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 x° 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]).