Defining $x \equiv \phi/M_{\text{Pl}}$, the Hubble flow functions are given by

$$ \epsilon_1 = \frac{2}{x^2} \left( \frac{1 - \alpha x^2 - 4\alpha x^2 \ln x}{1 - 2\alpha x^2 \ln x} \right)^2, \qquad (5.58) $$

$$ \epsilon_2 = \frac{4}{x^2} \frac{(1 + \alpha x^2)(1 + 2\alpha x^2) - 2\alpha x^2 \ln x (1 - \alpha x^2 - 4\alpha x^2 \ln x)}{(1 - 2\alpha x^2 \ln x)^2}, \qquad (5.59) $$

and

$$ \epsilon_3 = \frac{4}{x^2} \frac{1 - \alpha x^2 - 4\alpha x^2 \ln x}{(1 - 2\alpha x^2 \ln x)^2} \times \frac{1 - \alpha x^2 [\alpha x^2 (4\alpha x^2 + 9) + 1] - \alpha x^2 \ln x [4\alpha^2 x^4 \ln x (4 \ln x + 1) + (\alpha x^2 + 3)(6\alpha x^2 + 2)]}{(1 + \alpha x^2)(1 + 2\alpha x^2) - 2\alpha x^2 \ln x (1 - \alpha x^2 - 4\alpha x^2 \ln x)}. \qquad (5.60) $$

If $\alpha = 0$, one recovers the slow-roll parameters of the massive case (namely LFI with $p = 2$, see Section 5.2) as expected. The potential and Hubble flow functions have been represented in Fig. 13.

Let us now discuss the field domains in which inflation can take place. It is clear that the above potential is not positive definite for all field values. It becomes negative at the point

$$ x_{V=0} = \frac{\phi_{V=0}}{M_{\text{Pl}}} = \sqrt{\frac{1}{\alpha W_0 (1/\alpha)}}, \qquad (5.61) $$

where $W_0$ is the 0-branch of the Lambert function. The model is defined only in the regime $\phi < \phi_{V=0}$. On the other hand, the top of the potential, where $V' = 0$ (or equivalently $\epsilon_1 = 0$), is given by

$$ x_{\text{top}} = \frac{\phi_{\text{top}}}{M_{\text{Pl}}} = \sqrt{\frac{1}{2\alpha W_0 \left(\frac{\sqrt{e}}{2\alpha}\right)}}. \qquad (5.62) $$

As the model makes sense only if the logarithmic terms do not dominate the potential, the acceptable regime is $\phi < \phi_{\text{top}} < \phi_{V=0}$, and a large field region only exists for $\phi_{\text{top}}/M_{\text{Pl}} \gg 1$. From the above expression, this means that we must be in the regime $\alpha \ll 1$. For $\phi < \phi_{\text{top}}$ one can check from Eqs. (5.55) and (5.62) that the loop corrections never exceed $\alpha/e$.

Let us now turn to the slow-roll trajectory. It is given by

$$ N - N_{\text{end}} = -\frac{1}{2} \int_{\phi_{\text{end}}/M_{\text{Pl}}}^{\phi/M_{\text{Pl}}} \frac{x - 2\alpha x^3 \ln x}{1 - \alpha x^2 - 4\alpha x^2 \ln x} dx, \qquad (5.63) $$

an integral that cannot be performed analytically and which is numerically evaluated in ASPIC. For the purpose of this section, we can nevertheless make an expansion in $\alpha$ to obtain an approximate expression

$$ N - N_{\text{end}} = -\frac{x^2}{4} \left[ 1 + \alpha \frac{x^2}{4} (1 + 4 \ln x) \right] + \frac{x_{\text{end}}^2}{4} \left[ 1 + \alpha \frac{x_{\text{end}}^2}{4} (1 + 4 \ln x_{\text{end}}) \right] + \mathcal{O}(\alpha^2). \qquad (5.64) $$

Inflation stops close to the minimum of the potential when $\epsilon_1 = 1$. This last equation cannot be solved analytically but we can also perform an expansion at first order in $\alpha$ and one gets

$$ x_{\text{end}} = \frac{\phi_{\text{end}}}{M_{\text{Pl}}} \approx \frac{1}{\sqrt{2\alpha W_0 \left[ \frac{e^{1+1/(4\alpha)}}{2\alpha} \right]}} \approx \sqrt{2} - 2\sqrt{2\alpha}. \qquad (5.65) $$

In the limit $\alpha \to 0$, we recover the large field result for $p = 2$, i.e. $x_{\text{end}} \to \sqrt{2}$. The maximum total number of e-folds one can realize between $\phi = \phi_{\text{top}}$ and $\phi = \phi_{\text{end}}$ can be calculated from the previous expressions. It reads

$$ \begin{aligned} \Delta N_{\text{max}} = N_{\text{end}} - N_{\text{top}} &= \frac{5}{32\alpha W_0 \left(\frac{\sqrt{e}}{2\alpha}\right)} + \frac{1 + 2\alpha - 20\alpha W_0 \left[ \frac{e^{1+1/(4\alpha)}}{2\alpha} \right]}{128\alpha^2 W_0^2 \left[ \frac{e^{1+1/(4\alpha)}}{2\alpha} \right]} \\ &\approx -\frac{5}{32\alpha \ln(\alpha)}. \end{aligned} \qquad (5.66) $$

This is a decreasing function of $\alpha$, so that $\alpha$ has to be small enough if one wants a sufficiently high number of e-folds to take place. Indeed, if one wants at least $\Delta N_{\text{min}}$ e-folds to occur, one needs to work with

$$ \alpha < \frac{5}{32\Delta N_{\text{min}}} \frac{1}{\ln\left(\frac{32\Delta N_{\text{min}}}{10}\right)}. \qquad (5.67) $$

For example, $\Delta N_{\text{min}} = 50$ imposes $\alpha < 6 \times 10^{-4}$. The fact that $\alpha$ is bounded from above can be directly checked in Fig. 129. The field $\phi_*$ value at which the pivot mode crossed the Hubble radius during inflation is obtained from Eq. (3.48) whereas the corresponding e-fold number can be obtained from the trajectory.

Finally, the parameter $M$ can be determined from the amplitude of the CMB anisotropies, and one gets

$$ \left(\frac{M}{M_{\text{Pl}}}\right)^4 = \frac{2880\pi^2}{x_*^4} \frac{(1 - 2\alpha x_*^2 \ln x_*)^3}{(1 - \alpha x_*^2 - 4\alpha x_*^2 \ln x_*)^2} \frac{Q_{\text{rms-PS}}^2}{T^2}. \qquad (5.68) $$
The reheating consistent slow-roll predictions for the RCMI models are represented in Fig. 129. As expected, the LFI quadratic model case is properly recovered for $\alpha \to 0$. From this figure, we see that all models having $\alpha \gtrsim 10^{-3.7}$ lie outside the $2\sigma$ contour. Let us emphasize that the value of $\alpha$ cannot be infinitely small due to Eq. (5.57). At zero order, one has $\phi > \phi_{\text{end}} \simeq \sqrt{2}M_{\text{Pl}}$ such that Eq. (5.57) can be recast into

$$ \alpha > \frac{M^4}{8\pi^2 M_{\text{Pl}}^4} = \frac{m^2}{16\pi^2 M_{\text{Pl}}^2}. \qquad (5.69) $$

From the COBE normalization, and in the limit of small $\alpha$, one gets $M/M_{\text{Pl}} \gtrsim 10^{-3}$ and the lower bound reads $\alpha > 10^{-15}$.

### 5.5. Radiatively Corrected Quartic Inflation (RCQI)

This model is similar to RCMI discussed in Section 7.1 but implements radiative corrections due to fermion couplings over a quartic ($p=4$) large field model [310] (see Section 5.2). The potential is given by

$$ V = \lambda \phi^4 - \frac{g^4}{16\pi^2} \phi^4 \ln \left( \frac{\phi}{M_{\text{Pl}}} \right) = M^4 \left( \frac{\phi}{M_{\text{Pl}}} \right)^4 \left[ 1 - \alpha \ln \left( \frac{\phi}{M_{\text{Pl}}} \right) \right], \qquad (5.70) $$

where

$$ M^4 = \lambda M_{\text{Pl}}^4, \qquad \alpha \equiv \frac{g^4}{16\pi^2 \lambda}. \qquad (5.71) $$

Defining $x = \phi/M_{\text{Pl}}$, the Hubble flow functions in the slow-roll approximation read

$$ \epsilon_1 = \frac{8}{x^2} \left( \frac{1 - \frac{\alpha}{4} - \alpha \ln x}{1 - \alpha \ln x} \right)^2, \qquad \epsilon_2 = \frac{1 + \frac{\alpha}{4}(\alpha - 1) + \alpha \left( \frac{\alpha}{4} - 2 \right) \ln x + \alpha^2 \ln^2 x}{(1 - \alpha \ln x)^2}, \qquad (5.72) $$

and

$$ \epsilon_3 = \frac{8}{x^2} \frac{(1 - \frac{\alpha}{2} - \alpha \ln x)(1 - \frac{\alpha}{4} - \alpha \ln x) \left[ 1 + \frac{\alpha^2}{2} + \frac{\alpha}{4} - \alpha \left( 2 + \frac{\alpha}{4} - \alpha \ln x \right) \ln x \right]}{(1 - \alpha \ln x)^2 \left[ 1 + \frac{\alpha}{4}(\alpha - 1) - \alpha \left( 2 - \frac{\alpha}{4} - \alpha \ln x \right) \ln x \right]}. \qquad (5.73) $$

The shape of the potential and the Hubble flow functions are very similar to the ones of the RCMI model and have been represented in Fig. 14. In particular, the potential is vanishing and maximal at the field values

$$ x_{V=0} = \frac{\phi_{V=0}}{M_{\text{Pl}}} = e^{1/\alpha}, \qquad x_{\text{top}} = \frac{\phi_{\text{top}}}{M_{\text{Pl}}} = e^{1/\alpha - 1/4}, \qquad (5.74) $$

respectively. As the model makes sense only if the corrections are small compared to the quartic term, one should consider $\alpha \ll 1$ and not too large super-Planckian field values.

The slow-roll trajectory can be integrated analytically from Eqs. (3.11) and (5.70) and one gets

$$ N - N_{\text{end}} = -\frac{1}{16} \left[ 2x^2 - e^{-1/2+2/\alpha} \text{Ei}\left(\frac{1}{2} - \frac{2}{\alpha} + 2\ln x\right) - 2x_{\text{end}}^2 + e^{-1/2+2/\alpha} \text{Ei}\left(\frac{1}{2} - \frac{2}{\alpha} + 2\ln x_{\text{end}}\right) \right], \qquad (5.75) $$

where the exponential integral function is defined by

$$ \text{Ei}(x) \equiv - \int_{-x}^{+\infty} \frac{e^{-t}}{t} dt. \qquad (5.76) $$

The quartic limit $\alpha \to 0$ is recovered by noticing that

$$ \text{Ei}(-2/\alpha) \sim -\frac{\alpha}{2} e^{-2/\alpha}. \qquad (5.77) $$

Contrary to the RCMI model, the top of the potential is flat enough to support inflation. Indeed, one sees from Eq. (5.74) that the argument of the exponential integral function vanishes at $x = x_{\text{top}}$. Since for $y \to 0$, one has $\text{Ei}(y) \sim \gamma + \ln y$, whatever the value of $x_{\text{end}}$ the total number of e-folds is divergent. This means that it is always possible to realize the required $\Delta N_*$ number of e-folds provided inflation starts close enough to the top of the potential.

As for RCMI, inflation stops at $\epsilon_1 = 1$ but this equation can only be solved numerically. For illustrative purpose, one can nevertheless solve it at first order in $\alpha$ to get

$$ x_{\text{end}} = \frac{\phi_{\text{end}}}{M_{\text{Pl}}} \approx 2\sqrt{2} - \frac{\sqrt{2}}{2}\alpha. \qquad (5.78) $$

The link between $\phi_*$ and $\Delta N_*$ is given by the slow-roll trajectory with $\phi_*$ given by Eq. (3.48).

Finally, the parameter $M$ can be determined from the amplitude of the CMB anisotropies, and one gets

$$ \lambda = \frac{M^4}{M_{\text{Pl}}^4} = \frac{11520\pi^2 \left(1 - \frac{\alpha}{4} - \alpha \ln x_*\right)^2}{x_*^6 (1 - \alpha \ln x_*)^3} \frac{Q_{\text{rms-PS}}^2}{T^2}. \qquad (5.79) $$

The slow-roll predictions for RCQI are represented in Figs. 130 and 131. As expected, the quartic model case is properly recovered in the limit $\alpha \to 0$. From Fig. 130, we see that all the models seem to lie outside the $2\sigma$ contour for $\bar{w}_{\text{reh}} = 0$. As the reheating phase takes place at the bottom of a quartic-like potential, we have also represented the prediction for $\bar{w}_{\text{reh}} = 1/3$ in Fig. 131. For a radiation-dominated reheating, $\Delta N_*$ is fixed and for each value of $\alpha$ one has only a single point. In that situation, all these models are still disfavored at the two-sigma level.
<figure><img src="image_17.png" /><figcaption>Fig. 14. Radiatively Corrected Quartic Inflation (RCQI) for $\alpha = 0.8$. Top panels: the potential and its logarithm as a function of the field values. Bottom left panel: slow-roll parameter $\epsilon_1$. The shaded area indicates where inflation stops. Bottom right panel: slow-roll parameters $\epsilon_2$ (solid line) and $\epsilon_3$ (dotted line). The shaded region for $\epsilon_2$ and $\epsilon_3$ shows where the slow-roll approximation is violated for that value of $\alpha$.</figcaption></figure>

## 5.6. Natural Inflation (NI)

### 5.6.1. Theoretical justifications

Natural inflation was first proposed as an attempt to solve the so-called “fine-tuning” problem of inflation. In particular, in order to obtain sufficient inflation and the correct normalization for the microwave background anisotropies, the potential $V(\phi)$ of the inflaton must be sufficiently flat. It is usually argued that, on general grounds, such a flatness is not robust under radiative corrections, unless it is protected by some symmetry. This is the reason that has motivated Refs. [311,312] to put forward Natural Inflation, in which the inflaton potential is flat due to shift symmetries. The model makes use of Nambu-Goldstone bosons [313,314] which arise whenever a global symmetry is spontaneously broken. The main idea can be very simply illustrated with the following action

$$ S = - \int dx \sqrt{-g} \left[ g^{\mu\nu} \partial_{\mu} \Phi^{\dagger} \partial_{\nu} \Phi + i \bar{\Psi} \gamma^{\mu} \partial_{\mu} \Psi + \lambda \left( \Phi^{\dagger} \Phi - \frac{f^2}{2} \right)^2 + g_f \bar{\Psi}_L \Phi \Psi_R + g_f \bar{\Psi}_R \Phi^{\dagger} \Psi_L \right], \qquad (5.80) $$

where $\Phi$ is a complex scalar field, $\Psi$ a Dirac spinor and $\Psi_{LR} = (1 \pm \gamma_5)/2\Psi$. The quantity $f$ is the energy scale at which the symmetry is spontaneously broken, $\lambda$ is a dimensionless coupling constant and $g_f$ a dimensionless Yukawa coupling. This action is invariant under the U(1) transformation: $\Phi \to e^{i\alpha}\Phi$, $\Psi_L \to e^{i\alpha/2}\Psi_L$ and $\Psi_R \to e^{-i\alpha/2}\Psi_R$, where $\alpha$ is an arbitrary constant. Due to the “Mexican hat” potential for the scalar field, this symmetry is spontaneously broken below the scale $f$ and the scalar field acquires the vev $\langle\Phi\rangle = f/\sqrt{2}e^{i\phi/f}$. The field $\phi$ corresponds to an “angular variable” and is a Goldstone boson. Below the scale of broken symmetry, the effective Lagrangian can be expressed as

$$ \mathcal{L} = \frac{1}{2} \partial_{\mu} \Phi \partial^{\mu} \Phi + i \bar{\Psi} \gamma^{\mu} \partial_{\mu} \Psi + g_f \frac{f}{\sqrt{2}} (\bar{\Psi}_L \Psi_R e^{i\phi/f} + \bar{\Psi}_R \Psi_L e^{-i\phi/f}). \qquad (5.81) $$

It is now invariant under $\phi \to \phi + 2\pi f$, $\Psi_L \to e^{i\alpha/2}\Psi_L$ and $\Psi_R \to e^{-i\alpha/2}\Psi_R$. Then, we assume that an explicit symmetry breaking takes place, for instance through the appearance of a fermion condensate for which $\langle\bar{\Psi}\Psi\rangle \simeq M_s^3$ where $M_s < f$ is the scale at which this symmetry breaking occurs. As a consequence, the effective Lagrangian takes the form

$$ \mathcal{L} = \frac{1}{2} \partial_{\mu} \Phi \partial^{\mu} \Phi + 2 g_f M_s^3 \frac{f}{\sqrt{2}} \cos \left( \frac{\Phi}{f} \right). \qquad (5.82) $$

We see that the Nambu-Goldstone boson has acquired a cosine potential and the overall scale of the potential is given by $M^4 \simeq g_f M_s^3 f$. Therefore, if one takes $f \simeq M_{Pl}$, $M_s$ slightly below the GUT scale and a Yukawa coupling of order one, one can “naturally” generate a small ratio $M/f$. A last
remark is in order on this model. Suppose that quantum gravity effects generate non-renormalizable higher order terms in the action (5.80) like

$$ \Delta V = a_{mn} \frac{|\Phi|^{2m}}{M_{\text{Pl}}^{2m+n-4}} (\Phi^n + \Phi^{\dagger n}), \qquad (5.83) $$

where $a_{mn}$ are a priori unknown coefficients. After symmetry breaking, one would therefore obtain a correction of the form

$$ \Delta V = a_{mn} M_{\text{Pl}}^4 \left( \frac{f}{M_{\text{Pl}}} \right)^{2m+n} \cos \left( n \frac{\phi}{f} \right). \qquad (5.84) $$

If $f \gtrsim M_{\text{Pl}}$, as favored by current cosmological data (see below) these terms should dominate unless the coefficients $a_{mn}$ are fine-tuned to very small values. Notice that the overall scale of the potential is now given by $a_{mn} M_{\text{Pl}}^4$, which also demands that $a_{mn} \lesssim 10^{-15}$ in order to have the correct CMB normalization. These terms are therefore dangerous for the consistency and the natural character of the model. This model has been studied in more details in Refs. [315–329].

Many other types of candidates have subsequently been explored in order to produce scenarios similar to that of Natural Inflation. For example, in Ref. [330], it was suggested to use a pseudo-Nambu Goldstone boson as the rolling field in double field inflation. Then, NI potentials generated by radiative corrections in models with explicitly broken Abelian [331] and non-Abelian [332,333] symmetries were considered, showing that NI models with $f \simeq M_{\text{Pl}}$ and $f \ll M_{\text{Pl}}$ can both be generated. In Refs. [334], the field $\phi$ is considered to be a Polonyi field [335] and the model predicts that $f = M_{\text{Pl}}$. Refs. [336,337] have examined natural inflation in the context of extra dimensions and Ref. [338] has used pseudo-Nambu Goldstone bosons from little Higgs models to drive hybrid inflation. Also, Refs. [339,340] have used the natural inflation idea of pseudo-Nambu Goldstone bosons in the context of braneworld scenarios to drive inflation, Ref. [341] has studied the model in 5-D warped backgrounds. The same potential has also been obtained and studied in Ref. [342] when studying instantons in non-linear sigma models, and in Ref. [343] as providing quintessential inflation. In some of these references the potential is sometimes found with the minus sign in front of the cosine term, which is, up to a shift in the field $\text{vev } \phi/f \to \phi/f + \pi$, the same potential as already studied before. This last model has also been derived and studied in Refs. [336,337,344] in the context of orbifold GUT inflation, where the potential is given by

$$ V(\phi) = M^4 \left[ F\left(\frac{\phi}{\phi_0}\right) + F\left(2\frac{\phi}{\phi_0}\right) + \frac{F(0)}{2} \right], \qquad (5.85) $$

with

$$ F(x) = - \sum_{n=1}^{\infty} \frac{\cos(n\pi x)}{n^5}. \qquad (5.86) $$

This potential must be studied in its increasing branch, and in the small field limit. At leading order, one recovers the cosine potential.

Finally, an important question is whether a situation where $f > M_{\text{Pl}}$ makes sense from the high energy physics and effective field theory point of view. In fact, it was shown in Refs. [345–347] that $f \lesssim 10^{12}$ GeV in order for the corresponding energy density not to exceed the critical energy density. But this constraint applies to the post inflationary Universe and, during inflation, Ref. [348] has argued that it is not relevant. However, it remains the question of whether $f > M_{\text{Pl}}$ makes sense or not. To address this issue, an interesting mechanism has been proposed in Ref. [349] (see also Ref. [350]) which shows that two axion fields at sub-Planckian scales can have an effective dynamics similar to the one field Natural Inflation model with $f > M_{\text{Pl}}$.

Let us consider a model with two axions, $\theta$ and $\rho$ the effective Lagrangian of which is given by

$$ \mathcal{L} = \frac{1}{2} \partial_{\mu} \theta \partial^{\mu} \theta + \frac{1}{2} \partial_{\mu} \rho \partial^{\mu} \rho + M_1^4 \left[ 1 - \cos \left( \frac{\theta}{f} + \frac{\rho}{g_1} \right) \right] + M_2^4 \left[ 1 - \cos \left( \frac{\theta}{f} + \frac{\rho}{g_2} \right) \right], \qquad (5.87) $$

where $M_1$ and $M_2$, $f$, $g_1$ and $g_2$ are constant, a priori, arbitrary scales. The same model can be re-written in terms of the fields $\psi$ and $\xi$ defined by

$$ \psi = \frac{fg_1}{\sqrt{f^2 + g_1^2}} \left( \frac{\theta}{f} + \frac{\rho}{g_1} \right), \qquad \xi = \frac{fg_1}{\sqrt{f^2 + g_1^2}} \left( -\frac{\theta}{g_1} + \frac{\rho}{f} \right). \qquad (5.88) $$

It is easy to show that this leads to

$$ \mathcal{L} = \frac{1}{2} \partial_{\mu} \psi \partial^{\mu} \psi + \frac{1}{2} \partial_{\mu} \xi \partial^{\mu} \xi + M_1^4 \left[ 1 - \cos \left( \frac{\sqrt{f^2 + g_1^2}}{fg_1} \psi \right) \right] \\ + M_2^4 \left[ 1 - \cos \left( \frac{f^2 + g_1 g_2}{fg_2 \sqrt{f^2 + g_1^2}} \psi + \frac{g_1 - g_2}{g_2 \sqrt{f^2 + g_1^2}} \xi \right) \right]. \qquad (5.89) $$

Moreover, the mass of the two fields $\psi$ and $\xi$ can be expressed as

$$ m_{\psi}^2 = \left( \frac{1}{f^2} + \frac{1}{g_1^2} \right) M_1^4, \qquad m_{\xi}^2 = \frac{(g_1 - g_2)^2}{g_2^2 (f^2 + g_1^2)} M_2^4. \qquad (5.90) $$

If $g_1$ is very close to $g_2$, then the field $\xi$ will be light and, therefore, will have a non-trivial dynamics. In addition, if the field $\psi$ is sufficiently heavy (compared to the Hubble parameter), then its $\text{vev}$ will be frozen at $\psi = 0$. In this case, we see that the original two fields model effectively reduces to a one field NI model with a scale $f_{\xi}$ given by

$$ f_{\xi} = \frac{g_2 \sqrt{f^2 + g_1^2}}{g_1 - g_2}. \qquad (5.91) $$

But, since, $g_1$ is close to $g_2$, the scale $f_{\xi}$ will be large even if the fundamental scales $f$, $g_1$ and/or $g_2$ are sub-Planckian. In this way, one can generate super-Planckian values for the scale $f$ and, at the same time, have a theory which can be consistent from the effective field theory point of view.
<figure><img src="image_18.png" /><figcaption>Fig. 15. Natural Inflation (NI). Top left panel: potential for $f/M_{pl} = 1.5$. Top right panel: logarithm of the potential for the same value of $f$. Bottom left panel: slow-roll parameter $\epsilon_1$ for a potential with $f/M_{pl} = 1.5$. The shaded area indicates the breakdown of the slow-roll inflation (strictly speaking when the acceleration stops). Bottom right panel: slow-roll parameters $\epsilon_2$ (solid line) and $\epsilon_3$ (dotted line) for a potential with $f/M_{pl} = 1.5$.</figcaption></figure>

### 5.6.2. Slow-roll analysis

Summarizing the above discussion, the model that we consider in this section makes use of a potential that can be written as

$$V(\phi) = M^4 \left[ 1 + \cos \left( \frac{\phi}{f} \right) \right]. \qquad (5.92)$$

The scale $M$ is determined by the CMB normalization and the potential depends on one parameter: the a priori unknown scale $f$. The potential of Eq. (5.92) is displayed with its logarithm in Fig. 15. Since it is a periodic and even function of the field $\text{vev } \phi$, it is enough to study it in the range $\phi \in [0, \pi f]$ where inflation proceeds from the left to the right. If one lets $x \equiv \phi/f$, the slow-roll parameters can be expressed as

$$\epsilon_1 = \frac{M_{Pl}^2}{2f^2} \frac{\sin^2 x}{(1 + \cos x)^2}, \qquad \epsilon_2 = \frac{2M_{Pl}^2}{f^2} \frac{1}{1 + \cos x}, \qquad \epsilon_3 = 2\epsilon_1. \qquad (5.93)$$

They are displayed in Fig. 15, where one can see that they are all increasing functions of the field $\text{vev}$, which means that they all increase during inflation. Inflation stops at the position $x_{\text{end}}$ given by $\epsilon_1 = 1$ (see below), and one can see that $\epsilon_2$ and $\epsilon_3$ are already greater than one at this point. This means that the slow-roll approximation stops being valid slightly before the end of inflation, and the few last e-folds may not be properly described in this frame of approximations. Another remark to be made is the fact that one generically has

$$\epsilon_2 > \frac{M_{Pl}^2}{f^2}. \qquad (5.94)$$

This means that in order for the slow-roll approximation to be valid, one must require $f/M_{Pl} \gg 1$ which is not necessarily problematic from a high energy physics point of view (see the above discussion).

The end of inflation occurs when $\epsilon_1 = 1$, i.e. at a position given by

$$x_{\text{end}} = \arccos \left( \frac{1 - 2f^2/M_{Pl}^2}{1 + 2f^2/M_{Pl}^2} \right). \qquad (5.95)$$

From this expression, one can calculate the value of the other slow roll parameters at the end of inflation, namely $\epsilon_2^{\text{end}} = 2 + M_{Pl}^2/f^2$ and $\epsilon_3^{\text{end}} = 2\epsilon_2^{\text{end}}$, which confirms that the last few e-folds may not be described properly in the slow-roll approximation.

Let us now calculate the slow-roll trajectory. It is given by

$$N_{\text{end}} - N = \frac{f^2}{M_{Pl}^2} \ln \left( \frac{1 - \cos x_{\text{end}}}{1 - \cos x} \right), \qquad (5.96)$$
where $N_{\text{end}}$ is the number of e-folds at the end of inflation, and $N$ is the number of e-folds at some point when the scaled field $\text{vev}$ is $x$. This trajectory can be inverted and one obtains

$$x = \arccos \left\{ 1 - (1 - \cos x_{\text{end}}) \exp \left[ -\frac{M_{\text{Pl}}^2}{f^2} (N_{\text{end}} - N) \right] \right\}. \qquad (5.97)$$

Replacing $x_{\text{end}}$ by its value [see Eq. (5.95)] gives

$$x = \arccos \left\{ 1 - \frac{4f^2}{M_{\text{Pl}}^2 + 2f^2} \exp \left[ -\frac{M_{\text{Pl}}^2}{f^2} (N_{\text{end}} - N) \right] \right\}. \qquad (5.98)$$

Finally, the amplitude of the CMB anisotropies fixes the parameter $M$ to

$$\left(\frac{M}{M_{\text{Pl}}}\right)^4 = 720\pi^2 \frac{Q_{\text{rms}-\text{PS}}^2}{T^2} \frac{M_{\text{Pl}}^2}{f^2} \frac{\sin^2 x_*}{(1 + \cos x_*)^3}. \qquad (5.99)$$

If $f/M_{\text{Pl}} = \mathcal{O}(1)$, this expression simplifies to

$$\left(\frac{M}{M_{\text{Pl}}}\right)^4 \approx 720\pi^2 \frac{Q_{\text{rms}-\text{PS}}^2}{T^2} \frac{e^{-2M_{\text{Pl}}^2/f^2 \Delta N_*}}{1 + 2f^2/M_{\text{Pl}}^2}, \qquad (5.100)$$

which gives rise to $M/M_{\text{Pl}} \approx 10^{-13}$. On the contrary, if $f/M_{\text{Pl}} \gg 1$ one has

$$\left(\frac{M}{M_{\text{Pl}}}\right)^4 \approx 360\pi^2 \frac{Q_{\text{rms}-\text{PS}}^2}{T^2} \left(\frac{f}{M_{\text{Pl}}}\right)^2 \frac{1}{\Delta N_*^2}, \qquad (5.101)$$

and the potential energy scale goes up. For instance, if $f/M_{\text{Pl}} = 10^2$ one has $M/M_{\text{Pl}} \approx 10^{-2}$.

The reheating consistent slow-roll predictions for the natural inflation models are displayed in Fig. 132. The reheating equation of state parameter $\bar{w}_{\text{reh}}$ has been taken to 0 since the potential is quadratic close to its minimum. In the limit $f/M_{\text{Pl}} \to \infty$, the quadratic model predictions (LFI with $p=2$, see Section 5.2) seem to be recovered. Indeed, from the above formula, one can check that in this limit both $x_{\text{end}}$ and $x_*$ approach $\pi$ and the potential is, at leading order, a parabola. More precisely, one can check from Eq. (5.98) that in the limit $f/M_{\text{Pl}} \to \infty$, one has $\cos x_* \approx -1 + (1 + 2\Delta N_*) M_{\text{Pl}}^2/f^2$, from which one deduces that $\epsilon_{1*} \approx 1/(1 + 2\Delta N_*)$ and $\epsilon_{2*} \approx 2/(1 + 2\Delta N_*) \approx 2\epsilon_{1*}$. These relations are characteristic of the LFI quadratic models, see Eq. (5.42). However, one has $\epsilon_{3*} = 2\epsilon_{2*}$ which differs from the LFI quadratic relationship $\epsilon_{3*} = \epsilon_{2*}$, and therefore quantities sensitive to $\epsilon_3$, such as the running $\alpha_S$, would break the degeneracy between NI and the LFI quadratic model. As expected, large values of $f/M_{\text{Pl}}$ seem to be favored by the data (as well as high reheating temperatures), and in practice, $f/M_{\text{Pl}} < 4$ appears to be disfavored at the $2\sigma$ level by the Planck 2018 + Bicep-Keck data.

## 5.7. Exponential SUSY Inflation (ESI)

### 5.7.1. Theoretical justifications

This model has been discussed in Ref. [351] in the context of spin-driven inflation and derived in Ref. [352] in the context of supergravity and superstrings. The potential is given by $V(\phi) \propto (1 - e^{-q\phi/M_{\text{Pl}}})$. The same potential also appears in Ref. [353] in the context of brane inflation, in Ref. [354] in the context of type IIB string compactification as fiber inflation and more recently in Ref. [355] as unitarized Higgs inflation models. This type of models can be obtained under very general considerations. Suppose that one has a supergravity model with a Kähler potential depending on one field $\psi$ given by $K = -\beta/\kappa \ln(1 - \alpha\kappa\psi\psi^\dagger)$, where $\alpha$ and $\beta$ are two free parameters. This model leads to a scalar potential but for a field which is not canonically normalized. The canonically normalized field $\theta$ is given by

$$\kappa^{1/2}\theta \approx \frac{1}{\sqrt{\alpha}}(1 - 2e^{-\sqrt{2/\bar{\beta}}\kappa^{1/2}\psi}), \qquad (5.102)$$

where we have assumed that inflation takes place at relatively large $\psi$ vev's. Then, suppose that the superpotential leads to a given function $V = f(\theta)$. One can always expand $f$ such that

$$V(\phi) \approx V_0 (1 - e^{-\sqrt{2/\bar{\beta}}\kappa^{1/2}\phi}) + \dots, \qquad (5.103)$$

where $\kappa^{1/2}\phi \equiv \kappa^{1/2}\theta + \sqrt{\beta/2} \ln [2f_\theta/(\sqrt{\alpha}f)]$ and $V_0$ is just the function $f$ evaluated at $1/\sqrt{\alpha}$. We see that one obtains exactly the ESI potential with $q = \sqrt{2}/\bar{\beta}$. Preferred choices for $\beta$ are $\beta = 1$ or $\beta = 3$ leading to $q = \sqrt{2}$ or $q = \sqrt{2}/3$. In absence of any more further guidance, it seems reasonable to assume that $\beta$, and hence $q$, is just a number of order one.

### 5.7.2. Slow-roll analysis

Based on the previous considerations, we now study the following potential

$$V(\phi) = M^4 (1 - e^{-q\phi/M_{\text{Pl}}}), \qquad (5.104)$$

where $q$ is a positive dimensionless parameter and inflation proceeds at decreasing field values in the region where $\phi/M_{\text{Pl}} > 0$. Defining $x \equiv \phi/M_{\text{Pl}}$, the Hubble flow functions in the slow-roll approximation read

$$\epsilon_1 = \frac{q^2}{2} \frac{e^{-2qx}}{(1 - e^{-qx})^2}, \quad \epsilon_2 = 2q^2 \frac{e^{-qx}}{(1 - e^{-qx})^2}, \quad \epsilon_3 = q^2 \frac{e^{-qx}(1 + e^{-qx})}{(1 - e^{-qx})^2}. \qquad (5.105)$$

The potential and the Hubble flow functions with respect to the field values are represented in Fig. 16.

The slow-roll trajectory can be integrated analytically from Eq. (3.11) and one finds

$$N - N_{\text{end}} = -\frac{e^{qx} - qx}{q^2} + \frac{e^{qx_{\text{end}}} - qx_{\text{end}}}{q^2}. \qquad (5.106)$$
<figure><img src="image_19.png" /><figcaption>Fig. 16. Exponential SUSY Inflation (ESI) for $q = 0.1$. Top panels: the potential and its logarithm. Bottom left panel: slow-roll parameter $\epsilon_1$. The shaded area indicates where acceleration stops. Bottom right panel: slow-roll parameters $\epsilon_2$ (solid line) and $\epsilon_3$ (dotted line). For those, the shaded region signals the breakdown of the slow-roll approximation but not necessarily the end of the accelerated expansion.</figcaption></figure>

<figure><img src="image_20.png" /><figcaption>Fig. 17. Lambert functions $W_0(x)$ (dashed line) and $W_{-1}(x)$ (solid line). During Exponential SUSY inflation, inflation proceeds along the “-1” branch in the direction specified by the arrow on the figure.</figcaption></figure>

This equation can also be inverted in terms of the Lambert function to get the field value in terms of the number of e-folds:

$$x = q(N - N_{\text{end}}) - \frac{e^{qx_{\text{end}}} - qx_{\text{end}}}{q} - \frac{1}{q} W_{-1} \{ - \exp[q^2(N - N_{\text{end}}) - (e^{qx_{\text{end}}} - qx_{\text{end}})] \}. \quad (5.107)$$

The fact that one should choose the branch $W_{-1}$ is justified below. The argument of the Lambert function is always negative as the exponential is always positive. Moreover, since $x_{\text{end}} > 0$ and $N < N_{\text{end}}$, the maximal value of exponential argument is saturated for $x_{\text{end}} \to 0$, i.e. for a Lambert function argument equals to $-1/e$. As the result the Lambert function argument varies, at most, in $[-1/e, 0]$. Finally, since $x > 0$, we see directly from Eq. (5.107) that the Lambert function values have to be negative thereby ensuring that inflation proceeds only along the “-1”-branch (see Fig. 17).
<figure><img src="image_21.png" /><figcaption>Fig. 18. Power Law Inflation (PLI) for $\alpha = 0.3$. Top panels: power law potential (left) and its logarithm (right). Bottom left panel: slow-roll parameter $\epsilon_1$. Bottom right panel: slow-roll parameters $\epsilon_2 = \epsilon_3 = 0$. On these plots, the shaded area indicates the region where slow-roll is violated.</figcaption></figure>

With such a potential, inflation ends naturally at $\epsilon_1 = 1$, i.e. at the field value

$$x_{\text{end}} = \frac{1}{q} \ln \left( 1 + \frac{q}{\sqrt{2}} \right). \qquad (5.108)$$

From this equation and the trajectory, we have an explicit relation between the field value $\phi_*$ at which the pivot mode crossed the Hubble radius during inflation and the corresponding e-fold number $\Delta N_*$.

Finally, the parameter $M$ can be determined from the amplitude of the CMB anisotropies, and one gets

$$\left(\frac{M}{M_{\text{Pl}}}\right)^4 = 720q^2\pi^2 \frac{e^{-2qx_*} Q_{\text{rms-PS}}^2}{(1 - e^{-qx_*})^3 T^2}, \qquad (5.109)$$

where the value of $\phi_*$ (or $\Delta N_*$) is obtained from Eq. (3.48). The reheating consistent slow-roll prediction for the exponential Susy models are represented in Figs. 133 and 134. In the limit $q \to 0$, we recover the same prediction as a linear large field model. From Fig. 133, we see that all the models remains compatible with the current data. These figures correspond to $\bar{w}_{\text{reh}} = 0$, but one could argue that $\bar{w}_{\text{reh}} \gtrsim -1/3$ make more sense if a parametric reheating would feel the linear shape of the potential. This quite extreme situation is represented in Fig. 134. In that case, the low reheating temperatures are clearly disfavored.

## 5.8. Power Law Inflation (PLI)

These models refer to inflationary potentials of the form

$$V(\phi) = M^4 e^{-\alpha\phi/M_{\text{Pl}}}, \qquad (5.110)$$

where $\alpha$ is a dimensionless parameter. They have been intensively studied since they lead to an exact inflationary dynamics, of the power law form, hence their name. Moreover, the power spectrum can also be determined exactly in this case. The background solution reads $a \propto (t/t_0)^{2/\alpha^2}$ and $\phi = \phi_0 + 2M_{\text{Pl}}/\alpha \ln(t/t_0)$ with $t_0^2 = 2M_{\text{Pl}}^2/(\alpha^2 M^4)(6/\alpha^2 - 1)e^{\alpha\phi_0/M_{\text{Pl}}}$. We see that we have inflation provided $\alpha \in [0, \sqrt{2}]$.

This scenario was introduced in Ref. [356,357] where the two point correlation function of the cosmological fluctuations was calculated for the first time (see also Refs. [358,359]). The predictions of this model were recently compared to the Planck data in Ref. [360]. Soon after Ref. [357], it was also considered in Refs. [361,362] but in the context of quintessence, i.e. for models of dark energy in which the energy density of the scalar field redshifts as a power law of the scale factor $\rho \propto a^{-q}$. In that case, one has $\alpha = \sqrt{q}/2$. The same potential also arises in the case where large field inflation is considered (LFI, see Section 5.2) but with a non-minimal coupling of the inflaton to the gravity sector, see Refs. [363,364] (the exponential potential appears after the transformation to the Einstein frame). In Ref. [365], a cosmic no-hair theorem for Bianchi models was proven assuming that the potential of the inflaton is of the same type as in Eq. (5.110). It was shown that one must have $0 < \alpha < \sqrt{2}/3$ so that the isotropic power law solution is the unique attractor for any initially expanding Bianchi type model (except type IX). In Ref. [366], the
potential (5.110) has been studied in the Kantowski–Sachs metric, and it was found that the production of particles by the scalar field acts as viscous forces which enlarges the range of initial conditions leading to successful inflation. In Ref. [367], the nature of the potential V(φ) relevant to having inflation in presence of a minimally coupled scalar field together with a causal viscous fluid was investigated. It was shown that this leads to an exponential potential. In Refs. [368–370], the exponential potential was used to describe the dynamics of a tachyonic matter field (i.e. with a non-minimal kinetic term). In Ref. [371], the general transformations that leave unchanged the form of the field equations for Bianchi V cosmologies were investigated, and it was found that they admit asymptotic stable points that lead to power law solutions of the type (5.110). In Ref. [372], inflation was studied in the context of M-theory on S¹/Z₂ via the non-perturbative dynamics of M5-branes. The open membrane instanton interactions between the branes give rise to potentials of the type (5.110). Within the same framework, Ref. [373] has discussed a realization of cascade inflation as assisted inflation built upon a succession of power law inflationary eras. Ref. [374] has used the exponential potential (5.110) in the context of Randall–Sandrum type II Braneworld model. Finally, the general dynamics of power law inflation was studied in detail in Refs. [356,375–383], where various aspects of its phenomenology were highlighted.

The potential and its logarithm are displayed in Fig. 18. They are decreasing functions of the field, hence inflation proceeds from the left to the right. The slow-roll parameters take a simple form given by

$$ \epsilon_1 = \frac{\alpha^2}{2}, \quad \epsilon_{i>1} = 0. \qquad (5.111) $$

Since the first slow-roll parameter is constant, inflation cannot stop by slow-roll violation and one has to assume that, at some vev $\phi_{\text{end}}$, a tachyonic instability is triggered. A priori, this means that the model has in fact an additional new free parameter. However, because the slow-roll parameters do not depend on $\phi$, as well as all the other properties of the inflationary dynamics (even when the slow-roll approximation is not satisfied, see below), the observational predictions of the model cannot depend on $\phi_{\text{end}}$ and this parameter turns out to be irrelevant.

The Hubble flow hierarchy being almost trivial, the exact dynamics of the model can be worked out even if the slow-roll approximation is violated. Indeed, let us first notice that the slow-roll trajectory can be explicitly integrated, and gives

$$ \frac{\phi}{M_{\text{Pl}}} = \frac{\phi_{\text{end}}}{M_{\text{Pl}}} + \alpha (N - N_{\text{end}}). \qquad (5.112) $$

Then, one can remark that this trajectory is also a solution of the exact Klein–Gordon equation of motion, which reads in terms of the number of e-folds N,

$$ H^2 \frac{\partial^2 \phi}{\partial N^2} + \left( 3H^2 + H \partial \frac{\partial H}{\partial N} \right) \frac{\partial \phi}{\partial N} + \frac{dV}{d\phi} = 0. \qquad (5.113) $$

Indeed, the first term vanishes, and the second term requires

$$ H^2 = \frac{V + \phi^2/2}{3M_{\text{Pl}}^2} = \frac{V + \frac{H^2}{2} \left( \frac{\partial \phi}{\partial N} \right)^2}{3M_{\text{Pl}}^2} = \frac{V + \frac{H^2}{2} \alpha^2 M_{\text{Pl}}^2}{3M_{\text{Pl}}^2}, \qquad (5.114) $$

from which one gets

$$ H^2 = \frac{V}{3M_{\text{Pl}}^2} \frac{1}{1 - \alpha^2/6}. \qquad (5.115) $$

From there, one can evaluate all terms in the Klein–Gordon equation, and verify that Eq. (5.112) is indeed a solution of Eq. (5.113). Since it is a second order differential equation, other solutions exist, but it can be shown [361,362] that the exact solution is an attractor. Let us also notice that combining Eq. (5.115) with Eq. (5.112) gives rise to

$$ H = H_{\text{end}} \left( \frac{a_{\text{end}}}{a} \right)^{a^2/2}, \qquad (5.116) $$

which can be integrated and gives

$$ a(t) = a_{\text{end}} \left( \frac{t}{t_{\text{end}}} \right)^{2/a^2}. \qquad (5.117) $$

One recovers the solution mentioned at the beginning of this section. Finally, the equation of state $w = P/\rho$ can also be worked out exactly and one gets

$$ w = -1 + \frac{\alpha^2}{3}. \qquad (5.118) $$

Again, all the previous expressions are valid even if the slow-roll approximation is not satisfied. One can see that pure de Sitter corresponds to $\alpha = 0$. In this case the potential is constant, the equation of state is -1 and the scale factor expands exponentially.

Another nice feature of power-law inflation is that the spectrum of the perturbations can be computed exactly without relying on any approximation. Defining the parameter $\beta \le -2$ from $\alpha^2/2 = (\beta + 2)/(\beta + 1)$, the primordial scalar power spectrum is given by

$$ P_{\zeta} = \frac{H_*^2}{\pi \epsilon_1 (8\pi M_{\text{Pl}}^2)} f(\beta) \left(\frac{k}{k_*}\right)^{2\beta+4}, \qquad (5.119) $$

where

$$ f(\beta) \equiv \frac{1}{\pi} \left[ \frac{(1+\beta)^{1+\beta}}{2^{1+\beta}} \Gamma\left(\frac{1}{2}+\beta\right) \right]^2. \qquad (5.120) $$

In particular, $f(\beta = -2) = 1$. The power spectrum of gravitational waves can also be obtained remarking that we have $\mu_S = \mu_T$ for power law inflation. From

$$ P_{\zeta} = \frac{k^3}{8\pi^2} \left| \frac{\mu_S}{a\sqrt{\epsilon_1}} \right|^2, \qquad P_h = \frac{2k^3}{\pi^2} \left| \frac{\mu_T}{a} \right|^2, \qquad (5.121) $$
<figure><img src="image_22.png" /><figcaption>Fig. 19. Top left panel: Kähler moduli inflation (KMII) potential for $\alpha = 1.5$. The two arrows indicate the two regions of the potential where inflation can take place. Top right panel: logarithm of the potential for the same value of $\alpha$. Bottom left panel: slow-roll parameter $\epsilon_1$ for $\alpha = 0.5$ (solid green line), $\alpha = 1.5$ (solid blue line) and $\alpha = 2.5$ (solid pink line). Obviously, the number of solutions of the equation $\epsilon_1 = 1$ depends on the value of $\alpha$. Bottom right panel: slow-roll parameters $\epsilon_2$ (solid line) and $\epsilon_3$ (dotted line) for $\alpha = 1.5$.</figcaption></figure>

one gets

$$r \equiv \frac{P_h}{P_\zeta} = 16\epsilon_1 = \frac{16n_T}{n_T - 2}, \qquad (5.122)$$

since $n_T = n_S - 1 = 2\beta + 4$.

Finally, the overall amplitude of the CMB anisotropies leads to a determination of the scale $M$, namely

$$\left(\frac{M}{M_{\text{Pl}}}\right)^4 = 720\pi^2\alpha^2 e^{\alpha\phi_s/M_{\text{Pl}}} \frac{Q_{\text{rms-PS}}^2}{T^2}. \qquad (5.123)$$

Obviously, this normalization depends on the value of $\phi_{\text{end}}$, and it is more relevant to express it in terms of the potential energy, say, at the end of inflation:

$$\frac{V_{\text{end}}}{M_{\text{Pl}}^4} = 720\pi^2\alpha^2 e^{-\alpha^2\Delta N_*} \frac{Q_{\text{rms-PS}}^2}{T^2}, \qquad (5.124)$$

from which one typically gets $V_{\text{end}}^{1/4}/M_{\text{Pl}} \approx 10^{-4}$.

The reheating consistent slow-roll predictions for the power law inflation models are displayed in Fig. 135. Because the slow-roll parameters are constant during inflation, one can check that the predictions of the models do not depend on the energy scale at which the power law reheating ends. One has $n_S = 1 - \alpha^2$ and $r = 8\alpha^2$, and from the Planck 2018 + Bicep-Keck constraints, all the models are disfavored at more than two-sigma confidence level.

## 5.9. Kähler Moduli Inflation I (KMII)

These models are stringy models and arise when type IIB string theories via Calabi-Yau flux compactification are used. KMII scenarios have been derived and studied in Refs. [384–390]. More specifically, when internal spaces are weighted projective spaces, one of the Kähler moduli can play the role of an inflaton field and its potential, in the large field limit, reads

$$V(\phi) = M^4 \left( 1 - \alpha \frac{\phi}{M_{\text{Pl}}} e^{-\phi/M_{\text{Pl}}} \right), \qquad (5.125)$$

$\alpha$ being a positive dimensionless parameter. Actually, since we deal with a modulus, $\phi$ usually possesses a non-minimal kinetic term. Then, once the inflaton field has been canonically normalized, $\phi$ has to be replaced with $\propto \phi^{4/3}$. The corresponding corrected potential is studied as “Kähler Moduli Inflation II” (KMII) in Section 6.3. However, sometimes, the potential (5.125) (with $\phi$ already canonically normalized) is also studied as a toy model (notably in Ref. [390]), the hope being that it can give a simpler description of the physics that naturally appears in the context of moduli inflation. Therefore, in this section, we also consider this scenario.
The potential in Eq. (5.125) depends on one free parameter, $\alpha$. A priori, there does not exist any bound on its value. However, as explained below, in order for slow-roll inflation to occur, one must restrict the range of possible values for $\alpha$. Within this range, we will show that the predictions of the model turn out to be almost independent of $\alpha$ (in fact, they logarithmically depend on $\alpha$). The potential (5.125) and its logarithm are displayed in Fig. 19. It decreases from $\phi = 0$ (where it blows up), reaches a minimum at $\phi = M_{\text{Pl}}$, and then increases to the asymptotic value $V = M^4$ when $\phi \to +\infty$. Therefore, two regimes of inflation may a priori exist: either inflation proceeds from the left to the right in the decreasing $\phi < M_{\text{Pl}}$ branch of the potential (in this branch the vev $\phi$ increases during inflation) or it proceeds from the right to the left in the increasing $\phi > M_{\text{Pl}}$ branch of the potential (and the vev decreases during inflation). However, one should keep in mind that the potential is derived under the large field assumption and, consequently, only the second regime is in fact meaningful. As a toy model, one might nevertheless want to study both regimes but it turns out that, in the first one, inflation could not stop by violation of the slow-roll conditions. This is why we will mainly focus on the second regime in the rest of this section. Let us also notice that the minimum value of the potential is located at $\phi = M_{\text{Pl}}$ and is $V_{\text{min}} = M^4 (1 - \alpha/e)$. Therefore, if one requires the potential to be positive definite everywhere, then one must have $0 < \alpha < e \approx 2.72$. However, this condition may also be ignored if one considers that the potential (5.125) is in any case not valid at $\phi/M_{\text{Pl}} \lesssim 1$.

Defining $x \equiv \phi/M_{\text{Pl}}$, the three first slow-roll parameters can be expressed as

$$ \epsilon_1 = \frac{\alpha^2}{2} e^{-2x} \frac{(1-x)^2}{(1-\alpha e^{-x}x)^2}, \quad \epsilon_2 = \frac{2\alpha e^{-x}}{(1-\alpha e^{-x}x)^2} (\alpha e^{-x} + x - 2), \qquad (5.126) $$

and

$$ \epsilon_3 = \frac{\alpha e^{-x} (x-1)}{(1-\alpha e^{-x}x)^2 (\alpha e^{-x} + x - 2)} \left[ x - 3 + \alpha e^{-x} (x^2 - 3x + 6) - 2\alpha^2 e^{-2x} \right]. \qquad (5.127) $$

Let us now study in more detail how inflation stops in this model. As can be seen in Fig. 19, the number of solutions of $\epsilon_1 = 1$ depends on the value of $\alpha$. We now define the numbers $\alpha_1$ and $\alpha_2$ by

$$ \alpha_1 = \frac{\sqrt{2}}{\sqrt{2-1}} e^{\frac{2-\sqrt{2}}{\sqrt{2-1}}} \approx 0.83, \quad \alpha_2 = \frac{\sqrt{2}}{\sqrt{2+1}} e^{\frac{2+\sqrt{2}}{\sqrt{2+1}}} \approx 2.41. \qquad (5.128) $$

If $0 < \alpha < \alpha_1$, then there is no solution (this corresponds to the green line in the bottom left panel in Fig. 19). The inflaton field eventually oscillates around the minimum of its potential but remains in a region where inflation continues forever. In this case, in order to stop inflation, one must add an auxiliary field to the model such that a tachyonic instability is triggered at some value $x_{\text{end}}$. This of course increases the number of parameters of this model. If $\alpha_1 < \alpha < \alpha_2$ (which corresponds to the blue line in Fig. 19), then two solutions appear:

$$ x_{\epsilon_1=1|_{x<1}}^{-} = x_{\text{end}|_{x<1}} = \frac{1}{1-\sqrt{2}} - W_0 \left( \frac{\sqrt{2}}{1-\sqrt{2}} \frac{e^{\frac{1}{1-\sqrt{2}}}}{\alpha} \right) \approx -2.4 - W_0 \left( -\frac{0.3}{\alpha} \right), \qquad (5.129) $$

$$ x_{\epsilon_1=1|_{x<1}}^{+} = \frac{1}{1-\sqrt{2}} - W_{-1} \left( \frac{\sqrt{2}}{1-\sqrt{2}} \frac{e^{\frac{1}{1-\sqrt{2}}}}{\alpha} \right) \approx -2.4 - W_{-1} \left( -\frac{0.3}{\alpha} \right), \qquad (5.130) $$

where $W_0$ and $W_{-1}$ denotes the “0-branch” and the “-1-branch” of the Lambert function respectively. These two solutions are both smaller than one so that they both lie in the decreasing branch of the potential. Correspondingly, two regimes of inflation exist. The first one proceeds from the left to the right and stops at $x_{\text{end}|_{x<1}}$. However, using the expression for the slow-roll parameters (5.126), it is easy to see that $\epsilon_1$ is always larger than $1/2$ in this domain. Therefore, the slow-roll approximation breaks down in this case. The second regime takes place in the $\phi/M_{\text{Pl}} > 1$ branch of the potential but inflation cannot stop by slow-roll violation. Finally, if $\alpha_2 < \alpha$ (this situation corresponds to the pink line in the bottom left panel in Fig. 19), then four solutions exist: two were already given in Eqs. (5.129), (5.130) and the two new ones read

$$ x_{\epsilon_1=1|_{x>1}}^{-} = \frac{1}{1+\sqrt{2}} - W_0 \left( -\frac{\sqrt{2}}{1+\sqrt{2}} \frac{e^{\frac{1}{1+\sqrt{2}}}}{\alpha} \right) \approx 0.4 - W_0 \left( -\frac{0.9}{\alpha} \right), \qquad (5.131) $$

$$ x_{\epsilon_1=1|_{x>1}}^{+} = x_{\text{end}|_{x>1}} = \frac{1}{1+\sqrt{2}} - W_{-1} \left( -\frac{\sqrt{2}}{1+\sqrt{2}} \frac{e^{\frac{1}{1+\sqrt{2}}}}{\alpha} \right) \approx 0.4 - W_{-1} \left( -\frac{0.9}{\alpha} \right). \qquad (5.132) $$

The two new solutions are greater than one and therefore lie in the increasing branch of the potential. Thus two regimes exist in this situation. The first one is the same as before, proceeds again from the left to right, stops at $x_{\text{end}|_{x<1}}$ and suffers from the fact that $\epsilon_1$ is always larger than $1/2$. The second one proceeds from the right to the left and ends at $x_{\text{end}|_{x>1}}$. We conclude that this regime is the regime of interest for the KMII model and that we must therefore require $\alpha > \alpha_2$.

Let us now study the slow-roll trajectory. It can be integrated exactly and its expression can be written as

$$ N_{\text{end}} - N = x_{\text{end}} - \frac{e}{\alpha} \text{Ei}(x_{\text{end}} - 1) + \ln(x_{\text{end}} - 1) - x + \frac{e}{\alpha} \text{Ei}(x - 1) - \ln(x - 1), \qquad (5.133) $$

where $\text{Ei}$ is the exponential integral function [281,282]. At this point, a few remarks are in order. Firstly, let us notice that $N$ goes to $\infty$ when $x$ tends to 1. This means that, in the slow-roll approximation, the field can never cross the minimum of its potential. In particular, if $\alpha < \alpha_2$, that is to say if one starts from the $\phi/M_{\text{Pl}} < 1$ branch and rolls down from the left to the right, then one can never reach the physical $\phi/M_{\text{Pl}} > 1$ branch of the potential and inflation can never come to an end. Secondly, when $x \gg 1$, the trajectory can be approximated by

$$ N_{\text{end}} - N \approx \frac{e}{\alpha} \left( \frac{e^x}{x} - \frac{e^{x_{\text{end}}}}{x_{\text{end}}} \right). \qquad (5.134) $$
<figure><img src="image_23.png" /><figcaption>Fig. 20. Lambert functions $W_0(x)$ (dashed line) and $W_{-1}(x)$ (solid line). During Kähler moduli inflation, inflation proceeds along the “-1” branch in the direction specified by the arrow.</figcaption></figure>

Moreover, in this approximation, it can be inverted exactly and one obtains

$$x \approx -W_{-1} \left[ - \frac{1}{\alpha (N_{\text{end}} - N) / e + e^{x_{\text{end}}}/x_{\text{end}}} \right], \qquad (5.135)$$

in agreement with what was obtained in Ref. [390]. In the above expression, $W_{-1}$ is the -1 branch of the Lambert function. Let us also notice that, in Ref. [390], the branch of the Lambert function was in fact incorrectly chosen. The fact that the -1 branch of the Lambert function has to be considered comes from the following argument. When $N_{\text{end}} - N \to \infty$, the argument of the Lambert function goes to 0⁻ and, therefore, since $x$ must tend towards +∞ in this limit, the -1 branch must be chosen. In addition, if $N_{\text{end}} - N \to 0$, then one must have $x \to x_{\text{end}} > 1$ which is also the case if the -1 branch is retained. This is represented in Fig. 20 where the arrow indicates the direction along which inflation proceeds. In the third place, since, when $x \to \infty$, one has $N_{\text{end}} - N \to \infty$, a sufficient number of e-folds can always be realized in this model. Finally, it is inaccurate to assume that $x_{\text{end}} \gg 1$ and, therefore, the above approximated trajectory is not so useful. However, if one only assumes that $x \gg 1$ (which can be checked to be a good approximation, especially at $x = x_*$) but not $x_{\text{end}} \gg 1$, then one can write

$$N_{\text{end}} - N \approx \frac{e}{\alpha} \frac{e^x}{x} + x_{\text{end}} - \frac{e}{\alpha} \text{Ei}(x_{\text{end}} - 1), \qquad (5.136)$$

which, moreover, can be inverted into

$$x \approx -W_{-1} \left[ - \frac{1}{\alpha (N_{\text{end}} - N) e + \text{Ei}(x_{\text{end}} - 1) - \alpha x_{\text{end}}/e} \right], \qquad (5.137)$$

and which is valid whenever $x \gg 1$. However, one should keep in mind that, now, and contrary to the former approximated trajectory, taking the limit $N \to N_{\text{end}}$ in the above expression is meaningless.

The energy scale $M$ is, as before, given by the CMB normalization and one obtains the following expression

$$\left(\frac{M}{M_{\text{Pl}}}\right)^4 = 720\pi^2\alpha^2 \frac{(1-x_*)^2}{(1-\alpha x_* e^{-x_*})^3} e^{-2x_*} \frac{Q_{\text{rms-PS}}^2}{T^2}. \qquad (5.138)$$

If one uses the $x_* \gg 1$ approximation, then Eq. (5.137) tells us that $x_* \approx \ln(\alpha\Delta N_*)$ and Eq. (5.138) can be re-written as

$$\left(\frac{M}{M_{\text{Pl}}}\right)^4 = \mathcal{O}(1) 720 \frac{\pi^2}{\Delta N_*^2} \frac{Q_{\text{rms-PS}}^2}{T^2}. \qquad (5.139)$$

It is remarkable that this equation does not depend on $\alpha$. Using a fiducial value for $\Delta N_*$, one typically gets $M/M_{\text{Pl}} \sim 10^{-3}$.

The predictions of KMII models are displayed in Fig. 136, for $\alpha > \alpha_2$. The reheating equation of state parameter $\bar{w}_{\text{reh}}$ has been taken to 0 since the potential is quadratic close to its minimum [but, it should be reminded that, in principle, the potential Eq. (5.125) cannot be trusted close to its minimum]. One can see that, as announced at the beginning of this section, the predictions depend on $\alpha$ in a very mild way, a conclusion which is in agreement with Refs. [384,390]. This can be understood as follows. If one assumes that $x_* \gg 1$, then we have already noticed that Eq. (5.137) implies that $x_* \approx \ln(\alpha\Delta N_*)$. From this result, one obtains that

$$\epsilon_{1*} \approx \frac{1}{2\Delta N_*^2} \ln^2(\alpha\Delta N_*) , \quad \epsilon_{2*} \approx \frac{2}{\Delta N_*} \ln(\alpha\Delta N_*) , \quad \epsilon_{3*} \approx \frac{1}{\Delta N_*} \ln(\alpha\Delta N_*) . \qquad (5.140)$$

In these expressions, we notice that the slow-roll parameters (at Hubble crossing) logarithmically depend on $\alpha$. This explains the weak $\alpha$ dependence observed in Fig. 136. Of course, one can also calculate the corresponding expressions of the spectral index, tensor to scalar ratio and running. One arrives at

$$n_s \approx 1 - 2 \frac{\ln(\alpha \Delta N_*)}{\Delta N_*}, \quad r \approx 8 \frac{\ln^2(\alpha \Delta N_*)}{\Delta N_*^2}, \quad \alpha_s \approx -2 \frac{\ln^2(\alpha \Delta N_*)}{\Delta N_*^2}. \qquad (5.141)$$

These expressions are in accordance with the estimates derived in Refs. [384,390]. However, contrary to what is claimed in Refs. [390], the predicted value of the running is not excluded by the CMB observations since, according to the Planck results [209], one has $\alpha_s = -0.013 \pm 0.009$.
## 5.10. Horizon Flow Inflation at first order (HF1I)

The horizon flow models have been introduced in Ref. [391] and consist into designing field potentials to exactly produce a truncated Taylor expansion of the Hubble parameter with respect to the field. As such they constitute a whole class of phenomenological inflationary models. Here, we are considering a potential designed such that $H(\phi) = H_0(1+A_1\phi/M_{Pl})$, where $A_1$ is a free dimensionless parameter. The shape of the potential reads [391]

$$V(\phi) = M^4 \left( 1 + A_1 \frac{\phi}{M_{Pl}} \right)^2 \left[ 1 - \frac{2}{3} \left( \frac{A_1}{1 + A_1 \frac{\phi}{M_{Pl}}} \right)^2 \right]. \qquad (5.142)$$

Denoting $x \equiv \phi/M_{Pl}$, the potential admits a global minimum at $x_{V-min} = -1/A_1$, which is negative

$$V_{\min} = V(\phi_{V-\min}) = -\frac{2}{3} M^4 A_1^2 < 0. \qquad (5.143)$$

As a result, there are two disconnected field domains in which the potential remains definite positive, either $x > x_{V=0}^+$ or $x < x_{V=0}^-$ where $x_{V=0}^\pm$ are the two roots of $V(x_{V=0}^\pm) = 0$, i.e.

$$x_{V=0}^{+} = \sqrt{\frac{2}{3}} - \frac{1}{A_1}, \qquad x_{V=0}^{-} = -\sqrt{\frac{2}{3}} - \frac{1}{A_1}. \qquad (5.144)$$

An interesting consequence of the horizon flow approach is that the Hubble flow functions can be calculated exactly, i.e. without the slow-roll approximation because $H(\phi)$ is exactly known. As discussed in Refs. [46,392], one could compare them with the other hierarchy of parameters, $\epsilon_i^V$, that are defined by the successive logarithmic derivatives of the potential. In the slow-roll approximation, one precisely uses the potential derivatives to approximate the Hubble flow functions. From $H \propto 1 + A_1 x$, one gets the exact Hubble flow functions

$$\epsilon_1 = 2 \left( \frac{A_1}{1 + A_1 x} \right)^2, \qquad \epsilon_2 = \epsilon_3 = 2\epsilon_1, \qquad (5.145)$$

whereas the slow-roll functions associated with the potential are

$$\epsilon_1^V = \frac{18A_1^2(A_1x + 1)^2}{[3 + 6A_1x + A^2(3x^2 - 2)]^2}, \qquad \epsilon_2^V = \frac{12A_1^2[3 + 6A_1x + A_1^2(3x^2 + 2)]}{[3 + 6A_1x + A_1^2(3x^2 - 2)]^2}, \qquad (5.146)$$

and

$$\epsilon_3^V = \frac{108A_1^2(A_1x + 1)^2 [1 + 2A_1x + A_1^2(x^2 + 2)]}{[3 + 6A_1x + A_1^2(3x^2 - 2)]^2 [3 + 6A_1x + A_1^2(3x^2 + 2)]}. \qquad (5.147)$$

As shown in Ref. [46], the link between the two hierarchies can be made explicit and one has

$$\epsilon_1^V = \epsilon_1 \left( \frac{1 - \eta/3}{1 - \epsilon_1/3} \right)^2. \qquad (5.148)$$

The $\eta$ parameter is defined as

$$\eta \equiv \frac{2}{H} \frac{d^2 H}{dx^2}, \qquad (5.149)$$

and vanishes in our case. As a result, provided $\epsilon_1 \ll 1$, i.e. we are in the slow-roll approximation, both hierarchies give the same results at first order. In order to establish Eq. (5.148), one has to show first that

$$\eta = \epsilon_1 + \frac{1}{\sqrt{2\epsilon_1}} \frac{d\epsilon_1}{dx}, \qquad (5.150)$$

and then that<fnref n="6" />

$$\frac{d\epsilon_1}{dx} = (\epsilon_1 - 3) \left( \frac{d \ln V}{dx} - \sqrt{2\epsilon_1} \right). \qquad (5.151)$$

The potential and the exact Hubble flow functions have been represented in Fig. 21.

Inflation can take place inside the two positive definite domains of the potential, i.e. at negative or positive field values. However, the Hubble parameter has to be positive such that $H_0$ has to be chosen negative if $1 + A_1x < 0$ along the field trajectory. Since the potential is completely symmetric with respect to its minimum $x_{V-min}$, we can study in full generality only the $x > x_{V=0}^+$ branch. In particular, as the Hubble flow functions are exact, we can also derive the exact field trajectory

$$N - N_{\text{end}} = -\frac{1}{2A_1} \left( x + \frac{1}{2} A_1 x^2 - x_{\text{end}} - \frac{1}{2} A_1 x_{\text{end}}^2 - \frac{2}{3} A_1 \ln \left| \frac{1 + A_1 x}{1 + A_1 x_{\text{end}}} \right| \right). \qquad (5.152)$$

Let us notice that, in the slow-roll approximation, one would have derived the trajectory from $\epsilon_1^V$. Doing so, one would have obtained

$$N - N_{\text{end}} = -\frac{1}{2A_1} \left( x + \frac{1}{2} A_1 x^2 - x_{\text{end}} - \frac{1}{2} A_1 x_{\text{end}}^2 - \frac{2}{3} A_1 \ln \left| \frac{1 + A_1 x}{1 + A_1 x_{\text{end}}} \right| \right). \qquad (5.153)$$

<footnote n="6">⁶ A sign in these two equations differs from the ones typeset in Ref. [46], most probably due to a misprint.</footnote>
<figure><img src="image_24.png" /><figcaption>Fig. 21. Top left panel: Horizon Flow Inflation at first order potential for $A_1 = 0.1$. Top panels: the potential and its logarithm with respect to the field values. Bottom left panel: the first Hubble flow function $\epsilon_1$ (exact) and the corresponding shaded area where inflation stops. Bottom right panel: Hubble flow functions $\epsilon_2$ (solid line) and $\epsilon_3$ (dotted line) for the same potential. These two functions are equal to $2\epsilon_1$.</figcaption></figure>

It is amusing to remark that here, the simplest formula is not given by the slow-roll derived one, but rather by the exact one. From this remark one should keep in mind that, in order to simplify trajectories integration, one can always add factors of order $O(\epsilon_1)$. The exact trajectory (5.152) can be inverted and one finds

$$x = -\frac{1}{A_1} + \frac{1}{A_1} \sqrt{1 + 2A_1 x_{\text{end}} + A_1^2 [x_{\text{end}}^2 - 4(N - N_{\text{end}})]}. \qquad (5.154)$$

Along both the positive and negative branch of the potential, inflation ends naturally at $\epsilon_1 = 1$, that is at

$$x_{\epsilon_1=1}^{\pm} = \frac{-1 \pm \sqrt{2A_1}}{A_1}. \qquad (5.155)$$

Along the positive branch we are interested in, we therefore have

$$x_{\text{end}} = x_{\epsilon_1=1}^{+} = \frac{-1 + \sqrt{2A_1}}{A_1}. \qquad (5.156)$$

Plugging this expression into Eq. (5.154) gives the field value $x_*$ at which the pivot mode crossed the Hubble radius during inflation in terms of the e-fold number $\Delta N_* = N_{\text{end}} - N_*$. Let us remember that solving for $x_*$ (or $\Delta N_*$) is made through Eq. (3.48). From Eq. (5.145), one gets

$$\epsilon_{1*} = \frac{1}{1 + 2\Delta N_*} \qquad (5.157)$$

which, together with $\epsilon_2 = 2\epsilon_1$, yields

$$n_S - 1 = 2n_T, \qquad r = 4(1 - n_S). \qquad (5.158)$$

Notice that this relation is different from the power law case and consistent with Ref. [393]. In that reference, the authors mention that the horizon flow models predicts $r \simeq 4.8(1 - n_S)$ as a result of Monte-Carlo simulations.

Finally, the potential parameter $M$ can be determined from the CMB normalization

$$\left(\frac{M}{M_{\text{Pl}}}\right)^4 = 960\pi^2 \frac{A_1^2}{(1 + A_1 x_*)^4} \frac{Q_{\text{rms-PS}}^2}{T^2}. \qquad (5.159)$$

It is interesting to notice that the typical energy scale of inflation in these models does not depend on $A_1$. The previous equation indeed leads to

$$\frac{V(x_*)}{M_{\text{Pl}}^4} = \frac{480\pi^2}{1 + 2\Delta N_*} \frac{Q_{\text{rms-PS}}^2}{T^2} \left(1 - \frac{1}{3 + 6\Delta N_*}\right) \approx 10^{-9}. \qquad (5.160)$$
The reheating consistent (exact) predictions for the horizon flow inflation I models are represented in Fig. 137. As expected, the relation $\epsilon_2 = 2\epsilon_1$, which is the same as for the LFI quadratic case, is properly recovered. The predictions do not depend much on the potential parameter $A_1$.

## 5.11. Colemann-Weinberg Inflation (CWI)

### 5.11.1. Theoretical justifications

The potential of this model was first introduced by Coleman and Weinberg in Ref. [394], in the context of spontaneous symmetry breaking generated by radiative corrections. The starting point of this work is to calculate the effective potential for a massless charged meson minimally coupled to the electrodynamic field.

In that reference, the effective action is explicitly constructed from a Legendre transform of the partition function, and expanded into one-particle-irreducible Feynman diagrams with $n$ external lines (and summing up over $n$). The exact knowledge of the effective potential requires an infinite summation of all these Feynman diagrams, which is in practice intractable. It is thus made use of the one loop expansion method where all diagrams with no closed loops are first summed, then all diagrams with one closed loop are added, and all higher loops diagrams neglected. Starting with a quartic interacting scalar field, and requiring that the renormalized mass vanishes, one obtains a potential of the form

$$V(\phi) \propto 1 + \alpha \left(\frac{\phi}{Q}\right)^4 \ln\left(\frac{\phi}{Q}\right). \qquad (5.161)$$

Let us emphasize that another useful frame of approximation is the Gaussian effective potential method. The Gaussian effective potential is a non-perturbative approach to quantum field theory [395–403], originally developed in the context of quantum mechanics, and generalized to field theory afterwards. In quantum mechanics, when studying systems governed by Hamiltonians of the form $H = p^2/2 + V(\phi)$, the idea is to calculate an effective potential $V_{\text{GEP}}$ defined as

$$V_{\text{GEP}}(\phi_0) = \min_{\Omega} \left[ \langle \psi | H | \psi \rangle, \psi(\phi) = \left(\frac{\Omega}{\hbar\pi}\right)^{1/4} e^{-\Omega(\phi-\phi_0)^2/(2\hbar)} \right], \qquad (5.162)$$

i.e. the minimum possible quantum mean energy of a Gaussian wavefunction centered over $\phi_0$. Such an object turns out to be a powerful tool to addressing the effects of quantum fluctuations on the physical behavior of a system in a non-perturbative way. It can be easily generalized to quantum field theories, expanding the field operator $\Phi$ only over $\Omega$-massive excitations around the classical value $\phi_0$ in $d$ dimensions,

$$\Phi(t, \mathbf{x}) = \phi_0 + (2\pi)^{(1-d)/2} \int \frac{d^{d-1}\mathbf{k}}{\sqrt{2\sqrt{k^2 + \Omega^2}}} \left( a_k e^{-i\sqrt{k^2 + \Omega^2}t + i\mathbf{k}\cdot\mathbf{x}} + a_k^\dagger e^{i\sqrt{k^2 + \Omega^2}t - i\mathbf{k}\cdot\mathbf{x}} \right), \qquad (5.163)$$

where $a_k^\dagger$ and $a_k$ are the usual creation and annihilation operators, and minimizing the quantum mean value of the Hamiltonian density over $\Omega$. In Ref. [396], the quartic interacting scalar field has been worked out with this method, i.e. starting from $V(\phi) = m^2\phi^2/2 + \lambda\phi^4$. The Gaussian effective potential $V_{\text{GEP}}$ obtained in this way can be expanded in power of $\hbar$ to show that the first order terms match with the potential of Coleman and Weinberg. This is not surprising as this is equivalent of performing a one loop expansion over the effective action. However, it should be stressed that the Gaussian effective potential method provides a much more general expression for the potential, that is valid beyond this perturbative limit and that can address regimes where quantum diffusion dominates the dynamics of the scalar field.

The model is defined such that inflation ends by violation of the slow-roll conditions, and is followed by a preheating stage in which the inflaton field oscillates at the bottom of its potential. Therefore this potential minimum must be set to zero, which implies

$$\alpha = 4e. \qquad (5.164)$$

One is thus left with one mass parameter, $Q$, which sets the typical $\text{vev}$ at which inflation takes place. On the other hand, the value taken for $Q$ also depends on the underlying high energy model from which the CW potential emerges.

The CWI potential appears in various other contexts and, in fact, historically, it was the first model of inflation ever proposed [30] (also known as “old inflation”). The idea was that inflation occurs while the field is trapped in a false vacuum state $\langle\phi\rangle = 0$. Then, inflation comes to an end when the field tunnels from this state to the symmetry breaking true minimum. Unfortunately, this models was quickly realized to be ruled out since the above mentioned process is accompanied by bubble formation and these bubbles, while colliding, produce too large inhomogeneities. Then, this problem was solved by a modification of the old inflation scenario called “new inflation” [31,32]. The main idea is that inflation does not occur while the field is trapped but when the field is rolling down from the origin to its true minimum. Bubbles are also formed but there are so big that our entire universe is contained in one of them. As a consequence, we do not observe bubble collisions and our universe is extremely homogeneous as indicated by the observations. This new inflationary scenario was explicitly implemented in Ref. [31] where the SU(5) → SU(3) × SU(2) × U(1) phase transition in GUTs is investigated. The model makes use of a CWI potential that can be described by

$$V(\phi) = \frac{5625}{512\pi^2} g^4 \left[ \phi^4 \ln\left(\frac{\phi}{\phi_0}\right) - \frac{\phi^4}{4} + \frac{\phi_0^4}{4} \right], \qquad (5.165)$$

where $\phi_0 \approx 10^{14} - 10^{15}$ GeV, representing the GUT symmetry breaking scale, and $g^2 \approx 1/3$ is the SU(5) gauge coupling constant. However, as noticed afterwards in Refs. [404–408], this model has also a fatal flaw. Indeed, one sees in Eq. (5.165) that the overall normalization of the potential reads $M^4 = 5625g^4\phi_0^4/(2048\pi^2)$ and that, therefore, the amplitude of the fluctuations is in fact already fixed. Using the value of the SU(5) coupling constant and $Q/M_{\text{Pl}} = e^{1/4}\phi_0/M_{\text{Pl}} \approx 5 \times 10^{-5} - 5 \times 10^{-4}$, one arrives at $M^4 \approx (10^{-13} - 10^{-17}) M_{\text{Pl}}^4$. This turns out to be incompatible with the CMB normalization [see Eq. (5.173) below]. However, the same model was re-considered in Refs. [407,409] (see also Ref. [410]), but with additional fields and couplings. It was then shown that the scale $M$ acquires a different form and can scale as the inverse of the coupling constants. Since these ones are small, it becomes possible to obtain a higher value for $M$ and to correctly CMB normalize the model. In what follows, we will therefore consider the scale $M$ as a free parameter fixed by the overall amplitude of the cosmological fluctuations.

We also notice that, in Ref. [411], the CWI potential is obtained in the context of Kaluza-Klein inflation, i.e. in higher dimensions and with higher derivative terms and logarithmic dependence on the curvature scalar. Again, the typical value for $Q \approx 10^{15}$ GeV. The CWI potential appears
<figure><img src="image_25.png" /><figcaption>Fig. 22. Coleman–Weinberg Inflation (CWI) for $\alpha = 4e$. Top left panel: Coleman–Weinberg Inflation potential as a function of $\phi/Q$. Top right panel: logarithm of the potential for the same value of $\alpha$. Bottom left panel: normalized first slow-roll parameter $Q^2/M_{\text{Pl}}^2 \epsilon_1$. The shaded area indicates the where inflation stops if $Q = M_{\text{Pl}}$. Bottom right panel: normalized second and third slow-roll parameters $Q^2/M_{\text{Pl}}^2 \epsilon_2$ (solid line) and $Q^2/M_{\text{Pl}}^2 \epsilon_3$ (dotted line) for the same potential.</figcaption></figure>

also in Ref. [412], but the value used for $Q$ is rather different, $Q = 0.223 M_{\text{Pl}}$, and is fine-tuned in order to have two phases of inflation, a “chaotic inflationary” phase followed by a “new inflationary” phase. Finally, in Ref. [413], the Coleman–Weinberg potential is studied in the framework of Einstein–Brans–Dicke gravity, with the same typical value for $Q \simeq 10^{15}$ GeV and the same typical value for $M^4/M_{\text{Pl}}^4 \simeq 10^{-15}$ as in the original paper.

### 5.11.2. Slow-roll analysis

Considering the previous considerations, we take the potential to be

$$V(\phi) = M^4 \left[ 1 + \alpha \left( \frac{\phi}{Q} \right)^4 \ln \left( \frac{\phi}{Q} \right) \right], \qquad (5.166)$$

with a parameter $Q/M_{\text{Pl}}$ in the range $[10^{-5}, 10^{-3}]$ and $\alpha = 4e$. As already mentioned, the mass parameter $M$ will be viewed as free and fixed by the normalization to the amplitude of the CMB anisotropies. The potential is displayed Fig. 22. It starts decreasing with the inflaton vev at $\phi = 0$, reaches a minimum at $\phi/Q = e^{-1/4}$ where it vanishes, and then increases and diverges as $\phi$ goes to $\infty$. As mentioned above, inflation proceeds along the decreasing branch of the potential, in the direction specified by the arrow in the figure.

Let us now derive the first slow-roll parameters. Defining $x \equiv \phi/Q$, they are given by

$$\epsilon_1 = \frac{M_{\text{Pl}}^2 \alpha^2}{Q^2} x^6 \left( \frac{1 + 4 \ln x}{1 + \alpha x^4 \ln x} \right)^2, \qquad (5.167)$$

while

$$\epsilon_2 = 2 \frac{M_{\text{Pl}}^2}{Q^2} \alpha x^2 \frac{-7 - 12 \ln x + \alpha x^4 + \alpha x^4 \ln x + 4 \alpha x^4 \ln^2 x}{(1 + \alpha x^4 \ln x)^2}, \qquad (5.168)$$

and finally

$$\epsilon_3 = \frac{M_{\text{Pl}}^2}{Q^2} \left( \begin{aligned} & -26\alpha x^2 + 21\alpha^2 x^6 - 2\alpha^3 x^{10} - 128\alpha x^2 \ln x \\ & + 152\alpha^2 x^6 \ln x - 11\alpha^3 x^{10} \ln x - 96\alpha x^2 \ln^2 x \\ & + 368\alpha^2 x^6 \ln^2 x - 14\alpha^3 x^{10} \ln^2 x + 384\alpha^2 x^6 \ln^3 x \\ & - 16\alpha^3 x^{10} \ln^3 x - 32\alpha^3 x^{10} \ln^4 x \end{aligned} \right) (1 + \alpha x^4 \ln x)^{-2} \\ \times (7 - \alpha x^4 + 12 \ln x - \alpha x^4 \ln x - 4\alpha x^4 \ln^2 x)^{-1}. \qquad (5.169)$$
<figure><img src="image_26.png" /><figcaption>Fig. 23. End of inflation in Coleman–Weinberg inflation. The approximated formula of Eq. (5.170) for $x_{end}$ (red dashed line) is compared with the exact numerical solution of $\epsilon_1 = 1$ (blue solid line), for $\alpha = 4e$, in the physically relevant range of values for $Q/M_P$. The agreement is obviously excellent.</figcaption></figure>

The three of them have the same general behavior. They vanish at $x = 0$, increase with $x$ in the decreasing branch of the potential and diverge at the minimum of the potential. Then they decrease from infinity in the increasing branch of the potential, and reach asymptotically vanishing values when the field vev goes to infinity. Inflation stops by slow-roll violation when $\epsilon_1 = 1$. The value of $x$ at which this happens needs to be determined numerically, but in the limit $Q/M_{Pl} \ll 1$ (remember that $Q/M_{Pl} \simeq 10^{-4}$) where one expects $x_{end} \ll 1$, one can derive an analytic approximated formula, namely

$$x_{\text{end}} \approx e^{-1/4} \exp \left[ W_{-1} \left( -\frac{3\sqrt{2}}{4\alpha} \frac{Q}{M_{\text{Pl}}} e^{3/4} \right) \right], \qquad (5.170)$$

where $W_{-1}$ is the $-1$ branch of the Lambert function. A comparison between this approximated formula and the numerical solution for $x_{end}$ is displayed in Fig. 23. The agreement is excellent.

Let us now calculate the slow-roll trajectory from Eq. (3.11). It is given by

$$\begin{aligned} N_{\text{end}} - N = & \frac{Q^2}{M_{\text{Pl}}^2} \frac{\sqrt{e}}{4\alpha} \left[ \text{Ei}\left(-\frac{1}{2} - 2\ln x\right) - \text{Ei}\left(-\frac{1}{2} - 2\ln x_{\text{end}}\right) \right] \\ & + \frac{Q^2}{M_{\text{Pl}}^2} \frac{1}{16\sqrt{e}} \left[ \text{Ei}\left(\frac{1}{2} + 2\ln x_{\text{end}}\right) - \text{Ei}\left(\frac{1}{2} + 2\ln x\right) \right] \\ & + \frac{1}{8} \frac{Q^2}{M_{\text{Pl}}^2} (x^2 - x_{\text{end}}^2), \end{aligned} \qquad (5.171)$$

where $\text{Ei}$ is the exponential integral function, $N_{\text{end}}$ is the number of $e$-folds at the end of inflation and $N$ is the number of $e$-folds corresponding to the scaled field vev $x$. In the $Q/M_{Pl} \ll 1$ limit where $x \ll 1$, the first term of this expression dominates. Since $\alpha = 4e$, the previous expression can be slightly simplified:

$$\begin{aligned} N_{\text{end}} - N = & \frac{Q^2}{M_{\text{Pl}}^2} \frac{1}{16\sqrt{e}} \left[ \text{Ei}\left(-\frac{1}{2} - 2\ln x\right) - \text{Ei}\left(-\frac{1}{2} - 2\ln x_{\text{end}}\right) \right] \\ & + \text{Ei}\left(\frac{1}{2} + 2\ln x_{\text{end}}\right) - \text{Ei}\left(\frac{1}{2} + 2\ln x\right) + \frac{1}{8} \frac{Q^2}{M_{\text{Pl}}^2} (x_{\text{end}}^2 - x^2). \end{aligned} \qquad (5.172)$$

After having solved the above equation for $x_*$, the field value at which the pivot scale crossed the Hubble radius during inflation, $M$ is fixed by the amplitude of the CMB anisotropies to

$$\left(\frac{M}{M_{\text{Pl}}}\right)^4 = 720\pi^2\alpha^2 \frac{M_{\text{Pl}}^2}{Q^2} x_*^6 (1 + 4 \ln x_*)^2 (1 + \alpha x_*^4 \ln x_*)^{-3} \frac{Q^2_{\text{rms}} - \text{PS}}{T^2}. \qquad (5.173)$$

The reheating consistent slow-roll predictions of the Coleman–Weinberg models are displayed Fig. 138 in the physical range $Q/M_{Pl} \in [10^{-5}, 10^{-3}]$. The reheating equation of state parameter $\bar{\omega}_{\text{reh}}$ has been taken to 0 since the potential is quadratic close to its minimum $V(x) \simeq 2\alpha M^4 e^{-1/2} (x - e^{-1/4})^2$. The typical predicted amount of gravitational waves is extremely small, and a non-negligible deviation from $n_S = 1$ is noticed. Also, one could choose to relax the constraint on the parameter $Q$ and study the Coleman–Weinberg potential in general. This was done for instance in Ref. [409] where the Coleman–Weinberg potential predictions are compared with the WMAP observations on general grounds. It is found that the potential normalization should be of the order $M \simeq 10^{16}$ GeV, and that $Q \simeq 10 M_{Pl}$ in order to match $n_S \simeq 0.96$. For this reason the reheating consistent slow-roll predictions are displayed in Fig. 139 in the extended range $Q/M_{Pl} \in [1, 100]$. In the limit $Q/M_{Pl} \gg 1$, the model is well approximated by a quadratic potential around its minimum, and one asymptotically approaches the LFI predictions with $p = 2$ (see Section 5.2).
<figure><img src="image_27.png" /><figcaption>Fig. 24. Loop Inflation (LI). Top left panel: Loop Inflation potential for $\alpha = \pm 0.5$, the case $\alpha = 0.5$ being displayed in blue and the case $\alpha = -0.5$ being displayed in pink. Top right panel: logarithm of the potential for the same values of $\alpha$. Bottom left panel: slow-roll parameter $\epsilon_1$ with the same values of $\alpha$. The shaded area indicates where inflation stops. Bottom right panel: slow-roll parameters $\epsilon_2$ (solid line) and $\epsilon_3$ (dotted line) for the same values of $\alpha$.</figcaption></figure>

## 5.12. Loop Inflation (LI)

### 5.12.1. Theoretical justifications

The flatness of an inflationary potential is in general altered by radiative corrections. One loop order corrections generically take the form of a logarithmic function, $\ln(\phi/\mu)$, where $\mu$ is a renormalization scale. Starting from a perfectly flat potential, one obtains a potential of the form $V(\phi) = M^4 [1 + \alpha \ln(\phi/M_{Pl})]$ where $\alpha$ is a dimensionless parameter that tunes the strength of the radiative effects. Studying such potentials is therefore a simple way to discuss in which cases the quantum correction “spoil” the flatness of a potential, and how this happens.

In fact, this type of scenarios were invented in the context of F and D-term inflation in Refs. [414–417]. The original motivation was to build an inflationary model in supersymmetry but without the $\eta$-problem that appears in the F-term approach. Indeed, if one considers a simple superpotential $W = f/2X\phi^2 - \mu^2 X$ where $\phi$ and $X$ are two superfields, then it is easy to obtain the supersymmetric potential assuming a minimal Kähler potential: $V = |f\phi^2/2 - \mu^2|^2 + f^2|X|^2|\phi|^2$. There is a flat direction for $\phi = 0$ along the X direction with $V = \mu^4$. Lifting this direction with a one loop correction leads to the LI potential which is suitable for inflation. However, considering non-minimal term in the Kähler potential destroys the flatness of V. The D-term approach was shown to be a viable alternative. The idea is to consider a theory with a U(1) symmetry and three chiral superfields, $X$, $\phi_+$ and $\phi_-$ with charges 0, +1 and -1 respectively. It then follows that the superpotential has the form $W = \lambda X \phi_+ \phi_-$. If we compute the corresponding potential in global supersymmetry, one arrives at

$$V = \lambda^2 |X|^2 (|\phi_-|^2 + |\phi_+|^2) + \lambda^2 |\phi_+ \phi_-|^2 + \frac{g^2}{2} (|\phi_+|^2 - |\phi_-|^2 + \xi)^2, \qquad (5.174)$$

where the part proportional to g (g being the gauge coupling) represents the D-part of V. In this expression $\xi$ is a Fayet–Iliopoulos term. There is a unique supersymmetric vacuum at $X = \phi_+ = 0$ and $|\phi_-| = \sqrt{\xi}$ and a flat direction along the X direction with $\phi_+ = \phi_- = 0$ where the potential $V = g^2 \xi^2/2$ can drive inflation. Since supersymmetry is broken along the flat direction, this produces one loop corrections and we obtain

$$V = \frac{g^2}{2} \xi^2 \left[ 1 + \frac{g^2}{16\pi^2} \ln \left( \frac{\lambda^2 |X|^2}{\mu^2} \right) \right], \qquad (5.175)$$

where $\mu$ is a renormalization scale. We see that this potential has exactly the form of an LI potential where the scale M is related to the Fayet–Iliopoulos term $\xi$ and where $\alpha$ is in fact the square of the gauge coupling. In particular, this implies that $\alpha > 0$ in this context. One can also reproduce the above calculation in supergravity (with minimal Kähler potentials) and show that the D-part of the theory leads to the same potential which is free of the $\eta$ problem.

After these initial works on D-term inflation, many other papers addressing different issues were published. Observational constraints on this type of scenarios were discussed in Refs. [418,419]. Ref. [420] has discussed how to produce D-term inflation and to stabilize the moduli at the same time. Then, in Refs. [421–423], it was shown that the stringy implementation of D-term inflation is problematic. We have seen that the scale M is essentially controlled by the value of the Fayet–Iliopoulos term $\xi$. Therefore, the CMB normalization allows us to calculate the value
of $\xi$. Anticipating the calculation at the end of this section, if one uses the equation after Eq. (5.187) with $M^4 = g^2\xi^2/2$ and $\alpha = g^2/(8\pi^2)$ [from Eq. (5.175)], then one arrives at

$$ \xi \approx \left[ \left( \frac{90}{\Delta N_*} \right)^{1/4} \left( \frac{Q_{\text{rms}-\text{PS}}}{T} \right)^{1/2} M_{\text{Pl}} \right]^2 \approx (6.9 \times 10^{15} \text{ GeV})^2, \qquad (5.176) $$

where we have taken the fiducial value $\Delta N_* \approx 50$. As noticed in Refs. [421–423], in string theory, one typically obtains $\xi = (\text{Tr}Q)M_s^2/(192\pi^2)$ where $M_s$ is the string scale and $\text{Tr}Q \approx 100$ sums the U(1) charges of all massless states. This leads to $\xi \approx (\text{few} \times 10^{17} \text{GeV})^2$ and, therefore, does not match the CMB normalization (5.176). Then, Refs. [424,425] studied more complicated models in the supersymmetric context in order to fix the problem we have just discussed. Other scenarios were also investigated in Refs. [426–429]. D-term inflation in the context of string theory and brane inflation was also discussed in Ref. [309,430–435]. The same topic was also addressed in Refs. [436,437] but in the context where the Friedmann equations receives quadratic corrections. Finally, Ref. [438] studied LI potentials in the case of Wess–Zumino models. Let us emphasize again that, in all these models, the constant $\alpha$ is positive and given in terms of the square of a gauge coupling.

The LI potential was also derived in a different framework in Ref. [439]. This article uses the O’Raifeartaigh–Witten model that will be studied in more detail in Section 5.23. Therefore, we do not give the details here and only quote results that will be reviewed in that section. In particular, we will see in Eq. (5.338) that the only difference is that the parameter $\alpha$ is now given in terms of three coupling constants and has a rather involved form which allows for negative $\alpha$ values. For this reason we will not fix the sign of $\alpha$ in the following.

### 5.12.2. Slow-roll analysis

Let us now turn to the slow-roll study of loop inflation. We recall that the potential takes the following form

$$ V(\phi) = M^4 \left[ 1 + \alpha \ln \left( \frac{\phi}{M_{\text{Pl}}} \right) \right], \qquad (5.177) $$

where $\alpha$ is a dimensionless parameter, that can a priori be either positive or negative (see the above discussion). Let us define the quantity $x \equiv \phi/M_{\text{Pl}}$. The potential Eq. (5.177), as well as its logarithm, is displayed in Fig. 24. If $\alpha > 0$, it is an increasing function of the field $\text{vev}$, and vanishes at

$$ x_{V=0} = e^{-1/\alpha}. \qquad (5.178) $$

Hence inflation proceeds from the right to the left at $x > x_{V=0}$ in that case. If $\alpha < 0$ however, the potential is a decreasing function of the field, which vanishes at $x_{V=0}$, still given by Eq. (5.178), hence inflation proceeds from the left to the right at $x < x_{V=0}$.

The three first Hubble flow functions in the slow-roll approximation are given by

$$ \epsilon_1 = \frac{\alpha^2}{2} \frac{1}{x^2} (1 + \alpha \ln x)^{-2}, \quad \epsilon_2 = 2\alpha \frac{1}{x^2} \frac{1 + \alpha + \alpha \ln x}{(1 + \alpha \ln x)^2}, \qquad (5.179) $$

and

$$ \epsilon_3 = 2\alpha \frac{1}{x^2} (1 + \alpha \ln x)^{-2} (1 + \alpha + \alpha \ln x)^{-1} \times \left[ 1 + \frac{3\alpha}{2} + \alpha^2 + \left(2\alpha + \frac{3}{2}\alpha^2\right) \ln x + \alpha^2 \ln^2 x \right]. \qquad (5.180) $$

If $\alpha > 0$, the first slow-roll parameter is a decreasing function of the field $\text{vev}$, which diverges at $x_{V=0}$ and vanishes when $x \to \infty$. Therefore inflation stops by slow-roll violation, at the point $x_{\text{end}}$ satisfying $\epsilon_1 = 1$ and given by

$$ x_{\text{end}} = \frac{1}{\sqrt{2}} \left[ W_0 \left( \frac{e^{1/\alpha}}{\sqrt{2}} \right) \right]^{-1}, \qquad (5.181) $$

where $W_0$ is the 0-branch of the Lambert function. One can check that since $W_0(y) < y$ for any $y$, one always has $x_{\text{end}} > x_{V=0}$, as required. When $\alpha \ll 1$, one has $x_{\text{end}} \approx \alpha/\sqrt{2}$. If $\alpha < 0$ on the other hand, the first slow-roll parameter diverges at $x = 0$, decreases with $x$, reaches a minimum at $x_{\epsilon_2=0} = \exp(-1-1/\alpha)$, then increases with $x$ and diverges at $x_{V=0}$. The minimum value of $\epsilon_1$ equals $\epsilon_1(x_{\epsilon_2=0}) = \exp(2+2/\alpha)/2$ which is smaller than unity only if $\alpha > 2/(\ln 2-2) \approx -1.53$. Otherwise $\epsilon_1(x) > 1$ all over the domain and inflation cannot take place. If $\alpha > 2/(\ln 2-2)$, the inflationary domain lies between $x_{\epsilon_1=1}^{-}$ and $x_{\text{end}} = x_{\epsilon_1=1}^{+}$, with

$$ x_{\epsilon_1=1}^{-} = -\frac{1}{\sqrt{2}} \left[ W_{-1} \left( \frac{-e^{1/\alpha}}{\sqrt{2}} \right) \right]^{-1}, \quad x_{\text{end}} = x_{\epsilon_1=1}^{+} = -\frac{1}{\sqrt{2}} \left[ W_0 \left( \frac{-e^{1/\alpha}}{\sqrt{2}} \right) \right]^{-1}, \qquad (5.182) $$

and where $W_{-1}$ is the $-1$-branch of the Lambert function. When $|\alpha| \ll 1$, one has $x_{\text{end}} \approx e^{-1/\alpha} - 1/\sqrt{2} \gg 1$. Let us notice that the end of inflation occurs in the region $\phi \gg M_{\text{Pl}}$, where Eq. (5.177) may not be well defined. Therefore, depending on the underlying theoretical setting, the end of inflation by slow-roll violation may not be meaningful.

Let us now turn to the slow-roll trajectory. It can be integrated, giving rise to

$$ N_{\text{end}} - N = \frac{x^2}{2} \left( \ln x + \frac{1}{\alpha} - \frac{1}{2} \right) - \frac{x_{\text{end}}^2}{2} \left( \ln x_{\text{end}} + \frac{1}{\alpha} - \frac{1}{2} \right). \qquad (5.183) $$

When $|\alpha| \ll 1$, it approximately takes the form $2\alpha(N_{\text{end}} - N) = x^2 - x_{\text{end}}^2$. The trajectory Eq. (5.183) can be inverted making use of the Lambert function, and one obtains

$$ x^2 = \frac{4(N_{\text{end}} - N) - x_{\text{end}}^2 \left[ 1 - \frac{2}{\alpha} - \ln(x_{\text{end}}^2) \right]}{W_0(-1) \left\{ 4(N_{\text{end}} - N)e^{-(1-2/\alpha)} - \left[ 1 - \frac{2}{\alpha} - \ln(x_{\text{end}}^2) \right] \exp\left[-1 + \frac{2}{\alpha} + \ln(x_{\text{end}}^2)\right] \right\}}, \qquad (5.184) $$
<figure><img src="image_28.png" /><figcaption>Fig. 25. Left panel: Lambert functions $W_0(x)$ (dashed line) and $W_{-1}(x)$ (solid line). During loop inflation, inflation proceeds along the “0” branch in the direction specified by the green arrow on the figure if $\alpha > 0$, and along the “-1” branch in the direction specified by the pink arrow on the figure if $\alpha < 0$. Right panel: Maximal number of e-folds $\Delta N_{\text{max}}$ one can realize when $\alpha < 0$, between $x_{\epsilon_1=1}^-$ and $x_{\epsilon_1=1}^+$, as a function of $\alpha$.</figcaption></figure>

where the 0 branch of the Lambert function must be chosen if $\alpha > 0$, while the -1 branch must be chosen if $\alpha < 0$. The Lambert function is displayed in the left panel of Fig. 25, together with the regions in which inflation proceeds. Let us now comment and check that this expression is valid. Firstly, if $N = N_{\text{end}}$, the Lambert function is of the form $W(-z_{\text{end}}e^{-z_{\text{end}}}) = -z_{\text{end}}$, where $z \equiv (1 - 2/\alpha) - \ln(x^2)$, and this automatically cancels the numerator such that one has indeed $x = x_{\text{end}}$. Secondly, if $\alpha > 0$, the condition $x_{\text{end}} > x_{V=0}$ implies that $z_{\text{end}} < 1$, and the Lambert function at $N_{\text{end}}$ is equal to $-z_{\text{end}} > -1$. Therefore, at the end of inflation, one should use the zero branch of the Lambert function. Finally, as inflation is under way, the argument of the Lambert function is decreasing which implies that the whole inflationary stage takes place on the zero branch. On the other hand, if $\alpha < 0$ using similar arguments, the whole inflationary stage can be shown to take place on the -1 branch.

In this later case ($\alpha < 0$), it is also interesting to notice that the total number of e-folds is bounded, since inflation can only proceed between $x_{\epsilon_1=1}^-$ and $x_{\epsilon_1=1}^+$. The corresponding maximal number of e-folds $\Delta N_{\text{max}}$ is displayed, as a function of $\alpha$, in the right panel of Fig. 25. One can see that when $\alpha \lesssim -0.35$, not a sufficient number of e-folds can be realized. For such values of $\alpha$, one already has $x_{\text{end}} > 10$. Since inflation is supposed to take place at sub-Planckian vevs, it means that this regime of inflation is a priori forbidden. If one allows slightly super-Planckian field vevs, up to $x \simeq 100$ or $x \simeq 1000$, this implies that $\alpha < -0.1$. Therefore even in this case, $\alpha$ must lie in the rather narrow range $-0.3 < \alpha < -0.1$.

Making use of the approximated trajectories and expressions for $x_{\text{end}}$, some analytic predictions can be derived in the case $\alpha > 0$. The observable field value $x_*$, and its associated number of e-folds $\Delta N_* = N_{\text{end}} - N_*$ at which the pivot mode crossed the Hubble radius during inflation are obtained from the above equations together with Eq. (3.48). In the limit $\alpha \ll 1$, one obtains the approximate expressions

$$ \epsilon_{1*} \approx \frac{\alpha}{4\Delta N_*}, \quad \epsilon_{2*} \approx \epsilon_{3*} \approx \frac{1}{\Delta N_*}, \qquad (5.185) $$

hence

$$ r \approx \frac{\alpha}{64\Delta N_*}, \quad n_s - 1 \approx -\frac{1}{\Delta N_*}, \quad \alpha_s \approx \frac{1}{\Delta N_*^2}. \qquad (5.186) $$

Finally, the parameter $M$ can be determined from the amplitude of the CMB anisotropies, and one gets

$$ \left(\frac{M}{M_{\text{Pl}}}\right)^4 = 720\pi^2 \frac{\alpha^2 Q_{\text{rms-PS}}^2}{x_*^2 T^2} (1 + \alpha \ln x_*)^{-3}. \qquad (5.187) $$

In the small $|\alpha|$ limit, one obtains $M^4/M_{\text{Pl}}^4 \simeq 360\pi^2\alpha/\Delta N_* Q_{\text{rms-PS}}^2/T^2$ for $\alpha > 0$, and $M^4/M_{\text{Pl}}^4 \simeq 720\pi^2\alpha^2 e^{2/\alpha} Q_{\text{rms-PS}}^2/T^2$ for negative values of $\alpha$.

The reheating consistent slow-roll predictions of the loop inflation models are displayed in Fig. 140 for $\alpha > 0$, and in Fig. 141 for $\alpha < 0$. For $\alpha > 0$ and $\alpha \ll 1$, the approximations in Eqs. (5.185) give a good description of what is numerically obtained, namely a deviation from scale invariance which almost does not depend on $\alpha$, and an amount of gravitational waves which grows linearly with $\alpha$. For $\alpha < 0$, the predictions blow out of the observational one- and two-sigma contours when $\alpha$ approaches the upper bound derived above, as expected. Correspondingly, the parameter $\alpha$ does not seem to be much constrained when it is positive, whereas close-to-zero values are favored when it is negative.

### 5.13. ($R+R^{2p}$) Inflation (RpI)

This model is the Einstein frame description of a scalar-tensor theory equivalent to $f(R) = R + \epsilon R^{2p}/\mu^{4p-2}$ where $\mu$ is a mass scale, $\epsilon = \pm 1$, and $p > 1/2$ (otherwise the expansion is meaningless). It generalizes the original Starobinsky model [241] obtained for $p = 1$. Such theories are quite generic and appear as limiting cases of more general modified gravity theories [245,440–443] (see Ref. [246] for a review).

Following Refs. [245,246], one can introduce the scalar degree of freedom $\phi$ defined by

$$ \frac{\phi}{M_g} = \sqrt{\frac{3}{2}} \ln(|F(R)|), \qquad (5.188) $$

where $F(R) \equiv \partial f/\partial R$. For the sake of clarity, we identify the Lagrange multiplier field $\chi$ with its on shell value $\chi = R$ and drop the “tilde” over Einstein frame quantities, see Section 4.1.1 for a detailed discussion of this class of models.

The quantity $F \equiv \Omega^2$ is also the square of the conformal factor inducing the transformation from the Jordan frame to the Einstein frame. In the Einstein frame, the field $\phi$ evolves in a potential given by

$$ V(\phi) = \frac{M_g^2 |F| R F - f}{2 F F^2}. \qquad (5.189) $$
In the present case, one has

$$F(R) = 1 + 2\epsilon p \left(\frac{R}{\mu^2}\right)^{2p-1}, \qquad (5.190)$$

which, for small departures with respect to the Einstein-Hilbert action $R \ll \mu^2$, implies that $F(R) > 0$ as needed. Let us notice that in the opposite situation, accelerated (and super-accelerated) solutions have been shown to exist [246]. Defining the quantity $y$ by

$$y = \sqrt{\frac{2}{3} \frac{\phi}{M_g}}, \qquad (5.191)$$

and inserting Eq. (5.190) into Eq. (5.189) one obtains the Einstein frame potential

$$V = M^4 e^{-2y} |e^y - 1|^{2p/(2p-1)}. \qquad (5.192)$$

The normalization constant $M^4$ is related to the modified gravity scale $\mu$ through the following expression

$$M^4 = \frac{2p-1}{4p} \frac{M_g^2 \mu^2}{(2p)^{1/(2p-1)}}. \qquad (5.193)$$

For $F(R) > 0$, Eq. (5.188) implies that for $\epsilon = 1$, the model is defined in the domain $y > 0$, whereas for $\epsilon = -1$ one should consider the domain $y < 0$ only. Such a potential has also been studied in Ref. [444] for $p = 1$, in Refs. [245,445] for $p = 4$ and in Ref. [446] for $p = 2$. Let us notice that the case $p = 1$ corresponds to the Higgs inflation potential studied in Section 4.2. The case $p = 1/2$ is singular since one recovers $f(R) \propto R$. Taking the limit $p \to \infty$, the potential asymptotes $V \to M^4 e^{-2y} |e^y - 1|$ and varying $p$ allows us to explore different potential shapes.

Let us first consider the case $y > 0$ ($\epsilon = 1$). If $p > 1$, the potential admits a maximum at

$$y_{\max} = \ln \left( \frac{2p-1}{p-1} \right), \qquad (5.194)$$

such that inflation can proceed either for $0 < y < y_{\max}$ or $y > y_{\max}$. We respectively call these regimes Rpl1 and Rpl2. If $p < 1$, the potential is an increasing function of $y$, hence inflation proceeds from the right to the left. We call this regime Rpl3. The case $p = 1$ is singular and again, it corresponds to the Higgs inflation potential studied in Section 4.2.

The Hubble flow functions in the slow-roll approximation read

$$\epsilon_1 = \frac{4}{3} \frac{[1 + (p-1)e^y - 2p]^2}{(2p-1)^2 (e^y - 1)^2}, \qquad \epsilon_2 = \frac{8}{3} \frac{p e^y}{(2p-1)(e^y - 1)^2}, \qquad (5.195)$$

and

$$\epsilon_3 = -\frac{4}{3} \frac{(e^y + 1)[1 + (p-1)e^y - 2p]}{(2p-1)(e^y - 1)^2}. \qquad (5.196)$$

The potential and the Hubble flow functions for $y > 0$ have been represented in Fig. 26. As one can check on these figures, inflation never stops in the Rpl2 regime and one needs to complement the model with a mechanism that can end inflation, as for instance with an extra-field and a tachyonic instability. This adds one additional parameter $y_{\text{end}}$ to the model. When this parameter is large, all the three Hubble flow functions admit asymptotically constant values:

$$\lim_{y \to \infty} \epsilon_1 = \frac{4}{3} \left( \frac{p-1}{2p-1} \right)^2, \qquad \lim_{y \to \infty} \epsilon_2 = 0, \qquad \lim_{y \to \infty} \epsilon_3 = -\frac{4}{3} \frac{p-1}{2p-1}. \qquad (5.197)$$

If $p$ is an integer, except for the special case $p = 1$ (see Section 4.2), these values are always smaller than unity, but not particularly small. As such, all these models predict large deviation from scale invariance. Indeed, the spectral index at first order is given by

$$n_s - 1 \approx -\frac{8}{3} \left( \frac{p-1}{2p-1} \right)^2, \qquad (5.198)$$

which, for $p \ge 2$, remains always smaller than $-8/27 \approx -0.3$. This is strongly disfavored by current CMB measurements. Therefore, only the models such that $p$ is close enough to 1 are to be considered (i.e. non integer values of $p$.)

If inflation proceeds in the Rpl1 regime, then inflation stops naturally when $\epsilon_1 = 1$, i.e. at the field value

$$y_{\text{end}} = \ln \left[ (2p-1) \frac{1+2p(\sqrt{3}+1)}{8p^2-4p-1} \right]. \qquad (5.199)$$

However, the second Hubble flow function can only take relatively large value. From Eq. (5.195), since $y < y_{\max}$, one gets

$$\epsilon_2 > \epsilon_2(y_{\max}) = \frac{8}{3} \frac{p-1}{p}. \qquad (5.200)$$

For $p \ge 2$, we are in a situation where $\epsilon_2 > 4/3$ and again, the models are ruled out by a simple slow roll analysis. Therefore, as already noticed before, $p$ must take (non integer) close enough to 1 values for the models to be viable.

Finally, in the Rpl3 regime, inflation stops naturally when $\epsilon_1 = 1$, with $y_{\text{end}}$ still given by Eq. (5.199). This expression is defined only if $p > (1 + \sqrt{3})/2 \approx 0.68$ but the first slow roll parameter continuously decreases with $y$, and its asymptotic value is again given by Eq. (5.197). Therefore, this regime is viable only when $p$ is close enough to unity.

Let us now turn to the slow-roll trajectory. It is given by

$$N - N_{\text{end}} = \frac{3}{4} \left\{ \frac{p}{p-1} \ln \left[ \frac{(p-1)e^y + 1 - 2p}{(p-1)e^{y_{\text{end}}} + 1 - 2p} \right] + y - y_{\text{end}} \right\}. \qquad (5.201)$$

This expression is not properly defined for $p = 1$ but this case has already been considered in the section on the Higgs inflation model. When $p > 1$, if $y = y_{\max}$, the argument of the logarithm vanishes and the total number of $e$-folds diverges. As a result, provided inflation starts close enough to
<figure><img src="image_29.png" /><figcaption>Fig. 26. (R + R²p) Inflation (Rpl) in the Einstein frame for p = 2 (Rpl1 and Rpl2), and p = 0.9 (Rpl3) (assuming M_g ≈ M_n). Top panels: the potential and its logarithm. Bottom left panel: slow-roll parameter ε₁ with the region in which inflation stops (shaded area). In the Rpl2 regime, inflation never stops and one has to consider an extra-mechanism to end inflation. For this special case, one does not longer have M_g ≈ M_n and y is defined by Eq. (5.191). Bottom right panel: slow-roll parameters ε₂ (solid line) and ε₃ (dotted line).</figcaption></figure>

the top of the potential, it is always possible to find a long enough inflationary period. For p < 1, the number of e-folds diverges when y → ∞. The slow-roll trajectory cannot be analytically inverted, but using the same reheating model as in Section 4.2, one can solve for the field value y* at which the pivot mode crossed out the Hubble radius. The corresponding number of e-fold ΔN* = N*end - N* being given by Eq. (5.201).

Concerning the case ε = -1, i.e. the domain y < 0, all of the previous formula still apply but the potential is now a monotonic decreasing function of the field vev which is too steep to support inflation. In particular, over the whole negative domain, Eq. (5.195) implies that ε₁(y < 0) > ε₁(y → -∞) = 4/3, independently on whether p > 1 or p < 1.

Finally, the constant M can be determined from the amplitude of the CMB anisotropies. It follows that

$$ \frac{M^4}{M_g^4} = 1920\pi^2 \frac{[1 + (p-1)e^{y_*} - 2p]^2 e^{2y_*} Q_{\text{rms-PS}}^2}{(2p-1)^2 (e^{y_*} - 1)^{\frac{6p-2}{2p-1}}} \frac{1}{T^2}. \qquad (5.202) $$

The reheating consistent slow-roll predictions of the Rpl models are displayed in Fig. 142 for the Rpl1 regime, in Fig. 143 for the Rpl2 regime, and in Fig. 146 for the Rpl3 regime. In the Rpl1 regime, the Higgs inflation model predictions (see Fig. 125) are recovered when p → 1, and one can see that p < 1.02 is a necessary condition for the spectral index not to be too red. For Rpl2 the limit p → 1 is such that one does not reproduce the Higgs inflation results and for y*end → ∞ the predictions lie on the line ε₂* = 0. Moreover, one can see that when p > 1.1, the models predict too much gravity waves to be compatible with the CMB data. Let us stress that since y*end is not necessarily small for Rpl2, we are in a situation where the numerical value of M_g may be significantly different than M_gpl (see Section 4.2.2). Finally for the Rpl3 regimes, the Higgs inflation model predictions (see Fig. 125) are recovered when p → 1, and they remain compatible with the data within the two-sigma contours provided p > 0.99.

## 5.14. Double-Well Inflation (DWI)

In this section, we study the famous “Mexican hat” potential given by

$$ V(\phi) = M^4 \left[ \left( \frac{\phi}{\phi_0} \right)^2 - 1 \right]^2. \qquad (5.203) $$

Except for the mass M determined by the CMB normalization, it depends on one parameter, the vev φ₀. Historically, this potential was first introduced by Goldstone in Ref. [447] as a toy model for dynamical symmetry breaking. In cosmology, it is of course utilized to investigate the formation and the microscopic structure of topological defects [448–454]. In the context of inflation, it was first used to construct scenarios of topological inflation [455,456]. In this case, it is made use of the fact that the discrete Z₂ symmetry, φ → −φ, makes the state φ = 0 unstable. Therefore, the Universe will split into two different regions separated by a domain wall. One can then show that inflation takes place within this topological defect. More precisely, the potential is usually written as V = λ/4 (φ² − η²)² where η represents the position of the minima
<figure><img src="image_30.png" /><figcaption>Fig. 27. Top left panel: Double Well Inflation (DWI) potential as a function of $\phi/\phi_0$. Only the $\phi > 0$ region is displayed since the potential is symmetric under $\phi \to -\phi$. Top right panel: logarithm of the potential. The arrow indicates in which direction inflation can proceed. Bottom left panel: slow-roll parameter $\epsilon_1$, rescaled by the quantity $M_{\text{Pl}}^2/\phi_0^2$, such that the corresponding expression becomes universal, i.e. independent of $\phi_0$. Bottom right panel: slow-roll parameters $\epsilon_2$ (solid line) and $\epsilon_3$ (dotted line), rescaled by $M_{\text{Pl}}^2/\phi_0^2$ for the same reason as mentioned before.</figcaption></figure>

of the potential. Then, Refs. [455,456] show that topological inflation occurs if $\eta > M_{\text{Pl}}$. On the other hand, if one writes Eq. (5.203) as $V = M^4/\phi_0^4 (\phi^2 - \phi_0^2)^2$, one sees that one can identify $\eta$ with $\phi_0$. And we will precisely show that agreement with the CMB observations requires $\phi_0 > M_{\text{Pl}}$. The potential (5.203) was also used in Refs. [457,458] in the context of open inflation. In a rather different theoretical framework, Eq. (5.203) was studied in Refs. [459,460] where it was derived in $N = 1$ supergravity coupled to matter. It is also interesting to notice that it was obtained using various stringy constructions as early as the 80's, see Refs. [461,462]. More recently, this potential was found to be relevant in a large number of different physical situations [409,463–473]. The same potential was also obtained in the context of M-flation, see Refs. [224–226]. Let us finally mention that this model is sometimes viewed as a realistic version of Small Field Inflation (SFI) with $p = 2$ (see Section 6.1), the extra quartic term preventing the potential from becoming negative. However, as will be shown in the following, these two classes of models should actually be described separately since their predictions differ in the relevant range of parameters.

The parameter $\phi_0$ sets the typical vev at which inflation proceeds and depends on the symmetry breaking scale one considers. In principle, it could vary over a wide range of values, from $\phi_0 \sim 10^{15}$ GeV for GUT symmetry breaking schemes to super-Planckian vev in a stringy or supergravity context. As will be shown in the following, it is in fact constrained to be large (super-Planckian) in order for the predictions of the model to be compatible with the CMB data. The DWI potential is displayed in Fig. 27 together with its logarithm. One has represented the region $\phi > 0$ only because the potential is symmetric under $\phi \to -\phi$. We see that it decreases for $\phi < \phi_0$, vanishes at $\phi_0$ and then increases for $\phi > \phi_0$. As was already mentioned before, this potential is used to describe dynamical symmetry breaking and, as a consequence, inflation should proceed from the left to the right at $\phi < \phi_0$, in the direction specified by the arrow in Fig. 27.

Let us now calculate the slow-roll parameters. If one defines $x \equiv \phi/\phi_0$ they are given by

$$ \epsilon_1 = \left(\frac{M_{\text{Pl}}}{\phi_0}\right)^2 \frac{8x^2}{(x^2-1)^2}, \quad \epsilon_2 = \left(\frac{M_{\text{Pl}}}{\phi_0}\right)^2 \frac{8(1+x^2)}{(x^2-1)^2}, \quad \epsilon_3 = \left(\frac{M_{\text{Pl}}}{\phi_0}\right)^2 \frac{8(x^4+3x^2)}{(x^2-1)^2(x^2+1)}. \qquad (5.204) $$

The behavior of these parameters is represented in Fig. 27. The first slow-roll parameter $\epsilon_1$ is an increasing function of $\phi$ in the range $x \in [0, 1]$. It vanishes at $x = 0$ and blows up at $x = 1$. Then, for $x > 1$, it becomes a decreasing function going to zero when $x$ goes to infinity. We see in Fig. 27 that inflation stops by violation of the slow-roll conditions. The slow roll parameters $\epsilon_2$ and $\epsilon_3$ have similar behaviors, except that $\epsilon_2$ does not vanish when $x = 0$ but is equal to $\epsilon_2(x = 0) = 8(M_{\text{Pl}}/\phi_0)^2$. Therefore, in order for slow-roll to be valid, this last value should be less than one, which amounts to

$$ \frac{\phi_0}{M_{\text{Pl}}} > 2\sqrt{2}. \qquad (5.205) $$
<figure><img src="image_31.png" /><figcaption>Fig. 28. Lambert functions $W_0(x)$ (dashed line) and $W_{-1}(x)$ (solid line). In DWI, inflation proceeds along the negative part of the “0” branch in the direction specified by the arrow.</figcaption></figure>

This constraint on the parameter $\phi_0$ shows that the symmetry breaking scale needs to be super-Planckian. If this last condition is verified, then $\epsilon_2$ becomes greater than one during inflation at $\phi_{\epsilon_2=1}$ defined by

$$x_{\epsilon_2=1} = \sqrt{1 + 4 \left( \frac{M_{Pl}}{\phi_0} \right)^2 \left[ 1 - \sqrt{1 + \left( \frac{\phi_0}{M_{Pl}} \right)^2} \right]} \quad (5.206)$$

This happens before the end of inflation ($\epsilon_1 = 1$) which occurs at the following value of the field

$$x_{\text{end}} = \sqrt{2 + \left(\frac{\phi_0}{M_{\text{Pl}}}\right)^2} - \sqrt{2}. \quad (5.207)$$

Let us now turn to the slow-roll trajectory. It can be integrated exactly and yields the following formula

$$N_{\text{end}} - N = \frac{1}{4} \left( \frac{\phi_0}{M_{\text{Pl}}} \right)^2 \left[ \ln \left( \frac{x_{\text{end}}}{x} \right) - \frac{1}{2} (x_{\text{end}}^2 - x^2) \right], \quad (5.208)$$

where $N_{\text{end}}$ is the number of e-folds at the end of inflation. Using the 0-branch of the Lambert function $W_0$, this trajectory can be inverted. One obtains

$$x = \sqrt{-W_0 \left[ -x_{\text{end}}^2 e^{-x_{\text{end}}^2} e^{8 \left( \frac{M_{\text{Pl}}}{\phi_0} \right)^2 (N-N_{\text{end}})} \right]} \quad (5.209)$$

The fact that the 0-branch of the Lambert function should be chosen comes from the requirement that $x < 1$. The corresponding “trajectory” along the Lambert curve is displayed in Fig. 28, the arrow indicating in which direction inflation proceeds. This trajectory is remarkably similar to the one of SFI with $p = 2$, see Section 6.1 and Eq. (6.6), the only difference being that the factor 8 in front of $N - N_{\text{end}}$ is just 4 in the case of SFI. Therefore not only these two potentials coincide at small fields, but they also give rise to the same kind of slow-roll trajectory. This is why these two models are sometimes identified, DWI being considered as a realistic realization of SFI. However, as shown below, the observations favors super-Planckian values of $\phi_0$ and, in this limit, the two models are not equivalent (of course, this also has something to do with the debate about whether having super-Planckian vev is meaningful or not). In fact, in the regime $\phi_0/M_{\text{Pl}} \gg 1$, one can write

$$x_* \approx 1 - \sqrt{2 \frac{M_{\text{Pl}}}{\phi_0} \sqrt{1 + 2\Delta N_*} + \frac{1}{3} \left( \frac{M_{\text{Pl}}}{\phi_0} \right)^2 \left( 1 + 2\Delta N_* + \frac{2}{\sqrt{1 + 2\Delta N_*}} \right) + \dots} \quad (5.210)$$

From this expression it is clear that, for super-Planckian values of $\phi_0$, $\phi_*$ is close to the minimum of the potential where the quartic term plays an important role and, consequently, where the SFI potential is not a good approximation. A calculation of the Hubble flow parameters at Hubble crossing confirms this conclusion. They are given by

$$\epsilon_{1*} \approx \frac{1}{1 + 2\Delta N_*}, \quad \epsilon_{2*} \approx \frac{2}{1 + 2\Delta N_*}, \quad \epsilon_{3*} \approx \frac{2}{1 + 2\Delta N_*} \quad (5.211)$$

This allows us to establish the corresponding expressions of the tensor to scalar ratio, spectral index and running. One obtains

$$r \approx \frac{16}{1 + 2\Delta N_*}, \quad n_s - 1 \approx -\frac{4}{1 + 2\Delta N_*}, \quad \alpha_s \approx -\frac{8}{1 + 2\Delta N_*} \quad (5.212)$$

These expressions should be compared with Eqs. (6.17). We see that the first Hubble flow parameter for SFI and DWI differ by a factor close to 4 and that the $\epsilon_2$ roughly differ by a factor of 2. As a consequence, as can be checked in Fig. 147, the DWI predictions are such that $\epsilon_{2*} = 2\epsilon_{1*}$ [or
equivalently, $r = 4(1 - n_s)$], whereas, as can be checked in Fig. 169, we have $\epsilon_{2*} = 4\epsilon_{1*}$ for SFI [or equivalently, $r = 8/3(1 - n_s)$]. This explains why the two models can in fact lead to quite different predictions and why DWI cannot be simply viewed as a mere realistic continuation of SFI.

Finally, it is also interesting to constrain the energy scale M. For this purpose, we use the CMB normalization which gives

$$ \frac{M^4}{M_{\text{Pl}}^4} = 11520\pi^2 \left(\frac{M_{\text{Pl}}}{\phi_0}\right)^2 \frac{x_*^2}{(x_*^2 - 1)^4} \frac{Q_{\text{rms-PS}}^2}{T^2}. \qquad (5.213) $$

Then, using the approximated trajectory $x_* \approx 1 - \sqrt{2 + 4\Delta N_*} M_{\text{Pl}} / \phi_0$ in the above formula, one obtains the following expression

$$ \frac{M^4}{M_{\text{Pl}}^4} \approx 1440\pi^2 \left(\frac{\phi_0}{M_{\text{Pl}}}\right)^2 \frac{1}{(1 + 2\Delta N_*)^2} \frac{Q_{\text{rms-PS}}^2}{T^2}. \qquad (5.214) $$

Then, requiring that $M < M_{\text{Pl}}$ leads to the following upper bound on the value of $\phi_0$, $\phi_0 / M_{\text{Pl}} \lesssim 1.5 \times 10^5$. Combined with the lower limit of Eq. (5.205), we see that the possible range of variation of $\phi_0$ is quite large.

The reheating consistent slow-roll predictions for the DWI models are displayed in Fig. 147. The reheating equation of state parameter $\bar{w}_{\text{reh}}$ has been chosen to be 0 since the potential is quadratic close to its minimum $V(\phi) \approx 4M^4/\phi_0^2(\phi - \phi_0)^2$. As claimed before, one can check that only super-Planckian values of the symmetry breaking scale $\phi_0$ are compatible with the data. Actually, this is also true for the SFI models, see Section 6.1 and Fig. 169. As already mentioned before, in this regime, the two models differ while, as expected, they are very similar for sub-Planckian values of the field $\bar{w}_{\text{ev}}$.

## 5.15. Mutated Hilltop Inflation (MHI)

This model belongs to the class of hilltop models [474,475]. In this type of scenarios, inflation is supposed to occur at the top of the potential. In particular, it was shown in Refs. [474,475] that, by adding the contributions coming from higher order operators, F or D term inflation can be turned into hilltop models. Here, we consider mutated hilltop inflation which was first introduced and discussed in Refs. [476,477]. The potential is phenomenological only and given by

$$ V = M^4 \left[ 1 - \operatorname{sech} \left( \frac{\phi}{\mu} \right) \right], \qquad (5.215) $$

with $\operatorname{sech} x = 1 / \cosh x$. As argued in Refs. [476,477], it can be viewed as small field inflation (hilltop inflation) completed by an infinite number of higher order operators, these operators giving rise to a power series responsible for the appearance of the sech function. From an effective field theory point of view, reasonable values of the parameter $\mu$ seem to be such that $\mu < M_{\text{Pl}}$ but in other contexts such a restriction may not be necessary. This is why although the model is studied for any value of $\mu$, approximated formula will also be derived in the $\mu \ll M_{\text{Pl}}$ approximation.

Defining $x \equiv \phi / \mu$, the three first Hubble flow functions in the slow-roll approximation are given by

$$ \epsilon_1 = \frac{M_{\text{Pl}}^2}{2\mu^2} \coth^2\left(\frac{x}{2}\right) \operatorname{sech}^2 x, \qquad \epsilon_2 = \frac{M_{\text{Pl}}^2}{\mu^2} \left[ \operatorname{csch}^2\left(\frac{x}{2}\right) + 2 \operatorname{sech}^2 x \right], \qquad (5.216) $$

$$ \epsilon_3 = \frac{M_{\text{Pl}}^2}{\mu^2} \frac{\cosh x \coth^2\left(\frac{x}{2}\right) + 2 \tanh^2 x}{\cosh x + \sinh^2 x}. \qquad (5.217) $$

where $\operatorname{csch} x = 1 / \sinh x$. These three quantities are monotonic decreasing functions of the field values and inflation proceeds from large field values towards small field values. Together with the potential, they are represented as a function of $x$ in Fig. 29.

The slow-roll trajectory can be integrated exactly from Eq. (3.11) and reads

$$ N - N_{\text{end}} = \frac{\mu^2}{M_{\text{Pl}}^2} \left\{ 2 \ln \left[ \frac{\cosh(x/2)}{\cosh(x_{\text{end}}/2)} \right] - \cosh x + \cosh x_{\text{end}} \right\}. \qquad (5.218) $$

It can also be inverted analytically to give the field values in terms of the number of e-folds using the Lambert function $W_{-1}$. One obtains

$$ x = \operatorname{arccosh} \left( -1 - W_{-1} \left\{ - (1 + \cosh x_{\text{end}}) \exp \left[ \frac{M_{\text{Pl}}^2}{\mu^2} (N - N_{\text{end}}) - 1 - \cosh x_{\text{end}} \right] \right\} \right). \qquad (5.219) $$

Since $N - N_{\text{end}} < 0$ and the function $ye^{-y}$ has a global maximum equals to $1/e$, inflation proceeds along the $-1$ branch of the Lambert function as represented in Fig. 30. Note that in the $\mu \ll M_{\text{Pl}}$ limit, this trajectory simply becomes $N - N_{\text{end}} \approx \mu^2 / (2M_{\text{Pl}}^2) (e^{x_{\text{end}}} - e^x)$.

For MHI, inflation naturally stops when $\epsilon_1 = 1$, which has an unique solution given by

$$ x_{\text{end}} = \operatorname{arcsech} \left[ -\frac{1}{3} + \frac{1}{3} \left( 1 - 6 \frac{\mu^2}{M_{\text{Pl}}^2} \right) \left( -1 + 36 \frac{\mu^2}{M_{\text{Pl}}^2} + 3 \sqrt{6} \frac{\mu}{M_{\text{Pl}}} \sqrt{4 \frac{\mu^4}{M_{\text{Pl}}^4} + 22 \frac{\mu^2}{M_{\text{Pl}}^2} - 1} \right)^{-1/3} \right. \\ \left. + \frac{1}{3} \left( -1 + 36 \frac{\mu^2}{M_{\text{Pl}}^2} + 3 \sqrt{6} \frac{\mu}{M_{\text{Pl}}} \sqrt{4 \frac{\mu^4}{M_{\text{Pl}}^4} + 22 \frac{\mu^2}{M_{\text{Pl}}^2} - 1} \right)^{1/3} \right], \qquad (5.220) $$

and with $\operatorname{arcsech} x = \operatorname{arccosh}(1/x)$. One should note that the previous equation is always well defined, regardless of the sign of the square root argument by analytic continuation. Let us notice that from Eq. (5.216) one has

$$ \epsilon_2 - \epsilon_1 = \frac{1}{2} \operatorname{csch}^2 \left( \frac{x}{2} \right) + \operatorname{sech} x + \frac{5}{2} \operatorname{sech}^2 x > 0. \qquad (5.221) $$

Consequently, the slow-roll approximation may become inaccurate before the end of inflation because $\epsilon_2 > 1$ occurs just before $\epsilon_1 = 1$. However, one can check that this happens during a negligible number of e-folds and the observable predictions for MHI remain mostly unaffected. Also, in the limit $\mu \ll M_{\text{Pl}}$, Eq. (5.220) gives $x_{\text{end}} \approx \ln(\sqrt{2}M_{\text{Pl}}/\mu)$.
<figure><img src="image_32.png" /><figcaption>Fig. 29. Mutated Hilltop Inflation (MHI). The top panels show the potential and its logarithm as a function of $x = \phi/\mu$. Bottom left panel: Rescaled slow-roll parameter $\epsilon_1$ (divided by $M_*^2/\mu^2$). The shaded area represents the region in which inflation stops if $\mu = M_{pl}$. It should be accordingly rescaled for other values of $\mu$. Bottom right panel: slow-roll parameters $\epsilon_2$ (solid line) and $\epsilon_3$ (dotted line), again rescaled by $M_*^2/\mu^2$ together with the region of slow-roll violation for $\mu = M_{pl}$.</figcaption></figure>

<figure><img src="image_33.png" /><figcaption>Fig. 30. Lambert functions $W_0(x)$ (dashed line) and $W_{-1}(x)$ (solid line). During Mutated Hilltop inflation, inflation proceeds along the “-1” branch in the direction specified by the arrow on the figure.</figcaption></figure>

The value $x_* = \phi_*/\mu$ at which the pivot mode crossed the Hubble radius during inflation is obtained by solving Eq. (3.48) for a given reheating energy. In terms of $\Delta N_*$, and in the limit $\mu \ll M_{pl}$, one has $x_* \simeq \ln(2\Delta N_* M_{pl}^2/\mu^2)$. This enables to give estimates for the slow-roll parameters at Hubble crossing, namely

$$ \epsilon_{1*} \approx \frac{1}{2\Delta N_*^2} \left( \frac{\mu}{M_{pl}} \right)^2, \quad \epsilon_{2*} \approx \frac{2}{\Delta N_*}, \quad \epsilon_{3*} \approx \frac{1}{\Delta N_*} \qquad (5.222) $$

hence, at first order in slow-roll

$$ r \approx \frac{8}{\Delta N_*^2} \left( \frac{\mu}{M_{pl}} \right)^2, \quad n_s - 1 \approx -\frac{2}{\Delta N_*}, \quad \alpha_s \approx -\frac{2}{\Delta N_*^2}. \qquad (5.223) $$
<figure><img src="image_34.png" /><figcaption>Fig. 31. Radion Inflation (RGI) for $\alpha = 10^{-4}$. Top frames: the potential and its logarithm. Bottom left panel: slow-roll parameter $\epsilon_1$ and the shaded area in which inflation stops ($\epsilon_1 > 1$). Bottom right panel: slow-roll parameters $\epsilon_2$ (solid line) and $\epsilon_3$ (dotted line).</figcaption></figure>

One can see that for $\mu/M_{Pl} \ll 1$, the typical predicted amount of gravitational waves is very small, and the deviation from scale invariance almost does not depend on $\mu$.

Finally, the constant $M$ can be determined from the amplitude of the CMB anisotropies

$$ \frac{M^4}{M_{\text{Pl}}^4} = 90\pi^2 \frac{M_{\text{Pl}}^2}{\mu^2} \text{csch}^6\left(\frac{x_*}{2}\right) \sinh x_* \tanh x_* \frac{Q_{\text{rms-PS}}^2}{T^2}. \qquad (5.224) $$

In the $\mu/M_{Pl} \ll 1$ limit, one obtains

$$ \frac{M^4}{M_{\text{Pl}}^4} \approx \frac{720\pi^2}{\Delta N_*^2} \frac{\mu^2}{M_{\text{Pl}}^2} \frac{Q_{\text{rms-PS}}^2}{T^2}. \qquad (5.225) $$

Typically, for $\mu/M_{Pl} \approx 10^{-2}$, one has $M/M_{Pl} \approx 10^{-4}$.

The reheating consistent slow-roll predictions for MHI have been represented in Fig. 148. As expected, for small values of $\mu/M_{Pl}$, the predicted amount of gravitational waves is extremely small and the deviation from scale invariance almost does not depend on $\mu$.

## 5.16. Radion Gauge Inflation (RGI)

This model was studied in Ref. [478]. It is an extension of the gauge inflation scenario in which the radius modulus field around which the Wilson loop is wrapped assists inflation as it shrinks [311]. Assuming that the radion field value is such that the potential energy is minimal, for each value of the inflaton field $\phi$, one can derive an effective potential

$$ V(\phi) = M^4 \frac{(\phi/M_{\text{Pl}})^2}{\alpha + (\phi/M_{\text{Pl}})^2}, \qquad (5.226) $$

where $\alpha$ is a dimensionless positive parameter. In the context of Ref. [478], the model is natural for $\alpha < 1$ but larger than unity values are not forbidden. The same potential has been obtained in Ref. [479] in the context of S-dual superstring models. In that case, $\alpha$ represents a typical vev for the inflaton, in Planck units. Defining $x = \phi/M_{Pl}$, the first three slow-roll parameters read

$$ \epsilon_1 = \frac{2\alpha^2}{x^2 (\alpha + x^2)^2}, \quad \epsilon_2 = 4\alpha \frac{\alpha + 3x^2}{x^2 (\alpha + x^2)^2}, \quad \epsilon_3 = 4\alpha \frac{\alpha^2 + 3\alpha x^2 + 6x^4}{x^2 (\alpha + x^2)^2 (\alpha + 3x^2)}. \qquad (5.227) $$

The potential, its logarithm, and the Hubble flow functions are represented in Fig. 31.

The slow-roll trajectory can be integrated analytically from Eq. (3.11) to obtain

$$ N - N_{\text{end}} = \frac{x_{\text{end}}^2}{4} + \frac{x_{\text{end}}^4}{8\alpha} - \frac{x^2}{4} - \frac{x^4}{8\alpha}. \qquad (5.228) $$
Moreover, it can be inverted explicitly to give the field values in terms of the number of e-folds as

$$x = \sqrt{-\alpha + \sqrt{-8\alpha(N - N_{\text{end}}) + (\alpha + x_{\text{end}}^2)^2}}. \qquad (5.229)$$

The end of inflation naturally occurs for $\epsilon_1 = 1$, i.e., from Eq. (5.227), at the field value $x_{\text{end}}$ given by

$$x_{\text{end}} = \frac{-\sqrt{6}\alpha + [9\alpha + \sqrt{3\alpha^2(2\alpha + 27)}]^{2/3}}{162^{1/6}[9\alpha + \sqrt{3\alpha^2(2\alpha + 27)}]^{1/3}}. \qquad (5.230)$$

As for the MHI models, one should pay attention that

$$\epsilon_2 - \epsilon_1 = 2\alpha \frac{\alpha + 6x^2}{x^2(\alpha + x^2)^2} > 0, \qquad (5.231)$$

for any positive values of $\alpha$. As a result, slow-roll violation, i.e. $\epsilon_2 > 1$, occurs in RGI before inflation ends. However, since the first Hubble flow function is monotonic, this is not very problematic as it happens only during a negligible number of e-folds and only around $N_{\text{end}}$. The slow-roll observable predictions therefore remain accurate.

As before, the observable field value $x_*$ is obtained by solving Eq. (3.48) for a given reheating model and allows the determination of the parameter $M$ from the amplitude of the CMB anisotropies. One gets

$$\frac{M^4}{M_{\text{Pl}}^4} = \frac{2880\pi^2\alpha^2}{x_*^4(\alpha + x_*^2)} \frac{Q_{\text{rms-PS}}^2}{T^2}. \qquad (5.232)$$

The reheating consistent slow-roll predictions for these models are displayed in Fig. 149. Large values of $\alpha$ give back the same predictions as the large field models with $p=2$ (see Section 5.2) having $\epsilon_{2*} = 2\epsilon_{1*}$.

## 5.17. MSSM Inflation (MSSMI)

### 5.17.1. Theoretical justifications

The Minimal Supersymmetric Standard Model (MSSM) is an extension of the Standard Model of particle physics. Its Lagrangian is characterized by the following super potential

$$W_{\text{MSSM}} = \lambda_u^{ij} Q_i \cdot H_u U_j^c + \lambda_d^{ij} Q_i \cdot H_d D_j^c + \lambda_e^{ij} L_i \cdot H_d E_j^c + \mu H_u \cdot H_d. \qquad (5.233)$$

The quantity $Q_i$ denotes a doublet of left handed quarks superfields where $i$ is a family index. In practice this means that

$$Q_1 = \begin{pmatrix} U \\ D \end{pmatrix}, \quad Q_2 = \begin{pmatrix} C \\ S \end{pmatrix}, \quad Q_3 = \begin{pmatrix} T \\ B \end{pmatrix}. \qquad (5.234)$$

where the components of the doublets are superfields. For instance, the scalar part of $U$ is the $\bar{u}$ squark and its fermionic part is the ordinary $u$ quark. Of course, there is also a color index $a = 1, 2, 3$ and, in fact, one should write the corresponding doublet as $Q_{1a}$. Moreover, one can also introduce a third SU(2)$_L$ index $\alpha = 1, 2$ and write $Q_{1a\alpha}$ with, for instance, $Q_{1a1} = U$ and $Q_{1a2} = D$. On the other hand, the quantities $U_j^c$ and $D_j^c$ denote the right handed superfields where $j$ is the family index (and the color index has been ignored in order to simplify the notation): for instance, $U_2^c$ means the right handed charm quark superfield which is a singlet under SU(2)$_L$.

In the same fashion, $L_i$ denotes a doublet of left handed lepton superfields

$$L_1 = \begin{pmatrix} N_e \\ E_e \end{pmatrix}, \quad Q_2 = \begin{pmatrix} N_\mu \\ E_\mu \end{pmatrix}, \quad Q_3 = \begin{pmatrix} N_\tau \\ E_\tau \end{pmatrix}. \qquad (5.235)$$

where, for instance, $N_e$ denotes the electronic neutrino superfield (the scalar part being the neutralino and the fermionic part the electronic neutrino itself) while $E_e$ denotes the electron superfield. On the other hand, the quantities $E_j^c$ denote the right handed superfields that are singlet under SU(2)$_L$ (for instance, $E_2^c$ is the right handed muonic superfield). In the superpotential (5.233), there are two terms involving the quarks and only one involving the leptons because, as well-known, there is no right handed neutrinos in the standard model.

The last term in Eq. (5.233) describes the Higgs sector with two Higgs doublet $H_u$ and $H_d$. The quantity $\mu$ is a new dimensionful (of dimension one) parameter of the model. The dot indicates an SU(2) invariant product. Finally, $\lambda_u, \lambda_d, \lambda_e$ are the $3 \times 3$ Yukawa matrices.

From the superpotential (5.233), one can determine the scalar potential of the theory by means of the usual supersymmetric machinery. As is well-known, the scalar potential is made of two pieces, the $F$-term part and the $D$-term part. Clearly, given the number of fields in the theory, the scalar potential is a complicated object. For inflation, we are especially interested in the flat directions of this potential. A flat direction is a direction such that the $F$ and $D$-terms vanish, that is to say such that $V_F = 0$, $V_D = 0$ and, therefore, $V \equiv V_F + V_D = 0$. It was shown that the MSSM scalar potential contains nearly 300 gauge invariant flat directions [86,480,481]. Finding these directions is a non-trivial task and we now very briefly explain how this can be done. Usually, it consists in putting all the fields to zero except a few ones, these few ones being carefully chosen such that cancellations occur in such a way that the potential exactly vanishes. We now illustrate this method on a particular case. Let us first recall that the general formula giving the $D$-term potential is

$$V_D = \frac{1}{2} \sum_a g_a^2 D^a D^a, \qquad (5.236)$$

where $D^a = \phi^\dagger T^a \phi$, $T^a$ being the generator of the group and $\phi$ denoting a generic field (of course, the index $a$ should not be confused with the color index discussed above). For the standard model, we have the group SU(2)$_L \times U(1)_Y$ and, therefore, the explicit expression of the $D$-term reads

$$V_D = \frac{g^2}{2} (D_1^2 + D_2^2 + D_3^2) + \frac{g_Y}{2} D_Y^2, \qquad (5.237)$$
g and $g_Y$ being the coupling constants of the two groups. For the SU(2) group, the generators $T^a$ are nothing but the Pauli matrices and, therefore, $T^a = \sigma^a/2$. Following Refs. [480,482], let us consider a situation where all the fields in the MSSM are assumed to have a vanishing vev except $L_i$ and $E_i^c$ where we remind that i and j are family indices. If we write $L_i^\uparrow$ and $L_i^\downarrow$ as respectively the upper and lower component of the doublet $L_i$, then one has (i.e. we put $\phi = L_i$ in the general formula expressing $D^a$)

$$D_1 = \frac{1}{2} \sum_{i=1}^{3} (L_i^\dagger * L_i + L_i * L_i^\dagger), \quad D_2 = \frac{i}{2} \sum_{i=1}^{3} (L_i^\dagger * L_i - L_i * L_i^\dagger), \qquad (5.238)$$

$$D_3 = \frac{1}{2} \sum_{i=1}^{3} (|L_i^\uparrow|^2 - |L_i^\downarrow|^2). \qquad (5.239)$$

The quantity $E^c$ being a SU(2) singlet does not participate to the above expression. On the other hand, the contribution from the U(1) group reads

$$D_Y = \frac{1}{2} \sum_{i=1}^{3} (2|e_i|^2 - |L_i^\uparrow|^2 - |L_i^\downarrow|^2), \qquad (5.240)$$

where $e_i$ denotes the scalar field of the $E_i^c$ supersymmetric multiplet. We see that, if we take

$$L_i = \begin{pmatrix} \phi \\ 0 \end{pmatrix}, \quad L_j = \begin{pmatrix} 0 \\ \phi \end{pmatrix}, \quad e_k = \phi, \qquad (5.241)$$

then we have $V_D = 0$.

The next step consists in calculating the F-term for the choice made in Eq. (5.241). It is easy to check that $V_F = 0$. Therefore, we have identified a flat direction. It is denoted $L_i L_j e_k$ or LLe to recall that all family combination are possible. This direction is represented by a “composite operator $X_m$” formed by the product of the superfields making up the flat direction. In our case $X_3 = L_i L_j e_k = \phi^3$ and $m = 3$ since we have three operators participating to the definition of $X_3$. This direction has been proposed in Ref. [483] as a possible candidate for the inflaton field. Let us also remark that another choice put forward in that reference was udd.

We have just seen how to identify flat directions in the MSSM potential. However, this flatness is usually spoiled by the presence of higher order non-renormalizable operators appearing in the MSSM (viewed here as a low energy effective field) and by supersymmetry breaking [86,480,481]. Higher order operators are described by the following superpotential

$$W = \frac{\lambda_n}{n} \frac{X_m^k}{M_{\mathrm{Pl}}^{mk-3}}, \qquad (5.242)$$

where $\lambda_n$ is a coupling constant, $n \equiv mk$ and $k = 1$ or $k = 2$ depending on whether the flat direction is even or odd under R-parity. Recall that $Q$, $L$, $U^c$, $D^c$ and $E^c$ have R-parity $-1$ and $H_u$, $H_d$ have R-parity $+1$. It follows that LLe (for instance) has odd R-parity and, therefore, that $k = 2$. For the directions LLe (this is also true for uud), this means that

$$n \equiv mk = 6. \qquad (5.243)$$

The above superpotential (5.242) will produce a term $|\partial W/\partial \phi|^2 \propto \phi^{2(km-1)}$ in the scalar potential. Then, we have the contributions originating from supersymmetry breaking. They can be easily calculated if, for instance, we assume that we have an independent hidden sector where supersymmetry is broken and that this breaking is mediated by gravity only. This gives two types of soft terms, one proportional to $\phi^2$ and another, the so-called “A-term”, proportional to $(\phi\partial W/\partial\phi + \mathrm{cc})$ that is to say, given Eq. (5.242), proportional to $\phi^{mk}$.

More generally, if one starts from a flat direction with a given $n$, then the superpotential has the form $W = \lambda_n/n\Phi^n M_{\mathrm{Pl}}^{3-n}$, where $\Phi = \phi e^{i\theta}$ is the superfield which contains the flat direction. Then, the scalar potential takes the form

$$V(\phi) = \frac{1}{2} m_{\phi}^2 \phi^2 + A \cos(n\theta + \theta_0) \frac{\lambda_n}{n} \frac{\phi^n}{M_{\mathrm{Pl}}^{n-3}} + \lambda_n^2 \frac{\phi^{2(n-1)}}{M_{\mathrm{Pl}}^{2(n-3)}}, \qquad (5.244)$$

where the second term involves the angular part of the superfield via a term $\cos(n\theta+\theta_0)$, which in practice is fixed at $-1$ to maximize its contribution. As explained below, the fact that the second term appears with a negative coefficient plays a crucial role in making this scenario a credible inflationary one.

Together with the global minimum at $\phi = 0$, under the condition $A^2 \ge 8(n-1)m_{\phi}^2$, the potential has a secondary minimum at $\phi_0 \simeq (m_{\phi} M_{\mathrm{Pl}}^{n-3})^{1/(n-2)}$. If $A^2 \gg 8(n-1)m_{\phi}^2$, this secondary minimum becomes the deepest one and thus the true one. The curvature of the potential at this minimum is of the order $m_{\phi}^2$. If inflation occurs there, one gets $H \simeq m_{\phi}(m_{\phi}/M_{\mathrm{Pl}})^{1/(n-2)}$, which is much smaller than the potential curvature for $m_{\phi} \ll M_{\mathrm{Pl}}$. This implies that the potential is too steep for quantum effects during inflaton to kick $\phi$ out of the false minimum. Such a situation is similar to the old inflationary scenario. However, this barrier disappears if one saturates the previous inequality and takes

$$A^2 = 8(n-1)m_{\phi}^2. \qquad (5.245)$$

In that case, the potential has a flat inflection point at $\phi_0$ and inflation can proceed between this plateau and $\phi = 0$. This is the case we study in this section. This model (and its generalizations) has also been studied in Refs. [484–496]. Its generalizations will be investigated in more details in Sections 6.6 and 6.7. Let also us notice that when $n = 3$, the same potential appears in Refs. [497,498] as “Generalized Chaotic Inflation”, and later in Refs. [499–501] as “Punctuated Inflation”. In these references, it is shown that slow-roll inflation is briefly interrupted when the inflaton crosses the flat inflection point and this can produce step-like features in the primordial power spectra. These effects are outside the scope of the following slow-roll analysis as we will be dealing with the last slow-roll inflationary stage within this scenario.

### 5.17.2. Slow-roll analysis

We now turn to the slow-roll analysis of MSSM inflation. As discussed before, we assume that the inflaton is the flat direction LLe or uud. This implies that $n = 6$ in Eq. (5.244). Then, rewriting the potential (5.244) in a more convenient fashion, one arrives at

$$V(\phi) = M^4 \left[ \left( \frac{\phi}{\phi_0} \right)^2 - \frac{2}{3} \left( \frac{\phi}{\phi_0} \right)^6 + \frac{1}{5} \left( \frac{\phi}{\phi_0} \right)^{10} \right], \qquad (5.246)$$
<figure><img src="image_35.png" /><figcaption>Fig. 32. MSSM Inflation (MSSMI). Top left panel: MSSM Inflation potential Eq. (5.246) as a function of $\phi/\phi_0$. Top right panel: logarithm of the potential. Bottom left panel: slow-roll parameter $\epsilon_1$ scaled by $\phi_0^2/M_n^2$. Bottom right panel: slow-roll parameters $\epsilon_2$ (solid line) and $\epsilon_3$ (dotted line) scaled by $\phi_0^2/M_n^2$.</figcaption></figure>

where we have defined new parameters according to

$$M^8 = \frac{M_{Pl}^3 m_{\phi}^5}{4\sqrt{10}\lambda_6}, \qquad \phi_0^8 = \frac{M_{Pl}^6 m_{\phi}^2}{10\lambda_6^2}. \qquad (5.247)$$

These definitions and the value of the coefficients ensure that $\phi_0$ is the location of a flat inflection point. Since $m_{\phi}^2\phi^2$ is a soft SUSY breaking term, we typically expect that $m_{\phi} \simeq 1$ TeV and this is the reason why, in what follows, typical values of the field are taken to be

$$\phi_0 \approx 10^{14} \text{ GeV}, \qquad (5.248)$$

in agreement with the second of Eqs. (5.247) (the coupling constant $\lambda_6$ is taken to be of order one). An interesting feature of this model is that it provides inflation at sub-Planckian vev and at low scale $V \simeq (10^9 \text{ GeV})^4$. As noticed in Ref. [483], higher values than $n=6$ would produce too small amplitude for the scalar perturbations. This is why the model is commonly studied with $n=6$ (with $n=3$, this is RIPI, see Section 5.18).

The potential in Eq. (5.246) is displayed in Fig. 32, together with its logarithm. It is an increasing function of the field, the derivative of which vanishes at $\phi=0$ and at its second inflection point $\phi=\phi_0$, the position of the first inflection point being given by $\phi_{V''=0}^2 = \phi_0/\sqrt{3}$. Inflation proceeds in the region $\phi \in [0, \phi_0]$, in the direction specified by the arrow in Fig. 32.

Defining the dimensionless quantity $x$ by

$$x \equiv \frac{\phi}{\phi_0}, \qquad (5.249)$$

the first three Hubble flow functions in the slow-roll approximation are given by

$$\epsilon_1 = 450 \frac{M_{Pl}^2}{\phi_0^2} \frac{(x^4 - 1)^4}{x^2 (3x^8 - 10x^4 + 15)^2}, \qquad \epsilon_2 = 60 \frac{M_{Pl}^2}{\phi_0^2} \frac{3x^{16} - 58x^8 + 40x^4 + 15}{x^2 (3x^8 - 10x^4 + 15)^2}, \qquad (5.250)$$

and

$$\epsilon_3 = \frac{M_{Pl}^2}{\phi_0^2} \frac{60}{x^2} \left( -225 + 1575x^4 - 3165x^8 + 395x^{12} + 2605x^{16} - 1275x^{20} + 81x^{24} + 9x^{28} \right) \times (3x^8 - 10x^4 + 15)^{-2} \times (-15 - 55x^4 + 3x^8 + 3x^{12})^{-1}. \qquad (5.251)$$

These two slow-roll parameters diverge when the field vev goes to 0, and vanish when the field vev goes to infinity. The first slow roll parameter $\epsilon_1$ first decreases, vanishes at the flat inflection point where $\epsilon_2$ vanishes too, then increases to reach a local maximum where $\epsilon_2$ vanishes again, and eventually decreases again, to vanish at infinity where $\epsilon_2$ also goes to zero. Denoting by $x_{\epsilon_2=0}^+$ the position of the second extremum, one has

$$x_{\epsilon_2=0}^{+} = \left(\frac{1}{3}\right)^{1/4} \left[ 2^{4/3} \left(i\sqrt{685} - 1\right)^{1/3} + 14 \times 2^{2/3} \left(i\sqrt{685} - 1\right)^{-1/3} - 1 \right]^{1/4} \approx 1.41022. \qquad (5.252)$$
<figure><img src="image_36.png" /><figcaption>Fig. 33. Location of the slow-roll violation induced end of inflation $x_{\text{end}} = \phi_{\text{end}}/\phi$ for the MSSM inflation models, as a function of $\phi_0/M_{\text{Pl}}$. The blue solid curve represents a numerical solution of $\epsilon_1 = 1$, while the red dotted curve corresponds to the approximated analytic solution Eq. (5.254). For physical values $\phi_0 \simeq 10^{-4} M_{\text{Pl}}$, the agreement is obviously excellent.</figcaption></figure>

In between the two local extrema of $\epsilon_1$, the second slow-roll parameter $\epsilon_2$ is negative whereas it is positive elsewhere. The value of $\epsilon_1$ at its local maximum is given by

$$ \epsilon_1^{\max} = \epsilon_1(x_{\epsilon_2=0}^+) \approx 34.459 \frac{M_{\text{Pl}}^2}{\phi_0^2}. \qquad (5.253) $$

With the typical above-mentioned value for $\phi_0 \simeq 10^{14} \text{GeV}$, one has $M_{\text{Pl}}^2/\phi_0^2 \simeq 10^8$ and $\epsilon_1^{\max} > 1$. This means that if inflation proceeds for vev's larger than that of the flat inflection point, it can naturally stop by slow-roll violation. However, if this happens, inflation proceeds at $x \gg 1$ and the potential is effectively very close to a large field model one (LFI, see Section 5.2) with $p = 10$.

For this reason, we will be focused to the case in which inflation occurs for vev's smaller than that of the flat inflection point. In this case, the value of $x_{\text{end}}$ at which inflation stops by slow-roll violation must be determined numerically. In the limit $\phi_0/M_{\text{Pl}} \ll 1$ however, one has $x_{\text{end}} \simeq 1$ and an approximate analytic formula can be derived

$$ x_{\text{end}} \approx 1 - \frac{1}{2^{3/4} \sqrt{15}} \sqrt{\frac{\phi_0}{M_{\text{Pl}}}}. \qquad (5.254) $$

A comparison between this expression and the numerical solution of $\epsilon_1 = 1$ is displayed in Fig. 33. For physical values $\phi_0 \simeq 10^{-4} M_{\text{Pl}}$, the agreement is excellent.

Let us now turn to the slow-roll trajectory. It can be integrated from Eq. (3.11) and leads to

$$ N_{\text{end}} - N = \left( \frac{\phi_0}{M_{\text{Pl}}} \right)^2 \left\{ \frac{x^2 - x_{\text{end}}^2}{20} + \frac{1}{15} \left( \frac{x_{\text{end}}^2}{x_{\text{end}}^4 - 1} - \frac{x^2}{x^4 - 1} \right) - \frac{2}{15} \left[ \operatorname{arctanh}(x_{\text{end}}^2) - \operatorname{arctanh}(x^2) \right] \right\} $$

where $N_{\text{end}}$ is the number of e-folds at the end of inflation and $N$ is the number of e-folds at some point when the scaled field vev is $x$. A few remarks are in order. Firstly, when $x \simeq 1$, the second term of the previous expression dominates, and one has $N_{\text{end}} - N \simeq 1/15 (\phi_0/M_{\text{Pl}})^2 [1/(x_{\text{end}}^4 - 1) - 1/(x^4 - 1)]$, which can be inverted and gives

$$ x \approx 1 - \frac{1}{4} \left[ 2^{-5/4} \sqrt{15} \sqrt{\frac{M_{\text{Pl}}}{\phi_0}} + 15 \frac{M_{\text{Pl}}^2}{\phi_0^2} (N_{\text{end}} - N) \right]^{-1}. \qquad (5.255) $$

Secondly, one could wonder if a sufficient number of e-folds can be realized in the regime studied here. When $x \to 1$, the corresponding number of e-folds diverges, but in practice, the inflationary dynamics close to the flat inflection point is governed by the quantum diffusion and the classical equation of motion cannot be trusted in this domain.

If one introduces the ratio $\eta$ between the quantum kicks amplitude $H/(2\pi)$ and the classical drift $M_{\text{Pl}}^2 V_{\phi}/V$, when $x \simeq 1$, one has

$$ \eta \approx \frac{1}{90\sqrt{30\pi}} M^2 \phi_0 M_{\text{Pl}}^{-3} (x-1)^{-2} \approx \frac{4\sqrt{10}}{\pi\sqrt{3}} M^2 M_{\text{Pl}} \phi_0^{-3} (N_{\text{end}} - N)^2, \qquad (5.256) $$

where the last equality comes from the approximate trajectory. In order to estimate the value of $\eta$, one needs the value of $M$ which is fixed by the amplitude of the CMB anisotropies. With $x_*$ the observable field value associated with $\Delta N_* = N_{\text{end}} - N_*$, one gets

$$ \left(\frac{M}{M_{\text{Pl}}}\right)^4 = 2880\pi^2 \frac{M_{\text{Pl}}^2}{\phi_0^2} \frac{(1-x_*^4)^4}{x_*^4(1-\frac{2}{3}x_*^4+\frac{1}{5}x_*^8)^3} \frac{Q_{\text{rms-PS}}^2}{T^2}. \qquad (5.257) $$
In the $x_* \approx 1$ approximation, this gives

$$ \frac{M^4}{M_{\text{Pl}}^4} \approx \frac{3}{8} \pi^2 \frac{Q^2 \text{rms-PS}}{T^2} \frac{\phi_0^6}{M_{\text{Pl}}^6 (N_{\text{end}} - N_*)^4}, \qquad (5.258) $$

and thus

$$ \eta \approx \sqrt{20 \frac{Q^2 \text{rms-PS}}{T^2}} \left( \frac{N_{\text{end}} - N}{\Delta N_*} \right)^2. \qquad (5.259) $$

It is quite remarkable that this formula does not depend on $\phi_0$ anymore but only on the ratio $(N_{\text{end}} - N)/\Delta N_*$. From $Q_{\text{rms-PS}}/T \approx 6 \times 10^{-6}$, one has $N_{\text{end}} - N_{\text{min}} \approx 10^4$ in the classical regime [483]. For $\phi_0 \approx 10^{14}$ GeV, one obtains $M \approx 10^8$ GeV, in agreement with what was announced earlier.

Finally, it can be interesting to write down the approximated slow-roll parameters at Hubble crossing and in the limit $\phi_0/M_{\text{Pl}} \ll 1$. One obtains

$$ \epsilon_{1*} \approx \left( \frac{\phi_0}{M_{\text{Pl}}} \right)^6 \frac{1}{7200 \Delta N_*^4}, \quad \epsilon_{2*} \approx \frac{4}{\Delta N_*}, \quad \epsilon_{3*} \approx \frac{1}{\Delta N_*}, \qquad (5.260) $$

hence

$$ r \approx \left( \frac{\phi_0}{M_{\text{Pl}}} \right)^6 \frac{1}{450 \Delta N_*^4}, \quad n_S \approx 1 - \frac{4}{\Delta N_*}, \quad \alpha_S \approx -\frac{4}{\Delta N_*^2}. \qquad (5.261) $$

They are similar with the typical predictions of the RIPI models [see Eq. (5.277)].

The reheating consistent slow-roll predictions of the MSSMI models are displayed in Fig. 150. The reheating equation of state parameter $\bar{w}_{\text{reh}}$ has been taken to 0 since the potential is quadratic in the vicinity of its minimum. One can check that, in the limit $\phi_0/M_{\text{Pl}} \ll 1$, the first slow-roll parameter is indeed extremely small, while the second slow-roll parameter does not depend much on $\phi_0$. Remembering that $\phi_0/M_{\text{Pl}} \approx 10^{-4}$, one can see that these models seem to be disfavored by the data since they predict a too large deviation from scale invariance. In order to better reproduce the constraints on the spectral index, these models should be such that $\phi_0/M_{\text{Pl}} \gg 1$, for which they become similar to large field models (LFI, see Section 5.2). This can be seen from the previous formulas in the limit $x \gg 1$. Unfortunately, such values for $\phi_0$ are not compatible with the MSSM. Finally, comparing Fig. 150 with Fig. 151, one can see that the general features of MSSMI are very similar to the RIPI ones, and that the conclusions drawn here are rather robust against a change in $n$ appearing in Eq. (5.244).

## 5.18. Renormalizable Inflection Point Inflation (RIPI)

### 5.18.1. Theoretical justifications

In Section 5.17 inflation is implemented within the Minimal Supersymmetric Standard Model (MSSM) around a flat inflection point. Here, we consider a similar model but with $n=3$ instead of $n=6$. Such a scenario can emerge in the following situation, see Refs. [502,503]. Let us consider the MSSM with three additional superfields $N_i$ representing three right-handed neutrinos. These fields are singlet under the standard model gauge group but this one can be extended to $SU(3)_c \times SU(2)_L \times U(1)_Y \times U(1)_{B-L}$ and the $N_i$ are assumed to be charged under the extra $U(1)_{B-L}$. Then, we postulate the following superpotential

$$ W = W_{\text{MSSM}} + hN H_u L, \qquad (5.262) $$

where $h \lesssim 10^{-12}$ in order to explain the neutrino mass, $m_\nu \approx \mathcal{O}(0.1)$ eV. It follows that $NH_uL$ is a D-flat direction of the potential and we parametrize this direction by $\phi$. As a consequence, if one now calculates the corresponding potential, one finds that

$$ V = \frac{1}{2} m_{\phi}^2 \phi^2 - \frac{Ah}{6\sqrt{3}} \phi^3 + \frac{h^2}{12} \phi^4, \qquad (5.263) $$

where, as usual, we have included the soft supersymmetry breaking terms (since $W \propto \phi^3$, the A-term, proportional to $\phi \partial W / \partial \phi$ is, this time, cubic) and have minimized $V$ along the angular direction. If $A$ is chosen such that $A = 4m_\phi$, then we have a flat inflection point at $\phi_0 = \sqrt{3}m_\phi/h$. A discussion on the fine-tuning required to get a flat inflection point can be found in Section 6.7,

### 5.18.2. Slow-roll analysis

We now turn to the slow-roll analysis of the potential given in Eq. (5.263). For this purpose, it is more convenient to re-write it as

$$ V(\phi) = M^4 \left[ \left( \frac{\phi}{\phi_0} \right)^2 - \frac{4}{3} \left( \frac{\phi}{\phi_0} \right)^3 + \frac{1}{2} \left( \frac{\phi}{\phi_0} \right)^4 \right], \qquad (5.264) $$

where we have defined the quantities $M$ and $\phi_0$ by

$$ M^4 = \frac{1}{2} m_{\phi}^2 \phi_0^2, \quad \phi_0 = \sqrt{3} \frac{m_{\phi}}{h}. \qquad (5.265) $$

Relevant values of $m_\phi$ range from 100 GeV to 10 TeV and $h \approx 10^{-12}$. This means that [502,503]

$$ \phi_0 \approx 10^{14} \text{ GeV}, \qquad (5.266) $$

a value that turns out to be similar to the one considered in the MSSMI case (see Section 5.17).

Let us now define the quantity $x$ by the following expression

$$ x \equiv \frac{\phi}{\phi_0}. \qquad (5.267) $$
<figure><img src="image_37.png" /><figcaption>Fig. 34. Renormalizable Inflection Point Inflation (RIPI). Top left panel: renormalizable inflection point inflation potential as a function of $\phi/\phi_0$. Top right panel: logarithm of the potential, the required flatness of the potential close to its inflection point becomes obvious on this plot. Bottom left panel: slow-roll parameter $\epsilon_1$ normalized by $M_{Pl}^2/\phi_0^2$. The shaded area indicates the region in which $\epsilon_1 > 1$ and thus where inflation stops (this has to be rescaled for $\phi_0 \neq M_{Pl}$). Bottom right panel: slow-roll parameters $\epsilon_2$ (solid line) and $\epsilon_3$ (dotted line), normalized by $M_{Pl}^2/\phi_0^2$.</figcaption></figure>

The potential is an increasing function of the field $\text{vev}$, hence inflation proceeds from the right to the left. It has two inflection points $x_{V''=0}^{\pm}$, given by

$$x_{V''=0}^{\mp} = \frac{1}{3} \quad \text{and} \quad x_{V''=0}^{+} = 1, \qquad (5.268)$$

the second one being a flat inflection point [i.e. $V'(x_{V''=0}^{+}) = 0$], close to which inflation takes place. This potential is displayed in Fig. 34, together with its logarithm.

Let us now turn to the slow-roll parameters. The first three Hubble flow functions in the slow-roll approximation are given by

$$\epsilon_1 = 72 \frac{M_{Pl}^2}{\phi_0^2} \frac{(x-1)^4}{(3x^3 - 8x^2 + 6x)^2}, \quad \epsilon_2 = 24 \frac{M_{Pl}^2}{\phi_0^2} (x-1) \frac{3x^3 - 9x^2 + 10x - 6}{(3x^3 - 8x^2 + 6x)^2}, \qquad (5.269)$$

and

$$\epsilon_3 = 24 \frac{M_{Pl}^2}{\phi_0^2} (x-1) (36 - 144x + 246x^2 - 236x^3 + 144x^4 - 54x^5 + 9x^6) \times (6x - 8x^2 + 3x^3)^{-2} (10x - 9x^2 + 3x^3 - 6)^{-1}.$$

Both $\epsilon_1(x)$ and $\epsilon_2(x)$ diverge when the field $\text{vev}$ goes to 0, and vanish when the field $\text{vev}$ goes to infinity. The first slow-roll parameter $\epsilon_1$ first decreases, vanishes at $x_{V''=0}^{+}$ where $\epsilon_2$ vanishes too, $x_{\epsilon_2=0}^{-} = x_{V''=0}^{+}$, then increases to reach a local maximum at $x_{\epsilon_2=0}^{+}$ where $\epsilon_2$ vanishes again, and eventually decreases again. The value of $x_{\epsilon_2=0}^{+}$ is given by

$$x_{\epsilon_2=0}^{+} = 1 - \frac{1}{3(9 + \sqrt{82})^{1/3}} + \frac{1}{3}(9 + \sqrt{82})^{1/3} \approx 1.75. \qquad (5.270)$$

In between these two local extrema of $\epsilon_1$, the second slow roll parameter $\epsilon_2$ is negative, and it is positive elsewhere. The value of $\epsilon_1$ at its local maximum, $\epsilon_1^{\max}$, is given by

$$\epsilon_1^{\max} \approx 5.2753 \frac{M_{Pl}^2}{\phi_0^2}. \qquad (5.271)$$

Therefore, if $\phi_0/M_{Pl} \lesssim 2.3$, inflation can stop by slow-roll violation in the region corresponding to $\text{vev}$'s larger than that of the second inflection point $x_{\epsilon_2=0}^{+}$. Remembering that typically $\phi_0 \approx 10^{14}$ GeV $\approx 4 \times 10^{-5} M_{Pl}$, this condition is easily satisfied. In that case, an expression for the $\text{vev}$ at
which inflation ends, $x_{\epsilon_1=1}^+$, can be obtained but is does not add much to the discussion since for reasonable values of $\phi_0$, it is extremely far from the flat inflection point (e.g. for $\phi_0/M_{\text{Pl}} = 10^{-4}$, one has $x_{\epsilon_1=1}^+ \simeq 28285$). Since the potential is introduced in order to study inflation in the vicinity of the flat inflection point, it should be studied in the other regime, as it is the case for MSSM inflation (see Section 5.17), i.e. when inflation takes place between $x=0$ and the second inflection point $x_{\epsilon_2=0}^-$. In that situation, it ends at

$$x_{\text{end}} = x_{\epsilon_1=1}^{-} = \frac{1}{9} \frac{M_{\text{Pl}}}{\phi_0} \left[ \begin{aligned} & 6\sqrt{2} + 8 \frac{\phi_0}{M_{\text{Pl}}} + 2 \left( -36 + 6\sqrt{2} \frac{\phi_0}{M_{\text{Pl}}} - 5 \frac{\phi_0^2}{M_{\text{Pl}}^2} \right) \\ & \times \left( 216 \frac{\phi_0}{M_{\text{Pl}}} - 99\sqrt{2} \frac{\phi_0^2}{M_{\text{Pl}}^2} + 136 \frac{\phi_0^3}{M_{\text{Pl}}^3} - 432\sqrt{2} \right. \\ & \left. \quad + 27\sqrt{2} \sqrt{-72\sqrt{2} \frac{\phi_0^3}{M_{\text{Pl}}^3} + 33 \frac{\phi_0^4}{M_{\text{Pl}}^4} - 16\sqrt{2} \frac{\phi_0^5}{M_{\text{Pl}}^5} + 12 \frac{\phi_0^6}{M_{\text{Pl}}^6}} \right)^{-1/3} \\ & - \left( 216 \frac{\phi_0}{M_{\text{Pl}}} - 99\sqrt{2} \frac{\phi_0^2}{M_{\text{Pl}}^2} + 136 \frac{\phi_0^3}{M_{\text{Pl}}^3} - 432\sqrt{2} \right) \\ & + 27\sqrt{2} \sqrt{-72\sqrt{2} \frac{\phi_0^3}{M_{\text{Pl}}^3} + 33 \frac{\phi_0^4}{M_{\text{Pl}}^4} - 16\sqrt{2} \frac{\phi_0^5}{M_{\text{Pl}}^5} + 12 \frac{\phi_0^6}{M_{\text{Pl}}^6}} \right)^{1/3} \end{aligned} \right]. \quad (5.272)$$

For $\phi_0/M_{\text{Pl}} \ll 1$, one can numerically check that this expression is very close to the flat inflection point location $x_{\epsilon_2=0}^-$, namely

$$x_{\text{end}} \approx 1 - \sqrt{6\sqrt{2} \frac{\phi_0}{M_{\text{Pl}}}}. \quad (5.273)$$

The whole inflationary stage therefore proceeds in the vicinity of this point.

The slow-roll trajectory is obtained from Eq. (3.11) and reads

$$N_{\text{end}} - N = \frac{\phi_0^2}{M_{\text{Pl}}^2} \left[ \begin{aligned} & -\frac{x}{6} + \frac{x^2}{8} + \frac{1}{12(1-x)} - \frac{\ln(1-x)}{12} \\ & + \frac{x_{\text{end}}}{6} - \frac{x_{\text{end}}^2}{8} - \frac{1}{12(1-x_{\text{end}})} + \frac{\ln(1-x_{\text{end}})}{12} \end{aligned} \right]. \quad (5.274)$$

Several remarks are in order. Firstly, from this expression, one can see that the number of e-folds diverges when the field approaches the inflection point of the potential. This means that this point is never crossed and that, if inflation proceeds for vev's larger than that of this inflection point, then the field approaches it asymptotically but never actually reaches it. However, an exact numerical integration of the equations of motion reveals that, if the field approaches the inflection point in such a way that the slow-roll conditions are not satisfied, then it can cross it. This is typically the case if its speed is large enough. On the other hand, the field dynamics at the exact location of the inflection point is dominated by quantum diffusion, and a more careful study must be carried out to describe what exactly happens there. Following the considerations of Section 5.17, we focus on the inflationary regime only in the region where the vev of $\phi$ is smaller than that of the flat inflection and where deviations from slow-roll and quantum diffusion plays a negligible role. Since for $\phi_0/M_{\text{Pl}} \ll 1$ inflation takes place relatively close to the inflection point, the two last terms of Eq. (5.274) dominate over the two first ones. In this limit, the trajectory can be inverted to get

$$x_* \approx 1 - W_0^{-1} \left\{ \exp \left[ 12 \left( \frac{M_{\text{Pl}}}{\phi_0} \right)^2 \Delta N_* + \frac{1}{1-x_{\text{end}}} - \ln(1-x_{\text{end}}) \right] \right\}. \quad (5.275)$$

Making use of Eq. (5.273), and keeping only the dominant terms in $\phi_0/M_{\text{Pl}}$, one obtains

$$x_* \approx 1 - \frac{1}{12} \left( \frac{\phi_0}{M_{\text{Pl}}} \right)^2 \frac{1}{\Delta N_*}. \quad (5.276)$$

This expression can be useful to determine typical values for the slow-roll parameters evaluated at Hubble crossing. One obtains

$$\epsilon_{1*} \approx \frac{1}{288} \frac{\phi_0^6}{\Delta N_*^4 M_{\text{Pl}}^6}, \quad \epsilon_{2*} \approx \frac{4}{\Delta N_*} \quad \epsilon_{3*} \approx \frac{1}{\Delta N_*}, \quad (5.277)$$

hence

$$r \approx \frac{1}{18} \frac{1}{\Delta N_*^4 M_{\text{Pl}}^6}, \quad n_S - 1 \approx -\frac{4}{\Delta N_*}, \quad \alpha_S \approx -\frac{4}{\Delta N_*^2}. \quad (5.278)$$

One can see that these models typically predict a tiny amount of gravitational waves, but a substantial deviation from scale invariance $n_S - 1 \approx -4/\Delta N_* \approx 0.1$. The similarity with Eqs. (5.260) is obvious.

Finally, the parameter $M$ can be determined from the amplitude of the CMB anisotropies and the observable field value $x_* = x(N_*)$ by

$$\left(\frac{M}{M_{\text{Pl}}}\right)^4 = 622080 \frac{M_{\text{Pl}}^2}{\phi_0^2} \pi^2 \frac{(x_* - 1)^4}{x_*^4 (3x_*^2 - 8x_* + 3)^3} \frac{Q_{\text{rms}-\text{PS}}^2}{T^2}. \quad (5.279)$$
<figure><img src="image_38.png" /><figcaption>Fig. 35. Top left panel: Arctan Inflation (AI) potential as a function of $\phi/\mu$. Top right panel: logarithm of the potential. Bottom left panel: slow-roll parameter $\epsilon_1$ rescaled by $M_\phi^2/\mu^2$ which renders the corresponding expression “universal”, i.e. independent of the free parameter $\mu$. Bottom right panel: slow-roll parameters $\epsilon_2$ (solid line) and $\epsilon_3$ (dotted line) rescaled by $M_\phi^2/\mu^2$ (for the same reason as mentioned before).</figcaption></figure>

For $\phi_0/M_{Pl} \ll 1$, one can make use of Eq. (5.276) to get the approximate expression

$$ \left(\frac{M}{M_{Pl}}\right)^4 \approx 30 \frac{\pi^2}{\Delta N_*^4} \left(\frac{\phi_0}{M_{Pl}}\right)^6 \frac{Q_{\text{rms}-\text{PS}}^2}{T^2}. \qquad (5.280) $$

Using the typical value $\phi_0 \simeq 10^{14}$ GeV, one gets $M/M_{Pl} \simeq 5 \times 10^{-11}$.

The reheating consistent slow-roll predictions of the renormalizable inflection point models are displayed in Fig. 151. The reheating equation of state parameter $\bar{\phi}_{\text{reh}}$ has been taken to 0 since the potential is quadratic close to its minimum. One can check that in the limit $\phi_0/M_{Pl} \ll 1$, the first slow-roll parameter is indeed extremely small, while the second slow-roll parameter does not depend much on $\phi_0$. Remembering that $\phi_0/M_{Pl} \simeq 10^{-4}$, one can see that these models are disfavored by the CMB data since they predict a too large deviation from scale invariance. In order to remain inside the two-sigma confidence intervals, these models should be such that $\phi_0/M_{Pl} \gg 1$, for which they are close to the large field models (LFI, see Section 5.2). However, such values for $\phi_0$ are, a priori, outside the range of validity of the RIPI scenario. Finally, comparing Fig. 150 with Fig. 151, one can see that the general features of RIPI are very close to the MSSMI ones, and that the conclusions drawn before are therefore robust against the precise value of the power index $n$ in Eq. (5.244).

## 5.19. Arctan Inflation (AI)

This scenario was originally introduced in Ref. [504] as a toy model where the equation of state changes rapidly around $\phi = 0$. The potential reads

$$ V(\phi) = M^4 \left[ 1 - \frac{2}{\pi} \arctan \left( \frac{\phi}{\mu} \right) \right], \qquad (5.281) $$

and depends on one free parameter, $\mu$. This model was considered in order to test the reliability of different computational methods and schemes of approximation used in the calculations of the inflationary cosmological perturbations power spectrum, see Ref. [504]. More precisely, in Ref. [237], it was also used to study with which accuracy the first and second slow-roll order power spectra can approximate the actual power spectrum of the fluctuations in the case where the underlying model has both quite large tilt and running. This potential was considered again in Refs. [505,506] in order to study whether it can lead to the formation of long-lived primordial black holes. In the following slow-roll analysis, $\mu$ will be viewed as a free parameter with no restricted range of variation. Let us notice, however, that since it characterizes the typical vev at which inflation takes place, it could also be limited to the sub-Planckian regime if one wants inflation to proceed in a small field regime. As a matter of fact, it will be shown below that this needs to be the case to end inflation by slow-roll violation.

The potential (5.281), as well as its logarithm, are displayed in Fig. 35. They are decreasing functions of the field and, hence, inflation proceeds from the left to the right, in the direction specified by the arrow in Fig. 35.
Let us now compute the three first slow-roll parameters. If one defines $x \equiv \phi/\mu$, their expressions are given by

$$ \epsilon_1 = \frac{M_{\text{Pl}}^2}{\mu^2} \frac{2}{(1+x^2)^2 (\pi - 2 \arctan x)^2}, \quad \epsilon_2 = 8 \frac{M_{\text{Pl}}^2}{\mu^2} \frac{1 - \pi x + 2x \arctan x}{(1+x^2)^2 (\pi - 2 \arctan x)^2}, \qquad (5.282) $$

and

$$ \epsilon_3 = 2 \frac{M_{\text{Pl}}^2}{\mu^2} \left[ -4 + 6\pi x + \pi^2 (1 - 3x^2) + 4 (3\pi x^2 - 3x - \pi) \arctan x + 4 (1 - 3x^2) \arctan^2 x \right] \left[ (1 + x^2)^2 (\pi - 2 \arctan x)^2 (-1 + \pi x - 2x \arctan x) \right]^{-1}. \qquad (5.283) $$

They are displayed in Fig. 35. The first slow-roll parameter $\epsilon_1$ increases during inflation, reaches a maximum at $x_{\epsilon_1 \text{max}}$ and then decreases. Whether inflation can stop by violation of slow-roll or not depends on the value of $\epsilon_1$ at its maximum: $\epsilon_1^{\text{max}}$. This value is a solution of the following equation

$$ 2x_{\epsilon_1^{\text{max}}} \arctan (x_{\epsilon_1^{\text{max}}}) + 1 = \pi x_{\epsilon_1^{\text{max}}}, \qquad (5.284) $$

which can only be solved numerically. One gets $x_{\epsilon_1^{\text{max}}} \approx 0.428978$, from which one deduces that

$$ \epsilon_1^{\text{max}} \approx 0.262531 \frac{M_{\text{Pl}}^2}{\mu^2}. \qquad (5.285) $$

Therefore, in order for inflation to end by slow-roll violation, one needs to work under the assumption that $\mu/M_{\text{Pl}} < 0.512378$. In that case, inflation proceeds along the plateau located at values of $x$ such that $x < x_{\epsilon_1^{\text{max}}}$, in the direction specified by the arrow in Fig. 35 (i.e. from the left to the right). Otherwise, if one wants inflation to occur in other parts of the potential and/or for values of $\mu$ such that $\mu/M_{\text{Pl}} > 0.512378$, another mechanism needs to be considered in order to stop it (typically, we imagine a tachyonic instability in another direction in field space). This means that we also need to introduce an extra parameter $x_{\text{end}}$ which gives the location of the vev at which the tachyonic instability is triggered. Let us remark that we could also consider a model where the inflaton starts at $x < x_{\epsilon_1^{\text{max}}}$, then crosses the region where $\epsilon_1$ has its maximum and then causes the end of inflation by tachyonic instability. This case would give a bump in the power spectrum and, clearly, cannot be properly described in the slow-roll framework. In this article, we restrict ourselves to the first version of the scenario mentioned above. In this situation $x_{\text{end}}$ is given by the smallest solution of the equation $\epsilon_1 = 1$ and needs to be computed numerically. Before inflation stops, one can see in Fig. 35 that the second slow-roll parameter $\epsilon_2$ reaches a maximum, the location of which can be numerically computed to be $x_{\epsilon_2^{\text{max}}} \approx -0.28539 < x_{\epsilon_1^{\text{max}}}$.

At this point, one has $\epsilon_2^{\text{max}} \approx 1.02827 M_{\text{Pl}}^2 / \mu^2 > \epsilon_1^{\text{max}}$. As a consequence, the slow-roll approximation breaks down before the end of inflation. This conclusion is reinforced by the fact that $\epsilon_3$ diverges at $x_{\epsilon_1^{\text{max}}}$. This means that the last $e$-folds of inflation cannot be properly described in the slow-roll framework.

$$ N_{\text{end}} - N = \frac{\mu^2}{M_{\text{Pl}}^2} \left[ \frac{\pi x_{\text{end}}}{2} + \frac{x_{\text{end}}^2}{6} + \frac{\pi x_{\text{end}}^3}{6} - \left(1 + \frac{x_{\text{end}}^2}{3}\right) x_{\text{end}} \arctan x_{\text{end}} + \frac{1}{3} \ln(1+x_{\text{end}}^2) - \frac{\pi x}{2} - \frac{x^2}{6} - \frac{\pi x^3}{6} + \left(1 + \frac{x^2}{3}\right) x \arctan x + \frac{1}{3} \ln(1+x^2) \right], \qquad (5.286) $$

where $N_{\text{end}}$ is the number of $e$-folds at the end of inflation. In the vacuum dominated approximation where the potential is just given by $V(\phi) \approx M^4$, this trajectory can be approximated by $N_{\text{end}} - N = \mu^2/M_{\text{Pl}}^2(\pi x_{\text{end}} + x_{\text{end}}^2/6 + \pi x^3/3 - \pi x - x^2/6 - \pi x^3/3)$, which can be inverted exactly if needed.

This formula is valid if $\mu/M_{\text{Pl}} \ll 1$, since in that case, $x_{\text{end}} \approx -\sqrt{M_{\text{Pl}}/(\mu\pi\sqrt{2})} \ll -1$. Under this assumption, one has $x_*^3 \approx -3M_{\text{Pl}}^2/(\pi\mu^2) \Delta N_*$, from which one can approximate the values of the three first Hubble flow parameters at Hubble radius crossing

$$ \epsilon_{1*} = \frac{(\mu/M_{\text{Pl}})^{2/3}}{2(\pi\Delta N_*^2)^{2/3}}, \quad \epsilon_{2*} = \frac{4}{3\Delta N_*}, \quad \epsilon_{3*} = \frac{1}{\Delta N_*}, \qquad (5.287) $$

Then, one can calculate the tensor-to-scalar ratio, the spectral index and the running. One obtains the following expressions

$$ r = \frac{8 (\mu/M_{\text{Pl}})^{2/3}}{(\pi \Delta N_*^2)^{2/3}}, \quad n_s - 1 = -\frac{4}{3 \Delta N_*} \approx -0.03, \quad \alpha_s = -\frac{4}{3 \Delta N_*^2} \approx -5 \times 10^{-4}. \qquad (5.288) $$

These formulas are in agreement with the consistency relation $\alpha_s = -3/4(n_s - 1)^2$ obtained in Ref. [505].

Finally, it is interesting to estimate the energy scale $M$ from the CMB normalization. This leads to

$$ \left(\frac{M}{M_{\text{Pl}}}\right)^4 = \frac{2880 \pi^3 M_{\text{Pl}}^2 / \mu^2}{(1 + x_*^2)^2 (\pi - 2 \arctan x_*)^3} \frac{Q_{\text{rms-PS}}^2}{T^2}. \qquad (5.289) $$

Under the vacuum dominated approximation ($\mu/M_{\text{Pl}} \ll 1$), the above equation can be re-expressed as

$$ \left(\frac{M}{M_{\text{Pl}}}\right)^4 \approx \frac{40 \times 3^{2/3} \pi^{4/3}}{\Delta N_*} \left(\frac{\mu}{M_{\text{Pl}}}\right)^{2/3} \frac{Q_{\text{rms-PS}}^2}{T^2}. \qquad (5.290) $$

The requirement $M < M_{\text{Pl}}$ is always satisfied form sub-Planckian values of $\mu$. The typical value $M/M_{\text{Pl}} \simeq 10^{-3}$ corresponds to $\mu/M_{\text{Pl}} \simeq 10^{-2}$.

The slow-roll predictions of the AI models are displayed in Fig. 152, in the range $\mu/M_{\text{Pl}} < 0.512378$ (so that inflation can end by slow-roll violation). The reheating equation of state parameter $\bar{w}_{\text{reh}}$ has been taken to be 0 but since there is no potential minimum around which the inflaton field can oscillate at the end of inflation, this parameter is a priori unspecified. One can see that this model typically predicts a small amount of gravitational waves, and a deviation from scale invariance which is in accordance with the observations. The predictions in the planes $(n_s, r)$ are qualitatively well described by the vacuum dominated analysis presented before, see Eq. (5.288).
<figure><img src="image_39.png" /><figcaption>Fig. 36. Constant $n_s$ A Inflation (CNAI) potential and slow-roll parameters versus the vacuum expectation value of the inflaton field. Top left panel: Constant $n_s$ A Inflation potential for $\alpha = 1$. Top right panel: logarithm of the potential for the same value of $\alpha$. Bottom left panel: slow-roll parameter $\epsilon_1$ (same value of $\alpha$): inflation stops when $\epsilon_1 = 1$ in this model. Bottom right panel: slow-roll parameters $\epsilon_2$ and $\epsilon_3$ ($\alpha = 1$).</figcaption></figure>

## 5.20. Constant $n_s$ A Inflation (CNAI)

This class of models is designed in order to produce power spectra with constant spectral index. It was studied for the first time in Ref. [507]. The rational behind this approach is that, so far, no evidence for a significant running has been found in the cosmological data. Since, from a Bayesian point of view, one should avoid introducing parameters that are unnecessary in order to reproduce the observations, it makes sense to consider models which lead to exact power-law power spectra. This is of course the case for power-law inflation as discussed in Section 5.8 and we will see other examples in Sections 5.21, 6.15 and 7.6. In fact, in Ref. [507], a systematic analysis of potentials that yield constant spectral index was carried out. It was found that the following potential belongs to this category of models

$$V(\phi) = M^4 \left[ 3 - (3 + \alpha^2) \tanh^2 \left( \frac{\alpha}{\sqrt{2}} \frac{\phi}{M_{Pl}} \right) \right], \qquad (5.291)$$

where $\alpha$ is a positive massless parameter (denoted $n_0^2$ in Ref. [507]) and, in this section, we study this case. This potential is represented in Fig. 36 and, since it is symmetrical under the transformation $\phi \to -\phi$, only the $\phi > 0$ part is displayed. The potential is a decreasing function of the field $\phi$ and, therefore, inflation proceeds from the left to the right. It is positive provided $\phi < \phi_0$, where

$$\frac{\phi_0}{M_{Pl}} = \frac{\sqrt{2}}{\alpha} \operatorname{arctanh} \left( \sqrt{\frac{3}{3+\alpha^2}} \right). \qquad (5.292)$$

There is no value of $\alpha$ for which the potential is always positive. Defining $x = \phi/M_{Pl}$, the slow-roll parameters are given by

$$\epsilon_1 = \frac{4\alpha^2 (3 + \alpha^2)^2 \tanh^2\left(\frac{\alpha x}{\sqrt{2}}\right)}{\left[6 + \alpha^2 - \alpha^2 \cosh(\sqrt{2}\alpha x)\right]^2}, \qquad (5.293)$$

$$\epsilon_2 = \frac{2\alpha^2 (3 + \alpha^2) \left[ 12 + \alpha^2 - 2\alpha^2 \cosh(\sqrt{2}\alpha x) + \alpha^2 \cosh(2\sqrt{2}\alpha x) \right]}{\left[ 6 + \alpha^2 - \alpha^2 \cosh(\sqrt{2}\alpha x) \right]^2 \cosh^2\left(\frac{\alpha x}{\sqrt{2}}\right)}, \qquad (5.294)$$
$$ \epsilon_3 = 2\alpha^2 (3 + \alpha^2) \tanh^2\left(\frac{\alpha}{\sqrt{2}}x\right) \left[6(-24 + 2\alpha^2 - \alpha^4) + (120\alpha^2 + 7\alpha^4)\cosh(\sqrt{2}\alpha x)\right. \\ \left. - 2\alpha^2(\alpha^2 - 6)\cosh(2\sqrt{2}\alpha x) + \alpha^4\cosh(3\sqrt{2}\alpha x)\right] \\ \times \left[6 + \alpha^2 - \alpha^2\cosh(\sqrt{2}\alpha x)\right]^{-2} \left[12 + \alpha^2 - 2\alpha^2\cosh(\sqrt{2}\alpha x) + \alpha^2\cosh(2\sqrt{2}\alpha x)\right]^{-1}. \qquad (5.295) $$

These slow-roll parameters are displayed in Fig. 36. They all increase as inflation proceeds and diverge when the field approaches $\phi_0$. Hence inflation ends by slow-roll violation. Notice that the equation $\epsilon_1 = 1$ can be solved analytically. If we define $y \equiv \sinh^2(\alpha x/\sqrt{2})$, then one has to solve the following cubic equation $\alpha^4 y^3 + (\alpha^4 - 6\alpha^2)y^2 + [9 - 6\alpha^2 - \alpha^2(3 + \alpha^2)]y + 9 = 0$. The relevant solution reads

$$ y_{\text{end}} = \frac{6 - \alpha^2}{3\alpha^2} - \frac{1 - i\sqrt{3}}{3 \times 2^{1/3}} (3 + \alpha^2)^2 (1 + 3\alpha^2) P^{-1/3} - \frac{1 + i\sqrt{3}}{6 \times 2^{1/3} \alpha^4} P^{1/3}, \qquad (5.296) $$

where we have defined $P$ by

$$ P = -\alpha^6 (3 + \alpha^2)^2 (6 - 52\alpha^2 + 9\alpha^4) \\ + \sqrt{-27\alpha^{14} (3 + \alpha^2)^4 (36 - 60\alpha^2 + 96\alpha^4 + 25\alpha^6 + 4\alpha^8)}. \qquad (5.297) $$

The slow-roll parameters $\epsilon_1$ and $\epsilon_3$ both vanish when the field $\text{vev}$ goes to 0, whereas $\epsilon_2$ has a non-vanishing minimum value, given by $\epsilon_2 \to 2\alpha^2 (3 + \alpha^2)/3$ when $x=0$. Therefore, if $\alpha$ is larger than some maximum value

$$ \alpha_{\max} = \sqrt{\frac{1}{2} (\sqrt{15} - 3)} \approx 0.66, \qquad (5.298) $$

then $\epsilon_2$ is larger than 1 in the whole inflationary regime and the slow-roll approximation does not hold. It is therefore necessary to work under the assumption $\alpha < \alpha_{\max}$ which we assume in the following.

Let now us check that the spectral index $n_s - 1 = -2\epsilon_1 - \epsilon_2$ (at first order in slow-roll), can be made constant, as announced previously. Expanding the slow-roll parameters $\epsilon_1$ and $\epsilon_2$ in small values of $\alpha$, and crucially assuming that $\alpha x_*$ remains small, one obtains $\epsilon_1 = \mathcal{O}(\alpha^4)$ and $\epsilon_2 = 2\alpha^2 + \mathcal{O}(\alpha^4)$, so that $n_s - 1 = -2\alpha^2 + \mathcal{O}(\alpha^4)$. Therefore, the corresponding expression is indeed a constant (i.e. does no depend on $\phi_*$). Since we have $|n_s - 1| \ll 1$, this implies that $\alpha$ should be small which is consistent with the condition $\alpha < \alpha_{\max}$ derived above.

Let us now study the slow-roll trajectory of the system. This one can be integrated exactly leading to the following formula

$$ N - N_{\text{end}} = \frac{1}{\alpha^2 (3 + \alpha^2)} \left\{ \begin{aligned} & 3 \ln \left[ \sinh \left( \frac{\alpha}{\sqrt{2}} x \right) \right] - \frac{\alpha^2}{2} \sinh^2 \left( \frac{\alpha}{\sqrt{2}} x \right) \\ & - 3 \ln \left[ \sinh \left( \frac{\alpha}{\sqrt{2}} x_{\text{end}} \right) \right] + \frac{\alpha^2}{2} \sinh^2 \left( \frac{\alpha}{\sqrt{2}} x_{\text{end}} \right) \end{aligned} \right\}. \qquad (5.299) $$

Moreover, this trajectory can be inverted which allows us to explicitly express the $\text{vev}$ of the inflaton field in terms of the $e$-folds number. One obtains

$$ x = \frac{\sqrt{2}}{\alpha} \arcsinh \left[ -\frac{3}{\alpha^2} W_0 \left( -\frac{\alpha^2}{3} \exp \left\{ \frac{2}{3} \alpha^2 (3 + \alpha^2) (N - N_{\text{end}}) \right\} \right. \right. \\ \left. \left. + 2 \ln \left[ \sinh \left( \frac{\alpha}{\sqrt{2}} x_{\text{end}} \right) \right] - \frac{\alpha^2}{3} \sinh^2 \left( \frac{\alpha}{\sqrt{2}} x_{\text{end}} \right) \right] \right]^{1/2}, \qquad (5.300) $$

where $W_0$ is the 0 branch of the Lambert function as required since $x(N)$ is an increasing function of $N$. It is displayed in Fig. 37 where the CNAI trajectory takes place between $\phi/M_{\text{Pl}} = 0$ at the origin of the plot, and $x = \phi_0/M_{\text{Pl}}$ at the junction between the $-1$ branch and the 0 branch.

The slow-roll predictions of the CNAI models are displayed in Fig. 153. When $\alpha$ is small (but not too small), the value of $n_s$ is indeed constant (and compatible with the considerations presented above) but, unfortunately, too far from scale invariance to be compatible with CMB data. When $\alpha \ll 10^{-1}$, the predictions become roughly compatible with the data but, clearly, $n_s$ is no longer constant and no longer given by $-2\alpha^2$. At first sight, this is surprising since we expect the spectral index to tend towards $-2\alpha^2$ when $\alpha$ goes to zero (see above). In order to understand this point, let us remark that, in the limit where $\alpha$ vanishes, one can expand Eq. (5.296) to find $y_{\text{end}} \approx 3/\alpha^2 - 3/\alpha + \mathcal{O}(\alpha)$ (the term at order $\alpha^0$ is absent and this plays an important role in what follows). This leads to $x_{\text{end}} \approx (\sqrt{2}/\alpha) \ln(2\sqrt{3}/\alpha) - 1/\sqrt{2} + \mathcal{O}(\alpha)$. Notice that this last equation is compatible with the behavior of the first Hubble-flow parameter (5.293) in the vicinity of $\phi_0$: $\epsilon_1 \approx M_{\text{Pl}}^2/[2(\phi - \phi_0)^2]$. Therefore, the expression of $x_{\text{end}}$ found before corresponds in fact to writing $\epsilon_1 = 1$ with this approximated $\epsilon_1$. Then, using the slow-roll trajectory (5.300), one gets

$$ \sinh^2 \left( \frac{\alpha x_*}{\sqrt{2}} \right) = -\frac{3}{\alpha^2} W_0 \left( -\frac{\alpha^2}{3} e^{-2A/3} \right), \qquad (5.301) $$

where $A$ is given by the following expression

$$ A \equiv \alpha^2 (3 + \alpha^2) \Delta N_* - 3 \ln \left[ \sinh \left( \frac{\alpha x_{\text{end}}}{\sqrt{2}} \right) \right] + \frac{\alpha^2}{2} \sinh^2 \left( \frac{\alpha x_{\text{end}}}{\sqrt{2}} \right). \qquad (5.302) $$

This quantity can be expanded in $\alpha$ using the equation for $y_{\text{end}}$ derived above and, at leading order, one obtains

$$ -\frac{2}{3} A \approx -\frac{2}{3} \alpha^2 \Delta N_* + \ln \left( \frac{3}{\alpha^2} \right) - 1 - \frac{\alpha^2}{2}. \qquad (5.303) $$
<figure><img src="image_40.png" /><figcaption>Fig. 37. Lambert functions $W_0(x)$ (dashed line) and $W_{-1}(x)$ (solid line). During CNAI inflation, inflation proceeds along the “0” branch in the direction specified by the arrow on the figure.</figcaption></figure>

For simplicity, the last term in the previous expression can be ignored since $2\Delta N_* \gg 1/2$. It follows that, introducing the formula for $-2A/3$ into Eq. (5.301), one arrives at

$$ \sinh^2\left(\frac{\alpha x_*}{\sqrt{2}}\right) = -\frac{3}{\alpha^2} W_0\left(-\frac{1}{e}e^{-2\alpha^2 \Delta N_*}\right). \qquad (5.304) $$

If we ignore the exponential in the argument of the Lambert function (since $\alpha \ll 1$) and use the identity $\arcsinh(x) = \ln(x + \sqrt{x^2 + 1})$, one finally arrives at

$$ \alpha x_* \underset{\alpha \to 0}{\sim} \sqrt{2 \ln \left( \frac{2\sqrt{3}}{\alpha} \right)}. \qquad (5.305) $$

We now understand why, in the limit $\alpha \to 0$, the spectral index is no longer constant. The naive expression $n_s \simeq -2\alpha^2$ is obtained by expanding the expressions of $\epsilon_1$ and $\epsilon_2$ in $\alpha$, including the hyperbolic function of argument $\alpha x_*$. But we have just shown that, when $\alpha \ll 1$, $\alpha x_*$ is not small and, therefore, the Taylor expansion of those terms is no longer justified. This is why, in Fig. 153, we see a deviation from $n_s$ constant at very small values of $\alpha$. In fact, this questions the interest of this model since the condition of constant spectral index is obtained only for values of $n_s$ that are already ruled out by the CMB data. On the other hand, when $\alpha \ll 1$, the model seems compatible with the data and, therefore, represents a legitimate inflationary scenario even if the spectral index is not constant in this case.

Finally, it is also interesting to study the energy scale at which inflation takes place in this model. The CMB normalization gives

$$ \left(\frac{M}{M_{\text{Pl}}}\right)^4 = \frac{11520\pi^2\alpha^2(\alpha^2+3)^2\sinh^2\left(\frac{\alpha}{\sqrt{2}}x_*\right)Q_{\text{rms-PS}}^2}{[\alpha^2+6-\alpha^2\cosh(\sqrt{2}\alpha x_*)]^3T^2}. \qquad (5.306) $$

Since we have established the expression of $x_*$ above, it is sufficient to use it in the above formula. We have, however, to be careful about the calculation of the denominator. Indeed, if we neglect again the exponential in the argument of the Lambert function, Eq. (5.301), then $\sinh^2(\alpha x_*/\sqrt{2}) \simeq 3/\alpha^2$ and the denominator in Eq. (5.306) vanishes. Therefore, one needs to evaluate the Lambert function more precisely and to keep the corrections proportional to $\Delta N_*$. This can be done with the help of Eq. (33) of Ref. [508] which implies that $\sinh^2(\alpha x_*/\sqrt{2}) \simeq 3/\alpha^2 - 6\sqrt{\Delta N_*}/\alpha$. Using this expression, one arrives at

$$ \frac{M}{M_{\text{Pl}}} \simeq 0.016 \alpha^{-3/4} (\Delta N_*)^{-3/8}. \qquad (5.307) $$

For an order of magnitude estimate, one can use the fiducial value $\Delta N_* \simeq 55$. This leads to $M/M_{\text{Pl}} \simeq 0.0035 \alpha^{-3/4}$. Requiring $M < M_{\text{Pl}}$ puts a lower bound on the parameter $\alpha$, namely $\alpha \gtrsim 5 \times 10^{-4}$. This roughly corresponds to the range studied in Fig. 153.

## 5.21. Constant $n_s$ B Inflation (CNBI)

This model is another representative of the class of scenarios studied Section 5.20, it is designed such that the corresponding power spectrum has a constant spectral index. The potential is given by

$$ V(\phi) = M^4 \left[ (3 - \alpha^2) \tan^2 \left( \frac{\alpha}{\sqrt{2}} \frac{\phi}{M_{\text{Pl}}} \right) - 3 \right], \qquad (5.308) $$

where $\alpha$ is a positive dimensionless parameter [507]. Since the potential is periodic with period $\pi\sqrt{2}/\alpha$ and, moreover, invariant under $\phi \to -\phi$, one can restrict ourselves to the range $0 < \phi/M_{\text{Pl}} < \pi/(\sqrt{2}\alpha)$ without loss of generality. The potential is an increasing function of the field and,
<figure><img src="image_41.png" /><figcaption>Fig. 38. Top left panel: constant $n_s$ T Inflation (CNBI) potential for $\alpha = 0.1$, see Eq. (5.308). Top right panel: logarithm of this potential (for the same value of $\alpha$). Bottom left panel: slow-roll parameter $\epsilon_1$ still for $\alpha = 0.1$. Bottom right panel: slow-roll parameters $\epsilon_2$ and $\epsilon_3$ again for $\alpha = 0.1$.</figcaption></figure>

as a consequence, inflation proceeds from the right to the left. Finally, $V(\phi)$ is positive provided $\phi > \phi_0$, where

$$ \frac{\phi_0}{M_{Pl}} = \frac{\sqrt{2}}{\alpha} \arctan \left( \sqrt{\frac{3}{3-\alpha^2}} \right). \qquad (5.309) $$

Obviously, in order for the potential not to be negative everywhere, one needs to impose that $\alpha < \sqrt{3}$ and, as a result, the previous expression is well defined. The potential (and its logarithm) is displayed in Fig. 38, in the relevant range $\phi_0/M_{Pl} < \phi/M_{Pl} < \pi/(\sqrt{2}\alpha)$.

Then, defining $x = \phi/M_{Pl}$, the slow-roll parameters are given by

$$ \epsilon_1 = \frac{4\alpha^2 (\alpha^2 - 3)^2 \tan^2\left(\frac{\alpha}{\sqrt{2}}x\right)}{[\alpha^2 + (6-\alpha^2)\cos(\sqrt{2}\alpha x)]^2}, \qquad (5.310) $$

$$ \epsilon_2 = \frac{\alpha^2 (3-\alpha^2) [6+\alpha^2+2(6-\alpha^2)\cos(\sqrt{2}\alpha x)+(\alpha^2-6)\cos(2\sqrt{2}\alpha x)]}{2\cos^6\left(\frac{\alpha}{\sqrt{2}}x\right)[3+(\alpha^2-3)\tan^2\left(\frac{\alpha x}{\sqrt{2}}\right)]^2}, \qquad (5.311) $$

and

$$ \begin{aligned} \epsilon_3 = & 2\alpha^2 (\alpha^2 - 3) \tan^2\left(\frac{\alpha}{\sqrt{2}}x\right) [6(-72 + 14\alpha^2 - \alpha^4) + (\alpha^2 - 6)(7\alpha^2 + 78)\cos(\sqrt{2}\alpha x) \\ & - 2(\alpha^4 - 18\alpha^2 + 72)\cos(2\sqrt{2}\alpha x) + (\alpha^2 - 6)^2\cos(3\sqrt{2}\alpha x)] \\ & \times [\alpha^2 + (6 - \alpha^2)\cos(\sqrt{2}\alpha x)]^{-2} [6 + \alpha^2 + 2(6 - \alpha^2)\cos(\sqrt{2}\alpha x) \\ & + (\alpha^2 - 6)\cos(2\sqrt{2}\alpha x)]^{-1}. \end{aligned} \qquad (5.312) $$

These slow-roll parameters are displayed in Fig. 38 (bottom panels). The first slow-roll parameter $\epsilon_1$ first decreases as the field $v\epsilon$ increases and reaches a minimum value at $x_{\epsilon_2=0}$ where $\epsilon_2$ vanishes and then increases. The value of $x_{\epsilon_2=0}$ is given by

$$ x_{\epsilon_2=0} = \frac{1}{\alpha\sqrt{2}} \arccos \left[ \frac{\alpha^2 - 6 + \sqrt{\alpha^4 - 36\alpha^2 + 180}}{2(\alpha^2 - 6)} \right]. \qquad (5.313) $$

The second slow-roll parameter, $\epsilon_2$, always decreases as inflation proceeds, crossing $\epsilon_2 = 0$ at $x_{\epsilon_2=0}$. The third slow-roll parameter, $\epsilon_3$, is positive for $x < x_{\epsilon_2=0}$. In this domain, it decreases to reach a minimum and then increases and diverges when $x$ approaches $x_{\epsilon_2=0}$. On the contrary, for
<figure><img src="image_42.png" /><figcaption>Fig. 39. Lambert functions $W_0(x)$ (dashed line) and $W_{-1}(x)$ (solid line). During Constant $n_s$ T Inflation, inflation proceeds along the “-1” branch in the direction specified by the arrow.</figcaption></figure>

$x > x_{\epsilon_2=0}$, $\epsilon_3$ becomes negative. It first increases and reaches a local maximum, then decreases and goes to $-\infty$ at $x = \pi / (\sqrt{2\alpha})$. The three slow roll parameters diverge when $\phi$ goes to $\phi_0$ and to $M_{Pl}\pi / (\sqrt{2\alpha})$.

The minimum value of $\epsilon_1$ at $x_{\epsilon_2=0}$ turns out to be smaller than 1 only if $\alpha < \alpha_{max} \approx 0.2975$. A (rather long) analytic expression for $\alpha_{max}$ can be derived, but it does not provide much information to the present discussion. Therefore, one must require $\alpha < 0.2975$ in order to realize slow-roll inflation in this model. Then, assuming this is the case, it is clear from Fig. 38 and from the previous considerations that inflation ends by slow-roll violation. If we define $y \equiv \sin^2(\alpha x / \sqrt{2})$, then the condition $\epsilon_1 = 1$ is equivalent to $4(6 - \alpha^2)^2 y^3 - 4(12 - \alpha^2)(6 - \alpha^2)y^2 + 4(45 + 3\alpha^2 - 6\alpha^4 + \alpha^6)y - 36 = 0$. The relevant solution is given by

$$y_{\text{end}} = \frac{12 - \alpha^2}{3(6 - \alpha^2)} + \frac{4}{3} 2^{-2/3} (1 - i\sqrt{3}) \frac{(3\alpha^2 - 1)(18 - 9\alpha^2 + \alpha^4)^2}{(6 - \alpha^2)^2} P^{-1/3} \\ - (1 + i\sqrt{3}) \frac{2^{-1/3}}{24(6 - \alpha^2)^2} P^{1/3}, \qquad (5.314)$$

where we have defined the quantity $P$ by

$$P \equiv 64 (-6 + \alpha^2)^3 (-3 + \alpha^2)^2 \left( -6 + 110\alpha^2 - 9\alpha^4 + 3\alpha\sqrt{3} \times \sqrt{-36 + 408\alpha^2 - 12\alpha^4 - 25\alpha^6 + 4\alpha^8} \right). \qquad (5.315)$$

If $\alpha \ll 1$, then $y_{\text{end}} \approx 1/2$ and $x_{\text{end}} \approx \sqrt{2}/\alpha \arcsin(1/\sqrt{2}) = \pi/(2\sqrt{2\alpha})$.

As for the CNAI model, the spectral index $n_s - 1 = -2\epsilon_1 - \epsilon_2$, at first order in slow-roll, can be made constant in some limit. Expanding the slow-roll parameters in $\alpha$, while assuming $\alpha x$ to be small, gives $\epsilon_1 = x^2\alpha^4/2 + \mathcal{O}(\alpha^6)$ and $\epsilon_2 = 2\alpha^2 + \mathcal{O}(\alpha^4)$, so that $n_s - 1 = -2\alpha^2 + \mathcal{O}(\alpha^4)$. Therefore, approximate scale-invariance, $|n_s - 1| \ll 1$, implies $\alpha$ small.

Let us now turn to the slow-roll trajectory. This one can be integrated exactly, leading to the following formula

$$N - N_{\text{end}} = \frac{1}{\alpha^2 (3 - \alpha^2)} \left\{ 3 \ln \left[ \sin \left( \frac{\alpha}{\sqrt{2}} x \right) \right] - \frac{6 - \alpha^2}{2} \sin^2 \left( \frac{\alpha}{\sqrt{2}} x \right) \right. \\ \left. - 3 \ln \left[ \sin \left( \frac{\alpha}{\sqrt{2}} x_{\text{end}} \right) \right] + \frac{6 - \alpha^2}{2} \sin^2 \left( \frac{\alpha}{\sqrt{2}} x_{\text{end}} \right) \right\}. \qquad (5.316)$$

This formula can be inverted and $x$ can be expressed explicitly in terms of the $e$-folds number. One obtains

$$x = \frac{\sqrt{2}}{\alpha} \arcsin \left[ -\frac{3}{6 - \alpha^2} W_{-1} \left( -\frac{6 - \alpha^2}{3} \exp \left\{ \frac{2}{3} \alpha^2 (3 - \alpha^2) (N - N_{\text{end}}) \right. \right. \right. \\ \left. \left. \left. + 2 \ln \left[ \sin \left( \frac{\alpha}{\sqrt{2}} x_{\text{end}} \right) \right] - \frac{6 - \alpha^2}{3} \sin^2 \left( \frac{\alpha}{\sqrt{2}} x_{\text{end}} \right) \right\} \right]^{1/2}, \qquad (5.317)$$

where $W_{-1}$ is the $-1$ branch of the Lambert function. It is displayed in Fig. 39. When $x = \pi / (\sqrt{2\alpha})$, the argument of the Lambert function is $(\alpha^2 - 6) \exp((\alpha^2/3 - 2)/3)$ which is always larger than $-1/e$ for any value of $\alpha$ (this expression decreases with $\alpha$ when $\alpha < \sqrt{3}$), whereas when $x = \phi_0 / M_{Pl}$, the argument of the Lambert function is just given by $-1/e$. For $x > \phi_0 / M_{Pl}$, the value taken by the Lambert function must be less than $-1$ which indicates that the $-1$ branch is the relevant one. Therefore, inflation proceeds in the domain displayed in Fig. 39 in which one easily checks that the above trajectory is always well defined.

The slow-roll predictions of the CNBI models are displayed in Fig. 154 for the range $10^{-5} \lesssim \alpha \lesssim 10^{-1.3}$. For very small values of $\alpha$, the predictions are in agreement with the data with a value of $n_s$ centered around the constant value $n_s \approx 0.97$ and an amount of gravitational waves such that
$r \ge 0.07$. But one also notices that the spectral index is not really constant. In fact, it does not come as a surprise that the same phenomenon highlighted in Section 5.20 is at work here. Indeed, using the slow-roll trajectory (5.316), one has

$$ \sin^2\left(\frac{\alpha x_*}{\sqrt{2}}\right) = -\frac{3}{6-\alpha^2}W_{-1}\left(-\frac{6-\alpha^2}{3}e^{-2\Lambda/3}\right), \qquad (5.318) $$

where A is given by the following expression

$$ A \equiv \alpha^2 (3 - \alpha^2) \Delta N_* - 3 \ln \left[ \sin \left( \frac{\alpha x_{\text{end}}}{\sqrt{2}} \right) \right] + \frac{6 - \alpha^2}{2} \sin^2 \left( \frac{\alpha x_{\text{end}}}{\sqrt{2}} \right). \qquad (5.319) $$

Using the formula for $x_{\text{end}}$ derived above, one obtains, in the limit $\alpha \ll 1$ and at this order of approximation that $x_* \approx x_{\text{end}}$. Therefore, as in Section 5.20, $\alpha x_*$ is not a small quantity and one cannot always Taylor expand the trigonometric functions that appear in the expressions of the slow-roll parameters. This explains why, in the limit $\alpha \ll 1$, the spectral index is in fact not constant (see Section 5.20).

Finally, the CMB normalization gives

$$ \left(\frac{M}{M_{\text{Pl}}}\right)^4 = \frac{11520\pi^2\alpha^2(3-\alpha^2)^2\sin^2\left(\frac{\alpha}{\sqrt{2}}x_*\right)Q_{\text{rms-PS}}^2}{\left[(\alpha^2-6)\cos(\sqrt{2}\alpha x_*) - \alpha^2\right]^3 T^2}. \qquad (5.320) $$

In the limit $\alpha \ll 1$ we are interested in (since we have seen that, if $\alpha$ is not small, then the model is ruled out), the above expression takes the form $M/M_{\text{Pl}} \approx 0.02 \alpha^{-1/4} (\Delta N_*)^{-3/8}$. We obtain almost exactly the same result as for CNAI, see Eq. (5.306), except that the power of $\alpha$ is different. Taking the value $\Delta N_* = 55$, it follows that $M/M_{\text{Pl}} \approx 0.0044 \alpha^{-1/4}$ and requiring $M < M_{\text{Pl}}$, one obtains the following lower bound, $\alpha \gtrsim 3.8 \times 10^{-10}$.

## 5.22. Open String Tachyonic Inflation (OSTI)

### 5.22.1. Theoretical justifications

In this section, we consider tachyon inflation. It was shown in Refs. [509–512] that, in bosonic string theory, the four-dimensional action for a tachyon field $T$ on a D3-brane can be approximated as [511,512]

$$ S_T = T_3 \int d^4 x \sqrt{-g} \left[ \alpha' e^{-T/T_0} \partial_\mu \left( \frac{T}{T_0} \right) \partial^\mu \left( \frac{T}{T_0} \right) + \left( 1 + \frac{T}{T_0} \right) e^{-T/T_0} \right], \qquad (5.321) $$

where higher derivative terms have been ignored. In this stringy setting, $T_0$ is of the order of the string scale $T_0 \approx M_s = \ell_s^{-1} = 1/\sqrt{\alpha'}$, where $\ell_s$ is the string length. The constant $T_3$ is the brane tension which can be expressed as $T_3 \propto M_s^4/g_s$, $g_s$ being the string coupling. The tachyon is assumed to be minimally coupled to Einstein gravity and the Planck mass in four dimensions can be written as $M_{\text{Pl}}^2 = M_s^2 \nu/g_s^2$, where $\nu = (M_s r)^d/\pi$, $r$ being a radius of compactification and $d$ the number of compactified dimensions. This four dimensional approximation is valid provided $r \gg \ell_s$ or $\nu \gg 1$. The action (5.321) can be viewed as a truncated version of the action

$$ S_{\bar{T}} = \int d^4 x \sqrt{-g} V(\bar{T}) \sqrt{1 + \alpha' \partial_\mu \left(\frac{\bar{T}}{T_0}\right) \partial^\mu \left(\frac{\bar{T}}{T_0}\right)}. \qquad (5.322) $$

Indeed, following Refs. [368,513,514], redefining the field $\bar{T}$ by $\bar{T}/T_0 \equiv \sqrt{8(1+T/T_0)}$ with $V[\bar{T}(T)] \equiv T_3(1+T/T_0) \exp(-T/T_0)$, it is straightforward to show that the leading terms of Eq. (5.322) give back Eq. (5.321). Conversely, the full action of tachyonic inflation, under the assumptions discussed previously, can thus be described in terms of $\bar{T}$ by Eq. (5.322) with [513]

$$ V(\bar{T}) = \frac{T_3 e \bar{T}^2}{8 \bar{T}_0^2} e^{-T^2/(8\bar{T}_0^2)}. \qquad (5.323) $$

Because the action (5.322) is a particular case of k-inflation for which $S = \int d^4 x \sqrt{-g} P(T, X)$ with $X \equiv -g^{\mu\nu} \partial_\mu T \partial_\nu T/2$ and, here, $P(T, X) = \sqrt{1-2X}$, tachyonic inflation could produce observable non-Gaussianities. Therefore, one may wonder how accurate is the truncated action to describe the observable features of the model. On the theoretical point of view, knowing whether the truncated action is a faithful representation of the actual action is a complicated question since even an exact derivation of the complete action is still an open problem. On a more phenomenological point of view, non-Gaussianities are not observed by Planck [100]. More precisely, the parameter $f_{\text{NL}}$ (equilateral configuration) characterizing the amplitude of the bispectrum in Fourier space can be written as [169,515]

$$ f_{\text{NL}} = \frac{35}{108} \left( \frac{1}{c_s^2} - 1 \right) - \frac{5}{81} \left( \frac{1}{c_s^2} - 1 - 2\Lambda \right), \qquad (5.324) $$

where, in our case, $c_s^2 = 1-2X$ and $1/c_s^2 - 1 = 2\Lambda$ so that the last term in the above equation cancels out [515]. This leads to $f_{\text{NL}} = 35X/[54(1-2X)]$. In the range of interest $X \in [0, 1/2]$, the Planck constraint [100], $f_{\text{NL}} = -42 \pm 75$, yields $X \lesssim 0.495$. As a result, departures from the leading order action of Eq. (5.321) are, a priori, still allowed by the CMB data. We will see at the end of this section that tachyonic inflation has however other problems. For the moment, given that Eq. (5.321) can always be seen as a phenomenological model, we can continue to work with this action in order to see if, at least, this can lead to an inflationary scenario compatible with the CMB data.

### 5.22.2. Slow-roll analysis

The inflationary dynamics can be studied directly from Eq. (5.321) but since it is linear in $X$, the field can be canonically normalized. Performing the change of variable $e^{-T/T_0} \equiv (\phi/T_0)^2 / 8$, the Lagrangian can be re-written with an ordinary kinetic term, as a function of the field $\phi$ and with
<figure><img src="image_43.png" /><figcaption>Fig. 40. Top left panel: Open String Tachyonic Inflation (OSTI) potential as a function of $\phi/\phi_0$. Top right panel: logarithm of the potential. The arrow indicates in which direction inflation proceeds. Bottom left panel: slow-roll parameter $\epsilon_1$, rescaled by the quantity $M_{\phi_0}^2/\phi_0^2$, such that the corresponding expression becomes universal, i.e. independent of $\phi_0$. Bottom right panel: slow-roll parameters $\epsilon_2$ (solid line) and $\epsilon_3$ (dotted line), rescaled by $M_{\phi_0}^2/\phi_0^2$ for the same reason as mentioned before.</figcaption></figure>

a potential given by

$$V(\phi) = -M^4 \left(\frac{\phi}{\phi_0}\right)^2 \ln\left[\left(\frac{\phi}{\phi_0}\right)^2\right], \qquad (5.325)$$

where $M^4 \equiv eT_3$ and $\phi_0^2 \equiv 8eT_3^2$. We notice that it corresponds to a particular case of LPI discussed in Section 7.5, with $q=1$ and $p=2$. Such a potential was also introduced in Ref. [516] as a toy model of tachyon condensation. Let us also comment on the parameter $\phi_0$. In the original model $\phi_0 \simeq M_s$ and, as such, it is a zero-parameter scenario. Here, given the issues discussed before (see also the end of this section) we consider $\phi_0$ as a free parameter. If necessary, one can always recover the situation where $\phi_0$ is fixed to the string scale by assuming the corresponding prior $\phi_0 = M_s$.

The potential (5.325) is represented in Fig. 40, together with its logarithm (top panels), as a function of $x \equiv \phi/\phi_0$. Since it is invariant under $x \to -x$, and since it is positive definite only if $x^2 < 1$, it is only displayed in the range $0 < x < 1$. The potential vanishes at $x=0$, increases with $x$, reaches a maximum at $x_{V'=0} = e^{-1/2}$, then decreases with $x$ and vanishes at $x_{V'=0} = 1$. Inflation is supposed to take place between $x_{V'=0}$, where the effective mass of the inflaton is negative $m_{\phi}^2 = -4\phi_0^2$, and $x=0$, where the effective mass is positive and infinite $m_{\phi}^2 \to +\infty$. Hence it proceeds from the right to the left, at decreasing field values (see Fig. 40).

Let us now calculate the three first slow-roll parameters. They are given by

$$\epsilon_1 = 2 \left( \frac{M_{Pl}}{\phi_0} \right)^2 \left[ \frac{1 + \ln(x^2)}{x \ln(x^2)} \right]^2, \qquad (5.326)$$

$$\epsilon_2 = 4 \left( \frac{M_{Pl}}{\phi_0} \right)^2 \frac{2 + \ln(x^2) + \ln^2(x^2)}{x^2 \ln^2(x^2)}, \qquad (5.327)$$

and

$$\epsilon_3 = 4 \left( \frac{M_{Pl}}{\phi_0} \right)^2 \frac{1 + \ln(x^2) + 4 + 3 \ln(x^2) + \ln^2(x^2) + \ln^3(x^2)}{x^2 \ln^2(x^2) + 2 + \ln(x^2) + \ln^2(x^2)}. \qquad (5.328)$$

They are displayed in the bottom panels of Fig. 40. The first slow-roll parameter $\epsilon_1$ diverges when $x \to 0$, decreases with $x$, vanishes at $x_{V'=0}$ and then increases with $x$ and diverges when $x \to x_{V'=0}$. As a consequence, inflation stops by slow-roll violation at a point $x_{end}$ where $\epsilon_1 = 1$ that needs to be determined numerically. The second slow-roll parameter $\epsilon_2$ has the same kind of behavior, except that it has a non-vanishing minimum located at a point $x_{\epsilon_{min}}$, which is such that $0 < x_{\epsilon_{min}} < x_{V'=0}$. An analytic expression for $x_{\epsilon_{min}}$ can be derived but it does not add much to the discussion. It yields $\epsilon_2^{\min} \simeq 20.65 M_{Pl}^2/\phi_0^2$. This means that in order for a slow-roll inflationary regime to take place, $\epsilon_2^{\min} \ll 1$ requires that the parameter $\phi_0$ be