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*}
Hence
\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?
