Lemma. Each nonzero form is equivalent to a form as in Case 1 or Case 2:
Without loss of generality, we may assume that a term not involving
z is nonzero. We may then use the above Proposition on the
variables x,y, and t so that the terms in the form f involving
only these add up to
,
, or xyt. If it is xyt
replace t by t+x to get
. This means we can assume
Several times during the paper we
will transform by a diagonal matrix in G which amounts to transforming
each of x,y,z,t by various scalars. We refer to this as scaling.
In doing so, we will always without
reference be sure that the coefficients which have been fixed are
not changed.
Now scale to get (e.g., use the diagonal
matrix
).
In the sequel we
will denote transformations under G
by indicating the replacements to be performed. In particular, the
substitution of y by
will be given by
. Do this substitution
to obtain a new form in which
as well. The terms which were
assumed to be 0 are still 0;
the coefficients
, and
may have changed, the remaining terms have not.
If we are not yet in Case 1 or 2, we must have ,
,
and
.
Now scale to get
(e.g., by
).
Next apply
to obtain a new form in which the terms
that were assumed to be zero remain zero, except for
whose coefficient becomes
. Choose
so that this is not 0 and the form is as in Case 2.
A form f is in if all
submatrices of
have determinant 0. If some
submatrix is nonsingular,
then f is not in
. The method will be to take a form as
described above and find suitable
submatrices for which
the determinant 0 condition forces conditions on the coefficients.
For subsets J,K of of equal size,
we denote by
the determinant of the
submatrix of a square matrix A
obtained by excluding the rows J and
columns K. For example, the submatrix excluding rows 2,3
and columns 5,6 would have determinant denoted by
Case 1.
We take a form as described above in Case 1.
We have . We wish to
show
and so assume it is not.
However,
in order to do part of Case 2 we also must release
and
meaning they are not necessarily 0.
Lemma. If f satisfies
We can scale z (i.e., multiply z by the scalar ,
fixing the remaining variables),
so that the resulting coefficient
is 1. The coefficient
of
remains 1. We assume then that
.
Perform the substitution
to get a form in which
. No new terms appear and
and
are still
1.
Finally do the transformation
.
This sets
, adds no new terms and keeps the
nonzero terms as they were. When we transform this way,
we have always checked
that the transformation is invertible.
Lemma.
Let be as
in the previous lemma
and assume
.
Then Gf contains
.
Set equal to 1 by
scaling
,
,
,
.
Also,
by scaling
and multiplying by
we can
assume
as well.
Now
gives
and
gives
.
In particular,
.
Now factor
to get
.
The previous equation for
gives
and so
. If
, the transform
leads to
A word is in order here about how such tranformations are found.
For a given form f, its Lie algebra stabilizer can
be found as described in .1.
The Lie algebra for the form above consists of a 3-dimensional
commutative Lie algebra of linear tranformations all of which
have a unique eigenvector. The same is true of the Lie algebra of
the form
in Table 1. The transformation above is found by
producing an element of G which under conjugation transforms
one Lie algebra to the other.
Lemma.
Let
be as
in Lemma 5.2
and assume
,
.
Then Gf contains
.
As in the first paragraph of the proof of the previous lemma, we can
set .
Now
gives
.
Next,
gives
or
.
Here
gives
or
and so
.
Now
gives
.
Thus, the form is
We may now assume that . We are still assuming as well that
and
.
Lemma.
Suppose that is as
in Lemma 5.2.
If
,
then f is equivalent to one of
,
or
.
The equations and
give
and
, respectively.
But
gives
, so at least one of
,
must be 0. The form is now
If , scale by
to get
;
after interchanging
t and z this gives
, which was shown to be equivalent to
in the proof of Lemma 5.4.
If scale x and t by
to get
and the form is
which modulo cubics is
which is equivalent to
.
This is
.
Finally, if both and
are 0, it is
which becomes
equal to
by interchanging y and t.
These lemmas show that any form which satisfies the conditions of Lemma
is listed in the table. We can go back
to the beginning of Case 1 and assume
. We assume that
and
are 0 as our only need in Case 2 will
be when
.
Lemma.
Assume f is a form satisfying
and
.
Then f is equivalent to a form
with
and
.
Apply to get
. Notice
which had been 0 may now be nonzero.
Lemma.
Assume is as in
in Lemma 5.6.
If
,
then f is equivalent to
or
.
Set and then
gives
.
Now
gives
. Again
gives
. This means that
or
.
Suppose first that . Then
gives
and
the form is
. If
, this can be
scaled to
.
This, however, has rank 14 and so is not in
. It is easily seen to
be equivalent to
by the cyclic permutation
.
Otherwise,
and the form is
which
is
after interchanging t and z.
We can then assume and
. The form is now
If and
,
then f is equivalent to
or to
.
The equation gives
.
Now
gives
. If
, this is
. If not
scale to get
. This is
after interchanging x with z and t
with y.
To finish Case 1,
it remains to consider f as in Lemma 5.6 with
.
Lemma.
Assume f is as
in Lemma 5.6.
If
and
,
then f is equivalent to one of
,
,
,
,
.
First, assume . The substitution
gives
.
Scale to get
.
If , then f is a form in 3 variables, and hence known
by Propositon 4.1 to be equivalent to one of
,
,
.
If both and
are nonzero, scale by
If and
, then the form is
, which is readily seen to be
equivalent to
again.
If and
, then the form is
. Scaling to
and transforming via
, gives
after a permutation of variables.
It remains to consider the case .
The form is now
This completes Case 1.
Case 2.
In this case we assume ,
and
.
Lemma.
Assume f satisfies ,
and
.
Then f is equivalent to a form
with
and
.
First of all, we can set by scaling. Apply
to make
; this
keeps
, but releases
and
.
Lemma.
Suppose f is as in Lemma 5.10.
If
,
then f is equivalent to a form as in Lemma 5.2; in particular,
Gf contains a form in Case 1.
By scaling set .
Now make the transform
.
This gives a form in which
and
, that is, as required.
We may assume then
that .
At various stages during this case we will make the following
substitution, where
,
By scaling, set . The fact that
means we
can choose
and
as we wish and then choose
so that
after transforming by
,
the coefficient of
is 0. By choosing
and
and
, we can apply
to obtain a form in which
.
Now substitute
and then
. This
gives a form with
released but no other new nonzero
terms and
.
We could
proceed with this value for
but we will need to
do the case without this restriction anyway which we do below.
Lemma.
Suppose
is as in Lemma 5.10.
If
and
,
then f is equivalent to
.
By scaling set . Pick
and
and apply
to get a form in which
and
are 0. We must then choose
so that the coefficient of
is 0 which we may do.
Now
gives
and
gives
.
Moreover,
gives
. Now,
application of
and then
gives
.
After interchanging z and t, this becomes
of Table 1.
Lemma.
Suppose
is as
in Lemma 5.10.
If
then f is equivalent to a form in Case 1 or a form in three variables.
Again we distinguish between whether or not
. Suppose it is not and set it to 1. As
above we can make substitutions
with
,
and
to make
. Now let
be a cube root of
. Then transform by
. This gives
as well. But then, transforming by
gives a form in Case 1.
We can now
assume that
.
For the last time transform with
for
appropriately chosen
. As before
we can choose
to make
. We choose
,
so that the coefficient
of
remains 0.
The transformed coefficient
of
is
.
If
we can choose
so that
and
.
Then
gives the contradiction
.
This means we can assume but
is no longer restricted to 0.
Now
gives
. If
, the form is in 3 variables and we are done.
Assume not; we can then scale to .
Now apply
to get
. The form is now
This finishes Case 2 completely.
Thus, we have shown that all forms in are
as listed in Table 1.