Another question of index notation in tensor calculus

Question

I have this equation $$\nabla_{a}(g_{bc}\lambda^{c})=(\nabla_{a}g_{bc})\lambda^{c}+g_{bc}\nabla_{a}\lambda^{c}$$

And making some calculations $$ \lambda^{c} (\nabla_{a}g_{bc})= \nabla_{a}(g_{bc}\lambda^{c})- g_{bc}\nabla_{a}\lambda^{c} $$

$$ \lambda^{c} (\nabla_{a}g_{bc})= \nabla_{a}(\lambda_{b})- g_{bc}\nabla_{a}\lambda^{c} $$

And I want to get an expression for $\nabla_{a}g_{bc}$, so I make this

$$ \lambda^{c} (\nabla_{a}g_{bc})= \frac{\lambda}{\lambda} (\nabla_{a}(\lambda_{b})- g_{bc}\nabla_{a}\lambda^{c} ) $$

With $\lambda=\lambda^{e}\lambda_{e}$. Then $$ \lambda^{c} (\nabla_{a}g_{bc})= \frac{\lambda^{e}\lambda_{e}}{\lambda} (\nabla_{a}(\lambda_{b})- g_{bc}\nabla_{a}\lambda^{c} ) $$ Changing indexes $$ \lambda^{e} (\nabla_{a}g_{be})= \frac{\lambda^{e}\lambda_{e}}{\lambda} (\nabla_{a}(\lambda_{b})- g_{bc}\nabla_{a}\lambda^{c} ) $$ Then $$ (\nabla_{a}g_{be})= \frac{\lambda_{e}}{\lambda} (\nabla_{a}(\lambda_{b})- g_{bc}\nabla_{a}\lambda^{c} ) +k_{a} $$ With $k_{a}$ such that $\lambda^{a}k_{a}=0$

This is right? And if is not, there is a way that I can get an expression for $\nabla_{a}g_{be}$

Edit

Suppose that we are not working in $\nabla_{a}g_{bc}=0$

Answer 1

From what I can tell, you are simply rearranging your original expression

$$\nabla_{a}(g_{bc}\lambda^{c})=(\nabla_{a}g_{bc})\lambda^{c}+g_{bc}\nabla_{a}\lambda^{c}$$

and attempting to isolate an expression for $\nabla_a g_{bc}$. This is not possible, simply because every possible covariant derivative "candidate" satisfies the "product rule" exactly as this one does. Without additional information, you cannot determine how $\nabla$ acts on the metric.

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From an index manipulation point of view, in general one cannot "invert" contraction, in the same way that one cannot simply invert a dot product. Explicitly, your calculation fails here:

$$\lambda^{c} (\nabla_{a}g_{bc})= \frac{\lambda^{e}\lambda_{e}}{\lambda} (\nabla_{a}(\lambda_{b})- g_{bc}\nabla_{a}\lambda^{c} )$$

The index $e$ on the right hand side is contracted over, and therefore does not constitute a free index. When you relabel the left hand side $$\lambda^e(\nabla_a g_{be}) = \ldots$$

and then try to remove the $\lambda^e$ by contracting with $\lambda_e$, you are forgetting that $e$ is already a dummy index. In short, to re-use more familiar notation, you have that $$\vec \lambda \cdot \vec F = \ldots$$ and are trying to get an expression for $\vec F$, which you simply cannot do in the absence of additional information.

Answer 2

I want to get an expression for $\nabla_{a}g_{bc}$

If you are using the Levi-Civita connection, then the covariant derivative of the metric is zero by the definition of that connection.

See this PSE question.

If you don’t know which connection you’re supposed to be using, you’re almost certainly supposed to be using Levi-Cevita’s. It’s the unique torsion-free connection that preserves lengths and angles under parallel transport.