# Why can I use the Covariant Derivative in the Lie Derivative?

The Lie derivative is the change in the components of a tensor under an infinitesimal diffeomorphism. It seems that this definition does not depend on the metric: $$\mathcal{L}_X T^{\mu_1...\mu_p}_{\nu_1...\nu_q}= X^\lambda \partial_\lambda T^{\mu_1...\mu_p}_{\nu_1...\nu_q} - X^{\mu_1}\partial_\lambda T^{\lambda \mu_2...\mu_p}_{\nu_1...\nu_q} + {\rm upper\,indices} + X^\lambda\partial_{\nu_1}T^{\mu_1...\mu_p}_{\lambda\nu_2...\nu_q} + {\rm lower \,indices}\quad.$$

Now, for some reason if I replace all the derivatives by covariant derivatives $$\partial \to \nabla$$, then magically all the connection symbols $$\Gamma$$ cancel out! Why does that happen??

(A similar thing happens for exterior derivatives. If I take the $$d$$ of some $$p-$$form, I get an antisymmetric coordinate derivate, e.g. $$F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$$. If I replace the derivative with $$\nabla$$, the connection symbols cancel if they are assumed to be symmetric. What is happening?)

I suspect that some users might want to answer by saying that if the expression does not depend on the metric, then I can always choose a coordinate system where the connection vanishes and so the expression with the covariant derivative will be correct with that metric and therefore with any metric since the expression is independent of the metric. But if you have curvature, you can't make the connection vanish everywhere, right?

• I don't know the deeper reason, but notice that it's clearly related to the connection being torsionless. You could turn the question around and ask why the connection being torsionless implies this. – Javier Jun 14 at 13:21

The reason relies upon 4 facts. First of all, it turns out that, for vector fields, $${\cal L}_X Y = [X,Y] = \nabla_X Y - \nabla_Y X + T^{(\nabla)}(X,Y),$$ where $$T^{(\nabla)}$$ is (up to a sign) the torsion tensor of the connection $$\nabla$$. If you are dealing with the Levi-Civita connection $$T^{(\nabla)}=0$$ by definition. All that implies that, referring to vector fields, you can indifferently use the standard derivative or the covariant one (assumed to be either Levi-Civita or torsion-free) in coordinates to compute a Lie derivative.
Secondly, for scalar fields, $${\cal L}_X f = X(f) = \nabla_X(f)$$ by definition of Lie derivative and covariant derivative. (It can be used as a definition of Lie deriviative of a scalr field or it can be obtained by an autonomous definition based on the flow of $$X$$.)
As a third step, the Lie derivative of $$1$$-forms $$\omega$$ satisfies $${\cal L}_X\langle Y, \omega \rangle = \langle {\cal L}_X Y, \omega \rangle + \langle Y, {\cal L}_X \omega \rangle\:.$$ that is $${\cal L}_X\langle Y, \omega \rangle - \langle {\cal L}_X Y, \omega \rangle = \langle Y, {\cal L}_X \omega \rangle\:.$$ (Again, it can be used as a definition or it can be obtained by an autonomous definition based on the flow of $$X$$.) Since the terms in the left-hand side can be written using indifferently the standard derivative or the covariant derivative in view of the first two steps, $$\langle Y, {\cal L}_X \omega \rangle$$ and thus $${\cal L}_X \omega$$ itself can be expanded indifferently in terms of the standard derivative or of the covariant derivative.
In summary, for scalar fields, vector fields and $$1$$-forms, the use of standard or covariant derivative are completely equivalent in computing a Lie derivative.
Finally, the extension to tensor fields is implemented just by assuming the (pointwise) Leibnitz rule: $${\cal L}_X T \otimes T' = ({\cal L}_X T) \otimes T'+ T \otimes ({\cal L}_X T')$$ (Again, it can be used as a definition or it can be obtained by an autonomous definition based on the flow of $$X$$.) So that the result propagates to every tensor order.