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# Determination of

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 . Suppose f is an arbitrary form in V. Then

Our goal is to study the orbits under G of f. To this end, we may (and frequently shall) freely replace f by its image under an element .

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

If f is in , then . It is this kind of equation that we shall use throughout to find restrictions on the coefficients . We shall divide the work according to the two cases found in Lemma 5.1.

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

then Gf contains a form with

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 12 and is equivalent to after interchanging t and z. If not, scale by to get . Now apply and then again to get . Lemma. Assume is as in Lemma 5.6.

If and , then f is equivalent to or to .

First, gives .

Scale to make .

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 rank 12. It can be factored as where i is the square root of -1. The transform then yields , the form on the list.

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

If , this is in 3 variables and we are done by Proposition 4.1. If not scale to get . Then gives which is .

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 ,

It leaves the fixed coefficients the same with the exception of which has coefficient . The coefficients of xyz, yzt, and have become , and , respectively. We will make choices of the in such a way that the coefficient of remains 0 and some of the other coefficients become 0. Lemma. Suppose f is as in Lemma 5.10 with and . If , then f is equivalent to such a form with .

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.

This means we can assume .

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 .

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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