$L_X( \omega (Y_1,\ldots, Y_n)) = (L_X \omega) (Y_1, \ldots, Y_n) + \sum_{i = 1}^n \omega (Y_1, \ldots, Y_{i-1}, L_X Y_i, \ldots, Y_n)$

Question

Here $\omega$ is a smooth form on a manifold, $X, Y_1, \ldots, Y_p$ are smooth vector fields, and $L_X$ represents the Lie derivative.

I am having trouble proving this standard identity. All the books that I have checked in describe it as a trivial exercise, but I would appreciate a hint or a sketch of an argument. Even in the reduction to checking for $\omega = f \, dx_1 \vee dx_2$ in a local coordinate system I don't see an obvious way forward.

$$L_X( \omega (Y_1,\ldots, Y_n)) = (L_X \omega) (Y_1, \ldots, Y_n) + \sum_{i = 1}^n \omega (Y_1, \ldots, Y_{i-1}, L_X Y_i, \ldots, Y_n)$$

Answer 1

It is "trivial" - work in a local coordinate system and apply the usual tricks for computing the derivative of a product. Take advantage of the multilinearity of forms and the existence of a local rectification for $X$ (at nonzero points) to use the canonical identification between the tangent spaces on $R^n$.

Answer 2

A coordinate independent proof is possible if you use the definition $L_X = \frac{\mathrm d}{\mathrm d t}\big|_{t=0}\,\Phi_t^*$, where $\Phi$ is the flow of the field $X$, and note that:

1. Evaluation $\omega(Y_1,\dots,Y_n)$ of a tensor can be thought of as repeated contraction applied on $\omega\otimes Y_1\otimes Y_2 \otimes\dots\otimes Y_n$.
2. If $f : M\to M$ is a diffeomorphism, then $f^*$ commutes with contraction.
3. If $f$ is a diffeomorphism, and $A$ and $B$ are arbitrary tensors, then $f^*(A\otimes B) = f^*A\otimes f^*B$.
4. $\frac{\mathrm d}{\mathrm d t}\big|_{t=0}(A_t\otimes B_t) = \frac{\mathrm d A_t}{\mathrm d t}\big|_{t=0}\otimes B + A\otimes \frac{\mathrm d B_t}{\mathrm d t}\big|_{t=0}$.

The last part is the only bit where you really have to 'show' something that is related to the Lie derivative and not to general properties of diffeomorphisms, but of course the same trick that works for the Leibniz rule for classical calculus will work here as well.

From 3 and 4 together it follows that

$$ L_X(A\otimes B) = L_X(A)\otimes B + A\otimes L_X(B). $$

So that we have (say for $n=1$):

\begin{aligned} L_X(\omega(Y)) &= L_X(C(\omega\otimes Y)) \\ &= C(L_X(\omega\otimes Y)) \\ &= C(L_X\omega\otimes Y) + C(\omega\otimes L_XY) \\ &= (L_X\omega)(Y)+\omega(L_XY). \end{aligned}

(Where $C$ is a contraction.)

Finally, use the antisymmetrizer and linearity of $L_X$ to obtain the same result for differential forms.