Review

In the first lecture on General Relativity, the concept of parallel transport was introduced. The parallel transport operator acts like a derivative on sums and products (it is a "derivation"), with the additional properties assumed:

$$\delta_{\vert \vert}(\Phi,dx^{\sigma})=0,$$ (1)

for any scalar $\Phi$, and for a contravariant vector,

$$\delta_{\vert \vert}(A^{\mu},dx^{\sigma}) = -\Gamma_{\alpha \sigma}^{\mu} A^{\alpha}dx^{\sigma}.$$ (2)

Then the following result was proved by considering the parallel transport of the scalar $g_{\mu \nu}A^{\mu}B^{\nu}$:

$$\delta_{\vert \vert}(g_{\mu \nu},dx^{\sigma})=g_{\alpha \nu}\Gamm... ...ma}^{\alpha} dx^{\sigma}+g_{\mu \beta} \Gamma_{\nu \sigma}^{\beta} dx^{\sigma}.$$ (3)

The covariant derivative of the metric tensor is then defined in terms of the actual differential change, minus the change due to parallel transport:

$$g_{\mu \nu ; \sigma}dx^{\sigma}=(g_{\mu \nu}+g_{\mu \nu, \sigma}dx^{\sigma}) -(g_{\mu \nu}+\delta_{\vert \vert}(g_{\mu \nu},dx^{\sigma})$$ (4)

$$=g_{\mu \nu , \sigma}dx^{\sigma}-g_{\alpha \nu}\Gamma_{\mu \sigma}^{\alpha} dx^{\sigma}-g_{\mu \beta} \Gamma_{\nu \sigma}^{\beta} dx^{\sigma}.$$ (5)