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From mathworld.wolfram.com it states that the rank four permutation tensor satisfies:

$$ \epsilon_{\alpha \beta \gamma \delta} = -\epsilon^{\alpha \beta \gamma \delta}.\tag{6}$$

However, I am struggling to see how this follows from the definitions from the same source of

$$ \epsilon_{\alpha \beta ...\mu} = \sqrt{|g|} [\alpha, \beta,...,\mu]\tag{2}$$

$$\epsilon^{\alpha \beta ...\mu} = \frac{[\alpha, \beta,...,\mu]}{\sqrt{|g|}} \tag{3}$$

since it seems to require $\sqrt{|g|} = - \frac{1}{\sqrt{|g|}} \rightarrow |g| = -1$?

Additionally, by the usual raising/lowering of indices with the metric we can say,

$$ \epsilon_{\rho \sigma \mu \nu} = g_{\rho \alpha} g_{\sigma \beta} g_{\mu \lambda} g_{\nu \gamma} \epsilon^{\alpha \beta \lambda \gamma}$$

and so for a diagonal metric (e.g. Minkowski, Schwarzchild),

$$ \epsilon_{\rho \sigma \mu \nu} = g_{\rho \rho} g_{\sigma \sigma} g_{\mu \mu} g_{\nu \nu} \epsilon^{\rho \sigma \mu \nu}$$

Now for Minkowski, $g_{\rho \rho} g_{\sigma \sigma} g_{\mu \mu} g_{\nu \nu} = -1$, but this is not true. for e.g. Schwarzchild?

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  • $\begingroup$ Your last equation does not make sense. $\endgroup$ Commented Sep 2, 2022 at 10:26

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The first equation only holds for Minkowski spacetime, something that explicitly appears in Weinberg's "Gravitation and Cosmology" book (which the linked site references). In that case $\sqrt{-g} = 1$.

The minus sign comes from the fact that, on a Lorentzian manifold, contravariant and covariant tensors may exhibit a sign difference (since the metric is no longer positive-definite). In our case, a Minkowski spacetime will either have 1 or 3 coordinate directions with a negative sign, depending on the convention; $(- + + +)$ and $(+ - - -)$ respectively. Hence after raising the indices:

\begin{equation} \varepsilon ^{\mu \nu \rho \sigma} = g^{\mu \alpha} g^{\nu \beta} g^{\rho \gamma} g^{\sigma \delta} \, \varepsilon _{\alpha \beta \gamma \delta} \Rightarrow \, \varepsilon ^{\alpha \beta \gamma \delta} = -\varepsilon _{\alpha \beta \gamma \delta} \end{equation}

the contravariant Levi-Civita tensor picks up a minus sign. Bear in mind that the notation $[\alpha ,\, \beta \, , \ldots , \, \mu ]$ implies that the indices are lower/raised appropriately in each of the two cases. This means that between your equations (2) and (3), it is implied that there is a relative sign difference. It is not shown because I assume the text doesn't derive (3) from (2), so the indices are not necessarily the same (the identical naming is incidental).

To derive that relative sign difference, let's examine a more general case. For a general spacetime, notice how the product $g^{\mu \alpha} g^{\nu \beta} g^{\rho \gamma} g^{\sigma \delta}$ (assuming that the indices $\mu$, $\nu$, $\rho$ and $\sigma$ as such so that the contravariant Levi-Civita tensor in not 0) along with the implicit sum due to Einstein notation will produce the inverse of the determinant of the metric. This is especially easy to see in the case of a diagonal metric. Therefore one may write:

\begin{equation} \begin{array}{lr} \varepsilon _{\alpha \beta \gamma \delta} = \sqrt{-g} \, [\alpha , \, \beta ,\, \gamma ,\, \delta ] \\ \varepsilon ^{\alpha \beta \gamma \delta} = g^{-1} \, \varepsilon _{\alpha \beta \gamma \delta} \end{array} \Bigg \} \Rightarrow \, \varepsilon ^{\alpha \beta \gamma \delta} = -\frac{[\alpha , \, \beta ,\, \gamma ,\, \delta ]}{\sqrt{-g}} \end{equation}

Had we started backwards and defined:

\begin{equation} \varepsilon ^{\alpha \beta \gamma \delta} = \frac{[\alpha , \, \beta ,\, \gamma ,\, \delta ]}{\sqrt{-g}} \end{equation}

then it would be the covariant tensor that would pick up a minus sign in this convention.

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    $\begingroup$ Perfect answer. Thanks for taking the time! $\endgroup$ Commented Sep 2, 2022 at 15:08

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