\emph{Hence the vanishing of the Riemann-Christoffel tensor is a necessary condition for flat space-time.} This condition is also \emph{sufficient}---if the Riemann-Christoffel tensor vanishes space-time must be flat. This can be proved as follows--- We have found (\SecRef{34}) that if \[ B_{\mu\nu\sigma}^{\epsilon} = 0, \Tag{(36.1)} \] it is possible to construct a uniform vector-field by parallel displacement of a vector all over the region. Let $A_{(\alpha)}^{\mu}$~be four uniform vector-fields given by $\alpha = 1$, $2$, $3$,~$4$, so that \[ (A_{(\alpha)}^{\mu})_{\sigma} = 0 \] or by~\Eq{(29.4)} \[ \frac{\dd A_{(\alpha)}^{\mu}}{\dd x_{\sigma}} = -\{\epsilon\sigma, \mu\}\, A_{(\alpha)}^{\epsilon}. \Tag{(36.2)} \] Note that $\alpha$~is not a tensor-suffix, but merely distinguishes the four independent vectors. We shall use these four uniform vector-fields to define a new coordinate-system distinguished by accents. Our unit mesh will be the hyperparallelopiped contained by the four vectors at any point, and the complete mesh-system will be formed by successive parallel displacements of this unit mesh until the whole region is filled. One edge of the unit mesh, given in the old coordinates by \[ dx_{\mu} = A_{(1)}^{\mu}, \] has to become in the new coordinates \[ dx_{\alpha}' = (1, 0, 0, 0). \] \PageSep{77} Similarly the second edge, $dx_{\mu} = A_{(2)}^{\mu}$, must become $dx_{\alpha}' = (0, 1, 0, 0)$; etc. This requires the law of transformation \[ dx_{\mu} = A_{(\alpha)}^{\mu}\, dx_{\alpha}'. \Tag{(36.3)} \] Of course, the construction of the accented coordinate-system depends on the possibility of constructing uniform vector-fields, and this depends on \Eq{(36.1)} \index{Uniform!mesh-system}% being satisfied.