Hodge star

As explained in duality in exterior algebras, if we have a fixed volume form, for a simple $k$ -vector $w$ we have the $(n - k)$ -form

w \mapsto v o l (w, -)

Reciprocally, to every $k$ -form we can associate a $(n - k)$ -vector at this way.

In the presence of a metric, we can make use of this duality and then apply the musical isomorphism (raising and lowering indices). This is the Hodge star:
hodgedual_2.jpg|500

It turns out that given an isometry $f : M \to N$ of Riemannian manifolds, then the Hodge star commutes with the pullback $f^{*}$ of forms.

Alternative definition

The Hodge star operator in $R^{4}$ with a pseudo-Riemannian metric maps a $p$ -form $β$ to a $(4 - p)$ -form $* β$ , defined by:

α \land (* β) = ⟨ α, β ⟩ vol

where $vol$ is the volume form, and $⟨ α, β ⟩$ is the metric-induced inner product on forms (see @baez1994gauge page 88).

In an orthonormal frame

Exercise 69 in @baez1994gauge
Let $M$ be an oriented semi-Riemannian manifold of dimension $n$ and signature $(s, n - s)$ . Let $e^{μ}$ be an orthonormal basis of 1-forms on some chart. Define the Levi-Civita symbol for $1 \leq i_{j} \leq n$ by

ϵ_{i_{1} \dots i_{n}} = {\begin{cases} sign (i_{1}, \dots, i_{n}) & all i_{j} distinct \\ 0 & otherwise . \end{cases}

Show that for any $p$ -form

ω = \frac{1}{p!} ω_{i_{1} \dots i_{p}} e^{i_{1}} \land \dots \land e^{i_{p}},

⋆ ω = \frac{1}{p!} (⋆ ω)_{j_{1} \dots j_{n - p}} e^{j_{1}} \land \dots \land e^{j_{n - p}}

we have

(⋆ ω)_{j_{1} \dots j_{n - p}} = ϵ_{j_{1} \dots j_{n - p}}^{i_{1} \dots i_{p}} ω_{i_{1} \dots i_{p}} .

In coordinates

From here.
Let $X$ be a pseudo-Riemannian manifold of dimension $D$ and metric $g$ , and locally, on some open set $U \subset X$ , let

e^{1}, \dots, e^{D} \in Ω^{1} (U)

be a coframe. For example, if ${x^{i}}$ is a coordinate chart on $U$ , then $e^{i} : = d x^{i}$ can be such a coframe.

With this choice, any differential form $α \in Ω^{p} (U)$ has a component expansion

α = \frac{1}{p!} α_{i_{1} \dots i_{p}} e^{i_{1}} \land \dots \land e^{i_{p}}

where here and in the following we use the Einstein index notation.

In terms of these components, the Hodge dual $⋆ α$ of $α$ is expressed by the following formula:

\begin{aligned} ⋆ α & = \frac{1}{p! (D - p)!} \sqrt{| d e t ((g_{i j})) |} α_{j_{1} \dots j_{p}} g^{j_{1} i_{1}} \dots g^{j_{p} i_{p}} ϵ_{i_{1} \dots i_{p} i_{p + 1} \dots i_{D}} e^{i_{p + 1}} \land \dots \land e^{i_{D}} \\ = \frac{1}{p! (D - p)!} \sqrt{| d e t ((g_{i j})) |} α^{i_{1} \dots i_{p}} ϵ_{i_{1} \dots i_{p} i_{p + 1} \dots i_{D}} e^{i_{p + 1}} \land \dots \land e^{i_{D}} \end{aligned}

Examples

$R^{3}$ standard metric

The formula for the Hodge star operator might seem complicated, so consider an example. Take $d x, d y, d z$ as a basis of 1-forms on $R^{3}$ with its usual Euclidean metric and orientation. Then we have

⋆ d x = d y \land d z, ⋆ d y = d z \land d x, ⋆ d z = d x \land d y,

and conversely

⋆ (d x \land d y) = d z, ⋆ (d y \land d z) = d x, ⋆ (d z \land d x) = d y .

$R^{3}$ with general metric

Another example, given a 1-form $ω = ω_{a} d x^{a}$ in $R^{3}$ with a Riemannian metric $g_{a b}$ , $⋆ ω$ is a 2-form with components given by:

(⋆ ω)_{a b} = \frac{1}{2} ϵ_{a b c} g^{d c} ω_{d} \sqrt{| g |}

where $ϵ_{a b c}$ is the Levi-Civita symbol and $\sqrt{| g |}$ is the square root of the determinant of the metric tensor.

On the contrary, for a 2-form $ω = ω_{a b} d x^{a} \land d x^{b}$ we have

(⋆ ω)_{c} = \frac{1}{2} ϵ_{c a b} ω^{a b} \sqrt{| g |}

where $ω^{a b} = g^{a c} g^{b d} ω_{c d}$ .

$M^{4}$ , Minkowski space

For one-forms, the Hodge star operator $⋆$ acts as follows:

\begin{aligned} ⋆ d t & = - d x \land d y \land d z, \\ ⋆ d x & = - d t \land d y \land d z, \\ ⋆ d y & = - d t \land d z \land d x, \\ ⋆ d z & = - d t \land d x \land d y . \end{aligned}

For 2-forms, $⋆$ acts as:

\begin{aligned} ⋆ (d t \land d x) & = - d y \land d z, \\ ⋆ (d t \land d y) & = - d z \land d x, \\ ⋆ (d t \land d z) & = - d x \land d y, \\ ⋆ (d x \land d y) & = d t \land d z, \\ ⋆ (d z \land d x) & = d t \land d y, \\ ⋆ (d y \land d z) & = d t \land d x . \end{aligned}

These results can be summarized in index notation as:

\begin{aligned} ⋆ (d x^{μ}) & = η^{μ λ} ϵ_{λ ν ρ σ} \frac{1}{3!} d x^{ν} \land d x^{ρ} \land d x^{σ}, \\ ⋆ (d x^{μ} \land d x^{ν}) & = η^{μ κ} η^{ν λ} ϵ_{κ λ ρ σ} \frac{1}{2!} d x^{ρ} \land d x^{σ} . \end{aligned}

$R^{2}$ with Lorentzian metric, i.e., $M^{2}$

Consider $R^{2}$ with the metric $d t^{2} - d x^{2}$ . The volume form on this spacetime is given by:

vol = \sqrt{| det g |} d t \land d x = d t \land d x,

since $| det g | = 1$ for the metric $diag (1, - 1)$ .
For a 2-dimensional spacetime, a 2-form is mapped to a 0-form under the Hodge star operator. The relationship is:

α \land ⋆ α = (α, α) vol,

where $(α, α)$ is the inner product induced by the metric. For $d t \land d x$ , the inner product $(d t \land d x, d t \land d x)$ is computed using the determinant of the metric restricted to the directions of the 2-form.

In our case, the metric determinant is $- 1$ , so:

(d t \land d x, d t \land d x) = - 1.

Therefore:

⋆ (d t \land d x) = - 1.

Similarly:

⋆ d t = d x, ⋆ d x = - d t .

Hodge star

Alternative definition

In an orthonormal frame

In coordinates

Examples

R3 standard metric

R3 with general metric

M4, Minkowski space

R2 with Lorentzian metric, i.e., M2

$R^{3}$ standard metric

$R^{3}$ with general metric

$M^{4}$ , Minkowski space

$R^{2}$ with Lorentzian metric, i.e., $M^{2}$