Let $M$ be a manifold, $\alpha$ a two form and $x,y,z$ vector fields. How can we show that
$$L_x(\alpha(y,z))=(L_x \alpha)(y,z)+\alpha(L_x y,z)+\alpha(y,L_x z)$$
(where $L_x$ is the Lie derivative in the direction of $x$)? Does this property have a name?
Idea: We can get rid of the 2-form. Namely, we have to show that
$$ L_x \circ i_z \circ i_y = i_z \circ i_y \circ L_x + i_z \circ i_{L_x y} + i_{L_x z} \circ i_y $$
where $i_x$ means the contraction with $x$. I tried using Cartan's formula on the last expression to no success.
Best Answer
I'm going to use capital letters for the vector fields and $\mathcal{L}$ for the Lie derivative.
By Cartan's formula, we have
$$(\mathcal{L}_X\alpha)(Y, Z) = (d(i_X\alpha))(Y, Z) + (i_X(d\alpha))(Y, Z) = (d(i_X\alpha))(Y, Z) + (d\alpha)(X, Y, Z).$$
If $\beta$ is a one-form, $(d\beta)(Y, Z) = Y\beta(Z) - Z\beta(Y) - \beta([X, Y])$, so for $\beta = i_X\alpha$ we see that
\begin{align*} (d(i_X\alpha))(Y, Z) &= Y(i_X\alpha)(Z) - Z(i_X\alpha)(Y) - (i_X\alpha)([Y, Z])\\ &= Y\alpha(X, Z) - Z\alpha(X, Y) - \alpha(X, [Y, Z])\\ &= Y\alpha(X, Z) - Z\alpha(X, Y) + \alpha([Y, Z], X). \end{align*}
As for the other term, we have
\begin{align*} &\ (d\alpha)(X, Y, Z)\\ =&\ X\alpha(Y, Z) - Y\alpha(X, Z) + Z\alpha(X, Y) - \alpha([X, Y], Z) + \alpha([X, Z], Y) - \alpha([Y, Z], X). \end{align*}
Combining the two, we obtain
\begin{align*} (\mathcal{L}_X\alpha)(Y, Z) &= X\alpha(Y, Z) - \alpha([X, Y], Z) + \alpha([X, Z], Y)\\ &= X\alpha(Y, Z) - \alpha(\mathcal{L}_X Y, Z) + \alpha(\mathcal{L}_X Z, Y)\\ &= X\alpha(Y, Z) - \alpha(\mathcal{L}_X Y, Z) - \alpha(Y, \mathcal{L}_X Z). \end{align*}
Now note that if $f$ is a smooth function, then $\mathcal{L}_Xf = d(i_Xf) + i_X(df) = d(0) + df(X) = Xf$. In particular,
$$\mathcal{L}_X(\alpha(Y, Z)) = X\alpha(Y, Z).$$
Therefore
$$(\mathcal{L}_X\alpha)(Y, Z) = \mathcal{L}_X(\alpha(Y, Z)) - \alpha(\mathcal{L}_X Y, Z) - \alpha(Y, \mathcal{L}_X Z)$$
and hence
$$\mathcal{L}_X(\alpha(Y, Z)) = (\mathcal{L}_X\alpha)(Y, Z) + \alpha(\mathcal{L}_X Y, Z) + \alpha(Y, \mathcal{L}_X Z).$$
More generally, if $\alpha$ is a contravariant $k$-tensor (for example, a $k$-form),
$$\mathcal{L}_X(\alpha(Y_1, \dots, Y_k)) = (\mathcal{L}_X\alpha)(Y_1, \dots, Y_k) + \sum_{i=1}^k\alpha(Y_1, \dots, Y_{i-1}, \mathcal{L}_XY_i, Y_{i+1}, \dots, Y_k).$$
In fact, one usually defines the Lie derivative to act on tensors by precisely this rule.
An algebraic way of expressing the above equation is: the Lie derivative obeys the Leibniz rule with respect to contractions. In fact, this can be used as one of the axioms that uniquely determines the action of the Lie derivative on tensor fields - see here.