Page 298 - Advanced Linear Algebra
P. 298
282 Advanced Linear Algebra
We can now show that the symplectic transvections generate the symplectic
group.
Theorem 11.19 Every symplectic transformation on a nonsingular symplectic
geometry is the product of symplectic transvections.
=
Proof. Let be a symplectic transformation on . We proceed by induction on
=
~ dim ²= ³. If ~ , then = ~ / ~ span ²"Á '³ is a hyperbolic plane and
Theorem 11.18 implies that there is a product of symplectic transvections on
= for which
¢ ²"Á '³ ª ² "Á '³
This proves the result if ~ . Assume that the result holds for all dimensions
less than and let dim ² = ³ ~ .
Now,
=~ / p A
where /~ span ²"Á '³ and is a symplectic geometry of dimension less than
A
that of . As before, there is a product of symplectic transvections on for
=
=
which
¢ ²"Á '³ ª ² "Á '³
and so
/ O~ O /
Note that c /~ / and so Theorem 11.9 implies that c ²/ ³ ~ / .
Since dim²/ ³ dim²/³ , the inductive hypothesis applied to the symplectic
transformation c on / implies that there is a product of symplectic
transvections on / for which ~ c . As remarked earlier, is also a
product of symplectic transvections on that is the identity on and so
/
=
/ O~ / and ~ on /
Thus, ~ on both and on / and so ~ is a product of symplectic
/
transvections on .
=
The Structure of Orthogonal Geometries: Orthogonal Bases
)
We have seen that no interesting that is, not totally degenerate symplectic
(
geometries have orthogonal bases. By contrast, almost all interesting orthogonal
geometries have orthogonal bases.
=
To understand why, it is convenient to group the orthogonal geometries into two
classes: those that are also symplectic and those that are not. The reason is that
all orthogonal nonsymplectic geometries have orthogonal bases, as we will see.
However, an orthogonal symplectic geometry has an orthogonal basis if and
