Commuting of the covariant derivative: Menzel's Mathematical Physics

Question

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)?

Answer 1

I misunderstood what Menzel had intended by "covariant nature of the differential operator". He did not mean that the differential operation is synonymous with covariant differentiation. As Menzel is wont to do, he proceeded to expose a series of non-trivial equivalences, and then tacked on a final equivalence which does not obviously follow from the immediately preceding form, but does, however follow from the original expression (given at the top of the previous page).

$$D_{i}D_{j}\mathfrak{r}=D_{j}D_{i}\mathfrak{r}$$ follows directly from the fact that $\mathfrak{r}$ can be referred to a fixed orthonormal basis, and the self-commuting of partial differentiation.

$$\mathfrak{r}=x^{i}\hat{\mathfrak{e}}_{i}$$

$$\frac{\partial^{2}\mathfrak{r}}{\partial q^{j}\partial q^{k}}=\mathfrak{\hat{e}}_{i}\frac{\partial^{2}x^{i}}{\partial q^{j}\partial q^{k}}$$