Since $d^{2}$~is an invariant \begin{align*} g'_{\alpha\beta}\, dx_{\alpha}'\, dx_{\beta}' &= g_{\mu\nu}\, dx_{\mu}\, dx_{\nu} \\ &= g_{\mu\nu} A_{(\alpha)}^{\mu} A_{(\beta)}^{\nu}\, dx_{\alpha}'\, dx_{\beta}' \text{ by \Eq{(36.3)}.} \intertext{Hence} g'_{\alpha\beta} &= g_{\mu\nu} A_{(\alpha)}^{\mu} A_{(\beta)}^{\nu}. \end{align*} Accordingly, by differentiation, \begin{align*} \frac{\dd g_{\alpha\beta}'}{\dd x_{\sigma}} &= g_{\mu\nu} A_{(\alpha)}^{\mu}\, \frac{\dd A_{(\beta)}^{\nu}}{\dd x_{\sigma}} + g_{\mu\nu} A_{(\beta)}^{\nu}\, \frac{\dd A_{(\alpha)}^{\mu}}{\dd x_{\sigma}} + A_{(\alpha)}^{\mu} A_{(\beta)}^{\nu}\, \frac{\dd g_{\mu\nu}}{\dd x_{\sigma}} \\ &= -g_{\mu\nu} A_{(\alpha)}^{\mu} A_{(\beta)}^{\epsilon} \{\epsilon\sigma, \nu\} - g_{\mu\nu} A_{(\beta)}^{\nu} A_{(\alpha)}^{\epsilon} \{\epsilon\sigma, \mu\} + A_{(\alpha)}^{\mu} A_{(\beta)}^{\nu}\, \frac{\dd g_{\mu\nu}}{\dd x_{\sigma}} \end{align*} by~\Eq{(36.2)}. By changing dummy suffixes, this becomes \begin{align*} \frac{\dd g_{\alpha\beta}'}{\dd x_{\sigma}} &= A_{(\alpha)}^{\mu} A_{(\beta)}^{\nu} \left[-g_{\mu\epsilon} \{\nu\sigma, \epsilon\} - g_{\epsilon\nu} \{\mu\sigma, \epsilon\} + \frac{\dd g_{\mu\nu}}{\dd x_{\sigma}}\right]\displaybreak[0] \\ &= A_{(\alpha)}^{\mu} A_{(\beta)}^{\nu} \left[-[\nu\sigma, \mu] - [\mu\sigma, \nu] + \frac{\dd g_{\mu\nu}}{\dd x_{\sigma}}\right] \\ &= 0 \text{ by \Eq{(27.5)}.} \end{align*} Hence the $g_{\alpha\beta}'$ are constant throughout the region. We have thus constructed a coordinate-system fulfilling the condition that the $g$'s are constant, and it follows that the space-time is flat. It will be seen that a \emph{uniform} mesh-system, i.e.\ one in which the unit meshes are connected with one another by parallel displacement, is necessarily a Cartesian system (rectangular or oblique). Uniformity in this sense is impossible in space-time for which the Riemann-Christoffel tensor does not vanish, e.g.\ there can be no uniform mesh-system on a sphere.