# Lie Derivative of metric

We are given $$g = \sum_{i,j} g_{ij} dx_i \otimes dx_j$$ 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 $$h_{ij} = \sum_{k=1}^n (X^k \frac{\partial g_{ij}}{\partial x_k} + g_{kj} \frac{\partial X^k}{\partial x_i} + g_{ik}\frac{\partial X^k}{\partial x_j})$$

It is not clear to me how to do this expansion using the definition of the Lie Derivative. For instance if we take $$\mathcal{L}_X g = \frac{d}{dt}(\phi_t^{*}g)$$ evaluated at $$t=0$$, what are we do use as our $$\phi_t$$ given that the vector field $$X$$ is arbitrary?

• For a fixed point $p$, take any $\phi_t$ such that $\frac d{dt}\phi_t(0)=Xp$. – Berci Jul 12 at 11:22
• @Berci thanks for the feedback. I'm still not sure how to deal with the $h_{ij}$ arises from this though.. – DSS Jul 12 at 11:47

1. $$\mathcal{L} _X(A\otimes B) = \mathcal{L} _X(A) \otimes B +A\otimes \mathcal{L} _X(B)$$ for any tensor fields $$A$$ and $$B$$.
2. $$\mathcal{L} _X(f) = Xf$$ for any smooth funtion $$f$$.
3. $$\mathcal{L} _X(d\omega) = d\mathcal{L} _X(\omega)$$ for any differential form $$\omega$$.