Orthogonality relation four-vectors

Question

If $A,B,C,D$ are four-vectors and $A,B,C$ form an orthogonal hypersurface to $D$, and $det(g_{\mu\nu})=g$ where $g_{\mu\nu}$ are the metric components, then it is true that \begin{equation} \sqrt{-g}A^{\alpha}B^{\beta}C^{\gamma}\epsilon_{\alpha\beta\gamma\delta}=cte \cdot D_{\delta} \end{equation}

Where $cte$ is a constant. I dont understand how this equation is derived.

Making an analogy with 3D euclidean linear algebra, if $A,B$ are vectors that create a plane orthogonal to $C$ then $A \times B=C$, or, in components \begin{equation} A_{i}B_{j}\epsilon^{ijk}=C^{k} \end{equation}

Which looks very similar to the equation above. However, I don't understant where the $\sqrt{-g}$ comes from (I suppose it comes from the fact that this is in general a curved space, but I don't see it clearly enough). And I don't understand why a constant is added in the right side

Answer 1

Orthogonality is ensured even without the metric determinant on the left. However, it becomes necessary if you wish the length of the resultant vector $D$ to be preserved under coordinate changes.

To see this, contract the expression with $D^\delta$ and identify the lhs with the Riemannian volume form

$$ \sqrt{-g}\;\varepsilon_{\alpha\beta\gamma\delta}A^\alpha B^\beta C^\gamma D^\delta = \sqrt{-g}\,(dx^1\wedge dx^2\wedge dx^3\wedge dx^4)(A,B,C,D) = \mathrm{const}\,D^\delta D_\delta$$ where $\wedge$ denotes the exterior product of the coordinate one-forms.

The appearance of the metric determinant ensures the expression to be invariant under coordinate changes, i.e. the length of $D$ doesn't depend on the choice of basis.

The minus sign under the square root comes about because you are (presumably) dealing with a Lorentzian metric. More generally one would write $\sqrt{|g|}$.

The constant on the rhs is in my opinion meant to be a constant independent of the point on the manifold. You may choose it however to scale $D$.