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Metric Vector Spaces: The Theory of Bilinear Forms 277
To see why, suppose that is an isometry and 8¢= ¦ > ~ ²# Á ÃÁ# ³ is an
=
ordered basis for . Then ~ ² # Á 9 Ã Á # ³ is an ordered basis for > and
4 ²= ³ ~²º# Á # »³ ~²º # Á # »³ ~4 ²> ³
8
9
Thus, the congruence class of matrices representing = is identical to the
congruence class of matrices representing > .
Conversely, suppose that = and > are metric vector spaces with the same
congruence class of representing matrices. Then if 8 ~²# Á Ã Á # ³ is an
9
=
ordered basis for , there is an ordered basis ~ ²$ ÁÃÁ$ ³ for > for which
²º# Á # »³ ~ 4 ²= ³ ~ 4 ²> ³ ~ ²º$ Á $ »³
9
8
Hence, the map defined by # ~ $ is an isometry from to > .
¢= ¦ >
=
We have shown that two metric vector spaces are isometric if and only if they
have the same congruence class of representing matrices. Thus, we can
determine whether any two metric vector spaces are isometric by representing
each space with a matrix and determining whether these matrices are congruent,
using a set of canonical forms or a set of complete invariants.
Symplectic Geometry
We now turn to a study of the structure of orthogonal and symplectic geometries
)
(
and their isometries. Since the study of the structure and the structure itself of
symplectic geometries is simpler than that of orthogonal geometries, we begin
with the symplectic case. The reader who is interested only in the orthogonal
case may omit this section.
Throughout this section, let be a nonsingular symplectic geometry.
=
The Classification of Symplectic Geometries
Among the simplest types of metric vector spaces are those that possess an
orthogonal basis. However, it is easy to see that a symplectic geometry has an
=
orthogonal basis if and only if it is totally degenerate and so no “interesting”
symplectic geometries have orthogonal bases.
Thus, in searching for an orthogonal decomposition of = , we turn to two-
dimensional subspaces and this puts us in mind of hyperbolic spaces. Let be
<
the family of all hyperbolic subspaces of , which is nonempty since the zero
=
=
subspace ¸ ¹ is singular and so has a nonzero hyperbolic extension. Since is
finite-dimensional, < has a maximal member > . Since > is nonsingular, if
> £= , then
=~ > p >
where > . But then if # > £ ¸ ¹ is nonzero, there is a hyperbolic extension