Generalized derivations of Lie triple systems

([16]). A Lie triple system is a pair (T,[⋅, ⋅, ⋅]) consisting of a vector space T over a field 𝔽, a trilinear multiplication[⋅, ⋅, ⋅] : T × T × T → T such that for all x, y, z, u, v ∈ T,

End (T) denotes the set consisting of all linear maps of T. Obviously, End (T) is a Lie algebra over 𝔽 with the bracket [D₁, D₂] = D₁D₂ − D₂D₁, for all D₁, D₂ ∈ End (T).

([17]). Let (T, [⋅, ⋅, ⋅]) be a Lie triple system. A linear map D : T → T is said to be a derivation of T if it satisfies

for all x, y, z ∈ T.

We denote the set of all derivations by Der (T), then Der (T) provided with the commutator is a subalgebra of End (T) and is called the derivation algebra of T.

D ∈ End (T) is said to be a generalized derivation of T, if there exist D′, D″, D‴ ∈ End(T) such that

for all x, y, z ∈ T.

D ∈ End(T) is said to be a quasiderivation, if there exists D′ ∈ End(T) such that

for all x, y, z ∈ T.

Denote by GDer (T) and QDer (T) the sets of generalized derivations and quasiderivations, respectively.

([13]). If C(T) = {D ∈ End(T) ∣ [D(x), y, z] = [x, D(y), z] = [x, y, D(z)] = D([x, y, z])} for all x, y, z ∈ T, then C(T) is called a centroid of T.

QC (T) = {D ∈ End(T) ∣ [D(x), y, z] = [x, D(y), z] = [x, y, D(z)]} for all x, y, z ∈ T, then QC (T) is called a quasicentroid of T.

ZDer (T) = {D ∈ End(T) ∣ [D(x), y, z] = D([x, y, z]) = 0} for all x, y, z ∈ T, then ZDer(T) is called a central derivation of T.

It is easy to verify that

([13]). T is a Lie triple system and I is a non-empty subset of T. We call Z_T(I) = {x ∈ T ∣ [x, a, y] = [y, a, x] = 0, ∀a ∈ I, y ∈ T} the centralizer of I in T. In particular, Z_T(T) = {x ∈ T ∣ [x, y, z] = 0, ∀y, z ∈ T} is the center of T, denoted by Z(T)

Let T be a Lie triple system. Then the following statements hold.

1. GDer(T), QGer(T) and C(T) are subalgebras of End(T).

2. ZDer(T) is an ideal of Der(T).

(1) Assume that D₁, D₂ ∈ GDer(T). For all x, y, z ∈ T, we have

and

Thus for all x, y, z ∈ T, we have

From the definition of generalized derivation, one gets [D₁, D₂] ∈ GDer(T), so GDer(T) is a subalgebra of End(T).

Similarly, QGer (T) is a subalgebra of End(T).

Assume that D₁, D₂ ∈ C(T). ∀x, y, z ∈ T, note that

Similarly,

Then [D₁, D₂] ∈ C(T), C(T) is a subalgebra of End(T).

(2) Assume that D₁ ∈ ZDer(T). D₂ ∈ Der(T). For all x, y, z ∈ T, we have

and

Then [D₁, D₂] ∈ ZDer(T) and ZDer(T) is an ideal of Der(T).

Let T be a Lie triple system. Then

1. [Der(T), C(T)] ⊆ C(T);

2. [QDer(T), QC(T)] ⊆ QC(T);

3. C(T) ⋅ Der(T) ⊆ Der(T);

4. C(T) ⊆ QDer(T);

5. [QC(T), QC(T)] ⊆ QDer(T);

6. QDer(T) + QC(T) ⊆ GDer(T).

(1) Assume that D₁ ∈ Der(T), D₂ ∈ C(T). For all x, y, z ∈ T, we have

and

Hence,

Similarly,

Thus,[D₁, D₂] ∈ C(T) and we get[Der(T), C(T)] ⊆ C(T).

(2) Similar to the proof of (1).

(3) Assume that D₁ ∈ C(T), D₂ ∈ Der(T). For all x, y, z ∈ T, we have

So we have D₁D₂ ∈ Der(T).

(4) Assume that D ∈ QC(T). For all x, y, z ∈ T, we have

Hence,

Therefore, D ∈ QDer(T) since D′ = 3D ∈ C(T) ⊆ End(T).

(5) Assume that D₁, D₂ ∈ QC(T). For all x, y, z ∈ T, we have

and

Hence,

i.e. [D₁, D₂] ∈ QDer(T).

(6) It is obvious.

([17]). If T is a Lie triple system, I is an ideal of T, then Z_T (I) is also an ideal of T. Moreover, Z(T) = Z_T(T), Z(I) = Z_I (I) are ideals of T.

([17]). Let the Lie triple system T be decomposed into the direct sum of two ideals, i.e. T = A ⊕ B. Then we have

1. Z(T) = Z(A) ⊕ Z(B).

2. If Z(T) = 0, then Der(T) = Der (A) ⊕ Der (B).

If the Lie triple system T can be decomposed into the direct sum of two ideals, i.e. T = A ⊕ B and Z(T) = 0, then we have

1. GDer(T) = GDer (A) ⊕ GDer (B),

2. QDer(T) = QDer (A) ⊕ QDer (B),

3. C(T) = C (A) ⊕ C(B),

4. QC(T) = QC (A) ⊕ QC (B).

(1) For D′ ∈ GDer (A), extend it to a linear transformation on T by setting D′ (a + b) = D′(a), ∀a ∈ A, b ∈ B. Obviously, D′ ∈ GDer(T) and GDer(A) ⊆ GDer(T). Similarly, GDer(B) ⊆ GDer(T). Let a ∈ A, b₁, b₂ ∈ B and D ∈ Der(T). Then

Suppose D(a) = a′ + b′, where a ∈ A, b ∈ B, then

So [b′, b₁, b₂] = 0 and b′ ∈ Z(B). Since Z(T) = Z(A) ⊕ Z(B), b′ = 0. Hence D(a) = a′ ∈ A. Therefore D(A) ⊆A. Similarly, D(B) ⊆B.

Let D ∈ Der(T) and x = a + b ∈ A + B, where a ∈ A, b ∈ B. Define E, F ∈ End(T) by E(a + b) = D(a), F(a + b) = D(b), then E ∈ Der (A), F ∈ Der(B). Hence D = E + F ∈ Der (A) + Der (B). Since Der(A) ⋂ Der (B) = {0}, GDer(T) = GDer (A) + GDer (B) as a vector space.

Let E ∈ Der (A), F ∈ Der(B) and b ∈ B. Then [E, D] = (ED − DE)(b) = 0. Hence [E, D] ∈ Der (A) and Der (A) ⊲ Der(T). Similarly, Der (B) ⊲ Der(T).

(2), (3), (4) Similar to the proof of (1).

If T is a Lie triple system, then QC(T) + [QC(T), QC(T)] is a subalgebra of GDer(T).

By Lemma 3.2 (5) and (6), we have

and

It is easy to verify [QDer(L), [QC(L), QC(L)] ⊆[QC(L), QC(L)] by the Jacobi identity of Lie algebra. Thus

If T is a Lie triple system, then [C(T), QC(T)] ⊆ End (T, Z(T)). Moreover, if Z(T) = {0}, then[C(T), QC(T)] = {0}.

Assume that D₁ ∈ C(T), D₂ ∈ QC(T) and for all x, y, z ∈ T, we have

Hence [D₁, D₂] (x) ∈ Z(T) and [D₁, D₂] ∈ End (T, Z(T)) as desired. Furthermore, if Z(T) = {0}, it is clear that [C(T), QC(T)] = {0}.

([19]). Let L be an algebra over 𝔽 (char 𝔽 ≠ 2), if the multiplication satisfies the following identities:

for all x, y, z, w ∈ L, then we call L a Jordan algebra.

([19]). Let T be a Lie triple system over 𝔽(char 𝔽 ≠ 2), with the operation D₁ ∙ D₂ = D₁D₂ + D₂D₁, for all elements D₁, D₂ ∈ End(T). Then the pair. End(T), ∙) is a Jordan algebra.

Let T be a Lie triple system over 𝔽(char 𝔽 ≠ 2), with the operation D₁ ∙ D₂ = D₁D₂ + D₂D₁, for all elements D₁, D₂ ∈ QC(T). Then. QC(T) is a Jordan algebra.

We need only to show that D₁ ∙ D₂ ∈ QC(T), for all D₁, D₂ ∈ QC(T), we have

Similarly, [D₁ ∙ D₂(x), y, z] = [x, y, D₁ ∙ D₂(z)]. Then D₁ ∙ D₂ ∈ QC(T) and QC(T) is a Jordan algebra.

If T is a Lie triple system over 𝔽, then we have

1. If char 𝔽 ≠ 2, then QC(T) is a Lie algebra with [D₁, D₂] = D₁D₂ – D₂D₁if and only if QC(T) is also an associative algebra with respect to composition.

2. If char 𝔽 ≠ 3 and Z(T) = {0}, then QC(T) is a Lie algebra if and only if [QC(T), QC(T)] = 0.

(1). (⇐) For all D₁, D₂ ∈ QC(T), we have D₁D₂ ∈ QC(T) and D₂D₁ ∈ QC(T), so [D₁, D₂] = D₁D₂ – D₂D₁ ∈ QC(T). Hence, QC(T) is a Lie algebra.

(⇒) Note that D1D2=D1•D2+[D1,D2]2 and by Corollary 3.10, we have D₁ ● D₂ ∈ QC(T) ; [D₁, D₂] ∈ QC(T). It follows that D₁D₂ ∈ QC(T) as desired.

(2)(⇒) Assume that D₁, D₂ ∈ QC(T). For all x, y, z ∈ T, QC(T) is a Lie algebra, so [D₁, D₂] ∈ QC(T), then

From the proof of Lemma 3.2 (5), we have

Hence 3[[D₁, D₂](x), y, z] = 0. We have [[D₁, D₂](x), y, z] = 0, i.e. [D₁, D₂] = 0 since char 𝔽 ≠ 3.

(⇐) It is clear. □

Let V be a linear space and 𝒜 : V → V a linear map. f(x) denotes the minimal polynomial of 𝒜. If x²does not divide f(x), then V = Ker(𝒜) + Im(𝒜).

Obviously dim (V) = dim(Ker(𝒜))+dim(Im(𝒜)) because 𝒜 is a linear map. x² does not divide f(x) means that f(x) = x²g(x) + ax + b, a ≠ 0 or b ≠ 0.

Case 1: If b ≠ 0, then f(𝒜) = 𝒜²g(𝒜) + a𝒜 + bid = 0, so 𝒜(𝒜g(𝒜) + aid) = - bid and 𝒜 is invertible.

Hence Ker(𝒜) = {0}.

Case 2: If b = 0, that means a ≠ 0, f(𝒜) = 𝒜²g(𝒜) + a𝒜. Here we only prove Ker(𝒜) ∩ Im(𝒜) = {0}. Indeed, ∀ x ∈ Ker(𝒜) ∩ Im(𝒜), 𝒜(x) = 0 and there exists x′ ∈ V such that 𝒜(x) = x. So f(𝒜)(x′) = 𝒜²g(𝒜)(x′) + a𝒜(x′), which means a𝒜(x′) = ax = 0. Hence x = 0 since a ≠ 0.

Let T be a Lie triple system, D ∈ C(T). Then

1. Ker (D)and Im(D) are ideals in T

2. If T is indecomposable, D ∈ C(T) and D ≠ 0. Suppose x²does not divide the minimal polynomial of D, then D is invertible.

3. Suppose T perfect, i.e. T = [T, T, T]. If T is indecomposable and C(T) consists of semisimple elements, then C(T) is a field.

(1) Since D ∈ C(T), for all x ∈ Ker(D), y, z ∈ T, one gets D[x, y, z] = [D(x), y, z] = 0, that means [x, y, z] ∈ Ker(D).

Meanwhile, for all x ∈ Im(D), there exists x′ ∈ T, such that x = D(x′). So [x, y, z] = [D(x′), y, z] = D[x′, y, z] ∈ Im(D).

(2) From Lemma 3.12 and (1) there is an ideal sum T = Ker(D) ⊕ Im(D). Since T is indecomposable, one gets Ker(D) = 0 and Im(D) = T, which means D is invertible.

(3) For all semisimple element D ∈ C(T), since x² does not divide the minimal polynomial of D and T is indecomposable, from (2) one gets D is invertible. It is obvious that id ∈ C(T). If there exist D₁ ≠ 0, D₂ ≠ 0, D₁, D₂ ∈ C(T) such that D₁D₂ = 0, then D₁ = D₂ = 0, a contradiction. Hence C(T) has no zero divisor. Note that D₁D₂. [x, y, z]) = D₁([D₂(x), y, z]) = [D₂(x), D₁(y), z] = D₂([x, D₁(y), z]) = D₂D₁ ([x, y, z]) and T = [T, T, T]. Then D₁D₂ = D₂D₁, ∀ D₁, D₂ ∈ C(T). So C(T) is a field.

Let T be a Lie triple system with Z(T) = {0}. If D ∈ QC(T) and suppose x²does not divide the minimal polynomial of D, then T = Ker(D) ⊕ Im(D).

From Lemma 3.12, there is a vector space direct sum T = Ker(D) +̇ Im(D). Obviously, [Ker(D), D(T), T] = [D Ker(D)), T, T] = 0 and [T, D(T), Ker(D)]

= [T, T; D(Ker(D))] = 0, so Ker(D) ⊆ Z_T(Im(D)), Im(D) ⊆ Z_T(Ker(D)). Since Z_T. Im(D)) ∩ Z_T(Ker(D)) = Z(T) = {0}, we must have Ker (D) = Z_T(Im(D)), Im(D)

= Z_T(Ker(D)).

It is easy to get [[Ker(D), T, T], Im (D), T] = [T, Im(D), [Ker(D), T, T]] = 0, which means [Ker(D), T, T] ⊆ Z_T(Im(D)) = Ker(D).

Also [[Im(D), T, T], Ker(D), T] = [T, Ker(D), [Im(D), T, T]] = 0, [Im(D), T, T] ⊆ Z_T. Ker(D)) = Im(D). So Ker(D) and Im(D) are ideals.

Let (T,[⋅, ⋅, ⋅]) be a indecomposable Lie triple system over an algebraically field 𝔽 and Z(T) = 0. D ∈ QC(T) is semisimple, then D ∈ Z_C(T) (GDer(T)).

Let D ∈ QC(T), D has an eigenvalue λ since 𝔽 is an algebraically field. We denote the corresponding eigenspace by E_λ(D), it is easy to get (D – λid) ∈ QC(T) and Ker (D – λid) = E_λ(D) ≠ 0. From Lemma 3.14 and D is a semisimple element, Ker(D – λid ) is an ideal of T, so Ker (D – λid ) = T. That is D = λ id ∈ C(T) and [D, GDer(T)] = 0.

D ∈ QDer(T) if and only if D^*. Ker(ϕ)) ⊆ Ker(ϕ).

(⇒) For all Σx ⊗ y ⊗ z ∈ Ker(ϕ), we have Σ[x, y, z] = 0. Thus,

Since D ∈ QDer(T), we have

Hence D^* Ker(ϕ)) ⊆ Ker (ϕ).

(⇐) Since D^*. Ker(ϕ)) ⊆ Ker (ϕ), there exists an element D′ ∈ End(T) such that

for all D ∈ End(T). A direct computation shows that for all x, y, z ∈ T,

that is, D ∈ QDer(T). □

Suppose that End(T) acts on T ⊗ T ⊗ T via D ⋅ (x ⊗ y ⊗ z) = D^* (x ⊗ y ⊗ z] for all x, y, z ∈ T with D^*as above. Then (T ⊗ T ⊗ T)⁺and (T ⊗ T ⊗ T) – are two irreducible End(T)-modules.

Here we need only to show that (T ⊗ T ⊗ T)⁺ is an irreducible End(T)-module and the case of (T ⊗ T ⊗ T)^– is similar. For all x, y, z ∈ T and D ∈ End(T), from Eq.(3.1) we have

Hence D⋅(T ⊗ T ⊗ T)⁺ ⊆(T ⊗ T ⊗ T)⁺ and it is easy to check that

Therefore, (T ⊗ T ⊗ T)⁺ is an End(T)-module.

Suppose that there exists a nonzero End(T)-submodule V in (T ⊗ T ⊗ T)⁺. Choose a nonzero element Σ(x⊗y⊗z + y⊗x⊗z) in V for some x, y, z ∈ T. A direct computation shows that all elements in (T⊗T⊗T)⁺are obtainable by repeated application of elements of End(T) to Σ(x⊗y⊗z + y⊗x⊗z) and formation of linear combinations. Hence V is (T ⊗ T ⊗ T)⁺ itself. Thus, (T ⊗ T ⊗ T)⁺ as an End(T)-module is irreducible.□

Let T be a Lie triple system with[T, T, T] ≠0 and QDer(T) = End(T). Then T is a two-dimensional simple Lie triple system.

We consider the action of End(T) on T⊗T⊗T via D⋅x⊗y⊗z) = D*(x⊗y⊗z) for all x, y, z ∈ T with D* as in Eq. 3.1). By Lemma 4.1, QDer(T) = End(T) implies that End(T) ⋅ Ker(ϕ,) ⊆ Ker (ϕ,). Lemma 4.2 tells us that the only proper subspaces of T⊗T⊗T, invariant under this action of End(T), are (T⊗T⊗T)⁺ and (T⊗T⊗T)⁻. Thus we have ϕ, : T⊗T⊗T → T with kernel {0}, (T⊗T⊗T)⁺ and (T⊗T⊗T)⁻. Using

we have that n = 1, if Ker (ϕ) = {0} or (T⊗T⊗T)⁻; n ≤ 2, if Ker (ϕ) = (T⊗T⊗T)⁺.

We discuss the possibilities for n as follows:

1. If n = 1, then T is commutative, which is a contradiction with our assumption.

2. If n = 2, i.e. Ker (ϕ,) = (T⊗T⊗T)⁺, hence dim (Ker ϕ)) = 6 and dim [T, T, T]) = 2, so ϕ, must be surjective and we have [T, T, T] = T. So that T is the two-dimensional simple Lie triple system (See [12, Chapter 4.3 ]).

If T is a two-dimensional simple Lie triple system or an abelian Lie triple system, then QDer(T) = End(T).

By [12, Chapter 4.3 ], we have a basis e₁, e₂ such that [e₁, e₂, e₁] = −e₁, [e₁, e₂, e₂] = e₂. Then for all D ∈ End(T), one gets D(e1)=k1e1+k2′e2,k1,k1′,k1,k2′∈F. It obvious k′ = 0 since D is a linear map. Similarly, D(e₂) = k₂e₂. So we have

Let D′ ∈ End(T) such that D′(e₁) = −(2k₁ + k₂)/e₁ and D′(e₂) = (k₁ + 2k₂)e₂, thus D ∈ QDer(T). □

Let T be a Lie triple system over 𝔽 We defineT̆, ≔ {Σ(x⊗t + y⊗t³)|x, y ∈ T} ⊆ T ⊗(𝔽[t]/t⁴). Then T̆ is a Lie triple system with the operation [x⊗tⁱ, y⊗t^j, z⊗t^k] = [x, y, z]⊗t^i+j+k, for all x, y, z ∈ T, i, j, k ∈ {1, 3}

For all x, y, z, u; v ∈ T and i, j, k, m, n ∈ {1, 3}, we have

and

Hence T̆, is a Lie triple system. □

T, T̆ϕ are as defined above. Then

1. ϕ is injective and ϕ(D) does not depend on the choice of D′.

2. ϕ (QDer(T)) ⊆ Der T̆).

If ϕ(D₁) = ϕ(D₂), then for all a ∈ T, b ∈ [T, T, T] and u ∈ U, we have

that is

so D₁(a) = D₂(a). Hence D₁ = D₂, and ϕ is injective.

Suppose that there exists D″ such that

and

then we have

thus D′(b) = D″(b). Hence

which implies φ(D) is determined by D.

(2) We have [xtⁱ, yt^j, zt^k] = [x, y, z]t^i+j+k = 0, for all i + j + k ≥ 4. Thus, to show φ(D) ∈ Der. (Ť), we need only to check the validness of the following equation

For all x, y, z ∈ T, we have

Therefore, for all D ∈ QDer(T), we have φ(D) ∈ Der(Ť). ⎕

Let T be a Lie triple system. Z(T) = {0} and Ť, φ are as defined above. Then Der(Ť) = φ(QDer(T)) ∔ ZDer(Ť).

Since Z(T) = {0}, we have Z(Ť) = Tt³. For all g ∈ Der(Ť), we have g(Z(Ť)) ⊆ Z(Ť), hence g(Ut³) ⊆ g(Z(Ť) ⊆ Z(Ť) = Tt³. Now we define a map f : Tt + Ut³ + [T, T, T]t³ → Tt³ by

It is clear that f is linear. Note that

hence f ∈ ZDer(Ť). Since

and

there exist D, D′ ∈ End(T) such that for all a ∈ T, b ∈ [T, T, T],

Since (g-f) ∈ Der(Ť) and by the definition of Der(Ť), we have [g-f)(a₁t),a₂t,a₃t]+a₁t,(g-f)(a₂t),a₃t]+[a₁t,a₂t,(g-f)(a₃t)] = (g-f)([a₁t,a₂t,a₃t]), for all a₁,a₂,a₃ ∈ T. Hence

Thus D ∈ QDer(T). Therefore, g-f = φ(D) ∈ φ(QDer(T)), so Der(Ť) ⊆ φ(QDer(T)) + ZDer(Ť). By Proposition 5.2 (2) we have Der(Ť) = φ(QDer(T)) + ZDer(Ť).

For all f ∈ φ(QDer(T)) ∩ ZDer(Ť), there exists an element D ∈ QDer(T) such that f = φ(D). Then

for all a ∈ T, b ∈ [T, T, T].

On the other hand, since f ∈ ZDer(Ť), we have