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 .