Lie derivative along the commutator of two vector fields 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 $x^\mu$.
 A: One way to prove this result is to first show it for functions on the manifold, and use this to further prove it for vector fields, co-vector fields and finally general tensor fields. The advantage of this approach as it requires the minimum of formulas. 
We will need the following assumptions:
(a) The Lie derivative follows the Leibnitz rule when acting on a product of objects.
(b) $ \mathcal{L}_{X} (f) = X(f) $ 
Part 1 - Show for a function, f:
Using (b) we have 
$$ \mathcal{L}_X \mathcal{L}_Y f = \mathcal{L}_X ( Y(f)) = X(Y(f)) $$ 
In a coordinate basis this is:
$$ X^{\rho} \partial_{\rho} (Y^{\sigma} \partial_{\sigma} f) = X^{\rho} ( \partial_{\rho} Y^{\sigma}) (\partial_{\sigma} f) + X^{\rho} Y^{\sigma} (\partial_{\rho}  \partial_{\sigma} f) $$
Because $ \partial_{\rho}  \partial_{\sigma} f $ is symmetric in $\rho \leftrightarrow \sigma $ then the second term on the RHS is symmetric in $ X \leftrightarrow Y $. Therefore:
$$ \mathcal{L}_X \mathcal{L}_Y f - \mathcal{L}_Y \mathcal{L}_X f = X^{\rho} ( \partial_{\rho} Y^{\sigma}) (\partial_{\sigma} f) - Y^{\rho} ( \partial_{\rho} X^{\sigma}) (\partial_{\sigma} f) = (X^{\rho} ( \partial_{\rho} Y^{\sigma}) - X^{\rho} ( \partial_{\rho} Y^{\sigma}))(\partial_{\sigma} f) $$
$$ =  [X,Y]^{\sigma} \partial_{\sigma} f =  [X,Y](f) = \mathcal{L}_{[X,Y]} f $$
Part 2: Show for a vector field, $ T = V \in TM $
Using the result from Part 1 we have:
$$ \mathcal{L}_{[X,Y]} (V(f)) = \mathcal{L}_{X} \mathcal{L}_{Y} (V(f)) - \mathcal{L}_{Y} \mathcal{L}_{X} (V(f)) $$
Using the Leibnitz rule:
$$ \mathcal{L}_{X} \mathcal{L}_{Y} (V(f)) = \mathcal{L}_{X} (( \mathcal{L}_{Y} V)f) + \mathcal{L}_{X}( V(\mathcal{L}_{Y} f) ) = (\mathcal{L}_{X} \mathcal{L}_{Y} V)(f) + V(\mathcal{L}_{X} \mathcal{L}_{Y}f) + (\mathcal{L}_{X} V)(\mathcal{L}_{Y} f) + (\mathcal{L}_{Y} V)(\mathcal{L}_{X} f) $$
The last 2 terms taken together are symmetric in $ X \leftrightarrow Y $. We therefore have that
$$ \mathcal{L}_{[X,Y]} (V(f)) = \mathcal{L}_{X} \mathcal{L}_{Y} (V(f)) - \mathcal{L}_{Y} \mathcal{L}_{X} (V(f)) = (\mathcal{L}_{X} \mathcal{L}_{Y} V - \mathcal{L}_{Y} \mathcal{L}_{X} V)(f) +  V(\mathcal{L}_{X} \mathcal{L}_{Y} f - \mathcal{L}_{Y} \mathcal{L}_{X}f) $$
But we can also apply the Leibnitz rule to $\mathcal{L}_{[X,Y]} (V(f))$ to give:
$$ \mathcal{L}_{[X,Y]} (V(f)) = (\mathcal{L}_{[X,Y]} V)(f) + V(\mathcal{L}_{[X,Y]} f)) = (\mathcal{L}_{[X,Y]} V)(f) + V(\mathcal{L}_{X} \mathcal{L}_{Y} f - \mathcal{L}_{Y} \mathcal{L}_{X}f))$$
Comparing these 2 expressions we get that:
$$ (\mathcal{L}_{[X,Y]} V)(f) = (\mathcal{L}_{X} \mathcal{L}_{Y} V - \mathcal{L}_{Y} \mathcal{L}_{X} V)(f) $$
This is true for any function hence we have shown the identity for a vector field.
Part 3 - Show for a covector field, $ T = \eta \in T^{*}M $
We consider the action of $ \mathcal{L}_{[X,Y]} $ on the function $ \eta (V) $ where $ \eta $ is a covector field and V is a vector field. This works exactly the same as in part 2 giving us:
$$(\mathcal{L}_{[X,Y]} \eta)(V) = (\mathcal{L}_{X} \mathcal{L}_{Y} \eta - \mathcal{L}_{Y} \mathcal{L}_{X} \eta)(V) $$
This is true for any vector field hence the result is shown for covector fields.
Part 4 - General Tensor field, T
We will now consider the action of $ \mathcal{L}_{[X,Y]} $ on the function $T(X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s)$, where T is an (r,s) rank tensor, $ X_i $ is a vector field, and $ \eta_i $ is a covector field. This step proceeds similarly to before.
By the result from Part 1:
$$ \mathcal{L}_{[X,Y]}T(X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) = (\mathcal{L}_{X} \mathcal{L}_{Y} - \mathcal{L}_{Y} \mathcal{L}_{X})T(X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) $$
Similarly to Part 2 $ \mathcal{L}_{X} \mathcal{L}_{Y} $ acting on  $ T(X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) $ gives terms symmetric in $ X \leftrightarrow Y $ when the Lie derivatives hit seperate terms. These terms will cancel with terms contained in $ - \mathcal{L}_{Y} \mathcal{L}_{X} T(X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) $. Therefore we are left with:
$$ (\mathcal{L}_{X} \mathcal{L}_{Y} - \mathcal{L}_{Y} \mathcal{L}_{X})T(X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) = (\mathcal{L}_{X} \mathcal{L}_{Y}T - \mathcal{L}_{Y} \mathcal{L}_{X}T)(X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) + T((\mathcal{L}_{X}\mathcal{L}_{Y} - \mathcal{L}_{Y} \mathcal{L}_{X})X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) + T(X_1,(\mathcal{L}_{X}\mathcal{L}_{Y} - \mathcal{L}_{Y} \mathcal{L}_{X})X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) + \dots $$
Using the results from Parts 2 and 3 gives:
$$ \mathcal{L}_{[X,Y]}T(X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) =(\mathcal{L}_{X} \mathcal{L}_{Y}T - \mathcal{L}_{Y} \mathcal{L}_{X}T)(X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) + T(\mathcal{L}_{[X,Y]}X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) + T(X_1,\mathcal{L}_{[X,Y]}X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) + \dots $$
Lastly, the use of the Leibnitz rule gives 
$$ \mathcal{L}_{[X,Y]}T(X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) =  \mathcal{L}_{[X,Y]}T(X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) + T(\mathcal{L}_{[X,Y]}X_1,X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) + T(X_1,\mathcal{L}_{[X,Y]}X_2,...,X_r,\eta_1,\eta_2,...,\eta_s) + \dots $$
Comparing the 2 equalities and noting that they are true for any vector fields $ X_1,X_2,...,X_r $ and any covector fields $ \eta_1,\eta_2,...,\eta_s $ proves that the desired result holds for any general tensor field T.
A: 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.
A: 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.
