According to this site, the Lie derivative of a $(0,2)$-tensor is $$ \mathcal{L}_XT_{ab}=\partial_XT_{ab}+T_{cb}\partial_aX^c+T_{ac}\partial_bX^c $$

However, according the same website, the Lie derivative of the metric is $$ \mathcal{L}_Xg_{ab}=\nabla_Xg_{ab}+g_{cb}\nabla_aX^c+g_{ac}\nabla_bX^c $$

and if the connection is metric compatible, the first term vanish.

My question is, why do we change $\partial$ into $\nabla$ here? In fact, in a note, I see both two expressions for the Lie derivative of the metric. Are they really equivalent?


I tried to expand the second equation, and turns out it indeed is the first one. In fact, $$ \begin{align} &\nabla_Xg_{ab}+g_{cb}\nabla_aX^c+g_{ac}\nabla_bX^c\\ =&X^c(\partial_cg_{ab}-\Gamma^{d}_{ac}g_{db}-\Gamma^d_{cb}g_{ad})+g_{cb}\partial_aX^c+g_{ac}\partial_bX^c+g_{cb}\Gamma^{c}_{ad}X^d+g_{ac}\Gamma^c_{ad}X^d\\ =&\partial_Xg_{ab}+g_{cb}\partial_aX^c+g_{ac}\partial_bX^c \end{align} $$

All the $\Gamma$'s cancel out. But now my question is, is it generally true that when expanding $\mathcal{L}_XT$ for some tensor $T$, we can use either partial derivative or covarient derivative?

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    $\begingroup$ I believe if you would explicitly write the second equation in terms of the connection coefficients, then you would find that they cancel and you will get the first equation (but I'm not 100% sure). $\endgroup$ – Hunter May 12 '14 at 13:47
  • $\begingroup$ Assuming a metric compatible connection, the Lie derivative of our metric along a field $X$ is given by, $\mathcal{L}_X g_{ab}=\nabla_a X_b + \nabla_b X_a $ We can expand the expression by inserting the explicit covariant derivative with the Christofel symbols, $\mathcal{L}_X g_{ab} = \partial_a X_b + \Gamma^c_{ab}X_c + \partial_b X_a+\Gamma^c_{ba}X_c$. $\endgroup$ – JamalS May 12 '14 at 13:52
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    $\begingroup$ @JamalS, so? Sorry I didn't get your point.... $\endgroup$ – HanXu May 12 '14 at 13:54
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    $\begingroup$ @Hunter, exactly they are! Thanks!...However, I noticed it is true not only for the metric, but for any $(0,2)$-tensor. So is it a general fact that we can replace the partial derivative to covarient derivative here? $\endgroup$ – HanXu May 12 '14 at 13:55
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    $\begingroup$ @HanXu yeah I believe so, though again I'm not 100% sure. Sean Carroll's book has a really nice appendix about Lie derivatives and I vaguely remember that you can always make that replacement. But I don't have the book in front of me, so I can't double check right now. $\endgroup$ – Hunter May 12 '14 at 13:59

This is the first time I face this problem, I think that the answer is positive provided the connection is torsion free as, in fact, happens for the Levi-Civita connection associated with a metric. Here is my proof.

Let us indicate the Lie derivative with respect to the vector field $Z$ by ${\cal L}_Z$. The action on tensor fields is completely determined by the following requirements.

[0] ${\cal L}_Z(X) = [Z,X]$.

[1] It trivially acts on scalar fields: ${\cal L}_Z f = Z(f)$.

[2] It commutes with contractions: ${\cal L}_Z(\langle X, \omega \rangle) = \langle {\cal L}_Z(X), \omega \rangle + \langle X, {\cal L}_Z(\omega) \rangle$.

[3] It acts as a derivative with respect to the tensor product: ${\cal L}_Z (T\otimes U) = ({\cal L}_Z(T) )\otimes U + T \otimes {\cal L}_Z(U)$.

The requirements [0] and [2] completely define the actions on covariant vector fields in terms of the action of vector fields on scalar fields (components of covariant fields): $${\cal L}_Z(\omega) = \langle \partial_{x^a}, {\cal L}_Z(\omega) \rangle dx^a = Z(\langle \partial_{x^a}, \omega\rangle ) dx^a - \langle Z(\partial_{x^a}), \omega\rangle dx^a = Z(\omega_a) dx^a - \langle [Z, \partial_{x^a}], \omega \rangle dx^a\:,$$ i.e., $${\cal L}_Z(\omega)_a = Z(\omega_a) - \langle [Z, \partial_{x^a}], dx^b \rangle \omega_b\:. \tag{1}$$

From the definition of torsion we have that, if $\nabla$ is a torsion-free connection, then $$[Z,X] = \nabla_Z X - \nabla_X Z$$
since the torsion is nothing but the difference of the two sides. From [0] we conclude that $${\cal L}_Z(X) = \nabla_Z X - \nabla_X Z\:, \tag{2}$$ or, in components: $$({\cal L}_Z(X))^a = (Z^b\nabla_b X)^a - (\nabla_b Z)^aX^b\:,$$ and this identity just says that, for torsion-free connctions, we can replace derivatives for covariant ones when computing the Lie derivative of contravariant vector fields.

Let us pass to covariant vector fields. From (1) and (2) we have that $${\cal L}_Z(\omega)_a = \nabla_Z(\omega_a) - \langle \nabla_Z( \partial_{x^a}), dx^b \rangle \omega_b + \langle \nabla_{\partial_{x^a}}(Z), dx^b \rangle \omega_b\:,$$ that is $${\cal L}_Z(\omega)_a = (Z^b\nabla_b\omega)_a + \langle \nabla_{a}(Z), dx^b \rangle \omega_b\:,$$ namely $${\cal L}_Z(\omega)_a = (Z^b\nabla_b\omega)_a + (\nabla_{a}Z)^b\omega_b\:,$$ and this identity just says that, for torsion-free connctions, we can replace derivatives for covariant ones when computing the Lie derivative of covariant vector fields.

Since generic tensor fields can be constructed as a linear combination of tensor products of contravariant and covariant vector fields, from [3] the found results can easily be generalized to all types of tensor fields. In computing the Lie derivative of a tensor field we can always replace the standard derivative for the covariant one at every occurrence provided the connection is torsion free.

  • $\begingroup$ According to my book, the connection must indeed be torsion free. The book does not provide the above proof though which is really nice to see. $\endgroup$ – Hunter May 12 '14 at 16:19
  • $\begingroup$ Thanks, so my proof should be true... at least the statement :) $\endgroup$ – Valter Moretti May 12 '14 at 16:23
  • $\begingroup$ Yes, according to Sean Carroll the statement is definitely true ;). But he doesn't provide a formal proof, but introduces it more as an identity that you can convince yourself of. Also, as a sidenote, you can use \tag{1} if you want to number your equations. $\endgroup$ – Hunter May 12 '14 at 16:37
  • $\begingroup$ Yes, I know \tag{something} but I always forget to use it... $\endgroup$ – Valter Moretti May 12 '14 at 16:38
  • $\begingroup$ Ah this clears up a lot of things for me: in particular, once we adopt $\mathcal{L}_\xi g_{ab} = \nabla_a\xi_b + \nabla_b\xi_a$ where $\xi$ is a vector field, then we $\textit{cannot}$ undo the covariant into partial derivative, since OP and @ValterMoretti have shown that it's only interconvertible in its full form. $\endgroup$ – Everiana Apr 12 '19 at 5:37

For any vector field $X$, $\nabla_X$ is a derivation of the tensor algebra and thus can be decomposed as $$ \nabla_X = \mathcal L_Y + S $$ for some vector field $Y$ and an endomorphism $S$ of the tangent space (see eg Kobayashi/Nomizu, proposition 3.3).

For a function $f$, we have $$ \nabla_X f = \mathcal L_X f $$ and thus $Y=X$.

Given an arbitrary vector field $Z$, by definition of torsion $$ \nabla_X Z = \mathcal L_X Z + \nabla_Z X + T(X,Z) $$ In case of torsion-free connections, this yields $$ S: Z \mapsto \nabla_Z X $$ ie $$ \mathcal L_X = \nabla_X - \nabla_{(\cdot)} X $$ As $\partial$ is just a local connection on the coordinate patch, this actually yields both coordinate expressions mentioned in the question.

Note that in case of covariant tensors, the positive sign is due to the fact that covectors need to transform with the negative transpose of the endomorphism (see eg Kobayashi/Nomizu, proposition 2.13).

  • $\begingroup$ far from the most straight-forward way to prove this, but what the heck ;) $\endgroup$ – Christoph May 12 '14 at 17:03

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