Tensor Harmonics and Parity Decomposition

Spherical symmetry does not make a perturbation spherical: it organizes its angular dependence into independent angular-momentum blocks. Within each block, parity separates polar and axial fields when the equations and boundary conditions preserve that symmetry. We will construct the angular bases, derive the factors needed to project a field onto them, and identify the exceptional monopole and dipole sectors. This is an angular decomposition, not yet a solution of the gravitational constraints or a count of propagating modes.

Required background. PDEs and Green Functions for Gravitation supplies eigenfunction expansions and projection onto angular modes.

Helpful background. Black-Hole Geometry and Mechanics supplies spherical backgrounds and horizons; Formulations of General Relativity distinguishes gauge, constraints, and evolution; Killing Fields and Isometry Groups explains why rotations organize angular sectors.

Round-sphere geometry and scalar modes

On a spherically symmetric four-dimensional region with areal radius $r\gt0$, write

$$ds^2=g_{ab}(x)\,dx^a dx^b+r^2(x)\Omega_{AB}\,d\theta^A d\theta^B.$$

Here $g_{ab}$ is the Lorentzian metric on the two-dimensional orbit space; uppercase indices refer to the unit sphere. Throughout the angular calculation, raise those indices with $\Omega^{AB}$, not with $r^{-2}\Omega^{AB}$. On the usual angular chart,

$$\Omega_{AB}=\operatorname{diag}(1,\sin^2\theta), \qquad d\Omega=\sin\theta\,d\theta\,d\phi, \qquad \epsilon_{\theta\phi}=\sin\theta.$$

The sphere derivative $D_A$ preserves both $\Omega_{AB}$ and $\epsilon_{AB}$, and its Ricci tensor is $R_{AB}=\Omega_{AB}$. Coordinate components may be singular at a pole even when the tensor is smooth there. We work on the whole smooth round sphere, with no angular boundary or puncture.

Use the Condon–Shortley phase for complex scalar harmonics. Define $\lambda_\ell=\ell(\ell+1)$, with $\ell=0,1,\ldots$ and $-\ell\leq m\leq\ell$. Then

$$D^A D_A Y^{\ell m}=-\lambda_\ell Y^{\ell m}, \qquad \int_{S^2}\overline{Y^{\ell m}}Y^{\ell' m'}\,d\Omega =\delta_{\ell\ell'}\delta_{mm'}.$$

For a scalar $u$, its coefficient is $u_{\ell m}=\int\overline{Y^{\ell m}}u\,d\Omega$. The expansion converges in $L^2$ for square-integrable data; smooth data have rapidly decreasing coefficients and allow the finite differentiations used below. Reality is imposed on coefficients, not by discarding all nonzero $m$:

$$\overline{Y^{\ell m}}=(-1)^mY^{\ell,-m}, \qquad u_{\ell,-m}=(-1)^m\overline{u_{\ell m}}.$$

The same coefficient rule applies to the tensor bases below, whose defining angular operations are real. A real harmonic basis is equally valid, with the coefficient transformation made consistently. The broader spectral construction is developed in Scalar, Vector, and Tensor Spherical Harmonics.

Vector and trace-free tensor harmonics

For each scalar harmonic, define

$$Y_A^{\ell m}=D_A Y^{\ell m}, \qquad X_A^{\ell m}=-\epsilon_A{}^B D_B Y^{\ell m}.$$

The first follows the gradient; the second rotates it through a right angle in the tangent plane, with the displayed sign. These are covectors; the corresponding vectors are obtained by raising the index. The trace-free symmetric tensors are

$$\begin{aligned} Y_{AB}^{\ell m} &=D_AD_BY^{\ell m}+\frac{\lambda_\ell}{2}\Omega_{AB}Y^{\ell m},\\ X_{AB}^{\ell m} &=D_{(A}X_{B)}^{\ell m} =-\frac12\left(\epsilon_A{}^C D_B+\epsilon_B{}^C D_A\right)D_CY^{\ell m}. \end{aligned}$$

Parentheses mean half-weighted symmetrization: $D_{(A}X_{B)}=(D_AX_B+D_BX_A)/2$. The scalar eigenvalue removes the trace of $Y_{AB}$; antisymmetry of $\epsilon_{AB}$ removes that of $X_{AB}$. Thus $\Omega^{AB}Y_{AB}=\Omega^{AB}X_{AB}=0$.

These definitions follow Martel and Poisson 2005, §III, pp. 104003-3–104003-4, Eqs. (3.1)–(3.11), PDF. They fix a conventional normalization, not unit norm for every basis element. The Maxwell test-field page uses the opposite axial-vector sign: with the same sphere orientation, $X_A^{\rm here}=-X_A^{\rm Maxwell}$. A coefficient multiplying that vector must change sign too if it represents the same angular one-form. Nothing physical changes when the basis and coefficient are translated together.

The elementary divergence and curl identities are

$$\begin{aligned} D^A Y_A&=-\lambda_\ell Y,&D^A X_A&=0,\\ \epsilon^{AB}D_A Y_B&=0,&\epsilon^{AB}D_A X_B&=-\lambda_\ell Y. \end{aligned}$$

For the tensors, curvature is essential. Commuting a derivative through a scalar gradient gives

$$D^A D_A D_B Y=D_B(D^A D_A Y)+R_B{}^C D_CY =-(\lambda_\ell-1)D_BY.$$

It follows that

$$D^A Y_{AB}=-\frac{\lambda_\ell-2}{2}Y_B, \qquad D^A X_{AB}=-\frac{\lambda_\ell-2}{2}X_B.$$

For the second identity, use $D^AX_A=0$ in the divergence of $D_{(A}X_{B)}$, together with the same unit-sphere curvature commutator. Replacing these covariant derivatives by componentwise partial derivatives misses precisely these curvature terms.

Norms and coefficient projection

For angular tensors of the same rank, use the Hermitian inner product obtained by contracting every index with $\Omega^{AB}$ and integrating over $d\Omega$. Define

$$N_\ell=\frac12\lambda_\ell(\lambda_\ell-2) =\frac12(\ell-1)\ell(\ell+1)(\ell+2).$$

The table gives the squared norm of one basis element and its parity eigenvalue. Different $(\ell,m)$ modes are orthogonal within each listed family.

Angular basis	First nonzero multipole	Squared norm	Parity eigenvalue
$Y^{\ell m}$	$\ell=0$	$1$	$(-1)^\ell$
$Y_A^{\ell m}$	$\ell=1$	$\lambda_\ell$	$(-1)^\ell$
$X_A^{\ell m}$	$\ell=1$	$\lambda_\ell$	$(-1)^{\ell+1}$
$\Omega_{AB}Y^{\ell m}$	$\ell=0$	$2$	$(-1)^\ell$
$Y_{AB}^{\ell m}$	$\ell=2$	$N_\ell$	$(-1)^\ell$
$X_{AB}^{\ell m}$	$\ell=2$	$N_\ell$	$(-1)^{\ell+1}$

The vector norm follows from integration by parts on the closed sphere:

$$\int_{S^2}\overline{D_AY}D^AY\,d\Omega =-\int_{S^2}\overline Y D^2Y\,d\Omega=\lambda_\ell.$$

The rotation by $\epsilon_A{}^B$ preserves this norm. Its cross inner product vanishes because $\epsilon^{AB}D_AD_BY=0$ after integration by parts. This is an integrated orthogonality statement, not a claim that arbitrary polar and axial fields have zero pointwise contraction.

To derive the tensor normalization without guessing, write $Y_{AB}=D_AD_BY+\lambda_\ell\Omega_{AB}Y/2$. Its trace-free property and the divergence identity give

$$\begin{aligned} \int_{S^2}\overline{Y_{AB}}Y^{AB}\,d\Omega &=\int_{S^2}\overline{Y^{AB}}D_AD_BY\,d\Omega\\ &=-\int_{S^2}D_A\overline{Y^{AB}}D_BY\,d\Omega\\ &=\frac{\lambda_\ell-2}{2}\int_{S^2}\overline{Y^B}Y_B\,d\Omega=N_\ell. \end{aligned}$$

Since $X_{AB}=D_{(A}X_{B)}$ is symmetric, the same integration gives its norm $N_\ell$. Integrating a polar–axial tensor cross product reduces it to the vanishing vector cross product. The trace basis is pointwise orthogonal to either trace-free family. Its own norm is $2$, since $\Omega_{AB}\Omega^{AB}=2$. These factors agree with Martel and Poisson 2005, §III, pp. 104003-3–104003-4, Eqs. (3.3)–(3.11), PDF.

A smooth angular one-form and a smooth symmetric angular tensor therefore have expansions

$$\begin{aligned} v_A&=\sum_{\ell\geq1,m}\left(a_{\ell m}Y_A^{\ell m}+b_{\ell m}X_A^{\ell m}\right),\\ q_{AB}&=\sum_{\ell\geq0,m}k_{\ell m}\Omega_{AB}Y^{\ell m} +\sum_{\ell\geq2,m}\left(g_{\ell m}Y_{AB}^{\ell m}+h_{\ell m}X_{AB}^{\ell m}\right). \end{aligned}$$

There is no missing harmonic one-form on $S^2$: its one-forms have gradient and rotated-gradient parts. The trace-free tensor pair corresponds to the two spin-weight-two components of a symmetric tensor; see the explicit relation in Martel and Poisson 2005, Appendix A, p. 104003-11, PDF and the later spin-weighted construction. These whole-sphere completeness statements do not impose a radial equation or a spacetime boundary condition.

Orthogonality now provides a usable projection rule:

$$\begin{aligned} a_{\ell m}&=\frac1{\lambda_\ell}\int\overline{Y^{A\ell m}}v_A\,d\Omega, &b_{\ell m}&=\frac1{\lambda_\ell}\int\overline{X^{A\ell m}}v_A\,d\Omega,\\ k_{\ell m}&=\frac12\int\overline{Y^{\ell m}}\Omega^{AB}q_{AB}\,d\Omega,\\ g_{\ell m}&=\frac1{N_\ell}\int\overline{Y^{AB\ell m}}q_{AB}\,d\Omega, &h_{\ell m}&=\frac1{N_\ell}\int\overline{X^{AB\ell m}}q_{AB}\,d\Omega. \end{aligned}$$

Use the first line only for $\ell\geq1$, the last only for $\ell\geq2$, and the trace coefficient for every $\ell\geq0$. Unit-norm vector and trace-free tensor harmonics would divide these bases by $\sqrt{\lambda_\ell}$ and $\sqrt{N_\ell}$; their expansion coefficients would consequently be different.

Parity acts on the whole tensor

For the antipodal map $P:(\theta,\phi)\mapsto(\pi-\theta,\phi+\pi)$,

$$P^*\Omega=\Omega, \qquad P^*\epsilon=-\epsilon, \qquad P^*Y^{\ell m}=(-1)^\ell Y^{\ell m}.$$

The orientation sign is visible directly: $P^*(\sin\theta\,d\theta\wedge d\phi)=-\sin\theta\,d\theta\wedge d\phi$. Pullback commutes with the sphere covariant derivative because $P$ is an isometry. Each use of the orientation tensor therefore contributes an extra minus sign. This proves the parity column of the table.

“Polar” or “even-parity” means the family with eigenvalue $(-1)^\ell$; “axial” or “odd-parity” means the family with eigenvalue $(-1)^{\ell+1}$. Thus an even-parity dipole has parity eigenvalue $-1$, and an odd-parity dipole has eigenvalue $+1$. The family names do not assert a fixed sign independent of $\ell$.

Nor is parity obtained by substituting transformed angles into a component alone. A covariant tensor transforms as

$$(P^*q)_{AB}(\theta,\phi) =\frac{\partial P^C}{\partial\theta^A} \frac{\partial P^D}{\partial\theta^B} q_{CD}(\pi-\theta,\phi+\pi), \qquad DP=\operatorname{diag}(-1,1).$$

The off-diagonal component has one extra minus sign from this Jacobian. The quadrupole exercise below tests exactly that distinction.

Rotations alone do not separate polar and axial copies of the same angular-momentum representation. For example, the map $(\mathcal Jv)_A=\epsilon_A{}^Bv_B$ commutes with rotations but sends $Y_A$ to $-X_A$. It reverses parity rather than preserving it.

For a linear field operator $\mathcal L$ that preserves rotations, Schur’s lemma fixes the $(\ell,m)$ blocks, while allowing mixing among field components within a block. If in addition $[\mathcal L,P^*]=0$, the opposite parity eigenvalues at fixed $\ell$ forbid polar–axial mixing. To apply this statement to a boundary problem, its domain and boundary conditions must preserve the same symmetries. A nonspherical boundary or parity-breaking matter coupling can invalidate that conclusion even if an angular expansion is still useful. Spherical harmonics also generically cease to diagonalize the angular equation on Kerr; special parameter limits are exceptions, not a general finite-spin separation rule.

Metric amplitudes and exceptional multipoles

The metric slots explain how this angular calculation is used. For one mode with $\ell\geq2$, the conventions of the following metric pages are

$$\begin{aligned} p_{ab}&=H_{ab}Y^{\ell m},\\ p_{aB}&=j_aY_B^{\ell m}+h_aX_B^{\ell m},\\ p_{AB}&=r^2\left(K\Omega_{AB}Y^{\ell m}+G Y_{AB}^{\ell m}\right) +h_2X_{AB}^{\ell m}. \end{aligned}$$

The coefficients depend only on orbit-space coordinates. In particular, the odd convention leaves $h_2$ outside the displayed $r^2$ factor. The unit-sphere harmonics contain no hidden powers of $r$. Read Odd-Parity Metric Perturbations and Even-Parity Metric Perturbations for their gauge transformations and dynamical reductions.

Trace removal also changes a coefficient: if the polar angular metric is written instead as $r^2(K_{\rm Hess}\Omega_{AB}Y+G_{\rm Hess}D_AD_BY)$, then

$$G=G_{\rm Hess}, \qquad K=K_{\rm Hess}-\frac{\lambda_\ell}{2}G_{\rm Hess}.$$

This is a change of basis, not a gauge transformation; compare Martel and Poisson 2005, §IV.A, p. 104003-4, Eq. (4.3) and following discussion, PDF.

At $\ell=0$, the scalar is constant, so both vector harmonics and both trace-free tensor harmonics vanish. At $\ell=1$, the vectors survive but $N_1=0$ forces both trace-free tensors to vanish. The trace tensor $\Omega_{AB}Y^{\ell m}$ survives at both multipoles. Thus division by $N_\ell$ or by $(\ell-1)(\ell+2)$ is unavailable there; it cannot be repaired by substituting a low multipole into a radiative formula.

These are angular identities, not gauge choices. Interpreting a monopole as a mass shift, an axial dipole as angular momentum, or a polar dipole as removable gauge additionally requires the field equations, boundary conditions, and admissible gauge transformations. Low Multipoles, Parameter Shifts, and Gauge Modes supplies that separate analysis. No gravitational master equation, mode stability, or nonlinear backreaction follows from orthogonality alone.

Exercises

A projection factor that cannot be dropped. Let $v_A=3Y_A^{10}$, with $Y^{10}$ of unit scalar norm. Recover its polar coefficient by integration. What happens if the vector harmonic is mistakenly treated as unit normalized?

Solution

Here $\lambda_1=2$, so $\int\overline{Y^{A10}}v_A\,d\Omega=3\lambda_1=6$. The projection divides by $\lambda_1$ and returns $a_{10}=3$. Omitting the denominator returns $6$ and reconstructs twice the intended field. A unit vector basis would be $Y_A^{10}/\sqrt2$, whose coefficient is $3\sqrt2$, not $3$.

Separate the trace of a quadrupole Hessian. For $q_{AB}=D_AD_BY^{20}$, find the nonzero coefficients $k_{\ell m},g_{\ell m},h_{\ell m}$. Check the result by tracing and by the tensor projection formula.

Solution

Since $\lambda_2=6$ and $N_2=12$,

$$q_{AB}=Y_{AB}^{20}-3\Omega_{AB}Y^{20}, \qquad k_{20}=-3,\quad g_{20}=1,\quad h_{20}=0.$$

Its trace is $-6Y^{20}$, and the trace projection includes one half, giving $k_{20}=-3$. Contracting against $\overline{Y^{AB20}}$ eliminates the trace term and gives $12$; division by $N_2$ gives $1$. All other coefficients vanish by orthogonality. This example also checks the sign of the Hessian-to-trace-free coefficient map.

An unchanged component in an odd tensor. Set $C=\sqrt{5/(16\pi)}$ and $Y^{20}=C(3\cos^2\theta-1)$. Compute its axial vector and axial tensor. Is $X_{AB}^{20}$ even because its nonzero component function is unchanged by $\theta\mapsto\pi-\theta$?

Solution

The vector has $X_\theta^{20}=0$ and $X_\phi^{20}=-6C\sin^2\theta\cos\theta$. Use $\Gamma^\phi{}_{\theta\phi}=\cot\theta$ in its symmetric derivative:

$$X_{\theta\phi}^{20}=X_{\phi\theta}^{20} =\frac12\left(\partial_\theta X_\phi^{20}-2\cot\theta X_\phi^{20}\right) =3C\sin^3\theta, \qquad X_{\theta\theta}^{20}=X_{\phi\phi}^{20}=0.$$

The function $3C\sin^3\theta$ is unchanged at the antipodal arguments, but the mixed covariant component acquires $(-1)(+1)$ from the two Jacobian factors. Consequently $P^*X_{AB}^{20}=-X_{AB}^{20}$, exactly the axial quadrupole parity. Omitting the connection would give the wrong tensor; omitting the Jacobian would give the wrong parity. As a normalization check,

$$\int_{S^2}\overline{X_{AB}^{20}}X^{AB20}\,d\Omega =36\pi C^2\int_0^\pi\sin^5\theta\,d\theta=12=N_2.$$

The two equal off-diagonal entries both contribute to the contraction. The explicit component construction agrees with Martel and Poisson 2005, Appendix A, p. 104003-11, PDF.

References

• Martel, Karl, and Eric Poisson. “Gravitational Perturbations of the Schwarzschild Spacetime: A Practical Covariant and Gauge-Invariant Formalism.” Physical Review D 71 (2005): 104003. DOI. Open PDF.