its motion on the main
geodesic line, while the other one deviates orthogonally. At the second Poisson event the deviated
particle is separated into two pieces each of mass 1/2 2 ; one piece continues its motion on the same
geodesic line whereas the other one starts moving on the geodesic line orthogonal to that joining
its position with the origin O. In general, at the k-th Poisson event, only the deviating particle of
mass 1/2 k breaks into two fragments of equal mass 1/2 k+1 ; the first continues its motion on the
same geodesic line while the other one deviates orthogonally.
If up to time t, N(t) Poisson events have occurred, we have N(t) + 1 particles running, at a
constant hyperbolic velocity c, along different geodesic lines with a mass depending on the instant
of separation from the generating particle.
In the above branching process each particle can reproduce only once and the particle splitting
at time Tk of the k-th Poisson event and which never more disintegrates will preserve its mass,
equal to 1/2 k , for the successive time interval (Tk, t).
Our main result concerns the dynamics of the center of mass cm of the cloud of particles performing
the branching and diffusion process. In particular, we are able to give an exact expression
for the mean hyperbolic distance from the origin O, ηcm(t), of the center of mass cm at any time
t > 0
E{cosh ηcm(t)} = 23c2
3 −
e 22 λt
√
λ2 + 24c2 t −
e 22 √
λ2 +24c2 3 √ λ2 + 24c2 + 5λ +
+ λ + 2c
2(λ + 3c) ect λ − 2c
+
2(λ − 3c) e−ct .
.
e t
22 √
λ2 +24c2 3 √ λ 2 + 2 4 c 2 − 5λ
We give two different and independent proofs of the above result: our first technique is based on
Laplace transforms, while the other one brings about the following non-homogeneous second-order

differential equation
3 −
22 λt
d2 dt2 u − c2u = λc2e √
λ2 + 24c2 t −
e 22 √
λ2 +24c2 − e t
22 √
λ2 +24c2
.
We also examine the mean hyperbolic distance of each individual particle which stops changing
direction after the k-th Poisson event. Our main result shows that
where g(s; k, λ) = e−λs λ k s k−1
Γ(k)
E{cosh ηk(t)I {N(t)≥k}} = 1
2 k
t
0
cosh c(t − s)h(k, c, s)g(s; k, λ) ds
is a Gamma distribution and h(k, c, s) = k
r=0
viii
k cs(2Yr,k−1)
r E{e }
with Yr,k ∼ Beta(r, k − r) so that (2Yr,k − 1) ∈ (−1, 1) and with the assumption that Y0,k = 1 and
Yk,k = −1. Thus the last formula shows that the mean hyperbolic distance, at time t, of a particle
generated at the k-th Poisson event can be seen as the mean hyperbolic distance of a particle which
never deviates from the main geodesic line and which stops moving at a random time with law
h(k, c, s)g(s; k, λ). See also Cammarota and Orsingher (6).

Chapter 1
Tools of Riemannian Geometry
1.1 Settings
Definition 1.1.1. A topological space is a set X together with a collection O of subsets of X
satisfying the following properties
• ∅, X ∈ O
• the intersection of any finite collection of sets in O is also in O,
• the union of any collection of sets in O is also in O.
The sets in O are called open.
Definition 1.1.2. A topological space is Hausdorff if for any two distinct points x1, x2 ∈ X there
exists open sets O1, O2 ∈ O such that x1 ∈ O1, x2 ∈ O2 and O1 ∩ O2 = ∅.
The Hausdorff axiom is essential for the uniqueness of the limit of convergent sequences.
Definition 1.1.3. A map between two topological spaces is called continuos if the inverse image
of any open set is open. A bijective map which is continuous in both directions is called
homeomorphism.
Definition 1.1.4. A n-dimensional manifold is a Hausdorff topological space M such that for each
x ∈ M there is a neighborhood U of x that is homeomorphic to an open subset O of R n . Such a
homeomorphism ϕ : U → O is called coordinate chart.
The important point here is that locally the topology of a manifold is the same as that of R n .
Definition 1.1.5. Consider a family (Ui)i∈I of open sets such that M =
i∈I Ui. An atlas is the
family {Ui, ϕi}i∈I of charts.
Remark 1.1.1. Each ϕ −1
i (·) is referred to as parametrization and the set ϕi(Ui) is called parameter
domain.
Definition 1.1.6. A n-dimensional manifold M is C r differentiable if, for Ui ∩ Uj = ∅, the
composition
ϕj ◦ ϕ −1
i
: ϕi(Ui ∩ Uj) → ϕj(Ui ∩ Uj)

1.1 Settings 2
is C r differentiable for arbitrary i, j.
In what follows we will assume that M and N are two C ∞ differentiable manifolds.
Definition 1.1.7. Let M be a n-dimensional manifold and N be a m-dimensional manifold. A
function f : M → N is said to be differentiable if for all charts ϕi : Ui → Rn , ψi : Vi → Rm with
f(Ui) ⊂ Vi the composition ψi ◦ f ◦ ϕ −1
i is also differentiable.
The tangent space of a n-dimensional manifold M at a point p ∈ M is going to be thought of
as the n-dimensional set of “directions” which, starting at p, point in all directions of M. Without
making reference to the ambient space this notion has to be intrinsically defined as follows.
Definition 1.1.8. A tangent vector X(p) at p ∈ M is a derivative operator defined on the set of
germs of functions Fp(M) := {f : M → R|f differentiable}/ ∼, where the equivalence relation ∼
is defined by declaring f ∼ f ∗ if and only if f and f ∗ coincide in a neighborhood of p.
Definition 1.1.8 means more precisely that the tangent vector X(p) is a map X(p) : Fp(M) → R
with the two following properties
X(p)(αf + βg) = αX(p)f + βX(p)g, (1.1)
X(p)(fg) = g(p)X(p)f + f(p)X(p)g, (1.2)
where α, β ∈ R and f, g ∈ Fp(M). The value X(p)f is called directional derivative of f at p in the
direction X(p).
Definition 1.1.9. The tangent space TpM of M at p is defined as the set of all tangent vectors
at the point p.
Let M be a n-dimensional manifold, p ∈ U ⊂ M and let ϕ : U → R n be a chart. We denote by
(u 1 , . . . , u n ) the standard coordinates of R n and by (x 1 , . . . , x n ) the corresponding coordinates in
M. Thus x i (p) is the function given by the i-th coordinate of ϕ(p) in R n , that is
x i (p) = u i (ϕ(p)).
On the other hand it is possible to see the coordinates (x 1 , . . . , x n ) as variables with respect to
which we can form derivatives. For a function f : M → R, f ∈ C 1 , we set
∂f
∂xi
p
:= ∂(f ◦ ϕ−1 )
∂u i
In view of Definition 1.1.8, special tangent vectors at p are the partial derivatives ∂i(p), with
i = 1, . . . , n, defined by
∂i(p)f := ∂f
∂xi
.
It is possible to prove that the tangent space TpM at p of the n-dimensional manifold M is
a n-dimensional vector space on R spanned, in appropriately chosen coordinate (x 1 , . . . , x n ), in a
given chart by the tangent vectors ∂i(p), i = 1, . . . , n,
p
ϕ(p)
TpM = Span{∂1(p), . . . , ∂n(p)}.
.

3 Tools of Riemannian Geometry
For a proof see, for example, Kühnel (24) Theorem 5.6. Therefore for every tangent vector X(p) ∈
TpM one has
n
X(p) =
i=1
ξ i p ∂i(p), (1.3)
where ξ i p is the directional derivative of the coordinate function x i in the direction X(p). For a
function f : M → R, f ∈ C 1 we have
X(p)f =
n
i=1
ξ i p ∂i(p)f.
Definition 1.1.10. A differentiable vector field X on a n-dimensional manifold M is an association
p ∈ U ⊂ M → X(p) ∈ TpM such that in every chart ϕ : U → R n , with coordinates (x 1 , . . . , x n ), the
coefficients ξ i p : U → R in the representation (1.3) valid at the each point p ∈ U are differentiable
functions.
Common notations for the vector field and the directional derivative of a function f : M → R,
f ∈ C 1 in the direction of the vector field are
X =
n
ξ i ∂i, Xf =
i=1
n
ξ i ∂if.
The interpretation of X as a derivative operator permits us to consider the iterates of X. If
X and Y are two differentiable vector fields on M and f : M → R, f ∈ C 2 , we can consider the
functions X(Y f) and Y (Xf).
Proposition 1.1.1. Let X and Y be two differentiable vector fields on a C ∞ manifold M. Then
there exists a unique vector filed [X, Y ], which is referred to as the Lie bracket of X and Y , such
that for all f : M → R, f ∈ C 2 ,
i=1
[X, Y ]f = X(Y f) − Y (Xf).
For a proof see, for example, Do Carmo (14) Lemma 5.2.
We introduce now a metric structure on M.
Definition 1.1.11. An inner product on a vector space V over a field F is a bilinear form
such that is
• symmetric: 〈v, w〉 = 〈w, v〉
• positive definite: 〈v, v〉 > 0 for all v = 0.
(v, w) ∈ V × V → 〈v, w〉 ∈ F,
Definition 1.1.12. Given a C ∞ manifold M, define a Riemannian metric g on M to be a mapping
that associates with each p ∈ M an inner product
gp : MpT × MpT → R

1.1 Settings 4
satisfying the following differentiability property. If U is any open set in M and X, Y are two
differentiable vector fields on U, then the function g(X, Y ) : U → R given by
is differentiable on U.
g(X, Y )(p) := gp(X(p), Y (p))
Each tangent space is then equipped with an inner product g that varies smoothly from point to
point.
Definition 1.1.13. A Riemannian manifold is a C ∞ manifold M with a given Riemannian metric
g.
For a given Riemannian metric g we define
〈·, ·〉 := g(·, ·), gjk = gkj := 〈∂j, ∂k〉, G := (gjk), G −1 := (g jk ), g := det G,
where j, k = 1 . . . , n.
Definition 1.1.14. Let X and Y be two differentiable vector fields on a manifold M, a connection
is a mapping that associates to X and Y the differentiable vector field ▽XY that satisfies the
following properties
▽fX+gY Z = f▽XZ + g▽Y Z, (1.4)
▽X(Y + Z) = ▽XY + ▽XZ, (1.5)
▽X(fY ) = f▽XY + (Xf)Y. (1.6)
where X, Y and Z are differentiable vector fields, f, g are two differentiable functions on M and
fX denotes the vector field (fX)(p) := f(p)X(p).
The following theorem holds (for a proof see, e.g., Do Carmo (14) Theorem 3.6).
Theorem 1.1.1 (Levi-Civita). Given a Riemannian manifold M, there exists a unique connection
▽ on M satisfying the conditions
for all X, Y and Z differentiable vector fields on M.
▽XY − ▽Y X = [X, Y ], (1.7)
X〈Y, Z〉 = 〈▽XY, Z〉 + 〈Y, ▽XZ〉, (1.8)
The connection defined by Theorem 1.1.1 is referred to as the Levi-Civita (or Riemannian) connection.
Let X and Y be two differentiable vector fields on M such that
X =
n
ξ i ∂i, Y =
i=1
In view of (1.4), (1.5) and (1.6) we have
i
n
η j ∂j.
j=1
▽XY = ▽P i ξi
∂iY = ξ i ▽∂iY =
ξ i ▽∂i(η j ∂j) =
=
ξ i η j ▽∂i∂j +
ξ i (∂iη j )∂j =
ξ i η j Γ k ji∂k +
i,j
i,j
i,j
i,j,k
i,j
ξ i [η j ▽∂i∂j + (∂iη j )∂j]
i,j
ξ i (∂iη j )∂j, (1.9)

5 Tools of Riemannian Geometry
where the functions Γ k ij
defined by
From (1.9) if follows that
: U ⊂ M → R, i, j, k = 1, . . . , n, referred to as Christoffel symbols, are
▽∂j ∂i =
n
k=1
Γ k ij∂k.
▽XY − ▽Y X =
ξ i η j Γ k ji∂k +
ξ i (∂iη j )∂j −
η j ξ i Γ k ij∂k −
i,j,k
i,j
=
ξ i (∂iη j )∂j −
i,j
=
i,j
i,j
[ξ i (∂iη j ) − η i (∂iξ j )]∂j
η j (∂jξ i )∂i
i,j,k
i,j
η j (∂jξ i )∂i
(1.10)
In order to obtain an explicit expression of the Γ k ij , by setting X = ∂i and Y = ∂j, we have
from (1.10) that
▽∂i ∂j − ▽∂j ∂i = [∂i − ∂i]∂j = 0. (1.11)
But, since
▽∂i∂j − ▽∂j ∂i =
(Γ k ji − Γ k ij)∂k, (1.12)
we obtain from (1.11) and (1.12) that, for all i, j, k = 1, . . . , n,
If we set X = ∂i, Y = ∂j and Z = ∂k in (2.20) then it becomes
that is
k
Γ k ij = Γ k ji. (1.13)
∂i〈∂j, ∂k〉 = 〈▽∂i ∂j, ∂k〉 + 〈∂j, ▽∂i ∂k〉
∂igjk = 〈
Γ l ji∂l, ∂k〉 + 〈∂j,
Γ l ki∂l〉 =
Γ l ji glk +
l
l
l
l
Γ l ki gjl
=
[Γ l ji glk + gjl Γ l ki]. (1.14)
l
From (1.13) and (1.14) we deduce the following explicit expression for the Γ k ij
Γ k ij = 1
2
g kl {∂i glj + ∂j gil − ∂l gij} (1.15)
l

1.2 Laplace operator 6
in fact
g kl {∂i glj + ∂j gil − ∂l gij} =
g kl
[Γ m li gmj + glm Γ m ji] +
l
1.2 Laplace operator
l
−
m
m
[Γ m il gmj + gim Γ m jl ]
m
[Γ m ij gml + gim Γ m lj ]
=
g kl Γ m li gmj + glmΓ m ji + Γ m ij gml + gimΓ m lj − Γ m il gmj − gimΓ m jl
l,m
=
g kl glmΓ m ji + Γ m
ij gml = 2
l,m
= 2 Γ k ij.
l,m
g kl Γ m ij glm
Definition 1.2.1. Given a function f : M → R, f ∈ C 1 , the gradient of f, grad f, is the vector
field on M such that
〈grad f, X〉 = Xf (1.16)
for all differentiable vector fields X on M.
Given two functions f, h : M → R, f, h ∈ C 1 , it follows that
In fact, in view of (1.1), we have
grad (f + h) = grad f + grad h, (1.17)
grad (fh) = h(grad f) + f(grad h). (1.18)
〈grad (f + h), X〉 = X(f + h) = Xf + Xh = 〈grad f, X〉 + 〈grad h, X〉
and in view (1.2), we have
= 〈grad f + grad h, X〉,
〈grad (fh), X〉 = X(fh) = h(Xf) + f(Xh) = h〈grad f, X〉 + f〈grad h, X〉
= 〈h grad f + f grad h, X〉.
We now calculate the expression of gradf in local coordinates. Observing that, for j, k, l = 1, . . . , n,
we have
gjkg kl ∂l =
δjl∂l = ∂j,
k,l
where δjl is the Kronecker delta, and
gjkg kl ∂lf =
〈∂j, ∂k〉g kl ∂lf = 〈∂j,
g kl (∂lf)∂k〉,
k,l
k,l
l
k,l

7 Tools of Riemannian Geometry
we obtain
Xf =
n
ξ j ∂jf =
j=1
j,k,l
ξ j gjkg kl ∂lf =
Now, in view of Definition 1.2.1 and formula (1.19), we finally have
n
ξ j 〈∂j,
(g kl ∂lf)∂k〉 = 〈X,
g kl (∂lf)∂k〉. (1.19)
j=1
k,l
grad f =
g kl (∂lf)∂k =
k,l
k,l
g kl ∂(f ◦ ϕ−1 )
∂u l
∂
k,l
. (1.20)
∂xk From formula (1.20) and in view of the representation (1.3) it follows that, if f ∈ C k with k ≥ 1,
then grad f is a C k−1 vector field.
Definition 1.2.2. Given a differentiable vector field Y on M, define the real-valued function
divergence of Y , div Y , by
where X(p) ranges over MpT .
(divY )(p) = trace(X(p) → ▽ X(p)Y )
To obtain the expression of div Y in local coordinates we observe that, in view of (1.9), we have
therefore
▽XY =
ξ i [
div Y =
i,k
j
⎧
n ⎨
m=1
Inserting formula (1.15) in (1.21), we obtain
div Y =
j
= 1
2
l
η l Γ j j
lj + ∂jη
⎩
j
=
j,l
η j Γ k ji + ∂iη k ]∂k,
η j Γ m jm + ∂mη m
η l Γ j
lj
η j
g lk
{∂jgkl + ∂lgjk − ∂kgjl} +
∂jη j
j,l
k
j
= 1
η g
2
j,l k
lk ∂jgkl +
∂jη
j
j
=
⎧
⎨
1
ηj g
⎩2
j
k,l
kl ∂jglk + ∂jη j
⎫
⎬
⎭
=
1
2 ηj tr(G −1 ∂jG) + ∂jη j
,
j
⎫
⎬
. (1.21)
⎭
+ ∂jη j =
η j Γ l jl +
∂jη j
since g kl = g lk because G is symmetric and where ∂jG is the matrix obtained from G by differentiating
each of the entries with respect to the j-th coordinate. With the standard formula for
j
j
j,l
j

1.3 Sectional curvature 8
differentiating determinants we have that tr(G −1 ∂jG) = ∂j(ln detG) and then
div Y = div (
η
j
j ∂j) =
1
2
j
ηj tr(G −1 ∂jG) + ∂jη j
=
1
2
j
ηj ∂j(ln g) + ∂jη j
= j
η ∂j(ln √ g) + ∂jη j =
j 1
η √ ∂j(
g √ g) + ∂jη j
=
=
j
1
√ g
1
√ g
j
η ∂j( √ g) + (∂jη j ) √ g
j
j
∂j(η j√ g). (1.22)
j
Definition 1.2.3. For every function f : M → R, f ∈ C 2 , we define the Laplacian of f, ∆f, by
∆f = div(grad f).
By combining (1.20) and (1.22), we obtain the explicit expression of ∆f in local coordinates:
∆f = div(grad f) = div (
g kl
∂lf)∂k
=
=
1
√ g
1
√ g
∂k(
g kl ∂lf √ g)
k
l
k
l
∂k(g kl√ g ∂lf). (1.23)
Next we consider the following invariance property of the Laplace operator.
k,l
Theorem 1.2.1. Let Φ be a diffeomorphism of the Riemannian manifold M onto itself. Then Φ
leaves the Laplace operator ∆ invariant, that is
for all f ∈ C 2 , if and only if Φ is an isometry.
∆f = (∆(f ◦ Φ)) ◦ Φ −1
The proof of Theorem 1.2.1 follows by taking into account the representation in (1.23) and by
observing that, for p ∈ M, we have gij(p) = gij(Φ(p)) if and only if Φ is an isometry. For a proof
see Helgason (19) Proposition 2.4, Chapter 2.
1.3 Sectional curvature
Definition 1.3.1. Let X, Y and Z be three differentiable vector fields on a Riemannian manifold
M. The curvature R is a correspondence that associates to X, Y and Z the differentiable vector
field R(X, Y )Z given by
where ▽ is the Levi-Civita connection.
R(X, Y )Z = ▽Y ▽XZ − ▽X▽Y Z + ▽ [X,Y ]Z,

9 Tools of Riemannian Geometry
Since, if M = R n then R(X, Y )Z = 0 for all X, Y , Z, we are able to think of R as a way of
measuring how mach M deviates from being Euclidean. In view of (1.4), (1.5) and (1.6), we obtain
that
R(fX1 + gX2, Y )Z = fR(X1, Y )Z + gR(X2, Y )Z, (1.24)
R(X, fY1 + gY2)Z = fR(X, Y1)Z + gR(X, Y2)Z, (1.25)
R(X, Y )(fZ1 + gZ2) = fR(X, Y )Z1 + gR(X, Y )Z2. (1.26)
Given a coordinate system (U, ϕ) around a point p ∈ M and setting
X =
n
ξ i ∂i, Y =
i=1
from (1.24), (1.25) and (1.26) we have
n
η j ∂j, Z =
j=1
R(X, Y )Z =
ξ i R(∂i, Y )Z =
i
=
i,j,k
i,j
ξ i η j ζ k R(∂i, ∂j)∂k.
n
ζ k ∂k,
k=1
ξ i η j R(∂i, ∂j)Z
Since, in view of (1.7) and (1.11) we have [∂i, ∂j] = 0, applying Definition 1.3.1 we obtain that
R(∂i, ∂j)∂k = ▽∂j ▽∂i∂k − ▽∂i▽∂j ∂k = ▽∂j (
Γ l ik∂l) − ▽∂i(
and, if we put
we have that
l
l
Γ l jk∂l)
=
(Γ l ik▽∂j ∂l + (∂jΓ l ik)∂l) −
(Γ l jk▽∂i∂l + (∂iΓ l jk)∂l)
l
=
l
Γ l ik
Γ s jl∂s −
s
l
Γ l jk
R(∂i, ∂j)∂k =
R l ijk =
Γ s ikΓ l js −
s
s
l
Γ s il∂s +
(∂jΓ l ik)∂l −
(∂iΓ l jk)∂l
l
s
R l ijk∂l,
l
Γ s jkΓ l is + ∂jΓ l ik − ∂iΓ l jk. (1.27)
Theorem 1.3.1. Let σ ∈ TpM be a two-dimensional subspace of the tangent space TpM and let
x, y ∈ σ be two linearly independent vectors, then
K(x, y) =
does not depend on the choice of the vectors x, y ∈ σ.
〈R(x, y)x, y〉
|x| 2 |y| 2 − 〈x, y〉 2
For a proof of Theorem 1.3.1 see Do Carmo (14) Proposition 3.1. We can give now the following
definition.
Definition 1.3.2. Given a point p ∈ M and a two-dimensional subspace σ ∈ TpM where {x, y} is
any basis of σ, the real number K(σ) = K(x, y) is called sectional curvature of the two dimensional
section σ at p.
The importance of the sectional curvature comes from the fact that the knowledge of K(σ),
l

1.4 Hyperbolic space 10
for all σ ∈ TpM, determines the curvature R completely. Among the Riemannian manifolds those
with constant sectional curvature are the most simple and play a fundamental role.
1.4 Hyperbolic space
For a Riemmanian manifold M and a chart x : U → Rn , we write the Riemannian metric in the
chart as
ds 2 =
n
gjk(x)dx j dx k . (1.28)
j,k=1
If M is the Euclidean space with the usual inner product and x is the identity map, formula (1.28)
gives the usual Euclidean distance element
Consider the half-space of R n given by
ds 2 =
n
(dx j ) 2 = |dx| 2 .
j=1
H n = {(x 1 , . . . , x n ) ∈ R n : x n > 0},
with n ≥ 2, endowed with the Riemannian metric given by
We have that
gij = δij
(x n ) 2 , gij = (x n ) 2 δij, g =
ds 2 =
n
j=1
(dx j ) 2
(x n )
1
(xn .
) 2n
|dx|2
= 2 (xn . (1.29)
) 2
H n is a model for the n-dimensional hyperbolic space. We start by proving that the hyperbolic
space has constant sectional curvature equal to −1.
We put fi = ∂i(log x n ) and
Rijkm = 〈R(∂i, ∂j)∂k, ∂m〉 = 〈
R l ijk∂l, ∂m〉 =
We have to prove that for all i, j = 1, . . . , n,
K(∂i, ∂j) =
l
Rijij
|∂2 i ||∂j| 2 = −1.
− 〈∂i, ∂j〉 2
l
R l ijkglm.

11 Tools of Riemannian Geometry
To calculate Rijij, in view of (1.27), we have that
Rijij =
R l iji glj =
=
=
l
1
(x n ) 2
1
(x n ) 2
⎡
s
⎣
l
Γ s iiΓ j
js
R l iji
−
δlj
(xn Rj iji
=
) 2 (xn ) 2
s
Γ s jiΓ j j j
is + ∂jΓii − ∂iΓji (Γ
s=i,j
s iiΓ j
js − ΓsjiΓ j
is ) + ΓjiiΓjjj
− Γjji
Γjij
+ ΓiiiΓ j
ji − ΓijiΓ j j j
ii + ∂jΓii − ∂iΓji where, in view of (1.15), the Christoffel symbols for the hyperbolic metric are given by
Γ k ij = 1
2
g kl {∂iglj + ∂jgil − ∂lgij}
l
= 1
(x
2
l
n ) 2
δkl ∂i
(xn )
= 1
2 (xn ) 2
δkj
∂i
(xn δik
+ ∂j
) 2 (xn )
= −δkjfi − δikfj + δijfk.
δlj
+ ∂j 2
δil δij
− ∂l 2 (xn ) 2
(x n )
δij
− ∂k 2 (xn ) 2
The last formula implies that Γk ij = 0 if i = j = k, Γiij = −fj, Γ j
ij = −fi, Γk ii = fk and Γi ii = −fi
and thus
Rijij =
1
(xn ) 2
−
s
(fs) 2 + (fi) 2 + (fj) 2 + fii + fjj
Finally, observing that |∂i| 2 |∂j| 2 − 〈∂i, ∂j〉 2 = gii gjj, we have
K(∂i, ∂j) = Rijij
gii gjj
= (x n ) 4 Rijij = (x n ) 2
−
s
(fs) 2 + (fi) 2 + (fj) 2 + fii + fjj
and since fn = 1
x n and fi = 0 for i = n if follows that K(∂i, ∂j) = −1 for all i, j = 1, . . . , n.
In view of (1.23), the Laplace operator ∆ in the hyperbolic space is given by
∆ = (x n ) n
= (x n ) n
= (x n ) n
= (x n ) 2
n ∂
∂x
j=1
j
(x n ) 2 (x n −n ∂
)
∂xj
⎡
n−1
⎣ (x
j=1
n ) 2−n ∂2
∂(xj ) 2 + (2 − n)(xn )
⎡
n
⎣ (x n ) 2−n ∂2
∂(xj ) 2 + (2 − n)(xn )
j=1
n
j=1
= (x n ) n
n
j=1
2−n−1 ∂
1−n ∂
∂
∂xj
(x n 2−n ∂
)
∂xj
∂xn + (xn ) 2−n
∂(xn ) 2
⎦
⎤
∂xn ⎦
∂2 ∂(xj ∂
+ (2 − n)xn . (1.30)
) 2 ∂xn There are several models for the hyperbolic space, the ball model, for example, consists of the
.
∂ 2
⎤
⎤
⎦ ,

1.4 Hyperbolic space 12
open unit ball B n = {x ∈ R n : |x| < 1} endowed with the Riemannian metric
ds 2 = 4|dx|2
(1 − |x| 2 .
) 2
We will see in the next chapter that the fractional transformation z → z−i
z+i
maps isometrically H2 onto B2 .
of the complex plane

Chapter 2
Hyperbolic Geometry
2.1 Möbius transformations
The reflection (or inversion) in the sphere S(a, r) = {x ∈ R n : |x − a| = r} where a ∈ R n and
r > 0, is a function φ(·) such that, for a, x ∈ R n , x = a
φ(x) = a +
r2
(x − a), (2.1)
|x − a| 2
it is natural to define φ(a) = ∞ and φ(∞) = a. We observe that φ(x) = x if and only if x ∈ S(a, r).
We have that φ ◦ φ(x) = x and so φ(·) is a bijection of R n onto itself.
A plane P (a, t) in R n is a set of the form
P (a, t) = {x ∈ R n : 〈x, a〉 = t} ∪ {∞}
where a ∈ R n , a = 0, t ∈ R and 〈x, a〉 = aixi. The reflection φ(·) in the plane P (a, t) is defined,
for every x ∈ R n , by
φ(x) = x + λa
and φ(∞) = ∞. The parameter λ is chosen so that 1
2 (x + φ(x)) ∈ P (a, t), this gives the explicit
formula
φ(x) = x − 2a
(〈x, a〉 − t) .
|a| 2 (2.2)
In fact 1
λa
λ
2 (x + φ(x)) ∈ P (a, t) if and only if 〈x + 2 , a〉 = t that is 〈x, a〉 + 2 |a|2 =t. Again φ(x) = x
if and only if x ∈ P (a, t), we have φ ◦ φ(x) = x and so φ(·) is a bijection of R n onto itself.
It is possible to prove that the reflection in a sphere and the reflection in a plane are two continuous
functions in R n onto itself with respect to the chordal metric d(x, y) defined by
⎧
⎨
d(x, y) =
⎩
2|x−y|
(1+|x| 2 ) 1/2 (1+|y| 2 ) 1/2 , if x, y = ∞,
2
(1+|x| 2 ) 1/2 , if y = ∞.

2.1 Möbius transformations 14
The chordal metric in R n induces the same topology as does the Euclidean metric, this implies
that a function is continuous with respect to both or to neither of these metrics.
We have observed that for any reflection φ(·) we have that φ(x) = φ −1 (x), this implies that φ −1 (x)
is continuous on R n with respect to the chordal metric too and then each reflection φ(·) is a
homeomorphism of R n onto itself with respect to the chordal metric.
Remark 2.1.1. If φ(·) is the reflection in the sphere S(a, r), we have
|φ(y) − φ(x)| = r 2
y − a x − a
−
|y − a| 2 |x − a| 2
1 2〈x − a, y − a〉
= r2 −
|x − a| 2 |x − a| 2 1
+
|y − a| 2 |y − a| 2
1/2 =
r2 2 2
|y − a| − 2〈x − a, y − a〉 + |x − a|
|x − a||y − a|
1/2 This shows that
=
r2 |x − y|
. (2.3)
|x − a||y − a|
|φ(x + h) − φ(x)| r
lim
=
h→0 |h|
2
. (2.4)
|x − a| 2
Remark 2.1.2. Any reflection in a plane is a Euclidean isometry, in fact
|φ(x) − φ(y)| 2
2
=
2a
2a
x − (〈x, a〉 − t) − y + (〈y, a〉 − t)
|a| 2 |a| 2
2
=
2a
x − y − 〈x − y, a〉
|a| 2
= |x − y| 2
+ 2 − 2
〈x − y, a〉 〈x − y, a〉 +
|a| 2 4
|a| 4 〈x − y, a〉2 |a| 2
= |x − y| 2 . (2.5)
Definition 2.1.1. A Möbius transformation acting in R n is a finite composition of reflections in
spheres and planes.
It follows that each Möbius transformation is a homeomorphism of R n onto itself. The composition
of two Möbius transformations is again a Möbius transformation and so is the inverse of a
Möbius transformation. In fact if φ = φ1 ◦ · · · ◦ φm, where the φj, j =, 1 . . . , m, are reflections,
then φ −1 = φm ◦ · · · ◦ φ1. Finally, since for any Möbius transformation we have φ ◦ φ −1 (x) = x,
the identity map is a Möbius transformation. We can give the following definition.
Definition 2.1.2. The group of Möbius transformations acting in R n is called General Möbius
Group and is denoted by GM(R n ).
Example 2.1.1. The translation φ(x) = x + a, a, x ∈ Rn , is a Möbius transformation since it
is the reflection in the plane 〈x, a〉 = 0 followed by the reflection in the plane 〈x, a〉 = |a|2
2 . The
magnification φ(x) = kx, k > 0 is a Möbius transformation since it is the reflection in the sphere

15 Hyperbolic Geometry
S(0, 1) followed by the reflection in the sphere S(0, √ k). The transformations of the form
φ(x) =
ax + b
cx + d
where a, b, c, d are real and such that ad − bc = 0 are Möbius transformations. In fact φ(x) =
φ4 ◦ φ3 ◦ φ2 ◦ φ1(x) where φ1(x) = x + d
c is a translation, φ2(x) = 1
x is an inversion and a reflection
with respect to the real axis, φ3(x) = − ad−bc
c 2
is a translation.
(2.6)
x is a magnification and a rotation and φ4(x) = x+ a
c
In what follows we will denote with Σ either a sphere S(a, r) or a plane P (a, t) in R n .
Theorem 2.1.1. For every Möbius transformation φ(·) and every sphere or plane Σ we have that
φ(Σ) is also a sphere or a plane.
The inversive product (Σ, Σ ′ ), given in Definition 2.1.3, is a real expression which depends only
on Σ and Σ ′ and which is invariant under all Möbius transformations. When Σ and Σ ′ intersect
it is a function of their angle of intersection and when Σ and Σ ′ are disjoint it is a function of the
hyperbolic distance between them.
Since the equation defining a sphere S(a, r) is
and the equation defining a plane P (a, t) is
|x| 2 − 2〈x, a〉 + |a| 2 − r 2 = 0
−2〈x, a〉 + 2t = 0,
we have that every sphere or plane Σ can be written in a common form as
a0|x| 2 − 2〈x, a〉 + an+1 = 0,
where a = (a1, . . . , an) and the coefficient vector (a0, . . . , an+1) is determined within a real nonzero
multiple. Since for Σ = S(a, r) and Σ = P (a, t) we have that |a| 2 > a0an+1, we can give the
following definition.
Definition 2.1.3. Let Σ and Σ ′ have coefficient vectors (a0, . . . , an+1) and (b0, . . . , bn+1). The
inversive product (Σ, Σ ′ ) of Σ and Σ ′ is defined by
(Σ, Σ ′ ) =
|2〈a, b〉 − a0bn+1 − an+1b0|
2(|a| 2 − a0an+1) 1/2 (|b| 2 .
− b0bn+1)
1/2
Remark 2.1.3. It is helpful to obtain the explicit expression of (Σ, Σ ′ ) in the following three
cases.
• If Σ = S(a, r) and Σ ′ = S(b, t) then (Σ, Σ ′ ) = r2 +t 2 −|a−b| 2
2rt
• If Σ = S(a, r) and Σ ′ = P (b, t) then (Σ, Σ ′ ) = |〈a,b〉−t|
r|b| .
• If Σ = P (a, r) and Σ ′ = P (b, t) then (Σ, Σ ′ ) = |〈a,b〉|
|a||b| .
.

2.2 Poincaré extension of Möbius transformations 16
Theorem 2.1.2. For any Σ and Σ ′ and any Möbius transformation φ(·) we have that
(Σ, Σ ′ ) = (φ(Σ), φ(Σ ′ )).
Remark 2.1.4. It is simple to see that, if Σ and Σ ′ intersect, then (Σ, Σ ′ ) = cos θ where θ is one
of the angles of intersection between Σ and Σ ′ .
For example let Σ = S(a, r) and Σ ′ = S(b, t) be two spheres that meat at some point c ∈ R n ,
then there exists a translation φ(·) such that φ(c) = 0. It follows that |φ(a)| = r and |φ(b)| = t
and in view of Theorem 2.1.2 we have
(Σ, Σ ′ ) = (φ(Σ), φ(Σ ′
)) =
r
2 + t2 − |φ(a) − φ(a)| 2
2rt
= |〈φ(a), φ(b)〉|
|φ(a)||φ(b)|
= cos θ.
Theorem 2.1.3. Let Σ be any sphere or plane, σ the reflection in Σ and I the identity map. If
φ(·) is a Möbius transformation that fixes each x ∈ Σ then either φ = I or φ = σ.
2.2 Poincaré extension of Möbius transformations
Poincaré observed that each Möbius transformation φ(·) acting in R n has a natural extension to a
Möbius transformation ˜ φ(·) acting in R n+1 . This extension depends on the embedding
x = (x1, . . . , xn) ↦→ ˜x = (x1, . . . , xn, 0),
of R n into R n+1 . In particular if φ(·) is the reflection in S(a, r), a ∈ R n , then ˜ φ(·) is defined as the
reflection in S(ã, r) and in view of (2.1) we have that for every x ∈ R n+1
˜φ(x) = ã +
and in particular for x ∈ R n we have that
˜φ(˜x) = ˜ φ(x1, . . . , xn, 0) = ã +
r2
(x − ã) = φ(x),
|x − ã| 2 r2xn+1 |x − ã| 2
(2.7)
r2
|˜x − ã| 2 (˜x − ã) = (φ(x), 0) = φ(x). (2.8)
If φ(·) is the reflection in P (a, t), a ∈ R n , then ˜ φ(·) is defined as the reflection in P (ã, t) and for
every x ∈ R n+1 we have
For x ∈ R n it follows that
˜φ(x) = x − 2ã
|ã| 2 (〈x, ã〉 − t) = (φ(x), xn+1). (2.9)
˜φ(˜x) = ˜ φ(x1, . . . , xn, 0) = ˜x − 2ã
|ã| 2 (〈˜x, ã〉 − t) = (φ(x), 0) = φ(x). (2.10)
Since for every x ∈ R n and every reflection φ(·) we have that ˜ φ(˜x) = φ(x), we can see ˜ φ(·) as an
extension of φ(·), such extension is called Poincaré extension of φ(·).
Remark 2.2.1. In view of (2.7), (2.8), (2.9) and (2.10) we observe that each Poincaré extension
˜φ(·) leaves invariant the plane xn+1 = 0 and the half-spaces xn+1 > 0 and xn+1 < 0.

17 Hyperbolic Geometry
Since each Möbius transformation φ = φ1 ◦ · · · ◦ φm in R n is defined as the composition of a
finite number of reflections φj, j = 1, . . . , m, there is at last one Möbius transformation ˜ φ =
˜φ1 ◦ · · · ◦ ˜ φm which extends the action of φ(·) to R n+1 in the sense that ˜ φ(x, 0) = φ(x). This
Möbius transformation preserves the plane xn+1 = 0 and the half-spaces xn+1 > 0 and xn+1 < 0.
In fact there is at most one extension ˜ φ(·) of a Möbius transformation φ(·). If ˜ φ1 and ˜ φ2 are two
such extensions, then ˜ φ −1
2 ◦ ˜ φ1 fixes each point of the plane xn+1 = 0, then by Theorem 2.1.3 we
have that either ˜ φ −1
2 ◦ ˜ φ1 = I or ˜ φ −1
2 ◦ ˜ φ1 = σ where σ is the reflection in the plane xn+1 = 0. But
since ˜ φ −1
2 ◦ ˜ φ1 preserves the half-spaces xn+1 > 0 and xn+1 < 0 we must have that ˜ φ −1
2 ◦ ˜ φ1 = I
and then ˜ φ1 = ˜ φ2.
2.3 Hyperbolic metric
We now focus on the action of the Poincaré extension ˜ φ(·) of a Möbius transformation in the upper
half-plane
H n+1 = {x ∈ R n+1 : xn+1 > 0}.
If ˜ φ(·) is the reflection in the sphere S(ã, r), a ∈ R n , then by (2.3) for every y, x ∈ H n+1 we have
and from (2.7) it follows that
| ˜ φ(y) − ˜ φ(x)|
|y − x|
=
r2 , (2.11)
|x − ã||y − ã|
[ ˜ φ(x)]n+1 = r2xn+1 , (2.12)
|x − ã| 2
where [ ˜ φ(x)]n+1 denotes the (n + 1)-th component of ˜ φ(x). Formula (2.11) and formula (2.12)
imply that
and this shows that
is invariant under ˜ φ(·).
| ˜ φ(y) − ˜ φ(x)| 2
[ ˜ φ(y)]n+1[ ˜ =
φ(x)]n+1
r 4 |y−x| 2
|x−ã| 2 |y−ã| 2
r 2 xn+1
|x−ã| 2
|y − x| 2
xn+1yn+1
r 2 yn+1
|y−ã| 2
= |y − x|2
xn+1yn+1
(2.13)
Let ˜ φ(·) be the reflection in the plane P (ã, t), a ∈ R n , since in view of (2.5) every reflection in a
plane is a Euclidean isometry, we have that | ˜ φ(y) − ˜ φ(x)| = |y − x| and from (2.9) we have that
[ ˜ φ(x)]n+1 = xn+1, this implies that
| ˜ φ(y) − ˜ φ(x)| 2
[ ˜ φ(y)]n+1[ ˜ =
φ(x)]n+1
|y − x|2
xn+1yn+1
Thus (2.13) is invariant under all Poincaré extensions of any φ(·) in GM(R n ) and this implies that
such Poincaré extensions are isometry of H n+1 endowed with the Riemannian metric derived from
the differential
.
ds = |dx|
. (2.14)
xn+1

2.3 Hyperbolic metric 18
Given a positive continuous function λ(x) on D ∈ Rn , let ρ(x, y) be the function defined for
x, y ∈ D by
ρ(x, y) = inf λ(γ(t)) |γ
γ
′ (t)|dt,
where the infimum is taken over all curves γ(t) in D with derivative γ ′ (t), joining x to y in D.
The function ρ(x, y) is a metric in D since it is symmetric, non negative and satisfies the triangle
inequality. In view of (2.14) the function η(x, y), defined by
b
η(x, y) = inf
a
|γ ′ (t)|
dt,
[γ(t)]n+1
where γ : [a, b] → H n+1 is a differentiable curve that joins x to y in H n+1 , is the hyperbolic metric
on H n+1 . The couple (H n+1 , η) is a model of the hyperbolic space and
is the hyperbolic length of the curve γ.
||γ|| =
b
a
|γ ′ (t)|
dt
[γ(t)]n+1
We shall now map isometrically H n+1 onto B n+1 = {x ∈ R n+1 : |x| < 1} to obtain a second model
of the hyperbolic space.
Let φ0 denote the reflection in the sphere S(en+1, √ 2), for every x ∈ R n+1
then
|φ0(x)| 2 =
en+1 +
= |en+1| 2 +
= 1 +
φ0(x) = en+1 +
2(x − en+1)
|x − en+1| 2 , en+1 +
2(x − en+1)
,
|x − en+1|
2
2(x − en+1)
|x − en+1| 2
4
|x − en+1| 2 〈en+1,
4
2
x − en+1〉 +
|x − en+1|
|x − en+1|
4
4
|x − en+1| 2 (〈en+1,
4
x〉 − 〈en+1, en+1〉) +
|x − en+1| 2
= 1 + 4〈en+1, x〉 4xn+1
= 1 + . (2.15)
|x − en+1|
2 |x − en+1|
2
This shows that φ0 maps the lower half-space xn+1 < 0 onto B n+1 . Now let σ be the reflection in
the plane xn+1 = 0 and φ = φ0 ◦ σ. Since for x ∈ R n+1 we have
σ(x) = x − 2xn+1en+1 = (x1, . . . , xn, −xn+1),
we obtain that φ(·) maps the half-space H n+1 onto B n+1 . Observing that the reflection in a plane
is a Euclidean isometry and in view of (2.4), we find that
|φ(y) − φ(x)|
lim
y→x |y − x|
|φ0(σ(y)) − φ0(σ(x))| |φ0(σ(y)) − φ0(σ(x))|
= lim
= lim
y→x |y − x|
y→x |σ(y) − σ(x)|
2
=
.
|σ(x) − en+1|
2

19 Hyperbolic Geometry
Since, in view of (2.15), we have that
it follows that
1 − |φ(x)| 2 = 1 − |φ0(σ(x))| 2 = − 4[σ(x)]n+1 4xn+1
=
,
|σ(x) − en+1|
2 |σ(x) − en+1|
2
|φ(y) − φ(x)|
lim
=
y→x |y − x|
1 − |φ(x)|2
2xn+1
. (2.16)
In general, given a positive continuous function λ(x) on D ∈ Rn , a metric ρ(x, y) = inf
γ λ(γ(t)) |γ′ (t)|dt
and a bijection f of D onto D1 ∈ Rn such that
|f(y) − f(x)|
lim
= µ(x),
y→x |y − x|
it is possible to define a positive a continuous function λ1(x) = λ(x)/µ(x) on D1 and a metric
ρ1(x, y) = inf
γ λ1(γ(t)) |γ ′ (t)|dt on D1. It follows that f is an isometry of (D, ρ) onto (D1, ρ1)
(see e.g. Beardon (2) page 7).
Since φ0 ◦ σ is a bijection of H n+1 onto B n+1 that satisfies (2.16), taking into account (2.14), we
define λ(x) = 1
1−|φ(x)|2
2
, µ(x) = and λ1(x) =
xn+1 2xn+1
1−|φ(x)| 2 . It follows that φ = φ0◦σ is an isometry
between (H n+1 , η) and (B n+1 , η1) where η1 is the metric obtained from the differential
ds = 2|dx|
. (2.17)
1 − |x| 2
The upper-half plane H 2 = {z = x + iy ∈ C : Im[z] > 0} with the metric η derived from the
differential ds = |dz|
Im[z] and the disk B2 = {z = x+iy ∈ C : |z| < 1} with the metric η1 derived from
the differential ds = 2|du|
1−|u| 2 are two models for the hyperbolic plane. We give now a description of
the hyperbolic metric in the upper-half plane H 2 and in the disk B 2 .
Now let
g(z) =
az + b
cz + d
(2.18)
be a Möbius transformation of the form (2.6), such that a, b, c, d are real and ad − bc > 0. Since
for every z = x + iy ∈ H 2 we have
g(z) =
ax + b + iay
cx + d + icy
= (ax + b)(cx + d) + acy2
(cx + d) 2 + c 2 y 2
it follows that g(z) maps H 2 onto itself. Observing that
y(ad − bc)
+ i
(cx + d) 2 + c2y2 |g ′
(z)| =
a(cz + d) − c(az + b)
(cz + d) 2
=
ad − bc
(cz
+ d) 2
=
ad − bc
(cx + d) 2 + c2y2

2.3 Hyperbolic metric 20
we have that
|g ′ (z)|
Im[g(z)] =
ad − bc
(cx + d) 2 + c2y2 (cx + d) 2 + c2y2 =
y(ad − bc)
1 1
= . (2.19)
y Im[z]
From (2.19) it follows that, for z, w ∈ H 2 ,
η(g(z), g(w)) = inf
′ ′
|g (γ(t))| |γ (t)|
dt = inf
Im[g(γ(t))]
|γ ′ (t)|
dt (2.20)
Im[γ(t)]
= η(z, w) (2.21)
and then each g(z) of the form (2.18) is an isometry of (H 2 , η). The following theorem holds.
Theorem 2.3.1. The group of isometries of (H 2 , η) is precisely the group of the maps of the form
g(z) =
where a, b, c, d ∈ R and ad − bc > 0.
az + b
a(−¯z) + b
, q(z) =
cz + d c(−¯z) + d ,
This result will be used in the next theorem to obtain an explicit expression for the hyperbolic
metric η.
Theorem 2.3.2. For every z, w ∈ H 2 we have that
Proof
We have seen in (2.20) that
and from the invariance of (2.13) we have
it follows that is sufficient to prove that
cosh η(z, w) = 1 +
cosh η(z, w) = cosh η(g(z), g(w))
|z − w| 2 |g(z) − g(w)|2
=
Im[z]Im[w] Im[g(z)]Im[g(w)]
cosh η(g(z), g(w)) = 1 +
|z − w|2
. (2.22)
2Im[z]Im[w]
|g(z) − g(w)|2
2Im[g(z)]Im[g(w)]
for some convenient transformation g(z). Let l be the unique Euclidean circle or line which contains
z and w and is orthogonal to the real axis, if l is a Euclidean circle its center (x0, 0) is on the real
axis. l meets the real axis at some point α ∈ R. We choose
g1(z) = − 1
(z − α) + β and g2(z) = z − x0.
These are two η-isometries of the form (2.18). If l is a Euclidean line we choose g(z) = g1(z) with
β = 0 since it maps l onto the imaginary axis. If l is a Euclidean circle we choose β = − 1
2α and
g(z) = g2 ◦ g1(z) since it maps l onto the imaginary axis. We can now assume that g(z) = ip and
g(w) = iq for some p and q such that 0 < p < q and we have to prove that
η(ip, iq) = log(q/p)

21 Hyperbolic Geometry
since
1 +
|ip − iq|2 (p − q)2
= 1 +
2Im[ip]Im[iq] 2pq = p2 + q2 = cosh log(q/p).
2pq
But, if γ(t) = x(t) + iy(t) with t ∈ [0, 1] is any smooth curve joining z to w such that γ(1) = q and
γ(0) = p, we have that
||γ|| =
1
0
|x ′ (t) + iy ′ 1
(t)|
dt ≥
y(t)
0
In general, if we do not assume that p < q, we have that
y ′ (t)
dt = log(q/p). (2.23)
y(t)
η(ip, iq) = | log(p/q)|. (2.24)
Remark 2.3.1. Formula (2.22) can be easily generalized to the (n + 1)-dimensional case. For
z, w ∈ Hn+1 , we have
|z − w|2
cosh η(z, w) = 1 + . (2.25)
2zn+1wn+1
Remark 2.3.2. From Theorem 2.3.2 we obtain that the hyperbolic sphere with center y ∈ H n+1
and hyperbolic radius r is the Euclidean sphere with center (y1, . . . , yn, yn+1 cosh r) and radius
yn+1 sinh r. In fact {x ∈ Hn+1 : η(x, y) = r} = {x ∈ Hn+1 : cosh r = 1 + |x−y|2
} this implies
2xn+1yn+1
that a point x on the hyperbolic sphere satisfies the equations
2xn+1yn+1 cosh r = 2xn+1yn+1 + (x1 − y1) 2 + · · · + (xn+1 − yn+1) 2 ,
2xn+1yn+1 cosh r = (x1 − y1) 2 + · · · + (xn − yn) 2 + x 2 n+1 + y 2 n+1,
(yn+1 sinh r) 2 = (x1 − y1) 2 + · · · + (xn − yn) 2 + (xn+1 − yn+1 cosh r) 2 .
Remark 2.3.3. It is possible to prove that the following equations are equivalent to formula
(2.22).
sinh
cosh
tanh
η(z, w)
2
η(z, w)
2
η(z, w)
2
2 x
Starting from (2.22) and applying sinh 2
=
=
=
|z − w|
,
2(Im[z]Im[w]) 1/2
(2.26)
|z − ¯w|
,
2(Im[z]Im[w]) 1/2
z − w
z
− ¯w .
(2.27)
(2.28)
1
x
= 2 (cosh x − 1) and cosh2 2
1 = 2 (cosh x + 1) we obtain
easily (2.26) and (2.27). Formula (2.28) follows from (2.26) and (2.27). It is possible to obtain the
following explicit expression for the hyperbolic metric η
observing that
cosh log
η(z, w) = log
|z − ¯w| + |z − w|
= 1 +
|z − ¯w| − |z − w|
|z − ¯w| + |z − w|
, (2.29)
|z − ¯w| − |z − w|
2|z − w| 2
|z − ¯w| 2 |z − w|2
= 1 +
− |z − w| 2 2Im[z]Im[w] .

2.3 Hyperbolic metric 22
We now consider the disk model (B 2 , η1). The map
is a bijection of H 2 onto B 2 , in fact
f(z) =
z − i
z + i
|f(z)| 2 = x2 + (y − 1) 2
x2 4y
= 1 −
+ (y + 1) 2 x2 + y2 + 1 + 2y .
(2.30)
Thus the metric ρ ∗ (u, v) = η(f −1 (u), f −1 (v)), u, v ∈ B 2 , is a metric on B 2 . However, since we have
that
2|f ′ (z)|
=
1 − |f(z)| 2
2
1 − 1 +
2
x2 +(y+1) 2
4y
x2 +y2 +1+2y
= 1 1
=
y Im[z] ,
we can identify ρ ∗ with the metric η1 derived from the differential (2.17), then f is an isometry
between (H 2 , η) and (B 2 , η1). We can derive an explicit expression for the hyperbolic metric η1
in B 2 by simply rewriting Theorem 2.3.2 by means of f. In particular for u = f(z) ∈ B 2 and
v = f(w) ∈ B 2 we have that
sinh η1(u, v)
2
=
|u − v|
(1 − |u| 2 ) 1/2 (1 − |v| 2 ) 1/2
in fact, in view of (2.26), we have, for u = a + ib and v = α + iβ, that
sinh 2 η1(u, v)
2
= sinh 2 η(f −1 (z), f −1 (w))
=
=
=
4Im
−i u+1
u−1
−i u+1
u−1
2
v+1
+ i v−1
Im
2
−i v+1
v−1
|u−v| 2
[(a−1) 2 +b2 ][(α−1) 2 +β2 ]
1−a2−b2 (a−1) 2 +b2 1−α2−β2 (α−1) 2 +β2 |u − v| 2
(1 − |u| 2 )(1 − |v| 2 ) .
Starting from formula (2.31) we obtain that
in fact
cosh 2 η1(u, v)
2
cosh η1(u, v)
2
= 1 + sinh 2 η1(u, v)
2
=
|1 − u¯v| 2
(1 − |u| 2 )(1 − |v| 2 )
=
= 1 +
=
η
2
= sinh
=
|1 − u¯v|
(1 − |u| 2 ) 1/2 (1 − |v| 2 ) 1/2
−i u+1
u−1
v+1 , −i v−1
2
4|u−v| 2
|(u−1)(v−1)| 2
4 1−a2−b2 (a−1) 2 +b2 1−α2−β2 (α−1) 2 +β2 |u − v| 2
(1 − a 2 − b 2 )(1 − α 2 − β 2 )
(2.31)
(2.32)
|u − v| 2
(1 − |u| 2 )(1 − |v| 2 ) = (1 − |u|2 )(1 − |v| 2 ) + |u − v| 2
(1 − |u| 2 )(1 − |v| 2 )

23 Hyperbolic Geometry
since
|1 − u¯v| 2 = |1 − ūv| 2 = |ū| 2
2
1
− v
ū = |u| 2
2
1
− v
ū = |u| 2
1
ū
= 1 + |u| 2 |v| 2 − 2|u| 2
1
, v = 1 + |u|
ū 2 |v| 2 − 2 〈u, v〉
= 1 + |u| 2 |v| 2 − 2 〈u, v〉 ± |u| 2 ± |v| 2
= |u − v| 2 + (|u| 2 − 1)(|v| 2 − 1).
From (2.31) and (2.32) it follows that
and from (2.33) we obtain that
since
In the particular case u = 0 we have
tanh η1(u, v)
2
η1(u, v) = log
= |u − v|
|1 − u¯v|
|1 − u¯v| + |u − v|
|1 − u¯v| − |u − v|
1 |1 − u¯v| + |u − v|
tanh log =
2 |1 − u¯v| − |u − v|
|u − v|
|1 − u¯v| .
η1(0, v) =
1 + |v|
1 − |v| .
2
+ |v| 2
1
− 2 , v
ū
(2.33)
It follows that an hyperbolic circle in B 2 with center 0 and radius r is a Euclidean circle with center
0 and radius R = r−1
r+1 , in fact
v ∈ B 2 : η1(0, v) = r =
v ∈ B 2 :
1 + |v|
= r = v ∈ B
1 − |v| 2
r − 1
: |v| = . (2.34)
r + 1
In the next theorem is obtained the length of an hyperbolic circle and the area of an hyperbolic
disk.
Theorem 2.3.3. The area of a hyperbolic disk of hyperbolic radius r is 4π sinh 2 (r/2). The length
of a hyperbolic circle of hyperbolic radius r is 2π sinh r.
Proof
We use the model B2 . Let C = {v ∈ B2 : η1(0, v) = r} and D = {v ∈ B2 : η1(0, v) ≤ r} be the
circle and disk with center in the origin and hyperbolic radius r. In view of (2.31) and (2.34) we
have that
sinh 2 r
2 = sinh2 η1(0, v)
=
2
|v|2 R2
= .
1 − |v| 2 1 − R2 Then
h − area(D) = 4
D
= 4π sinh 2 r
2 .
1
(1 − |v| 2 ) 2 dv1dv2 = 4
2π
0
dθ
R
0
s
ds
(1 − s2 = 4(2π)
) 2 2(1 − R2 )
R 2

2.4 Geodesics 24
In view of (2.34) we have
then
h − length(C) = 2
C
2.4 Geodesics
1 + |v|
sinh r = sinh η1(0, v) = sinh log =
1 − |v|
2R
1 − R2 2π
|dv|
= 2
1 − |v| 2
0
R
R
dθ = 4π = 2π sinh r.
1 − R2 1 − R2 The circle of points at infinity is represented by the x-axis and denoted by ∂H 2 in the half-plane
model and by ∂B 2 = {u : |u| = 1} in the disk model.
Definition 2.4.1. An hyperbolic line is the intersection of the half-plane H 2 or the disk B 2 with
a Euclidean circle or a straight line which is orthogonal to the circle of the points at infinity.
The following facts follow easily from Definition 2.4.1.
1. There is a unique hyperbolic line through two distinct points of the hyperbolic plane.
2. Two distinct hyperbolic lines intersect in at most one point of the hyperbolic plan.
3. The reflection in a hyperbolic line is a hyperbolic isometry. In fact we have seen that (2.13) is
invariant under the action of Poincaré extensions of Möbius transformations and in particular
under reflections in spheres and planes of the form S(ã, r) = {z = x+iy ∈ H 2 : |x−a| 2 +y 2 =
r 2 } and P (ã, t) = {z = x + iy ∈ H 2 : ax = t} for a, t ∈ R and r > 0, but these are the
hyperbolic lines of H 2 .
4. Given two hyperbolic lines l1 and l2 there is a hyperbolic isometry g(·) such that g(l1) = l2.
In fact, as in the proof of Theorem 2.3.2, for every hyperbolic line li it is possible to find an
isometry gi(·) such that gi(li) maps li onto the imaginary axis of H2 . It follows that it is
sufficient to chose g = g −1
2 ◦ g1.
5. Given any hyperbolic line l and any point w ∈ H 2 , there is a unique hyperbolic line through
w orthogonal to l. Let g(z) be the η-isometry such that g(l) maps l onto the imaginary axis.
We observe that, since g(z) is Möbius, it preserves the angles, then {z ∈ H 2 : |z| = |g(w)|} is
the unique hyperbolic line which contains g(w) and is orthogonal to the positive imaginary
axis.
Given two distinct points z and w on an hyperbolic line l, we denote by [z, w] the closed hyperbolic
segment on l.
In the following theorem we prove that the hyperbolic lines are geodesics, that is curves of shortest
hyperbolic length.

25 Hyperbolic Geometry
Theorem 2.4.1. Let z and w be two points in H 2 . A curve γ joining z to w satisfies
||γ|| = η(z, w)
if and only if γ is a parametrization of [z, w] as a simple curve.
Proof
Let g(·) be a η-isometry such that g(ip) = z and g(iq) = w. Let γ1(t) = x1(t) + x1(t), t ∈ [0, 1],
be any curve joining ip to iq. In view of formula (2.23) we have that ||γ1|| = η(ip, iq) if and only
if x1(t) = 0 and y ′ 1(t) > 0, that is if and only if γ1 is a parametrization of [ip, iq] as a simple
curve. In fact, if ||γ1|| = η(ip, iq) this implies that x ′ 1(t) = 0 and y ′ 1(t) > 0 that is x1(t) = 0 and
y ′ 1(t) > 0 since x1(0) = 0. On the other side, if x1(t) = 0 and y ′ 1(t) > 0, than in view of (2.23)
we immediately have ||γ1|| = η(ip, iq). The theorem follows by taking into account that g(·) is a
η-isometry and then maps hyperbolic lines in hyperbolic lines.
Theorem 2.4.2. Let z and w be two distinct points in the hyperbolic plane. Then
if and only if ζ ∈ [z, w].
η(z, w) = η(z, ζ) + η(ζ, w)
Proof
If ζ ∈ [z, w] and g(·) is such that g(z) = ip, g(ζ) = iq and g(w) = iv than in view of (2.24) we
have that η(z, w) = η(z, ζ) + η(ζ, w). On the other side if ζ /∈ [z, w] and γ is a curve made of the
segments [z, ζ] and [ζ, w], then in view of Theorem 2.4.1 we have that ||γ|| > η(z, w).
Definition 2.4.2. Let l1 and l2 be two distinct geodesics. If l1 ∩ l1 = ∅, then l1 and l2 are intersecting.
If l1 ∩ l1 = ∅, then l1 and l2 are parallel.
In the hyperbolic geometry for any given hyperbolic line l and a point w not on l, there are infinitely
many hyperbolic lines through w which are parallel to l.
For each point ζ and any hyperbolic line l the distance of ζ from l is defined by
η(ζ, l) = inf η(ζ, w).
w∈l
Theorem 2.4.3. Let l1be the unique geodesic through ζ orthogonal to l and let ζ1 be the point of
intersection between l1 and l. We have that η(ζ, l) = η(ζ, ζ1).
Proof
We may assume that l is the positive imaginary axis, then l1 = {z ∈ H2 : |ζ| = |z|}. We need to
show that η(ζ, l) = η(ζ, i|ζ|). Since each point on l is of the form z = ip, p > 0, from Theorem
2.3.2 we have that
cosh η(ζ, ip) = x2 + y2 + p2 =
2yp
|ζ|
|ζ| p
+ ≥
2y p |ζ|
|ζ|
y ,

2.5 Hyperbolic trigonometry 26
for ζ = x + iy. The equality holds if and only if p = |ζ|.
Definition 2.4.3. A horocycle is a curve in the hyperbolic plane that cuts at right angles all the
parallels passing through the same point at infinity. In H 2 and B 2 the horocycles are represented
by Euclidean circles tangent to the circle of points at infinity.
2.5 Hyperbolic trigonometry
In this section we focus on some fundamental results of the hyperbolic trigonometry. With za, zb
and zc we denote the vertices of an hyperbolic triangle T , the sides opposite these vertices will
have lengths a, b and c respectively and the interior angles at the vertices will be α, β and γ. If,
for example, the vertex za is on the circle of points at infinity, then we have α = 0, b = c = ∞. As
isometries preserve lengths and angles, we note that the trigonometric formulae remain invariant
under isometries. In what follows we will use (H 2 , η) as model of the hyperbolic plane.
Theorem 2.5.1. Let T be a hyperbolic triangle with vertices za, zb, zc and let l be the geodesic
containing the longest side, say [zb, zc], of T . Then the geodesic l1 through za and orthogonal to l
meets l at a point w ∈ [zb, zc].
Proof
We may assume that l is the positive imaginary axis, it follows that w = i|za|. In view of Theorem
2.4.3 we have that η(za, zb) ≥ η(w, zb) and that η(za, zc) ≥ η(w, zc). As [zb, zc] is the longest side
of T we deduce that
max{η(za, w), η(zc, w)} ≤ max{η(za, zc), η(za, zb)} ≤ η(zb, zc).
Since the points zb, zc and w are collinear, from Theorem 2.4.2, we have that w must lie between
zb and zc or be equal to one of them.
Theorem 2.5.2. Let T be a right triangle with angles α = 0, β = 0 and γ = π/2. Then
cosh b sin α = 1, (2.35)
sinh b tan α = 1, (2.36)
tanh b sec α = 1. (2.37)
Proof
In H 2 we may assume that za = x + iy with x 2 + y 2 = 1 and zc = i. Since y = sin α, from Theorem
2.3.2 we obtain (2.35), in fact
cosh b = cosh η(i, x + iy) = 1 +
|i − x − iy|2
2y
= x2 + y 2 + 1
2y
= 1 1
=
y sin α .
Formulae (2.36) and (2.37) follow easily by applying sinh 2 x = cosh 2 x − 1 and tanh x =
sinh x
cosh x .

27 Hyperbolic Geometry
Remark 2.5.1. The famous Lobachevsky’s formula
cot α
2
follows easily from Theorem 2.5.2, in fact, in view of (2.37), we have tanh b = cos α, then
= eb
cot α
2 =
1 + cos α
1 − cos α =
eb e−b = eb .
Theorem 2.5.3 (Hyperbolic Pythagorean Theorem). For any triangle with angles α, β and γ =
π/2, we have
cosh c = cosh a cosh b.
Proof
For any right triangle with angles α, β and π/2, we may assume, by applying a suitable isometry,
that za = x + iy with x 2 + y 2 = 1, zb = it for t > 0 and zc = i. Using Theorem 2.3.2 we have
cosh c = cosh η(it, x + iy) = 1 +
cosh b = cosh η(i, x + iy) = 1 +
cosh a = cosh η(i, it) =
1 + t2
.
2t
|it − x − iy|2
2yt
|i − x − iy|2
2y
= 1 + t2
= 1
y ,
2yt ,

2.5 Hyperbolic trigonometry 28

Chapter 3
Hyperbolic Brownian Motion
3.1 Settings
The n-dimensional hyperbolic Brownian motion, n ≥ 2, is a diffusion process on the upper halfspace
H n = {z = (x, y) = (x1, · · · , xn−1, y) : x ∈ R n−1 , y > 0}
associated with the Laplace operator ∆n given by (see formula (1.30))
∆n = y 2
n−1
∂ 2
∂x
i=1
2 i
+ ∂2
∂y2
− y(n − 2) ∂
.
∂y
(3.1)
For a fixed z ′ = (x ′ , y ′ ) ∈ H n , for all t > 0 and z = (x, y) ∈ H n , we denote with Gn(z, z ′ , t) the
transition function of the hyperbolic Brownian motion in H n with starting point z ′ . The transition
function Gn(z, z ′ , t) is solution to the following heat-type equation
with initial condition
∂
∂t un(z, z ′ , t) = ∆n un(z, z ′ , t), (3.2)
un(z, z ′ , 0) =
n−1
i=1
δ(xi − x ′ i)δ(y − y ′ ).
We have seen in Theorem 1.2.1 that if Φ : H n → H n is a diffeomorphism, then ∆f = (∆(f ◦
Φ)) ◦ Φ −1 if and only if Φ is an isometry, this invariance property implies that
u(z, z ′ , t) = u(Φ(z), Φ(z ′ ), t).
Let η(z, z ′ ) be the hyperbolic distance between z and the starting point z ′ , given by formula
(2.25), since the Laplace operator commutes with rotations and translations, we have that the
heat kernel Gn(z, z ′ , t) is a function of η(z, z ′ ) and, without restrictions, we can assume that the
starting point z ′ is the origin O = (0, 1) of H n (see for example Davies (10) page 178 or Borthwick
(3) Section 4.1).
Since, for a fixed time t, the transition function Gn(z, z ′ , t) is in fact a function of the hyperbolic

3.1 Settings 30
distance η = η(z) = η(z, O), we denote it by Gn(η, t) and we take into account only the radial part
of the Laplace operator (3.2) with initial condition un(η, 0) = δ(η).
It is possible to prove that the following relation holds
∆n η(z) = (n − 1) coth η(z). (3.3)
The proof of relation (3.3) is postponed to Proposition 3.A.1 in the Appendix 3.A. In view of (3.3),
for a smooth function f on R, we deduce that
∆nf(η) = f ′′ (η) + (n − 1) coth η f ′′ (η). (3.4)
For a proof of (3.4) see Proposition 3.A.2 in the Appendix 3.A. Since we can rewrite formula (3.4)
in the following form
1
∆nf(η) =
sinh n−1
d
sinh
η dη
n−1 η d
dη f(η)
,
we finally note that the heat kernel Gn(η, t) is obtained by resolving the following Cauchy problem
⎧
⎨ ∂
∂t
⎩
un(η, t) =
un(η, 0) = δ(η).
1
sinh n−1 η
∂
∂η sinh n−1 η ∂
∂η un(η, t),
Since an hyperbolic sphere in H n with hyperbolic center O and hyperbolic radius η is a Euclidean
sphere with Euclidean center (0, cosh r) ∈ H n and Euclidean radius sinh r, we have that the volume
element in H n is given by
dv = sinh n−1 η dη dΩn,
where dΩn is the surface element of the n-dimensional unit sphere
n−1
dΩn = (sin θi) n−1−i dθi.
i=1
To obtain the normalized heat kernel we impose that
Gn(η, t) dv = 2πn/2
Γ( n
2 )
∞
0
(3.5)
Gn(η, t) sinh n−1 η dη = 1. (3.6)
The probability density pn(η, t) to be at time t at a distance η from the starting point, normalized
with respect to sinh n−1 η dη, is given by
pn(η, t) = 2πn/2
Γ( n
2 ) Gn(η, t).
We denote by (B (n) , B (n) , P (n) ) the n-dimensional standard Wiener space where B (n) is the
space of all R n valued continuous paths on [0, ∞), with (B1(s), . . . , Bn(s)) ∈ B (n) satisfying
(B1(0), . . . , Bn(0)) = 0 and (B (n) (s), s ≥ 0) is the canonical filtration (see Revuz and Yor (36)
page 35).
Taking into account the generator ∆n/2, where ∆n is given by (3.1), it follows that the Brownian

31 Hyperbolic Brownian Motion
motion on H n may be obtained by solving the stochastic differential equations
⎧
⎨dXi(s)
= Y (s)dBi(s), i = 1, ..., n − 1,
⎩dY
(s) = Y (s)dBn(s) − n−2Y
(s)ds.
By Itô’s formula we have that the solution {Z(t) = (X(t), Y (t)), t ≥ 0} satisfying Z(0) = z = (x, y)
may be represented as follows:
2
⎧
⎨Xi(t)
= xi + y
⎩
t
0 exp{Bµ (s)} dBi(s), i = 1, ..., n − 1,
Y (t) = y exp{B µ (t)},
where µ = (n − 1)/2 and B µ (t) = Bn(t) − µt has constant drift −µ. Since µ is assumed to be
positive, Y (t) is the usual geometric Brownian motion with negative drift and then {Z(t), t ≥ 0}
converges almost surely to the boundary of H n as t → ∞.
3.2 Hyperbolic Brownian motion in H 2
3.2.1 Transition function
The 2-dimensional hyperbolic Brownian motion is a diffusion process on the upper half-plane
H 2 = {z = (x, y) : x ∈ R, y > 0}
associated with the Laplace operator ∆2 given by
(3.7)
∆2 = y 2
2 ∂ ∂2
+
∂x2 ∂y2
. (3.8)
The position of points in H 2 can be given either in terms of Cartesian coordinates (x, y) or by
means of the hyperbolic coordinates (η, α). The coordinate η represents the hyperbolic distance
of (x, y) from the origin O = (0, 1) of H 2 . The coordinate α is the slope of the tangent in O to
the geodesic through O and (x, y) with Euclidean center (x0, 0) and Euclidean radius r (see Figure
5.3).
The formulas which relates the hyperbolic coordinate to the cartesian coordinates are
x =
y =
sinh η cos α
cosh η−sinh η sin α ,
1
cosh η−sinh η sin α ,
with η > 0 and α ∈ [− π π
2 , 2 ]. See, for example, Rogers and Williams (37), page 213 or Terras (41)
formula (3.7). For a proof of the relations in (3.9) see Proposition 3.A.3.
In Proposition 3.A.4 it is showed that the Laplace operator (3.8) in hyperbolic coordinates takes
the form
1 ∂
sinh η
sinh η ∂η
∂
1
+
∂η sinh 2 ∂
η
2
.
∂α2 (3.10)
Since, for a fixed t, the transition function p2(z, z ′ , t) is a function of the hyperbolic distance η,
we can take into account only the radial part of the Laplace operator (3.10).
(3.9)

3.2 Hyperbolic Brownian motion in H 2 32
Figure 3.1: Hyperbolic coordinates in H 2 .
In the next theorem, following the proof given in Lao and Orsingher (26), we give the explicit
form of the normalized heat kernel G2(η, t). In the literature there are several proofs of formula
(3.11), for example Buser obtain the heat kernel G2(η, t) by using the Abel transform, see Buser
(4) Theorem 7.4.1. An approach based on the Mehler transform is given in Chavel (9) Section X.2.
Other transform techniques are described in Terras (41) Section 3.2 and Helgason (19) page 29.
Theorem 3.2.1. The normalized heat kernel G2(η, t), t > 0, has the following explicit expression
G2(η, t) =
t − e 4
25/2 (πt) 3/2
∞
η
ϕ2
− ϕe 4t
√ dϕ, η > 0. (3.11)
cosh ϕ − cosh η
Proof
In order to obtain formula (3.11) we must solve the initial-value problem
⎧
⎨ ∂u 1 ∂
∂t = sinh η ∂η sinh η
⎩
∂
∂η u,
u(η, 0) = δ(η).
It is now convenient to perform the change of variable cosh η = y which yields
We now put
that gets the following equations
∂u
∂t =
(y 2 − 1) ∂2
∂
+ 2y u.
∂y2 ∂y
u(t, y) = T (t)F (y)
⎧
⎨T
⎩
′ (t) = −T (t) ω,
(y 2 − 1)F ′′ + 2yF ′ + ωF = 0,
(3.12)
(3.13)
where ω is an arbitrary positive constant. The method of separation of variables produces two
ordinary equations, one of which is strictly related to Legendre equation. The bounded solution of

33 Hyperbolic Brownian Motion
equation (3.12), because of its homogeneity and linearity, has thus the form
u(η, t) =
∞
0
T (t, ω)F (η, ω)Q(ω)dω, (3.14)
where Q(ω), ω > 0, is an arbitrary function which is determined by applying the initial condition.
The equation satisfied by F , for ω = −ν(ν + 1), becomes the classical Legendre equation
which is a special case of the hypergeometric equation
(1 − z 2 ) d2u du
− 2z + ν(ν + 1)u = 0 (3.15)
dz2 dz
t(1 − t) d2u du
+ [γ − (α + β + 1)t] − αβu = 0, (3.16)
dt2 dt
for z = 1 − 2t, α = −ν, β = ν + 1 and γ = 1. A solution to (3.16) is the hypergeometric function
2F1(α, β, γ; t) =
=
=
∞ (α)k(β)k
t
k!(γ)k
k
∞ α(α + 1)...(α + k − 1)β(β + 1)...(β + k − 1)
t
k!γ(γ + 1)...(γ + k − 1)
k
∞ Γ(α + k)Γ(β + k) Γ(γ)
k!Γ(γ + k) Γ(α)Γ(β) tk , |t| < 1.
k=0
k=0
k=0
This means that a solution to the Legendre equation (3.15) is the Legendre polynomial
Pν(z) = 2F1
=
∞
k=0
∞
k 1 − z (−ν)k(ν + 1)k 1 − z
−ν, ν + 1, 1; =
2
k!(1)k 2
k=0
(−ν)k(ν + 1)k
(k!) 2
k 1 − z
, |1 − z| < 2. (3.17)
2
By following Lebedev (27), page 165, it is possible to represent the Legendre polynomials in a more
suitable form by considering that
By substituting (3.18) in (3.17) we get that
Pν(z) =
π
2 2
sin
π 0
2k
1
2 k
ϕ dϕ = . (3.18)
Γ(k + 1)
∞ (−ν)k(ν + 1)k
1
k=0 2 k k!
k 1 − z 2
2 π
= 2
π ∞ 2 (−ν)k(ν + 1)k
π
1
0
2 k k!
sin 2 ϕ
k=0
= 2
π
2
2F1
π 0
−ν, ν + 1, 1
2 ; sin2 ϕ
π
2
0
1 − z
2
sin 2k ϕ dϕ
k 1 − z
dϕ
2
dϕ. (3.19)

3.2 Hyperbolic Brownian motion in H 2 34
The function in the integrand of (3.19) can be rewritten as
fν(ω) = 2F1 −ν, ν + 1, 1
; −ω
2
√ √ 2ν+1 √ √ −2ν−1 1 + ω + ω + 1 + ω + ω
=
2 √ ,
1 + ω
(3.20)
in fact fν(ω) solves the hypergeometric equation (3.16) for α = −ν, β = ν + 1, γ = 1
2 and t = −ω,
that is
ω(1 + ω) d2
fν 1 dfν
+ + 2ω
dω2 2 dω − ν(ν + 1)fν = 0,
which can also be rewritten in the more convenient form
√ √ω√ √ ′
ω 1 + ω 1 + ωfν
′
− ν + 1
2 fν = 0. (3.21)
2
It is now a relatively simple matter to show that (3.20) is a solution to (3.21) and also to check
that
fν(ω) = f−ν−1(ω).
where
By some calculations it can be seen that
fν
sin 2 ϕ
Therefore the Legendre polynomial (3.19)
Pν(cosh η) = 2
π
2
π
= 2
π
2
0
π
2
in view of (3.22), takes the following form
0
(cosh η − 1) = cosh θ(ν + 1
2 )
cosh θ
2
sinh θ η
= sinh sin ϕ. (3.22)
2 2
2F1
fν
Pν(cosh η) = 1
η
π −η
−ν, ν + 1, 1
2 ; sin2
ϕ
(1 − cosh η) dϕ
2
2
sin ϕ
(cosh η − 1) dϕ,
2
1
(ν+ e 2)θ
dθ. (3.23)
2(cosh η − cosh θ)
We skip the next step which consists in applying a contour integration to the complex-valued
function. The detailes of this transformation can be found in Lebedev (27), page 172.
1
(ν+ e 2)t
f(t) =
π , t ∈ C,
2(cosh η − cosh t)
and permits us to express the Legendre function (3.23) by means of an integral on [η, ∞) as follows
Pν(cosh η) = 2
cot π ν +
π 1
∞ sinh[
2 η
ν + 1
2 ϕ]
√ dϕ. (3.24)
2 cosh ϕ − 2 cosh η

35 Hyperbolic Brownian Motion
If we now choose ω = 1
4 + x2 in the second equation of (3.13) we note that it coincides with
(3.15) when ν = − 1
2 ± ix. From (3.14), by considering (3.24), we get that
u(η, t) =
= 2
π
∞
t −
e 4
0
−x2t P− 1
2 +ix(cosh η) x Q(x) dx (3.25)
∞
t −
x Q(x)e 4
0
−x2 ∞
t sin(xϕ)
coth(πx) dx √ dϕ.
η 2 cosh ϕ − 2 cosh η
We show now that if we choose Q(x) = k tanh(πx), for some real constant k, then the initial
condition u(η, 0) = δ(η) is satisfied. We have
u(η, t) =
=
=
√
2k
π
√
2k
π
∞
0
e− t
4
t − ke 4
√
π (2t) 3
t −
x tanh(πx) e 4 −x2 ∞
t sin(xϕ)
coth(πx) dx √ dϕ
cosh ϕ − cosh η
η
∞
∞
dϕ
√ xe
η cosh ϕ − cosh η 0
−x2t sin(xϕ)dϕ
∞
ϕ2
− ϕe 4t
√ dϕ (3.26)
cosh ϕ − cosh η
η
In order to check that (3.26) satisfies the initial condition it suffices to write that
lim
t→0
t→0
+ u(η, t) = lim
∞
k
+
= k
∞
η/ √ 2
=
η/ √ 2
ϕ2
− ϕe 2t
√
2πt3 δ(ϕ)
cosh( √ 2ϕ) − cosh η
dϕ
cosh( √ 2ϕ) − cosh η
dϕ
⎧
⎨0,
for η > 0,
⎩k
∞
0
δ(ϕ)
√ cosh( √ 2ϕ)−1 dϕ = ∞, for η = 0.
In the last step we took into account the behaviour of the distribution of the first-passage time of
a standard Brownian motion. We finally observe that if G2(η, t) = u(η, t), where u(η, t) is given
by (3.26) with k = 1
2π , then it satisfies the condition (3.6). In fact
G2(η, t) dv =
t − 1 e 4
√
π (2t) 3
= 1
t − e 4
√
π (2t) 3
=
t − e 4
√
π (2t) 3
t
2e− 4
= √
π (2t) 3
√
2
= √
π (2t) 3
= 1
t √
√2t
4πt
= 1
√ 2π
∞ ∞
ϕ2
− ϕe 4t
sinh η dη √ dϕ
0
η cosh ϕ − cosh η
∞ ϕ
ϕ2
− sinh η
ϕe 4t dϕ √ dη
0
0 cosh ϕ − cosh η
∞
η=ϕ 0
∞
ϕ2
−
ϕe 4t dϕ −2 cosh ϕ − cosh η
ϕe
ϕ2
− 4t
cosh ϕ − 1dϕ =
η=0
∞
t − 2e 4
√
π (2t) 3
(ϕ+t)2
−
ϕe 4t dϕ
ϕ2
− e
ϕe 4t
ϕ
ϕ
2 − − e 2
√ dϕ
2
0
∞
∞
(ϕ−t)2
−
ϕe 4t dϕ −
0
0
0
∞
− √ (
t
2
√ y2
−
2ty + t)e 2 dy − √ ∞
2t √ (
t
2
√ y2
−
2ty − t)e 2 dy
∞
y2
−
e 2 dy = 1.
−∞

3.2 Hyperbolic Brownian motion in H 2 36
Remark 3.2.1. In view of Theorem 3.2.1 we obtain that the probability density p2(η, t) to be at
time t at a distance η, normalized with respect to the area element sinh η dη, is given by
p2(η, t) = 2π G2(η, t) =
t − ∞
e 4
√
π (2t) 3
η
ϕ2
− ϕe 4t
√ dϕ. (3.27)
cosh ϕ − cosh η
Remark 3.2.2. Sometime, in the literature, instead of equation (3.2), it is considered the equation
∂ 1
∂t u = 2∆2u, in this case the probability density (3.27) must be replaced by
p2(η, t) =
t − ∞
e 8
√ √ 3
π t η
ϕ2
− ϕe 2t
√ dϕ.
cosh ϕ − cosh η
Remark 3.2.3. From (3.27) also the distribution function of η(t) can be obtained:
where ft(ϕ) =
Pr{η(t) > η} =
=
=
t − e 4
√
π (2t) 3
t − e 4
√
π (2t) 3
t − 2e 4
√
π (2t) 3
∞
η
∞
η
∞
η
= 2 √ ∞
t −
2e 4
η
∞
dη
η
ϕ2
−
ϕe 4t dϕ
ϕe
ϕ2
− 4t
ϕ2
− ϕe 4t sinh η
√ dϕ
cosh ϕ − cosh η
ϕ
η
sinh η
√ cosh ϕ − cosh η dη
cosh ϕ − cosh ηdϕ
ft(ϕ) cosh ϕ − cosh ηdϕ,
ϕ2
ϕe− 4t
√ , t > 0, ϕ > 0, as a function of t, is the distribution of the first-passage
2π(2t) 3
time for the standard Brownian motion, with infinitesimal variance equal to 2, through level ϕ.
3.2.2 Hitting distribution
In several papers (see Baldi et al. (1), Matsumoto and Yor (30)) it is considered the diffusion on
H 2 associated to the operator
L = y2
2
2 ∂ ∂2
+
∂x2 ∂y2
− νy ∂
∂x −
µ − 1
y
2
∂
∂y
where ν ∈ R and µ > 0. The coefficients µ and ν measure respectively the vertical and the
horizontal component of the drift. In view of (3.1) we obtain that, for µ = 1/2 and ν = 0, the
operator L coincides with ∆2/2 and in this case we have the standard two dimensional hyperbolic
Brownian motion.
Unlike ∆2, that is invariant under hyperbolic isometries, the differential operator L is invariant
under transformations of the form z → az + b with a > 0 and b ∈ R.
The diffusion process {Z(t) = (X(t), Y (t)), t > 0} associated to L and such that Z(0) = z =

37 Hyperbolic Brownian Motion
(x, y) ∈ H 2 may be realized as the solution of the stochastic differential equations
⎧
⎨dX(t)
= Y (t)dB1(t) − νY (t)dt,
⎩dY
(t) = Y (t)dB2(t) − µ − 1
Y (t)dt,
where B1(t) and B2(t) are two independent one-dimensional Brownian motions. Z(t) is represented
as follows ⎧
⎨X(t)
= x + y
⎩
t
0 exp{Bµ 2 (s)} dBν 1 (s),
Y (t) = y exp{B µ
2 (t)},
where B ν 1 (t) = B1(t)−νt and B µ
2 (t) = B2(t)−µt. Since µ is assumed to be positive, Y (t) converges
to 0 as t tends to ∞. It is possible to show that the distribution of the horizontal component X(t)
converges weakly as t tends to infinity to a given limiting distribution.
Let ha = {z ∈ H2 : Im{z} = a}, with 0 ≤ a < y, be the horocycle through ∞. If a = 0 we
have that h0 is the boundary of the hyperbolic half plane H2. The hitting distribution on ha, with
a > 0, of the diffusion process associated to L is given by the law of the random variable
τa
Xτa = x + y exp{B
0
µ
2 (s)} dBν 1 (s)
where τa = inf{t ≥ 0 : Y (t) = a} is the first hitting time on ha. With a slight abuse of terminology
the law of the random variable
is the hitting distribution on h0.
Hitting distribution on h0
X∞ = x + y
∞
0
exp{B µ
2 (s)} dBν 1 (s)
Let pz(ξ) be the density of the hitting distribution on h0 for the process associated to the operator
L and starting at z = (x, y). We point out that pz(ξ) is the Poisson kernel of L in the domain H 2
and satisfies the following conditions:
⎧
Lpz(ξ) = 0, ∀ξ ∈ h0,
⎪⎨ pz(ξ) > 0, ∀ξ ∈ h0,
∞
−∞
⎪⎩
pz(ξ)dξ = 1,
limy→0 + pz(ξ) = 0, if ξ = x.
Since L is invariant under the action of the map z → az + b then so is pz(ξ)dξ on R. That is
pz(ξ) dξ = paz+b(aξ + b) d(aξ + b) = a paz+b(aξ + b) dξ.
Setting a = 1 x
y , b = − y and f(x) = pi(x) we have
pz(ξ) = px+iy(ξ) = 1
y pi
ξ − x
y
2
= 1
y f
ξ − x
y
(3.28)
(3.29)

3.2 Hyperbolic Brownian motion in H 2 38
then the hitting distribution with arbitrary starting point z is obtained by rescaling the one with
starting point z = i.
Theorem 3.2.2. The horizontal component X(t), with starting point z = i, converges in distribution,
as t → ∞, to the random variable X∞ with density
f(x) = 22µ−1 Γ( 1
2 + µ − iν) 2
πΓ(2µ)
−2ν arctan x e
(1 + x2 .
) µ+1/2
Proof
The first condition in (3.28), Lpz(ξ) = 0, ∀ξ ∈ h0, can be rewritten for f(x) using formula (3.29).
We have
∂ 1
∂x y f
ξ − x
y
∂2 ∂x2
1
y f
ξ − x
y
∂ 1
∂y y f
ξ − x
y
∂2 ∂y2
1
y f
ξ − x
y
= − 1
y2 f ′ (u)| ξ−x ,
y
= 1
y3 f ′′ (u)| ξ−x ,
y
= − 1
ξ − x ξ − x
f −
y2 y y3 f ′ (u)| ξ−x ,
y
= (ξ − x)2
y 5
f ′′ (u)| ξ−x
y
ξ − x
+ 4
y4 f ′ (u)| ξ−x
y
+ 2
f
y3 ξ − x
then we obtain
Lpz(ξ) = y2
2 ∂
2 ∂x2
1
y f
ξ − x
+
y
∂2
∂y2
1
y f
ξ − x
− νy
y
∂
1
∂x y f
ξ − x
y
− µ − 1
y
2
∂
1
∂y y f
=
ξ − x
y
1
2y f ′′ (u)| ξ−x +
y
(ξ − x)2
2y3 f ′′ ξ − x
(u)| ξ−x + 2
y y2 f ′ (u)| ξ−x +
y
1
y f
ξ − x
+
y
ν
y f ′ (u)| ξ−x
y
− µ − 1
−
2
1
y f
ξ − x ξ − x
−
y y2 f ′
(u)| ξ−x .
y
Setting y = 1 and x = 0 we have that f(ξ) solves the second-order linear differential equation
Mf(ξ) = 0 where
Mf(ξ)
= 1
2 f ′′ (ξ) + ξ2
2 f ′′ (ξ) + 2ξf ′ (ξ) + f(ξ) + νf ′
(ξ) − µ − 1
[−f(ξ) − ξf
2
′ (ξ)]
= 1 + ξ2
f
2
′′
(ξ) + ν + µ + 3
ξ f
2
′
(ξ) + µ + 1
f(ξ) (3.30)
2
= d
1 + ξ2
ξf(ξ) + f
dξ
2
′
(ξ) + νf(ξ) + µ − 1
ξf(ξ)
2
= d
2 d 1 + ξ
f(ξ) + νf(ξ) + µ −
dξ dξ 2
1
ξf(ξ) .
2
With the change of variable ξ = tan φ
2
and by imposing that f(ξ) = 2
1+ξ 2
y
.
g(2 arctan ξ) =

39 Hyperbolic Brownian Motion
2 φ
2 cos 2 g(φ), we obtain
2
d d 1 + ξ
f(ξ) + νf(ξ) + µ −
dξ dξ 2
1
ξf(ξ)
2
2 φ d 2 φ d
= 2 cos 2 cos g(φ) + ν + µ −
2 dφ 2 dφ 1
tan
2
φ
2 φ
2 cos
2 2 g(φ)
2 φ d d
= 4 cos g(φ) + ν + µ −
2 dφ dφ 1
tan
2
φ
2 φ
g(φ) cos .
2
2
since d
dξ
= 2 cos2 φ
2
d
dφ
2 φ d
4 cos
2 dφ
and is proportional to g1(φ) that satisfies
. The function g(φ) is solution to the differential equation
d
g(φ) + ν + µ −
dφ 1
tan
2
φ
g(φ) cos
2
= 0
2
2 φ
d
dφ g1(φ)
+ ν + µ − 1
tan
2
φ
−2 φ
g1(φ) = k cos
2
2 .
The function g(φ) has the following form
−F (φ)
g(φ) = c1g1(φ) = c1e
where c, c1 are two real constants and
Then
F (φ) =
c
2 +
φ
0
F (ψ) k
e
cos2 ψ
2
ν + µ − 1
tan
2
φ
dφ = νφ − 2 µ −
2
1
log cos
2
φ
.
2
g(φ) = c1e −νφ
2µ−1 φ c
cos + k
2 2
φ
0
dψ
e νψ −2µ−1 ψ
cos
2 dψ
c, c1, k ∈ R. Since µ > 0 and φ ∈ (−π, π) the integral takes values of either sign. Taking into
account the second condition in (3.28) we have that k = 0 and
g(φ) = c2
2 e−νφ 2µ−1 φ
cos
2
for some real constant c2. Recalling that φ = 2 arctan ξ and f(ξ) = 2
1+ξ 2 g(2 arctan ξ) we have
and finally
g(φ) =
1 + ξ2
2
= c2 arctan ξ
e−2ν
2
f(ξ) = c2
2 e−νφ cos
1
1 + ξ2 f(ξ) = c2
2µ−1 φ
2
2µ−1
−2ν arctan ξ e
(1 + ξ2 1
) µ+ 2
= c2
2 e−2ν arctan ξ cos 2µ−1 (arctan ξ)
By imposing the third condition in (3.28) that is π
g(φ)dφ = 1 we obtain the value of the constant
−π
.

3.2 Hyperbolic Brownian motion in H 2 40
c2. With the change of variable u = φ
2
obtain
1 =
π
−π
g(φ)dφ = c2
2
=
π
−π
and by Gradshteyn and Ryzhik (16) formula 3.892.2. we
e −νφ 2µ−1 φ
cos dφ = c2
2
c2 π
2 2µ−1 2µB 2µ+i2ν+1
2
, 2µ−i2ν+1.
2
π/2
−π/2
e −2νu cos 2µ−1 u du
It follows that the value of the constant c2 is (see Gradshteyn and Ryzhik (16) formula 8.384.1 and
8.326.1)
c2 = 22µ−1
2µ
B µ +
π
1
1
+ iν, µ + − iν =
2 2 22µ−1 Γ(µ +
π
1
2
= 22µ−1
π
|Γ(µ + 1
2 − iν)|2
Γ(2µ)
For a different proof of Theorem 3.2.2 see Baldi et al. (1) page 595.
1
+ iν)Γ(µ + 2 − iν)
Γ(2µ)
. (3.31)
Remark 3.2.1. We note that if ν = 0 and µ = 1
2 , that is if we consider the hyperbolic Brownian
motion on H2 without drift, the limiting distribution is the Cauchy distribution as in the case
of the hitting distribution on the x-axis for the Euclidean Brownian motion on R2 . In fact f(x)
becomes
f(x) =
1
π(1 + x 2 ) .
In general f(x) belongs to the type IV family of Pearson distributions (see Johnson et al. (20)).
Characteristic function of the hitting distribution on h0
In view of (3.29) the characteristic function uz(λ) of X z ∞ for arbitrary starting point z = (x, y) ∈ H 2
can be derived easily from the special case u(λ) = ui(λ) of starting point i. In fact
uz(λ) =
∞
−∞
= e iλx u(λy).
e iλξ pz(ξ)dξ =
∞
−∞
iλξ 1
e
y pi
ξ − x
y
dξ = e iλx
∞
−∞
e iλyw pi(w)dw
Theorem 3.2.3. The characteristic function of the random variable X z ∞, with starting point z = i,
is given by
u(λ) =
1 Γ( 2 + µ + iν)
e
Γ(2µ)
−λ
1
Φ − µ + iν, 1 − 2µ; 2λ
2
where Φ is the confluent hypergeometric function of the second kind.
Proof
Since f(ξ) is a real-valued function we have u(−λ) = u(λ). Starting from the differential equation
(3.30)
1 + ξ 2
2
f ′′
(ξ) + ν + µ + 3
ξ f
2
′
(ξ) + µ + 1
f(ξ) = 0
2

41 Hyperbolic Brownian Motion
and applying the inverse Fourier transform to both members, we get
∞
−∞
e iλx
2 1 + ξ
f
2
′′
(ξ) + ν + µ + 3
ξ f
2
′
(ξ) + µ + 1
f(ξ) dξ = 0.
2
Since, for k = 0, 1, 2, the function xkf (k) (x) is integrable, we have that the function λku (k) (λ) exists,
is continuous in R and vanishes at infinity. In particular u(λ) is twice continuously differentiable
outside 0. We have
∞
e
−∞
iλx f (k) (x)dx = (−iλ) k u(λ) = (−1) k u (k) (λ),
∞
−∞
∞
−∞
e iλx xf ′ (x)dx = −i ∂
∞
e
∂λ −∞
iλx f ′ (x)dx = −i ∂
∂λ [−iλu(λ)] = −u(λ) − λu′ (λ),
e iλx x 2 f ′′ (x)dx = − ∂2
∂λ2 ∞
e
−∞
iλx f ′′ (x)dx = − ∂2
∂λ2 [(iλ)2u(λ)] = 2u(λ) + 4λu ′ (λ) + λ 2 u ′′ (λ).
It follows that u(λ) is in the kernel of the operator N given by
Nu(λ) = λ2
2 u′′
(λ) − µ − 1
λu
2
′
2 λ
(λ) − + iνλ u(λ).
2
Taking into account the third condition in (3.28) we have u(0) = ∞
−∞ pz(ξ)dξ = 1. With the
change of variable u(λ) = e −λ v(2λ) and ω = 2λ we get
ωv ′′ (ω) + (1 − 2µ − ω)v ′
1
(ω) − − µ + iν v(ω) = 0.
2
Since the confluent hypergeometric differential equation ωv ′′ (ω) + (b − ω)v ′ (ω) − av(ω) = 0 has
solution
Φ(a, b; ω) = 21−b Γ(1 − a)e ω
2
π
π
2
0
ω
cos tan θ + (2a − b)θ cos
2 −b θdθ
where Φ(a, b; ω) is the confluent hypergeometric function of the second kind with Re{b} < 1 and a
not a positive integer (see Gradshteyn and Ryzhik (16) formula 9.216.1), the characteristic function
u(λ) we are looking for is given by
u(λ) = e −λ v(2λ) = Ke −λ
1
Φ − µ + iν, 1 − 2µ; 2λ .
2
for some real constant K. In particular we note that Φ(a, b; ω) has a finite non-zero limit for
ω → 0 + ω − and that limω→∞ e 2 Φ(a, b; ω) = 0. To obtain the value of the constant K we observe
that
1
1 = u(0) = KΦ − µ + iν, 1 − 2µ; 0 = K
2 22µ
π Γ
π
1
2
+ µ − iν cos(2iνθ) cos
2 0
2µ−1 θ dθ.

3.2 Hyperbolic Brownian motion in H 2 42
Recalling that e −2νθ = 2 cos(2iνθ) − e 2νθ , we have
π/2
−π/2
e −2νθ cos 2µ−1 θdθ =
= 2
π/2
−π/2
π/2
then in view of formula (3.31) we obtain
0
[2 cos(2iνθ) − e 2νθ ] cos 2µ−1 π/2
θdθ = 2
cos(2iνθ) cos 2µ−1 θdθ,
−π/2
1 = K 22µ
π Γ
π
1
2
+ µ − iν cos(2iνθ) cos
2 0
2µ−1 θ dθ
= K 22µ
π Γ
π/2
1
1
+ µ − iν e
2 2 −π/2
−2νθ cos 2µ−1 θdθ
= K 22µ
π Γ
1
1 π
+ µ − iν
2 2 22µ−1 Γ(2µ)
.
|Γ(1/2 + µ − iν)| 2
cos(2iνθ) cos 2µ−1 θdθ
It is possible to obtain the characteristic function u(λ) by means of the Feynman-Kac formula too,
see Baldi et al. (1) Section 3.
Hitting distribution on ha with a > 0
It is possible to show that the characteristic function of the hitting distribution on ha, with a > 0,
can be derived from the characteristic function of the hitting distribution on h0. We know that
u(λy) =
∞
−∞
= Eiy
e iλyω
f(ω)dω = E e iλyXi
∞ = E e iλy R ∞
0 exp{Bµ 2 (s)}dBν 1 (s)
e iλ R ∞
0
Y (s)dBν
1 (s)
where Y (s) = y exp{B µ
2 (s)} is the y-component of the hyperbolic Brownian motion starting at
z = (0, y). By conditioning with respect to the σ-algebra Fτa and using the strong Markov
property one has, for a < y, that
u(λy) = Eiy Eiy
= Eiy e iλ R τa
0
It follows that
e iλ R ∞
Y (s)dBν
0 1 (s)
Fτa
Y (s)dBν 1 (s)
Eia
= Eiy e iλ R τa
0
e iλ R ∞
0
Y (s)dBν
1 (s)
= Eiy
Eiy[e iλXτa ] = u(λy)
u(λa) .
Y (s)dBν 1 (s) Eiy
iλXτa e u(λa).
e iλ R ∞
Y (s)dB(ν)
τa 1 (s)
Fτa
iλXτa
Then the characteristic function Eiy e of the hitting distribution on ha of the hyperbolic
Brownian motion starting at iy is given by the ratio of the characteristic functions u(λy) and u(λa)
of the hitting distribution on h0 for the hyperbolic Brownian motion starting at i.
The densities of the random variables X∞ and Xτa belong to the domain of attraction of a
stable law of which it is possible to determine the parameters, see Baldi et al. (1) Section 6.

43 Hyperbolic Brownian Motion
3.2.3 Hyperbolic Brownian bridge
Let {Bz1 (t), t ≥ 0} be a Euclidean Brownian motion in Rn starting from z1 ∈ R n , n ≥ 2, then the
Euclidean Brownian bridge {W (t), t ∈ [0, 1]} between z1 and z2 ∈ R n is given by
W (t) = Bz1 (t) + t[z2 − Bz1 (t)]
for every t ∈ [0, 1]. For the Brownian bridge in the Euclidean space we have that the probability
that it stays uniformly close to the line joining z1 and z2 does non depend on the distance between
z1 and z2.
Let U(t) be the hyperbolic Brownian bridge in H n , n ≥ 2, between the origin O and z ′ ∈ H n ,
Eberle (see (13)) proved that on the hyperbolic space H n the sample paths of the hyperbolic
Brownian bridge actually concentrate, according to some exponential rate, around the geodesic
through O and z ′ when z ′ tends to infinity in finite time. Simon in (39) has found the exact
exponential rate of concentration in the case of the hyperbolic Brownian bridge on H 2 .
Let lz ′ be the unique geodesic segment joining O to z′ and define the following function on H 2
d O z ′(·) = inf{η(·, z), z ∈ lz ′}.
For every a > 0, it is possible to prove the following
lim
η(O,z ′ )→∞ −
1
η(O, z ′ ) log
P sup d
0≤t≤1
O
z ′(U(t)) > a = 2 log cosh a.
The proof of the above result is based on the comparison between the law of the hyperbolic bridge
and that of the hyperbolic Brownian motion conditioned to tend toward infinity in the direction
of z ′ .
3.2.4 Fractional hyperbolic Brownian motion
A generalization of the classical hyperbolic Brownian motion in H 2 is given in Lao and Orsingher
(26) where it is considered the time-fractional equation
with α ∈ (0, 1], subject to the initial condition
∂α ∂tα uα = 1
sinh η
sinh η
∂
uα, (3.32)
∂η
uα(η, 0) = δ(η).
We assume that the fractional derivative appearing in (3.32) is understand in the sense of Dzerbashyan-
Caputo that is, for f ∈ C m ,
dαf =
dtα t
1
Γ(m − α) 0
f (m) (s)
ds, m − 1 < α ≤ m.
(t − s) α+1−m

3.3 Multidimensional hyperbolic Brownian motion 44
Proceeding as in the proof of Theorem 3.2.1, we get
⎧
⎨
⎩
d α T
dtα = −ωT,
(y2 − 1)F ′′ + 2yF ′ + ωF = 0.
For the first equation, by applying the Laplace transform to both members, we obtain that
where Eα,1 = ∞
k=0
x k
Γ(αk+1)
T (t, ω) = Eα,1(ωt α )
is the Mittag-Leffler function. By making use of the representation
(3.25) and choosing again Q(x) = tanh(πx), we have
uα(η, t) = 2
π
∞
0
xEα,1
− tα
4 − x2t α
∞
sin(xϕ)
dx √ dϕ, (3.33)
η 2 cosh ϕ − 2 cosh η
for 0 < α ≤ 1. We observe that the classical result (3.11) can be derived from (3.33) as a special
case for α = 1 and that for α = 1/2 we get that
∞
u1/2(η, t) = 2
0
τ2 − e 4t
√ p2(η, τ)dτ. (3.34)
2π2t
Formula (3.34) implies that the fractional hyperbolic Brownian motion with α = 1/2 is equal in
distribution to the classical hyperbolic Brownian motion at a reflected Brownian time.
3.3 Multidimensional hyperbolic Brownian motion
3.3.1 Transition function in H 3
We consider now the behavior of the transition function of the hyperbolic brownian motion in H 3 .
We follow the proof given in Orsingher and De Gregorio (32).
Theorem 3.3.1. The normalized heat kernel G3(η, t), t > 0, has the following explicit expression
G3(η, t) =
e −t
2 3 (πt) 3/2
Proof
We show that (3.35) satisfies the Cauchy problem (3.5). Let us write
We start by observing that
h(η, t) = e−t
t 3/2
∂
∂t h(η, t) = e−tt −3/2 η2
−
e 4t
η2
− ηe 4t
, η > 0. (3.35)
sinh η
η2
− ηe 4t
sinh η .
η
−1 −
sinh η
3 η2
+
2t 4t2
.

45 Hyperbolic Brownian Motion
On the other hand,
and we have
sinh 2 η ∂
∂
∂η h(η, t) = e−tt −3/2 η2
−
e 4t
∂η
∂
sinh
∂η
2 η ∂
∂η
1
sinh η
cosh η
1 − η
sinh η
η2
− ,
2t
h(η, t) = e −t t −3/2
η2
−
e 4t sinh η − η cosh η − η2
h(η, t) = e −t t −3/2
η2
−
e 4t η sinh η −1 − 3 η2
+
2t 4t2 2t
sinh η
Since lim t→0 + h(η, t) = δ(η), we obtain that the function h(η, t) a is solution to the Cauchy problem
(3.5). By imposing the condition (3.6) we obtain that the normalized heat kernel G3(η, t) is given
by
in fact
G3(η, t)dv = 4π
=
∞
1
2 √ π
= t−3/2
4 √ π
0
e−tt−3/2 ∞
G3(η, t) =
G3(η, t) sinh 2 η dη
2
∞
0
= t−3/2 √ 2t
4 √ π
= t−3/2 √ 2t
4 √ π
0
h(η, t)
23 ,
π3/2 η2
−
ηe 4t (e η − e −η )dη
(η−2t)2
(η+2t)2
−
η(e 4t −
− e 4t )dη
∞
− √ (2t + w
2t
√ w2 −
2t)e
∞
w2 −
2t e 2 dw = 1.
−∞
2 dw −
∞
√ 2t
.
(w √ 2t − 2t)e
− w2
2 dw
The probability density p3(η, t) to be at time t at a distance η from the starting point is then
given by
p3(η, t) = 2π3/2 h(η, t)
√ G3(η, t) =
π
2
2
√ e−t
=
π 2 √ πt3/2 η2
− ηe 4t
sinh η .
In what follows we will see that in the odd-dimensional hyperbolic spaces it is always possible
to express the heat kernel Gn(η, t) in a closed form.
3.3.2 The Millson recursive formula
The following recursive formula holds for all n = 1, 2, . . .
Gn+2(η, t) = − e−nt ∂
2π sinh η ∂η Gn(η, t). (3.36)
Formula (3.36) is attributed to Millson (unpublished) by Debiard, Gaveau and Mazet in (12)
page 369, it is also reported in Chavel, see (9) page 151, in the general setting of Riemannian

3.3 Multidimensional hyperbolic Brownian motion 46
manifolds and in Grigor’yan and Noguchi, see (17) page 664, and Davies, see (10) page 178, for the
hyperbolic spaces. For a proof of Theorem (3.3.2) see Davies (11) Theorem 2.1 and Matsumoto
(28) Corollary 3.2.
Theorem 3.3.2. For n = 1, 2, . . . , the following relation holds
Gn+2(η, t) = − e−nt ∂
2π sinh η ∂η Gn(η, t).
Remark 3.3.1. We note that, if Gn(η, t) satisfies ∂
∂t Gn =
function
satisfies the following differential equation
We first observe that
1
sinh n−1 η
Gn+2(η, t) = − e−nt ∂
2π sinh η ∂η Gn(η, t),
∂
∂t Gn+2 =
1
sinh n+1 η
∂
sinh
∂η
n+1 η ∂
Gn+2.
∂η
∂
∂η sinh n−1 η ∂
∂η Gn, then the
∂
∂t Gn+2
= 1
∂
−
2π ∂t
e−nt ∂
sinh η ∂η Gn
= 1 ne
2π
−nt ∂
sinh η ∂η Gn − 1 e
2π
−nt
∂ ∂
sinh η ∂η ∂t Gn
= 1 ne
2π
−nt ∂
sinh η ∂η Gn − 1 e
2π
−nt
∂ 1
sinh η ∂η sinh n−1
∂
sinh
η ∂η
n−1 η ∂
∂η Gn
= 1 ne
2π
−nt ∂
sinh η ∂η Gn − 1 e
2π
−nt
∂ cosh η ∂
(n − 1)
sinh η ∂η sinh η ∂η Gn + ∂2
Gn
∂η2 = 1 ne
2π
−nt ∂
sinh η ∂η Gn − 1 e
2π
−nt
1
(1 − n)
sinh η sinh 2 ∂
η ∂η Gn
cosh η ∂
+ (n − 1)
sinh η
2
∂η2 Gn + ∂3
Gn
∂η3 = 1 ne
2π
−nt ∂
sinh η ∂η Gn
n − 1 e
+
2π
−nt
sinh 3 ∂
η ∂η Gn
n − 1 cosh η
− e−nt
2π sinh 2 ∂
η
2
∂η2 Gn − 1 e
2π
−nt ∂
sinh η
3
Gn.
∂η3 On the other side we have
1
sinh n+1
∂
sinh
η ∂η
n+1 η ∂
Gn+2
∂η
= 1 1
2π sinh n+1
∂
sinh
η ∂η
n+1 η ∂
−
∂η
e−nt ∂
sinh η ∂η Gn
= 1 1
2π sinh n+1
∂
e
η ∂η
−nt sinh n−1 η cosh η ∂
= n − 1
2π e−nt cosh2 η
sinh 3 η
− e−nt
2π
= n e−nt
2π
1
sinh η
cosh 2 η
sinh 3 η
∂
∂η Gn + e−nt
2π
∂3 ∂η3 Gn − n e−nt
2π
= 1 ne
2π
−nt ∂
sinh η ∂η Gn
n − 1
+
2π
cosh η
sinh 2 η
1
sinh η
∂η Gn − e −nt sinh n η ∂2
Gn
∂η2 ∂
∂η Gn + e−nt
2π
∂2 Gn
∂η2 ∂
∂η Gn − e−nt
2π sinh 3 ∂
η ∂η Gn − (n − 1) e−nt
2π
e −nt
sinh 3 η
∂
∂η Gn −
cosh η
sinh 2 η
cosh η
sinh 2 η
n − 1 cosh η
e−nt
2π sinh 2 η
∂2 Gn
∂η2 ∂2 ∂η2 Gn − e−nt ∂
2π sinh η
3
Gn
∂η3 ∂ 2
∂η 2 Gn − 1
2π
e −nt
sinh η
∂3 Gn.
∂η3

47 Hyperbolic Brownian Motion
We finally note that, if lim t→0 + Gn(η, 0) = δ(η), then
lim
t→0 + Gn+2(η, t) = lim
t→0
2π sinh η
+ − e−nt
∂
∂η Gn(η, t) = δ(η).
In view of Theorem 3.2.1 and Theorem 3.3.1 we have the explicit form for G2(η, t) and G3(η, t),
now, applying recursively the Millson formula in (3.36), we can obtain an explicit form for Gn(η, t)
for all n = 1, 2, . . . We start by considering the odd dimensional case.
• n = 2k + 1
G2k+1(η, t) = − e−[2(k−1)+1]t ∂
2π sinh η ∂η G2(k−1)+1 = − e−(2k−1)t
∂
−
2π sinh η ∂η
e−[2(k−2)+1]t ∂
2π sinh η ∂η G
2(k−2)+1
= e−2(2k−2)t
(2π) 2
− 1
2 ∂
G2(k−2)+1 sinh η ∂η
= e−2(2k−2)t
(2π) 2
− 1
2
∂
−
sinh η ∂η
e−[2(k−3)+1]t ∂
2π sinh η ∂η G
2(k−3)+1
= e−3(2k−3)t
(2π) 3
− 1
3 ∂
G2(k−3)+1 sinh η ∂η
= e−(k−1)[2k−(k−1)]t
(2π) k−1
− 1
k−1 ∂
G2[k−(k−1)]+1 sinh η ∂η
= e−(k2 −1)t
(2π) k−1
− 1
k−1 ∂
G3. (3.37)
sinh η ∂η
In the even dimensional case we obtain
• n = 2k + 2
− 1
sinh η
2 ∂
G2(k−2)+2 ∂η
G2k+2(η, t) = − e−2kt ∂
2π sinh η ∂η G2k = e−2(2k−1)t
(2π) 2
= e−3(2k−2)t
(2π) 3
− 1
3 ∂
G2(k−3)+2 sinh η ∂η
= e−k(2k−(k−1))t
(2π) k
− 1
k ∂
G2
sinh η ∂η
= e−(k2 +k)t
(2π) k
− 1
k ∂
G2. (3.38)
sinh η ∂η
Formulas (3.37) and (3.38) are obtained with a different approach in Grigor’yan (17), Theorem
1.1, where the proof is obtained by investigating the relation between the heat equation and the
wave equation in H n .
3.3.3 Bounds for the hyperbolic heat kernel
Davies and Mandouvalos in (11) obtained sharp upper and lower bounds for the heat kernel Gn(η, t)
on the hyperbolic space H n for all n ≥ 2.
Theorem 3.3.3. For all n ≥ 2 there exists a positive constant cn such that
c −1
n hn(η, t) ≤ Gn(η, t) ≤ cn hn(η, t)

3.3 Multidimensional hyperbolic Brownian motion 48
for all t > 0 and η > 0, where
n −
hn(η, t) = t 2 exp − (n − 1)2t (n − 1)η
− −
4 2
η2
(1 + η + t)
4t
n−3
2 (1 + η).
The proof of Theorem 3.3.3 is based on the Millson formula (3.36) that relates the heat kernel on
hyperbolic spaces of different dimension. In particular, for n = 3, we obtain the uniform estimate
3 −
G3(η, t) ∼ t 2 exp −t − η − η2
(1 + η)
4t
that is consistent with (3.35) because of the uniform equivalence
η
sinh η ∼ (1 + η)e−η . (3.39)
Recalling that dv = sinh n−1 η dη dΩn and since, in view of (3.39), we have
we obtain that
sinh n−1 η Gn(η, t) ∼
=
sinh n−1 η ∼
If we fix η > 0, Theorem 3.3.3 implies that
n−1 η
e
1 + η
(n−1)η ,
n−1
η
n −
t 2 exp −
1 + η
(n − 1)2t +
4
(n − 1)η
−
2
η2
(1 + η + t)
4t
n−3
2 (1 + η)
n−1
η
n − [η − (n − 1)t]2
t 2 exp − (1 + η + t)
1 + η
4t
n−3
2 (1 + η). (3.40)
n
3
−
Gn(η, t) ∼ t 2 −
if t → 0 and Gn(η, t) ∼ t 2 e − (n−1)2t 4 if t → ∞.
n − For large t, since t 2 (1 + η + t) n−3
2 goes to zero, we have that the dominant term in formula (3.40)
is exp{−[η − (n − 1)t] 2 /4t} and it has its maximum around η = (n − 1)t. Intuitively formula (3.40)
states that the diffusion on the hyperbolic space behaves like a motion with asymptotic velocity
η/t = n − 1 directed away from the starting point. This behavior is totally different from that of
the Brownian motion in the Euclidean space.
This result is in agreement with Theorem 2.1 in Matsumoto (29) where a central limit theorem
for the radial component of the hyperbolic Brownian motion is shown. Starting from the stochastic
representation of the hyperbolic distance {η(Z(t), z), t ≥ 0}, where {Z(t), t ≥ 0} is the hyperbolic
Brownian motion starting at z ∈ H n , Matsumoto proved the following theorem.
Theorem 3.3.4. The probability distribution of the random variable
η(Z(t), z) − (n − 1)t
√ 2t
converges weakly, as t → ∞, to the standard Normal distribution.

49 Hyperbolic Brownian Motion
For the particular case n = 2, in Lao and Orsingher (26), it is proved that
E{cosh η(Z(t), z)} = e t .
3.3.4 Gruet’s formula for the heat kernel
While the explicit expressions for Gn(η, t) given in (3.37) and (3.38) have different forms for odd
and even dimensions, Gruet in (18) gives a unique integral representation for the normalized heat
kernel Gn(η, t) that holds for every dimension n (see Matsumoto (28) and Matsumoto and Yor
(31)). This representation is based on the Yor’s formula on the distribution of the integral of
geometric Brownian motion (see Yor (43)).
Let {B(t), t ≥ 0} be a standard one-dimensional Brownian motion starting from the origin and
defined on the probability space (Ω, F, P ), set B µ (t) = B(t) + µt the corresponding Brownian
motion with constant drift µ ∈ R. We consider the exponential functional A µ (t) defined by
A µ (t) =
t
0
exp{2B µ (s)}ds =
t
0
exp{2B(s) + 2µs}ds.
Yor’s formula gives an explicit expression for the density of the joint distribution of (A µ (t), B µ (t)).
For r > 0 and t > 0 we have
where
P{A µ (t) ∈ du, B µ (t) ∈ dx} = e µx− µ2 x
t 1 + e2x e dudx
2 exp − θ , t , (3.41)
2u u u
θ(r, t) =
re π2
2t
√ 2π 3 t
∞
0
ξ2
−
e 2t −r cosh ξ sinh(ξ) sin
πξ
dξ. (3.42)
t
Yor (see Yor (44) or Matsumoto and Yor (30)) has shown that the following relation holds
Iα(r) =
∞
where Iα(r) is the modified Bessel function.
0
e − α2 t
2 θ(r, t)dt, α > 0, (3.43)
The proof of the Gruet’s formula (3.44) is based on the explicit expression of the joint density
of (A µ (t), B µ (t)). In this section we will assume that the heat kernel Gn(η, t) is solution to
∂ 1
∂t u = 2∆2u. Theorem 3.3.5. For every n ≥ 2 and t > 0 it holds that
Gn(η, t) =
e−(n−1)2 t
8
π(2π) n
2
∞
n + 1
√ Γ
t 2
0
e π2−ξ 2
2t sinh ξ sin πξ
t
(cosh η + cosh ξ) n+1 dξ. (3.44)
2
Proof
We denote by ˜ δz ′ and δz ′ the Dirac delta function concentrated at z′ with respect to the volume
element dv = dxdy
yn and the Lebesgue measure dz = dxdy respectively. Applying the generalized
expectation in the sense of Watanabe (see (42) page 16), it is possible to represent the heat kernel

3.3 Multidimensional hyperbolic Brownian motion 50
as
Gn(z, z ′
, t) =
B (n)
˜δz ′(Z(t))dP (n) = (y ′ ) n
B (n)
δz ′(Z(t))dP (n) , (3.45)
where Z(t) = (X(t), Y (t)) is given by (3.7). Let F µ (t) be the R n−1 -valued random variable
F µ (t) =
t
from (3.7) and (3.45) it follows that
Gn(z, z ′ , t) = (y ′ ) n
0
=
exp{B µ (s)}dB1(s), ...,
B (n)
′ n
y
y
B (n)
t
0
exp{B µ
(s)}dBn−1(s) ,
δz ′(x + yF µ (t), y exp{B µ (t)})dP (n)
δ “ x ′ −x
y , y′
” µ µ (n)
(F (t), exp{B (t)})dP . (3.46)
y
We note that the conditional distribution of F µ (t) given {B(s), 0 ≤ s ≤ t} is the (n−1)-dimensional
centered Gaussian distribution with covariance matrix A µ (t)In−1, where In−1 is the (n − 1)dimensional
unit matrix. Therefore, taking the conditional expectation on the last member of
(3.46), we obtain, using the same notation for the Dirac delta function on R n and R, that
Gn(z, z ′ , t) =
=
′ n
y
y
B (n)
′ n−1
y
y
B (n)
e − ||x′ −x|| 2
2y 2 A µ (t)
(2πA µ (t)) n−1
2
e − ||x′ −x|| 2
2y 2 A µ (t)
(2πA µ (t)) n−1
2
δ y ′ (exp{B
y
µ (t)})dP (n)
δ y
log ′ (B
y
µ (t))dP (n)
where it is used the relation δc(ex ) = c−1δlog c(x), for c > 0. The last generalized expectation is
the conditional expectation under the condition B µ (t) = log y′
y and can be computed by applying
formula (3.41) and recalling that µ = (n − 1)/2, we have
P A µ (t) ∈ du| B µ ′ y
(t) = log =
y
It follows that
Gn(z, z ′ ′ n−1 ∞
y
, t) =
y
0
e − ||x′ −x|| 2
2y2u (2πu) n−1
2
by means of the change of variable v = y′
yu
Gn(z, z ′ , t) =
= e−(n−1)2 t
∞
0
v
n−1
2
exp −
2π
v
2y ′
8
(2π) n−1
2
∞
0
y
y ′
n−1
2
e − (n−1)2
t
8 exp − 1 + (y′ /y) 2
′ y du
θ , t
2u yu u .
y
y ′
n−1
2
e − (n−1)2
t
8 exp − 1 + (y′ /y) 2
′ y du
θ , t
2u yu u ,
and recalling formula (2.25), we obtain
||x ′ − x|| 2 + y 2 + y ′2
y
− (n − 1)2
t θ(v, t)
dv
8 v
exp {−v cosh η(z, z ′ )} v n−3
2 θ(v, t)dv. (3.47)
Finally, taking into account formula (3.42) and changing the order of integration it is possible to

51 Hyperbolic Brownian Motion
obtain the representation (3.44), in fact
Gn(η, t) = e− (n−1)2t π2
8 + 2t
(2π) n−1
2
√ 2tπ 3
= e− (n−1)2t π2
8 + 2t
= e−(n−1)2 t
∞
0
∞
(2π) n−1 √
2 2tπ3 0
∞
8 n + 1
√ Γ
t 2
π(2π) n
2
ξ2
−
e 2t sinh(ξ) sin
ξ2
−
e 2t sinh(ξ) sin
0
πξ
t
πξ
t
dξ
∞
0
dξ Γ
e π2−ξ 2
2t sinh ξ sin πξ
t
(cosh η + cosh ξ) n+1 dξ.
2
v n−1
2 exp{−v[cosh η + cosh ξ]}dv
n + 1
2
n+1
−
(cosh η + cosh ξ) 2
It is no simple to see directly the coincidence of Gruet’s formula with the classical formulas in
(3.11) and (3.35) obtained for the particular cases n = 2 and n = 3, nevertheless it is simple to see
the coincidence of their Laplace transforms in t.
We first compute the Laplace transform of Gruet’s formula (3.44). Starting from formula (3.47)
we have
∞
0
α2 −
e 2 t Gn(η, t)dt =
=
recalling formula (3.43) we get
∞
0
∞
0
1
2 t e− (n−1)2t ∞
8
α2 −
e
(2π) n−1
2
α2 −
e 2 t Gn(η, t)dt =
=
=
∞
0
1
(2π) n−1
2
1
(2π) n−1
2
2
(2π) n
2
(2π) n−1
2
0
exp{−v cosh η}v n−3
2 θ(v, t)dv dt
v n−3
2 exp{−v cosh η}dv
∞
0
2
π
n−2
−iπ e 2
sinh n−2
2 r
∞
v n−3
2 exp{−v cosh η}dv I q
n−2
−iπ e 2
sinh n−2
2 r
n−2
2 Qq n−2
2 Qq α 2 + (n−1)2
4
0
α 2 + (n−1)2
4
1 − 2
e −
„
α 2
«
(n−1)2
2 + 8 t
θ(v, t)dt,
1 − 2
(cosh η)
α 2 + (n−1)2
4
(cosh η)
where Iα(v) is the modified Bessel function and Q m n (x) is the associated Legendre function of the
second kind. In the second equality we have used formula 6.622.3 in Gradshteyn and Ryzhik (16).
For n = 2 we obtain
∞
For n = 3 we have the simple expression
0
α2 −
e 2 t G2(η, t)dt = 1
∞
0
π Q√ α 2 + 1
4
α2 −
e 2 t G3(η, t)dt = e−η√ α2 +1
2π sinh η
1 (cosh η).
− 2
since, in view of Gradshteyn and Ryzhik (16) formula 8.754.4, we have Q 1
2
ν− 1
2
(v)
(cosh r) = π
2 sinh ν ie−νr .
On the other hand, starting from the representation of G2(η, t) given in (3.11) and replacing t

3.3 Multidimensional hyperbolic Brownian motion 52
with t/2, we obtain
∞
0
α2 −
e 2 t G2(η, t)dt =
where we have used that ∞
0
=
√
2
(2π) 3/2
√
2
(2π) 3/2
1
=
π √ 2
= 1
e −ta− b t
t 3/2
∞
η
∞
η
∞
η
∞
ϕ
√ dϕ
cosh ϕ − cosh η
√
ϕ
2π
√ dϕ
cosh ϕ − cosh η ϕ e−2
e −ϕ√ α 2 + 1
4
√ cosh ϕ − cosh η dϕ
π Q√ α2 + 1 1 (cosh η)
4 − 2
0
t −
e 2(α 2 + 1
4) − e ϕ2
2t
t 3/2
r “ α 2
2
”
1 ϕ2 + 8 2
dt = π
b e−2√ ab and the last step follows from formula 3.544 of
Gradshteyn and Ryzhik (16) and Qn(x) is the Legendre function of the second kind. Concerning
the Laplace transform of G3(η, t) given in formula (3.35), when t is replaced by t/2, we have
∞
0
α2 −
e 2 t G3(η, t)dt =
1
(2π) 3/2
η
∞ t − e
sinh η 0
= e−η√ α 2 +1
2π sinh η .
2 (α2 +1)− η2
2t
Remark 3.3.2. We note that the Millson formula (3.36) follows from the representation (3.44) of
the heat kernel. If we differentiate both sides of (3.44) with respect to η we have
∂
∂η Gn(η, t)
= e−(n−1)2 t
8
π √ t(2π) n
2
∞
n + 1
Γ
2
= − sinh η 2πe nt
2 Gn+2(η, t).
0
e π2 −ξ 2
2t sinh ξ sin
πξ
−
t
3.3.5 Hyperbolic branching Brownian motion
t 3/2
n + 1
2
dt
sinh η
dt
(cosh η + cosh ξ) n+3
2
The hyperbolic branching Brownian motion is a branching diffusion in which individual particles
follow independent Brownian paths in the hyperbolic space H n and undergo binary fission at
exponentially distributed random times independent of the motion.
Lalley and Sellke (25) considered the hyperbolic branching Brownian motion in the hyperbolic
plane when the rate λ of fission is assumed to be constant. Unlike the branching Brownian motion
in the Euclidean plane, the hyperbolic branching Brownian motion in H 2 exhibits a phase transition
in λ. For λ ≤ 1/8 the number of particles in any compact region of H 2 is eventually zero with
probability one, but for λ > 1/8 the number of particles in any open set grows to infinity with
probability one. This phase transition manifests itself in the behavior of the process at infinity.
Let Λ be the set of all limit points in the boundary of H 2 (it is irrelevant whether ∂H 2 is viewed
as the real axis of the half-plane model or the unit circle of the disk model) of particle trails. In
(25), Theorem 1 and Proposition 10, it is shown that for λ ∈ (0, 1/8] the Hausdorff dimension of
Λ is, with probability one, equal to
1
2 (1 − √ 1 − 8λ)
dξ

53 Hyperbolic Brownian Motion
and that for λ > 1/8 the complement of Λ in ∂H 2 has Lebesgue measure zero. Note that as λ
goes to 1/8 the Hausdorff dimension increases continuously to 1/2 and not to 1, so the Hausdorff
dimension of Λ is discontinuous at the critical value λ = 1/8.
Karpelevich, Pechersky and Suhov (21) generalized the Lalley and Sellke result for an hyperbolic
branching Brownian motion in H n , with n ≥ 2. Let v = λ(χ − 1) be the fission potential where
χ > 1 is the mean number of offspring in a single fission. The Hausdorff dimension h(Λ) of the
limiting set Λ on ∂H n is a function of v:
⎧
⎨ 1
2
h(Λ) =
⎩
(n − 1 − (n − 1) − 8v), if 0 ≤ v ≤ (n−1)2
8 ,
n − 1, if v > (n−1)2
.
It follows that h(Λ) monotonically increases with v from 0 to (n − 1)/2 and then jumps to n − 1,
we have that Λ = ∂H n for v > (n − 1) 2 /8. Kelbert and Suhov in (22) considered the case of a
variable fission mechanism.
3.A Appendix
Proposition 3.A.1. If z = (x, y) and z ′ = (x ′ , y ′ ) are in H n , ∆n is given by formula (3.1) and
η(z, z ′ ) is given by (2.25), we have that
∆n η(z, z ′ ) = (n − 1) coth η(z, z ′ ).
Proof
Since coth(arcosh(x)) = x
√ x 2 −1 and since, in view of formula (2.25), we have
we have to prove that
∆n η(z, z ′ ) = (n − 1)
η(z, z ′ ) = arcosh |x − x′ | 2 + y2 + y ′2
2yy ′ ,
|x − x ′ | 2 + y 2 + y ′2
|x − x ′ | 2 + (y + y ′ ) 2 |x − x ′ | 2 + (y − y ′ .
) 2
Using the following notation δ+ = |x − x ′ | 2 + (y + y ′ ) 2 and δ− = |x − x ′ | 2 + (y − y ′ ) 2 , we obtain
that
∂
∂y η(z, z′ ) = y2 − y ′2 − |x − x ′ | 2
y ,
δ− δ+
∂2 ∂y2 η(z, z′ ) = 2y2δ− δ+ − (y2 − y ′2 − |x − x ′ | 2 ) ∂
y2δ−δ+ =
2
−
δ− δ+
y2 − y ′2 − |x − x ′ | 2
y2δ− δ+
8
∂y (y δ− δ+)
− y2 − y ′2 − |x − x ′ | 2
δ 3/2
− δ 1/2
+
+ y2 − y ′2 − |x − x ′ | 2
y δ 3/2
− δ 1/2
y
+
′ − y2 − y ′2 − |x − x ′ | 2
δ 1/2
− δ 3/2
−
+
y2 − y ′2 − |x − x ′ | 2
y δ 1/2
− δ 3/2
+
y ′ ,

3.A Appendix 54
since
∂
∂y (y δ− δ+) = δ− δ+ + y(y − y′ ) δ+
δ−
On the other side, for i = 1, . . . , n − 1, we have
since
∂
∂xi
∂ 2
∂x 2 i
η(z, z ′ ) = 2(xi − x ′ i )
δ− δ+
η(z, z ′
δ− δ+ − (xi − x
) = 2
′ ∂
i ) ∂xi (δ− δ+)
∂
∂xi
So, finally, we obtain that
∆nη(z, z ′ ) =
δ−δ+
( δ− δ+) = δ1/2 +
2(n − 1)y2
δ− δ+
2δ 1/2
−
=
2(xi − x ′ i) + δ1/2 −
2δ 1/2
+
+ y(y + y′ ) δ−
.
δ+
2
δ− δ+
2(xi − x ′ i).
− 2y 2 δ+
n−1
+ δ−
(xi − x ′ i) 2 + 2y2
δ 3/2
+ δ 3/2
−
i=1
δ− δ+
− 2(xi − x ′ i) 2 δ+ + δ−
,
δ 3/2
− δ 3/2
+
− y2 − y ′2 − |x − x ′ | 2
δ− δ+
− y2 − y ′2 − |x − x ′ | 2
δ 3/2
− δ 1/2
y
+
2 + y2 − y ′2 − |x − x ′ | 2
δ 3/2
− δ 1/2
yy
+
′ − y2 − y ′2 − |x − x ′ | 2
δ 1/2
− δ 3/2
y
+
2
− y2 − y ′2 − |x − x ′ | 2
δ 1/2
− δ 3/2
yy
+
′ − (n − 2) y2 − y ′2 − |x − x ′ | 2
δ− δ+
= 2(n − 1)y2 + (n − 1)(−y 2 + y ′2 + |x − x ′ | 2 ) + 2y 2
δ− δ+
+ −y2 (y2 − y ′2 − |x − x ′ | 2 ) + yy ′ (y2 − y ′2 − |x − x ′ | 2 ) − 2y2 |x − x ′ | 2
δ 3/2
− δ 1/2
+
− −y2 (y2 − y ′2 − |x − x ′ | 2 ) + yy ′ (y2 − y ′2 − |x − x ′ | 2 ) − 2y2 |x − x ′ | 2
δ 3/2
+ δ 1/2
−
= (n − 1) |x − x′ | 2 + y 2 + y ′2
δ− δ+
and the proposition is proved.
Proposition 3.A.2. If f is a smooth function on R, we have
Proof
We have
∆nf(η(z, z ′ )) = y 2
n−1
∆nf(η) = f ′′ (η) + (n − 1) coth η f ′′ (η).
∂ 2
∂x
i=1
2 i
f(η(z, z ′ )) + ∂2
∂y 2 f(η(z, z′ ))
− (n − 2)y ∂
∂y f(η(z, z′ ))

55 Hyperbolic Brownian Motion
where
∂ ∂f ∂η
f(η) =
∂y ∂η ∂y ,
We have than
∆nf(η(z, z ′ )) = y 2
In fact
n−1
∂η
i=1
∂xi
∂ 2
∂y 2 f(η) = ∂2 f
∂η 2
n−1
= ∂2f y2
∂η2 = ∂2f y2
∂η2
∂ 2 f
∂η
i=1
2
n−1
∂η
∂xi
∂η
i=1
n−1
∂xi
∂η
i=1
= ∂2f ∂f
+
∂η2 ∂η ∆nη.
2
+
2 ∂η
∂y
=
∂xi
2
2 ∂η
+
∂y
∂f
∂η
2
2
∂2η ,
∂y2 ∂ 2
∂x 2 i
f(η) = ∂2f ∂η2 2 ∂η
+
∂xi
∂f
∂η
∂2η ∂x2 .
i
+ ∂f ∂
∂η
2η ∂x2
+
i
∂2f ∂η2 2 ∂η
+
∂y
∂f ∂
∂η
2η ∂y2
− (n − 2)y ∂f ∂η
∂η ∂y
2
∂η
+ +
∂y
∂f
y
∂η
2
n−1 ∂
i=1
2η ∂x2 +
i
∂2η ∂y2
− (n − 2)y ∂η
∂y
2
∂η
+ +
∂y
∂f
∂η ∆nη
n−1
4
(xi − x
δ+δ−
i=1
′ i) 2 + (y2 − y ′2 − |x − x ′ | 2 ) 2
y2δ+δ− = y4 + y ′4 + |x − x ′ | 4 − 2y 2 y ′2 + 2y 2 |x − x ′ | 2 + 2y ′ |x − x ′ | 2
y 2 δ+δ−
= 1
.
y2 In view of Proposition 3.A.1 we obtain
and the proposition is proved.
∆nf(η(z, z ′ )) = ∂2f ∂f
+
∂η2 ∂η ∆nη = ∂2f ∂f
+ (n − 1) coth η
∂η2 ∂η
Proposition 3.A.3. The hyperbolic coordinates (η, α) and the cartesian coordinates (x, y) in H 2
satisfy the following relationships
Proof
We first observe that
x =
y =
and from these relations we obtain
sinh η cos α
cosh η−sinh η sin α , η > 0,
1
cosh η−sinh η sin α , π
π
− 2 < α ≤ 2 .
(x − x0) 2 + y 2 = r 2 , x 2 0 + 1 = r 2 , x0 = r sin α,
tan α = x2 + y 2 − 1
2x
Since, in view of (2.25), we have
(3.48)
and x = tan α ±
tan2 α + 1 − y2 . (3.49)
cosh η = x2 + y2 − 1
, (3.50)
2y

3.A Appendix 56
substituting (3.49) in (3.50) we have
that is
which implies
2y cosh η = y 2
+ 1 + tan α ± tan2 α − y2 2 + 1
= 2 + 2 tan 2
α ± 2 tan α tan2 α1 − y2 ,
y cosh η − 1 − sin2 α
cos2 α
=
sin α
± tan
cos α
2 α + 1 − y2 ,
(y cosh η cos 2 α − 1) 2 = sin 2 α cos 2 α(tan 2 α + 1 − y 2 ),
y 2 [cosh 2 η cos 4 α + sin 2 α cos 2 α] + cos 2 α − 2y cosh η cos 2 α = 0,
y 2 [cosh 2 + sin 2 (1 − cos 2 η)] + 1 − 2y cosh η = 0,
[y cosh η − 1] 2 − y 2 sin 2 α sinh 2 η = 0,
(y cosh η − 1 − y sin α sinh η)(y cosh η − 1 + y sin α sinh η) = 0.
The last equation and formula (3.49) imply both relations in (3.9).
Proposition 3.A.4. The hyperbolic Laplacian in cartesian coordinates
∆2 = y 2
2 ∂ ∂2
+
∂x2 ∂y2
is expressed in hyperbolic coordinates (η, α) by means of the differential operator
Proof
By deriving (3.9) we get
and
∂2 ∂2
=
∂η2 ∂x2 ∂ ∂
=
∂η ∂x
∆2 = 1
∂
sinh η
sinh η ∂η
∂
+
∂η
1
sinh 2 η
∂2 .
∂α2 cos α
∂ − sinh η + cosh η sin α
2 +
(cosh η − sinh η sin α) ∂y
2 , (3.51)
(cosh η − sinh η sin α)
∂ ∂ sinh
=
∂α ∂x
2 η − sinh η cosh η sin α
(cosh η − sinh η sin α) 2
∂ cos α sinh η
+
∂y (cosh η − sinh η sin α)
cos 2 α
(cosh η − sinh η sin α)
∂2
4 +
∂y2 +2 ∂2 cos α (− sinh η + cosh η sin α)
∂x∂y (cosh η − sinh η sin α) 4
+ ∂ −2 cos α (sinh η − cosh η sin α)
∂x (cosh η − sinh η sin α) 3 + ∂
∂y
(− sinh η + cosh η sin α) 2
(cosh η − sinh η sin α) 4
− cos2 α + (sinh η − cosh η sin α) 2
(cosh η − sinh η sin α) 3 ,(3.52)
2 ,

57 Hyperbolic Brownian Motion
∂2 ∂2
=
∂α2 ∂x2 sinh 2 η (sinh η − cosh η sin α) 2
(cosh η − sinh η sin α) 4 + ∂2
∂y2 sinh 2 η cos2 α
(cosh η − sinh η sin α) 4
+2 ∂2 sinh
∂x∂y
2 η cos α (sinh η − cosh η sin α)
(cosh η − sinh η sin α) 4
+ ∂ sinh η cos α
∂x
sinh 2 η − 1 − sinh η cosh η sin α
(cosh η − sinh η sin α) 3
+ ∂ sinh η
∂y
− sin α cosh η + sinh η + sinh η cos2 α
(cosh η − sinh η sin α) 3 . (3.53)
By multiplying (3.52) by sinh 2 η and then summing (3.53) we obtain that
∂ 2
∂η2 sinh2 η + ∂2
∂α
2 = ∂2
∂x2 sinh2 η cos2 α + (sinh η − cosh η sin α) 2
(cosh η − sinh η sin α) 4
+ ∂2
∂y2 sinh2 η cos2 α + (sinh η − cosh η sin α) 2
(cosh η − sinh η sin α) 4
+ ∂ sinh η cos α
∂x
− cosh 2 η + sinh η cosh η sin α
(cosh η − sinh η sin α) 3
+ ∂ sinh η cosh η (sinh η − cosh η sin α)
∂y (cosh η − sinh η sin α) 2
= sinh 2 η y 2
2 ∂ ∂2
+
∂x2 ∂y2
2 ∂
− sinh η cosh η cos α y
∂x
2 ∂
+ sinh η cosh η(sinh η − cosh η sin α) y
∂y
= sinh 2 η y 2
2 ∂ ∂2
+
∂x2 ∂y2
− sinh η cosh η ∂
∂η .
Where in the last equality we have used formula (3.51).

3.A Appendix 58

Chapter 4
Travelling Randomly on H 2 with a
Pythagorean Compass
4.1 Description of the planar random motion on H 2
We start our analysis by considering a particle located at the origin O of H 2 . The particle initially
moves on the half-circumference with center at (0, 0) and radius 1. The motion of the particle
develops on the geodesic lines represented by half-circles with the center located on the x axis.
Changes of direction are governed by a homogeneous Poisson process of rate λ.
At the occurrence of the first Poisson event, the particle starts moving on the circumference
orthogonal to the previous one.
After having reached the point P2, where the second Poisson event happened, the particle
continues its motion on the circumference orthogonal to that joining O with P2 (see Figure 4.1).
In general, at the n-th Poisson event, the particle is located at the point Pn and starts moving
on the circumference orthogonal to the geodesic curve passing through Pn and the origin O (consult
again Figure 4.1).
At each Poisson event the particle moves from the reached position P clockwise or counterclockwise
(with probability 1
2 ) on the circumference orthogonal to the geodesic line passing through
P and O.
The hyperbolic length of the arc run by the particle during the inter-time between two successive
changes of direction, occurring at tk−1 and tk respectively, is given by c(tk − tk−1), with k ≥ 1 and
t0 = 0. The velocity c is assumed to be the constant hyperbolic velocity
c = ds
dt
1 dx2 + dy2 =
y dt2 .
The Cartesian coordinates of the points Pk, where the changes of direction occur, can be explicitly
evaluated, but they are not important in our analysis because we study only the evolution
of the hyperbolic distance from the origin of the moving particle.
The construction outlined above shows that the arcs OPk−1, Pk−1Pk, and OPk form right

4.1 Description of the planar random motion on H 2 60
Figure 4.1: In the first three figures a sample path where the particle chooses the outward direction
is depicted. In the last one a trajectory with one step moving towards the origin is depicted.
triangles with the vertex of the right angle at Pk−1.
In force of the Hyperbolic Pythagorean Theorem we have that
cosh d(OPk) = cosh d(OPk−1) cosh d(Pk−1Pk).
The hyperbolic distance η(t) of the moving point Pt after n changes of direction is thus given by
cosh η(t) = cosh d(OPt)
= cosh d(PnPt) cosh d(OPn)
n
= cosh c(t − tn) cosh c(tk − tk−1)
=
n+1
k=1
k=1
cosh c(tk − tk−1), (4.1)
where t0 = 0 and tn+1 = t. The instants tk, k = 0, 1, . . . , n are uniformly distributed in the set
T = {0 = t0 < t1 < · · · < tk < · · · < tn < tn+1 = t}.
This means that cosh η(t), defined in (4.1), can be viewed as the hyperbolic distance from O of
the moving particle for fixed time points of the underlying Poisson process and for a fixed number
N(t) = n of changes of direction.
We remark that the geodesic distance (4.1) depends on how much time the particle spends on
each geodesic curve (but not on the chosen direction). Of course, (4.1) depends on the number n
of changes of direction and on the speed c of the moving particle, as well.
The set of possible positions at different times t is depicted in Figure 4.2. The vertices A and
B are reached when the particle never changes direction, whereas C and D are reached if the
deviation occurs immediately after the start.
The ensemble of points having the same hyperbolic distance from O at time t, forms the circle

61 Travelling Randomly on H 2 with a Pythagorean Compass
0.8 1.0 1.2 1.4
0 2 4 6 8
A
C
D
−0.4 −0.2 0.0 0.2 0.4
(a) N=2, t=6
C
A D
−4 −2 0 2 4
B
(c) N=2, t=40
B
0 1 2 3 4
0 2 4 6 8 10 12 14
A
C
D
−2 −1 0 1 2
(b) N=2, t=25
C
A D
−6 −4 −2 0 2 4 6
B
(d) N=2, t=50
Figure 4.2: The set of all possible points reachable by the process for different values of t is drawn.
In each domain a trajectory of the process, with c = 0.05 and N(t) = 2, is depicted.
with center C = (0, cosh η(t)) and radius sinh η(t). Since
cosh η(t) =
n+1
k=1
cosh c(tk − tk−1),
the ordinate of the center C is obtained by successively multiplying the ordinates of the centers
of equally distant points at each step. For the radius, however, such a fine interpretation is not
possible (the radii do not exhibit the same multiplicative behavior) but nevertheless we will study
their product
because
n+1
k=1
sinh η(t) ≥
sinh c(tk − tk−1) (4.2)
n+1
k=1
sinh c(tk − tk−1)
and (4.2) represents a lower bound for the circle of equally distant points at time t.
B

4.2 Equations related to the mean hyperbolic distance 62
4.2 Equations related to the mean hyperbolic distance
In this section we study the conditional and unconditional mean values of the hyperbolic distance
η(t). Our first result concerns the derivation of the equations satisfied by the mean values
where
and by
En(t) = E{cosh η(t)|N(t) = n} (4.3)
= n!
t n
t
0
dt1
= n!
In(t),
tn In(t) =
t
0
t
t1
t
dt1 · · ·
dt2 · · ·
tn−1
t
dtn
tn−1
n+1
k=1
dtn
n+1
k=1
cosh c(tk − tk−1)
cosh c(tk − tk−1),
E(t) = E{cosh η(t)} (4.4)
∞
= E{cosh η(t)|N(t) = n}P{N(t) = n}
n=0
= e −λt
∞
λ n In(t).
n=0
At first, we state the following result concerning the evaluation of the integrals In(t), n ≥ 1.
Lemma 4.2.1. The functions
In(t) =
t
0
dt1
t
t1
t
dt2 · · ·
tn−1
dtn
n+1
with t0 = 0 and tn+1 = t, satisfy the difference-differential equations
where t > 0, n ≥ 1, and I0(t) = cosh ct.
Proof
We first note that
d
dt In =
t
t
dt1 · · ·
0
t
+c
0
= In−1 + c
tn−2
t
dt1 · · ·
t
0
k=1
d 2
dt 2 In = d
dt In−1 + c 2 In
dtn−1
dtn
n
cosh c(tk − tk−1)
k=1
tn−1 k=1
t
dt1 · · ·
tn−1
cosh c(tk − tk−1), (4.5)
n
cosh c(tk − tk−1) sinh c(t − tn)
dtn
n
cosh c(tk − tk−1) sinh c(t − tn)
k=1
(4.6)

63 Travelling Randomly on H 2 with a Pythagorean Compass
and therefore
d 2
dt 2 In = d
dt In−1 + c 2
t
0
t
dt1 · · ·
tn−1
dtn
n+1
k=1
cosh c(tk − tk−1)
= d
dt In−1 + c 2 In. (4.7)
In view of Lemma 4.2.1 we can prove also the following:
Theorem 4.2.2. The mean value E(t) = E{cosh η(t)} satisfies the second-order linear homogeneous
differential equation
d2 d
E(t) = −λ
dt2 dt E(t) + c2E(t) with initial conditions
E(0) = 1,
d
dt E(t) t=0 = 0.
The explicit value of the mean hyperbolic distance is therefore
λt −
E(t) = e 2
Proof
From (4.4), it follows that
cosh t√ λ 2 + 4c 2
2
+
d
E(t) = −λE(t) + e−λt
dt
(4.8)
λ
√
λ2 + 4c2 sinh t√λ2 + 4c2
. (4.9)
2
∞
n=0
and thus, in view of (4.7) and by letting I−1 = 0, we have that
d2
d
d
E(t) = −λ E(t) − λ E(t) + λE(t)
dt2 dt dt
n d
λ
dt In
+ e −λt
= −2λ d
dt E(t) − λ2 E(t) + c 2 E(t) + e −λt
= −2λ d
dt E(t) − λ2 E(t) + c 2 E(t) + λ
∞
n=0
∞
n=0
λ n
d
dt In−1 + c 2
In
n d
λ
dt In−1
d
E(t) + λE(t)
dt
(4.10)
= −λ d
dt E(t) + c2 E(t). (4.11)
While it is straightforward to see that the first condition in (4.8) is verified, the second one
needs some explanations: if we write
d
dt E(t)
t=0
E(∆t) − 1
= lim
∆t↓0 ∆t
(4.12)

4.2 Equations related to the mean hyperbolic distance 64
and observe that
E(∆t) = (1 − λ∆t) cosh c∆t + λ
∆t
= (1 − λ∆t) cosh c∆t + λ∆t
2
0
cosh ct1 cosh c(∆t − t1)dt1 + o(∆t)
λ
cosh c∆t + sinh c∆t + o(∆t),
2c
by substituting (4.13) in (4.12), the second condition emerges. The integral in (4.13) represents
the mean value E{cosh η(∆t)|N(∆t) = 1} and is in fact evaluated by applying the Pythagorean
hyperbolic theorem, as in (4.3), for k = 1 and t = ∆t.
The general solution to equation (4.11) has the form
λt −
E(t) = e 2
Ae t
2
√ λ 2 +4c 2
+ Be
t − 2
√ λ 2 +4c 2
. (4.13)
By imposing the initial conditions, the constants A and B can be evaluated and coincide with:
From (4.13) and (4.14) we obtain
E(t) =
e− λt
2
so that (4.9) emerges.
2
A = λ + √ λ2 + 4c2 2 √ λ2 + 4c2 √
λ2 + 4c2 − λ
, B =
2 √ λ2 + 4c2 . (4.14)
λ + √ λ 2 + 4c 2
√ λ 2 + 4c 2
e t
√
√
2 λ2 +4c2 λ2 + 4c2 − λ
+ √ e
λ2 + 4c2 √
t − 2 λ2 +4c2 Remark 4.2.1. The mean value E(t) tends to infinity as t → ∞ so that the moving particle, in
the long run, either reaches the x axis or moves away towards the infinity.
Of course, if c = 0 we have that E(t) = 1, and for λ → ∞ we have again that E(t) = 1 because
in both cases the particle cannot leave the starting point.
If λ → 0 we get E(t) = cosh ct because the particle will simply move on the basic geodesic line
and its hyperbolic distance grows linearly with t.
We note that the hyperbolic distance itself tends to infinity as t → ∞ because
lim cosh η(t) =
t→∞
∞
cosh d(PkPk−1) = ∞ (4.15)
k=1
and (4.15) is the infinite product of terms bigger than one.
Remark 4.2.2. By taking into account the difference-differential equation (4.6), or directly from
(4.3), it follows that the conditional mean values En(t) satisfy the following equation with nonconstant
coefficients
d 2
dt2 En + 2n
t
d
dt En − n d
t dt En−1 + n2 − n
t2 (En − En−1) − c 2 En = 0. (4.16)
In order to obtain the explicit value of the conditional mean value En(t) it is convenient to
perform a series expansion of E(t), instead of solving the difference-differential equation (4.16). In
this way we can prove the following result.

65 Travelling Randomly on H 2 with a Pythagorean Compass
Theorem 4.2.3. The conditional mean values En(t), n ≥ 1, can be expressed as
En(t) =
[ n
2 ]
r=0
+
1
2 n
[ n−1
2 ]
r=0
n!
(n − 2r)!
1
2 n
∞
2j
r + j (ct)
j (2r + 2j)!
j=0
n!
(n − 2r − 1)!
∞
r + j (ct)
j
2j
(2r + 2j + 1)!
j=0
Proof
By expanding the hyperbolic functions in (4.9) we have that
λt −
E(t) = e 2
+
∞
k=0
1
(2k)!
λ
√ λ 2 + 4c 2
∞
k=0
t
2k λ2 + 4c2 2
1
(2k + 1)!
t
2k+1
λ2 + 4c2 .
2
(4.17)
(4.18)
By applying the Newton binomial formula to the terms in the round brackets and by expanding
e λt
2 it follows that
E(t) = e −λt
+ λ
∞
m=0
∞
k=0
1
m!
1
(2k + 1)!
m ∞
λt
2
k=0
t
2
1
(2k)!
2k+1 k
r=0
t
2
Finally, interchanging the summation order, it results that
Since
E(t) = e −λt
⎡
∞ ∞
2r+m ⎣
1 λt (2r + m)!
m!r! 2 (2r + m)!
+
∞
m=0 r=0
∞
m=0 r=0
1
m!r!
2k k
r=0
k
λ
r
2r (2c) 2k−2r
∞
j=0
2r+m+1 λt (2r + m + 1)!
2 (2r + m + 1)!
E(t) = e −λt
∞ (λt) n
n=0
k
λ
r
2r (2c) 2k−2r
(r + j)!
j!
∞
j=0
(r + j)!
j!
.
(ct) 2j
(2r + 2j)!
(ct) 2j
⎤
⎦ .
(2r + 2j + 1)!
n! En(t), (4.19)

4.2 Equations related to the mean hyperbolic distance 66
from (4.19) and (4.19), we have that
En(t) =
=
m, r: 2r+m=n
+
[ n
2 ]
r=0
+
m, r: 2r+m+1=n
1
2 n
[ n−1
2 ]
r=0
n!
(n − 2r)!
1
2 n
1
22r+m ∞
(2r + m)! r + j 2j (ct)
m! j (2r + 2j)!
j=0
1
22r+m+1 ∞
(2r + m + 1)! r + j (ct)
m!
j
2j
(2r + 2j + 1)!
∞
2j
r + j (ct)
j (2r + 2j)!
j=0
n!
(n − 2r − 1)!
j=0
∞
r + j (ct)
j
2j
(2r + 2j + 1)! ,
and this represents the explicit form of the conditional mean values.
j=0
Remark 4.2.3. We check formula (4.17) by evaluating the mean value En(t) for n = 0, 1, 2, 3.
It can be noted that for n = 0 only the term r = 0 of the first sum in (4.17) must be considered,
so that
∞ (ct)
E{cosh η(t)|N(t) = 0} =
2j
= cosh ct.
(2j)!
For n = 1 both sums of (4.17) contribute to the mean value with the r = 0 term
E{cosh η(t)|N(t) = 1} = 1
2
j=0
∞
j=0
(ct) 2j
(2j)!
+ 1
2
∞
j=0
= 1 1
cosh ct + sinh ct.
2 2ct
(ct) 2j
(2j + 1)!
For n = 2 we have two terms in the first sum (corresponding to r = 0, 1) and the term r = 0 in
the second sum, so that
E{cosh η(t)|N(t) = 2} = 1
2 2
+ 1
2
∞
j=0
(ct) 2j
(2j)!
= 1
cosh ct +
22 For n = 3 we need to consider two terms in both sums
+ 1
2
∞
2j
j + 1 (ct)
j (2j + 2)!
j=0
∞ (ct)
j=0
2j
(2j + 1)!
1
22
1
+ sinh ct.
ct 2ct
E{cosh η(t)|N(t) = 3} = 1
23 ∞ (ct)
j=0
2j 3!
+
(2j)! 23 ∞
2j
j + 1 (ct)
j (2j + 2)!
j=0
+ 3
23 ∞ (ct)
j=0
2j 3
+
(2j + 1)! 22 ∞
2j
j + 1 (ct)
j (2j + 3)!
j=0
1 3
= +
23 23 (ct) 2
6
cosh ct +
23 3
−
ct (2ct) 3
sinh ct.

67 Travelling Randomly on H 2 with a Pythagorean Compass
The same results can be obtained directly from (4.3) by successive integrations.
For each step the ensemble of points with hyperbolic distance equal to c(tk − tk−1) forms a
Euclidean circumference Ck with radius sinh c(tk −tk−1) and center located at (0, cosh c(tk −tk−1)).
At time t, if n steps have occurred, the set of points Ct with hyperbolic distance equal to η(t) is a
circumference with center at (0, cosh η(t)) and radius sinh η(t). Clearly
cosh η(t) =
n+1
k=1
cosh c(tk − tk−1)
so that the ordinate of the center of Ct is equal to the product of the ordinates of Ck. However
sinh η(t) =
=
≥
1 + cosh 2 η(t)
n+1
1 + cosh 2 c(tk − tk−1)
n+1
k=1
k=1
sinh c(tk − tk−1)
and this shows that the quantity n+1
k=1 sinh c(tk − tk−1) represents a lower bound of the radius of
the circle Ct.
Theorem 4.2.4. The functions
Jn(t) =
t
0
dt1
t
t1
t
dt2 · · ·
tn−1
where t0 = 0, tn+1 = t > 0, and n ≥ 1, take the form
where J0(t) = sinh ct.
Jn(t) = t2n+1 c n+1
n!
∞
r=0
dtn
(n + r)!
r!
n+1
k=1
sinh c(tk − tk−1),
(ct) 2r
, (4.20)
(2r + 2n + 1)!
Proof
We first note that the functions Jn(t), n ≥ 1, t > 0 satisfy the difference-differential equations
Since
we have that
d
dt Jn = c
d 2
dt 2 Jn = cJn−1 + c 2 Jn, n ≥ 1, t > 0. (4.21)
t
0
d 2
dt 2 Jn = c
t
dt1 · · ·
t
+c 2
tn−1
0
t
dtn
n
sinh c(tk − tk−1) cosh c(t − tn),
k=1
t
dt1 · · ·
0
tn−2
t
dt1 · · ·
= c Jn−1 + c 2 Jn.
dtn−1
tn−1
dtn
n
sinh c(tk − tk−1)
k=1
n+1
k=1
sinh c(tk − tk−1)

4.2 Equations related to the mean hyperbolic distance 68
From (4.21), we have that the generating function
satisfies the differential equation
In fact, by (4.21), we have
∞
n=0
G(s, t) =
n d2
s
dt2 Jn = cs
∞
n=0
s n Jn
(4.22)
d2 G = c(s + c)G. (4.23)
dt2 ∞
s n−1 Jn−1 + c 2
∞
n=0
n=0
s n Jn
and this easily yields (4.23). Considering that the general solution to (4.23) is
and that G(s, t) satisfies the initial conditions
it follows that
G(s, t) = Ae t√ c(s+c) + Be −t √ c(s+c)
G(s, t) =
G(s, 0) = 0,
d
dt G(s, t) t=0 = c,
By expanding the sinh function in (4.26) we obtain that
G(s, t) =
=
=
c
s + c
k=0 j=0
(4.24)
(4.25)
√ c
√ s + c sinh t c(s + c). (4.26)
∞ (t
k=0
c(s + c)) 2k+1 ∞
=
(2k + 1)!
(2k + 1)!
k=0
∞ k
k
s
j
j c k−j t2k+1ck+1 (2k + 1)! =
∞
s
j=0
j
⎧
⎨ ∞
k
⎩ j
k=j
∞
s j
t2j+1cj+1 ∞ (j + r)! (ct)
j! r!
2r
(2r + 2j + 1)!
j=0
r=0
and, in view of (4.22), formula (4.20) appears.
Remark 4.2.4. We consider the quantity
∞
n=0
n!
t n Jn(t)P{N(t) = n} = e −λt
∞
n=0
t 2k+1 c k+1 (s + c) k
c k−j t2k+1 c k+1
(2k + 1)!
⎫
⎬
⎭
λ n Jn(t) = e −λt G(λ, t) (4.27)
= e −λt
√
c
√ sinh t
λ + c c(λ + c)
which represents a lower bound for mean values of the radius of the circle C of points with equal
hyperbolic distance from the origin at time t. We note that the bound (4.27) increases if
c 2 + cλ − λ 2 > 0.

69 Travelling Randomly on H 2 with a Pythagorean Compass
For large values of λ the radius of the circle C tends to decrease because the particle often changes
direction and hardly leaves the starting point O.
4.3 About the higher moments of the hyperbolic distance
In this section we study the conditional and unconditional higher moments of the hyperbolic
distance η(t). Our first results concern the derivation of the equations satisfied by the secondorder
moments
where
and by
Mn(t) = E{cosh 2 η(t)|N(t) = n}
= n!
t n
t
0
dt1
= n!
Un(t),
tn Un(t) =
t
0
t
t1
t
dt1 · · ·
dt2 · · ·
tn−1
t
dtn
tn−1
n+1
k=1
dtn
n+1
k=1
cosh 2 c(tk − tk−1)
cosh 2 c(tk − tk−1),
M(t) = E{cosh 2 η(t)} (4.28)
=
∞
E{cosh 2 η(t)|N(t) = n}P{N(t) = n}
n=0
= e −λt
∞
λ n Un(t).
n=0
At first, we state the following results concerning the evaluation of the integrals Un(t), n ≥ 1.
Lemma 4.3.1. The functions
Un(t) =
t
0
dt1
t
t1
t
dt2 · · ·
tn−1
dtn
n+1
k=1
cosh 2 c(tk − tk−1), (4.29)
with t0 = 0 and tn+1 = t, satisfy the following third-order difference-differential equations
where t > 0, n ≥ 1, and U0(t) = cosh 2 ct.
Proof
d3 dt3 Un = d2
dt2 Un−1
2 d
+ 4c
dt Un − 2c 2 Un−1, (4.30)

4.3 About the higher moments of the hyperbolic distance 70
We first note that
d
dt Un =
t
0
+2c
= Un−1
+2c
t
dt1 · · ·
t
0
t
A further derivation yields
0
tn−2
t
dt1 · · ·
dtn−1
tn−1
t
dt1 · · ·
tn−1
d 2
dt 2 Un = d
dt Un−1 + 2c 2
= d
+2c 2
t
0
dtn
dtn
n
cosh 2 c(tk − tk−1)
k=1
t
0
t
dt1 · · ·
n
cosh 2 c(tk − tk−1) cosh c(t − tn) sinh c(t − tn)
k=1
n
cosh 2 c(tk − tk−1) cosh c(t − tn) sinh c(t − tn).
k=1
t
dt1 · · ·
tn−1
dt Un−1 + 2c 2 Un
t t
+2c 2
0
dt1 · · ·
tn−1
dtn
dtn
tn−1
dtn
n+1
k=1
cosh 2 c(tk − tk−1)
n
cosh 2 c(tk − tk−1) sinh 2 c(t − tn)
k=1
n
cosh 2 c(tk − tk−1) sinh 2 c(t − tn).
Since it is not possible to express the integral in (4.31) in terms of Un and its first two derivatives, a
further derivation is necessary, that, in view of (4.31), leads to the following third-order differencedifferential
equation
d3 dt3 Un = d2
dt2 Un−1
2 d
+ 2c
dt Un + 2 2 c 3
× sinh c(t − tn) cosh c(t − tn)
k=1
t
0
t
dt1 · · ·
tn−1
= d2
dt2 Un−1
2 d
+ 2c
dt Un
2 d
+ 2c
dt Un − 2c 2 Un−1.
In view of Lemma 4.3.1 we can prove also the following:
dtn
n
cosh 2 c(tk − tk−1)
Theorem 4.3.2. The function M(t) = E{cosh 2 η(t)} satisfies the third-order linear differential
equation
d3 d2
M(t) = −2λ
dt3 dt2 M(t) + (4c2 − λ 2 ) d
dt M(t) + 2c2λM(t), (4.31)
with initial conditions ⎧⎪ ⎨
Proof
⎪⎩
M(0) = 1,
d
dtM(t) = 0, t=0
d 2
= 2c2 .
dt2 M(t)
t=0
k=1
(4.32)

71 Travelling Randomly on H 2 with a Pythagorean Compass
By multiplying both members of (4.30) by λ n and summing up we have that
and also
d 3
dt 3
∞
n=0
λ n Un = λ d2
dt 2
∞
n=1
λ n−1 2 d
Un−1 + 4c
dt
∞
λ n Un − 2c 2 ∞
λ
n=0
n=1
λ n−1 Un−1,
d3 dt3 λt d
e M(t) = λ 2
dt2 λt 2
e M(t) + 4c d λt 2 λt
e M(t) − 2c λe M(t),
dt
so that, after some manipulations, equation (4.31) appears. While the first condition in (4.32) is
obvious, the second one can be inferred from (4.31) as follows
and also
i.e.,
d λt
e M(t)
dt
=
∞ d
λ
dt
n=0
n Un
=
∞
λ n ∞
Un−1 + 2c λ n
n=0
n=0
t
× cosh c(t − tn) sinh c(t − tn)
λe λt λt d
M(t) + e
dt M(t) = λeλtM(t) + 2c
0
∞
λ n
n=0
t
dt1 · · ·
t
× cosh c(t − tn) sinh c(t − tn),
d
dt M(t)
t=0
0
tn−1
t
dt1 · · ·
dtn
tn−1
n
cosh 2 c(tk − tk−1)
k=1
dtn
n
cosh 2 c(tk − tk−1)
k=1
= 0 (4.33)
since 2c cosh ct sinh ct| t=0 =0. By differentiating twice (4.28) and by taking into account (4.31), we
have that
λ 2 e λt λt d
M + 2λe
= e λt
dt
λ 2 M + λ d
dt M
+2c 2 e λt M + 2c 2
d2
M + eλt M 2
∞
λ n
n=0
t
0
dt
t
dt1 · · ·
tn−1
dtn
n
cosh 2 c(tk − tk−1) sinh 2 c(t − tn),
and therefore, by considering (4.33), we obtain the second condition of (4.32).
In order to solve the differential equation (4.31) we need to first solve the related third-order
algebraic equation
r 3 + 2λr 2 − (4c 2 − λ 2 )r − 2c 2 λ = 0
which can be reduced to the standard form by means of the change of variable
This leads to
k=1
s = r + 2λ
. (4.34)
3
s 3
2 λ
− s + 4c2 +
3 2λ
c
3
2 − λ2
32
= 0 (4.35)

4.3 About the higher moments of the hyperbolic distance 72
to which the well-known Cardano formula can be applied. In fact, for the third-order equation
the solution can be expressed as
s = 3
− q
2 +
s 3 + ps + q = 0, (4.36)
p 3
By comparing (4.35) and (4.36) it results
p 3
3
3 + q2
3
3 + q2
3
+
22 − q
2 −
p 3
3
3 + q2
c2
= −
22 33 3 2 2 2 2 2 2
(2 c + λ ) + λ (λ − 3c ) ,
− q
= −λ c
2 3
2 − λ2
32
.
. (4.37)
22 The simplest case is that of c = λ
3 for which the solutions of (4.35) are s1 = 0, s2 = √ 7c and
s3 = − √ 7c. After some calculations we get that
E{cosh 2 η(t)} = e−2ct
7
1 + 6 cosh √ 7ct + 2 √ 7 sinh √
7ct .
Following Lemma (4.3.1) we can prove a more general result:
Theorem 4.3.3. The functions
K m t
n (t) =
0
t
dt1 · · ·
tn−1
dtn
n+1
k=1
cosh m c(tk − tk−1),
with t0 = 0 and tn+1 = t, are solutions of difference-differential equations of order m + 1.
Proof
For m = 1 and m = 2 this statement has already been shown above since, in Theorem 4.2.2 and
Theorem 4.3.2, we have obtained that
and
We easily see that
d
dt Km n = K m n−1 + c m
and
d 2
dt 2 K1 n − d
dt K1 n−1 − c 2 K 1 n = 0,
d3 dt3 K2 n − d2
dt2 K2 2 d
n−1 − 4c
dt K2 n + 2c 2 K 2 n−1 = 0.
t
0
t
dt1 · · ·
tn−1
dtn
n
k=1
d 2
dt 2 Km n = d
dt Km n−1 + c 2 mK m n + c 2 m(m − 1)
× cosh m−2 c(t − tn) sinh 2 c(t − tn).
cosh m c(tk − tk−1) cosh m−1 c(t − tn) sinh c(t − tn),
t
0
t
dt1 · · ·
tn−1
dtn
n
cosh m c(tk − tk−1)
k=1

73 Travelling Randomly on H 2 with a Pythagorean Compass
In view of (4.38) it also results
d 3
dt 3 Km n = d2
dt 2 Km n−1 + c 2 m d
+c 3 m(m − 1)(m − 2)
dt Km n + 2c 2
d
(m − 1)
t
0
dt1 · · ·
t
tn−1
dt Km n − K m n−1
dtn
n
cosh m c(tk − tk−1)
× cosh m−3 c(t − tn) sinh 3 c(t − tn). (4.38)
After (m − 1) derivatives the following equation is obtained
dm−1 dtm−1 Km n = dm−2
dtm−2 Km n−1 + c 2 m dm−3
dtm−3 Km n + · · · + c m−1 m(m − 1) · · · (m − (m − 1) + 1)
t t
×
0
dt1 · · ·
and the next derivative gives
tn−1
dtn
n
cosh m c(tk − tk−1)
k=1
× cosh c(t − tn) sinh m−1 c(t − tn), (4.39)
dm dtm Km n = dm−1
dtm−1 Km n−1 + c 2 m dm−2
dtm−2 Km n + · · · + c m m(m − 1) · · · 2
t t n
× dt1 · · · dtn cosh
0
tn−1 k=1
m c(tk − tk−1) sinh m c(t − tn)
k=1
+c m m(m − 1) · · · 2 · (m − 1)
t t n
× dt1 · · · dtn cosh m c(tk − tk−1)
0
tn−1
k=1
× cosh 2 c(t − tn) sinh m−2 c(t − tn). (4.40)
The second integral of (4.40) can be expressed in terms of the derivatives of order (m − 2) and
lower.
By further differentiating equation (4.40) it turns out that, because of (4.39), the derivative of the
first integral in (4.40) can be expressed in terms of the derivatives of order (m − 1) and lower. The
theorem is thus proved.
Likewise Theorem 4.2.4, the following theorem holds:
Theorem 4.3.4. The function
Vn(t) =
t
0
dt1
t
t1
t
dt2 · · ·
tn−1
dtn
n+1
k=1
sinh 2 c(tk − tk−1)
with t0 = 0 and tn+1 = t, satisfies the third-order difference-differential equation
where t > 0, n ≥ 1, and V0(t) = sinh 2 ct.
Proof
d3 dt3 Vn
2 d
= 4c
dt Vn + 2c 2 Vn−1

4.4 Motions with jumps backwards to the starting point 74
We first note that
d
dt Vn = 2c
and therefore
and
t
0
t
dt1 · · ·
tn−1
d 2
dt 2 Vn = 2c 2 Vn + 2c 2
t
dtn
0
n
sinh 2 c(tk − tk−1) sinh c(t − tn) cosh c(t − tn) (4.41)
k=1
t
dt1 · · ·
tn−1
d3 dt3 Vn
2 d
= 2c
dt Vn + 2c 2 Vn−1 + 4c 3
dtn
t
0
n
sinh 2 c(tk − tk−1) cosh 2 c(t − tn),
k=1
t
dt1 · · ·
tn−1
dtn
n
sinh 2 c(tk − tk−1)
× sinh c(t − tn) cosh c(t − tn). (4.42)
Finally, by substituting (4.41) in (4.42), we obtain
d3 dt3 Vn
2 d
= 4c
dt Vn + 2c 2 Vn−1.
4.4 Motions with jumps backwards to the starting point
We here examine the planar motion dealt with so far assuming now that, at the instants of changes
of direction, the particle can return to the starting point and commence its motion from scratch.
The new motion and the original one are governed by the same Poisson process so that changes
of direction occur simultaneously in the original as well as in the new motion starting afresh from
the origin. This implies that the arcs of the original sample path and those of the new trajectories
have the same hyperbolic length. However, the angles formed by successive segments differ in order
to make the Hyperbolic Pythagorean Theorem applicable to the trajectories of the new motion.
In order to make our description clearer, we consider the case where, in the interval (0, t),
N(t) = n Poisson events (n ≥ 1) occur and we assume that the jump to the origin happens at the
first change of direction, i.e., at the instant t1. The instants of changes of direction for the new
motion are
t ′ k = tk+1 − t1
where k = 0, · · · , n with t ′ 0 = 0 and t ′ n = t − t1 and the hyperbolic lengths of the corresponding
arcs are
c(t ′ k − t ′ k−1) = c(tk+1 − tk).
Therefore, at the instant t, the hyperbolic distance from the origin of the particle performing the
k=1

75 Travelling Randomly on H 2 with a Pythagorean Compass
motion which has jumped back to O at time t1 is
n
k=1
cosh c(t ′ k − t ′ k−1) =
=
n
cosh c(tk+1 − tk)
k=1
n+1
k=2
cosh c(tk − tk−1) (4.43)
where 0 = t ′ 0 < t ′ 1 < · · · < t ′ n = t − t1 and tk+1 = t ′ k + t1. Formula (4.43) shows that the new
motion has an hyperbolic distance equal to that of the original motion where the first step has been
deleted. However, the distance between the position Pt and the origin O of the moving particle
which jumped back to O after having reached the position P1, is different from the distance of
Pt from P1 since the angle between successive steps must be readjusted in order to apply the
Hyperbolic Pythagorean Theorem.
If we denote by T1 the random instant of the return to the starting point (occurring at the first
Poisson event), we have that
E{cosh η1(t)I {N(t)≥1}|N(t) = n} = E{cosh η(t − T1)I {T1≤t}|N(t) = n}
By observing that
and that
=
=
t
0
t
0
E{cosh η(t − T1)I {T1∈dt1}|N(t) = n}dt1
E{cosh η(t − T1)|T1 = t1, N(t) = n}P{T1 ∈ dt1|N(t) = n}dt1.
E{cosh η(t − T1)|T1 = t1, N(t) = n} = E{cosh η(t − t1)|N(t) = n − 1}
=
(n − 1)!
(t − t1) n−1 In−1(t − t1),
with 0 < t1 < t, formula (4.44) becomes
P{T1 ∈ dt1|N(t) = n} = n!
tn (t − t1) n−1
(n − 1)! dt1
E{cosh η1(t)I {N(t)≥1}|N(t) = n} = n!
t n
t
0
In−1(t − t1)dt1. (4.44)
From (4.44) we have that the mean hyperbolic distance for the particle which returns to O at time
T1 has the form:
E{cosh η1(t)|N(t) ≥ 1} =
=
e −λt
P{N(t) ≥ 1}
λe −λt
P{N(t) ≥ 1}
∞
λ n
n=1
t
0
t
0
In−1(t − t1)dt1
e λ(t−t1) E(t − t1)dt1
We give here a general expression for the mean value of the hyperbolic distance of a particle
which returns to the origin for the last time at the k-th Poisson event Tk. We shall denote the
distance by the following equivalent notation η(t−Tk) = ηk(t) where the first expression underlines
that the particle starts from scratch at time Tk and then moves away for the remaining interval of

4.4 Motions with jumps backwards to the starting point 76
length t − Tk. In the general case we have the result stated in the next theorem:
Theorem 4.4.1. If N(t) ≥ k, then the mean value of the hyperbolic distance ηk is equal to
E{cosh ηk(t)|N(t) ≥ k} =
where E(t) is given by (4.9).
Proof
We start by observing that
E{cosh ηk(t)|N(t) ≥ k} =
=
=
=
λ k e −λt
P{N(t) ≥ k}
t
0
t
dt1 · · ·
tk−1
λke−λt t
e
P{N(t) ≥ k}(k − 1)! 0
λ(t−tk) k−1
tk ∞
E{cosh ηk(t)I {N(t)=n}|N(t) ≥ k}
n=k
n=k
e λ(t−tk) E(t − tk)dtk
E(t − tk)dtk,
∞
P{N(t) = n}
E{cosh ηk(t)I {N(t)≥k}|N(t) = n}
P{N(t) ≥ k}
∞
E{cosh ηk(t)I {N(t)≥k}|N(t) = n}P{N(t) = n|N(t) ≥ k}.
n=k
Since Tk = inf{t : N(t) = k}, the conditional mean value inside the sum can be developed as
follows
E{cosh ηk(t)I {N(t)≥k}|N(t) = n} = E{cosh η(t − Tk)I {Tk≤t}|N(t) = n}
In view of (4.3) we have that
=
=
t
0
t
0
E{cosh η(t − tk)I {Tk∈dtk}|N(t) = n}dtk
E{cosh η(t − tk)|Tk = tk, N(t) = n}P{Tk ∈ dtk|N(t) = n}dtk.
E{cosh η(t − Tk)|Tk = tk, N(t) = n} = E{cosh η(t − tk)|N(t − tk) = n − k}
=
(n − k)!
(t − tk) n−k In−k(t − tk),
and on the base of well-known properties of the Poisson process we have that
where 0 < tk < t. In conclusion we have that
P{Tk ∈ dtk|N(t) = n} = n!
tn (t − tk) n−k t
(n − k)!
k−1
k
(k − 1)! dtk
E{cosh ηk(t)I {N(t)≥k}|N(t) = n} = n!
t n
1
(k − 1)!
t
0
t k−1
k
In−k(t − tk)dtk

77 Travelling Randomly on H 2 with a Pythagorean Compass
and, from this and (4.45), it follows that
E{cosh ηk(t)|N(t) ≥ k} =
=
∞
n=k
n!
t n (k − 1)!
t
0
t k−1
k
λ k e −λt
P{N(t) ≥ k}(k − 1)!
e −λt (λt) n
In−k(t − tk)dtk
n!P{N(t) ≥ k}
t
e
0
λ(t−tk) k−1
tk Finally, in view of Cauchy formula of multiple integrals, we obtain that
P{N(t) ≥ k}(k − 1)!
=
λke−λt t
e
0
λ(t−tk) k−1
tk λke−λt t t
dt1 · · ·
P{N(t) ≥ k} 0
tk−1
E(t − tk)dtk
e λ(t−tk) E(t − tk)dtk.
E(t − tk)dtk.
Theorem 4.4.2. The mean of the hyperbolic distance of the moving particle returning to the origin
at the k-th change of direction is
where
E{cosh ηk(t)|N(t) ≥ k}
=
λ k e −λt
√ λ 2 + 4c 2 P{N(t) ≥ k}
e At
A
k−1 − eBt
k−1
1 1
+ −
Bk−1 Bi Ai
k−i−1 t
(k − i − 1)!
i=1
A = 1
2 (λ + λ 2 + 4c 2 ), B = 1
2 (λ − λ 2 + 4c 2 ). (4.45)
For k = 1, the sum in (4.45) is intended to be zero.
Proof
We can prove (4.45) by applying both formulas in (4.45). We start our proof by employing the
first one:
E{cosh ηk(t)|N(t) ≥ k} (4.46)
=
λ k e −λt
P{N(t) ≥ k}
t
0
dt1 · · ·
Therefore, in view of (4.15), formula (4.46) becomes
E{cosh ηk(t)|N(t) ≥ k}
=
λ k e −λt
P{N(t) ≥ k}
t
× e (t−tk ) √
2 λ2 +4c2 +
0
dt1 · · ·
t
tk−1
t
e λ(t−tk)
−λ + √ λ 2 + 4c 2
√ λ 2 + 4c 2
tk−1
e λ(t−tk) E(t − tk)dtk.
λ − e 2 (t−tk)
2
e − (t−tk ) √
2 λ2 +4c2 λ + √ λ2 + 4c2
√
λ2 + 4c2
dtk.

4.4 Motions with jumps backwards to the starting point 78
By introducing A and B as in (4.45), we can easily determine the k-fold integral
E{cosh ηk(t)|N(t) ≥ k}
=
=
=
λ k e −λt
√ λ 2 + 4c 2 P{N(t) ≥ k}
λ k e −λt
√ λ 2 + 4c 2 P{N(t) ≥ k}
t
0
λke−λt t
√
λ2 + 4c2P{N(t) ≥ k} 0
t
At the j-th stage the integral becomes
E{cosh ηk(t)|N(t) ≥ k} =
0
t
dt1 · · ·
dt1 · · ·
dt1 · · ·
tk−1
t
tk−2
t
tk−3
λ k e −λt
√ λ 2 + 4c 2 P{N(t) ≥ k}
i=1
At the k − 1-th stage the integral becomes
E{cosh ηk(t)|N(t) ≥ k} =
Ae A(t−tk) − Be B(t−tk)
dtk
e A(t−tk−1) − e B(t−tk−1)
dtk−1
e A(t−tk−2)
t
0
A
t
dt1 · · ·
j−1
1 1
+ −
Bi Ai
(t − tk−j) j−i−1
.
(j − i − 1)!
λ k e −λt
√ λ 2 + 4c 2 P{N(t) ≥ k}
i=1
t
− eB(t−tk−2)
B
tk−j−1
dt1
0
k−2
1 1
+ −
Bi Ai
(t − t1) k−i−2
.
(k − i − 2)!
+ 1
1
− dtk−2.
B A
e A(t−tk−j)
e A(t−t1)
A j−1
eB(t−t1)
−
Ak−2 Bk−2 eB(t−tk−j)
−
Bj−1 At the k-th integration we obtain formula (4.45). By means of the second formula in (4.45) and
by repeated integrations by parts we can obtain again result (4.45).
Remark 4.4.1. For k = 1 we have that
E{cosh η1(t)|N(t) ≥ 1} =
λ
√ λ 2 + 4c 2
sinh t
√
2 λ2 + 4c2 sinh λt
2
. (4.47)
It is clear that the mean value (4.47) tends to infinity as t → ∞. Furthermore, if λ, c → ∞
(so that c2
λ → 1) then E{cosh η1(t)I {N(t)≥1}} → e t . It can also be checked that if c = 0 then
E{cosh η1(t)I {N(t)≥1}} = 1, since the particle never leaves the starting point.
For k = 2 formula (4.45) yields
E{cosh η2(t)|N(t) ≥ 2} =
=
λ2 λt − e 2
c2
P{N(t) ≥ 2}
−
λ
√ λ 2 + 4c 2
cosh t
λ2 + 4c2 2
t
sinh λ2 + 4c2 −
− e
2
λt
2
λ 2
c 2 P{N(t) ≥ 2} [E{cosh η(t)} (4.48)
− P{N(t) ≥ 1}E{cosh η1(t)|N(t) ≥ 1} − e −λt

79 Travelling Randomly on H 2 with a Pythagorean Compass
Since
lim
c→0
1
c2
cosh t
2
we have, as expected, that
λ 2 + 4c 2 −
λ
√ λ 2 + 4c 2
sinh t
2
λ2 + 4c2 −
− e λt
2 = e λt
2
P{N(t) ≥ 2},
λ2 lim
c→0 E{cosh η2(t)|N(t) ≥ 2} = 1. (4.49)
Also, when λ → ∞, we obtain the same limit as in (4.49). The expression (4.48) suggests the
following decomposition
E{cosh η(t)} = c2
λ 2 P{N(t) ≥ 2}E{cosh η2(t)|N(t) ≥ 2}
+P{N(t) ≥ 1}E{cosh η1(t)|N(t) ≥ 1} + e −λt
= c2
λ 2 E{cosh η2(t)I {N(t)≥2}} + E{cosh η1(t)I {N(t)≥1}} + e −λt
Remark 4.4.2. The result in (4.45) appears as the mean hyperbolic distance of a motion starting
from the origin and running, without returns, until time t − Tk, where Tk has a truncated Gamma
distribution (Erlang distribution) with density
P{Tk ∈ dtk} =
In other words we can write (4.45) as
λ k e −λtk t k−1
k
(k − 1)!P{Tk ≤ t} dtk
0 < tk < t.
E{cosh ηk(t)|N(t) ≥ k} = E{E{cosh η(t − Tk)}} (4.50)
=
t
0
E{cosh η(t − Tk)}P{Tk ∈ dtk}.
Furthermore the expression (4.45) contains a fractional integral of order k for the function g(s) =
e λs E(s)
E{cosh ηk(t)|N(t) ≥ k} =
λ k
∞ (λt)
j=k
j
j!
If the mean value (4.50) is taken with respect to
then we have that
P{Tν ∈ ds} = λν e −λs s ν−1
E{E{cosh η(t − Tν)}} = λν e −λt
1
Γ(k)
t
(t − s)
0
k−1 e λs E(s)ds
ds 0 < s < t,
Γ(ν)P{Tν ≤ t}
P{Tν ≤ t}
1
Γ(ν)
t
(t − s)
0
ν−1 e λs E(s)ds
that also contains a fractional integral of order ν in the sense of Riemann-Liouville. The expression
(4.51) can be interpreted as the mean hyperbolic distance at time t where the particle can jump
back to the origin at an arbitrary instant (different from the instants of change of direction).
.

4.5 Motion at finite velocity on the surface of a three-dimensional sphere 80
4.5 Motion at finite velocity on the surface of a three-dimensional
sphere
Let P0 be a point on the equator of a three-dimensional sphere. Let us assume that the particle
starts moving from P0 along the equator in one of the two possible directions (clockwise or counterclockwise)
with velocity c.
At the first Poisson event (occurring at time T1) it starts moving on the meridian joining the
north pole PN with the position reached at time T1 (denoted by P1) along one of the two possible
directions (see Figure 4.3).
At the second Poisson event the particle is located at P2 and its distance from the starting point
P0 is the length of the hypothenuse of a right spherical triangle with cathetus P0P1 and P1P2; the
hypothenuse belongs to the equatorial circumference through P0 and P2.
Now the particle continues its motion (in one of the two possible directions) along the equatorial
circumference orthogonal to the hypothenuse through P0 and P2 until the third Poisson event
occurs.
In general, the distance d(P0Pt) of the point Pt from the origin P0 is the length of the shortest
arc of the equatorial circumference through P0 and Pt and therefore it takes values in the interval
[0, π]. Counter-clockwise motions cover the arcs in [−π, 0] so that the distance is also defined in
[0, π] or in [−π/2, π/2] with a shift that avoids negative values for the cosine.
P0
Pt
PN
Figure 4.3: Motion on the surface of a three-dimensional sphere.
By means of the spherical Pythagorean relationship we have that the Euclidean distance d(P0P2)
satisfies
cos d(P0P2) = cos d(P0P1) cos d(P1P2)
and, after three displacements,
cos d(P0P3) = cos d(P0P2) cos d(P2P3)
P2
P1
= cos d(P0P1) cos d(P1P2) cos d(P2P3).
After n displacements the position Pt on the sphere at time t is given by
cos d(P0Pt) =
n
cos d(PkPk−1) cos d(PnPt).
k=1

81 Travelling Randomly on H 2 with a Pythagorean Compass
Since d(PkPk−1) is represented by the amplitude of the arc run in the interval (tk, tk−1), it results
d(PkPk−1) = c(tk − tk−1).
The mean value E{cos d(P0Pt)|N(t) = n} is given by
where t0 = 0, tn+1 = t, and
En(t) = E{cos d(P0Pt)|N(t) = n}
= n!
t n
t
0
dt1
= n!
Hn(t),
tn Hn(t) =
t
0
t
t1
dt2 · · ·
t
dt1 · · ·
The mean value E{cos d(P0Pt)} is given by
tn−1
t
dtn
tn−1
n+1
k=1
dtn
n+1
k=1
cos c(tk − tk−1)
cos c(tk − tk−1).
E(t) = E{cos d(P0Pt)}
∞
= E{cos d(P0Pt)|N(t) = n}P{N(t) = n}
n=0
= e −λt
∞
λ n Hn(t).
n=0
By steps similar to those of the hyperbolic case we have that Hn(t), t ≥ 0, satisfies the differencedifferential
equation
d2 dt2 Hn = d
dt Hn−1 − c 2 Hn,
where H0(t) = cos ct, and therefore we can prove the following:
Theorem 4.5.1. The mean value E(t) = E{cos d(P0Pt)} satisfies
with initial conditions
and has the form
Proof
⎧
⎪⎨
E(t) =
⎪⎩
λt − e 2 cosh t
λt − e 2
λt − e 2
√
2 λ2 − 4c2 +
λt
1 + 2
√
4c2 − λ2 +
cos t
2
d2 d
E = −λ
dt2 dt E − c2E (4.51)
E(0) = 1,
d
dt E(t) t=0 = 0,
λ
√ λ 2 −4c 2
λ
√ 4c 2 −λ 2
√
t sinh 2 λ2 − 4c2 √
t sin 2 4c2 − λ2 0 < 2c < λ,
λ = 2c > 0,
2c > λ > 0.
(4.52)
(4.53)

4.5 Motion at finite velocity on the surface of a three-dimensional sphere 82
The solution to the problem (4.51)-(4.52) is given by
E(t) =
so that (4.53) emerges.
e− λt
2
+
2
e t
√
2 λ2−4c2 λ
√ λ 2 − 4c 2
+ e
e t
√
2 λ2−4c2 √
t − 2 λ2−4c2
− e
t − 2
√ λ 2 −4c 2
,
For large values of λ the first expression furnishes E(t) ∼ 1 and therefore the particle hardly
leaves the starting point.
If λ
2 < c, the mean value exhibits an oscillating behavior; in particular, the oscillations decrease as
time goes on, and this means that the particle moves further and further reaching in the limit the
poles of the sphere.
Remark 4.5.1. By assuming that c is replaced by ic in (4.53) we formally extract from the first
and the third expression in (4.53) the hyperbolic mean distance (4.9). This is because the space
H 2 can be regarded as a sphere with imaginary radius. Clearly the intermediate case λ = 2c has
no correspondence for the motion on H 2 because the Poisson rate must be a real positive number.

Chapter 5
Cascades of Particles Moving at
Finite Velocity in H 2
5.1 Description of the randomly moving and branching model
We assume that a unit-mass particle is placed at time t = 0 at the origin O of H 2 and starts
moving on the main geodesic line represented in H 2 by the half-circle of radius 1 passing through
O. This particle chooses with probability 1/2 one of the two possible directions and moves with
constant hyperbolic velocity equal to c (see Figure 5.1 (a)). The hyperbolic velocity
c = ds
dt
dx
1
=
y dt
2
+
2 dy
dt
is assumed to be constant. For an Euclidean observer, the closer to the x-axis is the moving particle
the slower it moves.
A Poisson process of rate λ governs the changes of direction. At the first Poisson event the
particle splits into two pieces of equal mass: one continues its motion on the same geodesic line
while the other one starts moving (in one of the two possible directions) on the geodesic line
orthogonal to the previous one (see Figure 5.1 (b)). In general, at the k-th Poisson event, the
deviating particle of mass 1/2 k undergoes a further decomposition: one splinter of mass 1/2 k+1
continues undisturbed its motion, while the other one, also of mass 1/2 k+1 , is forced to move onto
the geodesic line orthogonal to that joining O with the position it occupied at the time where the
splitting took place.
Therefore, if no Poisson event occurs (i.e., {N(t) = 0}, where N(t) is the number of Poisson
events in [0, t]), the unit-mass particle is located, at time t, on the first geodesic line at an hyperbolic
distance from O equal to η0(t) = ct (see Figure 5.1 (a)).
If one Poisson event happens, at time S1 < t (i.e., {N(t) = 1}), then, at time t, one fragment
of mass 1/2 will be at distance η0(t) = ct on the first geodesic line, while the other splinter, also
of mass 1/2, will be located at hyperbolic distance η1(t) given by
cosh η1(t) = cosh c S1 cosh c(t − S1), (5.1)

5.1 Description of the randomly moving and branching model 84
1
1
1 2 3
●
O
●
0
(a) N(t)=0
1 2 3
●
O
●
0
(d) N(t)=3
● 1
●
1 2
●
1 2 2
1
1
● 1 2 4
1 2 4
●
O
●
0
(b) N(t)=1
1 2 3
●
O
●
0
(e) N(t)=4
●
●
1 2
1 2
●
1 2
●
1 2 2
1
1
4
∏ coshc(Sj − Sj−1)
j=1
5
∏
j=1
coshc(S j − Sj−1)
●
1 2 2
●
O
●
0
(c) N(t)=2
3
∏ coshc(Sj − Sj−1)
j=1
●
●
0
(f) N(t)=4
●
●
1 2
coshct
●
1 2 2
coshcS 1coshc(t − S1)
Figure 5.1: In (a), the trajectory of the unit-mass particle initially placed at the origin O of H 2
and moving on the main geodesic line is shown. When N(t) = 0, no disintegration occurs. In (b),
(c), (d), and (e), the trajectories of the particles generated by N(t) = 1, 2, 3, 4 Poisson events are
plotted. The relevant mass associated with each particle is also indicated by a suitable label. In
(f), for each particle the relevant hyperbolic cosine of the hyperbolic distance from the origin is
also reported.
where in formula (5.1) the Hyperbolic Pythagorean Theorem 2.5.3 has been applied (see Figure
5.1 (b)).
If N(t) = n, the process described above produces n + 1 splinters. The particle generated at
the k-th Poisson event, with k = 0, · · · n − 1, (and which will not break up after the k-th event)
has mass 1/2 k+1 while the last splinter, which has changed direction at all fission events, has mass
1/2 n . For every k = 0, · · · n, the particle generated at the k-th Poisson event is located, at time t,
at the hyperbolic distance ηk(t) from O. Such a distance, in force of the Hyperbolic Pythagorean
Theorem, reads
cosh ηk(t) =
k+1
j=1
cosh c(Sj − Sj−1),
where S0 = 0, Sk+1 = t and the random times Sj with j = 1, · · · k represent the instants where
the Poisson events happen and the deviations of motion occur. (see Figure 5.1 (f)).
In general, if the number of splits recorded is N(t), the number of particles is N(t) + 1 and each
runs on a different geodesic line of H 2 . The hyperbolic distance of the center of mass cm of the
O
●
●

85 Cascades of Particles Moving at Finite Velocity in H 2
cloud of moving particles, at time t > 0, is denoted by ηcm(t) and is represented by
cosh ηcm(t) =
N(t)−1
k=0
1
2k+1 k+1
j=1
cosh c(Sj − Sj−1)I {N(t)>0} + 1
2N(t) N(t)+1
j=1
cosh c(Sj − Sj−1), (5.2)
where S1, S2 . . . S N(t) are the random times at which the disintegrations occur, and S0 = 0,
S N(t)+1 = t. The second term in (5.2) refers to the splinter which underwent all disintegrations
occurred until time t, while the first one is related to those particles produced during the
branching process.
The assumption that the particles deviate on geodesic lines orthogonal to those joining the
origin O with their current position is crucial since it makes the Hyperbolic Pythagorean Theorem
applicable. Otherwise it should be necessary to apply the Carnot hyperbolic formula and this
would make the analytic treatment of the problem extremely difficult.
Under the condition that N(t) = n, the hyperbolic distance of the center of mass of the cloud
of particles at time t is the random variable
n−1
k=0
1
2k+1 k+1
j=1
cosh c(Sj − Sj−1)I {n>0} + 1
2n n+1
cosh c(Sj − Sj−1). (5.3)
We observe that the n instants S1, · · · , Sn where the fissions take place are uniformly distributed
under the condition that N(t) = n, and possess density
j=1
P{S1 ∈ ds1, · · · , Sn ∈ dsn} = n!
t n ds1 · · · dsn
for 0 = s0 < s1 < · · · < sn+1 = t. Therefore, under the condition that the number of splitting
events is N(t) = n, the hyperbolic distance ηk(t) of the k-th splinter is for k = 0, · · · n − 1
E{cosh ηk(t)|N(t) = n} = n!
t n
and for k = n
= n!
t n
t
0
t
E{cosh ηn(t)|N(t) = n} = n!
t n
0
t
ds1 · · ·
sk−1
t
ds1 · · ·
sk−1
t
dsk · · ·
sn−1
dsn
k+1
j=1
(t − sk)
dsk
n−k k+1
(n − k)!
j=1
cosh c(sj − sj−1)
cosh c(sj − sj−1)
= n!
t n Gn,k(t), (5.4)
t
0
t
ds1 · · ·
sn−1
dsn
n+1
j=1
cosh c(sj − sj−1)
= n!
t n Gn,n(t). (5.5)
The branching process described above can be adapted to the Poincaré disk B: since the mapping
(2.30) preserves the hyperbolic distance, the trajectories of splitting and moving particles can be
conveniently depicted in B as well as H 2 (see Figure 5.1 and 5.2).
We restrict ourselves to the mean hyperbolic distance because this leads to fine explicit results.
The analysis of the distribution of the hyperbolic distance, even in the case of a random motion of

5.2 Mean hyperbolic distance: the Laplace transform approach 86
an individual particle that changes direction at all Poisson events, implies a much more complicated
analysis and is almost intractable since multiple integrals of the form
must be evaluated.
0
0
E{e iα cosh η(t) |N(t) = n} = n!
t n
●
O
0
(a) N(t)=0
● 1 2 3
●
1 2 3
●
O
0
(d) N(t)=3
●
1 2 2
1
●
1 2
●
0
0
t
0
t
ds1 · · ·
●
O
0
(b) N(t)=1
1 2
●
3
1 2 3
●
●
1 2 4
●
O
0
(e) N(t)=4
sn−1
●
●
1 2
1 2 2
e iα Q n+1
j=1 cosh c(sj−sj−1) dsn
1 2
●
1 2
●
0
0
1 2 2
●
●
O
0
(c) N(t)=2
3
∏ coshc(Sj − Sj−1) 4
j=1
∏ coshc(Sj − Sj−1) ●
j=1 5
●
∏ coshc(Sj − Sj−1) j=1
0
(f) N(t)=4
●
1 2 2
1 2
●
●
coshcS1coshc(t − S1) Figure 5.2: Same trajectories as in Figure 5.1 represented through the Poincaré disk model.
5.2 Mean hyperbolic distance: the Laplace transform ap-
proach
We are able to obtain the explicit form of the mean-value of (5.2) by means of two different and
independent approaches. In this section we present the Laplace-transform derivation which leads
to our first theorem.
Theorem 5.2.1. The mean-value of the hyperbolic distance (5.2) of the center of mass is
E{cosh ηcm(t)} = 23c2
3 −
e 22 λt
√
λ2 + 24c2 t −
e 22 √
λ2 +24c2 3 √ λ2 + 24c2 + 5λ +
e t
22 √
λ2 +24c2 3 √ λ 2 + 2 4 c 2 − 5λ
+ λ + 2c
2(λ + 3c) ect λ − 2c
+
2(λ − 3c) e−ct , t > 0.
●
O
coshct
●
●
(5.6)

87 Cascades of Particles Moving at Finite Velocity in H 2
Proof
In view of (5.3), (5.4) and (5.5), we first note that
E{cosh ηcm(t)|N(t) = n} = n!
tn n−1
k=0
=
+ n!
tn 1
2n ⎧
⎪⎨
⎪⎩
1
2 k+1
t
0
n!
tn n−1 k=0
t
0
t
ds1 · · ·
t
ds1 · · ·
Our task is therefore to study the Laplace transform,
∞
0
e −µt E{cosh ηcm(t)}dt =
=
sn−1
sn−1
dsn
1
2k+1 Gn,k(t) + n!
tn dsn
n+1
j=1
k+1
j=1
cosh c(sj − sj−1)
cosh c(sj − sj−1)
1
2 n Gn,n(t), n ≥ 1,
G0,0(t), n = 0.
∞
0
∞
e −µt
λ n
n−1
∞
E{cosh ηcm(t)|N(t) = n}P{N(t) = n}dt
n=0
1
2k+1 n=1 k=0
∞ λ
+
n=0
n
2n ∞
0
∞
0
e −(λ+µ)t Gn,k(t)dt
e −(λ+µ)t Gn,n(t)dt, (5.7)
where µ > 0. We evaluate the Laplace transform appearing in (5.7) in the following way. If
γ = λ + µ and c < γ, then we have, by successively inverting the inner integrals, that
∞
e −γt Gn,k(t)dt
∞
=
0
e −γt
⎧
⎨ t
⎩ 0
∞ ∞
= ds1 e −γt ⎧
⎨
dt
⎩
0
=
=
=
0
∞
0
∞
ds1
0
∞
×
0
∞
0
s1
∞
s1
∞
ds1 · · ·
t
ds1 · · ·
t
sk−1
s1
∞
ds2 · · ·
sk−1
k−1
dsk−1
sk−2 j=1
−γw wn−k
e
⎫
n−k k+1
(t − sk)
⎬
dsk
cosh c(sj − sj−1)
(n − k)!
⎭
j=1
dt
⎫
t
n−k k+1
(t − sk)
⎬
ds2 · · · dsk
cosh c(sj − sj−1)
sk−1 (n − k)!
⎭
j=1
k
∞
n−k
−γt (t − sk)
dsk cosh c(sj − sj−1) e cosh c(t − sk)dt
(n − k)!
j=1
cosh c(sj − sj−1)
∞
sk−1
cosh cw dw
(n − k)!
e −γw k ∞
−γw wn−k
cosh cw e cosh cw dw.
(n − k)!
0
sk
e −γsk cosh c(sk − sk−1)dsk
In the last step above the change of variable sj − sj−1 = wj applied k times leads to the final
expression. For k = n, from the previous calculations, we obtain
∞
0
e −γt Gn,n(t)dt =
∞
0
e −γw n+1 cosh cw dw .

5.2 Mean hyperbolic distance: the Laplace transform approach 88
Since
we have that
∞
0
∞
0
∞
e
0
−γw γ
cosh cw dw =
γ2 ,
− c2
−γw wn−k
1 1
1
e cosh cw dw = +
(n − k)! 2 (γ − c) n−k+1 (γ + c) n−k+1
,
e −µt E{cosh ηcm(t)}dt
∞
= λ
n=1
n
n−1 1
2
k=0
k+1
γ
γ2 − c2 k
1 1
1
+
2 (γ − c) n−k+1 (γ + c) n−k+1
∞
n
λ γ
+
2 γ
n=0
2 − c2 n+1 = 1
22 ∞
λ n
1
(γ − c) n+1
n−1
k γ
1
+
2(γ + c) (γ + c) n+1
n−1
k
γ
2(γ − c)
n=1
k=0
2γ
+
2γ2 − 2c2 , (5.8)
− γλ
where the last sum converges if µ satisfies the inequality 2c 2 < λ 2 + 2µ 2 + 3λµ. The double sum
in (5.8) can be calculated by inverting the order of summation in the following way:
∞
λ
n=1
n
1
(γ − c) n+1
n−1
k γ
1
+
2(γ + c) (γ + c)
k=0
n+1
n−1
k
γ
2(γ − c)
k=0
∞
k ∞ n 1 γ
λ
=
+
γ − c 2(γ + c) γ − c
k=0
n=k+1
1
∞
k ∞ n γ
λ
γ + c 2(γ − c) γ + c
k=0
n=k+1
∞
k k+1 ∞ r 1 γ λ
λ
=
γ − c 2(γ + c) γ − c γ − c
k=0
r=0
+ 1
∞
k k+1 ∞ r γ λ
λ
γ + c 2(γ − c) γ + c γ + c
k=0
r=0
λ
=
(γ − c) 2
∞
γλ
2(γ
k=0
2 − c2 k γ − c
) γ − c − λ +
λ
(γ + c) 2
∞
γλ
2(γ
k=0
2 − c2 k γ + c
) γ + c − λ
1 1 1 1 2(γ
= λ
+
γ − c γ − c − λ γ + c γ + c − λ
2 − c2 )
2γ2 − 2c2 . (5.9)
− γλ
The inversion of the Laplace transform is made possible by suitably rearranging the expression
(5.9) as follows:
∞
e
0
−µt E{cosh ηcm(t)}dt
= λ
1 1 1 1 γ
+
2 γ − c γ − c − λ γ + c γ + c − λ
2 − c2 2γ2 − 2c2 − γλ +
2γ
2γ2 − 2c2 − γλ
= λ
1 1
2 λ + µ − c µ − c +
1 1 (λ + µ)
λ + µ + c µ + c
2 − c2 λ2 + 2µ 2 2λ + 2µ
+
+ 3λµ − 2c2 λ2 + 2µ 2 + 3λµ − 2c2 = λ
λ + µ + c λ + µ − c
1
+
2 µ − c µ + c λ2 + 2µ 2 2λ + 2µ
+
+ 3λµ − 2c2 λ2 + 2µ 2 . (5.10)
+ 3λµ − 2c2 k=0

89 Cascades of Particles Moving at Finite Velocity in H 2
By means of the decomposition
2µ 2 + 3λµ + λ 2 − 2c 2 = 2
µ + 3λ − √ λ 2 + 2 4 c 2
2 2
the expression (5.10) can be further worked out by writing
∞
0
e −µt E{cosh ηcm(t)}dt
= λ
2 2
=
+
+
+
2
µ + 3λ−√ λ 2 +2 4 c 2
2 2
3
2
µ + 3λ−√ λ 2 +2 4 c 2
2 2
1
µ + 3λ−√λ2 +24c2 22
1
1
λ λ + 2 + 2µ
−
µ + 3λ−√ λ 2 +2 4 c 2
2 2
1
µ + 3λ−√ λ 2 +2 4 c 2
2 2
µ + 3λ+√ λ 2 +2 4 c 2
2 2
µ + 3λ+√λ2 +24c2 22 1
µ + 3λ+√ λ 2 +2 4 c 2
2 2
−
+
1
µ + 3λ+√ λ 2 +2 4 c 2
2 2
1
µ + 3λ+√ λ 2 +2 4 c 2
2 2
λ + 2c λ − 2c
2 + +
µ − c µ + c
λ 1
√ +
λ2 + 24c2 2
λ
2 √ λ2 + 24c2
λ + 2c
µ − c
λ
22 (2µ + 3
2λ). µ + 3λ + √ λ 2 + 2 4 c 2
2 2
1
µ + 3λ−√λ2 +24c2 22 + λ − 2c
µ + c
+
,
1
µ + 3λ+√ λ 2 +2 4 c 2
2 2
It is now a simple matter to invert the Laplace transform (5.11) and we arrive at the mean
hyperbolic distance of the center of mass in an integral form
E{cosh ηcm(t)}
=
1
2 +
λ
√ λ 2 + 2 4 c 2
+ λ(λ + 2c)
2 √ λ2 + 24c2 + λ(λ − 2c)
2 √ λ2 + 24c2 + λ
2 3
t
0
t
√
3λ− λ2 +24c2 −t
e 22
1
+
2 −
e cs √
3λ− λ2 +24c2 −(t−s)
e 22 ds −
√
λ
3λ+ λ2 +24c2 −t
√ e 2
λ2 + 24c2 2
t
0
0
t
e
0
−cs √
3λ− λ2 +24c2 −(t−s)
e 22 t
ds −
0
3 −
e 22 λs √
3λ− λ2 +24c2 −(t−s)
e 22 t
ds +
0
e cs √
3λ+ λ2 +24c2 −(t−s)
e 22
ds
e −cs √
3λ+ λ2 +24c2 −(t−s)
e 22
ds
3 −
e 22 λs 3λ+
−(t−s)
e
√
λ2 +24c2 22 ds
. (5.12)
The expression (5.12) can be further developed and simplified by observing that, after some simple
calculations, we have that
λ(λ + 2c)
2 √ λ 2 + 2 4 c 2
=
t
0
e cs √
3λ− λ2 +24c2 −(t−s)
e 22 ds −
t
λ(λ + 2c)
2 √ λ2 + 24c2 22 22c + 3λ − √ λ2 + 24c2
e ct 3λ−
−t
− e
λ(λ + 2c)
−
2 √ λ2 + 24c2 22 22c + 3λ + √ λ2 + 24c2
e ct 3λ+
−t
− e
0
e cs √
3λ+ λ2 +24c2 −(t−s)
e 22
ds
√ λ 2 +2 4 c 2
2 2
√ λ 2 +2 4 c 2
2 2
. (5.13)
(5.11)

5.2 Mean hyperbolic distance: the Laplace transform approach 90
Similar manipulations yield
and also
λ(λ − 2c)
2 √ λ 2 + 2 4 c 2
=
t
0
e −cs √
3λ− λ2 +24c2 −(t−s)
e 22 ds −
t
λ(λ − 2c)
2 √ λ2 + 24c2 22 −22c + 3λ − √ λ2 + 24c2
e −ct 3λ−
−t
− e
λ(λ − 2c)
−
2 √ λ2 + 24c2 22 −22c + 3λ + √ λ2 + 24c2
e −ct 3λ+
−t
− e
λ
23 t
3 −
e 2
0
2 λs √
3λ− λ2 +24c2 −(t−s)
e 22 ds +
=
λ
2 √ λ 2 + 2 4 c
3
e− 22 λt
2
e t
0
t
0
√ λ 2 +2 4 c 2
2 2 − e −t
e −cs √
3λ+ λ2 +24c2 −(t−s)
e 22
ds
√ λ 2 +2 4 c 2
2 2
√ λ 2 +2 4 c 2
2 2
, (5.14)
3 −
e 22 λs √
3λ+ λ2 +24c2 −(t−s)
e 22
ds
√ λ 2 +2 4 c 2
2 2
. (5.15)
By inserting results (5.15), (5.13), and (5.14) into (5.12) we now obtain the final formula
E{cosh ηcm(t)}
= e− 3
22 λt
3λ
1 + √ e
2
λ2 + 24c2 t
22
√
λ2 +24c2 3λ
t −
+ 1 − √ e 2
λ2 + 24c2 2
√
λ2 +24c2 + λ + 2c
2(λ + 3c) ect √
2λ
3λ− λ2 +24c2 + √ e−t 2
λ2 + 24c2 2
λ − 2c
22c − 3λ + √ λ2 + 24 λ + 2c
−
c2 22c + 3λ − √ λ2 + 24c2
+ λ − 2c
2(λ − 3c) e−ct √
2λ
3λ+ λ2 +24c2 + √ e−t 2
λ2 + 24c2 2
λ + 2c
22c + 3λ + √ λ2 + 24 λ − 2c
−
c2 22c − 3λ − √ λ2 + 24c2
3 −
= e 22 λt e t
22
√
λ2 +24c2 1
2 +
3λ
2 √ λ2 + 24
2 3λ
− √
c2 λ2 + 24c2 2 − λ √ λ2 + 24c2 − 23c2 5λ − 3 √ λ2 + 24c2
3 −
+e 22 λt t −
e 22
√
λ2 +24c2 1
2 −
3λ
2 √ λ2 + 24
2 3λ
+ √
c2 λ2 + 24c2 2 + λ √ λ2 + 24c2 − 23c2 5λ + 3 √ λ2 + 24c2
+ λ + 2c
2(λ + 3c) ect λ − 2c
+
2(λ − 3c) e−ct
= 23c2 3
t
−
e 22 λt −
e 2
√
λ2 + 24c2 2
√
λ2 +24c2 5λ + 3 √ λ2 + 24 e
−
c2 t
22 √
λ2 +24c2 5λ − 3 √ λ2 + 24c2
+ λ + 2c
2(λ + 3c) ect λ − 2c
+
2(λ − 3c) e−ct .
Remark 5.2.1. Apparently a critical point in formula (5.6) is λ = 3c. We show that the mean
hyperbolic distance (5.6) is finite for λ = 3c. We first write ε = λ−3c and r(ε) = √ ε 2 + 6cε + 5 2 c 2
and evaluate the limit
lim
λ→3c
3 −
22 λt
e t
22 √
λ2 +24c2 λ − 2c
2(λ − 3c) e−ct − 23c2e √
λ2 + 24c2 5λ − 3 √ λ2 + 24c2 ε + c
= lim
ε→0 2ε e−ct −
23c2 t
e− 2
r(15c + 5ε − 3r) 2 (9c+3ε−r) , (5.16)
which refers to the components of (5.6) with diverging coefficients. We expand the second expo-

91 Cascades of Particles Moving at Finite Velocity in H 2
1.04
1.03
1.02
1.01
* * * * * * * * * * * * * * * * * * * * * * * * *
1.00
*
*
*
*
*
*
*
*
*
*
λ = 0.1c
λ = 4c
λ = 1c
λ = 5c
λ = 2c
λ = 6c
* λ = 3c
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
0.10 0.15 0.20 0.25 0.30
(a)
*
*
*
*
1.036
1.034
1.032
1.030
1.028
*
*
1.026
*
*
*
*
0.250 0.260 0.270
Figure 5.3: In (a) and (b) the mean-value E{cosh ηcm(t)} is plotted for c = 1 and different values
of λ, the curve corresponding to λ = 3c is obtained by plotting the limit value E{cosh ηcm(t)} =
e−ct 53
2252 + 3
10ct ct 5 + e 22 7ct
3 + e− 2
22
523 .
nential in (5.16) as
t −
e 22 (9c+3ε−r) = e −ct t −
e 22 (5c+3ε−r) = e −ct
1 − t
t2
(5c + 3ε − r) +
22 25 (5c + 3ε − r)2 + o(ε 3
) ,
where we have taken into account that r(ε) → 5c for ε → 0. It is now convenient to write (5.16)
in the form
e −ct
ε + c
lim
ε→0 2ε −
23c2 r(15c + 5ε − 3r) + 2tc2 (5c + 3ε − r)
r(15c + 5ε − 3r) − t2c2 (5c + 3ε − r) 2
22r(15c + 5ε − 3r) + o(ε2
) .
By considering that r(ε) ∼ 5c + 3
5ε for ε → 0, we can easily realize that
ε + c
lim
ε→0 2ε −
23c2
=
r(15c + 5ε − 3r)
2 · 7
2tc
, lim
52 ε→0
2 (5c + 3ε − r) 3 t
= ct, lim
r(15c + 5ε − 3r) 2 · 5 ε→0
2c2 (5c + 3ε − r) 2
22 = 0.
r(15c + 5ε − 3r)
This suffices to show that near λ = 3c the mean hyperbolic distance is finite. Considering also the
two additional terms of (5.6) leads to the following asymptotic estimate of the mean hyperbolic
distance of the center of mass for large values of t and near λ ∼ 3c. In particular, we have that
E{cosh ηcm(t)} ∼ 5
12 ect .
In Figure 5.3, the function E{cosh ηcm(t)} is plotted for c = 1 and different values of λ in the
neighborhood of λ = 3c (including the limiting case λ = 3c).
Remark 5.2.2. We now show that the center of mass (as well as each individual particle) goes
further and further away from the origin O of H 2 (or equivalently it migrates towards the frontier
of the Poincaré disc B) as time t increases. In order to show this result we work on (5.12) and,
(b)
*
*
*
*

5.3 Equation governing the hyperbolic distance 92
after some manipulations, we obtain that
d
dt E{cosh ηcm(t)} = 22c2 3 −
e 22 λt
t
√ sinh
λ2 + 24c2 22
λ2 + 24c2 2cλ
+
2
√
λ2 + 24c2 + 22 c 2 λ
√ λ 2 + 2 4 c 2
t
0
t
0
3 −
e 22 λs sinh c(t − s) sinh s
22 3 −
e 22 λs cosh c(t − s) sinh s
22 λ 2 + 2 4 c 2 ds
λ 2 + 2 4 c 2 ds. (5.17)
The first term in the right hand side of (5.17) is produced by the derivatives of all terms of (5.12)
while the integrals stem from the third and fourth term only. From (5.17) it is easy to check that
d 2
dt 2 E{cosh ηcm(t)} > 0 for all t so that the center of mass gets off from the starting point with
positive acceleration.
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Figure 5.4: In (a), (b), and (c) the trajectories and the positions of the splinters at time t = 40,
when c = 0.05 and N(t) = 20, are shown. In (d), (e) and (f) only the positions of the splinters at
time t = 50, when c = 0.05 and N(t) = 300, are drawn. In (a) and (d) each splinter chooses the
clockwise direction, in (b) and (e) each splinter chooses the counterclockwise direction, in (c) and
(f) the clockwise and counterclockwise directions are alternatively chosen.
5.3 Equation governing the hyperbolic distance
We are able to confirm result (5.6) by a completely different method based on a system of differential
equations governing all the quantities appearing in the mean hyperbolic distance of the center of

93 Cascades of Particles Moving at Finite Velocity in H 2
mass. In view of formulas (5.4) and (5.5) we can write the mean hyperbolic distance of the center
of mass in the following form
E{cosh ηcm(t)} = e −λt
∞
n=1
λ n
n−1
k=0
1
2 k+1 Gn,k(t) + e −λt
∞
n=0
λ n
2 n Gn,n(t). (5.18)
We start by deriving the difference-differential equations governing the functions Gn,k(t), with
t > 0 and 0 ≤ k ≤ n. As it has been pointed out in the previous chapter, the j-th moment of
the conditional hyperbolic distance of a single particle leads to a difference-differential equation of
order equal to j +1. The same type of phenomenon occurs in the present case where the hyperbolic
distance of a cloud of points is envisaged.
Lemma 5.3.1. The functions represented by the following multiple integrals
Gn,k(t) =
t
0
t
ds1 · · ·
sk−1
dsk
are solutions to the difference-differential equations
⎧
⎪⎨
⎪⎩
d 2
n−k
(t − sk) k
cosh c(sj − sj−1) cosh c(t − sk), 0 ≤ k ≤ n,
(n − k)!
j=1
dt2 Gn,k = 2 d
dtGn−1,k − Gn−2,k + c2Gn,k, k ≤ n − 2,
d 2
dt2 Gn,n−1 = 2 d
dtGn−1,n−1 − Gn−2,n−2 + c2Gn,n−1, k = n − 1,
d 2
dt 2 Gn,n = d
dt Gn−1,n−1 + c 2 Gn,n, k = n.
Proof
We start by considering the case k ≤ n − 2 where the first derivative reads
d
dt Gn,k = Gn−1,k + c
t
0
t
ds1 · · ·
sk−1
and, from (5.20), the second derivative becomes
d 2
dt 2 Gn,k = d
dt Gn−1,k + c 2 Gn,k
+c
t
0
ds1 · · ·
t
sk−1
(t − sk)
dsk
n−k
(n − k)!
(t − sk)
dsk
n−k−1
(n − k − 1)!
k
j=1
(5.19)
cosh c(sj − sj−1) sinh c(t − sk), (5.20)
k
cosh c(sj − sj−1) sinh c(t − sk)
= 2 d
dt Gn−1,k − Gn−2,k + c 2 Gn,k. (5.21)
The expression (5.20), for k = n − 1, is qualitatively different and must be handled carefully. In
this case we have that
d
dt Gn,n−1 = Gn−1,n−1
t
+c
0
ds1 · · ·
t
sn−2
j=1
n−1
dsn−1(t − sn−1)
j=1
cosh c(sj − sj−1) sinh c(t − sn−1), (5.22)

5.3 Equation governing the hyperbolic distance 94
and thus, from (5.22),
d 2
dt 2 Gn,n−1 = d
dt Gn−1,n−1 + c 2 Gn,n−1
+c
t
0
ds1 · · ·
= d
dt Gn−1,n−1 +
t
n−1
dsn−1
sn−2 j=1
d
dt Gn−1,n−1 − Gn−2,n−2
= 2 d
dt Gn−1,n−1 − Gn−2,n−2 + c 2 Gn,n−1.
cosh c(sj − sj−1) sinh c(t − sn−1)
+ c 2 Gn,n−1
We omit the derivation of the last formula in (5.19) because it coincides with the result of Lemma
4.2.1.
The results of Lemma 5.3.1 permit us to obtain the equation governing the mean-value of
the hyperbolic distance of the center of mass of the randomly moving splinters produced by the
disintegration process.
Theorem 5.3.2. The mean-value E{cosh ηcm(t)} in (5.2) is solution to the non-homogeneus
second-order linear equation
3 −
22 λt
d2 dt2 u − c2u = λc2e √
λ2 + 24c2 Proof
We begin by successively deriving the expression (5.18) as follows
and
d
u = −λu + e−λt
dt
∞
n=1
d2 d
u = −2λ
dt2 dt u − λ2u + e −λt
λ n
n−1
k=0
∞
n=1
t −
e 22 √
λ2 +24c2 − e t
22 √
λ2 +24c2
. (5.23)
1
2 k+1
λ n
n−1
k=0
d
dt Gn,k + e −λt
1
2 k+1
∞
n=0
λ n
2 n
d 2
dt 2 Gn,k + e −λt
d
dt Gn,n,
∞
n=0
λn 2n d2 dt2 Gn,n. (5.24)
We must now insert all the expressions of Lemma 5.3.1 into (5.24) by accurately taking into account

95 Cascades of Particles Moving at Finite Velocity in H 2
the constrains on n and k. It is convenient to write the last two terms of (5.24) as follows
∞
n=1
=
=
λ n
n−1
k=0
∞
n=2
+
∞
n=2
+
+
1
2 k+1
λ n
n−2
∞
n=1
k=0
λ n
2 n
λ n
n−2
∞
n=2
∞
n=1
k=0
λ n
2 n
λ n
2 n
d 2
dt 2 Gn,k +
1
2 k+1
d 2
∞
n=0
d 2
dt 2 Gn,k +
λn 2n d2 Gn,n
dt2 ∞
n=2
dt2 Gn,n + d2
G0,0
dt2 λ n
2 n
d2 dt2 Gn,n−1 + λ d
2
2
2 d
dt Gn−1,k − Gn−2,k + c 2
Gn,k
G1,0
dt2 1
2k+1
2 d
dt Gn−1,n−1 − Gn−2,n−2 + c 2
Gn,n−1 + λ d
dt G0,0 + λ
2 c2G1,0
d
dt Gn−1,n−1 + c 2
Gn,n + c 2 G0,0. (5.25)
By regrouping the above terms we notice that (5.25) can take the form
c 2
∞
n=1
λ n
n−1
−λ 2
1
∞
2
k=0
k+1 Gn,k +
n=0
∞
n=1
λ n
n−1
k=0
1
2 k+1 Gn,k +
λn
Gn,n + 2λ
2n d
dt
∞
n=0
= c 2 e λt u + 2λ d
dt [eλtu] − λ 2 e λt u − λ
2
λn Gn,n
2n ∞
n=0
∞
n=1
− λ
2
λ n
n−1
∞
n=0
k=0
λn 2n d
dt Gn,n + λ2
22 Multiplying by e −λt the right hand side of (5.26), we have that
e −λt
∞
n=1
λ n
n−1
k=0
1
2 k+1
d 2
dt 2 Gn,k + e −λt
∞
n=0
= c 2 u + 2λ 2 u + 2λ d
dt u − λ2 u − λ
2 e−λt
By inserting result (5.27) into (5.24) we finally obtain
d 2
dt 2 u − c2 u = − λ
2 e−λt
∞
n=0
λn 2n d2 Gn,n
dt2 ∞
n=0
1
2 k+1 Gn,k +
λn 2n d
dt Gn,n + λ2
22 ∞
n=0
∞
n=0
∞
n=0
λn
Gn,n
2n λn Gn,n
2n λ n
2 n Gn,n. (5.26)
∞
λn 2n d
dt Gn,n + λ2 λ
e−λt
22 n=0
n
2n Gn,n. (5.27)
∞
λn 2n d
dt Gn,n + λ2 λ
e−λt
22 n=0
n
2n Gn,n. (5.28)
λt − ∞
The series e 2
n=0 λn
2n Gn,n has been studied in the previous chapter and represents the mean
hyperbolic distance of the particle moving in H2 and changing direction at each Poisson event
when the rate of the Poisson process is λ/2. We have that
λt −
e 2
Since
∞
n=0
λ n
2 n Gn,n =
λt −
e 2
e− λt
4
∞
n=0
2
λ n
2 n
1 +
λ
√
λ2 + 24c2 d
dt Gn,n = d
λt −
e 2
dt
e t
√
4 λ2 +24c2 + 1 −
∞
n=0
λn
Gn,n
2n + λ λt
e− 2
2
λ
√ e
λ2 + 24c2 ∞
n=0
λn Gn,n,
2n
√
t − 4 λ2 +24c2 .

5.3 Equation governing the hyperbolic distance 96
the right-hand side of (5.28) can be rewritten as
− λ
2 e−λt
∞ λ
n=0
n
2n d
dt Gn,n + λ2
∞ λ
e−λt
22 n=0
n
2n Gn,n = − λ
∞
λt d λt
e− 2 − λ
e 2
2 dt
n=0
n
Gn,n
2n = − λ
λt
e− 2 −
2 λ λt
e− 2
23 2
λ
1 + √ e
λ2 + 24c2 t
22
√
λ2 +24c2 λ
t −
+ 1 − √ e 2
λ2 + 24c2 2
√
λ2 +24c2 √
λ2 + 24c2 +
23 λt −
e 22
λ
1 + √ e
λ2 + 24c2 t
22
√
λ2 +24c2 λ
t −
− 1 − √ e 2
λ2 + 24c2 2
√
λ2 +24c2
= λ2 3λt
e− 2
24 2
λ
1 + √ e
λ2 + 24c2 t
22
√
λ2 +24c2 λ
t −
+ 1 − √ e 2
λ2 + 24c2 2
√
λ2 +24c2
λ
1 + √ e
λ2 + 24c2 t
22
√
λ2 +24c2 λ
t −
− 1 − √ e 2
λ2 + 24c2 2
√
λ2 +24c2 − λ√λ2 + 24c2 24 3λt −
e 22 3λt −
= −e 22 e t
22 √
λ2 +24c2 λc2 3λt
√ + e− 2
λ2 + 24c2 2 t −
e 22 √
λ2 +24c2 λc2 √
λ2 + 24c2 .
This lets equation (5.23) appear.
The expression (5.6) of the mean-value E{cosh ηcm(t)} derived in Theorem 5.2.1 satisfies the
non-homogeneous second-order linear equation (5.23) as shown in the next theorem. This proves
that the mean hyperbolic distance (5.6) is obtained by two different and independent methods.
Theorem 5.3.3. The mean hyperbolic distance E{cosh ηcm(t)} in (5.6) is the solution to the
following Cauchy problem
⎧
⎪⎨
⎪⎩
d 2
dt2 u − c2u = λc2e u(0) = 1,
d
dtu(t) = 0.
t=0
− 3
22 λt
√
λ2 +24c2 t −
e 22 √
λ2 +24c2 − e t
22 √
λ2 +24c2
,
Proof
In order to perform the calculations it is convenient to write (5.6) as
where
K =
E{cosh ηcm(t)} = KA(t)B(t) +
23c2 3
√ , A(t) = e− 2
λ2 + 24c2 2 λt , B(t) =
and B(t) has derivatives
B ′ (t) = −
√
λ2 + 24c2 2 2
t −
e 22 √
λ2 +24c2 5λ + 3 √ λ2 + 24 +
c2 λ + 2c
2(λ + 3c) ect λ − 2c
+
2(λ − 3c) e−ct
t
e− 22 √
λ2 +24c2 5λ + 3 √ λ2 + 24 −
c2 e t
22 √
λ2 +24c2 5λ − 3 √ λ 2 + 2 4 c 2
e t
22 √
λ2 +24c2 5λ − 3 √ λ2 + 24 ,
c2 , B ′′ (t) = λ2 + 24c2 24 B(t).

97 Cascades of Particles Moving at Finite Velocity in H 2
Therefore
d2 dt2 E{cosh ηcm(t)} − c 2 E{cosh ηcm(t)} = λKA
5λ
B − 3B′
2 22 = 22λc2
3 −
e 22 λt
5λ
√
λ2 + 24c2 22
t −
e 22 √
λ2 +24c2 5λ + 3 √ λ2 + 24 e
−
c2 t
22 √
λ2 +24c2 5λ − 3 √ λ2 + 24c2
√
λ2 + 24c2 + 3
22
t −
e 22 √
λ2 +24c2 5λ + 3 √ λ2 + 24 e
+
c2 t
22 √
λ2 +24c2 5λ − 3 √ λ2 + 24c2
= λc2
3
√
−
e 22 λt
t − e 4 λ2 +24c2 √
λ2 + 24c2 5λ + 3 √ λ2 + 24c2 (5λ + 3λ2 + 24c2 ) +
3 −
22 λt
= λc2e √
λ2 + 24c2
e
√
t − 4 λ2 +24c2 − e t
√
4 λ2 +24c2
.
e t
√
4 λ2 +24c2 5λ − 3 √ λ2 + 24c2 (−5λ + 3λ2 + 24c2 )
We can easily check that E{cosh ηcm(0)} = 1 and we have that d
dt E{cosh ηcm(t)} t=0 = 0 since
d
dt E{cosh ηcm(t)}
= KA − 3λ
λ + 2c
B + B′ + c
22 2(λ + 3c) ect λ − 2c
− c
2(λ − 3c) e−ct
= 2c2
3 −
e 22 λt
e
√
λ2 + 24c2 t
22 √
λ2 +24c2 5λ − 3 √ λ2 + 24c2 (3λ − λ2 + 24c2 t
e− 2
) − 2
√
λ2 +24c2 5λ + 3 √ λ2 + 24c2 (3λ + λ2 + 24c2
)
λ + 2c
+c
2(λ + 3c) ect λ − 2c
− c
2(λ − 3c) e−ct .
Remark 5.3.1. From Theorem 5.3.3 an interesting relationship between the hyperbolic mean
distance of the center of mass and its acceleration can be extracted
3 −
22 λt
c 2 E{cosh ηcm(t)} − d2
dt2 E{cosh ηcm(t)} = 2λc2e √
λ2 + 24c2 t
sinh
22
λ2 + 24c2 . (5.29)
For large values of t, if √ λ 2 + 2 4 c 2 − 3λ > 0, the mean distance and its second derivative tend to
coincide while in the opposite case they tend to diverge. In view of Remark 5.2.2 this means that
there is a different rate of growth for the mean distance and its acceleration only if √ λ 2 + 2 4 c 2 −
3λ < 0.
5.4 Mean hyperbolic distance of the k-th splinter
If N(t) = n ≥ k, the k-th splinter, at time t, is located at the hyperbolic distance ηk(t) from O
equal to
cosh ηk(t) =
k+1
j=1
cosh c(Sj − Sj−1), 0 ≤ k ≤ n, (5.30)
The expression (5.30) refers to sample paths of particles which change direction until the k-th
Poisson event and then remain on the same geodesic line until time t. Clearly η0(t) represents the
distance of the particle which never changed direction and ηn(t) is the distance of the particle that
changed direction at all Poisson events.

5.4 Mean hyperbolic distance of the k-th splinter 98
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Figure 5.5: The position of the splinters in the hyperbolic half-plane H 2 at time t = 50, when
c = 0.05 and N(t) = 300, are drawn. In (a) each splinter chooses the clockwise direction, in (b)
each splinter chooses the counterclockwise direction and in (c) the clockwise and counterclockwise
directions are alternatively chosen.
The mean-value of (5.30) becomes
E{cosh ηk(t)I {N(t)≥k}} =
=
∞
E{cosh ηk(t)I {N(t)=n}}
n=k
●
●
●
● ●
∞
E{cosh ηk(t)|N(t) = n}P{N(t) = n}. (5.31)
n=k
We now evaluate the Laplace transform of (5.31). By letting γ = λ + µ and by considering (5.4)
we have that
∞
e
0
−µt E{cosh ηk(t)I {N(t)≥k}}dt =
∞
λ n
n=k
∞
0
e −γt Gn,k(t)dt
∞
= λ
n=k
n
γ
γ2 − c2 k
1 1
1
+
2 (γ − c) n−k+1 (γ + c) n−k+1
= 1
γλ
2 γ2 − c2
k ∞
n−k 1 λ
+
γ − c γ − c
n=k
1
∞
n−k
λ
γ + c γ + c
n=k
= 1
γλ
2 γ2 − c2 k
1
γ − c − λ +
1
γ + c − λ
= 1
λ(λ + µ)
2 (λ + µ) 2 − c2 k
1 1
+
µ + c µ − c
= 1
∞
e
2
−µt t
dt e −c(t−s) t
fk(s)ds + e c(t−s)
fk(s)ds
=
0
∞
0
e −µt dt
t
0
0
0
cosh c(t − s)fk(s)ds (5.32)
●
●
●

99 Cascades of Particles Moving at Finite Velocity in H 2
where fk(s) is the inverse Laplace transform of the function
λ(λ + µ)
(λ + µ) 2 − c2 k = λk
2k
1
λ + µ − c +
k 1
=
λ + µ + c
λk
2k
λ(λ+µ)
(λ+µ) 2−c2 k . Let us write
k
r=0
k
r
1
(λ + µ − c) r
j
u If k = 0 we have f0(s) = δ(s). When k = 1, 2, . . . , recalling that L j! eαu
1+(u) =
s > α and j = 0, 1, 2 . . . , we obtain the inverse Laplace transform
fk(s) = λk
2 k
sk−1 (k − 1)! es(c−λ)
k−1
s
k w
r
r−1 (s − w)k−r−1
ew(c−λ)
(r − 1)! (k − r − 1)! e−(s−w)(λ+c) dw
+ λk
2 k
+ λk
2 k
r=1
0
s k−1
(k − 1)! e−s(c+λ) .
1
.
(λ + µ + c) k−r
1
(s−α) j+1 with
By inverting the Laplace transform (5.32) we obtain for k = 0 that E{cosh η0(t)I {N(t)≥0}} = cosh ct,
while for k = 1, 2, . . . we have
E{cosh ηk(t)I {N(t)≥k}}
t
cosh c(t − s)e
0
cs e−λsλksk−1 k−1
= 1
2k (k − 1)! ds
+ 1
2k
t
k
cosh c(t − s)e
r 0
−λs λ k s k−1
+ 1
2k = 1
2 k
1
r=1
t
cosh c(t − s)e
0
−cs e−λsλksk−1 (k − 1)! ds
t
cosh c(t − s)e
0
cs e−λsλksk−1 (k − 1)! ds
k−1
cosh c(t − s)E{e cs(2Yr,k−1) e
} −λsλksk−1 + 1
2 k
+ 1
2k = 1
2 k
t
k
r 0
r=1
t
cosh c(t − s)e
0
−cs e−λsλksk−1 t
k−1
0
r=1
0
e cs(2y−1) yr−1 (1 − y) k−r−1
dy ds
(r − 1)!(k − r − 1)!
Γ(k)
(k − 1)! ds
cosh c(t − s) e cs
k
+ E{e
r
cs(2Yr,k−1)
−cs
} + e g(s; k, λ) ds (5.33)
where g(s; k, λ) = e−λsλ k s k−1
Γ(k) , Yr,k ∼ Beta(r, k − r) and (2Yr,k − 1) ∈ (−1, 1). If we adopt the
convention that Y0,k = 1 and Yk,k = −1 it is possible to simplify the last expression rewriting it as
E{cosh ηk(t)I {N(t)≥k}} = 1
2 k
where h(k, c, s) = k
r=0
= 1
2 k
t
0
t
0
cosh c(t − s)
k
r=0
ds
k
E{e
r
cs(2Yr,k−1)
}g(s; k, λ) ds
cosh c(t − s) h(k, c, s) g(s; k, λ) ds
k cs(2Yr,k−1)
r E{e }. The last result suggests regarding the mean hyperbolic
distance of a particle generated at the k-th Poisson event as the mean hyperbolic distance of a
particle moving on the first geodesic line and stopping at a randomly distributed time; the law of
the stopping time is a suitable combination of a Beta and Gamma distributions.

5.4 Mean hyperbolic distance of the k-th splinter 100
Remark 5.4.1. In some particular cases it is possible to check easily that formula (5.33) gives
exactly the same expression of the mean hyperbolic distance of the k-th particle as that obtained
starting from formula (5.31). In particular for k = 1 we have
E{cosh η1(t)I {N(t)≥1}}
= 1
2
t
For k = 2 we obtain
E{cosh η2(t)I {N(t)≥2}}
= 1
22 = 1
2
t
0
t
0
cosh c(t − s) e cs + e −cs g(s; 1, λ)ds =
cosh c(t − s)
where Y1,2 ∼ Beta(1, 1).
0
t
e cs + 2E{e cs(2Y1,2−1) } + e −cs
g(s; 2, λ) ds
cosh c(t − s) cosh cs λ2 se −λs
Γ(2)
ds + 1
2
0
cosh c(t − s) cosh cs λe −λs ds.
t
cosh c(t − s) E{e
0
cs(2Y1,2−1) } λ2se−λs Γ(2)
ds

102

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