12
I. AUTOMORPHIC FORMS AND ABELIAN VARIETIES
y G (U(r]n)A H Pj^ )(U(r)n)A H C). Since p is diagonal, we obtain the desired
result for GU(r)n).
1.10. The notation being as in §1.8, put C\}] = Cfe
-1
, 3]. Then GL
m + n
(K")
A
=
PA C[%] by the above lemma. Therefore every element x of GLm+n(K)A can
be written x = yz with y G P^
m , n )
and z G C[a]. We then define an t-ideal il3(x)
by
(1.19) il3(x) = det(dy)t,
where the right-hand side is an r-ideal defined in §1.6. We easily see that this is
well-defined. (But it depends on m, n, and 3.)
Next, given x G (KA)^
w e
define an integral r-ideal uo(x) and a positive
integer i/(x) by
(1.20) i/o(a;)v = ^o(^) for every v ek,
(1.21) i/(x)= W(i/0(aO).
Here
VQ(XV)
is defined by (1.15). In other words, take g G
GLU(K)A
and ft G
(.KA)™
S O
that
x^
=
9Zlhy 1S a
reduced expression for every u G k . Then ^(x ) =
det(#)t.
1.11. Lemma. For x G GLm+n(K)A the following assertions hold.
(1) If a = diag[lm, «ln] with K G if£, then
ilKi{oLxa~l)
= il3(a:).
(2) If dx G GLn(K)A and 3 = /it with \i G if£, then det^Jil^x)"" 1 =
v^~ld~lcx).
(3) If x G GL
m+n
(K)nP]
i
m
'
n)
D[t), 3] and 93 ^ t, then det{dx) ^ 0 and det(dx)
•il3(x)_1 is prime to 1)3.
For the proof see [S97, Lemma 9.4].
1.12. By a CM-field we mean a totally imaginary quadratic extension of a totally
real algebraic number field of finite degree. Given a CM-field K and an absolute
equivalence class r of representations of K by complex matrices, we call (K, r)
a CM-type if the direct sum of r and its complex conjugate is equivalent to the
regular representation of K over Q. If (K, r) is a CM-type and [K : Q] = 2n, then
r is the class of diag[ri, ... , rn] with n isomorphic embeddings r* of K into C
such that { T\, ... , r
n
,
TIO;,
... ,
TUUJ
} is exactly the set of all embeddings of K into
C, where u denotes complex conjugation. In this setting we write r =
{TJ}^
= 1
.
If F is the totally real field over which K is quadratic and p is the generator of
Gal(K/F), then we have
TIUJ
= prt, because of the following easy fact:
(1.22) If a is an isomorphism of K onto a subfield of C, then xpo is the complex
conjugate of for every x G K.
Therefore, if X is a matrix with entries in K, then (X*)a = t(XT), where the bar
is complex conjugation. Thus putting Y* = lY for a complex matrix Y, we have
(x*y
= {x*y.
Given (K, r) as above, let K' be the field generated over Q by YTi=\
flTi f°r a
^
a G K. We call K' the reflex Geld of (K, r). It can easily be shown that K' is a
CM-field and contains [JlLi aTi f o r all a G K (see [S98, pp.62-63, p.122, Lemma
18.2]).
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