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My problem is question 12-11 of "introduction to smooth manifolds" by lee. I have to prove that $$\mathcal L_V \mathcal L_W A - \mathcal L_W \mathcal L_V A=\mathcal L_{[V,W]}A$$ A is smooth covariant tensor field on M $$V,W\in \mathscr X(M)$$ I don't know how hint works about this problem.

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    What characterisation of the Lie derivative are you using?2017-01-03

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Well, $\mathcal{L}_V$ for any $V$ is a derivation of the tensor algebra. So first check that the commutator of two derivations is once again a derivation. This can be checked by a direct calculation. In fact, the commutator of a grade $k$ and grade $l$ derivations is a derivation of grade $k+l$. Since the Lie-derivative is a derivation of grade $0$, the commutator is again a derivation of grade $0$, so $[\mathcal{L}_V,\mathcal{L}_W]$ is a candidate for being a Lie-derivative itself.

If $f\in C^\infty(M)$ a smooth function (grade 0 tensor field), $$ [\mathcal{L}_V,\mathcal{L}_W]f=\mathcal{L}_V\mathcal{L}_Wf-\mathcal{L}_W\mathcal{L}_Vf=\mathcal{L}_V(Wf)-\mathcal{L}_W(Vf)=VWf-WVf=[V,W]f=\mathcal{L}_{[V,W]}f. $$

If $X\in\Gamma(TM)$ is a smooth vector field (grade 1 tensor field), we have $$ [\mathcal{L}_V,\mathcal{L}_W]X=\mathcal{L}_V\mathcal{L}_WX-\mathcal{L}_W\mathcal{L}_VX=\mathcal{L}_V[W,X]-\mathcal{L}_W[V,X]=[V,[W,X]]-[W,[V,X]]. $$

The Jacobi identity is $$ [V,[W,X]]+[W,[X,V]]+[X,[V,W]]=0, $$ from which we have by skew symmetry $$ [[V,W],X]=[V,[W,X]]+[W,[X,V]]=[V,[W,X]]-[W,[V,X]]. $$

The RHS is $[\mathcal{L}_V,\mathcal{L}_W]X$, but the LHS is $[[V,W],X]=\mathcal{L}_{[V,W]}X,$ so we have $$ [\mathcal{L}_V,\mathcal{L}_W]X=\mathcal{L}_{[V,W]}X. $$

Now, check for yourself that any derivation of the tensor algebra that commutes with contractions is uniquely determined by its action on grade 0 and grade 1 (the tensor algebra is a bigraded algebra, but since $\Gamma(T^*M)$ is related to $\Gamma(TM)$ by duality and the Lie-derivative commutes with contractions, it is sufficient to know the action of the Lie-derivative on $C^\infty(M)$ and $\Gamma(TM)$), which means that if $[\mathcal{L}_V,\mathcal{L}_W]=\mathcal{L}_{[V,W]}$ on functions and vector fields (and it is so), then by that accord, this identity will hold on all of the tensor algebra.