Let L and A be Lie triple systems, and let θA be a representation of L on A. We first construct the third-order cohomology classes by derivations of A and L, then obtain a Lie algebra GθA with a representation Φ on H3(L,A), where θA is given by an abelian extension
0⟶A⟶˜Lπ→L⟶0.
We study obstruction classes for extensibility of derivations of A and L to those of ˜L. An application of Φ is discussed.
Citation: Xueru Wu, Yao Ma, Liangyun Chen. Abelian extensions of Lie triple systems with derivations[J]. Electronic Research Archive, 2022, 30(3): 1087-1103. doi: 10.3934/era.2022058
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Let L and A be Lie triple systems, and let θA be a representation of L on A. We first construct the third-order cohomology classes by derivations of A and L, then obtain a Lie algebra GθA with a representation Φ on H3(L,A), where θA is given by an abelian extension
0⟶A⟶˜Lπ→L⟶0.
We study obstruction classes for extensibility of derivations of A and L to those of ˜L. An application of Φ is discussed.
Lie triple systems were introduced by E. Cartan in his studies on Riemannian geometry. Since then, it has been studied by many scholars. For example, Jacobson studied Lie triple systems by an algebraic method in [1,2]. The cohomology theories have been given by Yamaguti in [3] and some problems about cohomology have been solved, see [4,5,6]. Note that Lie triple systems are closely related to Lie algebras, so it is natural to generalize some properties of Lie algebras to Lie triple systems.
Automorphisms are very important subjects in the research of algebras. In [7], the authors studied extension of a pair of automorphisms of Lie algebras, and they gave a necessary and sufficient condition for a pair of automorphisms to be extensible. Since derivations are infinitesimals of automorphisms, it is interesting to study extension of a pair of derivations. Recently, extension of a pair of derivations on Lie algebras, 3-Lie algebras, Leibniz algebras and associative algebras have been studied, refer to [8,9,10,11]. We attempt to consider the same problems on Lie triple systems. Inspired by [11], we define a Lie algebra GθA, where θA is a representation of a Lie triple system L on a Lie triple system A, using compatible derivations of L and A. Then we show that a certain representation of GθA can characterize the extensibility of above compatible derivations.
This paper is organized as follows. In Section 2, we recall some basic definitions and properties of Lie triple systems. Then we construct a Lie algebra GθA and consider its representation on H3(L,A) in Section 3. In Section 4, we investigate the abelian extension and extensibility of a pair of derivations. We prove that a compatible pair (∂a,∂l) is extensible if and only if [Ob˜L(∂a,∂l)]∈H3(L,A) is trivial and (∂a,∂l) is extensible in every reversible extension if and only if Φ(∂a,∂l)=0.
In this paper, all Lie triple systems L are defined over a fixed but arbitrary field F.
In this section, we first recall some basic definitions and properties of Lie triple systems, then we show that H3(L,A)=0 for an abelian extension implies split property.
Definition 2.1. [2] A Lie triple system is a vector space L endowed with a trilinear operation [⋅,⋅,⋅]:L×L×L⟶L satisfying
[a,a,b]=0,[a,b,c]+[b,c,a]+[c,a,b]=0,[a,b,[c,d,e]]=[[a,b,c],d,e]+[c,[a,b,d],e]+[c,d,[a,b,e]], |
for all a,b,c,d,e∈L.
For a Lie triple system L, a linear map ∂:L⟶L is called a derivation of L, if for all a,b,c∈L,
∂([a,b,c])=[∂(a),b,c]+[a,∂(b),c]+[a,b,∂(c)]. |
Denote by Der(L) the space of derivations of L.
Definition 2.2. [3] An L-module is a vector space V with a bilinear map
θV:L×L⟶End(V)(a,b)⟶θV(a,b) |
such that the following conditions hold:
θV(c,d)θV(a,b)−θV(b,d)θV(a,c)−θV(a,[b,c,d])+DV(b,c)θV(a,d)=0,θV(c,d)DV(a,b)−DV(a,b)θV(c,d)+θV([a,b,c],d)+θV(c,[a,b,d])=0, |
where DV(a,b)=θV(b,a)−θV(a,b), for all a,b,c,d∈L. Also θV is called a representation of L on V.
Definition 2.3. [3] Let θV be a representation of a Lie triple system L. An n-linear map f:L×⋯×L⏟ntimes→V satisfying
f(x1,x2,...,xn−3,x,x,xn)=0 |
and
f(x1,x2,...,xn−3,x,y,z)+f(x1,x2,...,xn−3,y,z,x)+f(x1,x2,...,xn−3,z,x,y)=0 |
is called an n-cochain of L on V. Denote by Cn(L,V) the set of all n-cochains, for n≥0.
Definition 2.4. [3] Let L be a Lie triple system and θV a representation of L on V. The coboundary operator δ:C2n−1(L,V)→C2n+1(L,V) is given by
(δf)(x1,x2,...,x2n+1)=θV(x2n,x2n+1)f(x1,x2,...,x2n−1)−θV(x2n−1,x2n+1)f(x1,x2,...,x2n)+n∑i=1(−1)i+1DV(x2i−1,x2i)f(x1...,x2i−2,x2i+1,...,x2n+1)+n∑i=12n+1∑j=2i+1(−1)i+n+1f(x1...,x2i−2,x2i+1,...,xj,[x2i−1,x2i,xj],...,x2n+1) | (2.1) |
for any x1,x2,...,x2n+1∈L, f∈C2n−1(L,V).
Let L be a Lie triple system and V an L-module. The set
Z2n+1(L,V)={f∈C2n+1(L,V)|δf=0} |
is called the space of (2n+1)-cocycles of L on V.
The set
B2n+1(L,V)={δf|f∈C2n−1(L,V)} |
is called the space of (2n+1)-coboundaries of L on V.
The n-th cohomology group is
Hn(L,V):=Zn(L,V)/Bn(L,V). |
Note that, by Eq 2.1, for f∈C3(L,V) we have
δf(x1,x2,x3,x4,x5)=θV(x4,x5)f(x1,x2,x3)−θV(x3,x5)f(x1,x2,x4)+DV(x1,x2)f(x3,x4,x5)−DV(x3,x4)f(x1,x2,x5)+f([x1,x2,x3],x4,x5)+f(x3,[x1,x2,x4],x5)+f(x3,x4,[x1,x2,x5])−f(x1,x2,[x3,x4,x5]). | (2.2) |
Definition 2.5. [6] Suppose that L and A are Lie triple systems. If
0⟶A⟶˜Lπ→L⟶0 |
is an exact sequence of Lie triple systems, and [˜L,A,A]=0 (which implies that [A,˜L,A]=[A,A,˜L]=0), then we call ˜L an abelian extension of L by A. A linear map s:L→˜L is called a section of ˜L if it satisfies π∘s=idL. If there exists a section which is also a homomorphism between Lie triple systems, we say that the abelian extension is split.
The following properties have been proved in [6], which will be used in Sections 3 and 4.
Let ˜L be an abelian extension of L by A. We construct a representation of L on A. Fix any section s:L⟶˜L of π and define θA:L×L⟶End(A) by
θA(x,y)(v)=[v,s(x),s(y)]˜L, | (2.3) |
for all x,y∈L, v∈A. In particular, DA(x,y)(v)=(θA(y,x)−θA(x,y))(v)=[s(x),s(y),v]˜L. Note that θA is independent on the choice of s. Moreover, since
[s(x),s(y),s(z)]˜L−s([x,y,z]L)∈A, |
for any x,y,z∈L, we have a map ω:L×L×L⟶A given by
ω(x,y,z)=[s(x),s(y),s(z)]˜L−s([x,y,z]L)∈A, | (2.4) |
for all x,y,z∈L.
Lemma 2.6. Let ˜L be an abelian extension of a Lie triple system L by A. Then
(1) θA given by Eq 2.3 is a representation of L on A;
(2) ω given by Eq 2.4 is a 3-cocycle associated to θA.
Corollary 2.7. Let ˜L be an abelian extension of a Lie triple system L by A. Keep the same notations as in Lemma 2.6. Then the cohomology class [ω] does not depend on the choice of s.
Proposition 2.8. If θA is a representation of L on A and ω is a 3-cocycle i.e., δω=0 in Eq 2.2 then L⊕A is a Lie triple system with the bracket given by
[x+u,y+v,z+w]L⊕A=[x,y,z]L+ω(x,y,z)+DA(x,y)(w)+θA(y,z)(u)−θA(x,z)(v), |
where x,y,z∈L, u,v,w∈A.
Corollary 2.9. Retain all the notations and assumptions in Proposition 2.8. Let π:L⊕A⟶L be the canonical projection. Then there is an abelian extension ˜L=L⊕A of Lie triple systems L by A.
Based on the previous notations, we have the following
Proposition 2.10. Let θA be a representation of a Lie triple system L on a Lie triple system A. If H3(L,A)=0 then any abelian extension of L by A is split.
Proof. It suffices to show that there is a section s of π which is a homomorphism. Recall that θA given by Eq 2.3 is independent on the choice of s. Consider the 3-cocycle ω given by Lemma 2.6, since H3(L,A)=0, there exists an α∈C1(L,A) such that ω=δα. For any x,y,z∈L, it holds that
ω(x,y,z)=−α([x,y,z]L)+DA(x,y)(α(z))+θA(y,z)(α(x))−θA(x,z)(α(y)). |
Define a linear map s′:L⟶˜L by s′(x)=s(x)−α(x). Note that s′ is also a section of π. Then for any x,y,z∈L, we have
[s′(x),s′(y),s′(z)]˜L=[s(x)−α(x),s(y)−α(y),s(z)−α(z)]˜L=[s(x),s(y),s(z)]˜L−[s(x),s(y),α(z)]−[α(x),s(y),s(z)]−[s(x),α(y),s(z)]=[s(x),s(y),s(z)]˜L−DA(x,y)(α(z))+θA(x,z)(α(y))−θA(y,z)(α(x))=s([x,y,z]L)+ω(x,y,z)−DA(x,y)(α(z))+θA(x,z)(α(y))−θA(y,z)(α(x))=s([x,y,z]L)−α([x,y,z]L)=s′([x,y,z]L). |
Therefore, s′ is a homomorphism.
In this section, L and A denote Lie triple systems. We choose derivations of L and A, and use these derivations to construct third-order cohomology classes. Based on these preparations, we construct a Lie algebra GθA and its representation on H3(L,A).
Given a representation θA of L on A. Suppose ω∈C3(L,A). For any pair (∂a,∂l)∈Der(A)×Der(L), define a 3-cochain Obω(∂a,∂l)∈C3(L,A) as
Obω(∂a,∂l)=∂aω−ω(∂l⊗id⊗id)−ω(id⊗∂l⊗id)−ω(id⊗id⊗∂l), | (3.1) |
or equivalently,
Obω(∂a,∂l)(x,y,z)=∂aω(x,y,z)−ω(∂l(x),y,z)−ω(x,∂l(y),z)−ω(x,y,∂l(z)), |
for all x,y,z∈L.
Lemma 3.1. Let θA be a representation of L on A and ω∈C3(L,A) with respect to the representation θA. Assume that a pair (∂a,∂l)∈Der(A)×Der(L) satisfies that
∂aθA(x,y)−θA(x,y)∂a=θA(∂l(x),y)+θA(x,∂l(y)), | (3.2) |
for all x,y∈L. If ω is a 3-cocycle then Obω(∂a,∂l) given by Eq 3.1 is also a 3-cocycle.
Proof. We only need to prove that δObω(∂a,∂l)=0. Since ω is a 3-cocycle, δω=0, by Eq 2.2 it follows that, for any x1,x2,x3,x4,x5∈L,
0=θA(x4,x5)ω(x1,x2,x3)−θA(x3,x5)ω(x1,x2,x4)+DA(x1,x2)ω(x3,x4,x5)−DA(x3,x4)ω(x1,x2,x5)+ω([x1,x2,x3],x4,x5)+ω(x3,[x1,x2,x4],x5)+ω(x3,x4,[x1,x2,x5])−ω(x1,x2,[x3,x4,x5]). | (3.3) |
Then for any x1,x2,x3,x4,x5∈L, we have
(δObω(∂a,∂l))(x1,x2,x3,x4,x5)=θA(x4,x5)Obω(∂a,∂l)(x1,x2,x3)⏟(1)−θA(x3,x5)Obω(∂a,∂l)(x1,x2,x4)⏟(2)+DA(x1,x2)Obω(∂a,∂l)(x3,x4,x5)⏟(3)−DA(x3,x4)Obω(∂a,∂l)(x1,x2,x5)⏟(4)+Obω(∂a,∂l)([x1,x2,x3],x4,x5)⏟(5)+Obω(∂a,∂l)(x3,[x1,x2,x4],x5)⏟(6)+Obω(∂a,∂l)(x3,x4,[x1,x2,x5])⏟(7)−Obω(∂a,∂l)(x1,x2,[x3,x4,x5])⏟(8). | (3.4) |
Since (∂a,∂l)∈Der(A)×Der(L) and Obω(∂a,∂l) satisfies Eq 3.1, we have
(1)=θA(x4,x5)∂aω(x1,x2,x3)−θA(x4,x5)ω(∂l(x1),x2,x3)−θA(x4,x5)ω(x1,∂l(x2),x3)−θA(x4,x5)ω(x1,x2,∂l(x3)), | (3.5) |
(2)=−θA(x3,x5)∂aω(x1,x2,x4)+θA(x3,x5)ω(∂l(x1),x2,x4)+θA(x3,x5)ω(x1,∂l(x2),x4)+θA(x3,x5)ω(x1,x2,∂l(x4)), | (3.6) |
(3)=DA(x1,x2)∂aω(x3,x4,x5)−DA(x1,x2)ω(∂l(x3),x4,x5)−DA(x1,x2)ω(x3,∂l(x4),x5)−DA(x1,x2)ω(x3,x4,∂l(x5)), | (3.7) |
(4)=−DA(x3,x4)∂aω(x1,x2,x5)+DA(x3,x4)ω(∂l(x1),x2,x5)+DA(x3,x4)ω(x1,∂l(x2),x5)+DA(x3,x4)ω(x1,x2,∂l(x5)), | (3.8) |
(5)=∂a(ω([x1,x2,x3],x4,x5))−ω([∂l(x1),x2,x3],x4,x5)−ω([x1,∂l(x2),x3],x4,x5)−ω([x1,x2,∂l(x3)],x4,x5)−ω([x1,x2,x3],∂l(x4),x5)−ω([x1,x2,x3],x4,∂l(x5)), | (3.9) |
(6)=∂a(ω(x3,[x1,x2,x4],x5))−ω(∂l(x3),[x1,x2,x4],x5)−ω(x3,[∂l(x1),x2,x4],x5)−ω(x3,[x1,∂l(x2),x4],x5)−ω(x3,[x1,x2,∂l(x4)],x5)−ω(x3,[x1,x2,x4],∂l(x5)), | (3.10) |
(7)=∂a(ω(x3,x4,[x1,x2,x5]))−ω(∂l(x3),x4,[x1,x2,x5])−ω(x3,∂l(x4),[x1,x2,x5])−ω(x3,x4,[∂l(x1),x2,x5])−ω(x3,x4,[x1,∂l(x2),x5])−ω(x3,x4,[x1,x2,∂l(x5)]), | (3.11) |
(8)=−∂a(ω(x1,x2,[x3,x4,x5]))+ω(∂l(x1),x2,[x3,x4,x5])+ω(x1,∂l(x2),[x3,x4,x5])+ω(x1,x2,[∂l(x3),x4,x5])+ω(x1,x2,[x3,∂l(x4),x5])+ω(x1,x2,[x3,x4,∂l(x5)]). | (3.12) |
For Eqs 3.5–3.12, by suitable combination and with the aid of Eq 3.3, we get the following
−θA(x4,x5)ω(∂l(x1),x2,x3)+θA(x3,x5)ω(∂l(x1),x2,x4)−ω([∂l(x1),x2,x3],x4,x5)+DA(x3,x4)ω(∂l(x1),x2,x5)−ω(x3,[∂l(x1),x2,x4],x5)−ω(x3,x4,[∂l(x1),x2,x5])+ω(∂l(x1),x2,[x3,x4,x5])=DA(∂l(x1),x2)ω(x3,x4,x5), | (3.13) |
−θA(x4,x5)ω(x1,∂l(x2),x3)+θA(x3,x5)ω(x1,∂l(x2),x4)−ω([x1,∂l(x2),x3],x4,x5)+DA(x3,x4)ω(x1,∂l(x2),x5)−ω(x3,[x1,∂l(x2),x4],x5)−ω(x3,x4,[x1,∂l(x2),x5])+ω(x1,∂l(x2),[x3,x4,x5])=DA(x1,∂l(x2))ω(x3,x4,x5), | (3.14) |
−θA(x4,x5)ω(x1,x2,∂l(x3))−DA(x1,x2)ω(∂l(x3),x4,x5)−ω([x1,x2,∂l(x3)],x4,x5)−ω(∂l(x3),[x1,x2,x4],x5)−ω(∂l(x3),x4,[x1,x2,x5])+ω(x1,x2,[∂l(x3),x4,x5])=−θA(∂l(x3),x5)ω(x1,x2,x4)−DA(∂l(x3),x4)ω(x1,x2,x5), | (3.15) |
θA(x3,x5)ω(x1,x2,∂l(x4))−DA(x1,x2)ω(x3,∂l(x4),x5)−ω([x1,x2,x3],∂l(x4),x5)−ω(x3,∂l(x4),[x1,x2,x5])+ω(x1,x2,[x3,∂l(x4),x5])−ω(x3,[x1,x2,∂l(x4)],x5)=θA(∂l(x4),x5)ω(x1,x2,x3)−DA(x3,∂l(x4))ω(x1,x2,x5), | (3.16) |
−DA(x1,x2)ω(x3,x4,∂l(x5))−ω([x1,x2,x3],x4,∂l(x5))−ω(x3,[x1,x2,x4],∂l(x5))+DA(x3,x4)ω(x1,x2,∂l(x5))−ω(x3,x4,[x1,x2,∂l(x5)])+ω(x1,x2,[x3,x4,∂l(x5)])=θA(x4,∂l(x5))ω(x1,x2,x3)−θA(x3,∂l(x5))ω(x1,x2,x4). | (3.17) |
Next, let us substituting Eqs 3.13–3.17 into Eq 3.4 having the following
(δObω(∂a,∂l))(x1,x2,x3,x4,x5)=θA(x4,x5)∂aω(x1,x2,x3)−θA(x3,x5)∂aω(x1,x2,x4)+DA(x1,x2)∂aω(x3,x4,x5)−DA(x3,x4)∂aω(x1,x2,x5)+∂aω([x1,x2,x3],x4,x5)+∂aω(x3,[x1,x2,x4],x5)+∂aω(x3,x4,[x1,x2,x5])−∂aω(x1,x2,[x3,x4,x5])+DA(∂l(x1),x2)ω(x3,x4,x5)+DA(x1,∂l(x2))ω(x3,x4,x5)−θA(∂l(x3),x5)ω(x1,x2,x4)−DA(∂l(x3),x4)ω(x1,x2,x5)−DA(x3,∂l(x4))ω(x1,x2,x5)+θA(∂l(x4),x5)ω(x1,x2,x3)+θA(x4,∂l(x5))ω(x1,x2,x3)−θA(x3,∂l(x5))ω(x1,x2,x4). | (3.18) |
By Eq 3.3, we have
∂aω([x1,x2,x3],x4,x5)+∂aω(x3,[x1,x2,x4],x5)+∂aω(x3,x4,[x1,x2,x5])−∂aω(x1,x2,[x3,x4,x5])=−∂aθA(x4,x5)ω(x1,x2,x3)+∂aθA(x3,x5)ω(x1,x2,x4)−∂aDA(x1,x2)ω(x3,x4,x5)+∂aDA(x3,x4)ω(x1,x2,x5). | (3.19) |
Here, inserting Eq 3.19 into Eq 3.18 gives that
(δObω(∂a,∂l))(x1,x2,x3,x4,x5)=θA(x4,x5)∂aω(x1,x2,x3)−θA(x3,x5)∂aω(x1,x2,x4)+DA(x1,x2)∂aω(x3,x4,x5)−DA(x3,x4)∂aω(x1,x2,x5)−∂aθA(x4,x5)ω(x1,x2,x3)+∂aθA(x3,x5)ω(x1,x2,x4)−∂aDA(x1,x2)ω(x3,x4,x5)+∂aDA(x3,x4)ω(x1,x2,x5)+DA(∂l(x1),x2)ω(x3,x4,x5)+DA(x1,∂l(x2))ω(x3,x4,x5)−θA(∂l(x3),x5)ω(x1,x2,x4)−DA(∂l(x3),x4)ω(x1,x2,x5)+θA(∂l(x4),x5)ω(x1,x2,x3)−DA(x3,∂l(x4))ω(x1,x2,x5)+θA(x4,∂l(x5))ω(x1,x2,x3)−θA(x3,∂l(x5))ω(x1,x2,x4)=(−∂aθA(x4,x5)+θA(x4,x5)∂a+θA(∂l(x4),x5)+θA(x4,∂l(x5)))ω(x1,x2,x3)+(∂aθA(x3,x5)−θA(x3,x5)∂a−θA(∂l(x3),x5)−θA(x3,∂l(x5)))ω(x1,x2,x4)+(−∂aDA(x1,x2)+DA(x1,x2)∂a+DA(∂l(x1),x2)+DA(x1,∂l(x2)))ω(x3,x4,x5)+(∂aDA(x3,x4)−DA(x3,x4)∂a−DA(∂l(x3),x4)−DA(x3,∂l(x4)))ω(x1,x2,x5). |
Since (∂a,∂l) satisfies Eq 3.2, we only need to show that
∂aDA(x,y)−DA(x,y)∂a=DA(∂l(x),y)+DA(x,∂l(y)), |
byDA(x,y)=θA(y,x)−θA(x,y), we have
∂aDA(x,y)−DA(x,y)∂a=∂a(θA(y,x)−θA(x,y))−(θA(y,x)−θA(x,y))∂a=(∂aθA(y,x)−θA(y,x)∂a)−(∂aθA(x,y)−θA(x,y)∂a)=(θA(∂l(y),x)+θA(y,∂l(x)))−(θA(∂l(x),y)+θA(x,∂l(y)))=(θA(∂l(y),x)−θA(x,∂l(y)))+(θA(y,∂l(x))−θA(∂l(x),y))=DA(x,∂l(y))+DA(∂l(x),y). |
Then we have δObω(∂a,∂l)=0 as required.
Definition 3.2. Let θA be a representation of L on A. A pair (∂a,∂l)∈Der(A)×Der(L) is called compatible with respect to θA if Eq 3.2 holds.
Based on the previous works, we are ready to construct a Lie algebra and its representation on the third cohomology group. Set
GθA={(∂a,∂l)∈Der(A)×Der(L)|(∂a,∂l)iscompatiblewithrespecttoθA}. |
Lemma 3.3. There is a linear map Φ:GθA⟶End(H3(L,A)) given by
Φ(∂a,∂l)([ω])=[Obω(∂a,∂l)],foranyω∈Z3(L,A), | (3.20) |
where Obω(∂a,∂l) is given by Eq 3.1.
Proof. By Lemma 3.1 and (∂a,∂l) is compatible with respect to θA it follows that Obω(∂a,∂l) is a 3-cocycle whenever ω is a 3-cocycle. Therefore, we only need to show that if δλ is a 3-coboundary, then Φ(∂a,∂l)(δλ)=0, which means that Φ is well-defined.
(Φ(∂a,∂l)(δλ))(x,y,z)=(∂a(δλ)−(δλ)(∂l⊗id⊗id)−(δλ)(id⊗∂l⊗id)−(δλ)(id⊗id⊗∂l))(x,y,z)=∂a(−λ([x,y,z])+θA(y,z)(λ(x))−θA(x,z)(λ(y))+DA(x,y)(λ(z)))−(−λ([∂l(x),y,z])+θA(y,z)(λ(∂l(x)))−θA(∂l(x),z)(λ(y))+DA(∂l(x),y)(λ(z))))−(−λ([x,∂l(y),z])+θA(∂l(y),z)(λ(x))−θA(x,z)(λ(∂l(y)))+DA(x,∂l(y))(λ(z)))−(−λ([x,y,∂l(z)])+θA(y,∂l(z))(λ(x))−θA(x,∂l(z))(λ(y))+DA(x,y)(λ(∂l(z)))). |
Since ∂l is a derivation, we have
λ([∂l(x),y,z])+λ([x,∂l(y),z])+λ([x,y,∂l(z)])=λ(∂l([x,y,z])). |
Then
(Φ(∂a,∂l)(δλ))(x,y,z)=∂aθA(y,z)(λ(x))−∂aθA(x,z)(λ(y))+∂aDA(x,y)(λ(z))−θA(y,z)(λ(∂l(x)))+θA(∂l(x),z)(λ(y))−DA(∂l(x),y)(λ(z))−θA(∂l(y),z)(λ(x))+θA(x,z)(λ(∂l(y)))−DA(x,∂l(y))(λ(z))−θA(y,∂l(z))(λ(x))+θA(x,∂l(z))(λ(y))−DA(x,y)(λ(∂l(z)))−∂a(λ([x,y,z]))+λ(∂l([x,y,z]))=−θA(y,z)(λ(∂l(x)))+θA(x,z)(λ(∂l(y)))−DA(x,y)(λ(∂l(z)))+θA(y,z)∂a(λ(x))−θA(x,z)∂a(λ(y))+DA(x,y)∂a(λ(z))−∂a(λ([x,y,z]))+λ(∂l([x,y,z]))=δ(∂a∘λ−λ∘∂l)(x,y,z)=0, |
it is obvious that Φ is a linear map.
We will end this section with the following conclusion.
Theorem 3.4. Let θA be a representation of L on A. Then GθA is a Lie subalgebra of Der(A)×Der(L) and the map Φ given by Eq 3.20 is a Lie algebra homomorphism.
Proof. Note that GθA is a subalgebra of Der(A)×Der(L). First, we prove that if (∂a1,∂l1), (∂a2,∂l2)∈GθA, then the commutator [(∂a1,∂l1),(∂a2,∂l2)]∈GθA. For all x,y∈L, we have
(∂a1∂a2−∂a2∂a1)θA(x,y)−θA(x,y)(∂a1∂a2−∂a2∂a1)=∂a1(∂a2θA(x,y))−∂a2(∂a1θA(x,y))−θA(x,y)∂a1∂a2+θA(x,y)∂a2∂a1=∂a1(θA(x,y)∂a2+θA(∂l2(x),y)+θA(x,∂l2(y)))⏟I1−∂a2(θA(x,y)∂a1+θA(∂l1(x),y)+θA(x,∂l1(y)))⏟I2−θA(x,y)∂a1∂a2+θA(x,y)∂a2∂a1. | (3.21) |
By Eq 3.2 it follows that
I1=θA(x,y)∂a1∂a2+θA(∂l1(x),y)∂a2+θA(x,∂l1(y))∂a2+θA(∂l2(x),y)∂a1+θA(∂l1∂l2(x),y)+θA(∂l2(x),∂l1(y))+θA(x,∂l2(y))∂a1+θA(∂l1(x),∂l2(y))+θA(x,∂l1∂l2(y)), | (3.22) |
I2=θA(x,y)∂a2∂a1+θA(∂l2(x),y)∂a1+θA(x,∂l2(y))∂a1+θA(∂l1(x),y)∂a2+θA(∂l2∂l1(x),y)+θA(∂l1(x),∂l2(y))+θA(x,∂l1(y))∂a2+θA(∂l2(x),∂l1(y))+θA(x,∂l2∂l1(y)). | (3.23) |
Then taking Eqs 3.22 and 3.23 into Eq 3.21 gives that
(∂a1∂a2−∂a2∂a1)θA(x,y)−θA(x,y)(∂a1∂a2−∂a2∂a1)=θA((∂l1∂l2−∂l2∂l1)(x),y)+θA(x,(∂l1∂l2−∂l2∂l1)(y)), |
which implies that [(∂a1,∂l1),(∂a2,∂l2)] is compatible.
Next, for any (∂a1,∂l1), (∂a2,∂l2)∈GθA, [ω]∈H3(L,A), we have
[Φ(∂a1,∂l1),Φ(∂a2,∂l2)]([ω])=Φ(∂a1,∂l1)Φ(∂a2,∂l2)([ω])−Φ(∂a2,∂l2)Φ(∂a1,∂l1)([ω]). | (3.24) |
By Eqs 3.1 and 3.20, we have
Φ(∂a1,∂l1)Φ(∂a2,∂l2)([ω])=Φ(∂a1,∂l1)(∂a2ω−ω(∂l2⊗id⊗id)−ω(id⊗∂l2⊗id)−ω(id⊗id⊗∂l2))=[∂a1(∂a2ω−ω(∂l2⊗id⊗id)−ω(id⊗∂l2⊗id)−ω(id⊗id⊗∂l2))−(∂a2ω−ω(∂l2⊗id⊗id)−ω(id⊗∂l2⊗id)−ω(id⊗id⊗∂l2))(∂l1⊗id⊗id)−(∂a2ω−ω(∂l2⊗id⊗id)−ω(id⊗∂l2⊗id)−ω(id⊗id⊗∂l2))(id⊗∂l1⊗id)−(∂a2ω−ω(∂l2⊗id⊗id)−ω(id⊗∂l2⊗id)−ω(id⊗id⊗∂l2))(id⊗id⊗∂l1)]. |
Similarly, one obtains
Φ(∂a2,∂l2)Φ(∂a1,∂l1)([ω])=Φ(∂a2,∂l2)(∂a1ω−ω(∂l1⊗id⊗id)−ω(id⊗∂l1⊗id)−ω(id⊗id⊗∂l1))=[∂a2(∂a1ω−ω(∂l1⊗id⊗id)−ω(id⊗∂l1⊗id)−ω(id⊗id⊗∂l1))−(∂a1ω−ω(∂l1⊗id⊗id)−ω(id⊗∂l1⊗id)−ω(id⊗id⊗∂l1))(∂l2⊗id⊗id)−(∂a1ω−ω(∂l1⊗id⊗id)−ω(id⊗∂l1⊗id)−ω(id⊗id⊗∂l1))(id⊗∂l2⊗id)−(∂a1ω−ω(∂l1⊗id⊗id)−ω(id⊗∂l1⊗id)−ω(id⊗id⊗∂l1))(id⊗id⊗∂l2)]. |
Inserting the above two identities into Eq 3.24, we deduce
[Φ(∂a1,∂l1),Φ(∂a2,∂l2)]([ω])=[(∂a1∂a2−∂a2∂a1)ω−ω((∂l1∂l2−∂l2∂l1)⊗id⊗id)−ω(id⊗(∂l1∂l2−∂l2∂l1)⊗id)−ω(id⊗id⊗(∂l1∂l2−∂l2∂l1))]=[Obω(∂a1∂a2−∂a2∂a1,∂l1∂l2−∂l2∂l1)]=Φ(∂a1∂a2−∂a2∂a1,∂l1∂l2−∂l2∂l1)([ω])=Φ[(∂a1,∂l1),(∂a2,∂l2)]([ω]) |
as desired.
In this section, we construct obstruction classes for extensibility of derivations by Lemma 3.1. Also, for GθA we give a representation in terms of extensibility of derivations.
Lemma 4.1. Keep notation as above. The cohomology class [Obω(∂a,∂l)]∈H3(L,A) does not depend on the choice of sections of π. (Hence we will denote [Obω(∂a,∂l)] by [Ob˜L(∂a,∂l)] ).
Proof. Suppose that s1 and s2 are sections of π and ω1, ω2 are defined by Eq 2.4, while Obω1(∂a,∂l), Obω2(∂a,∂l) are defined by Eq 3.1 with respect to ω1, ω2. Then
Obω1(∂a,∂l)(x,y,z)−Obω2(∂a,∂l)(x,y,z)=∂a(ω1(x,y,z))−ω1(∂l(x),y,z)−ω1(x,∂l(y),z)−ω1(x,y,∂l(z))−∂a(ω2(x,y,z))+ω2(∂l(x),y,z)+ω2(x,∂l(y),z)+ω2(x,y,∂l(z))=∂a(ω1(x,y,z)−ω2(x,y,z))⏟I1−(ω1(∂l(x),y,z)−ω2(∂l(x),y,z))⏟I2−(ω1(x,∂l(y),z)−ω2(x,∂l(y),z))⏟I3−(ω1(x,y,∂l(z))−ω2(x,y,∂l(z)))⏟I4. | (4.1) |
A map λ:L⟶A is defined by λ(x)=s1(x)−s2(x), for all x∈L. Recall the proof of Corollary 2.7, we have
ω1(x,y,z)−ω2(x,y,z)=−λ([x,y,z])+θA(y,z)(λ(x))−θA(x,z)(λ(y))+DA(x,y)(λ(z)), |
for any x,y,z∈L. So
I1=∂a(−λ([x,y,z])+θA(y,z)(λ(x)))−θA(x,z)(λ(y))+DA(x,y)(λ(z))). |
Similarly, we have
I2=−λ([∂l(x),y,z])+θA(y,z)(λ(∂l(x)))−θA(∂l(x),z)(λ(y))+DA(∂l(x),y)(λ(z)), |
I3=−λ([x,∂l(y),z])+θA(∂l(y),z)(λ(x))−θA(x,z)(λ(∂l(y)))+DA(x,∂l(y))(λ(z)), |
and
I4=−λ([x,y,∂l(z)])+θA(y,∂l(z))(λ(x))−θA(x,∂l(z))(λ(y))+DA(x,y)(λ(∂l(z))). |
By I1, I2, I3, I4 and Eq 4.1 have
Obω1(∂a,∂l)(x,y,z)−Obω2(∂a,∂l)(x,y,z)=(∂aθA(y,z)−θA(∂l(y),z)−θA(y,∂l(z)))(λ(x))−(∂aθA(x,z)−θA(∂l(x),z)−θA(x,∂l(z)))(λ(y))+(∂aDA(x,y)−DA(∂l(x),y)−DA(x,∂l(y)))(λ(z))−θA(y,z)(λ(∂l(x)))+θA(x,z)(λ(∂l(y)))−DA(x,y)(λ(∂l(z)))−∂a(λ([x,y,z]))+λ(∂l([x,y,z])). |
Since (∂a,∂l) satisfies Eq 3.2 it follows that
Obω1(∂a,∂l)(x,y,z)−Obω2(∂a,∂l)(x,y,z)=θA(y,z)∂a(λ(x))−θA(x,z)∂a(λ(y))+DA(x,y)∂a(λ(z))−θA(y,z)(λ(∂l(x)))+θA(x,z)(λ(∂l(y)))−DA(x,y)(λ(∂l(z)))−∂a(λ([x,y,z]))+λ(∂l([x,y,z]))=δ(∂a∘λ−λ∘∂l)(x,y,z), |
i.e., [Obω1(∂a,∂l)]=[Obω2(∂a,∂l)]∈H3(L,A), so we have done the proof.
Next, we will define the extensibility of derivations, and we will give a necessary and sufficient condition that (∂a,∂l) is extensible.
Definition 4.2. Let ˜L be an abelian extension of a Lie triple system L by A. A pair (∂a,∂l)∈Der(A)×Der(L) is called extensible if there is a derivation ∂˜l∈Der(˜L) such that the diagram
![]() |
(4.2) |
is commutative, where ι:A⟶˜L is the inclusion map.
Proposition 4.3. Assume that ˜L is an abelian extension of a Lie triple system L by A. If (∂a,∂l)∈Der(A)×Der(L) is extensible, then (∂a,∂l) is compatible with respect to θA.
Proof. Since (∂a,∂l) is extensible, there is a derivation ∂˜l∈Der(˜L) such that the diagram (4.2) is commutative, i.e., ι∘∂a=∂˜l∘ι, hence there is a map μ:L⟶A given by
μ(x)=∂˜l(s(x))−s(∂l(x)). | (4.3) |
Since ι∘∂a=∂˜l∘ι, (equivalent, ∂˜l|A=∂a) and ∂˜l∈Der(˜L), we obtain
∂a(θA(x,y)(v))−θA(x,y)(∂a(v))=∂a([v,s(x),s(y)]˜L)−[∂a(v),s(x),s(y)]˜L=∂˜l([v,s(x),s(y)]˜L)−[∂a(v),s(x),s(y)]˜L=[∂˜l(v),s(x),s(y)]˜L+[v,∂˜l(s(x)),s(y)]˜L+[v,s(x),∂˜l(s(y))]˜L−[∂a(v),s(x),s(y)]˜L=[∂˜l(v),s(x),s(y)]˜L+[v,s(∂l(x)),s(y)]˜L+[v,μ(x),s(y)]˜L+[v,s(x),s(∂l(y))]˜L+[v,s(x),μ(y)]˜L−[∂a(v),s(x),s(y)]˜L=[v,s(∂l(x)),s(y)]˜L+[v,s(x),s(∂l(y))]˜L=θA(∂l(x),y)(v)+θA(x,∂l(y))(v), |
hence (∂a,∂l) is compatible.
Theorem 4.4. Let ˜L be an abelian extension of a Lie triple system L by A and (∂a,∂l)∈Der(A)×Der(L) compatible with respect to θA. Then (∂a,∂l) is extensible if and only if [Ob˜L(∂a,∂l)]∈H3(L,A) is trivial.
Proof. (⇒) Fix any linear section s of π. Since (∂a,∂l) is extensible, there exists a derivation ∂˜l∈Der(˜L) such that the associated diagram (4.2) is commutative. By π∘∂˜l=∂l∘π, we have ∂˜l(s(x))−s(∂l(x))∈A, for x∈L, then there exists a map μ:L⟶A given byEq 4.3. It sufficient to show that
Ob˜L(∂a,∂l)(x,y,z)=(δμ)(x,y,z), | (4.4) |
for all x,y,z∈L. Next, we compute both hand sides of the following identity
∂˜l([s(x1)+v1,s(x2)+v2,s(x3)+v3]˜L)=[∂˜l(s(x1)+v1),s(x2)+v2,s(x3)+v3]˜L+[s(x1)+v1,∂˜l(s(x2)+v2),s(x3)+v3]˜L+[s(x1)+v1,s(x2)+v2,∂˜l(s(x3)+v3)]˜L, | (4.5) |
for any x1,x2,x3∈L, v1,v2,v3∈A.
At first, since ˜L is an abelian extension of L by A, we have [˜L,A,A]=0 and
[s(x1)+v1,s(x2)+v2,s(x3)+v3]˜L=[s(x1),s(x2),s(x3)]˜L+[s(x1),s(x2),v3]˜L+[s(x1),v2,s(x3)]˜L+[v1,s(x2),s(x3)]˜L=[s(x1),s(x2),s(x3)]˜L+θA(x2,x3)(v1)−θA(x1,x3)(v2)+DA(x1,x2)(v3), |
by Eq 2.4 the left-hand side of Eq 4.5 is
LHSofEq4.5=∂˜l([s(x1),s(x2),s(x3)]˜L+θA(x2,x3)(v1)−θA(x1,x3)(v2)+DA(x1,x2)(v3))=∂˜l(s([x1,x2,x3]L)+ω(x1,x2,x3)+θA(x2,x3)(v1)−θA(x1,x3)(v2)+DA(x1,x2)(v3)). |
Since the diagram (4.2) is commutative, we have
LHSofEq4.5=s(∂l([x1,x2,x3]L))+μ([x1,x2,x3]L)+∂a(ω(x1,x2,x3))+∂a(θA(x2,x3)(v1))−∂a(θA(x1,x3)(v2))+∂a(DA(x1,x2)(v3))=s([∂l(x1),x2,x3]L)+s([x1,∂l(x2),x3]L)+s([x1,x2,∂l(x3)]L)+μ([x1,x2,x3]L)+∂a(ω(x1,x2,x3))+∂a(θA(x2,x3)(v1))−∂a(θA(x1,x3)(v2))+∂a(DA(x1,x2)(v3)). | (4.6) |
Now we compute the right-hand side of Eq 4.5. Note that, since ∂˜l|A=∂a, it holds that
∂˜l(s(xi)+vi)=∂˜l(s(xi))+∂a(vi)=∂˜l(s(xi))−s(∂l(xi))+s(∂l(xi))+∂a(vi)=s(∂l(xi))+μ(xi)+∂a(vi)∈s(L)⊕A. |
By this the right-hand side of Eq 4.5 is
RHSofEq4.5=[s(∂l(x1))+μ(x1)+∂a(v1),s(x2)+v2,s(x3)+v3]˜L+[s(x1)+v1,s(∂l(x2))+μ(x2)+∂a(v2),s(x3)+v3]˜L+[s(x1)+v1,s(x2)+v2,s(∂l(x3))+μ(x3)+∂a(v3)]˜L=[s(∂l(x1)),s(x2),s(x3)]˜L+[s(∂l(x1)),s(x2),v3]˜L+[s(∂l(x1)),v2,s(x3)]˜L+[μ(x1),s(x2),s(x3)]˜L+[s(x1),μ(x2),s(x3)]˜L+[∂a(v1),s(x2),s(x3)]˜L+[s(x1),s(∂l(x2)),v3]˜L+[v1,s(∂l(x2)),s(x3)]˜L+[s(x1),∂a(v2),s(x3)]˜L+[s(x1),s(x2),s(∂l(x3))]˜L+[s(x1),s(x2),μ(x3)]˜L+[s(x1),s(x2),∂a(v3)]˜L+[s(x1),v2,s(∂l(x3))]˜L+[v1,s(x2),s(∂l(x3))]˜L+[s(x1),s(∂l(x2)),s(x3)]˜L. | (4.7) |
By Eqs 4.6 and 4.7, we have
s([∂l(x1),x2,x3]L)+s([x1,∂l(x2),x3]L)+s([x1,x2,∂l(x3)]L)+μ([x1,x2,x3]L)+∂a(ω(x1,x2,x3))+∂a(θA(x2,x3)(v1))−∂a(θA(x1,x3)(v2))+∂a(DA(x1,x2)(v3))=[s(∂l(x1)),s(x2),s(x3)]˜L+DA(∂l(x1),x2)(v3)−θA(∂l(x1),x3)(v2)+θA(x2,x3)(μ(x1))+θA(x2,x3)(∂a(v1))+[s(x1),s(∂l(x2)),s(x3)]˜L+DA(x1,∂l(x2))(v3)−θA(x1,x3)(μ(x2))−θA(x1,x3)(∂a(v2))+θA(∂l(x2),x3)(v1)+[s(x1),s(x2),s(∂l(x3))]˜L+DA(x1,x2)(μ(x3))+DA(x1,x2)(∂a(v3))−θA(x1,∂l(x3))(v2)+θA(x2,∂l(x3))(v1). |
Then
0=−ω(∂l(x1),x2,x3)−ω(x1,∂l(x2),x3)−ω(x1,x2,∂l(x3))+∂a(ω(x1,x2,x3))−DA(x1,x2)(μ(x3))−θA(x2,x3)(μ(x1))+θA(x1,x3)(μ(x2))+μ([x1,x2,x3]L)+(∂aDA(x1,x2)−DA(x1,x2)∂a−DA(∂l(x1),x2)−DA(x1,∂l(x2)))(v3)+(∂aθA(x2,x3)−θA(x2,x3)∂a−θA(∂l(x2),x3)−θA(x2,∂l(x3)))(v1)−(∂aθA(x1,x3)−θA(x1,x3)∂a−θA(∂l(x1),x3)−θA(x1,∂l(x3)))(v2). |
Since (∂a,∂l) is compatible and by the proof of Lemma 3.1,
(∂aDA(x1,x2)−DA(x1,x2)∂a−DA(∂l(x1),x2)−DA(x1,∂l(x2)))(v3)=0,(∂aθA(x2,x3)−θA(x2,x3)∂a−θA(∂l(x2),x3)−θA(x2,∂l(x3)))(v1)=0,(∂aθA(x1,x3)−θA(x1,x3)∂a−θA(∂l(x1),x3)−θA(x1,∂l(x3)))(v2)=0. | (4.8) |
Thus we have
∂a(ω(x1,x2,x3))−ω(∂l(x1),x2,x3)−ω(x1,∂l(x2),x3)−ω(x1,x2,∂l(x3))−DA(x1,x2)(μ(x3))−θA(x2,x3)(μ(x1))+θA(x1,x3)(μ(x2))+μ([x1,x2,x3]L)=0, | (4.9) |
which is exactly Eq 4.4 due to Eq 3.1 and the definition of 3-cohomology group. So [Ob˜L(∂a,∂l)]=0 as required.
(⇐) Assume that [Ob˜L(∂a,∂l)] is trivial. Then there is a map μ:L⟶A such that Ob˜L(∂a,∂l)=δμ. For any s(x)+v∈˜L, define ∂˜l:˜L⟶˜L by
∂˜l(s(x)+v)=s(∂l(x))+μ(x)+∂a(v), |
then the associated diagram in (4.2) is commutative: for any x∈L, v∈A,
(π∘∂˜l)(s(x)+v)=π(s(∂l(x))+μ(x)+∂a(v))=∂l(x)=(∂l∘π)(s(x)+v);∂˜l∘ι(v)=∂˜l(v)=∂a(v)=ι∘∂a(v). |
Moreover, since (∂a,∂l) satisfies Eq 3.2, by Eqs 4.8 and 4.9, it follows that Eq 4.5 holds by Eqs 4.6 and 4.7, that is, ∂˜l∈Der(˜L).
Hence, for a compatible pair (∂a,∂l)∈Der(A)×Der(L), the cohomology class [Ob˜L(∂a,∂l)] can be regarded as an obstruction class to extensibility.
By Proposition 4.3 and Theorem 4.4 we have the following result.
Corollary 4.5. Let ˜L be an abelian extension of a Lie triple system L by A. If H3(L,A)=0, then any pair (∂a,∂l)∈Der(A)×Der(L) is compatible if and only if it is extensible.
We know that the condition H3(L,A)=0 is in general not equivalent to split property of extensions. However, we still have the following result.
Corollary 4.6. Let ˜L be a split abelian extension of a Lie triple system L by A. Then any pair (∂a,∂l)∈Der(A)×Der(L) is compatible if and only if it is extensible.
Proof. (⇐) It holds due to Proposition 4.3.
(⇒) Since the extension is split there exists a section s′, which is a homomorphism. Let θs′, (resp. ωs′) be defined by Eq 2.3 (resp. Eq 2.4) with respect to s′. Then we get ωs′=0. By Eq 3.1, we have Obωs′(∂a,∂l)=0. In view of Lemmas 3.1 and 4.1, one has [Ob˜L(∂a,∂l)]=[Obωs′(∂a,∂l)]=0. Then by Theorem 4.4, we deduce that (∂a,∂l) is extensible as required.
Let us end this section with the relation between the representation Φ of GθA and extensibility of derivations.
Definition 4.7. Let θA be a representation of L on A and ˜L an abelian extension of Lie triple systems L by A. So there exists a section s of π such that the representation θs defined by Eq 2.3 is the same as θA, then the extension is called reversible with respect to θA.
Example 4.8. Let θA be a representation of L on A. By Proposition 2.8 we have a Lie triple system ˜L:=L⊕A with the bracket given by
[x+u,y+v,z+w]˜L=[x,y,z]L+θA(y,z)(u)−θA(x,z)(v)+DA(x,y)(w), |
where x,y,z∈L, u,v,w∈A. By Corollary 2.9, we have an abelian extension
0⟶A⟶˜Lp→L⟶0, |
and p is the canonical projection. Choose a section s of p given by s(x)=x, x∈L. Then s:L⟶˜L is a homomorphism, and hence the representation θs given by Eq 2.3 is the same as θA which means that the extension
0⟶A⟶˜Lp→L⟶0 |
is reversible with respect to θA.
Theorem 4.9. Let θA be a representation of L on A. The pair (∂a,∂l)∈GθA is extensible in every reversible extension if and only if Φ(∂a,∂l)=0.
Proof. (⇒) For any [φ]∈H3(L,A), by Corollary 2.9, there is an abelian extension
0⟶A⟶˜Lp→L⟶0, |
where p is the canonical projection and the bracket on ˜L:=L⊕A is given by
[x+u,y+v,z+w]˜L=[x,y,z]L+φ(x,y,z)+θA(y,z)(u)−θA(x,z)(v)+DA(x,y)(w), |
for any x,y,z∈L, u,v,w∈A. Choose a section s of p defined by s(x)=x, for any x∈L. Then the representation θs is given by Eq 2.3. Let Hnθs(L,A) denote the cohomology group with respect to θs. We first show that θs is the same as θA for x,y∈L, v∈A,
θs(x,y)(v)=[v,s(x),s(y)]˜L=[v,x,y]˜L=θA(x,y)(v). |
Hence we obtain that the cohomology group Hnθs(L,A) associated to θs is the same as Hn(L,A). Since we define ωs by Eq 2.4 (resp. φ) is a 3-cocycle in Hnθs(L,A) (resp. in Hn(L,A)), we have [ωs]=[φ]. Then
Φ(∂a,∂l)([φ])=Φ(∂a,∂l)([ωs])(byLemma3.3)=[Ob˜L(∂a,∂l)]=0.(byTheorem4.4) |
(⇐) Suppose Φ(∂a,∂l)=0. For any reversible abelian extension
0⟶A⟶˜Lp→L⟶0, |
there exists a section s of π such that the representation θs is the same as θA. Therefore, ωs defined by Eq 2.4 is a 3-cocycle in H3(L,A). Then we have
[Ob˜L(∂a,∂l)]=Φ(∂a,∂l)([ωs])=0. |
By Theorem 4.4, (∂a,∂l) is extensible. This completes the proof.
The following corollary is straightforward.
Corollary 4.10. Let θA be a representation of L on A. Then any pair (∂a,∂l)∈GθA is extensible in every reversible extension if and only if Φ≡0.
The second author is supported by NNSF of China (Nos. 11801066, 11771410). The third author is supported by NNSF of China (Nos.11771069, 12071405).
The authors declare there is no conflicts of interest.
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