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The Lie derivative gives a way to define the derivative of a tensor field in the direction of a vector field.

Let $A$ be a tensor field on a smooth manifold $M$, and let $V$ be a smooth vector field on $M$. Then the Lie derivative of $A$ with respect to $V$ is the tensor field given by

$$\mathcal{L}_VA = \left.\frac{d}{dt}\phi_t^*A\right|_{t=0}$$

where $\phi_t$ is the flow of $V$. In the case of a vector field $X$, we have $\mathcal{L}_VX = [V, X]$.

If $A$ is a $k$-contravariant, $l$-covariant tensor field, $X_1, \dots, X_k$ are smooth vector fields, and $\omega^1, \dots, \omega^l$ are smooth one-forms, then an equivalent definition of $\mathcal{L}_VA$ is

\begin{align*} &(\mathcal{L}_VA)(X_1, \dots, X_k, \omega^1, \dots, \omega^l)\\ =&\ V(A(X_1, \dots, X_k, \omega^1, \dots, \omega^l)) - \sum_{i=1}^kA(X_1, \dots, X_{i-1}, \mathcal{L}_VX_i, X_{i+1}, \dots, X_k, \omega^1, \dots, \omega^l)\\ &\ - \sum_{j=1}^lA(X_1, \dots, X_k, \omega^1, \dots, \omega^{j-1}, \mathcal{L}_V\omega^j, \omega^{j+1}, \dots, \omega^l)\\ =&\ V(A(X_1, \dots, X_k, \omega^1, \dots, \omega^l)) - \sum_{i=1}^kA(X_1, \dots, X_{i-1}, [V, X_i], X_{i+1}, \dots, X_k, \omega^1, \dots, \omega^l)\\ &\ - \sum_{j=1}^lA(X_1, \dots, X_k, \omega^1, \dots, \omega^{j-1}, \mathcal{L}_V\omega^j, \omega^{j+1}, \dots, \omega^l). \end{align*}

Acting on differential forms (skew contravariant tensor fields), the Lie derivative is related to interior multiplication and the exterior derivative via Cartan's magic formula:

$$\mathcal{L}_V = i_V\circ d + d\circ i_V.$$