# Commuting of the covariant derivative: Menzel's Mathematical Physics

Menzel defines covariant differentiation as equivalent to partial differentiation with respect to the general coordinates. “To indicate the covariant nature of the differential operator, set $$\frac{\partial}{\partial q^{i}}\equiv D_{i} \tag{1}$$"

He then provides the following which I have greatly abbreviated. Consider $$D_{i}\mathfrak{u}=D_{i}\left(\mathfrak{e}_{k}u^{k}\right)=\mathfrak{e}_{k}D_{i}u^{k}+u^{k}D_{i}\mathfrak{e}_{k}=\mathfrak{e}_{k}\frac{\partial u^{k}}{\partial q^{i}}+u^{k}D_{i}\mathfrak{e}_{k} \tag{2}.$$

“The operation $D_{i}u^{k}$ amounts to ordinary differentiation of the function $q^{i}$.” But we still have $D_{i}\mathfrak{e_{j}}$ to contend with. To that end define the following:

$$D_{i}\mathfrak{e}_{j}\equiv\Gamma_{ij}^{k}\mathfrak{e}_{k} \tag{3}$$

$$\Gamma_{ij}^{k}\mathfrak{e}_{k}=\Gamma_{ij}^{k}\mathfrak{e}^{l}g_{kl}\equiv\Gamma_{ij,k}\mathfrak{e}^{k} \tag{4}$$

Dot-multiply the first of these through by $\mathfrak{e}^{k}$

$$\Gamma_{ij}^{k}=\mathfrak{e}^{k}\cdot D_{i}\mathfrak{e}_{j} \tag{5}$$

Take the covariant derivative of $\delta_{\ j}^{k}$

$$D_{i}\delta_{\ j}^{k}=D_{i}(\mathfrak{e}^{k}\cdot\mathfrak{e}_{j})=\mathfrak{e}^{k}\cdot D_{i}\mathfrak{e}_{j}+\mathfrak{e}_{j}\cdot D_{i}\mathfrak{e}^{k}=0 \tag{6}$$

Therefore

$$\Gamma_{ij}^{k}=-\mathfrak{e}_{j}\cdot D_{i}\mathfrak{e}^{k} \tag{7}$$

and

$$D_{i}\mathfrak{e}^{k}=-\Gamma_{ij}^{k}\mathfrak{e}^{j} \tag{8}.$$

My reasoning to justify the relationship immediately above is that (7) expresses the projection of the $D_{i}\mathfrak{e}^{k}$ onto the $\mathfrak{e}_{i}$ basis. If $D_{i}\mathfrak{e}^{k}$ is replaced in (7) by the right-hand side of (8), the equivalence holds. The two sides of (8) are therefore component equivalent.

Menzel defines the covariant basis vectors as $$\mathfrak{e}_{i}\equiv\frac{\partial\mathfrak{r}}{q^{i}}=D_{i}\mathfrak{r} \tag{9}$$

Where the $\mathfrak{r}$ is the position.

He then asserts that (7) leads to $$\Gamma_{ij}^{k}=\mathfrak{e}^{k}\cdot D_{i}D_{j}\mathfrak{r}=\mathfrak{e}^{k}\cdot D_{j}D_{i}\mathfrak{r}=\mathfrak{e}^{k}\cdot D_{j}\mathfrak{e}_{i}=\Gamma_{ji}^{k} \tag{10}$$

The gammas are therefore symmetrical on the lower indices.

I am not following his reasoning leading to (10). What does (7) have to do with (10)?

$$D_{i}D_{j}\mathfrak{r}=D_{j}D_{i}\mathfrak{r}$$ follows directly from the fact that $\mathfrak{r}$ can be referred to a fixed orthonormal basis, and the self-commuting of partial differentiation.
$$\mathfrak{r}=x^{i}\hat{\mathfrak{e}}_{i}$$
$$\frac{\partial^{2}\mathfrak{r}}{\partial q^{j}\partial q^{k}}=\mathfrak{\hat{e}}_{i}\frac{\partial^{2}x^{i}}{\partial q^{j}\partial q^{k}}$$