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I'm following the formula for the Hodge dual as follows $$ (\star \omega)_{\mu_1\cdots\mu_{m-p}} =\frac{1}{p!} \sqrt{|g|} \varepsilon_{\mu_1\cdots\mu_{m-p}\nu_1\cdots\nu_p} \omega^{\nu_1\cdots\nu_p} $$ As the case for $\mathrm dy$ in 4D Minkowski space with signature ($1,-1,-1,-1$), I got $$ (\star \mathrm d y)_{013}=\varepsilon_{0132} (\mathrm dy)^2=\varepsilon_{0132} g^{22}(\mathrm dy)_2=-1\times(-1) =1. $$

Thus $$ \star \mathrm dy=\mathrm dt\wedge \mathrm dx\wedge \mathrm dz. $$ But the correct answer should be $-1$ in the front.

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The cleanest notation that works for all metric signatures in 4D is found in Wikipedia: $$\tag{1} \star\,(dx^\mu)=\eta^{\mu\lambda}\varepsilon_{\lambda\nu\rho\sigma}\frac1{3!}dx^\nu\wedge dx^\rho \wedge dx^\sigma\,. $$ I just learned here that there might be different conventions for the sign of the Levi-Civita symbol but thankfully Wikipedia mentions clearly that in (1) they use the convention $\varepsilon_{0123}=1\,.$

  • Now for the metric $\eta^{\mu\lambda}$ assume West coast convention $(+---)$ which is popular in particle physics.

By the anti symmetry of $\varepsilon_{\lambda\nu\rho\sigma}$ and the anti symmetry of $dx^\nu\wedge dx^\rho \wedge dx^\sigma$ in all indices it is clear that when summing over $\nu,\rho,\sigma$ we can just take one term where $\nu<\rho<\sigma$ and drop the factor $\frac1{3!}\,.$ That is because any permutation in an index pair leads to a sign change of the form $(-1)(-1)=1\,.$ Another drastic simplification arises because $\eta^{\mu\lambda}$ is diagonal.

For $\mu=2\,,$ we have $dx^\mu=dy$ and therefore \begin{align}\tag{2} \star\,dy&= \underbrace{\eta^{22}\varepsilon_{2013}}_{(-1)(+1)}\, dt \wedge dx\wedge dz=-\,dt\wedge dx\wedge dz \end{align} which is the answer that you said is correct.

It is quite obvious, too, that with the East coast convention $(-+++)$ and Levi-Civita symbol normalized to $\varepsilon_{0123}=1$ we get a plus sign: $$\tag{3} \star\,dy=dt\wedge dx\wedge dz $$ which agrees with WP because they write this as $$\tag{4} \star\,dy=-\,dt\wedge\color{red}{dz}\wedge\color{red}{dx}\,. $$

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