The Arithmetic of Quaternion Algebra

More documents

Recommendations

Info

94 CHAPTER 4. APPLICATIONS TO ARITHMETIC GROUPS Proof. The assertion for Γ comes from Example 2.1. As for G, it suffices to verify that the possible values of the order of the cyclic groups which are contained in G are 1, 2, 4, 6, 8, 12. It can be obtained at once from the structure of G\Γ and the order of the cyclic groups in Γ. The formulae for eq(G) are not so simple as for eq(Γ) but can be obtained by elementary method. The following table gives the list of all surfaces barΓ\H of genus 0,1,2. here is Table 1 !!! Using the results of Ogg about the hyperelliptic Riemann surfaces of genus g ≥ 2, we can determine the surfaces ¯ Γ\H of genus g ≥ 2 which are hyperelliptic. In all these cases, the hyperelliptic involution is induced by an element of G. We denote by πi the element of O of reduced norm pi (1 ≤ i ≤ 2m) and gd the element of G defined by gd = d −1/2 π ε1 1 ...πε2m 2m for d = πε1 1 ...πε2m 2m , εi = 0 or 1 The table below gives the list of the hyperelliptic surfaces with their genus and the element of G which induces the hyperelleptic involution: Here is a talbe!!! C The construction of a fundamental domain for Γ and G in The case of D = 15.(Michon [1]). The quaternion algebra is generated by i, j satisfying The order O generated over Z by i 2 = 3, j 2 = 5, ij = −ji. 1, i, (i + j)/2, (i + k)/2 is maximal. It has the matrix representation O = { 1 � √ x 2 5¯y √ � 5y , |x, y ∈ Q( ¯x √ 3) are integer, and x ≡ y mod (2)}. The group Γ = O 1 is consists of the above matrices such that (1) n(x) − 5n(y) = 4. The group G normalizing Γ is consists of the matrices satisfying (2) n(x) − 5n(y) = 4, 12, 20, or 60 divided by the square root of their determinants. The fixed points in C of an element of G are distinct and given by z = b√ 3 ∓ √ a 2 − 4 √ 5 ¯ (y) if x = a + b √ 3, a, b ∈ Z. The elliptic fixed points corresponds to a = −1, or 1. It can be restricted to a = 0 or 1, since the change of sign of the matrix does not change the
4.3. EXAMPLES AND APPLICATIONS 95 homography. The elliptic points are distributed on the ray staring from origin and with slope b −1 . All the elliptic points which locate on an admissible ray obtained by solving the following equation (3) − 5n(y) = 4 − N(x), y is integer in Q( √ 3). If z0 is an elliptic point, we see that ε n z0, n ∈ Z is also an elliptic point if ε is fundamental unit of Q( √ 3). Let η be the fundamental unit of Q √ 5, namely 1 2 (1 + √ 5). It has norm −1. Consider its square η 2 included in Γ, with image k = 1 2 � 3 √ 5 √ 5 3 In view of the symmetry, k n (z0), n ∈ Z is also an elliptic point. The first values of b such that the equation (3) having a solution are b = ∓2, ∓8. for b = 2, it becomes −n(y) = 3, y is integer in Q( √ 3). For b = 8, it becomes Denote � . −n(y) = 37, y is integer in Q( √ 3). A = 1 2 + i √ 5 2 − √ 3 , C = 1√ 8 + i 5 4 + √ 3 , C′ = 1 8 + i 5 4 − √ 3 . The set of elliptic points on the line of slope 1/2 is {ε n , n ∈ Z}; on the line on slope 1/8, it is {ε n C, ε n C ′ , n ∈ Z}. Denote by B, B ′ the symmetry of A, A ′ with respect to the imaginary axis with A ′ = ε 2 A. Lemma 4.3.2. The hyperbolic hexagon BACC ′ A ′ B ′ is a fundamental domain of Γ. Proof. Let We have h = � � ε 0 , l = 0 ε 1 � √ √ � −4 √ + 3 − 15 √ . 2 15 −4 − 3 A ′ = h(A), B ′ = h(B) C = k(B), C ′ = k(B ′ ) A = l(A ′ ), C = l(C ′ ) The hexagon has for the angles at vertices B, B ′ C, C ′ are π/6, and π/3 at A, A ′ . It is a fundamental domain for the group < l, h, k > generated by l.h.k. It has two cycles {A, A ′ }, {B, B ′ C, C ′ } each of order 3. Its hyperbolic volume is (6 − 2)π − 2 2π 3 = 8π 3 . On the other hand, for the hyperbolic measure the volume of ¯ Γ\H from the first table of the precedent exercise is equal to 8π/3. Therefore, Γ =< l, h, k > and the polygon is fundamental. The same procedure allows to treat by the same way the case of G. .
Page 1:
The Arithmetic of Quaternion Algebr
Page 4 and 5:
ii CONTENTS
Page 6 and 7:
2 CHAPTER 1. QUATERNION ALGEBRA OVE
Page 8 and 9:
Page 10 and 11:
Page 12 and 13:
Page 14 and 15:
10 CHAPTER 1. QUATERNION ALGEBRA OV
Page 16 and 17:
Page 18 and 19:
Page 20 and 21:
Page 22 and 23:
Page 24 and 25:
Page 26 and 27:
Page 28 and 29:
Page 30 and 31:
Page 32 and 33:
Page 34 and 35:
Page 36 and 37:
Page 38 and 39:
Page 40 and 41:
Page 42 and 43:
Page 44 and 45:
Page 46 and 47:
Page 48 and 49: 44 CHAPTER 3. QUATERNION ALGEBRA OV
Page 84 and 85: 80 CHAPTER 4. APPLICATIONS TO ARITH
Page 104 and 105: 100 CHAPTER 4. APPLICATIONS TO ARIT
Page 110 and 111: 106CHAPTER 5. QUATERNION ARITHMETIC
Page 122: 118CHAPTER 5. QUATERNION ARITHMETIC
show all

The Arithmetic of Quaternion Algebra

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?