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In this section we show any nonzero form in is equivalent under the action of to one of the forms, , in Table 1 with . We do this in a series of lemmas which themselves provide an algorithm for transforming an arbitrary form in into the appropriate .

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

which is the case . Thus, we may assume . But now the transformation gives the form .
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

Finally, transform this form according to . This gives , that 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

Scale by to get . Now apply to get the form . If , the form is which has rank
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

to get . Now apply to get the form which has rankIf 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

If , this is inThis 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 ,

It leaves the fixed coefficients the same with the exception of which has coefficient . The coefficients of
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

Again set to get which is a form as in Case 1.This finishes Case 2 completely. Thus, we have shown that all forms in are as listed in Table 1.

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