(1) From $A$ to $B$, the absolute change is $\PadTo[r]{-A_{\mu\sigma}\, dx_{\sigma}}{A_{\mu\nu}\, dx_{\nu}}$, calculated for~$x_{\sigma}$\footnotemark.\footnotetext {We suspend the summation convention since $dx_{\nu}$ and $dx_{\sigma}$ are edges of a particular mesh. The convention would give correct results; but it goes too fast, and we cannot keep pace with it.}% (2) From $B$ to $C$, the absolute change is $\PadTo[r]{-A_{\mu\sigma}\, dx_{\sigma}}{A_{\mu\sigma}\, dx_{\sigma}}$, calculated for~$x_{\nu} + dx_{\nu}$. (3) From $C$ to $D$, the absolute change is $\PadTo[r]{-A_{\mu\sigma}\, dx_{\sigma}}{-A_{\mu\nu}\, dx_{\nu}}$, calculated for~$x_{\sigma} + dx_{\sigma}$. (4) From $D$ to $A$, the absolute change is $-A_{\mu\sigma}\, dx_{\sigma}$, calculated for~$x_{\nu}$. \\ Combining (2) and (4) the net result is the difference of the changes $A_{\mu\sigma}\, dx_{\sigma}$, at $x_{\nu} + dx_{\nu}$ and at $x_{\nu}$~respectively. We might be tempted to set this difference down as \[ \frac{\dd}{\dd x_{\nu}} (A_{\mu\nu}\, dx_{\sigma})\, dx_{\nu}. \] But as already explained that would give only the difference of the mathematical components and not the ``absolute difference.'' We must take the covariant derivative instead, obtaining (since $dx_{\sigma}$~is the same for (2) and~(4)) \[ A_{\mu\sigma\nu}\, dx_{\sigma}\, dx_{\nu}. \] Similarly (3) and (1) give \[ -A_{\mu\nu\sigma}\, dx_{\nu}\, dx_{\sigma}, \] so that the total absolute change round the circuit is \[ (A_{\mu\sigma\nu} - A_{\mu\nu\sigma})\, dx_{\nu}\, dx_{\sigma}. \Tag{(33.2)} \] We should naturally expect that on returning to our starting point the absolute change would vanish. How could there have been any absolute change on balance, seeing that the vector is now the same~$A_{\mu}$ that we started with? Nevertheless in general $A_{\mu\nu\sigma} \neq A_{\mu\sigma\nu}$, that is to say the order of covariant differentiation is not permutable, and \Eq{(33.2)} does not vanish. \PageSep{70} That this result is not unreasonable may be seen by considering a two-dimensional space, the surface of the ocean. If a ship's head is kept straight on the line of its wake, the course is a great circle. Now suppose that the ship sails round a circuit so that the final position and course are the same as at the start. If account is kept of all the successive changes of course, and the angles are added up, these will not give a change zero (or~$2\pi$) on balance. For a triangular course the difference is the well-known ``spherical excess.'' Similarly the changes of velocity do not cancel out on balance. Here we have an illustration that the absolute changes of a vector do not cancel out on bringing it back to its initial position.