Under the above correspondence every ``ruled" geometric object in is
represented
by a set of points on
in
. Thus a linear complex
is
represented
by the intersection of a hyperplane
of
with
(cf.
2.1).
will be tangent to
if
is special, otherwise
will be a non-degenerate quadric of
. As for a linear
congruence of
, one has to intersect the hyperplanes
representing the corresponding complexes and then intersect this 3-dimensional
subspace of
with
. Thus a linear congruence
is represented
by an ``ordinary" quadric lying on
. According to
being
degenerate, parabolic, hyperbolic or elliptic, the quadric will be degenerate
into two planes, a cone, or a non-degenerate quadric of index 2 and 1
respectively. In the same way, a regulus of a ruled quadric of
can be
seen as the intersection with
of a plane of
, i.e. as a conic
lying on
.
Let be the form defining the Klein's quadric
, let
and
be the
exterior square of
. Then, for every
,
.
Thus the map
induces a homomorphism from
to
.
Let
,
be the projective transformations corresponding to
and
respectively. It is easy to check that
iff
is a square. It follows that
is isomorphic
to the subgroup of
consisting of the elements having determinant a
square. From this we get an isomorphism between
and
.
Now
preserves the non-special linear complex
consisting of the
isotropic lines of
, which is represented in
by the
intersection of a hyperplane
with
. This is
a non-degenerate quadric of
. Hence we get a monomorphism of
into
, i.e. an isomorphism between
and
.
(For details see [VdW] and [Dieu].)