Scalar, Vector, and Tensor Spherical Harmonics

The angular basis must match the tensor type of the field. Scalar harmonics diagonalize the sphere Laplacian; their gradients and curls span one-forms; their trace, trace-free Hessian, and symmetrized curl span symmetric two-tensors. With fixed normalization and parity conventions, projecting a rotationally invariant operator onto these bases turns a four-dimensional field equation into independent orbit-space equations labeled by $(\ell,m)$, while the norm factors expose the exceptional monopole and dipole.

Required background. Killing Fields and Isometry Groups supplies the rotation-group action and conserved labels; Events, Intervals, and Minkowski Geometry supplies the local Lorentz and tensor language used to interpret field components.

Helpful background. Black-Hole Geometry and Mechanics provides the spherical backgrounds; PDEs and Green Functions for Gravitation reviews eigenfunction expansions; Formulations of General Relativity separates constraints and evolution; Frames, Coframes, and Anholonomy prepares the dyad description used on the next page.

Scalar modes of the sphere

Let $\Omega_{AB}$ be the unit-sphere metric and $D_A$ its covariant derivative. Choose the orientation $\epsilon_{\theta\phi}=\sin\theta$. The scalar phase is the Condon–Shortley convention:

$$Y_{\ell m}(\theta,\phi) =\left[\frac{2\ell+1}{4\pi} \frac{(\ell-m)!}{(\ell+m)!}\right]^{1/2} P_\ell^m(\cos\theta)e^{im\phi},$$

where, for $m\ge0$, $P_\ell^m(x)=(-1)^m(1-x^2)^{m/2}d^mP_\ell/dx^m$; negative-$m$ harmonics are fixed by $Y_{\ell,-m}=(-1)^m\overline{Y_{\ell m}}$. Then

$$D^2Y_{\ell m}=-\ell(\ell+1)Y_{\ell m}, \qquad \int_{S^2}\overline{Y_{\ell m}}Y_{\ell' m'}\,d\Omega =\delta_{\ell\ell'}\delta_{mm'}.$$

For every square-integrable scalar, completeness gives $f=\sum_{\ell m}f_{\ell m}Y_{\ell m}$ in $L^2(S^2)$, with $f_{\ell m}=\int\overline{Y_{\ell m}}f\,d\Omega$. Smoothness makes the coefficients decay faster than any power of $\ell$; a discontinuity slows the decay and makes a truncated expansion ring. Completeness is a functional statement, not permission to exchange arbitrary singular limits with the sum.

Vector harmonics diagonalize exact and coexact parts

For $\ell\ge1$, 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 is even (polar) and the second odd (axial). Under the antipodal map their complete tensorial parities are $(-1)^\ell$ and $(-1)^{\ell+1}$. They satisfy

$$D^AY_A^{\ell m}=-\ell(\ell+1)Y_{\ell m}, \qquad D^AX_A^{\ell m}=0,$$ $$\epsilon^{AB}D_AY_B^{\ell m}=0, \qquad \epsilon^{AB}D_AX_B^{\ell m}=-\ell(\ell+1)Y_{\ell m}.$$

Integration by parts gives equal norms $\ell(\ell+1)$ and zero even–odd inner product. The Hodge decomposition on $S^2$ has no harmonic one-form, so these exact and coexact bases are complete for square-integrable one-forms. The factor $\ell(\ell+1)$ vanishes at $\ell=0$, correctly diagnosing the absence of a vector monopole.

Symmetric tensor harmonics and exceptional modes

A symmetric two-tensor uses the trace harmonic $\Omega_{AB}Y_{\ell m}$ and, for $\ell\ge2$, the trace-free pair

$$Y_{AB}^{\ell m} =\left[D_AD_B+\frac12\ell(\ell+1)\Omega_{AB}\right]Y_{\ell m},$$ $$X_{AB}^{\ell m} =-\frac12\left(\epsilon_A{}^CD_B+\epsilon_B{}^CD_A\right) D_CY_{\ell m}.$$

Their squared norm is

$$\frac12(\ell-1)\ell(\ell+1)(\ell+2).$$

The two zeros at low $\ell$ are physical bookkeeping information. A monopole has neither vector nor trace-free tensor pieces. A dipole has vector pieces but no trace-free tensor piece. Therefore a radiative metric master field whose definition divides by $(\ell-1)(\ell+2)$ is restricted to $\ell\ge2$. The low modes must be solved from their own projected equations.

These definitions, including the sign of $X_A$ and the trace-free tensor basis, follow Martel and Poisson 2005, § III, Eqs. (3.1)–(3.11), pp. 104003-3–4; Appendix A, p. 104003-11. Other sources reverse the axial sign or use $D_AD_BY$ without splitting its trace; amplitudes must be translated with the basis.

Projection, completeness, and the convention bridge

On a spherically symmetric background, an operator built from $\Omega_{AB}$, $D_A$, and $\epsilon_{AB}$ preserves the rotation representations. Its matrix is block diagonal in $(\ell,m)$, but parity is a separate condition: the map $v_A\mapsto\epsilon_A{}^Bv_B$ commutes with rotations while exchanging polar and axial vectors. Parity blocks decouple only when the operator and its domain or boundary conditions also preserve antipodal parity. Projection does not by itself diagonalize fields of different tensor type in one block, nor does it remove gauge constraints. The tensor-harmonic decomposition gives the explicit pullback and coefficient factors.

Projecting coefficients with their norms

Define $L_\ell=\ell(\ell+1)$ and $N_\ell=L_\ell(L_\ell-2)/2$. For a one-form and a symmetric two-tensor, write

$$\begin{aligned} v_A&=\sum_{\ell\ge1,m} (a_{\ell m}Y_A^{\ell m}+b_{\ell m}X_A^{\ell m}),\\ T_{AB}&=\sum_{\ell\ge0,m}k_{\ell m}\Omega_{AB}Y_{\ell m}\\ &\quad+\sum_{\ell\ge2,m} (g_{\ell m}Y_{AB}^{\ell m}+h_{\ell m}X_{AB}^{\ell m}). \end{aligned}$$

Taking the inner product with each orthogonal basis tensor gives

$$\begin{aligned} a_{\ell m}&=\frac1{L_\ell}\int_{S^2}\overline{Y^{A\ell m}}v_A\,d\Omega, &b_{\ell m}&=\frac1{L_\ell}\int_{S^2}\overline{X^{A\ell m}}v_A\,d\Omega,\\ k_{\ell m}&=\frac12\int_{S^2}\overline{Y_{\ell m}} \Omega^{AB}T_{AB}\,d\Omega,\\ g_{\ell m}&=\frac1{N_\ell}\int_{S^2}\overline{Y^{AB\ell m}}T_{AB}\,d\Omega, &h_{\ell m}&=\frac1{N_\ell}\int_{S^2}\overline{X^{AB\ell m}}T_{AB}\,d\Omega. \end{aligned}$$

Every raised angular index uses the inverse unit-sphere metric. The trace basis has squared norm two, not one. Divide by $L_\ell$ only for $\ell\ge1$ and by $N_\ell$ only for $\ell\ge2$; the trace basis still exists at $\ell=0,1$. These factors follow from the norms above, without renormalizing the harmonics.

A smooth tensor reconstructed at two cutoffs

Take $f=Y_{10}+Y_{20}$ and $T_{AB}=D_AD_Bf$. The definitions give the complete finite expansion

$$T_{AB}=-\Omega_{AB}Y_{10}-3\Omega_{AB}Y_{20}+Y_{AB}^{20}.$$

The only nonzero coefficients are $k_{10}=-1$, $k_{20}=-3$ and $g_{20}=1$. The dipole has no trace-free tensor harmonic, but its Hessian’s trace survives. Keeping all tensor modes through cutoff $\ell_{\mathrm{cut}}=1$ gives $T^{[1]}_{AB}=-\Omega_{AB}Y_{10}$; cutoff two recovers $T$ exactly.

Measure the field and its divergence, not merely its values at quadrature nodes. For the residual $R_{AB}=T_{AB}-T^{[\ell_{\mathrm{cut}}]}_{AB}$, define

$$\begin{aligned} \|R\|_2^2&=\int_{S^2}\overline{R_{AB}}R^{AB}\,d\Omega,\\ \|\operatorname{div}R\|_2^2 &=\int_{S^2}\overline{D^AR_{AB}}D_CR^{CB}\,d\Omega. \end{aligned}$$

For a unit scalar harmonic with $D^2Y=-LY$, the sphere’s curvature gives

$$D^AD_AD_BY=(1-L)D_BY, \qquad \int_{S^2}\overline{D_AD_BY}D^AD^BY\,d\Omega=L(L-1).$$

The first identity follows by commuting the derivatives and using $\operatorname{Ric}_{AB}=\Omega_{AB}$; integration by parts then gives the second. At cutoff one the residual is $D_AD_BY_{20}$, with $L=6$. Thus the checks are

Cutoff	Tensor error norm	Divergence error norm
$\ell_{\mathrm{cut}}=1$	$\sqrt{30}$	$5\sqrt6$
$\ell_{\mathrm{cut}}=2$	$0$	$0$

The full Hessian’s squared norm is $30$, not the trace-free norm $12$: its trace contributes $18$. Dropping that contribution leaves a plausible tensor but fails both reconstruction and derivative checks. This finite example verifies the projection factors; it is not a proof of completeness or convergence rates for every smooth field. In larger truncations, quadrature-node agreement alone can hide aliasing, so monitor invariant norms and derivative residuals independently.

For later pages we reserve spin weight $s$ for dyad rotation, define the eth signs explicitly before use, take the spheroidal parameter to be $c=a\omega$ with $e^{-i\omega t}$ time dependence, and use the $S^2$ inner product above. The spheroidal eigenvalue is called $A_{s\ell m}(c)$ and satisfies $A_{s\ell m}(0)=\ell(\ell+1)-s(s+1)$. These conventions prevent the common confusion among $A$, $E=A+s(s+1)$, and the radial Teukolsky constant $\lambda=A+c^2-2mc$.

General representation theory and proofs of the full Hodge spectral theorem require further mathematical study. Exact Solutions develops the geometry of hidden symmetries; here the harmonic bases organize perturbation equations.

Exercise: reconstruct a vector dipole

Let $f=Y_{10}$ and $V_A=D_Af+2\epsilon_A{}^BD_BY_{11}$. Compute its nonzero harmonic coefficients and verify that its divergence and curl recover the two scalar sources. Why can no $\ell=0$ vector coefficient be added?

Solution

By definition, the exact coefficient is $1$ in the $(\ell,m)=(1,0)$ sector and the coexact coefficient is $-2$ in the convention $X_A=-\epsilon_A{}^BD_BY$: thus $V_A=Y_A^{10}-2X_A^{11}$. Orthogonality makes all other coefficients vanish. The identities above give

$$D^AV_A=-2Y_{10}, \qquad \epsilon^{AB}D_AV_B=4Y_{11}.$$

The first term has zero curl and the second zero divergence. A vector monopole would have to be a gradient or dual gradient of $Y_{00}$, but that harmonic is constant, so both candidates vanish. The missing mode is a basis degeneracy, not a choice of gauge.

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.