<figure><img src="image_71.png" /><figcaption>Fig. 68. Left panel: $x_{c_1=0}^{\pm}$ and $x_{V'=0}^{\pm}$ [defined in Eq. (6.121)] together with $x_{c_1^{\text{min}}}^{\pm}$ [see Eq. (6.126)] as a function of $\alpha$. Right panel: minimal value of the slow-roll parameter $\epsilon_1$, i.e. $\epsilon_1(x_{c_1^{\text{min}}}^{\pm})$, rescaled by $\phi_0^2/M_n^2$, as a function of $\alpha$. When it is greater than unity, inflation cannot occur.</figcaption></figure>

However, the constraints on $\alpha$ are tighter to get a sufficient number of e-folds. Let us therefore now turn to the determination of the slow-roll trajectory. It can be integrated exactly to give

$$N_{\text{end}} - N = \frac{\phi_0^2}{M_{\text{Pl}}^2} \left\{ \frac{5-4\alpha}{12\sqrt{\alpha(1-\alpha)}} \arctan\left(\frac{x-1}{\sqrt{1/\alpha-1}}\right) + \frac{x}{2}\left(\frac{x}{4} - \frac{1}{3}\right) \right. \\ \left. + \left(\frac{1}{8\alpha} - \frac{1}{6}\right) \ln\left[1+\alpha x(x-2)\right] - \frac{5-4\alpha}{12\sqrt{\alpha(1-\alpha)}} \arctan\left(\frac{x_{\text{end}}-1}{\sqrt{1/\alpha-1}}\right) \right. \\ \left. - \frac{x_{\text{end}}}{2}\left(\frac{x_{\text{end}}}{4} - \frac{1}{3}\right) - \left(\frac{1}{8\alpha} - \frac{1}{6}\right) \ln\left[1+\alpha x_{\text{end}}(x_{\text{end}}-2)\right] \right\} \quad (6.129)$$

Exactly the same remarks we have made for the GMSSMI model also applies here (see Section 6.6). In particular, for $\alpha < 1$, and contrary to the RIPI models ($\alpha = 1$), the number of e-folds never diverges at a given point $x$. Therefore, the total number of e-folds is bounded by some maximal finite value. From Eq. (6.129) when $\alpha \to 1$, one has

$$N_{\text{end}} - N_{\text{ini}} \le \left( \frac{\phi_0}{M_{\text{Pl}}} \right)^2 \frac{\pi}{24} \frac{1}{\sqrt{1-\alpha}}. \quad (6.130)$$

Therefore, if one require at least $\Delta N = N_{\text{end}} - N_{\text{ini}}$ e-folds, one has to fine-tune $\alpha$ to

$$\alpha > 1 - \left(\frac{\phi_0}{M_{\text{Pl}}}\right)^4 \frac{\pi^2}{576 \Delta N^2}. \quad (6.131)$$

Remembering that the small parameter here is $\phi_0/M_{\text{Pl}}$, one can see that it is a much tighter constraint than the one of Eq. (6.128). Taking $\phi_0/M_{\text{Pl}} \simeq 10^{-1}$ and $\Delta N \simeq 50$, one obtains $\alpha > 1 - 10^{-10}$. This makes the fine-tuning quite important and, as explained below, the same condition $|\alpha - 1| < \phi_0^4/M_{\text{Pl}}^4/\Delta N^2$ also applies to the case $\alpha > 1$ to maintain an acceptable deviation from scale invariance, making the whole class of models fine-tuned. However, as already mentioned above, the value $\phi_0 \simeq 10^{17}$GeV makes the fine-tuning issue easier to accept than the value $\phi_0 \simeq 10^{14}$GeV. Finally, the amplitude of the CMB anisotropies fixes the parameter $M$ to

$$\left(\frac{M}{M_{\text{Pl}}}\right)^4 = 622080\pi^2 \frac{M_{\text{Pl}}^2}{\phi_0^2} \frac{(1-2\alpha x_* + \alpha x_*^2)^2}{x_*^4(6-8\alpha x_* + 3\alpha x_*^2)^3} \frac{Q_{\text{rms}-\text{PS}}^2}{T^2}. \quad (6.132)$$

As explained in Section 5.17, this leads to $M/M_{\text{Pl}} \simeq 10^{13}$ GeV for $\phi_0/M_{\text{Pl}} \simeq 10^{-4}$.

The reheating consistent slow-roll predictions of the RIPI models are displayed in Figs. 195, 197, for $\alpha > 1$ and $\alpha < 1$ respectively, and for values of $\phi_0$ such that $\phi_0 \simeq 10^{17}$ GeV: $\phi_0/M_{\text{Pl}} = 10^{-2}, 10^{-1.5}, 10^{-1}, 10^{-0.5}, 1$. 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 both cases, one can see that in the limit $\alpha \to 1$, the standard RIPI predictions are recovered, see Fig. 151. The amount of gravitational waves $r$ seems to be quite independent on $\alpha$ while the spectral index $n_s$ strongly depends on it. In the case $\alpha > 1$, the fine-tuning is as important as in the case $\alpha < 1$ as mentioned above. Considering values of $\alpha$ very different from 1 worsens the spectral index problem, already present in standard RIPI. These models are therefore strongly disfavored by the data. In the case $\alpha < 1$ however, there is a very narrow range of acceptable values for $\alpha$. They are well inside the $|\alpha - 1| < \phi_0^4/M_{\text{Pl}}^4/\Delta N^2$ condition and the spectral index is inside the two-sigma confidence intervals. But as can be seen in Fig. 197, the spectral index varies so quickly with $\alpha$ that, even if the fine-tuning is less problematic than in the GMSSMI case (due to the different value of $\phi_0$), it is still very important.

## 6.8. Brane SUSY Breaking Inflation (BSUSYBI)

This model has been studied in Ref. [576] in the context of superstrings models.<fnref n="9" /> The potential is a sum of two exponential terms

$$V(\phi) = M^4 \left( e^{\sqrt{6}\frac{\phi}{M_{\text{Pl}}}} + e^{\sqrt{6}\gamma\frac{\phi}{M_{\text{Pl}}}} \right), \quad (6.133)$$

<footnote n="9">⁹ see Eq. (1.1) and Eq. (2.9) in that reference.</footnote>
<figure><img src="image_72.png" /><figcaption>Fig. 69. Brane SUSY Breaking Inflation (BSUSYBI) for $\gamma = 0.1$. Upper panels: the potential and its logarithm. Bottom left panel: the first slow-roll parameter $\epsilon_1$ as a function of the field value, the shaded area indicates where inflation stops. Bottom right panel: slow-roll parameter $\epsilon_2$ and $\epsilon_3$.</figcaption></figure>

one is a “hard” exponential brought about by a SUSY breaking mechanism and the other is a “slow-roll term” having $0 < \gamma < 1/\sqrt{3}$ and that dominates the eventual inflationary dynamics. It was shown in Ref. [576] that the inflationary dynamics can also generate superimposed oscillations in the primordial power spectrum but we will not focus on this case since, obviously, slow-roll is not satisfied in this situation [577–579]. Let us also notice that if the term in $\sqrt{6}$ in the first exponential function is relaxed to be a free parameter, the potential becomes as in Ref. [580], i.e. a general exponential brane potential. Defining

$$x \equiv \frac{\phi}{M_{\text{Pl}}}, \qquad (6.134)$$

the first three Hubble flow functions in the slow-roll approximation read

$$\epsilon_1 = 3 \left( \frac{e^{\sqrt{6x}} + \gamma e^{\sqrt{6\gamma x}}}{e^{\sqrt{6x}} + e^{\sqrt{6\gamma x}}} \right)^2, \qquad \epsilon_2 = -12(\gamma - 1)^2 \frac{e^{\sqrt{6(\gamma+1)x}}}{(e^{\sqrt{6x}} + e^{\sqrt{6\gamma x}})^2}, \qquad (6.135)$$

and

$$\epsilon_3 = 6(1-\gamma) \frac{(e^{\sqrt{6x}} - e^{\sqrt{6\gamma x}})(e^{\sqrt{6x}} + \gamma e^{\sqrt{6\gamma x}})}{(e^{\sqrt{6x}} + e^{\sqrt{6\gamma x}})^2}. \qquad (6.136)$$

These functions together with the potential are displayed in Fig. 69. The two exponential components are clearly visible on the plot of the logarithm of the potential. The required flatness of the potential is realized only along the $\gamma$ branch and for negative values of $x$. The first Hubble flow function $\epsilon_1$ is an increasing function of $x$ which varies between its asymptotic values:

$$\lim_{x \to -\infty} \epsilon_1 = 3\gamma^2, \qquad \lim_{x \to +\infty} \epsilon_1 = 3. \qquad (6.137)$$

For $\gamma$ small enough ($\gamma < 1/\sqrt{3}$), there is a regime where it is less than unity. This regime is given by the condition $x < x_{\epsilon_1=1}$ with

$$x_{\epsilon_1=1} = \frac{1}{\sqrt{6}(\gamma-1)} \ln \left( \frac{\sqrt{3}-1}{1-\gamma\sqrt{3}} \right). \qquad (6.138)$$

As a result, inflation can only proceed in the domain $x < x_{\epsilon_1=1}$ and it never stops. Hence the need for an extra-parameter $x_{\text{end}}$ encoding the field value at which some unspecified mechanism (such as a tachyonic instability) is triggered and stops inflation. Let us notice that the slow-roll parameter $\epsilon_2$ is always negative and goes to zero at large $|x|$ with a local minimum in $x = 0$ equals to $\epsilon_2^{\min} = -3(\gamma - 1)^2$. Finally, the slow-roll parameter $\epsilon_3$ vanishes when $x = 0$ and shares the same sign as $x$. Its asymptotic values are

$$\lim_{x \to -\infty} \epsilon_3 = 6\gamma(\gamma - 1), \qquad \lim_{x \to +\infty} \epsilon_3 = 6(1 - \gamma). \qquad (6.139)$$
<figure><img src="image_73.png" /><figcaption>Fig. 70. Maximum value of $x_{end}$ in order to realize N e-folds of inflation between $x_{c_1=1}$ and $x_{end}$ as a function of $0 < \gamma < 1/\sqrt{3}$. This condition defines a prior for the model parameter $x_{end}$, which is the region lying under the curves on the figure.</figcaption></figure>

The slow-roll trajectory can be integrated and gives

$$N - N_{\text{end}} = -\frac{1}{\sqrt{6}}(x - x_{\text{end}}) + \frac{1}{6\gamma}\ln\left[\frac{1 + \gamma e^{\sqrt{6}(\gamma-1)x}}{1 + \gamma e^{\sqrt{6}(\gamma-1)x_{\text{end}}}}\right]. \qquad (6.140)$$

This equation cannot be analytically inverted but since inflation requires $x < x_{c_1=1}$, it shows that $x_{end}$ should not be too close to $x_{c_1=1}$ in order to realize enough e-folds of inflation. This puts some upper bound on $x_{end}$, that can be computed numerically and that is displayed in Fig. 70. This value $x_{end}^{\text{max}}$ defines a prior for the model parameter $x_{end}$, which is the region lying under the curves on the figure.

Integrating Eq. (3.48) finally gives the field value $x_*$ at which the pivot mode crossed the Hubble radius during inflation. The parameter $M$ being fixed by the amplitude of the CMB anisotropies

$$\left(\frac{M}{M_{\text{Pl}}}\right)^4 = 4320\pi^2 \frac{\left(e^{\sqrt{6}x_*} + \gamma e^{\sqrt{6}\gamma x_*}\right)^2}{\left(e^{\sqrt{6}x_*} + e^{\sqrt{6}\gamma x_*}\right)^3} \frac{Q_{\text{rms-PS}}^2}{T^2}. \qquad (6.141)$$

The reheating consistent slow-roll predictions of the BSUSYBI models have been plotted in Fig. 199. The parameter $x_{end}$ varies between $2x_{end}^{\text{max}} < x_{end} < x_{end}^{\text{max}}$ with $x_{end}^{\text{max}} < 0$, under which the predictions of the model coincide with those of PLI (see Section 5.8). Large values for the parameter $\gamma$ are disfavored and it has to be smaller than $\lesssim 5 \times 10^{-2}$ to generate a reasonable amount of gravitational waves.

## 6.9. Tip Inflation (TI)

### 6.9.1. Theoretical justifications

This model is a scenario based on string theory in which the motion of branes in extra-dimensions causes the four-dimensional spacetime to inflate, see for instance Refs. [207,290,581–586]. Let us assume string theory with flux compactification. In this situation, the six-dimensional Calabi–Yau space has generically the shape of a bulk with warped throat(s) attached to it. The metric in the bulk is usually not known but, along the throat, explicit examples are available. A representative case is the Klebanov–Strassler throat [587] for which one can write the metric as

$$ds^2 = h^{-1/2}(r)\eta_{\mu\nu}dx^\mu dx^\nu + h^{1/2}(r)(dr^2 + r^2ds_5^2). \qquad (6.142)$$

The function $h(r)$ describes the warping along the radial coordinate $r$ of the throat. We see that the throat is in fact a cone with five-dimensional sections given by the metric $ds_5^2$. For a conifold, these sections are two spheres $S_2 \times S_3$ which shrink to zero at the tip of the cone [588]. Let us recall that a conifold can also be defined by the equation $\sum_{A=1}^4 (Z_A)^2 = 0$, i.e. a six-dimensional (or three complex dimension) surface in $\mathbb{C}^4$. However, if one has a deformed conifold, then, at the tip the $S_2$ sphere shrinks to zero but the $S_3$ remains finite [588]. A deformed conifold can similarly be defined by the equation $\sum_{A=1}^4 (Z_A)^2 = \epsilon^2$ and, at the tip, one has $\sum_{A=1}^4 |Z_A|^2 = \epsilon^2$. Usually brane inflation takes place when a brane is moving along the radial direction of the throat, see Section 6.19. Here, following Ref. [585], we will consider a different situation, namely the case of a brane moving at the tip of the deformed conifold. In addition, we will not only consider radial motion only but also angular motion.

Technically, the above model can be described in the framework of supergravity (viewed, in this context, as a low energy effective field theory). Let us assume that there is a D3-brane moving at the tip and that complex structure moduli and the dilaton are stabilized, thanks to the presence of fluxes. Furthermore, following Ref. [585], we suppose that there is only one volume modulus, $\rho$, plus three fields $z_i$, $i = 1, \dots, 3$ describing the D3-brane position. It follows that the corresponding Kähler potential is given by

$$K(\rho, z_i, z_i^\dagger) = -3M_{\text{Pl}}^2 \ln \left[ \rho + \rho^\dagger - \gamma k (z_i, z_i^\dagger) \right], \qquad (6.143)$$

where $k$ is a function of the brane coordinates and $\gamma$ is a constant (of mass dimension $-2$) related to the brane tension $T_3$, an approximate expression of which will be given below. In the vicinity of the deformed conifold tip, the function $k$ takes the form

$$k(z_i, z_i^\dagger) = k_0 + c\epsilon^{-2/3} \left( \sum_{A=1}^{4} |Z_A|^2 - \epsilon^2 \right). \qquad (6.144)$$
Here c is a numerical constant $c = 2^{1/6}/3^{1/3} \approx 0.77$ and $k_0$ stands for the value of the function k at the tip. The quantity $\epsilon^{2/3} = r_{\text{tip}}$ can be viewed as the radius of the tip as illustrated in Figs. 1 and 2 of Ref. [585].

The last ingredient of the model is a stack of n D7-branes placed far from the tip. Then, the superpotential (Kuperstein embedding [589]) can be written as

$$W = W_0 + A(z_1)e^{-a\rho} = W_0 + A_0 \left(1 - \frac{z_1}{\mu}\right)^{1/n} e^{-a\rho}. \qquad (6.145)$$

In this expression, $\mu^{2/3}$ represents the distance between the stack of D7-branes and the tip (see Fig. 2 of Ref. [585] for an illustration). We always assume that this distance is much larger than the size of the tip, i.e. $\epsilon/\mu \ll 1$. The quantities $W_0$, $A_0$ and $a$ are constants. It is interesting to remark that the above superpotential only depends on $z_1$ and therefore breaks the symmetry of the tip.

We are now in a position where the potential and the kinetic term can be calculated for the fields $z_i$ and $\rho$. The F-term potential reads

$$V(\sigma, x_1) = \frac{2ae^{-a\sigma}}{M_{Pl}^2 U^2} \left( \frac{aU}{6} |A|^2 e^{-a\sigma} + |A|^2 e^{-a\sigma} - |W_0 A| \right) + \frac{e^{-2a\sigma}}{3M_{Pl}^2 U^2 n^2 \mu^2} \frac{|A|^2 \epsilon^{2/3}}{c} \left( 1 - \frac{x_1^2}{\epsilon^2} \right) \left( 1 - \frac{x_1}{\mu} \right)^{-2} + \frac{D}{U^b}, \qquad (6.146)$$

where we have taken, from the definition $z_i = x_i + iy_i$, $z_1 = x_1$ at the tip. Because of our choice of the superpotential, V no longer depends on $x_2$, $x_3$. In the above expression, we have defined $\rho = \sigma + i\tau$ and $\tau$ is chosen such that V is minimal. The quantity U is defined by $U = \rho + \rho^\dagger - k = 2\sigma - k_0$ at the tip. Finally, the last term $D/U^b$, with D and b constant, is an uplifting term which is added in order to avoid having an anti-de Sitter minimum. In practice, uplifting potentials generically have $b = 3$ [590].

The calculation of the kinetic term is difficult since the Kähler matrix mixes all the fields $z_i$. For this reason, it is easier to use another parametrization such where $z_1 = \epsilon \cos \varphi$, $z_2 = \epsilon \sin \varphi \cos \theta$, $z_3 = \epsilon \sin \varphi \sin \theta \cos \psi$ and $z_4 = \epsilon \sin \varphi \sin \theta \sin \psi$, as appropriate since the tip of the deformed conifold is $S_3$. In this case, the Kähler matrix becomes diagonal and expanding everything in the small parameter $\epsilon/\mu \ll 1$, one obtains

$$V(\sigma, \varphi) = \Lambda(\sigma) + B(\sigma) \cos \varphi + C(\sigma) \sin^2 \varphi + \dots, \qquad (6.147)$$

where

$$\Lambda(\sigma) = \frac{2a|A_0|e^{-a\sigma}}{M_{Pl}^2 U^2} \left( \frac{aU}{6} |A_0| e^{-a\sigma} + |A_0| e^{-a\sigma} - |W_0| \right) + \frac{D}{U^b}, \qquad (6.148)$$

$$B(\sigma) = \frac{2a|A_0|e^{-a\sigma}}{M_{Pl}^2 U^2 n} \frac{\epsilon}{\mu} \left( -\frac{aU|A_0|}{3} e^{-a\sigma} - 2|A_0|e^{-a\sigma} + |W_0| \right), \qquad (6.149)$$

$$C(\sigma) = \frac{|A_0|^2 e^{-2a\sigma}}{3M_{Pl}^2 U^2 \gamma \mu^2 n^2} \frac{\epsilon^{2/3}}{c}. \qquad (6.150)$$

Let us now discuss this result. If one ignores, for the moment, all terms depending on the brane position, it remains only the term $\Lambda(\sigma)$ which is nothing but the Kachru-Kallosh-Linde-Trivedi (KKLT) potential for the volume modulus [590]. We see that in absence of the uplifting term $D/U^b$, its minimum given by $\partial\Lambda/\partial\sigma = 0$ would be located at $\sigma = \sigma_0$, solution of the implicit equation

$$W_0 = -A_0 \left[ 1 + \frac{a}{3} (2\sigma_0 - k_0) \right] e^{-a\sigma_0}. \qquad (6.151)$$

The corresponding value of the potential would actually be negative (anti-de Sitter) and given by

$$\Lambda(\sigma_0) = - \frac{a^2 |A_0|^2}{3 M_{Pl}^2 U} e^{-2a\sigma_0} < 0. \qquad (6.152)$$

Hence the required uplifting term from which one can find a new minimum at which V is positive. This is precisely how KKLT managed to find a de Sitter minimum instead of an anti-de Sitter one for the first time in string theory [590].

If the position of the minimum were not changed by adding the uplifting term, one would obtain a vanishing value of V for

$$D_0 = \frac{a^2 |A_0|^2 U^{b-1} (\sigma_0)}{3 M_{Pl}^2} e^{-2a\sigma_0}. \qquad (6.153)$$

This suggests to introduce a new parameter $\beta$, defined by

$$\beta \equiv D \frac{3 M_{Pl}^2}{a^2 |A_0|^2 U^{b-1}(\sigma_0)} e^{2a\sigma_0}, \qquad (6.154)$$

such that one can trade D for $\beta$ in all the uplifting terms. Therefore, $\beta = 1$ represents a situation in which the potential is uplifted while the position of its minimum is unchanged. In general, as expected in presence of the brane, the KKLT minimum $\sigma_0$ of $\Lambda(\sigma)$ will be shifted. The correction due to the uplifting terms can be evaluated perturbatively and one obtains the following expression

$$\sigma_{\min} = \sigma_0 + \frac{b\beta}{2a^2\sigma_0} + \dots, \qquad (6.155)$$

valid provided $b\beta/(2a^2\sigma_0) \ll 1$. For $\beta = 0$, one recovers that $\sigma_{\min} = \sigma_0$ as expected without uplifting terms (and with a negative minimum for V). There are other corrections to the position of the minimum due to the presence of the brane but one can show that they do not play an important role (they are calculated in Ref. [585]). The final argument consists in considering that the modulus is stabilized at this minimum. Then, one obtains a single field model $V(\varphi) = V(\sigma_{\min}, \varphi)$ where the coefficients in Eq. (6.147) are now given by

$$\Lambda(\sigma_{\min}) \equiv \Lambda \approx \frac{a^2 |A_0|^2 e^{-2a\sigma_0}}{6 M_{Pl}^2 \sigma_0} [(\beta - 1) + \dots], \qquad (6.156)$$
$$B(\sigma_{\min}) \equiv B \approx \frac{a|A_0|^2 \epsilon e^{-2a\sigma_0}}{6M_{\text{Pl}}^2 n\mu\sigma_0^2} \left[ (b\beta - 3) + \frac{b\beta}{4a\sigma_0}(14 - 3b\beta) + \dots \right], \qquad (6.157)$$

$$C(\sigma_{\min}) \equiv C \approx \frac{|A_0|^2 \epsilon^{2/3} e^{-2a\sigma_0}}{12 M_{\text{Pl}}^2 n^2 \mu^2 \sigma_0^2 \gamma c} + \dots, \qquad (6.158)$$

The above relations express the parameters of the potential in terms of the stringy parameters. We see that, if $\beta > 1$, we have that the KKLT potential is positive at the minimum that could account for a cosmological constant today for $\beta - 1 = \mathcal{O}(\sigma_0^{-2})$ [585].

Finally, the kinetic term for $\phi$ remains to be calculated. Using the explicit form of the Kähler metric, one obtains

$$K_{I\bar{J}} \partial_{\mu} z^I \partial^{\mu} z^J \approx \frac{3 M_{\text{Pl}}^2}{U} \gamma c \epsilon^{4/3} \partial_{\mu} \varphi \partial^{\mu} \varphi + \dots, \qquad (6.159)$$

where, at the minimum, one has

$$\gamma \approx \frac{\sigma_0 T_3}{3 M_{\text{Pl}}^2}, \qquad (6.160)$$

$T_3$ being the brane tension. Therefore, in the large volume limit, the canonical field $\phi$ is $\phi = \sqrt{T_3 c \epsilon^{2/3}} \phi$. As a consequence, the final form of the potential reads

$$V(\phi) = \Lambda + B \cos \left( \frac{\phi}{\sqrt{T_3 c \epsilon^{2/3}}} \right) + C \sin^2 \left( \frac{\phi}{\sqrt{T_3 c \epsilon^{2/3}}} \right). \qquad (6.161)$$

To end this section, it is interesting to discuss the orders of magnitude of the parameters appearing in the above potential. For this purpose, it is useful to recall that $\sigma_0$, being a volume modulus, is related to the size (or volume) of the extra-dimensions, $V_6 \approx \sigma_0^{3/2} \alpha'^3$. The brane tension can be written as $T_3 = (2\pi)^{-3} g_s^{-1} \alpha'^{-2}$ while the Planck mass takes the form $M_{\text{Pl}}^2 = (2\pi)^{-7} V_6 g_s^{-2} \alpha'^{-4}$ (g_s is the string coupling). As already mentioned, the distance $\mu^{2/3}$ can be viewed as the distance between the stack of D7-branes and the tip. It is therefore of the order of the size of the throat which allows us to write that $\mu \approx (27\pi g_s N \alpha'^2/4)^{3/8}$ where the positive integer N is the total background Ramond–Ramond charge.

In order to have a successful slow-roll scenario, we must assume that the potential vanishes at its minimum. This amounts to take $\Lambda = B$ which can always be achieved by choosing $\beta = \beta_{sr}$ such that (with $b = 3$, see before)

$$\beta_{sr} = 1 + \frac{45\epsilon}{4n\mu a^2 \sigma_0^2} + \dots, \qquad (6.162)$$

where we have performed a large volume expansion. Then, at the top of the potential, one has $\partial^2 V / \partial \phi^2 \approx 2C - \Lambda$ and if one wants a flat potential $2C - \Lambda = 2C - B$ must be a very small quantity, i.e. $C/B \approx 1/2$. Using the equations established above, one can write

$$\frac{C}{B} = Y \frac{\sigma_0^{3/2}}{g_s(g_s \pi N)^{3/8}} \left( \frac{r_{\text{tip}}}{\ell_s} \right)^{-1/2}, \qquad (6.163)$$

where the numerical factor $Y = (12/15) \times (4/27)^{3/8} / [(2\pi)^4 n c] \approx 5 \times 10^{-5}$ and $r_{\text{tip}} \equiv \epsilon^{2/3}$. The string length is given by $\ell_s = \sqrt{\alpha'}$. Let us also recall that we have taken $b = 3$. We see in the above expressions, especially Eq. (6.157), that this case is special because $\beta_{sr} \approx 1$ and we have an additional suppression. It is also interesting to discuss the mass scale which appears in the arguments of the trigonometric functions. Straightforward calculations lead to

$$\frac{\sqrt{T_3 c \epsilon^{2/3}}}{M_{\text{Pl}}} = (2\pi)^2 \sqrt{\frac{c}{2}} g_s^{1/2} \sigma_0^{-3/4} \left( \frac{r_{\text{tip}}}{\ell_s} \right). \qquad (6.164)$$

For fixed $g_s$ and $N$, the two inflationary parameters $C/B$ and $\sqrt{T_3 c \epsilon^{2/3}}/M_{\text{Pl}}$ are in fact controlled by the radius of the tip and the volume of the extra-dimensions.

Finally, if one requires $C/B = 1/2$, as appropriate in a slow-roll analysis, then the above equations imply that

$$\frac{\sqrt{T_3 c \epsilon^{2/3}}}{M_{\text{Pl}}} \approx 2 \times 10^8 \sigma_0^{9/4}. \qquad (6.165)$$

This equation is relevant for the question of the priors that should be put on the model parameters.

### 6.9.2. Slow-roll analysis

We now turn to the slow-roll analysis of the model. For the canonically normalized inflaton field, we have just seen that the potential is given by

$$V = M^4 \left( 1 + \cos \frac{\phi}{\mu} + \alpha \sin^2 \frac{\phi}{\mu} \right), \qquad (6.166)$$

where inflation proceeds in the region $0 < \phi/\mu < \pi$. Here, we have written $\Lambda = M^4$, $C/B = \alpha$ and $\mu = \sqrt{T_3 c \epsilon^{2/3}}$ (not to be confused with the scale $\mu$ introduced above and related to the distance between the stack of branes and the tip). When $\alpha \ll 1$, the potential reduces to the natural inflation (NI) one. Yet, it was shown in Section 5.6 that only super-Planckian decay constants $\mu/M_{\text{Pl}} > \mathcal{O}(1)$ could make the natural inflation models compatible with observations (see e.g. Fig. 132). As noticed in Ref. [585], this means that tip inflation models with $\alpha \ll 1$ are not viable. On the other hand, as was discussed in detail in the previous sub-section, if $\alpha$ is fine-tuned to $\alpha \approx 1/2$, then the potential of Eq. (6.166) becomes very flat at the top and a phenomenologically successful slow-roll inflationary stage could occur. This is why, in the following, these models are studied with $\alpha \approx 1/2$.

Defining

$$x \equiv \frac{\phi}{\mu}, \qquad (6.167)$$
<figure><img src="image_74.png" /><figcaption>Fig. 71. Tip Inflation (TI). Upper panels: Tip Inflation potential and its logarithm for $\alpha = 0.1$ (blue line) and $\alpha = 1$ (pink line), as a function of $\phi/\mu$. Bottom left panel: slow-roll parameter $\epsilon_1$ normalized by $M_{Pl}^2/\mu^2$. The shaded area indicates the breakdown of the slow-roll inflation if $\mu = M_{Pl}$ (strictly speaking when the acceleration stops). Bottom right panel: slow-roll parameter $\epsilon_2$ (solid line) and $\epsilon_3$ (dotted line), again rescaled by $M_{Pl}^2/\mu^2$.</figcaption></figure>

the potential of Eq. (6.166) and its logarithm with respect to $x$ are displayed in Fig. 71. Its general shape depends on the value of $\alpha$. If $\alpha < 1/2$, it is a decreasing function of the field $\text{vev}$, hence inflation proceeds from the left to the right, and it has a vanishing minimum at $x = \pi$. Its first derivative vanishes at the top of the potential for $x = 0$ while its second derivative $V''(x = 0) \propto 2\alpha - 1$. It vanishes there when $\alpha = 1/2$ and the potential becomes flat enough to support inflation. If $\alpha > 1/2$, the potential maximum is not located at $x = 0$ anymore but at $x = \arccos[1/(2\alpha)]$. Let us thus define

$$x_{V'=0} = \begin{cases} 0 & \text{if } \alpha < 1/2, \\ \arccos\left(\frac{1}{2\alpha}\right) & \text{if } \alpha > 1/2. \end{cases} \qquad (6.168)$$

If $\alpha > 1/2$, the potential decreases with the field $\text{vev}$ in the range $x_{V'=0} < x < \pi$, where inflation proceeds from the left to the right. Again, the first derivative of the potential vanishes at the top of the potential while its second derivative $V''(x = x_{V'=0}) \propto 1/(2\alpha) - 2\alpha$ again vanishes when $\alpha = 1/2$. This is why $\alpha$ must be close enough to 1/2 in order for a viable slow-roll inflationary regime to take place.

Let us calculate the Hubble flow functions within the slow-roll approximation. They read

$$\epsilon_1 = \frac{M_{Pl}^2}{\mu^2} \frac{(1 - 2\alpha \cos x)^2 \sin^2 x}{2(1 + \cos x + \alpha \sin^2 x)^2}, \qquad (6.169)$$

$$\epsilon_2 = \frac{M_{Pl}^2}{\mu^2} \frac{2 \cos^2 \frac{x}{2}}{(1 + \cos x + \alpha \sin^2 x)^2} [2 + \alpha(3 + 4\alpha) - 2\alpha(3 + 2\alpha) \cos x - \alpha \cos(2x)], \qquad (6.170)$$

and

$$\epsilon_3 = \frac{M_{Pl}^2}{\mu^2} \left\{ -2 - \frac{2+4\alpha}{(1+\alpha-\alpha\cos x)^2} + \frac{5+3\alpha}{1+\alpha-\alpha\cos x} + \frac{1}{\cos^2\left(\frac{x}{2}\right)} + \frac{4(1+\alpha+3\alpha^2)-2\alpha(7+4\alpha)\cos x}{\alpha[\cos(2x)+(6+4\alpha)\cos x-3-4\alpha]-2} \right\}. \qquad (6.171)$$

They are displayed in Fig. 71 and are increasing functions of the field $\text{vev}$ in the inflationary domain $x_{V'=0} < x < \pi$. Notice that they diverge when $x \to \pi$. The first and third slow-roll parameters $\epsilon_1$ and $\epsilon_3$ vanish at the potential maximum. However, the second slow-roll parameter $\epsilon_2$ takes a
non-vanishing positive value given by

$$ \epsilon_2 (x = x_{V'=0}) = \begin{cases} \frac{M_{\text{Pl}}^2}{\mu^2} (1 - 2\alpha) & \text{if } \alpha < 1/2, \\ 4 \frac{M_{\text{Pl}}^2}{\mu^2} \frac{2\alpha - 1}{2\alpha + 1} & \text{if } \alpha > 1/2. \end{cases} \qquad (6.172) $$

Requiring $|\epsilon_2| < 1$ implies again to adjust $\alpha$ close to $1/2$ such that $|\alpha - 1/2| \ll \mu^2/M_{\text{Pl}}^2 \ll 1$. Inflation stops when $\epsilon_1 = 1$ at the position $x_{\text{end}}$ given by

$$ x_{\text{end}} = \arccos \left[ \Sigma + \frac{(1 + i\sqrt{3})\sigma}{3 \times 2^{2/3} (\delta + \sqrt{\Delta})^{1/3}} - \frac{(1 - i\sqrt{3})\sigma'}{6 \times 2^{1/3}} (\delta + \sqrt{\Delta})^{1/3} \right]. \qquad (6.173) $$

In this formula, we have defined

$$ \Delta = -864\alpha^6 (2\alpha + 1)^3 \frac{\mu^2}{M_{\text{Pl}}^2} \left( \frac{\mu^2}{M_{\text{Pl}}^2} + 2 \right)^2 \times \left\{ (2\alpha - 1)^3 + 2(2\alpha + 1)[(\alpha - 10)\alpha - 2] \frac{\mu^2}{M_{\text{Pl}}^2} - 4(2\alpha + 1)^2 \frac{\mu^4}{M_{\text{Pl}}^4} \right\}, \qquad (6.174) $$

and

$$ \delta = 8\alpha^3 \left[ 2(2\alpha - 1)^3 - 3(1 + 2\alpha)(5 + 2\alpha)(1 + 4\alpha) \frac{\mu^2}{M_{\text{Pl}}^2} - 15(1 + \alpha)(1 + 2\alpha)^2 \frac{\mu^4}{M_{\text{Pl}}^4} - 2(1 + 2\alpha)^3 \frac{\mu^6}{M_{\text{Pl}}^6} \right], \qquad (6.175) $$

together with

$$ \sigma = 3 + 4\alpha(1 - \alpha) - 2 \frac{\mu^2}{M_{\text{Pl}}^2} (1 + 2\alpha)^2 - \frac{8}{2 + \frac{\mu^2}{M_{\text{Pl}}^2}}, \qquad \sigma' = \frac{1}{2\alpha^2 \left( 2 + \frac{\mu^2}{M_{\text{Pl}}^2} \right)}. \qquad (6.176) $$

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

$$ N_{\text{end}} - N = \frac{\mu^2}{M_{\text{Pl}}^2} \frac{1}{2\alpha - 1} \ln \left( \frac{1 - \cos x}{1 - \cos x_{\text{end}}} \right) - \frac{\mu^2}{2M_{\text{Pl}}^2} \frac{2\alpha + 1}{2\alpha - 1} \ln \left( \frac{1 - 2\alpha \cos x}{1 - 2\alpha \cos x_{\text{end}}} \right). \qquad (6.177) $$

For $\alpha = 1/2$, this expression is singular, and one has

$$ N_{\text{end}} - N = \frac{\mu^2}{M_{\text{Pl}}^2} \left[ \frac{1}{1 - \cos x} - \frac{1}{1 - \cos x_{\text{end}}} - \frac{1}{2} \ln \left( \frac{1 - \cos x}{1 - \cos x_{\text{end}}} \right) \right]. \qquad (6.178) $$

Finally, the parameter $M$ can be determined from the amplitude of the CMB anisotropies and the observable field value $x_*$ [see Eq. (3.48)], and one gets

$$ \left( \frac{M}{M_{\text{Pl}}} \right)^4 = 720 \pi^2 \frac{M_{\text{Pl}}^2 (1 - 2\alpha \cos x_*)^2 \sin^2 x_* Q_{\text{rms-PS}}^2}{\mu^2 (1 + \cos x_* + \alpha \sin^2 x_*)^3 T^2}. \qquad (6.179) $$

The reheating consistent slow-roll predictions of the TI models are displayed in Fig. 201 for $\alpha < 1/2$ and in Fig. 204 for $\alpha > 1/2$, with $\mu/M_{\text{Pl}} = 10^{-6}$, $10^{-4}$ and $10^{-2}$. In both cases, one can see that $\alpha$ needs to be sufficiently adjusted to $1/2$, namely $|2\alpha - 1| \ll \mu^2/M_{\text{Pl}}^2$, otherwise the deviation from scale invariance is too important. The typical amount of gravitational waves is very small. To see how $\mu/M_{\text{Pl}}$ is constrained, the slow-roll predictions are displayed for $\alpha = 1/2$ in Fig. 207, and with $\mu$ varying. One can see that even if one allows values of $\mu$ larger than the typical ones ($\mu/M_{\text{Pl}} \simeq 10^{-4}$) these models are disfavored by the observations since they deviate too much from scale invariance.

## 6.10. $\beta$ Exponential Inflation (BEI)

This model was introduced and studied in Ref. [591] as a phenomenological generalization of the PLI exponential potential (see Section 5.8). The potential is given by

$$ V(\phi) = M^4 \exp_{1-\beta} \left( -\lambda \frac{\phi}{M_{\text{Pl}}} \right), \qquad (6.180) $$

where the generalized exponential function $\exp_{1-\beta}$ is defined by

$$ \exp_{1-\beta}(f) = \begin{cases} (1 + \beta f)^{1/\beta} & \text{for } 1 + \beta f > 0, \\ 0 & \text{otherwise.} \end{cases} \qquad (6.181) $$

As discussed in Ref. [591], for $f > 0$ and $g > 0$, this function satisfies the following identities:

$$ \exp_{1-\beta}[\ln_{1-\beta}(f)] = f, \qquad \ln_{1-\beta}(f) + \ln_{1-\beta}(g) = \ln_{1-\beta}(fg) - \beta[\ln_{1-\beta}(f)\ln_{1-\beta}(g)], \qquad (6.182) $$
<figure><img src="image_75.png" /><figcaption>Fig. 72. $\beta$ Exponential Inflation (BEI) for $\beta = 0.1$. Upper panels: the potential and its logarithm. Bottom left panel: slow-roll parameter $\epsilon_1$ with respect to the field values. The shaded area indicates where inflation stops if $\lambda = 1$. Bottom right panel: slow-roll parameters $\epsilon_2 = \epsilon_3$.</figcaption></figure>

where $\ln_{1-\beta}(f) = (f^{\beta} - 1)/\beta$ is the generalized logarithmic function. In the limit $\beta \to 0$, all the above expressions reproduce the usual exponential and logarithm properties. Therefore, the limit $\beta \to 0$ reproduces the PLI potential (see Section 5.8). However, as discussed below, this is not the case for the observable predictions which remain different. Defining the quantity $x$ by

$$x \equiv \frac{\phi}{M_{Pl}}, \qquad (6.183)$$

the range of field $\nu\nu$ for which inflation occurs depends on the sign of $\beta$. For $\beta > 0$, the field values are such that $x < 1/(\beta\lambda)$, whereas if $\beta < 0$, the potential is defined for $x > 1/(\beta\lambda)$. In both cases, inflation proceeds from the left to the right. The first three Hubble flow functions in the slow-roll approximation are given by

$$\epsilon_1 = \frac{\lambda^2}{2(1 - \beta\lambda x)^2}, \qquad \epsilon_2 = \frac{2\beta\lambda^2}{(1 - \beta\lambda x)^2} = 4\beta\epsilon_1, \qquad \epsilon_3 = \epsilon_2. \qquad (6.184)$$

Together with the potential, they are represented in Fig. 72.

One immediately sees that $\epsilon_1$ is an increasing function of $x$ only for the case where $\beta > 0$. Therefore inflation can naturally stop at $x_{end}$ such that $\epsilon_1(x_{end}) = 1$. In the opposite situation, namely $\beta < 0$, inflation has to be ended by some additional mechanism and $x_{end}$ would become an extra-parameter. Since this model is purely phenomenological, in the following, we restrict ourselves to the case $\beta > 0$ for which

$$x_{end} = \frac{1}{\beta} \left( \frac{1}{\lambda} - \frac{1}{\sqrt{2}} \right). \qquad (6.185)$$

The next step consists in determining the slow-roll trajectory. It can be integrated explicitly and the result reads

$$N - N_{end} = \frac{1}{\lambda} (x - x_{end}) - \frac{\beta}{2} (x^2 - x_{end}^2). \qquad (6.186)$$

It can also be inverted and one obtains the following expression for $x$ as a function of the $e$-folds number

$$x = \frac{1}{\lambda\beta} - \sqrt{\left(x_{end} - \frac{1}{\lambda\beta}\right)^2 - \frac{2}{\beta}(N - N_{end})}. \qquad (6.187)$$

Using these expressions, the observable field value $x_*$ can be related to the number of $e$-folds $\Delta N_* = N_{end} - N_*$ at which the pivot scale crossed out the Hubble radius during inflation. Making use of Eq. (6.185), one gets

$$x_* = \frac{1}{\lambda\beta} - \sqrt{\frac{1}{2\beta^2} + \frac{2}{\beta}\Delta N_*}. \qquad (6.188)$$
Inserting this expression into the slow-roll parameters formulas yields

$$ \epsilon_{1*} = \frac{1}{1 + 4\beta\Delta N_*}, \quad \epsilon_{2*} = \epsilon_{3*} = 4\beta\epsilon_{1*}. \qquad (6.189) $$

Therefore, the slow-roll predictions of these models do not depend on the parameter $\lambda$. Moreover, the limit $\beta \to 0$ does not give the same observable predictions as for the PLI models due to the singular behavior of $x_{end}$. These models can therefore be viewed as a completely different class.

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

$$ \left(\frac{M}{M_{\text{Pl}}}\right)^4 = 720\pi^2 \lambda^2 (1 - \beta \lambda x_*)^{-2-\frac{1}{\beta}} \frac{Q^2_{\text{rms}-\text{PS}}}{T^2}. \qquad (6.190) $$

Notice that, from Eq. (6.188), the above expression can be written in terms of $\Delta N_*$ and that it does not depend on $\lambda$ anymore. The reheating consistent slow-roll predictions for the BEI models are displayed in Fig. 208. The parameter $\beta$ must be such that $\beta \gtrsim 0.6$ in order for the predictions of the model to remain inside the two-sigma confidence intervals, while the parameter $\lambda$ remains totally unconstrained.

## 6.11. Pseudo Natural Inflation (PSNI)

### 6.11.1. Theoretical justifications

Pseudo Natural Inflation (PSNI) was introduced and studied in Ref. [337]. This model has common points with NI, see Section 5.6. Indeed, in PSNI, the inflaton field is also a pseudo-Nambu Goldstone boson which appears after symmetry breaking. The corresponding potential is nearly flat which is well-suited for inflation. The main ideas behind this construction are reviewed in Section 5.6. The main difference with respect to natural inflation, for which the broken symmetry is a shift symmetry, is that in pseudo natural inflation the broken symmetry is now a $U(1)$ one. A concrete implementation of this idea has been proposed in Ref. [337] and starts with the following supersymmetric hybrid superpotential

$$ W(S, X, \varphi, \psi_1, \psi_2) = \lambda_0 S (\psi_1^2 + \psi_2^2 - f^2) + \frac{\lambda_1}{2} \psi_1 \varphi^2 + \lambda_2 X (\varphi^2 - \psi^2), \qquad (6.191) $$

with $\lambda_1^2 f^2 > 2\lambda_2^2 \psi^2$, where $S, X, \psi_1, \psi_2$ and $\varphi$ are scalar fields and $\lambda_0, \lambda_1$ and $\lambda_2$ are coupling constants. We see that the $U(1)$ symmetry is explicitly broken by the term proportional to $\lambda_1$. The corresponding potential can be written as

$$ V = \lambda_0^2 |\psi_1^2 + \psi_2^2 - f^2|^2 + 2|\lambda_0 S \psi_1 + \frac{\lambda_1}{2} \varphi^2|^2 + 4\lambda_0^2 |S \psi_2|^2 + |\varphi|^2 |\lambda_1 \psi_1 + 2\lambda_2 X|^2 + \lambda_2^2 |\varphi^2 - \psi^2|^2. \qquad (6.192) $$

The flat directions of this superpotential can be reparametrized as

$$ \psi_1 + i\psi_2 \equiv (f + \sigma)e^{i\phi/f}, \quad \psi_1 - i\psi_2 \equiv (f - \sigma)e^{-i\phi/f}, \qquad (6.193) $$

where $\phi$ is the Nambu–Goldstone boson associated to the broken $U(1)$ symmetry and $\sigma$ is a modulus. One can assume that $\sigma$ is stabilized and sits at $\sigma = 0$, the minimum of a potential originating from supersymmetry breaking. The field $\phi$ plays the role of the inflaton. Using the above expressions and the condition $\sigma = 0$, one obtains that $\psi_1 = f \cos(\phi/f)$ and $\psi_2 = f \sin(\phi/f)$. In that case, a flat direction for $\phi$ is obtained for $\phi = 0$ and $S = 0$ since then we have

$$ V = \lambda_2^2 \psi^4. \qquad (6.194) $$

Notice that SUSY is broken because $F_X \equiv \langle \partial W / \partial X \rangle = \lambda_2 \psi^4 \neq 0$. As a consequence, the corresponding vacuum energy density is indeed given by $V_0 \simeq |F_X|^2 = \lambda_2^2 \psi^4$.

This tree level potential is corrected by two kind of contributions. First, supergravity induces a soft SUSY breaking mass of order $H$ for every scalar, but since $\phi$ is a pseudo Nambu–Goldstone boson, it only receives a potential due to the explicit breaking term proportional to $\lambda_1$. The corresponding contribution is loop suppressed, $m_\phi^2 \simeq 3\lambda_1^2 H^2 / (16\pi^2)$, as soon as $\lambda_1 \lesssim 1$ which will be assumed. Second, the potential receives a direct Yukawa mediated contribution through a $\phi$ loop and Ref. [337] has shown that it takes the form

$$ V(\phi) \simeq V_0 \left( 1 + \frac{\lambda_2^2}{4\pi^2} \ln \frac{\lambda_1 \psi_1}{\mu} \right) = V_0 \left[ 1 + \frac{\lambda_2^2}{4\pi^2} \ln \frac{\cos(\phi/f)}{\mu/f} \right]. \qquad (6.195) $$

where $\mu$ is some renormalization scale. The above formula gives rise to a new type of potential that we study in the next sub-section.

### 6.11.2. Slow-roll analysis

We now turn to the slow-roll analysis of the PSNI model. Using more friendly notations, the potential (6.195) can be re-expressed as

$$ V = M^4 \left[ 1 + \alpha \ln \left( \cos \frac{\phi}{f} \right) \right], \qquad (6.196) $$

with the following definitions

$$ M^4 = \lambda_2^2 \psi^4 \left[ 1 + \frac{\lambda_2^2}{4\pi^2} \ln \left( \frac{\lambda_1 f}{\mu} \right) \right], \quad \alpha = \frac{\lambda_2^2 / (4\pi^2)}{1 + \lambda_2^2 / (4\pi^2) \ln \left( \frac{\lambda_1 f}{\mu} \right)}. \qquad (6.197) $$

Therefore, one typically has $\alpha \ll 1$, and the scale $f$ should a priori be such that $f \lesssim M_{\text{Pl}}$ in order to avoid the usual problems of natural inflation.

The potential (6.196) as well as its logarithm are displayed in Fig. 73. Since $\phi$ is assumed to be such that $\phi \simeq 0$ initially, the potential must be studied in the range $\phi/f \in [0, \pi/2]$. It is positive definite in the range $\phi/f \in [0, \arccos(e^{-1/\alpha})]$. We see that it is a decreasing function of the inflaton $\text{vev}$, which means that inflation proceeds from the left to the right in the direction specified by the arrow in Fig. 73.

Let us now turn to the slow-roll parameters. If one defines $x \equiv \phi/f$, then the three first Hubble flow parameters are given by

$$ \epsilon_1 = \frac{M_{\text{Pl}}^2}{2f^2} \frac{\alpha^2 \tan^2 x}{(1 + \alpha \ln \cos x)^2}, \quad \epsilon_2 = 2\alpha \frac{M_{\text{Pl}}^2}{f^2} \frac{1 + \alpha + \alpha \ln \cos x - \alpha \cos^2 x}{\cos^2 x (1 + \alpha \ln \cos x)^2}, \qquad (6.198) $$
<figure><img src="image_76.png" /><figcaption>Fig. 73. Top left panel: Pseudo Natural Inflation (PSNI) potential, for $\alpha = 0.1$, as a function of $\phi/f$. Top right panel: logarithm of the potential for the same value of $\alpha$. Bottom left panel: slow-roll parameter $\epsilon_1$, rescaled by the quantity $M_{\text{Pl}}^2/f^2$ such that it acquires a universal form, for the same value of $\alpha$. Bottom right panel: slow-roll parameter $\epsilon_2$ (solid line) and $\epsilon_3$ (dotted line), rescaled by the quantity $M_{\text{Pl}}^2/f^2$, still for the same value of $\alpha$.</figcaption></figure>

$$ \epsilon_3 = \alpha \frac{M_{\text{Pl}}^2}{f^2} (\tan x)^2 \frac{2 + 3\alpha + \alpha^2 - \alpha^2 \cos(2x) + (4 + 3\alpha)\alpha \ln \cos x + 2\alpha^2 \ln^2 \cos x}{(1 + \alpha \ln \cos x)^2 (1 + \alpha \ln \cos x + \alpha \sin^2 x)} \quad (6.199) $$

They are displayed in Fig. 73. We see on this plot that the slow-roll parameters $\epsilon_1$ and $\epsilon_3$ vanish when $x$ goes to 0 and diverge when $x$ goes to $\pi/2$. On the other hand, the slow-roll parameter $\epsilon_2$ has a non-zero limit when $x$ goes to 0, namely

$$ \lim_{x \to 0} \epsilon_2 = 2 \frac{M_{\text{Pl}}^2}{f^2} \alpha. \quad (6.200) $$

This quantity should be small in order for slow-roll to be valid. This means that, at a fixed scale $f$, the parameter $\alpha$ needs to be smaller than $f^2/M_{\text{Pl}}^2$. From the monotonic behavior of $\epsilon_1$, one also notices that inflation naturally stops at $\epsilon_1 = 1$. Unfortunately, this equation cannot be solved exactly and the solution needs to be determined numerically. However, since we are in a regime where $f/M_{\text{Pl}} \ll 1$ and $\alpha M_{\text{Pl}}^2/f^2 \ll 1$, $x_{\text{end}}$ must be close to $\pi/2$. One can derive a better approximation by solving the equation $\epsilon_1 = 1$ using an expansion in the small quantities of the problem. One arrives at

$$ x_{\text{end}} \approx \frac{\pi}{2} - \frac{\alpha M_{\text{Pl}}}{\sqrt{2} f}, \quad (6.201) $$

that is to say the first correction to $\pi/2$ is linear in $\alpha M_{\text{Pl}}/f$ and, as expected, negative. As usual, the ASPIC code makes use of the complete slow-roll solution.

Let us now turn to the slow-roll trajectory. It can be integrated exactly in terms of the dilogarithm function $Li_2$ (also referred to as Spence's function, or Jonquière function). This function was already used in this paper, for instance in Section 5.1. The explicit expression of the trajectory reads

$$ N_{\text{end}} - N = \frac{f^2}{\alpha M_{\text{Pl}}^2} \left[ \begin{aligned} & (1 + \alpha \ln \cos x_{\text{end}}) \ln \sin x_{\text{end}} + \frac{\alpha}{4} \text{Li}_2 (\cos^2 x_{\text{end}}) \\ & - \frac{f^2}{\alpha M_{\text{Pl}}^2} \left[ (1 + \alpha \ln \cos x) \ln \sin x + \frac{\alpha}{4} \text{Li}_2 (\cos^2 x) \right] \end{aligned} \right] \quad (6.202) $$

where $N_{\text{end}}$ is the number of e-folds at the end of inflation. Unfortunately, this trajectory cannot be inverted analytically. However, if one uses the two conditions $f/M_{\text{Pl}} \ll 1$ and $\alpha M_{\text{Pl}}^2/f^2 \ll 1$, one can simplify a lot its expression. In particular, at Hubble crossing, one can write

$$ \Delta N_* \approx \frac{f^2}{2\alpha M_{\text{Pl}}^2} \left[ \left(x_* - \frac{\pi}{2}\right)^2 - \left(x_{\text{end}} - \frac{\pi}{2}\right)^2 \right], \quad (6.203) $$
from which one can obtain an explicit formula for $x_*$

$$x_* \approx \frac{\pi}{2} - \sqrt{2\alpha\Delta N_*} \frac{M_{\text{Pl}}}{f}. \qquad (6.204)$$

Then, this also allows us to derive useful approximated equations for the first three Hubble flow parameters, namely

$$\epsilon_{1*} \approx \frac{\alpha}{4\Delta N_*}, \quad \epsilon_{2*} \approx \epsilon_{3*} \approx \frac{1}{\Delta N_*}. \qquad (6.205)$$

The expressions of the tensor-to-scalar ratio, spectral index and running are

$$r \approx \frac{4\alpha}{\Delta N_*}, \quad n_s - 1 \approx \alpha_s \approx -\frac{1}{\Delta N_*}, \qquad (6.206)$$

These formulas are in agreement with the estimates given in Ref. [337]. Interestingly enough, we see that these predictions are independent of the scale $f$ and that the spectral index (and the running) is even independent of $\alpha$.

The last step consists in using the CMB normalization in order to extract the mass scale $M$. Straightforward manipulations lead to

$$\left(\frac{M}{M_{\text{Pl}}}\right)^4 = 720\pi^2\alpha^2 \frac{M_{\text{Pl}}^2}{f^2} \frac{\tan^2 x_*}{(1 + \alpha \ln \cos x_*)^3} \frac{Q_{\text{rms-PS}}^2}{T^2}. \qquad (6.207)$$

Under the two conditions $f/M_{\text{Pl}} \ll 1$ and $\alpha M_{\text{Pl}}^2/f^2 \ll 1$ and using the same method as before, this leads to

$$\left(\frac{M}{M_{\text{Pl}}}\right)^4 \approx \frac{360\pi^2\alpha}{\Delta N_*} \frac{Q_{\text{rms-PS}}^2}{T^2}. \qquad (6.208)$$

Requiring $M < M_{\text{Pl}}$ is easily achieved since, for the fiducial value $\Delta N_* \approx 55$, this is equivalent to $\alpha \lesssim 2580$ whereas we have $\alpha \ll 1$. Taking the more realistic value $\alpha \approx 10^{-6}$ and $\Delta N_* \approx 55$, one typically obtains that $M/M_{\text{Pl}} \approx 10^{-3}$.

The predictions of the PSNI models are displayed in Fig. 210 for $f/M_{\text{Pl}} = 10^{-3}, 10^{-1}, 10$ respectively (although this last value is considered just for the purpose of illustration since super-Planckian values of $f$ are not very physical). The reheating equation of state parameter $\bar{w}_{\text{reh}}$ has been taken to 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 and can take different values (in the ASPIC code, this parameter can be freely chosen). One can see that the rough description provided by Eqs. (6.205) is correct: when $\alpha M_{\text{Pl}}^2/f^2 \ll 1$, the deviation from scale invariance does not depend on the model parameters and is of the order of $n_s \approx 1 - 1/\Delta N_* \approx 0.975$, while $r \approx 4\alpha/\Delta N_*$ is typically very small.

## 6.12. Non Canonical Kähler Inflation (NCKI)

### 6.12.1. Theoretical justifications

This model was introduced and studied in Ref. [474] as a way to model hilltop inflation. The idea is to consider $F$ or $D$ term inflation in which we have a flat direction lifted by one loop corrections. This gives rise to loop inflation as discussed in Section 5.12. The LI potential has been obtained, however, under the assumption of a minimal Kähler potential. Now, corrections originating from higher order operators, always present in the Kähler potential, should typically produce a mass term and, therefore, the scalar potential gets modified and takes the form

$$V(\phi) \approx V_0 + \alpha \ln \left(\frac{\phi}{Q}\right) + b\phi^2, \qquad (6.209)$$

where $Q$ is a renormalization scale. This is the model we study in this section. Let us notice that the coefficient $b$ can be positive or negative. The case $b > 0$ has been investigated in Refs. [592,593] as “hybrid inflation with quasi-canonical supergravity” and the case $b < 0$ was studied in Ref. [474]. For $b > 0$, the potential (6.209) can be viewed as a valley hybrid potential [VHI, see Section 7.2 and Eq. (7.29)] plus logarithmic radiative corrections. Therefore, a consistency check of our calculations will be that, when $\alpha \to 0$, all the formulas derived below must reproduce those derived in Section 7.2. Finally, let us mention that the potential (6.209) has also been studied in Ref. [594] for $b < 0$ under the name “SUSY breaking potential” and in Ref. [595] in the context of supersymmetric hybrid inflation.

### 6.12.2. Slow-roll analysis

In this sub-section, we now turn to the slow-roll analysis of the NCKI scenario. For this purpose, it is convenient to re-write the potential (6.209) under the following form

$$V = M^4 \left[ 1 + \alpha \ln \left( \frac{\phi}{M_{\text{Pl}}} \right) + \beta \left( \frac{\phi}{M_{\text{Pl}}} \right)^2 \right], \qquad (6.210)$$

where $\alpha$ is a small positive dimensionless parameter and $\beta$ a dimensionless parameter of order $\mathcal{O}(1)$ which can be either positive or negative. Notice that the coefficient $\alpha$ has been redefined and that $\beta$ is directly related to $b$.

The potential (6.210), as well as its logarithm, are displayed in Fig. 74. We now describe its shape. For this purpose, let us first define the quantity $x \equiv \phi/M_{\text{Pl}}$. If $\beta > 0$, the potential is definite positive provided $x > x_{V=0}^-$, where

$$x_{V=0}^- = \left[ \frac{\alpha}{2\beta} W_0 \left( \frac{2\beta}{\alpha} e^{-2/\alpha} \right) \right]^{1/2}, \qquad (6.211)$$

and where $W_0$ is the “0”-branch of the Lambert function. In this case, the potential is an increasing function of the field $\text{vev}$ and, therefore, inflation proceeds from the right to the left in the direction indicated by the arrow in Fig. 74. Let us also notice that, in this case, the potential has an inflection point located at $x_{V''=0} = \sqrt{\alpha/(2\beta)}$. If $\beta < 0$, we must have $2\beta/\alpha \exp(1-2/\alpha) > -1$ in order to avoid the situation where the potential is everywhere negative. This implies that either $\beta > -1$ or $\beta < -1$ and, in this last case, $\alpha < -2/W_{-1} [1/(e\beta)]$ or $\alpha > -2/W_0 [1/(e\beta)]$. If one of these
<figure><img src="image_77.png" /><figcaption>Fig. 74. Top left panel: Non Canonical Kähler Inflation (NCKI) potential for $\alpha = 0.1$ and $\beta = \pm 1$. The solid blue line represents the case $\beta = -1$ while the solid pink line represents the case $\beta = 1$. Top right panel: logarithm of the potential for the same values of $\alpha$ and $\beta$. Bottom left panel: slow-roll parameter $\epsilon_1$, for a potential with the same values of $\alpha$ and $\beta$ and the same color code. The shaded area indicates the region where inflation is not possible. Bottom right panel: slow-roll parameters $\epsilon_2$ (solid blue and pink lines) and $\epsilon_3$ (dotted blue and pink lines), for a potential with the values of $\alpha$ and $\beta$ already considered in the other panels.</figcaption></figure>

conditions is satisfied (which is generically the case when $\alpha \ll 1$), the potential is positive provided $x_{V=0}^{-} < x < x_{V=0}^{+}$, where $x_{V=0}^{-}$ is defined in Eq. (6.211) and where

$$x_{V=0}^{+} = \left[ \frac{\alpha}{2\beta} W_{-1} \left( \frac{2\beta}{\alpha} e^{-2/\alpha} \right) \right]^{1/2}, \qquad (6.212)$$

$W_{-1}$ being the $-1$ branch of the Lambert function. In this case, the potential is a concave function of the field $\text{vev}$, with a maximum located at $x_{V'=0} = \sqrt{-\alpha/(2\beta)}$. Typically, inflation proceeds from the right to the left at small values of the field $\text{vev}$ compared to the Planck mass. The Hubble flow functions in the slow-roll approximation are given by

$$\epsilon_1 = \frac{(\alpha + 2\beta x^2)^2}{2x^2 (1 + \alpha \ln x + \beta x^2)^2}, \qquad (6.213)$$

$$\epsilon_2 = 2 \frac{\alpha(\alpha + 1) + (5\alpha - 2)\beta x^2 + 2\beta^2 x^4 + \alpha(\alpha - 2\beta x^2) \ln x}{x^2 (1 + \alpha \ln x + \beta x^2)^2}, \qquad (6.214)$$

and

$$\epsilon_3 = \frac{1}{x^2} \left[ \frac{2(\alpha + 2\beta x^2)^2}{(1 + \alpha \ln x + \beta x^2)^2} + \frac{\alpha - 2\beta x^2}{1 + \alpha \ln x + \beta x^2} + \frac{\alpha^2 + 8\alpha\beta x^2 - 4\beta^2 x^4}{\alpha(\alpha + 1) + (5\alpha - 2)\beta x^2 + 2\beta^2 x^4 + \alpha(\alpha - 2\beta x^2) \ln x} \right]. \qquad (6.215)$$

The are displayed in the bottom panels in Fig. 74. If $\beta > 0$, the first slow-roll parameter $\epsilon_1$ diverges when $x \to x_{V=0}^{-}$. For $x > x_{V=0}^{-}$, it first decreases, then reaches a minimum, then increases and reaches a local maximum. Finally, from this maximum, it decreases again and vanishes at infinity. Therefore, inflation stops at a $\text{vev}$ $x_{\text{end}}$ solution of $\epsilon_1(x_{\text{end}}) = 1$, which cannot be solved analytically. It can be noticed that the value of $\epsilon_1$ as its local maximum increases when $\alpha$ decreases. In the limit $\alpha \ll 1$, one has

$$\epsilon_1^{\text{max}} \approx \frac{\beta}{2}, \qquad (6.216)$$

which is reached at $x_{\epsilon_1^{\text{max}}} \approx 1/\sqrt{\beta}$ (still in the limit of very small $\beta$). This sets an upper bound on $\beta$ in order for this local maximum to satisfy $\epsilon_1 \ll 1$. If not, inflation would proceed in the part of the potential beyond its inflection point, corresponding to “large values” of the field $\text{vev}$ and the model would formally be equivalent to a quadratic model (LFI$_2$, see Section 5.2).
If $\beta < 0$, the first slow-roll parameter diverges when $x \to x_{V=0}^-$. For $x > x_{V=0}^-$, $\epsilon_1$ decreases, vanishes at the potential local maximum $x_{V'=0}$, and then increases to blow up when $x \to x_{V=0}^+$. At the same time, the second slow-roll parameter $\epsilon_2$ decreases in the inflationary range $x_{V=0}^- < x < x_{V'=0}^-$. Let us also notice that, since $\epsilon_2(x_{V'=0}) \propto 2\alpha - \alpha^2 + \alpha^2 \ln[-\alpha/(2\beta)]$, one has $\epsilon_2 > 0$, thanks to the condition $2\beta/\alpha \exp(1-2/\alpha) > -1$. Therefore the minimum value of $\epsilon_2$ in the increasing branch of the potential is reached at the potential maximum and is given by

$$ \epsilon_2^{\min} = \frac{-16\beta}{2 - \alpha \left[ 1 + \ln(-2\frac{\beta}{\alpha}) \right]} \qquad (6.217) $$

For $\alpha < -2\beta/e$ (which is generically the case since $\alpha \ll 1$), this number is such that $\epsilon_2^{\min} > -8\beta$, which puts a lower bound on $\beta$ in order for $\epsilon_2$ to remain small and slow-roll to be satisfied. As it was the case for $\beta > 0$, inflation also ends when $\epsilon_1 = 1$. Notice that the exact calculations are implemented in the ASPIC routines.

Let us now turn to the slow-roll trajectory. It can be analytically integrated using the dilogarithm function $Li_2$ and the corresponding expression reads

$$ \begin{aligned} N_{\text{end}} - N = & \left(1 - \frac{\alpha}{2} + \alpha \ln x\right) \frac{\ln\left(\alpha + 2\beta x^2\right)}{4\beta} + \frac{x^2}{4} - \frac{\alpha}{4\beta} \ln \alpha \ln x + \frac{\alpha}{8\beta} \text{Li}_2\left(-2\frac{\beta}{\alpha}x^2\right) \\ & - \left(1 - \frac{\alpha}{2} + \alpha \ln x_{\text{end}}\right) \frac{\ln\left(\alpha + 2\beta x_{\text{end}}^2\right)}{4\beta} - \frac{x_{\text{end}}^2}{4} + \frac{\alpha}{4\beta} \ln \alpha \ln x_{\text{end}} - \frac{\alpha}{8\beta} \text{Li}_2\left(-2\frac{\beta}{\alpha}x_{\text{end}}^2\right), \end{aligned} \qquad (6.218) $$

where $N_{\text{end}}$ is the number of $e$-folds at the end of inflation. An approximate and simpler expression can be derived in the limit $\alpha \ll 1$. In that limit, one obtains $N_{\text{end}} - N = x^2/4 + \ln(x)/(2\beta) - x_{\text{end}}^2/4 - \ln(x_{\text{end}})/(2\beta)$, which is precisely the slow-roll trajectory for the VHI models with $\mu = M_{\text{Pl}}/\sqrt{\beta}$ and $p=2$, see Eq. (7.35). For $\alpha \neq 0$, the exact trajectory cannot be inverted analytically.

Finally, the parameter $M$ can be determined from the CMB normalization. One obtains the following expression

$$ \left(\frac{M}{M_{\text{Pl}}}\right)^4 = 720\pi^2 \frac{(\alpha + 2\beta x_*^2)^2}{x_*^2 (1 + \alpha \ln x_* + \beta x_*^2)^3} \frac{Q_{\text{rms}-\text{PS}}^2}{T^2}. \qquad (6.219) $$

The slow-roll predictions of the NCKI models are displayed in Figs. 213 and 217 for $\beta > 0$ and $\beta < 0$, respectively. 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 in fact unspecified. Some remarks are in order at this point. Firstly, when $\beta > 0$, we notice that $\epsilon_2$ at Hubble crossing is either positive or negative while, when $\beta < 0$, it is always positive. This is in agreement with what we have discussed before. Secondly, when $\beta > 0$ and $\alpha \ll 1$, one can check that the predictions of the models are similar to the VHI ones with $p=2$ (compare with Fig. 345). Again, this is consistent with the previous considerations. Thirdly, when $|\beta| \gtrsim \mathcal{O}(1)$, the predictions of the models do not depend much on $\beta$. Finally, as expected, when $\beta \to 0$, one recovers the predictions of the LI models, see Section 5.12 and Fig. 140. Now, in the regime $|\beta| = \mathcal{O}(1)$ and $\alpha \ll 1$, Figs. 213 and 217 indicate that the case $\beta > 0$ is disfavored by the observations. The situation is even worst for $\beta < 0$, the deviation from scale invariance being clearly too important to satisfy the observational constraints.

## 6.13. Constant Spectrum Inflation (CSI)

This potential belongs to the class of models discussed in Ref. [596] and is constructed in order to produce a power spectrum $P(k) \propto k^0$ for the primordial density fluctuations, i.e. a power spectrum with constant spectral index such that $n_s = 1$ (exact scale invariance). It reads

$$ V(\phi) = \frac{M^4}{\left(1 - \alpha \frac{\phi}{M_{\text{Pl}}}\right)^2}. \qquad (6.220) $$

There is a symmetry for $\phi/M_{\text{Pl}} \to 2/\alpha - \phi/M_{\text{Pl}}$ and inflation can proceed indifferently in the branch $\phi/M_{\text{Pl}} < 1/\alpha$ or in the branch $\phi/M_{\text{Pl}} > 1/\alpha$, leading to the same physical predictions. For this reason, in the following, we will be interested in the branch $\phi/M_{\text{Pl}} < 1/\alpha$. Defining the quantity $x$ by

$$ x \equiv \frac{\phi}{M_{\text{Pl}}}, \qquad (6.221) $$

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

$$ \epsilon_1 = \frac{2\alpha^2}{(\alpha x - 1)^2}, \qquad \epsilon_2 = \epsilon_3 = -2\epsilon_1. \qquad (6.222) $$

The previous relation $\epsilon_2 = -2\epsilon_1$ means that, at first order in slow-roll, the spectral index is indeed equals to unity, $n_s - 1 = 0$. Recall that the potential of this model is precisely constructed in order for this relation to be true. Let us notice, however, that, at second order in slow-roll, $\epsilon_2 = \epsilon_3 = -2\epsilon_1$ yields $n_s - 1 = 4\epsilon_1^2 > 0$. One should note that another way to realize $n_s - 1 = 0$ at first order in slow-roll is to take the large field inflation potential LFI (see Section 5.2) with a negative power index $p = -2$. In that case one also has $\epsilon_2 = \epsilon_3 = -2\epsilon_1$ and, at second order, $n_s - 1 = 4\epsilon_1^2$ is also verified. However, since the explicit expressions of $\epsilon_1$ for CSI and LFI ($p = -2$) are different, the actual value of the spectral index at second order is also different. The potential and the Hubble flow functions have been represented in Fig. 75.

As can be checked in this figure, $\epsilon_1$ is a monotonic function of $x$ in both branches of the potential. It diverges at $x = 1/\alpha$ and vanishes for $x \to \pm\infty$. Inflation can therefore take place in the region $x < x_{\epsilon_1=1}^-$ for the branch $x < 1/\alpha$ (or $x > x_{\epsilon_1=1}^+$ for the branch $x > 1/\alpha$), where $x_{\epsilon_1=1}^{\pm}$ are the field values at which $\epsilon_1 = 1$:

$$ x_{\epsilon_1=1}^{\pm} = \frac{1 \pm \sqrt{2\alpha}}{\alpha}. \qquad (6.223) $$
<figure><img src="image_78.png" /><figcaption>Fig. 75. Constant Spectrum Inflation (CSI) for $\alpha = 0.1$. Upper panels: the potential and its logarithm along the branch $x < 1/\alpha$. Bottom left panel: slow-roll parameter $\epsilon_1$ together with the region in which it is larger than unity and in which inflation cannot occur (shaded). Bottom right panel: slow-roll parameter $\epsilon_2 = \epsilon_3$ along the same branch $x < 1/\alpha$.</figcaption></figure>

Since the field evolution proceeds from the right to the left from $x_{\epsilon_1=1}^{\pm}$, inflation does not stop by slow-roll violation and an extra mechanism parametrized by $x_{\text{end}}$ should be considered in order to end it. For this reason, CSI is in fact a two parameters model. Let us also notice that the slow-roll parameters $\epsilon_2 = \epsilon_3$ are negative monotonic functions of $x$ in both branches of the potential and cross the line $\epsilon_2 = \epsilon_3 = -1$ at

$$x_{\epsilon_2=-1}^{\pm} = x_{\epsilon_3=-1}^{\pm} = \frac{1 \pm 2\alpha}{\alpha}. \qquad (6.224)$$

As a result, there is a small domain $x_{\epsilon_2=-1}^{-} < x < x_{\epsilon_1=1}^{-}$ where we have inflation but where the slow-roll approximation is violated (this is also true for the other branch). This is not problematic since the system is driven away from this regime towards a situation in which all the Hubble flow functions become small (see Fig. 75).

The slow-roll trajectory can be integrated explicitly and reads

$$N_{\text{end}} - N = -\frac{x^2}{4} + \frac{x}{2\alpha} + \frac{x_{\text{end}}^2}{4} - \frac{x_{\text{end}}}{2\alpha}. \qquad (6.225)$$

It can also be inverted analytically and it follows that

$$x = \frac{1 \pm \sqrt{1 - 2\alpha x_{\text{end}} + \alpha^2 x_{\text{end}}^2 + 4\alpha^2 (N - N_{\text{end}})}}{\alpha}. \qquad (6.226)$$

The sign $\mp$ depends on whether one works in the $x < 1/\alpha$ branch or in the $x > 1/\alpha$ branch, respectively. A consequence of this formula is the fact that, if one requires $N_{\text{end}} - N_{\text{ini}}$ e-folds during inflation, then $x_{\text{end}}$ should be smaller than some value $x_{\text{end}}^{\text{max}}$ given by

$$x_{\text{end}}^{\text{max}} = \frac{1}{\alpha} - \sqrt{2 + 4(N_{\text{end}} - N_{\text{ini}})}, \qquad (6.227)$$

in the $x < 1/\alpha$ branch. Equivalently, taking the minus sign in this expression would lead to $x_{\text{end}}^{\text{min}}$ for the branch $x > 1/\alpha$.

Finally, the observable field value $x_*$ is obtained by solving Eq. (3.48) while the amplitude of the CMB anisotropies fixes the parameter $M$ to

$$(\frac{M}{M_{\text{Pl}}})^4 = 2880\pi^2\alpha^2 \frac{Q^2}{T^2}. \qquad (6.228)$$

Interestingly enough, it only depends on $\alpha$, and not on $x_*$ (i.e. it has no explicit dependence on the reheating). The reheating consistent slow-roll predictions for the CSI models are represented in Figs. 221 and 222 for $\alpha = 10^{-3}$ and $\alpha = 1$, respectively.
## 6.14. Orientifold Inflation (OI)

### 6.14.1. Theoretical justifications

The model is based on the following considerations. Let us start with a $N = 1$ supersymmetric Yang-Mills gauge theory the Lagrangian of which can be written as

$$ \mathcal{L} = -\frac{1}{4} F_{\mu\nu}^a F^{a\mu\nu} + \frac{i}{2} \bar{\lambda}^a \mathcal{D}_{ab} \lambda^b, \qquad (6.229) $$

with $a = 1, ..., N_c^2$, $N_c$ being the number characterizing the group SU($N_c$). $F_{\mu\nu}^a$ is the field strength, $\lambda^a$ a spinor field and $\mathcal{D}$ a covariant derivative. A is a composite scalar field, i.e. a bound state denoted by $\varphi \simeq \lambda\bar{\lambda}$, can actually appear in the theory if a strongly interacting regime takes place. The effective Lagrangian aimed at describing its dynamics has been derived in Ref. [597] and reads

$$ \mathcal{L}_{\text{YV}} = -\frac{N_c^2}{\alpha_{\text{OI}}} (\varphi\varphi^\dagger)^{-2/3} \partial_\mu \varphi \partial^\mu \varphi^\dagger - \frac{4\alpha_{\text{OI}} N_c^2}{9} (\varphi\varphi^\dagger)^{2/3} \ln\left(\frac{\varphi}{\Lambda^3}\right) \ln\left(\frac{\varphi^\dagger}{\Lambda^3}\right), \qquad (6.230) $$

where $\alpha_{OI}$ is a constant and $\Lambda$ a mass scale. This class of theories are discussed in more detail in Section 7.5. However, in Ref. [598], it was argued that in “orientifold theories”, the above Lagrangian can be slightly deformed and now takes the form

$$ \mathcal{L}_{\text{OI}} = -\frac{N_c^2}{\alpha_{\text{OI}}} (\varphi\varphi^\dagger)^{-2/3} \partial_\mu \varphi \partial^\mu \varphi^\dagger - \frac{4\alpha_{\text{OI}} N_c^2}{9} (\varphi\varphi^\dagger)^{2/3} \left[ \ln\left(\frac{\varphi}{\Lambda^3}\right) \ln\left(\frac{\varphi^\dagger}{\Lambda^3}\right) - \beta \right], \qquad (6.231) $$

where $\beta = \mathcal{O}(1/N_c)$. Ref. [598] raised the possibility that $\varphi$ (or, rather, its canonically conjugated version) could be the inflaton. In fact, in order to study this question, one must also specify the gravitational coupling. In Ref. [598], the scalar field $\varphi$ is non-minimally coupled to gravity such that, in the Jordan frame,

$$ S = \int d^4 x \sqrt{-g} \left[ -\frac{M^2 + N_c^2 \xi (\varphi \varphi^\dagger)^{1/3}}{2} R + \mathcal{L}_{\text{OI}} \right], \qquad (6.232) $$

where $M$ is a mass scale. There is a new parameter in the problem, $\xi$, which describes the strength of the non-minimal coupling to gravity (as it was the case for Higgs inflation, see Section 4.2). Then, in the Einstein frame, one can write the above model as Ref. [598]

$$ S = \int d^4 x \sqrt{-g} \left\{ -\frac{1}{2} M_{\text{Pl}}^2 R - \frac{N_c^2}{\alpha_{\text{OI}}} \Omega^{-2} \left[ 1 + \frac{\alpha_{\text{OI}} N_c^2 \xi^2}{3 M_{\text{Pl}}^2} \Omega^{-2} (\varphi \varphi^\dagger)^{1/3} \right] (\varphi \varphi^\dagger)^{-2/3} \partial_\mu \varphi \partial^\mu \varphi^\dagger - \Omega^{-4} V_{\text{OI}} \right\}. \qquad (6.233) $$

In this expression, $V_{OI}$ refers to the second term in Eq. (6.231) and

$$ \Omega^2 = \frac{M^2 + N_c^2 \xi (\varphi \varphi^\dagger)^{1/3}}{M_{\text{Pl}}^2}. \qquad (6.234) $$

In the following, we consider two situations: the case where $\xi \neq 0$ such that $\Omega^2 \simeq N_c^2 \xi \varphi^{2/3}/M_{\text{Pl}}^2$, i.e. the second term in the definition of $\Omega^2$ dominates (the large field limit) and the case $\xi = 0$. In the first case, taking $\varphi = \varphi^\dagger$ and canonically normalizing the field one finds

$$ V(\varphi) = \frac{4\alpha_{\text{OI}} M_{\text{Pl}}^4}{9 N_c^2 \xi^2} \left[ \left( \ln \frac{\varphi}{\Lambda^3} \right)^2 - \beta \right]. \qquad (6.235) $$

The canonically normalized field is $\phi/M_{\text{Pl}} \propto \ln \varphi$. Since $\beta$ is a small number, it can be neglected and this model is in fact a LFI model with $V(\phi) \propto \phi^2$ which was already studied in Section 5.2. For the second case, it is sufficient to restart from Eq. (6.231). Then, the canonically normalized field reads

$$ \frac{\varphi}{\Lambda^3} = \left( \frac{\phi}{\phi_0} \right)^3, \qquad (6.236) $$

$$ \text{with} \qquad \phi_0 = 3N_c \left( \frac{2}{\alpha_{\text{OI}}} \right)^{1/3} \Lambda. \qquad (6.237) $$

It follows that the potential can be written as

$$ V = \alpha_{\text{OI}} N_c^2 \Lambda^4 \left( \frac{\phi}{\phi_0} \right)^4 \left[ \ln^2 \left( \frac{\phi}{\phi_0} \right) - \frac{\beta}{9} \right]. \qquad (6.238) $$

This model is studied in detail in the next subsection. The case $\beta = 0$ will also be investigated in Section 7.5.

### 6.14.2. Slow-roll analysis

We now turn to the slow-roll study of the potential derived previously in Eq. (6.238). This one can be re-written as

$$ V(\phi) = M^4 \left( \frac{\phi}{\phi_0} \right)^4 \left[ \left( \ln \frac{\phi}{\phi_0} \right)^2 - \alpha \right], \qquad (6.239) $$

where we have defined

$$ M^4 = \alpha_{\text{OI}} N_c^2 \Lambda^4, \qquad \alpha \equiv \frac{\beta}{9}. \qquad (6.240) $$
<figure><img src="image_79.png" /><figcaption>Fig. 76. Orientifold Inflation (OI) for $\alpha = 0.1$. Upper panels: the potential and its logarithm. Bottom left panel: slow-roll parameter $\epsilon_1$, rescaled by the factor $\phi_0^2/M_{pl}^2$. The shaded area indicates where inflation cannot occur (for $\phi_0 = M_n$). Bottom right panel: rescaled slow-roll parameter $\epsilon_2$.</figcaption></figure>

One should be careful that $\alpha_{01}$ appearing in the first of the two above equations stems from the Lagrangian used in the previous subsection while the observable constant $\alpha$ only refers to the quantity $\beta/9 = \mathcal{O}(1/N_c) \ll 1$. The scale $\phi_0$ is defined in Eq. (6.237) and will be chosen such that $\phi_0 \simeq 10^{16}$ GeV. The potential as well as its logarithm are displayed in Fig. 76.

Defining the quantity $x$ by the following expression

$$x = \frac{\phi}{\phi_0}, \qquad (6.241)$$

the potential remains positive provided $x < x_{V=0}^{-}$ or $x > x_{V=0}^{+}$, where

$$x_{V=0}^{\pm} = e^{\pm\sqrt{\alpha}}. \qquad (6.242)$$

It vanishes at $x = 0$, then increases to reach a local maximum at $x_{V'=0}^{-}$, decreases again to become negative at $x_{V=0}^{-}$, reaches a local minimum at $x_{V'=0}^{+}$, then increases again to become positive at $x_{V=0}^{+}$ and diverges asymptotically. The values of $x_{V'=0}^{-}$ and $x_{V'=0}^{+}$ are given by

$$x_{V'=0}^{\pm} = e^{-\frac{1}{4}\pm\sqrt{\frac{1}{16}+\alpha}}. \qquad (6.243)$$

A priori three regimes of inflation may exist: $x < x_{V'=0}^{-}$ and inflation proceeds from the right to the left, $x_{V'=0}^{-} < x < x_{V=0}^{-}$ and inflation proceeds from the left to the right, $x_{V=0}^{+} < x$ and inflation proceeds from the right to the left in the direction specified by the arrow in Fig. 76. As explained below, only the third possibility allows us to have a slow-roll inflationary regime.

Let us now calculate the quantities $\epsilon_n$. The first three Hubble flow functions in the slow-roll approximation are given by

$$\epsilon_1 = 2 \frac{M_{Pl}^2}{\phi_0^2} \left( \frac{2 \ln^2 x + \ln x - 2\alpha}{x \ln^2 x - \alpha x} \right)^2, \qquad (6.244)$$

$$\epsilon_2 = 4 \frac{M_{Pl}^2}{\phi_0^2} \frac{2 \ln^4 x + \ln^3 x + (1 - 4\alpha) \ln^2 x - \alpha \ln x + \alpha + 2\alpha^2}{(x \ln^2 x - \alpha x)^2}, \qquad (6.245)$$

and

$$\epsilon_3 = 2 \frac{M_{Pl}^2}{\phi_0^2} \left[ \begin{aligned} & 8\alpha^4 + 6\alpha^3 - \alpha^2 (8\alpha + 15) \ln x + 2\alpha (3 - 16\alpha^2 - 2\alpha) \ln^2 x \\ & + 8\alpha (3\alpha + 1) \ln^3 x + 2 (24\alpha^2 - 5\alpha + 1) \ln^4 x + (7 - 24\alpha) \ln^5 x + 8 (1 - 4\alpha) \ln^6 x \\ & + 8 \ln^7 x + 8 \ln^8 x \end{aligned} \right] (x \ln^2 x - \alpha x)^{-2} \\ \times [2\alpha^2 + \alpha - \alpha \ln x + (1 - 4\alpha) \ln^2 x + \ln^3 x + 2 \ln^4 x]^{-1}. \qquad (6.246)$$
They have been represented in Fig. 76. One can see that the slow-roll regime can only take place in the $x > x_{V=0}^+$ region, where $\epsilon_1$ continuously increase as inflation proceeds from the right to the left, and diverges at $x_{V=0}^+$. In the other domains, $\epsilon_2$ remains too large to support slow-roll inflation. Within the $x > x_{V=0}^+$ domain, inflation naturally ends by slow-roll violation, but the field value $x_{end}$ at which this occurs has to be determined numerically. However, since $\phi_0 \simeq 10^{16}$ GeV, one can derive an approximated formula for $x_{end}$ in the $\phi_0 \ll M_{Pl}$ limit, namely

$$x_{\text{end}} \approx 2\sqrt{2} \frac{M_{\text{Pl}}}{\phi_0}. \qquad (6.247)$$

The next step is to derive the slow-roll trajectory. It can be obtained from Eq. (3.11) and reads

$$N_{\text{end}} - N = -\frac{\phi_0^2}{M_{\text{Pl}}^2} \left\{ \frac{x_{\text{end}}^2 - x^2}{8} + \frac{\ln^2(x_{V'=0}^+) - \alpha}{2\sqrt{1+16\alpha}} (x_{V'=0}^+)^2 \right[ \text{Ei}\left(2\ln\frac{x_{\text{end}}}{x_{V'=0}^+}\right) \right. \\ \left. - \text{Ei}\left(2\ln\frac{x}{x_{V'=0}^+}\right) \right] - \frac{\ln^2(x_{V'=0}^-) - \alpha}{2\sqrt{1+16\alpha}} (x_{V'=0}^-)^2 \left[ \text{Ei}\left(2\ln\frac{x_{\text{end}}}{x_{V'=0}^-}\right) - \text{Ei}\left(2\ln\frac{x}{x_{V'=0}^-}\right) \right] \right\}, \qquad (6.248)$$

where $\text{Ei}$ is the exponential integral function, and where $x_{V=0}^+ = 0$ have been defined in Eq. (6.243). In the $\phi_0 \ll M_{Pl}$ limit, this trajectory reduces to $\Delta N_* \simeq \phi_0^2/(8M_{Pl}^2(x_*^2 - x_{end}^2)$, where we have introduced the observable field value $x_*$ at which the pivot scale crossed the Hubble radius during inflation. It can be inverted to give $x_*$ in terms of $\Delta N_* = N_{end} - N_*$ and one gets

$$x_* \approx 2\sqrt{2} \frac{M_{\text{Pl}}}{\phi_0} \sqrt{\Delta N_* + 1}. \qquad (6.249)$$

Plugging this into Eqs. (6.244), (6.245) and (6.246) gives the approximated expressions

$$\epsilon_{1*} \approx \epsilon_{2*} \approx \epsilon_{3*} \approx \frac{1}{\Delta N_* + 1}, \qquad (6.250)$$

hence

$$r \approx \frac{16}{\Delta N_* + 1}, \qquad n_s - 1 \approx -\frac{3}{\Delta N_* + 1}, \qquad \alpha_s \approx -\frac{3}{(\Delta N_* + 1)^2}. \qquad (6.251)$$

From $x_*$, the parameter $M$ is fixed by the amplitude of the CMB anisotropies and one obtains

$$\left(\frac{M}{M_{\text{Pl}}}\right)^4 = \frac{2880\pi^2 (2 \ln^2 x_* + \ln x_* - 2\alpha)^2 M_{\text{Pl}}^2 Q_{\text{rms-PS}}^2}{x_*^6 (\ln^2 x_* - \alpha)^3 T^2}. \qquad (6.252)$$

In the $\phi_0 \ll M_{Pl}$ limit, the previous expression reduces to the following formula

$$\left(\frac{M}{M_{\text{Pl}}}\right)^4 \approx \frac{45\pi^2}{2(\Delta N_* + 1)^3} \left(\frac{\phi_0}{M_{\text{Pl}}}\right)^4 \frac{1}{\ln^2\left(2\sqrt{2}\frac{M_{\text{Pl}}}{\phi_0}\sqrt{\Delta N_* + 1}\right)} \frac{Q_{\text{rms-PS}}^2}{T^2}. \qquad (6.253)$$

With $\phi_0 \simeq 10^{16}$ GeV, this typically gives $M/M_{Pl} \simeq 5 \times 10^{-4}$.

The reheating consistent slow-roll predictions for the orientifold inflation models are displayed in Fig. 223, for $\phi_0/M_{Pl} = 10^{-4}$, $10^{-2}$, and 1. Let us recall that natural values are around $\phi_0 \simeq 10^{16}$ GeV and $\alpha \in [10^{-3}, 1]$. The reheating equation of state parameter has been fixed to $\bar{w}_{reh} = 0$ since the potential is quadratic in the vicinity of its minimum. According to the rough picture provided by Eq. (6.250), the predictions of these models almost do not depend on its parameters $\phi_0$ and $\alpha$, which is why all the points in Fig. 223 are superimposed. In particular, one can see that these models generically predict an important amount of gravitational waves which is disfavored by the observations.

## 6.15. Constant $n_s$ C Inflation (CNCI)

This model has been obtained in Ref. [507] and is the third example of a class of scenarios already studied in Sections 5.20 and 5.21. As explained in those sections, the corresponding potential is designed in order to produce a power spectrum with constant spectral index. The potential studied in this section reads

$$V(\phi) = M^4 \left[ (3 + \alpha^2) \coth^2 \left( \frac{\alpha}{\sqrt{2}} \frac{\phi}{M_{\text{Pl}}} \right) - 3 \right], \qquad (6.254)$$

where $\alpha$ is a positive dimensionless parameter (denoted $n_0$ in Ref. [507]). The potential being symmetrical in $\phi \to -\phi$, only the $\phi > 0$ part is displayed in Fig. 77. It is a decreasing function of the field $v\phi$, and its asymptotic value when $\phi/M_{Pl}$ goes to infinity is given by $\alpha^2 M^4$, hence the potential is always positive.

Defining $x = \phi/M_{Pl}$, the three first slow-roll parameters are given by

$$\epsilon_1 = \frac{4\alpha^2 (3 + \alpha^2)^2 \coth^2\left(\frac{\alpha x}{\sqrt{2}}\right)}{\left[6 + \alpha^2 + \alpha^2 \cosh\left(\sqrt{2}\alpha x\right)\right]^2}, \qquad (6.255)$$

$$\epsilon_2 = -\frac{2\alpha^2 (3 + \alpha^2) \left[12 + \alpha^2 + 2\alpha^2 \cosh\left(\sqrt{2}\alpha x\right) + \alpha^2 \cosh\left(2\sqrt{2}\alpha x\right)\right]}{\left[6 + \alpha^2 + \alpha^2 \cosh\left(\sqrt{2}\alpha x\right)\right]^2 \sinh^2\left(\frac{\alpha x}{\sqrt{2}}\right)}, \qquad (6.256)$$
<figure><img src="image_80.png" /><figcaption>Fig. 77. Top left panel: Constant $n_s$ C inflaton potential for $\alpha = 0.1$. Inflation proceeds from the left to the right as indicated by the arrow. Top right panel: logarithm of the potential for the same value of $\alpha$. Bottom left panel: the first slow-roll parameter $\epsilon_1$ for $\alpha = 0.1$. Bottom right panel: slow-roll parameters $\epsilon_2$ and $\epsilon_3$, still for $\alpha = 0.1$.</figcaption></figure>

and

$$ \epsilon_3 = -2\alpha^2 (3 + \alpha^2) \left[ 6(24 - 2\alpha^2 + \alpha^4) + (120\alpha^2 + 7\alpha^4) \cosh(\sqrt{2\alpha x}) \right. \\ \left. + 2\alpha^2 (\alpha^2 - 6) \cosh(2\sqrt{2\alpha x}) + \alpha^4 \cosh(3\sqrt{2\alpha x}) \right] \coth^2\left(\frac{\alpha}{\sqrt{2}}x\right) \\ \times \left[ 6 + \alpha^2 + \alpha^2 \cosh(\sqrt{2\alpha x}) \right]^{-2} \left[ 12 + \alpha^2 + 2\alpha^2 \cosh(\sqrt{2\alpha x}) + \alpha^2 \cosh(2\sqrt{2\alpha x}) \right]^{-1}. \quad (6.257) $$

These slow-roll parameters are displayed in Fig. 77 (bottom panels). We see that the first slow-roll parameters monotonically decrease during inflation. It blows up as the field $\mathrm{vev}$ approaches zero and tends to zero when the field $\mathrm{vev}$ goes to infinity. On the contrary, the second and third slow-roll parameters monotonically increase from $-\infty$ to zero as inflation proceeds.

Given the above described behavior of $\epsilon_1$, it is clear that inflation cannot stop by slow-roll violation. Therefore, it should be stopped by instability which means that an extra parameter $x_{\mathrm{end}}$ should be added to the model.

As for CNAI and CNBI, 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 $\epsilon_1$ and $\epsilon_2$ in $\alpha$, assuming that $x\alpha$ remains small, one obtains $\epsilon_1 = 2/x^2 + 2\alpha^2/3 + \mathcal{O}(\alpha^4)$ and $\epsilon_2 = -4/x^2 + 2\alpha^2/3 + \mathcal{O}(\alpha^4)$, so that $n_s - 1 = -2\alpha^2 + \mathcal{O}(\alpha^4)$. As for the similar calculations performed in Sections 5.20 and 5.21, one should remark that, if $x_{\mathrm{end}}$ is such that $\alpha x_* \gtrsim 1$, the previous expansion can be inaccurate and some deviations from constant $n_s$ may appear.

Let us now consider the slow-roll trajectory. It can be integrated analytically and is given by the following formula

$$ N - N_{\mathrm{end}} = \frac{1}{\alpha^2 (3 + \alpha^2)} \left\{ 3 \ln \left[ \cosh \left( \frac{\alpha}{\sqrt{2}} x \right) \right] + \frac{\alpha^2}{2} \cosh^2 \left( \frac{\alpha}{\sqrt{2}} x \right) \right. \\ \left. - 3 \ln \left[ \cosh \left( \frac{\alpha}{\sqrt{2}} x_{\mathrm{end}} \right) \right] - \frac{\alpha^2}{2} \cosh^2 \left( \frac{\alpha}{\sqrt{2}} x_{\mathrm{end}} \right) \right\}. \quad (6.258) $$

Moreover, this expression can be explicitly inverted. As a consequence, the function $x(N)$ can be written as

$$ x = \frac{\sqrt{2}}{\alpha} \operatorname{arccosh} \left[ \frac{3}{\alpha^2} W_0 \left( \frac{\alpha^2}{3} \exp \left\{ \frac{2}{3} \alpha^2 (3 + \alpha^2) (N - N_{\mathrm{end}}) \right\} \right) \right. \\ \left. + 2 \ln \left[ \cosh \left( \frac{\alpha}{\sqrt{2}} x_{\mathrm{end}} \right) \right] + \frac{\alpha^2}{3} \cosh^2 \left( \frac{\alpha}{\sqrt{2}} x_{\mathrm{end}} \right) \right]^{1/2}, \quad (6.259) $$

where $W_0$ is the Lambert function. The fact that we deal with the 0-branch is obvious since the argument of this function is positive definite.

The predictions of the CNCI models are displayed in Fig. 225, for $\alpha = 10^{-3}$, 0.1 and 0.2. The thin black solid lines are the lines such that $n_s - 1 = -2\alpha^2$. We see that, for very small values of $\alpha$, the predictions are indeed such that the spectral index is constant. For $\alpha$ not too small,
however, we also notice deviations from this law and the larger $\alpha$ the stronger these deviations. This is reminiscent with the phenomenon observed in Sections 5.20 and 5.21 but now $x_{\text{end}}$ is a free parameter and, for a given value of $\alpha$, the deviations from $n_s - 1 = -2\alpha^2$ become larger when $x_{\text{end}}$ increase (i.e. when the line becomes redder in Fig. 225). In this case, the Taylor expansion of the trigonometric functions which appear in the expressions of the slow-roll parameters is no longer valid because a larger $x_{\text{end}}$ implies a larger $x_*$. This has for consequence that CNCI inflation is only marginally consistent with the data. Indeed, it is precisely in the region where $n_s - 1 = -2\alpha^2$ would be compatible with the observations that the deviations play an important role and push the predictions away from the allowed contours. In fact, these properties can be better illustrated by deriving explicitly $x_*$. Using Eq. (6.258), one gets

$$ \cosh^2\left(\frac{\alpha x_*}{\sqrt{2}}\right) = \frac{3}{\alpha^2} W_0\left(\frac{\alpha^2}{3} e^{2\Lambda/3}\right), \qquad (6.260) $$

where we have defined the quantity $A$ by

$$ A \equiv -\alpha^2 (3 + \alpha^2) \Delta N_* + 3 \ln \left[ \cosh \left( \frac{\alpha x_{\text{end}}}{\sqrt{2}} \right) \right] + \frac{\alpha^2}{2} \cosh^2 \left( \frac{\alpha x_{\text{end}}}{\sqrt{2}} \right). \qquad (6.261) $$

In the regime where both $\alpha \ll 1$ and $\alpha x_{\text{end}} \ll 1$, the previous expression reduces to $x_*^2 \simeq x_{\text{end}}^2 - 4\Delta N_*$. This last formula is identical to the slow-roll trajectory for LFI provided $p = -2$, see Eq. (5.36). At the beginning of this section, we have shown that, at leading order $\epsilon_1 \simeq 2/x^2$ and $\epsilon_2 \simeq -4/x^2$ and, comparing with Eqs. (5.35), we notice that these are also the slow-roll parameters for LFI with $p = -2$. In fact, expanding Eq. (6.254), one sees that $V(\phi) \propto \phi^{-2}$ which confirms the previous considerations. In the regime where $\alpha \ll 1$ and $\alpha x_{\text{end}} \ll 1$, the model is very close to LFI with $p = -2$. On the contrary, if $\alpha x_{\text{end}}$ is not small, then the above relation does not hold anymore and one does not recover a constant spectral index.

Finally, we conclude this section by discussing how the mass scale $M$ can be chosen. The CMB normalization gives

$$ \left(\frac{M}{M_{\text{Pl}}}\right)^4 = \frac{11520\pi^2\alpha^2(3+\alpha^2)^2\cosh^2\left(\frac{\alpha}{\sqrt{2}}x_*\right) Q_{\text{rms-PS}}^2}{\left[6+\alpha^2+\alpha^2\cosh\left(\sqrt{2}\alpha x_*\right)\right]^3 T^2}. \qquad (6.262) $$

From Eq. (6.260), one deduces that $\cosh^2(\alpha x_*/\sqrt{2}) \simeq 1 - 2\alpha^2\Delta N_* + \alpha^2 x_{\text{end}}^2/2 \simeq 1$. Inserting this formula into Eq. (6.262), and taking the leading order in $\alpha$, one obtains $M/M_{\text{Pl}} \simeq 0.02\sqrt{\alpha}$. This implies that $M < M_{\text{Pl}}$ if $\alpha \lesssim 2420$, which is largely the case for the predictions displayed in Fig. 225.

## 6.16. Supergravity Brane Inflation (SBI)

### 6.16.1. Theoretical justifications

This model can emerge in different contexts. Following Ref. [310], let us consider a model with a scalar field and a massive fermion interacting through a Yukawa type term (with a coupling constant $g$). The corresponding Lagrangian can be written as

$$ -\mathcal{L} = \frac{1}{2} \partial_{\mu} \phi \partial^{\mu} \phi + \frac{i}{2} \bar{\psi} \gamma^{\mu} \partial_{\mu} \psi + \frac{1}{2} m^2 \phi^2 + \frac{\lambda}{4!} \phi^4 + m_f \bar{\psi} \psi + \frac{1}{2} g \phi \bar{\psi} \psi, \qquad (6.263) $$

where we have assumed the most general renormalizable scalar potential. At one loop level, the potential takes the form

$$ V(\phi) = V_0 + \frac{1}{2}m^2\phi^2 + \frac{\lambda}{4!}\phi^4 + \frac{1}{64\pi^2}\left(m^2 + \frac{\lambda}{2}\phi^2\right)^2 \ln\left(\frac{m^2 + \frac{\lambda\phi^2}{2}}{\mu^2}\right) \\ - \frac{2}{64\pi^2}\left(g\phi + m_f\right)^4 \ln\left[\frac{\left(g\phi + m_f\right)^2}{\mu^2}\right], \qquad (6.264) $$

where $\mu$ is a renormalization scale. Then, assuming that, for some reason, the bosonic and fermionic massive terms are negligible, the potential can be expressed as

$$ V(\phi) \approx V_0 + \left[ \frac{\lambda}{4!} + \frac{\lambda^2}{256\pi^2} \ln\left(\frac{\lambda}{2}\right) - \frac{g^4}{16\pi^2} \ln g \right] \phi^4 + \frac{1}{64\pi^2} \left( \frac{\lambda^2}{2} - \frac{g^4}{4} \right) \phi^4 \ln\left(\frac{\phi}{\mu}\right). \qquad (6.265) $$

This is the type of potential that we study in this section. Notice that a change in the renormalization scale $\mu$ is in fact equivalent to a change in the coefficient of the terms $\propto \phi^4$ and $\propto \phi \ln(\phi/\mu)$. This potential was also studied in Ref. [599] but the coefficient of the $\phi^4$ term was chosen such that, at its minimum, the potential exactly vanishes. This particular case will also be treated in what follows. Finally, it is interesting to remark that this model was also proposed in Refs. [600,601] in the context of brane cosmology within a supergravity bulk spacetime.

### 6.16.2. Slow-roll analysis

Let us now turn to the slow-roll analysis of the potential given by Eq. (6.265). It is more convenient to write it under the following form

$$ V(\phi) = M^4 \left\{ 1 + \left[ -\alpha + \beta \ln \left( \frac{\phi}{M_{\text{Pl}}} \right) \right] \left( \frac{\phi}{M_{\text{Pl}}} \right)^4 \right\}, \qquad (6.266) $$

where $\alpha$ and $\beta$ are dimensionless quantities that must be considered as small quantities since they are typically proportional to coupling constants, see Eq. (6.265). It is worth noticing that setting $\alpha = 0$ in the above expression allows us to recover the Coleman-Weinberg CWI models already studied in Section 5.11. Defining the quantity $x$ by the following expression

$$ x \equiv \frac{\phi}{M_{\text{Pl}}}, \qquad (6.267) $$

one sees that the potential decreases from $x = 0$ to reach a minimum located at $x = x_{V'=0}$, then increases and diverges when $x$ goes to infinity. The value of $x_{V'=0}$ is given by

$$ x_{V'=0} = \exp\left(\frac{\alpha}{\beta} - \frac{1}{4}\right). \qquad (6.268) $$
Since the logarithm terms in Eq. (6.266) are one loop corrections, they should not dominate the leading order terms. As a result, inflation can take place only in the domain $x < x_{V'=0}$ if one wants the model to be such that additional corrections to $V(\phi)$ are negligible. The value of the potential at the minimum reads

$$V_{\min} = V(x_{V'=0}) = M^4 \left( 1 - \frac{\beta}{4} e^{4\alpha/\beta - 1} \right), \qquad (6.269)$$

which is negative or vanishing if the following condition is satisfied

$$\alpha \ge \alpha_{\min}(\beta) = \frac{\beta}{4} \left[ 1 - \ln \left( \frac{\beta}{4} \right) \right]. \qquad (6.270)$$

Inflation proceeds from the left to the right in the range $0 < x < x_{V=0} < x_{V'=0}$ where $x_{V=0}$ is the value at which the potential vanishes. It is given by

$$x_{V=0} = \left[ \frac{-4/\beta}{W_{-1}(-4/\beta e^{-4\alpha/\beta})} \right]^{1/4}, \qquad (6.271)$$

where $W_{-1}$ is the $-1$ branch of the Lambert function. In this situation, inflation stops by slow-roll violation at $x = x_{V=0}$. As noticed above, the case $\alpha = \alpha_{\min}(\beta)$ is also interesting. It corresponds to tuning the parameters $\alpha$ and $\beta$ such that the minimum of the potential exactly vanishes. When this condition is satisfied the previous formula reduces to $x_{V=0} = x_{V'=0} = (\beta/4)^{-1/4}$. Then, the first slow roll parameter $\epsilon_1$ diverges at this point (see below) and, as a consequence, inflation also ends by slow roll violation.

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

$$\epsilon_1 = \frac{x^6 (-4\alpha + \beta + 4\beta \ln x)^2}{2 (1 - \alpha x^4 + \beta x^4 \ln x)^2}, \qquad (6.272)$$

$$\epsilon_2 = 2 \frac{(12\alpha - 7\beta - 12\beta \ln x)x^2 + (4\alpha^2 - \alpha\beta + \beta^2 + \beta^2 \ln x - 8\alpha\beta \ln x + 4\beta^2 \ln^2 x)x^6}{[1 + x^4(-\alpha + \beta \ln x)]^2}, \qquad (6.273)$$

$$\epsilon_3 = \frac{8}{x^2} + 2 \frac{(-4 + \beta x^4)^2}{x^2 (1 - \alpha x^4 + \beta x^4 \ln x)^2} + \frac{1}{x^2} \frac{-52 + 9\beta x^4}{1 - \alpha x^4 + \beta x^4 \ln x} + \frac{144\alpha - 84\beta + (28\alpha - 11\beta)\beta x^4 - 4\beta (36 + 7\beta x^4) \ln x}{(12\alpha - 7\beta - 12\beta \ln x)x^2 + (4\alpha^2 - \alpha\beta + \beta^2 - 8\alpha\beta \ln x + \beta^2 \ln x + 4\beta^2 \ln^2 x)x^6}. \qquad (6.274)$$

Together with the potential, they are represented in Fig. 78 for the physical branch $0 < x < x_{V=0}$.

As already mentioned, inflation stops by violation of the slow-roll conditions. This happens when $x = x_{end}$ where $x_{end}$ is the solution of $\epsilon_1(x_{end}) = 1$. We see in Eq. (6.272) that there is no simple analytic solution for $x_{end}$ and this equation must in fact be solved numerically. We have, however, already stressed that, when $\alpha \le \alpha_{\min}(\beta)$, $\epsilon_1$ diverges for $x \to x_{V=0}$, and therefore one already knows that $x_{end} < x_{V=0}$.

Let us now consider the slow-roll trajectory. It can be integrated analytically and one obtains the following expression

$$N - N_{\text{end}} = \frac{e^{\frac{2\alpha}{\beta}} - \frac{1}{2}}{16} \left[ \text{Ei} \left( \frac{1}{2} - 2\frac{\alpha}{\beta} + 2 \ln x \right) - \text{Ei} \left( \frac{1}{2} - 2\frac{\alpha}{\beta} + 2 \ln x_{\text{end}} \right) \right] - \frac{e^{\frac{1}{2} - 2\frac{\alpha}{\beta}}}{4\beta} \left[ \text{Ei} \left( -\frac{1}{2} + 2\frac{\alpha}{\beta} - 2 \ln x \right) - \text{Ei} \left( -\frac{1}{2} + 2\frac{\alpha}{\beta} - 2 \ln x_{\text{end}} \right) \right] - \frac{x^2 - x_{\text{end}}^2}{8}. \qquad (6.275)$$

The field value $x_*$ at which the pivot scale crossed the Hubble radius during inflation is obtained by solving Eq. (3.48). Clearly, it must also been done numerically and those calculations are implemented in the corresponding ASPIC routines.

Finally, the parameter $M$ is fixed by the amplitude of the CMB anisotropies and one obtains

$$\left(\frac{M}{M_{\text{PI}}}\right)^4 = \frac{720\pi^2 (4\alpha - \beta - 4\beta \ln x_*)^2 Q_{\text{rms-PS}}^2}{(1 - \alpha x_*^4 + \beta x_*^4 \ln x_*)^3 T^2}. \qquad (6.276)$$

The reheating consistent slow-roll predictions for the SBI models are displayed in Figs. 228 and 229, for $\beta = 5 \times 10^{-5}$ and $\beta = 10^{-3}$, respectively, and with $\alpha \le \alpha_{\min}(\beta)$. These plots show that the larger values of $\beta$, the more negligible the amount of gravitational waves. The predictions for the special case $\alpha = \alpha_{\min}(\beta)$ are also displayed in Fig. 230, where it is clear that smaller values of $\beta$ are preferred.

## 6.17. Spontaneous Symmetry Breaking Inflation (SSBI)

### 6.17.1. Theoretical justifications

The potential that we study in this section is given by the following expression

$$V(\phi) = V_0 + a\phi^2 + b\phi^4, \qquad (6.277)$$

where $a$ and $b$ are constant coefficients the sign of which is not a priori determined. Before turning to the slow-roll analysis, it is interesting to study in which context such a potential can arise.

First of all, it is clear that this potential is very general since it is just made of the three first terms of a general Taylor expansion. Therefore, it can just be considered as a phenomenological description of a generic inflaton potential. This view was for instance adopted in Ref. [408], where this potential was used as a toy model to implement “new inflation”. In the same fashion, it was also considered in Ref. [602] (with the assumptions $a < 0$ and $b > 0$) in the framework of models with spontaneous symmetry breaking where $\phi$ represents one of the components of a Higgs field. In Ref. [603], it was also studied in the context of “mixmaster inflation”.

However, there are also models where this specific shape explicitly arises and, here, when necessary, we also briefly review them.
<figure><img src="image_81.png" /><figcaption>Fig. 78. Supergravity Brane Inflation (SBI) for $\beta = 0.7$ and $\alpha = 0.13 > \alpha_{min}(\beta)$, $\alpha = \alpha_{min}(\beta)$, and $\alpha = 0.09 < \alpha_{min}(\beta)$ (where $\alpha_{min}$ is defined in Eq. (6.270)). Upper panels: the potential and its logarithm. Inflation proceeds in the place and direction labeled by the arrow. 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), only displayed in the branch of the potential where inflation proceeds.</figcaption></figure>

The first example is given by Refs. [604,605]. In these articles, inflation was investigated in the context of gauge mediated SUSY breaking scenarios. One of the basic idea of this approach is that the inflaton field should not be an extra field added to the theory on purpose but rather a field which is already present in known high energy theories. In the MSSM, see also Section 5.17, we know that the Higgs sector superpotential contains the term $\mu H_u \cdot H_d$ where $\mu$ should be of the order of the electroweak scale, that is to say far from the Planck scale. This is the so-called $\mu$-problem. One possible solution is to consider that this term dynamically arises due to the presence of another superfield (usually a singlet), $S$, in the theory. Refs. [604,605] take advantage of this fact and build a model where $S$ can also play the role of the inflaton. Since the model is also formulated in the framework of gauge-mediated supersymmetry breaking scenarios, there is an additional superfield $X$ such that its scalar component (also denoted $X$) and auxiliary component $F_X$ acquire non-vanishing $\text{vev}$. Let us now consider the following super-potential

$$W = -\beta \frac{X S^4}{M_{\text{Pl}}^2} + \frac{S^5}{M_{\text{Pl}}^2} + \lambda \frac{S^2}{M_{\text{Pl}}^2} H_u \cdot H_d + \bar{W}, \qquad (6.278)$$

where the function $\bar{W}$ describes all the other extra terms in $W$ and, crucially, is assumed to be independent of $S$. The quantities $\lambda$ and $\beta$ are constant coefficients. As argued in Refs. [604,605], this form of $W$ can be enforced by discrete symmetries. In particular, we notice the absence of a term $SH_u \cdot H_d$. Another important ingredient of the model is the assumption that the $\text{vev } F_X$ comes from the extra-terms in the above superpotential, i.e. $F_X \approx \partial \bar{W}/\partial X$. Then, the scalar potential reads

$$V = \left( F_X - \beta \frac{S^4}{M_{\text{Pl}}^2} \right)^2 + \left( 5 \frac{S^4}{M_{\text{Pl}}^2} - 4 \beta \frac{X}{M_{\text{Pl}}^2} S^3 \right)^2. \qquad (6.279)$$

Taking into account supergravity corrections, which are typically of the form $(\partial W/\partial X)/M_{\text{Pl}}^2$, i.e. $m^2 = a F_X^2/M_{\text{Pl}}^2$, where $a$ is a coefficient of order one we are led to

$$V \approx F_X^2 - a \frac{F_X^2}{M_{\text{Pl}}^2} S^2 - 2\beta F_X \frac{S^4}{M_{\text{Pl}}^2} + 16\beta^2 \frac{X^2}{M_{\text{Pl}}^4} S^6 - 40\beta \frac{X}{M_{\text{Pl}}^4} S^7 + (25 + \beta^2) \frac{S^8}{M_{\text{Pl}}^4}. \qquad (6.280)$$

In addition, making the reasonable assumption that the field $X$ is stabilized at a $\text{vev}$ such that $X/M_{\text{Pl}} \ll 1$, one can neglect higher order terms in this expression. Then, we see that $S$ can play the role of the inflaton with a potential of the form given by Eq. (6.277), namely

$$V \approx F_X^2 \left( 1 - a \frac{S^2}{M_{\text{Pl}}^2} - \frac{2\beta M_{\text{Pl}}^2}{F_X} \frac{S^4}{M_{\text{Pl}}^4} \right). \qquad (6.281)$$

At the minimum of the potential, $S^4 \approx M_{\text{Pl}}^2 F_X$ and this implies a $\mu$ term for the MSSM of the form $\mu \approx \lambda \sqrt{F_X}$. As explained before, this model dynamically produces the $\mu$ term while obtaining a candidate for the inflaton field. Finally, let us remark that the CMB normalization will determine the scale $F_X$ and that the spectrum of the superparticles depends on the ratio $F_X/X$. Therefore, given a value of $F_X$, one can always choose $X$ in order to obtain reasonable values for the superparticle masses.
The SSBI potential was also used, as a toy model, in Refs. [606,607] to study a model of “Spinodal Inflation”. After the 90’s, it was considered again several times: in the context of the Randall-Sundrum model in Ref. [608] (but within the framework of Brans-Dicke theories), in the context of the little Higgs model in Ref. [337] and in the context of induced gravity inflation in Ref. [609]. In this last reference, a potential of the form (6.277) was considered but in the Jordan frame. Since the potential is different in the Einstein frame, in fact, this model does not belong to the class of scenarios studied here. Finally, it was also considered in the context of electroweak inflation in Ref. [610].

In Ref. [611], an inflationary scenario was studied in which the superpartner of the right-handed neutrino plays the role of the inflaton field. Let us denote by N the singlet neutrino superfield, $\phi$ the super waterfall field (that can be put to zero during inflation) and S another singlet superfield (which can also be put to zero during inflation). Then, on very general grounds, the Kähler potential can be written as

$$ K = |S|^2 + |\phi|^2 + |N|^2 + \kappa_S \frac{|S|^4}{4M_{\text{Pl}}^2} + \kappa_N \frac{|N|^4}{4M_{\text{Pl}}^2} + \kappa_\phi \frac{|\phi|^4}{4M_{\text{Pl}}^2} + \kappa_{S\phi} \frac{|S|^2|\phi|^2}{M_{\text{Pl}}^2} + \kappa_{SN} \frac{|S|^2|N|^2}{M_{\text{Pl}}^2} \\ + \kappa_{N\phi} \frac{|N|^2|\phi|^2}{M_{\text{Pl}}^2} + \dots, \qquad (6.282) $$

where the dimensionless coefficients $\kappa$ are a priori of order one. The superpotential can be expressed as

$$ W = \kappa S \left( \frac{\phi^4}{M'^2} - M^2 \right) + \frac{\lambda}{M_*} N^2 \phi^2 + \dots, \qquad (6.283) $$

where $M, M'$ and $M_*$ are three mass scales and $\kappa$ and $\lambda$ are coupling constants. Since the three fields introduced before are singlets the potential does not contain D-term contributions. As a consequence, for $S \simeq 0$ and $\phi \simeq 0$, we are left with the F-term potential only and this one can be written as

$$ V(N) \approx \kappa^2 M^4 \left[ 1 + (1 - \kappa_{SN}) \frac{N^2}{M_{\text{Pl}}^2} + \left( \frac{1}{2} + \frac{\kappa_N}{4} - \kappa_{SN} + \kappa_{SN}^2 \right) \frac{N^4}{M_{\text{Pl}}^4} + \dots \right]. \qquad (6.284) $$

We see that it has the form of Eq. (6.277). Ref. [611] also discusses how to stop inflation by tachyonic instability. Since the field $\phi$ is viewed as the waterfall field, one has to calculate his mass to see when the instability is triggered. This can be done by evaluating the quadratic correction in $\phi$ to the potential calculated before. This leads to

$$ m_{\phi}^2 = \left( 1 + \kappa_{N\phi} \frac{N^2}{M_{\text{Pl}}^2} - \kappa_{S\phi} \right) \frac{\kappa^2 M^4}{M_{\text{Pl}}^2} + 4 \frac{\lambda^2}{M_*^2} N^4. \qquad (6.285) $$

Neglecting the term $N^2/M_{\text{Pl}}^2 \ll 1$ in this expression, the effective mass vanishes for

$$ N_{\text{cri}} \approx \frac{\kappa M^2 M_*}{2 \lambda M_{\text{Pl}}} \sqrt{-(1 - \kappa_{S\phi})}. \qquad (6.286) $$

We see that this requires $1 - \kappa_{S\phi} < 0$. On the other hand, this model also provides an expression for the coefficients $a$ and $b$ in terms of the fundamental coefficients of the Kähler potential. Except from the above mentioned condition, there is no other constraint on the coefficients $\kappa$ and, as a consequence, the sign of $a$ and $b$ is, a priori, not fixed in this scenario.

Another context in which Eq. (6.277) arises is “racetrack inflation” [612,613]. Racetrack inflation is a string inspired inflationary scenario where the inflaton is a volume modulus. Therefore, this model belongs to the same class as KMIII, see Section 6.3. The Kähler and super potentials are given by standard formulas, namely

$$ K = -\frac{3}{\kappa} \ln(T + T^{\dagger}), \qquad W = W_0 + Ae^{-aT} + Be^{-bT}. \qquad (6.287) $$

Writing $T = X + iY$, it follows that the scalar F-term potential reduces to

$$ V(X, Y) = \frac{\kappa}{6X^2} \left\{ \begin{aligned} & aA^2 (3 + aX) e^{-2aX} + bB^2 (3 + bX) e^{-2bX} + 3aA W_0 e^{-aX} \cos(aY) \\ & + 3bBW_0 e^{-bX} \cos(bY) + AB [2abX + 3(a+b)] e^{-(a+b)X} \cos[(a-b)Y] \end{aligned} \right\} + \frac{E}{X^a}, \qquad (6.288) $$

where an uplifting term $\propto X^{-a}$ has been added. Let us mention that $X$ and $Y$ are not canonically normalized and their kinetic term reads $3[(\partial_{\mu}X)^2 + (\partial_{\mu}Y)^2]/(4\kappa X^2)$. The above potential has a very rich structure and for $W_0 = 0$ and $a = b$, we have a flat direction in $Y$. Moreover, for $Y = 0$, one can find a minimum in the $X$ direction. If we then combine the two above remarks, then it is clear that there exists a choice of parameters such that one has a saddle point around $Y = 0$ (a specific example was exhibited in Ref. [612]). This point seems suitable for inflation. Around such a point, it is argued in Ref. [613] that one can write

$$ V(Y) = V_0 \left( 1 + \frac{\eta_0}{2} y^2 + \frac{C}{4} y^4 + \dots \right), \qquad (6.289) $$

where $y$ is now the canonically normalized field when $X$ is stabilized. This is again a potential of the type given by Eq. (6.277). In order to phenomenologically reproduce racetrack inflation, one should have $\eta_0$ small and negative and $C$ large and positive.

The potential of Eq. (6.277) was also used, as a toy model, in the context of minimal left-right symmetric models with spontaneous D-parity breaking in Ref. [614] and in the context of hilltop supernatural inflation in Refs. [615–617]. A justification based on high energy physics was offered and the idea is to assume that the full potential has a SUSY flat direction. The approach is therefore similar to what was already investigated in Section 5.17. In that situation, one can write $V(\phi)$ as

$$ V = V_0 + \frac{1}{2} m^2 \phi^2 - A \frac{\lambda_p \phi^p}{p M_{\text{Pl}}^{p-3}} + \lambda_p^2 \frac{\phi^{2p-2}}{M_{\text{Pl}}^{2p-6}}, \qquad (6.290) $$

where the term $V_0$ is added by hand. If one chooses $p = 4$ and neglects the last term (for instance if $\phi \ll M_{\text{Pl}}$), then one arrives at

$$ V(\phi) \approx V_0 + \frac{1}{2} m^2 \phi^2 - \frac{\lambda_4 A}{4 M_{\text{Pl}}^2} \phi^4, \qquad (6.291) $$
<figure><img src="image_82.png" /><figcaption>Fig. 79. Spontaneous Symmetry Breaking Inflation (SSBI) potential and the corresponding Hubble flow parameter $\epsilon_1$ for the two cases $\alpha > 0$, $\beta > 0$ (SSBI1), and $\alpha < 0$, $\beta < 0$ (SSBI2). The values of the parameters are chosen to be $\alpha, \beta = \pm 1$. The four other possibilities, namely SSBI3, SSBI4, SSBI5, SSBI6 are displayed in Fig. 80.</figcaption></figure>

which is of the form of Eq. (6.277). In this framework, $m$ and $A$ are SUSY soft terms and, therefore, should be taken of $\mathcal{O}(\text{TeV})$. The term $V_0 = M_s^4$ where $M_s$ is the SUSY breaking scale, $M_s \simeq 10^{11}$GeV.

Finally, let us mention that SSBI was also considered in the context of a supersymmetric B-L extension of the standard model in Refs. [618,619] and in the context of Kähler-driven “tribrid inflation” in Ref. [620]. In this last case, one obtains a situation very similar to the one discussed above for sneutrino inflation. In particular, the coefficients $a$ and $b$ can be expressed in terms of the coefficients appearing in the Kähler potential. To end this part, let us notice that the potential (6.277) also arises in the context of Higgs inflation in a false vacuum, as shown in Refs. [621–623].

As already mentioned above, these works differ on the signs of $\alpha$ and $\beta$. Summarizing, Refs. [603,611] require $\alpha > 0$, $\beta > 0$ while Refs. [337,408,602,606,607,609,610,613,614] assume $\alpha < 0$, $\beta > 0$. On the other hand, Refs. [615–617] consider that $\alpha > 0$ and $\beta < 0$ and Refs. [604,605,621–623] have $\alpha < 0$, $\beta < 0$. We see that the four possible combinations have all been studied. Also, in Refs. [618,619], one has $\alpha, \beta \lesssim \mathcal{O}(1)$ and inflation only takes place in the increasing branches of the potential (see below). Finally, in Refs. [608,620], $\beta$ is taken to be positive and the sign of $\alpha$ is left unspecified.

### 6.17.2. Slow-roll analysis

Let us now turn to the slow-roll analysis of SSBI. For this purpose, it is more convenient to rewrite the potential (6.277) as

$$V(\phi) = M^4 \left[ 1 + \alpha \left( \frac{\phi}{M_{Pl}} \right)^2 + \beta \left( \frac{\phi}{M_{Pl}} \right)^4 \right], \qquad (6.292)$$

where $\alpha$ and $\beta$ are two dimensionless parameters. Based on the previous brief review of the literature, we conclude that it is necessary to study the model in full generality and, therefore, in what follows, we investigate all possible situations. As mentioned above, four cases should be distinguished: $\alpha > 0$, $\beta > 0$; $\alpha < 0$, $\beta < 0$; $\alpha > 0$, $\beta < 0$ and $\alpha < 0$, $\beta > 0$, with two possible domains of inflation in the two latter cases. Therefore we have six regimes of inflation that we label SSBI1, SSBI2, SSBI3, SSBI4, SSBI5 and SSBI6. The different potentials and inflationary regimes are displayed and defined in Figs. 79 and 80. Since the potential is symmetric under $\phi/M_{Pl} \to -\phi/M_{Pl}$, it is only displayed and studied for $\phi > 0$.

Let us now calculate the slow-roll parameters. If one defines $x$ by $x \equiv \phi/M_{Pl}$, then the three first Hubble parameters are given by the following expressions

$$\epsilon_1 = \frac{2 (\alpha x + 2\beta x^3)^2}{(1 + \alpha x^2 + \beta x^4)^2}, \qquad \epsilon_2 = \frac{4 [-\alpha + (\alpha^2 - 6\beta) x^2 + \alpha\beta x^4 + 2\beta^2 x^6]}{(1 + \alpha x^2 + \beta x^4)^2}, \qquad (6.293)$$

and

$$\epsilon_3 = \frac{4x^2 (\alpha + 2\beta x^2) [-3\alpha^2 + 6\beta + \alpha (\alpha^2 - 12\beta) x^2 + 3(\alpha^2 - 8\beta)\beta x^4 + 2\beta^3 x^6]}{(1 + \alpha x^2 + \beta x^4)^2 [-\alpha + (\alpha^2 - 6\beta) x^2 + \alpha\beta x^4 + 2\beta^2 x^6]} \qquad (6.294)$$

The first slow-roll parameter $\epsilon_1$ is displayed in the right panels of Figs. 79 and 80 while the second and third slow-roll parameters $\epsilon_2$ and $\epsilon_3$ are displayed in Fig. 81. Let us describe the behavior of these slow-roll parameters, for the six models under consideration. For SSBI1, $\epsilon_1$ vanishes at
<figure><img src="image_83.png" /><figcaption>Fig. 80. Spontaneous Symmetry Breaking Inflation (SSBI) potential and the corresponding Hubble flow parameter $\epsilon_1$ for the two cases $\alpha > 0$, $\beta < 0$ (corresponding to SSB13 to SSB14) and $\alpha < 0$, $\beta > 0$ (corresponding to SSB15 and to SSB16). In each of these cases, the direction in which inflation proceeds is indicated by the arrow.</figcaption></figure>

$x=0$, reaches a maximum at $x_{\epsilon_2=0}^{SSBI1}$ (where $\epsilon_2$ vanishes and $\epsilon_3$ diverges) and then decreases to asymptotically vanish when $x$ goes to infinity. The value of $x_{\epsilon_2=0}^{SSBI1}$ is given by

$$x_{\epsilon_2=0}^{SSBI1\&3\&6} = \left\{ -\frac{\alpha}{6\beta} + \frac{1}{6\beta} \left[ 8\alpha^3 + \sqrt{64\alpha^6 + (5\alpha^2 - 36\beta)^3} \right]^{1/3} + \frac{36\beta - 5\alpha^2}{6\beta} \left[ 8\alpha^3 + \sqrt{64\alpha^6 + (5\alpha^2 - 36\beta)^3} \right]^{-1/3} \right\}^{1/2} . \quad (6.295)$$

Whether the maximum of $\epsilon_1$ at this point is larger or smaller than 1 depends on $\alpha$ and $\beta$. In the following, we restrict ourselves to the physical regime where $\alpha, \beta \le O(1)$. For each value of $\beta$, there is a minimum value of $\alpha$, denoted $\alpha_{min}$, above which the maximum is larger than 1. The line $\alpha_{min}(\beta)$ is displayed in Fig. 82 and the shaded area in this plot represents the region in the parameter space where inflation stops by slow-roll violation. When $\beta \ll 1$, $\alpha_{min}(\beta)$ approaches 2 as can be noticed in the figure. In addition, for $\beta \gtrsim 0.25$, the maximum value for $\epsilon_1$ becomes larger than 1 for any value of $\alpha$.

For SSB12, the three first slow-roll parameters are monotonic increasing functions of the field $\nuev$ and diverge when the potential vanishes at

$$x_{\nu=0}^{SSBI2\&4\&5} = \sqrt{-\frac{\alpha + \sqrt{\alpha^2 - 4\beta}}{2\beta}}. \quad (6.296)$$

Hence inflation ends by slow-roll violation at $x_{end}$. Unfortunately, the corresponding $\nuev$ cannot be found exactly and one has to rely on numerical calculations. Let us also notice that, while the first and third slow-roll parameters $\epsilon_1$ and $\epsilon_3$ vanish at $x=0$, $\epsilon_2$ is equal to $\epsilon_2^{min} = -4\alpha$ at this point. Therefore, in order for the slow-roll approximation to be valid, one needs to work with $|\alpha| \ll 1$.

For SSB13, the first slow-roll parameter $\epsilon_1$ vanishes at $x=0$ and at $x = \sqrt{-\alpha/(2\beta)}$. In between, it reaches a maximum located at

$$x_{\epsilon_2=0}^{SSBI3} = x_{\epsilon_2=0}^{SSBI1}, \quad (6.297)$$

a point where $\epsilon_2$ vanishes and $\epsilon_3$ diverges. Whether the maximum of $\epsilon_1$ at this point is larger or smaller than 1 depends again on $\alpha$ and $\beta$. For each value of $\beta$, there is a minimum value for $\alpha$ above which inflation stops by slow-roll violation, similarly to the SSB11 case. This corresponds to the green dotted line in Fig. 82 (top right panel). One way to estimate whether a slow roll regime of inflation can occur in the decreasing branch of $\epsilon_1$ is to look at the value of $\epsilon_2$ at the top of the potential. It is given by

$$\epsilon_2^{\text{top}} = \frac{-32\alpha\beta}{\alpha^2 - 4\beta}. \quad (6.298)$$

This number is smaller than one when $\beta < -1/64$, or when $\alpha$ lies outside the range with limits given by $-16\beta \pm \sqrt{\beta(1+64\beta)}$, displayed in Fig. 82 with the red and cyan dotted lines. Therefore, requiring that $\epsilon_2^{top} < 1$ and that inflation stops by slow roll violation leads to the allowed space $\alpha > \alpha_{min}$, represented by the shaded region in Fig. 82.
<figure><img src="image_84.png" /><figcaption>Fig. 81. Second slow-roll parameter $\epsilon_2$ (solid line) and third slow-roll parameter $\epsilon_3$ (dotted line), for the six SSB1 models studied in this section. The free parameters of the models are chosen to be $\alpha, \beta = \pm 1$.</figcaption></figure>

<figure><img src="image_85.png" /><figcaption>Fig. 82. The black solid line gives the minimum value of $|\alpha|$, denoted here by $\alpha_{min}$, as a function of $\beta$ in order for inflation to stop by slow-roll violation for SSB1 (top left panel), SSB15 (bottom left panel) and SSB16 (bottom right panel). For SSB13 (top right panel), the green dotted line denotes the minimum value of $\alpha$ for inflation to stop by slow-roll violation, and the cyan and red dotted line restrict the values of $\alpha$ for which $\epsilon_2^{\text{op}} > 1$ (defined only for $\beta < -1/64$). In the bottom panels, the dotted lines correspond to $\alpha^2 = 4\beta$, see the discussion in the text. In all the panels, the region above the black solid curve (shaded region) represents the allowed region (i.e. the one where a slow roll regime of inflation stops because $\epsilon_1$ reaches one). For SSB11, when $\tilde{\beta} \gtrsim 0.25$, this is always the case. For SSB11 and SSB13, $\alpha_{min}$ approaches the asymptotic value $\alpha_{min} = 2$ when $|\tilde{\beta}| \ll 1$. For SSB15 and SSB16, inflation stops by slow-roll violation when $\alpha < -|\alpha_{min}|$.</figcaption></figure>
For SSB14, the three first slow-roll parameters are monotonic increasing functions of the field *vev* and diverge when the potential vanishes at $x_{V=0}^{SSB12\&4}$. The first and third slow-roll parameters $\epsilon_1$ and $\epsilon_3$ vanish when $x = \sqrt{-\alpha/(2\beta)}$ while $\epsilon_2$ has a non-zero value $\epsilon_2^{\min} = 8\alpha\beta/(\beta^2 - \alpha^2/4)$ at this point. From the above discussion, it is clear that, in this version of the scenario, inflation also stops by violation of the slow-roll condition. As for SSB12, however, the corresponding *vev* cannot be determined exactly and a numerical calculation is needed.

For SSB15, the behavior of the slow-roll parameters depend on $\alpha^2/\beta$. If $\alpha^2/\beta \ge 4$, the minimum of the potential at $x = \sqrt{-\alpha/(2\beta)}$ is negative. The potential vanishes at $x_{V=0}^{SSB12\&4\&5}$ and the three first slow-roll parameters continuously increase between $x = 0$ where they vanish (except $\epsilon_2$ for which $\epsilon_2^{\min} = -4\alpha$) and $x_{V=0}^{SSB12\&4\&5}$ where they diverge. Inflation ends by slow-roll violation at some point $x_{end}$ that needs to be determined numerically. On the other hand, if $\alpha^2/\beta \le 4$, $\epsilon_1$ vanishes at $x = 0$, reaches a maximum at $x_{\epsilon_2=0}^{SSB15}$ (where $\epsilon_2$ vanishes and $\epsilon_3$ diverges), then decreases and finally vanishes at $x = \sqrt{-\alpha/(2\beta)}$. The value of $x_{\epsilon_2=0}^{SSB15}$ is given by

$$x_{\epsilon_2=0}^{SSB15} = \left\{ -\frac{\alpha}{6\beta} - \frac{1+i\sqrt{3}}{12\beta} \left[ 8\alpha^3 + \sqrt{64\alpha^6 + (5\alpha^2 - 36\beta)^3} \right]^{1/3} + \frac{5\alpha^2 - 36\beta}{12\beta} (1 - i\sqrt{3}) \left[ 8\alpha^3 + \sqrt{64\alpha^6 + (5\alpha^2 - 36\beta)^3} \right]^{-1/3} \right\}^{1/2} \quad (6.299)$$

Whether the maximum of $\epsilon_1$ at this point is larger or smaller than 1 depends on $\alpha$ and $\beta$ and is again similar to what has already been discussed before. The region in the parameter space where inflation ends by slow-roll violation is displayed in Fig. 82 and corresponds to the points such that $\alpha < -|\alpha_{min}|$. In this plot, the dotted line represents the curve $\alpha^2 = 4\beta$, above which one is sure that inflation ends by slow-roll violation since the minimum of the potential is negative in this case. For values of $\beta \ll 1$, one can see that $|\alpha_{min}| \approx 2\sqrt{\beta}$ and the allowed region becomes negligible.

Finally the case SSB16 remains to be treated. The behavior of the slow roll parameters depend on $\alpha^2/\beta$ in the same way as before. If $\alpha^2/\beta \ge 4$, the minimum of the potential at $x = \sqrt{-\alpha/(2\beta)}$ is negative. The potential vanishes at $x_{V=0}^{SSB16}$ and the slow-roll parameters continuously decrease from this value (where they blow up) and go to zero at infinity. The value of $x_{V=0}^{SSB16}$ can be expressed as

$$x_{V=0}^{SSB16} = \sqrt{\frac{-\alpha + \sqrt{\alpha^2 - 4\beta}}{2\beta}} \quad (6.300)$$

On the other hand, if $\alpha^2/\beta \le 4$, $\epsilon_1$ vanishes at $x = \sqrt{-\alpha/(2\beta)}$, reaches a maximum at $x_{\epsilon_2=0}^{SSB16}$ and then decreases. At infinity, it goes to zero. The value of $x_{\epsilon_2=0}^{SSB16}$ is given by

$$x_{\epsilon_2=0}^{SSB16} = x_{\epsilon_2=0}^{SSB13} = x_{\epsilon_2=0}^{SSB11} \quad (6.301)$$

Whether the maximum of $\epsilon_1$ at this point is larger or smaller than 1 depends on $\alpha$ and $\beta$. The corresponding region in the parameter space is displayed in Fig. 82 and corresponds to the inequality $\alpha < -|\alpha_{min}|$. The dotted line represents the law $\alpha^2 = 4\beta$. Above this line, one is sure that inflation can stop by slow-roll violation since, in this case, the potential becomes negative at some point. It is also interesting to notice that, when $\beta \gtrsim 1.48$, the maximum value of $\epsilon_1$ is larger than 1 for any value of $\alpha$. On the other hand, if $\beta \ll 1$, the allowed region shrinks to zero.

Let us now turn to the slow-roll trajectory. This one can be integrated analytically to get

$$N_{\text{end}} - N = -\frac{1}{2\alpha} \ln \left( \frac{x_{\text{end}}}{x} \right) - \frac{x_{\text{end}}^2 - x^2}{8} - \frac{\alpha^2 - 4\beta}{16\alpha\beta} \ln \left( \frac{1 + \frac{2\beta}{\alpha} x_{\text{end}}^2}{1 + \frac{2\beta}{\alpha} x^2} \right), \quad (6.302)$$

where $N_{\text{end}}$ is the number of e-folds at the end of inflation. It is important to notice that the argument of the logarithm is always positive. This trajectory cannot be inverted analytically. But, numerically, it is easy to use this expression in order to determine $x_*$, the value of $x$ at Hubble radius crossing.

Finally, it is interesting to constrain the value of the scale $M$ with the CMB normalization. It follows that

$$\left(\frac{M}{M_{\text{Pl}}}\right)^4 = \frac{2880 (\alpha x_* + 2\beta x_*^3)^2 \pi^2 Q_{\text{rms-PS}}^2}{(1 + \alpha x_*^2 + \beta x_*^4)^3 T^2} \quad (6.303)$$

We are now in a position where we can discuss the predictions of the six versions of this model. The reheating consistent slow-roll predictions for the SSB11 models are displayed in Figs. 231, 232 and 233 for $\beta = 10^{-3}$, $\beta = 10^{-1}$ and $\beta = 10$, respectively. SSB11 seems to be disfavored by the observations. The predictions of SSB12 models are displayed in Fig. 234 for different values of $\beta$ and $\alpha$. We notice that they depend on the parameter $\alpha$ quite strongly. The spectral index is clearly red and, for values of $\beta$ of order one, the contribution of gravity waves becomes very small. For SSB13, the predictions are presented in Figs. 238, 239 and 240 for $\beta = -10^{-3}$, $\beta = -5 \times 10^{-3}$ and $\beta = -10^{-2}$, respectively. As we increase $\beta$, the points start spreading in the plane $(n_s, r)$. For this class of models, the spectrum is red and the level of gravity waves quite important. The predictions for the SSB14 models are displayed in Figs. 241, 242, and 243 for $\beta = -10^{-5}$, $\beta = -10^{-4}$, $\beta = -10^{-3}$, respectively. One can notice that the typical predicted values for $\epsilon_1$ decrease with the absolute value of $\beta$. As before the spread of the points increases with $\beta$. The tilt is still red and the contribution of gravity waves is small for small values of $\alpha$. The predictions for the SSB15 models are displayed in Figs. 244, 245 and 246 for $\beta = 10^{-6}$, $\beta = 10^{-5}$ and $\beta = 10^{-4}$, respectively. Once again, for $\mathcal{O}(1)$ values of $\beta$, one can see that the model predict a small amount of gravitational waves but has a deviation from scale invariance strongly disfavored by the observational constraints. Finally, the reheating consistent slow-roll predictions for the SSB16 models are displayed in Figs. 247, 248 and 249 for $\beta = 10^{-6}$, $\beta = 10^{-1}$ and $\beta = 1$, respectively. When $\beta \ll 1$ the predictions of the model do not depend on $\beta$. Moreover, for values of $\beta$ of order one, the predictions become almost independent of the two parameters of the model.

## 6.18. Inverse Monomial Inflation (IMI)

These models are characterized by the inverse monomial potential given by

$$V(\phi) = M^4 \left(\frac{\phi}{M_{\text{Pl}}}\right)^{-p}, \quad (6.304)$$
<figure><img src="image_86.png" /><figcaption>Fig. 83. Top left panel: Inverse Monomial Inflation (IMI) potential for $p = 2$. Top right panel: logarithm of the potential for the same value of $p$. Bottom left panel: slow-roll parameter $\epsilon_1$ for $p = 2$. Bottom right panel: slow-roll parameters $\epsilon_2$ and $\epsilon_3$ for $p = 2$. Only one line appears because $\epsilon_2 = \epsilon_3$. On these plots, the shaded region represents the region where the slow-roll approximation breaks down.</figcaption></figure>

where $p$ is a positive number. This scenario has been studied in many different situations: in Refs. [361,624,625] it was considered in the context of quintessential inflation, in Refs. [626–629] in the context of tachyon inflation, in Refs. [555,557] in the context of intermediate inflation and in Ref. [374] in the context of Randall-Sundrum braneworld models. In all these articles, the potential was just postulated. An attempt to derive this potential from high energy considerations was made in Refs. [630,631] in the context of supersymmetric QCD. Let us, however, notice that this was done in order to build a model of quintessence and not of inflation. The model uses the group SU($N_c$) and has $N_f$ flavors. The quarks $Q^i$, $i = 1, ..., N_f$ are placed in the fundamental representation of SU($N_c$) and the anti-quarks $Q_i^\dagger$ in the conjugate representation [630]. At scales below the gauge breaking scale $\Lambda$, the relevant degrees of freedom are the pions $\pi_j^i = Q^i Q_j^\dagger$ and one can show that the corresponding superpotential is given by [632,633]

$$W = (N_c - N_f) \frac{\Lambda^{3(N_c - N_f)/(N_c - N_f)}}{(\det \pi)^{1/(N_c - N_f)}}. \qquad (6.305)$$

The potential (6.304) then follows from the F-term associated to the above superpotential.

The potential is represented in Fig. 83 for $p = 2$. It is a decreasing function of the field $\text{vev}$ and, hence, inflation proceeds from the left to the right, in the direction specified by the arrow in the figure.

The three Hubble flow functions are straightforwardly obtained from Eqs. (3.4), (3.5) and (3.6). Defining $x \equiv \phi/M_{\text{Pl}}$, one gets

$$\epsilon_1 = \frac{p^2}{2x^2}, \qquad \epsilon_2 = -\frac{2p}{x^2}, \qquad \epsilon_3 = \epsilon_2. \qquad (6.306)$$

These functions are represented in the two bottom panels in Fig. 83. The first slow-roll parameter is a monotonic decreasing function of $\phi$ while $\epsilon_2$ and $\epsilon_3$ are negative increasing functions. From these expressions, one can also immediately deduce that, for a given $p$, the model in the plane $(\epsilon_1, \epsilon_2)$ is represented by the line $\epsilon_1 = -(p/4)\epsilon_2$. Since inflation proceeds from the left to the right, it cannot stop by slow-roll violation. As a consequence, an extra-mechanism, such as e.g. tachyonic instability, must be implemented to end inflation. Let us denote $x_{\text{end}}$ the position at which such a process occurs. The model has therefore two free parameters: $p$ and $x_{\text{end}}$.

The slow-roll trajectory can be obtained by quadrature from Eq. (3.11), and one obtains

$$N - N_{\text{end}} = \frac{1}{2p} (x^2 - x_{\text{end}}^2). \qquad (6.307)$$

This expression can be inverted and reads

$$x = \sqrt{x_{\text{end}}^2 + 2p(N - N_{\text{end}})}. \qquad (6.308)$$

Let us now derive some prior condition on $x_{\text{end}}$. One can notice that when $x < x_{\epsilon_1=1} = p/\sqrt{2}$, one has $\epsilon_1 > 1$ and inflation cannot take place. This means that inflation can only proceed between $x_{\epsilon_1=1}$ and $x_{\text{end}}$, where the maximum number of e-folds is, using Eq. (6.307), $\Delta N_{\text{max}}(x_{\text{end}}) =$
$(x_{\text{end}}^2 - x_{\epsilon_1=1}^2)/(2p)$. Put it differently, if one wants to realize at least $\Delta N$ e-folds, then one has to work with $x_{\text{end}} > x_{\text{end}}^{\text{min}}$ where

$$x_{\text{end}}^{\text{min}}(\Delta N) = \sqrt{p^2/2 + 2p\Delta N}. \qquad (6.309)$$

This defines a prior condition on $x_{\text{end}}$.

Finally, the parameter $M$ can be determined from the amplitude of the CMB anisotropies, and it follows that

$$(\frac{M}{M_{\text{Pl}}})^4 = 720\pi^2 p^2 x_*^{p-2} \frac{Q_{\text{rms}-\text{PS}}^2}{T^2}. \qquad (6.310)$$

The reheating consistent slow-roll predictions for the IMI models are displayed in Fig. 250. For a given value of $p$, they lie along the line $(1-2/p)r = 8(1-n_s)$, i.e. $\epsilon_1 = -(p/4)\epsilon_2$. As expected, large values of $x_{\text{end}}$, or small values of the reheating temperature (these two parameters being degenerate), are preferred.

## 6.19. Brane Inflation (BI)

### 6.19.1. Theoretical justifications

This section is devoted to brane inflation, a class of models widely discussed in the literature [207,417,433,464,634–637,637–647]. The idea is that inflation is caused by branes moving in the extra dimensions as it was already the case in TI, see Section 6.9. For this reason, the setup is very similar to the one considered in that section. One starts from type IIB superstring theory where six dimensions are compactified. The effective, low energy, description of the model contains various fields among which are the dilaton, the axion and the (tensorial) gravitational field. One also has anti-symmetric fields with their corresponding field strength. The compact dimensions form a Calabi–Yau space and, generically, this Calabi–Yau space is made of a bulk plus throats attached to it. Along a given throat, a solution for the ten-dimensional metric is given by the conifold already discussed in Section 6.9 whose metric is given in Eq. (6.142). In this equation, the metric $ds_5^2$ lives on the five-dimensional section $\Sigma_5$ and $r$ is the “radial” coordinate. In the following, we will denote by $r_{\text{UV}}$ the radial coordinate at which the cone is glued to the bulk and $r_0$ the coordinate at the tip of the cone. The volume of the cone section is denoted by $\text{Vol}(\Sigma_5)$ and will be measured in terms of the volume of the five-dimensional sphere, namely

$$v \equiv \frac{\text{Vol}(\Sigma_5)}{\text{Vol}(S_5)}. \qquad (6.311)$$

The geometry of the section $\Sigma_5$ depends on the background fluxes, denoted by $M$ and $K$, that are quantities related to the values of the anti-symmetric fields. If these fluxes vanish then the five-dimensional sections are simply given by $S_2 \times S_3$. In that case, the conifold can be written as $\sum_{i=1}^4 w_i^2 = 0$ where $w_i$ are four complex coordinates, see also Section 6.9. Moreover, an exact expression for the warp function $h(r)$ can be found and reads

$$h(r) = C_2 + \frac{C_1}{r^4}, \qquad (6.312)$$

$C_1$ and $C_2$ being constants. On the other hand, if the fluxes are turned on, then the background geometry responses accordingly and, as a consequence, the geometry of the cone is modified. It is now given by a “deformed conifold”, $\sum_{i=1}^4 w_i^2 = z$, where $z$ is a number which depends on $M$ and $K$. The warp function acquires a more complicated form and, obviously, becomes $z$-dependent, i.e. $h(r, z)$. The explicit form of this warp function is not needed here but it is interesting to notice that, far from the tip, one has $h(r, z) \simeq h(r)$. In other words, the modification of the extra-dimensional geometry due to the fluxes is significant only in the vicinity of the tip. Notice that, provided the depth of the throat is comparable to its width, the radial coordinate $r_{\text{UV}}$ can be expressed in terms of the quantity $N \equiv MK$. One obtains [648]

$$r_{\text{UV}}^4 = 4\pi g_s \alpha'^2 \frac{N}{v}, \qquad (6.313)$$

where $g_s$ is the string coupling and $\alpha' \equiv \ell_s^2$, $\ell_s$ being the string length.

Finally, an anti-$D3$ brane is placed at the tip of the conifold, i.e. at the bottom of the throat. This brane is heavy and is supposed to slightly disturb the geometry of the throat in a way that has been calculated for instance in Refs. [207,646,649]. Then, in this geometry, one studies the motion of a light $D3$ brane with tension

$$T_3 = \frac{1}{(2\pi)^3 g_s \alpha'^2}. \qquad (6.314)$$

This brane is attracted by the anti-$D3$ brane and as a consequence moves radially along the throat. In principle it possesses a DBI kinetic term but one can show that, in the regime considered here, it always reduces to an ordinary, minimal, kinetic term, see Ref. [207]. If $r$ represents the distance between the two branes, then the effective Lagrangian of the system can be expressed as

$$\mathcal{L} = -\frac{1}{2} \left( \frac{\partial \phi}{\partial t} \right)^2 - \frac{2T_3 r_0^4}{r_{\text{UV}}^4} \left( 1 - \frac{r_0^4 T_3^2}{N} \frac{1}{\phi^4} \right), \qquad (6.315)$$

where $\phi \equiv \sqrt{T_3}r$. The shape of the potential is now completely fixed and the behavior $\propto \phi^{-4}$ is of course due to the particular scaling $\propto r^{-4}$ of the warp function given by Eq. (6.312).

In order to be valid, the effective model described above must satisfy some conditions that we now discuss in more detail. Defining $\phi_0 \equiv \sqrt{T_3}r_0$ and $\phi_{\text{UV}} \equiv \sqrt{T_3}r_{\text{UV}}$, it is clear that the presence of the brane in the throat implies that $\phi_0 < \phi < \phi_{\text{UV}}$. In addition, as discussed for instance in Ref. [207], from the trivial fact that the volume of the throat, $V_6^{\text{throat}} = 2\pi^4 g_s N \alpha'^2 r_{\text{UV}}^2$, cannot be bigger than the volume of the total Calabi-Yau manifold $V_6^{\text{tot}}$, one can derive the bound

$$\phi_{\text{UV}} < \frac{m_{\text{Pl}}}{\sqrt{2\pi N}}, \qquad (6.316)$$
where the Planck mass can be expressed as $m_{\text{Pl}}^2 = 8\pi V_6^{\text{tot}}/\kappa_{10}$ and $\kappa_{10} = (2\pi)^7 g_s^2 \alpha'^4/2$. Another constraint comes from the fact that the effective model is valid only if the proper distance between the two branes is larger than the Planck length. One can show, see Ref. [207], that this means $r > r_{\text{stg}}$ where

$$r_{\text{stg}} \equiv r_0 e^{\sqrt{\alpha'} / r_{\text{UV}}}. \qquad (6.317)$$

In particular, as will be seen in the following, the value of $r_{\text{stg}}$ plays an important role regarding the mechanism ending inflation. In the next section, we carry out the slow-roll analysis of this model.

Let us also mention that the same potential arises in the context of tachyon inflation [650,651], in the context of SQCD inflation [652] and in the context of the strong coupling limit of twisted models of SQCD inflation, (see TWI, Section 6.5 and Ref. [568]). It is also worth noticing that the same kind of inverse power law potential is sometimes used in quintessence models [361,624,625]. The brane inflation potential can also receive power law corrections [653] with either positive (UV models) or negative sign (IR models). The UV case is similar to RIPI models while the IR corresponds to SFI models.

### 6.19.2. Slow-roll analysis

We now turn to the slow-roll analysis of BI. For this purpose, it is more convenient to re-write the potential appearing in Eq. (6.315) in the following way

$$V(\phi) = M^4 \left[ 1 - \left( \frac{\phi}{\mu} \right)^{-p} \right], \qquad (6.318)$$

where $\mu$ and $p$ are free parameters. Compared to Eq. (6.315), we have generalized by hand the expression of $V(\phi)$ by considering an arbitrary $p$. In such a way, this potential can be viewed as a generalization of the small field models to negative values of $p$ (see Section 6.1). In the following, we will also consider the non-approximated KKLT potential

$$V(\phi) = \frac{M^4}{1 + \left(\frac{\phi}{\mu}\right)^{-p}}, \qquad (6.319)$$

from which (6.318) is the $\mu \ll M_{\text{Pl}}$ limit.

In the context of the brane inflationary scenario, the value $p = 4$ is special in the sense that, as explained above, it corresponds to the motion of a test D3 brane in a warped throat and is, therefore, a case of physical interest. Let us notice that the parameters of the potential are related to their stringy counterparts by

$$M^4 = \frac{2T_3 r_0^4}{r_{\text{UV}}^4} = \frac{4\pi^2 \nu}{\mathcal{N}^4} \phi_0^4, \qquad \mu^4 = \frac{T_3^2 r_0^4}{\mathcal{N}^4} = \frac{M^4}{4\pi^2 \nu}. \qquad (6.320)$$

Moreover, brane inflation proceeds under the condition $\mu/M_{\text{Pl}} \ll 1$. Indeed, using the formulas established in the previous subsection, it is easy to show that

$$\frac{\mu^4}{M_{\text{Pl}}^4} = \frac{1}{\mathcal{N}^4} \left(\frac{\phi_0}{M_{\text{Pl}}}\right)^4 < \frac{1}{\mathcal{N}^4} \left(\frac{\phi_{\text{UV}}}{M_{\text{Pl}}}\right)^4 < \frac{16}{\mathcal{N}^3} \ll 1, \qquad (6.321)$$

where we have used the condition $\phi_0 < \phi_{\text{UV}}$ and Eq. (6.316). Finally, let us stress that the brane motion in the throat ends by a tachyonic instabilities at $\phi = \phi_{\text{stg}}$. As we discuss below, the observable predictions of the model crucially depends on whether the universe is still inflating at $\phi \gtrsim \phi_{\text{stg}}$, or not. Therefore, in the context of string theory, we necessarily have $\mu/M_{\text{Pl}} \ll 1$, $p = 4$ and an additional model parameter $\phi_{\text{stg}}$.

In the following, we will first consider arbitrary values for $\mu$ and $p$ viewing Eq. (6.318) as a phenomenological potential in which $\phi_{\text{stg}}$ has no meaning, and then, the discussion will be focused on the stringy scenario. BI is another proto-typical case exemplifying how two models having exactly the same potential can lead to different observable predictions. Here this will be due to the mechanism ending inflation.

The potential (6.318), as well as its logarithm, are displayed in Fig. 84. It is an increasing function of the field, hence inflation proceeds from the right to the left. It vanishes for $\phi/\mu = 1$ and, hence, it should be studied in the $\phi/\mu > 1$ region only. Let us calculate the slow-roll parameters. Defining the quantity $x$ by the following expression

$$x \equiv \frac{\phi}{\mu}, \qquad (6.322)$$

one can express the first three Hubble flow functions in the slow-roll approximation as

$$\epsilon_1 = \left( \frac{M_{\text{Pl}}}{\mu} \right)^2 \frac{p^2}{2x^2 (1 - x^p)^2}, \qquad \epsilon_2 = 2p \left( \frac{M_{\text{Pl}}}{\mu} \right)^2 \frac{(1+p)x^p - 1}{x^2 (1 - x^p)^2}, \qquad (6.323)$$

and

$$\epsilon_3 = p \left( \frac{M_{\text{Pl}}}{\mu} \right)^2 \frac{2 + (p-4)(p+1)x^p + (1+p)(2+p)x^{2p}}{x^2 (1-x^p)^2 [(1+p)x^p - 1]}, \qquad (6.324)$$

These functions are displayed in Fig. 84. They become very small at large fields $x \gg 1$, and diverge when the potential vanishes at $x \to 1$. Therefore inflation can naturally end with slow-roll violation at a field value $x_{\text{end}}$, solution of $\epsilon_1(x_{\text{end}}) = 1$, i.e., verifying

$$x_{\text{end}}^{p+1} - x_{\text{end}} = \frac{p}{\sqrt{2}} \frac{M_{\text{Pl}}}{\mu}. \qquad (6.325)$$

Unless $p$ takes integer values, this equation has to be solved numerically (see also Section 6.1).
<figure><img src="image_87.png" /><figcaption>Fig. 84. Brane Inflation (BI) for $p = 2$. Upper panels: the potential and its logarithm as a function of $\phi/\mu$. Bottom left panel: slow-roll parameter $\epsilon_1$ divided by $M_{\text{Pl}}^2/\mu^2$. The shaded area indicates the region in which inflation cannot occur for $\mu = M_{\text{Pl}}$. Bottom right panel: slow-roll parameters $\epsilon_2$ (solid line) and $\epsilon_3$ (dotted line), divided by $M_{\text{Pl}}^2/\mu^2$.</figcaption></figure>

However, in the limits $\mu/M_{\text{Pl}} \ll 1$ and $\mu/M_{\text{Pl}} \gg 1$ we can find an approximate expression for $x_{\text{end}}$. Solving perturbatively the equation $\epsilon_1 = 1$, one obtains

$$x_{\text{end}} \underset{\mu \ll M_{\text{Pl}}}{\approx} \left( \frac{p M_{\text{Pl}}}{\sqrt{2} \mu} \right)^{\frac{1}{p+1}} + \frac{1}{p+1} \left( \frac{p M_{\text{Pl}}}{\sqrt{2} \mu} \right)^{\frac{1-p}{1+p}}, \quad x_{\text{end}} \underset{\mu \gg M_{\text{Pl}}}{\approx} 1 + \frac{1}{\sqrt{2}} \frac{M_{\text{Pl}}}{\mu} - \frac{p+1}{4} \frac{M_{\text{Pl}}^2}{\mu^2}. \qquad (6.326)$$

It is also interesting to find the solution of $\epsilon_2 = 1$. As before, this cannot be done exactly but, perturbatively, one obtains

$$x_{\epsilon_2=1} \underset{\mu \ll M_{\text{Pl}}}{\approx} \left[ 2p(1+p) \left( \frac{M_{\text{Pl}}}{\mu} \right)^2 \right]^{\frac{1}{p+2}}, \quad x_{\epsilon_2=1} \underset{\mu \gg M_{\text{Pl}}}{\approx} 1 + \sqrt{2} \frac{M_{\text{Pl}}}{\mu}. \qquad (6.327)$$

From the above expressions, we deduce that slow-roll violation always occurs before the end of inflation, that is to say $\epsilon_2$ becomes unity before $\epsilon_1$. This has not effect on the observable predictions since only a few e-folds of inflation are spent in this regime (see Fig. 84).

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

$$N_{\text{end}} - N = \frac{\mu^2}{2p M_{\text{Pl}}^2} \left( x_{\text{end}}^2 - \frac{2}{p+2} x_{\text{end}}^{p+2} - x^2 + \frac{2}{p+2} x^{p+2} \right), \qquad (6.328)$$

an expression which cannot be inverted in general. However, in the $\mu \ll M_{\text{Pl}}$ and $\mu \gg M_{\text{Pl}}$ limits, one has $x \gg 1$ and $x \simeq 1$ respectively and the previous equation can be approximately inverted leading to the following expressions

$$x_* \underset{\mu \ll M_{\text{Pl}}}{\approx} \left[ p(p+2) \frac{M_{\text{Pl}}^2}{\mu^2} \Delta N_* + x_{\text{end}}^{p+2} \right]^{\frac{1}{p+2}}, \quad x_* \underset{\mu \gg M_{\text{Pl}}}{\approx} 1 + \frac{M_{\text{Pl}}}{\mu} \sqrt{\frac{1}{2} + 2\Delta N_*}, \qquad (6.329)$$

where use has been made of Eq. (6.326). Also, making use of the full KKLT potential (6.319), the slow roll trajectory reads

$$N_{\text{end}} - N = \frac{\mu^2}{2p M_{\text{Pl}}^2} \left( -x_{\text{end}}^2 - \frac{2}{p+2} x_{\text{end}}^{p+2} + x^2 + \frac{2}{p+2} x^{p+2} \right), \qquad (6.330)$$

which coincides with (6.328) in the limit $\mu \ll M_{\text{Pl}}$.

The mass scale M is given by the CMB normalization and verifies

$$\left( \frac{M}{M_{\text{Pl}}} \right)^4 = 720 \pi^2 p^2 \left( \frac{M_{\text{Pl}}}{\mu} \right)^2 \frac{x_*^{p-2}}{(x_*^p - 1)^3} \frac{Q_{\text{rms-PS}}^2}{T^2}. \qquad (6.331)$$

which can be further simplified in the appropriate limits using Eqs. (6.326) and (6.329).
The reheating consistent slow-roll predictions for the phenomenological models are displayed in Figs. 254, 255, 256 for $p = 2$, $p = 3$ and $p = 4$, respectively, and with $\mu/M_{\text{pl}} \in [10^{-3}, 10^3]$. The reheating equation of state parameter $\bar{w}_{\text{reh}} = 0$ but since the shape of the potential is unknown at $x < 1$, this parameter is a priori unspecified and could take different values. For small values of $\mu$, we see that $n_s \simeq 0.96$ and $r \ll 1$. In the opposite case, $\mu \gg M_{\text{pl}}$, the model predictions lie around $\epsilon_2 \simeq 4\epsilon_1$ with $n_s \simeq 0.97$ and $r \simeq 0.08$. These behaviors can be recovered by plugging the approximated expressions given in Eqs. (6.326) and (6.329) into the Hubble flow functions. For $\mu \ll M_{\text{pl}}$, one obtains

$$ \epsilon_{1*} \approx \frac{p^2}{2} \left[ p(p+2) \Delta N_* \right]^{-\frac{2p+2}{p+2}} \left( \frac{\mu}{M_{\text{pl}}} \right)^{\frac{2p}{p+2}}, \quad \epsilon_{2*} \approx \frac{2}{\Delta N_*} \frac{p+1}{p+2}, \quad \epsilon_{3*} \approx \frac{1}{\Delta N_*}, \qquad (6.332) $$

and the spectral index is of the order $n_s \simeq 1 - 2/\Delta N_*(p+1)/(p+2) \sim 0.96$ with $r \ll 1$. Similarly, for $\mu \gg M_{\text{pl}}$ limit, the Hubble flow parameters at Hubble crossing behave as

$$ \epsilon_{1*} \approx \frac{1}{4\Delta N_*}, \quad \epsilon_{2*} \approx \frac{1}{\Delta N_*}, \quad \epsilon_{3*} \approx \frac{1}{\Delta N_*}. \qquad (6.333) $$

Therefore, the predicted level of gravity waves is now of the order $r \simeq 4/\Delta N_* \simeq 0.08$ and the spectral index is $n_s \simeq 1 - 3/(2\Delta N_*) \simeq 0.97$, which is again in agreement with the numerical results.

Finally, the predictions for the KKLTI models, i.e. using the full potential (6.319), are displayed in Figs. 258, 259, 260 for the same parameters. One can see that they deviate from the ones of brane inflation only when $\mu \gg M_{\text{pl}}$.

### 6.19.3. Slow-roll analysis of the stringy scenario

In the case where the model is interpreted as a stringy scenario, with $p = 4$, we have seen before that the low energy description is valid provided $r > r_{\text{stg}}$, or $x > x_{\text{stg}}$ with

$$ x_{\text{stg}} \equiv \frac{\sqrt{T_3} r_{\text{stg}}}{\mu} = \mathcal{N}^{1/4} \exp \left[ \left( 4\pi g_s \frac{\mathcal{N}}{\nu} \right)^{-1/4} \right]. \qquad (6.334) $$

If slow-roll violation occurs before the system reaches $x_{\text{stg}}$, then the effective string description is always valid and the observable predictions will be exactly the same as those derived in the previous paragraph (for $p = 4$ and $\mu \ll M_{\text{pl}}$). However, if, on the contrary, slow-roll violation occurs after the field crosses the value $x_{\text{stg}}$, then inflation stops by instability at $x_{\text{stg}}$ instead of the naively expected $x_{\text{end}}$. Indeed, in this case, a tachyon appears and triggers the process of branes annihilation. Therefore, the mechanism ending inflation in this model depends on whether slow-roll violation occurs in a regime where the distance between the branes is larger or smaller than the string length. And this question depends on the value of the parameters characterizing BI. One can determine the two regimes by evaluating the ratio

$$ \frac{x_{\epsilon_2=1}}{x_{\text{stg}}} = 40^{1/6} \left( \frac{M}{M_{\text{pl}}} \right)^{-1/3} \mathcal{N}^{-1/4} (4\pi^2\nu)^{1/12} \exp \left[ - \left( 4\pi g_s \frac{\mathcal{N}}{\nu} \right)^{-1/4} \right], \qquad (6.335) $$

in which we have used Eqs. (6.320), (6.327) and (6.334) (with $p = 4$ and $\mu \ll M_{\text{pl}}$). If this ratio is larger than one, inflation stops by slow-roll violation and if it is smaller than one by instability. The complicated part of the analysis lies in the fact that the above equation depends on the mass scale $M$. In order to have an explicit expression of $M$ in terms of the parameters of the model, one must first CMB normalize the model which, in turn, requires the knowledge of the mechanism ending inflation. However, we are interested in calculating the frontier where $x_{\epsilon_2=1} = x_{\text{stg}}$ and, therefore, the two possible mechanisms for stopping inflation coincide in that case. Replacing $x_{\text{end}}$ by $x_{\text{stg}} = x_{\epsilon_2=1}$ in Eq. (6.329) yields

$$ x_*^f \approx \left[ 24 \frac{M_{\text{pl}}^2}{\mu^2} \left( \Delta N_* + \frac{5}{3} \right) \right]^{1/6}, \qquad (6.336) $$

from which one can obtain an explicit formula for the first slow-roll coefficient (6.323) at Hubble radius crossing

$$ \epsilon_{1*}^f \approx 8 \left[ 24 \left( \Delta N_* + \frac{5}{3} \right) \right]^{-5/3} \left( \frac{\mu}{M_{\text{pl}}} \right)^{4/3}. \qquad (6.337) $$

Comparing this expression to Eq. (6.332), we see that there is a very small shift by $5/3$ in $\Delta N_*$. It accounts for the difference of e-folds between the time at which slow-roll violations occur, i.e. for $x = x_{\epsilon_2=1}$, and the end of inflation at $x_{\text{end}}$. As argued before, we see that these effects are too small to be observable and completely degenerated with the reheating duration. Plugging this expression into the CMB normalization, and using the relation $M^4 = 4\pi^2\nu\mu^4$, one arrives at the following expression for $M$

$$ \frac{M}{M_{\text{pl}}} = C (4\pi^2 \nu)^{-1/8} \left( \Delta N_* + \frac{5}{3} \right)^{-5/8}, \qquad (6.338) $$

where we have defined

$$ C \equiv 3^{-5/8} (8\pi^2 Q_*)^{3/8}, \quad Q_* \equiv 45 \frac{Q_{\text{rms}-\text{PS}}}{T^2} = 2700 P_* \qquad (6.339) $$

We can now insert this expression of $M$ in Eq. (6.335) to get the equation defining the frontier in the string parameter space, namely

$$ \left. \frac{x_{\epsilon_2=1}}{x_{\text{stg}}} \right|_f = 1 = \left( \frac{40}{C^2} \right)^{1/6} \left( \Delta N_* + \frac{5}{3} \right)^{5/24} (4\pi^2\nu)^{1/8} \mathcal{N}^{-1/4} \exp \left[ - \left( 4\pi g_s \frac{\mathcal{N}}{\nu} \right)^{-1/4} \right]. \qquad (6.340) $$

Following Ref. [207], if one defines the two following rescaled stringy parameters

$$ y \equiv 4\pi g_s \frac{\mathcal{N}}{\nu}, \quad \bar{v} \equiv \frac{\nu}{(4\pi g_s)^2}, \qquad (6.341) $$

then the frontier (6.340) is defined by the following “universal” form

$$ y^{1/4} e^{y^{-1/4}} \bar{v}^{1/8} - \left( \frac{40}{C^2} \right)^{1/6} \left( \Delta N_* + \frac{5}{3} \right)^{5/24} (4\pi^2)^{1/8} = 0, \qquad (6.342) $$
<figure><img src="image_88.png" /><figcaption>Fig. 85. Theoretical prior space for the stringy scenario of brane inflation [207] in the plane of the “universal” coordinates (y, v̄). The solid blue line is the frontier above which inflation ends by tachyonic pre-heating triggered by brane annihilation (light green region). Only in the region enclosed by this curve (light blue region), inflation ends by slow-roll violation. The upper thick red line is the volume bound of Eq. (6.343). The lower black straight line is the “UV” limit given by (6.346) and is relevant only if inflation stops by slow-roll violation. The solid green curve is given by (6.350) and also represents the “UV” limit but, this time, in the regime where inflation stops when the two branes collide. As a consequence, the admissible region is the one shaded in light black. We see that, even in this allowed region, inflation can either end by tachyonic instability or slow-roll violation depending on the string parameter values. In principle, the blue, black and green lines should cross at a single point. Due to the approximations used here, we see that this is true only approximately. In order to give a more faithful description of the allowed region, the light black area has been slightly deformed around the crossing point (see Ref. [207] for an exact determination of these frontiers).</figcaption></figure>

which is independent of the string coupling $g_s$. As represented in Fig. 85, in the plane (y, v̄), this relation is a curve that separates the region where inflation stops by slow-roll violation (below the curve) and the region where inflation stops by instability due to brane annihilation (above the curve).

The requirement of having the throat contained within the Calabi-Yau manifold can equally be written in terms of the universal variables. From Eqs. (6.316) and (6.341), one gets

$$y^{3/2} \bar{v} < 8\pi^2 M_{\text{pl}}^2 \ell_s^2, \qquad (6.343)$$

which therefore depends on the string length $\ell_s = \sqrt{\alpha'}$ but not on the string coupling $g_s$.

Finally, the last theoretical prior comes from requiring that the brane motion remains located inside the throat, i.e. $x < x_{\text{UV}}$ with

$$x_{\text{UV}} \equiv \frac{\sqrt{T_3} r_{\text{UV}}}{\mu} = \frac{M_{\text{pl}}}{M} \left( \frac{\mathcal{N}}{4\pi^3 \alpha'^2 g_s} \right)^{1/4}. \qquad (6.344)$$

Since during inflation $x$ decreases, this condition gives an upper limit on the admissible initial field values. However, the initial field values depends on the total number of e-folds of inflation, say $\Delta N_{\text{tot}}$, and on the field value at which inflation ends, i.e. either $x_{\text{stg}}$ or $x_{\epsilon_2=1}$ depending on if brane annihilation occurs before slow-roll violations.

Let us first assume that brane annihilation occurs well after the end of inflation, i.e. we are in lower part of the string parameter space (y, v̄) separated by Eq. (6.342). For the relevant limit, $\mu \ll M_{\text{pl}}$, the initial field value is given by

$$x_{\text{ini}}^{\epsilon_2} \approx \left[ 24 \frac{M_{\text{pl}}^2}{\mu^2} \left( \Delta N_{\text{tot}} + \frac{5}{3} \right) \right]^{1/6}. \qquad (6.345)$$

This expression involves $\mu$ and therefore $M$ through Eq. (6.320). Again, one has to determine $M$ using the CMB normalization and we are assuming that inflation ends at $x_{\epsilon_2=1}$, i.e. exactly Eq. (6.338). Plugging everything together and making use of the universal variables, one gets

$$y \bar{v} \stackrel{x_{\text{stg}} > x_{\epsilon_2=1}}{\ge} C^{8/3} \pi^2 M_{\text{pl}}^2 \ell_s^4 \left[ 24 \left( \Delta N_{\text{tot}} + \frac{5}{3} \right) \right]^{2/3} \left( \Delta N_* + \frac{5}{3} \right)^{-5/3}. \qquad (6.346)$$

If inflation ends by brane annihilation at $x = x_{\text{stg}}$, i.e. the string parameters (y, v̄) lie above the curve given by Eq. (6.338), then $x_{\text{ini}}$ and $x_*$ are accordingly modified. For $\mu \ll M_{\text{pl}}$, their new expressions are however still given by Eq. (6.329), up to the replacement $x_{\text{end}} \to x_{\text{stg}}$, i.e.

$$x_{\text{ini}}^{\text{stg}} \approx \left( 24 \frac{M_{\text{pl}}^2}{\mu^2} \Delta N_{\text{tot}} + x_{\text{stg}}^6 \right)^{1/6}, \quad x_{*}^{\text{stg}} \approx \left( 24 \frac{M_{\text{pl}}^2}{\mu^2} \Delta N_{*} + x_{\text{stg}}^6 \right)^{1/6}. \qquad (6.347)$$

As before, $x_{\text{ini}}^{\text{stg}}$ and $x_{*}^{\text{stg}}$ depend on $\mu$ and therefore on $M$, which is determined by the CMB normalization. However, since inflation now ends by tachyonic instability this one has to be re-determined by plugging $x_{*}^{\text{stg}}$ into Eq. (6.331). Doing so gives an implicit expression for $M$

$$\frac{M}{M_{\text{pl}}} \approx C(4\pi^2 v)^{-1/8} \left( \Delta N_* + \frac{\mu^2}{M_{\text{pl}}^2} \frac{x_{\text{stg}}^6}{24} \right)^{-5/8} = C(4\pi^2 v)^{-1/8} \left[ \Delta N_* + \frac{5}{3} \left( \frac{x_{\text{stg}}}{x_{\epsilon_2=1}} \right)^6 \right]^{-5/8}, \qquad (6.348)$$
where use has been made of Eq. (6.327), for $\mu \ll M_{\text{pl}}$. This equation cannot be analytically solved for $M$ because $\mu$, and $x_{\epsilon_2=1}$, depends on $M$. However, if brane annihilation occurs well before slow-roll violation, one has $x_{\text{stg}} \gg x_{\epsilon_2=1}$ such that the term in $\Delta N_*$ can be neglected. In that situation, from $\mu^4 = M^4/(4\pi^2\nu)$, one gets the approximate expression

$$ \frac{M}{M_{\text{pl}}} \underset{x_{\text{stg}} \gg x_{\epsilon_2=1}}{\approx} 24^{5/18} C^{4/9} (4\pi^2\nu)^{1/12} x_{\text{stg}}^{-5/3}. \qquad (6.349) $$

Requiring $x_{\text{ini}}^{\text{stg}} < x_{\text{UV}}$ finally yields

$$ y^{19/6} \bar{v}^{7/3} \exp\left(\frac{20}{3}y^{-1/4}\right) \underset{x_{\text{stg}} \gg x_{\epsilon_2=1}}{>} \left(8\pi^2 \ell_s^2\right)^3 Q_* \left[y^{2/3} \bar{v}^{1/3} \exp\left(\frac{8}{3}y^{-1/4}\right) + \frac{6\Delta N_{\text{tot}}}{Q_*^{1/3}}\right], \qquad (6.350) $$

which completes the bounds coming from $x_{\text{UV}}$.

Brane inflation within the string scenario has therefore a rather involved set of priors. In addition to have $p=4$ and $\mu \ll M_{\text{pl}}$, the model parameters should simultaneously verify Eq. (6.343) and either Eq. (6.346), or Eq. (6.350), according to the sign of the left hand side of Eq. (6.342). All these equations involve the amplitude of the CMB anisotropies, which is well measured, the total number of e-folds $\Delta N_{\text{tot}}$, which is an unknown quantity, and the number of e-folds $\Delta N_*$ before the end of inflation at which the pivot mode crossed the Hubble radius. As discussed in Section 3.2, $\Delta N_*$ can only be obtained by solving Eq. (3.45), i.e. after having specified the reheating parameter. As the result, the reheating slow-roll predictions for the string scenario can only be sorted out numerically, paying attention that for a given reheating history, all of the previous theoretical constraints are satisfied. As an illustration, we have plotted in Fig. 85 the bounds for the typical values $\Delta N_* = 50$ and $\Delta N_{\text{tot}} = 60$ with $\alpha' M_{\text{pl}}^2 \approx 1/4$ [207,654].

The reheating consistent slow-roll predictions for the string models are displayed in Figs. 257 for a set of realistic fundamental parameters. Also, making use of the full potential (6.319), the predictions of the corresponding KKLT inflation models are displayed in Figs. 261. One can check that they match perfectly.

## 6.20. String Axion Inflation I (SAII)

The model emerges from geometrical compactifications on Calibi-Yau manifold, in presence of fluxes, and in the framework of type IIB superstring theory. It has been proposed in Ref. [655] and shares some similarities with the KKLT construction of Section 6.19. However, here, the inflaton is identified with an axion of the complex structure moduli while its potential comes from worldsheet instanton effects [656] that have been derived in Ref. [655]. The potential reads

$$ V(\phi) = M^4 \left[ 1 - \cos\left(\frac{\phi}{\mu}\right) + \alpha \frac{\phi}{\mu} \sin\left(\frac{\phi}{\mu}\right) \right], \qquad (6.351) $$

where $\mu$ is a vev and $\alpha$ a dimensionless constant that is not required to be small. The first two terms in Eq. (6.351) match, up to a field redefinition, the potential of Natural Inflation (NI) in Section 5.6. Therefore, the potential of SAII can be viewed as a modulated addition to NI, a situation also discussed in Ref. [657]. Let us stress, however, that, depending on the values of $\alpha$, the predictions of SAII can be quite different from the ones of NI. Ref. [655] also considers higher-order terms in the instanton effects and, under some assumptions, these ones can generate an additional mass term in the potential. This case corresponds to the model String Axion Inflation II (SAIII), which is analyzed in Section 7.7.

The potential in Eq. (6.351) depends on two parameters, $\mu$ and $\alpha$, that can take any value. It is symmetric with respect to $\phi=0$, and we can therefore restrict the analysis to the $\phi \ge 0$ region. As soon as $\alpha$ is non-vanishing, the potential becomes negative in some regions, and for $\alpha < -1/2$, this occurs around the origin (see below). As a result, slow-roll inflation can take place only within some limited field range, that depends on $\alpha$, and around the maxima of the potential. The potential and its logarithm are displayed in the top panels of Fig. 86.

Defining $x \equiv \phi/\mu$, the smallest local maximum of the potential, denoted $x = x_{\text{V-max}}$, is a solution of

$$ (1 + \alpha) \sin(x) + \alpha x \cos(x) = 0. \qquad (6.352) $$

This is a transcendental equation, which has to be solved numerically for each value of $\alpha$. Here, we are interested in the smallest positive solution of this equation for which $V(x_{\text{V-max}}) > 0$. Expanding Eq. (6.351) around the origin, one gets

$$ \frac{V(x)}{M^4} = \left(\alpha + \frac{1}{2}\right) x^2 + \mathcal{O}(x^4), \qquad (6.353) $$

which implies that, for $\alpha > -1/2$, the potential is a positive increasing function of $x$ in a neirboorhood of $x=0$, up to its first local maximum at $x = x_{\text{V-max}}$. Therefore, inflation can take place within the domain $x \in [0, x_{\text{V-max}}]$, at decreasing field values, and this regime will be referred to as SAII1. The potential is minimal at the origin, with $V(x=0) = 0$, such that reheating after inflation ends up in a Minkowski vacuum. For $\alpha < -1/2$, Eq. (6.353) shows that the potential is decreasing towards a negative minimum around the origin, and then becomes positive to reach its first local maximum at $x = x_{\text{V-max}}$. In that situation, the SAII1 inflationary domain lies within $x \in [x_{\bar{V}=0}^-, x_{\text{V-max}}]$, where $x_{\bar{V}=0}$ is the smallest strictly positive solution of $V(x) = 0$, i.e.

$$ 1 - \cos(x) + \alpha x \sin(x) = 0. \qquad (6.354) $$

This equation is again transcendental and has to be solved numerically. However, there are some trivial solutions, namely $x = 2\pi n$ where $n$ is an integer number. Unfortunately, for $\alpha < -1/2$, $x_{\bar{V}=0}^+ < 2\pi$ does not belong to this subset of solutions. Let us also notice that, in this situation, the reheating would proceed after inflation around an anti-de Sitter minimum, which should thus be lifted somehow. If one is ready to accept to rely on such a mechanism, then one should also consider the inflating regime at $x > x_{\text{V-max}}$, where inflation proceeds at increasing field values and for which reheating also occurs around an anti-de Sitter minimum. This regime will be referred to as SAII2, see Fig. 86. Strictly speaking, there are an infinite numbers of negative minima for the potential at larger values of $x$, but the value of $V$ at the minimum becomes negatively larger for each of them. We will be therefore ignore these in the following.

For all values of $\alpha$, SAII2 occurs in the range $x \in [x_{\text{V-max}}, x_{\bar{V}=0}^+]$ where $x_{\bar{V}=0}^+$ is the smallest solution of Eq. (6.354) satisfying $x_{\bar{V}=0}^+ > x_{\text{V-max}}$. This time, if $\alpha \le 0$, one has $x_{\bar{V}=0}^+ = 2\pi$, but for $\alpha > 0$, Eq. (6.354) has to be solved numerically in order to determine $x_{\bar{V}=0}^+$. The situation is summarized in Fig. 87 where we have plotted $x_{\text{V-max}}(\alpha)$, $x_{\bar{V}=0}^+(\alpha)$ and $x_{\bar{V}=0}^+(\alpha)$.
<figure><img src="image_89.png" /><figcaption>Fig. 86. String Axion Inflation I (SAII) for $\alpha = -0.2$ (blue curve) and $\alpha = -0.8$ (red curve). Top panels: the potential and its logarithm, for comparison the potential of Natural Inflation ($\alpha = 0$, black curve) is represented. Bottom left panel: slow-roll parameter $\epsilon_1$ for $\alpha = -0.2$ and $\mu = 10 M_{\text{pl}}$, with the different inflationary regimes of SAII annotated with an arrow indicating the direction to which the field evolves. Bottom right panel: slow-roll parameters $\epsilon_2$ (solid line) and $\epsilon_3$ (dotted line) for the same parameters value. When $\epsilon_3$ becomes negative, the plot shows $|\epsilon_3|$ as a cyan dotted line, the blue dotted line corresponds to positive values.</figcaption></figure>

<figure><img src="image_90.png" /><figcaption>Fig. 87. Field domains in which SAI1 and SAI2 are defined. The black central curve shows the field value $x_{\text{vmax}}$ (in unit of $\mu$) at which the potential is maximal. SAI1 takes place at decreasing field value from $x_{\text{vmax}}$ while SAI2 inflates from $x_{\text{vmax}}$ at increasing field values. The curves $x_{\text{vmax}}^-(\alpha) > 0$ and $x_{\text{vmax}}^+(\alpha)$ are the separatrix under which (and, respectively, above which) the potential becomes negative. In both regimes, inflation gracefully ends before the separatrix is reached at $x = x_{\text{end}}$, where $x_{\text{end}}$ is the relevant solution of $\epsilon_1(x) = 1$. The dashed lines show the value of $x_{\text{end}}(\alpha)$ for SAI1 (lower dashed curve) and SAI2 (top dashed curve) when $\mu = 10 M_{\text{pl}}$.</figcaption></figure>

The first two slow-roll parameters read

$$ \epsilon_1 = \frac{1}{2\mu^2} \left[ \frac{(1+\alpha)\sin(x) + \alpha x \cos(x)}{1 - \cos(x) + \alpha x \sin(x)} \right]^2, \qquad (6.355) \\ \epsilon_2 = \frac{1}{\mu^2} \frac{2 + 2\alpha x \sin(x) - 2(1 + 2\alpha)\cos(x) - \alpha^2 \cos(2x) + \alpha(4 + \alpha + 2\alpha x^2)}{[1 - \cos(x) + \alpha x \sin(x)]^2}, $$
while the third one is given by

$$ \epsilon_3 = -\frac{1}{\mu^2[1 - \cos(x) + \alpha x \sin(x)]^2} \times \frac{(\alpha + 1) \sin(x) + \alpha x \cos(x)}{\alpha^2 + 4\alpha + 2\alpha^2 x^2 - \alpha^2 \cos(2x) + 2\alpha x \sin(x) - 2(2\alpha + 1) \cos(x) + 2} \times \left[ \begin{aligned} & -4\alpha^2 x^2 \sin(x) - \alpha^2 x^2 \sin(2x) - 2\alpha x (6\alpha + 2\alpha^2 x^2 + 1) \cos(x) - 3\alpha^3 \sin(x) \\ & + \alpha^3 \sin(3x) + 9\alpha^2 x - 12\alpha^2 \sin(x) + 6\alpha^2 \sin(2x) - 6\alpha \sin(x) + 3\alpha \sin(2x) \\ & + (3\alpha + 2)\alpha x \cos(2x) - 2 \sin(x) + \sin(2x) \end{aligned} \right]. \quad (6.356) $$

The denominator of $\epsilon_1(x)$ in Eq. (6.355) diverges for $x \to 0$, $x \to x_{V=0}^{-}$ and $x \to x_{V=0}^{+}$, which ensures that inflation gracefully ends before the potential becomes negative. Moreover, all three slow-roll parameters feature an overall scaling $\propto \mu^{-2}$. This implies that, for any value of $\alpha$, at fixed $x$, the slow-roll parameters can be made arbitrarily small by choosing large values of $\mu$. Conversely, for small values of $\mu$, one expects inflation to be of the hilltop kind and confined around $x_{V,\text{max}}$. The three slow-roll parameters have been plotted for $\alpha = -0.2$ in the bottom panels of Fig. 86.

Both SAI1 and SAI2 inflationary regimes end for $\epsilon_1(x_{end}) = 1$ at a field value $x_{end}$ that is solution of

$$ (1 + \alpha) \sin(x) + \alpha x \cos(x) = \pm \mu \sqrt{2} [1 - \cos(x) + \alpha x \sin(x)], \quad (6.357) $$

which needs to be solved in the range $x_{end} \in [x_{V=0}^{-}, x_{V,\text{max}}]$ for SAI1 and $x_{end} \in [x_{V,\text{max}}, x_{V=0}^{+}]$ for SAI2. Again, Eq. (6.357) being transcendental, it has to be solved numerically and the solutions have been represented as dashed curves in Fig. 87.

The slow-roll trajectory also requires a numerical integration and reads

$$ N_{\text{end}} - N = \mu^2 \int_{x_{\text{end}}}^{x} \frac{1 - \cos(y) + \alpha y \sin(y)}{(1 + \alpha) \sin(y) + \alpha y \cos(y)} dy, \quad (6.358) $$

with $x_{end}$ the relevant solution of Eq. (6.357), depending on the regime under consideration. This expression diverges for $x \to x_{V,\text{max}}$ and the total number of e-folds is therefore unbounded.

Finally, the normalization of the potential $M^4$ can be obtained from the amplitude of the CMB anisotropies once one has determined the field value $x_*$ at which the pivot mode crossed the Hubble radius. It reads

$$ \left(\frac{M}{M_{\text{Pl}}}\right)^4 = \frac{720\pi^2}{\mu^2} \frac{[(1 + \alpha) \sin(x_*) + \alpha x_* \cos(x_*)]^2}{[1 - \cos(x_*) + \alpha x_* \sin(x_*)]^3}. \quad (6.359) $$

The reheating consistent slow-roll predictions for SAI1 and SAI2 are represented in Figs. 262 to 267 for various values of $\alpha$ and $\mu$.

## 6.21. Mukhanov Inflation (VFMI)

This model has been proposed by Mukhanov in Ref. [658] and relies on a hydrodynamical representation of the inflationary background evolution. Instead of specifying a potential, one postulates the form of the equation-of-state parameter $w$ as a function of the number of e-folds elapsed before the end of inflation $N_{end} - N$. VFMI is defined by

$$ W(N) \equiv 1 + w(N) \equiv \frac{\beta}{(C + N_{\text{end}} - N)^{\alpha}}, \quad (6.360) $$

where $\alpha > 0$ and $\beta > 0$ are the model parameters and $C$ is a constant that has been set to unity in Ref. [658]. Because inflation ends at $N = N_{end}$ and, by definition, when the universe stops accelerating, i.e., for $w(0) = -1/3$, one has in fact

$$ C = \left(\frac{3\beta}{2}\right)^{1/\alpha}. \quad (6.361) $$

At the perturbative level, cosmological fluctuations are still assumed to be of quantum-mechanical origin, adiabatic, and conserved on super-Hubble scales, i.e. coupled with a single scalar field.

### 6.21.1. Equivalence with a scalar field theory

On general grounds, giving the functional form of $W(N) = 1 + w(N)$ is strictly equivalent to specifying a parametric potential for a canonically normalized scalar field $\phi$. This can be seen from the hydrodynamical Friedmann-Lemaître equations

$$ H^2 = \frac{\rho}{3M_{\text{Pl}}^2}, \qquad \frac{\ddot{a}}{a} = -\frac{\rho}{6M_{\text{Pl}}^2} \left(1 + 3\frac{P}{\rho}\right). \quad (6.362) $$

From $P = w\rho$, and in terms of the number of e-folds $N$, the second equation reads

$$ H \frac{dH}{dN} + H^2 = -\frac{\rho}{6M_{\text{Pl}}^2}(1 + 3w). \quad (6.363) $$

Plugging the first Friedmann-Lemaître equation $\rho = 3M_{\text{Pl}}^2 H^2$, and dividing Eq. (6.363) by $H^2$, one gets

$$ -\frac{1}{H} \frac{dH}{dN} = \frac{3}{2}(1+w). \quad (6.364) $$

By definition of the Hubble-flow functions in Eq. (3.3), this equation reduces to

$$ \epsilon_1(N) = \frac{3}{2} [1 + w(N)] = \frac{3\beta}{2(C + N_{\text{end}} - N)^{\alpha}}. \quad (6.365) $$
As a result, postulating the equation of state during inflation is exactly equivalent to postulating the first Hubble-flow function $\epsilon_1(N)$. Therefore, the complete hierarchy of Hubble-flow functions is exactly obtained by taking the successive logarithmic derivatives of Eq. (6.365). For instance,

$$ \epsilon_2 = \frac{1}{W} \frac{dW}{dN} = \frac{\alpha}{C + N_{\text{end}} - N}, \qquad \epsilon_3 = \frac{1}{(dW/dN)} \frac{d^2W}{dN^2} - \frac{1}{W} \frac{dW}{dN} = \frac{1}{C + N_{\text{end}} - N}. \qquad (6.366) $$

Comparing these expressions with the ones associated with a homogeneous scalar field, see Eqs. (3.7) and (3.8), one obtains

$$ \left\{ \begin{array}{l} \left( \frac{d\phi}{dN} \right)^2 = 3M_{\text{Pl}}^2 W, \\ \\ \frac{d\ln V}{dN} = -3W + \frac{d\ln(2-W)}{dN}. \end{array} \right. \qquad (6.367) $$

They can be formally integrated into the exact field trajectory and parametric potential

$$ \phi(N) = \pm \sqrt{3} M_{\text{Pl}} \int^{N} W^{1/2}(x) dx + \phi_0, \qquad (6.368) \\ V(N) = M^4 \left[ 1 - \frac{W(N)}{2} \right] \exp \left[ -3 \int^{N} W(x) dx \right], $$

where $\phi_0$ and $M^4$ are two expected integration constants. Because specifying the equation of state is equivalent to postulate $\epsilon_1$, which is also the field velocity in e-folds, there is a hard-coded shift symmetry in the problem and the field values are defined up to a constant. The ambiguity of sign in the trajectory is reminiscent with the problems associated with horizon-flow inflation [659]: one of the two exact solutions would actually climb up the potential and is strongly unstable. The integration constant $M^4$ fixes the energy scale of inflation, which remains obviously unspecified by postulating only the equation of state parameter.

### 6.21.2. Exact field trajectory and potential

The field trajectory and potential of Eq. (6.368) can be exactly integrated for the VFMI equation of state given in Eq. (6.360). Defining

$$ x \equiv \frac{\phi}{M_{\text{Pl}}}, \qquad (6.369) $$

one gets

$$ x = \frac{\sqrt{3\beta}}{1 - \alpha/2} \left[ (C + N_{\text{end}} - N)^{1-\alpha/2} - 1 \right], \qquad (6.370) $$

in which the integration constant $\phi_0$ has been chosen such that the limiting case $\alpha = 2$ takes the simple form

$$ x_{(\alpha=2)} = \sqrt{3\beta} \ln (C + N_{\text{end}} - N). \qquad (6.371) $$

The field value at the end of inflation can be immediately read off for $N = N_{\text{end}}$

$$ x_{\text{end}} = \frac{\sqrt{3\beta}}{1 - \alpha/2} \left[ \left( \frac{3\beta}{2} \right)^{\frac{2-\alpha}{2\alpha}} - 1 \right], \qquad (6.372) $$

where Eq. (6.361) has also been used. Similarly, the parametric potential $V(N)$ reads

$$ V(N) = M^4 \left[ 1 - \frac{\beta}{2(C + N_{\text{end}} - N)^\alpha} \right] \exp \left\{ \frac{3\beta}{1-\alpha} \left[ (C + N_{\text{end}} - N)^{1-\alpha} - 1 \right] \right\}, \qquad (6.373) $$

which, reduces to

$$ V_{(\alpha=1)}(N) = M^4 \left[ 1 - \frac{\beta}{2(C + N_{\text{end}} - N)} \right] (C + N_{\text{end}} - N)^{3\beta} \qquad (6.374) $$

in the limiting case $\alpha = 1$. Inverting the field trajectory in Eq. (6.370) gives

$$ C + N_{\text{end}} - N = \left( 1 + \frac{2 - \alpha}{2\sqrt{3\beta}} x \right)^{\frac{2}{2-\alpha}}, \qquad (6.375) $$

which can be plugged into Eq. (6.373) to obtain the exact field potential for VFMI

$$ V(\phi) = M^4 \left[ 1 - \frac{\beta}{2 \left( 1 + \frac{2-\alpha}{2\sqrt{3\beta}} \frac{\phi}{M_{\text{Pl}}} \right)^{\frac{2(1-\alpha)}{2-\alpha}}} \right] \exp \left\{ \frac{3\beta}{1-\alpha} \left[ \left( 1 + \frac{2-\alpha}{2\sqrt{3\beta}} \frac{\phi}{M_{\text{Pl}}} \right)^{\frac{2(1-\alpha)}{2-\alpha}} - 1 \right] \right\}. \qquad (6.376) $$

According to the values of $\alpha$, the potential smoothly interpolates between various typical inflationary regimes [658]. For $\alpha \le 1$, the potential is unbounded for $x \to \infty$, which is reminiscent with Large Field Inflation (LFI). For $1 < \alpha \le 2$, the potential is of the plateau-type and takes a finite value at large $x$. For $\alpha > 2$ inflation takes place at the top of the potential, around $x = x_{\text{V-max}}$ with

$$ x_{\text{V-max}} = \frac{2\sqrt{3\beta}}{\alpha - 2}, \qquad (\alpha > 2), \qquad (6.377) $$
<figure><img src="image_91.png" /><figcaption>Fig. 88. Mukhanov Inflation (VFMI) for three typical values of $\alpha \le 1$, $1 < \alpha \le 2$ and $\alpha > 2$ and $\beta = 0.8$. Upper panels: the potential $V(\phi)$ and its logarithm as a function of the normalized field values $\phi/M_{\text{Pl}}$. Bottom left panel: first Hubble-flow function $\epsilon_1$, for the same three values of $\alpha$. Bottom right panels: second Hubble-flow function $\epsilon_2$ (solid lines) and third Hubble flow functions $\epsilon_3$ (dotted lines). In all cases inflation proceeds towards small field values. For the cases in which $\alpha > 2$, only the small-field inflationary regime $x < x_{\text{v-max}}$ is plotted, $x > x_{\text{v-max}}$ being a symmetric case.</figcaption></figure>

which is similar to Small Field Inflation (SFI). Let us notice that for $\alpha > 2$, the left hand side of Eq. (6.375) becomes infinite for $x \to x_{\text{v-max}}$ and the maximal number of e-folds in VFMI for $\alpha > 2$ is unbounded. Within our choice of sign for the field trajectory (6.370), inflation always proceeds from large field values towards small field values and stops at $x = x_{\text{end}}$.

The exact Hubble flow functions have been derived in Eqs. (6.365) and (6.366) in terms of the number of e-folds. From Eq. (6.375), they can be expressed in terms of field values and read

$$ \epsilon_1 = \frac{3\beta}{2\left(1 + \frac{2-\alpha}{2\sqrt{3\beta}}x\right)^{\frac{2\alpha}{2-\alpha}}}, \quad \epsilon_2 = \frac{\alpha}{\left(1 + \frac{2-\alpha}{2\sqrt{3\beta}}x\right)^{\frac{2}{2-\alpha}}}, \quad \epsilon_{n \ge 3} = \frac{\epsilon_2}{\alpha}. \qquad (6.378) $$

Together with the potential, they have been represented in Fig. 88 for three typical values of $\alpha$.

The field value $x_*$, or the e-folds number $\Delta N_*$, at which the observable pivot scale crossed the Hubble radius during inflation are obtained by solving the reheating equations Eq. (3.48) and Eq. (3.45), respectively. The observed values of the Hubble-flow functions are immediately given by Eqs. (6.365) and (6.366), i.e.,

$$ \epsilon_{1*} = \frac{3\beta}{2(C + \Delta N_*)^\alpha}, \quad \epsilon_{2*} = \frac{\alpha}{C + \Delta N_*}, \quad \epsilon_{n \ge 3} = \frac{\epsilon_{2*}}{\alpha}. \qquad (6.379) $$

Finally, the integration constant $M^4$ fixing the energy scale of inflation is inferred from the amplitude of the CMB anisotropies. One gets

$$ \left(\frac{M}{M_{\text{Pl}}}\right)^4 = \frac{4320\pi^2\beta}{2(C + \Delta N_*)^\alpha - \beta} \exp\left\{\frac{3\beta}{1-\alpha}[(C + \Delta N_*)^{1-\alpha} - 1]\right\} \frac{Q_{\text{rms-PS}}^2}{T^2}. \qquad (6.380) $$

Let us remark that within any equation of state parametrization of the inflationary background, $M^4$ being an integration constant, its value cannot be a theoretical output of the model. This has to be contrasted with the more usual specification of a field potential in which the value of $M^4$ may very well be predicted, as it is the case in Higgs/Starobinsky inflation (HI), the original Coleman-Weinberg model (CWI) and Dual Inflation (DI). The reheating consistent predictions for VMFI have been represented in Fig. 268 for various values of $\alpha$ and $\beta$.
<figure><img src="image_92.png" /><figcaption>Fig. 89. Top left panel: Fiber Inflation potential for $n = 0$ and $\delta = 10^{-5}$. Top right panel: logarithm of the potentials for the same values of $n$ and $\delta$. Bottom left panel: slow-roll parameter $\epsilon_1$. Bottom right panel: slow-roll parameters $\epsilon_2$ (solid line) and $\epsilon_3$ (dotted line) for the same values of $n$ and $\delta$.</figcaption></figure>

## 6.22. Fiber Inflation (FI)

This model was proposed in Ref. [660] in the context of string theory, where inflation is driven by a closed string modulus that parameterizes the size of the extra dimensions. This imposes that $\phi > 0$, and the potential is given by

$$V(\phi) = M^4 \left[ \left(1 + \frac{2}{3}\delta\right) e^{-\frac{4}{\sqrt{3}}\frac{\phi}{M_{Pl}}} - 4\left(1 + \frac{\delta}{6}\right) e^{-\frac{1}{\sqrt{3}}\frac{\phi}{M_{Pl}}} + \frac{\delta}{1+n}e^{\frac{2(1+n)}{\sqrt{3}}\frac{\phi}{M_{Pl}}} + 3 - \frac{\delta}{1+n} \right], \quad (6.381)$$

where $n$ is a non-negative integer number (when $n = 0$, the model was studied in Ref. [354]), and $\delta$ is a parametrically small positive number that is related to the string coupling $q$ via $\delta \propto q^{4(1+n/3)}$.

The potential is displayed in Fig. 89. It vanishes at the origin $\phi = 0$, where the derivative of the potential vanishes too, and it is a monotonic increasing function of the field. Indeed, the equation $V'(\phi) = 0$ reduces to

$$(1 + \frac{\delta}{6}) e^{\frac{\sqrt{3}}{M_{Pl}}} + \frac{\delta}{2} e^{\frac{2(3+n)}{\sqrt{3}}\frac{\phi}{M_{Pl}}} = 1 + \frac{2}{3}\delta, \quad (6.382)$$

which can be satisfied only if $\phi = 0$ (since the left-hand side of Eq. (6.382) is a manifestly increasing function of $\phi$). When $\delta = 0$, the potential asymptotes a constant, so it has a plateau shape. When $\delta > 0$, this plateau is broken at some field value (which can be estimating by equating the constant term with the exponential growing term in the potential function) above which the potential grows exponentially.

Defining

$$x \equiv \frac{\phi}{M_{Pl}}, \quad (6.383)$$

the first Hubble flow function in the slow-roll approximation reads

$$\epsilon_1 = \frac{16}{6} \left[ \frac{-1 - \frac{2}{3}\delta + \left(1 + \frac{\delta}{6}\right) e^{\sqrt{3}x} + \frac{\delta}{2} e^{\frac{6+2n}{\sqrt{3}}x}}{1 + \frac{2}{3}\delta - 4 \left(1 + \frac{\delta}{6}\right) e^{\sqrt{3}x} + \frac{\delta}{n+1} e^{\frac{6+2n}{\sqrt{3}}x} + \left(3 - \frac{\delta}{n+1}\right) e^{\frac{4}{\sqrt{3}}x}} \right]^2, \quad (6.384)$$
while the second is given by

$$ \epsilon_2 = \frac{4}{9(1+n)} e^{-\frac{8}{\sqrt{3}}x} \left[ \begin{aligned} & 3(1+n)(6+\delta)(3+2\delta)e^{\sqrt{3}x} - 2\delta(3+2\delta)(3+n)^2 e^{\frac{2(3+n)}{\sqrt{3}}x} \\ & + \delta(6+\delta)(3+2n)^2 e^{\frac{9+2n}{\sqrt{3}}x} + 8(3+2\delta)(\delta-3-3n)e^{\frac{4}{\sqrt{3}}x} - (6+\delta)(\delta-3-3n)e^{\frac{7}{\sqrt{3}}x} \\ & + 6(n+1)\delta(\delta-3-3n)e^{\frac{2(5+n)}{\sqrt{3}}x} \end{aligned} \right] \times \left[ 3 + \left(1+\frac{2\delta}{3}\right) e^{-\frac{4}{\sqrt{3}}x} - \frac{2}{3}(6+\delta)e^{-\frac{1}{\sqrt{3}}x} - \frac{\delta}{1+n} + \frac{\delta}{1+n} e^{\frac{2(1+n)}{\sqrt{3}}x} \right]^{-2} \quad (6.385) $$

and, finally, the third one reads

$$ \epsilon_3 = \left[ 2(1+n)e^{-\frac{8}{\sqrt{3}}x} \left\{ 6 \left( e^{\sqrt{3}x} - 1 \right) + \delta \left[ -4 + e^{\sqrt{3}x} + 3e^{\frac{2(3+n)}{\sqrt{3}}x} \right] \right\} \right. \\ \times \left( 8 \left\{ 6 \left( e^{\sqrt{3}x} - 1 \right) + \delta \left[ -4 + e^{\sqrt{3}x} + 3e^{\frac{2(3+n)}{\sqrt{3}}x} \right] \right\}^3 - \frac{6}{1+n} \left\{ 6 \left( e^{\sqrt{3}x} - 1 \right) \right. \right. \\ \left. + \delta \left[ -4 + e^{\sqrt{3}x} + 3e^{\frac{2(3+n)}{\sqrt{3}}x} \right] \right\} \left\{ 3(1+n) \left( -1 - 3e^{\frac{4}{\sqrt{3}}x} + 4e^{\sqrt{3}x} \right) \right. \\ \left. + \delta \left[ 3e^{\frac{4}{\sqrt{3}}x} - 3e^{\frac{2(3+n)}{\sqrt{3}}x} - 2(1+n) + 2(1+n)e^{\sqrt{3}x} \right] \right\} \left\{ 6 \left( -4 + e^{\sqrt{3}x} \right) \right. \\ \left. + \delta \left[ -16 + e^{\sqrt{3}x} - 6(1+n)e^{\frac{2(3+n)}{\sqrt{3}}x} \right] \right\} + 9e^{\frac{8}{\sqrt{3}}x} \left[ 3 + \left(1+\frac{2\delta}{3}\right) e^{-\frac{4}{\sqrt{3}}x} \right. \\ \left. - \frac{2}{3}(6+\delta)e^{-\frac{1}{\sqrt{3}}x} - \frac{\delta}{1+n} + \frac{\delta}{1+n} e^{\frac{2(1+n)}{\sqrt{3}}x} \right] \left( 2 \left\{ 6 \left( -16 + e^{\sqrt{3}x} \right) + \delta \left[ -64 + e^{\sqrt{3}x} \right. \right. \right. \\ \left. \left. + 12(1+n)^2 e^{\frac{2(3+n)}{\sqrt{3}}x} \right] \right\} \left( 81 \left[ 3 + \left(1+\frac{2\delta}{3}\right) e^{-\frac{4}{\sqrt{3}}x} - \frac{2}{3}(6+\delta)e^{-\frac{1}{\sqrt{3}}x} - \frac{\delta}{1+n} \right. \right. \\ \left. \left. + \frac{\delta}{1+n} e^{\frac{2(1+n)}{\sqrt{3}}x} \right] \right) \left[ - \left( 1 + \frac{2\delta}{3} \right) e^{-\frac{4}{\sqrt{3}}y} + \left( 1 + \frac{\delta}{6} \right) e^{-\frac{1}{\sqrt{3}}y} + \frac{\delta}{1+n} e^{\frac{2(1+n)}{\sqrt{3}}y} + 3 \right. \right. \\ \left. \left. - \frac{\delta}{1+n} \left[ - \left( 1 + \frac{2\delta}{3} \right) e^{-\frac{4}{\sqrt{3}}y} + \left( 1 + \frac{\delta}{6} \right) e^{-\frac{1}{\sqrt{3}}y} + \frac{\delta}{2} e^{\frac{2(1+n)}{\sqrt{3}}y} \right] \right]^{-1} \right) \quad (6.386) $$

They are displayed in the lower panels of Fig. 89. One can see that the first slow-roll parameter diverges at $x = 0$, decreases until a field value that we denote $x_{\epsilon_2=0}$ and that, in practice, needs to be computed numerically, and then increases again to reach the asymptotic value

$$ \epsilon_{1,\infty} = \frac{2}{3} (1+n)^2. \quad (6.387) $$

This indicates that inflation stops by violation of the slow-roll conditions, at a field value $x_{\text{end}}$ that needs to be determined numerically by solving the equation $\epsilon_1 = 1$.

The slow-roll trajectory

$$ N_{\text{end}} - N = \frac{\sqrt{3}}{4} \int_{x_{\text{end}}}^{x} \left[ \left(1 + \frac{2\delta}{3}\right) e^{-\frac{4}{\sqrt{3}}y} - 4\left(1 + \frac{\delta}{6}\right) e^{-\frac{1}{\sqrt{3}}y} + \frac{\delta}{1+n} e^{\frac{2(1+n)}{\sqrt{3}}y} + 3 \right. \\ \left. - \frac{\delta}{1+n} \left[ - \left(1 + \frac{2\delta}{3}\right) e^{-\frac{4}{\sqrt{3}}y} + \left(1 + \frac{\delta}{6}\right) e^{-\frac{1}{\sqrt{3}}y} + \frac{\delta}{2} e^{\frac{2(1+n)}{\sqrt{3}}y} \right] \right]^{-1} dy, \quad (6.388) $$

also needs to be integrated numerically.

Let us notice that, when $n \ge 1$, $\epsilon_{1,\infty} > 1$ so there is a finite range of field value where the first slow-roll parameter is below unity. As a consequence, only a finite number of $e$-folds can be realized in such cases, between the two field values where $\epsilon_1 = 1$. This number is displayed in Fig. 90, where one can check that, for a sufficiently long inflationary phase to take place, $\delta$ needs to be small enough, and the upper bound on $\delta$ is smaller for larger $n$. These considerations are based on a slow-roll analysis and one may wonder whether, in the large-field region, inflation can take place outside the slow-roll regime. This is not the case since in the large-field limit, the potential function becomes approximately exponential, and is of the same form as in Power Law Inflation (PLI, see Section 5.8). In this model, the dynamics can be solved exactly, i.e. without resorting to the slow-roll approximation, and in Section 5.8 it is shown that inflation requires the coefficient in the exponential to be smaller than $\sqrt{2}$ (when the field is expressed in reduced Planck mass units). This is not the case for $n \ge 1$ in the present model, which confirms the validity of the above discussion beyond the slow-roll approximation.
<figure><img src="image_93.png" /><figcaption>Fig. 90. Maximum number of inflationary e-folds realized in Fiber Inflation, as a function of δ and for a few values of n. When n ≥ 1, the first Hubble-flow parameter is smaller than one across a finite field range, and ΔNmax corresponds to the number of e-folds, Eq. (6.388), realized between the two boundaries of that interval, the location of which are obtained by numerically solving the equation e1 = 1.</figcaption></figure>

Finally, the normalization of the potential M⁴ can be obtained from the amplitude of the CMB anisotropies once one has determined the field value x* at which the pivot mode crossed the Hubble radius. It reads

$$ \left(\frac{M}{M_{\text{Pl}}}\right)^4 = 3840\pi^2 \frac{e^{\frac{4}{\sqrt{3}}x_*} \left[-1 - \frac{2}{3}\delta + \left(1 + \frac{\delta}{6}\right)e^{\sqrt{3}x} + \frac{\delta}{2}e^{\frac{6+2n}{\sqrt{3}}x}\right]^2}{\left[1 + \frac{2}{3}\delta - 4\left(1 + \frac{\delta}{6}\right)e^{\sqrt{3}x} + \frac{\delta}{n+1}e^{\frac{6+2n}{\sqrt{3}}x} + \left(3 - \frac{\delta}{n+1}\right)e^{\frac{4}{\sqrt{3}}x}\right]^3} \frac{Q_{\text{rms-PS}}^2}{T^2}. \quad (6.389) $$

The reheating consistent predictions for FI have been represented in Figs. 274 and 275 for n = 0 and n = 1 respectively. One can see that, in the small-δ limit, the model is in good agreement with the data since the potential has a plateau shape. One should also note that the potential (hence the predictions of the model) is independent of n in that limit.

## 6.23. Hyperbolic Inflation (HBI)

### 6.23.1. Theoretical justifications

In Ref. [661], the cosmological evolution driven by a scalar field φ in the presence of a perfect fluid (denoted “f” hereafter) is discussed, in the case where both the scalar field and the perfect fluid follow scaling solutions, i.e. ρφ = Cφa⁻ⁿₚ and ρf = Cf a⁻ⁿₚ. In these expressions, Cφ, Cf, nₚ and nₚ are constant, and nₚ > nₚ for the scalar field to dominate at early time.<fnref n="10" /> In that case, the inflaton field φ still follows the Klein-Gordon equation (3.2), but the Friedmann-Lemaître equation (3.1) becomes

$$ H^2 = \frac{\rho_{\phi} + \rho_{f}}{3M_{\text{Pl}}^2}, \quad (6.390) $$

where ρφ = V(φ) + φ²/2. The scalar-field potential V(φ) leading to such a solution is derived in Ref. [661], and the inflationary model associated to that potential is compared with the Planck 2015 data in Ref. [662].

Let us see how the potential function V(φ) can be obtained. By making use of the Klein-Gordon equation (3.2), the time derivative of the energy density associated to the inflaton field reads ˙ρφ = −3Hφ². Since ρφ = Cφa⁻ⁿₚ, one also has ˙ρφ = −HnₚCφa⁻ⁿₚ, hence the above implies that φ² = nₚCφa⁻ⁿₚ/3, so the kinetic energy scales in the same way as the total energy density associated to φ. This is simply because, for the inflaton to follow a scaling solution, its equation-of-state parameter must be constant, hence the ratio between its potential energy and its kinetic energy is constant too, and both the potential and the kinetic energy follow the same scaling solution.

Since dφ/da = φ/(aH), using the modified Friedmann-Lemaître equation (6.390), one has

$$ \frac{d\phi}{da} = \pm \frac{M_{\text{Pl}}}{a} \sqrt{\frac{n_{\phi}}{1 + \frac{C_f}{C_{\phi}} a^{n_{\phi} - n_f}}}. \quad (6.391) $$

In the absence of perfect fluid, i.e. if Cf = 0, this can be readily integrated as φ/MPl = ±√nφ(N - Nend) + φend, which is the dynamics of Power Law Inflation (PLI), see Eq. (5.112) in Section 5.8. Indeed, it is shown that PLI (where no perfect fluid is considered) yields scaling solutions. Even if Cf ≠ 0, Eq. (6.391) can still be integrated, and one obtains

$$ \phi(a) = \phi_{\text{end}} \pm \frac{2\sqrt{n_{\phi}}}{n_{\phi} - n_{f}} M_{\text{Pl}} \left[ \arcsinh\left(\sqrt{\frac{C_{\phi}}{C_{f} a^{n_{\phi} - n_{f}}}}\right) - \arcsinh\left(\sqrt{\frac{C_{\phi}}{C_{f} a^{n_{\phi} - n_{f}}}}\right) \right]. \quad (6.392) $$

<footnote n="10">¹⁰ This is the choice made in Ref. [662], but the opposite assumption can also be made, if the scalar field accounts for late-time acceleration rather than inflation, see Ref. [661].</footnote>
<figure><img src="image_94.png" /><figcaption>Fig. 91. Hyperbolic Inflation potential (HBI) for $n = 1.1$. Top left panel: the potential as a function of $\phi/\mu$. Top right panel: logarithm of the potential. Bottom left panel: rescaled slow-roll parameter $\epsilon_1\mu^2/M_{\text{pl}}^2$. Bottom right panel: rescaled slow-roll parameters $\epsilon_2\mu^2/M_{\text{pl}}^2$ (solid line) and $\epsilon_3\mu^2/M_{\text{pl}}^2$ (dotted line). HBI can only inflate for Planckian and super-Planckian values of $\mu$.</figcaption></figure>

This can be inverted to yield the function $a(\phi)$. Since $V = \rho_{\phi} - \dot{\phi}^2/2$, one has $V(a) = (1 - n/6)C_{\phi}a^{-n\phi}$ and an explicit form of the potential can be obtained as

$$V(\phi) = \left(1 - \frac{n\phi}{6}\right) C_{\phi} \left(\frac{C_f}{C_{\phi}}\right)^{\frac{n\phi}{n\phi - n_f}} \sinh^{\frac{2n\phi}{n\phi - n_f}} \left[ \frac{n\phi - n_f}{2\sqrt{n\phi}} \frac{\phi - \phi_{\text{end}}}{M_{\text{pl}}} + \arcsinh\left(\sqrt{\frac{C_{\phi}}{C_f a^{n\phi - n_f}}}\right) \right]. \quad (6.393)$$

For $n_{\phi} > n_{\text{f}}$, we have chosen a growing potential with respect to $\phi$, implying a rolling down trajectory, and thus a negative sign in Eq. (6.391). One can check that when $C_{\text{f}}$ goes to zero (i.e. in the absence of perfect fluid), the argument of the $\arcsinh()$ function in Eq. (6.392) becomes very large, hence the scale factor $a(\phi)$ approaches an exponential function, and one recovers the potential of Power Law Inflation.

Using the shift symmetry of the problem, one can absorb the constant term in the argument of the hyperbolic sine function in Eq. (6.393) and rewrite the potential as $V \propto \sinh^n(\phi/\mu)$ where

$$n = \frac{2n_{\phi}}{n_{\phi} - n_{\text{f}}}, \quad \mu = 2M_{\text{pl}}\frac{\sqrt{n_{\phi}}}{n_{\phi} - n_{\text{f}}}. \qquad (6.394)$$

Since $n_{\phi}$ and $n_{\text{f}}$ are of order one, with $n_{\phi} > n_{\text{f}}$, the consistency of the model demands $n > 2$. This also implies that the typical vev $\mu$ of the field is of the order of the Planck mass. In Ref. [661], this potential is studied in the context of a single scalar field theory, i.e. by neglecting completely the presence of the perfect fluid and making use of the Friedmann-Lemaître equation (3.1) rather than Eq. (6.390). At early times, with $n_{\phi} > n_{\text{f}}$, the contribution of the perfect fluid may indeed be negligible with respect to the one of the field. However, as inflation proceeds, the fluid energy density becomes more and more important and such an assumption will eventually break down. As such, without an explicit description of the fluid, one cannot guarantee that the usage of the non-modified FL equations is justified till the end of inflation. Let us also notice that the problem could be alleviated by assuming $C_{\text{f}} \to 0$, but, in that case, the model predictions would be undistinguishable from PLI.

In the next section, these issues will be ignored and we will take a phenomenological approach to explore the observable predictions of the HBI potential.

### 6.23.2. Slow-roll analysis

From the previous discussion, we take the potential of Hyperbolic Inflation as

$$V = M^4 \sinh^n \left( \frac{\phi}{\mu} \right), \qquad (6.395)$$

where $M$ is a mass scale. Because $n$ is not necessarily an even integer, we will consider inflation to occur only in the branch $\phi > 0$. The potential is a monotonic increasing function of the field values, so inflation proceeds at decreasing field value. It is represented in Fig. 91.

Defining

$$x \equiv \frac{\phi}{\mu}, \qquad (6.396)$$
the Hubble-flow functions in the slow-roll approximation are given by

$$ \epsilon_1 = \frac{n^2 M_{\text{Pl}}^2}{2\mu^2 \tanh^2(x)}, \quad \epsilon_2 = \frac{2n M_{\text{Pl}}^2}{\mu^2 \sinh^2(x)}, \quad \epsilon_3 = \frac{2n M_{\text{Pl}}^2}{\mu^2 \tanh^2(x)}, \qquad (6.397) $$

and are displayed in the lower panels of Fig. 91. They all decrease with the field value $x$, hence they increase as inflation proceeds, and diverge in the limit $x \to 0$. In the opposite limit, when $x \to +\infty$, $\epsilon_1$ approaches a constant value

$$ \epsilon_1^{\min} = \frac{n^2 M_{\text{Pl}}^2}{2\mu^2}. \qquad (6.398) $$

In order for slow-roll inflation to occur, one must have $\epsilon_1^{\min} < 1$ and this gives a lower bound on $\mu$ given by<fnref n="11" />

$$ \mu_{\min} = \frac{n}{\sqrt{2}} M_{\text{Pl}}. \qquad (6.399) $$

For $\mu > \mu_{\min}$, Hyperbolic Inflation gracefully ends when $\epsilon_1 = 1$, at a field value given by

$$ x_{\text{end}} = \arctanh \left( \frac{n M_{\text{Pl}}}{\sqrt{2} \mu} \right). \qquad (6.400) $$

The slow-roll trajectory can be integrated, and one obtains

$$ N_{\text{end}} - N = \frac{\mu^2}{n M_{\text{Pl}}^2} \ln \left[ \frac{\cosh(x)}{\cosh(x_{\text{end}})} \right]. \qquad (6.401) $$

This trajectory can be explicitly inverted to get the field values $x$ as a function of the number of e-folds as

$$ x = \operatorname{arccosh} \left[ \cosh(x_{\text{end}}) e^{-n \frac{M_{\text{Pl}}^2}{\mu^2} (N_{\text{end}} - N)} \right] = \operatorname{arccosh} \left( \frac{e^{-n \frac{M_{\text{Pl}}^2}{\mu^2} \Delta N}}{\sqrt{1 - \frac{n^2 M_{\text{Pl}}^2}{2 \mu^2}}} \right), \qquad (6.402) $$

with $\Delta N = N_{\text{end}} - N$ and where Eq. (6.400) has been used to express $x_{\text{end}}$. The trajectory (6.401), combined with the reheating equation (3.48), allows us to determine $x_*$, the field value at which the pivot mode crossed the Hubble radius during inflation. In turn, this determines the mass scale $M$ of the potential from the CMB normalization and one finds

$$ \left(\frac{M}{M_{\text{Pl}}}\right)^4 = 720\pi^2 \frac{n^2 M_{\text{Pl}}^2}{\mu^2 \tanh^2(x_*) \sinh^n(x_*)} \frac{Q_{\text{rms-PS}}^2}{T^2}. \qquad (6.403) $$

The reheating-consistent slow-roll predictions of HBI are displayed in Figs. 276 to 278, for $n = 0.5$, $n = 1$ and $n = 1.5$, and various values of $\mu$. As these plots suggest, HBI produces a significant amount of tensor modes putting it under pressure.

The model predictions can be analytically understood by plugging Eq. (6.402) into the Hubble flow functions in Eqs. (6.397). One obtains

$$ \epsilon_{1*} = \frac{\frac{n^2 M_{\text{Pl}}^2}{2 \mu^2}}{1 - \left(1 - \frac{n^2 M_{\text{Pl}}^2}{2 \mu^2}\right) e^{-2n \frac{M_{\text{Pl}}^2}{\mu^2} \Delta N_*}}, \quad \epsilon_{2*} = 2n \frac{M_{\text{Pl}}^2}{\mu^2} \frac{1 - \frac{n^2 M_{\text{Pl}}^2}{2 \mu^2}}{e^{-2n \frac{M_{\text{Pl}}^2}{\mu^2} \Delta N_*} - 1 + \frac{n^2 M_{\text{Pl}}^2}{2 \mu^2}}, \qquad (6.404) $$

and $\epsilon_{3*} = (4/n)\epsilon_{1*}$. As stressed above, for slow-roll inflation to take place, $\mu$ must be super-Planckian, which is why it is interesting to evaluate these formulas in the limit $\mu/M_{\text{Pl}} \gg \sqrt{\Delta N_*}$, where one has

$$ \epsilon_{1*} \approx \frac{n}{4 \left( \Delta N_* + \frac{n}{4} \right)}, \quad \epsilon_{2*} \approx \epsilon_{3*} \approx \frac{1}{\Delta N_* + \frac{n}{4}}. \qquad (6.405) $$

One notices that these expressions are the same as the ones obtained for Large Field Inflation (LFI), see Eq. (5.42), where the parameter denoted $\rho$ in LFI is identified with $n$ in HBI. This is because, in the regime $\mu \gg M_{\text{Pl}}$, the last e-folds of inflation proceed at $\phi \ll \mu$, where the potential function (6.395) can be Taylor expanded, yielding $V \simeq M^4(\phi/\mu)^\rho$. Note that the condition $n > 2$ therefore implies that the original fluid model is disfavored.

## 6.24. Smeared Higgs Inflation (SHI)

### 6.24.1. Theoretical justifications

In Ref. [663], an extension of the Coleman–Weinberg model is considered, see Section 5.11, in the context of SU(5) GUT. The two fields in presence are a Higgs gauge singlet $\phi$ and a field $\chi$ that breaks the SU(5) group. After taking radiative corrections into account, the potential is given by

$$ V = -\frac{m^2}{2}\phi^2 + \frac{\lambda}{4}\phi^4 - \frac{\beta^2}{2}\phi^2\chi^2 + \frac{a}{4}\chi^4 + A\phi^4 \left[ \ln\left(\frac{\phi}{\phi_0}\right) + C \right] + V_0. \qquad (6.406) $$

The situation where $m^2 = 0$ was considered in Ref. [408] and, as we will see below, it leads to Coleman–Weinberg inflation (CWI). In that case, the fields $\phi$ and $\chi$ have vanishing masses and quartic dimensionless self-coupling $\lambda$ and $a$ respectively. The parameter $\beta$ is also dimensionless

<footnote n="11">¹¹ From Eq. (6.394), the bound $\mu > \mu_{\min}$ implies that $n_\phi < 2$. Since $n_f < n_\phi$, this also implies that $n_f < 2$ and, as noted in Ref. [661], this excludes the possibility that the perfect fluid is pressureless matter ($n_f = 3$) or radiation ($n_f = 4$).</footnote>
<figure><img src="image_95.png" /><figcaption>Fig. 92. Smeared Higgs Inflation potential (SHI) for $\alpha = 1$. Top left panel: the potential as a function of $\phi/\phi_0$. Top right panel: logarithm of the potential. Bottom left panel: rescaled slow-roll parameter $\epsilon_1\phi_0^2/M_n^2$. Bottom right panel: rescaled slow-roll parameters $\epsilon_2\phi_0^2/M_n^2$ (solid line) and $\epsilon_3\phi_0^2/M_n^2$ (dotted line).</figcaption></figure>

and couples the two fields, while $A \sim \beta^4/(16\pi^2)$, $C$ and $\phi_0$ are renormalization parameters. The potential energy at vanishing field configurations is denoted $V_0$. The additional contribution considered in Ref. [663] is a negative squared mass for $\phi$, and we now explain how it modifies the Coleman–Weinberg potential.

The first step is to set the field $\chi$ at the minimum of its effective potential, so $\chi = \beta\phi/\sqrt{a}$. Then, the parameter $C$ can be set such that the resulting effective potential for $\phi$ is minimum at $\phi = \phi_0$. This leads to $4AC = m^2/\phi_0^2 + \beta^4/a - \lambda - A$. Finally, $V_0$ is set such that the potential vanishes at this minimum, which gives rise to $4V_0 = A\phi_0^4 + m^2\phi_0^2$. Only three parameters remain, namely $m$, $\phi_0$ and $A$, in terms of which the potential reads

$$V = \frac{m^2 \phi_0^2}{4} \left[ 1 - \left( \frac{\phi}{\phi_0} \right)^2 \right]^2 + A \phi^4 \left[ \ln \left( \frac{\phi}{\phi_0} \right) - \frac{1}{4} \right] + \frac{A \phi_0^4}{4}. \qquad (6.407)$$

As announced above, when $m^2 = 0$, one recovers the Coleman–Weinberg potential, see Eq. (5.165). In the opposite limit where $m^2$ is very large, one obtains the Higgs tree-level potential, see Eq. (4.78), which corresponds to the double-well inflation model (DWI) studied in Section 5.14. One therefore expects SHI to interpolate between these two limits, CWI and DWI. It can either be viewed as a generalization of CWI, or as a radiatively-corrected version of DWI.

### 6.24.2. Slow-roll analysis

The potential (6.407) can be more conveniently rewritten as

$$V = M^4 \left\{ \left[ 1 - \left( \frac{\phi}{\phi_0} \right)^2 \right]^2 + \alpha \left( \frac{\phi}{\phi_0} \right)^4 \left[ \ln \left( \frac{\phi}{\phi_0} \right) - \frac{1}{4} \right] + \frac{\alpha}{4} \right\}, \qquad (6.408)$$

where we have introduced $M \equiv m^2\phi_0^2/4$ and $\alpha \equiv A\phi_0^4/M^4$. It is a two-parameter potential, defined for $\phi > 0$, where the DWI-limit now corresponds to $\alpha \to 0$ and the CWI-limit to $\alpha \to \infty$. It is represented in Fig. 92. Starting from $\phi = 0$ where $V' = 0$, it decreases until $\phi = \phi_0$ where it vanishes, and then it increases with $\phi$ at $\phi > \phi_0$. As a consequence, there are a priori two relevant regimes for inflation: a hilltop regime at $\phi < \phi_0$, and a large-field regime at $\phi > \phi_0$. However, since the model was introduced in Ref. [663] as a hilltop model, we will focus on the first regime. Moreover, in the large-field regime, the potential is approximately quartic, which is strongly disfavored by CMB observations (see the discussion regarding LFI$_4$ in Section 5.2 and Fig. 127).

Defining

$$x \equiv \frac{\phi}{\phi_0}, \qquad (6.409)$$
the Hubble-flow functions in the slow-roll approximation are given by

$$ \epsilon_1 = \frac{8 M_{\text{Pl}}^2}{\phi_0^2} x^2 \frac{[-1 + x^2 + \alpha x^2 \ln(x)]^2}{\left( (1 - x^2)^2 + \alpha x^4 \left[ \ln(x) - \frac{1}{4} \right] + \frac{\alpha}{4} \right)^2}, \qquad (6.410) $$

$$ \epsilon_2 = \frac{32 M_{\text{Pl}}^2}{\phi_0^2} \left\{ (1 - x^2) \left[ 4 + \alpha - \alpha(6 + \alpha)x^2 - (4 - \alpha + \alpha^2)x^4 \right] \right. \\ \left. - \alpha x^2 \left[ (\alpha - 8)x^4 + 4x^2 + 3(4 + \alpha)\ln(x) + 4\alpha^2 x^6 \ln^2(x) \right] \right\} \\ \times \left[ 4 + \alpha - (\alpha - 4)x^4 - 8x^2 + 4\alpha x^4 \ln(x) \right]^{-2}, \qquad (6.411) $$

$$ \epsilon_3 = 16 \frac{M_{\text{Pl}}^2}{\phi_0^2} x^2 \left[ -1 + x^2 + \alpha x^2 \ln(x) \right] \left\{ \begin{aligned} & (4 + \alpha)(8 - 6\alpha + 5\alpha^2)x^8 + 32\alpha(\alpha + 5)x^6 \\ & - 2(96 + 96\alpha + 30\alpha^2 + 5\alpha^3)x^4 - 32(\alpha^2 + \alpha - 8)x^2 + (4 + \alpha)(6 + \alpha)(5\alpha - 4) \\ & + 2\alpha \ln(x) \left[ (48 - 16\alpha + 7\alpha^2)x^8 + 80\alpha x^6 - 2(144 + 68\alpha + 5\alpha^2)x^4 + 48(4 + \alpha)x^2 + 3(4 + \alpha)^2 \right] \\ & + 16\alpha^2 x^4 \ln^2(x) \left[ -6(\alpha + 4) + (6 - \alpha)x^4 + 2\alpha x^4 \ln(x) \right] \end{aligned} \right\} \left\{ \begin{aligned} & \left[ (\alpha - 4)x^4 + 8x^2 \right] \\ & - (\alpha + 4) - 4\alpha x^4 \ln(x) \right] \left[ (4 - \alpha + \alpha^2)x^6 + (7\alpha - 4)x^4 - (4 + 7\alpha + \alpha^2)x^2 \right. \\ & \left. + 4 + \alpha + (8 - \alpha)\alpha x^6 \ln(x) - 4\alpha x^4 \ln(x) - 3\alpha(4 + \alpha)x^2 \ln(x) + 4\alpha^2 x^6 \ln^2(x) \right] \end{aligned} \right\}^{-1}, \qquad (6.412) $$

and are displayed in the lower panels of Fig. 92. When $\phi < \phi_0$, i.e. $x < 1$, they all increase with the field value $x$, hence they increase as inflation proceeds, and diverge in the limit $x \to 1$. In the opposite limit, when $x \to 0$, $\epsilon_1$ and $\epsilon_3$ vanish, while $\epsilon_2$ approaches a constant value

$$ \epsilon_2^{\min} = \frac{32 M_{\text{Pl}}^2}{(\alpha + 4)\phi_0^2}. \qquad (6.413) $$

As a consequence, slow-roll inflation requires $\phi_0^2/M_{\text{Pl}}^2 \gg 1/(\alpha + 4)$. Inflation ends when $\epsilon_1 = 1$, at a field value $\phi_{\text{end}}$ that needs to be determined numerically. The slow-roll trajectory,

$$ N_{\text{end}} - N = \frac{\phi_0^2}{M_{\text{Pl}}^2} \int_{\phi_{\text{end}}/\phi_0}^{\phi/\phi_0} dx \frac{(1-x^2)^2 + \alpha x^4 (\ln x - 1/4) + \frac{\alpha}{4}}{4x(-1+x^2+\alpha x^2 \ln x)}, \qquad (6.414) $$

also needs to be integrated and inverted numerically. Combined with the reheating equation (3.48), this allows us to determine $x_*$, the field value at which the pivot mode crosses out the Hubble radius during inflation. In turn, this determines the mass scale $M$ of the potential from the CMB normalization and one finds

$$ \left(\frac{M}{M_{\text{Pl}}}\right)^4 = 11520\pi^2 \frac{M_{\text{Pl}}^2}{\phi_0^2} x_*^2 \frac{\left[-1 + x_*^2 + \alpha x_*^2 \ln(x_*)\right]^2}{\left\{ (1 - x_*^2)^2 + \alpha x_*^4 \left[\ln(x_*) - \frac{1}{4}\right] + \frac{\alpha}{4} \right\}^3} \frac{Q_{\text{rms}-\text{PS}}^2}{T^2}. \qquad (6.415) $$

Let us note that inflation necessarily explores the regime where $x$ is of order one, such that one cannot use Taylor expansions in $x$ in order to approximate the slow-roll trajectory. Indeed, if one assumes that $x_{\text{end}} \ll 1$, Eq. (6.410) gives rise to $x_{\text{end}} \simeq (4 + \alpha)\phi_0/(8\sqrt{2}M_{\text{Pl}})$, which is much smaller than one provided $\phi_0^2/M_{\text{Pl}}^2 \ll 1/(4 + \alpha)^2$. This leads to $\epsilon_2^{\min} \gg 1$, hence it discards this possibility.

The reheating-consistent slow-roll predictions of SHI are displayed in Figs. 279 to 281, for $\phi_0/M_{\text{Pl}} = 10, 15, 20$ and 25 respectively, and various values of $\alpha$. One notices that when $\phi_0/M_{\text{Pl}} \gg 1$, the model's predictions approach the ones of LFI$_2$ (see Section 5.2 and Fig. 127). This is because, in that regime, the last $e$-folds of inflation are realized close to the quadratic minimum of the potential at $x = 1$. Indeed, from Eq. (6.410), one can check that inflation ends at $\phi_{\text{end}} = \phi_0 - \sqrt{2}M_{\text{Pl}}$ in this limit (so $x_{\text{end}} \simeq 1 - \sqrt{2}M_{\text{Pl}}/\phi_0$ is close to one), which coincides with Eq. (5.38) with $p = 2$ and when the field value is displayed by $\phi_0$. The slow-roll trajectory (6.414) can then be integrated as

$$ x_* \approx 1 - 2 \frac{M_{\text{Pl}}}{\phi_0} \sqrt{\frac{1}{2} + \Delta N_*}, \qquad (6.416) $$

which is again close to one when $\phi_0 \gg M_{\text{Pl}}$. The slow-roll parameters at Hubble crossing of the pivot scale are given by

$$ \epsilon_{1*} \approx \frac{1}{2(\Delta N_* + 1/2)}, \qquad \epsilon_{2*} \approx \frac{1}{\Delta N_* + 1/2}, \qquad \epsilon_{3*} \approx \epsilon_{2*}, \qquad (6.417) $$

which coincides with Eq. (5.42) when $p = 2$.