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 $x^\mu$.
Best Answer
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.