# Hodge dual of $4$-form in Minkowski spacetime

In the Wikipedia article about the Hodge dual, I'm clear on how to compute the Hodge star of $1$-forms, $2$-forms, and $3$-forms in the $4$-dimensional Minkowski spacetime of metric signature $(+---)$. What I'm not so clear on is if there is a way to compute the Hodge star of the $4$-form $dt \wedge dx \wedge dy \wedge dz$. I know that by definition the Hodge star sends $k$-forms to $(n-k)$-forms, where $n$ is the dimension of the underlying vector space. So in that regard $4$-forms would be sent to $0$-forms under the Hodge star. My question is would that be a smooth (scalar) function $\phi(t, x, y, z)$?

• In fact, $dt \wedge dx \wedge dy \wedge dz$ is (up to a choice of sign) the volume form of the Minkowski metric, so $\ast(dt \wedge dx \wedge dy \wedge dz) = \pm 1$. – Travis Jun 3 '17 at 18:04
• What do you mean exactly? It was computed explicitly in my first comment. – Travis Jun 3 '17 at 18:07
• You only computed it up to a sign. But to determine the sign we need to know the orientation. The notation suggests an orientation which gives the answer as +1 – ziggurism Jun 3 '17 at 18:09
• The orientation is an additional datum that was not specified---it's not usually taken to be part of the data of the metric. This is the 'choice' in "up to a choice of sign". Fixing the order $(t, x, y, z)$ for the variables determines an orientation ov $V$, namely the one for which $dt \wedge dx \wedge dy \wedge dz$ is positively oriented. – Travis Jun 3 '17 at 18:16
• It follows from the definition of $\ast$. If we regard $1$ as a $0$-form, we have $\ast 1 = 1 \wedge \ast 1 = \langle 1, 1 \rangle \Omega = \Omega$, where $\Omega$ is the volume form of the metric (which depends on the orientation). On the other hand, on $k$-forms we have $\ast^2 \alpha = s (-1)^{k (n - k)} \alpha$, so in our case ($k = 0$, and Lorenztian signature gives $s = -1$) $\ast \Omega = \ast^2 1 = -1$. Thus, $\pm$ is $-$ if the given $4$-form is positively oriented (in which case it is $\Omega$) and $+$ if not. – Travis Jun 3 '17 at 18:24

Regard the scalar $1$ as a $0$-form: By the definition of the Hodge star operator $\ast$, we have $$\ast 1 = 1 \wedge \ast 1 = \langle 1, 1 \rangle \Omega = \Omega ,$$ where $\Omega$ is the volume form of the metric (which depends on a choice orientation).
On the other hand, on $k$-forms $\alpha$ the square of the Hodge star operator satisfies the duality identity $$\ast^2 \alpha = s (-1)^{k (n - k)} \alpha ,$$ where $s \in \{ \pm 1 \}$ depends on the signature of the metric. In our case, Lorentzian signature in dimension $4$, $s = -1$. Thus, $$\ast \Omega = \ast^2 1 = -1 .$$
Now, $(dt, dx, dy, dz)$ is a (pseudo-)orthonormal coframe, $\Omega = \pm dt \wedge dx \wedge dy \wedge dz$---the choice of sign is the same as the choice of the orientation. Thus, $$\ast (dt \wedge dx \wedge dy \wedge dz) = \ast(\pm \Omega) = \mp 1 ,$$ where $\mp$ is $-$ if $dt \wedge dx \wedge dy \wedge dz$ is positive and $+$ if not.