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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? $$ $$

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  • $\begingroup$ The worst thing in studying during dawn is the long wait I must endure until someone comments/answers :P $\endgroup$ – Marra Jul 28 '13 at 7:00
  • $\begingroup$ Is $\mathcal L_X$ the Lie derivative, right? $\endgroup$ – Avitus Aug 23 '13 at 8:37
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Let us consider the following general setting. We need it to prove the statement.

  • The dg Lie algebra of poly-vector fields

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.

  • Statement in the OP

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.

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  • $\begingroup$ While I believe this is correct, it is totally out of my context :( $\endgroup$ – Marra Aug 23 '13 at 16:12
  • $\begingroup$ What is your context? The idea in the above answer can be summarized as follows: if you consider the identity you want to prove, on the l.h.s. you have the Lie derivative applied to a 2-vector field. Question: what is the Lie derivative applied to such an object? One starts with the Lie derivative applied to vector fields in diff. geometry, am I right? In order to define the l.h.s. of the identity you introduce multi vector fields, i.e. functions, vector fields, 2-vector fields, etc... and define the Lie derivative on such space, called space of poly vector fields. The Lie derivative... $\endgroup$ – Avitus Aug 23 '13 at 16:33
  • $\begingroup$ ...on such space is called Schouten bracket; from its very definition the identity you want to prove is shown to hold. The end :-) $\endgroup$ – Avitus Aug 23 '13 at 16:34
  • $\begingroup$ Yeah, I meant no harm, is just that the proof you gave me is somewhere beyond my reading, I'd have to spend some time to understand it. But thanks! At least I know now that the identity holds :-) $\endgroup$ – Marra Aug 23 '13 at 16:41
  • $\begingroup$ No harm has been felt! :-) Try the Schouten bracket for $i,l$ equal to $(1,1)$, $(2,1)$ etc...it helped me a lot! If you are interested in what comes from this area of diff. geo+ dg algebras you could have a look at Vaismann's book on symplectic and Poisson geometry. $\endgroup$ – Avitus Aug 23 '13 at 17:17
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@Marra: Avirus gave a nice explanation of the Schouten-Nijenhuis bracket (useful in Poisson geometry). If you want just to know the Lie derivative of the exterior product $\mathscr{L}_X(Y\wedge Z)$, then you can start from the definition, for any tensor field $T$: $$\mathscr{L}_X T=\frac{d}{dt}\Big|_{t=0}(\exp tX)^*T$$ where $\exp tX$ the local flow of $X$. From there you can prove, for any tensor fields $S$ and $T$, that $$\mathscr{L}_X(S\wedge T)=\mathscr{L}_X S\wedge T+S\wedge\mathscr{L}_X T.$$ Finally, it's useful to keep in mind that a Lie derivative is always a derivation: $$\mathscr{L}_X(fg)=\mathscr{L}_X(f)g+f\mathscr{L}_X(g),\ \forall f,g\in C^\infty(M).$$ $$\mathscr{L}_X(\alpha\wedge\beta)=\mathscr{L}_X\alpha\wedge\beta+\alpha\wedge\mathscr{L}_X\beta,\ \forall\alpha,\beta\in\Omega^*(M).$$ $$\mathscr{L}_X<\alpha,Y>=<\mathscr{L}_X\alpha,Y>+<\alpha,\mathscr{L}_X Y>,\ \forall\alpha,\in\Omega^1(M),\ \forall Y\in\mathcal{X}(M).$$ and the general definition of a derivation: if $(\mathcal{A},\cdot)$ is an algebra, a derivation on $\mathcal{A}$ is linear map $\delta : \mathcal{A}\to\mathcal{A}$ such that, $$\delta(x\cdot y)=(\delta x)\cdot y+x\cdot(\delta y).$$ P.S: Sorry for my bad English!

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