Motivated by [the answer to this question](https://mathoverflow.net/questions/215587/manifold-lie-algebra-compatibility) we ask:



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


