Tutorial

V. Jungic, L. Robinson

This tutorial is designed to accompany result given by Arjeh M. Cohen and David B. Wales in [2] and [3]. The basic definitions and some details and examples necessary for understanding the result and its proof are given as well as a sketch of the proof. The reader will find a few suggestions as to how Maple could be used for proving and checking some parts of the proof.

Let k be an algebraically closed field of characteristic 3 and let be the general linear group of degree 4 over k (i.e. the group of invertible matricies of order 4 with entries in k). Let be the set of all homogeneus polynomials in four variables over k, of degree n. Thus

and

Let

and let

Thus

if and only if:

(We identify M with any

, usually with f such that, without adding new monomials, f can not be written in the form

) V is a 16 - dimensional vector space over k. An action of G on V , i.e. a function

is defined in the following way. If

and

, let

be such that

Then

is given by

We say that V is a kG - module.For

, the G - orbit of f is the set

. Let

Theorem 1

There are exactly ten G - orbits of vectors in . Representative vectors for these orbits are listed above.

We can formulate the Theorem in the following way:

Chen Zhijie [1] found eight of these orbits and conjuctured that there were no other G - orbits. The remaining two, and , were found by Cohen and Wales.The above kG - modul V was one of the open cases in a prospective classification of irreducible modules for almost simple algebraic groups over an algebraically closed field of positive characteristic for which there are a finite number of points. P is the space of all cubic hypersurfaces on projective space. Considering hypersurfaces in the same -orbit as "equivalent" means working modulo a choice of coordinates. For example, a view on the real points in the hyperplane t=1 of the cubic hypersurface

is given on the following picture.

We will follow the idea of the proof given in [2].

Let be the general linear algebra (i.e. the algebra of all matricies of order 4 with entries in k).We consider g as a Lie algebra. A Lie algebra g is a vector space together with a bilinear map

such that

and

In our case

. We shall use as basis

, where

and, for

. If

and

, the action of g on

is given by

where

For example, since

we have

This then determines the action of g on

, and hence the action on V, in the following way. If

and

where

, then

For example, let

be such that

As a representative of M we take

Let

Then

Hence,

and its representative is

Notice that

and, in general, if

As a basis for V, we shall use the 16 monomials in x, y, z, t which are not cubes. Let us denote by the l - th monomial in the second row of the array below. The action of g on V can now be explicitly given in terms of the following matrix where is written as . Here is matrix whose ij - th row consists of the coefficients of written out on the basis of V just given. In the array below, the row beginning with represents the ij - th row of , that is the vector such that

For example, Hence, and for , . From the forth row ( which begins with ) we have and, for , . Hence

For

, the Lie stabilizer of f is

.We compute

in the following way (and Maple can do this). {

.
Form: .
Solve: (mod 3);.
Then .

For example:

If , then

.
:

.
.
.

is a Lie subalgebra of g .

The proof of the fact that

is based on the following Lemma.

Lemma 2

Let and and let h=Yf. Then .

Proof Let

and let

. Then

Obviously

Let

Then

and

Let

Then

and

Hence

and

Therefore

and the Lemma is proved.

Hence, if f and h belong to the same orbit then and are isomorphic as Lie algebras. To prove

the idea is to compute

, and to verify that, for

, the Lie algebras

and

are not isomorphic. The list of

follows:

Since dim

, dim

dim

, dim

dim

, dim

, and dim

, it remains to consider two cases:

and

. It is easy to check that

(i. e.

is abelian). But,for

and

we have

and

. Hence,

is not abelian. Since a Lie algebra isomorphism preserves the property of being abelian, it follows that

and

are not isomorphic. Our way of proving that

has two steps. First, ask Maple to find the set of all

such that

. The answer is

If there is any

such that

, then

. Now, ask Maple to calculate

, where

is written in general form. The answer is

Since

we have

Therefore

To prove , Cohen and Wales consider two cases. If and 16 - dim then there is such that . An algorithm how to construct such that is given. Secondly, it is shown, in a rather complicated manner, that