I'm doing an exercise where I needed to find, for an arbitary $4$-vector $V^\mu$ $$ V_{r;\theta;\phi}-V_{r;\phi;\theta}=\ ???$$ for the space-time metric $$ ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}=\left(1-\frac{r_g}{r}\right)c^2dt^2-\left(1+\frac{r_g}{r}\right)dr^2-r^2g_\Omega $$ Note that here $;$ denote the covariant differential so that $$V_{r;\theta;\phi}=\frac{D }{Dx^\phi}\left(\frac{D V_r}{Dx^\theta}\right)$$
I'm able to show that the required quantity turn out to be
$$V_{r;\theta;\phi}-V_{r;\phi;\theta}=\frac{1}{r}\left(\frac{\partial V_\phi}{\partial x^\theta}-\frac{\partial V_\theta}{\partial x^\phi}\right)$$
It's asked to interpret this result which I'm not able to do. I can understand that the right-hand side is a radial component of the curl. But I'm not able to see any geometrical interpretation of why this should be equal to left-hand side. Please help me with this.