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Write $K=k_1+k_2+k_3>0$ and $C=ϵ+1-c_s^2$. The inflationary scalar Fourier mode has real late-time value $A_k=H/\sqrt{4ϵM_{\mathrm{Pl}}^2{c}_s{k}^3}$ and $u_k^{\prime \ast }(\tau )=A_k{c}_s^2{k}^2\tau e^{ik{c}_s\tau }$. Thus the time integrand, before the factor $-2i$, isThe Bunch-Davies vacuum prescription supplies damping in the past. Equivalently, use a positive regulator $d$ and then take the limit:The real part of $-2i$ times this integral is $4/(c_s^3{K}^3)$. Including the six Wick contractions yields the cubic time-derivative curvature bispectrumThe resulting primordial bispectrum is a homogeneous function of degree $-6$ in the wavenumbers, as expected for scale-independent background parameters.

The printed limit $c_s\ll 0$ is not a physical sound-speed limit for these modes. The normalization and positive-frequency choice assume $c_s>0$; a negative value is outside the positive sound-speed convention used by these modes. The likely intended limit is $0<c_s\ll 1$, which is analyzed explicitly here rather than silently substituted for the PDF.

To compare primordial non-Gaussianity, normalize by the power spectrum, not only by the raw three-point amplitude. HereAt an equilateral bispectrum configuration, the exact normalized amplitude isThus $c_s\to 1$ gives an amplitude of order $ϵ$, consistent with weak non-Gaussianity in single-field slow-roll inflation. For $0<c_s\ll 1$, this vertex gives an amplitude of order one, enhanced relative to the slow-roll limit but not diverging as $c_s^{-2}$ after power-spectrum normalization. At fixed $H,ϵ,M_{\mathrm{Pl}}$ the raw bispectrum does scale as $c_s^{-2}$, but $P_{\zeta }^2$ has the same scaling. The often-quoted larger small-sound-speed amplitudes of more general inflationary actions require their actual interaction coefficients and other vertices; they do not follow from the lone supplied vertex. These conclusions concern its tree contribution within the stated weak-perturbation approximation.

The bispectrum shape function isAt fixed perimeter, the arithmetic-geometric mean inequality bounds this by $1/27$, attained only at an equilateral bispectrum configuration. In a squeezed bispectrum configuration with $k_1=q\ll k_2\simeq k_3=k$, it behaves as $q/(8k)\to 0$. At a flattened bispectrum configuration $(k,k,2k)$ it is finite, $1/32$, with no boundary singularity. Thus the shape prefers equilateral triangles and suppresses squeezed triangles, with a broad finite flattened contribution. This is a shape preference, not a complete signal-to-noise calculation: actual observational weighting also depends on cosmological transfer functions, covariance and accessible triangles. If $C=0$, this vertex produces no bispectrum at any configuration; if $C<0$, the signed shape reverses while its absolute triangle preference is unchanged.

When $V$ is approximately constant, $dV/d\varphi \simeq 0$, and exponential expansion gives the constant Hubble parameter $H=H_{\inf }$. Take the spatial Fourier transformSinceeach independent inflationary scalar Fourier mode satisfies