# Derivation of the contravariant form for the Levi-Civita tensor

• I
AndersF
TL;DR Summary
If the covariant form for the Levi-Civita is defined as ##\varepsilon_{i,j,k}:=\sqrt{g}\epsilon_{i,j,k}##, how could be shown from this definition that it's contravariant form is given by ##\varepsilon^{i,j,k}=\frac{1}{\sqrt{g}}\epsilon^{i,j,k}##?
The covariant form for the Levi-Civita is defined as ##\varepsilon_{i,j,k}:=\sqrt{g}\epsilon_{i,j,k}##. I want to show from this definition that it's contravariant form is given by ##\varepsilon^{i,j,k}=\frac{1}{\sqrt{g}}\epsilon^{i,j,k}##.

My attempt

What I have tried is to express this tensor ##\varepsilon^{i j k}## through the contraction with the metric tensor of ##\varepsilon_{i j k}## the contravariant form, and then to replace the definition of ##\varepsilon_{i j k}##:

##\varepsilon^{i j k}=g^{i p} g^{j q} g^{k r} \varepsilon_{p q r}=g^{i p} g^{j q} g^{k r} \sqrt{g} \epsilon_{p q r}##

This expression reminds me of the the expression of the determinant of the dual metric tensor, ##g^{-1}=\det (g^{ij})##, through the Levi-Civita symbol:

##g^{-1}=g^{1 p} g^{2 q} g^{3 r} \epsilon_{p q r}##

But I'm stuck here, as I don't know how to match these expressions... Would there be any way to achieve this? Would this be a good way to prove the theorem?

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I guess you talk about a 3D Riemannian manifold here. Then let's define ##\Delta_{jkl}=\Delta^{jkl}## as the totally antisymmetric symbol with ##\Delta_{123}=\Delta^{123}=1##.

Then
$$\epsilon_{jkl}=\sqrt{g} \Delta_{jkl} \quad \text{with} \quad g=\mathrm{det}(g_{\mu \nu})$$
are covariant (pseudo) tensor components.

Indeed, using the transformation law for covariant tensor components you get
$$\epsilon_{abc}'=\sqrt{g} \Delta_{jkl} \frac{\partial q^j}{\partial q^{\prime a}} \frac{\partial q^k}{\partial q^{\prime b}} \frac{\partial q^l}{\partial q^{\prime c}}=\sqrt{g} J \Delta_{abc},$$
where
$$J=\mathrm{det} (\partial_a' q^j)$$
is the Jacobian of the coordinate transformation.

On the other hand
$$g' = \mathrm{det} g_{ab}'=\mathrm{det} (\partial_a' q^j \partial_b' q^k g_{jk})=J^2 g.$$
So if ##J>0##, then indeed
$$\epsilon_{abc}'=\sqrt{g'} \Delta_{abc}.$$
If ##J<0## you get an additional minus sign. That's why ##\epsilon_{jkl}## are the components of a pseudo-tensor rather than a true tensor.

For the contravariant components you get
$$\epsilon^{jkl}=g^{ja} g^{kb} g^{lc} \epsilon_{abc} = \sqrt{g} g^{ja} g^{kb} g^{lc} \Delta_{abc} = \sqrt{g} \mathrm{det}(g^{ab}) \Delta^{jkl}=\frac{1}{g} \sqrt{g} \Delta^{jkl} = \frac{1}{\sqrt{g}} \Delta^{jkl},$$
where I've used that in a Riemannian manifold ##\mathrm{det}(g_{ab})>0##.

AndersF