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The Arithmetic of Quaternion Algebra

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4.2. RIEMANN SURFACES 87<br />

Corollary 4.2.2. Let N be a positive real number. For every point z 0 ∈ H <strong>of</strong><br />

its real part zero, we have<br />

log N = d(z0, Nz0) = inf d(z, Nz).<br />

z∈H<br />

Pro<strong>of</strong>. d(z, Nz) = arccosh(1 + (N−1)2 x2<br />

2N (1 + y2 )) is minimal for x = 0 and equal<br />

to log N.<br />

Proposition 4.2.3. <strong>The</strong> area <strong>of</strong> a triangle whose vertices are at the infinity is<br />

equal to π.<br />

Pro<strong>of</strong>. (A picture here!!!)<br />

� �<br />

y −2 dxdy =<br />

� π<br />

0<br />

� ∞<br />

− sin θdθ y<br />

sin θ<br />

−2 dy = π.<br />

<strong>The</strong> common area <strong>of</strong> these triangles can be taken as the definition <strong>of</strong> the value<br />

π.<br />

Proposition 4.2.4. <strong>The</strong> area <strong>of</strong> a hyperbolic triangle <strong>of</strong> the angles at the vertices<br />

being θ1, θ2, θ3 is equal to π − θ1 − θ2 − θ3,<br />

Pro<strong>of</strong>. <strong>The</strong> formula is true if every vertex is at infinity. We use the Green formula<br />

if no any vertex is at the infinity: if Ci, i = 1, 2, 3 are the edges <strong>of</strong> the<br />

triangle , then � � y−2dxdy = � �<br />

i Ci dx/y<br />

Here are two pictures!!!<br />

�<br />

C dx/y = � θ 2<br />

r sin u/(−r sin u)du = α.<br />

θ1<br />

<strong>The</strong> area is thus I = α1 + α2 + α3. <strong>The</strong> total rotation <strong>of</strong> the turning normal<br />

along is the triangle 2π, and that around a vertex <strong>of</strong> angle θ is π − θ. It follows<br />

2π = �<br />

i (π − θi) + �<br />

i αi, from this we have I = π − θ1 − θ2 − θ3. It turns back<br />

to one <strong>of</strong> these two cases when one <strong>of</strong> the angles is zero (its vertex corresponds<br />

to the infinity). By the triangulation we can compute the area <strong>of</strong> a polygon.<br />

Corollary 4.2.5. <strong>The</strong> area <strong>of</strong> a hyperbolic polygon with the angles at vertices<br />

θ1, ..., θn equals (n − 2)π − (θ1 + ... + θ2).<br />

EXAMPLE. A fundamental domain <strong>of</strong> P Sl(2, Z). <strong>The</strong> group P SL(2, Z) is<br />

generated by the homographies t = z + 1 and t = −1/z. We show that the<br />

hatching domain in the picture is a fundamental set<br />

F = {z ∈ C|Imz > o, |z| ≥ 1, −1/2 ≤ Rez ≤ 1/2}.<br />

It ia a triangle with one <strong>of</strong> its vertices being at the infinity. Its area is π−2π/3 =<br />

π/3. It equals the area <strong>of</strong> the triangle without hatching, which is also a fundamental<br />

set <strong>of</strong> SL(2, Z) in H.<br />

Here is a picture!!!

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