I am having trouble here. The context is: Let $X$, $Y$ and $S$ be vector fields ina a manifold (we can assume it's $\mathbb{C}^2$ though I'm pretty sure this should work in any manifold), and we can work fine with the exterior product or wedge product of vector fields, since they are tensors.
I need to know if the formula
$$
\mathcal{L}_X(S\wedge Y)=\mathcal{L}_X(S)\wedge Y+S\wedge\mathcal{L}_X(Y)
$$
I know this is true when $X$, $Y$ and $S$ are differential forms. The demonstration is basd solely on the property that says that, for tensors fields, we have
$$
\mathcal{L}_X(S\otimes Y)=\mathcal{L}_X(S)\otimes Y+S\otimes\mathcal{L}_X(Y)
$$
I don't think I can say that since it is correct for the tensor product, it would be for the exterior product. I guess I must use the fact that vector fields are antissimetric 1-linear forms and use the operator (in my reference it is called "anti-simetrization operator")
$$
\mathcal{\alpha}(X)=\sum_{s\in \mathcal{G}_p} \epsilon (s)s\circ X
$$
where $\mathcal{G}_p$ is the set or permitations of $p$ indexes and the composition means a permutation on the indexes of the base elements of $X$. The application $\alpha$ turns linear p-forms into antissimetric forms and then we have exterior product of those. If $X$ is already antissimetric, then $\alpha(X)=p!X $.
Now, we also have the definition
$$
X\wedge Y=\dfrac{1}{p!q!}\alpha (X\otimes Y)
$$
So I'm guessing I can argue that $\alpha (X\otimes Y)=(p+q)!(X\otimes Y)$, and the calculations work, that is, I get the expression $\mathcal{L}_X(S\wedge Y)=\mathcal{L}_X(S)\wedge Y+S\wedge\mathcal{L}_X(Y)$ as I wanted. But I don't know for sure if this is correct. I am trying to self-learn somethings on tensors.
Can someone tell me if it's correct and, if not, point me my mistakes?
$$
$$
Best Answer
Let us consider the following general setting. We need it to prove the statement.
Let $M$ be a real manifold of dimension $n$ over the ground field $\mathbb K$. Let $$\operatorname{T}^{\bullet}_{poly}(M):=\mathcal C^{\infty}(M)\otimes_{\mathbb K}\wedge^{\bullet+1}\operatorname{T}(M) $$
be the algebra of poly vector fields on $M$. Note the shift of grading: for example, vector fields are polyvectors of degree $0$.
There exists a structure of differential graded Lie algebra on $\operatorname{T}^{\bullet}_{poly}(M)$ given as follows. The differential is equal to $0$. The Lie bracket is the Schouten bracket $[\cdot,\cdot]_\mathcal{S} $ given by
$$[e_1 ∧ ... ∧ e_k, \eta_1 ∧ ... ∧ \eta_l]_\mathcal{S} = \\ \sum_{i=1}^k\sum_{j=1}^l(−1)^{i+j}\mathcal L_{e_i}(\eta_j)\wedge e_1 \wedge\dots\wedge\hat{e}_i\wedge\dots\wedge e_k\wedge \eta_1\wedge\dots\wedge\hat{\eta}_l\wedge\dots\wedge\eta_l, $$ for all $e_{\bullet}$ and $\eta_{\bullet}$ in $\operatorname{T}^{0}_{poly}(M)$ and denoting omission by $\hat{\cdot}$. Note that the Schouten bracket reduces to the Lie bracket
$$\mathcal L_X(Y):=[X,Y],$$
on $\operatorname{T}^{0}_{poly}(M)$.
In summary, using some lengthy but straightforward computations, one can prove that
$$(\operatorname{T}^{\bullet}_{poly}(M),0,[\cdot,\cdot]_\mathcal{S}) $$
is a dg Lie algebra (I do not want to introduce the exact definition and further discuss the gradings). Note that we have also an associative product, i.e. the wedge product. In other words, the structure on $\operatorname{T}^{\bullet}_{poly}(M)$ is even richer, but let us skip the discussion about Gerstenhaber algebras.
In the above setting, the original statement is equivalent to
For all $X,Y,S\in \operatorname{T}^{0}_{poly}(M)$ the identity
$$[X,S\wedge Y]_\mathcal{S}=[X,S]\wedge Y+S\wedge[X,Y]~~(*) $$
holds.
On the l.h.s. of $(*)$ it is necessary to consider the Schouten bracket because $S\wedge Y\in \operatorname{T}^{1}_{poly}(M)$.
Let us prove it; by definition of the Schouten bracket
$$[X,S\wedge Y]_\mathcal{S}=(−1)^{1+1}\mathcal L_{X}(S)\wedge Y+(−1)^{1+2}\mathcal L_{X}(Y)\wedge S=\mathcal L_{X}(S)\wedge Y-\mathcal L_{X}(Y)\wedge S, $$
or
$$[X,S\wedge Y]_\mathcal{S}=\mathcal L_{X}(S)\wedge Y+S\wedge\mathcal L_{X}(Y), $$
as claimed. This ends the proof.