I am trying to understand (rather than memorise) the derivation of the Christoffel symbols from the vanishing covariant derivative of the metric, the very first step is

\begin{equation} \label{eq:first} \nabla_\sigma g_{\mu\nu}=\partial_\sigma g_{\mu\nu}-\Gamma^\lambda_{\sigma\mu}g_{\lambda\nu}-\Gamma^\lambda_{\sigma\nu}g_{\mu\lambda}=0. \end{equation}

I'm wondering what is a good way to remember (or better yet work out) how to index the $\Gamma$ coefficients? How to know which indexes are subscripted on the $\Gamma$ and which index on the metric is involved in the summation with the upstairs index on the $\Gamma$? Is there a consistent way to work these out? Because at the moment I take at face value what is written in notes which is not very satisfying and feels like keeping the training wheels on.

The second issue I have is in the Leibniz when taking the covariant derivative of the metric. The metric is a rank-2 tensor and so can be written in the form

\begin{equation} g_{\mu\nu}\tilde e^\mu \otimes \tilde e^\nu. \end{equation}

When we take the covariant derivative we apply the product rule first (I assume) and then use Leibniz rule to take the derivatives of the dual basis vectors, i.e:

\begin{equation} \nabla_\sigma g_{\mu\nu}\tilde e^\mu \otimes \tilde e^\nu=(\partial g_{\mu\nu})\tilde e^\mu \otimes \tilde e^\nu+g_{\mu\nu}\bigl((\nabla_\sigma\tilde e^\mu)\otimes\tilde e^\nu\bigr)+g_{\mu\nu}\bigl((\nabla_\sigma\tilde e^\nu)\otimes\tilde e^\mu\bigr)\end{equation}

My question is, is this the correct method? It feels wrong to be taking the tensor product of the connection coefficient this way given that it isn't a true tensor. And if this is correct, where does the tensor product fit into Eq.1?

  • 2
    $\begingroup$ Wikipedia has a description of how to take the covariant derivative of any tensor. Look for the sentence that starts “Or, in words:”. $\endgroup$
    – G. Smith
    Commented Mar 5, 2020 at 1:34
  • $\begingroup$ That's very useful I hadn't read that section thank you. $\endgroup$
    – Charlie
    Commented Mar 5, 2020 at 1:39

1 Answer 1


1. I believe you have read all the corresponding info about Christoffel symbols on wikipedia and all you need is a good mnemonic. We need to deal with two things: indices and sign. The index part is actually simple. If you have a covector: $$ \nabla_iu_j=\partial_iu_j-\Gamma^{\square}_{\square\square}u_{\square} $$ then it's obvious that $\Gamma$ should contract with $u$, in other words the dummy index should go to $u$ and to top of $\Gamma$. The two lower free indices should go to bottom of $\Gamma$: $$ \nabla_iu_j=\partial_iu_j-\Gamma^{k}_{ij}u_{k} $$ For vector: $$ \nabla_iv^j=\partial_iv^j +\Gamma^{\square}_{\square\square}v^{\square} $$ it's again obvious that $v$ should have a dummy index and it doesn't matter to which of the lower indices of $\Gamma$ it goes, since $\Gamma$ is symmetric on lower indices (however, traditionally dummy index comes first). Free indices should go to corresponding places: one up, one down $$ \nabla_iv^j=\partial_iv^j +\Gamma^j_{ki}v^k. $$

Generally, if you have an arbitrary tensor: $$ \nabla_iT_{\color{red}{l}\color{green}m\ldots}^{\color{magenta}p\color{orange}q\ldots}=\partial_iT_{lm\ldots}^{pq\ldots} \color{red}{-\Gamma^k_{il}T}_{\color{red}km\ldots}^{pq\ldots} \color{green}{-\Gamma^k_{im}T}_{l\color{green}k\ldots}^{pq\ldots} \ldots \color{magenta}{+ \Gamma^p_{ki}T}_{lm\ldots}^{\color{magenta}kq\ldots} \color{orange}{+ \Gamma^q_{ki}T}_{lm\ldots}^{p\color{orange}k\ldots} \ldots $$ for each of the index you add a term as if this thing was just a vector or a covector.

Finally, the sign part is easy to remember since “vector-covector” and “plus-minus” are both naturally ordered.

2. That's a right approach. When you substitute $$ \frac{\partial\mathbf e^\mu}{\partial x^\sigma} = -\Gamma^\mu_{\sigma\nu}\mathbf e^\nu, $$ you will obtain the equation above.


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