Page 297 - Advanced Linear Algebra

Metric Vector Spaces: The Theory of Bilinear Forms 281

We now wish to prove that any symplectic transformation on a nonsingular
symplectic geometry is the product of symplectic transvections. The proof is
=
not difficult, but it is a bit lengthy, so we break it up into parts. Our first goal is
to show that we can get from any hyperbolic pair to any other hyperbolic pair
using a product of symplectic transvections.

Let us say that two hyperbolic pairs ²%Á &³ and ²$Á '³ are connected if there is a
product of symplectic transvections that carries to and to and write
'
&
$
%

¢²%Á&³ ª ²$Á'³
or ²%Á &³ © ²$Á '³ . It is clear that connectedness is an equivalence relation on
the set of hyperbolic pairs.
Theorem 11.18 In a nonsingular symplectic geometry = , every pair of
hyperbolic pairs are connected.
Proof. Note first that if º Á !» £ , then £ ! and so
Á

!cÁ ~b º c ! c !³ ~c º ! c !³
»
»
²
Á
²

Taking ~ °º Á » gives c Á ! ~ ! . Therefore, if ² Á "³ is hyperbolic, then we
!
can always find a vector for which
%
² Á "³ © ²!Á %³
namely, %~ c!Á ". Also, if both ² Á "³ and ²!Á "³ are hyperbolic, then
² Á "³ © ²!Á "³
since º c !Á "» ~ and so c!Á " ~ . "
Actually, these statements are still true if º Á !» ~ . For in this case, there is a
&
nonzero vector for which º Á & » £ and ! º Á & » £ . This follows from the fact
!
that there is an  = i for which £ and £ and so the Riesz vector 9

is such a vector. Therefore, if ² Á "³ is hyperbolic, then
² Á "³©²&Á "³©²!Á c&Á &c!Á c&Á "³

Z
and if both ² Á "³ and ²!Á "³ are hyperbolic, then
² Á"³ © ²&Á"³ © ²!Á"³
Hence, transitivity gives the same result as in the case º Á !» £ .

Finally, if ²" Á " ³ and ²#Á #³ are hyperbolic, then there is a & for which

²" Á " ³ © ²# Á &³ © ²# Á # ³

and so transitivity shows that ²" Á " ³ © ²# Á # ³.