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 $\phi(t, x, y, z)$?
[Math] Hodge dual of $4$-form in Minkowski spacetime
differential-forms
Related Solutions
So, what is the precise expression of $⟨v,w⟩$?
Let $(M,g)$ be an orientable Riemannian manifold of dimension $n$. For any $0 \le k \le n$, there is a unique extension of the metric $g$ to the metric $⟨ \, ,\, ⟩: \Omega^k(M) \times \Omega^k(M) \to \mathbb{R}$ such that for any O.N.B. $ \{e^1, ...,e^n \} $ of one-forms, the basis of $\Omega^k(M)$ is an O.N.B. with respect to $⟨ \, ,\, ⟩$
$$\text{i.e. } ⟨ e^{i_1} \wedge...\wedge e^{i_k} \, ,\, e^{j_1} \wedge...\wedge e^{j_k}⟩ = \delta^{i_1 j_1}...\delta^{i_k j_k} \, .\tag4$$
More generally, for $v^1,...,v^k,w^1,...,w^k \in \Omega(M)$,
$$ ⟨v^1\wedge...\wedge v^k,w^1\wedge...\wedge w^k⟩ := \mathrm{det}(g(v^i,w^j)) \,. \tag5$$
It is easy to check that this form is bilinear, symmetric and positive-definite.
Now suppose $ v := v_{i_1...i_k} e^{i_1} \wedge...\wedge e^{i_k}$ and $w:=w_{j_1...j_k}e^{j_1} \wedge...\wedge e^{j_k}$ are the coordinate representations of the $k$-forms $v$ and $w$. Therefore,
\begin{align*} ⟨v,w⟩ &= v_{i_1...i_k} \ w_{j_1...j_k} \,\delta^{i_1 j_1}...\delta^{i_k j_k} \\ &= v_{i_1...i_k} \ w^{i_1...i_k} \,. \tag8 \\ \end{align*}
How to derive the coordinate expression of the Hodge dual?
To begin with, I will take your definition and write it down as follows.
The Hodge dual is the unique isomorphism
\begin{align*} \star:\Omega^k(M) &\to \Omega^{n-k}(M), \\ \omega &\mapsto \star \omega \\ \end{align*}
such that the following holds:
$$ \forall \omega, \eta \in \Omega^k(M): \omega \wedge \star \eta = ⟨\omega,\eta⟩ \ vol$$
where $vol := \sqrt{g} \ \mathrm{d} x^1 \wedge ... \wedge \mathrm{d}x^n$ is the volume form.
I have purposefully deviated from the notation used in (1) to avoid confusion between $w$ and $\omega$. Moreover, I believe that (2) is not correct at all. I will derive the correct coordinate representation below. Since the Hodge star operator is a $C^\infty (M)$-linear map, it suffices to evaluate it on the basis elements.
Proposition:
$$ \star (\mathrm{d}x^{i_1}\wedge...\wedge \mathrm{d}x^{i_k}) = \frac{\sqrt{g}}{(n-k)!} g^{i_1 j_1} ... g^{i_k j_k} \ \varepsilon_{j_1 ... j_n} \ \mathrm{d}x^{j_{k+1}} \wedge ... \wedge \mathrm{d}x^{j_n}\,.\tag9$$
This immediately implies
$$ \boxed{(\star \omega)_{j_{k+1} ... j_n} = \frac{\sqrt{g}}{(n-k)!} g^{i_1 j_1} ... g^{i_k j_k} \ \varepsilon_{j_1 ... j_n} \ \omega_{i_1 ... i_k}} \,. \tag{10}$$
Note the structural difference between this (10) and your claim (2).
Lemma: The permutation symbol obeys
$$ \varepsilon^{i_1 ... i_k} = \frac{1}{(n-k)!} g^{i_1 j_1} ... g^{i_k j_k} \ \varepsilon_{j_1 ... j_n} \ \varepsilon^{j_{k+1} ... j_n} \,.\tag{11}$$
Proof of Prop.: Let $\eta := \mathrm{d}x^{i_1}\wedge...\wedge \mathrm{d}x^{i_k}$ and $\widetilde{\eta} := \frac{\sqrt{g}}{(n-k)!} g^{i_1 j_1} ... g^{i_k j_k} \ \varepsilon_{j_1 ... j_n} \ \mathrm{d}x^{j_{k+1}} \wedge ... \wedge \mathrm{d}x^{j_n}$. In order that $\widetilde{\eta} = \star \eta$, we have to check if
$$ \forall \omega \in \Omega^k(M): \omega \wedge \widetilde{\eta} = ⟨\omega,\eta⟩ \ vol$$
Since both sides are linear in $\omega$, it suffices to check on a basis. Let $\omega := \mathrm{d}x^{1}\wedge...\wedge \mathrm{d}x^{k}$. Then, one immediately notices that $\eta = \varepsilon^{i_1 ... i_k} \omega$.
Observe that \begin{align*} ⟨\omega,\eta⟩ \ vol &= \varepsilon^{i_1 ... i_k} \ ⟨\omega,\omega⟩ \sqrt{g} \ \mathrm{d} x^1 \wedge ... \wedge \mathrm{d}x^n\\ &= \sqrt{g} \ \varepsilon^{i_1 ... i_k} \ g^{1 j_1} ... g^{k j_k} \ \varepsilon_{j_1 ... j_k} \mathrm{d} x^1 \wedge ... \wedge \mathrm{d}x^n\\ &= \sqrt{g} \ \varepsilon^{i_1 ... i_k} \ \mathrm{d} x^1 \wedge ... \wedge \mathrm{d}x^n\\ &= \sqrt{g} \ \Big( \frac{1}{(n-k)!} g^{i_1 j_1} ... g^{i_k j_k} \ \varepsilon_{j_1 ... j_n} \ \varepsilon^{j_{k+1} ... j_n} \Big) \ \mathrm{d} x^1 \wedge ... \wedge \mathrm{d}x^n\\ &= \mathrm{d} x^1 \wedge ... \wedge \mathrm{d} x^k \wedge \Big( \frac{\sqrt{g}}{(n-k)!} g^{i_1 j_1} ... g^{i_k j_k} \ \varepsilon_{j_1 ... j_n} \Big) \ \mathrm{d} x^{j_{k+1}} \wedge ... \wedge \mathrm{d}x^{j_n}\\ &= \omega \wedge \widetilde{\eta} \end{align*}
This concludes our proof. $\tag{Q.E.D.}$
First of all, I don't think there is any other reasonable interpretation using the slide besides that $$ \| \alpha \|^2 = (\alpha, \alpha).$$ This is used in no.34$\to$no.35 and no.39$\to$no.40. And
$$(\alpha, \alpha) = \int_M \alpha \wedge \star \alpha$$
are used when applying the formal adjoint $\delta$ in several places. So 1 is the interpretation used in the slide.
To show that $\|\alpha\| = 0\Rightarrow \alpha = 0$, we use 2.1, that is
$$ \alpha \wedge \star \alpha = g(\alpha,\alpha)\operatorname{ vol}_{(M,g)},$$
so if $\|\alpha\|^2 = 0$, by 1 we have
$$0 = \int_M g(\alpha,\alpha)\operatorname{ vol}_{(M,g)},$$
and since $g(\alpha, \alpha)$ is non-negative, $g(\alpha, \alpha) = 0$ everywhere and thus $\alpha(x) =0$ for all $x\in M$. So $\alpha = 0$.
Best Answer
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.