Tensor Calculus – Sign of the Totally Anti-Symmetric Levi-Civita Tensor When Raising Indices

conventionstensor-calculus

I am confused with the sign we get when we want to raise or lower all indices of the totally anti-symmetric tensor of any rank. Take the metric to be mostly plus ($-+\ldots+$). Then is it
$$\varepsilon^{ijk}=\varepsilon_{ijk}$$
or
$$\varepsilon^{ijk}=-\varepsilon_{ijk}?$$
so I am confused to as which one is true.
And if we consider higher rank does something change? For example
$$\varepsilon^{ijkl}=\varepsilon_{ijkl}$$
or
$$\varepsilon^{ijkl}=\varepsilon_{ijkl}?$$

Best Answer

In flat spacetime, the isomorphism between contra- and covariant components is furnished by the Minkowski metric $\eta_{\mu\nu}$. The Minkowski metric in $n$ spacetime dimensions is just $\operatorname{diag}(-1,1,\dots,1)$. Thus for all dimensions, $$\operatorname{det}\eta=-1$$ For any $n\times n$ matrix $A_{\mu\nu}$, there is a well-known theorem $$\epsilon_{\mu_1\cdots\mu_n}\operatorname{det}A=\sum_{i}\sum_{\nu_i}\epsilon_{\nu_1\cdots\nu_n}A_{\nu_1\mu_1}\cdots A_{\nu_n\mu_n}$$ where we have made no distinction between covariance and contravariance. But suppose $A=\eta$. Then all the multiplications by $\eta$ will raise the indices on $\epsilon$. And since $\operatorname{det}\eta=-1$ in any spacetime dimension, we get $$\epsilon^{\cdots}=-\epsilon_{\cdots}$$ where the dots represent any number of indices.

Related Question