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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?

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?

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 we obtain an abelian Lie algeba at each fiber $T_x(S^2)$?

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?

Motivated by 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 we obtain an abelian Lie algeba at each fiber $T_x(S^2)$?

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 we obtain an abelian Lie algeba at each fiber $T_x(S^2)$?