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

  • $\begingroup$ The constant is presumably there because you haven’t imposed any normalization conditions on the vectors. And I don’t see the point of the $\sqrt{-g}$. The four vectors live at some point of the manifold, and the value of $\sqrt{-g}$ at that point is just a number that can be absorbed into the constant. If you stated where you found this equation, I suspect we might discover that there is more to this than you have described. $\endgroup$
    – G. Smith
    Commented Aug 5, 2019 at 5:01

1 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 $D$.

  • $\begingroup$ Thank you... do you know why the minus sign inside the square root? And I didnt understand the $^$ notation, is it the outer product between the 1-forms? $\endgroup$ Commented Aug 5, 2019 at 11:31
  • $\begingroup$ @JuanPabloArcila See the edit. $\endgroup$
    – Nephente
    Commented Aug 5, 2019 at 12:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.