- #1

Jufa

- 101

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- TL;DR Summary
- I am struggling to proving the following identity

First, we shall mention that it is known that the covariant derivative of the metric vanishes, i.e ##\nabla_i g_{mn} = 0##.

Now I want tro prove the following:

$$ \nabla_i A_k = g_{kn}\nabla_i A^n$$

The demonstration I encounter takes advantage of the Leibniz rule:

$$ \nabla_i A_k = \nabla_i \big(g_{kn}A^n \big) = \nabla_i\big(g_{kn})A^n + g_{kn}\nabla_iA^n = 0 + g_{kn}\nabla_i A^n$$

Nevertheless, we must be careful at the last step. To my undersand, when writing ##\nabla_i\big(g_{kn})##, we are not referring (in this case) to taking the covariant derivative of the whole covariant tensor ##g_{kn}## and evaluating its ##kn## component, but instead we are fixing the value of ##n##. That is, it is not the same to compute the covariant derivative of the #mn# component of a certain tensor ##T^{mn}##, which will be just the partial derivative of this component, than to compute the ##mn## component of the whole tensor's covariant derivative, which will involve christoffel symbols. Following this reasoning, when I compute ##\nabla_i\big(g_{kn})A^n## I get the following:

$$ \nabla_i\big(g_{kn})A^n = \Big( \partial g_{kn} - \Gamma^l_{ik} g_{ln} \Big) A^n $$

Which clear does not vanish in general, since the quantity that we know that actually vanishes is:

$$ 0 = \nabla_i\big(g_{kn})= \partial g_{kn} - \Gamma^l_{ik} g_{ln} - \Gamma^l_{in} g_{kl} $$

Note that in the latter the covariant derivative is taken in the whole tensor.

Can someone help?

Thanks in advance.

Now I want tro prove the following:

$$ \nabla_i A_k = g_{kn}\nabla_i A^n$$

The demonstration I encounter takes advantage of the Leibniz rule:

$$ \nabla_i A_k = \nabla_i \big(g_{kn}A^n \big) = \nabla_i\big(g_{kn})A^n + g_{kn}\nabla_iA^n = 0 + g_{kn}\nabla_i A^n$$

Nevertheless, we must be careful at the last step. To my undersand, when writing ##\nabla_i\big(g_{kn})##, we are not referring (in this case) to taking the covariant derivative of the whole covariant tensor ##g_{kn}## and evaluating its ##kn## component, but instead we are fixing the value of ##n##. That is, it is not the same to compute the covariant derivative of the #mn# component of a certain tensor ##T^{mn}##, which will be just the partial derivative of this component, than to compute the ##mn## component of the whole tensor's covariant derivative, which will involve christoffel symbols. Following this reasoning, when I compute ##\nabla_i\big(g_{kn})A^n## I get the following:

$$ \nabla_i\big(g_{kn})A^n = \Big( \partial g_{kn} - \Gamma^l_{ik} g_{ln} \Big) A^n $$

Which clear does not vanish in general, since the quantity that we know that actually vanishes is:

$$ 0 = \nabla_i\big(g_{kn})= \partial g_{kn} - \Gamma^l_{ik} g_{ln} - \Gamma^l_{in} g_{kl} $$

Note that in the latter the covariant derivative is taken in the whole tensor.

Can someone help?

Thanks in advance.

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