# Invariance of ##\epsilon^{\mu \nu \alpha \beta}##

Gold Member
Homework Statement:
Show that the components of the totally antisymmetric symbol ##\epsilon^{\mu \nu \alpha \beta}## are invariant under transformation belonging to SO(3,1) group.
Relevant Equations:
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Hi, I'm reading some introductory notes about SR and I'm completely stuck at this problem. I imagine I should consider a transformation ##L## such that
$$\hat \epsilon^{\mu \nu \alpha \beta} = L^{\mu}_{\delta}L^{\nu}_{\gamma}L^{\alpha}_{\theta}L^{\beta}_{\psi} \hat \epsilon^{\delta \gamma \theta \psi}$$
and somehow play around with the LHS to show it is equal to ##\epsilon^{\mu \nu \alpha \beta}##. Am I right ? Problem is I'm completely out of ideas.

Thanks
Ric

JD_PM

$$\bar{\epsilon}^{\mu\nu\rho\sigma} = \Lambda^{\mu}{}_{\alpha} \Lambda^{\nu}{}_{\beta} \Lambda^{\rho}{}_{\gamma} \Lambda^{\sigma}{}_{\tau} \epsilon^{\alpha\beta\gamma\tau}.$$ The determinant of any $4 \times 4$ matrix, such as $\Lambda$, is given by $$\Lambda^{\mu}{}_{\alpha} \Lambda^{\nu}{}_{\beta} \Lambda^{\rho}{}_{\gamma} \Lambda^{\sigma}{}_{\tau} \epsilon^{\alpha\beta\gamma\tau} = \mbox{det}(\Lambda) \epsilon^{\mu\nu\rho\sigma}.$$ But for $\Lambda \in \mbox{SO}^{\uparrow}(1,3)$, $\mbox{det}(\Lambda) = 1$. Thus, it follows that $\bar{\epsilon}^{\mu\nu\rho\sigma} = \epsilon^{\mu\nu\rho\sigma}$, is invariant.

JD_PM, dRic2, vanhees71 and 2 others
Gold Member
Thank you very much. I didn't know how to evaluate the determinant of a nxn matrix. Where did you learn such trick ?

JD_PM
$$\bar{\epsilon}^{\mu\nu\rho\sigma} = \Lambda^{\mu}{}_{\alpha} \Lambda^{\nu}{}_{\beta} \Lambda^{\rho}{}_{\gamma} \Lambda^{\sigma}{}_{\tau} \epsilon^{\alpha\beta\gamma\tau}.$$ The determinant of any $4 \times 4$ matrix, such as $\Lambda$, is given by $$\Lambda^{\mu}{}_{\alpha} \Lambda^{\nu}{}_{\beta} \Lambda^{\rho}{}_{\gamma} \Lambda^{\sigma}{}_{\tau} \epsilon^{\alpha\beta\gamma\tau} = \mbox{det}(\Lambda) \epsilon^{\mu\nu\rho\sigma}.$$ But for $\Lambda \in \mbox{SO}^{\uparrow}(1,3)$, $\mbox{det}(\Lambda) = 1$. Thus, it follows that $\bar{\epsilon}^{\mu\nu\rho\sigma} = \epsilon^{\mu\nu\rho\sigma}$, is invariant.

Where did you learn such trick ?

A magician never reveals his secrets... (?)

Antarres
That formula for determinant is just a conveniently written form of the formula that you already know(the permutation sum). You basically have, by contracting the identity he has given over all free indices:
$$\det A = \frac{1}{4!}A^\alpha_{\hphantom{a}\mu} A^\beta_{\hphantom{a}\nu} A^\gamma_{\hphantom{a}\rho} A^\delta_{\hphantom{a}\sigma} \epsilon_{\alpha\beta\gamma\delta}\epsilon^{\mu\nu\rho\sigma}$$
for some ##4 \times 4## matrix ##A## (this is Leibniz formula). For ##n## dimensional case you'd have ##n## indices above.

We can try to check it like this:

First we remember the actual Leibniz formula for expanding the determinant:

$$\det A = \sum_\sigma \text{sgn}(\sigma_i) \prod_{i=1}^{n} A_{i,\sigma_i}$$
where ##\sigma## is a permutation of indices, and ##\text{sgn}(\sigma)## is its sign.
Now we notice that this can be rewritten in the form:
$$\det A = \sum_{i_1, \dots, i_n} \epsilon_{i_1 \dots i_n} A^1_{i_1}\dots A^n_{i_n}$$

This is because ##\epsilon_{i_1 \dots i_n}## is nonzero only if ##i_1, \dots, i_n## is a permutation of indices, and it equals the sign of this permutation.
In ##4## dimensions, this formula is written as(using Einstein summation convention for repeated indices):
$$\det A = \epsilon^{\mu\nu\rho\sigma} A^0_{\hphantom{a}\mu} A^1_{\hphantom{a}\nu} A^2_{\hphantom{a}\rho} A^3_{\hphantom{a}\sigma}$$

Now if we return to the first formula for determinant which we wanted to explain, if we formally sum that formula over indices ##\alpha##, ##\beta##, ##\gamma##, ##\delta##, we find the formula above(in that we use that epsilon is totally antisymmetric, and we use that ##\epsilon_{0123} = 1##, if that's the convention we use).
Finally, we would like to see if the transformation law that @samalkhaiat mentioned holds. For that, we just notice that the formula above is antisymmetric with respect to the indices marked by ##0##, ##1##, ##2##, ##3##. So if we want to make the equivalent formula to this with a random permutation of ##(0123)##, we would find the same formula, just multiplied by the sign of this permutation. But this sign is, as we said before, denoted exactly by ##\epsilon^{\alpha\beta\gamma\delta}##, where ##(\alpha\beta\gamma\delta)## is the chosen random permutation of ##(0123)##. Thus we have for every value of free indices:
$$(\det A) \epsilon^{\alpha\beta\gamma\delta} = \epsilon^{\mu\nu\rho\sigma} A^\alpha_{\hphantom{a}\mu} A^\beta_{\hphantom{a}\nu} A^\gamma_{\hphantom{a}\rho} A^\delta_{\hphantom{a}\sigma}$$

This explains the transformation law of ##\epsilon## symbol.
All in all, it's not a trick, it's just the normal determinant formula written in the form of a contraction.
It's a very convenient formula, and you'll probably find it useful on numerous occassions, so it's good to know.

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dRic2
Gold Member
Thank you very much

Gold Member
2022 Award
You only must be careful with the signs. If you are in Minkowski space it depends on the convention you use to define the Levi-Civita symbol. That's why if you use textbooks, papers, manuscripts you have to look for the definition of the Levi-Civita tensor's sign.

In my community (HEP/relativistic nuclear physics/heavy-ion collisions) the usual convention is that ##\epsilon^{\mu \nu \rho \sigma}## is totally antisymmetric under exchange of its indices, and ##\epsilon^{0123}=+1##, i.e., the Levi-Civita tensor with upper indices is the signature of the permutation of ##(\mu,\nu,\rho,\sigma)## relative to ##(0,1,2,3)##.

Then since ##\eta_{\mu \nu}=\mathrm{diag}(1,-1,-1,-1)## (note that you can use equally well the other sign convention here, but that doesn't change much in what follows) you have ##\mathrm{det} \hat{\eta}=-1## and thus
$$\epsilon_{\mu \nu \rho \sigma}=\mathrm{det} \hat{\eta} \epsilon^{\mu \nu \rho \sigma}=-\epsilon^{\mu \nu \rho \sigma}.$$
A related symbol are the generalized Kronecker-##\delta##'s defined on the space of totally antisymmetric tensors. It's a really invariant symbol, i.e., it's invariant under general basis changes, not only under those with determinant +1.

E.g., ##\delta_{\alpha \beta \gamma \delta}^{\mu \nu \rho \sigma}## is antisymmetric under exchange of both the lower and the upper indices, and it's ##+1## if ##(\alpha,\beta,\gamma,\delta)=(\mu,\nu,\rho,\sigma)##. With the above defined sign conventions you have
$$\delta_{\alpha \beta \gamma \delta}^{\mu \nu \rho \sigma}=-\epsilon_{\alpha \beta \gamma \delta} \epsilon^{\rho \mu \nu \sigma}.$$
For more details, see

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

Appendix A.4 ff.

dRic2 and Antarres
$$\epsilon_{\mu \nu \rho \sigma}=\mathrm{det} \hat{\eta} \epsilon^{\mu \nu \rho \sigma}=-\epsilon^{\mu \nu \rho \sigma}.$$
What you wrote is numerically correct (by choice), but covariantly horrible. In flat space-time, $\epsilon^{\mu\nu\rho\sigma}$ is a (Lorentz) tensor. Therefore, we can lower its indices using the (Lorentz) metric $\eta_{\mu\nu}$: $$\epsilon^{\mu\nu\rho\sigma}\ \eta_{\mu\alpha} \ \eta_{\nu\beta} \ \eta_{\rho\gamma} \ \eta_{\sigma\tau} = \epsilon_{\alpha\beta\gamma\tau} . \ \ \ \ \ \ (1)$$ This covariant expression ensures that the choice $\epsilon_{0123} = -1$ implies (and is implied by) $\epsilon^{0123} = +1$.

Now, with a simple trick, we can transform (1) into an extremely important equation. Using the fact that $\mbox{det}(\eta) = - 1$, we rewrite (1) as $$\epsilon^{\mu\nu\rho\sigma} \ \eta_{\mu\alpha} \ \eta_{\nu\beta} \ \eta_{\rho\gamma} \ \eta_{\sigma\tau} = - \mbox{det}(\eta) \ \epsilon_{\alpha\beta\gamma\tau} . \ \ \ \ \ \ (2)$$ In curved spacetime [ $\eta_{\mu\nu} \to g_{\mu\nu}(x), \ \mbox{det}(\eta) \to g(x)$] our symbol $\epsilon^{\mu\nu\rho\sigma}$ is no loger a tensor, i.e., it does not transform as a tensor under general coordinate transformations. However, the curved space version of (2) is still valid: $$\epsilon^{\mu\nu\rho\sigma} \ g_{\mu\alpha} \ g_{\nu\beta} \ g_{\rho\gamma} \ g_{\sigma\tau} = - g(x) \ \epsilon_{\alpha\beta\gamma\tau} .$$ If we rewrite is as $$\frac{1}{\sqrt{-g}} \epsilon^{\mu\nu\rho\sigma} \ g_{\mu\alpha} \ g_{\nu\beta} \ g_{\rho\gamma} \ g_{\sigma\tau} = \sqrt{- g} \ \epsilon_{\alpha\beta\gamma\tau} ,$$ we see that the corresponding (curved-space) totally antisymmetric tensor is given by $$\varepsilon_{\alpha\beta\gamma\tau} \equiv \sqrt{- g} \ \epsilon_{\alpha\beta\gamma\tau}, \ \ \ \varepsilon^{\mu\nu\rho\sigma} \equiv \frac{1}{\sqrt{- g}} \ \epsilon^{\mu\nu\rho\sigma} .$$

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dRic2 and vanhees71
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Sure, but everywhere where you write "tensor" you should have written "tensor components".

Gold Member
Some time ago, when I started learning a little bit of tensors autonomously (I still don't know much though), a professor of mine told me to beware the fact that ##\epsilon## is not a tensor per se, but it becomes a tensor only if I multiply it by ##\sqrt{ |g|}## (is this formula right?), where ##g## is the metric tensor. In Euclidean and Minkowski space this might go unnoticed.

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2022 Award
Of course, if you consider general tensors, i.e., objects that are invariant under general coordinate transformations, which are not unimodular, then you must use
$$\Delta^{\mu \nu \rho \sigma}=\frac{\sqrt{-g}} \epsilon^{\mu \nu \rho \sigma}$$
and also
$$\Delta_{\mu \nu \rho \sigma} = g_{\mu \alpha} g_{\nu \beta} g_{\rho \gamma} g_{\sigma \delta} \Delta^{\mu \nu \rho \sigma}=\sqrt{-g} \epsilon_{\mu \nu \rho \sigma},$$
where ##g=\mathrm{det} (g_{\mu \nu})<0##.

dRic2