Assume that I have the expression for a Christoffel symbol:

$$ \Gamma^\mu_{\alpha \beta}=\frac{1}{2}g^{\mu \lambda}(\partial_\alpha g_{\beta \lambda}+\partial_\beta g_{\alpha \lambda} - \partial_\lambda g_{\alpha \beta}).\tag{1}$$

If the metric $g_{\mu\nu}$ is diagonal then the identity $$g^{\mu\lambda}g_{\lambda\nu}=\delta^\mu_\nu\tag{2}$$ simplifies to the expression $$g^{\mu\mu}g_{\mu\mu}=1.\tag{3}$$

Therefore is the following expression for the Christoffel symbol notationally correct?

$$\Gamma^\mu_{\alpha \beta}=\frac{1}{2g_{\mu \mu}}(\partial_\alpha g_{\beta \mu}+\partial_\beta g_{\alpha \mu} - \partial_\mu g_{\alpha \beta}).\tag{4}$$

Does it obey the Einstein summation convention correctly as the repeated $\mu$ index is not summed over?

If the expression is not correct how should it be written?


Ok I see the correct manipulation to get an all covariant form using Einstein notation:

\begin{eqnarray}\tag{5} \Gamma_{\gamma\alpha\beta}&=&g_{\gamma\mu}\Gamma^\mu_{\alpha\beta}\\ &=&\frac{1}{2}g_{\gamma\mu}g^{\mu\lambda}(\partial_\alpha g_{\beta\lambda} + \partial_\beta g_{\alpha \lambda} - \partial_\lambda g_{\alpha\beta})\\ &=&\frac{1}{2}\delta^\lambda_\gamma(\partial_\alpha g_{\beta\lambda} + \partial_\beta g_{\alpha \lambda} - \partial_\lambda g_{\alpha\beta})\\ &=&\frac{1}{2}(\partial_\alpha g_{\beta\gamma} + \partial_\beta g_{\alpha \gamma} - \partial_\gamma g_{\alpha\beta}) \end{eqnarray}

That's correct isn't it?


1 Answer 1


Your last expression is not valid, for two reasons: first, any given index can only occur twice per term in the Einstein convention, once as an upper index and once as a lower index. Remember that when an index is repeated, it means you sum over it with the metric: $$T^a T_a = \sum_{a,b} g_{ab} T^a T^b$$ You have terms like $\frac{1}{2g_{\mu\mu}}\partial_\alpha g_{\beta\mu}$ which contain $\mu$ three times. That's meaningless in Einstein notation.

The other reason is that the metric doesn't necessarily commute with the partial derivative: $g\partial \neq \partial g$. So you can't convert a factor like $g^{\mu\lambda} \partial_{\alpha}g_{\beta \lambda}$ into $\partial_{\alpha}g^{\mu\lambda} g_{\beta\lambda}$, which I think is what you had in mind there. (You could put in an extra term to account for the commutator if you wanted to: $g\partial = \partial g + [g,\partial]$.)

By the way, it's not true that contracting the result of $g^{\mu\lambda}g_{\lambda\nu}$ gives you $1$. You get $g^{\mu\lambda}g_{\lambda\mu} = \delta^{\mu}_{\ \mu} = \sum_{\mu} 1$, which is the dimensionality of the space. In normal 3+1D space, this gives 4.

  • 1
    $\begingroup$ I'd be interested to know what someone thinks is wrong with this answer.... $\endgroup$
    – David Z
    Commented Nov 1, 2015 at 13:50
  • $\begingroup$ Thanks for your very helpful comments but I still think $g^{\mu\lambda}g_{\lambda\nu}=\delta^\mu_\nu$ is correct (source Sean Carroll's grtiny document). $\endgroup$ Commented Nov 1, 2015 at 14:03
  • 1
    $\begingroup$ @JohnEastmond He never said $g^{\mu\lambda}g_{\lambda\nu}=\delta^\mu_\nu$ wasn't correct... $\endgroup$
    – jinawee
    Commented Nov 1, 2015 at 14:11
  • $\begingroup$ ok - I see - sorry I was wrong! $\endgroup$ Commented Nov 1, 2015 at 14:16

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