an:06484529 Zbl 1350.17004 B??r??, C??me J. A.; Ouedraogo, M. Fran??oise; Pilabr??, Nakelgbamba B. On the existence of ad-nilpotent elements EN Afr. Mat. 26, No. 5-6, 813-823 (2015). 00348126 2015
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17A32 17B30 Lie algebras; Leibniz algebras; ad-nilpotent; derivation; anti-derivation; module For an arbitrary element $$x$$ of a Leibniz algebra $$L$$ consider two operators: $\mathrm{ad}_x: L\rightarrow L,\quad y \rightarrow [y,x]$ and $\mathrm{Ad}_x:L \rightarrow L,\quad y\rightarrow [x,y].$ Clearly, $$\mathrm{ad}_x$$ and $$\mathrm{Ad}_x$$ are derivation and anti-derivation of $$L$$, respectively. For these operators the following relations hold true: \begin{aligned} \mathrm{ad}_{[x,y]} & =\mathrm{ad}_y \mathrm{ad}_x -\mathrm{ad}_x \mathrm{ad}_y,\\ \mathrm{Ad}_{[x,y]} & =\mathrm{ad}_y \mathrm{Ad}_x - \mathrm{Ad}_x \mathrm{ad}_y,\\ \mathrm{Ad}_{[x,y]} & =\mathrm{ad}_y \mathrm{Ad}_x + \mathrm{Ad}_x \mathrm{Ad}_y,\\ 0 & =\mathrm{Ad}_x \mathrm{Ad}_y +\mathrm{Ad}_x \mathrm{ad}_y.\end{aligned} A (bi)module $$M$$ over a Leibniz algebra $$L$$ is a vector space with two (left $$l$$ and right $$r$$) actions, satisfying the above relations. Let $$L$$ be a Leibniz algebra and $$M$$ be $$L$$-(bi)module. We denote by $$\mathrm{Ess}(M)$$ the subspace of $$M$$ spanned by elements of the type $$l_x(v)+r_x(v)=xv+vx$$ for all $$(x,v)\in L \times M.$$ In the present paper the authors prove the invariance of $$\mathrm{Ess}(L)$$ under derivations of $$L$$ and $$\mathrm{Ess}(L)\subseteq \mathrm{Ker } \widetilde{D}$$ for any anti-derivation $$\widetilde{D}$$ of $$L$$. Moreover, the embedding $$[L_\lambda, L_\mu]\subseteq L_{\lambda+\mu}+\mathrm{Ess}(L)$$, where $$L_\lambda, L_\mu$$ are weight spaces with respect to a given anti-derivation, is established. The main results of the paper is the following: Theorem. Let $$L$$ be a Leibniz algebra over an algebraically closed field. Let $$X$$ be a non-empty subset of $$L$$ such that for every $$x\in X$$, all eigenvectors of $$\mathrm{ad}_x$$ (correspondingly, of $$\mathrm{Ad}_x$$) lie in $$X$$. Then $$\mathrm{ad}_y$$ (correspondingly, $$\mathrm{Ad}_x$$) is nilpotent for some $$y\in X$$. Sh. A. Ayupov (Tashkent)