On generators of the algebra of polynomial invariants of group F_{4}

A.V.Iltyakov School of Mathematics and Statistics University of Sydney 2006, N.S.W. Australia

Abstract

We describe the algebra of polynomial invariants of a system of 3 vectors of an irreducible 26-dimensional representation of a simple algebraic group of type F_{4} (the ground field is algebraically closed of characteristic zero). Generators are given in the form of trace polynomials of the Albert algebra (an exceptional simple Jordan algebra); a system of homogeneous parameters and a basis over it are also found.

1 Introduction

The problem of finding generators of the algebra of invariants of a system of vectors with respect to a linear algebraic group is classical ( see, e.g., [17]); this paper is devoted to the problem of extending the results of [14, 15] on invariants of a simple algebraic group of type G_{2} to that of F_{4}.

In the sequel the ground field F is algebraically closed of characteristic zero.

Notations.

∘ End_{F} V is the algebra of linear transformations of a vector space V;

∘ vect_{F} X is the subspace spanned by a subset X of a vector space over F;

∘ alg_{F} X is the subalgebra generated by a subset X of an algebra over F;

∘ F [V] (respectively, F (V)) is the algebra (field) of polynomial (rational) functions from a vector space V to F;

∘ V : G is a G-module V, i.e., V is a vector space and G is closed subgroup of the group GL(V) of non-singular linear transformations of V;

∘ F [V]^{G} (F (V)^{G}) is the subalgebra (the subfield) of polynomial (rational) invariants of V : G;

∘ kV = \underbrace{V ⊕ … ⊕ V}_{k} is the direct sum of k copies of a vector space V; ∘ Pol(V, W) ≃ F [V] ⊗_{F} W is the space of polynomial mappings from a vector

space V to a vector space W.

It is well-known that the minimal linear representations of exceptional algebraic groups are connected with some non-associative algebras. Thus, the automorphism group G = Aut(O) of the Cayley-Dickson algebra O is a simple algebraic group of type G_{2} [13]. It is known that every element x ∈ O satisfies the quadratic identity

x^{2} − t(x)x + n(x) = 0 ,

where t(x) is the trace of x. As a G-module O is the direct sum of a 1-dimensional trivial submodule and an irreducible 7-dimensional submodule O′ = {a ∈ O| t(a) = 0}.

In [14, 15] it was proven that the algebra of invariants F [kO]^{G} with respect to the diagonal action (the system of k vectors of O : G) is generated by trace polynomials, which can be defined in the following way.

Let X_{i} is the projector of kO onto the i-th summand (a generic element of the algebra O), then X_{1}, …, X_{k} lie in the algebra Pol(kO, O) with respect to the pointwise operations. The subalgebra F_{k}(O) generated by X_{1}, …, X_{k} is called the algebra of generic elements of O of rank k; for every mapping u ∈ F_{k}(A) we can consider a polynomial function t(u) ∈ F [kO] which is the composition of u with t(x). Moreover, t(u) is a G-invariant, and the subalgebra t[kO] generated by 1 and all such polynomials is called the algebra of trace polynomials of O and it is equal to F [kO]^{G} by the result cited above. Respectively, F [kO′]^{G} is generated by trace polynomials in generic elements X′_{i} with t(X′_{i}) = 0.

A similar conjecture makes sense in the case of the Albert algebra A (for definition see, e.g., [8, 16] or section 3); it is known that G = Aut(A) is a simple algebraic group of type F_{4} (e.g., [7]); also, every element x ∈ A satisfies a cubic equation

x^{3} − tr(x)x^{2} + S(x)x − N(x) = 0 ,  (1)

where tr(x) is the trace of x. The G-module A is decomposed into the sum of a 1-dimensional one and 26-dimensional irreducible module A′ = {a ∈ A| tr(a) = 0}.

Denote by tr[kA] (tr[kA′]) the subalgebra of F [kA]^{G} (F [kA′]^{G}) of trace polynomials, i.e., the subalgebra generated by tr(u), where u is a monomial in X_{1}, … X_{k} (X′_{1}, …, X′_{k}). It is known that that F [kA]^{G} (F [kA′]^{G}) is integral over tr[kA] (tr[kA′]), their spectra are homeomorphic and their fields of fractions coincide; in particular, for k = 1, 2 they are equal [4].

The case k = 3 is special: the action of G on 3A is almost free, that is, the stabilizer of a point in general position is trivial. Indeed, the algebra A can be generated by 3 elements [16, p.58,ex.12], hence, the set of points (x, y, z) ∈ 3A such that x, y, z generate A is non-empty and, obviously, it is open in 3A. Besides, if ϕ ∈ G fixes x, y, z, then ϕ is the identity mapping.

The main result of this paper is the following theorem.

Theorem 1 Let A be the Albert algebra over an algebraically closed field F of characteristic 0, let G be the automorphism group of A. Then the algebra of polynomial invariants F [3A]^{G} coincides with the algebra of trace polynomials tr[3A].

Recall that the subspace A′ of zero trace elements is an irreducible G-module of minimal dimension and 3A′ : G is almost free (A can be generated by 3 elements in A′ [16, p.58, ex.12]).

Corollary 1 The algebra F [3A′]^{G} is generated by 1 and trace polynomials tr(u), where u ranges over all monomials in the generic elements with zero trace.

As a matter of fact, to prove it we need first to study the structure of tr[3A] in detail; in particular, we find a system of homogeneous parameters and a basis over it for tr[3A] (that is, tr[3A] is a Cohen-Macauley algebra) and then we prove theorem 1; thus, they become eventually such a system and a basis for F [3A]^{G} (theorem 2); finally, it yields a similar decomposition for F [3A′]^{G}.

2 System of homogeneous parametres

Let x, y, z be the generic elements of A; denote by Q the subalgebra of tr[3A] generated by 1 and the following polynomials:

(G1) tr(x^{i}), tr(y^{i}), tr(z^{i}), where i = 1, 2, 3;

(G2) tr(x^{i}y^{j}), tr(y^{i}z^{j}), tr(z^{i}x^{j}), where 1 ≤ i, j ≤ 2;

(G3) tr(x^{i}y^{j} · z^{k}), where 1 ≤ i, j, k ≤ 2

The following statement is due to I.P.Shestakov (an oral communication); also, cf. [9, lemma 2].

Proposition 1 The algebra of trace polynomials tr[3A] is generated by Q and

ω = tr(((zx)y)^{2} · (zy)x)  (2)

We will divide the proof into several steps.

Recall that the Albert algebra A is a Jordan one, i.e., it is commutative and it satisfies the Jordan identity:

(a^{2}b)a = a^{2}(ba) ;  (3)

from the definition it follows that F_{k}(A) is also a Jordan algebra for all k.

Now let A be a Jordan algebra, for a ∈ A denote by R_{a} the operator of the (right) multiplication : b → ba; therefore, R_{a} ∈ End_{F} (A). Let B be a subalgebra of A, put R^{A}(B) = alg_{F} {R_{b}| b ∈ B} ⊆ End(A).

From the commutative law and (3) one can get the following identities [8, p.34]:

R_{a}R_{b}R_{c} + R_{c}R_{b}R_{a} + R_{(ac)b} = R_{ab}R_{c} + R_{bc}R_{a} + R_{ca}R_{b} ;  (4)
[R_{ab}, R_{c}] + [R_{bc}, R_{a}] + [R_{ca}, R_{b}] = 0 ,  (5)

where [s, t] is the commutator st − ts.

Lemma 1 [16, p.67] Let A be a Jordan algebra and let B be the subalgebra generated by a subset X ⊆ A. Then R^{A}(B) = alg_{F} {R_{a}, R_{ab}| a, b ∈ X}.

Let A be a Jordan algebra satisfying the identity

a^{3} = 0 .  (6)

Lemma 2 The following identities hold in A:

R_{a^{2}} = −2R_{a}^{2} ; R_{ab} = −R_{a}R_{b} − R_{b}R_{a} ;  (7)
R_{a^{2}}R_{a} = 0 ; R_{ab}R_{c} + R_{bc}R_{a} + R_{ca}R_{b} = 0 ;  (8)
R_{a}R_{b}R_{a} = −1/2R_{ba^{2}} ;  (9)
R_{a^{2}}R_{b} = −2R_{ab}R_{a} ;  (10)
2(ab)^{2} = −a^{2}b^{2}  (11)
2(ab)(ac) = −a^{2}(bc)  (12)

Proof. The identity (7) is a consequence of the linearized (6). The first identity in (8) follows from (3) and (7); the second one is its linearization. Identity (9) follows from (4) and (8); next, (10) is a consequence of (8). Finally, (11) follows from the previous one and (12) is a linearization of (11). □

Notation: let v_{1}, …, v_{m}, u_{1}, …, u_{n} be elements in a vector space, we write

v_{1}, …, v_{m} → u_{1}, …, u_{n}

if v_{1}, …, v_{m} belong to the subspace spanned by u_{1}, …, u_{n}.

Now we suppose that A is a free algebra in the variety of Jordan algebras defined by (6) in the free generators x, y, z [16, chapter 1].

Lemma 3 Let u = u(x, y, z) ∈ A, then

(i)

u → xU_{1}, yU_{2}, zU_{3} ,

where U_{1}, U_{2}, U_{3} ∈ alg_{F} {R_{x}, R_{y}, R_{z}}, (ii) if deg(u) ≥ 4, then

u → x^{2}U_{1}, y^{2}U_{2}, z^{2}U_{3} ,

where U_{1}, U_{2}, U_{3} ∈ alg_{F} {R_{x}, R_{y}, R_{z}}.

Proof. Part (i) follows from (7). Next, thanks to (7) we may assume that the initial part of u is equal to xR_{y}R_{z}R_{x}, hence, by (7), (12)

xR_{y}R_{z}R_{x} → xR_{y}R_{x}R_{z}, (zx)(xy) → x^{2}R_{y}R_{z}, x^{2}R_{zy} → x^{2}R_{y}R_{z}, x^{2}R_{z}R_{y} .

□

Now we are going to describe homogeneous generators of the vector space A. Let U(k) be the subspace of A spanned by all monomials in x, y, z of the total degree k. For polynomials u, v we write u ≃ v, if u is equal to v where some variables are swept. Denote by V (i, j, k) the subspace of homogeneous polynomials f, where deg_{x}(f) = i, deg_{y}(f) = j, deg_{z}(f) = k. Also, let

V (r) = ∑_{i+j+k=r,i≥j≥k} V (i, j, k) .

Notice that for every monomial u ∈ U(k) there is v ∈ V (k) such that u ≃ v.

Lemma 4 The subspace V (k) is spanned by a set of monomials M(k), where (a1) M(1) = {x}; (a2) M(2) = {x^{2}, xy}; (a3) M(3) = {x^{2}y, xR_{y}R_{z}, xR_{z}R_{y}} ; (a4) M(4) = M(2, 2, 0) ∪ M(2, 1, 1), where

M(2, 2, 0) = {x^{2}R_{y^{2}}} , M(2, 1, 1) = {x^{2}R_{y}R_{z}, x^{2}R_{z}R_{y}} ;

(a5) M(5) = M(3, 1, 1) ∪ M(2, 2, 1), where

M(3, 1, 1) = {x^{2}R_{y}R_{z}R_{x} } , M(2, 2, 1) = vect{x^{2}R_{y^{2}}R_{z} , x^{2}R_{z}R_{y^{2}}} ;

(a6) M(6) = M(3, 2, 1) ∪ M(2, 2, 2), where

M(3, 2, 1) = {x^{2}R_{z}R_{y^{2}}R_{x}},
M(2, 2, 2) = {x^{2}R_{y^{2}}R_{z^{2}}, x^{2}R_{z^{2}}R_{y^{2}}} ;

(a7) M(7) = M(3, 3, 1) ∪ M(3, 3, 2), where

M(3, 3, 1) = {zR_{x^{2}y^{2}}R_{x}R_{y}} , M(3, 2, 2) = vect{z^{2}R_{x^{2}y^{2}}R_{x}, y^{2}R_{x^{2}z^{2}}R_{x}} ;

(a8) M(8) = {z^{2}R_{x^{2}y^{2}}R_{x}R_{y}, zR_{x^{2}y^{2}}R_{x}R_{y}R_{z}} ; (a9) M(9) = {z^{2}R_{x^{2}y^{2}}R_{x}R_{y}R_{z}, y^{2}R_{x^{2}z^{2}}R_{x}R_{z}R_{y}}. (a10) M(k) = 0 for all k ≥ 10.

Proof. We will check it by the induction on k. Notice that (a1), (a2) are trivial and (a3) follows easily from (7). By lemma 3, the subspace U(k + 1) is spanned by

v ∈ {uR_{x}, uR_{y}, uR_{z}},

where u runs over monomials in U(k). Also, all such v satisfying

deg_{x}(v) ≥ deg_{y}(v) ≥ deg_{z}(v)  (13)

generate V (k + 1). To get all them it suffices to pick u ∈ M(k) (by induction, M(k) generates V (k)), then to multiply it by t ∈ {x, y, z} and to check that every monomial v ≃ uR_{t} with property (13) belong to M(k). It is equivalent to the following conditions: (*) for every v ∈ M(k + 1) and for every permutation σ of the variables if σ(v)

satisfies (13), then σ(v) ∈ vect_{F} M(k + 1); (**) for all u ∈ M(k) and t ∈ {x, y, z} there is a suitable permutation of the

variables x, y, z which takes the monomial uR_{t} to vect_{F} M(k + 1).

For the rest of the proof of lemma 4 the relation ≡ means the equality modulo vect_{F} M(k + 1).

(a4) We have k = 3; the condition (*) for M(4) is evident (it follows from the com-

mutative law).

Next, let us check (**): consider u = x^{2}R_{y}; by (8) we have

uR_{x} = x^{2}yR_{x} = 0; uR_{y} = x^{2}R_{y}R_{y} = −1/2x^{2}y^{2} ≡ 0 ; uR_{z} = x^{2}yR_{z} ≡ 0 ;

if u = (xy)z, then by (9), (7)

uR_{x} = (xy)zR_{x} → (x^{2}z)y ≡ 0; (xy)zR_{y} ≃ (xy)zR_{x} ≡ 0 ;
(xy)zR_{z} → z^{2}(xy) ≃ x^{2}(zy) → x^{2}R_{y}R_{z}, x^{2}R_{z}R_{y} ≡ 0 .

In a similar way one can check the case u = (xz)y.

(a5) Now k = 4; for M(5) condition (*) follows from (7), (8):

x^{2}R_{z}R_{y}R_{x} = −x^{2}R_{yz}R_{x} − x^{2}R_{y}R_{z}R_{x} = −x^{2}R_{y}R_{z}R_{x} ≡ 0;
y^{2}R_{z}R_{x^{2}} → x^{2}R_{z}R_{y^{2}} ≡ 0, x^{2}R_{y^{2}}R_{z} ≡ 0 .

Next, let u = x^{2}y^{2}, then, by (8),

x^{2}y^{2}R_{x} = 0 ; x^{2}y^{2}R_{y} = 0 ; x^{2}y^{2}R_{z} ≡ 0 ;

if u = x^{2}R_{y}R_{z}, then by (9),

uR_{x} = x^{2}R_{y}R_{z}R_{x} ≡ 0; uR_{y} = x^{2}R_{y}R_{z}R_{y} → x^{2}y^{2}R_{z} ≡ 0 ;
uR_{z} = x^{2}R_{y}R_{z}R_{z} → x^{2}R_{y}R_{z^{2}} ≃ x^{2}R_{z}R_{y^{2}} ≡ 0 ;

for u = x^{2}R_{z}R_{y} it can be checked similarly. (a6) We have k = 5; the property (*) follows from (7). Next, let u = x^{2}R_{y}R_{z}R_{x},

then

uR_{x} = x^{2}R_{y}R_{z}R_{x}R_{x} → x^{2}R_{y}R_{z}R_{x^{2}} = yR_{x^{2}}R_{z}R_{x^{2}} = 0;
uR_{y} = x^{2}R_{y}R_{z}R_{x}R_{y} → x^{2}R_{y}R_{zx}R_{y} → x^{2}R_{y^{2}}R_{zx}, x^{2}R_{zx}R_{y^{2}} →
→ x^{2}R_{y^{2}}R_{z}R_{x}, x^{2}R_{z}R_{x}R_{y^{2}} → x^{2}R_{z}R_{y^{2}}R_{x} ≡ 0 ;
uR_{z} = x^{2}R_{y}R_{z}R_{x}R_{z} → x^{2}R_{y}R_{z^{2}}R_{x} ≃ x^{2}R_{z}R_{y^{2}}R_{x} ≡ 0 .

Now let u = x^{2}y^{2}R_{z}:

uR_{x} = x^{2}y^{2}R_{z}R_{x} → x^{2}R_{z}R_{y^{2}}R_{x} ≡ 0 ;
uR_{y} = x^{2}y^{2}R_{z}R_{y} ≃ x^{2}R_{y^{2}}R_{z}R_{x} → x^{2}R_{z}R_{y^{2}}R_{x} ≡ 0 ;
uR_{z} = x^{2}R_{z}R_{y^{2}}R_{z} → x^{2}R_{y^{2}}R_{z^{2}} ≡ 0, x^{2}R_{z^{2}}R_{y^{2}} ≡ 0 .

(a7) The case k = 6; let us check (*): by (7) and (11),

zR_{y^{2}x^{2}}R_{y}R_{x} → zR_{(xy)^{2}}R_{xy} , zR_{x^{2}y^{2}}R_{x}R_{y} ≡ 0 .

Next, let u = x^{2}R_{z}R_{y^{2}}R_{x}, then

uR_{x} = x^{2}R_{z}R_{y^{2}}R_{x}R_{x} → zR_{x^{2}}R_{y^{2}}R_{x^{2}} = 0;
uR_{y} = x^{2}R_{z}R_{y^{2}}R_{x}R_{y} → zR_{x^{2}}R_{y^{2}}R_{x}R_{y} → zR_{x^{2}y^{2}}R_{x}R_{y} ≡ 0;
uR_{z} = x^{2}R_{z}R_{y^{2}}R_{x}R_{z} → x^{2}R_{z}R_{y^{2}}R_{z}R_{x}, x^{2}R_{z}R_{y^{2}}R_{zx} →

→ x^{2}R_{y^{2}z^{2}}R_{x} = 0, xR_{y^{2}(zx)^{2}} → xR_{y^{2}(x^{2}z^{2})} = z^{2}R_{x^{2}}R_{y^{2}}R_{x} → z^{2}R_{x}R_{y^{2}}R_{x^{2}} ≡ 0 ;

for u = x^{2}R_{y^{2}}R_{z^{2}}:

uR_{x} = x^{2}R_{y^{2}}R_{z^{2}}R_{x} → z^{2}R_{x^{2}y^{2}}R_{x} ≡ 0 ;
uR_{y} ≃ x^{2}R_{y^{2}}R_{z^{2}}R_{x} ≡ 0 ;
uR_{z} = 0 ;

the case u = x^{2}R_{y^{2}}R_{z^{2}} can be checked in a similar way.

(a8) Now k = 7; the condition (*) follows from (12) and (7). Let u = zR_{x^{2}y^{2}}R_{x}R_{y},

then

uR_{x} = zR_{x^{2}y^{2}}R_{x}R_{y}R_{x} → zR_{x^{2}}R_{y^{2}}R_{yx^{2}} → zR_{x^{2}}R_{y^{2}}R_{y}R_{x^{2}} = 0, zR_{x^{2}}R_{y^{2}}R_{x^{2}}R_{y} = 0;

uR_{y} = zR_{x^{2}y^{2}}R_{x}R_{y}R_{y} → zR_{x^{2}}R_{y^{2}}R_{x}R_{y^{2}} = 0;

uR_{z} = zR_{x^{2}y^{2}}R_{x}R_{y}R_{z} → zx^{2}R_{y^{2}}R_{x}R_{y}R_{z} → zx^{2}R_{y^{2}}R_{z}R_{y}R_{x} , zx^{2}R_{y^{2}}R_{y(zx)} →

→ z^{2}R_{y^{2}}R_{x^{2}}R_{y}R_{x} , (zx)xR_{y(zx)}R_{y^{2}} → z^{2}R_{y^{2}}R_{x^{2}}R_{y}R_{x}, (zx)^{2}(yx)R_{y^{2}} →

→ z^{2}R_{y^{2}}R_{x^{2}}R_{y}R_{x}, z^{2}R_{x^{2}}R_{y^{2}}R_{x}R_{y} → z^{2}R_{x^{2}y^{2}}R_{y}R_{x} → z^{2}R_{x^{2}y^{2}}R_{x}R_{y} ≡ 0 .

Let u = z^{2}R_{x^{2}y^{2}}R_{x}, then

uR_{x} = z^{2}R_{x^{2}y^{2}}R_{x}R_{x} = 0;
uR_{y} = z^{2}R_{x^{2}y^{2}}R_{x}R_{y} ≡ 0;
uR_{z} = z^{2}R_{x^{2}y^{2}}R_{x}R_{z} ≃ y^{2}R_{x^{2}z^{2}}R_{x}R_{y} ≡ 0 .

The case u = y^{2}R_{x^{2}z^{2}}R_{x} is similar.

(a9) Now k = 8; let us check condition (*). It suffices to show that V′(9) is invariant

under two transpositions: (yz) and (xy). The first case is trivial; the latter one: by (7)

z^{2}R_{x^{2}y^{2}}R_{y}R_{x}R_{z} ≡ 0 ;

next, by (7) and (9),

x^{2}R_{y^{2}z^{2}}R_{y}R_{z}R_{x} = z^{2}R_{y^{2}}R_{x^{2}}R_{y}R_{z}R_{x} → z^{2}R_{x^{2}y^{2}}R_{y}R_{z}R_{x} →
→ z^{2}R_{x^{2}y^{2}}R_{z(xy)}, z^{2}R_{x^{2}y^{2}}R_{x}R_{z}R_{y} ≡ 0 → z^{2}R_{(xy)^{2}}R_{z(xy)} = 0 .

Now, let u = z^{2}R_{x^{2}y^{2}}R_{x}R_{y}, then

uR_{x} = z^{2}R_{x^{2}y^{2}}R_{x}R_{y}R_{x} → z^{2}R_{x^{2}}R_{y^{2}}R_{x}R_{y}R_{x} →
→ z^{2}R_{x^{2}}R_{y^{2}}R_{yx^{2}} → z^{2}R_{x^{2}}R_{y^{2}}R_{x^{2}}R_{y} = 0 ;
uR_{y} = z^{2}R_{x^{2}y^{2}}R_{x}R_{y}R_{y} → z^{2}R_{x^{2}}R_{y^{2}}R_{x}R_{y^{2}} = 0;
uR_{z} = z^{2}R_{x^{2}y^{2}}R_{x}R_{y}R_{z} ≡ 0 .

(a10)

Finally, k = 9; since V′(10) = 0, condition (*) is trivial.

Next, let u = z^{2}R_{x^{2}y^{2}}R_{x}R_{y}R_{z}, then

uR_{x} = z^{2}R_{x^{2}y^{2}}R_{x}R_{y}R_{z}R_{x} → (z(xy))^{2}R_{x}R_{y}R_{z}R_{x} → (z(xy))^{2}R_{x}R_{z(xy)} = 0 ;

the case uR_{y} is similar; further,

uR_{z} = z^{2}R_{x^{2}y^{2}}R_{x}R_{y}R_{z}R_{z} → z^{2}R_{x^{2}y^{2}}R_{x}R_{y}R_{z^{2}} → (z^{2}x^{2})R_{y^{2}}R_{x}R_{y}R_{z^{2}} →

→ (z^{2}x^{2})R_{y^{2}}R_{z^{2}}R_{y}R_{x} = 0, (z^{2}x^{2})R_{y^{2}}R_{y(z^{2}x)} → (zx)^{2}y^{2}R_{y((zx)z)} = zR_{x_{1}}R_{y}R_{x_{1}^{2}y^{2}}

where x_{1} = zx; by (7), (9) (or by (a7)), we have

zR_{x_{1}}R_{y}R_{x_{1}^{2}y^{2}} → zR_{x_{1}^{2}}R_{y^{2}}R_{x_{1}}R_{y} = 0 ,

because zR_{x_{1}^{2}} = z(zx)^{2} → z(z^{2}x^{2}) = 0. So, lemma 4 is proven. □

Proof of proposition 1. Let B = F_{3}(A); denote by P the subalgebra of tr[3A] generated by Q and ω. We need to show that tr(u) ∈ P for every monomial u ∈ B; we will carry out the induction on the degree of u.

If u ∈ {x, y, z}, then it is trivial; next, u = vt, where v, t are monomials of lesser degree. Also, since

tr((a, b, c)) = 0 ,  (14)

where (a, b, c) = (ab)c − a(bc), we may assume that t is a generator in {x, y, z}.

Now if u, v are homogeneous elements in B of equal degree, then tr(u) ≡ tr(v) means that for some non-zero α ∈ F the element tr(u) − αtr(v) belongs to P modulo the subalgebra generated by elements of the form tr(w), where w runs over all monomials in B of degree being less than deg(u) = deg(v).

Let I be the ideal of B generated by a^{3}, where a ranges over all elements in B; from (1) it follows that for all c ∈ I and b ∈ B we have tr(cb) ≡ 0. Hence, by lemma 4, it suffices to show that tr(vt) ≡ 0, where v runs over the generators in (a1)-(a9).

The cases (a1), (a2) are trivial; consider (a3): for example, if v = xR_{y}R_{z} and t = x, then, by (13) and (a2),

tr(vt) = tr((xy)zR_{x}) = tr(((xy)x)z) ≡ tr((x^{2}y)z) ≡ 0 .

Let us check some more cases (the others are left as an exercise for the reader). (a5) Let v = x^{2}R_{y}R_{z}R_{x}; if t = x, then

tr(vt) = tr(x^{2}R_{y}R_{z}R_{x^{2}}) = tr(x^{2}R_{z}R_{x^{2}}R_{z}) ≡ 0 ;

if t = y, then

tr(vt) = tr(x^{2}R_{y}R_{z}R_{x}R_{y}) ≡ tr(x^{2}R_{z}R_{y}R_{x}R_{y}) = tr(x^{2}R_{z}R_{y}R_{y}R_{x}) ≡
≡ tr(x^{2}R_{z}R_{y^{2}}R_{x}) = tr(zR_{x^{2}}R_{x}R_{y^{2}}) ≡ 0 .

(a7) Let v = zR_{x^{2}y^{2}}R_{x}R_{y} and t = z, then

tr(vt) = tr((x^{2}y^{2})R_{z}R_{x}R_{y}R_{z}) = tr((x^{2}y^{2})R_{z}R_{x}R_{z}R_{y}) ≡
≡ tr((x^{2}y^{2})R_{xz^{2}}R_{y}) = tr((x^{2}y^{2})R_{y}R_{xz^{2}}) ≡ 0 .

(a8) Let v = z^{2}R_{x^{2}y^{2}}R_{x}R_{y} and t = z, then

tr(vt) = tr(z^{2}R_{x^{2}y^{2}}R_{x}R_{y}R_{z}) ≡ tr(z^{2}R_{x^{2}}R_{y^{2}}R_{x}R_{y}R_{z}) ≡
≡ tr(((zx)y)^{2}R_{x}R_{y}R_{z}) = tr((zx)y)^{2}((zy)x)) ≡ 0 .

Let v = zR_{x^{2}y^{2}}R_{x}R_{y}R_{z}; if t = x, then

tr(vt) ≡ tr(zR_{x^{2}y^{2}}R_{y}R_{x}R_{z}R_{x}) = tr(zR_{x^{2}y^{2}}R_{y}R_{x}R_{x}R_{z}) ≡
≡ tr(zy^{2}R_{x^{2}}R_{y^{2}}R_{x^{2}}R_{z}) ≡ 0 ;

if t = z, then

tr(vt) = tr(zR_{x^{2}y^{2}}R_{x}R_{y}R_{z}R_{z}) = tr(z^{2}R_{y}R_{x}R_{x^{2}y^{2}}R_{z}) ≡ tr(z^{2}R_{x^{2}y^{2}}R_{x}R_{y}R_{z}) ,

by the previous case, the latter element belongs to P.

Finally, let v = z^{2}R_{x^{2}y^{2}}R_{x}R_{y}R_{z} and let t = x, then

tr(vt) = tr(z^{2}R_{x^{2}}R_{y^{2}}R_{x}R_{y}R_{x}R_{z}) ≡ tr(z^{2}R_{x^{2}}R_{y^{2}}R_{x^{2}y}R_{z}) ≡
≡ tr(z^{2}R_{x^{2}}R_{y^{2}}R_{x^{2}}R_{y}R_{z}) ≡ 0 .

□

Proposition 2 The element (2) is integral over Q of degree 2.

Proof. Put a = (zx)y and b = (zy)x, then ω = tr(a^{2}b). We will check first that tr(a^{2}b) + tr(b^{2}a) ∈ Q. Notice that tr(u) ∈ Q for every monomial u ∈ B of degree less then 9; also, tr(a^{3}), tr(b^{3}) ∈ Q. Denote by ≡ the equality modulo Q; from (10) and (7) we have

tr(a^{2}b) = tr(((zx)y)^{2} · (zy)x) ≡ 1/4tr((z^{2}x^{2})y^{2} · (zy)x) ≡

≡ −1/4tr(z^{2}(x^{2}y^{2}) · (zy)x) ≡ 1/4tr(z^{2}(x^{2}y^{2}) · (zx)y) ≡ −1/4tr((z^{2}y^{2})x^{2} · (zx)y) ≡

≡ −tr(((zy)x)^{2} · (zx)y) = −tr(b^{2}a) .

Now consider f = tr(a^{2}b · b^{2}a); by (8) and (1)

f = tr(a^{2} · (b · b^{2}a)) ≡ tr(a^{2}b)tr(b^{2}a) .

Furthermore,

f = tr(a^{2} · b^{2}(ba)) = tr(a^{2}b^{2}(ab)) ;

by lemma 4, (a6)

(ab) → (x^{2}y^{2})z^{2}, (x^{2}z^{2})y^{2}

modulo identity (6), hence, modulo Q+Qω the polynomial f is equal to tr(w(x^{2}, y^{2}, z^{2})), where w = w(x, y, z) is a polynomial in x, y, z. We claim that

tr(w(x^{2}, y^{2}, z^{2})) = ∑_{j} tr(u_{j})h_{j} ,  (15)

where deg(u_{j}) ≤ 8 and h_{j} ∈ tr[3A] for all j.

Indeed, by lemma 1 we may assume that w is equal to w_{0}U, where w_{0} is a monomial in x, y, z of degree 4 or 5 and U is a product of operators of right multiplication. If deg(w_{0}) = 4, then we make use of lemma 3; othewise, w_{0} = (pq)rR_{st}, where p, q.r, s, t ∈ {x, y, z}. If the initial segment of w_{0} is of the form x^{2} or xR_{yx} (up to the order of the variables), then the claimed statement (15) follows from (7). Hence, we may assume that w_{0} = (xy)zR_{yz} or w_{0} = (xy)zR_{xy}; the first case is reduced to the previous ones by (3); next, thanks to (7) we may change the order of the first three multiples and get also in the previous situation.

So, (15) is proven; since deg(f) = 18 we get f ∈ Q + ωQ; on the other hand, ω^{2} = ω(tr(a^{2}b) + tr(b^{2}a)) − tr(a^{2}b)tr(b^{2}a), hence,

ω^{2} = p + qω  (16)

for suitable p, q ∈ Q; so, the proposition is proven. □.

From this proposition it follows that the transcendence degree of Q over F is equal to that of K; also, since F (3A)^{G} is the field of fractions of K, we have tr. deg_{F} Q = 29. Comparing it with the number of elements in (G1), (G2) and (G3), we get the following statement.

Corollary 2 The elements in (G1),(G2),(G3) are algebraically independent generators of the algebra Q.

3 Proof of the main statement

It is known (e.g., [11, sect.3.1]) that the field of fractions of F [3A]^{G} is equal to the field of rational invariants F (3A)^{G}. By [5, 4], it is generated by trace polynomials, therefore, for every f ∈ F [3A]^{G} there are g, h ∈ tr[3A] such that h ≠ 0 and f = g/h. Now we will show that h of a special form can be found.

Proposition 3 For every f ∈ F [3A]^{G} there is a nonzero trace polynomial h ∈ Q such that f · h ∈ tr[3A] and h does not depend on z.

The underlying idea of the proof comes from the slice method (e.g., [2]).

Let V be a G-module and let E be a subspace of V; denote by N_{G}(E) the normalizer of E in G:

N_{G}(E) = {g ∈ G| g(E) ⊆ E} .

Denote by f|_{E} the restriction to E of a function f defined on V, then the mapping

ϕ(f) → f|_{E}  (17)

is a homomorphism from F [V]^{G} to F [E]^{N(E)}. It is common that ϕ is not an isomorphism; however, if the orbit G(E) = {g(v)| g ∈ G , v ∈ E} is a dense subset of the affine space V = F^{dim V}, then ϕ is obviously a monomorphism, hence, it can be extended to a homomorphism from the field F (V)^{G} to F (E)^{N(E)}. Moreover, this mapping is an isomorphism, if E is a slice, that is, the following conditions are satisfied [2]:

(b1) the orbit G(E) is a dense subset of the affine space V = F^{dim V};

(b2) there is an open subset E′ ⊆ E such that for every g ∈ G and every v ∈ E′ if g(v) ∈ E, then g ∈ N_{G}(E).

Now, let E be a slice of 3A : G and let f ∈ F [3A]^{G}; since the field F (3A)^{G} is generated by trace polynomials, ϕ(f) ∈ F [E]^{N(E)} is equal to ϕ(g/h) = ϕ(g)/ϕ(h), where g, h are trace polynomials. Our task is to find E where ϕ(h) does not depend z.

Since we make use of some arguments from [6], we will keep notations of that paper.

The elements of a Cayley-Dickson algebra O have the form

( \begin{matrix}α & v \\ v^{∗} & β \end{matrix} ) ,

where α, β ∈ F and v, v^{∗} ∈ F^{3} = F × F × F. The multiplication is defined as follows:

( \begin{matrix}α_{1} & v_{1} \\ v_{1}^{∗} & β_{1} \end{matrix} ) · ( \begin{matrix}α_{2} & v_{2} \\ v_{2}^{∗} & β_{2} \end{matrix} ) = ( \begin{matrix}α_{1}α_{2} + (v_{1}, v_{2}^{∗}) & α_{1}v_{2} + β_{2}v_{1} − v_{1}^{∗} × v_{2}^{∗} \\ α_{2}v_{1}^{∗} + β_{1}v_{2}^{∗} + v_{1} × v_{2} & (v_{1}^{∗}, v_{2}) + β_{1}β_{2} \end{matrix} ) ,

where (v, w), v×w are the scalar and vector product in F^{3} respectively. The algebra O is equipped with the involution

\overline{( \begin{matrix}α & v \\ v^{∗} & β \end{matrix} )} = ( \begin{matrix}β & −v \\ −v^{∗} & α \end{matrix} ) ,

where the trace t(x) of a matrix x ∈ O equals to x + x̄ ∈ 1 · F ≃ F.

Let O_{3} = O ⊗_{F} F_{3} be the algebra of 3 × 3-matrices over O; a linear mapping a⊗b → ā⊗b^{⊤}, where b → b^{⊤} is the matrix transposition, is an involution of O_{3}; the subspace H(O_{3}) of symmetric elements with the multiplication a ∘ b = 1/2(ab + ba) is a central simple Jordan algebra A of dimension 27 [8, 16], which is called an Albert algebra.

Let e_{i} = e_{ii}, e_{ij} (i, j ∈ {1, 2, 3}, i ≠ j) be the elementary matrices in O_{3} (we identify 1 ⊗ F_{3} with F_{3}), then an arbitrary element in A can be written in the form

∑_{i=1}^{3} α_{i}e_{i} + ∑_{(ijk)}[c_{i}]_{jk},  (18)

where α_{i} ∈ F, c_{i} ∈ O and [c]_{jk} means c · e_{jk} + c̄ · e_{kj}, and (ijk) runs over the subgroup A_{3} of even permutations.

The subspace of elements (18), where

c_{1}, c_{2}, c_{3} ∈ ( \begin{matrix}F & 0 \\ 0 & F \end{matrix} ) ⊆ O ,

is a subalgebra of A isomorphic to F_{3}^{(+)} (the matrix algebra F_{3} with the symmetrized multiplication ∘b). Also, denote by C the subalgebra of diagonal matrices F e_{1} + F e_{2} + F e_{3}.

The automorphism group G = Aut(A) contains the following subgroups:

(c1) the symmetric subgroup S_{3} whose action is induced by permutation of the idempotents e_{1}, e_{2}, e_{3} [8, p.389,ex.3];

(c2) the subgroup Z_{2} generated by an automorphism of the second order [6, sect.2]:

σ : e_{i} → e_{i} ∀i ∈ {1, 2, 3} ;

σ : [ \begin{matrix}α & v \\ v^{∗} & β \end{matrix} ]_{ij} → [ \begin{matrix}β & −v^{∗} \\ −v & α \end{matrix} ]_{ij}

for all (ijk) ∈ A_{3};

(c3) the subgroup T_{2} = {ξ = (ξ_{1}, ξ_{2}, ξ_{3}) ∈ F^{∗} × F^{∗} × F^{∗}| ξ_{1}ξ_{2}ξ_{3} = 1} acting in the

following way:

ξ : e_{i} → e_{i} ∀i ∈ {1, 2, 3} ;

ξ : [ \begin{matrix}α & v \\ v^{∗} & β \end{matrix} ]_{ij} → [ \begin{matrix}ξ_{i}ξ_{j}^{−1}α & ξ_{k}^{−1}v \\ ξ_{k}v^{∗} & ξ_{i}^{−1}ξ_{j}β \end{matrix} ]_{ij}

for all (ijk) ∈ A_{3};

(c4) the subgroup M of automorphisms of A which keep the elements in F_{3}^{(+)} being fixed; by [8, p.386, th.8], it is isomorphic to SL_{3}(F). More precisely, if ψ ∈ SL_{3}(F), then we can define an automorphism ψ̂ ∈ M in following way (it suffices to consider only elements in the complement of F_{3}^{(+)} to A):

ψ̂ : [ \begin{matrix}0 & v \\ v^{∗} & 0 \end{matrix} ]_{ij} → [ \begin{matrix}0 & ψ(v) \\ (ψ^{⊤})^{−1}(v^{∗}) & 0 \end{matrix} ]_{ij} ,

where (ijk) ∈ A_{3}. Then ψ → ψ̂ is an isomorphism of SL_{3}(F) onto M.

Note that the product H of the subgroups S_{3} and Z_{2} is direct (cf. [6]); also, it is easy to check that T_{2} normalizes M and M ∩ T_{2} = {1}; finally, H normalizes the (internal) semidirect product H_{0} = M T_{2} and H_{0} ∩ H = {1}.

Put E = C ⊕ F_{3}^{(+)} ⊕ A ⊆ 3A.

Lemma 5 (cf. [1, lemma 4.5] ) The subspace E is a slice of 3A : G, where N_{G}(E) = H_{0} H.

Proof. Let U be the subset of points (x, y, z) in the affine space 3A such that

1) the characteristic polynomial of x has 3 distinct eigen-values,

2) x, y generate a subalgebra isomorphic to F_{3}^{(+)} .

From [9, lemma 2] it follows that U is open. In addition, by [8, p.370, th.3] for every (x, y, z) ∈ M there is g_{1} ∈ G such that g_{1}(x), g_{1}(y) ∈ F_{3}^{(+)}. The subalgebra generated by g_{1}(x) is spanned by three orthogonal idempotents, hence, thanks to the Coordinatization Theorem [8, p.137] there is an automorphism g_{2} of F_{3}^{(+)} such that g_{2}(g_{1}(x)) ∈ C. By [8, p.370, th.3] we may assume that g_{2} ∈ G, so, g = g_{2} ∘ g_{1} sends (x, y, z) into E, therefore, the orbit of E is dense.

Next, let E′ = U ∩E, then for every g ∈ G and (x, y, z) ∈ E′ if (g(x), g(y), g(z)) ∈ E, then the element g(x) generates C and g(x), g(y) generate F_{3}^{(+)}, hence, since g is an automorphism, we have g(C) = C and g(F_{3}^{(+)}) = F_{3}^{(+)}, therefore, g(E) = E. □

Also, from the proof it follows that the subspace E_{0} = C ⊕ F_{3}^{(+)} ⊂ E is a slice of 2A : G, where the normalizer N_{G}(E_{0}) is equal to N_{G}(E). Hence, the restriction of ϕ to F (2A)^{G} is an isomorphism onto K = F (E_{0})^{N_{G}(E)} ⊆ F (E)^{N_{G}(E)}. On the other hand, by [6, lemma 9], F (2A)^{G} is generated by trace polynomials in x, y. Put S = ϕ^{∗}(tr[3A]) ⊆ F [E]^{N_{G}(E)} and let S̃ be the subalgebra over K generated by S. Obviously, S̃ = vect_{K}S and its pre-image in F [3A]^{G} lies in the subspace over F (2A)^{G} spanned by tr[3A]; hence, to prove proposition 3 it suffices to show that

F [E]^{N_{G}(E)} ⊆ S̃ .  (19)

We wiil do it in the following way: we will extend S̃ by some elements in F (E_{0}) such that the extension will contain F [E]^{N_{G}(E)} and, on the other hand, we will show that the subalgebra of all elements in the extension which are fixed under the action of a subgroup of N_{G}(E) coincides with S̃.

The restriction f → f|_{E} defines a homomorphism from F_{3}(A) to the algebra Pol(E, A) of polynomial mappings from E to A. We also denote it by ϕ; by [6, lemma 1]), ϕ is a monomorphism; denote by B_{3} the image alg_{F} {ϕ(x), ϕ(y), ϕ(z)}.

The algebra vect_{F (E)}A ≃ F (E) ⊗_{F} A is a scalar extension of A, where the trace function is extended by linearity. Hence, the algebra S is generated by 1 and tr(u), where u ranges over elements in B_{3}.

In the standard coordinates (18) of A the images of the generic elements have the following form:

ϕ(x) = ∑_{i=1}^{3} x_{i}e_{i} ,

ϕ(y) = ∑_{i=1}^{3} y_{i}e_{i} + ∑_{(ijk)}[y_{i}]_{jk},

ϕ(z) = ∑_{i=1}^{3} z_{i}e_{i} + ∑_{(ijk)}[z_{i}]_{jk},

where

y_{i} = ( \begin{matrix}y_{jk} & 0 \\ 0 & y_{kj} \end{matrix} ) , z_{i} = ( \begin{matrix}z_{jk} & v_{i} \\ v_{i}^{∗} & z_{kj} \end{matrix} )

and v_{i} = (v_{i1}, v_{i2}, v_{i3}), v_{i}^{∗} = (v_{i1}^{∗}, v_{i2}^{∗}, v_{i3}^{∗}).

Let us extend the field K: consider K̂ = K(x_{1}, x_{2}, x_{3}, α) and let Ŝ = vectK̂ S̃ = vectK̂S. Respectively, let B̂_{3} = Ŝ · B_{3} ⊆ vect_{F (E)}A, Notice that tr(u) ∈ Ŝ for all u ∈ B̂_{3} and, therefore, Ŝ = alg_{F} {tr(u)| u ∈ B̂_{3}}. Now we will show that F [E]^{N_{G}(E)} ⊆ Ŝ; in fact, it will be more convinient to prove the same relation for some extension of F [E]^{N_{G}(E)}.

The next statement follows from a classical result on invariants of SL_{n}(F) (e.g., [17, 11]).

Lemma 6 The subalgebra D_{1} = F [E]^{M} is generated by 1 and the following elements:

(d1) x_{i}, y_{i}, z_{i} (i ∈ {1, 2, 3});

(d2) y_{ij}, z_{ij} (i, j ∈ {1, 2, 3}, i ≠ j);

(d3) (v_{i}, v_{j}^{∗}) (i, j ∈ {1, 2, 3});

(d4) Δ := det(v_{1}, v_{2}, v_{3}), Δ̄ := − det(v_{1}^{∗}, v_{2}^{∗}, v_{3}^{∗}) .

Obviously, D_{2} = F [E]^{M T_{2}} = (F [E]^{M})^{T_{2}} = D_{1}^{T_{2}}. Now, let L be the subalgebra of F (E) generated over K̂ by 1 and the following elements :

(e1) z_{i} (i ∈ {1, 2, 3});

(e2) z_{ij}y_{ji} (i, j ∈ {1, 2, 3}, i ≠ j);

(e3) (v_{i}, v_{i}^{∗}) (i ∈ {1, 2, 3});

(e4) (v_{i}, v_{j}^{∗})y_{ij}, (i, j ∈ {1, 2, 3}, i ≠ j).

(e5) Δ, Δ̄;

Lemma 7 L ⊆ Ŝ .

Proof. Arguing like in [6, lemma 3 and 4], we will show simultaneously the following:

(i) the field K̂ includes y_{k}, t_{k} = y_{ij}y_{ji} ((ijk) ∈ A_{3}) and ᾱ; (ii) elements (e1)-(e5) belong to Ŝ; (iii) the algebra B̂_{3} contains the following elements:

(f1) e_{1}, e_{2}, e_{3};

(f2)

[ \begin{matrix}y_{ij} & 0 \\ 0 & 0 \end{matrix} ]_{ij} , [ \begin{matrix}0 & 0 \\ 0 & y_{ji} \end{matrix} ]_{ji} ,

where (ijk) ∈ A_{3};

(f3)

[ \begin{matrix}0 & v_{i} \\ 0 & 0 \end{matrix} ]_{jk} , [ \begin{matrix}0 & 0 \\ v_{i}^{∗} & 0 \end{matrix} ]_{jk} ,

where (ijk) ∈ A_{3};

(f4)

[ \begin{matrix}0 & v_{j}y_{ji} \\ 0 & 0 \end{matrix} ]_{jk} , [ \begin{matrix}0 & 0 \\ v_{j}^{∗}y_{ij} & 0 \end{matrix} ]_{jk}

where (ijk) ∈ S_{3}.

First, since ϕ(x) ∈ B_{3} and x_{1}, x_{2}, x_{3} ∈ K̂ are the roots of its characteristic polynomial, we have e_{1}, e_{2}, e_{3} ∈ B̂_{3}. Applying the Peirce decomposition [8, 16], we get y_{i} ∈ K̂, then [y_{i}]_{jk} ∈ B̂_{3} and, therefore, y_{jk}y_{kj} = tr([y_{i}]_{jk}^{2}) ∈ K̂ for all (ijk) ∈ A_{3}. Also, z_{i} ∈ Ŝ and [z_{i}]_{jk} ∈ B̂_{3}.

Next, ᾱ = y_{13}y_{32}y_{21} = y_{12}y_{21} · y_{23}y_{32} · y_{31}y_{13}/α ∈ K̂. Hence,

[ \begin{matrix}y_{ij} & 0 \\ 0 & 0 \end{matrix} ]_{ij} = \frac{1}{α − ᾱ}(α[y_{k}]_{ij} − 2(y_{ij}y_{ji})([y_{i}]_{jk} ∘ [y_{j}]_{ki}))

belongs to B̂_{3}. Also, [ \begin{matrix}y_{ji} & 0 \\ 0 & 0 \end{matrix} ]_{ji} = [y_{k}]_{ij} − [ \begin{matrix}y_{ij} & 0 \\ 0 & 0 \end{matrix} ]_{ij} ∈ B̂_{3}. Further, z_{ij}y_{ji} =

tr([z_{k}]_{ij} ∘ [ \begin{matrix}y_{ji} & 0 \\ 0 & 0 \end{matrix} ]_{ji}) ∈ Ŝ; therefore, z_{ij}/y_{ij} = (z_{ij}y_{ji})/(y_{ij}y_{ji}) ∈ Ŝ and

[ \begin{matrix}0 & v_{i} \\ v_{i}^{∗} & 0 \end{matrix} ]_{jk} = [z_{i}]_{jk} − \frac{z_{jk}}{y_{jk}} [ \begin{matrix}y_{jk} & 0 \\ 0 & 0 \end{matrix} ]_{jk} − \frac{z_{kj}}{y_{kj}} [ \begin{matrix}0 & 0 \\ 0 & y_{kj} \end{matrix} ]_{jk}

lies in B̂_{3}. Hence, (v_{i}, v_{i}^{∗}) = −\frac{1}{2}tr([ \begin{matrix}0 & v_{i} \\ v_{i}^{∗} & 0 \end{matrix} ]_{jk}^{2}) ∈ Ŝ. Next,

[ \begin{matrix}0 & 0 \\ −v_{j}^{∗}y_{ij} & 0 \end{matrix} ]_{jk} = 2 [ \begin{matrix}0 & v_{j} \\ v_{j}^{∗} & 0 \end{matrix} ]_{ki} ∘ [ \begin{matrix}y_{ij} & 0 \\ 0 & 0 \end{matrix} ]_{ij} ∈ Ŝ ,

and all elements (f4) can be obtained in this way. Furthermore,

(v_{j}, v_{k}^{∗})y_{ji}y_{ik} = −tr([ \begin{matrix}0 & v_{j}y_{ji} \\ 0 & 0 \end{matrix} ]_{jk} ∘ [ \begin{matrix}0 & 0 \\ v_{k}^{∗}y_{ik} & 0 \end{matrix} ]_{jk}) ∈ Ŝ ,

and since y_{ji}y_{ik}/y_{jk} = ᾱ/t_{i} ∈ K̂, we have also (v_{j}, v_{k}^{∗})y_{jk} ∈ Ŝ. Using an appropriate pair of elements (f4), we prove all elements (e4) to lie in Ŝ.

Finally,

[ \begin{matrix}0 & v_{i}y_{ik} \\ 0 & 0 \end{matrix} ]_{ij} ∘ [ \begin{matrix}y_{ki} & 0 \\ 0 & 0 \end{matrix} ]_{ki} = \frac{−y_{ik}y_{ki}}{2} [ \begin{matrix}0 & v_{i} \\ 0 & 0 \end{matrix} ]_{jk} ,

So, [ \begin{matrix}0 & v_{i} \\ 0 & 0 \end{matrix} ]_{jk} ∈ B̃_{3}. Moreover, [ \begin{matrix}0 & 0 \\ v_{j} × v_{k} & 0 \end{matrix} ]_{jk} = −2 [ \begin{matrix}0 & v_{j} \\ 0 & 0 \end{matrix} ]_{ki} ∘ [ \begin{matrix}0 & v_{k} \\ 0 & 0 \end{matrix} ]_{ij}

is an element in B̂_{3}, and

Δ = (v_{j} × v_{k}, v_{i}) = −tr([ \begin{matrix}0 & 0 \\ v_{j} × v_{k} & 0 \end{matrix} ]_{jk} ∘ [ \begin{matrix}0 & v_{i} \\ 0 & 0 \end{matrix} ]_{jk}) ∈ Ŝ.

The lemma is proven. □

From the next lemma it follows that F [E]^{N_{G}(E)} is contained in Ŝ.

Lemma 8 D_{2} ⊆ L.

Proof. Indeed, let L̃ = alg_{F} L ∪ {y_{12}, y_{23}, y_{12}^{−1}, y_{23}^{−1}}, then y_{ij}, y_{ij}^{−1} ∈ L̃ for all i, j ∈ {1, 2, 3}, i ≠ j; hence, D_{1} ⊆ L̃. On the other hand, L̃^{T_{2}} = L, hence, D_{2} ⊆ L. □

Recall that Ŝ = vectK̂S; also, using the argument of [6, lemma 6], one can show that K̂^{H} = K, hence, since S consists of invariants of H ⊂ N_{G}(E), we have

Ŝ^{H} = vectK̂^{H} S = S̃ .

Therefore, by lemma 8, F [E]^{N_{G}(E)} = D_{2}^{H} ⊆ L^{H} ⊆ Ŝ^{H} = S̃ and proposition 3 is proven. □

Lemma 9 The element (2) does not belong to the field of fractions of Q.

Proof. Let T be the subfield of F (E) generated by ϕ(Q) and x_{1}, x_{2}, x_{3}, α, then, arguing like in the proof of lemma 8 (or [6, lemma 3 and 4]), one can show that K_{0} contains the elements

(g1) x_{i}, y_{i}, z_{i}, i ∈ {1, 2, 3}; (g2) t_{i}, i ∈ {1, 2, 3}; (g3) y_{ij}z_{ji}, i, j ∈ {1, 2, 3}, i ≠ j; (g4) (v_{i}, v_{i}^{∗}), i ∈ {1, 2, 3}; (g5) (v_{i}, v_{j}^{∗})y_{ij}, i, j ∈ {1, 2, 3}, i ≠ j; (g6) Δ + Δ̄.

(g7) α .

Moreover, one can easily check that every generator of ϕ(Q) is a polynomial in (g1)-(g7), hence, these elements generate T over F. The total number of them is 29; it is precisely the transcendence degree of F (3A)^{G} over F [6]. Recall that F [3A]^{G} is integral over Q and T is a finite field extension of ϕ(Q), hence, the trascendence degree of T over F is also equal to 29, therefore, (g1)-(g7) are algebraically independent.

Suppose that ω ∈ Q, then Q must be equal to P, and, by lemma 8 (or [6, lemma 3]), Δ ∈ T.

Let B be the subalgebra of T generated over F by (g1)-(g7). Notice that

Δ̄Δt_{1}t_{2}t_{3}α = det((v_{i}, v_{j}^{∗})|_{i,j=1}^{3}) = (v_{1}, v_{1}^{∗})(v_{2}, v_{2}^{∗})(v_{3}, v_{3}^{∗})t_{1}t_{2}t_{3}α+
+((v_{1}, v_{2}^{∗})y_{12})((v_{2}, v_{3}^{∗})y_{23})((v_{3}, v_{1}^{∗})y_{31})t_{1}t_{2}t_{3}+
+((v_{2}, v_{1}^{∗})y_{21})((v_{3}, v_{2}^{∗})y_{32})((v_{1}, v_{3}^{∗})y_{13})α^{2}−
−((v_{3}, v_{1}^{∗})y_{31})(v_{2}, v_{2}^{∗})((v_{1}, v_{3}^{∗})y_{13})αt_{3}t_{1}−
−((v_{1}, v_{2}^{∗})y_{21})(v_{3}, v_{3}^{∗})((v_{2}, v_{1}^{∗})y_{21})αt_{1}t_{2}−
−((v_{2}, v_{3}^{∗})y_{23})(v_{1}, v_{1}^{∗})((v_{3}, v_{2}^{∗})y_{32})αt_{2}t_{3},

belongs to B. Therefore, a = Δαt_{1}t_{2}t_{3} satisfies a quadratic equation over B:

a^{2} − a(Δ + Δ̄)αt_{1}t_{2}t_{3} + Δ̄Δα^{2}t_{1}^{2}t_{2}^{2}t_{3}^{2} = 0 ;

since B is integrally closed, we get a ∈ B. On the other hand, comparing degrees in the coordinates of E, we see that it is impossible (Δ would be equal to β(Δ + Δ̄) for some β ∈ F). □

We immediately get the following.

Corollary 3 The P is a free Q-module with basis 1, ω.

Proof of theorem 1. From [11, th.3.19] it follows that the algebra N = F [3A]^{G} is a Cohen-Macaulay one. Since the homogeneous generators (G1)-(G3) of Q are algebraically idependent and N is a finitely generated module over Q, the elements (G1)-(G3) are homogeneous parameters of N and, therefore, N is a free Q-module. Let 1, u_{1}, …, u_{m} be a basis of N over Q, we may assume that every u_{i} is homogeneous (cf. [11, p.178]). Moreover, by the preceding statement, P is a free Q-module of rank 2, on the other hand, from proposition 3 it follows that ∀i u_{i}h ∈ Q for some nonzero h ∈ F [2A]^{G} ⊆ Q, hence, m = 1.

The following assesment is evident.

Lemma 10 Let u ∈ N\Q be a homogeneous polynomial of degree deg(u_{1}). then u = a + αu_{1}, where a ∈ Q and α ∈ F, α ≠ 0.

Since Q is homogeneous in all variables x, y, z, we may assume that u_{1} is homogeneous in x, y, z (we may replace u_{1} with a homogeneous component of it, by the previous lemma). Moreover, every permutation of the variables x, y, z induces an automorphism τ of N which keeps Q being invariant; by the previous lemma, τ(u_{1}) = a + αu_{1}, hence, deg_{x}(u_{1}) = deg_{y}(u_{1}) = deg_{z}(u_{1}).

Next, ω = a + bu_{1} for some a, b ∈ Q; recall that ω and u_{1} are homogeneous in x, y, z, hence, we may assume that a, b are the same and, since deg_{x}(ω) = deg_{y}(ω) = deg_{z}(ω) = 3, for every t ∈ {x, y, z}

deg_{t}(b) = deg_{t}(ω) − deg_{t}(u_{1}) = deg_{z}(ω) − deg_{z}(u_{1}) = deg_{z}(b) .

On the other hand, by proposition 3, u_{1}h = c + dω, where h, c, d ∈ Q, h ≠ 0 and h does not depend on z. Since ω and u_{1} are homogeneous in x, y, z, we may assume that h, c, d have the same property and deg_{z}(h) = 0. Hence, ωh = ah + b(u_{1}h) = ah + bc + bdω; this yields h = bd and, since h ≠ 0, we have deg_{z}(b) = 0. Hence, deg_{x}(b) = deg_{y}(b) = deg_{z}(b) = 0, that is, b is a non-zero scalar in F. Therefore, u_{1} ∈ tr[3A] and the theorem is proven. □

From proposition 2 and corollories 2 3 we get the following statement.

Theorem 2 The elements (G1)-(G2) form a system of homogeneous parametries of F [3A]^{G}, where 1 and polynomial (2) are a basis over it.

Note that the elements p, q in (16) can be found explicitly as polynomials in (G1)- (G3) and this yields a syzygy for the generators of F [3A]^{G}; obviously, it is unique. Also, from the proof of proposition 2 it follows that this relation is a consequence of commutivity, the Jordan identity (3) and the Cayley-Hamilton identity (1) (cf. Second Main Theorem for G_{2} [15]).

If we replace x, y, z in (G1)-(G3) and (2) by the generic elements with zero trace, then we will get generators of F [3A′]^{G}. Moreover, since the elements tr(x), tr(y) and tr(z) vanish, the rest of elements obtained from (G1)-(G3) will form a system of homogeneous parameters of the algebra F [3A′]^{G}. In the same way from (15) on can get the unique syzygy for this set of generators.

Acknowledgments: the author thanks G.Lehrer for attention to this work.

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