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 $\phi(t, x, y, z)$?

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.