Scalar, Vector, and Tensor Decomposition

Linear perturbations of an FLRW universe fall into scalar, vector, and tensor sectors. This is not merely a naming scheme: spatial isotropy makes the three helicity sectors evolve independently at first order. The decomposition turns ten metric components and the matter perturbations into smaller systems with different physical content, while also exposing the zero modes and boundary conditions that a formal inverse Laplacian can hide.

Helpful background. Einstein Equations and relativistic matter introduce the metric and stress–energy fields being decomposed. Formulations of general relativity distinguishes gauge freedom from constraints; the cosmology overview locates the FLRW background; and PDEs and Green functions explains the Fourier modes and boundary conditions used below.

The perturbation fields

Work at first order about a spatially flat FLRW background in conformal time,

$$d\bar s^2=a^2(\eta)\left(-d\eta^2+\delta_{ij}dx^i dx^j\right), \qquad \mathcal H\equiv\frac{a'}a.$$

A prime denotes $d/d\eta$. We use the spatial Fourier convention

$$X(\eta,\mathbf x) =\int\frac{d^3k}{(2\pi)^3}X(\eta,\mathbf k)e^{i\mathbf k\cdot\mathbf x}, \qquad \partial_i\longleftrightarrow ik_i, \qquad \nabla^2\longleftrightarrow-k^2.$$

The most general first-order metric perturbation may be written

$$\begin{aligned} ds^2=a^2\Big\{&-(1+2\phi)d\eta^2 +2(\partial_iB+S_i)d\eta\,dx^i\\ &+\left[(1-2\psi)\delta_{ij} +2\partial_i\partial_jE +\partial_iF_j+\partial_jF_i+h_{ij}\right]dx^i dx^j\Big\}, \end{aligned}$$

where

$$\partial^iS_i=0, \qquad \partial^iF_i=0, \qquad \partial^ih_{ij}=0, \qquad h^i{}_i=0.$$

Thus $\phi,B,\psi,E$ are scalar fields; $S_i$ and $F_i$ are transverse vectors; and $h_{ij}$ is transverse and traceless. The signs and factors in this definition are conventions, so later equations use this whole package rather than importing one variable at a time from another source. Malik and Wands give the corresponding metric split and degree count in Malik and Wands 2009, §2.2, printed pp. 3–6, eqs. (2.3)–(2.17) (Open PDF).

The ten components are still all present. Four scalar functions contribute four components. Each transverse three-vector has two components, contributing four more. A transverse-traceless symmetric three-tensor has two components. Hence

$$4+2+2+2=10.$$

This is a kinematic decomposition, not a removal of coordinate freedom. Four first-order coordinate functions still act on these variables; the next page derives that gauge action.

Helmholtz and tensor decompositions

For a spatial vector $V_i$, the Helmholtz split is

$$V_i=\partial_iV+V_i^{\rm T}, \qquad \partial^iV_i^{\rm T}=0, \qquad V=\nabla^{-2}\partial^iV_i.$$

For a symmetric spatial tensor $X_{ij}$, a useful trace-separated form is

$$\begin{aligned} X_{ij}={}&\frac13\delta_{ij}X +\left(\partial_i\partial_j-\frac13\delta_{ij}\nabla^2\right)X_L\\ &+2\partial_{(i}X^{\rm T}_{j)}+X^{\rm TT}_{ij}, \end{aligned}$$

with $\partial^iX_i^{\rm T}=0$ and $\partial^iX_{ij}^{\rm TT}=X^{{\rm TT}\,i}{}_i=0$. Its six components split as two scalars, one two-component transverse vector, and one two-component transverse-traceless tensor.

Here $X=\delta^{ij}X_{ij}$ and $2\partial_{(i}X^{\rm T}_{j)}=\partial_iX^{\rm T}_j+\partial_jX^{\rm T}_i$. The word “tensor” in the sector name means the transverse-traceless part; the scalar and vector pieces are still components of the original spatial tensor. The classification concerns how the pieces transform under rotations, not whether their indices have disappeared.

The symbols $\nabla^{-2}$ and $k^{-2}$ require a domain. On $\mathbb R^3$ one may require sufficient decay and choose the corresponding Green function. In a periodic box one removes or separately fixes the mean mode. This page assumes $k\ne0$; a homogeneous perturbation can instead redefine the FLRW background, and harmonic zero modes or physical boundaries need their own boundary data. Curved spatial slices require scalar, vector, and tensor harmonics rather than plane waves.

A direct Fourier check makes the split concrete. Choose $\mathbf k=k\hat{\mathbf z}$. For a symmetric $X_{ij}$, the combinations $X_{zz}$ and $X_{xx}+X_{yy}$ are scalar; $X_{xz}$ and $X_{yz}$ form the transverse-vector sector; and

$$X_+=\frac12(X_{xx}-X_{yy}), \qquad X_\times=X_{xy}$$

are the two tensor polarizations. Equivalently, choose plus/cross tensors $e_{ij}^{\lambda}(\widehat{\mathbf k})$ so that

$$\begin{aligned} e_{ij}^{\lambda}e^{\lambda'ij} &=2\delta_{\lambda\lambda'},\\ X_{ij}^{\rm TT}(\mathbf k) &=\sum_{\lambda=+,\times}e_{ij}^{\lambda}(\widehat{\mathbf k}) X_{\lambda,\mathbf k},\\ X_{\lambda,\mathbf k} &=\frac12e_{\lambda}^{ij}(\widehat{\mathbf k}) X_{ij}^{\rm TT}(\mathbf k). \end{aligned}$$

For $\mathbf k=k\widehat{\mathbf z}$ this inverse projection gives exactly $X_+=(X_{xx}-X_{yy})/2$ and $X_\times=X_{xy}$. The factor $1/2$ is therefore part of the norm-two basis, not an optional source normalization. No component is lost, and rotations about $\mathbf k$ assign helicities $0$, $\pm1$, and $\pm2$ to these groups.

Matter decomposes in the same way

The background density and pressure are spatial scalars, so their first-order perturbations $\delta\rho$ and $\delta p$ belong to the scalar sector. Define the momentum and trace-free stress by

$$\delta T^0{}_i=\partial_i\delta q+\delta q_i^{\rm T},$$ $$\begin{aligned} \delta T^i{}_j={}&\delta p\,\delta^i{}_j +\left(\partial^i\partial_j-\frac13\delta^i{}_j\nabla^2\right)\Pi\\ &+\partial^i\Pi_j^{\rm T}+\partial_j\Pi^{{\rm T}i} +\Pi^{{\rm TT}i}{}_j. \end{aligned}$$

The transverse and trace-free conditions are the same as for the metric. This notation does not assume a perfect fluid: free-streaming particles, for example, can supply scalar, vector, and tensor anisotropic stress. The fluid velocity and anisotropic-stress decompositions are developed in Malik and Wands 2009, §4 and §4.1, printed pp. 11–13, eqs. (4.3)–(4.17) (Open PDF).

Why the sectors decouple

The flat FLRW background is invariant under spatial translations and rotations. After Fourier transformation, a linear background operator can depend on $\mathbf k$ only through rotationally covariant combinations. It therefore commutes with rotations about $\mathbf k$ and cannot map one helicity representation into another. Scalar matter sources scalar metric perturbations, transverse-vector sources drive the vector sector, and transverse-traceless stress drives gravitational waves.

“Decouple” means no linear mixing between helicities; it does not mean that a sector is source-free or dynamically trivial. Nor does it survive unchanged beyond first order. Products such as two scalar gradients can contain vector and tensor pieces, so nonlinear mode coupling mixes sectors. Nonlinear structure evolution is outside this chapter, as are perturbations of theory-specific modifications of general relativity.

Exercises

Let $V_i(\mathbf k)$ be a Fourier-space vector with $k\ne0$. Construct its longitudinal and transverse projections and verify that their sum is $V_i$.

Solution

Define

$$P_{ij}=\delta_{ij}-\frac{k_i k_j}{k^2}, \qquad V_i^{\rm T}=P_i{}^jV_j, \qquad V_i^{\rm L}=\frac{k_i k^j}{k^2}V_j.$$

Then $k^iV_i^{\rm T}=0$ and $V_i^{\rm L}+V_i^{\rm T}=V_i$. With $V_i^{\rm L}=ik_iV$, the scalar potential is $V=-i k^jV_j/k^2$, consistent with $V=\nabla^{-2}\partial^iV_i$ under the declared Fourier convention. The construction fails at $k=0$, which is why that mode was excluded rather than silently divided by zero.

References

• Malik, K. A., and D. Wands, “Cosmological Perturbations,” Physics Reports 475, 1–51 (2009), DOI, Open PDF.