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Using the approximate large-angle CMB relation $T/S\simeq -7n_T$ gives $T/S\simeq 7{\sigma }^4/(96M^4{N}^3)$ in quartic hilltop inflation. It is suppressed by the fourth power of the small-field scale. This large-angle temperature ratio is not the primordial tensor-to-scalar ratio $r=16{ϵ}_V$; their transfer functions and normalization differ.

// Target: cosmology.bigb

The given large-angle temperature transfer relation gives the low-multipole tensor fraction in quartic hilltop inflationThis is not the primordial tensor-to-scalar ratio $r=16{ϵ}_V$, because the scalar and tensor contributions to low CMB multipoles have different transfer functions. With $\sigma \ll m_{\mathrm{pl}}$, the tensor contribution is far below the scalar temperature signal and its cosmic variance, and the corresponding primordial gravitational-wave or polarization signal is extremely small.

For COBE normalization of quartic hilltop inflation, the historical scalar horizon amplitude is of order ${\delta }_H\simeq 2\times 10^{-5}$, or ${\mathcal{P}}_S\simeq 25{\delta }_H^2/4\sim 2.5\times 10^{-9}$ in the usual matter-era convention. The slow-roll spectrum simplifies toHenceLate-time transfer and tilt corrections to the COBE normalization change the order-one coefficient, not the conclusion that the dimensionless coupling must be extraordinarily small. The leading normalization is independent of $\sigma$: reducing the height of the potential also changes the slope and field value at fixed $N$, causing the scale factors to cancel.

For the monomial inflation potential on $\varphi >0$, $V_{,\varphi }/V=n/\varphi$ and $V_{,\varphi \varphi }/V=n(n-1)/{\varphi }^2$. Thus, to leading slow-roll approximation,For $n=2,3,4$, $2n\ge n^2/2$. Therefore the loose order-one requirements $ϵ\le 1$ and $|\eta |\le 1$ giveThis is the interpretation under which the printed threshold follows. It is not a sufficient criterion for genuine slow roll, which requires much smaller parameters and hence ${\varphi }^2/M_{\mathrm{Pl}}^2\gg 2n$ for these fixed powers. Nor is its equality the actual end of acceleration for $n=2,3$: there $ϵ\simeq n/4<1$. For $n=4$ the leading estimate has $ϵ\simeq 1$. The potential-based order-one criterion $|{\eta }_V|<1$ also differs from the Hubble-flow criterion, especially at $n=4$. We use the endpoint the paper explicitly stipulates, rather than treating this boundary as a controlled slow-roll approximation.

Let ${\varphi }_e^2=2n{M}_{\mathrm{Pl}}^2$ by that prescription. The slow-roll e-fold count, using $\dot{\varphi }/H\simeq -n{M}_{\mathrm{Pl}}^2/\varphi$, isFor $N=60$, the monomial inflation with a prescribed endpoint result isApplying the leading scalar spectral index in Hubble slow-roll parameters and tensor-to-scalar ratio formulas givesThese are predictions at the chosen 60-e-fold initial point, normally identified with the pivot scale's horizon exit. A physical endpoint determined by $ϵ=1$ would instead have the leading estimate ${\varphi }_e^2\simeq n^2{M}_{\mathrm{Pl}}^2/2$; its change to the e-fold count and predictions must not be silently mixed with the stipulated $122n$ calculation.

Using the leading predictions just obtained givesThe quadratic and cubic monomial inflation potentials satisfy the quoted interval for the scalar spectral index, whereas the quartic model lies below its lower endpoint. All three exceed the quoted bound on the tensor-to-scalar ratio. ThusThis conclusion uses the exam's numerical bounds and the specified 60-e-fold endpoint convention, not a substitution of later observational constraints. Even a modest endpoint adjustment does not reduce the predicted $r$ by the roughly factor of two or more needed here.

For $V=\lambda {\varphi }^4$,Thus $n_s-1=-6{ϵ}_V+2{\eta }_V=-3{ϵ}_V$. Combining this with the tensor-to-scalar ratio $r=16{ϵ}_V$ gives the quartic-inflation tilt-tensor relationThe measured tilt $n_s-1=-0.04$ would require $r\simeq 0.213$, whereas $r=0.001$ predicts $n_s-1\simeq -1.88\times 10^{-4}$. Therefore quartic slow-roll inflation is inconsistent with the stated measurements.

The expressions in part c implySince the tensor-to-scalar ratio is $r=16ϵ$,For positive $\alpha$, this class predicts a red scalar tilt andA blue tilt with positive $r$, or $r\ge 8(1-n_s)$, is incompatible. An allowed pair fixes $\alpha =2r/[8(1-n_s)-r]$, subject to slow roll and observational bounds.

Taking the ratio of the supplied primordial tensor power spectrum and scalar spectrum cancels the common factor $(H/2\pi {)}^2$:To first slow-roll order $ϵ={ϵ}_V$, so the tensor-to-scalar ratio is

For canonical slow-roll inflation with $V(\varphi )=\lambda {\varphi }^4$, the leading scalar spectral index and tensor-to-scalar ratio obeyIt is a parameter-free consistency check of the quartic potential at leading slow-roll order.