Hodge Dual

Prerequisites: Exterior Product. Full path (2 pages)

Intuitive
Concrete
Abstract

Intuitive

The Hodge dual, also known as the Hodge star operator ( $\star$ ), in $n$ -dimensional space takes an antisymmetric rank- $p$ tensor and maps it to an antisymmetric rank-( $n-p$ ) tensor. If we picture the rank- $p$ tensor as a $p$ -dimensional volume element, the Hodge dual can be pictured as its orthogonal complement.

Example #1: In Euclidean 3D space ( $n=3$ ), the Hodge dual of the antisymmetric rank-2 tensor $a \wedge b$ is a vector, thus mapping rank $p=2$ to $p=3-2=1$ . In this case, the resulting vector is just the ordinary cross product: $\star (a \wedge b) = a \times b$ . (For the meaning of $a \wedge b$ , see Exterior Product.)

Example #2: In Euclidean 3D space ( $n=3$ ), the Hodge dual of the antisymmetric rank-3 tensor $a \wedge b \wedge c$ is a scalar, thus mapping rank $p=3$ to $p=3-3=0$ . In this case, the resulting scalar is just the scalar triple product: $\star (a \wedge b \wedge c) = a \times b \cdot c$ .

Example #3: In 4D space ( $n=4$ ), the Hodge dual of an antisymmetric rank-2 tensor is again an antisymmetric rank-2 tensor ( $p=4-2=2$ ). This is the reason why self-dual ( $\star T=T$ ) and anti-self-dual ( $\star T = -T$ ) tensors exist in 4D (as real tensors in Euclidean signature; in Lorentzian signature $\star^2=-1$ and one needs complex tensors with $\star T=\pm i T$ ).

Concrete

If we increase the rank of a general tensor it acquires more and more components. For example in 4D space a rank-0 tensor (= scalar) has 1 component, a rank-1 tensor (= vector) has 4 components, a rank-2 tensor has 4x4=16 components, a rank-3 tensor has 4x4x4=64 components, and a rank-4 tensor has 4x4x4x4=256 components.

The situation changes completely if we consider (totally) antisymmetric tensors instead of general tensors. Now the number of independent components increases at first but then decreases again. For example in 4D space a rank-0 tensor has 1 component, a rank-1 tensor has 4 components, a rank-2 tensor has 6 components, a rank-3 tensor has 4 components, and a rank-4 tensor has just 1 component. Notice the pattern: 1 4 6 4 1.

This pattern suggests a relationship between antisymmetric rank- $p$ tensors and and antisymmetric rank-( $4-p$ ) tensors in 4D space. This is the idea behind the Hodge dual! The diagram below shows this pattern for 2, 3, and 4-dimensional spaces.

For a more detailed explanation of this picture see Fun with Symmetry.

Abstract

Let $(M,g)$ be an oriented $n$ -dimensional (pseudo-)Riemannian manifold, and let $s=\operatorname{sign}\det g$ ( $s=+1$ for Riemannian, $s=-1$ for Lorentzian signature). The metric induces a pointwise inner product on $p$ -forms, $\langle \alpha,\beta\rangle = \frac{1}{p!}\,\alpha_{\mu_1\dots\mu_p}\beta^{\mu_1\dots\mu_p},$ and the orientation gives the volume form $\mathrm{vol} = \sqrt{|g|}\,dx^1\wedge\dots\wedge dx^n$ in positively oriented coordinates. The Hodge star is the unique linear map $\star:\Omega^p(M)\to\Omega^{n-p}(M)$ with $\alpha\wedge\star\beta = \langle\alpha,\beta\rangle\,\mathrm{vol}\quad\text{for all } \alpha\in\Omega^p(M).$ In components, with $\epsilon_{12\dots n}=+1$ , $(\star\omega)_{\nu_1\dots\nu_{n-p}} = \frac{\sqrt{|g|}}{p!}\,\omega^{\mu_1\dots\mu_p}\,\epsilon_{\mu_1\dots\mu_p\nu_1\dots\nu_{n-p}}.$ For $\mathbb{R}^3$ with the Euclidean metric this gives $\star(e^1\wedge e^2)=e^3$ , which is the cross product rule of the Intuitive layer. The star depends on the metric and on the orientation; reversing the orientation flips its sign.

Applying it twice gives $\star\star = (-1)^{p(n-p)}\,s \quad\text{on } \Omega^p(M).$ On 2-forms in four dimensions, $p(n-p)=4$ , so $\star^2=+1$ in Euclidean signature and $\star^2=-1$ in Lorentzian signature. Real self-dual and anti-self-dual 2-forms (eigenvalues $\pm1$ ) therefore exist in Euclidean signature, which is the setting of instantons. In Minkowski space the eigenvalues are $\pm i$ , and the split of the field strength into self-dual and anti-self-dual parts requires complex forms.

With $\star$ one defines the codifferential $\delta = \pm\star d\star:\Omega^p\to\Omega^{p-1}$ , where the sign depends on $n$ , $p$ and $s$ and is fixed by requiring $\delta$ to be the formal adjoint of $d$ with respect to $(\alpha,\beta)=\int_M\alpha\wedge\star\beta$ . The Laplace-de Rham operator is $\Delta = d\delta+\delta d$ . On a compact oriented Riemannian manifold the Hodge theorem states that every de Rham cohomology class contains exactly one harmonic form ( $\Delta\omega=0$ ), and that $\Omega^p(M) = \mathcal{H}^p\oplus d\Omega^{p-1}\oplus\delta\Omega^{p+1}$ is an orthogonal decomposition. Since $\star$ commutes with $\Delta$ and maps harmonic $p$ -forms to harmonic $(n-p)$ -forms, it gives Poincaré duality $b_p=b_{n-p}$ for the Betti numbers.

In physics the star turns the source-free half of Maxwell's equations into the half with sources: $dF=0$ and $d\star F=\star J$ (up to a sign and factors that depend on conventions). On $p$ -forms with $n=2p$ the star is invariant under conformal rescalings $g\to e^{2\sigma}g$ , because the factor $\sqrt{|g|}$ and the $p$ inverse metrics used to raise the indices cancel. This is why the Maxwell action $\int F\wedge\star F$ is conformally invariant in four dimensions.

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Authors: Eduard Sackinger