Random Processes in Hyperbolic Spaces Hyperbolic Brownian ...

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
⎤
⎤
⎦ ,