1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $\omega$.
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.