Again \[ E^{\alpha\beta\gamma\delta}\, E^{\epsilon\zeta\eta\theta}\, g_{\alpha\epsilon}\, g_{\beta\zeta}\, g_{\gamma\eta}\, g_{\delta\theta} \] is seen by inspection to be an invariant. But this is equal to \begin{align*} &E^{2} \epsilon_{\alpha\beta\gamma\delta}\, \epsilon_{\mu\nu\sigma\tau}\, g_{\alpha\epsilon}\, g_{\beta\zeta}\, g_{\gamma\eta}\, g_{\delta\theta} \\ = {} &4!\, E^{2} g. \end{align*} Hence \[ \text{$E^{2} g$ is an invariant.} \Tag{(48.65)} \] Accordingly \[ E^{2} g = E'^{2} g' = (EJ)^{2} g',\quad\text{ by \Eq{(48.6)}} \] giving another proof that \[ g = J^{2} g'. \Tag{(48.7)} \] \emph{Corollary.} If $a$~is the determinant formed from the components~$a_{\mu\nu}$ of \emph{any} covariant tensor, $E^{2} a$~is an invariant and \[ a = J^{2} a'. \Tag{(48.8)} \] \Section{49.}{Element of volume. Tensor-density} In \SecRef{32} we found that the surface-element corresponding to the parallelogram contained by two displacements, $\delta_{1} x_{\mu}$, $\delta_{2} x_{\mu}$, is the antisymmetrical tensor \[ dS^{\mu\nu} = \left|\begin{array}{@{}cc@{}} \delta_{1} x_{\mu} & \delta_{1} x_{\nu} \\ \delta_{2} x_{\mu} & \delta_{2} x_{\nu} \\ \end{array}\right|. \] Similarly we define the volume-element (four-dimensional) corresponding to \index{Volume-element}% the hyperparallelopiped contained by four displacements, $\delta_{1} x_{\mu}$, $\delta_{2} x_{\mu}$, $\delta_{3} x_{\mu}$, $\delta_{4} x_{\mu}$, as the tensor \[ dV^{\mu\nu\sigma\tau} = \left|\begin{array}{@{}cccc@{}} \delta_{1} x_{\mu} & \delta_{1} x_{\nu} & \delta_{1} x_{\sigma} & \delta_{1} x_{\tau} \\ \delta_{2} x_{\mu} & \delta_{2} x_{\nu} & \delta_{2} x_{\sigma} & \delta_{2} x_{\tau} \\ \delta_{3} x_{\mu} & \delta_{3} x_{\nu} & \delta_{3} x_{\sigma} & \delta_{3} x_{\tau} \\ \delta_{4} x_{\mu} & \delta_{4} x_{\nu} & \delta_{4} x_{\sigma} & \delta_{4} x_{\tau} \\ \end{array}\right|. \Tag{(49.1)} \] It will be seen that the determinant is an antisymmetrical tensor of the fourth rank, and its $256$~components accordingly have one or other of the three values \[ +dV,\quad 0,\quad -dV, \] where $dV = ±dV^{1234}$. It follows from \Eq{(48.65)} that $(dV)^{2} g$~is an invariant, so that \[ \text{$\sqrt{-g} · dV$ is an invariant.} \Tag{(49.2)} \] %\PageSep{110} Since the sign of $dV^{1234}$ is associated with some particular cycle of enumeration of the edges of the parallelopiped, which is not usually of any importance, the single positive quantity~$dV$ is usually taken to represent the volume-element fully. Summing a number of infinitesimal volume-elements, \index{Volume!physical and geometrical}% we have \[ \iiiint \sqrt{-g} · dV\quad\text{is an invariant,} \Tag{(49.3)} \] the integral being taken over any region defined independently of the coordinates.