The Arithmetic of Quaternion Algebra

5.3. EXAMPLES 113
isomorphic to
Q ×
A H× ∞O × p
� O × q \H ×
A /H× = O × p Qp\H × p /O (p)× = X/Γ.
Precisely, if (Ii) is a system of the representatives of classes of the ideals to the
left of O, the right orders the ideals Ii , denoted by O(i), constitute a system
of representatives of the quotient graph X/Γ. An order O ′ , vertices of the tree
X, is Γ-equivalent to O (i) if it is joined to O by an ideal I which is equivalent
to Ii.
The Brandt matrix can be explained geometrically as the homomorphism of the
free group Z[X/Γ] generated by the vertices of the quotient graph X/Γ. Let
f : Z[X] → Z[X/Γ] be the homomorphism induced by the surjection X → X/Γ.
For every integer n ≥ 1, let Pn be the homomorphism od Z[X/Γ] such that
Pnf = fTn where Tn is the homomorphism of Z[X] defined by the relations,
Ch.II,§1,
T0(O ′ ) = O ′ , T1(O ′ ) = �
O ′′ , T1Tn = Tn+1 + qTn−1.
d(O ′ ,O ′′ )=1
The Brandt matrices P (p n ) are the matrices of homomorphism Pn on the basis
of Z[X/Γ] consisting of the vertices xi , the images of the maximal order Oi.
In fact, it suffices to prove for n = 0, 1, since the last relation of Tn is true
for Pn and P (p n ). For n = 0, It is evident because P (1) is the identity matrix.
For n = 1, the coefficient aij of the matrix P1 on the basis xi defined by
P1(xi) = � aijxj, is:
aij = Card{O ′′ |f(O ′′ ) = Oj, d(Oi, O ′′ ) = 1}.
this is the number of the integral ideals I to the left of Oi with reduced norm pZ
such that IiI is equivalent to Ij. We do discover here the definition of Brandt
matrix P (p).
The group Γ (O ′ ,O ′′ ) of the isometries of Γ which fix an edge (O ′ , O ′′ ) with starting
point O ′ and end point O ′′ is (O ′ × ∩ O ′′ ×)/Z × . The number CardΓ(O ′ ,O ′′ )
is called the order of the edge of quotient graph X/Γ,which is the image of
(O ′ , O ′′ ). For every vertex x of the quotient graph X/Γ we denote by A(x),
resp. S(x) the set of the edges y of X/Γ with the starting point x, resp. of the
end points of that edges with starting point x, and e(x), resp. e(y), the order
of vertex x, resp. the order of the edge y. We have
q + 1 = e(x) �
and the homomorphism P1 is given by
P1(x) = e(x) �
x ′ ∈S(x)
y∈A(x)
e(y) −1
e(y) −1 x ′ , where x ′ is the end point of y.
We see thus immediately that the matrix of P1 is symmetric with respect to the
basis (e(x) −1/2 x), where x runs through the vertices of X/Γ. This is simply the
matrix (a(x, x ′ )), where a(x, x ′ ) = e(x, x ′ ) −1 if the vertices x, x ′ are joined by
an edge y = (x, x ′ ) and a(x, x ′ ) = 0 if there is no any edge joining x to x ′ .