Page 313 - Advanced Linear Algebra
P. 313
Metric Vector Spaces: The Theory of Bilinear Forms 297
It is not hard to see that the maximum dimension ²= ³ of a hyperbolic subspace
of = is $²= ³ , where $²= ³ is the Witt index of = . First, the nonsingular
extension of a maximal totally degenerate subspace < of is a hyperbolic
= $
space of dimension $²= ³ and so ²= ³ $²= ³ . On the other hand, there is a
contained in any hyperbolic space > and so
totally degenerate subspace <
$²=³, that is, dim ²> ³ $²=³. Hence ²=³ $²=³ and so
²=³ ~ $²=³.
Theorem 11.37 Let be a nonsingular orthogonal geometry over a field ,
=
-
with char²-³ £ .
)
1 All maximal hyperbolic subspaces of have dimension $ ² = . ³
=
)
2 Any hyperbolic subspace of dimension $²= ³ must be maximal.
)
3 The Witt index of a hyperbolic space > is .
The Anisotropic Decomposition of an Orthogonal Geometry
If is a maximal hyperbolic subspace of , then
=
>
=~ > p >
Since is maximal, > is anisotropic, for if " > were isotropic, then the
>
nonsingular extension of > p ² span " ³ would be a hyperbolic space strictly
larger than .
>
Thus, we arrive at the following decomposition theorem for orthogonal
geometries.
Theorem 11.38 The anisotropic decomposition of an orthogonal geometry)
(
Let =~ < p rad ²= ³ be an orthogonal geometry over , with char ²-³ £ . Let
-
> , where > be a maximal hyperbolic subspace of < ~ ¸ ¹ if < has no
isotropic vectors. Then
=~ : p > p rad ²= ³
where is anisotropic, is hyperbolic of dimension $ ² = ³ and rad ² => ³ is
:
totally degenerate.
Exercises
1. Let <Á > be subspaces of a metric vector space . Show that
=
a ) < > ¬ > <
)
b < <
c ) < ~ <
=
2. Let <Á > be subspaces of a metric vector space . Show that
a ) ²< b > ³ ~ < q >
)
b ²< q > ³ ~ < b >
3. Prove that the following are equivalent:
a = is nonsingular
)
)
b º"Á %» ~ º#Á %» for all % = implies " ~ #
