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In this question we try to improve some parts of this post as follows:

What is an example of a manifold $M$ and a Lie algebra $L$ (with the same dimension) such that $M$ does not admit an smooth $(1,2)$ tensor $\alpha$ which satisfies the Jacobi identity at each fibre $T_x(M)$ and the fibrewise Lie algebra structure $\alpha_x$ on $T_x (M)$ is isomorphic to $L$?

What type of obstructions would appear?

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A simple example is to let $M=S^2$ and let $L$ be the nonabelian Lie algebra of dimension $2$. If such an $\alpha$ existed, its range would be a rank-1 subbundle $L\subset TS^2$, but this cannot exist for topological reasons.

The general obstruction is whether the manifold $M$ admits an $\mathrm{Aut}(L)$-structure, and this can be determined by homotopy-theoretical methods.

Another good example is to take $M=S^4$ and $L$ any nonabelian Lie algebra of dimension~$4$. Since $\mathrm{Aut}(L)$ always preserves a nontrivial subalgebra of $L$ in this case, such a structure $\alpha$ on $M$ would induce a nontrivial subbundle of $TS^4$, and this does not exist, so $\alpha$ cannot exist.

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  • $\begingroup$ Fibrewise the range is a subvector space of tangent space. But why is it a subbundle? $\endgroup$
    – Ali Taghavi
    Commented Aug 25, 2015 at 11:08
  • $\begingroup$ Thank you for the answer. Does the question automatically imply that "We have a Bundle of Lie algebras"? $\endgroup$
    – Ali Taghavi
    Commented Aug 25, 2015 at 11:13
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    $\begingroup$ If $\alpha$ is smooth, then, for any linearly independent vector fields $X$ and $Y$, $\alpha(X,Y)$ will be a smooth nonvanishing vector field that takes values in the subspace at each point, so the range will be a smooth subbundle. For your second comments question, the answer is 'yes'. $\endgroup$
    – Robert Bryant
    Commented Aug 25, 2015 at 11:13
  • $\begingroup$ I am sorry if my question is elementary: Is it obvious that the conditions in my initial question automatically implies that the structure group can be reduced to Aut(L)? $\endgroup$
    – Ali Taghavi
    Commented Aug 25, 2015 at 12:44
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    $\begingroup$ @AliTaghavi: Yes, it is obvious: If such an $\alpha$ exists, let $F\to M$ be the bundle of coframes (i.e., linear isomorphisms) $u:T_xM\to L$ that are isomorphisms of $(T_xM,\alpha_x)$ with $(L,[,])$ as Lie algebras. Then $F\to M$ is a reduction of the $\mathrm{GL}(n,\mathbb{R})$ coframe bundle to a principal right $\mathrm{Aut}(L)$-subbundle. $\endgroup$
    – Robert Bryant
    Commented Aug 25, 2015 at 14:22

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