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Considering a partial derivative of a vector field $w^a$ in x-direction (also called here 1-direction) I can write it as $$\frac{ \partial w^a}{\partial x^1 } = \partial_1 w^a - \Gamma^a_{1c} w^c + \Gamma^a_{b1} w^b = v^b \nabla_b w^a - w^b \nabla_b v^a \tag{1}$$

where $\Gamma$ is assumed to be the Levi-Civita connection which is symmetric on the last 2 indices and $v^1=1$. $v^{a \neq 1}=0$. $v$ is to be considered as a vector on a manifold. So on the left side there is an expression which apparently not coordinate covariant whereas the right side is covariant. How does this fit together respectively where am I misled ? This post is inspired by the appendix C.2 of Wald's book of General Relativity (1983).

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  • $\begingroup$ I changed some parts of the question in order to avoid misunderstandings. $\endgroup$ Sep 4, 2023 at 15:39
  • $\begingroup$ I edited the answer to match the new definition $\endgroup$
    – LolloBoldo
    Sep 4, 2023 at 16:20

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Your expression has an issue since $$ \mathcal{L}_v w \neq v^b \nabla_b w^a-w^b\nabla_bv^a$$

As you can clearly see, your RHS does have a free index, and hence is not covariant exactly like $\partial_1 w^a$, while $\mathcal{L}_v w$ is index-free hence covariant.

The correct equivalence is:

$$\mathcal{L}_v w = e_a \big(\mathcal{L}_v w\big)^a = e_a\big(v^b \nabla_b w^a-w^b\nabla_bv^a\big) =e_a\big[v^b (\partial_b w^a +\Gamma^a_{bc}w^c) -w^b(\partial_b v^a +\Gamma^a_{bc}v^c)\big] = e_a(v^b\partial_bw^a -w^b\partial_bv^a) $$

Given your definition of $v^a$ it is clear that:

$$\partial_b v^a =0$$ and

$$\frac{\partial w^a}{\partial x^1} = v^b \partial_b w^a $$

Hence for your choice of $v^a$ you have:

$$\frac{\partial w^a}{\partial x^1} = v^b \partial_b w^a = (\mathcal{L}_v w)^a$$

Where the RHS is in fact non covariant.

If you want your expression to equal the Lie derivative then it should become:

$$(v^b\partial_bw^a) e_a = \mathcal{L}_v w$$


Important note: The correct nomenclature in the OP post should be covariant for indexed espressions and invariant for index free expressions, instead of not covariant for the former and covariant for the latter

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  • $\begingroup$ Let's the vocabulary. My point is that in equation (1) the very right side transforms like vector whereas the very left side does not. How is it possible? $\endgroup$ Sep 4, 2023 at 18:03
  • $\begingroup$ Your new edited version transform as a vector both the RHS and the LHS. $v^b \partial_b w^a$ transform as a vector and $(\mathcal{L}_v w)^a$ also transform as a vector $\endgroup$
    – LolloBoldo
    Sep 4, 2023 at 18:06
  • $\begingroup$ Under coordinate transformations the both transform as $X^a \rightarrow \Lambda^a_bX^b$ $\endgroup$
    – LolloBoldo
    Sep 4, 2023 at 18:07
  • $\begingroup$ Well in order to transform $\partial_1 w^a$ one has commute $\Lambda _b^a$ with the partial derivative $\partial_1$ without creating an additional term. I fear that that is not possible. $\endgroup$ Sep 4, 2023 at 18:51
  • $\begingroup$ This is true, im not sure if they commute. Nonetheless the RHS and the LHS are still the same expression (see the proof above) so if the LHS doesn't transform as a vector, neither does the RHS. I'll check the Commutation property as soon as I have time and I will edit my answer $\endgroup$
    – LolloBoldo
    Sep 4, 2023 at 20:40

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