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

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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
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Thank you very much

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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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2022 Award
Sure, but everywhere where you write "tensor" you should have written "tensor components".

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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.

$$\Delta^{\mu \nu \rho \sigma}=\frac{\sqrt{-g}} \epsilon^{\mu \nu \rho \sigma}$$
$$\Delta_{\mu \nu \rho \sigma} = g_{\mu \alpha} g_{\nu \beta} g_{\rho \gamma} g_{\sigma \delta} \Delta^{\mu \nu \rho \sigma}=\sqrt{-g} \epsilon_{\mu \nu \rho \sigma},$$
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