The Arithmetic of Quaternion Algebra

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100 CHAPTER 4. APPLICATIONS TO ARITHMETIC GROUPS F Hyperbolic space of dimension 3 . We want to extend a complex homography to a transformation of R 3 . Every complex homography is an even product of inversions with respect to the cicles in plane which is identified with C. Consider now the spheres which have the same circle and same rays as that circles, and the operation of R 3 consistent with the performance of the product of the inversion with respect to these spheres. extend then a complex homography to R 3 . We now verify the consistence of this definition (Poincarè[1]). It remains to find the equations of the transformation. We identify the points of R 3 with the points u = (z, v) ∈ C × R or with the matrices � z u = v � −v . ¯z � a The operation of R3 extending the homography associated with g = c � b ∈ d SL(2, C) is u ↦→ U = (au + b)(cu + d) −1 . Set U = ¯g(u) = (Z, V ). We verify the following formulae: Z = ((az + b)(cz + d) + a¯cv 2 )(|cz + d| 2 + |c| 2 v 2 ) −1 , V = v(|cz + d| 2 + |c| 2 v 2 ) −1 . Differentiating the formula U = g(u) we see that V −1 dU = v −1 du. We equip H3 = {u ∈ R 3 |v > 0} the upper half-space the hyperbolic metric v −2 (dx 2 + dy 2 + dv 2 ), u = (x + iy, v). The group SL(2, C) acts on the hyperbolic half-space by isometries. Its action is transitive. the isotropic group of (1, 0) is equal to SU(2, C) and SL(2, C)/SU(2, C) is homeomorphic to H3. The group of all the isomorphisms of H3 is generated by the mapping (z, v) ↦→ (¯z, v) and the group is isomorphic to P SL(2, C) of isometries associated with SL(2, C).The geodesics are the circles(or the straight lines) orthogonal to plane C. Definition 4.11. The volume element deduced from the hyperbolic metric is v −3 dxdydv. Definition 4.12. Milnor (Thurston, [1]) introduced a function, i.e. the Lobachevski function, L(θ) = − � θ 0 log |2 sin u|du. The function allows to express the volume of a tetrahedron. The function is related to the values of the zeta functions of the (complex) number field at point 2, hence we have the relation (1) L(θ) = 1/2 � sin(2nθ)/n 2 , 0 ≤ θ ≤ π n≥1
4.3. EXAMPLES AND APPLICATIONS 101 deduced from the relation between L(θ) and the dilogarithm � ψ(z) = − log(1 − w)dw/w = � z n /n 2 , for |z|leq1, |w|leq1, obtained by setting z = e 2iθ : 0 z n≥1 ψ(e 2iθ − ψ(1) = −θ(π − θ) + 2iL(θ). From this and using the Fourier transformation we have (2) � kmod(D) We also have the relations ( −D k )L(πk/D) = √ D � ( −D n≥1 (3) L(θ) is periodic of period π and odd. (4) L(nθ) = � nL(θ + j/n), for every integer n �= 0. jmod(n) n )n−2 = √ Dζ Q( √ −D) (2)/ζQ(2) = 6π −2√ Dζ Q( √ −D) (2). The relation (3) is immediate, the relation (4) is deduced from the trigonometric identity 2 sin u = � jmod(n) 2 sin(u+jπ/n) which can be proved by factoring the polynomial Xn − 1. Milnor conjecture that every rational linear relation among the real numbers L(θ), for the angles which are the rational multiples of π, is a consequence of (3) and (4). See also Lang [1]. Volume of a tetrahedron of which one vertices is at the infinity. A picture here!!! The base of such a tetrahedron is a sphere with center on C. The projection on C from the tetrahedron is a triangle, of which the angles are the dihedral angles of their edges meeting at the infinity:= α, β, γ. Therefore, α + β + γ = π. Suppose γ = π/2 and A is projected to (0, 0), and let V be the volume of the tetrahedron � � � V = v −3 � � dxdydv = dxdy/2(1 − x 2 − y 2 ) where t = {(x, y)|0 < x ≤ x tan α}, we obtain by setting x = cos θ, then � δ V = −1/4 log(sin(θ + α)/ sin(θ − α))dθ, π/2 V = 1/4(L(α + δ) + L(α − δ) + 2L(π/2 − α)). If the vertex B is in C (two vertices are at the infinity), we have δ = α and V = 1/2L(α). t
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The Arithmetic of Quaternion Algebra

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