In Dirac's “General Theory of Relativity”, he derives the covariant derivative of a contravariant vector (his Eq. (10.7)):

$$ A^\mu_{: \sigma} = A^\mu_{, \sigma} + \Gamma^\mu_{\alpha \sigma}A^\alpha $$

from the definition of the covariant derivative of a covariant vector (his Eq. (10.1)):

$$ A_{\mu : \nu} = A_{\mu , \nu} - \Gamma^\alpha_{\mu\nu} A_\alpha $$

by applying the covariant derivative to the inner product $A^\mu B_\mu$.

I wanted to derive the covariant derivative of a contravariant vector directly, as follows. First, note that

\begin{eqnarray*} g^{\mu\nu} A_{\mu,\sigma} &=& (g^{\mu\nu} A_\mu)_{,\sigma} - g^{\mu\nu}_{,\sigma} A_\mu \\ &=& A^\nu_{,\sigma} - g^{\mu\nu}_{,\sigma} A_\mu \qquad (*) \end{eqnarray*}


\begin{eqnarray*} A^\nu_{:\sigma} &=& (g^{\mu\nu} A_\mu)_{:\sigma} \\ &=& g^{\mu\nu} A_{\mu :\sigma} \qquad (\text{using }g^{\mu\nu}_{:\sigma}=0) \\ &=& g^{\mu\nu} ( A_{\mu ,\sigma} - \Gamma^\alpha_{\mu\sigma} A_\alpha ) \\ &=& A^\nu_{ , \sigma} - g^{\mu\nu}_{, \sigma}A_\mu - g^{\mu\nu}\Gamma^\alpha_{\mu\sigma} A_\alpha \qquad \text{using}\, (*). \end{eqnarray*}

It remains, therefore, to show that

$$ - g^{\mu\nu}_{, \sigma}A_\mu - g^{\mu\nu}\Gamma^\alpha_{\mu\sigma} A_\alpha = \Gamma^\nu_{\alpha\sigma} A^\alpha$$

or equivalently

$$ - g^{\alpha\nu}_{, \sigma} A_\alpha - g^{\mu\nu}\Gamma^\alpha_{\mu\sigma} A_\alpha = \Gamma^\nu_{\beta\sigma} g^{\alpha\beta} A_\alpha$$

or equivalently

$$ - g^{\alpha\nu}_{, \sigma} - g^{\mu\nu}\Gamma^\alpha_{\mu\sigma} = g^{\mu\alpha}\Gamma^\nu_{\mu\sigma}$$

or equivalently

$$ - g^{\alpha\nu}_{, \sigma} = g^{\mu\nu}\Gamma^\alpha_{\mu\sigma} + g^{\mu\alpha}\Gamma^\nu_{\mu\sigma} \qquad (**) $$

Now, the right-hand side of $(**)$ is

\begin{eqnarray*} &=& (g^{\mu\nu}g^{\alpha\phi} + g^{\mu\alpha}g^{\nu\phi})\Gamma_{\phi\mu\sigma} \\ &=& (g^{\mu\nu}g^{\alpha\phi} + g^{\mu\alpha}g^{\nu\phi}) \frac{1}{2} (g_{\phi\mu,\sigma} + g_{\phi\sigma,\mu} - g_{\mu\sigma,\phi}) \\ &=& \frac{1}{2}(g^{\mu\nu}g^{\alpha\phi} + g^{\mu\alpha}g^{\nu\phi}) g_{\phi\mu,\sigma} + \frac{1}{2} (g^{\mu\nu}g^{\alpha\phi} + g^{\mu\alpha}g^{\nu\phi}) (g_{\phi\sigma,\mu} - g_{\mu\sigma,\phi}) \qquad (***). \end{eqnarray*}

Now we use Dirac’s Eq. (7.9) which provides a relationship between the derivatives of the metric tensor and inverse metric tensor:

$$ g^{\alpha\nu}_{,\sigma} = - g^{\alpha\mu} g^{\nu\lambda} g_{\mu\lambda,\sigma}.$$

Then the first term of $(***)$ is $-g^{\alpha\nu}_{,\sigma}$, which is precisely the required term on the left-hand side of $(**)$.

But what of the other four terms:

$$\frac{1}{2}(g^{\mu\nu}g^{\alpha\phi} + g^{\mu\alpha}g^{\nu\phi}) (g_{\phi\sigma,\mu} - g_{\mu\sigma,\phi}) ?$$

How does one show that this expression vanishes identically? I am having trouble demonstrating that. But it must be true. Any help?

  • 2
    $\begingroup$ Interesting you denote covariant derivatives with ${}_:$. I've only ever seen ${}_;$. $\endgroup$
    – J.G.
    Dec 23, 2022 at 16:34
  • $\begingroup$ I'm just following Dirac's notation. It may be archaic. He also refers to the metric tensor as the "fundamental tensor". $\endgroup$
    – Khun Chang
    Dec 23, 2022 at 17:13
  • 1
    $\begingroup$ How interesting. $\endgroup$
    – J.G.
    Dec 23, 2022 at 17:19
  • $\begingroup$ I think it's worth mentioning that what OP is asking requires a (pseudo-)Riemannian manifold and a connection compatible with the metric (which is the Levi-Civita connection). A connection does not need a metric to be defined, though and so the method of finding the result applying the covariant derivative to the covariant vector field (which is a 1-form) evaluated (what you've called "inner product) on the contravariant field is a more general method. $\endgroup$ Dec 23, 2022 at 23:20

1 Answer 1


That's actually the easy part of your calculation. Note that $g^{\mu\nu}g^{\alpha\phi} + g^{\mu\alpha}g^{\nu\phi}$ is $\mu\leftrightarrow\phi$-symmetric (because $g^{\kappa\lambda}$ is $\kappa\leftrightarrow\lambda$-symmetric), while $g_{\phi\sigma,\mu} - g_{\mu\sigma,\phi}$ is $\mu\leftrightarrow\phi$-antisymmetric. This implies their product vanishes when contracted as usual under $\mu,\,\phi$.

  • $\begingroup$ Damn, you're right it's almost obvious. Many thanks. $\endgroup$
    – Khun Chang
    Dec 23, 2022 at 17:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.