Timeline for Sign of the totally anti-symmetric Levi-Civita tensor $\varepsilon^{\mu_1 \ldots}$ when raising indices

5 events

when toggle format	what		by	license	comment
Feb 22, 2017 at 20:30	vote	accept	Marion
Feb 22, 2017 at 20:38
Jan 6, 2015 at 12:15	comment	added	user10851		But your first formula and my sign change from ${}^{**}H$ to $H$ depend on the dimension of the space, the number of indices on $H$, and the number of negative signs in the metric signature, so I don't guarantee this works if those things change.
Jan 6, 2015 at 12:14	comment	added	user10851		@Marion That's very confusing notation without Hodge stars, since $H^{\mu\nu}$ should by all rights be $g^{\mu\alpha} g^{\nu\beta} H_{\alpha\beta}$. In any case, we have $({}^H)^{\mu\nu} = -(1/2)\epsilon^{\mu\nu\rho\sigma}H_{\rho\sigma}$. Then $(1/2)\epsilon_{\mu\nu\rho\sigma}({}^H)^{\rho\sigma} = (1/2)g_{\mu\alpha}g_{\nu\beta}\epsilon^{\alpha\beta\rho\sigma}({}^H)_{\rho\sigma} = -g_{\mu\alpha}g_{\nu\beta}({}^{*}H)^{\alpha\beta} = g_{\mu\alpha}g_{\nu\beta}H^{\alpha\beta} = H_{\mu\nu}$.
Jan 6, 2015 at 10:47	comment	added	Marion		So if the dual tensor is defined as $H^{\mu \nu} = -\frac{1}{2}\varepsilon^{\mu \nu \rho \sigma} H_{\rho \sigma} $ does the above mean that with indices down we have $H_{\mu \nu} = \frac{1}{2}\varepsilon_{\mu \nu \rho \sigma} H^{\rho \sigma} $?
Jan 6, 2015 at 4:47	history	answered	user10851	CC BY-SA 3.0