The Hodge Dual in Lorentzian Signature

The Hodge star converts a $p$-form into an $(n-p)$-form by combining a nondegenerate metric with an orientation. Its square is not always $+1$: on an oriented pseudo-Riemannian $n$-manifold of index $s$, it obeys $\star^2=(-1)^{p(n-p)+s}$ on real $p$-forms. In four-dimensional Lorentzian signature, $s=1$, so the square is $-1$ on two-forms but $+1$ on one- and three-forms.

Required background. Tensor Fields and Natural Operations supplies metric contractions and tensor transformations.

Helpful background. Calculus and Linear Algebra for Relativity supplies determinant signs; Exterior Algebra and Differential Forms supplies wedge products and orientations; Events, Intervals, and Minkowski Geometry supplies a Lorentzian orthonormal basis.

Metric, orientation, and degree

Let $(M,g)$ be oriented, $n$-dimensional, and pseudo-Riemannian of index $s$. The chosen orientation fixes a metric volume form $\mathrm{vol}_g$. The metric induces an inner product on $p$-forms,

$$\langle\alpha,\beta\rangle_g =\frac{1}{p!} \alpha_{\mu_1\cdots\mu_p} \beta^{\mu_1\cdots\mu_p}.$$

For each degree $p$, the Hodge operator $\star:\Omega^p(M)\to\Omega^{n-p}(M)$ is defined by

$$\alpha\wedge\star\beta =\langle\alpha,\beta\rangle_g\,\mathrm{vol}_g$$

for all $p$-forms $\alpha,\beta$. This equation displays every ingredient: dimension fixes the target degree, the metric raises indices and supplies the inner product, and the orientation fixes the sign of the volume form. Reversing orientation reverses $\star$ once, while changing the metric signature changes the norm signs used in its definition.

In a positively oriented chart on a four-dimensional Lorentzian spacetime with signature $(-,+,+,+)$,

$$\mathrm{vol}_g =\sqrt{-g}\,dx^0\wedge dx^1\wedge dx^2\wedge dx^3, \qquad \epsilon_{0123}=+\sqrt{-g}.$$

The component expression for a $p$-form is

$$(\star\alpha)_{\mu_{p+1}\cdots\mu_n} =\frac{1}{p!} \alpha^{\mu_1\cdots\mu_p} \epsilon_{\mu_1\cdots\mu_n}.$$

This index ordering follows from the displayed wedge definition. A useful source comparison is Tong 2019, §3.1.3, eqs. (3.5)–(3.6), p. 97, PDF: Tong puts the complementary indices first in the alternating tensor. Consequently, for the same orientation and metric,

$$\star_{\mathrm{here}}\alpha =(-1)^{p(n-p)}\star_{\mathrm{Tong}}\alpha.$$

The two conventions agree on two-forms in four dimensions but differ on one- and three-forms. They give the same square rule because applying the conversion twice cancels the extra sign. A source’s $\star^2$ formula alone therefore does not determine its single-star convention.

Deriving the square

Choose an oriented pseudo-orthonormal coframe $e^1,\ldots,e^n$, with $s$ negative norm squares. For a basis $p$-form $e^I=e^{i_1}\wedge\cdots\wedge e^{i_p}$, the first star supplies the complementary ordered form and the product of the metric signs in $I$. Applying the star again produces two signs. Moving the complementary $(n-p)$ indices back past the $p$ original indices contributes $(-1)^{p(n-p)}$; the product of all $n$ metric signs contributes $(-1)^s$. Therefore

$$\star\star\alpha=(-1)^{p(n-p)+s}\alpha, \qquad \alpha\in\Omega^p(M).$$

This proof is pointwise, so it requires neither flatness nor a coordinate basis. Calegari derives the metric Hodge construction in the Riemannian case in Calegari 2022, §§9.1–9.2, pp. 83–85, PDF; the additional factor $(-1)^s$ is precisely the pseudo-Riemannian correction.

Four-dimensional Lorentzian basis check

Take Minkowski space with positive orientation $dt\wedge dx\wedge dy\wedge dz$ and signature $(-,+,+,+)$. For degree one, the defining wedge identity gives

$$\begin{aligned} \star dt&=-dx\wedge dy\wedge dz,& \star dx&=-dt\wedge dy\wedge dz,\\ \star dy&=\phantom{-}dt\wedge dx\wedge dz,& \star dz&=-dt\wedge dx\wedge dy. \end{aligned}$$

For degree two under the same dimension, signature, and orientation,

$$\begin{aligned} \star(dt\wedge dx)&=-dy\wedge dz,& \star(dt\wedge dy)&=\phantom{-}dx\wedge dz,\\ \star(dt\wedge dz)&=-dx\wedge dy,& \star(dx\wedge dy)&=\phantom{-}dt\wedge dz,\\ \star(dx\wedge dz)&=-dt\wedge dy,& \star(dy\wedge dz)&=\phantom{-}dt\wedge dx. \end{aligned}$$

Applying the two-form table twice yields $-\alpha$. For a one-form, the defining wedge identity gives, for example, $\star(dx\wedge dy\wedge dz)=-dt$, hence $\star^2dt=+dt$. Both agree with $(-1)^{p(4-p)+1}$. In particular, real Lorentzian two-forms cannot have eigenvalue $+1$ or $-1$ under $\star$; after complexification the eigenvalues are $\pm i$.

What the Hodge star does not preserve

The star is metric dependent. A change of orientation flips its sign, and a conformal rescaling changes it by a degree-dependent power. In dimension $n$, for $\widetilde g=\Omega^2g$ with $\Omega>0$,

$$\widetilde\star\alpha=\Omega^{n-2p}\star\alpha$$

on $p$-forms. Thus the star on two-forms is conformally invariant in four dimensions, while the star on one-forms is not. Degenerate bilinear forms admit no inverse metric and hence no Hodge star by this definition.

The formula also does not choose an orientation or a boundary orientation. Those must be supplied independently. When an identity uses a source with the opposite signature, translating only the displayed $\star^2$ sign is unsafe: the raised Levi-Civita tensor and the norms of basis forms must be translated together.

Two further checks expose common sign errors. In dimension four and Lorentzian index one, with the stated orientation,

$$\star1=\mathrm{vol}_g, \qquad \star\mathrm{vol}_g=-1.$$

The second identity follows because the volume form has squared norm $-1$. It verifies the degree-zero case of the square rule: $\star^2 1=-1$. By contrast, for a one-form $\alpha$, $\star^2\alpha=+\alpha$, and for a two-form $F$, $\star^2F=-F$. A sign table that reports the same square in every degree has therefore lost either the permutation factor or the metric-index factor.

The defining pairing is also indefinite. Although $\alpha\wedge\star\alpha$ resembles a squared norm times volume, its coefficient may be negative or zero for a nonzero form. For example, $dt\wedge\star dt=-\mathrm{vol}$, while $dx\wedge\star dx=+\mathrm{vol}$. This is not a defect: the induced inner product on forms inherits the pseudo-Riemannian signature.

On an oriented manifold with boundary, the Hodge star does not set the boundary orientation. The latter still follows the outward-normal-first rule. Nor does $\star$ commute with every pullback: an isometry preserving orientation intertwines Hodge stars, while a general immersion or conformal map introduces the metric, dimension, and degree effects stated above. Keeping these structures separate prevents a correct algebraic star formula from being paired with an inconsistent integration sign.

Exercises

In four-dimensional Minkowski space with the orientation and signature used above, let $F=dt\wedge dx+dy\wedge dz$. Compute $\star F$ and $\star^2F$.

Solution

The basis table gives

$$\star F=-dy\wedge dz+dt\wedge dx.$$

Applying the star again,

$$\star^2F =-\star(dy\wedge dz)+\star(dt\wedge dx) =-dt\wedge dx-dy\wedge dz=-F.$$

This is the degree-$p=2$, dimension-$n=4$, index-$s=1$ case of $\star^2=(-1)^{p(n-p)+s}$.

References

• Calegari, Danny. Introduction to Differential Geometry. Preliminary version, May 26, 2022, University of Chicago, §§9.1–9.2, pp. 83–85. Course PDF.
• Tong, David. General Relativity. University of Cambridge lecture notes, 2019, §3.1, pp. 91–99. Official course PDF.