6 I. GENERALIZE D UNITAR Y GROUP S
1.9. Lemma, Suppose that the characteristic of K is not 2. Put
Z={zeIQ\ l + zeGL n(K)},
Y = G*nz, x = {xez\xif = -px*}.
Put also f(z) = (l z)(l +
z)~1
for z e Z. Then the following assertions hold:
(1) / gives a bijection of Z onto itself, and its inverse is f.
(2) / gives a bijection of X onto Y.
(3) Suppose K is commutative and p = idx', then det(y) = 1 for every y E Y,
and det( l + w) = 0 if w G G^ and det(w) ^ 1.
Proof. Sinc e 1 z commutes with 1 + z, w e can write also f(z) = (l +
z)~1(l
z).
Put h(x) = x 1 and g(x) = 2x~
x
fo r x G GLn(K). The n h i s a bijectio n o f
GLn(K) ont o Z an d f(h(x)) = ( 2 x)x
_ 1
= h(g(x)). Sinc e # i s a bijectio n o f
GLn(K) ont o itself an d g(g(x)) = x, w e obtain (1). Next , i f x f(y) wit h y G Z,
then ( 1 + i/) x = 1 y, s o tha t
(l + y)(xp + yx*)( i + y* ) = ( i - y)v( i + /* ) + ( l + y)y( i - y*)
= 2((p-ytpy*).
This shows that x G l i f and only if y G Y, which proves (2) . Finally , in the settin g
of (3) , fo r x G X w e have ( 1 x)p = ip(l + *#) , and s o det( l x) = det( l + a;) .
Thus det[/(x) ] = 1, which prove s the firs t par t o f (3) . I f w G G^ an d det(w ) ^ 1,
then w £ Z, so that det( l + u ) = 0 .
1.10. Suppos e no w K i s a field and ip is nondegenerate; pu t
F={xeK\xp = x}.
Then F i s a subfield o f K , an d [i( T : F] = 1 or 2 according a s p = id/ : or p ^ idx .
Let £ - N
K/F
(KX) ii K ^ F and E = {a 2 \a e F x } if K = F. Let p
0
b e a s i n
§1.7. Sinc e det(fp
0
f*) = det(y
0
)det(f)det(f)p G det(^0 )^, th e cose t det(ip
0
)E i n
i^
x
/i£ is completely determined b y (p. We call the coset det(ipo)E the discriminant
of ip and denote it by d((p). I f e = 1, then clearly det(y?o ) £ ^
x
»
a n
d d(/? ) belong s
to
Fx
/i£. If n i s odd and e = 1, then for c G
Fx
w e have d(cy ) = c
nd(p)
= cd(p).
Therefore, give n y with od d n an d £ = 1, w e ca n alway s tak e c G
Fx
s o tha t
d(op) i s represented b y a given elemen t o f F
x.
\i K ^ F an d £ = —1, w e ca n reduc e th e proble m t o th e cas e e = 1. Indeed ,
take c K
x
s o tha t c
p
= —c . The n cy ? i s hermitia n an d G* = G
C(p.
However ,
as we shall se e later, i t i s sometimes convenien t t o conside r a skew-hermitian for m
instead o f a hermitian form .
1.11. Lemma. Suppose that K is a held, dim(V ) = 2 , and ip is nondegenerate.
Then (p is isotropic if and only if d(p) is represented by —e.
Proof. If (p i s isotropic , the n Lemm a 1.6 show s tha t (V , (p) is isomorphi c t o
(Hi, 7/i ) an d s o d(p) i s represente d b y —e. To prov e th e converse , suppos e tha t
d(ip) i s represente d b y e an d ip(x, x) ^ 0 fo r ever y x G V, ^ 0 . The n th e las t
part o f th e proo f o f Lemm a 1.8 show s tha t V = Kw + Ku wit h som e w an d u
such tha t p(w, u) = 0 . Pu t a = p(w, w) an d 6 = ip(u, u). The n a & = —ecc
p
with c G
ifx.
Le t x bw + cw ; the n x ^ 0 . Sinc e fr
p
= efr , w e hav e /?(# , x) =
abb9
+ 6cc
p
= b(eab + cc
p)
= 0 , a contradiction. Thi s complete s th e proof .
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