I am considering a 4D-Lorentzian Manifold $M$ with metric $g$ and a Spacelike hypersurface $\Sigma$ with the induced metric $\gamma$.
Now i look at the following Volume-Element on $\Sigma$:
$f_b=\sqrt{-det(g)} \ \epsilon_{brst}\ dx^{r} \wedge dx^{s} \wedge dx^{t}$
now in the book "Relativity on Curved Manifolds" by De Felice and Clark they state that this is equal to:
$f_{b}=n_b \sqrt{det(\gamma)} \ \epsilon_{\alpha \beta \gamma} \ dx^{\alpha} \wedge dx^{\beta} \wedge dx^{\gamma}$
where $n$ is the unit normal vectorfield to $\Sigma$ and $\alpha, \beta, \gamma \neq b$.
I would like to understand where this identity comes from. I think it is a direct consequence of
$i_n vol_{M}=vol_{\Sigma} $
but i don't really see how one follows from the other.
I don't know Physics, but mathematically it is rather simple.
If the manifold $M$ is oriented, of dimension $n$ and Riemannian, i.e., equiped with a metric, there is only a volume form $\omega$ such that $\omega_x(v_1, v_2, \ldots, v_n) = 1$ for any positively oriented orthonormal bases $v_1, v_2, \ldots, v_n \in T_xM,$ i.e., the volume of the positively oriented unit cube should be equal to 1.
Thanks to this fact, we can construct the volume form of the hypersurface if we know the volume of the ambient manifold, as your anticipation above.