Proof of the invariance of the Levi-Civita tensor

Question

My question is related with the proof of the following: the Levi Civita tensor, $\epsilon _{\mu \nu \rho \sigma}$ is an invariant tensor, that is, if we make a change between one reference frame with some coordinates to another one in the way is expected from a (0,4) tensor, that is

$$\epsilon' _{\mu \nu \rho \sigma} = \dfrac{\partial x^a}{\partial x'^{\mu}} \dfrac{\partial x^b}{\partial x'^{\nu}} \dfrac{\partial x^c}{\partial x'^{\rho}} \dfrac{\partial x^d}{\partial x'^{\sigma}} \epsilon _{a b c d},$$

and apply its properties we should arrive at the result

$$\epsilon' _{\mu \nu \rho \sigma} = \epsilon _{a b c d}.$$

So the tensor doesn't change between reference frames. After going around the problem for a while I haven't be able to prove it. Could anyone give me a helping hand?

Answer 1

I suspect that what you are really after is that $\epsilon$ is invariant under Lorentz transformations between inertial observes. In fact, we have a well known formula for the determinat of an operator: $$ \epsilon_{i_1...i_n}A^{i_1}_{\ \ j_1}...A^{i_n}_{\ \ j_n} = \mathrm{det} A \epsilon_{j_1...j_n}. $$ All Lorentz transformations have $\mathrm{det} A=\pm 1$. If you restrict attention to proper ortochronous transformations (basically exclude various reflections) you get that $\epsilon$ is invariant. We say that it is pseudoscalar with respect to Lorentz group. On the other hand, as pointed by others in this thread, it is not a tensor with respect to general curvilinear transformations.

Answer 2

If we are referring to the same Levi-Civita symbol, i.e. the completely antisymmetric symbol, then this is not a tensor. It is a tensor density of weight $-1$, which means that under a general coordinate transformation it gets multiplied by the inverse Jacobian of the transformation. So it is normal that you can't prove what you say.

What it is usually done in General Relativity is to define a quantity $$E_{\mu\nu\sigma\rho}=\sqrt{\det[g_{ab}]}\epsilon_{\mu\nu\sigma\rho}$$ which is a real tensor, since the determinant of the metric tensor is a scalar density of weight $+1$ and as a consequence the Jacobian and inverse Jacobian cancel out. This is related to differential forms and invariant volume elements, so you can look these up if you need more information.