Lie derivative of a metric

Question

This question has been asked before but the answers were not clear to me so I am asking again.

We have $g = \sum_{i,j} g_{ij} dx_i \otimes dx_j$ as a smooth $(0,2)$-tensor and asked to show that given a smooth vector field $X=\sum X^i \frac{\partial}{\partial x_i}$ $$\mathcal{L}_X g = \sum_{i,j} = h_{ij} dx_i \otimes dx_j$$ where $$h_{ij} = \sum_{k=1}^n (X^k \frac{\partial g_{ij}}{\partial x_k} + g_{kj} \frac{\partial X^k}{\partial x_i} + g_{ik}\frac{\partial X^k}{\partial x_j})$$

I don't know how to do this expansion using the definition of the Lie Derivative.

Answer 1

Following Wikipedia, if you take as definition these 4 axioms:

1. $\mathcal{L}_Yf = Y(f)$
2. $\mathcal{L}_Y(S\otimes T)=(\mathcal{L}_YS)\otimes T+S\otimes (\mathcal{L}_YT)$
3. $\mathcal{L}_X (T(Y_1, \ldots, Y_n)) = (\mathcal{L}_X T)(Y_1,\ldots, Y_n) + T((\mathcal{L}_X Y_1), \ldots, Y_n) + \cdots + T(Y_1, \ldots, (\mathcal{L}_X Y_n))$
4. $[\mathcal{L}_X, d] = 0$

then: $$ \begin{align} (\mathcal {L}_X g) &= (\mathcal {L}_X g)_{ab} dx^a\otimes dx^b\\ &= X(g_{ab})dx^a\otimes dx^b + g_{cb} \mathcal{L}_X (dx^c) \otimes dx^b + g_{ac} dx^a \otimes \mathcal{L}_X (dx^c)\\ &= (X^c \partial_c g_{ab}+g_{cb}\partial_a X^c+g_{ac}\partial_b X^c)dx^a\otimes dx^b\\ \end{align} $$

which gives your desired result.