The Arithmetic of Quaternion Algebra

3.2. ZETA FUNCTION, TAMAGAWA NUMBER 55
Proof. When X is a field, the computation of Tamagawa number is implicitly
contained in the theorem 2.2 of the functional equation. If X = M(2, K) , it
can compute directly. The theorem 2.3 can be extended to the central simple
algebra X. In this case we have τ(X) = τ(X1) = τ(H 1 ) = 1 and τ(G) = n,
if [X : K] = n 2 . Reference: Weil [2]. By the definition of Tamagawa measure,
τ(X) = 1. We shall prove τ(G) = 2τ(H 1 ) and τ(H1) = τ(H 1 ), after that then
τ(H 1 ) = 1. The proof is analytic, and the Poisson formula is involved. The
exact sequence which is compatible with thee Tamagawa measures
1 −−−−→ H 1 A /H×
K
−−−−→ HA,1/H ×
K
n
−−−−→ KA,1/K × −−−−→ 1
proves that τ(H 1 ) = τ(H1)τ(K −1
1 ). The theorem 2.2 shows τ(H1) = τ(K1) = 1
if H is a field because of the definition of Tamagawa measure itself. Therefore
τ(H 1 = τ(H1) for every quaternion algebra H/K. It follows from the proof of
Theorem 2.2 that
�
2
K ×
A /K×
H ×
A /H×
f(||h||H)dhA
K
·τ(H1 )
f(||k||K)dk ∗ A =
K ×
A /K×
�
K ×
A /K×
f(||k|| K 2)dk ∗ A
for every function f such that the integrals converge absolutely. Applying
||h||H = ||n(h)|| K2 if h ∈ H times
A , we see that
�
�
f(||k||K)dk ∗ �
A = τ(G)
K ×
A /K×
f(||k||K)dkA∗, it follows τ(G) = 2τ(H 1 ). The Theorem has been proved when X = H or K is
a field.
It remains to prove τ(SL(2, K)) = 1. The starting point is the formula
�
�
(2) f(x)dx =
[
�
f(ua)]τ(u),
A 2
SL(2,A)/SL(2,K)
a∈K 2 −
where f is an admissible function on A 2 , cf. II,§2, and τ(u) is a Tamagawa
measure on SL(2, A)/SL(2, K), and where A 2 is identified with the column
vectors of two elements in A on which SL(2, A) operates by
� � � � � �
a b x ax + by
=
.
c d y cx + dy
� �
0
The orbit of is A
1
2 � �
� �
0
1 x
− and its isotropic group is NA = { |x ∈ A}.
0
0 1
We apply Poisson formula,
�
f(ua) = �
f ∗ ( t u −1 a)
a∈K 2
a∈K 2
because of det(u) = 1. Here we give an other expression for the integral (2) in
function of f ∗ . In fact, it prefers to write the integral in f ∗ into the function on
f ∗∗ = f(−x). Since τ( tu−1 ) = τ(u), we obtain
(3)
�
A 2
f ∗ �
(x)dx =
SL(2,A)/SL(2,K)
a∈K2 ⎛
⎝0 0
⎞
⎠
[ �
f(ux) − f ∗ (0)]τ(u).