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For plane-parallel pressureless matter in an Einstein-de Sitter universe, before shell crossing, the symmetrized growing-mode standard perturbation theory density kernel isIn particular $F_{2,s}=(q_1+q_2{)}^2/(2q_1{q}_2)$. An unsymmetrized representation of the one-dimensional beta mode-coupling kernel is $(q_1+q_2)/q_1$; its symmetric part is the expression above.

The background Einstein-de Sitter universe satisfies $a^{\prime }=\mathcal{H}a$ and ${\mathcal{H}}^{\prime }=-{\mathcal{H}}^2/2$. Introduce ${\mathcal{F}}_n$ and ${\mathcal{G}}_n$ for the spatial convolutions in the Fourier transform representation with kernels $F_n$ and $G_n$, respectively, so ${\delta }^{(n)}=a^n{\mathcal{F}}_n$ and ${\theta }^{(n)}=-\mathcal{H}{a}^n{\mathcal{G}}_n$. Differentiation givesAt first order the cosmological continuity equation yields $F_1-G_1=0$, while the cosmological Euler equation gives $3F_1/2-3G_1/2=0$. The growing-mode normalization therefore gives $\boxed{G_1=F_1=1}$.

At second order, take the unsymmetrized order of the inputs printed in the paper. Matching the alpha mode-coupling kernel and beta mode-coupling kernel terms givesHenceWith the paper's unsymmetrized convention $\alpha =\beta =(q_1+q_2)/q_1$, one possible unsymmetrized representation isOnly the symmetric part enters a convolution of two identical linear fields. Symmetrizing givesThis is the one-dimensional cosmological density kernel. The physical symmetric one-dimensional Euler kernel is also $(q_1+q_2{)}^2/(2q_1{q}_2)$: symmetrizing the printed $\beta$ reproduces it, so its apparently different unsymmetrized form is not a contradiction. Unsymmetrized kernels themselves are not unique, since antisymmetric parts integrate to zero. Zero input momenta require the usual separate treatment of the homogeneous background.

The cosmological continuity equation expresses conservation of dark-matter mass: $\delta$ is the density contrast, $\mathbf{v}$ is the peculiar velocity, and a prime denotes a conformal time derivative. The cosmological Euler equation expresses momentum conservation: $\mathcal{H}=a^{\prime }/a$ is the conformal Hubble rate, $\varphi$ is the peculiar gravitational potential, $\rho$ is the density, and ${\sigma }_{ij}$ is the velocity-dispersion tensor of collisionless matter. The terms are respectively Hubble drag, convective acceleration, gravity, and velocity-dispersion stress.

For curl-free flow, introduce the peculiar-velocity divergence $\theta =\nabla \cdot \mathbf{v}$ and set ${\sigma }_{ij}=0$. A Fourier transform of the nonlinear continuity term giveswith the alpha mode-coupling kernelTaking the divergence of the Euler equation gives the quadratic velocity kernelThis is the beta mode-coupling kernel; its symmetry follows from the two velocity factors.

Let $\theta =\nabla \cdot \mathbf{v}$. Vanishing vorticity implies in Fourier spaceFourier transforming $\nabla \cdot (\delta \mathbf{v})$ in the continuity equation giveswith the alpha mode-coupling kernelTaking the divergence of $(\mathbf{v}\cdot \nabla )\mathbf{v}$ and symmetrizing its two velocity arguments giveswhere the beta mode-coupling kernel is

Use the Fourier convention $f(\mathbf{x})={\int }_{\mathbf{k}}f(\mathbf{k})e^{i\mathbf{k}\cdot \mathbf{x}}$ with ${\int }_{\mathbf{k}}=\int d^{3}k/(2\pi {)}^3$. Neglect anisotropic stress and assume the pressureless velocity is irrotational, soFourier transforming $\nabla \cdot (\delta \mathbf{v})$ in the nonlinear continuity equation giveswhere the alpha mode-coupling kernel and its symmetrization areTaking the divergence of the Euler equation giveswith the symmetric beta mode-coupling kernel

In the Einstein-de Sitter universe, $a^{\prime }=\mathcal{H}a$ and ${\mathcal{H}}^{\prime }=-{\mathcal{H}}^2/2$. At linear order the ansatz gives ${\tilde{\theta }}^{(1)}={\tilde{\delta }}^{(1)}$. At second order, the continuity and Euler equations becomewhere each integral includes the momentum-conserving measure above. Eliminating ${\tilde{\theta }}^{(2)}$ yieldswith the standard perturbation theory density kernel