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Let M be a manifold.

To what extent all Lie algebra structures with tensorial property on $\chi^{\infty}(M)$ are studied?

That is a Lie algebra structure for which $[X,fY]=f[X,Y]$.

(For every parallelizable $n$-manifold $M$ and every Lie algebra structure on $\mathbb{R}^{n}$, one can introduce a Lie algebra structure on $\chi^{\infty}(M)$ with this tensorial property. So what is an example of non parallelizable case?)

Any such structure on a manifold is a vector valued 2-form, that is an element of $\Omega_{1,2}(M)$. Such vector valued 2-forms are called Jacobi 2-form. The Jacoby identity on $\Omega_{1,2}(M)$ is denoted by $I_{2}$

What identities $I_{1}$ and $I_{3}$ can be defined on $\Omega_{1,1}(M)$ and $\Omega_{1,3}(M)$ such that they are invariant under the differentiation? That is: The differential of every $I_{2}$- form, satisfies $I_{3}$ or the differential of every $I_{1}$-form satisfy $I_{2}$(The Jacoby I dentity). Can one extend these (possible) identities for arbitrary $k$ on $\Omega_{1,k}$?(Then study the subcomplex and its cohomology of vector valued differential forms which satisfy $I_{k}$, for all $k$? May be Such cohomology can be named "Jacobi cohomology")

For compact manifolds we may pose some other questions as follows:

  1. Are there some obstructions for existence of a Jacobi 2-form whose fibre wise structure is a simple Lie algebra?

2)Is there an example of a Jacobi 2-form such that we obtain different(perhaps infinit number of) Lie algebra structures on the tangent spaces?

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    $\begingroup$ The condition on "tensorial Lie algebra structures" means that the structure comes from point-wise operations, i.e. your bundle is a locally trivial bundle of Lie algebras. I am not aware of a natural example of this situation, though there are cases in which the cotangent bundle $T^*M$ naturally is a bundle of Lie algebras. There is no natural notion of "differentiation" on vector valued forms unless you choose an additional structure there, e.g. a linear connection on $TM$. The main natural operation there is the Froelicher-Nijenhuis bracket. $\endgroup$ Commented Jun 24, 2015 at 12:49
  • $\begingroup$ @AndreasCap Thank you very much for your very interesting comment. As I wrote in the question, such tensorial Lie algebra struct exists, when M is parallelizable. As you said we need a connection. For example put $M=\mathbb{R}^{3}$ then the vector valued two form $P(x,y,z)(dy\wedge dz, dz\wedge dx, dx\wedge dy)$ satisfies the jacoby identity.It is the natural Lie algebra of $\mathbb{R}^{3}$. The differential of this two form(with flat connection) is the standard volum form muliplied with a gradient vector field. $\endgroup$ Commented Jun 24, 2015 at 21:26
  • $\begingroup$ As you said a manifold which fibres of the tangent bundle are equiped with a given(fixed)Lie algebra structure possess 2-forms which satisfies the jacoby identity. but it would be possible that a 2-form satisfies Jacoby identity but the structure of Lie algebras is not fixed, when the base point varies. I think that for vector bundle which fibres are matrix ALGEBRA, there is an invariant which is introduced by Douady. But I do not know whether there is a Lie algebra analogy? Any way, in my question, I do not fix a given lie algebra structure. I just consider 2-forms with Jacoby identity. $\endgroup$ Commented Jun 24, 2015 at 21:37
  • $\begingroup$ Finally, according to your comment, we need a connection. So we can associate to a Riemannian manifold(with LC connection), such type of cohomology(provided we can extent the Jacoby identity to higher order forms. Does this idea leads one to triviality? Thanks again for your comment. $\endgroup$ Commented Jun 24, 2015 at 21:40
  • $\begingroup$ @AndreasCap In line of this post one can ask the following particular question: Is there a vector valued 3-form $(\alpha, \beta, \gamma)$ on $\mathbb{R}^{3}$ which is not the differential of a jacobi 2-form? $\endgroup$ Commented Jun 25, 2015 at 20:45

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