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Random Processes in Hyperbolic Spaces Hyperbolic Brownian ...

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23 <strong>Hyperbolic</strong> Geometry<br />

s<strong>in</strong>ce<br />

|1 − u¯v| 2 = |1 − ūv| 2 = |ū| 2<br />

2<br />

<br />

<br />

1 <br />

− v<br />

ū = |u| 2<br />

2<br />

<br />

<br />

1 <br />

− v<br />

ū = |u| 2<br />

<br />

1 <br />

<br />

ū<br />

= 1 + |u| 2 |v| 2 − 2|u| 2<br />

<br />

1<br />

, v = 1 + |u|<br />

ū 2 |v| 2 − 2 〈u, v〉<br />

= 1 + |u| 2 |v| 2 − 2 〈u, v〉 ± |u| 2 ± |v| 2<br />

= |u − v| 2 + (|u| 2 − 1)(|v| 2 − 1).<br />

From (2.31) and (2.32) it follows that<br />

and from (2.33) we obta<strong>in</strong> that<br />

s<strong>in</strong>ce<br />

In the particular case u = 0 we have<br />

tanh η1(u, v)<br />

2<br />

η1(u, v) = log<br />

= |u − v|<br />

|1 − u¯v|<br />

|1 − u¯v| + |u − v|<br />

|1 − u¯v| − |u − v|<br />

<br />

1 |1 − u¯v| + |u − v|<br />

tanh log =<br />

2 |1 − u¯v| − |u − v|<br />

|u − v|<br />

|1 − u¯v| .<br />

η1(0, v) =<br />

1 + |v|<br />

1 − |v| .<br />

2<br />

+ |v| 2 <br />

1<br />

− 2 , v<br />

ū <br />

(2.33)<br />

It follows that an hyperbolic circle <strong>in</strong> B 2 with center 0 and radius r is a Euclidean circle with center<br />

0 and radius R = r−1<br />

r+1 , <strong>in</strong> fact<br />

v ∈ B 2 : η1(0, v) = r =<br />

<br />

v ∈ B 2 :<br />

<br />

1 + |v|<br />

= r = v ∈ B<br />

1 − |v| 2 <br />

r − 1<br />

: |v| = . (2.34)<br />

r + 1<br />

In the next theorem is obta<strong>in</strong>ed the length of an hyperbolic circle and the area of an hyperbolic<br />

disk.<br />

Theorem 2.3.3. The area of a hyperbolic disk of hyperbolic radius r is 4π s<strong>in</strong>h 2 (r/2). The length<br />

of a hyperbolic circle of hyperbolic radius r is 2π s<strong>in</strong>h r.<br />

Proof<br />

We use the model B2 . Let C = {v ∈ B2 : η1(0, v) = r} and D = {v ∈ B2 : η1(0, v) ≤ r} be the<br />

circle and disk with center <strong>in</strong> the orig<strong>in</strong> and hyperbolic radius r. In view of (2.31) and (2.34) we<br />

have that<br />

s<strong>in</strong>h 2 r<br />

2 = s<strong>in</strong>h2 η1(0, v)<br />

=<br />

2<br />

|v|2 R2<br />

= .<br />

1 − |v| 2 1 − R2 Then<br />

<br />

h − area(D) = 4<br />

D<br />

= 4π s<strong>in</strong>h 2 r<br />

2 .<br />

1<br />

(1 − |v| 2 ) 2 dv1dv2 = 4<br />

2π<br />

0<br />

dθ<br />

R<br />

0<br />

s<br />

ds<br />

(1 − s2 = 4(2π)<br />

) 2 2(1 − R2 )<br />

R 2

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