TITLE: Lie derivative of a (0,2)-tensor
QUESTION [0 upvotes]: Let $\alpha$ be a (0,2)-tensor and X a vector field.
I want to find $(L_X\alpha)$ so I have tried to do the following steps:
Let Y, Z be other vector fields. Then:
$$(L_X\alpha)(Y,Z) = X \cdot \alpha_{ij}dx^i \otimes dx^j(Y,Z) + [\alpha_{ij}L_X(dx^i) \otimes dx^j] (Y,Z) + \alpha_{ij}dx^i \otimes L_X(dx^j)](Y,Z), $$
using the general expression of Lie derivative.
Now I would use that X and Y are $X_k\dfrac{\partial}{\partial x^k}$ and $Y_l\dfrac{\partial}{\partial y^l}, Z_m\dfrac{\partial}{\partial z^m}$ locally and how acts the Lie derivative on a 1-form.
Nevertheless, I'm stuck and I don't know how to continue. How shall I follow?

REPLY [3 votes]: Just have to use the definition $$(\mathcal{L}_X\alpha)(Y,Z) = X(\alpha(Y,Z)) - \alpha([X,Y],Z) - \alpha(Y,[X,Z]),$$taking $Y=\partial_i$ and $Z = \partial_j$, together with the general relation $[X,\partial_k] = [X^i\partial_i,\partial_k] = -(\partial_kX^i)\partial_i$. Thus: $$\begin{align}(\mathcal{L}_X\alpha)_{ij} &= X(\alpha(\partial_i,\partial_j)) - \alpha([X,\partial_i],\partial_j) - \alpha(\partial_i,[X,\partial_j]) \\ &= X(\alpha_{ij}) + \alpha((\partial_iX^k)\partial_k, \partial_j) + \alpha(\partial_i, (\partial_jX^k)\partial_k) \\ &= X^k(\partial_k\alpha_{ij}) + \alpha_{kj}(\partial_iX^k) + \alpha_{ik}(\partial_jX^k). \end{align}$$
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