The Arithmetic of Quaternion Algebra

10 CHAPTER 1. QUATERNION ALGEBRA OVER A FIELD
We now show the rotation (r, 2α) of the space R (identifying with the Hamilton
quaternion of zero reduced trace) of angle 2α, and of the axis being carried out
by a vector of r units, is induced by the inner automorphism associated with
q = cos α + r sin α. In fact, we have r 2 = −1, and by the theorem 2.1 of automorphisms
we can choose a quaternion s ∈ H such that s 2 = −1 and rs = −sr.
The pure quaternions form the R-vector space with basis r, s, rs. Under this
basis we shall verify the restriction of the inner automorphism induced by q to
the quaternions with zero trace is the rotation defined above. We have:
(cos α + r sin α)r(cos α − rsinα) = r,
(cos α + r sin α)s(cos α − r sin α) = cos 2α.s + sin 2α.rs,
(cos α + r sin α)s(cos α − r sin α) = cos 2α.rs − sin 2α.s.
We derive from this that H 1 /∓1 is isomorphic to SO(3, R). We now show H 1
is a non-trivial covering of SO(3, R). Otherwise, H 1 would contain a subgroup
of index 2, hence distinguished. There would exist a surjective homomorphism
c of H 1 to ∓1. We shall see it is impossible. Since −1 is a square in H 1 , we
have c(−1) = 1. All of the elements of square −1 being conjugated by an inner
automorphism of H 1 , we have c(i) − c(j) = c(ij) or i, j are defined as that in
§1. From it we obtain c(i) = 1, and c(x) = 1 for every quaternion of square −1.
Because every quaternion of H 1 is the product of two quaternions of square
−1 and of a sign, we then deduce c equals 1 on H 1 identically. We note that
H 1 /∓1 being isomorphic to SO(3, R) is a simple group. It is well-known that
P SL(2, K) = SL(2K)/∓1 is a simple group if the field K is not a finite field
consisting of 2 or 3 elements (Dieudonné,[1]). the property can not be generalized.
The group H 1 /∓1 is not always simple. It can be found in Dieudonné an
infinite number of examples where the groups are not simple. Put the following
question: if K is a global field, and H/K a quaternion algebra such that for the
completion of Kv of K, the group Hv/∓1 is simple( where Hv = H ⊗ Kv), is
Hv/∓1 a simple group? The group of commutator of a group G is the group
generated by the elements of G of the form uvu −1 v −1 , u, v ∈ G. The group of
commutator of H × is then contained in H 1
Proposition 1.3.5. The group of the commutators of H × equals H 1 .
Proof. Let h be a quaternion of reduced norm 1. If the algebra K(h) is a
separable quadratic algebra over K, the theorem 90 of Hilbert indicates there
exists an element x ∈ K(h) × such that h = xx −1 . We can moreover verify this
property directly: if K(h) is a field, we choose x = h+1 if h �= −1, and x ∈ H × 0 ;
if K(h) is not a field, it is isomorphic to K + K, and if h = (a, b) ∈ K + K is of
norm ab = 1, we then choose x = (c, d) with cd −1 = a. Since x, x are conjugate
by an inner automorphism (they satisfy the same minimal polynomial), we
deduce from this that h is a commutator.
If K(h)/K is not separable quadratic extension, we then have h = h,hence (h −
1) 2 = 0. If H is a field, then h = 1, otherwise H is isomorphic to M(2, K), and
we would have the assertion: SL(2K) is the group of commutator of GL(2, K),
cf. Dieudonneé[1].
The explanation of the group H 1 /∓1 as the group of rotations of R 3 permits
us to determine the structure of the finite groups of real quaternions as that of
finite rotation groups(Coxeter,[1]).We start from recalling this well-known result
about the structure of finite rotation groups.