# which Lie algebra admit a left symmetric algebra structure?

Hello, I have a some questions, but first I must point out some definitions. A left symmetric algebra LSA is given a a vector space $V$ equipped with a bilinear product $\cdot$, which is not necessarily associative but satisfies: $$(x\cdot y)\cdot z-x\cdot(y\cdot z)=(y\cdot x)\cdot z-y\cdot(x\cdot z)$$ to such algebra is associated a Lie algebra structure given by the bracket $$[x,y]=x\cdot y-y\cdot x$$ This algebraic structure of LSA is important in geometry: 1) Affine structures: if a manifold $M$ is equipped with a linear connection $\nabla$ which is flat and torsion-free then the space of its vector fields $\mathcal{X}(M)$ is a LSA for the product $$X\cdot Y=\nabla_XY.$$ 2) Symplectic Lie groups: if $(G,\omega)$ is a Lie group with a left invariant symplectic form so we can bring $G$ an affine connection (linear, flat, torsion-free) which is left invariant $$\omega(\nabla_xy,z)=-\omega(y,[x,z])$$ and its Lie algebra $\mathcal{G}$ is an LSA for the product $x\cdot y=\nabla_xy$. My question is: 1) What are the Lie algebras that admit a structure of LSA? Only solvable ones, why not semi-simple?

• LSA = what Gerstenhaber calls a pre-Lie algebra? – darij grinberg Jul 31 '11 at 13:00
• @darij: yes, more precisey a left pre-Lie algebra. Here's the nLab link de rigueur: ncatlab.org/nlab/show/pre-Lie+algebra – José Figueroa-O'Farrill Jul 31 '11 at 14:58
• Have you looked into the various papers by Dietrich Burde on the subject? For instance: Simple left-symmetric algebras with solvable Lie algebra. Manuscripta Math. 95 (1998), no. 3, 397–411 (plus later correction in the journal). – Jim Humphreys Aug 1 '11 at 22:14
• Many thanks, José and Jim for your comments! You can see the related topic, (mathoverflow.net/users/16578), Which Lie algebra admit symplectic forms, mathoverflow.net/questions/71766 (version: 2011-07-31) – amine Aug 3 '11 at 8:23
• See the paper of J. Helmstetter, Radical d'une algèbre symétrique a gauche, Ann. Inst. Fourier, 29, 4(1979), 17-35. He defines the radical R(A) of an LSA A to be the largest left ideal contained in the kernel of a→trR(a) where R(a) is the right multiplication operator, and gives several equivalent characterizations of this ideal. If A=[A,A] this forces R(A)=A, which forces the underlying Lie algebra to be solvable. That no semisimple Lie algebra admits a compatible LSA structure follows. – Dan Fox Mar 5 '13 at 13:26

As said, semisimple Lie algebras over a field of characteristic zero do not admit an LSA-structure. There are several proofs. Basically it relies on Whitehead's lemma for semisimple Lie algebras, saying that $H^1(\mathbb{g},M)=0$ for all finite-dimensional $\mathbb{g}$-modules $M$.
On the other hand, not all solvable Lie algebras admit a compatible LSA-structure. There are even nilpotent Lie algebras not admitting such structures, see Y. Benoist's paper "A nilvariete non-affine", and this paper here. The proof is difficult and shows that such counterexamples can be constructed by $n$-dimensional Lie algebras not admitting a faithful linear representation of degree $n+1$.