# Levi Civita covariance and contravariance

I read some older posts about this question, but I don't know if I'm getting it. I'm working with a Lagrangian involving some Levi Civita symbols, and when I calculate a term containing $\epsilon^{ijk}$ I obtain the contrary sign using $\epsilon_{ijk}$. I always apply the normal rules: $\epsilon_ {ijk}=\epsilon^{ijk}=1$; $\epsilon_ {jik}=\epsilon^{jik}=-1$ etc. I believed that there is no difference between covariant and contravariant Levi-Civita symbol. What do you know about this?

• You might be interested by formulae $(2.42)$ and $(2.43)$, p.$52$ in this Sean Carroll Lecture Notes on General Relativity paper Jul 5, 2014 at 12:10
• Comment to the post (v2): Presumably $ijk$ should be $123$. Oct 5 at 9:28

1. On one hand, there is the Levi-Civita symbol with upper (lower) indices, whose entries are only $0$s and $\pm 1$s; it is a contravariant (covariant) pseudotensor density, respectively.
2. On the other hand, there is the Levi-Civita tensor with upper (lower) indices, whose definition differs from the Levi-Civita symbol by a factor of $\sqrt{|\det(g_{\mu\nu})|}$; it is a contravariant (covariant) pseudotensor, respectively.
• I think the most common trend in the literature I see is for the lower-indexed symbol defined with signs given by the parity of the index permutation in the normal way, while the upper-indexed symbol differs from the lower-indexed one by $\mathrm{sgn}(\det g)$. But the OP seems to be in Euclidean $\mathbb{R}^3$ anyway.
• Hi there, without going into the exact reasons is $\epsilon_ {ijk}=\epsilon^{ijk}=1$ or $\epsilon_ {ijk}=\epsilon^{ijk}=-1$? I'm only asking as I need to perform calculations with the contravariant and covariant forms so I need to know which one is correct, thanks! Oct 5 at 1:57