aa r X i v : . [ m a t h . K T ] A p r A SUBCOMPLEX OF LEIBNIZ COMPLEX
TEIMURAZ PIRASHVILI
Abstract.
Using the free graded Lie algebras we introduce a natural subcom-lex of the Loday’s complex of a Leibniz algebra. Introduction
This note, except for the introduction and the second half of Section 3, werewritten in 2002 during my stay in Nantes, where I was invited by Vincent Franjou.Let h be a Leibniz algebra and let CL ∗ ( h ) be the chain compex constructed byLoday [1]. The aim is to intriduce a natural subcomplex L ( h , ∗ of CL ∗ ( h ), whichconjecturaly computes the Quillen derived functors of the left adjoint functor to theinclusion Lie Algebras ⊂ Leibniz Algebras . The paper also includes a conjecture of Jean-Luis Loday (see Section 3).We refer the reader to [6] about other conjectures on Leibniz homology.2.
Definition of the subcomplex
We are working over a field K of characteristic zero.A Leibniz algebra h is a vector space equipped with a bilinear operation {− , −} : h ⊗ h → h satisfying the Leibniz identity { x, { y, z }} = {{ x, y } , z } − {{ x, z } , y } . It is clear that Lie algebras are Leibniz algebras for which the antisymmetry condi-tion { x, y } + { y, x } = 0 holds.We refer the reader to [3], [5] and [4] for more on Leibniz algebras, Leibnizhomology and Leibniz representations.For any Leibniz algebra h we let h Lie be the quotient of h by the relation { x, y } + { y, x } = 0. In this way one obtains the functor( − ) Lie : Leibniz algebras → Lie algebras which is the left adjoint of the inclusion functor
Lie algebras ⊂ Leibniz algebras . For Leibniz algebras we have
Leibniz homology introduced by Jean-Louis Lodayin [1], see also [3]. For a Lie algebra h , the Leibniz homology (denoted by HL ∗ ( h ))is defined as the homology of the complex CL ∗ ( h ). By definition CL n ( h ) = h ⊗ n ,while the Loday boundary is given by d ( x ⊗ · · · ⊗ x n ) = X i 1) denote the tensor algebra on V , where V is concentrated in the degree 1. In other words T ( V, 1) = M n ≥ T ( V, n , T ( V, n = V ⊗ n , n ≥ . We let L ( V, 1) = L n ≥ L ( V, n , be the graded Lie subalgebra of T ( V, 1) gener-ated by V . The Lie algebra L ( V, 1) is called free graded Lie algebra generatedby V concentrated in the degree 1 . It is clear that L ( V, ∼ = V and the map x ⊗ y [ x, y ] yields the isomorphism S V ∼ = L ( V, , where S denotes the secondsymmetric power. The vector space L ( V, n is spanned by the elements of the form[ x , [ x , [ · · · [ x n − , x n ] · · · ] with x , · · · , x n ∈ V . For simplicity we denote this typeof element by [[ x , · · · , x n ]].We let i n : L ( V, n ⊂ T ( V, n = V ⊗ n be the inclusion. The main observationof this note is the fact that if h is a Leibniz algebra, then L ( h , ∗ is closed underLoday boundary map, and therefore L ( h , ∗ is a subcomplex of CL ∗ ( h ). In otherwords one has the following Proposition 2.1. Let h be a Leibniz algebra. Then d ( i n ( L ( h , n )) ⊂ i n − ( L ( h , n − ) . In order to describe the induced boundary map δ : L ( h , n → L ( h , n − , weneed the linear map p n : T ( V, n → L ( V, n given by p n ( x ⊗ · · · ⊗ x n ) = [[ x , · · · , x n ]] . Proposition 2.2. Let h be a Leibniz algebra. Then, for any element ω ∈ L ( h , n +1 one has di n +1 ( ω ) = ( − n p n ◦ f n ◦ i n +1 ( ω ) , where f n : h ⊗ n +1 → h ⊗ n is the linear map given by f n ( x ⊗ · · · ⊗ x n ⊗ x n +1 ) = x ⊗ · · · ⊗ x n − ⊗ { x n , x n +1 } . We give the proof of Proposition 2.1 and Proposition 2.2 in Section 5.As a corollary we get for any Leibniz algebra g a chain complex L ( h , ∗ := ( · · · → L ( h , n +1 δ n −→ L ( h , n δ n − −−−→ · · · δ −→ L ( h , δ −→ L ( h , ) SUBCOMPLEX OF LEIBNIZ COMPLEX 3 The first nontrivial boundary maps are given by δ ([ x, y ]) = { x, y } + { y, x } ,δ ([ x, [ y, z ]]) = [ x, { y, z } ] + [ x, { z, y } ] − [ y, { z, x } ] − [ z, { y, x } ] δ ([ x, [ y, [ z, u ]]]) = [ x, [ y, { z, u } ]] + [ x, [ y, { u, z } ]] − [ x, [ u, { z, y } ]] − [ x, [ z, { u, y } ]]+ [ y, [ u, { z, x } ]] + [ y, [ z, { u, x } ]] − [ z, [ u, { y, x } ]] − [ u, [ z, { y, x } ]]We let Lie n ( h ), n ≥ L ( h , ∗ . Lemma 2.3. For any Leibniz algebra h the group Lie ( h ) is isomorphic to theLiezation of h : Lie ( h ) ∼ = h Lie Two conjectures The fisrt one says: Conjecture 1 . If h is a free Leibniz algebra, then Lie n ( h ) = 0 for n > Lie ∗ ( h ) computes the Quillen derived functors ofthe functor ( − ) Lie : Leibniz Algebras → Lie Algebras . Since 2002, I have had the oportunity to discuss the conjecture with severalpeople, including Jean-Luis Loday, Michael Robinson and Christine Vespa, butunfortunately, our attemts to prove the conjecture failed.The second one claims that our chain complex is only a part of a big pichture: Conjecture 2 . (J.-L. Loday. 2006 [2] ). Let e ( i ) n be the image of the i -th Eulerianidempotent on g ⊗ n . Then d ( e ( i ) n ) ⊂ e (1) n − ⊕ · · · ⊕ e ( i ) n − . For definition of the Eulerian idempotents see [1, pp 142-144].4. A generalization of Wigner’s identity In this section V denotes a graded vector space and ⊗ denotes the tensor productof graded vector spaces. Any linear map D : V → V has an extension as a linearmap D n : T ( V ) → T ( V ) given by D n ( v ⊗ · · · ⊗ v n ) = n X i =1 v ⊗ · · · ⊗ D ( v i ) ⊗ · · · ⊗ v n . In particular D (1) = 0. Moreover D n ( ω ) = nω , provided D = Id . Here ω ∈ T n ( V ) . Let us recall that T ( V ) is a graded Hopf algebra with following coproduct:∆( v ) = v ⊗ ⊗ v, v ∈ V. We let µ be the multiplication map T ( V ) ⊗ T ( V ) → T ( V ) and we let L ( V ) be theLie subalgebra of T ( V ) generated by V . Let p D : T n ( V ) → L ( V ) n be the map given by p D ( v ⊗ · · · ⊗ v n ) = [ v , [ v , · · · , [ v n − , Dv n ] · · · ] . Proposition 4.1. For any ω ∈ T n ( V ) one has µ ◦ ( p D ⊗ Id T ( V ) ) ◦ ∆( ω ) = D n ( ω ) TEIMURAZ PIRASHVILI Let us observe that Proposition 4.1 for D = Id is a result of [7]. The proof inour setup is completely similar to one given in loc.cit.Proof . The result is obvious for n = 1 and therefore we can use the induction on n . For an element ω of T n ( V ) we use the notation ω = v · · · v n , v , · · · , v n ∈ V . Wemay write ∆( v · · · v n ) = 1 ⊗ v · · · v n + P i a i ⊗ b i with each a i ∈ L i ≥ T i ( V ). Theinduction hypothesis gives P i p D ( a i ) b i = P k ≥ v · · · D ( v k ) · · · v n . On the otherhand, we have p D ( vτ ) = [ v, p D ( τ )] = vp D ( τ ) − ( − | τ | p D ( τ ) v. We have also∆( ω ) = ( v ⊗ ⊗ v )(1 ⊗ v · · · v n + X i a i ⊗ b i )= v ⊗ v · · · v n + X i v a i ⊗ b i + 1 ⊗ ω + ( − | a i | X i a i ⊗ v b i . Thus µ ( p D ⊗ Id )∆( ω ) = ( Dv ) v · · · v n + X i p D ( v a i ) b i + +( − | a i | X i p D ( a i ) v b i = ( Dv ) v · · · v n + X i v p D ( a i ) b i = ( Dv ) v · · · v n + v X k ≥ v · · · D ( v k ) · · · v n ✷ Corollary 4.2. For any ω ∈ L ( V ) n one has p D ( ω ) = D n ( ω ) . Proof . If ω ∈ L ( V ) n then ∆( ω ) = ω ⊗ ⊗ ω . Thus the result is a directconsequence of Proposition 4.1. ✷ Proof of Propositions 2.1 and 2.2 In this section we let δ = p n ◦ f n ◦ i n +1 : L ( h , n +1 → L ( h , n . We will provethe following Proposition 5.1. di n +1 = ( − n i n δ. Since i k is injective for all k , this implies both Propositions 2.1 and 2.2.We start with general remarks. Let g be a Lie algebra and let V be a left g -module. For x ∈ V and g ∈ g we let { x, g } be the result of the action of g on x . Itis well known that g acts also on V ⊗ n by { x ⊗ · · · ⊗ x n , g } := n X i =1 x ⊗ · · · ⊗ { x i , g } ⊗ · · · ⊗ x n . Lemma 5.2. One has the following identity: { [[ x , · · · , x n ]] , g } = n X i =1 [[ x , · · · , { x i , g } , · · · , x n ]] Proof . For n = 2 we have { [ x , x ] , g } = { x ⊗ x + x ⊗ x , g } = { x , g } ⊗ x + x ⊗ { x , g } + { x , g } ⊗ x + x ⊗ { x , g } = [ { x , g } , x ] + [ x , { x , g } ] . For n ≥ x , · · · , x n ]] = [ x , [[ x , · · · , x n ]] and Lemma follows by theinduction. Corollary 5.3. For any n ≥ the vector space L ( V, n is a g -submodule of T ( V, n = V ⊗ n and the map p n is g -linear. SUBCOMPLEX OF LEIBNIZ COMPLEX 5 Let h be a Leibniz algebra. Since { x, { y, z } + { y, z }} = 0, it follows that the Liealgebra g := h Lie acts on h via { x, ¯ y } := { x, y } . Here ¯ y denotes the class of y ∈ h in g . Let us observe that f ( x, y ) = { x, y } defies a g -linear map f : h ⊗ h → h . Thus f n is also a g -linear map and as a consequence δ = p n ◦ f n ◦ i n +1 is aslo a g -linear. Lemma 5.4. For any x ∈ h and ω ∈ L ( h , n − one has δ ([ x, ω ]) = [ x, δω ] − ( − | ω | { ω, ¯ x } Proof . We have δ ([ x, ω ]) − [ x, δω ] = − ( − | ω | p n − f n − ( i n − ω ⊗ x ) . We let D : h → h be the map given by D ( v ) = { v, x } . Since f n − = Id h ⊗ n − ⊗ f ,we have p n − ( Id h ⊗ n − ⊗ f )( i n − ω ⊗ x ) = p D ( ω )and Lemma is a direct consequence of Corollary 4.2. ✷ For a linear map D : V → V we let ˜ D n : V ⊗ n → V ⊗ n be the map given by˜ D n ( v ⊗ · · · ⊗ v n ) = n X i =1 ( − i D ( v i ) ⊗ v ⊗ · · · ⊗ ˆ v i ⊗ · · · ⊗ v n . Lemma 5.5. For any ω ∈ L ( V, n , n ≥ and any linear map D : V → V one has ˜ D n ( ω ) = 0 Proof . We work by induction. If n = 2 we have˜ D ( x ⊗ y + y ⊗ x ) = − D ( x ) ⊗ y + D ( y ) ⊗ x − D ( y ) ⊗ x + D ( x ) ⊗ y = 0 . If n > ω = [ x, τ ], where τ ∈ L ( V, n − . Thus ω = x ⊗ τ − ( − | τ | τ ⊗ x and therefore we have˜ D n ( ω ) = − D ( x ) ⊗ τ − j ( x )( ˜ D n − ( τ )) − ( − | τ | ˜ D n − ( τ ) ⊗ x + D ( x ) ⊗ τ = 0thanks to the induction assumption. Here for any x ∈ V we let j ( x ) : V ⊗ n − → V ⊗ n be the map given by y ⊗ · · · y n − 7→ ⊗ y ⊗ x ⊗ · · · ⊗ y n − . ✷ The following lemma is an immediate consequence of the definitions (comparealso with formula (10.6.3.1) of [1]). Lemma 5.6. For any x ∈ h and ω ∈ V ⊗ n one has d ( ω ⊗ x ) = d ( ω ) ⊗ x + ( − | ω | { ω, ¯ x } and d ( x ⊗ ω ) = − x ⊗ d ( ω ) + ˜ D | ω | ( ω ) , for D = { x, −} : h → h . Lemma 5.7. For any x ∈ h and ω ∈ L ( h , n one has d ([ x, ω ]) = − [ x, dω ] − { ω, ¯ x } . Proof . Since [ x, ω ] = x ⊗ ω − ( − n ω ⊗ x , we have d ([ x, ω ]) = d ( x ⊗ ω − ( − n ω ⊗ x )By the previous Lemma, this expresion can be rewriten: − x ⊗ dω + ˜ D n ( ω ) − ( − n dω ⊗ x − ( − n ( − n { ω, ¯ x } Thanks to Lemma 5.5 ˜ D ( ω ) = 0 and we get the result. ✷ Now we are in position to prove Proposition 5.1. Proof Proposition 5.1 . We use the induction on n . First consider the case n = 1.So we may assume that ω = [ x, y ] = x ⊗ y + y ⊗ x . Hence di ( ω ) = d ( x ⊗ y + y ⊗ x ) = −{ x, y } − { y, x } = − δ ([ x, y ]) . TEIMURAZ PIRASHVILI If n > 1, then we may assume that ω = [ x, τ ], with τ ∈ L ( V, n . Thanks to Lemma5.7 we have di n +1 ( ω ) = d ([ x, τ ]) = − [ x, dτ ] − { τ, ¯ x } and the induction assumption gives di n +1 ( ω ) = ( − n [ x, δτ ] − { τ, ¯ x } and the result follows from Lemma 5.4. References [1] J.-L. Loday . Cyclic homology, Grundl. Math. Wiss. Bd. , 2 nd edition, Springer1998.[2] J.-L. Loday Emeil dated 07.07.06.[3] J. L. Loday and T. Pirashvili . 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