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 $\nabla_{a}g_{be}$


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

  • $\begingroup$ $\nabla_{a}g_{be}=0$ is an independent calculation. It ensures the connection is metric compatible, i.e., it ensures vector angles and magnitudes are invariant when $g$ is parallel transported. If $\nabla_{a}g_{be}\neq0$ then it implies your metric may degenerate. The connection can be affine or Levi-Cevita - and you can use the Christoffel symbols either way - but there are potentially an infinite number of affine connections which may be compatible with the metric - and the Christoffel symbols may not posses the same symmetry properties as the Levi-Cevita connection. $\endgroup$ Commented Apr 30, 2020 at 0:56

2 Answers 2


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


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.

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.


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.

  • $\begingroup$ Oh that is the problem, I'm not using levi civita conecction. I know that in $\nabla_{a}g_{bc}=0$ when torsion is $0$. $\endgroup$
    – Nothing
    Commented Apr 29, 2020 at 18:05
  • $\begingroup$ Your previous questions indicate that you have just started to study differential geometry, so I find it somewhat... surprising!... that you are using a connection with torsion. Please ignore my answer. It is the right answer for the other 99.99% of readers. $\endgroup$
    – G. Smith
    Commented Apr 29, 2020 at 19:20

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