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?

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

Just use the following facts (not so hard to show with the definition you have):

  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 $\omega$.
  • $\begingroup$ Thank you for this guide. I am making progress with it now. $\endgroup$ – DSS Jul 13 at 2:15
  • $\begingroup$ @DSS If you get stuck, just let me know. $\endgroup$ – TheGeekGreek Jul 13 at 10:41

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