I was reading about Lorentz transformations and frequently I hear the notion of Lorentz transforming quantities like $\epsilon^{\mu \nu \rho \sigma}$. But I have never heard an explanation as to why that is even something that should be done.

My problem with this notion is that even $\epsilon^{\mu \nu \rho \sigma}$ has indices that suggest the transformation behaviour is not an observable at all. Rather just defined as being either 1,-1 or 0 in certain cases. The transformation behaviour of 4 vectors like $x^{\mu}$ or $A^{\mu}$ is a physical necessity because it corresponds to how those quantities will look like if you measure them in either frame.

But the Levi Civita tensor is not something that is measured, it's just a mathematical object. And it doesn't make sense to me that a purely abstract non physical object should at all be subject to change under Lorentz transformation which are a set of rule for transforming physical observations from one frame to another.

Just the explanation of: "Look it has indices!" doesn't cut it for me. Of course the Levi Civita Symbol is invariant anyway for certain Lorentz transformations and differs at most by a sign but that's beside the point.

Any thoughts on this?


2 Answers 2


What's "actually going on" is that if you have some tensor object $A_{\alpha \beta \gamma \delta}$, it transforms in the tensor representation $$\left(\frac{1}{2}, \frac{1}{2}\right) \otimes \left(\frac{1}{2}, \frac{1}{2}\right) \otimes \left(\frac{1}{2}, \frac{1}{2}\right) \otimes \left(\frac{1}{2}, \frac{1}{2}\right) = 4(0,0) + 6(1,0) + 6(0,1) + 9(1,1) + 2(0,2) + 2(2,0) + 3(1,2) + 3(2,1) + (2,2)$$ of the Lorentz group, which is $SU(2)_L \otimes SU(2)_R$. But, $\frac{1}{2} \otimes \frac{1}{2} = 1 \oplus 0$, so there is a singlet of the Lorentz group hiding in this tensor representation. You see that we have four singlets, three of these dimensions are spanned by contractions like: $${A^{\alpha \beta}}_{\alpha \beta}, {{{A^{\alpha}}_\alpha}^\beta}_\beta, {{A^{\alpha\beta}}_{\beta\alpha}}$$

The $\epsilon^{\alpha \beta \gamma \delta}$ is what pulls out the last singlet, so that you end up with a scalar. To me, its a bit of abuse of notation, I'm not sure its meaningful to think of the indices on the epsilon tensor as "actually being" lorentz indices.


if it's a tensor, it transforms like a tensor, which means, the lorentz transformation matrices can act on it. If nothing else, the levi-civita symbol appears in things like the definition of the determinant of the metric, and the algebra of forms, both of which are pretty fundamental objects in doing any sort of tensor analysis in spacetime. Like, there's a Levi-Civita tensor in the definition of the Maxwell tensor, certainly we need to know how the Maxwell tensor transforms.

And yes, the discussion is moot anyway, because the Levi-Civita Tensor just transforms into itself when you apply the Lorentz matrices.

  • 3
    $\begingroup$ The Levi-Cita symbol $\epsilon^{ijkl}$ has only $\pm 1$ as its components. The Levi-Civita tensor $ \varepsilon^{ijkl}= \epsilon^{ijkl}/\sqrt{-g}$ has different components in different frames because the $\sqrt{-g}$ that is necessary for it to transform as a tensor changes as you change frame. $\endgroup$
    – mike stone
    Jun 5, 2022 at 20:17
  • $\begingroup$ Okay you say that it's a tensor but a tensor is a map from products of vector space and its dual to the real numbers. What I was talking about so far is just the "Levi Citiva Symbol" (title was changed) that is not a tensor at all but just a mapping of the form (i,j,k,l) -> {-1,0,1} But I see now that there is a distinction to me made here between the definition of the symbol and the tensor that's associated with it. And indeed I see that the tensor that the symbol induces has to transform as it does. $\endgroup$ Jun 5, 2022 at 20:23
  • $\begingroup$ @mikestone, you're of course right. $\endgroup$ Jun 5, 2022 at 20:59

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