LAPLACE OPERATOR AND POLYNOMIAL INVARIANTS

A.V.ILTYAKOV

Abstract. Let G be a simple linear algebraic group of type F_{4} over an algebraically closed field of characteristic zero. We describe generators of the algebra of invariants of a system of vectors of a minimal irreducible representatioin of G in terms of trace polynomials of an exceptional simple Jordan algebra (the Albert algebra) and Laplace operators.

1. Preliminaries

Let F denote an algebraically closed field of characteristic zero.

Notations.

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

∘ F [V] (respectively, F (V)) is the algebra (field) of polynomial (rational) functions from

a vector space V to F; all subalgebras of F [V] are supposed to be with 1; ∘ V : G is a module V over a group G; ∘ 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; ∘ R_{a} (L_{a}) the operator of right (left) multiplication by an element a, i.e., R_{a} : x → xa

(L_{a} : x → ax). It is well-known that minimal linear representations of exceptional algebraic groups are con-

nected 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} (cf. [12, Prop. 1.10]). It is known that every element x ∈ O satisfies the quadratic identity

x^{2} − t(x)x + n(x) = 0 ,  (1)

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

In [12, 13] G.Schwarz proved 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)} be 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 all 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 polynomial G-invariant, and the subalgebra t_{k}[O] generated by all such polynomials is called the algebra of trace polynomials of O of rank k; it is equal to F [kO]^{G} by Schwarz’s theorem. 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 (an exceptional simple Jordan algebra H(O_{3}) [9, ch. IX]); it is known that G = Aut(A) is a simple algebraic group of type F_{4} (e.g., [8]); also, every element x ∈ A satisfies a cubic equation

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

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} [8].

Denote by tr_{k}[A] (tr_{k}[A′]) 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 F [kA]^{G} (F [kA′]^{G}) is integral over tr_{k}[kA] (tr_{k}[A′]), their spectra are homeomorphic and their fields of fractions coincide; in particular, for k ≤ 3 they are equal [5, 6]. However, for k ≥ 5 we exhibit an invariant which does not belong to tr_{k}[A] (proposition 8).

Let V be a vector space over F; pick a ∈ Pol[kV, V] and let x = x^{(r)} ∈ Pol[kV, V] be a projector from kV onto the r-th summand (we also call it a generic element ); denote by D_{x}(a) the following derivation of F [kV] [15]:

f · D_{x}(a) = ∑_{i=1}^{n} \frac{∂f}{∂x_{i}} a_{i}

where x_{i}, a_{i} are the coordinates with respect to a basis of V. In fact, the defintion does not depend on the choice of the basis; besides, if V is an algebra and a is another generic element, then D_{x}(a) can be regarded as an operator of partial linearization in a more general context [14, ch.1]. Thus, if u ∈ F_{k}(A) is a monomial of degree n in x, then tr(u)D_{x}(a) = ∑_{i=1}^{n} tr(u_{i}), where u_{i} is obtained from u by replacing the i-th occurrence of x with a.

Now suppose that V is endowed with a nondegenerate symmetric bilinear form ⟨x, y⟩; fix an orthonormal basis E_{1}, …, E_{n} and for generic elements x = ∑_{i=1}^{n} x_{i}E_{i}, y = ∑_{i=1}^{n} y_{i}E_{i} define a linear operator Δ(x, y) on F [kV]: for all f ∈ F [kV] put

fΔ(x, y) = ∑_{i=1}^{n} f · D_{x}(E_{i})D_{y}(E_{i}) = ∑_{i=1}^{n} \frac{∂^{2}f}{∂x_{i}∂y_{i}} .  (3)

We call it a Laplace operator, because the case x = y will be the most important one in the proof.

By lemma 7, the definition of Δ(x, y) does not depend on the choice of orthonormal basis. Besides, if V is a G-module and the form ⟨x, y⟩ is G-invariant, then the algebra of polynomial invariants F [kV]^{G} is closed under Laplace operators .

It is convinient to consider the algebra of invariants with an arbitrary number of vector variables. Namely, let V⃗ denote the direct sum of a countable number of copies of a G-module V; we consider it as lim_{k→∞} kV. Therefore, F [V⃗] (F [V⃗]^{G}) is the sum of F [kV] (F [kV]^{G} ).

In the case V = A and G = Aut(A) we put ⟨x, y⟩ = tr(xy), it is a nondegenerate associative symmetric bilinear form of A [9, p.240, th.5]. We also denote by tr[A] the sum of tr_{k}[A], k ∈ N and F(A) = lim_{k→∞} F_{k}(A) is generated by x^{(r)}, r ∈ N.

Theorem 1. Let S be a set of homogeneous in each x^{(r)} generators of tr[A], then the algebra of invariants F [A⃗]^{Aut(A)} is generated by elements

fΔ_{1} … Δ_{t} ,  (4)

where f ∈ S, t ≥ 0 and Δ_{r} = Δ(x^{(i_{r})}, x^{(j_{r})}) for all r ∈ {1, …, t}.

By lemma 6 we can estimate the degree of generators.

Corollary 1. The algebra F [A⃗]^{Aut(A)} is generated by homogeneous invariants of degree ≤ 85.

Probably, this estimate is not precise; besides, it is not clear how many Δ-s are necessary in (4), the example from section 6 shows that t ≥ 1.

Using the projection from A onto the submodule A′, we get similar statements on generators of F [A⃗′]^{G}. Actually, the restriction of tr(xy) to A′ is nondegenerate, hence, we may choose E_{p} ∈ A′ for all p > 1 and E_{1} = 1/\sqrt{3}. Next, for a generic element x^{(r)} the projector x_{∗}^{(r)} is equal to x^{(r)} − 1/3tr(x^{(r)}). Put

f · D_{x_{∗}}(a) = ∑_{i=2}^{27} \frac{∂f}{∂x_{i}} a_{i}

then D_{x_{∗}}(a) = D_{x_{∗}}(a_{∗}), where a_{∗} = ∑_{i=2}^{27} a_{i}E_{i}. Also, for all f ∈ F [A⃗′]

f · D_{x}(a) = f · D_{x_{∗}}(a) = f · D_{x_{∗}}(a_{∗}) ,

hence, a Laplace operator Δ(x, y) acts on F [A⃗′] as

Δ(x_{∗}, y_{∗}) = ∑_{i=2}^{27} D_{x_{∗}}(E_{i})D_{y_{∗}}(E_{i}) .  (5)

Corollary 2. Suppose that tr[A′] is generated by a subset S′ of homogeneous polynomials. Then the algebra of invariants F [A⃗′]^{Aut(A)} is generated by

fΔ_{1} … Δ_{t} ,  (6)

where f ∈ S′, t ≥ 0 and Δ_{r} = Δ(x_{∗}^{(i_{r})}, x_{∗}^{(j_{r})}) for all r ∈ {1, …, t}.

Also, F [A⃗′]^{Aut(A)} is generated by polynomials of degree ≤ 85.

Proof. Put S_{0} = {tr(x^{(r)})| r ∈ N}, then tr[A] is generated by tr[A′] and S_{0}. Hence, S = S′ ∪ S_{0} is a set of homogeneous generators of tr[A] and, therefore, F [A⃗]^{Aut(A)} is generated by elements (4). Obviously, Δ(x, y) kills tr(x^{(r)}), hence, by (5) the set of elements (4) consists of (6) and tr(x^{(r)}); using the projection on A′, we prove the statement. □

The underlying idea of the proof was taken from Luna’s slice method (e.g., [2]) that is very efficient in birational Invariant Theory.

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

N_{G}(W) = {σ ∈ G| W · σ ⊆ W } .

Let f|_{W} be the restriction to W of a function f defined on V, then the mapping

ϕ : f → f|_{W}  (7)

is a homomorphism from F [V]^{G} to F [W]^{N_{G}(W)}. It is common that ϕ is not an isomorphism; however, if the orbit W · G = {w · σ| w ∈ W, σ ∈ G} of W is a dense subset of the affine space V = F^{dim V}, then ϕ is obviously a monomorphism and it extends to a monomorphism from the field of fractions to F (W)^{N_{G}(W)}. Moreover, this mapping is an isomorphism between F (V)^{G} and F (W)^{N_{G}(W)}, if W is a slice, that is, the following conditions are satisfied [2]:

1. the orbit W · G is a dense subset of the affine space V = F^{dim V};

2. there is an open subset W′ ⊆ W such that for every σ ∈ G and every v ∈ W′ if v · σ ∈ W, then σ ∈ N_{G}(W).

Now, let G = Aut(A) and let W be a “good” slice of (k + 2)A : G. Pick f ∈ F [kA]^{G}, since the field F ((k + 2)A)^{G} is generated by trace polynomials, f = g/h for some g, h ∈ tr_{k+2}[A]. In fact, h can be found in a special form if we pass to F [W]^{N_{G}(W)} and consider ϕ(f). In particular, h does not depend on x^{(r)}, r ≤ k. Then we apply the operators Δ(x^{(k+1)}, x^{(k+1)}), Δ(x^{(k+2)}, x^{(k+2)}) several times to the equalty fh = g; since f does not depend on x^{(k+1)}, x^{(k+2)}, the operators act only on h and make it a nonzero constant. On the other hand, g becomes an element of the algebra generated by (4).

In the next section we use this method in the case of G_{2} and give another proof of Schwarz’s theorem. The rest of the paper is devoted to invariants of F_{4}; in section 3 using the slice technique, we find a suitable representation of invariants in the form of a quotient of trace polynomials; then, in section 4, we describe some properties of the Laplace operator and reduce the proof of theorem 1 to some technical statements proven in section 5. In the last section we give an example of an invariant of degree 5 which does not belong to tr[A].

2. Invariants of G_{2}

The Cayley-Dickson algebra O can be represented as the set of matrices (e.g., [14, ch.2])

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

where α, β ∈ F and v, w ∈ F^{3} = F × F × F. The multiplication is defined in the following way:

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

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

where the trace t(x) of a matrix x ∈ O is equal to x + x̄ ∈ 1 · F ≃ F. One can easily check that O satisfies the Cayley-Hamilton identity (1), where δ(x) = x̄x = 1/2(t(x)^{2} − t(x^{2})).

Besides, O is an alternative algebra, this means that

x^{2}y = x(xy) ; yx^{2} = (yx)x  (8)

for all x, y ∈ O.

Throughout this section G denotes Aut(O).

Theorem 2. ([12, 13]) The algebra of polynomial invariants F [kO]^{G} is equal to t_{k}[O].

We start the proof with the following lemma.

Lemma 1. The algebra t[kO] is generated by the following elements (a1) t(a), t(a^{2}), (a2) t(ab), t(ab · c), t((ab)c · d) , where a, b, c, d are different generic elements from {x^{(1)}, …, x^{(k)}}.

Proof. Put Q = t_{k}[O] and let A be the Q-module generated by F_{k}(O) (in Pol[kO, O]). Denote by Q_{+} the subspace spanned by homogeneous elements of positive degree; it is an ideal of Q and I = Q_{+} · F_{k}(O) is an ideal of the Q-algebra A.

By (1) Ā = A/I is an alternative algebra of nil index 2. Hence, by (8) Ā satisfies the identities

ab = −ba ; (ab)c = −(ac)b .

Hence, it is spanned by u = x_{i}R_{x_{j}} … R_{x_{r}} and u is skew symmetric in x_{i}, x_{j}, … x_{r}; in particular, a nonzero u is multilinear. Observe that xy = 1/2[x, y], hence, by (8) v = (xy)z·t = −z(xy)·t = (zt)(xy) = 1/8[[x, y], [z, t]]. Also, since u is skew symmetric,

v = \frac{1}{8 · 4!} ∑_{σ∈S_{4}} sgn(σ)σ · [[x, y], [z, t]]

where σ acts by permutation of the variables. Obviously, the sum is equal to 0 (it is skew symmetric under an even permutation (12)(34)).

Thus, Ā is spanned by skew-symmetric monomials of degree ≤ 3. Let u be a monomial from A, we may assume that deg(u) ≥ 3, hence, u = vx, where x is a variable. If deg(u) > 4, then v ∈ I, that is, v = ∑_{p} v_{p}ϕ_{p}, where ϕ_{p} ∈ Q_{+}. We may assume that u_{p}, ϕ_{p} are homogeneous and deg(u_{p})+deg(ϕ_{p}) = deg(v). Therefore, to prove the lemma we can use the induction on deg(u). Also, v does not depend on x : if v = ay, where a depends on x, then t(u) = t((xa)y), hence, xa ∈ I and we can reduce the degree of u. Thus, y must be equal to x and t(u) = t(vx^{2}) = t(vx)t(x) − t(v)n(x) = t(vx)t(x) − 1/2t(v)(t(x)^{2} − t(x^{2})). □

Let R(x) denote the descriminant of the characteristic polynomial (1),

R(x) = 2t(x^{2}) − t(x)^{2} .  (9)

Proposition 1. For all f ∈ F [kO]^{G} there exists m ∈ N such that f · R(x_{k+1})^{m} belongs to t_{k+1}[O].

Proof. Consider a subspace W = kO ˙+F^{2} ⊆ (k + 1)O, where F^{2} is the subalgebra of the diagonal matrices. The mapping θ : f → f|_{W} is an algebra homomorphism from F [(k+1)O] → F [W]. Moreover, it is monic on the subalgebra of invariants F [(k + 1)O]^{G}. Indeed, the subset U of elements from O which are conjugate to a diagonal matrix is Zariski dense (U is defined by the condition R(x) ≠ 0). Hence, the orbit of every element from the Zariski dense subset kO × U ⊆ (k + 1)O meets W and if f|_{W} = 0 for an invariant f, then f = 0 everywhere.

Similarly, f → f|_{W} defines a monomorphism from the algebra of generic elements F_{k+1}(O) to Pol[W, O]; we also denote it by θ.

Thus x = θ(x_{k+1}) is a diagonal matrix

( \begin{matrix}α & 0 \\ 0 & β \end{matrix} )

and

θ(x^{(r)}) = x^{(r)} = ( \begin{matrix}α_{r} & v_{r} \\ w_{r} & β_{r} \end{matrix} ) ,

for all r ≤ k.

The normalizer N_{G}(W) = {σ ∈ G| W σ ⊆ W } is a semidirect product of the group Z_{2} and the special linear group SL_{3}. The generator of Z_{2} is an automorphism

σ : ( \begin{matrix}α & v \\ w & β \end{matrix} ) → ( \begin{matrix}β & −w \\ −v & α \end{matrix} ) ;

the action of SL_{3} is defined in the following way: let P be a matrix from SL_{3}, then

( \begin{matrix}α & v \\ w & β \end{matrix} ) · P = ( \begin{matrix}α & vP \\ w(P^{⊤})^{−1} & β \end{matrix} ) .

Indeed, for all ψ ∈ N_{G}(W) one can find i ∈ {0, 1} and P ∈ SL_{3} such that ψσ^{i}P = id.

By the classical result on invariants of SL_{3} (cf. [10, p. 254]) the algebra F [W]^{SL_{3}} is generated by (b1) α, β, α_{r}, β_{r}; (b2) (v_{p}, w_{q}), (b3) det(v_{p}, v_{q}, v_{r}), det(w_{p}, w_{q}, w_{r}) .

Let B = θ(F_{k+1}(O)), it is the subalgebra of Pol[W, O] generated by x^{(r)}, r ≤ k and x; also, consider the subalgebra B̃ generated by B and the idempotents

e_{1} = ( \begin{matrix}1 & 0 \\ 0 & 0 \end{matrix} ) ; e_{2} = ( \begin{matrix}0 & 0 \\ 0 & 1 \end{matrix} ) .

Lemma 2. The subalgebra P ⊆ F [W] generated by t(u), u ∈ B̃, contains F [W]^{SL_{3}}.

Proof. For all r we have α_{r} = t(e_{1}x^{(r)}e_{1}) ∈ B̃; similarly, β_{r}, α, β ∈ B̃. Also,

(v^{(p)}, w^{(q)}) = t((e_{1}x^{(p)} · e_{2})(e_{2}x^{(q)} · e_{1})) ∈ P

and

det(v^{(p)}, v^{(q)}, v^{(r)}) = t((e_{1}x^{(p)} · e_{2})(e_{1}x^{(q)} · e_{2}) · (e_{1}x^{(r)} · e_{2})) ∈ P;

similarly, det(w^{(p)}, w^{(q)}, w^{(r)}) ∈ P. □

Proof of proposition 1. Put R = R(x^{(k+1)}) and pick f ∈ F [(k + 1)O]^{G}, we may assume that f is homogeneous of positive degree. Since θ is monic on F [(k + 1)O]^{G}, it extends to an isomorphism from the field of fractions (in fact, it is equal to the field of invariants F ((k+1)O)^{G} [10, p.165]) into F (W). Hence, it suffices to show that θ(f)θ(R)^{m} ∈ θ(t_{k+1}[O]) for some m.

By the previous lemma, θ(f) ∈ P, hence θ(f) is equal to a trace polynomial g ∈ t[O] where some variables are replaced by e_{1}, e_{2} and x. Observe that

e_{1} = \frac{(x − β · 1)(α − β)}{(α − β)^{2}} , e_{2} = \frac{(x − α · 1)(β − α)}{(α − β)^{2}} ;

we replace each occurrence of e_{i} in θ(f) by the corresponding element and get

θ(f) = \frac{1}{(α − β)^{2m}} ∑_{p} θ(f_{p})h_{p}

for some m ≥ 0, f_{p} ∈ t_{k+1}[O] and h_{p} ∈ F [α, β].

Since θ(f), θ(f_{p}) are Z_{2}-invariants, we may assume that h_{p} is a symmetric polynomial in two variables α, β. Hence, it is a polynomial in α + β = θ(t(x^{(k+1)})) and ω. Finally,

θ(R) = 2t(x^{2}) − t(x)^{2} = 2(α^{2} + β^{2}) − (α + β)^{2} = (α − β)^{2}

hence, ∑_{p} θ(f_{p})h_{p} ∈ θ(t_{k+1}[O]) and θ(f)θ(R)^{m} ∈ θ(t_{k+1}[O]). □

The next lemma is quite trivial but very useful.

Lemma 3. Let V be a vector space over F with a nondegenerate symmetric bilinear form ⟨x, y⟩ and an orthonormal basis e_{1}, …, e_{n}. Then

∑_{i=1}^{n}⟨ae_{i}⟩⟨be_{i}⟩ = ⟨a, b⟩  (10)

for all a, b ∈ V

Proof. Indeed, a = ∑_{i=1}^{n} a_{i}e_{i} and b = ∑_{j=1}^{n} b_{j}e_{j}, hence

⟨a, b⟩ = ⟨∑_{i=1}^{n} a_{i}e_{i}, ∑_{j=1}^{n} b_{j}e_{j}⟩ =

= ∑_{i,j=1}^{n} a_{i}b_{j}⟨e_{i}, e_{j}⟩ = ∑_{i=1}^{n} a_{i}b_{i} = ∑_{i=1}^{n}⟨a, e_{i}⟩⟨b, e_{i}⟩ .

□

It is well-known (e.g.[14, ch.2]) that the form tr(xy) is nondegenerate. Let e_{1}, e_{2}, …, e_{8} be an orthonormal basis of O, by (10)

∑_{i=1}^{8} t(ae_{i})t(be_{i}) = t(ab)  (11)

for all a, b ∈ O.

Now put x = x^{(k+1)} and

Δ = ∑_{i=1}^{8} D_{x}(e_{i})^{2} ,  (12)

it is the Laplace operator Δ(x, x).

Lemma 4. The algebra of trace polynomials t_{k+1}[O] is closed under Δ.

Proof. By lemma 1, t_{k+1}[O] is spanned by 1 and f = t(u_{1}) … t(u_{m}), where t(u_{p}) are polynomials (a1)-(a2).

Since D_{x}(e_{i}) is a derivation, we have

f · D_{x}^{2}(e_{i}) = ∑_{p,q=1; p≠q}^{m} t(u_{1}) … (t(u_{p}) · D_{x}(e_{i})) … (t(u_{q}) · D_{x}(e_{i})) … t(u_{m})+

+ ∑_{p=1}^{m} t(u_{1}) … (t(u_{p}) · D_{x}^{2}(e_{i})) … t(u_{m}) .

The lemma will be proven, if we show that

∑_{i=1}^{8} (t(u_{p}) · D_{x}(e_{i})) (t(u_{q}) · D_{x}(e_{i}))

and

∑_{i=1}^{8} t(u_{p}) · D_{x}^{2}(e_{i})

belong to t_{k+1}[O].

Consider the first sum; we may assume that u_{p} = v_{p}x, where deg_{x}(v_{p}) = 0 or v_{p} = x. Anyway, by (11) the first sum lies in t_{k+1}[O].

The latter sum is not equal to zero only if u_{p} = x^{2}, in this case it is a constant. □

Proof of theorem 2.

Pick f ∈ F [kO]^{G} ⊆ F [(k + 1)O]^{G}, by proposition 1 f · R(x)^{m} ∈ t_{k+1}[O] for some m.

If m = 0, then f ∈ t[kO]; otherwise we shall show that f · R(x)^{m−1} ∈ t_{k+1}[O].

By lemma 4, (f · R(x)^{m}) · Δ ∈ t_{k+1}[O]. On the other hand, since f does not depend on x,

(f · R(x)^{m}) · Δ = f · (R(x)^{m} · Δ) .

Denote D = D_{x}(e_{i}), a = t(x) and b = t(x^{2}), then R = R(x) = 2b − a^{2} and

R^{m}D^{2} = m(m − 1)R^{m−2}(RD)^{2} + mR^{m−1}(RD^{2})
RD = 2bD − 2a(aD) ;
RD^{2} = 2bD^{2} − 2(aD)^{2} − 2a(aD^{2}) = 2bD^{2} − 2(aD)^{2},

because aD^{2} = 0. Hence, by (11)

∑_{i=1}^{8} R · D_{x}(e_{i})^{2} = 4 ∑_{i=1}^{8} t(e_{i}^{2}) − 2 ∑_{i=1}^{8} t(e_{i})^{2} = 4 × 8 − 2 ∑_{i=1}^{8} t(e_{i} · 1)tr(e_{i} · 1) = 32 − 2t(1) = 28 .

Also

(R · D)^{2} = 4a^{2}(aD)^{2} − 8a(aD)(bD) + 4(bD)^{2}

and by (11)

∑_{i=1}^{8}(R · D_{x}(e_{i}))^{2} = 4 ∑_{i=1}^{8} t(x)^{2}t(e_{i})^{2} − 8 ∑_{i=1}^{8} t(x)t(e_{i}) · 2t(xe_{i}) + 4 ∑_{i=1}^{8}(2t(xe_{i}))^{2} =

= 4t(x)^{2}t(1) − 16 · t(x)^{2} + 16t(x^{2}) = 8t(x)^{2} − 16t(x)^{2} + 16t(x^{2}) = 8R .

Thus,

R^{m} · Δ = ∑_{i=1}^{8} R^{m}D_{x}(e_{i})^{2} = m(m − 1)R^{m−2} ∑_{i=1}^{8}(RD_{x}(e_{i}))^{2} + mR^{m−1} ∑_{i=1}^{8}(RD_{x}(e_{i})^{2})

= m(m − 1)R^{m−2}8R + 28mR^{m−1} = m(8m + 20)R^{m−1} .

Theorem 2 is proven. □

3. Slices of kA : F_{4}.

Recall the definition of Albert algebra [9, 14]. Let O_{3} = O ⊗_{F} F_{3} be the algebra of 3 × 3matrices 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 simple Jordan algebra A of dimension 27 called an Albert algebra.

Let e_{i} = e_{ii}, e_{ij} (i, j ∈ {1, 2, 3}, i ≠ j) denote 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},  (13)

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

The next formulae will be used very often

[a]_{ij} ∘ [b]_{jk} = [\overline{ab}]_{ki} ;  (14)
[a]_{ij} ∘ [b]_{ij} = \frac{1}{2}t(ab̄)(e_{i} + e_{j}) .  (15)

If c = ( \begin{matrix}α & v \\ w & β \end{matrix} ), then [ \begin{matrix}α & v \\ w & β \end{matrix} ]_{jk} denotes [c]_{jk}. In fact, the subspace of such elements is the

Pierce component A_{ij} = AU_{e_{i},e_{j}}, where U_{x,y} ∈ End_{F} A, zU_{x,y} = 1/2((zx)y + (zy)x) − z(xy)); also F e_{i} = A_{ii} = AU_{e_{i},e_{i}} (cf. [9, p.118]).

The subspace of elements (13), where

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

is a subalgebra of A isomorphic to the Jordan algebra F_{3}^{(+)} (the associative algebra F_{3} with the symmetrized multiplication a ∘ b). For all j, k ∈ {1, 2, 3}, j ≠ k, put

ϵ_{jk} = [ \begin{matrix}1 & 0 \\ 0 & 0 \end{matrix} ]_{jk} ,  (16)

then e_{i}, ϵ_{jk} form a basis of F_{3}^{(+)}. In particular, [c_{i}]_{jk} = c_{jk}ϵ_{jk} + c_{kj}ϵ_{kj} for c_{i} ∈ F^{2}. Also, denote by F^{3} the subalgebra of diagonal matrices F e_{1} + F e_{2} + F e_{3} ⊆ F_{3}^{(+)} ⊆ A. It is known that the automorphism group G = Aut(A) is a simple linear algebraic group of

type F_{4} (e.g.[8]). We need certain subgroups of G: (c1) the symmetric group S_{3} whose action is induced by permutation of the idempotents

e_{1}, e_{2}, e_{3} [9, p.389,ex.3]; (c2) the subgroup Z_{2} generated by an automorphism of the second order (cf. [7, sect.2]):

σ : e_{i} → e_{i} ∀i ∈ {1, 2, 3} ;

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

for all (ijk) ∈ A_{3}; (c4) by [9, p.386, th.8], the stabilizer M of F_{3}^{(+)} is isomorphic to the special linear group SL_{3}.

More precisely, if Q ∈ SL_{3}, then we can define an automorphism Q̂ ∈ M in following way (it suffices to consider only elements in the complement of F_{3}^{(+)} to A):

ψ̂ : [ \begin{matrix}0 & v \\ w & 0 \end{matrix} ]_{ij} → [ \begin{matrix}0 & vQ \\ w(Q^{⊤})^{−1} & 0 \end{matrix} ]_{ij}

for all (ijk) ∈ A_{3}. Then Q → Q̂ is an isomorphism of SL_{3} onto M; (c5) the stabilizer of F^{3} is isomorphic to a spinor group Spin_{7} [9, p.388,ex.2]; note that

SL_{3}, T_{2}, Z_{2} ⊆ Spin_{7}.

Put W = kA⊕F_{3}^{(+)}⊕F^{3} ⊆ (k+2)A; obviously, subgroups (c1)-(c4) are contained in N_{G}(W). Also, it is easy to check that Z_{2} normalizes T_{2} and Z_{2} ∩ T_{2} = {1}, therefore, H_{0} = T_{2}Z_{2} is a semidirect product T_{2} Z_{2}. Besides, H_{0} normalizes SL_{3} and H_{0} ∩ SL_{3} = {1}, hence, H = SL_{3} H_{0} ≤ N_{G}(W). Finally, S_{3} normalizes H and the semidirect product H S_{3} is isomorphic to a subgroup H · S_{3} of N_{G}(W). Moreover, one can check that they coincide and W is a slice (e.g. [1, lemma 4.5],[6, lemma 5]).

Similarly, let W_{0} = (k + 1)A ⊕ F^{3} ⊂ (k + 2)A, then it is also a slice and N_{G}(W_{0}) is a semidirect product of S_{3} and the stabilizer of e_{1}, e_{2}, e_{3}, i.e., Spin_{7}.

Denote by θ (θ_{0}) the algebra homomorphism from F [(k +2)A] to F [W] (respectively, F [W_{0}]) which takes f to the restriction f|_{W} (f|_{W_{0}}). The corresponding mapping from Pol[(k + 2)A, A] to Pol[W, A] (Pol[W_{0}, A]) will be denoted by the same symbol.

Recall that W (W_{0}) is a slice, in particular, θ (θ_{0}) restricts to a monomorphism from F [(k + 2)A]^{G} to F [W] (F [W_{0}]).

Denote α_{i} = α_{i}^{(k+1)}, β_{i} = α_{i}^{(k+2)}, x_{ij} = x_{ij}^{(k+1)} for all i, j. Notice that θ(x^{(r)}) = θ_{0}(x^{(r)}) = x^{(r)} for all r ≤ k; also θ_{0}(x^{(k+1)}) = x^{(k+1)}. Next, put x = θ(x^{(k+1)}) = ∑_{i=1}^{3} α_{i}e_{i} + ∑_{(ijk)}[x_{i}]_{jk}, where

[x_{i}]_{jk} = [ \begin{matrix}x_{jk} & 0 \\ 0 & x_{kj} \end{matrix} ]_{jk} .

Finally, put y = θ(x^{(k+2)}) = θ_{0}(x^{(k+2)}) = ∑_{i=1}^{3} β_{i}e_{i}, By the classical result on invariants of SL_{n} the algebra K_{1} = F [W]^{SL_{3}} is generated by

(d1) α_{i}^{(r)}, α_{i}, β_{i}; (d2) x_{ij}^{(r)}, x_{ij}, i ≠ j; (d3) (v_{i}^{(p)}, w_{j}^{(q)}); (d4) det(v_{i}^{(p)}, v_{j}^{(q)}, v_{k}^{(r)}), det(w_{i}^{(p)}, w_{j}^{(q)}, w_{k}^{(r)}), where r, p, q ∈ {1, …, k}, i, j, k ∈ {1, 2, 3}.

Let B denote the Jordan subalgebra generated by x^{(r)}, r ≤ k and x, y in Pol[W, A], i.e., B = θ(F_{k+2}(A)) (in fact, θ is an isomorphism from F_{k+2}(A) onto B [7, lemma 1]).

Also, denote by B̃ the Jordan subalgebra generated by B and e_{i}, ϵ_{ij} and let P be the subalgebra of F [W] generated by tr(u) , where u ∈ B̃.

Proposition 2. P contains K_{1}.

Proof. It suffices to show that the generators (d1)-(d4) belong to P.

First of all, the elements α_{i}^{(r)}e_{i} = x^{(r)}U_{e_{i},e_{i}}, [x_{k}^{(r)}]_{ij} = x^{(r)}U_{e_{i},e_{j}}, α_{i}e_{i}, β_{i}e_{i}, [x_{k}]_{ij} lie in B̃. Hence, α_{i}^{(r)} = tr(α_{i}^{(r)}e_{i}) ∈ P and similarly, α_{i}, β_{i} ∈ P.

Besides, x_{ij} = tr(ϵ_{ij} ∘ [x_{k}]_{ij}) ∈ P for all i, j. Hence,

[ \begin{matrix}0 & v_{k}^{(r)} \\ w_{k}^{(r)} & 0 \end{matrix} ]_{ij} = [x_{k}]_{ij} − x_{ij}ϵ_{ij} − x_{ji}ϵ_{ji} ∈ B̃

and

[ \begin{matrix}0 & v_{k}^{(r)} \\ 0 & 0 \end{matrix} ]_{ij} = ([ \begin{matrix}0 & v_{k}^{(r)} \\ w_{k}^{(r)} & 0 \end{matrix} ]_{ij} ∘ ϵ_{kj}) ∘ ϵ_{jk} ∈ B̃ .

Multiplying these elements by ϵ_{ij}, we get

[ \begin{matrix}0 & v_{k}^{(r)} \\ 0 & 0 \end{matrix} ]_{st} , [ \begin{matrix}0 & 0 \\ w_{k}^{(r)} & 0 \end{matrix} ]_{st} ∈ B̃

for all k, r, s, t. Therefore,

(v_{i}^{(p)}, w_{j}^{(q)}) = tr ([ \begin{matrix}0 & v_{i}^{(p)} \\ 0 & 0 \end{matrix} ]_{st} ∘ [ \begin{matrix}0 & 0 \\ w_{j}^{(q)} & 0 \end{matrix} ]_{st}) ∈ P

and

det(v_{i}^{(p)}, v_{j}^{(q)}, v_{k}^{(r)}) = tr (([ \begin{matrix}0 & v_{i}^{(p)} \\ 0 & 0 \end{matrix} ]_{12} ∘ [ \begin{matrix}0 & v_{j}^{(q)} \\ 0 & 0 \end{matrix} ]_{23})) ∘ [ \begin{matrix}0 & v_{k}^{(r)} \\ 0 & 0 \end{matrix} ]_{31}) ∈ P ;

similarly, det(w_{i}^{(p)}, w_{j}^{(q)}, w_{k}^{(r)}) ∈ P. So, all generators of K_{1} belong to P. □

The image B_{0} = θ_{0}(F_{k+2}(A)) ⊆ Pol[W_{0}, A] is generated by x^{(r)}, r ≤ k+1, and y; consider the Jordan algebra B̃_{0} generated by B_{0} and e_{1}, e_{2}, e_{3}. Let P_{0} be the subalgebra of F [W_{0}] generated by tr(u), where u runs over B̃_{0}. Since Spin_{7} leaves fixed e_{i} and y, we have P_{0} ⊆ F [W_{0}]^{Spin_{7}}.

Put t = t(x_{1}^{(k+1)}x_{2}^{(k+1)} · x_{3}^{(k+1)}), t_{i} = t(x_{i}^{(k+1)}x̄_{i}^{(k+1)}) and

ω = t^{2} − \frac{1}{2}t_{1}t_{2}t_{3} ;  (17)

since [x_{i}^{r}]_{jk} = x^{(r)}U_{e_{i},e_{j}} ∈ B̃_{0}, by (15) t, t_{i}, ω ∈ P_{0}.

Proposition 3. The algebra F [W_{0}]^{Spin_{7}} is contained in the algebra

P_{0}ω^{−1} = {f/ω^{i}| f ∈ P_{0}; i = 0, 1, … }

Proof. Consider a homomorphism θ_{1} from Pol[W_{0}, A] (F [W_{0}]) to Pol[W, A] (F [W]) which takes f to f|_{W}.

Notice that θ_{1} restricts to a monomorphism from F [W_{0}]^{Spin_{7}} to F [W]. Actually, there is a dense subset U of W_{0} = (k + 1)A ⊕ F^{3} such that for all a ∈ U there is an automorphism σ ∈ G which takes a to W (cf.[7, lemma 1]). Using a suitable element from S_{3}, we can choose such σ leaving fixed all e_{i}, i.e., σ ∈ Spin_{7}. This means that the ordit W · Spin_{7} is a dense subset of W_{0}, hence, θ_{1} is a monomorphism.

Since P_{0} ⊆ F [W_{0}]^{Spin_{7}}, θ_{1} extends to an isomorphism from P_{0}ω^{−1} onto θ_{1}(P_{0})θ_{1}(ω)^{−1}. Therefore, it suffices to show that

θ_{1}(F [W_{0}]^{Spin_{7}}) ⊆ θ_{1}(P_{0})θ(ω)^{−1} .

Pick f ∈ θ_{1}(F [W_{0}]^{Spin_{7}}), then it belongs to (F [W]^{SL_{3}})^{T_{2} Z_{2}} ⊆ P^{T_{2} Z_{2}}. Let B̃′ be the Jordan subalgebra of Pol[W, A] generated by B and e_{1}, e_{2}, e_{3}; obviously, θ_{1}(B̃_{0}) = B̃′ ⊆ B̃ and θ_{1}(P_{0}) is generated by tr(u), where u ∈ B̃′.

Put μ = x_{12}x_{23}x_{31}, μ̄ = x_{21}x_{13}x_{32}, then θ_{1}(t_{i}) = 2x_{jk}x_{kj}, θ_{1}(t) = μμ̄ and therefore

θ_{1}(ω) = (μ − μ̄)^{2} ;

also, one can check that

ϵ_{ij} = \frac{1}{μ − μ̄}([x_{k}]_{ij}x_{jk}x_{ki} − [x_{i}]_{jk} ∘ [x_{j}]_{ki}x_{ij})

hence, for all f ∈ F [W]^{SL_{3}} ⊆ P replacing all occurrences of ϵ_{ij} (i, j ∈ {1, 2, 3}, i ≠ j) in f, we get f to belong to the algebra generated by θ_{1}(P_{0}), x_{ij}, i, j = 1, 2, 3; i ≠ j and θ_{1}(ω)^{−1} . Hence,

f = θ_{1}(ω)^{−d} ∑_{q} f_{q}h_{q}

where f_{q} ∈ θ_{1}(P_{0}) and h_{q} ∈ F [x_{ij}] (all polynomials in x_{ij}). Moreover, T_{2} Z_{2} is a linearly reductive nilpotent group, hence h_{p} is equal to a sum of semi-invariants of T_{2} Z_{2}. Since f_{q} ∈ θ_{1}(P_{0}) ⊆ K_{1}^{T_{2} Z_{2}}, we may assume that h_{p} ∈ F [x_{ij}]^{T_{2} Z_{2}}

Lemma 5. F [x_{ij}]^{T_{2} Z_{2}} is generated by θ_{1}(ω), θ_{1}(t_{i}), θ_{1}(t).

Proof. Recall that (ξ_{1}, ξ_{2}, ξ_{3}) ∈ T_{2} takes x_{ij} to ξ_{i}^{−1}ξ_{j}x_{ij}; using this fact one can easily check that F [x_{ij}]^{T_{2}} is generated by θ_{1}(t_{i}) = x_{jk}x_{kj}, μ and μ̄. In addition, Z_{2} swops μ and μ̄ and leaves fixed θ_{1}(t_{i}), hence, F [x_{ij}]^{T_{2} Z_{2}} = (F [x_{ij}]^{T_{2}})^{Z_{2}} is generated by θ_{1}(t_{i}), μ + μ̄ = θ_{1}(t) and θ_{1}(ω) = (μ − μ̄)^{2}. □

Obviously, lemma 5 completes the proof of proposition 3. □

Recall that W_{0} is a slice of (k + 2)A : G, hence, θ_{0} restricts to a monomorphism from F [(k + 2)A]^{G} and this restriction extends to a field isomorphism from F ((k + 2)A)^{G} onto F (W_{0})^{N_{G}(W_{0})}, where N_{G}(W_{0}) = Spin_{7} S_{3}. Hence, P_{0}^{S_{3}} ⊆ F (W_{0})^{N_{G}(W_{0})}; denote by Q_{0} the pre-image θ_{0}^{−1}(P_{0}^{S_{3}}) in F ((k + 2)A)^{G}; we shall describe it explicitly.

Let R(x) denote the descriminant of the polynomial (2), then R(x) is a polynomial in tr(x^{i}), i = 1, 2, 3.

Proposition 4. The algebra Q_{0} is contained in

tr_{k+2}[A] · R^{−1}

where R = R(x^{(k+2)}).

Proof. In Pol[W_{0}, A] consider a system of linear equations

∑_{i=1}^{3} β_{i}^{p}e_{i} = θ_{0}(x^{(k+2)})^{p} = y^{p} ,

where p = 0, 1, 2 (we assume that (x^{(k+2)})^{0} = 1). The square of the determinant of this system det(β_{i}^{p}) is equal to θ_{0}(R), hence, e_{i} is equal to a linear combination of powers of θ_{0}(x^{(k+2)}) with coefficients in F [β_{i}]θ_{0}(R)^{−1}.

Now pick f ∈ P_{0}, like in the proof of proposition 3 we replace all occurrences of e_{i} with the corresponding linear combination of powers of θ_{0}(x^{(k+2)}) and get

f = θ_{0}(R)^{−d} ∑_{q} f_{q}h_{q} ,

where d ∈ N, f_{q} ∈ θ_{0}(tr_{k+2}[A]) and h_{q} = h_{q}(β_{1}, β_{2}, β_{3}). Since S_{3} is a finite group, we may assume that h_{q} is an S_{3}-invariant, i.e., it is a symmetric polynomial. It is well-known that h_{q} is a polynomial in tr(y^{i}) = θ_{0}(tr((x^{(k+2)})^{i})), where i = 1, 2, 3. Thus, f = θ_{0}(R)^{−d}θ(h) for some h ∈ tr_{k+2}[A]. Recall that θ_{0} is monic on F ((k + 2)A)^{G}, thus this yields proposition 4 □

Recall that ω ∈ P_{0}, denote by Ω ∈ Q_{0} the pre-image of ω.

Corollary 3. For all f ∈ F [(k + 2)A]^{G} there exist p, q ≥ 0 such that fΩ^{p}R^{q} ∈ tr_{k+2}[A].

Proof. The mapping θ_{0} takes F [(k + 2)A]^{G} to F [W_{0}]^{N_{G}(W_{0})} = (F [W_{0}]^{Spin_{7}})^{S_{3}} which lies in (P_{0}ω^{−1})^{S_{3}} = P_{0}^{S_{3}}ω^{−1} by proposition 3.

Since θ_{0} is monic on F ((k + 2)A)^{G} , the pre-image of F [W_{0}]^{N_{G}(W_{0})} is contained in Q_{0}Ω^{−1}, hence, F [(k + 2)A]^{G} ⊆ Q_{0}Ω^{−1}; so, the statement follows from proposition 4. □

4. Differentiation

In the case of G_{2} the algebra of trace polynomials t[O] is closed under the action of Laplace operators; it happens because t[O] is generated by polynomials of small degree (≤ 4). The situation becomes more complicated in the case of F_{4}.

Lemma 6. The algebra tr[A] is generated by homogeneous invariants of degree ≤ 85.

Proof. Recall that A is the direct sum of submodules F · 1 and A′, dim_{F} A′ = 26. Using a standard argument based on representation of the symmetric group (or that of the general linear group) one can show that tr[A] is generated by linearization (i.e., polarization) of generators of tr_{26}[A].

Thanks to the cubic identity (2) it suffices to show that a Jordan nil algebra A of nil-index 3 generated by 26 variables is nilpotent of index ≤ 84.

The linearization of the identity x^{3} = 0 in a commutative algebra is the Jacobi identity

(x_{1}x_{2})x_{3} + (x_{2}x_{3})x_{1} + (x_{3}x_{1})x_{2} = 0;

hence, the product of two ideals of A is also an ideal and the power J^{k} of an ideal J ◁ A is equal to JR_{J}^{k−1} (like in a Lie algebra). Let I = (A^{2})^{k}/(A^{2})^{k+1}, then I is an A-module and IR_{A}^{27} = 0. By [3, lemma 3] A^{2}R_{A^{2}}^{3}R_{A} = 0, hence, A^{2}R_{A}^{82} = 0, that is A^{84} = 0. □

The next statements says that Laplace operators might give new invariants. Probably, it is a well-known fact, because the action of a Laplace operator on a multilinear polynomial can be viewed as a tenzor contraction.

Lemma 7. Let V be a vector space of dimension n with a nondegenerate symmetric bilinear form ⟨x, y⟩. Then the definition (3) of Laplace operator Δ(x, y) does not depend on the choice of orthonormal basis.

Besides, if ⟨a, b⟩ is G-invariant then

F [kV]^{G}Δ(x, y) ⊆ F [kV]^{G} .

Proof. Let f ∈ F [kV], we shall show that fΔ(x, y) does not depend on the choice of orthonormal basis and if f is G-invariant, then fΔ(x, y) ∈ F [V⃗]^{G} (therefore, fΔ(x, y) ∈ F [kV]^{G} = F [V⃗]^{G} ∩ F [kV]). We may assume that f is homogeneous in each generic element and Δ(x, y) acts nontrivially (hence, f depends on x, y as well).

Pick different generic elements x′, y′ ∉ {x^{(1)}, …, x^{(k)}}, then for all a fD_{x}(a) = fD_{x}(x′)D_{x′}(a), hence

fΔ(x, y) = fD_{x}(x′)D_{y}(y′)Δ(x′, y′).  (18)

Observe that fD_{x}(x′)D_{y}(y′) is linear in x′ and y′ and belongs to F [V⃗]^{G}, if f ∈ F [kV]^{G}. So, we may assume that f = f(x, y, x^{(1)}, …, x^{(k)}) is linear in x, y.

Since ⟨a, b⟩ is a nondegenerate symmetric bilinear form, there exists

Φ = Φ(x^{(1)}, …, x^{(k)}) ∈ Pol[V⃗ , End_{F} V]

such that

f = ⟨y · Φ, x⟩ .

Hence, the trace of Φ is equal to

∑_{i=1}^{n}⟨E_{i} · Φ , E_{i}⟩ =

= ∑_{i=1}^{n} f(E_{i}, E_{i}, x^{(1)}, …, x^{(k)}) = f · Δ(x, y) ,

thus f · Δ(x, y) does not depend on the choice of E_{1}, …, E_{n}.

Put Φσ = Φ(x^{(1)} · σ^{−1}, …, x^{(k)} · σ^{−1}), then for all σ ∈ GL(V) we have

⟨y · Φ, x⟩ = f = f(x · σ^{−1}, y · σ^{−1}, x^{(1)} · σ^{−1}, …, x^{(k)} · σ^{−1}) =
= ⟨y · σ^{−1}(Φσ) , (x · σ^{−1})⟩ = ⟨y · σ^{−1}(Φσ)(σ^{−1})^{⊤} , x⟩ .

where σ → σ^{⊤} is the transposition with respect to ⟨a, b⟩. Hence,

Φσ = σΦσ^{⊤} .

If σ ∈ G, then σ^{⊤} = σ^{−1}, hence, fΔ(x, y) = tr(Φ) is a G-invariant. □

Let tr[A]^{∘} denote the minimal subalgebra which contains tr[A] and is closed under Laplace operators.

Proposition 5. The algebra tr[A]^{∘} is generated by elements (4).

Proof. Let A be the subalgebra of F [A⃗] generated by (4); then A contains tr[A] and A ⊆ tr[A]^{∘}. To prove the reciprocal inclusion it suffices to show that A is closed under Laplace operators, i.e., fΔ(x, y) ∈ A for all f ∈ A.

We may assume that f is a product of elements (4), i.e., f = ∏_{i=1}^{q} f_{i}, where f_{i} = g_{i}Δ_{1}^{(i)} … Δ_{t_{i}}^{(i)} and g_{i} ∈ S.

Using (18) and a sufficient number of different generic elements, we can replace all Δ_{j}^{(i)} with another ones Δ̄_{j}^{(i)} depending on different generic elements (which differ x, y as well); at the same time, each g_{i} is replaced with some linearization (it lies in tr[A]). Moreover, we choose the generic elements so that Δ̄_{j}^{(i)} = Δ(x^{(r)}, x^{(l)}) and all g_{k}, k ≠ i do not depend on x^{(r)}, x^{(l)}, therefore, D_{x^{(r)}}(a), D_{x^{(l)}}(a) kill g_{k}. Hence, f can be written in the form gΔ_{1} … Δ_{t}, where Δ_{1} … Δ_{t} is the product of all Δ̄_{j}^{(i)} and g ∈ tr[A] is multilinear in all generic elements whose the operators depend on. Recall that Δ(x, y) commutes with all Δ_{i}, hence, by (18) fΔ(x, y) can be written in the same form (we linearize g and attach Δ(x′, y′)). Finally, since S generates tr[A], we can replace g with a monomial in the generators, hence it suffices to show that the following element belong to A:

(∏_{i=1}^{q} h_{i})Δ_{1} … Δ_{d}  (19)

where h_{i} ∈ S, Δ_{i} = Δ(x^{(i)}, y^{(i)}), all x^{(i)}, y^{(j)} (i ∈ {1, …, d}) are different and ∏_{i=1}^{q} h_{i} is linear in every x^{(i)}, y^{(i)}.

We carry out the induction on d; the case d = 0 is trivial. Now, suppose that there are i ∈ {1, …, d}, s, p ∈ {1, …, q} such that and h_{s} (h_{p}) depends on x^{(i)} (respectively, y^{(i)}). Since Δ-s commute, we may assume i = 1; besides, by [9, p.227, cor. 4], tr((ab)c−(ba)c)) = 0, hence, h_{s} = tr(w_{s}x^{(1)}), h_{p} = tr(w_{p}y^{(1)}) for some

w_{s}, w_{p} ∈ {∑_{j} u_{j}t_{j}| u_{j} ∈ F(A) ∪ {1}; t_{j} ∈ tr[A]}

and using (10), we reduce d. If such i, s, p do not exist, Δ-s are distributed on multiples h_{i}, thus (19) becomes a product of elements (4). □

Proof of theorem 1.

By lemma 7, S ⊆ F [A⃗]^{G}; hence, by lemma 5 we need to show only F [A⃗]^{G} ⊆ tr[A]^{∘}.

Pick a homogeneous f ∈ F [A⃗]^{G} depending on k vector variables, i.e., f ∈ F [kA]^{G}; by corollary 3 there are p, q ∈ N such that fΩ^{p}R^{q} ∈ tr[A]. Put Δ_{1} = Δ(x^{(k+1)}, x^{(k+1)}), Δ_{2} = Δ(x^{(k+2)}, x^{(k+2)}); recall that Ω is equal to a fraction g/h, where h = R^{n} depends on x^{(k+2)}, hence, Ω^{p}Δ_{1} is defined (it is equal to (g^{p}Δ_{1})/h^{p}). Moreover, since the mapping θ_{0} does not change the first k + 1 generic elements of (k + 2)A, Δ_{1} is defined on F [W_{0}] and commutes with θ_{0}.

We may assume that p, q are even; in the next section, proposition 6, we show that ω^{p}Δ_{1}^{3p} is a non-zero constant c. Hence,

fΩ^{p}R^{q}Δ_{1}^{3p} = f(Ω^{p}Δ_{1}^{3p})R^{q} = fθ_{0}(ω^{p}Δ_{1}^{3p})R^{q} = c(fR^{q}) ,

therefore, fR^{q} ∈ tr[A]^{∘}. A similar fact about R and Δ_{2} from proposition 7 enables us to reduce R^{q} and prove theorem 1.

5. Cancellation of ω and R

Recall that E_{p}, p = 1, …, 27, is an orthonormal basis of A with respect to the bilinear form tr(xy). Moreover, by (14), the Pierce components are pairwise orthogonal, hence, we may find such E_{p} in A_{ij}; in particular, E_{p} = e_{p} for p = 1, 2, 3.

Given a ∈ A and (ijk) ∈ A_{3}, denote by (a)_{k} the element c ∈ O such that [c]_{ij} = aU_{e_{i},e_{j}}; it is a function from A to O and in these notations,

(x^{(r)})_{k} = x_{k}^{(r)}  (20)

for all r.

By (15) for all a, b ∈ O

tr([a]_{ij} ∘ [b]_{ij}) = tr(\frac{1}{2}t(ab̄)(e_{i} + e_{j})) = t(ab̄) .  (21)

Let x, y be generic elements of A; by (21) and (10) for all (ijk):

∑_{p=1}^{27} t(x̄_{k}(E_{p})_{k})t(ȳ_{k}(E_{p})_{k}) = ∑_{p=1}^{27} tr([x_{k}]_{ij} ∘ E_{p})tr([y_{k}]_{ij} ∘ E_{p}) = tr([x_{k}]_{ij} ∘ [y_{k}]_{ij}) = t(x̄_{k}y_{k}) . (22)

Proposition 6. For all s ∈ N ω^{2s} · Δ_{1}^{6s} ≠ 0.

Proof. Recall that x = θ_{0}(x^{(k+1)}) and t_{i} = t(x_{i}x̄_{i}), t = t(x_{1}x_{2} · x_{3}).

Put D_{p} = D_{x}(E_{p}); recall that (E_{p})_{q} ≠ 0 for some q and (E_{p})_{i} = 0 for all i ≠ q. Fix q ∈ \{1, 2, 3\}, since x_{j}x_{k} = \overline{([x_{k}]_{ij} ∘ [x_{j}]_{ki})_{i}}, we have (∑_{p} denotes ∑_{p=1}^{27})

∑_{p}(tD_{p})(t_{q}D_{p}) = ∑_{p} ∑_{(ijk)} t((E_{p})_{i} · x_{j}x_{k})2t((E_{p})_{q}x̄_{q}) =  (23)

= ∑_{(ijk), i=q} ∑_{p} t((E_{p})_{i} · x_{j}x_{k})2t((E_{p})_{q}x̄_{q}) = 2t ;

Similarly,

∑_{p}(t_{i}D_{p})^{2} = 4t_{i} ;  (24)

also, by (22) and t(āa) = 2āa

∑_{p}(tD_{p})^{2} = ∑_{p}( ∑_{(ijk)} t((E_{p})_{i} · x_{j}x_{k}))^{2} =  (25)

= ∑_{p} ∑_{(ijk)} t((E_{p})_{i} · x_{j}x_{k})^{2} = ∑_{(ijk)} t(\overline{x_{j}x_{k}} · x_{j}x_{k}) = ∑_{(ijk)} 1/2 · t_{i}t_{j} ;

∑_{p}(t_{i}D_{p}^{2}) = ∑_{p} 2t((E_{p})_{i}\overline{(E_{p})_{i}}) = 2 ∑_{(E_{p})_{i}≠0} t((E_{p})_{i}\overline{(E_{p})_{i}}) = 16 .  (26)

For all i ≠ j

tD_{p}^{2} = (t_{i}D_{p})(t_{j}D_{p}) = 0 .  (27)

Hence, by (27)

t^{m}t_{1}^{n_{1}}t_{2}^{n_{2}}t_{3}^{n_{3}}D_{p}^{2} = m(m − 1)t^{m−2}t_{1}^{n_{1}}t_{2}^{n_{2}}t_{3}^{n_{3}} · (tD_{p})^{2}+  (28)

+ ∑_{(ijk)} 2mn_{i} · t^{m−1}t_{i}^{n_{i}−1}t_{j}^{n_{j}}t_{k}^{n_{k}}(tD_{p})(t_{i}D_{p}) + ∑_{(ijk)} n_{i} · t^{m}t_{i}^{n_{i}−1}t_{j}^{n_{j}}t_{k}^{n_{k}}(t_{i}D_{p}^{2})+

+ ∑_{(ijk)} n_{i}(n_{i} − 1) · t^{m}t_{i}^{n_{i}−2}t_{j}^{n_{j}}t_{k}^{n_{k}}(t_{i}D_{p})^{2} .

Next, by (23)-(26)

t^{m}t_{1}^{n_{1}}t_{2}^{n_{2}}t_{3}^{n_{3}}Δ_{1} = \frac{1}{2}m(m − 1)t^{m−2} ∑_{(ijk)} t_{i}^{n_{i}+1}t_{j}^{n_{j}+1}t_{k}^{n_{k}}+  (29)

+ ∑_{(ijk)}(4mn_{i} + 16n_{i} + 4n_{i}(n_{i} − 1)) · t^{m−1}t_{i}^{n_{i}−1}t_{j}^{n_{j}}t_{k}^{n_{k}} .

For m = 2r we have

t^{2r}t_{1}^{n_{1}}t_{2}^{n_{2}}t_{3}^{n_{3}}Δ_{1} =  (30)

= r(2r − 1) ∑_{(ijk)} t^{2(r−1)}t_{i}^{n_{i}+1}t_{j}^{n_{j}+1}t_{k}^{n_{k}}+

+ ∑_{(ijk)} 4n_{i}(n_{i} + 2r + 3) · t^{2r}t_{i}^{n_{i}−1}t_{j}^{n_{j}}t_{k}^{n_{k}} .

Put

γ_{r} = ∏_{q=1}^{r} \frac{4(q + 3)}{2q − 1}  (31)

for all r ≥ 1 and γ_{0} = 1.

Lemma 8. For all r, n_{1}, n_{2}, n_{3} ≥ 0

(t_{1}^{n_{1}+r}t_{2}^{n_{2}+r}t_{3}^{n_{3}+r} − γ_{r}t^{2r}t_{1}^{n_{1}}t_{2}^{n_{2}}t_{3}^{n_{3}})Δ_{1}^{3k+n_{1}+n_{2}+n_{3}} = 0

Proof. We shall prove it by induction on r. If r = 0, then it is trivial; suppose that for all s < r and for all n_{1}, n_{2}, n_{3} ≥ 0 it is proven , we carry out the second induction on n = n_{1} + n_{2} + n_{3}.

Let n = 0, then n_{1} = n_{2} = n_{3} = 0; by (30)

t_{1}^{r}t_{2}^{r}t_{3}^{r}Δ_{1} = ∑_{(ijk)} 4r(r + 3)t_{i}^{r−1}t_{j}^{r}t_{k}^{r}

and

t^{2r}Δ_{1} = ∑_{(ijk)} r(2r − 1)t^{2(r−1)}t_{i}t_{j}

Denote by V (d) the subspace of polynomials which vanish under Δ_{1}^{d}; by the induction hypothesis on r modulo V (3r − 1) the latter sum is equal to

∑_{(ijk)} \frac{r(2r − 1)}{γ_{r−1}}t_{i}^{r}t_{j}^{r}t_{k}^{r−1} = ∑_{(ijk)} \frac{4r(r + 3)}{γ_{r}}t_{i}^{r−1}t_{j}^{r}t_{k}^{r} ,

therefore,

t^{2r}Δ_{1} − γ_{r}t_{1}^{r}t_{2}^{r}t_{3}^{r}Δ_{1} ∈ V (3r − 1) .

and the statement holds for n = 0. Now let us show it for n > 0.

By (30)

t_{1}^{n_{1}+r}t_{2}^{n_{2}+r}t_{3}^{n_{3}+r}Δ_{1} = ∑_{(ijk)} 4(n_{i} + r)((n_{i} + r) + 3)t_{i}^{n_{i}+r−1}t_{j}^{n_{j}+r}t_{k}^{n_{k}+r}

on the other hand, modulo V (3r + n_{1} + n_{2} + n_{3})

t^{2r}t_{1}^{n_{1}}t_{2}^{n_{2}}t_{3}^{n_{3}}Δ_{1} ≡ \frac{r(2r − 1)}{γ_{r−1}} ∑_{(ijk)} t_{i}^{n_{i}+1+(r−1)}t_{j}^{n_{j}+1+(r−1)}t_{k}^{n_{k}+(r−1)}+

+ ∑_{(ijk)} \frac{4n_{i}(n_{i} + 2r + 3)}{γ_{r}}t_{i}^{n_{i}−1+r}t_{j}^{n_{j}+r}t_{k}^{n_{k}+r} =

= \frac{r(2r − 1)}{γ_{r−1}} ∑_{(ijk)} t_{i}^{n_{i}+r}t_{j}^{n_{j}+r}t_{k}^{n_{k}+r−1}+

+\frac{1}{γ_{r}} ∑_{(ijk)} 4n_{i}(n_{i} + 2r + 3)t_{i}^{n_{i}+r−1}t_{j}^{n_{j}+r}t_{k}^{n_{k}+r}

Hence, by (31)

(t_{1}^{n_{1}+r}t_{2}^{n_{2}+r}t_{3}^{n_{3}+r} − γ_{r}t^{2r}t_{1}^{n_{1}}t_{2}^{n_{2}}t_{3}^{n_{3}})Δ_{1} ≡

≡ ∑_{(ijk)}(4r(r + 3) − \frac{γ_{r}r(2r − 1)}{γ_{r−1}})t_{i}^{n_{i}+r−1}t_{j}^{n_{j}+r}t_{k}^{n_{k}+r} = 0

□.

Recall that ω = t^{2} − 1/2t_{1}t_{2}t_{3}, hence,

ω^{2s} = ∑_{r=0}^{2s} C_{2s}^{r}(−2)^{−2s+r}t^{2r}t_{1}^{2s−r}t_{2}^{2s−r}t_{3}^{2s−r}

By lemma 8, modulo V (6s) it is equal to

2^{−2s}t_{1}^{2s}t_{2}^{2s}t_{3}^{2s} ∑_{r=0}^{2s} C_{2s}^{r}2^{r} \frac{1}{γ_{r}} (−1)^{2s−r} ,

hence, by (30) it follows that t_{1}^{2s}t_{2}^{2s}t_{3}^{2s}Δ_{1}^{6s} is a positive integer, hence, to prove the proposition it suffices to show that

T = ∑_{r=0}^{2s} C_{2s}^{r}2^{r} \frac{1}{γ_{r}} (−1)^{2s−r} ≠ 0

Lemma 9.

T = \frac{3}{2π} ∫_{0}^{1} ∫_{0}^{x_{3}} ∫_{0}^{x_{2}} ∫_{0}^{π/2}(x_{1}Sin^{2}x − 1)^{2s}dxdx_{1}dx_{2}dx_{3} .

Obviously, this yields T > 0.

Proof. We make use of the following well-known formulae

∫_{0}^{1} ∫_{0}^{x_{3}} ∫_{0}^{x_{2}} x_{1}^{r}dx_{1}dx_{2}dx_{3} = \frac{1}{(r + 3)(r + 2)(r + 1)} ;

∫_{0}^{π/2} Sin^{2r}(x)dx = \frac{π}{2} ∏_{q=1}^{r} \frac{2q − 1}{2q} .

This gives

T = ∑_{r=0}^{2s} C_{2s}^{r} \frac{2^{r}}{γ_{r}} (−1)^{2s−r} =

= ∑_{r=0}^{2s} C_{2s}^{r}2^{r} ∏_{q=1}^{r} \frac{2q − 1}{4(q + 3)}(−1)^{2s−r} = ∑_{r=0}^{2s} C_{2s}^{r} ∏_{q=1}^{r} \frac{2q − 1}{2(q + 3)}(−1)^{2s−r}

= ∑_{r=0}^{2s} C_{2s}^{r} \frac{3!}{2^{3}(r + 3)(r + 2)(r + 1)} ∏_{q=1}^{r} \frac{2q − 1}{2q}(−1)^{2s−r} =

= ∑_{r=0}^{2s} C_{2s}^{r}(−1)^{2s−r} \frac{3}{2π} ∫_{0}^{1} ∫_{0}^{x_{3}} ∫_{0}^{x_{2}} ∫_{0}^{π/2} x_{1}^{r}Sin^{2r}(x)dxdx_{1}dx_{2}dx_{3} =

= \frac{3}{2π} ∫_{0}^{1} ∫_{0}^{x_{3}} ∫_{0}^{x_{2}} ∫_{0}^{π/2} (∑_{r=0}^{2s} C_{2s}^{r}(−1)^{2s−r}x_{1}^{r}Sin^{2r}(x)) dxdx_{1}dx_{2}dx_{3} =

= \frac{3}{2π} ∫_{0}^{1} ∫_{0}^{x_{3}} ∫_{0}^{x_{2}} ∫_{0}^{π/2}(x_{1}Sin^{2}x − 1)^{2s}dxdx_{1}dx_{2}dx_{3} .

□.

Now let us prove an analog of proposition 6 for R.

Proposition 7. For all s ∈ N R^{2s} · Δ_{2}^{6s} ≠ 0.

Denote y = x^{(k+2)}, then R(y)D_{y}(1) = 0. Indeed, if y is diagonal matrix ∑_{i=1} β_{i}e_{i}, then R(y) = det(β_{i}^{j}|_{i=1,j=0}^{3,2}) and one may check it easily. On the other hand, R(y)D_{y}(1) is a Ginvariant, hence it is equal to 0 for all y ∈ A.

Put z = y − 1/3tr(y), then tr(z) = 0 and R = R(y) = R(z) = αa^{3} + βb^{2}, where a = tr(z^{2}), b = tr(z^{3}), because R(y) is a polynomial in tr(y), tr(y^{2}), tr(y^{3}) of degree 6 with respect to y. Substituting z = 2e_{2} − e_{1} − e_{3} and z = e_{1} − e_{2} we get

R(z) = 1/2a^{3} − 3b^{2} .

We fix an orthonormal basis G_{p}, p = 1, …, 27, of A where G_{1} = 1/\sqrt{3} and tr(G_{p}) = 0 for all p ≥ 2.

By (5) for any polynomial f = f(z)

f(z)Δ_{2} = f(z)Δ(z, z) .

Now ∑_{p} denotes ∑_{p=2}^{27}; for all u, v ∈ A′ tr(uG_{1}) = tr(vG_{1}) = 0, hence, by (10) we have

∑_{p} tr(uG_{p})tr(vG_{p}) = ∑_{p=1}^{27} tr(uG_{p})tr(vG_{p}) = tr(uv) .  (32)

Put D_{p} = D_{y}(G_{p}), then

∑_{p}(aD_{p})^{2} = 4 ∑_{p} tr(zG_{p})^{2} = 4tr(z^{2}) = 4a .  (33)

Next,

∑_{p} aD_{p}^{2} = 2 ∑_{p} tr(G_{p}^{2}) = 52 .  (34)

Also,

∑_{p}(aD_{p})(bD_{p}) = ∑_{p} 2tr(zG_{p}) · 3tr(z^{2}G_{p}) = 6b  (35)

and by (2)

∑_{p}(bD_{p})^{2} = 9 ∑_{p} tr(z^{2}G_{p})^{2} = 9tr(z^{4}) =  (36)

= 9tr(\frac{1}{2}tr(z^{2})z^{2} + \frac{1}{3}tr(z^{3})z) = 9/2a^{2} .

Besides, since tr(z) = 0 and G_{1}^{2} = 1/3, we have (cf. lemma 7)

∑_{p} bD_{p}^{2} = 6 ∑_{p} tr(zG_{p}^{2}) = 6 ∑_{p=1}^{27} tr(zG_{p}^{2}) = 6Tr_{A}(R_{z}) = 0  (37)

because Tr_{A}(R_{x}) = 9tr(x) for all x ∈ A.

Using the previous formulae, we get

a^{m}b^{n}D_{p}^{2} = ma^{m−1}b^{n}(aD_{p}^{2}) + m(m − 1)a^{m−2}b^{n}(aD_{p})^{2}+
+2mna^{m−1}b^{n−1}(aD_{p})(bD_{p}) + na^{m}b^{n−1}(bD_{p}^{2}) + n(n − 1)a^{m}b^{n−2}(bD_{p})^{2}

hence,

a^{m}b^{n}Δ_{2} = 52ma^{m−1}b^{n} + 4m(m − 1)a^{m−1}b^{n}+  (38)
+12mna^{m−1}b^{n} + 0 + 9/2n(n − 1)a^{m+2}b^{n−2} =
a^{m}b^{n}Δ_{2} = 4m(m + 3n + 12)a^{m−1}b^{n} + 9/2n(n − 1)a^{m+2}b^{n−2} .

Put

γ_{r} = ∏_{q=1}^{r} \frac{4(q + 4)}{2q − 1}

for all r ≥ 1 and γ_{0} = 1; using the induction on r, one may check that for all r, m ≥ 0

(a^{m+3r} − γ_{r}a^{m}b^{2r})Δ^{2m+3r} = 0 .

This gives

R^{2s} = 2^{−2s} ∑_{r=0}^{2s} C_{2s}^{r}6^{r}b^{2r}a^{3(2s−r)}(−1)^{2s−r} ≡

2^{−2s}(∑_{r=0}^{2s} C_{2s}^{r} \frac{6^{r}}{γ_{r}} (−1)^{2s−r})a^{6s}

modulo V (6s) = {f| fΔ_{2}^{6s} = 0}. By (38), a^{6s} ∉ V (6s), hence, the next lemma (the proof is similar to lemma 9 and we leave it as an exercise) yields proposition 7 and, therefore, completes the proof of theorem 1.

Lemma 10.

(∑_{r=0}^{2s} C_{2s}^{r} \frac{6^{r}}{γ_{r}} (−1)^{2s−r}) =

= \frac{3}{4π} ∫_{0}^{1} ∫_{0}^{x_{4}} ∫_{0}^{x_{3}} ∫_{0}^{x_{2}} ∫_{0}^{π/2}(3x_{1}Sin^{2}x − 1)^{2s}dxdx_{1}dx_{2}dx_{3}dx_{4} .

6. Example

In this section we show that Laplace operators do extend the algebra tr[A].

Proposition 8. tr[A] ≠ tr[A]^{∘}.

Proof. First of all, observe that tr[A] has no multilinear polynomials of degree 5 which are skew-symmetric in 4 variables.

Actually, let u = u(x_{1}, …, x_{n}) ∈ F(A) be a multilinear monomial; since A is commutative, the alternation of u in x_{1}, …, x_{n−1} might be nonzero only if (up to the enumeration of the first n − 1 variables)

u = x_{n}R_{x_{1}} … R_{x_{n−1}} .

Suppose that n ≥ 3, put v = x_{n}R_{x_{1}} … R_{x_{n−3}}, then tr(vx_{n−2} · x_{n−1}) − tr(vx_{n−1} · x_{n−2}) = 0 in A (cf. [9, p.229]). Hence, the alternation of tr(u) in any n − 1 variables makes it equal to 0.

Next, any polynomial from tr[A] of degree 5 is a linear combination of polynomials

f = tr(u_{1}) … tr(u_{t})

where t ≤ 5. Consider the alternation of any 4 variables in f; if t ≥ 4, then 3 of u-s are variables, hence f vanishes. Otherwise, there is u_{i} with deg(u_{i}) ≥ 3, hence f vanishes too, or there are u_{i}, u_{j} with deg(u_{i}), deg(u_{j}) ≥ 2, hence, f also becomes zero under the alternation.

Now we shall exhibit a multilinear polynomial f ∈ tr[A]^{∘} of degree 5 which is skew-symmetric in 4 variables.

Let Tr_{A}(Φ) denote the trace of a linear operator Φ ∈ End_{F} A. Put

Q = ∑_{σ∈S_{4}} sgn(σ)R_{x_{σ(1)}}R_{x_{σ(2)}}R_{x_{σ(3)}}R_{x_{σ(4)}}R_{x_{5}} ,

then Tr_{A}(Q) = fΔ(x, y), where f = tr(yQ · x). Thus, we only need to show that Tr_{A}(Q) ≠ 0. Note that invariants of this type are not new (cf.[4, 11]).

Put

x_{1} = e_{12} , x_{2} = e_{23}, x_{3} = e_{31}, x_{4} = e_{1} = e_{11}, x_{5} = e_{2} = e_{22} ;  (39)

all these elements belong to F_{3}^{(+)}. Consider a basis of A consisting of e_{i}, ϵ_{ij} (a basis of F_{3}^{(+)})

and [ \begin{matrix}0 & s_{r} \\ 0 & 0 \end{matrix} ]_{jk} , [ \begin{matrix}0 & 0 \\ s_{r} & 0 \end{matrix} ]_{jk}, where s_{1}, s_{2}, s_{3} is a basis of the 3-dimensional vector space. Then

for any element a of the basis, the subspace F_{3}^{(+)} + F a lies in an associative subalgebra of O_{3} = F_{3}⊗_{F} O; it follows by the definition of multiplication (in fact, this subalgebra is isomorphic to Q_{3}, where Q is the algebra of quaternions).

Let λ_{x} (ρ_{x}) denote the operator of left (right) multiplication in O_{3}, then the operator R_{x} = 1/2(ρ_{x} + λ_{x}) and we get the following (as usual, (ijk) runs over A_{3} and [a, b] = ab − ba):

Q = 2 ∑_{(ijk)}[R_{x_{i}}, R_{x_{j}}] ∘ [R_{x_{k}}, R_{x_{4}}]R_{x_{5}} =

= 1/16 ∑_{(ijk)}[ρ_{x_{i}} + λ_{x_{i}}, ρ_{x_{j}} + λ_{x_{j}}] ∘ [ρ_{x_{k}} + λ_{x_{k}}, ρ_{x_{4}} + λ_{x_{4}}](ρ_{x_{5}} + λ_{x_{5}}) .

The associativity law can be written in the operator form as ρ_{x}ρ_{y} = ρ_{xy}, λ_{x}λ_{y} = λ_{yx}, [ρ_{x}, λ_{y}] = 0, hence

16 · Q = ∑_{(ijk)}([ρ_{x_{i}}, ρ_{x_{j}}] + [λ_{x_{i}}, λ_{x_{j}}]) ∘ ([ρ_{x_{k}}, ρ_{x_{4}}] + [λ_{x_{k}}, λ_{x_{4}}])(ρ_{x_{5}} + λ_{x_{5}})

= ∑_{(ijk)} ρ_{[x_{i},x_{j}]∘[x_{k},x_{4}]x_{5}} + ρ_{[x_{i},x_{j}]∘[x_{k},x_{4}]}λ_{x_{5}} + λ_{x_{5}[x_{i},x_{j}]∘[x_{k},x_{4}]} + λ_{[x_{i},x_{j}]∘[x_{k},x_{4}]}ρ_{x_{5}}+

−ρ_{[x_{i},x_{j}]}λ_{x_{5}[x_{k},x_{4}]} − ρ_{[x_{i},x_{j}]x_{5}}λ_{[x_{k},x_{4}]} − λ_{[x_{i},x_{j}]}ρ_{[x_{k},x_{4}]x_{5}} − λ_{x_{5}[x_{i},x_{j}]}ρ_{[x_{k},x_{4}]} .

Using 39) and the multiplication table for matrix units, we get

16Q = \frac{1}{2}(−ρ_{e_{2}} − λ_{e_{2}} + ρ_{e_{3}}λ_{e_{2}} + λ_{e_{3}}ρ_{e_{2}} − 2λ_{e_{2}}ρ_{e_{2}} + 2ρ_{e_{12}}λ_{e_{21}}) .

Observe that the vector space A is closed under those operators and Tr_{A}(ρ_{e_{i}}) = Tr_{A}(λ_{e_{i}}) = 9, Tr_{A}(ρ_{e_{i}}λ_{e_{j}}) = 4, Tr_{A}(ρ_{e_{i}}λ_{e_{i}}) = 1 for all i ≠ j and Tr_{A}(ρ_{e_{ij}}λ_{e_{pq}}) = 0 if i ≠ j or p ≠ q. Therefore,

Tr_{A}(16Q) = −6.

The proposition is proven. □

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School of Mathematics and Statistics, University of Sydney, N.S.W., 2006, Australia (current address); Sobolev Institute of Mathematics, Siberian Branch of Russian Academy of Sciences, Novosibirsk, 630090, Russia