Let a displacement~$A^{\mu}$ be carried by parallel displacement round a small circuit~$C$. The condition for parallel displacement is by~\Eq{(91.1)} \[ \frac{\dd A^{\mu}}{\dd x_{\nu}} = -\Gamma_{\nu\alpha}^{\mu} A^{\alpha}. \Tag{(92.1)} \] Hence the difference of the initial and final values is \begin{align*} \delta A^{\mu} &= \int_{C} \frac{\dd A^{\mu}}{\dd x_{\nu}}\, dx_{\nu} \\ &= -\int_{C} \Gamma_{\nu\alpha}^{\mu} A^{\alpha}\, dx_{\nu} \\ &= \frac{1}{2} \iint \left\{ \frac{\dd}{\dd x_{\sigma}} (\Gamma_{\nu\alpha}^{\mu} A^{\alpha}) - \frac{\dd}{\dd x_{\nu}} (\Gamma_{\sigma\alpha}^{\mu} A^{\alpha})\right\} dS^{\nu\sigma} \end{align*} by Stokes's theorem~\Eq{(32.3)}. \index{Stokes's theorem!application of}% The integrand is equal to \begin{align*} &\phantom{{}={}} A^{\alpha} \left(\frac{\dd}{\dd x_{\sigma}} \Gamma_{\nu\alpha}^{\mu} - \frac{\dd}{\dd x_{\nu}} \Gamma_{\sigma\alpha}^{\mu}\right) + \Gamma_{\nu\alpha}^{\mu}\, \frac{\dd A^{\alpha}}{\dd x_{\sigma}} - \Gamma_{\sigma\alpha}^{\mu}\, \frac{\dd A^{\alpha}}{\dd x_{\nu}} \\ &= A^{\epsilon} \left(\frac{\dd}{\dd x_{\sigma}} \Gamma_{\nu\epsilon}^{\mu} - \frac{\dd}{\dd x_{\nu}} \Gamma_{\sigma\epsilon}^{\mu}\right) - \Gamma_{\nu\alpha}^{\mu} \Gamma_{\sigma\epsilon}^{\alpha} A^{\epsilon} + \Gamma_{\sigma\alpha}^{\mu} \Gamma_{\nu\epsilon}^{\alpha} A^{\epsilon} \quad\text{by~\Eq{(92.1)}} \\ &= -\Star{B}_{\epsilon\nu\sigma}^{\mu} A^{\epsilon}, \end{align*} where \[ \Star{B}_{\epsilon\nu\sigma}^{\mu} = -\frac{\dd}{\dd x_{\sigma}} \Gamma_{\nu\epsilon}^{\mu} + \frac{\dd}{\dd x_{\nu}} \Gamma_{\sigma\epsilon}^{\mu} + \Gamma_{\nu\alpha}^{\mu} \Gamma_{\sigma\epsilon}^{\alpha} - \Gamma_{\sigma\alpha}^{\mu} \Gamma_{\nu\epsilon}^{\alpha}. \Tag{(92.2)} \] \PageSep{215} Hence \[ \delta A^{\mu} = -\frac{1}{2} \iint \Star{B}_{\epsilon\nu\sigma}^{\mu} A^{\epsilon}\, dS^{\nu\sigma}. \Tag{(92.31)} \] As in \SecRef{33} the formula applies only to infinitesimal circuits. In evaluating the integrand we assumed that $A^{\alpha}$~satisfies the condition of parallel displacement~\Eq{(92.1)} not only on the boundary but at all points within the circuit. No single value of~$A^{\alpha}$ can satisfy this, since if it holds for one circuit of displacement it will not hold for a second. But the discrepancies are of order proportional to~$dS^{\nu\sigma}$, and another factor~$dS^{\nu\sigma}$ occurs in the integration; hence \Eq{(92.31)}~is true when the square of the area of the circuit can be neglected.