Hence interchanging $\nu$ and~$\sigma$, and subtracting, the complete expression is \begin{multline*} \Star{B}_{\mu\nu\sigma}^{\epsilon} = B_{\mu\nu\sigma}^{\epsilon} + g_{\mu}^{\epsilon} \left(\frac{\dd\kappa_{\nu}}{\dd x_{\sigma}} - \frac{\dd\kappa_{\sigma}}{\dd x_{\nu}}\right) + (g_{\nu}^{\epsilon} \kappa_{\mu\sigma} - g_{\sigma}^{\epsilon} \kappa_{\mu\nu}) + (g_{\mu\sigma} \kappa_{\nu}^{\epsilon} - g_{\mu\nu} \kappa_{\sigma}^{\epsilon}) \\ + (g_{\nu}^{\epsilon} \kappa_{\mu} \kappa_{\sigma} - g_{\sigma}^{\epsilon} \kappa_{\mu} \kappa_{\nu}) + (g_{\sigma}^{\epsilon} g_{\mu\nu} - g_{\nu}^{\epsilon} g_{\mu\sigma}) \kappa_{\alpha} \kappa^{\alpha} + (g_{\mu\sigma} \kappa_{\nu} - g_{\mu\nu} \kappa_{\sigma}) \kappa^{\epsilon}. \Tag{(87.4)} \end{multline*} Next set $\epsilon = \sigma$. We obtain the contracted in-tensor \begin{multline*} \Star{G}_{\mu\nu} = G_{\mu\nu} - F_{\mu\nu} + (\kappa_{\mu\nu} - 4\kappa_{\mu\nu}) + (\kappa_{\mu\nu} - g_{\mu\nu} \kappa_{\alpha}^{\alpha}) + (\kappa_{\mu} \kappa_{\nu} - 4\kappa_{\mu} \kappa_{\nu}) \\ + (4g_{\mu\nu} - g_{\mu\nu}) \kappa_{\alpha} \kappa^{\alpha} + (\kappa_{\mu} \kappa_{\nu} - g_{\mu\nu} \kappa_{\alpha} \kappa^{\alpha}) \\ = G_{\mu\nu} - 2F_{\mu\nu} - (\kappa_{\mu\nu} + \kappa_{\nu\mu}) - g_{\mu\nu} \kappa_{\alpha}^{\alpha} - 2\kappa_{\mu} \kappa_{\nu} + 2g_{\mu\nu} \kappa_{\alpha}^{\alpha}.\footnotemark \Tag{(87.5)} \end{multline*} \PageSep{205} Finally multiply by~$g^{\mu\nu}$. We obtain the co-invariant \footnotetext{The unit of~$\kappa_{\mu}$ is arbitrary; and in the generalised theory in Part~II the $\kappa_{\mu}$ there employed corresponds to twice the $\kappa_{\mu}$ of these formulae. This must be borne in mind in comparing, for example, \Eq{(87.5)} and~\Eq{(94.3)}.}% \[ \Star{G} = G - 6\kappa_{\alpha}^{\alpha} + 6\kappa_{\alpha} \kappa^{\alpha}. \Tag{(87.6)} \] The multiplication by~$g^{\mu\nu}$ reintroduces the unit of gauge, so that $\Star{G}$~becomes multiplied by~$\lambda^{-2}$ when the gauge is transformed. If the suffix~$\epsilon$ is lowered in~\Eq{(87.4)} the only part of~$\Star{B}_{\mu\nu\sigma\epsilon}$ which is symmetrical in $\mu$~and $\epsilon$ is $g_{\mu\nu} (\dd\kappa_{\nu}/\dd x_{\sigma} - \dd\kappa_{\sigma}/\dd x_{\nu}) = g_{\mu\epsilon} F_{\nu\sigma}$, which agrees with the condition~\Item{(a)} of Weyl's geometry (\SecRef{84}). \Section{88.}{The in-invariants of a region} \index{In-invariants}% There are no functions of the~$g_{\mu\nu}$ and~$\kappa_{\mu}$ at a point which are in-invariants; but functions which are in-invariant-densities may be found as follows---