Raising and lowering indices with the metric tensor

The metric tensor plays the extremely important role in tensor analysis, of allowing one to lower or raise indices on vectors and tensors. For example, starting with an arbitrary contravariant vector $A^{\mu}$, one can obtain a covariant vector that represents the same physical quantity by summing as follows:

$$A_{\mu} = g_{\mu \nu}A^{\nu}.$$ (6)

We also introduce the inverse of the metric tensor, the matrix $g^{\mu \nu}$, as follows:

$$g_{\mu \nu}g^{\nu \alpha} = \delta_{\mu}^{\alpha},$$ (7)

where the quantity on the right side of the above equation is the Kronecker delta. Then also

$$A^{\mu} = g^{\mu \nu}A_{\nu}.$$ (8)

Now we carry out the following process. (1) Carry a contravariant vector by parallel transport along $dx^{\sigma}$, then lower the index. (2) First lower the index,

These results must be equal. Keeping only terms of first order in $dx^{\sigma}$, gives

$$g_{\nu \mu}A^{\mu}+ g_{\nu \mu , \sigma}A^{\mu}dx^{\sigma}-g_{\nu... ...^{\sigma})A^{\mu}-g_{\nu \mu}\Gamma_{\alpha \sigma}^{\mu}A^{\alpha}dx^{\sigma}.$$ (9)

Therefore:

$$\delta_{\vert \vert}(g_{\nu \mu},dx^{\sigma})A^{\mu}= g_{\nu \mu , \sigma}A^{\mu}dx^{\sigma},$$ (10)

and so we obtain the following simple relation for the parallel transport of the metric tensor, since both $A^{\mu}$ and $dx^{\sigma}$ are arbitrary:

$$\delta_{\vert \vert}(g_{\nu \mu},dx^{\sigma})=g_{\nu \mu,\sigma}dx^{\sigma}.$$ (11)

Combining this result with the previous results implies that the covariant derivative of the metric tensor vanishes:

$$g_{\nu \mu;\sigma}=g_{\nu \mu , \sigma}-g_{\alpha \nu}\Gamma_{\mu \sigma}^{\alpha} -g_{\mu \beta} \Gamma_{\nu \sigma}^{\beta}=0.$$ (12)

It is not difficult to show also that the metric tensor inverse has zero covariant derivative, since the Kronecker delta is a scalar having the same values (0 or 1) in all reference frames. Therefore under covariant differentiation, the metric tensor acts like a constant.