Lie derivative along the commutator of two vector fields

Question

I would like to know how to show that the Lie derivative on a differentiable manifold satisfies

\begin{equation*} \mathcal{L}_{[X, Y]} = \mathcal{L}_X \mathcal{L}_Y - \mathcal{L}_Y \mathcal{L}_X \end{equation*}

for any tensor field on which the derivative is applied, where $[X, Y] = XY - YX$ is the commutator of $X$ and $Y$, which are arbitrary vector fields.

Edit:

In component notation, $[X, Y]^\mu = X^\lambda \partial_\lambda Y^\mu - Y^\lambda \partial_\lambda X^\mu = X^\lambda \nabla_\lambda Y^\mu - Y^\lambda \nabla_\lambda X^\mu$, where $\partial_\mu$ and $\nabla_\mu$ are respectively the partial and the covariant derivatives with respect to the variable $x^\mu$.

Answer 1

What I did in the end is something that goes along the lines of a proof by induction. In this sense, it is similar to Markus Heinrich's answer. First, the Lie derivative of an arbitrary $(k, l)$ tensor is

\begin{align*} \mathcal{L}_V {T^{\mu_1 \mu_2 \ldots \mu_k}}_{\nu_1 \nu_2 \ldots \nu_l} &= V^\sigma \partial_\sigma{T^{\mu_1 \mu_2 \ldots \mu_k}}_{\nu_1 \nu_2 \ldots \nu_l} \\ &- (\partial_\lambda V^{\mu_1}){T^{\lambda \mu_2 \ldots \mu_k}}_{\nu_1 \nu_2 \ldots \nu_l} - (\partial_\lambda V^{\mu_2}){T^{\mu_1 \lambda \ldots \mu_k}}_{\nu_1 \nu_2 \ldots \nu_l} - \ldots \\ &+ (\partial_{\nu_1} V^\lambda){T^{\mu_1 \mu_2 \ldots \mu_k}}_{\lambda \nu_2 \ldots \nu_l} + (\partial_{\nu_2} V^\lambda){T^{\mu_1 \mu_2 \ldots \mu_k}}_{\nu_1 \lambda \ldots \nu_l} + \ldots \end{align*}

For a function $f: M \rightarrow \mathbb{R}$ on the manifold, this reduces to

\begin{align*} \mathcal{L}_V f = V^\mu \partial_\mu f \end{align*}

while for a one-form field $\omega: M \rightarrow T^*_p M$, this reduces to

\begin{align*} \mathcal{L}_V \omega_\mu = V^\nu \partial_\nu f\omega_\mu + (\partial_\mu V^\nu) \omega_\nu \end{align*}

For a vector field, the formula $\mathcal{L}_V U^\mu = [V, U]^\mu$ is simpler than the above formula.

First, I showed that $\mathcal{L}_{[X, Y]} = \mathcal{L}_X \mathcal{L}_Y - \mathcal{L}_Y \mathcal{L}_X$ is true when applied to a function, to a vector field, and to a covector field. It is straightforward using the above formulae. This step stood as the base case for my "proof by induction". Then I used the fact that any tensor can be expressed as the tensor product of vectors and covectors and that the Lie derivative obeys the Leibniz rule: $\mathcal{L}_V (S \otimes T) = (\mathcal{L}_V S) \otimes T + S \otimes (\mathcal{L}_V T)$. As my induction hypothesis, I claimed that $\mathcal{L}_{[X, Y]} T = \mathcal{L}_X \mathcal{L}_Y T - \mathcal{L}_Y \mathcal{L}_X T$ is true for any tensor T. Then, taking two tensors $S$ and $T$ obeying this rule, I proved that a more general tensor $S \otimes T$ also obeyed it:

\begin{align*} \mathcal{L}_{[X, Y]} (S \otimes T) = \mathcal{L}_X \mathcal{L}_Y (S \otimes T) - \mathcal{L}_Y \mathcal{L}_X (S \otimes T) \end{align*} Therefore, by induction, I had what I wanted. $\blacksquare$

NB: The first formula is also true if the partial derivatives are replaced by covariant derivatives.

Answer 2

To show this for vector fields, we use the relation $$ \mathcal L_XY=[X,Y], $$ and hence the statement follows from the properties of the Lie bracket of vector fields.

For forms, first show that the interior product fulfills $$ i_{[X,Y]}=[\mathcal L_X,i_y], $$ (easy if you use Cartan's formula) and then a quick calculation using Cartan's formula and the fact that the Lie and exterior derivative commute gives $$ \mathcal L_{[X,Y]} \alpha = i_{[X,Y]}d\alpha + di_{[X,Y]}\alpha = \mathcal L_X(i_Y\alpha+di_y\alpha) - (i_y d - di_y)\mathcal L_X\alpha = [\mathcal L_X,\mathcal L_Y]\alpha. $$ and by the Leibniz rule this generalises to arbitrary tensors.