# Hodge dual of dy in Minkowski space

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.

## 1 Answer

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 $$\tag{4} \star\,dy=-\,dt\wedge\color{red}{dz}\wedge\color{red}{dx}\,.$$