On split Lie color triple systems

Definition 2.1

[14] Let Γ be an abelian group. A bi-character on Γ is a map ε : Γ × Γ → 𝕂 ∖ {0} satisfying

1. ε(α, β)ε(β, α) = 1,

2. ε(α, β + γ) = ε(α, β)ε(α, γ),

3. ε(α + β, γ) = ε(α, γ)ε(β, γ)

for all α, β, γ ∈ Γ.

It is clear that ε(α, 0) = ε(0, α) = 1 for any α ∈ Γ, where 0 denotes the identity element of Γ.

Definition 2.2

[14] Let L = ⊕_g∈ΓL_g be a Γ-graded 𝕂-vector space. For a nonzero homogeneous element v ∈ L, denote by v̄ the unique group element in Γ such that v ∈ L_v̄, which will be called the homogeneous degree of v. We shall say that L is a Lie color algebra if it is endowed with a 𝕂-bilinear map [⋅, ⋅] : L × L → L satisfying

1. [v, w] = −ε(v̄, w̄)[w, v], (Skew- symmetry)

2. [v, [w, t]] = [[v, w], t] + ε(v̄, w̄)[w, [v, t]] (Jacobi identity)

for all homogeneous elements v, w, t ∈ L.

Lie superalgebras are examples of Lie color algebras with Γ = ℤ₂ and ε(i, j) = (−1)^ij, for any i, j ∈ ℤ₂. We also note that L₀ is a Lie algebra.

Definition 2.3

[13] A Lie triple system is a vector space T endowed with a trilinear map {⋅, ⋅, ⋅} : T × T × T → T satisfying

1. {x, y, z} = −{y, x, z},

2. {x, y, z} + {y, z, x} + {z, x, y} = 0,

3. {x, y, {a, b, c}} = {{x, y, a}, b, c} + {a, {x, y, b}, c} + {a, b, {x, y, c}}

   for x, y, z, a, b, c ∈ T.

Definition 2.4

Let T = ⊕_g∈ΓT_g be a Γ-graded 𝕂-vector space. For a nonzero homogeneous element v ∈ T, denote by v̄ the unique group element in Γ such that v ∈ T_v̄, which will be called the homogeneous degree of v. We shall say that T is a Lie color triple system if it is endowed with a 𝕂-trilinear map

1. {x, y, z} = −ε(x̄, ȳ){y, x, z},

2. ε(z̄, x̄){x, y, z} + ε(x̄, ȳ){y, z, x} + ε(ȳ, z̄){z, x, y} = 0,,

3. {x, y, {a, b, c}} = {{x, y, a}, b, c} + ε(x̄ + ȳ, ā){a, {x, y, b}, c}

+ε(x̄ + ȳ, ā + b̄){a, b, {x, y, c}}

for all homogeneous elements x, y, z, a, b, c ∈ T.

Example 2.5

Lie triple systems are examples of Lie color triple systems with Γ = {0} and ε(0,0) = 1.

Example 2.6

Lie triple supersystems are examples of Lie color triple systems with Γ = {ℤ₂} = {0, 1} and ε(ᾱ, β̄) = (−1)^ᾱβ̄, for any ᾱ, β̄ ∈ ℤ₂.

Example 2.7

If L is a Lie color algebra with product [⋅, ⋅], then L becomes a Lie color triple system by putting {x, y, z} = [[x, y], z].

Definition 2.8

Let I = ⊕_g∈ΓI_g be a graded subspace of a Lie color triple system T. Then I is called a subsystem of T, if {I, I, I} ⊆ I; I is called an ideal of T, if {I, T, T} ⊆ I.

Definition 2.9

The standard embedding of a Lie color triple system T is the ℤ₂-graded Lie color algebra L = L⁰ ⊕ L¹, L⁰ being the 𝕂-span of {L(x, y) : x, y ∈ T}, where L(x, y) denotes the left multiplication operator in T, L(x, y)(z) := {x, y, z}; L¹ := T and where the product is given by

Let us observe that L⁰ with the product induced by the one in L = L⁰ ⊕ L¹ becomes a Lie color algebra.

Let us recall the concept of split Lie color algebra. Denote by H = ⊕_g∈ΓH_g a maximal abelian (graded) subalgebra, (MAGSA), of a Lie color algebra L. For a linear functional

we define the root space of L (with respect to H) associated to α as the subspace

The elements α : H₀ → 𝕂 satisfying L_α ≠ 0 are called roots of L with respect to H. We denote Λ := {α ∈ H0∗ ∖ {0} : L_α ≠ 0}. We say that L is a split Lie color algebra, with respect to H, if L = H ⊕ (⊕_α∈ΛL_α). We also say that Λ is the root system of L.

Definition 2.10

Let T be a Lie color triple system, L = L⁰ ⊕ L¹ be its standard embedding, and H⁰ = ⊕_g∈Γ Hg0 be a maximal abelian (graded) subalgebra (MAGSA) of L⁰. For a linear functional α ∈ (H00)∗ , we define the root space of T (with respect to H⁰) associated to α as the subspace T_α := {t_α ∈ T : [h, t_α] = α(h)t_α for any h ∈ H00 }. The elements α ∈ (H00)∗ satisfying T_α ≠ 0 are called roots of T with respect to H⁰ and we denote Λ¹ := {α ∈ (H00)∗ ∖ {0} : T_α ≠ 0}.

Let us observe that T₀ := {t₀ ∈ T : [h, t₀] = 0 for any h ∈ H00 }. In the following, we shall denote by Λ⁰ the set of all nonzero α ∈ (H00)∗ such that Lα0:={vα0∈L0:[h,vα0]=α(h)vα0 for any h ∈ H00 } ≠ 0.

Lemma 2.11

Let T be a Lie color triple system, L = L⁰ ⊕ L¹ be its standard embedding, and H⁰ be a MAGSA of L⁰. For α, β, γ ∈ Λ¹ ∪ {0} and ξ, q ∈ Λ⁰ ∪ {0}, the following assertions hold.

1. If [T_α, T_β] ≠ 0, then α + β ∈ Λ⁰ ∪ {0} and [T_α, T_β] ⊆ Lα+β0.

2. If [Lξ0,Tα]≠0, then ξ + α ∈ Λ¹ ∪ {0} and [Lξ0,Tα]⊆Tξ+α.

3. If [Lξ0,Lq0]≠0, then ξ + q ∈ Λ⁰ ∪ {0} and [Lξ0,Lq0]⊆Lξ+q0.

4. If {T_α, T_β, T_γ} ≠ 0, then α + β + γ ∈ Λ¹ ∪ {0} and {T_α, T_β,T_γ} ⊆ T_α+β+γ.

Proof

1. For any x ∈ T_α, y ∈ T_β and h ∈ H00 , by Definition 2.2 (2), one has [h, [x, y]] = ε(0, x̄)[x, [h, y]] + [[h, x], y] = [x, β(h)y] + [α(h)x, y] = (α + β)(h)[x, y].

2. For any x ∈ Lξ0 , y ∈ T_α and h ∈ H00 , by Definition 2.2 (2), one has [h, [x, y]] = [x, [h, y]] + [[h, x], y] = ε(0, x̄)[x, α(h)y] + [ξ(h)x, y] = (ξ + α)(h)[x, y].

3. For any x ∈ Lξ0 , y ∈ Lq0 and h ∈ H00 , by Definition 2.2 (2), one has [h, [x, y]] = ε(0, x̄)[x, [h, y]] + [[h, x], y] = [x, q(h)y] + [ξ(h)x, y] = (ξ + q)(h)[x, y].

4. It is a consequence of Lemma 2.11 (1) and (2).□

Definition 2.12

Let T be a Lie color triple system, L = L⁰ ⊕ L¹ be its standard embedding, and H⁰ = ⊕_g∈Γ Hg0 be a MAGSA of L⁰. We shall call that T is a split Lie color triple system (with respect to H⁰) if T = T₀ ⊕ (⊕_α∈Λ¹ T_α). We say that Λ¹ is the root system of T.

We observe that it is straightforward to prove that if T is a split Lie color triple system with respect to H⁰ and we let L = L⁰ ⊕ L¹ be its standard embedding algebra, then L⁰ is a split Lie color algebra with respect to the splitting Cartan subalgebra H⁰, with set of nonzero roots Λ⁰. We also note that the facts H⁰ ⊂ L⁰ = [T, T] and T = T₀ ⊕ (⊕_α∈Λ¹ T_α) imply

Lemma 2.13

Let T = ⊕_g∈ΓT_g be a split Lie color triple system with corresponding root space decomposition T = T₀ ⊕ (⊕_α∈Λ¹ T_α). If we denote by T_α,g = T_α ∩ T_g, the following assertions hold.

1. T_α = ⊕_g∈Γ T_α,g for any α ∈ Λ¹ ∪ {0}.

2. Hg0=∑g1+g2=g,g1,g2∈Γ[T0,g1,T0,g2]+∑α∈Λ1g1+g2=g,g1,g2∈Γ[Tα,g1,T−α,g2].

3. T₀ is a split Lie triple system, with respect to H00 , with root space decomposition T₀ = T_0,0 ⊕ (⊕_α∈Λ¹T_α,0).

Proof

1. By the Γ-grading of T we may express any v_α ∈ T_α, α ∈ Λ¹ ∪ {0}, in the form v_α = v_α,g1 + ⋯ + v_α,gn with v_α,gi ∈ T_gi for distinct g₁, ⋯ ,g_n ∈ Γ. If h₀ ∈ H00 then [h₀, v_α,gi] = α(h₀)v_α,gi for i = 1, ⋯, n. Hence T_α = ⊕_g∈Γ(T_α ∩ T_g) and we can write T_α = ⊕_g∈ΓT_α,g for any α ∈ Λ¹ ∪ {0}.

2. Consequence of Eq. (2.1) and Lemma 2.13 (1).

3. By considering g = 0 we get T₀ = T_0,0 ⊕ (⊕_α∈Λ¹T_α,0). Hence, the direct character of the sum and the fact that α ≠ 0 for any α ∈ Λ¹ give us that H00 is a maximal abelian subalgebra of the Lie algebra L00 . Hence T₀ is a split Lie triple system respect to H00 .□

Definition 2.14

A root system Λ¹ of a split Lie color triple system T is called symmetric if it satisfies that α ∈ Λ¹ implies −α ∈ Λ¹.

A similar concept applies to the set Λ⁰ of nonzero roots of L⁰.

In the following, T denotes a split Lie color triple system with a symmetric root system Λ¹, and T = T₀ ⊕ (⊕_α∈Λ¹ T_α) the corresponding root decomposition. We begin the study of split Lie color triple system by developing the concept of connections of roots.

Definition 2.15

Let α and β be two nonzero roots, we shall say that α and β are connected if there exists a family {α₁, α₂, ⋯, α_2n, α_2n+1} ⊂ Λ¹ ∪ {0} of roots of T such that

1. {α₁, α₁ + α₂ + α₃, α₁ + α₂ + α₃ + α₄ + α₅, ⋯, α₁ + ⋯ + α_2n + α_2n+1} ⊂ Λ¹;

2. {α₁ + α₂, α₁ + α₂ + α₃ + α₄, ⋯, α₁ + ⋯ + α_2n} ⊂ Λ⁰;

3. α₁ = α and α₁ + ⋯ + α_2n + α_2n+1 ∈ ±β.

   We shall also say that {α₁, α₂, ⋯, α_2n, α_2n+1} is a connection from α to β.

   We denote by

   Λα1:={β∈Λ1:αandβareconnected},

   we can easily get that {α} is a connection from α to itself and to −α. Therefore ±α ∈ Λα1 .

Definition 2.16

A subset Ω¹ of a root system Λ¹, associated to a split Lie color triple system T, is called a root subsystem if it is symmetric, and for α, β, γ ∈ Ω¹ ∪ {0} such that α + β ∈ Λ⁰ and α + β + γ ∈ Λ¹ then α + β + γ ∈ Ω¹.

Let Ω¹ be a root subsystem of Λ¹. We define

and V_Ω¹ := ⊕_α∈Ω¹T_α. Taking into account the fact that {T₀, T₀, T₀} = 0, it is straightforward to verify that T_Ω¹ := T_0,Ω¹ ⊕ V_Ω¹ is a subsystem of T. We will say that T_Ω¹ is a subsystem associated to the root subsystem Ω¹.

Proposition 2.17

If Λ⁰ is symmetric, then the relation ∼ in Λ¹, defined by α ∼ β if and only if β ∈ Λα1 , is of equivalence.

Proof

{α} is a connection from α to itself and therefore α∼ α.

If α∼ β and {α₁, α₂, ⋯, α_2n, α_2n+1} is a connection from α to β, then

is a connection from β to α in the case α₁ + ⋯ + α_2n + α_2n+1 = β, and

in the case α₁ + ⋯ + α_2n + α_2n+1 = −β. Therefore β∼ α.

Finally, suppose α∼ β and β∼ γ, {α₁, α₂, ⋯, α_2n, α_2n+1} is a connection from α to β and {β₁, ⋯, β_2m+1} is a connection from β to γ. If m ≠ 0, then

is a connection from α to γ in the case α₁ + ⋯ + α_2n + α_2n+1 = β, and

in the case α₁ + ⋯ + α_2n + α_2n+1 = −β. If m = 0, then γ ∈ ± β and so

is a connection from α to γ. Therefore α∼ γ and ∼ is of equivalence.□

Proposition 2.18

Let α be a nonzero root and suppose Λ⁰ is symmetric. Then Λα1 is a root subsystem.

Proof

If β ∈ Λα1 then there exists a connection {α₁, α₂, ⋯, α_2n, α_2n+1} from α to β. It is clear that {α₁, α₂, ⋯, α_2n, α_2n+1} also connects α to −β and therefore −β ∈ Λα1 . Let β₁, β₂, β₃ ∈ Λα1 ∪ {0} be such that β₁ + β₂ ∈ Λ⁰ and β₁ + β₂ + β₃ ∈ Λ¹. If β₁ = 0, as β₁ + β₂ ∈ Λ⁰ then β₂ ≠ 0 and there exists a connection {α₁, α₂, ⋯, α_2n, α_2n+1} from α to β₂. We have {α₁, α₂, ⋯, α_2n+1, 0, β₃} is a connection from α to β₂ + β₃ in case α₁ + ⋯ + α_2n + α_2n+1 = β₂ and {α₁, α₂, ⋯, α_2n+1, 0, −β₃} in case α₁ + ⋯ + α_2n + α_2n+1 = −β₂. So β₁ + β₂ + β₃ = β₂ + β₃ ∈ Λα1 . Suppose β₁ ≠ 0, then there exists a connection {α₁, α₂, ⋯, α_2n, α_2n+1} from α to β₁. Hence, {α₁, α₂, ⋯, α_2n+1,β₂, β₃} is a connection from α to β₁ + β₂ + β₃ in case α₁ + ⋯ + α_2n + α_2n+1 = β₁ and {α₁, α₂, ⋯, α_2n+1, −β₂, −β₃} in case α₁ + ⋯ + α_2n + α_2n+1 = −β₁. Therefore β₁ + β₂ + β₃ ∈ Λα1 .□

Lemma 3.1

The following assertions hold.

1. If α, β ∈ Λ¹ with [T_α, T_β] ≠ 0, then α is connected with β.

2. If α, β ∈ Λ¹, α ∈ Λ⁰ and [Lα0,Tβ]≠0, then α is connected with β.

3. If α, β ∈ Λ¹, α, β ∈ Λ⁰ and [Lα0,Lβ0]≠0, then α is connected with β.

4. If α, β ∈ Λ¹ such that α is not connected with β, then [T_α, T_β] = 0, [Lα0,Tβ¯]=0 and [Tβ¯,Lα0]=0 if furthermore α ∈ Λ⁰. If α, β ∈ Λ¹ such that α is not connected with β, then [Lα0,Lβ¯0]=0 if furthermore α, β ∈ Λ⁰.

Proof

1. Suppose [T_α, T_β] ≠ 0, by Lemma 2.11 (1), one gets α + β ∈ Λ⁰ ∪ {0}. If α + β = 0, then β = −α and so α is connected with β. Suppose α + β ≠ 0. Since α + β ∈ Λ⁰, one gets {α, β, −α} is a connection from α to β.

2. Suppose [Lα0,Tβ]≠0, by Lemma 2.11 (2), one gets α + β ∈ Λ¹ ∪ {0}. If α + β = 0, then β = −α and so α is connected with β. Suppose α + β ≠ 0. Since α + β ∈ Λ¹, we obtain {α, 0, −α−β} is a connection from α to β.

3. Suppose [Lα0,Lβ0]≠0, by Lemma 2.11 (3), one has α + β ∈ Λ⁰ ∪ {0}. If α + β = 0, then β = −α and so α is connected with β. Suppose α + β ≠ 0. Since α + β ∈ Λ⁰, one gets {α, β, −α} is a connection from α to β.

4. It is a consequence of Lemma 3.1 (1), (2) and (3).□

Lemma 3.2

If α, β ∈ Λ¹ are not connected, then {T_α, T_−α, T_β} = 0.

Proof

If [T_α, T_−α] = 0, it is clear. One may suppose that [T_α, T_−α] ≠ 0 and {T_α, T_−α, T_β} ≠ 0. By the definition of Lie color triple systems, one has {T_−α, T_β, T_α} ≠ 0 or {T_β, T_α, T_−α} ≠ 0, contradicting Lemma 3.1 (4). Hence, {T_α, T_−α, T_β} = 0.□

Lemma 3.3

Fix α₀ ∈ Λ¹ and suppose Λ⁰ is symmetric. For α ∈ Λα01 and β, γ ∈ Λ¹ ∪ {0}, then the following assertions hold.

1. If {T_α, T_β, T_γ} ≠ 0, then β, γ, α + β + γ ∈ Λα01 ∪ {0}.

2. If {T_γ, T_α, T_β} ≠ 0, then γ, β, γ + α + β ∈ Λα01 ∪ {0}.

3. If {T_β, T_γ, T_α} ≠ 0, then β, γ, β + γ + α ∈ Λα01 ∪ {0}.

Proof

1. It is easy to see that [T_α, T_β] ≠ 0, for α ∈ Λα01 and β ∈ Λ¹ ∪ {0}. By Lemma 3.1 (1), one gets α ∼ β in the case β ≠ 0. From here, β ∈ Λα01 ∪ {0}. In order to complete the proof, we will show γ, α + β + γ ∈ Λα01 ∪ {0}. We distinguish two cases.

   1. Suppose α + β + γ = 0. It is clear that α + β + γ ∈ Λα01 ∪ {0}. The fact that {T₀, T₀, T₀} = 0 and {T_α, T_−α, T₀} = 0 for α ∈ Λ¹ gives us γ ≠ 0. By Lemma 2.11 (1), one gets α + β ∈ Λ⁰. As α + β = −γ, {α, β, 0} would be a connection from α to γ and we conclude γ ∈ Λα01 ∪ {0}.

   2. Suppose α + β + γ ≠ 0. We treat separately two cases.

      Suppose α + β ≠ 0. By Lemma 2.11 (1), one gets α + β ∈ Λ⁰ and so {α, β, γ} is a connection from α to α + β + γ. Hence α + β + γ ∈ Λα01 ∪ {0}. In the case γ ≠ 0, {α, β, −α−β−γ} is a connection from α to γ. So γ ∈ Λα01 . Hence γ ∈ Λα01 ∪ {0}.

      Suppose α + β = 0. Then necessarily γ ∈ Λα01 ∪ {0}. Indeed, if γ ≠ 0 and α is not connected with γ, by Lemma 3.2, {T_α, T_β, T_γ} = {T_α, T_−α, T_γ} = 0, a contradiction. We also have α + β + γ = γ ∈ Λα01 ∪ {0}.

2. The fact that [T_γ, T_α] ≠ 0 implies by Lemma 3.1 (1) that α ∼ γ in the case γ ≠ 0. From here, γ ∈ Λα01 ∪ {0}. In order to complete the proof, we will show β, γ + α + β ∈ Λα01 ∪ {0}. We distinguish two cases.

   1. Suppose γ + α + β = 0. It is clear that γ + α + β ∈ Λα01 ∪ {0}. The fact that {T₀, T₀, T₀} = 0 and {T_α, T_−α, T₀} = 0 for α ∈ Λ¹ gives us β ≠ 0. By Lemma 2.11 (1), one has γ + α ∈ Λ⁰. As γ + α = −β, {α, γ, 0} would be a connection from α to β and we conclude β ∈ Λα01 ∪ {0}.

   2. Suppose γ + α + β ≠ 0. We treat separately two cases.

      Suppose γ + α ≠ 0. By Lemma 2.11 (1), one gets γ + α ∈ Λ⁰ and so {α, γ, β} is a connection from α to γ + α + β. Hence γ + α + β ∈ Λα01 ∪ {0}. In the case β ≠ 0, we have {α, γ, −α−γ−β} is a connection from α to β. So β ∈ Λα01 . Hence β ∈ Λα01 ∪ {0}.

      Suppose γ + α = 0. Then necessarily β ∈ Λα01 ∪ {0}. Indeed, if β ≠ 0 and α is not connected with β, by Lemma 3.2, {T_γ, T_α, T_β} = {T_−α, T_α, T_β} = 0, a contradiction. We also have γ + α + β = β ∈ Λα01 ∪ {0}.

3. By the definition of Lie color triple systems, one has

   ε(α¯,β¯){Tβ,Tγ,Tα}⊂ε(γ¯,α¯){Tα,Tβ,Tγ}+ε(β¯,γ¯){Tγ,Tα,Tβ}.

   So either {T_α, T_β, T_γ} ≠ 0 or {T_γ, T_α, T_β} ≠ 0. By Lemma 3.3 (1) and (2), one gets β, γ ∈ Λα01 ∪ {0}. Next we will show that β + γ + α ∈ Λα01 ∪ {0}. We treat separately three cases.

   1. Suppose β ≠ 0, then β ∈ Λα01 . By Lemma 3.3 (1), one has β + γ + α ∈ Λα01 ∪ {0}.

   2. Suppose β = 0 and γ ≠ 0, then γ ∈ Λα01 . By Lemma 3.3 (2), one has β + γ + α ∈ Λα01 ∪ {0}.

   3. Suppose β = 0 and γ = 0, then β + γ + α = α ∈ Λα01 , we also have β + γ + α ∈ Λα01 ∪ {0}.

Lemma 3.4

Fix α₀ ∈ Λ¹ and suppose Λ⁰ is symmetric. For α, β, γ ∈ Λα01 ∪ {0} with α + β + γ = 0 and τ, ϵ ∈ Λ¹ ∪ {0}, the following assertions hold.

1. If {{T_α, T_β, T_γ}, T_τ, T_ϵ} ≠ 0, then τ, ϵ, τ + ϵ ∈ Λα01 ∪ {0}.

2. If {T_ϵ, {T_α, T_β, T_γ}, T_τ} ≠ 0, then τ, ϵ, ϵ + τ ∈ Λα01 ∪ {0}.

3. If {T_τ, T_ϵ, {T_α, T_β, T_γ}} ≠ 0, then τ, ϵ, τ + ϵ ∈ Λα01 ∪ {0}.

Proof

1. From the fact that α + β + γ = 0, {T₀, T₀, T₀} = 0 and {T_α, T_−α, T₀} = 0 whenever α ∈ Λ¹, one may suppose that at least two distinct elements in {α, β, γ} are nonzero and one may consider the case {T_α, T_β, T_γ} ≠ 0, α + β ≠ 0 and γ ≠ 0. Since

   0≠{{Tα,Tβ,Tγ},Tτ,Tϵ}⊂{Tα,Tβ,{Tγ,Tτ,Tϵ}}−ε(α¯+β¯,γ¯){Tγ,{Tα,Tβ,Tτ},Tϵ}−ε(α¯+β¯,γ¯+τ¯){Tγ,Tτ,{Tα,Tβ,Tϵ}},

   any of the above three summands is nonzero. In order to complete the proof, we first will show τ, ϵ ∈ Λα01 ∪ {0}. We distinguish three cases.

   1. Suppose {T_α, T_β, {T_γ, T_τ, T_ϵ}} ≠ 0. As γ ≠ 0 and {T_γ, T_τ, T_ϵ} ≠ 0, Lemma 3.3 (1) shows that τ, ϵ are connected with γ in the case of being nonzero roots and so τ, ϵ ∈ Λα01 ∪ {0}.

   2. Suppose {T_γ, {T_α, T_β, T_τ}, T_ϵ} ≠ 0. As α + β ≠ 0 and γ ≠ 0. So either α ≠ 0 or β ≠ 0. By Lemma 3.3 (1) and (2), one has τ, ϵ ∈ Λα01 ∪ {0}.

   3. Suppose {T_γ, T_τ, {T_α, T_β, T_ϵ}} ≠ 0. As α + β ≠ 0 and γ ≠ 0. So either α ≠ 0 or β ≠ 0. By Lemma 3.3 (1) and (2), one has τ, ϵ ∈ Λα01 ∪ {0}.

      Finally, we will show τ + ϵ ∈ Λα01 ∪ {0}. From the fact that α + β + γ = 0, {T₀, T₀, T₀} = 0 and {{T_α, T_β, T_γ}, T_τ, T_ϵ} ≠ 0, let us suppose that at least one element in {τ, ϵ} is nonzero. So either τ ∈ Λα01 or ϵ ∈ Λα01 . Then {{T_α, T_β, T_γ}, T_τ, T_ϵ} ⊂ {T₀, T_τ, T_ϵ}. By Lemma 3.3 (2) and (3), one has τ + ϵ ∈ Λα01 ∪ {0}.

2. From the fact that α + β + γ = 0, {T₀, T₀, T₀} = 0 and {T_α, T_−α, T₀} = 0 whenever α ∈ Λ¹, one may suppose that at least two distinct elements in {α, β, γ} are nonzero and one may consider the case {T_α, T_β, T_γ} ≠ 0, α + β ≠ 0 and γ ≠ 0. Since

   0≠ε(α¯+β¯,ϵ¯){Tϵ,{Tα,Tβ,Tγ},Tτ}⊂{Tα,Tβ,{Tϵ,Tγ,Tτ}}−ε(α¯+β¯,ϵ¯+γ¯){Tϵ,Tγ,{Tα,Tβ,Tτ}}−{{Tα,Tβ,Tϵ},Tγ,Tτ},

   any of the above three summands is nonzero. In order to complete the proof, we firstly will show τ, ϵ ∈ Λα01 ∪ {0}. We distinguish three cases.

   1. Suppose {T_α, T_β, {T_ϵ, T_γ, T_τ}} ≠ 0. As γ ≠ 0 and {T_ϵ, T_γ, T_τ} ≠ 0, Lemma 3.3 (2) shows that ϵ, τ are connected with γ in the case of being nonzero roots and so ϵ, τ ∈ Λα01 ∪ {0}.

   2. Suppose {T_ϵ, T_γ, {T_α, T_β, T_τ}} ≠ 0. As α + β ≠ 0 and γ ≠ 0. So either α ≠ 0 or β ≠ 0. By Lemma 3.3 (1) and (2), one has ϵ, τ ∈ Λα01 ∪ {0}.

   3. Suppose {{T_α, T_β, T_ϵ}, T_γ, T_τ} ≠ 0. As α + β ≠ 0 and γ ≠ 0. So either α ≠ 0 or β ≠ 0. By Lemma 3.3 (1) and (2), one has ϵ, τ ∈ Λα01 ∪ {0}.

      Finally, we will show ϵ + τ ∈ Λα01 ∪ {0}. From the fact that α + β + γ = 0, {T₀, T₀, T₀} = 0 and {T_ϵ, {T_α, T_β, T_γ}, T_τ} ≠ 0, let us suppose that at least one element in {ϵ, τ} is nonzero. So either ϵ ∈ Λα01 or τ ∈ Λα01 . Then {T_ϵ, {T_α, T_β, T_γ}, T_τ} ⊂ {T_ϵ, T₀, T_τ}. By Lemma 3.3 (1) and (3), one has ϵ + τ ∈ Λα01 ∪ {0}.

3. By the definition of Lie color triple systems, one has

   0≠{Tτ,Tϵ,{Tα,Tβ,Tγ}}⊂{{Tα,Tβ,Tγ},Tτ,Tϵ}−ε(τ¯,ϵ¯){Tϵ,{Tα,Tβ,Tγ},Tτ}.

   Suppose {{T_α, T_β, T_γ}, T_τ, T_ϵ} ≠ 0, by Lemma 3.4 (1), one has τ, ϵ, τ + ϵ ∈ Λα01 ∪ {0}. Suppose {T_ϵ, {T_α, T_β, T_γ}, T_τ} ≠ 0, by Lemma 3.4 (2), one has τ, ϵ, ϵ + τ ∈ Λα01 ∪ {0}. Therefore, in these two cases, we get τ, ϵ, τ + ϵ ∈ Λα01 ∪ {0}.□

Lemma 3.5

Fix α₀ ∈ Λ¹and suppose Λ⁰ is symmetric. If α₁, α₂, α₃ ∈ Λα01 ∪ {0} with α₁ + α₂ + α₃ = 0 and ϵ ∈ Λ¹ ∖ Λα01 , then the following assertions hold.

1. [{T_α1, T_α2, T_α3}, T_ϵ] = 0.

2. In case ϵ ∈ Λ⁰, then [{T_α1, T_α2, T_α3}, Lϵ¯0 ] = 0.

3. [[{T_α1, T_α2, T_α3}, T₀], T_ϵ] = 0.

Proof

1. From the fact α₁ + α₂ + α₃ = 0, {T₀, T₀, T₀} = 0 and {T_α, T_−α, T₀} = 0 for α ∈ Λ¹, it is clear that if α₃ = 0 one gets [{T_α1, T_α2, T_α3}, T_ϵ] = 0. Let us consider the case α₃ ≠ 0. By the definition of Lie color triple systems, we have

   [{Tα1,Tα2,Tα3},Tϵ¯]⊂[[Tα1,Tα2],[Tα3,Tϵ¯]]−ε(α1¯+α2¯,α3¯)[Tα3,[[Tα1,Tα2],Tϵ¯]]. (3.2)

   Let us consider the first summand in Eq. (3.2). As α₃ ≠ 0, one has α₃ ∈ Λα01 . For ϵ ∈ Λ¹ ∖ Λα01 and Lemma 3.1 (4), one easily gets [T_α3, T_ϵ] = 0. Therefore [[T_α1, T_α2], [T_α3, T_ϵ]] = 0.

   Let us now consider the second summand in Eq. (3.2), it is sufficient to verify that

   [Tα3,[[Tα1,Tα2],Tϵ¯]]=0.

   To do so, we first assert that [[T_α1, T_α2], T_ϵ] = 0. Indeed, by the definition of Lie color algebras, we have

   [[Tα1,Tα2],Tϵ¯]⊂[Tα1,[Tα2,Tϵ¯]]−ε(α1¯,α2¯)[Tα2,[Tα1,Tϵ¯]], (3.3)

   where α₁, α₂ ∈ Λα01 ∪ {0}, ϵ ∈ Λ¹ ∖ Λα01 . In the following, we distinguish three cases.

   1. α₁ ≠ 0 and α₂ ≠ 0. As α₁ ∈ Λα01 and ϵ ∈ Λ¹ ∖ Λα01 , by Lemma 3.1 (1), one gets [T_α1, T_ϵ] = 0. As α₂ ∈ Λα01 and ϵ ∈ Λ¹ ∖ Λα01 , by Lemma 3.1 (1), one gets [T_α2, T_ϵ] = 0. Therefore by Eq. (3.3), one can show that [[T_α1, T_α2], T_ϵ] = 0.

   2. α₁ ≠ 0 and α₂ = 0. As α₁ ∈ Λα01 and ϵ ∈ Λ¹ ∖ Λα01 , by Lemma 3.1 (1), one gets [T_α1, T_ϵ] = 0. That is [T_α2,[T_α1, T_ϵ]] = 0. As α₂ = 0, [T_α2, T_ϵ] = [T₀, T_ϵ] ⊂ Lϵ¯0. . By Lemma 3.1 (4), one gets [T_α1, [T_α2, T_ϵ]] = 0. Therefore by Eq. (3.3), one can show that [[T_α1, T_α2], T_ϵ] = 0.

   3. α₁ = 0 and α₂ ≠ 0. As α₂ ∈ Λα01 and ϵ ∈ Λ¹ ∖ Λα01 , by Lemma 3.1 (1), one gets [T_α2, T_ϵ] = 0. That is [T_α1, [T_α2, T_ϵ]] = 0. As α₁ = 0, [T_α1, T_ϵ] = [T₀, T_ϵ] ⊂ Lϵ¯0. By Lemma 3.1 (5), we get [ T_α2,[T_α1, T_ϵ]] = 0. Therefore by (3.3), one can show that [[T_α1, T_α2], T_ϵ] = 0.

      So [T_α3,[[T_α1, T_α2], T_ϵ]] = 0 is a consequence of [[T_α1, T_α2], T_ϵ] = 0. By (3.2), one gets [{T_α1, T_α2, T_α3}, T_ϵ] = 0. The proof is complete.

2. From the fact α₁ + α₂ + α₃ = 0, {T₀, T₀, T₀} = 0 and {T_α, T_−α, T₀} = 0 for α ∈ Λ¹, one gets if α₃ = 0 then it is clear that [{T_α1, T_α2, T_α3}, Lϵ¯0 ] = 0. Let us consider the case α₃ ≠ 0. Note that

   [{Tα1,Tα2,Tα3},Lϵ¯0]⊂[[Tα1,Tα2],[Tα3,Lϵ¯0]]−ε(α1¯+α2¯,α3¯)[Tα3,[[Tα1,Tα2],Lϵ¯0]]. (3.4)

   Let us consider the first summand in Eq. (3.4). As α₃ ≠ 0, one gets [[T_α1, T_α2], [T_α3, Lϵ¯0 ]] = 0 by Lemma 3.1 (4). Let us now consider the second summand in Eq. (3.4). As either α₁ ≠ 0 or α₂ ≠ 0, the definition of Lie color algebras, the fact [T₀, Lϵ¯0 ] ⊂ T_ϵ and Lemma 3.1 (4), we obtain that [T_α3,[[T_α1, T_α2], Lϵ¯0 ]] = 0. So, the second summand in Eq. (3.4) is also zero and then [{T_α1, T_α2, T_α3}, Lϵ¯0 ] = 0.

3. It is a consequence of Lemma 3.5 (1), (2) and

   [[{Tα1,Tα2,Tα3},T0],Tϵ¯]⊂[[T0,Tϵ¯],{Tα1,Tα2,Tα3}]+[[Tϵ¯,{Tα1,Tα2,Tα3}],T0].

   □

Definition 3.6

A Lie color triple system T is said to be simple, if {T, T, T} ≠ 0 and its only ideals are {0} and T.

Theorem 3.7

Suppose Λ⁰ is symmetric, the following assertions hold.

1. For any α₀ ∈ Λ¹, the subsystem

   TΛα01=T0,Λα01⊕VΛα01

   of T associated to the root subsystem Λα01 is an ideal of T.

2. If T is simple, then there exists a connection from α to β for any α, β ∈ Λ¹.

Proof

1. Recall that

   T0,Λα01:=spanK{{Tα,Tβ,Tγ}:α+β+γ=0;α,β,γ∈Λα01∪{0}}⊂T0

   and VΛα01:=⊕γ∈Λα01Tγ. In order to complete the proof, it is sufficient to show that

   {TΛα01,T,T}⊂TΛα01.

   We first check that {TΛα01,T,T}⊂TΛα01. It is easy to see that

   {TΛα01,T,T}={T0,Λα01⊕VΛα01,T,T}={T0,Λα01,T,T}+{VΛα01,T,T}.

   Next, we will show that {T0,Λα01,T,T}⊂TΛα01. Note that

   {T0,Λα01,T,T}={T0,Λα01,T0⊕(⊕α∈Λ1Tα),T0⊕(⊕α∈Λ1Tα)}={T0,Λα01,T0,T0}+{T0,Λα01,T0,⊕α∈Λ1Tα}+{T0,Λα01,⊕α∈Λ1Tα,T0}+{T0,Λα01,⊕α∈Λ1Tα,⊕β∈Λ1Tβ}.

   Here, it is clear that { T0,Λα01, T₀, T₀} ⊂ {T₀, T₀, T₀} = 0. Taking into account { T0,Λα01, T₀, T_α}, for α ∈ Λ¹, Lemma 3.4 (1) and the fact that either α ∈ Λα01 or α ∉ Λα01 , give us that { T0,Λα01, T₀, T_α} ⊂ VΛα01 or { T0,Λα01, T₀, T_α} = 0. Similarly, one gets that { T0,Λα01, T_α, T₀} ⊂ VΛα01 or { T0,Λα01, T_α, T₀} = 0. Next, we will consider { T0,Λα01, T_α, T_β}, where α, β ∈ Λ¹. We treat five cases.

   1. If α ∈ Λα01 , β ∈ Λα01 and α + β = 0. One has

      {T0,Λα01,Tα,Tβ}⊂T0,Λα01.

   2. If α ∈ Λα01 , β ∈ Λα01 and α + β ≠ 0. By Λα01 is a root subsystem, one gets

      {T0,Λα01,Tα,Tβ}⊂VΛα01.

   3. If α ∈ Λα01 and β ∉ Λα01 . By Lemma 3.4 (1), one has

      {T0,Λα01,Tα,Tβ}=0.

   4. If β ∈ Λα01 and α ∉ Λα01 . By Lemma 3.4 (1), one has

      {T0,Λα01,Tα,Tβ}=0.

   5. If β ∉ Λα01 and α ∉ Λα01 . By Lemma 3.4 (1), one has

      {T0,Λα01,Tα,Tβ}=0.

      Therefore, { T0,Λα01, T, T} ⊂ TΛα01 .

      Next, we will show that { VΛα01 , T, T} ⊂ TΛα01 . It is obvious that

      {VΛα01,T,T}={⊕γ∈Λα01Tγ,T0⊕(⊕α∈Λ1Tα),T0⊕(⊕α∈Λ1Tα)}={⊕γ∈Λα01Tγ,T0,T0}+{⊕γ∈Λα01Tγ,T0,⊕α∈Λ1Tα}+{⊕γ∈Λα01Tγ,⊕α∈Λ1Tα,T0}+{⊕γ∈Λα01Tγ,⊕α∈Λ1Tα,⊕β∈Λ1Tβ}.

      Here, it is clear that {T_γ, T₀, T₀} ⊂ VΛα01 , for γ ∈ Λα01 . Next, we will consider {T_γ, T₀, T_α}, for γ ∈ Λα01 , α ∈ Λ¹. We treat three cases.

   1. If γ ∈ Λα01 , α ∉ Λα01 . By Lemma 3.3 (1), one has

      {Tγ,T0,Tα}=0.

   2. If γ ∈ Λα01 , α ∈ Λα01 and γ + α ≠ 0. By Λα01 is a root subsystem, one has

      {Tγ,T0,Tα}⊂VΛα01.

   3. If γ ∈ Λα01 , α ∈ Λα01 and γ + α = 0. It is clear that

      {Tγ,T0,Tα}⊂T0,Λα01.

      Hence, {T_γ, T₀, T_α} ⊂ TΛα01 , for γ ∈ Λα01 , α ∈ Λ¹. Similarly, it is easy to get {T_γ, T_α, T₀} ⊂ TΛα01 , for γ ∈ Λα01 , α ∈ Λ¹. At last, we will consider {⊕γ∈Λα01 T_γ, ⊕_α∈Λ¹ T_α, ⊕_β∈Λ¹ T_β}, for γ ∈ Λα01 , α ∈ Λ¹ and β ∈ Λ¹. We treat five cases.

   1. If γ ∈ Λα01 , α ∈ Λα01 , β ∈ Λα01 and γ + α + β = 0, one gets

      {Tγ,Tα,Tβ}⊂T0,Λα01.

   2. If γ ∈ Λα01 , α ∈ Λα01 , β ∈ Λα01 and γ + α + β ≠ 0, one gets

      {⊕γ∈Λα01Tγ,⊕α∈Λ1Tα,⊕β∈Λ1Tβ}⊂VΛα01.

   3. If γ ∈ Λα01 , α ∈ Λα01 and β Λα01 . By Lemma 3.3 (1) and (2), one gets

      {Tγ,Tα,Tβ}=0.

   4. If γ ∈ Λα01 , α ∉ Λα01 and β ∈ Λα01 . By Lemma 3.3 (1) and (3), one gets

      {Tγ,Tα,Tβ}=0.

   5. If γ ∈ Λα01 , α ∉ Λα01 and β ∉ Λα01 . By Lemma 3.3 (1), one gets

      {Tγ,Tα,Tβ}=0.

      So, { VΛα01 , T, T} ⊂ TΛα01 . Therefore { TΛα01 , T, T} ⊂ TΛα01 is a consequence of { T0,Λα01, T, T} ⊂ TΛα01 and { VΛα01 , T, T} ⊂ TΛα01 . Consequently, this proves that TΛα01 is an ideal of T.

2. The simplicity of T implies TΛα01 = T. Hence Λα01 = Λ¹.□