# Alternate derivation of the covariant derivative of a contravariant vector

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?

• Interesting you denote covariant derivatives with ${}_:$. I've only ever seen ${}_;$.
– J.G.
Commented Dec 23, 2022 at 16:34
• I'm just following Dirac's notation. It may be archaic. He also refers to the metric tensor as the "fundamental tensor". Commented Dec 23, 2022 at 17:13
• How interesting.
– J.G.
Commented Dec 23, 2022 at 17:19
• 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. Commented Dec 23, 2022 at 23:20

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 $$\mu,\,\phi$$.