A counterexample to a Proposition of Feldvoss-Wagemann and Burde-Wagemann
aa r X i v : . [ m a t h . K T ] A ug A COUNTEREXAMPLE TO A PROPOSITION OFFELDVOSS-WAGEMANN AND BURDE-WAGEMANN
T. PIRASHVILI Introduction
In what follows g is a finite dimensional Lie algebra over the field of complexnumbers. Around 1990 J.-L. Loday discovered that the the Chevalley-Eilenberg[2] boundary operator d : Λ n ( g ) → Λ n − ( g ) has a canonical lift as the boundaryoperator δ : g ⊗ n → g ⊗ n − , such that the canonical quotient map g ⊗ n → Λ n g defines a morphism of chain complexes · · · δ / / g ⊗ (cid:15) (cid:15) δ / / g ⊗ δ / / (cid:15) (cid:15) g id (cid:15) (cid:15) · · · d / / Λ ( g ) d / / Λ ( g ) d / / g Here d and δ are given by d ( x ∧ · · · ∧ x n ) = X ≤ i
0. Whether the converse is also true is ourweak conjecture [10]. The strong conjecture of [10] says that g is a semi-simple iff HL i ( g ) = 0 for all i >> Notations
By abuse of notation, the adjoint representation of g is also denoted by g . Denoteby r the solvable radical of g and put s = g / r . So, s is either trivial or semi-simpleLie algebra. In any case, one has a split short exact sequence of Lie algebras0 → r → g → s → . By choosing such a section, we obtain an action of s on r . By functoriality s actalso on H ∗ ( r ) and H ∗ ( r , r ). In general these actions are highly nontrivial.For a vector space V , we let V ♯ the dual of V . Lemma 1.
For a Lie algebra g the following conditions are equivalent:(i) The restriction map g ♯ → r ♯ yields an isomorphism HL ∗ ( g , g ♯ ) → HL ∗ ( g , r ♯ ) (ii) The canonical map HL ∗ ( g , r ) → HL ∗ ( g , g ) is an isomorphism(iii) One has HL ∗ ( g , s ) = 0 .(iv) One has H ∗ ( g , s ) = 0 .(v) One has H ( s , H ∗ ( r ) ⊗ s )) = 0 . Proof.
The equivalence (i) ⇐⇒ (ii) follows from the fact that the cohomologies aredual vector spaces of homologies.The equivalence (ii) ⇐⇒ (iii) follows from the homological long exact sequenceassociated to the short exact sequence of g -modules 0 → r → g → s → ⇒ (iv). Assume HL ∗ ( g , s ) = 0. Then by Theorem A of [9]we have H rel ∗ ( g , s ) = 0 and thus by Proposition 1 of [9] we have HL i ( g , s ) = H i ( g , s )for i ≥
2. Since these groups are also isomorphic for i = 0 , H ∗ ( g , s ) = 0.To show (iv) = ⇒ (iii), assume H ∗ ( g , s ) = 0. We will prove by induction on i that HL n ( g , s ) = 0. This is obvious for n = 0 , HL i ( g , s ) = 0 for all0 ≤ i ≤ n. By Theorem A [9] we obtain HL reli ( g , s ) = 0 for all 0 ≤ i ≤ n . Hence byProposition 1 [9] the canonical map HL i ( g , s ) → H i ( g , s ) is an isomorphism for all0 ≤ i ≤ n + 2 and the induction step works. Thus HL ∗ ( g , s ) = 0.To show (iv) ⇐⇒ (v), We consider the Lie algebra extension 0 → r → g → s → s is semi-simple or zero, we can use the homological version of Theorem 13of [4] to obtain H n ( g , s ) = M p + q = n H p ( s ) ⊗ H ( s , H q ( r , s ))Since the action of r on s is trivial, we can rewrite H n ( g , s ) = M p + q = n H p ( s ) ⊗ H ( s , H q ( r ) ⊗ s )) . This show that v) imply iv). Since H ( s ) one-dimensional, we see that the conditioniv) also imply v). (cid:3) A counterexample
Proposition 3.1 of [1] claims that for any g and p ≥ HL p ( g ) ∼ = HL p − ( g , r ♯ ) . We will introduce an example, which shows that this holds not always.Since HL ∗ +1 ( g ) ∼ = HL ∗ ( g , g ♯ ) holds always, Proposition 3.1 of [1] is equivalentto the claim that the equivalent conditions of Lemma 1 holds for all g . Now weconsider the six dimensional Lie algebra, for which this is not the case.We take s = sl and r to be the abelian Lie algebra, which as a module over s isthe adjoint representation. Thus g is the semi-direct product r ⋊ s . Since H ( r ) = s , COUNTEREXAMPLE TO A PROPOSITION OF FELDVOSS-WAGEMANN AND BURDE-WAGEMANN3 the Killing form defines a nontrivial s -module map H ( r ) ⊗ s → C . Hence g doesnot satisfy the condition v). References [1]
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