1. ALGEBRAIC PRELIMINARIES

11

We now consider our setting to be that of §1.4, with an algebraic number field

as F. Given x G (Fv)™ with v G h, we can naturally define uo(x) to be an integral

gv-ideal, taking gv to be A. Then we put

(1.16) u{x) = N(v0{x)) = [fl„ : i/0(*)].

If x = d~1c is a left reduced expression for x, then v{x) = | det(d)|~1. Moreover,

if uxv =

n n

with u G GLm(gv), v G GLn(gv), and a = diag[ai, ... , a

r

], then

i/(x)

- 1

is the product of \ai\v for all i such that a* ^ gv (see [S97, Lemma 3.8

(2)])-

1.8. Take our setting to be the same as in Cases I and II in §1.1 with an algebraic

number field as F. Namely, K = F or K is a quadratic extension of F. We denote

by r the ring of algebraic integers in K and by k the set of all nonarchimedean

primes of K. (Thus r = g and k = h if K = F.) Given v G k, an tv-ideal a, and

a matrix x with entries in Kv, we write x - a if all the entries of x belong to a.

Similarly, for a matrix y with entries in KA and an r-ideal b, we write y - b if

all the entries of yv belong to bv for every v G k.

Take two positive integers m and n. For x

c d

G GLm+n(K)A with

a G

(.KA)™

and d G (K/i)n write a = ax, b = 6X, c = cx, and d = dx. With fixed

t-ideals rj and 3 such that 93 C r, we put

(1.17) Cfo, 3] = {x G GLm+n{K)A I det(x)h G F U h tf

ax ""

r

frx - t), cx - 3, dx - r | .

We easily see that this is a subgroup of GL

m

+

n

(if )A (see [S97, §9.1]). We also note

that if x G C[t), 3] and t/|t)3, then (det(arc) det(dx) — det(#)) G t)v$v, and hence

we see that

(1.18) The map x »— • ((ax, 6x)u) 1 defines a homomorphism of C[t), 3] into

FIvloj [CLm(^/9i;3v) X GLn^/t)^^] .

1.9. Lemma. Define a subgroup p(

m

'

n

) of GLm+n(K) by

p(m n) = {xeGLm+n{K)\cx = 0}.

Let C denote the group C[tj, 3] 0/ (1.16). Then

p

{jn,n)c

=

1 ^ ^

G j L m + n (

^ )

A

J ( ^

€

GLn(ift;) and

(d"

1

^),, - iv for every v\t)} } .

Moreover, assuming m = n, /e£ G denote GU{r)n), U(rjn), or SU(nn) with r\n of

(1.8); putP = GH P("'

n

), and D = GAnC. Then

PAD = GAnP£'n)C

= { x G GA I (dx)v G GLn(K)v and (d~1cx)v - 3^forevery i/|tj3 } .

in particular, GLrnjrn (K)A = P^n)C and GA = PAD if 93 = r.

PROOF .

The assertions for GLm+n(K) and U(rjn) are proved in [S97, Lemma

9.2]. Combining the result for U(rjn) with [S97, Lemma 9.10 (2)], we obtain the

assertion for SU(rjn). As for GU(rjn), let x G GU(nn)A, P = diag[ln, v(a)ln],

and y = p~xx. Then y G U(r)n). Suppose (dx)v G GLn(K)v and (d" 1 ^)^ - 3^

for every v|t)3- Then we easily see that y satisfies the same conditions, and so