Can an equation containing a specific tensor be Lorentz invariant?

Let $A_u$ be a vector field in spacetime. If we restrict to a $2+1$ spacetime, and define the Levi-Civita tensor $\epsilon^{uvp}$ by $\epsilon^{123}=1$, then is the following equation Lorentz invariant?

\begin{equation*} \epsilon^{uvp}\partial_uA_p+mA^v=0. \end{equation*}

I checked that it transforms covariantly under a Lorentz transformation. Thus, because the specific tensor $\epsilon^{uvp}$ is invariant, the whole equation can be claimed to be Lorentz invariant. Is my argument correct?

• Yes, it's correct. The only caveat is that it is invariant only under proper Lorentz transformations. – Prahar Mar 27 '18 at 17:40
• @Prahar Can you expand on how an equation can be invariant when there are free indices? Invariant means it doesn't change under transformation, so if it is covariant - i.e. if it changes in the right way under transformation - how can it be invariant. – ZeroTheHero Mar 27 '18 at 17:59
• Because its a very special tensor. You can ask the same question about the metric tensor $\eta_{\mu\nu}$. How does it remain invariant if its a tensor? It does because it is a special type of tensor. Not all tensors have this property. But the metric $\eta_{\mu\nu}$ and Levi-Civita tensor $\varepsilon_{\mu_1\cdots\mu_d}$ have such invariance properties. – Prahar Mar 27 '18 at 18:01
• @Prahar right, $\eta_{\mu\nu}$ is invariant but $\eta_{\mu\nu}A^\nu$ isn't. – ZeroTheHero Mar 27 '18 at 18:03
• @ZeroTheHero - I never said it was. – Prahar Mar 27 '18 at 18:03

The metric tensor is a special tensor that is invariant under Lorentz transformations. This is of course trivial to see due to the very definition. Under Lorentz transformations $$\eta_{\mu\nu} \to \eta_{\rho\sigma} \Lambda^\rho{}_\mu \Lambda^\sigma{}_\nu.$$ But due to the property of Lorentz transformations, the above is just $\eta_{\mu\nu}$ so that $$\eta_{\mu\nu} \to \eta_{\rho\sigma} \Lambda^\rho{}_\mu \Lambda^\sigma{}_\nu = \eta_{\mu\nu}.$$ Thus $\eta$ is Lorentz invariant.

Similarly, the Levi-Civita tensor is invariant. Under Lorentz transformations $$\varepsilon_{\mu_1 \cdots \mu_d} \to \varepsilon_{\nu_1 \cdots \nu_d} \Lambda^{\mu_1}{}_{\nu_1} \cdots \Lambda^{\mu_d}{}_{\nu_d}$$ But now, due to the definition of the epsilon tensor and the definition of determinant of matrix (see this link), the above is equal to $$\varepsilon_{\mu_1 \cdots \mu_d} \to \varepsilon_{\nu_1 \cdots \nu_d} \Lambda^{\mu_1}{}_{\nu_1} \cdots \Lambda^{\mu_d}{}_{\nu_d} = (\det \Lambda)\varepsilon_{\mu_1 \cdots \mu_d}.$$ However, we know that $\det \Lambda = \pm 1$. Thus, $\varepsilon$ is invariant under proper Lorentz transformations (which have $\det \Lambda =1$) but picks up a sign under improper Lorentz transformations.

Thus, the equation you have in the problem is invariant under proper Lorentz transformations but not under the improper ones such as parity or time reversals.

PS - It looks like you got that equation trying to solve massive CS theories and the CS action is famously parity non-invariant.

• In many cases just SO(1,3) seems to be considered as the lorentz group, that is boosts and rotations. In that case is there no problem? – Keith Mar 27 '18 at 18:44
• If that's all you're interested in then there is no problem. That's not to say that there is a problem if you are talking about parity. It's just that in that case, these equations are not parity invariant. – Prahar Mar 27 '18 at 18:45
• man I'm sorry to bother you with this... I don't have a problem with the $\epsilon$ tensor being invariant but clearly $mA^\nu$ is NOT Lorentz-invariant, it's "merely" covariant (unless all $A^\nu$ are $0$). So given this, how can $\epsilon^{u\nu p}\partial_u A_p$ be invariant when it's just equal to -$mA^\nu$, which is NOT invariant? – ZeroTheHero Mar 27 '18 at 19:03
• I am not saying each individual term is invariant. What I said was "The equation is invariant" which means "In a Lorentz transformed frame, the equation takes the same form as it does here". This is what those words mean. Each term in that equation itself is covariant. The entire equation is invariant because it is equal to 0. – Prahar Mar 27 '18 at 19:14
• I suddenly wonder what the solution of this equation would be. What kind of particle does this equation describe? – Keith Mar 28 '18 at 2:32

Perhaps it's helpful to note that the equation can be rewritten in terms of differential forms without any indices at all.

Write $A$ for the $1$-form $A = A_\mu dx^\mu$. Denote by $\star$ the Hodge star map which sends $p$-forms to $3-p$ forms. $\star$ acts on $2$-forms by replacing $dt \wedge dx$ with $dy$, $dx \wedge dy$ with $dt$, and so forth.

In this notation, your equation is $*dA + mA = 0$.