Does an analytic tensorial Lie structure on $S^2$ gives a fiberwise Abelian Lie algebra structure?

Question

Motivated by the answer to this question we ask:

Is it true to say that for every real analytic tensorial Lie algebra structure $\alpha$ on $\chi^{\infty}(S^2)$, all fibers are necessarily Abelian Lie algebra? In the other word:Assume that $\alpha$ is a real analytic $(1,2)$ tensor on $S^2$. Moreover assume that the restriction $\alpha_x$ of $\alpha$ to each fiber satisfies the Jacobi identity. Does this imply that $\alpha$ gives us an abelian Lie algeba at each fiber $T_x(S^2)$?That is; Is $\alpha$ identically zero?

Answer 1

No. A skew-symmetric bilinear map $V\times V\to V$ satisfies the Jacobi identity automatically if $\dim V=2$ since the Jacobi identity is skew-symmetric in its arguments. So if you take a generic skew-symmetric $\alpha$, then on most fibres it will create a two-dimensional non-Abelian Lie algebra, and on some fibres it will be Abelian. Essentially it is the Hairy Ball Theorem in disguise.

Answer 2

No. there is an analytic $(1,2)$ tensor on $S^2$ which satisfies the Jacobi identity. Its restriction to the equator is abelian. It is non abelian at points out of equator.

The construction is similar to the idea of poincare compactification of polynomial vector fields.

$\mathbb{R}^2$ is diffeomorphic to each of upper and lower hemi sphere via $$ \phi{_\pm}(x,y)=(\frac{x}{\sqrt{1+x^2+y^2}}, \frac{y}{\sqrt{1+x^2+y^2}}, \frac{\pm 1}{\sqrt{1+x^2+y^2}})$$

With this diffeomorphism, we pull back the standard non abelian structure $[\partial/\partial x, \partial/ \partial y]=\partial / \partial x$ of the plane to these hemi spheres. The resulting tensor, denoted by $\alpha$, is real analytic on $S^2 \setminus \text{equator}$. Now, $z^k \alpha$ for $k$ sufficiently large, is real analytic on $S^2$.Moreover it vanish on equator. So it is the desired counter example. (here $z$ is the third coordinate on sphere).

Obviously this can be generalized to arbitrary $S^{n}.$