The Centroid of a Lie Triple Algebra

Definition 1 (see [1].)A Lie triple algebra (also called a Lie-Yamaguti algebra or a general Lie triple system) 𝔤 is a vector space over an arbitrary field F with a bilinear map denoted by xy of 𝔤 × 𝔤 into 𝔤 and a ternary map denoted by [x, y, z] of 𝔤 × 𝔤 × 𝔤 into 𝔤 satisfying the following axioms:

• (a)

  xx = 0,

• (b)

  [x, x, y] = 0,

• (c)

  (xy)z + (yz)x + (zx)y + [x, y, z]+[y, z, x]+[z, x, y] = 0,

• (d)

  [xy, z, w]+[yz, x, w]+[zx, y, w] = 0,

• (e)

  [x, y, zw] = ([x, y, z])w + z([x, y, w]),

• (f)

  ([x, y, v])([z, w, v]) = [[x, y, z], w, v]+[z[x, y, w], v],

where x, y, z, w, v ∈ 𝔤.

Remark 2. Any Lie algebra is a Lie triple algebra relative to xy = [x, y] and D(x, y)z = [[x, y], z], for x, y, z ∈ 𝔤. If D(x, y)z = 0, for all x, y, z ∈ 𝔤, the axioms stated earlier reduce to that of Lie algebra, and if xy = 0, for all x, y ∈ 𝔤, the axioms stated earlier reduce to that of Lie triple system. In this sense, the Lie triple algebra is a more general concept than that of the Lie algebra and Lie triple system.

Definition 3 (see [1], [2].)A derivation of a Lie triple algebra 𝔤 is a linear transformation D of 𝔤 into 𝔤 satisfying the following conditions:

• (1)

  D(xy) = (Dx)y + x(Dy),

• (2)

  D([x, y, z]) = [Dx, y, z]+[x, Dy, z]+[x, y, Dz], for all x, y, z ∈ 𝔤.

Let Der(𝔤) be the set of all derivation of 𝔤; then Der(𝔤) is regarded as a subalgebra of the general Lie algebra gl(𝔤) and is called the derivation algebra of 𝔤.

Definition 4. A Lie triple subalgebra I of 𝔤 is called an ideal if 𝔤I⊆I and [𝔤, I, 𝔤]⊆I.

Definition 5. A Lie triple algebra 𝔤 is perfect if 𝔤 = 𝔤𝔤 = [𝔤, 𝔤, 𝔤].

Definition 6. Let I be a nonempty subset of 𝔤; we call Z_𝔤(I) = {x ∈ 𝔤∣xa = [x, a, y] = 0,  for all  a ∈ I, y ∈ 𝔤} the centralizer of I in 𝔤. In particular, Z(𝔤) = {x ∈ 𝔤∣x𝔤 = [x, 𝔤, 𝔤] = 0} is the center of 𝔤.

Definition 7. Suppose that I is an ideal of the Lie triple algebra 𝔤, on the quotient vector space

Define the operator: $$, where x, y, z ∈ g. Then g/I is also a Lie triple algebra, and it is called a quotient algebra of 𝔤 by I.

Definition 8. Let 𝔤 be a Lie triple algebra over a field F. The centroid of 𝔤 is the transform on 𝔤 given by Γ(𝔤) = {ψ ∈ End_F(𝔤)∣ψ([x, y, z]) = [ψ(x), y, z], ψ(xy) = (ψ(x))y,    for all  x, y, z, ∈𝔤}.

By (a)–(f), we can conclude that if ψ ∈ Γ(𝔤), then we have ψ([x, y, z]) = [ψ(x), y, z] = [x, ψ(y), z] = [x, y, ψ(z)], ψ(xy) = (ψ(x))y = x(ψ(y)), for all x, y, z, ∈𝔤.

From the definition, it is clear that the scalars will always be in the centroid.

Proposition 9. If 𝔤 is perfect, then the centroid Γ(𝔤) is commutative.

Proof. For all x, y, z ∈ 𝔤, ϕ, ψ ∈ Γ(𝔤), we have ϕψ([x, y, z]) = [ψ(x), ϕ(y), z] = ψϕ([x, y, z]),  ϕψ(xy) = ψ(x)ϕ(y) = ψϕ(xy).

Proposition 10. Let 𝔤 be a Lie triple algebra over a field F and B a subset of 𝔤. Then;

• (1)

  Z_𝔤(B) is invariant under Γ(𝔤),

• (2)

  every perfect ideal of 𝔤 is invariant under Γ(𝔤).

Proof. (1) For any ψ ∈ Γ(𝔤),  x ∈ Z_𝔤(B),  y ∈ B,  z ∈ 𝔤, we have [ψ(x), y, z] = [ψ(x), y, z] = ψ[x, y, z] = 0,  (ψ(x))z = ψ(xz) = 0, and [z, y, ψ(x)] = ψ[z, y, x] = 0,  z(ψ(x)) = ψ(zx) = 0. Therefore, ψ(x) ∈ Γ(𝔤), which implies that Z_𝔤(B) is invariant under Γ(𝔤).

(2) Let J be any perfect ideal of 𝔤; then J = JJ = [J, J, J]. For any y ∈ J, there exist a, b, c, d, e ∈ J, such that y = ab = [c, d, e], and then we have ψ(y) = ψ(ab) = (ψ(a))b ∈ J𝔤⊆J, ψ(y) = ψ([c, d, e]) = [c, ψ(d), e]∈[J, 𝔤, 𝔤]⊆J. Hence J is invariant under Γ(𝔤).

Definition 11. Let φ(𝔤)⊆Z(𝔤) and φ(𝔤𝔤) = φ([𝔤, 𝔤, 𝔤]) = 0; then φ is called a central derivation.

Proposition 12. If 𝔤 has no nonzero ideals I and J with [𝔤, I, J] = 0,   IJ = 0, then Γ(𝔤) is an integral domain.

Proof. Clearly, id ∈ Γ(𝔤). If there exist ψ, φ ∈ Γ(𝔤), ψ ≠ 0, ϕ ≠ 0 such that ψφ = 0, then there exist x, y ∈ 𝔤 such that ψ(x) ≠ 0 and ϕ(y) ≠ 0. Then, [𝔤, ψ(x), ϕ(y)] = ψϕ([𝔤, x, y]) = 0, (ψ(x))(ϕ(y)) = ψϕ(xy) = 0. Therefore, ψ(x) and ϕ(y) can span two nonzero ideals I, J of 𝔤 such that [𝔤, I, J] = IJ = 0, which is a contradiction. Hence, Γ(𝔤) has no zero divisor; it is an integral domain.

Theorem 13. If 𝔤 is a simple Lie triple algebra over an algebraically closed field F, then Γ(𝔤) = F id .

Proof. Let ϕ ∈ Γ(𝔤)⊆End_F(𝔤). Since F is algebraically closed, ϕ has an eigenvalue λ. We denote the corresponding eigenspace to be E_λ(ϕ) = {v ∈ 𝔤∣ϕ(v) = λ(v)}, so E_λ(ϕ) ≠ 0, for any v ∈ E_λ(ϕ), x, y ∈ 𝔤, and we have ϕ([v, x, y]) = [ϕ(v), x, y] = λ[v, x, y], ϕ((vx)) = ϕ(v)x = (λv)x = λ(vx), so [v, x, y] ∈ E_λ(ϕ), vx ∈ E_λ(ϕ). It follows that E_λ(ϕ) is an ideal of 𝔤. But, 𝔤 is simple, so E_λ(ϕ) = 𝔤; that is, E_λ(ϕ) = λid_𝔤. This proves the theorem.

Proposition 14. Let 𝔤 be a Lie triple algebra over a field F; then

• (1)

  𝔤 is indecomposable if and only if  Γ(𝔤) does not contain idempotents except 0 and id ;

• (2)

  if 𝔤 is perfect, then every ϕ ∈ Γ(𝔤) is symmetric (f([a, b, c], d) = f(a[d, c, b]),  f(ab, c) = f(a, bc)) with respect to any invariant form on 𝔤.

Proof. (1)  If there exists ϕ ∈ Γ(𝔤) which is an idempotent and satisfies ϕ ≠ 0, id, then ϕ²(x) = ϕ(x), for all x ∈ 𝔤. We assert that Ker ϕ and Im ϕ are ideals of 𝔤. In fact, for any x ∈ Ker ϕ and y, z ∈ 𝔤, we have ϕ([x, y, z]) = [ϕ(x), y, z] = 0,   ϕ(xy) = (ϕ(x))y = 0y = 0, which implies [x, y, z] ∈ Ker ϕ, xy ∈ Ker ϕ. For any x ∈ Im ϕ, there exists a ∈ 𝔤 such that x = ϕ(a). Then, we have

This proves our assertion. Moreover, Ker ϕ∩Im ϕ = 0. Indeed, if x ∈ Ker ϕ∩Im ϕ, then there exists y ∈ 𝔤 such that x = ϕ(y) and 0 = ϕ(x) = ϕ²(y) = ϕ(y) = x. We have a decomposition x = ϕ(x) + y, for all x ∈ 𝔤, where ϕ(y) = 0. So, we have 𝔤 = Ker ϕ ⊕ Im  ϕ, which is a contradiction.

On the other hand, suppose 𝔤 has a decomposition 𝔤 = 𝔤₁ ⊕ 𝔤₂. Then for any x ∈ 𝔤, we have x = x₁ + x₂, x_i ∈ 𝔤_i,  i = 1,2. We choose ϕ ∈ Γ(𝔤) such that ϕ(x₁) = x₁ and ϕ(x₂) = 0. Then, ϕ²(x) = ϕ(x₁) = x₁ = ϕ(x). Hence, ϕ is an idempotent. By assumption, we have ϕ = 0 or ϕ = id. If ϕ = 0, then x₁ = 0, implying 𝔤 = 𝔤₂. If ϕ = id, then x₂ = 0, implying 𝔤 = 𝔤₁.

(2)  Let f be an invariant F-bilinear form on 𝔤. Then f([a, b, c], d) = f(a[d, c, b]),  f(ab, c) = f(a, bc),  for all  a, b, c, d ∈ 𝔤. Since 𝔤 is perfect, for ϕ ∈ Γ(𝔤), we have

The result follows.

Remark 15. Set L(x)(y) = xy, L(x, y)(z) = [x, y, z], R(x)(y) = yx, R(x, y)(z) = [z, x, y], for  all  x, y, z, ∈𝔤; then we call L(x), and L(x, y) the left multiplication operator of 𝔤;  R(x) and R(x, x) right multiplication operator of 𝔤. Denote by Mult(𝔤) = Mult_F(𝔤) the subalgebra of End_F(𝔤) generated by the left and right multiplication operators of 𝔤. Then, Γ(𝔤) is the centralizer of Mult(𝔤).

Theorem 16. Let π : 𝔤₁ → 𝔤₂ be an epimorphism of Lie triple algebra; for any f ∈ End _F(𝔤₁; Ker π): = {g ∈ End _F(𝔤₂)∣g(Ker π)⊆Ker π}, there exists a unique $$ satisfying $$. Moreover, the following results hold.

• (1)

  The map π _End : End _F(𝔤₁; Ker π) → End _F(𝔤₂), $$ is a homomorphism with the following properties:

•  

  π _End ( Mult (𝔤₁)) = Mult (𝔤₂), π _End (Γ(𝔤₁)∩ End _F(𝔤₁; Ker π))⊆Γ(𝔤₂). There is a homomorphism $$.

If Ker π = Z(𝔤₁), then every ϕ ∈ Γ(𝔤₁) leaves Ker π invariant; that is, π_Γ is defined on all of Γ(𝔤₁).

• (2)

  Suppose 𝔤₁ is perfect and Ker π⊆Z(𝔤₁); then $$ is injective.

• (3)

  If 𝔤₁ is perfect, Z(𝔤₂) = 0, and Ker π⊆Z(𝔤₁), then π_Γ : Γ(𝔤₁) → Γ(𝔤₂) is a monomorphism.

Proof. (1) It is easy to see that π_End is a homomorphism. Since Ker π is an ideal of 𝔤₁ and all left and right multiplication operators of 𝔤₁ leave Ker π invariant, Mult(𝔤₁)⊆End_F(𝔤₁; Ker π). Furthermore, for the left multiplication operator L(x, y), L(x) on 𝔤₁, we have π∘L(x, y) = L(π(x), π(y))∘π,    π∘L(x) = L(π(x))∘π, so π_End(L(x, y)) = L(π(x), π(y)), π_End(L(x)) = L(π(x)). For the right multiplication, we have the analogous formula π_End(R(x, y)) = R(π(x), π(y)), π_End(R(x)) = R(π(x)). Moreover, π is an epimorphism, so π_End(Mult(𝔤₁)) = Mult(𝔤₂). Now, we show that π_End(Γ(𝔤₁)∩End_F(𝔤₁; Ker π))⊆Γ(𝔤₂). Let ϕ ∈ Γ(𝔤₁)∩End_F(𝔤₁; Ker π). For any x^′, y^′, z^′ ∈ 𝔤₂, there exist x, y, z ∈ 𝔤₁ such that π(x) = x^′, π(y) = y^′, π(z) = z^′. Then, we have

which proves $$.

(2) If $$ for, ϕ ∈ Γ(𝔤₁)∩End_F(𝔤₁; Ker π), then $$, which means that ϕ(𝔤₁)⊆Ker π⊆Z(𝔤₁). Hence, ϕ([x, y, z]) = [ϕ(x), y, z] = 0, ϕ(xy) = (ϕ(x)y) = 0, for  all  x, y, z ∈ 𝔤₁. Furthermore, since 𝔤₁ is a perfect ideal, we can get ϕ = 0.

(3) We can see that π(Z(𝔤₁))⊆Z(𝔤₂) = 0, which follows that Z(𝔤₁)⊆Ker π. So, Ker π⊆Z(𝔤₁). From (1), we know that π_Γ : Γ(𝔤₁) → Γ(𝔤₂) is a well-defined homomorphism, which is an injection by (2).

Proposition 17. If the characteristic of  F is not 2, then

Proof. If ψ ∈ Γ(𝔤)∩Der(𝔤), then by the definition of Γ(𝔤) and Der(𝔤), for all x, y, z ∈ 𝔤, we have ψ(xy) = (ψ(x))y + x(ψ(y)), ψ([x, y, z]) = [ψ(x), y, z]+[x, ψ(y), z]+[x, y, ψ(z)], and ψ(xy) = (ψ(x))y = x(ψ(y)), ψ([x, y, z]) = [ψ(x), y, z] = [x, ψ(y), z] = [x, y, ψ(z)], so ψ(𝔤𝔤) = ψ([𝔤, 𝔤, 𝔤]) = 0 and ψ(𝔤)⊆C(𝔤). It follows easily that Γ(𝔤)∩Der(𝔤)⊆C(𝔤).

To show the inverse inclusion, let ψ ∈ C(𝔤); then 0 = ψ([x, y, z]) = [ψ(x), y, z] = [x, ψ(y), z] = [x, y, ψ(z)], 0 = ψ(xy) = (ψ(x))y = x(ψ(y)). Thus, ψ ∈ Γ(𝔤)∩Der(𝔤).

This implies that C(𝔤) = Γ(𝔤)∩Der(𝔤).

Lemma 18. Let I be a nonzero Γ(𝔤)-invariant ideal of 𝔤, V(I) = {ψ ∈ Γ(𝔤)ψ(I) = 0}, and let Hom(𝔤/I, Z_𝔤(I)) be the vector space of all linear maps from 𝔤/I to Z_𝔤(I) over F. Define $$, where $$. Then one has the following.

• (1)

  T(I) is a subspace of Hom(𝔤/I, Z_𝔤(I)) and V(I)≅T(I) as vector spaces.

• (2)

  If  Γ(I) = F id _I, then Γ(𝔤) = F id _𝔤 ⊕ V(I) as vector spaces.

Proof. (1) It is easily seen that V(I) is an ideal of the associative algebra Γ(𝔤). To prove (1) consider the following map α : V(I) → T(I) given by

where ψ ∈ V(I) and $$. The map α is well defined. If $$, then y − y₁ ∈ I, and so ψ(y − y₁) = 0. It follows easily that α is injective. We now show that α is onto. For every f ∈ T(I), set $$, for all x ∈ 𝔤. It follows from the definition of T(I) that, for all x, y, z ∈ 𝔤, $$. Thus, ψ_f ∈ Γ(𝔤), and so ψ_f ∈ V(I), ψ_f(I) = 0. But, α(ψ_f) = f implies that α is onto. It is fairly easy to see that α preserves operations on vector spaces from 𝔤/I to Z_𝔤(I). This proves (1).

We now prove (2). If Γ(I) = F id _I, then for all ψ ∈ Γ(𝔤), ψ∣_I = λid_I, for some λ ∈ F. If ψ ≠ λid_𝔤, let φ(x) = λx, for all x ∈ 𝔤; then ψ ∈ Γ(𝔤) and ψ − φ ∈ V(I). Clearly, ψ = φ + (ψ − φ). Furthermore, Fid_𝔤∩V(I) = 0, and so (2) is proved.

Lemma 19. Let 𝔤 be a Lie triple algebra; then φD is a derivation for φ ∈ Γ(𝔤), D ∈ Der(𝔤).

Proof. If x, y, z ∈ 𝔤, then

Thus, φD is a derivation.

Theorem 20. Let 𝔤 be a Lie triple algebra, and for any φ ∈ Γ(𝔤), D ∈ Der(𝔤), then one has the following.

• (1)

  Der(𝔤) is contained in the normalizer of Γ(𝔤) in End (𝔤).

• (2)

  Dφ is contained in Γ(𝔤) if and only if φD is a central derivation of 𝔤.

• (3)

  φD is a derivation of 𝔤 if and only if [D, φ] is a central derivation of 𝔤.

Proof. For any φ ∈ Γ(𝔤), D ∈ Der(𝔤), and x, y, z ∈ 𝔤,

Then, we get (Dφ − φD)(xy) = ((Dφ − φD)(x))y. On the other hand, we have

So, (Dφ − φD)([x, y, z]) = [(Dφ − φD)(x), y, z]; this is [D, φ] = Dφ − φD ∈ Γ(𝔤). This proves (1). From Lemma 19 and (1), Dφ is an element of Γ(𝔤) if and only if φD ∈ Γ(𝔤)∩Der(𝔤). Thanks to Proposition 17, we get the result (2). It follows from (1), Proposition 17, and Lemma 19 that the result holds.

Theorem 21. Suppose that 𝔤 is a Lie triple algebra over F and 𝔤 = 𝔤₁ ⊕ 𝔤₂ with 𝔤₁ and 𝔤₂ being ideals of 𝔤; then

as vector spaces, where C(1) = {φ ∈ Hom(𝔤₁, 𝔤₁)∣φ(𝔤₁)⊆Z(𝔤₂), φ(𝔤₂𝔤₂) = φ[𝔤₂, 𝔤₂, 𝔤₂] = 0}, C(2) = {φ ∈ Hom(𝔤₂, 𝔤₂)∣φ(𝔤₂)⊆Z(𝔤₁), φ(𝔤₁𝔤₁) = φ[𝔤₁, 𝔤₁, 𝔤₁] = 0}.

Proof. Let π_i : 𝔤 → 𝔤_i be canonical projections for i = 1,2; then π₁, π₂ ∈ Γ(𝔤) and π₁ + π₂ = id_𝔤. So we have for φ ∈ Γ(𝔤),

Note that π_iφπ_j ∈ Γ(𝔤) for i, j = 1,2. We claim that It suffices to show that π₁Γ(𝔤)π₁∩π₁Γ(𝔤)π₂ (other cases are similar). For any φ ∈ π₁Γ(𝔤)π₁∩π₁Γ(𝔤)π₂, there exist f_i ∈ Γ(𝔤), i = 1,2, such that φ = π₁f₁π₁ = π₁f₂π₂. Then, φ(x) = π₁f₂π₂(x) = π₁f₂π₂(π₂(x)) = π₁f₁(0) = 0, for all x ∈ 𝔤, and so φ = 0. Let Γ(𝔤) _ij = π_iΓ(𝔤)π_j, i, j = 1,2. We now prove that Since φ(𝔤₂) = 0 for φ ∈ Γ(𝔤) ₁₁, we have $$. On the other hand, one can regard Γ(𝔤₁) as a subalgebra of Γ(𝔤) by extending any φ₀ ∈ Γ(𝔤₁) on 𝔤₂ being equal to zero; that is, φ₀(x₁) = φ₀(x₁), φ₀(x₂) = 0, for all x₁ ∈ 𝔤₁, x₂ ∈ 𝔤₂. Then φ₀ ∈ Γ(𝔤) and φ₀ ∈ Γ(𝔤) ₁₁. Therefore, Γ(𝔤) ₁₁≅Γ(𝔤₁) with isomorphism $$, for all φ ∈ Γ(𝔤) ₁₁. Similarly, we have Γ(𝔤) ₂₂≅Γ(𝔤₂).

Next, we prove that Γ(𝔤) ₁₂≅C₂. If φ ∈ Γ(𝔤) ₁₂ there exists φ₀ in Γ(𝔤) such that φ = π₁φ₀π₂. For $$, where $$ and k = 1,2, 3, we have

Then, φ(𝔤)⊆Z(𝔤) and φ(𝔤𝔤) = φ[𝔤, 𝔤, 𝔤] = 0. It follows that $$, and so $$.

Conversely, for φ ∈ C₂, expending φ on 𝔤 (also denoted by φ) or by φ(𝔤₁𝔤₁) = φ[𝔤₁, 𝔤₁, 𝔤₁] = 0, we have π₁φπ₂ = φ and φ ∈ Γ(𝔤) ₁₂. This proves that Γ(𝔤) ₁₂ is isomorphic to C₂ with the following isomorphism:

for all φ ∈ Γ(𝔤) ₁₂. Similarly, we can prove that Γ(𝔤) ₂₁≅C₁. Summarizing the aforesaid discussion, we have

The proof is completed.

Theorem 22. Suppose that 𝔤 is a Lie triple algebra over F and with a decomposition of ideals 𝔤 = 𝔤₁ ⊕ 𝔤₂ ⊕ ⋯⊕𝔤_m. Then, one has

as vector spaces, where C_ij = {φ ∈ Hom(𝔤_i, 𝔤_j)∣φ(𝔤_i)⊆Z(𝔤_j), φ(𝔤_i𝔤_i) = φ[𝔤_i, 𝔤_i, 𝔤_i] = 0, for 1 ≤ i ≠ j ≤ m}.

Definition 23. Let A be an associative algebra over a field F, where the centroid of A is the space of F-linear transforms on A given by Γ(A) = {φ ∈ End_F(A)φ(ab) = aφ(b) = φ(a)b, for  all  a, b ∈ A}; then Γ(A) is an associative algebra of End_F(A). If 𝔤 is be a finite-dimensional Lie triple algebra over F, let 𝔤 ⊗ A be a tensor product of the underlying vector spaces A and 𝔤. Then, 𝔤 ⊗ A over a field F with respect to the following 2-ary and 3-ary multilinear operation:

where x_i ∈ 𝔤,   a_i ∈ A,   i = 1,2, 3. This Lie triple algebra 𝔤 ⊗ A is called the tensor product of A and 𝔤. For f ∈ End_F(𝔤) and φ ∈ End_F(A), there exists a unique map $$ such that The map should not be confused with the element f ⊗ φ of the tensor product End_F(𝔤) ⊗ End_F(A). Of course, we have a canonical map as the following: It is easy to see that if ϕ ∈ Γ(𝔤) and ψ ∈ Γ(A), then $$. Hence, $$, where $$ is the F-span of all endomorphisms $$.

Definition 24. The transformation ϕ ∈ Γ(𝔤 ⊗ A) is said to have finite 𝔤-image, if for any a ∈ A, there exist finitely many a₁, …, a_n ∈ A such that

It is easy to see that $$-image}. In addition, ϕ ∈ Γ(𝔤 ⊗ A) has finite 𝔤-image, if ϕ(𝔤 ⊗ 1)⊆𝔤 ⊗ Fa₁ + ⋯+𝔤 ⊗ Fa_n for suitable a_i ∈ A.

Lemma 25. Let 𝔤 be a Lie triple algebra over F and A a unitary commutative associative over F. Let {a_r} _r∈ℜ be a basis of A, and let ϕ ∈ Γ(𝔤 ⊗ A). Define ϕ_r ∈ End _F(𝔤) by

then all ϕ_r ∈ Γ(𝔤).

Proof. For x₁, x₂, x₃ ∈ 𝔤, we have

Hence, one has for all x_i ∈ 𝔤. So all ϕ_r ∈ Γ(𝔤).

Proposition 26. Let 𝔤 be a perfect Lie triple algebra over F, and let A be a unitary commutative associative over F which is free as a F-module. Then, one has the following.

• (1)

  If 𝔤 is perfect, then 𝔤 ⊗ A is perfect too.

• (2)

  If 𝔤 is finitely generated as a Mult (𝔤)-module (or as a Γ(𝔤)-module), or if 𝔤 is a central and a torsion-free F-module, then every ϕ ∈ Γ(𝔤 ⊗ A) has finite 𝔤-image.

• (3)

  If Γ(𝔤) is a free F-module and the map ω of Definition 23 is injective, then $$  has finite  𝔤-image}.

Proof. The definition of perfect shows that (1) holds. Let M = Mult(𝔤). It follows that $$ from A  is unital. Since 𝔤 is finitely generated as a Mult(𝔤)-module, we can suppose 𝔤 = Mx₁ + ⋯+Mx_n for x₁, …, x_n ∈ 𝔤. Fix ϕ ∈ Γ(𝔤 ⊗ A) and a ∈ A. There exist finite families {x_ij}⊆𝔤 and {a_ij}⊆A such that

for 1 ≤ i ≤ n. Hence, Since 𝔤  ⊗  A is perfect, the centroid Γ(𝔤  ⊗  A) is commutative. By replacing M with Γ(𝔤) we can use the same argument aforementione to show that every ϕ ∈ Γ(T ⊗ A) has finite 𝔤-image, if 𝔤 is a finitely generated Γ(𝔤)-module.

Now, suppose that Γ(𝔤) = Fid and that 𝔤 is a torsion-free F-module. Then there exist scalars k_r ∈ F such that ϕ_r = k_rid. Hence,

Fix x ∈ 𝔤; then almost all k_rx = 0 and almost all k_r = 0, which in turn implies that ϕ has finite 𝔤-image.

From the aforementione discussion above, we get

So, it suffices to prove that $$. We suppose that ϕ ∈ Γ(𝔤 ⊗ A) has finite 𝔤-image, and then there exists a finite subset 𝔍⊆ℜ such that equation in Lemma 25 becomes as follows: For x₁, x₂, x₃ ∈ 𝔤 and a ∈ A, we get Since 𝔤 is perfect, we can get where λ_r is the left multiplication in A by a_r. Let {ψ_s∣s ∈ 𝔖} be a basis of Γ(𝔤). Then, there exist a finite subset 𝔅⊆𝔖 and scalars k_rs ∈ F (r ∈ 𝔍, s ∈ 𝔅) such that We then get Now we show that ϕ_s ∈ Γ(A). For any x₁, x₂, x₃ ∈ 𝔤, a₁, a₂ ∈ A, we have Since [𝔤𝔤, 𝔤] = 𝔤𝔤 = 𝔤, we have for  all  x ∈ 𝔤 and a₁, a₂ ∈ A. It follows that where μ_s ∈ End_F(A) is defined by μ_s(a₂) = ϕ_s(a₁a₂)  −  ϕ_s(a₁)a₂, for all a₂ ∈ A. Since ω is injective, we also have So by the linear independence of the ψ_s, we get that μ_s = 0. Then ψ_s ∈ Γ(A). Hence $$.

Theorem 27. Let Ψ denote the subspace of Γ(𝔤) spanned by {φ_j}. Then {e_k, φ_j} is a basis of Γ(𝔤) and Γ(𝔤) = Ψ ⊕ C(𝔤) as vector spaces.

Proof. Since {φ_j∣_𝔤𝔤} and {φ_j∣_{[𝔤,𝔤,𝔤]}} are linear independent, {φ_j} is linear independent in Γ(𝔤). By definition of {e_k, φ_j}, the {e_k, φ_j} is independent in Γ(𝔤). For φ ∈ Γ(𝔤) since {φ_j∣_𝔤𝔤} is a basis of vector spaces {φ∣_𝔤𝔤∣φ ∈ Γ(𝔤)}, and {φ_j∣_{[𝔤,𝔤,𝔤]}} is a basis of vector spaces {φ∣_{[𝔤,𝔤,𝔤]}∣φ ∈ Γ(𝔤)}, there exist g_s ∈ F, s ∈ J (J is a finite set of positive integers) such that

We then have If y₁, y₂, y₃ ∈ 𝔤, then It follows that is a central derivation. So there exist r_i ∈ F, i ∈ I (I is a finite set of positive integers) such that Therefore, The proof is completed.

Lemma 28. Let 𝔤 be a simple Lie triple algebra, R = F[x₁, …, x_n], and $$; then one has $$.

Proof. Let ψ ∈ Γ(𝔤), m ∈ R. Then, we have

Thus $$. For any d ∈ C(g) and $$, we have Hence $$.

Theorem 29. Let 𝔤 be a simple Lie triple algebra, R = F[x₁, …, x_n], and $$. Then $$.

Proof. From Lemma 28, we get $$. Now we prove the opposite inclusion. Let {m_i} be a basis for $$, and let η_i(−, p) be suitable maps in End_F(𝔤) such that

Then, while So, for each i and p, we have η_i(p[x, y, z]) = [η_i(x, p), y, z] = [x, η_i(y, p), z] = [x, y, η_i(z, p)], η_i(p, (xy)) = (η_i(x, p), y) = (x, η_i(y, p)), for  all  x, y, z ∈ 𝔤. Hence, η_i(−, p) ∈ Γ(𝔤). But Γ(𝔤) = Fid. Therefore, η_i(x, p) = λ_i(p)x, for all x ∈ 𝔤, for suitable scalars λ_i(p).

Thus, we may write

Since the right-hand side is in 𝔤 ⊗ R, for each p, we get λ_i(p) = 0 for all except for a finite number of i. That is, Then, the map is well defined. Hence, we have ψ(x ⊗ p) = x ⊗ ρ(p). Thus, Choose x, y, z ∈ 𝔤 such that xy ≠ 0, [x, y, z] ≠ 0. Then we can conclude that ρ(p) = pρ(1), for  all  p ∈ R. So ρ is determined by its action on 1. Therefore, ρ ∈ R. Thus, That is, ψ ∈ id ⊗ R≅Γ(𝔤) ⊗ R. So, we have $$.