The Arithmetic of Quaternion Algebra

116CHAPTER 5. QUATERNION ARITHMETIC IN THE CASE WHERE THE EICHLER CONDITI
. The Leech lattice is the lattice in R 24 equipped with the above scalar product
and defined by one of the following equivalent ways:
(a) L = {x ∈ O 3 |ex1 ≡ ex2 ≡ ex3 ≡ � ximod(2)},
(b) L is the free O-module of basis f = (1, 1, e), g = (0, ē, ē), h = (0, 0, 2).
we shall prove that we obtain rightly the Leech lattice. Actually the lattice L is
– even, since x, y ∈ Z, and x · x ∈ 2Z, it is evident.
– unimodular, since if x ∈ H3 the equality x·L ⊂ Z is equivalent to h(x, L) ⊂ 2R
and the definition (b) shows the last inclusion is equivalent to x ∈ L.
– not contains any element x such that x · x = 2. Otherwise x ∈ L, x · x = 2,
then put ri = n(xi). It follows that � ri = 2, and since the elements ri are
totally positive, (1) implies that one of them at least should be annihilated. The
definition (a) of lattice implies then that exi ∈ 2O for every 1 ≤ i ≤ 3, From it
we have 2n(xi) ∈ 4R and xi ∈ 2O. Taking again the same reason, we see that
at most one of xi is nonzero and ri ∈ 4R, It leads to a contradiction.
E Tables.
If H is a quaternion algebra totally defined over Q, i.e. HR = H the Hamilton
quaternion field, of reduced discriminant D = �
p∈Ram(H) p, the class number and the type number of t
are given by the formulae:
h = h(D, N) = 1 �
(p−1)
12
�
(p+1)+ 1
4 f(D, N)(1) + 1
3 f(D, N)(3) �
−r
, t = 2
p|D
p|N
where r is the number of prime divisors of DN,
f(D, N) (m) = �
(1 − ( d(−m)
))
p
�
(1 + ( d(−m)
)
p
p|D
d(−m), h(−m) are the discriminant and the class number of Q( √ m) respectively.
�
−m, if m ≡ −1mod(4)
d(−m) =
, d(−1) = −4, d(−3) = −3,
−4m, if m ≇ −1mod(4)
g(D, N) (m) = 2 �
(1 − ( d(−m)
))
p
�
(1 + ( d(−m)
)), defined if N is even.
p
Set:
p|D
p|(N/2
p|N
⎧
⎪⎨ 1, if m ≇ −1mod4
a(m) = 2,
⎪⎩
4,
�
if m ≡ 7mod8 or m = 3 ,
if m ≡ 3mod8 and m �= 3
b(m) =
a(m),
3,
if m ≇ 3mod8 or m = 3
if m ≡ 3mod8 and m �= 3 ,
the number tr(m) are the traces of the Brandt matrices P (Zm) for m|DN:
⎧
⎪⎨ f(D, N)
2tr(m) =
⎪⎩
(m) h(−m), if D is even
f(D, N) (m) h(−m)a(m),
g(D, N)
if DN is odd
(m) h(−m)b(m),
.
if N is even
m|DN
tr(m)