No. there is an analytic $(1,2)$ tensor on $S^2$ tor 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 hei 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 thirth coordinate on sphere).

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 $S^{n}.$