homework and exercises - Proof of the invariance of the Levi-Civita tensor

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

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.