sufficiently super-Planckian. Finally, the third slow-roll parameter has the same behavior as the two previous ones, except that it has a negative minimum $\epsilon_3^{\min} \simeq -0.2733 M_{\text{Pl}}^2/\phi_0^2$, located between $x_{\phi_2}^{\min}$ and $x_{V'=0}^{\min}$ where it vanishes.

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

$$N_{\text{end}} - N = \frac{1}{4} \left( \frac{\phi_0}{M_{\text{Pl}}} \right)^2 \left[ x^2 - \frac{1}{e} \text{Ei}(1 + \ln x^2) - x_{\text{end}}^2 + \frac{1}{e} \text{Ei}(1 + \ln x_{\text{end}}^2) \right], \qquad (5.329)$$

where Ei is the exponential integral function [281,282] and $N_{\text{end}}$ is the number of e-folds at the end of inflation. This trajectory can only be inverted numerically to obtain $\phi(N)$.

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

$$\left(\frac{M}{M_{\text{Pl}}}\right)^4 = 2880\pi^2 \left(\frac{M_{\text{Pl}}}{\phi_0}\right)^2 \frac{[1 + \ln(x_*)^2]^2 Q_{\text{rms-PS}}^2}{x_*^4 |\ln(x_*)|^3 T^2}. \qquad (5.330)$$

The reheating consistent slow-roll predictions of the open string tachyonic inflation models are displayed in Fig. 155. It is interesting to notice that, as expected, these models are compatible with the CMB data only for super-Planckian values of $\phi_0$, $\phi_0/M_{\text{Pl}} \gg 1$. In this limit, one has $x_{\text{end}} \simeq \sqrt{2}M_{\text{Pl}}/\phi_0$, the quadratic terms in the slow roll trajectory Eq. (5.329) dominate over the exponential integral ones, such that one has $x_* \simeq 2M_{\text{Pl}}/\phi_0\sqrt{\Delta N_* + 1/2}$. It follows that

$$\epsilon_{1*} \approx \frac{1}{2\Delta N_* + 1}, \qquad \epsilon_{2*} \approx \epsilon_{3*} \approx 2\epsilon_{1*}, \qquad (5.331)$$

hence

$$r \approx \frac{16}{2\Delta N_* + 1}, \qquad 1 - n_s \approx \frac{4}{2\Delta N_* + 1}, \qquad \text{and} \qquad \alpha_s \approx -\frac{8}{(2\Delta N_* + 1)^2}. \qquad (5.332)$$

One can check that indeed, in the $\phi_0/M_{\text{Pl}} \gg 1$ limit, the prediction points lie in the line $\epsilon_2 = 2\epsilon_1$, or equivalently, $1 - n_s = r/4$.

Finally, let us close this section by some additional considerations on the difficulties that tachyonic inflation faces [513]. Using the above equations, it is easy to show that

$$\left(\frac{M}{M_{\text{Pl}}}\right)^4 \approx \frac{2880\pi^2 Q_{\text{rms-PS}}^2}{16\Delta N_* T^2} \frac{\phi_0^2 \left[5 - 2\ln(\phi_0/M_{\text{Pl}})\right]^2}{M_{\text{Pl}}^2 \left[4 - 2\ln(\phi_0/M_{\text{Pl}})\right]^3} \ll 1. \qquad (5.333)$$

Given that $T_3 \simeq M^4$, this implies that $g_s^3 \ll v^2$. On the other hand, we have seen that the model is compatible with the CMB data only if $\phi_0/M_{\text{Pl}} = (g/v)^{1/2} \gg 1$. This last inequality is consistent with $g_s^3 \ll v^2$ only if $v \ll 1$. But $v \ll 1$ is in contradiction with the assumption that $r \gg \ell_s$, which implies that $v \gg 1$. Therefore, it seems that the constraints obtained from the CMB data invalidates the use of an effective four-dimensional approach to describe tachyonic inflation [513]. On the other hand, this can also justify our approach which just considers this scenario as a phenomenological model.

## 5.23. Witten–O’Raifeartaigh Inflation (WRI)

### 5.23.1. Theoretical justifications

This model arises in different contexts and we now briefly review one of its theoretical motivation. The first situation originates from supersymmetric theories aimed at explaining the gauge hierarchy problem (that is to say why the GUT scale differs so much from the weak scale). In the supersymmetric scenario of Ref. [517], three chiral superfields A, X and Y are considered in a superpotential of the O’Raifeartaigh type [518],

$$W = \lambda X(A^2 - m^2) + gY A, \qquad (5.334)$$

where m and g are constant of mass dimension. The corresponding (global) supersymmetric potential can be expressed as

$$V = \lambda^2 |A^2 - m^2|^2 + g^2 |A|^2 + |2\lambda X A + gY|^2. \qquad (5.335)$$

The minimum of this potential is given by $\langle Y \rangle = -2\lambda\langle X \rangle\langle A \rangle/g$ and $\langle A \rangle = 0$ [there is also another minimum at $\langle A \rangle = \sqrt{m^2 - g^2/(2\lambda^2)}$]. Clearly, the potential is minimized regardless of $\langle X \rangle$, that is to say we have a flat direction along X. Along that direction, $V = \lambda^2 m^4$ and supersymmetry is broken since $F_X \equiv \partial W/\partial X \neq 0$. As a consequence, the mass of the real part and imaginary parts of A are split and are given by $4\lambda^2|X|^2 + g^2 \pm 2m^2\lambda^2$. The mass of the fermion field $\psi_A$ is $4\lambda^2|X|^2 + g^2$. The fact that supersymmetry is broken implies that the potential will receive corrections: as is well-known, if supersymmetry is preserved, the corrections originating from bosons and fermions exactly cancel out. Here, this is not the case and the amplitude of the corrections will be determined by the split between the bosonic and fermionic masses that we have just evaluated before. A simple calculation leads to

$$V = \lambda^2 m^4 \left[ 1 + \frac{\lambda^2}{8\pi^2} \ln \left( \frac{|X|^2}{\mu^2} \right) \right], \qquad (5.336)$$

where $\mu$ is the renormalization scale. Therefore, one obtains an increasing function of the field $\text{vev}$ and this implies that X cannot become large because it cannot climb its potential. As a consequence, one cannot generate a large hierarchy in this scenario. In fact, as explained in Ref. [517], this is due to the fact that the one loop correction is positive, as appropriate in a theory with scalars and fermions. This can also be understood from the renormalization group perspective where the appearance of the logarithm in the above expression of $V(X)$ can be viewed as the renormalization of the coupling constant such that $\lambda^2 \to \lambda^2 [1 + \lambda^2/(8\pi^2) \ln(|X|^2/\mu^2)]$. The conclusion of Ref. [517] is that if m is the small scale (the weak scale) and $\langle X \rangle$ the large one (the GUT scale), a large hierarchy cannot be achieved in this approach.
However, it is well-known that asymptotic freedom is possible in non-Abelian gauge theories. This means that the renormalization group equations have to produce negative one loop corrections. In such a situation, the field could run to infinity, in the non-perturbative regime. For this reason, it is interesting to re-consider the previous model in the framework of a non-Abelian gauge group such as in Grand Unified SU(5) theories. Refs. [519,520] consider two matter fields $A_a^b$ and $Z_a^b$ in the adjoint representation of SU(5) and one singlet X in a superpotential given by

$$W = \lambda_1 \mathrm{Tr}(ZA^2) + \lambda_2 X [\mathrm{Tr}(A^2) - m^2], \qquad (5.337)$$

which is the non-Abelian generalization of Eq. (5.334). One can show that supersymmetry is again necessarily broken<fnref n="7" /> and that the potential exhibits a flat direction with the value $V = \frac{\lambda_1^2 \lambda_2^2 m^4}{30(\lambda_2^2 + \lambda_1^2)}$. As it was the case in the first simple example presented above, and since supersymmetry is broken, quantum corrections modify the potential. At the one loop order, one obtains the following expression [519]

$$V(X) = \frac{\lambda_1^2 \lambda_2^2 m^4}{30 \lambda_2^2 + \lambda_1^2} \left( 1 + \frac{\lambda_2^2}{\lambda_2^2 + \lambda_1^2/30} \frac{29 \lambda_1^2 - 50 g^2}{80 \pi^2} \ln |X|^2 \right), \qquad (5.338)$$

where g is the SU(5) gauge coupling constant. If $29\lambda_1^2 < 50g^2$, the correction is negative contrary to the case studied before. Again, this is precisely because we deal with non-Abelian gauge interaction. The field X will grow and can reach a point where the perturbative approach is no longer valid. However, asymptotic freedom tells us that the potential could develop a minimum in this regime in which X could be stabilized, hence the original motivation for this scenario: the scale m can be taken to be relatively small while $\langle X \rangle$ can now be very large thereby addressing the gauge hierarchy problem.

This class of model was considered in Ref. [521] in order to build a new inflationary scenario. The idea is to start from a potential of the form derived above, namely $V(\phi) = M^4(1 + \bar{b}\ln\phi)$ with a negative coefficient $\bar{b}$. Therefore, the field is driven towards a regime where higher corrections must become important. Typically, one expects $\bar{b}$ to acquire a logarithmic dependence in $\phi$ and the potential to develop a minimum at, say $\phi = m_{\mathrm{GUM}}$. Therefore, this leads to $V(\phi) = M^4[1 + b\ln^2(\phi/m_{\mathrm{GUM}})]$ where b is a constant. Moreover, if one requires the potential to vanish at the minimum, we are led to $V(\phi) \propto \ln^2(\phi/m_{\mathrm{GUM}})$ and this is the potential studied in this section. In Ref. [521], it is argued that $m_{\mathrm{GUM}} \simeq M_{\mathrm{Pl}}$ and that, initially, $\phi \simeq \mu \simeq (m_{\mathrm{weak}}m_{\mathrm{GUM}})^{1/2} \simeq 10^{12}\mathrm{GeV}$. We will come back to these conditions in what follows.

Another way to obtain the same potential is based on Ref. [522,523] in which one considers the following action

$$S = - \int d^4 x \sqrt{-g} \left[ \hat{g}_{AB} (z^C, \bar{z}^C) g^{\mu\nu} \partial_\mu z^A \partial_\nu \bar{z}^B - V (z^C, \bar{z}^C) \right]. \qquad (5.339)$$

The $z^A$'s are complex scalar fields and $\hat{g}_{AB}$ is the Kähler metric. The corresponding equations of motion can be expressed as

$$g^{\mu\nu} \nabla_{\mu} \nabla_{\nu} \bar{z}^{\bar{D}} + \Gamma^{\bar{D}}_{AB} g^{\mu\nu} \partial_{\mu} \bar{z}^{A} \partial_{\nu} \bar{z}^{B} - \hat{g}^{CD} \frac{\partial V}{\partial z^{C}} = 0, \qquad (5.340)$$

where $\Gamma^{\bar{D}}_{AB} \equiv \hat{g}^{CD} \partial_{\bar{A}} \hat{g}_{C\bar{B}}$. If we restrict ourselves to cosmological spacetimes, the above equation becomes $\bar{z}^{\bar{D}} + 3H\dot{z}^{\bar{D}} + \Gamma^{\bar{D}}_{AB} \dot{z}^{\bar{A}} \dot{z}^{\bar{B}} + \hat{g}^{CD} \partial V / \partial z^C = 0$, where $H$ is the Hubble parameter. Then, for simplicity, we assume that there is only one field Z and we denote its real part as u and its imaginary part as v. We also assume that the potential is flat in the v-direction and take $V = V(z + \bar{z})$, $\hat{g}_{ZZ} \equiv \hat{g}(Z + \bar{Z})$. It follows that

$$\ddot{u} + 3H\dot{u} + \Gamma(u) (\dot{u}^2 - \dot{v}^2) + \partial_u V / (2\hat{g}) = 0, \qquad (5.341)$$

$$\ddot{v} + 3H\dot{v} + 2\Gamma(u)\dot{u}v = 0, \qquad (5.342)$$

with $\Gamma = \partial_u \hat{g} / (2\hat{g})$. The second differential equation can be integrated and one obtains $\dot{v} = Qa^{-3}/\hat{g}$, where Q is a constant. The next step consists in defining the field $\phi$ by $\dot{\phi} \equiv \sqrt{\hat{g}}u$. As a consequence, the first differential equation can be re-written as $\dot{\phi} + 3H\dot{\phi} + \partial_\phi [V + Q^2/(\hat{g}a^6)] = 0$, that is to say $\phi$ is now canonically normalized and its evolution is controlled by the effective potential $V(\phi) + Q^2/(\hat{g}a^6)$. One can show that the presence of the additional term proportional to $Q^2$ is not crucial [522,523]. Initially, it dominates because a is small but, quickly, since it is proportional to $a^{-6}$, it goes to zero as the universe expands. As a consequence, one is left with $V(\phi)$ only. A specific version of this scenario has been studied in details in Ref. [522]. In that article, it is assumed that $\hat{g} = e^{-2u}/2$ and $V = 0$. This corresponds to the bosonic action of a model which is superconformal invariant [524]. Then, this invariance is softly broken by adding a term $m^2u^2/2$ and, through the redefinition of the field, one can check that this leads to a potential proportional to $m^2(\ln\phi)^2$, that is to say of the type studied in this section. Moreover, one can also verify that, in the regime discussed above where the term $Q^2/(\hat{g}a^6)$ dominates, an exact solution can be found and reads: $a = a_0 t^{1/3}$ and $\phi^2(t) = E^2(\ln t + C)^2 + 4Q^2/(a_0^2 E^2)$, where E and C are two integration constants. As a consequence, when the universe expands, $Q^2/(\hat{g}a^6)$ goes to zero and one is left with the logarithmic potential only.

### 5.23.2. Slow-roll analysis

Based on the previous considerations, we study the WRI potential

$$V(\phi) = M^4 \ln^2 \left( \frac{\phi}{\phi_0} \right), \qquad (5.343)$$

where $\phi_0$ is viewed as a free parameter but we also keep in mind that a natural prior is $\phi_0 = M_{\mathrm{Pl}}$. The potential Eq. (5.343) is displayed in Fig. 41, together with its logarithm (top panels). The arrow indicates that inflation proceeds from the right to the left. Let us now calculate the Hubble

<footnote n="7">⁷ For this purpose, it is convenient to write that $A_d^c = (\phi_A)_a^b (T_b^\mu)_d^c$ and $Z_d^c = (\phi_Z)_a^b (T_b^\mu)_d^c$, where $T_a^b$, $a,b = 1, ..., 5$ is a basis of SU(5) generators. Concretely, one has $(T_b^a)_d^c = \delta_b^c \delta_d^a - \delta_b^a \delta_d^c / 5$. As a consequence, the three F-term can be expressed as $F_X = \lambda_2 [\mathrm{Tr}(\phi_A^2) - m^2]$, $F_Z = \lambda_1 [\phi_A^2 - \mathrm{Tr}(\phi_A^2) \perp / 5]$ and $F_A = \lambda_1 [\phi_Z \phi_A + \phi_A \phi_Z - 2\mathrm{Tr}(\phi_Z \phi_A) \perp / 5] + 2\lambda_2 \phi_X \phi_A$. These expressions are obtained by explicitly writing the superpotential in terms of the components $(\phi_A)_a^a$ and $(\phi_Z)_b^a$ and differentiating W with respect to them. From $F_X = 0$ it follows that $\mathrm{Tr}(\phi_A^2) = m^2$ and, therefore, $F_Z = 0$ implies that $\phi_A^2 = m^2 \perp / 5$. This last relation is compatible with $\mathrm{Tr}(\phi_A^2) = m^2$ but not with $\mathrm{Tr}(\phi_A) = 0$ in five dimensions. The conditions $F_X = 0$ and $F_Z = 0$ are thus incompatible and supersymmetry is spontaneously broken in this model.</footnote>
<figure><img src="image_44.png" /><figcaption>Fig. 41. Witten-O’Raifeartaigh Inflation (WRI) potential as a function of $\phi/\phi_0$. Top right panel: logarithm of the potential. The arrow indicates in which direction inflation proceeds. Bottom left panel: slow-roll parameter $\epsilon_1$, rescaled by the quantity $M_{\text{Pl}}^2/\phi_0^2$, such that the corresponding expression becomes universal, i.e. independent of $\phi_0$. Bottom right panel: slow-roll parameters $\epsilon_2$ (solid line) and $\epsilon_3$ (dotted line), rescaled by $M_{\text{Pl}}^2/\phi_0^2$ for the same reason as mentioned before.</figcaption></figure>

flow parameters. If one defines $x \equiv \phi/\phi_0$, they are given by

$$ \epsilon_1 = 2 \frac{M_{\text{Pl}}^2}{\phi_0^2} \frac{1}{x^2 \ln^2 x}, \qquad (5.344) $$

$$ \epsilon_2 = 4 \frac{M_{\text{Pl}}^2}{\phi_0^2} \frac{1 + \ln x}{x^2 \ln^2 x}, \qquad (5.345) $$

and

$$ \epsilon_3 = 2 \frac{M_{\text{Pl}}^2}{\phi_0^2} \frac{2 + 3 \ln x + 2 \ln^2 x}{x^2 \ln^2 x (1 + \ln x)}. \qquad (5.346) $$

They are displayed in the bottom panels of Fig. 41. One can see that they all vanish when $x \to \infty$, that they increase as inflation proceeds, diverging when $x \to 1$. At this stage, a remark is in order about Ref. [521]. As already mentioned above, a natural prior is $\phi_0 = M_{\text{Pl}}$. This means that if, initially, one has $\phi \simeq \mu$, one is in fact in the decreasing branch of the potential and, as a matter of fact, one cannot have inflation since $\epsilon_1 > 1$ always. Clearly, the only way to have inflation in this branch is to assume that $\phi_0 \gg M_{\text{Pl}}$, a case which appears to be difficult to justify in this context. Here, we do not consider this case. In the increasing branch of the potential, inflation stops by slow-roll violation when $\epsilon_1 = 1$, at a vev $x_{\text{end}}$ given by

$$ x_{\text{end}} = \exp \left[ W_0 \left( \sqrt{2} \frac{M_{\text{Pl}}}{\phi_0} \right) \right], \qquad (5.347) $$

where $W_0$ is the 0-branch of the Lambert function, which must be chosen in order to have $x > 1$.

Let us now turn to the slow-roll trajectory. It can be integrated exactly and this leads to the following expression

$$ N_{\text{end}} - N = \frac{1}{4} \frac{\phi_0^2}{M_{\text{Pl}}^2} \left( x^2 \ln x - \frac{x^2}{2} - x_{\text{end}}^2 \ln x_{\text{end}} + \frac{x_{\text{end}}^2}{2} \right), \qquad (5.348) $$

where $N_{\text{end}}$ is the number of e-folds at the end of inflation. Interestingly enough, this trajectory can be inverted, and one obtains

$$ x = \exp \left\{ \frac{1}{2} W_0 \left[ \frac{8 M_{\text{Pl}}^2}{e \phi_0^2} (N_{\text{end}} - N) + \frac{2}{e} x_{\text{end}}^2 \ln x_{\text{end}} - \frac{x_{\text{end}}^2}{e} \right] + \frac{1}{2} \right\}, \qquad (5.349) $$

where $W_0$ is still the 0-branch of the Lambert function. It is displayed in Fig. 42, together with the region where inflation proceeds.
<figure><img src="image_45.png" /><figcaption>Fig. 42. Lambert functions $W_0(x)$ (dashed line) and $W_{-1}(x)$ (solid line). During Witten–O’Raifeartaigh inflation, inflation proceeds along the “0” branch in the direction specified by the arrow.</figcaption></figure>

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

$$ \left(\frac{M}{M_{\text{Pl}}}\right)^4 = 2880\pi^2 \left(\frac{M_{\text{Pl}}}{\phi_0}\right)^2 \frac{1}{x_*^2 \ln^4 x_*} \frac{Q_{\text{rms-PS}}^2}{T^2}. \qquad (5.350) $$

The reheating consistent slow-roll predictions of the Witten- O’Raifeartaigh inflation models are displayed in Fig. 156. One should remember that in principle, $\phi_0 \simeq M_{\text{Pl}}$, even if a wider range of values for $\phi_0$ is displayed in order to understand how the predictions depend on this parameter. In particular, when $\phi_0 \gg M_{\text{Pl}}$, the predictions lie along the line $\epsilon_2 = 2\epsilon_1$. Indeed, in this limit, Eq. (5.347) shows that $x_{\text{end}} \to 1$ while Eq. (5.349) indicates that $x_* \to 1$. As a consequence, one obtains $\epsilon_{2*} \simeq \epsilon_{1*}$ from Eqs. (5.344) and (5.345).

## 5.24. Dual Inflation (DI)

### 5.24.1. Theoretical justifications

This model finds its roots in the $N=2$ supersymmetry SU(2) Yang-Mills theories à la Seiberg-Witten [525,526]. If $\phi^i$ ($i \in \{1, 2, 3\}$) denote the scalars belonging to the $N=2$ vector multiplets $\mathcal{A}^i$ in the adjoint representation, the classical potential of the theory is given by [525]

$$ V = \frac{1}{g^2} \text{Tr} [\phi, \phi^\dagger]^2. \qquad (5.351) $$

It exhibits a set of inequivalent vacua defined by the vanishing commutator $[\phi, \phi^\dagger] = 0$. Up to a gauge transformation, the minima of Eq. (5.351) can be chosen along $\phi \equiv \frac{1}{2}a\sigma_3$, where $\sigma_3 = \text{diag}(1, -1)$ and $a$ is a complex scalar. A gauge invariant representation of this moduli space can be restored by using the complex variable $u = \frac{1}{2}a^2 = \text{Tr}\phi^2$, such that for $u \neq 0$, SU(2) is spontaneously broken to a residual U(1). Integrating out the multiplets which acquire a mass for $u \neq 0$ gives a low-energy effective theory describing a single $N=2$ vector multiplet $\mathcal{A}$ whose scalar component is the order parameter $a$. As shown in Refs. [525,526], expressed in terms of $N=1$ supersymmetry, $\mathcal{A}$ contains a $N=1$ chiral multiplet $\mathcal{A}$ and a vector multiplet $W_\alpha$ whose dynamics in superspace is described by the Lagrangian

$$ \mathcal{L} = \frac{1}{4\pi} \Im \left[ \int d^4\theta \frac{\partial F(A)}{\partial A} \bar{A} + \int d^2\theta \frac{1}{2} \frac{\partial^2 F(A)}{\partial A^2} W_\alpha W^\alpha \right]. \qquad (5.352) $$

From an exhaustive analysis of the monodromies that must exist within the moduli space of the quantum vacua, Refs. [525,526] were able to give an explicit expression for the prepotential $F(A)$. Remarkably, it includes perturbative and non-perturbative quantum corrections as a function of the only free parameter of the model: the dynamical generated mass scale $\mathcal{A}$ (which naturally appears at one loop). The spectrum of the theory contains various magnetic and electric charges, dyons and monopoles. As shown in Ref. [525], adding a $N=1$ preserving-mass term to the Lagrangian softly breaks $N=2$ to $N=1$ supersymmetry and triggers monopole condensation. Because the strongly coupled regime of the theory is dual to weakly coupled monopoles [527], this condensation has been shown to explicitly describe confinement of the electric charges [525,526].

In this context, and motivated by the actual QCD vacuum structure, Álvarez-Gaumé et al. have shown in Ref. [528] that the only possible way to soft-break $N=2$ supersymmetry directly to $N=0$ while preserving the analyticity properties of the Seiberg-Witten model is to promote the dynamic scale $\mathcal{A}$ to a function of a new $N=2$ vector multiplet $\mathcal{S}$. Once the scalar and auxiliary components of this superfield are frozen, $N=2$ is softly broken to $N=0$ while the non-perturbative scalar potential of the low energy effective action still remains uniquely determined by the analyticity properties of the moduli space of quantum vacua. Denoting $a_0 = s$ and $a_1 = a$ the order parameters of the two multiplets, up to small terms, the low-energy effective potential for the vacuum with $\mathcal{S}$ frozen reads [528]

$$ V(a) = -\frac{2}{b_{11}} \rho^4 - \frac{\det(b)}{b_{11}} f_0^2, \qquad \rho^2 = \sup \left\{ -b_{11} |a|^2 + \frac{|b_{01}| f_0}{\sqrt{2}} , 0 \right\}. \qquad (5.353) $$
In this expression, $f_0 < \Lambda$ stands for the vev of S's auxiliary field once frozen. The $\rho^4$ term is non vanishing only when the monopoles acquire a non-vanishing vev, which then lowers the vacuum energy ($b_{11} > 0$, see below) and shows that confinement is favored [528]. The $b_{ij}$ are the imaginary parts of the coupling matrix $\tau_{ij}$

$$b_{ij} \equiv \frac{1}{4\pi} \mathfrak{F}(\tau_{ij}) = \frac{1}{4\pi} \mathfrak{F}\left(\frac{\partial^2 \mathfrak{F}}{\partial a_i \partial a_j}\right), \qquad (5.354)$$

and everything can be expressed in terms of the complex variable $u$. One gets [525,526,528]

$$a = \frac{4i\Lambda}{\pi k}(E' - K'), \quad \tau_{11} = i \frac{K}{K'}, \quad \tau_{01} = i \frac{2\Lambda}{kK'} \quad \tau_{00} = i \frac{8\Lambda^2}{\pi} \left( \frac{E'}{k^2 K'} - \frac{1}{2} \right), \qquad (5.355)$$

where $E(k)$, $K(k)$, $E'(k)$ and $K'(k)$ are the first and second kind complete elliptic integrals [282]. The elliptic modulus $k$ is related to $u$ by

$$k^2 \equiv \frac{2}{1 + \frac{u}{\Lambda^2}}. \qquad (5.356)$$

The potential of Eq. (5.353) is thus defined over a two-dimensional Kähler manifold, which in terms of $u$, has a non-canonical Kähler metric

$$ds^2 = \mathfrak{F}\left(\frac{\partial^2 \mathfrak{F}}{\partial a^2}\right) da d\bar{a} = 4\pi b_{11} \left|\frac{da}{du}\right|^2 du d\bar{u}. \qquad (5.357)$$

As remarked by Garcia-Bellido in Ref. [529], far from the monopole condensation region, the potential $V(u)$ exhibits a flat valley along the $\mathfrak{R}(u)$-direction and can support an inflationary period. Because inflation ends within the monopole condensation region, reheating can naturally occur by exciting the monopoles which are the confined states associated with the electric charges of theory. In this picture, inflation appears as a consequence of a Yang-Mills phase transition from asymptotic freedom to confinement and the inflaton is the order parameter. As discussed in Ref. [529], the potential (5.353) is not yet completely satisfactory for cosmological purposes as it admits a negative minimum and has to be uplifted. Assuming $f_0 \ll \Lambda$, the minimum occurs at $k^2 \approx 1$ and the uplifting constant to add to the potential is

$$V_0 \approx 1. \qquad (5.358)$$

Under these assumptions, dual inflation has been shown in Refs. [529,530] to generically yield a spectral index in the range $n_s = 0.9 \pm 0.1$ while being compatible with the measured amplitude of the CMB anisotropies. In the next section, we extend these papers to second order in slow roll, without any other approximation, and then calculate how the model predictions are affected by all the possible reheating histories. As detailed in Section 5.24.3, Dual Inflation shares with Starobinsky/Higgs Inflation the remarkable feature to predict the overall amplitude of the CMB anisotropies such that some care should be taken when solving for the reheating equations.

### 5.24.2. Slow-roll analysis

Focussing on the inflationary valley defined to be on the real axis of the complex moduli plane $u$ with $|u| > |\Lambda^2|$, and assuming without loss of generality that $\Lambda = |\Lambda|$ is also real, some simplifications can be made. As explicit in Eq. (5.356), the elliptic parameter $m \equiv k^2$, is bounded to $0 < m < 1$ and all the complete elliptic integrals are real. Plugging Eqs. (5.355) and (5.354) into Eq. (5.353) yields, after some algebra

$$V(m) = \frac{f_0^2 \Lambda^2}{\pi^2} \left\{ 1 + V_0 - 2 \frac{K - E}{mK} - \frac{\pi}{mK K'} v^2(m) \Theta[v(m)] \right\}, \qquad (5.359) \\ v(m) \equiv 1 - \frac{8\sqrt{2}}{\pi^2} \frac{\Lambda}{f_0} \frac{K(E' - K')^2}{m^{1/2}},$$

where $\Theta(x)$ stands for the Heaviside function, $V_0$ is the uplifting constant, and the terms in $v^2(m)$ correspond to the monopole condensation previously discussed. They are non-vanishing for large values of $m > m_{mon}$ where $v(m_{mon}) = 0$. All elliptic integrals are implicitly assumed to be evaluated at the modulus $k = \sqrt{m}$. This potential alone does not encode the inflationary dynamics because $m$ is not the canonical scalar degree of freedom. From the Kähler metric in Eq. (5.357), along the $u$ (and $m$) real direction, the canonical scalar field $\phi$ can be defined as

$$\left(\frac{d\phi}{dm}\right)^2 = \frac{8\Lambda^2}{\pi^2} \frac{KK'}{m^3} \Leftrightarrow \phi(m) = \Lambda \frac{2\sqrt{2}}{\pi} \int_m^1 \frac{\sqrt{K(p^{1/2})K'(p^{1/2})}}{p^{3/2}} dp, \qquad (5.360)$$

in which $\phi(1) = 0$ and $\phi(m) \to \infty$ for $m \to 0$. The above expression cannot be explicitly integrated, neither inverted, such that the potential of dual inflation is only known parametrically as $\phi(m)$ and $V(m)$. Nonetheless, Eqs. (5.359) and (5.360) show that dual inflation depends on two parameters $f_0$ and $\Lambda$. They appear in the definition of $v(m)$ as the ratio $f_0/\Lambda$ such that the shape of the potential actually depends only on $f_0/\Lambda$. The combination $f_0^2 \Lambda^2$ is an overall multiplicative constant rescaling the potential as a whole whereas $\Lambda$ alone rescales the field values. For these reasons, it is more convenient to introduce the equivalent set of parameters, and dimensionless field values, defined by

$$f \equiv \frac{f_0}{\Lambda} < 1, \qquad M^4 \equiv \frac{f^2 \Lambda^4}{\pi^2}, \qquad x \equiv \frac{\phi}{\Lambda}. \qquad (5.361)$$

Because the parameter $M^4$ can be determined by the amplitude of the CMB anisotropies, see Eq. (3.32), dual inflation is a one-parameter model parametrized by $f$. Once $f$ is chosen, the value of $\Lambda$ can be obtained from $M^4$ using the above equation. Let us notice that the uplifting term $V_0$ must be such that the minimum of the potential is exactly vanishing. As such, although $V_0$ is of order unity for small values of $f$, it is a non-trivial function of the parameter $f$ given by

$$V_0(f) = -1 + \left( 2 \frac{K - E}{mK} + \frac{\pi v^2}{mK K'} \right) \Big|_{m=m_{min}} , \qquad (5.362)$$

where $m_{min}$ is the elliptic parameter at which $V(m)$ is minimal. The logarithmic derivative of Eq. (5.359) reads

$$\frac{d \ln V}{d m} = \frac{\pi v \Theta(v) \{-4 m(m-1)\dot{v} K K' - v [K'(E+K) - E'K]\} + 2 K'^2 [E^2 + (m-1)K^2]}{2m(m-1)K K' \{K'[2E+K(mV_0 + m-2)] - \pi v^2 \Theta(v)\}}, \qquad (5.363)$$
<figure><img src="image_46.png" /><figcaption>Fig. 43. Dual Inflation (DI) for $f = 0.5$. Upper panels: the parametric potential $V(m)$ and field values $\phi(m)/\Lambda$ as a function of the elliptic parameter $m \equiv k^2$. Bottom panels: the potential $V(\phi)$ and its logarithm as a function of the canonically normalized field values $\phi/\Lambda$.</figcaption></figure>

and $m_{\min}$ is obtained by (numerically) solving

$$ \pi \nu \{-4m(m-1)\dot{\nu}KK' - \nu[K'(E+K) - E'K]\} + 2K'^2[E^2 + (m-1)K^2] = 0, \quad (5.364) $$

where $\dot{\nu}$ stands for the derivative of $\nu(m)$ with respect to $m$:

$$ \dot{\nu}(m) = \frac{4\sqrt{2}(E' - K')(EE' - K'[E + 2(m-1)K])}{\pi^2 f (m-1)m^{3/2}}. \quad (5.365) $$

The potentials $V(m)$ and $V(\phi)$, as well as the field values $x(m)$ have been represented in Fig. 43 for $f = 0.5$. Inflation proceeds from large field values $x$ to small field values, or, equivalently, from small elliptic parameter values $m$ to values close to unity.

In order to calculate the slow-roll parameters, one can use the chain rule to introduce the parametric trajectory as a proxy. From the definition of the Hubble flow functions in Eqs. (3.3) and (3.8), one has

$$ \epsilon_1 = \frac{1}{2M_{\text{Pl}}^2} \left(\frac{d\phi}{dm}\right)^2 \left(\frac{dm}{dN}\right)^2 = \frac{4\Lambda^2}{\pi^2 M_{\text{Pl}}^2} \frac{KK'}{m^3} \left(\frac{dm}{dN}\right)^2, \\ \epsilon_{n+1} = \frac{d \ln |\epsilon_n|}{dN} = \frac{d \ln |\epsilon_n|}{dm} \frac{dm}{dN}, \quad (5.366) $$

such that the derivation of all $\epsilon_n$ requires only the determination of the parametric trajectory $m(N)$. Assuming slow-roll for the inflaton $\phi$, one has, at leading order, from Eq. (3.10)

$$ \frac{d\phi}{dN} = -M_{\text{Pl}}^2 \frac{d\ln V}{d\phi} \Rightarrow \frac{dm}{dN} = -\frac{M_{\text{Pl}}^2 \pi^2}{8\Lambda^2} \frac{m^3}{KK'} \frac{d\ln V}{dm}, \quad (5.367) $$

such that the parametric slow-roll expressions for $\epsilon_n(m)$ simplify to

$$ \epsilon_1 = \frac{M_{\text{Pl}}^2 \pi^2}{16 \Lambda^2} \frac{m^3}{KK'} \left( \frac{d \ln V}{dm} \right)^2, \\ \epsilon_n = -\frac{M_{\text{Pl}}^2 \pi^2}{8 \Lambda^2} \frac{m^3}{KK'} \left( \frac{d \ln V}{dm} \right) \left( \frac{d \ln |\epsilon_n|}{dm} \right). \quad (5.368) $$

Let us notice that all the slow-roll parameters have the same dependency in the parameter $\Lambda$, namely $\epsilon_n \propto 1/\Lambda^2$. Because $\Lambda$ also enters into the definition of $M^4$, it means that the field value at which inflation ends depends on how the potential normalization matches the amplitude of the CMB anisotropies, which itself depends on the reheating, and the reheating quantities depend on the field value at which inflation ends. The consistent way to simultaneously solve these conditions is discussed in Section 5.24.3. For the time being, plugging Eqs. (5.359) into Eqs. (5.368)
gives an explicit expression for the parametric slow-roll functions. For the first one, one gets

$$ \epsilon_1(m) = \frac{M_{\text{Pl}}^2 \pi^2}{16 \Lambda^2} \frac{m^3}{K K'} \times \left( \frac{\pi \nu \Theta(\nu) \{-4 m(m-1)\ddot{\nu} K K' - \nu [K'(E+K) - E'K]\} + 2 K'^2 [E^2 + (m-1)K^2]}{2m(m-1)K K' \{K'[2E + K(mV_0 + m-2)] - \pi \nu^2 \Theta(\nu)\}} \right)^2, \quad (5.369) $$

in which $\nu(m)$ and $\dot{\nu}(m)$ are given in Eqs. (5.359) and (5.365), respectively. The second Hubble flow function is a bit more involved and reads

$$ \epsilon_2(m) = \frac{M_{\text{Pl}}^2 \pi^2}{8 \Lambda^2} \frac{m^4}{4(m-1)^2 K^4 K'^4 \{K'[2E + K(mV_0 + m-2)] - \pi\Theta(\nu)\nu^2\}^2} \\ \times (4KK' \{K'[2E + K(mV_0 + m-2)] - \pi\Theta(\nu)\nu^2\} \\ \times (\pi\Theta(\nu) \{K'[K(m-1)\{\nu[4\ddot{\nu}(m-1)m^2 - \nu] + 4\dot{\nu}^2(m-1)m^2 + 4\dot{\nu}m^2\nu\} - E\nu^2\} \\ + E\nu^2E'\} + 2K' \{K'[E + K(m-1)]^2 - E'[E^2 + K^2(m-1)]\}) \\ + \{K[(4m-5)K' - 3E'] + 3EK'\} \\ \times \{\pi\Theta(\nu)\nu(K \{K'[4\ddot{\nu}(m-1)m + \nu] - \nu E'\} + E\nu K') - 2K'^2 [E^2 + K^2(m-1)]\} \\ \times \{K'[2E + K(mV_0 + m-2)] - \pi\Theta(\nu)\nu^2\} + 2KK' \\ \times \{\pi\Theta(\nu)\nu(K \{K'[4\ddot{\nu}(m-1)m + \nu] - \nu E'\} + E\nu K') - 2K'^2 [E^2 + K^2(m-1)]\} \\ \times \{4\pi\ddot{\nu}\Theta(\nu)(m-1)m\nu - E'[2E + K(mV_0 + m-2)] + mK'[E(V_0 + 1) + K(V_0 - 1)]\}, \quad (5.370) $$

in which $\ddot{\nu}$ stands for the double derivative of $\nu(m)$. It reads

$$ \ddot{\nu}(m) = \frac{2\sqrt{2}}{\pi^2 f} \frac{1}{(m-1)^2 m^{5/2}} \left\{ 2E'K'[E(4m-2) + Km(4m-5) + K] + E'^2[-2Em - 3K(m-1)] + K'^2[E(4-6m) - K(m-1)(10m-7)] \right\}. \quad (5.371) $$

Let us notice the appearance of Heaviside functions $\Theta(\nu)$ alone in the expression of $\epsilon_2(m)$ showing that it is discontinuous, but finite, when the monopole terms become non-vanishing. Finally, after some algebra, one straightforwardly gets the third Hubble flow function:

$$ \begin{aligned} \epsilon_3(m) = & -\frac{M_{\text{Pl}}^2 \pi^2}{8 \Lambda^2} \frac{m^3}{K K'} \\ & \times \left( 4KK' \{[2E + K(V_0m + m - 2)]K' - \nu^2 \pi \Theta(\nu)\} \{2K' \{[E + K(m-1)]^2 K' \\ & - [E^2 + K^2(m-1)]E'\} + \pi \Theta(\nu) [E E' \nu^2 + (K(m-1) \{4\dot{\nu}^2(m-1)m^2 + 4\dot{\nu}v m^2 \\ & + [4\ddot{\nu}(m-1)m^2 - \nu] \nu] - E \nu^2) K'\} \right) + 2KK' \{4\dot{\nu}(m-1)m \nu \pi \Theta(\nu) \\ & - [2E + K(V_0m + m - 2)]E' + m[K(V_0 - 1) + E(V_0 + 1)]K' \} \\ & \times \{\nu \pi \Theta(\nu) (E \nu K' + K \{[4\dot{\nu}(m-1)m + \nu]K' - \nu E'\}) - 2 [E^2 + K^2(m-1)] K'^2 \} \\ & + [2E + K(V_0m + m - 2)]K' - \nu^2 \pi \Theta(\nu) \{3E K' + K \{[4m-5]K' - 3E'\} \} \\ & \times \{\nu \pi \Theta(\nu) (E \nu K' + K \{[4\dot{\nu}(m-1)m + \nu]K' - \nu E'\}) - 2 [E^2 + K^2(m-1)] K'^2 \} \\ & \times \left( 2K(m-1)m K' \{[2E + K(V_0m + m - 2)] K' - \nu^2 \pi \Theta(\nu) \} \right. \\ & \times \{2 [E^2 + K^2(m-1)] K'^2 - \nu \pi \Theta(\nu) (E \nu K' + K \{[4\dot{\nu}(m-1)m + \nu]K' - \nu E'\}) \} \\ & \times \left\{ \{[2E + K(V_0m + m - 2)]K' - \nu^2 \pi \Theta(\nu)\} \{3E K' + K \{[4m-5]K' - 3E'\} \} \right. \\ & \times \{-2(6K'E^4 + K \{[m(5V_0 + 13) - 24]K' - 2E'\} E^3 - K^2 \{(V_0m + m - 2)E' \\ & + [-4(V_0 + 1)m^2 + (7V_0 + 29)m - 28] K'\} E^2 - K^3(m-1) \} \\ & \times \{2E' + [3m(V_0 + 1) - 8]K'\} E - K^4(m-1) \{(V_0m + m - 2)E' \\ & - [m(V_0 - 3) + 2]K'\} K'^3 + \pi \Theta(\nu) \{12\dot{\nu}^2 K'^2 E^3 + K \nu K' \{[8\ddot{\nu}(m-1)m \\ & + \nu(5V_0m + 21m - 36)]K' - 6\dot{\nu}E'\} E^2 + 2K^2 \{[5\dot{\nu}^2 E'^2 + \nu \{4 - m(3V_0 + 7)\} \\ & - 20\dot{\nu}(m-1)m\} K' E' + (16\dot{\nu}^2(m-1)^2 m^2 + 2\dot{\nu}(m-1)\nu[m(5V_0 + 21) - 16]m \\ & + \nu \{16\ddot{\nu}(m-1)^2 m^2 + \nu [2(V_0 + 1)m^2 - (V_0 + 12)m + 10]\} K'^2 \} E \\ & + K^3 \{5\dot{\nu}^2(V_0m + m - 2)E'^2 - 2\nu \{10\dot{\nu}(m-1)m(V_0m + m - 2) \\ & + \nu [2(V_0 + 1)m^2 + (V_0 - 6)m + 1]\} K' E' + (16\dot{\nu}^2(m-1)^2(V_0m + m - 2)m^2 \\ & + 4\dot{\nu}(m-1)\nu [8(V_0 + 1)m^2 - 3(V_0 + 9)m + 14] m \\ & + \nu \{16\ddot{\nu}(m-1)^2(V_0m + m - 2)m^2 + \nu[m(V_0 - 5) + 4]\} K'^2 \} K' + \nu^2 \pi^2 \Theta(\nu)^2 \right. \\ & \times \left[ (-3\dot{\nu}^2 E'^2 + 2\nu[2\dot{\nu}(m-1)m + (2m-1)\nu]K' E' + \{16\dot{\nu}^2(m-1)^2 m^2 \\ & - 4\dot{\nu}(8m^2 - 15m + 7)\nu m + \nu [\nu - 16\ddot{\nu}(m-1)^2 m^2]\} K'^2) K^2 + 2E \nu K' \right. \\ & \times \left. \{\nu E' + [(3 - 2m)\nu - 2\dot{\nu}(m-1)m]K'\} K - 3E^2 \nu^2 K'^2 \right] \end{aligned} $$
$$
\begin{align}
& - 2KK' \{ -4\dot{v}(m-1)m\nu\pi\Theta(v) + [2E + K(V_0m + m - 2)]E' \} \nonumber \\
& - m[K(V_0 - 1) + E(V_0 + 1)]K' \} \{ 4KK' \{ [2E + K(V_0m + m - 2)]K' - v^2\pi\Theta(v) \} \} \nonumber \\
& \times \{ 2K' \{ [E + K(m-1)]^2K' - [E^2 + K^2(m-1)]E' \} \nonumber \\
& \quad + \pi\Theta(v) [E'E'v^2 + (K(m-1) \{ 4\dot{v}^2(m-1)m^2 + 4\dot{v}vm^2 + [4\ddot{v}(m-1)m^2 - v]v \} \nonumber \\
& \quad - E^2)K' \} \} + 2KK' \{ 4\dot{v}(m-1)m\nu\pi\Theta(v) - [2E + K(V_0m + m - 2)]E' \} \nonumber \\
& + m[K(V_0 - 1) + E(V_0 + 1)]K' \} \{ \nu\pi\Theta(v) \{ E\nu K' + K \{ [4\dot{v}(m-1)m + v]K' - vE' \} \} \nonumber \\
& - 2 [E^2 + K^2(m-1)] K'^2 \} + \{ [2E + K(V_0m + m - 2)]K' - v^2\pi\Theta(v) \} \nonumber \\
& \times (3EK' + K \{ [4m-5]K' - 3E' \} \} \{ \nu\pi\Theta(v) \{ E\nu K' + K \{ [4\dot{v}(m-1)m + v]K' - vE' \} \} - 2 [E^2 + K^2(m-1)] K'^2 \} \} \nonumber \\
& \times (8K^4(m-1)^3 K'^4 \{ [2E + K(V_0m + m - 2)]K' - v^2\pi\Theta(v) \}^3)^{-1} \nonumber \\
& + \{ m \{ 4KK' \{ [2E + K(V_0m + m - 2)]K' - v^2\pi\Theta(v) \} \{ 4K(m-1)^2m^2\pi\delta(v)K'\dot{v}^3 \} \} \nonumber \\
& + [m(m-1)]^{-1} \{ -[E^2 + K^2(m-1)]E'^2 + 2 \{ -(m-2)E^2 + 4K(m-1)E + K^2(m^2-3m+2) \} K' E' - [E^2 + 2K(m-1)E - K^2(m-1)]mK'^2 \} \nonumber \\
& + (m-1)\pi\Theta(v) \{ (E\nu(2\dot{v}m + v) + K \{ 2\dot{v}^2(m-1)m^2 + 2\dot{v}vm^2 + [2\ddot{v}(m-1)m^2 - v]v \} \} E' + 2m \{ K \{ m(8m^2 - 11m + 3)\dot{v}^2 + [v(1-2m)^2 + 6\dot{v}(m-1)^2m^2]\dot{v} + (m-1)m[2\ddot{v}(m-1)m + \ddot{v}(8m-3)]v \} \nonumber \\
& - E \{ (m-1)m\dot{v}^2 + (m+1)v\dot{v} + \ddot{v}(m-1)mv \} K' \} \} + [2(m-1)]^{-1} (m^{-1} \nonumber \\
& \times [4KK' \{ -4\dot{v}(m-1)m\nu\pi\Theta(v) + [2E + K(V_0m + m - 2)]E' \} \nonumber \\
& - m[K(V_0 - 1) + E(V_0 + 1)]K' \} \{ 2K' \{ [E + K(m-1)]^2K' - [E^2 + K^2(m-1)]E' \} \nonumber \\
& + \pi\Theta(v) [E'E'v^2 + (K(m-1) \{ 4\dot{v}^2(m-1)m^2 + 4\dot{v}vm^2 + [4\ddot{v}(m-1)m^2 - v]v \} \nonumber \\
& - E^2)K' \} \} + (4KK' \{ 4\dot{v}(m-1)m\nu\pi\Theta(v) - [2E + K(V_0m + m - 2)]E' \} \nonumber \\
& + m[K(V_0 - 1) + E(V_0 + 1)]K' \} \{ 2K' \{ [E + K(m-1)]^2K' - [E^2 + K^2(m-1)]E' \} \nonumber \\
& + \pi\Theta(v) [E'E'v^2 + (K(m-1) \{ 4\dot{v}^2(m-1)m^2 + 4\dot{v}vm^2 + [4\ddot{v}(m-1)m^2 - v]v \} \nonumber \\
& - E^2)K' \} \} + (2 \{ [2E + K(V_0m + m - 2)]K' - v^2\pi\Theta(v) \} \{ 3EK' + K \nonumber \\
& \times \{ [4m-5]K' - 3E' \} \} \{ 2K' \{ [E + K(m-1)]^2K' - [E^2 + K^2(m-1)]E' \} + \pi\Theta(v) \nonumber \\
& \times [E'E'v^2 + (K(m-1) \{ 4\dot{v}^2(m-1)m^2 + 4\dot{v}vm^2 + [4\ddot{v}(m-1)m^2 - v]v \} \nonumber \\
& - E^2)K' \} \} + 4KK' \{ 4(m-1) \{ (m-1)m\dot{v}^2 + (2m-1)v\dot{v} + \ddot{v}(m-1)mv \} \pi\Theta(v) \nonumber \\
& + [E(V_0 - 1) - K(m-1)(V_0 + 1)]E' + E(V_0m + m - 2V_0)K' \} \{ \nu\pi\Theta(v) \{ E\nu K' + K \{ [4\dot{v}(m-1)m + v]K' - vE' \} \} - 2 [E^2 + K^2(m-1)] K'^2 \} + m^{-1} \nonumber \\
& \times \{ \{ -4\dot{v}(m-1)m\nu\pi\Theta(v) + [2E + K(V_0m + m - 2)]E' - m[K(V_0 - 1) \nonumber \\
& + E(V_0 + 1)]K' \} \{ 3EK' + K \{ (4m-5)K' - 3E' \} \} \{ \nu\pi\Theta(v) \{ E\nu K' + K \{ [4\dot{v}(m-1)m + v]K' - vE' \} \} - 2 [E^2 + K^2(m-1)] K'^2 \} \nonumber \\
& + \{ [2(-2E + K(m-1)]K' \{ [2E + K(V_0m + m - 2)]K' - v^2\pi\Theta(v) \} \nonumber \\
& \times \{ 2K' \{ [E + K(m-1)]^2K' - [E^2 + K^2(m-1)]E' \} + \pi\Theta(v) \nonumber \\
& \times [E'E'v^2 + (K(m-1) \{ 4\dot{v}^2(m-1)m^2 + 4\dot{v}vm^2 + [4\ddot{v}(m-1)m^2 - v]v \} \nonumber \\
& - E^2)K' \} \} + 2K \{ (E' - mK') \{ [2E + K(V_0m + m - 2)]K' - v^2\pi\Theta(v) \} \nonumber \\
& \times \{ 2K' \{ [E + K(m-1)]^2K' - [E^2 + K^2(m-1)]E' \} + \pi\Theta(v) [E'E'v^2 + (K(m-1) \{ 4\dot{v}^2(m-1)m^2 + 4\dot{v}vm^2 + [4\ddot{v}(m-1)m^2 - v]v \} - E^2)K' \} \} + K \{ (E' - mK') \nonumber \\
& \times \{ -4\dot{v}(m-1)m\nu\pi\Theta(v) + [2E + K(V_0m + m - 2)]E' - m[K(V_0 - 1) + E(V_0 + 1)] \nonumber \\
& \times K' \} \{ 2 [E^2 + K^2(m-1)] K'^2 - \nu\pi\Theta(v) \{ E\nu K' + K \{ [4\dot{v}(m-1)m + v]K' - vE' \} \} \nonumber \\
& - \nu E' \} \} + [E + K(m-1)]K' \{ 4\dot{v}(m-1)m\nu\pi\Theta(v) - [2E + K(V_0m + m - 2)]E' \nonumber \\
& + m \{ K(V_0 - 1) + E(V_0 + 1) \} K' \} \{ 2 [E^2 + K^2(m-1)] K'^2 - \nu\pi\Theta(v) \nonumber \\
& \times (E\nu K' + K \{ [4\dot{v}(m-1)m + v]K' - vE' \} \} \} + \{ [3E + 2K(m-2)]E' \nonumber \\
& + [E(1-2m) + K(3m-1)]K' \} \{ [2E + K(V_0m + m - 2)]K' - v^2\pi\Theta(v) \} \nonumber \\
& \times \{ \nu\pi\Theta(v) \{ E\nu K' + K \{ [4\dot{v}(m-1)m + v]K' - vE' \} \} \nonumber \\
& - 2 [E^2 + K^2(m-1)] K'^2 \} \} \} \} (4K^3(m-1)^2 K'^3 \nonumber \\
& \times \{ [2E + K(V_0m + m - 2)]K' - v^2\pi\Theta(v) \}^2)^{-1} \} \tag{5.372}
\end{align}
$$
<figure><img src="image_47.png" /><figcaption>Fig. 44. Left panel: rescaled parametric slow-roll function $\epsilon_1(m)$ (multiplied by $\Lambda^2/M_P^2$) with respect to the elliptic parameter $m$ for dual inflation (the potential is represented in Fig. 43). Right panel: rescaled slow-roll functions $\epsilon_2(m)$ (solid line) and $\epsilon_3(m)$ (dotted lines), both multiplied by $\Lambda^2/M_P^2$. Although $\epsilon_2(m)$ is discontinuous at $m = m_{mon}$, it remains finite. The parameter $\epsilon_3(m)$ is however singular at $m = m_{mon}$ and changes sign (negative values have been represented in red on the logarithmic scale). Because $\Lambda < M_P$, inflation always occurs in the domain $m < m_{mon}$ for which $\epsilon_1(m) < 1$.</figcaption></figure>

<figure><img src="image_48.png" /><figcaption>Fig. 45. Elliptic parameter $m_{end}$ at which inflation ends (solid lines), and $m_{mon}$ at which monopole condensation occurs (dotted line), as a function of $f$. The upper red curve shows $m_{end}$ for $\Lambda/M_P = 0.2$ while the lower blue one for $\Lambda/M_P = 0.1$ ($m_{mon}$ does not depend on $\Lambda$). For small values of $f$, inflation always ends before monopole condensation $m_{end} < m_{mon}$. However, for Planckian-like values of $\Lambda$, inflation may end in the monopole condensation if $f$ is sufficiently large. As these curves show, in that situation, inflation does no last much longer, the potential with monopole terms switched on being steep, one has $m_{end} \gtrsim m_{mon}$ independently of $f$ (see also $\epsilon_1$ in Fig. 44).</figcaption></figure>

In this expression, $\ddot{v}$ is the third derivative of $v(m)$ with respect to $m$ and reads

$$ \ddot{v} = \frac{\sqrt{2}}{\pi^2 f} \frac{1}{(m-1)^3 m^{7/2}} \left( -2E' K' \{E[m(26m-23)+5] + K(m-1)[m(24m-13)-3]\} \right. \\ \left. + E'^2 \{E[m(8m+7)-7] + 4K(m-1)(7m-5)\} \right. \\ \left. + K'^2 [E(50m^2-65m+23) + 2K(m-1)(36m^2-51m+19)] \right). \quad (5.373) $$

As expected, the expression of $\epsilon_3(m)$ contains a Dirac distribution $\delta(v)$ and is therefore singular at the elliptic parameter value $m_{mon}$ for which $v(m_{mon}) = 0$. As explicit in Eq. (5.359), the potential contains a feature for $m = m_{mon}$ and the slow-roll approximation breaks down, in a way similar to the Starobinsky model [215]. There are however two differences. The first one is that this feature generically occurs after, or at the end of, inflation for $\Lambda < M_P$ (see Fig. 44), namely during the reheating stage. Therefore, it is not observable and cannot affect the slow-roll predictions which are confined within the inflationary valley at $m \ll m_{mon}$. The second difference with respect to Ref. [215] is that the singular behavior induced by the feature only appears at the perturbative level. Indeed, since both $\epsilon_1(m)$ and $\epsilon_2(m)$ remain finite at $m = m_{mon}$, slow roll is not necessarily violated for the background field trajectory [see Eq. (3.10)]. However, the singularity in $\epsilon_3(m)$ necessarily implies that the perturbations generated at $m = m_{mon}$ cannot be of the slow-roll kind. Again, this is not problematic for our purpose as this concerns very small length scales but such a feature might have some interesting consequences concerning reheating microphysics and black-hole formation [531], especially if the model is viewed as a toy realization of the QCD-like phase transition [532]. The parametric Hubble flow functions have been represented in Fig. 44 as a function of the elliptic parameter $m$.

Owing to the parametric expression of the slow-roll functions given in Eqs. (5.369), (5.370) and (5.372), observable predictions can be directly extracted from the parametric trajectory $m(N)$. From Eq. (5.367), one gets

$$ N_{\text{end}} - N = \frac{16\Lambda^2}{M_{\text{Pl}}^2 \pi^2} [I(m, f) - I(m_{\text{end}}, f)], \quad (5.374) $$
where

$$I(m, f) = \int^{m} \frac{(p-1)(KK')^2 \{K'[2E + K(pV_0 + p - 2)] - \pi v^2 \Theta(v)\}}{\pi v \Theta(v) p^2 \{-4p(p-1)\dot{v} K K' - v[K'(E + K) - E'K]\} + 2p^2 K'^2 [E^2 + (p-1)K^2]} dp. \quad (5.375)$$

In practice, provided one remains focused on the inflationary dynamics for $m < m_{\text{mon}}$, this expression can be simplified into

$$I(m < m_{\text{mon}}, f) = \int_{m_{\text{mon}}}^{m} \frac{(p-1)(KK')^2 \{K'[2E + K(pV_0 + p - 2)]\}}{2p^2 K'^2 [E^2 + (p-1)K^2]} dp. \quad (5.376)$$

There is no explicit analytic solutions for these two integrals and $I(m, f)$ has to be evaluated numerically. One should notice the overall $\Lambda^2$ factor in Eq. (5.374) showing that the number of e-folds, at given elliptic parameter $m$, indirectly depends on the amplitude of the CMB anisotropies (see Section 5.24.3). This expression has to be completed by the value of the elliptic parameter $m_{\text{end}}$ at which inflation ends. As can be seen in Fig. 44, dual inflation gracefully exits as soon as $\epsilon_1(m_{\text{end}}) = 1$, i.e. for

$$\frac{4\Lambda}{\pi M_{\text{Pl}}^4} = \frac{m_{\text{end}}^{1/2} (K_{\text{end}} K'_{\text{end}})^{-3/2}}{2(m_{\text{end}} - 1) \{K'_{\text{end}}[2E_{\text{end}} + K_{\text{end}}(m_{\text{end}}V_0 + m_{\text{end}} - 2)] - \pi v_{\text{end}}^2 \Theta(v_{\text{end}})\}} \\ \times (\pi v_{\text{end}} \Theta(v_{\text{end}}) \{ -4m_{\text{end}}(m_{\text{end}} - 1)\dot{v}_{\text{end}} K_{\text{end}} K'_{\text{end}} \\ - v_{\text{end}} [K'_{\text{end}}(E_{\text{end}} + K_{\text{end}}) - E'_{\text{end}} K_{\text{end}}] \} + 2K'_{\text{end}} [E_{\text{end}}^2 + (m_{\text{end}} - 1)K_{\text{end}}^2]). \quad (5.377)$$

There is neither analytical solution to this equation and it has to be solved for $m_{\text{end}}(\Lambda, f)$ numerically. In Fig. 45, we have represented the values of $m_{\text{end}}(\Lambda, f)$ and $m_{\text{mon}}(f)$ for $\Lambda = 0.1 M_{\text{Pl}}$ and $\Lambda = 0.2 M_{\text{Pl}}$. At small $f$, inflation always ends before monopole condensation ($m_{\text{end}} < m_{\text{mon}}$). Only for Planckian-like values of $\Lambda$ there exists a domain, corresponding to values of $f$ or order unity, in which $m_{\text{end}} \gtrsim m_{\text{mon}}$. As the solid curves of Fig. 45 show, in this situation, $m_{\text{end}}$ still remains very close to $m_{\text{mon}}$ because the potential for $m > m_{\text{mon}}$ is very steep and inflation ends almost instantaneously. From an observational point of view, the value of $f$ at which the two curves intersect correspond to a transition between two typical paradigms. For small values of $f$, $\epsilon_1$ smoothly increases during inflation to reach unity before the reheating starts. Such a situation is reminiscent with the parametric reheating paradigm. On the contrary, for large values of $f$, corresponding to $m_{\text{end}} \gtrsim m_{\text{mon}}$, the monopole condensation stops inflation almost instantaneously, i.e., the first Hubble flow function $\epsilon_1(m)$ can still be small just before the end of inflation. The latter situation actually mimics the effect of a tachyonic reheating.

In the next section, we discuss how to determine $m_*$, the elliptic parameter value at which the pivot mode crosses the Hubble radius during inflation, in a consistent way, by taking into account the CMB normalization of the parameter $M^4$.

### 5.24.3. Reheating consistent observable predictions

From Eqs. (3.32) and (5.361), at a given parameter $f$, the value of $\Lambda$ matching the amplitude of the CMB anisotropies verifies

$$\frac{\Lambda^4}{M_{\text{Pl}}^4} = \frac{24\pi^4 P_* \epsilon_1(m_*)}{f^2 v(m_*)}, \quad (5.378)$$

where, because $m_* < m_{\text{mon}}$, one has

$$v(m_*) \equiv \frac{V(m_*)}{M^4} = 1 + V_0(f) - 2 \frac{K(m_*^{1/2}) - E(m_*^{1/2})}{m_* K(m_*^{1/2})}. \quad (5.379)$$

From the expression of $\epsilon_1$ given in Eq. (5.369), using again the fact that $m_* < m_{\text{mon}}$, this formula can be simplified and one gets

$$\frac{\Lambda^6}{M_{\text{Pl}}^6} = \frac{3\pi^6 P_*}{2f^2} \frac{m_* [E_*^2 + (m_* - 1)K_*^2]^2}{(m_* - 1)^2 K_*^3 K_*' [2E_* + K_*(m_*V_0 + m_* - 2)]^2 [1 + V_0(f) - 2 \frac{K_* - E_*}{m_* K_*}]}, \quad (5.380)$$

in which all the elliptic integrals have a “*” because they are evaluated at $m = m_*$. Difficulties arise because $m_*$ must solve the reheating equation (3.45). Plugging Eq. (5.374) into Eq. (3.45) gives a new “parametric” reheating equation to solve,

$$\frac{16\Lambda^2}{M_{\text{Pl}}^2 \pi^2} [I(m_*, f) - I(m_{\text{end}}, f)] = \ln R_{\text{rad}} - N_0 - \frac{1}{4} \ln \left[ \frac{9}{2\epsilon_1(m_*)} \frac{v(m_{\text{end}})}{v(m_*)} \right] + \frac{1}{4} \ln (8\pi^2 P_*) , \quad (5.381)$$

where $I(m, f)$ is given in Eq. (5.375), $\epsilon_1(m)$ in Eq. (5.369) and $m_{\text{end}}(\Lambda, f)$ is the solution of Eq. (5.377). In this expression, although $v(m_*)$ is given by Eq. (5.379), $v(m_{\text{end}})$ requires the full expression of Eq. (5.359)

$$v(m_{\text{end}}) = 1 + V_0(f) - 2 \frac{K_{\text{end}} - E_{\text{end}}}{m_{\text{end}} K_{\text{end}}} - \frac{\pi}{m_{\text{end}} K_{\text{end}} K'_{\text{end}}} v_{\text{end}}^2 \Theta(v_{\text{end}}). \quad (5.382)$$

The parametric reheating equation (5.381) may be compared to the ordinary one in Eq. (3.48). We see that the price to pay for having worked with the non-canonical normalized variable $m$ is an explicit dependence on $\Lambda/M_{\text{Pl}}$. Let us emphasize that this factor is not a mere rescaling of $M_{\text{Pl}}^2$ but an explicit function of $m_*$, thereby rendering Eq. (5.381) fundamentally different than the usual reheating equation.

Plugging the expression of $\Lambda/M_{\text{Pl}}$ given in Eqs. (5.380) into Eqs. (5.377) and (5.381) yields a closed algebraic system of two equations whose unknowns are $m_{\text{end}}$ and $m_*$. For an input value of $f$ and of the reheating parameter $\ln R_{\text{rad}}$, this system can be numerically solved to get the values of $m_{\text{end}}$ and $m_*$. From $m_*$, one can finally use Eq. (5.380) to get the actual value of $\Lambda/M_{\text{Pl}}$. In Fig. 46, we have represented $m_{\text{end}}$, $m_*$ and $\Lambda/M_{\text{Pl}}$ as a function of $f$ and for all possible values of the reheating parameters $\ln R_{\text{rad}}$. The left panel of this figure shows that, in order to match the amplitude of the CMB anisotropies, the dynamic scale $\Lambda/M_{\text{Pl}}$ has to take larger values for smaller values of $f$. This is expected as the overall
<figure><img src="image_49.png" /><figcaption>Fig. 46. Left panel: CMB normalized values of $\Lambda/M_{\text{pl}}$ with respect to the parameter $f$ for all possible reheating histories. The color bar encodes the values of $\ln R_{\text{rad}}$. Right panel: CMB normalized values of $m_*$ (diamonds) and $m_{\text{end}}$ (plus signs) as a function of $f$ and for various values of $\ln R_{\text{rad}}$ (color bar). For sub-Planckian values of $\Lambda$, the parameter $f$ cannot take arbitrarily small values, i.e. $f > \mathcal{O}(10^{-5})$.</figcaption></figure>

potential normalization scales as $M^4 = f^2 A^4 / \pi^2$. Requiring that $\Lambda/M_{\text{pl}} < 1$ imposes a lower bound on $f$. As can be seen in this plot, the reheating modulates this bound by, typically, a factor of five showing that $f > \mathcal{O}(10^{-5})$.

Let us finally remark that, from $m_*$, the observable values of the slow-roll parameters, spectral index, tensor-to-scalar ratio and running can be immediately read-off from the expression of the $\epsilon_n(m_*)$ given earlier. The reheating consistent slow-roll predictions for dual inflation are represented in Fig. 157.

## 5.25. Cubically Corrected Starobinsky Inflation (CCSI)

### 5.25.1. Theoretical justifications

We have already encounter one class of corrections to the Starobinsky model [241] with Rpl in Section 5.13. Another possibility which has been discussed in Refs. [9,533–535] is to consider Starobinsky Inflation as the leading correction of a Taylor-expanded $f(R)$ modified gravity theory [245,246,442,443]. As such, the next-to-leading order correction would include a cubic dependency in $R$ such that

$$f(R) = R + \frac{R^2}{\mu^2} + \alpha \frac{R^3}{\mu^4}, \qquad (5.383)$$

where $\mu$ is the mass scale introduced in the Starobinsky model and $\alpha$ a new dimensionless parameter encoding the strength of the next-to-leading order corrections.

Following the same notation and methodology as for Rpl, one can introduce the scalar degree of freedom $\phi$ defined by

$$\frac{\phi}{M_g} = \sqrt{\frac{3}{2}} \ln(|F(R)|), \qquad (5.384)$$

where $F(R) \equiv \partial f / \partial R$. This is also the square of the conformal factor allowing to recast all the equations in the Einstein frame and one finds the associated potential for the field $\phi$ to be

$$V(\phi) = \frac{M_g^2 |F|}{2} \frac{RF - f}{F^2}. \qquad (5.385)$$

This is the same expression as for Rpl, see Eq. (5.189), but the function $F$ now reads

$$F = 1 + 2 \frac{R}{\mu^2} + 3\alpha \frac{R^2}{\mu^4}. \qquad (5.386)$$

Defining the quantity $y$ by

$$y = \sqrt{\frac{2}{3} \frac{\phi}{M_g}}, \qquad (5.387)$$

one gets from Eq. (5.384) that $F = e^y$. Solving Eq. (5.386) for $R$ gives a quadratic equation with two solutions, but only one allows us to recover the Starobinsky model in the limit $\alpha \to 0$. It reads

$$\frac{R}{\mu^2} = \frac{\sqrt{1 + 3\alpha(e^y - 1)} - 1}{3\alpha}. \qquad (5.388)$$

Plugging this expression into Eq. (5.385) yields a closed expression for the potential of CCSI in the Einstein frame

$$V(\phi) = \frac{M_g^2 \mu^2}{2} (1 - e^{-y})^2 \frac{1 + \sqrt{1 + 3\alpha(e^y - 1) + 2\alpha(e^y - 1)}}{\left[1 + \sqrt{1 + 3\alpha(e^y - 1)}\right]^3}. \qquad (5.389)$$

The multiplicative constant terms can be absorbed into the potential normalization and we define

$$M^4 \equiv \frac{1}{2} M_g^2 \mu^2. \qquad (5.390)$$
<figure><img src="image_50.png" /><figcaption>Fig. 47. Cubically Corrected Starobinsky Inflation (CCSI) in the Einstein frame for $\alpha = \pm 10^{-4}$. Top panels: the potential and its logarithm. Bottom left panel: slow-roll parameter $\epsilon_1$ with the different inflationary regimes of CCSI annotated with an arrow indicating the direction to which the field evolves. Notice that, for the CCSI2 regime, inflation never stops and one has to consider an extra-mechanism to end inflation. Bottom right panel: slow-roll parameters $\epsilon_2$ (solid line) and $\epsilon_3$ (dotted line). For CCSI2, $\epsilon_3$ becomes negative at large field values and this is represented as a cyan dotted line while the blue dotted line corresponds to positive values.</figcaption></figure>

The limiting case $\alpha \to 0$ gives

$$ \lim_{\alpha \to 0} V(\phi) = \frac{M^4}{4} (1 - e^{-y})^2, \qquad (5.391) $$

which is, up to a different definition of $M^4$, the potential of the Starobinsky model and also matches the one of Higgs inflation, see Eq. (4.21) and (4.88).

### 5.25.2. Slow-roll analysis

In terms of the canonically normalized field $\phi$, the potential of CCSI depends on only one parameter $\alpha$ and reads

$$ V(\phi) = M^4 \left(1 - e^{-\sqrt{\frac{2}{3}} \frac{\phi}{M_g}}\right)^2 \frac{1 + \sqrt{1 + 3\alpha \left(e^{\sqrt{\frac{2}{3}} \frac{\phi}{M_g}} - 1\right) + 2\alpha \left(e^{\sqrt{\frac{2}{3}} \frac{\phi}{M_g}} - 1\right)}}{\left[1 + \sqrt{1 + 3\alpha \left(e^{\sqrt{\frac{2}{3}} \frac{\phi}{M_g}} - 1\right)}\right]^3}. \qquad (5.392) $$

Because the sign of $\alpha$ changes the shape and the domain of definition of $V(\phi)$, this leads us to consider three different regimes. For $\alpha > 0$, the potential is defined over all field values and exhibits a maximum at

$$ \phi_{V \max} = \sqrt{\frac{3}{2}} M_g \ln \left( \frac{2 + 4\sqrt{\alpha}}{\sqrt{\alpha}} \right). \qquad (5.393) $$

For $0 < \phi < \phi_{V \max}$, inflation can proceed at decreasing field values and this regime will be referred to as CCSI1. For $\phi > \phi_{V \max}$, it proceeds at increasing field values and we call this regime CCSI2. Finally, if $\alpha < 0$, the potential is defined only in the domain $\phi < \phi_{UV}$ where

$$ \phi_{UV} = \sqrt{\frac{3}{2}} M_g \ln \left(1 - \frac{1}{3\alpha}\right). \qquad (5.394) $$

Inflation here proceeds at decreasing field values and will be referred to as CCSI3. The potential and its logarithm have been represented in the top panels of Fig. 47.
In terms of the dimensionless field y, the Hubble flow functions in the slow-roll approximation read

$$
\begin{align}
\epsilon_1 &= \frac{1}{3} \left\{ \frac{2 - 8\alpha - e^y \left[ 1 + 2\alpha (e^y - 5) - \sqrt{1 + 3\alpha (e^y - 1)} \right]}{(e^y - 1) \left[ 1 + 4\alpha (e^y - 1) \right]} \right\}^2, \\
\epsilon_2 &= \frac{2e^y}{3(e^y - 1)^2 [1 + 4\alpha (e^y - 1)]^2 \sqrt{3\alpha (e^y - 1) + 1}} \nonumber \\
&\quad \times \left( 12\alpha^2 (e^y - 1)^2 \left[ 4\sqrt{3\alpha (e^y - 1) + 1} + e^y + 2 \right] + 2 \left[ \sqrt{3\alpha (e^y - 1) + 1} + 1 \right] \right. \nonumber \\
&\quad \left. - \alpha (e^y - 1) \left\{ e^y \left[ 4\sqrt{3\alpha (e^y - 1) + 1} - 5 \right] - 2 \left[ 10\sqrt{3\alpha (e^y - 1) + 1} + 7 \right] \right\} \right), \tag{5.395}
\end{align}
$$

and

$$
\begin{equation}
\begin{aligned}
\epsilon_3 = & - \left[ 3(e^y - 1)^2 [3\alpha(e^y - 1) + 1] [4\alpha(e^y - 1) + 1]^2 \left( 2\left[\sqrt{3\alpha(e^y - 1) + 1}\right] \right. \right. \\
& \left. \left. \quad + 12\alpha^2(e^y - 1)^2 \left[4\sqrt{3\alpha(e^y - 1) + 1} + e^y + 2\right] - \alpha(e^y - 1) \left\{e^y\left[4\sqrt{3\alpha(e^y - 1) + 1} - 5\right]\right. \right. \\
& \left. \left. \quad - 2\left[10\sqrt{3\alpha(e^y - 1) + 1} + 7\right]\right) \right) \right]^{-1} \\
& \times \left\{8\alpha + e^y\left[2\alpha(e^y - 5) - \sqrt{3\alpha(e^y - 1) + 1}\right] - 2\right\} \\
& \times \left\{ 144\alpha^4 e^{6y} + 24\alpha^3 e^{5y} \left\{12\alpha\left[4\sqrt{3\alpha(e^y - 1) + 1} + 3\right] - 4\sqrt{3\alpha(e^y - 1) + 1} + 3\right\} \right. \\
& \left. - \alpha^2 e^{4y} \left(24\alpha\left\{36\alpha\left[4\sqrt{3\alpha(e^y - 1) + 1} + 5\right] - 48\sqrt{3\alpha(e^y - 1) + 1} - 55\right\} \right. \right. \\
& \left. \left. + 8\sqrt{3\alpha(e^y - 1) + 1} + 23\right) + 3(4\alpha - 1)\alpha e^{2y} \left(\alpha\left\{12\alpha\left[16\sqrt{3\alpha(e^y - 1) + 1} - 15\right] \right. \right. \\
& \left. \left. \quad - 16\sqrt{3\alpha(e^y - 1) + 1} + 165\right) - 4\left[3\sqrt{3\alpha(e^y - 1) + 1} + 8\right]\right) \right. \\
& \left. + 2\alpha e^{3y} \left[2\alpha\left(6\alpha\left\{48\alpha\left[2\sqrt{3\alpha(e^y - 1) + 1} + 5\right] - 72\sqrt{3\alpha(e^y - 1) + 1} - 155\right\} \right. \right. \\
& \left. \left. \quad + 68\sqrt{3\alpha(e^y - 1) + 1} + 155\right) + 4\sqrt{3\alpha(e^y - 1) + 1} - 5\right] \right. \\
& \left. + 4(1 - 4\alpha)^2(3\alpha - 1) \left\{\alpha\left[6\sqrt{3\alpha(e^y - 1) + 1} + 3\right] - \sqrt{3\alpha(e^y - 1) + 1} - 1\right\} \right. \\
& \left. - 2(4\alpha - 1)e^y \left[\alpha\left(3\alpha\left\{36\alpha\left[4\sqrt{3\alpha(e^y - 1) + 1} + 1\right] - 72\sqrt{3\alpha(e^y - 1) + 1} - 11\right\} \right. \right. \\
& \left. \left. \quad + 20\sqrt{3\alpha(e^y - 1) + 1} - 7\right) + 2\sqrt{3\alpha(e^y - 1) + 1} + 2\right] \right\}.
\end{aligned}
\tag{5.396}
\end{equation}
$$

They have been represented in the bottom panel of Fig. 47. Notice that $\epsilon_3(y)$ changes sign, which is represented by the cyan blue dotted line in the figure (logarithmic scale). One also remarks that in the regime CCSI2, obtained for $\alpha > 0$ and $\phi > \phi_{v\text{max}}$, inflation does not end and one has to introduce another mechanism to end the accelerated expansion, as for instance a tachyonic instability triggered by another field. Therefore, CCSI2 has an additional parameter, say $y_{\text{end}}$ (or $\phi_{\text{end}}$), the field value at which inflation ends. For the two other models, CCSI1 and CCSI3, inflation naturally stops for $y_{\text{end}}$ solution of $\epsilon_1(y_{\text{end}}) = 1$, namely

$$
y_{\text{end}} = \ln \left( \frac{-15 - 14\sqrt{3} + 176\alpha + 132\sqrt{3}\alpha + \sqrt{813 + 420\sqrt{3} + 4444\alpha + 2728\sqrt{3}\alpha}}{242\alpha} \right). \qquad (5.397)
$$

This formula is also valid for CCSI3 ($\alpha < 0$), although the equation $\epsilon_1 = 1$ admits a second root in that case, $y_{\epsilon_1=1}$, which bounds the inflationary domain to $y_{\text{end}} < y < y_{\epsilon_1=1}$ ($y_{\epsilon_1=1} < y_{\text{UV}}$) with

$$
y_{\epsilon_1=1} = \ln \left( \frac{-15 - 14\sqrt{3} + 176\alpha + 132\sqrt{3}\alpha - \sqrt{813 + 420\sqrt{3} + 4444\alpha + 2728\sqrt{3}\alpha}}{242\alpha} \right). \qquad (5.398)
$$
<figure><img src="image_51.png" /><figcaption>Fig. 48. Maximal number of e-folds of inflation within the CCSI3 regime as a function of $\alpha < 0$. In order to have at least 60 e-folds of inflation, $|\alpha|$ should be smaller than $|\alpha| < 8 \times 10^{-4}$.</figcaption></figure>

Let us now turn to the slow-roll trajectory, it can be integrated analytically and one gets

$$ 
\begin{aligned}
N_{\text{end}} - N = & -\frac{3}{4}(y - y_{\text{end}}) - \frac{9}{8} \log \left[ \frac{4 - \alpha (e^y - 4)^2}{4 - \alpha (e^{y_{\text{end}}}-4)^2} \right] \\
& + \frac{3+9\sqrt{\alpha}}{4\sqrt{\alpha}} \left\{ \text{arccoth} \left[ \frac{3\sqrt{\alpha+1}}{\sqrt{3\alpha(e^y-1)+1}} \right] - \text{arccoth} \left[ \frac{3\sqrt{\alpha+1}}{\sqrt{3\alpha(e^{y_{\text{end}}}-1)+1}} \right] \right\} \\
& + \frac{3}{4\sqrt{\alpha}} \left\{ \text{arctanh} \left[ \frac{1}{2}\sqrt{\alpha}(e^y-4) \right] - \text{arctanh} \left[ \frac{1}{2}\sqrt{\alpha}(e^{y_{\text{end}}}-4) \right] \right\} \\
& - \frac{3}{4\sqrt{\alpha}} \left\{ \text{arccoth} \left[ \frac{1-3\sqrt{\alpha}}{\sqrt{3\alpha(e^y-1)+1}} \right] - \text{arccoth} \left[ \frac{1-3\sqrt{\alpha}}{\sqrt{3\alpha(e^{y_{\text{end}}}-1)+1}} \right] \right\} \\
& - \frac{9}{4} \left\{ \text{arccoth} \left[ \frac{3\sqrt{\alpha-1}}{\sqrt{3\alpha(e^y-1)+1}} \right] - \text{arccoth} \left[ \frac{3\sqrt{\alpha-1}}{\sqrt{3\alpha(e^{y}-1)+1}} \right] \right\}.
\end{aligned} 
\quad (5.399) $$

This expression implicitly assumes complex numbers, as for instance in the CCSI3 regime where $\alpha < 0$, but the result is a real number. Moreover, because the inflationary domain for CCSI3 is bounded within $y_{\text{end}} < y < y_{\text{el}}=1$, the total number of e-folds of inflation $\Delta N_{\text{max}}$ is also bounded and a function of $\alpha$ only. Its value can be obtained by the formal replacement $y \to y_{\text{el}}=1$ in Eq. (5.399) and using Eqs. (5.397) and (5.398). The expression being not particularly illuminating, we have plotted $\Delta N_{\text{max}}$ as a function of $\alpha$ in Fig. 48. As this plot shows, $|\alpha|$ should be small for inflation to be long enough, typically $|\alpha| < \Theta(10^{-3})$ to get more than 60 e-folds of accelerated expansion.

The slow-roll trajectory giving $y$ as a function of $\Delta N = N_{\text{end}} - N$ cannot be analytically inverted from Eq. (5.399), but as for RPI, $y_*$, the dimensionless field value at which the pivot mode crossed the Hubble radius, is uniquely determined from the reheating model described in Section 4.2. The corresponding number of e-fold $\Delta N_* = N_{\text{end}} - N_*$ is given by Eq. (5.399).

Finally, the potential normalization $M$ is determined from the amplitude of the CMB anisotropies and satisfies

$$ 
\begin{aligned}
\left(\frac{M}{M_g}\right)^4 = & 480\pi^2 \frac{\left\{2 - 8\alpha - e^{y_*} \left[1 + 2\alpha(e^{y_*} - 5) - \sqrt{1 + 3\alpha(e^{y_*} - 1)}\right]\right\}^2}{e^{-2y_*} (e^{y_*} - 1)^4 [1 + 4\alpha(e^{y_*} - 1)]^2} \\
& \times \frac{\left[1 + \sqrt{1 + 3\alpha(e^{y_*} - 1)}\right]^3}{1 + \sqrt{1 + 3\alpha(e^{y_*} - 1)} + 2\alpha(e^{y_*} - 1)} \frac{Q^2_{\text{rms-PS}}}{T^2}.
\end{aligned} 
\quad (5.400) $$

The reheating consistent slow-roll predictions for the CCSI model in its different regimes are represented in Figs. 158 to 162. For CCSI1 and CCSI3, the limit $\alpha \to 0$ gives back the model predictions of Starobinsky Inflation (and Higgs Inflation), but not for the CCSI2 regime for which the model predictions strongly depend on the value of the new parameter $y_{\text{end}}$. Such a situation is reminiscent with the RPI2 case in Section 5.13 and, here as well, for CCSI2 one does not longer have numerical equality between $M_g$ and $M_{\text{Pl}}$.

## 5.26. Symmetry Breaking Kähler Inflation (SBKI)

### 5.26.1. Theoretical justifications

This model was proposed in Ref. [536], in the context of the supergravity constructions of Large Field Inflation (see Section 5.2.1). In supergravity, the scalar potential $V$ of a chiral multiplet $\phi^i$ is determined by the Kähler potential $K(\phi^i, \phi^{*i})$ and the superpotential $W(\phi^i)$ according to

$$ V = e^K [K^{li} (W_i + K_i W) (W_i^* + K_i W^*) - 3 |W|^2], \quad (5.401) $$
<figure><img src="image_52.png" /><figcaption>Fig. 49. Symmetry Breaking Kähler Inflation (SBKI) for $\alpha = 0.01$ (blue curves) and $\alpha = -0.01$ (red curves). Upper panels: the potential and its logarithm as a function of $\phi/M_{\text{Pl}}$. Bottom left panel: slow-roll parameter $\epsilon_1$. Bottom right panel: slow-roll parameters $\epsilon_2$ (solid lines) and $\epsilon_3$ (dotted lines).</figcaption></figure>

where $\phi^{*T}$ is the conjugate multiplet and the D-term contributions are omitted. As explained in Section 5.2, Large-Field Inflation with $p = 2$ (LFI2) can be achieved by introducing two chiral multiplets $\Phi$ and $X$ and by considering that the Kähler potential and the superpotential are respectively given by

$$K = XX^* + \frac{1}{2}(\Phi + \Phi^*)^2, \qquad (5.402)$$

$$W = m\Phi X. \qquad (5.403)$$

The Kähler potential enjoys a shift symmetry under the transformation $\Phi \to \Phi + ic$, so if the inflaton $\phi$ is identified with the imaginary part of $\Phi$, its potential does not receive any exponential contribution [that would otherwise be present, due to the prefactor $e^K$ in Eq. (5.401)]. It ends up being of the quadratic form $V(\phi) = m^2\phi^2/2$.

In Ref. [536], it is pointed out that the shift symmetry, which is here broken by the superpotential, could also be broken at the level of the Kähler potential by means of a non-holomorphic spurious field $\mathcal{E}$. The Kähler potential is then expanded around the origin $\mathcal{E} = 0$ according to

$$K = XX^* + \frac{1}{2}(\Phi + \Phi^*)^2 - \frac{\mathcal{E}}{2}(\Phi - \Phi^*)^2 + \frac{\mathcal{E}^2}{4!}(\Phi - \Phi^*)^4 + \dots, \qquad (5.404)$$

where the dots denote higher-order terms in the $\mathcal{E}$ expansion. These extra-terms modify the potential of LFI2 according to

$$V(\phi) = \exp\left(\mathcal{E}\phi^2 + \frac{\mathcal{E}^2}{6}\phi^4 + \dots\right) \frac{m^2}{2}\phi^2, \qquad (5.405)$$

and this expression will define the potential of SBKI.

### 5.26.2. Slow-roll analysis

It is more convenient to rewrite the potential as

$$V = M^4 \left(\frac{\phi}{M_{\text{Pl}}}\right)^2 \exp\left[\alpha \left(\frac{\phi}{M_{\text{Pl}}}\right)^2 + \frac{\alpha^2}{6} \left(\frac{\phi}{M_{\text{Pl}}}\right)^4\right], \qquad (5.406)$$

where $M^4 = m^2 M_{\text{Pl}}^2/2$ and $\alpha = \mathcal{E}/M_{\text{Pl}}$, and where we have restored the Planck masses. The parameter $\alpha$ can be positive or negative. Since it is an even function of the field value, it is enough to restrict the analysis to the region $\phi > 0$. The derivative of the potential is proportional to $3+3\alpha\phi^2+\alpha^2\phi^4$. Seen as a polynomial in $\phi$, the discriminant of that expression is equal to $-3\alpha^2$ and it is always negative, so the polynomial [hence $V'(\phi)$] is always positive. This shows that the potential is an increasing function of the field value, regardless of the sign of $\alpha$, and that inflation proceeds at decreasing field value. The potential and its logarithm are displayed in Fig. 49
Let us define the dimensionless field value

$$x \equiv \frac{\phi}{M_{Pl}}. \qquad (5.407)$$

The first and second Hubble-flow functions, in the slow-roll approximation, are given by

$$\epsilon_1 = 2 \left[ \frac{1}{x} + \alpha x + \frac{\alpha^2}{3} x^3 \right]^2, \quad \epsilon_2 = \frac{4}{x^2} - 4\alpha - 4\alpha^2 x^2, \qquad (5.408)$$

while the third reads

$$\epsilon_3 = - \frac{4(1 + \alpha^2 x^4) [\alpha x^2 (\alpha x^2 + 3) + 3]}{3x^2 (\alpha^2 x^4 + \alpha x^2 - 1)}. \qquad (5.409)$$

They are also displayed in the lower panels of Fig. 49. The first Hubble-flow parameter diverges when $x \to 0$ and also for $x \to \infty$. In between, it reaches a minimum at a field value where the second Hubble-flow parameter vanishes and that we denote $x_{\epsilon_2=0}$. Its expression depends on the sign of $\alpha$, and one finds

$$x_{\epsilon_2=0}^{+} = \sqrt{\frac{-1 + \sqrt{5}}{2\alpha}} \quad \text{for } \alpha > 0, \qquad x_{\epsilon_2=0}^{-} = \sqrt{\frac{1 + \sqrt{5}}{2|\alpha|}} \quad \text{for } \alpha < 0. \qquad (5.410)$$

The corresponding values of $\epsilon_1$ can be evaluated at its minimum, and one obtains

$$\epsilon_{1+}^{\min} = \frac{4}{9}(5\sqrt{5} + 11)\alpha, \quad \epsilon_{1-}^{\min} = \frac{4}{9}(5\sqrt{5} - 11)|\alpha|. \qquad (5.411)$$

For inflation to proceed, one must impose that $\epsilon_1^{\min} < 1$, which requires that $\alpha$ lies in the range $[\alpha_{\min}, \alpha_{\max}]$ with

$$\alpha_{\min} \equiv -\frac{9}{16}(5\sqrt{5} + 11), \quad \alpha_{\max} \equiv \frac{9}{16}(5\sqrt{5} - 11). \qquad (5.412)$$

In order to identify where inflation may proceed, and ends, one has to find the roots of the polynomial equation $\epsilon_1 = 1$, i.e.

$$\frac{\alpha^2}{3} x^4 + \alpha x^2 - \frac{x}{\sqrt{2}} + 1 = 0. \qquad (5.413)$$

This equation can be solved exactly but the explicit form of the solution (which only depends on $\alpha$) is not especially illuminating. Here, we just remark that the above equation has four solutions, two of which being positive. We denote these two solutions as $x_{\epsilon_1=1}^{+}(\alpha)$ for the largest one, and as $x_{\epsilon_1=1}^{-}(\alpha)$ for the smallest positive. Having $x_{\epsilon_1=1}^{+} > x_{\epsilon_1=1}^{-}$, this implies that inflation starts for $x < x_{\epsilon_1=1}^{+}$ and ends at $x_{\text{end}} = x_{\epsilon_1=1}^{-}$. In the small coupling limit, $|\alpha| \ll 1$, one has the limiting forms

$$x_{\text{end}} = \sqrt{2} + 2\sqrt{2\alpha} + \frac{28}{3}\sqrt{2\alpha^2} + \mathcal{O}(\alpha^3), \qquad (5.414)$$

$$x_{\epsilon_1=1}^{+} = \frac{3^{1/3}}{2^{1/6}|\alpha|^{2/3}} - \text{sign}(\alpha) \frac{2^{1/6}}{3^{1/3}|\alpha|^{1/3}} - \frac{\sqrt{2}}{3} + \mathcal{O}(\alpha^{1/3}), \qquad (5.415)$$

while the exact expressions can be found in the ASPIC library.

The slow-roll trajectory can be integrated explicitly and reads

$$N_{\text{end}} - N = \frac{\sqrt{3}}{2\alpha} \left[ \arctan \left( \frac{3 + 2\alpha x^2}{\sqrt{3}} \right) - \arctan \left( \frac{3 + 2\alpha x_{\text{end}}^2}{\sqrt{3}} \right) \right], \qquad (5.416)$$

and it can be inverted to get the field value as a function of the number of e-folds as

$$x = \sqrt{-\frac{3}{2\alpha} + \frac{\sqrt{3}}{2\alpha}} \tan \left[ \frac{2\alpha}{\sqrt{3}}(N_{\text{end}} - N) + \arctan \left( \frac{3 + 2\alpha x_{\text{end}}^2}{\sqrt{3}} \right) \right]. \qquad (5.417)$$

Since the first Hubble-flow parameter is below unity only over a finite field range of field values, the total number of inflationary e-folds that can be realized in SBK1, which we denote $\Delta N_{\text{max}}$, is bounded. It can be estimated by evaluating Eq. (5.416) with $x_{\text{end}} = x_{\epsilon_1=1}^{-}$ and $x_{\text{ini}} = x_{\epsilon_1=1}^{+}$. In practice, one must check that this number is large enough to solve the FLRW problems. Although it is possible to give an exact expression, for clarity, we here only report the result expanded at first order in $\alpha$. The condition $\Delta N_{\text{max}} > \Delta N_{\text{min}}$, where $\Delta N_{\text{min}}$ is the minimum number of e-folds required, translates into

$$-\frac{5\pi\sqrt{3}}{12\Delta N_{\min}} \lesssim \alpha \lesssim \frac{\pi\sqrt{3}}{12\Delta N_{\min}}. \qquad (5.418)$$

With $\Delta N_{\min} = 60$, this implies that $-0.038 < \alpha < 0.0075$, which is a more stringent condition than Eq. (5.412) and justifies, a posterior, the Taylor expansion in $\alpha$. The exact expression has been plotted in Fig. 50 as a function of $\alpha$ (assumed to be in the interval $|\alpha_{\min}, \alpha_{\max}|$). As expected from the above estimate, $\alpha$ should be small for $\Delta N_{\max}$ to be large.

Solving for the reheating equation (3.48), together with the field value $x_{\text{end}}$ at which inflation ends, Eq. (5.417) uniquely determines $x_*$, the field value at which the pivot mode crossed the Hubble radius during inflation. The CMB normalization fixes the overall scale of the potential to

$$\left(\frac{M}{M_{Pl}}\right)^4 = 2880\pi^2 \frac{\left[\frac{1}{x_*^2} + \alpha + \frac{\alpha^2}{3}x_*^2\right]^2}{\exp\left(\alpha x_*^2 + \frac{\alpha^2}{6}x_*^4\right)} \frac{Q_{\text{rms-PS}}^2}{T^2}. \qquad (5.419)$$

The reheating-consistent slow-roll predictions of the model are shown in Fig. 163. For $\alpha = 0$, one recovers the same predictions as LFI2, as already stressed. When $\alpha$ increases away from 0, the first Hubble-flow function, hence the tensor-to-scalar ratio $r$, increases, which makes the model
<figure><img src="image_53.png" /><figcaption>Fig. 50. Maximal total number of e-folds $\Delta N_{\text{max}}$ achievable within SBK1, from $x_{\epsilon_1=1}^+$ to $x_{\epsilon_1=1}^-$, as a function of the parameter $\alpha$. This one has to be small for inflation to last long enough.</figcaption></figure>

disfavored by the data. However, when $\alpha$ decreases away from 0, the tensor-to-scalar ratio decreases, making the model in better agreement with the data, before the spectral index becomes too blue.

The behavior displayed in these figures can also be analytically recovered by expanding both Eqs. (5.414) and (5.417) in $\alpha$. One obtains the approximate expression

$$x_* = 2\sqrt{\Delta N_*} + 2\Delta N_*^{3/2}\alpha + \mathcal{O}(\alpha^2), \qquad (5.420)$$

where we have also used that $\Delta N_* \gg 1$. Expanding Eqs. (5.408) in the same manner, one obtains the approximate expressions of the Hubble flow functions

$$\epsilon_{1*} \approx \frac{1}{2\Delta N_*} + 3\alpha, \quad \epsilon_{2*} \approx \frac{1}{\Delta N_*} - 6\alpha. \qquad (5.421)$$

As expected, the leading order in $\alpha$ gives back the predictions of LFI2, see Eq. (5.42).

## 5.27. Axion Hilltop Inflation (AHI)

### 5.27.1. Theoretical justifications

This model is a generalization of Natural Inflation (NI), see Section 5.6, where two oscillatory functions in the potential are considered [537]. A supergravity realization of this idea was proposed in Ref. [538]. It introduces an axion chiral superfield $\Phi$ having Kähler potential and superpotential respectively given by

$$K = \frac{\lambda^2}{2} (\Phi + \Phi^*)^2, \quad W = W_0 + Ae^{-a\Phi} + Be^{-b\Phi}. \qquad (5.422)$$

The scalar component of $\Phi$ is $\sigma+i\varphi$ where the axion $\varphi$ will play the role of the inflaton. The choice of the Kähler potential ensures a shift symmetry in the axion direction ($K$ is invariant under $\varphi \to \varphi + C$) and gives a mass to the saxon $\sigma$. This one is then stabilized and will remain a spectator field during inflation. Indeed, in Planck units, provided $|A| \ll |W_0|$, $|B| \ll |W_0|$, the shift symmetry is only weakly broken by the superpotential and one can show that the axion mass remains much smaller than the saxon mass. Therefore, taking the saxon at vanishing expectation value, the potential for the canonically normalized axion $\Phi = \sqrt{2\lambda\varphi}$ reads, up to an overall constant [538],

$$V(\phi) \approx 6|A||W_0| \left[ 1 - \cos\left(\frac{\phi}{\lambda_1}\right) \right] + 6|B||W_0| \left[ 1 - \cos\left(\frac{\phi}{\lambda_2} + \theta\right) \right] \\ - 2|A||B| \left( \frac{2}{\lambda_1\lambda_2} - 3 \right) \left[ 1 - \cos\left(\frac{\lambda_2 - \lambda_1}{\lambda_1\lambda_2}\phi - \theta\right) \right], \qquad (5.423)$$

where $B = |B|e^{i\theta}$, $\lambda_1 = \sqrt{2\lambda/a}$ and $\lambda_2 = \sqrt{2\lambda/b}$. Having a saxon more massive than the Hubble scale during inflation and forging a plateau-like shape for this potential requires some adjustment between the parameters. As discussed in Ref. [538], this amounts to fixing the axion value at which the potential is maximal at $\phi_{\text{max}} = \pi\lambda_1$, having $A/\lambda_1^2 = B/\lambda_2^2$ and fixing $\theta = -\pi\lambda_1/\lambda_2$. A first possible choice is to set $\lambda_1 = 2\lambda_2$, the leading terms in Eq. (5.423) reduce to $V \approx V_0 + C\phi^4$, which is a small-field inflation model with a power law index $p = 4$ (SFI4, see Section 6.1). The other choice considered in Ref. [538] is to consider the limit $\lambda_1 \to \lambda_2$, for which the leading terms of the potential simplify to

$$V = 6|A||W_0| \frac{\lambda_1 - \lambda_2}{\lambda_2} \left[ v_0 - 2 \cos \left( \frac{\phi}{\lambda_1} \right) + \left( \pi - \frac{\phi}{\lambda_1} \right) \sin \left( \frac{\phi}{\lambda_1} \right) \right] + \dots, \qquad (5.424)$$

where $v_0 \approx 4.82$ is a constant ensuring that the potential vanishes at its minimum. The remaining terms are, at most, of order $(\lambda_1 - \lambda_2)^2/\lambda_1^2$ and will be neglected in the following.
<figure><img src="image_54.png" /><figcaption>Fig. 51. Axion Hilltop Inflation. Top panels: the potential and its logarithm as a function of $\phi/f$. Bottom left panel: slow-roll parameter $\epsilon_1 f^2/M_{Pl}^2$. There are two inflationary domains corresponding to the field running away from the top of the potential. Because they are symmetrical, we only study the regime annotated with an arrow in which inflation proceeds at decreasing field value. Bottom right panel: slow-roll parameters $\epsilon_2 f^2/M_{Pl}^2$ (solid line) and $\epsilon_3 f^2/M_{Pl}^2$ (dotted line).</figcaption></figure>

<figure><img src="image_55.png" /><figcaption>Fig. 52. The dimensionless field value $x_{end} = \phi_{end}/f$ at which inflation ends as a function of $f/M_{Pl}$. For $f \ll M_{Pl}$, inflation is confined around the top of the potential ($x_{max} = \pi$).</figcaption></figure>

### 5.27.2. Slow-roll analysis

From the previous discussion, we can rewrite the potential of Axion Hilltop Inflation as

$$V(\phi) = M^4 \left[ v_0 - 2 \cos\left(\frac{\phi}{f}\right) + \left(\pi - \frac{\phi}{f}\right) \sin\left(\frac{\phi}{f}\right) \right], \qquad (5.425)$$

where we have introduced the parameter $f \equiv \lambda_1$ and $M^4 \equiv 6|A||W_0|(\lambda_1 - \lambda_2)/\lambda_2$. This is a periodic function and, in the following, the analysis is restricted to the region nearby the origin where the potential exhibits a plateau-like maximum.

As already mentioned, the constant $v_0$ is chosen such that the potential vanishes at its minimum. Therefore, the zeros of the potential are also solutions of $V'(x) = 0$, i.e.,

$$\tan(x) = x - \pi. \qquad (5.426)$$
where we have defined

$$x \equiv \frac{\phi}{f} \qquad (5.427)$$

There is one obvious solution to this equation at $x = \pi$, but the others have to be determined numerically. The geometrical interpretation is however clear. The solutions are the intersections between the line $y = x - \pi$ and the function $y = \tan(x)$ in the Euclidean plane $(x, y)$. Among the three solutions closest to the origin, two of them corresponds to the minima of the potential, $x_{V=0}^{\pm}$, the other, $x_{Vmax}$, is the field value at which the potential is maximal. Their numerical values read

$$x_{V=0}^{\mp} \approx -1.35, \quad x_{V=0}^{+} \approx 7.63, \quad x_{Vmax} = \pi. \qquad (5.428)$$

Let us notice that the actual values of $x_{V=0}^{\pm}$ uniquely determine the value of $v_0$. Enforcing $V(x_{V=0}^{\pm}) = 0$, one gets

$$v_0 = 2 \cos(x_{V=0}^{\pm}) + (x_{V=0}^{\pm} - \pi) \sin(x_{V=0}^{\pm}), \qquad (5.429)$$

which is numerically $v_0 \approx 4.82$, matching the value given above. Because $x_{V=0}^{-} < x_{Vmax} < x_{V=0}^{+}$, there are two inflationary domains. Either $x_{Vmax} < x < x_{Vmax}^{+}$ and inflation proceeds at increasing field value, or $x_{V=0}^{-} < x < x_{Vmax}$ and inflation proceeds at decreasing field values. However the potential is symmetric with respect to $x_{Vmax}$ and both of these regimes lead to identical observable predictions. For this reason, in the following, we focus the analysis on the first domain.

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

$$\epsilon_1 = \frac{M_{\mathrm{Pl}}^2}{2f^2} \left[ \frac{\sin(x) + (\pi - x) \cos(x)}{v_0 - 2 \cos(x) + (\pi - x) \sin(x)} \right]^2, \qquad (5.430)$$

while the second Hubble-flow function is given by

$$\epsilon_2 = \frac{M_{\mathrm{Pl}}^2}{f^2} \frac{1 + 2(\pi - x)^2 - \cos(2x) + 2v_0(\pi - x) \sin(x)}{[v_0 - 2 \cos(x) + (\pi - x) \sin(x)]^2}, \qquad (5.431)$$

and the third one by

$$\epsilon_3 = \frac{M_{\mathrm{Pl}}^2}{f^2} \frac{(\pi - x) \cos(x) + \sin(x)}{[v_0 - 2 \cos(x) + (\pi - x) \sin(x)]^2 [1 + 2(\pi - x)^2 - \cos(2x) + 2v_0(\pi - x) \sin(x)]} \times \left\{ 9v_0(\pi - x) - [8 + 2v_0^2 - 4(\pi - x)^2] (\pi - x) \cos(x) - v_0(\pi - x) \cos(2x) + [5 + 2v_0^2 + 8(\pi - x)^2] \sin(x) - v_0 [4 - (\pi - x)^2] \sin(2x) + \sin(3x) \right\}. \qquad (5.432)$$

They have been plotted, along with the potential and its logarithm, in Fig. 51. We have also represented the inflationary regime for $x < x_{Vmax}$ with an arrow.

Because the potential vanishes at its minimum, this guarantees that inflation gracefully ends at a field value for which $\epsilon(x) = 1$ in the domain $]x_{V=0}^{-}, x_{Vmax}[$, i.e., for which

$$[\sin(x) + (\pi - x) \cos(x)]^2 - \frac{2f^2}{M_{\mathrm{Pl}}^2} [v_0 - 2 \cos(x) + (\pi - x) \sin(x)]^2 = 0. \qquad (5.433)$$

This equation has to be solved numerically and we denote the solution $x_{end}$. Its dependence with respect to $f/M_{Pl}$ is been represented in Fig. 52.

The slow-roll trajectory cannot be integrated analytically and one has to numerically solve the following equation

$$N_{\mathrm{end}} - N = \frac{f^2}{M_{\mathrm{Pl}}^2} \int_{x_{\mathrm{end}}}^{x} \frac{v_0 - 2 \cos(y) + (\pi - y) \sin(y)}{\sin(y) + (\pi - y) \cos(y)} dy. \qquad (5.434)$$

The denominator of this equation vanishes at the top of the potential, for $x \to x_{Vmax} = \pi$. This ensures that a sufficient number of $e$-folds can always be made in that region. As can be seen in Fig. 52, for $f \ll M_{Pl}$ on has $x_{end} \to x_{Vmax}$ such that the inflationary domain is actually confined around the top of the potential. For $x \to x_{Vmax}$, one can derive an approximate expression for the trajectory by expanding both the numerator and denominator of Eq. (5.434). One gets

$$N_{\mathrm{end}} - N \approx \frac{6 + 3v_0}{2} \left[ \frac{1}{(x - \pi)^2} - \frac{1}{(x_{\mathrm{end}} - \pi)^2} \right], \qquad (5.435)$$

which can be inverted into

$$x \approx \pi - \frac{1}{\sqrt{\frac{1}{(x_{\mathrm{end}} - \pi)^2} + \frac{2}{6 + 3v_0} \frac{M_{\mathrm{Pl}}^2}{f^2} (N_{\mathrm{end}} - N)}}. \qquad (5.436)$$

The trajectory of Eq. (5.434) combined with the reheating equation (3.48), numerically determine $x_*$, the field value at which the pivot mode crossed the Hubble radius during inflation. The mass scale $M$ of the potential is then determined from the CMB normalization and one finds

$$\left(\frac{M}{M_{\mathrm{Pl}}}\right)^4 = 720\pi^2 \frac{M_{\mathrm{Pl}}^2}{f^2} \frac{[\sin(x_*) + (\pi - x_*) \cos(x_*)]^2}{[v_0 - 2 \cos(x_*) + (\pi - x_*) \sin(x_*)]^3} \frac{Q_{\mathrm{rms}-\mathrm{PS}}^2}{T^2}. \qquad (5.437)$$

The reheating consistent observable predictions have been represented in Fig. 164. For $f/M_{Pl} \ll 1$, $\epsilon_1$ becomes very small while $\epsilon_2$ reaches a constant value. It is possible to understand this limit by using the approximate trajectory of Eq. (5.436). This equation needs as an input $x_{end}$. In
<figure><img src="image_56.png" /><figcaption>Fig. 53. Pure Arctan Inflation. Top panels: the potential and its logarithm as a function of $\phi/\mu$. Bottom left panel: slow-roll parameter $\epsilon_1\mu^2/M_{Pl}^2$. Bottom right panel: slow-roll parameters $\epsilon_2\mu^2/M_{Pl}^2$ (solid line) and $\epsilon_3\mu^2/M_{Pl}^2$ (dotted line).</figcaption></figure>

the limit $f \ll M_{Pl}$, we have shown that $x_{end} \to x_{vmax} = \pi$ such that an approximate solution can be found by expanding Eq. (5.433) around $\pi$. One finds

$$x_{\text{end}} \approx \pi - \left[ \sqrt{2(3v_0 + 6)} \right]^{1/3} \left( \frac{f}{M_{\text{Pl}}} \right)^{1/3}, \qquad (5.438)$$

and $(x_{end} - \pi)^{-2} \propto (M_{Pl}/f)^{2/3}$ remains negligible in Eq. (5.436) for small-enough $f$. Therefore, one has

$$x_* \approx \pi - \sqrt{\frac{6 + 3v_0}{2\Delta N_*}} \frac{f}{M_{\text{Pl}}}, \qquad (5.439)$$

which gives, from Eqs. (5.430) and (5.431),

$$\epsilon_1 \approx \frac{6 + 3v_0}{16} \frac{f^4}{M_{\text{Pl}}^4 \Delta N_*^3}, \qquad \epsilon_2 \approx \frac{3}{\Delta N_*} - \frac{6 + 3v_0}{4} \frac{f^2}{M_{\text{Pl}}^2 \Delta N_*^4}. \qquad (5.440)$$

At fixed reheating history, the model therefore asymptotes to a constant spectral index with vanishing tensor-to-scalar ratio.

## 5.28. Pure Arctan Inflation (PAI)

This model has been proposed and discussed in Refs. [539,540] in the context of brane inflation within a five-dimensional bulk (see also Section 6.19). In this reference, it is argued that the interaction of bulk particles with a four-dimensional domain wall, assumed to be our universe, can trigger an accelerated expansion. From a four-dimensional point of view, inflation is driven by the so-called radion field, associated with the position of the wall in the fifth dimension, and the effective potential reads

$$V(\phi) = M^4 \arctan \left( \frac{\phi}{\mu} \right). \qquad (5.441)$$

The functional shape of this potential can be obtained from the one of Arctan Inflation (AI), see Section 5.19, by the transformation $\phi \to 1/\phi$. However, such a transformation on the field cannot be reabsorbed into a redefinition of some constants and Pure Arctan Inflation is a different model than Arctan Inflation. The potential is negative for $\phi < 0$ where it does not describe a physical situation in the context of brane inflation. We therefore restrict our analysis to the positive domain only.

The Hubble-flow functions associated with the potential (5.441), in the slow-roll approximation, are

$$\epsilon_1 = \frac{M_{\text{Pl}}^2}{2\mu^2} \frac{1}{\arctan^2(x) (1+x^2)^2}, \qquad \epsilon_2 = \frac{2M_{\text{Pl}}^2}{\mu^2} \frac{1+2x\arctan(x)}{\arctan^2(x) (1+x^2)^2}, \qquad (5.442)$$
and

$$ \epsilon_3 = \frac{2M_{\text{Pl}}^2}{\mu^2} \frac{1 + 3x \arctan(x) - (1 - 3x^2) \arctan^2(x)}{\arctan^2(x) (1 + x^2)^2 [1 + 2x \arctan(x)]} \quad (5.443) $$

where we have defined the dimensionless field

$$ x \equiv \frac{\phi}{\mu}. \qquad (5.444) $$

They have been represented, together with the potential, as a function of x in Fig. 53. The potential vanishes at x = 0 and this triggers a divergence of $\epsilon_1$ ensuring that inflation gracefully ends within the positive domain. Notice that, in the region close to the origin, one also has $\epsilon_2$ and $\epsilon_3$ larger than unity showing that slow-roll violations may also occur just before the end of inflation. The value of the inflaton field at which inflation ends, that we denote by $x_{\text{end}}$, is the positive root of $\epsilon_1(x) = 1$ and can be obtained by solving the following equation:

$$ \arctan(x) (1 + x^2) = \frac{M_{\text{Pl}}}{\sqrt{2\mu}}. \qquad (5.445) $$

There is not analytical solution to this equation, which has to be solved numerically. However, in the limits $\mu \ll M_{\text{Pl}}$ and $\mu \gg M_{\text{Pl}}$, the solution can be approximated as

$$ x_{\text{end}} \approx \begin{cases} \sqrt{\frac{2M_{\text{Pl}}}{\pi\mu}} & \text{if } \mu \ll M_{\text{Pl}} \\ \frac{M_{\text{Pl}}}{\sqrt{2\mu}} & \text{if } \mu \gg M_{\text{Pl}} \end{cases}. \qquad (5.446) $$

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

$$ N_{\text{end}} - N = \frac{\mu^2}{6M_{\text{Pl}}^2} \left[ 2x(x^2 + 3) \arctan(x) - 2x_{\text{end}}(x_{\text{end}}^2 + 3) \arctan(x_{\text{end}}) + (x_{\text{end}}^2 - x^2) + 2 \ln \left( \frac{1 + x_{\text{end}}^2}{1 + x^2} \right) \right]. \qquad (5.447) $$

This trajectory, together with the reheating equation (3.48) and $x_{\text{end}}$ from Eq. (5.445), allow us to determine the field value $x_*$ at which the pivot mode crossed the Hubble radius during inflation. This also fixes the energy scale of the potential by the CMB normalization and one gets

$$ \left(\frac{M}{M_{\text{Pl}}}\right)^4 = 720\pi^2 \frac{M_{\text{Pl}}^2}{\mu^2} \frac{1}{\arctan^3(x_*) (1 + x_*^2)^2} \frac{Q_{\text{rms-PS}}^2}{T^2}. \qquad (5.448) $$

The reheating consistent slow-roll prediction for Pure Arctan inflation have been represented in Fig. 165. The two regimes $\mu \ll M_{\text{Pl}}$ and $\mu \gg M_{\text{Pl}}$ can be understood as follows. In these two limits, the trajectory (5.447) can be inverted as

$$ x_* \approx \begin{cases} \left( x_{\text{end}}^3 + \frac{6\Delta N_*}{\pi} \frac{M_{\text{Pl}}^2}{\mu^2} \right)^{1/3} & \mu \ll M_{\text{Pl}} \\ \sqrt{x_{\text{end}}^2 + 2\Delta N_* \frac{M_{\text{Pl}}^2}{\mu^2}} & \mu \gg M_{\text{Pl}} \end{cases}. \qquad (5.449) $$

Using the approximate expressions for $x_{\text{end}}$ given in Eq. (5.446), and plugging the resulting expressions of $x_*$ into Eqs. (5.442) and (5.443), properly expanded in the relevant limit for $\mu$, one finally gets

$$ \epsilon_{1*} \approx 2 \left( \frac{\mu}{\pi M_{\text{Pl}}} \right)^{2/3} \frac{1}{(6\Delta N_*)^{4/3}}, \quad \epsilon_{2*} \approx \frac{4}{3\Delta N_*}, \quad \epsilon_{3*} \approx \frac{1}{\Delta N_*} \quad \text{if } \mu \ll M_{\text{Pl}}, \qquad (5.450) $$

$$ \epsilon_{1*} \approx \frac{8\mu^2}{\pi^2 (1 + 4\Delta N_*)^2 M_{\text{Pl}}^2}, \quad \epsilon_{2*} \approx \frac{16\sqrt{2}\mu}{(1 + 4\Delta N_*)^{3/2} \pi M_{\text{Pl}}}, \quad \epsilon_{3*} \approx \frac{3\epsilon_{2*}}{4} \quad \text{if } \mu \gg M_{\text{Pl}}. \qquad (5.451) $$

One can see that, in the $\mu \ll M_{\text{Pl}}$ limit, $\epsilon_1$ becomes very small while $\epsilon_2$ remains constant at fixed $\Delta N_*$, while in the limit $\mu \gg M_{\text{Pl}}$, the first three slow-roll parameters become large, and the model is excluded.

## 5.29. Superconformal $\alpha$-Attractor A Inflation (SAAI)

### 5.29.1. Theoretical justifications

The model is based on the vector multiplet Lagrangian introduced in (4.22) which depends on one arbitrary function, $J(C) = 3/2 \ln \Phi$. In Section 4.1, it was shown that the choice given by (4.23), namely $\Phi = -Ce^C$, leads to the Starobinsky model. In Refs. [254,541], an extension based on the choice

$$ \Phi(C) = (-C)^a e^{\beta C}, \qquad (5.452) $$

where $\alpha$ and $\beta$ are two new free parameters, was considered. It leads to the so-called “$\alpha - \beta$ model”. Using (4.22), it is easy to show that

$$ V(C) = \frac{9}{8} g^2 \left( \beta + \frac{\alpha}{C} \right)^2, \qquad (5.453) $$
<figure><img src="image_57.png" /><figcaption>Fig. 54. Superconformal α-attractor A Inflation (SAAI). Top left panel: the potential as a function of $\phi/M_g$. Top right panel: logarithm of the potential. Bottom left panel: the first slow-roll parameter $\epsilon_1$. Bottom right panel: slow-roll parameters $\epsilon_2$ (solid line) and $\epsilon_3$ (dotted line).</figcaption></figure>

where the field C is not canonically normalized. In terms of the canonically normalized field $\phi$, $C = -\exp[\sqrt{2/(3\alpha)}\phi/M_g]$, the potential acquires the following form

$$V(\phi) = \frac{9}{8}g^2 \left( \beta - \alpha e^{\sqrt{\frac{2}{3\alpha}} \frac{\phi}{M_g}} \right)^2. \qquad (5.454)$$

Writing $M^4 = 9g^2\beta^2/8$ and shifting the field $\phi - \phi_0 \to \phi$, where $\exp(-\sqrt{\frac{2}{3\alpha}}\frac{\phi_0}{M_g}) = \alpha/\beta$, one arrives at the potential (5.455). It is also interesting to notice that the potential (5.455) interpolates between the Starobinsky model ($\alpha = 1$) and the quadratic LFI model. Indeed, when $\alpha \to +\infty$, one has $V \sim 2M^4/(3\alpha)(\phi/M_g)^2$.

Let us finally notice that the above potential can also be obtained in the context of the models discussed in Section 6.29.

### 5.29.2. Slow-roll analysis

The potential of α-attractor models can be written as

$$V(\phi) = M^4 \left( 1 - e^{-\sqrt{\frac{2}{3\alpha}} \frac{\phi}{M_g}} \right)^2. \qquad (5.455)$$

It clearly bears resemblance with the Starobinsky/Higgs models. In fact, if $\alpha = 1$, it exactly reduces to these models. It can thus be seen as a generalization of these scenarios. The potential (5.455) is represented in Fig. 54 for different values of $\alpha$.

The three Hubble flow functions can be easily calculated and, defining $x \equiv \phi/M_g$, one obtains

$$\epsilon_1 = \frac{4}{3\alpha} \left( 1 - e^{\sqrt{\frac{2}{3\alpha}} x} \right)^{-2}, \quad \epsilon_2 = \frac{2}{3\alpha} \left[ \sinh \left( \frac{x}{\sqrt{6\alpha}} \right) \right]^{-2}, \qquad (5.456)$$

$$\epsilon_3 = \frac{2}{3\alpha} \left[ \coth \left( \frac{x}{\sqrt{6\alpha}} \right) - 1 \right] \coth \left( \frac{x}{\sqrt{6\alpha}} \right). \qquad (5.457)$$

Evidently, when $\alpha = 1$, these expressions reduce to Eqs. (4.48).

In this scenario, inflation ends by violation of the slow-roll conditions. Inflation stops when $x = x_{end}$ with

$$x_{end} = \sqrt{\frac{3\alpha}{2}} \ln \left( 1 + \frac{2}{\sqrt{3\alpha}} \right). \qquad (5.458)$$
However, as it was the case for Higgs inflation, see Section 4.2, violation of the slow-roll conditions can occur before. Indeed, one has $\epsilon_2 = 1$ for

$$x_{\epsilon_2=1} = \sqrt{6\alpha} \operatorname{arcsinh}\left(\sqrt{\frac{2}{3\alpha}}\right), \qquad (5.459)$$

and $\epsilon_3 = 1$ if

$$x_{\epsilon_3=1} = \sqrt{6\alpha} \operatorname{arctanh}\left(\frac{2}{1+\sqrt{1+6\alpha}}\right). \qquad (5.460)$$

In fact as inflation proceeds, the field reaches first the value $x_{\epsilon_2=1}$, then $x_{\epsilon_3=1}$ and, finally, $x_{\text{end}}$. It is interesting to notice that this hierarchy does not depend on $\alpha$.

The next step consists in calculating the slow-roll trajectory. Straightforward manipulations lead to the following expression

$$N_{\text{end}} - N = \frac{1}{2} \sqrt{\frac{3\alpha}{2}} (x_{\text{end}} - x) + \frac{3\alpha}{4} \left( e^{\sqrt{\frac{2}{3\alpha}}x} - e^{\sqrt{\frac{2}{3\alpha}}x_{\text{end}}} \right). \qquad (5.461)$$

Of course, one can check that, for $\alpha = 1$, the above formula reduces to the trajectory found in the Higgs scenario. For large values of $x$, $x \gg 1$, the last term is dominant. This trajectory can be inverted and expressed in term of the “-1-branch” of the Lambert function $W_{-1}$. One obtains

$$x = \sqrt{\frac{3\alpha}{2}} \left\{ -\frac{4}{3\alpha}\Delta N + \sqrt{\frac{2}{3\alpha}}x_{\text{end}} - e^{\sqrt{\frac{2}{3\alpha}}x_{\text{end}}} - W_{-1} \left[ -\exp \left( -\frac{4}{3\alpha}\Delta N + \sqrt{\frac{2}{3\alpha}}x_{\text{end}} - e^{\sqrt{\frac{2}{3\alpha}}x_{\text{end}}} \right) \right] \right\}, \qquad (5.462)$$

where, as usual, $\Delta N = N_{\text{end}} - N$. The reason that inflation proceeds along the -1 branch of the Lambert function can be understood by means of arguments similar to those presented in Section 4.2. The Lambert function in Eq. (5.462) can be written as $W_{-1}\{\exp[-4\Delta N/(3\alpha)]ye^y\}$ with $y = -\exp[\sqrt{2/(3\alpha)}x_{\text{end}}]$. At the end of inflation, by definition, one has $\Delta N = 0$. Using the property that $W_{-1}(ye^y) = -e^y$ (if $y < -1$), one has therefore that the Lambert function equals $-\exp(\sqrt{2/(3\alpha)}x_{\text{end}})$, which is smaller than -1; and, as can be seen in Fig. 6, a value of the Lambert function smaller than -1 necessarily corresponds to the -1 branch. The previous argument is valid at the end of inflation only. On more general grounds, for N's before the end of inflation, $\Delta N > 0$ becomes large and, therefore, the argument of the Lambert function becomes small. In order, for $x$ to say positive in Eq. (5.462), the Lambert function must be large and negative in this limit. This immediately implies that the branch -1 is the relevant one.

Finally, the value of $x_*$, at which the pivot mode crossed out the Hubble radius during inflation can be expressed as

$$x_* = \sqrt{\frac{3\alpha}{2}} \left( -\frac{4}{3\alpha}\Delta N_* + \ln\left(1 + \frac{2}{\sqrt{3\alpha}}\right) - \left(1 + \frac{2}{\sqrt{3\alpha}}\right) - W_{-1} \left\{ -\exp\left[-\frac{4}{3\alpha}\Delta N_* + \ln\left(1 + \frac{2}{\sqrt{3\alpha}}\right) - \left(1 + \frac{2}{\sqrt{3\alpha}}\right)\right] \right\} \right). \qquad (5.463)$$

where, in this expression, we have used the value of $x_{\text{end}}$ derived above. From the knowledge of $x_*$, the energy scale $M$ of the potential can be inferred and one obtains

$$\frac{M^4}{M_g^4} = 1920 \frac{\pi^2}{\alpha} \left(1 - e^{\sqrt{\frac{2}{3\alpha}}x_*}\right)^{-4} e^2 \sqrt{\frac{2}{3\alpha}} x_* \frac{Q_{\text{rms}-\text{PS}}^2}{T^2}. \qquad (5.464)$$

The reheating consistent slow-roll prediction for Superconformal $\alpha$-attractor A Inflation have been represented in Fig. 166.

## 5.30. T-Model Inflation (TMI)

### 5.30.1. Theoretical justifications

The theoretical motivations underlying these scenarios, which were named “T-models” in Ref. [262], find their roots in the superconformal context already presented for Starobinsky Inflation (SI) in Section 4.1.2. As discussed in that section, the corresponding scenarios have some attractive features. In particular, they are conformally and SO(1, 1) invariant. If the conformal invariance is broken in the rapidity gauge, $\chi^2 - \phi^2 = 6$, it was shown that the model reduces to the standard action of a scalar field, minimally coupled to gravity, with a constant potential, see Eq. (4.39). The corresponding solution is de Sitter spacetime, thus showing a connection between the action of Eq. (4.35) and the theory of inflation. Let us add that the way conformal invariance is broken is a choice and other instances are possible, but leading to the same conclusions. For instance, the conformal gauge $\chi = \sqrt{6}$ is a good illustration of the above claim since it is particularly simple. Indeed, in that case, the action (4.35) takes the form

$$S(g_{\mu\nu}, \phi) = \frac{M_g^2}{2} \int d^4 x \sqrt{-g} \left[ \left(1 - \frac{\phi^2}{6}\right) R - g^{\mu\nu} \partial_\mu \phi \partial_\nu \phi - \frac{\lambda}{2} (\phi^2 - 6)^2 \right]. \qquad (5.465)$$

Using the notations presented in Section 4.2.1, while dropping the “bar” over Jordan frame quantities here, we see that this is the action of a scalar tensor theory with $F(\phi) = 1 - \phi^2/6$, $Z(\phi) = 1$ and $U = \lambda(\phi^2 - 6)^2/4$, see Eq. (4.72). Expressed in the Einstein frame, the action exactly reduces to Eq. (4.39) since the potential reads $V = M_g^2 U / F^2 = 9\lambda M_g^2$. Therefore, this confirms that the final result is the same regardless of the gauge is used to break conformal invariance.

The above preliminary considerations help us to understand the context in which the TMI model is designed. Indeed, let us now consider the following action which is a generalization of the Eq. (4.35),

$$S(g_{\mu\nu}, \chi, \phi) = \frac{M_g^2}{2} \int d^4 x \sqrt{-g} \left[ \chi^2 R + g^{\mu\nu} \partial_\mu \chi \partial_\nu \chi - \frac{\phi^2}{6} R - g^{\mu\nu} \partial_\mu \phi \partial_\nu \phi \right]$$
$$ -\frac{1}{36} F_{\text{T}} \left(\frac{\phi}{\chi}\right) (\phi^2 - \chi^2)^2], \qquad (5.466) $$

where $F_T(\cdot)$, the new ingredient of the model, is a priori an arbitrary function. Given that $F_T = 1$ leads to de Sitter, deviations from this case should lead to non-trivial models of inflation. It is also important to notice that the action (5.466) is still conformally invariant, thanks to the dependence of $F_T(\cdot)$ in $\phi/\chi$. However, the symmetry SO(1, 1) is broken, unless $F_T$ is a constant which was of course precisely the case in Eq. (4.35). Notice also that, in Section 4.1, it was shown that the above model with a term $\sim \phi^2(\phi - \chi)^2$, or, equivalently, $F_T \sim (\phi/\chi)^2/(1+\phi/\chi)^2$ leads to the Starobinsky model. This important model is therefore included in the class of models studied in this section.

Then, in order to proceed, one needs to choose a gauge. Taking $\chi = \sqrt{6}$, Eq. (5.466) takes the form

$$ S(g_{\mu\nu}, \chi, \phi) = \frac{M_g^2}{2} \int d^4\mathbf{x} \sqrt{-g} \left[ \left(1 - \frac{\phi^2}{6}\right) R - g^{\mu\nu} \partial_\mu \phi \partial_\nu \phi - \frac{1}{36} F_T \left(\frac{\phi}{\sqrt{6}}\right) (\phi^2 - 6)^2 \right]. \qquad (5.467) $$

Again, one recognizes a scalar tensor theory with $F(\phi) = 1 - \phi^2/6$, $Z(\phi) = 1$ and $2U = F_T(\phi/\sqrt{6})(\phi^2/6 - 1)^2$. The potential, $V = M_g^2 U / F^2$, is then given by $V = M_g^2 F_T(\phi/\sqrt{6})/2$. In addition, the relationship between the fields $\phi$ and $\tilde{\phi}$ is given by Eq. (4.74) and leads to

$$ \frac{\tilde{\phi}}{M_g} = -\sqrt{\frac{3}{2}} \ln \left( \frac{\sqrt{6} - \phi}{\sqrt{6} + \phi} \right). \qquad (5.468) $$

This relation can be inverted and one arrives at

$$ \phi = \sqrt{6} \tanh \left( \frac{\tilde{\phi}}{M_g \sqrt{6}} \right). \qquad (5.469) $$

As a consequence, in the Einstein frame, the action takes the form

$$ S(\tilde{g}_{\mu\nu}, \tilde{\phi}) = \int d^4\mathbf{x} \sqrt{-\tilde{g}} \left\{ \frac{M_g^2}{2} \tilde{R} - \frac{1}{2} \tilde{g}^{\mu\nu} \partial_\mu \tilde{\phi} \partial_\nu \tilde{\phi} - \frac{M_g^2}{2} F_T \left[ \tanh \left( \frac{\tilde{\phi}}{\sqrt{6} M_g} \right) \right] \right\}. \qquad (5.470) $$

One can also use the rapidity condition $\chi^2 - \phi^2 = 6$. Then, using the parametrization defined in Eq. (4.38) for this gauge, one can check that the action (5.470) is directly recovered.

The next question is of course which function $F_T(\cdot)$ should be chosen? In absence of any deep reasons, Ref. [262] argues that a reasonable and interesting choice is simply to take a power-law, namely

$$ F_T \left( \frac{\phi}{\chi} \right) = \lambda \left( \frac{\phi}{\chi} \right)^{2n}. \qquad (5.471) $$

This leads to the potential studied in the next section. One remarks that for large values of the field, $F_T\{\tanh[\tilde{\phi}/(\sqrt{6}M_g)]\}$ tends to a constant and, in this regime, the SO(1, 1) symmetry is restored.

In Ref. [262], the model is also discussed in the framework of conformal supergravity, the action of which was given by Eq. (4.41). In that case, the embedding Kähler potential can be chosen as

$$ \mathcal{N}(X^I, \tilde{X}^I) = -|X^0|^2 + |X^1|^2 + |S|^2 - 3\zeta \frac{(S\tilde{S})^2}{|X^0|^2 - |X^1|^2}. \qquad (5.472) $$

Here, $X^0$ is a conformon, $X^1 = \Phi$ is the inflaton and $X^3 = S$ is a Goldstino and $\zeta$ is a dimensionless parameter. The above Kähler potential has a SU(1,1) symmetry between the conformon and the inflaton. As it was the case in Section 4.1, the last term, proportional to the parameter $\zeta$, is introduced to stabilize the inflationary trajectory. The superpotential is taken to be

$$ \mathcal{W} = S \left[ (X^0)^2 - (X^1)^2 \right] f \left( \frac{X^1}{X^0} \right). \qquad (5.473) $$

where, at this stage, the function $f$ is arbitrary. It is interesting to emphasize the difference with the superconformal model of Section 4.1. In that section, the embedding and superpotential were both given in terms of an exponential of the fields, leading to a power-law Kähler potential while, here, the embedding potential is expressed directly in terms of powers of the fields and, as a consequence, the corresponding Kähler potential will be given by a logarithm, a structure reminiscent of no scale supergravity.

Then, as usual, we fix the conformon field and choose the gauge where $X^0 = \tilde{X}^0 = \sqrt{3}M_g$. As a consequence, looking at Eqs. (4.42), the above choices imply that a (ordinary) supergravity description of the model can be expressed, as announced above, in terms of the following logarithmic Kähler potential

$$ K = -3M_g^2 \ln(1+k), \qquad (5.474) $$

where $k$ is a function given by

$$ k = -\frac{\Phi \tilde{\Phi}}{3 M_g^2} - \frac{S \tilde{S}}{3 M_g^2} + \frac{\zeta}{M_g^2} \frac{S^2 \tilde{S}^2}{3 M_g^2 - \Phi \tilde{\Phi}}. \qquad (5.475) $$

The exact scalar potential corresponding to the theory we have just described is quite complicated. Indeed, as already mentioned several times, in general, it can be written as

$$ V = M_g^4 e^G (G^{\tilde{B}C} G_{\tilde{B}} G_C - 3), \qquad (5.476) $$
with

$$G = \frac{K}{M_g^2} + \ln \left( \frac{W W^\dagger}{M_g^6} \right). \qquad (5.477)$$

Expanding this formula, one obtains the following complicated expression

$$V = \frac{e^{K/M_g^2}}{M_g^2} G^{\bar{B}C} \left( \frac{W W^\dagger}{M_g^4} \frac{\partial K}{\partial \bar{X}^\bar{B}} \frac{\partial K}{\partial X^C} + \frac{W^\dagger}{M_g^2} \frac{\partial K}{\partial \bar{X}^\bar{B}} \frac{\partial W}{\partial X^C} + \frac{W}{M_g^2} \frac{\partial W^\dagger}{\partial \bar{X}^\bar{B}} \frac{\partial K}{\partial X^C} \right. \\ \left. + \frac{\partial W^\dagger}{\partial \bar{X}^\bar{B}} \frac{\partial W}{\partial X^C} - 3 W W^\dagger \right). \qquad (5.478)$$

However, a crucial aspect of this scenario is that the superpotential, after symmetry breaking, can be expressed as

$$W = S(3M_g^2 - \Phi^2) f \left( \frac{\Phi}{\sqrt{3} M_g} \right), \qquad (5.479)$$

and we notice that, as it was the case in Section 4.1, this superpotential is again of the form $W = Sf(\Phi)$. Moreover, a fundamental remark made in Ref. [262] is that inflation takes place at $S = 0$ and in the direction where $\Phi$ is real. This greatly simplifies the calculations. Indeed, the simplification comes from the fact that, if $S = 0$, then $W = 0$ and the previous expression for the scalar potential can be reduced to a compact formula, namely

$$V = \frac{e^{K/M_g^2}}{M_g^2} G^{\bar{B}C} \frac{\partial W^\dagger}{\partial \bar{X}^\bar{B}} \frac{\partial W}{\partial X^C} = \frac{e^{K/M_g^2}}{M_g^2} G^{\bar{S}S} \frac{\partial W^\dagger}{\partial \bar{S}} \frac{\partial W}{\partial S}. \qquad (5.480)$$

To go further, we must now calculate the Kähler matrix. Using Eq. (5.474), one obtains that

$$G_{A\bar{B}} = -\frac{3}{1+k} \frac{\partial^2 k}{\partial X^A \partial \bar{X}^B} + \frac{3}{(1+k)^2} \frac{\partial k}{\partial X^A} \frac{\partial k}{\partial \bar{X}^B}. \qquad (5.481)$$

It follows that, along the inflationary direction, the Kähler matrix can be simplified as

$$G_{A\bar{B}} = \frac{1}{M_g^2(1+k)^2} \begin{pmatrix} 1 & 0 \\ 0 & 1+k \end{pmatrix}. \qquad (5.482)$$

Given that, along the inflationary direction, $k = -\Phi^2/(3M_g^2)$, the above form of the Kähler matrix implies that the canonically normalized field $\varphi$ is related to $\Phi$ by the following formula

$$\Phi = \sqrt{6} M_g \tanh \left( \frac{\varphi}{\sqrt{6} M_g} \right). \qquad (5.483)$$

Finally, by noticing that $\partial W / \partial S = 3 M_g^2 (1+k) f$, it follows that the potential of the canonically normalized field $\varphi$ reads

$$V(\varphi) = 9 M_g^4 \left| f \left[ \tanh \left( \frac{\varphi}{\sqrt{6} M_g} \right) \right] \right|^2. \qquad (5.484)$$

As a consequence, the T-model can be reproduced by choosing the function $f$ appropriately, namely proportional to $F_T$ in Eq. (5.470).

### 5.30.2. Slow-roll analysis

Going back to our usual notation for the inflaton, we now denote by $\varphi$ the canonically normalized field (noted $\varphi$ in the previous section, and not be confused with the Jordan frame field of Section 5.30.1).

The potential of the TMI can therefore be written as [541]

$$V(\varphi) = M_g^4 \left[ \tanh \left( \frac{\varphi}{\sqrt{6} M_g} \right) \right]^{2n}, \qquad (5.485)$$

and describes a one parameter model, the parameter being $n$.

From the above potential, we can obtain the Hubble flow functions. Defining $x \equiv \varphi/M_g$, one gets

$$\epsilon_1 = \frac{4n^2}{3} \sinh^{-2} \left( \frac{2x}{\sqrt{6}} \right), \quad \epsilon_2 = \frac{8n}{3} \frac{\cosh \left( \frac{2x}{\sqrt{6}} \right)}{\sinh^2 \left( \frac{2x}{\sqrt{6}} \right)}, \quad \epsilon_3 = \frac{3 + \cosh \left( \frac{4x}{\sqrt{6}} \right)}{\sinh^2 \left( \frac{2x}{\sqrt{6}} \right) \cosh \left( \frac{2x}{\sqrt{6}} \right)}. \qquad (5.486)$$

The potential and the Hubble-flow functions have been plotted in Fig. 55.

In this scenario, inflation stops by violation of the slow-roll conditions. This happens when $\epsilon_1 = 1$ corresponding to following vacuum expectation value of the field

$$x_{\text{end}} = \frac{\sqrt{6}}{2} \arcsin h \left( \frac{2n}{\sqrt{3}} \right). \qquad (5.487)$$
<figure><img src="image_58.png" /><figcaption>Fig. 55. T-Model Inflation (TMI). Top left panel: the potential as a function of $\phi/M_g$. Top right panel: logarithm of the potential. Bottom left panel: the first slow-roll parameter $\epsilon_1$. Bottom right panel: slow-roll parameters $\epsilon_2$ (solid line) and $\epsilon_3$ (dotted line).</figcaption></figure>

The slow-roll trajectory can be integrated and one gets

$$N_{\text{end}} - N = \frac{3}{4n} \left[ \cosh\left(\frac{2x}{\sqrt{6}}\right) - \cosh\left(\frac{2x_{\text{end}}}{\sqrt{6}}\right) \right]. \qquad (5.488)$$

This trajectory can be inverted leading to an explicit formula for $\phi/M_g$ during slow-roll inflation

$$x = \frac{\sqrt{6}}{2} \operatorname{arccosh} \left( \sqrt{1 + \frac{4n^2}{3}} + \frac{4n}{3} \Delta N \right), \qquad (5.489)$$

where $\Delta N = N_{\text{end}} - N$. The value of $x_*$, namely the value of the field when the pivot scale crosses out the Hubble radius during inflation is given just given by the above expression with $\Delta N = \Delta N_*$. Finally, the mass scale $M$ that normalizes the potential can be expressed as

$$\frac{M_*^4}{M_g^4} = \frac{1920\pi^2 n^2}{\sinh^2\left(\frac{2x_*}{\sqrt{6}}\right) \left[\tanh\left(\frac{x_*}{\sqrt{6}}\right)\right]^{2n}} \frac{Q_{\text{rms-PS}}^2}{T^2}. \qquad (5.490)$$

The reheating consistent observable predictions for TMI have been represented in Fig. 167 for various values of $n$. One notices that the dependence on $n$ of the spectral index and of the tensor-to-scalar ratio is very small. Indeed, provided the quantity $n\Delta N$ dominates in Eq. (5.489), one has

$$x_* \approx \frac{\sqrt{6}}{2} \operatorname{arccosh} \left( \frac{4n}{3} \Delta N_* \right). \qquad (5.491)$$

Plugging this approximation into Eq. (5.486) gives

$$\epsilon_{1*} \approx \frac{3}{4\Delta N_*^2}, \quad \epsilon_{2*} \approx \frac{2}{\Delta N_*}, \quad \epsilon_{3*} \approx \frac{1}{\Delta N_*}, \qquad (5.492)$$

and the Hubble-flow functions are independent of $n$ in the large $\Delta N_*$ limit.

# 6. Two parameters models

## 6.1. Small Field Inflation (SFI)

This model is proto-typical of inflation occurring at the top of a flat-enough potential. As such it appears in very different contexts. It has been introduced in Ref. [31,460] and derived in Ref. [32] in the context of radiatively induced symmetry breaking. It appears within superstring
<figure><img src="image_59.png" /><figcaption>Fig. 56. Small Field Inflation (SFI) for $p = 4$ and $\mu = M_{\text{Pl}}$. Upper panels: the potential and its logarithm as a function of $\phi/\mu$. 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).</figcaption></figure>

models [542], low scale symmetry breaking [332,543], supersymmetry [418,544] and supergravity [311,312,316,331,545–549]. It is also obtained in non-linear sigma models [342] or using moduli as inflatons [550]. It has been discussed in braneworld cosmology in Refs. [551–553] and is more recently referred to as “hilltop inflation” from Ref. [474,475]. The potential is given by

$$V(\phi) = M^4 \left[ 1 - \left( \frac{\phi}{\mu} \right)^p \right], \qquad (6.1)$$

and has two parameters in addition to the overall normalization $M$: a typical $\text{vev } \mu$ and the power index $p$. As this potential can be associated with very different physical frameworks, $\mu$ can take any values while $p > 0$ for being at the top of a potential (in the small field limit, namely $\phi \ll \mu$). In particular, we will allow super-Planckian values for $\mu$ even though, in the supergravity context, one would require $\mu < M_{\text{Pl}}$. Let us stress that Eq. (6.1) is defined only in the domain $\phi < \mu$ as one assumes that the small field potential describes only the field dynamics during inflation. The equation of state during reheating is thus not specified by Eq. (6.1). Defining

$$x \equiv \frac{\phi}{\mu}, \qquad (6.2)$$

the first three Hubble flow functions read

$$\epsilon_1 = \frac{p^2}{2} \left( \frac{M_{\text{Pl}}}{\mu} \right)^2 \frac{x^{2p-2}}{(1-x^p)^2}, \qquad \epsilon_2 = 2p \left( \frac{M_{\text{Pl}}}{\mu} \right)^2 \frac{p-1+x^p}{(1-x^p)^2}, \qquad (6.3)$$

and

$$\epsilon_3 = p \left( \frac{M_{\text{Pl}}}{\mu} \right)^2 \frac{x^{p-2} [2x^{2p} + (p-1)(p+4)x^p + (p-1)(p-2)]}{(1-x^p)^2 (p-1+x^p)}. \qquad (6.4)$$

They are monotonic functions of the field value but also decreasing functions of the $\text{vev } \mu$. The potential, its logarithm and the Hubble flow functions are represented in Fig. 56.

The slow-roll trajectory is obtained by integrating Eq. (3.11) to get

$$N - N_{\text{end}} = \frac{1}{2p} \frac{\mu^2}{M_{\text{Pl}}^2} \left[ -x^2 + x_{\text{end}}^2 + \frac{2}{2-p} (x^{2-p} - x_{\text{end}}^{2-p}) \right]. \qquad (6.5)$$

This equation seems to be well-defined only for $p \neq 2$. However, the particular case $p = 2$ can be directly obtained from Eqs. (3.11) and (6.1) to get

$$N - N_{\text{end}} = \frac{1}{4} \frac{\mu^2}{M_{\text{Pl}}^2} \left[ -x^2 + x_{\text{end}}^2 + 2 \ln \left( \frac{x}{x_{\text{end}}} \right) \right]. \qquad (6.6)$$
This expression can also be viewed as the limit of Eq. (6.5) for $p \to 2$. In general, the trajectory cannot be analytically inverted to give the field value $x(N)$ but one can find some analytic form for almost all integer values of $p$ (e.g. for $p=1, p=2, p=3, p=4, p=6$) that we do not write down for the sake of clarity.

From the potential Eq. (6.1), inflation can stop naturally at $\epsilon_1(x_{end}) = 1$ with $x_{end} < 1$. This condition gives the algebraic equation

$$x_{\text{end}}^p + \frac{p}{\sqrt{2}} \frac{M_{\text{Pl}}}{\mu} x_{\text{end}}^{p-1} = 1, \qquad (6.7)$$

which cannot be solved analytically in full generality. As for the trajectory, there are however explicit solutions for almost all integer values of $p$, the first two being

$$x_{\text{end}}^{(p=1)} = 1 - \frac{M_{\text{Pl}}}{\sqrt{2}\mu}, \qquad x_{\text{end}}^{(p=2)} = \frac{M_{\text{Pl}}}{\sqrt{2}\mu} \left( -1 + \sqrt{1 + 2\frac{\mu^2}{M_{\text{Pl}}^2}} \right). \qquad (6.8)$$

Together with Eq. (3.48), these equations are enough to allow the determination of the field value $x_*$ at which the observable modes crossed the Hubble radius during inflation. This fixes the value of the parameter $M$ to match the observed amplitude of the CMB anisotropies at

$$\frac{M^4}{M_{\text{Pl}}^4} = 720\pi^2 p^2 \frac{M_{\text{Pl}}^2}{\mu^2} \frac{x_*^{2p-2}}{(1-x_*^p)^3} \frac{Q_{\text{rms}-\text{PS}}^2}{T^2}. \qquad (6.9)$$

The reheating consistent slow-roll predictions for the small field models are represented in Figs. 168 to 170 for $p=1, p=2$ and $p=4$. The $p=1$ case is trivial since one then has $\epsilon_{2*} = 4\epsilon_{1*}$. For $p=2$ or $p=4$, one sees that the reheating temperature is limited from below to fit in the observable range. For instance, with $p=2$, values of $\mu$ such that $\mu/M_{\text{Pl}} < 10$ are clearly disfavored. Let us notice that the relation $\epsilon_{2*} = 4\epsilon_{1*}$ is recovered in the limit $\mu/M_{\text{Pl}} \gg 1$ whereas one clearly observes a systematic shift in $n_s$ (or $\epsilon_2$) when $\mu \ll M_{\text{Pl}}$. These behaviors can in fact be understood analytically.

Small field models in the supergravity context are commonly studied in the limit $\mu \ll M_{\text{Pl}}$. In this situation it is possible to find some approximate solution to both the trajectory and $x_{end}$. Keeping only the dominant term in Eq. (6.7), one gets

$$x_{\text{end}}^{(p \neq 1)} \approx \left( \frac{\sqrt{2}}{p} \frac{\mu}{M_{\text{Pl}}} \right)^{1/(p-1)}, \qquad (6.10)$$

the case $p \le 1$ being incompatible with the limit $\mu \ll M_{\text{Pl}}$ and the consistency requirement that $x_{end} < 1$. The small $\text{vev}$ limit can also be used to invert Eq. (6.5). Assuming $\mu \ll M_{\text{Pl}}$ and $x_{end} \ll 1$, neglecting the quadratic terms for $p > 1$, the approximate trajectory reads

$$N - N_{\text{end}} \approx \frac{\mu^2}{M_{\text{Pl}}^2} \frac{x^{2-p} - x_{\text{end}}^{2-p}}{p(2-p)}, \qquad (6.11)$$

which can be inverted to

$$x \approx \left[ x_{\text{end}}^{2-p} - \frac{M_{\text{Pl}}^2}{\mu^2} p(2-p) (N_{\text{end}} - N) \right]^{1/(2-p)}. \qquad (6.12)$$

Notice that far from the end of inflation, i.e. $N \ll N_{\text{end}}$, the first term can be neglected (for $p > 2$) since $x_{end} < 1$ and $M_{\text{Pl}}/\mu \gg 1$. Defining $\Delta N_* = N_{\text{end}} - N_*$, one can now plug this expression for $x_*$ into the Hubble flow functions of Eqs. (6.3) and (6.4) to get their observable values:

$$\epsilon_{1*} \approx \frac{p^2}{2} \left( \frac{M_{\text{Pl}}}{\mu} \right)^2 \left[ \Delta N_* p(p-2) \left( \frac{M_{\text{Pl}}}{\mu} \right)^2 \right]^{\frac{2(p-1)}{p-2}}, \qquad \epsilon_{2*} \approx \frac{2}{\Delta N_*} \frac{p-1}{p-2}, \qquad \epsilon_{3*} \approx \frac{1}{\Delta N_*}. \qquad (6.13)$$

It is crucial to keep in mind that the above formulas are valid only in the limit $\mu \ll M_{\text{Pl}}$ and $p > 2$. As before, the limiting case $p \to 2$ has to be taken with care and, starting with Eq. (6.6), one obtains

$$\epsilon_{1*}^{(p=2)} = \exp\left(-4\frac{M_{\text{Pl}}^2}{\mu^2}\Delta N_*\right), \qquad \epsilon_{2*}^{(p=2)} = 4\frac{M_{\text{Pl}}^2}{\mu^2}, \qquad \epsilon_{3*}^{(p=2)} = 6\epsilon_{1*}^{(p=2)}. \qquad (6.14)$$

Both Eqs. (6.13) and (6.14) describes the observed behavior in Figs. 168 to 170 when $\mu/M_{\text{Pl}} \to 0$ but they do fail in the intermediate region as we have discussed in the introduction (see Fig. 3).

If the theoretical motivations underlying the potential (6.1) do not require the $\text{vev}$ to be small, one can similarly derive approximate expressions for the observables in the limit $\mu/M_{\text{Pl}} \gg 1$ (but still with $x < 1$). Defining $\epsilon \equiv M_{\text{Pl}}/\mu$, one has $x_{\text{end}}(\epsilon)$ and we can search for a Taylor expanded solution of Eq. (6.7) to get

$$x_{\text{end}} = 1 - \frac{\epsilon}{\sqrt{2}} + \frac{p-1}{4}\epsilon^2 + \mathcal{O}(\epsilon^3). \qquad (6.15)$$

Similarly one can search for a Taylor expanded solution for the trajectory Eq. (6.5), plugging in the previous expression for $x_{end}$. Doing so yields

$$x_* = 1 - \epsilon \sqrt{\frac{1}{2} + 2\Delta N_* + \mathcal{O}(\epsilon^2)}. \qquad (6.16)$$

From this, one gets the corresponding Hubble flow functions

$$\epsilon_{1*} \approx \frac{1}{4\Delta N_* + 1}, \qquad \epsilon_{2*} \approx 4\epsilon_{1*}, \qquad \epsilon_{3*} \approx \epsilon_1. \qquad (6.17)$$

This result is quite remarkable since the observable slow-roll parameters become $\mu$ and $p$ independent. Performing the same calculation in the singular case $p \to 2$ yields exactly the same result. The spectral index, tensor-to-scalar ratio and running are immediately obtained from Eq. (6.17) with $r = 16\epsilon_{1*}$, $n_s - 1 \approx -3r/8$ and $\alpha \approx -r$. Again, these expressions match with Figs. 168 to 170 when $\mu/M_{\text{Pl}} \to \infty$.
## 6.2. Intermediate Inflation (II)

This model was introduced in Refs. [554–557] as an implementation of an equation of state of the form

$$ \rho + p = \gamma \rho^{\lambda}, \qquad (6.18) $$

where $\rho$ stands for the energy density and $p$ the pressure. Both $\gamma > 0$ and $\lambda > 1$ are dimensionless constants. As will be made explicit, this equation of state leads to a scale factor which is given by $a(t) \propto \exp(A t^\gamma)$ where $0 < f < 1$. In some sense the expansion is thus faster than power law but slower than de Sitter, hence the name of the model. The pure de Sitter case corresponds to $f = 1$. Inserting the Friedmann-Lemaître equation, $3M_{\text{Pl}}^2 H^2 = \rho$ as well as the equation of state Eq. (6.18) into the equation of conservation $\dot{\rho} + 3H(\rho + p) = 0$, one obtains a closed equation for $\rho$ which is solved by

$$ \rho = \rho_0 \left[ 3\gamma(\lambda - 1) \ln\left(\frac{a}{a_0}\right) \right]^{1/(1-\lambda)}, \qquad (6.19) $$

where $\rho_0$ and $a_0$ are positive constants. Making use of the Friedmann-Lemaître equation again, one deduces the behavior for $a$,

$$ \ln\left(\frac{a}{a_0}\right) = 3^{\lambda/(1-2\lambda)} \gamma^{1/(1-2\lambda)} \frac{\left(\lambda - \frac{1}{2}\right)^{(1-\lambda)/(1-2\lambda)}}{\lambda - 1} \left(\frac{t}{t_0}\right)^{(1-\lambda)/(1-2\lambda)}, \qquad (6.20) $$

i.e. the announced form $a(t) \propto \exp(A t^\gamma)$, with $f = 2(1 - \lambda)/(1 - 2\lambda)$. Since $\lambda > 1$, this means that $0 < f < 1$. Then, one can notice that it is possible to reinterpret the matter source as that of a scalar field with the potential $V(\phi)$ given by

$$ V(\phi) = 3A^2 f^2 M_{\text{Pl}}^4 \left[ \frac{\phi - \phi_0}{M_{\text{Pl}} \sqrt{8A (f^{-1} - 1)}} \right]^{4(1-1/f)} \\ - M_{\text{Pl}}^4 A f (1-f) \left[ \frac{\phi - \phi_0}{M_{\text{Pl}} \sqrt{8A (f^{-1} - 1)}} \right]^{-4/f}. \qquad (6.21) $$

Indeed, starting from this potential, the Klein-Gordon equation with $H = A f t^{-1}$, has an exact non-trivial solution given by

$$ \phi = \phi_0 + M_{\text{Pl}} \sqrt{8A(f^{-1} - 1)} \left(\frac{t}{t_0}\right)^{f/2}. \qquad (6.22) $$

It is then straightforward to calculate $\rho = \dot{\phi}^2/2 + V$ and $p = \dot{\phi}^2/2 - V$, and to show that they satisfy the equation of state Eq. (6.18). The potential can be recast in the form

$$ V(\phi) = M^4 \left( \frac{\phi - \phi_0}{M_{\text{Pl}}} \right)^{-\beta} - M^4 \frac{\beta^2}{6} \left( \frac{\phi - \phi_0}{M_{\text{Pl}}} \right)^{-\beta-2}, \qquad (6.23) $$

with $\beta = 4(1/f - 1)$. The constraint $0 < f < 1$ means that $\beta > 0$. Defining

$$ x \equiv \frac{\phi - \phi_0}{M_{\text{Pl}}}, \qquad (6.24) $$

it is shown below that the model predictions do not depend on $\phi_0$. Therefore Intermediate Inflation is a priori a one parameter family of models, but as explained below, one needs an extra parameter $x_{\text{end}}$ specifying the field value at which an unspecified mechanism is triggered to end of inflation. It is thus a two parameters model.

This potential appears in the earlier work of Ref. [558] as a solution for a cosmological model containing a string creation term. It is also discussed in the context of tachyon fields in Refs. [559,560]. Warm intermediate inflation was considered in Refs. [561,562], intermediate inflation within a Gauss-Bonnet braneworld was studied in Ref. [563], and with Jordan-Brans-Dicke theory in Refs. [564,565].

The potential (6.23), as well as its logarithm, are displayed in Fig. 57. It is positive definite for $x > x_{V=0} \equiv \beta/\sqrt{6}$. Therefore, one must restrict the inflaton vev to lie beyond this value. The potential increases with $x$, reaches a maximum at $x_{V'=0} \equiv \sqrt{\beta(\beta+2)}/6$, then decreases with $x$ to asymptotically vanish when $x$ goes to infinity. Therefore, a priori, two regimes of inflation exist. Either inflation proceeds at $x < x_{V'=0}$ from the right to the left, either it proceeds at $x > x_{V'=0}$ from the left to the right. However, in Eq. (6.22), one can see that the inflaton vev has to increase with time. Therefore only the branch $x > x_{V'=0}$ can produce an equation of state of the form of Eq. (6.18), which is where the model will be studied in the following.

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

$$ \epsilon_1 = \frac{1}{2} \left[ \frac{\beta^2(\beta + 2) - 6\beta x^2}{-\beta^2 x + 6x^3} \right]^2, \qquad \epsilon_2 = \frac{-2\beta x^4 + \frac{\beta^2}{3}(2\beta + 6)x^2 - \frac{\beta^4}{18}(\beta + 2)}{\left(x^3 - \frac{\beta^2 x}{6}\right)^2}, \qquad (6.25) $$

and

$$ \epsilon_3 = \frac{\beta [6x^2 - \beta(2+\beta)] \left[ \frac{\beta^5}{18}(2+\beta) - \beta^3(2+\beta)x^2 + 6\beta(4+\beta)x^4 - 12x^6 \right]}{\left(x^3 - \frac{\beta^2}{6}x\right)^2 \left[ \beta^3(\beta+2) - 12\beta(\beta+3)x^2 + 36x^4 \right]} \qquad (6.26) $$
<figure><img src="image_60.png" /><figcaption>Fig. 57. Intermediate Inflation (II). Upper panels: the potential and its logarithm for $\beta = 2.5$. Bottom left panel: slow-roll parameter $\epsilon_1$ for a potential with $\beta = 2.5$ and $\beta = 12$. The position of the maximum of $\epsilon_1$ with respect to one depends on $\beta$. The shaded area indicates where inflation stops. Bottom right panel: slow-roll parameters $\epsilon_2$ (solid line) and $\epsilon_3$ (dotted line) for a potential with $\beta = 2.5$.</figcaption></figure>

They are displayed in Fig. 57. The first slow-roll parameter diverges where the potential vanishes at $x_{V'=0}$, decreases from here and vanishes at the maximum of the potential $x_{V'=0}$. Then it increases again, reaches a local maximum at $x_{\epsilon_1 \text{max}}$, and decreases to asymptotically vanish when $x$ goes to infinity. The location $x_{\epsilon_1 \text{max}}$ is given by

$$x_{\epsilon_1 \text{max}} = \sqrt{\frac{\beta}{2} \left(1 + \frac{\beta}{3} + \sqrt{1 + \frac{4\beta}{9}}\right)}. \qquad (6.27)$$

At this point, the maximum value of $\epsilon_1$ is

$$\epsilon_1^{\text{max}} = \frac{\beta}{9} \frac{\left(1 + 3\sqrt{1 + \frac{4\beta}{9}}\right)^2}{\left(1 + \sqrt{1 + \frac{4\beta}{9}}\right)^2 \left(1 + \frac{\beta}{3} + \sqrt{1 + \frac{4\beta}{9}}\right)}. \qquad (6.28)$$

If $\beta < 9/2(1 + \sqrt{2}) \approx 10.86$, this maximum value is smaller than one. In this case inflation cannot stop by slow-roll violation in the decreasing branch of the potential and an extra parameter $x_{\text{end}}$ must be added to the model to specify the location where another mechanism such as e.g. tachyonic instability could trigger the end of inflation. If $\beta > 9/2(1 + \sqrt{2}) \approx 10.86$, the local maximum value of $\epsilon_1$ is higher than one and in the decreasing branch of the potential, either inflation takes place between $x_{V'=0}$ and the first solution of $\epsilon_1 = 1$, either it takes place between the second solution of $\epsilon_1 = 1$ and $x = \infty$. As will be shown below, only the latter case is consistent with the exact trajectory Eq. (6.22) which allows for an equation of state of the form of Eq. (6.18).

The slow-roll trajectory of the model can be obtained from Eq. (3.11). However, as already mentioned, a non-trivial and exact field evolution is given by Eq. (6.22). Written in terms of the number of e-folds $N - N_0 = \ln(a/a_0) = A(t^f - t_0^f)$, one obtains

$$x = \sqrt{x_{\text{end}}^2 + 2\beta(N - N_{\text{end}})}. \qquad (6.29)$$

This expression is exact and does not involve any approximations. It can be compared to slow-roll trajectory which reads

$$N_{\text{end}} - N = \frac{1}{2\beta} (x_{\text{end}}^2 - x^2) + \frac{1}{6} \ln \left[ x_{\text{end}}^2 - \frac{\beta(\beta + 2)}{6} \right] - \ln \left[ x^2 - \frac{\beta(\beta + 2)}{6} \right], \qquad (6.30)$$

where $N_{\text{end}}$ is the number of e-folds at the end of inflation and $N$ is the number of e-folds at some point when the scaled field $\text{vev}$ is $x$. As mentioned above, the slow-roll trajectory should match the exact one in the decreasing branch of the potential. For $x \to \infty$, one can neglect the logarithmic terms in Eq. (6.30) and one indeed recovers Eq. (6.29). This is expected since in this limit, the slow-roll parameters all go to zero and the slow-roll approximation becomes increasingly accurate. As a result, the domain of validity lies at $x \gg x_{V'=0}$, i.e. between the second solution of $\epsilon_1 = 1$ and $x = \infty$ and inflation cannot stop by slow-roll violation. This justifies the need of the extra-parameter $x_{\text{end}}$. This parameter is thus constrained to
<figure><img src="image_61.png" /><figcaption>Fig. 58. Prior space on $x_{end}$ derived from Eq. (6.34) with $N_{end} - N_{ini} = 60$, as a function of $\beta > 9/2(1 + \sqrt{2})$ (black solid line). The black dotted line corresponds to $x_{V'=0}$. For $\beta < 9/2(1 + \sqrt{2})$, provided some fine-tuning on the initial conditions, $x_{end}$ can take any values. The dashed area corresponds to parameters for the model which produce at least the required number of e-folds.</figcaption></figure>

$x_{end} > x_{V'=0}$ and should be large enough to allow for a sufficient number of e-folding. In order to get $N_{end} - N_{ini}$ e-folds, making use of Eq. (6.29), one gets

$$x_{\text{end}} = \sqrt{x_{\text{ini}}^2 + 2\beta(N_{\text{end}} - N_{\text{ini}})}. \qquad (6.31)$$

If $\beta > 9/2(1 + \sqrt{2}) \approx 10.86$, $x_{ini}$ is bounded from below by the highest solution of the equation $\epsilon_1 = 1$. This equation admits three solutions which, from the smallest to the biggest, are given by

$$x_{\epsilon_1=1}^0 = -\frac{\beta}{3\sqrt{2}} + \frac{\sqrt{2}}{3} \frac{\beta^{4/3}}{\sqrt[3]{9+2\beta+i\sqrt{-81-36\beta+4\beta^2}}} + \frac{\beta^{2/3}}{3\sqrt{2}} \sqrt[3]{9+2\beta+i\sqrt{-81-36\beta+4\beta^2}}, \qquad (6.32)$$

$$x_{\epsilon_1=1}^{\mp} = \frac{\beta}{3\sqrt{2}} + \frac{1 \mp i\sqrt{3}}{3\sqrt{2}} \frac{\beta^{4/3}}{\sqrt[3]{9+2\beta+i\sqrt{-81-36\beta+4\beta^2}}} + (1 \pm i\sqrt{3}) \frac{\beta^{2/3}}{6\sqrt{2}} \sqrt[3]{9+2\beta+i\sqrt{-81-36\beta+4\beta^2}}. \qquad (6.33)$$

The first solution is located below the maximum of the potential $x_{\epsilon_1=1}^0 < x_{V'=0}$, while the two others are located beyond it $x_{\epsilon_1=1}^{\mp} > x_{V'=0}$. Using the larger solution as a lower bound for $x_{ini}$, one gets

$$x_{\text{end}} > \sqrt{(x_{\epsilon_1=1}^{+})^2 + 2\beta(N_{\text{end}} - N_{\text{ini}})}. \qquad (6.34)$$

If $\beta < 9/2(1 + \sqrt{2})$, only one solution to $\epsilon_1 = 1$ exists,

$$x_{\epsilon_1=1} = -\frac{\beta}{3\sqrt{2}} + \frac{\sqrt{2}}{3} \frac{\beta^{4/3}}{\sqrt[3]{9+2\beta+\sqrt{81+36\beta-4\beta^2}}} + \frac{\beta^{2/3}}{3\sqrt{2}} \sqrt[3]{9+2\beta+\sqrt{81+36\beta-4\beta^2}}, \qquad (6.35)$$

which is located below the maximum of the potential $x_{\epsilon_1=1}^0 < x_{V'=0}$. In principle $x_{ini}$ is now only bounded from below by $x_{V'=0}$ and one can check from Eq. (6.30) that the total number of e-folds diverges close to $x_{V'=0}$. As a result, provided $x_{ini}$ is fine-tuned to the top of the potential, there is no bound on $x_{end}$. The prior space described by these relations is displayed in Fig. 58.

According to the previous discussion, the observable field value, at which the pivot mode crossed the Hubble radius during inflation, is such that $x_* \gg 1$. In this limit, it is possible to approximate the slow-roll parameters at Hubble crossing with

$$\epsilon_1^* \approx \frac{\beta^2}{2x_*^2}, \qquad \epsilon_2^* \approx \epsilon_3^* \approx -\frac{2\beta}{2x_*^2}, \qquad (6.36)$$

hence

$$r \approx \frac{8\beta^2}{x_*^2}, \qquad n_s - 1 \approx \frac{\beta(2-\beta)}{x_*^2}, \qquad \alpha_s = \frac{2\beta^2(\beta-2)}{x_*^4}. \qquad (6.37)$$

These estimates match with those of Ref. [557]. Finally, the parameter M is obtained from the amplitude of the CMB anisotropies

$$(\frac{M}{M_{\text{Pl}}})^4 = 720\pi^2 \left[ \frac{\beta^2 (\beta + 2)}{6} - \beta x_*^2 \right]^2 \left( x_*^3 - \frac{\beta^2 x_*}{6} \right)^{-2} \left( x_*^{-\beta} - \frac{\beta^2}{6} x_*^{-\beta-2} \right) \frac{Q_{\text{rms-PS}}^2}{T^2}. \qquad (6.38)$$
In the $x_* \gg 1$ limit, this gives

$$ \frac{M^4}{M_{\text{Pl}}^4} \approx 720\pi^2\beta^2 x_*^{2-\beta} \frac{Q^2}{T^2}, \qquad (6.39) $$

which yields $M/M_{\text{Pl}} \lesssim 10^{-2}$.

The reheating consistent slow-roll predictions for the intermediate inflation models are displayed in Fig. 171, for different values of $\beta > 0$, and for $x_{\text{end}}$ describing the prior space displayed in Fig. 58. 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 any case the reheating temperature is fully degenerate with the parameter $x_{\text{end}}$, and therefore these two parameters cannot be constrained independently. However one can see that $x_{\text{end}}$ is clearly limited from below as expected. The black solid lines represent the locus of the points such that $\epsilon_1^* = -\beta/4\epsilon_2^*$, or equivalently, $n_s - 1 = (1/\beta - 1/2)r/4$, these consistency relations arising from Eqs. (6.36). One can check that they provide a good qualitative description of the model predictions. In particular, they explain why, for $\beta < 2$, one has a blue tilt $n_s > 1$.

## 6.3. Kähler Moduli Inflation II (KMIII)

### 6.3.1. Theoretical justifications

These models are string motivated scenarios. They arise in the context of type IIB string theory via Calabi-Yau flux compactification. They have been derived and studied in Refs. [384–390], and a two-field generalization of this model has been investigated in Refs. [385–389]. They can be understood in the context of supergravity, viewed as an effective theory. In this framework, one starts with the following superpotential for the moduli $T_i$

$$ W = W_0 + \sum_{i=2}^{n} A_i e^{-a_i T_i}, \qquad (6.40) $$

where $a_i = 2\pi/(g_s N)$, $N$ being a positive integer (not to be confused with the e-fold number), $g_s$ the string coupling, and $W_0$ and $A_i$ are model dependent constants. The Kähler potential can be written as

$$ K = -2 M_{\text{Pl}}^2 \ln \left( \frac{\nu}{2\epsilon_s^6} + \frac{\xi}{2} \right), \qquad (6.41) $$

where the constant $\xi$ is given by $\xi = -\zeta(3)\chi(M)/[2(2\pi)^2]$, $\chi(M)$ being the Euler characteristic of the compactification manifold. The quantity $\mathcal{V}$ represents the overall volume of the Calabi-Yau manifold and can be taken to be

$$ \mathcal{V} = \frac{\gamma \epsilon_s^6}{2\sqrt{2}} \left[ (T_1 + T_1^\dagger)^{3/2} - \sum_{i=2}^{n} \lambda_i (T_i + T_i^\dagger)^{3/2} \right], \qquad (6.42) $$

where $\gamma$ and $\lambda_i$ are positive constants and depend on the details of the model. From the expression of the Kähler and superpotentials, it is then straightforward to calculate the corresponding F-term potential which is a relatively complex expression that can be found in Ref. [388]. If, however, one consider the limit $\mathcal{V} \gg 1$ (and $T_1 \gg T_i$), then the F-term simplifies a lot and gives rise to the following equation

$$ V(\tau_i) \approx \frac{3\xi W_0^2}{4M_{\text{Pl}}^2 \mathcal{V}_s^3} + \sum_{i=2}^{n} \left[ \frac{4W_0 a_i A_i}{M_{\text{Pl}}^2 \mathcal{V}_s^2} \tau_i e^{-a_i \tau_i} \cos(a_i \theta_i) + \frac{8(a_i A_i)^2}{3M_{\text{Pl}}^2 \gamma \lambda_i \mathcal{V}_s} \sqrt{\tau_i} e^{-2a_i \tau_i} \right], \qquad (6.43) $$

where we have written $T_i = \tau_i + i\theta_i$ and $\mathcal{V}_s \equiv \mathcal{V}/\epsilon_s^6$. We see that all the constants introduced before, namely $a_i$, $A_i$, $W_0$, $\xi$, $\gamma$ and $\lambda_i$ participate to the expression of the potential. From Eq. (6.43), solving $\partial\mathcal{V}/\partial\tau_i = 0$, one can estimate the value of each $\tau_i$ at the global minimum of the potential. In the following, we denote this quantity by $\tau_i^{\text{min}}$. Then, one can also calculate the value of the potential at this minimum. One finds [where, as usual, we have taken $\cos(a_i \theta_i) = -1$]

$$ V_{\text{min}} \approx \frac{3\xi W_0^2}{4M_{\text{Pl}}^2 \mathcal{V}_s^3} - \frac{3W_0^2 \gamma}{2M_{\text{Pl}}^2 \mathcal{V}_s^3} \sum_{i=2}^{n} \frac{\lambda_i}{a_i^{3/2}} (a_i \tau_i^{\text{min}})^{3/2}. \qquad (6.44) $$

As a consequence, if for one of the fields, say $\tau_n$, one has $(\lambda_n/a_n^{3/2}) / [\sum_{i=2}^{n-1}(\lambda_i/a_i^{3/2})] \ll 1$, then the value of $V_{\text{min}}$ is not modified even if one displaces $\tau_n$ from $\tau_n^{\text{min}}$. In other words, we have an inflationary valley along the $\tau_n$ direction and one can use it to produce inflation. In that case, the potential can be re-written as

$$ V(\tau_n) \approx \frac{BW_0^2}{M_{\text{Pl}}^2 \mathcal{V}_s^3} - \frac{4W_0 a_n A_n}{M_{\text{Pl}}^2 \mathcal{V}_s^2} \tau_n e^{-a_n \tau_n}, \qquad (6.45) $$

where the second exponential in Eq. (6.43) has been neglected, thanks to the condition $a_n \tau_n \gg 1$ and $B$ is a constant that includes the constant term in Eq. (6.43) as well as the contributions of the other fields at their minimum, i.e. $B = 3\xi/4 + \cdots$. It is important to notice that the assumption of large volume translates into a condition on the vev of $\tau_n$. The above potential is of the form of the toy model studied as “Kähler Moduli Inflation I (KMII)” in Section 5.9. The field is however not canonically normalized since it is a modulus. It is therefore necessary to first canonically normalize it and, then, re-derive the corresponding potential. Using the form of the Kähler potential given above, denoting by $\phi$ the canonical field, one arrives at

$$ \tau_n = \left( \frac{3\mathcal{V}_s}{4\gamma\lambda_n} \right)^{2/3} \left( \frac{\phi}{M_{\text{Pl}}} \right)^{4/3}. \qquad (6.46) $$

As a consequence, the final form of the inflaton's potential is given by

$$ V(\phi) = \frac{BW_0^2}{M_{\text{Pl}}^2 \mathcal{V}_s^3} - \frac{4W_0 a_n A_n}{M_{\text{Pl}}^2 \mathcal{V}_s^2} \left( \frac{3\mathcal{V}_s}{4\gamma\lambda_n} \right)^{2/3} \left( \frac{\phi}{M_{\text{Pl}}} \right)^{4/3} \exp \left[ -a_n \left( \frac{3\mathcal{V}_s}{4\gamma\lambda_n} \right)^{2/3} \left( \frac{\phi}{M_{\text{Pl}}} \right)^{4/3} \right]. \qquad (6.47) $$
<figure><img src="image_62.png" /><figcaption>Fig. 59. Top left panel: Kähler moduli inflation II (KMIII) potential for $\alpha = 4$ and $\beta = 1$. These parameters are not physical but they are used for display convenience. Top right panel: logarithm of the potential for the same value of $\alpha$ and $\beta$. Bottom left panel: slow-roll parameter $\epsilon_1$ for a potential with $\alpha = 4$ and $\beta = 1$. The shaded area indicates the breakdown of the slow-roll inflation (strictly speaking when the acceleration stops). Bottom right panel: slow-roll parameters $\epsilon_2$ (solid line) and $\epsilon_3$ (dotted line) for $\alpha = 4$ and $\beta = 1$.</figcaption></figure>

Let us now see what are the typical values that the parameters appearing in the above potential can take. As already mentioned, the quantity $V_s$ represents the Calabi-Yau volume and is supposed to be such that $V_s \gg 1$ or $V \gg \epsilon_s^6$. In Ref. [390] the typical value $V_s \approx 3 \times 10^6$ was chosen. The parameter $A_n$ depends on the complex structure moduli and is typically of order $\mathcal{O}(\epsilon_s^3)$. This is also the case for $W_0$. One has $a_n = 2\pi/N$, where $N$ is a positive integer (for D3-brane instantons, one has $N = 1$). The dimensionless parameter $\lambda_n$ is model dependent but is considered to be of order $\mathcal{O}(1)$. The quantity $\xi = \zeta(3)\chi / [2(2\pi)^3]$, where $\chi$ is the Euler number of the internal Calabi-Yau space, is also of order $\mathcal{O}(1)$ as well as the coefficient $\gamma$. This means that $B$ is of order $\mathcal{O}(1)$.

### 6.3.2. Slow-roll analysis

We now study the inflationary scenario based on the potential derived above. Re-writing $V(\phi)$ in a more convenient way, we have

$$V(\phi) = M^4 \left[ 1 - \alpha \left( \frac{\phi}{M_{Pl}} \right)^{4/3} e^{-\beta(\phi/M_{Pl})^{4/3}} \right]. \qquad (6.48)$$

where we have defined the parameters $M$, $\alpha$ and $\beta$ by

$$M^4 = \frac{BW_0^2}{M_{Pl}^2 \epsilon_s^3}, \quad \alpha = \frac{16V_s a_n}{3} \frac{A_n}{W_0} \left( \frac{3V_s}{4\gamma\lambda_n} \right)^{2/3}, \quad \beta = a_n \left( \frac{3V_s}{4\gamma\lambda_n} \right)^{2/3}. \qquad (6.49)$$

Making use of the typical orders of magnitude for the various quantities entering these expression, one sees that

$$\alpha = \mathcal{O}(\mathcal{V}_s^{5/3}), \quad \beta = \mathcal{O}(\mathcal{V}_s^{2/3}), \qquad (6.50)$$

with $\mathcal{V}_s \gg 1$.

The potential (6.48) and its logarithm are displayed in Fig. 59. $V(\phi)$ decreases from $V/M^4 = 1$ at $\phi = 0$, reaches a minimum at $\phi/M_{Pl} = \beta^{-3/4}$, and then increases to the asymptotic value $V/M^4 = 1$ when $\phi/M_{Pl} \to +\infty$. However, since the potential is derived under the large field assumption, only the increasing branch of the potential is relevant. Inflation proceeds from the right to the left along this branch. The minimum value of the potential at $\phi = M_{Pl}\beta^{-3/4}$ is given by $V_{min} = M^4 [1 - \alpha/(\beta e)]$. Therefore, if one wants the potential to be definite positive everywhere, one must have $\alpha/\beta < e$. However, from Eq. (6.50), we see that this condition cannot be satisfied since $\alpha/\beta = \mathcal{O}(V_s) \gg 1$. This means that the potential necessarily vanishes at some point. In the increasing branch of the potential, this occurs for a vev given by

$$x_{V=0} \equiv \frac{\phi_{V=0}}{M_{Pl}} = \left[ -\frac{1}{\beta} W_{-1} \left( -\frac{\beta}{\alpha} \right) \right]^{3/4}. \qquad (6.51)$$

Anyway, since the potential (6.48) is only valid in the large field region, this criterion does not play an important role in what follows.
<figure><img src="image_63.png" /><figcaption>Fig. 60. Comparison between the exact numerical value of $x_{\text{end}}(\alpha, \beta)$ (blue solid line), and the approximated formula given by Eq. (6.55) (red dotted line) for $\alpha = \nu^{5/3}$ and $\beta = \nu^{2/3}$. The agreement is excellent but a numerical calculation is used in ASPIC anyway.</figcaption></figure>

Let us now calculate the three first Hubble flow parameters. Defining $x \equiv \phi/M_{\text{Pl}}$, they are given by

$$ \epsilon_1 = \frac{8\alpha^2}{9} x^{2/3} e^{-2\beta x^{4/3}} \left( \frac{1 - \beta x^{4/3}}{1 - \alpha x^{4/3} e^{-\beta x^{4/3}}} \right)^2, \qquad (6.52) $$

$$ \epsilon_2 = \frac{8\alpha}{9} x^{-2/3} e^{-2\beta x^{4/3}} \frac{3\alpha x^{4/3} + \alpha\beta x^{8/3} + e^{\beta x^{4/3}}(1 - 9\beta x^{4/3} + 4\beta^2 x^{8/3})}{(1 - \alpha x^{4/3} e^{-\beta x^{4/3}})^2}, \qquad (6.53) $$

and

$$ \epsilon_3 = \left\{ 8\alpha (1 - \beta x^{4/3}) \left[ \alpha^2 x^{8/3} (9 + \beta x^{4/3}) - 2\alpha e^{\beta x^{4/3}} x^{4/3} \left( -4 + 19\beta x^{4/3} - 9\beta^2 x^{8/3} + 4\beta^3 x^4 \right) - e^{2\beta x^{4/3}} (1 + 11\beta x^{4/3} - 30\beta^2 x^{8/3} + 8\beta^3 x^4) \right] \right\} \left\{ 9x^{2/3} (e^{\beta x^{4/3}} - \alpha x^{4/3})^2 \times \left[ \alpha x^{4/3} (3 + \beta x^{4/3}) + e^{\beta x^{4/3}} (1 - 9\beta x^{4/3} + 4\beta^2 x^{8/3}) \right] \right\}^{-1}. \qquad (6.54) $$

Inflation stops when $\epsilon_1(x_{\text{end}}) = 1$. As can be seen in Fig. 59, for $\alpha/\beta \gg 1$, the first slow-roll parameter $\epsilon_1$ starts increasing from $\epsilon_1 = 0$ at $x = 0$, diverges at a vev that we do not need to compute here, and then decreases to vanish at $x = \beta^{-3/4}$. Then, it increases again, blows up at $x_{V=0}$ and, finally, asymptotically vanishes when $x \to \infty$. Since inflation proceeds at $x > x_{V=0}$ it always stops by violation of the slow-roll conditions. Unfortunately is not possible to find an analytic expression for $x_{\text{end}}$ but one can provide the following approximated formula,

$$ x_{\text{end}} \approx \left[ -\frac{5}{4\beta} W_{-1} \left( -\frac{4 \times 9^{2/5}}{5 \times 8^{2/5}} \alpha^{-4/5} \beta^{1/5} \right) \right]^{3/4}, \qquad (6.55) $$

where $W_{-1}$ is the Lambert function. It is compared to the numerical solution for $x_{\text{end}}$ implemented in the ASPIC code in Fig. 60. The agreement is excellent.

Let us now turn to the slow-roll trajectory. Unfortunately, KMIII is one of the rare cases for which it cannot be integrated by quadrature. As such, in the ASPIC library, the slow-roll trajectory is numerically integrated. However, in the large field limit $x \gg \beta^{-3/4}$, one can obtain an approximate analytic formula given by

$$ N_{\text{end}} - N \approx \frac{9}{16\alpha\beta^2} \left( \frac{e^{\beta x^{4/3}}}{x^2} - \frac{e^{\beta x_{\text{end}}^{4/3}}}{x_{\text{end}}^2} \right), \qquad (6.56) $$

from which one deduces that

$$ x \approx \left( -\frac{3}{2\beta} W_{-1} \left\{ -\frac{2}{3}\beta \left[ \frac{e^{\beta x_{\text{end}}^{4/3}}}{x_{\text{end}}^2} + \frac{16\alpha\beta^2}{9} (N_{\text{end}} - N) \right]^{-2/3} \right\} \right)^{3/4}. \qquad (6.57) $$

This approximation is in agreement with what was obtained in Ref. [390], up to an incorrect choice of the Lambert function branch. The Lambert function is displayed in Fig. 61 and the part of the curve where inflation proceeds is indicated by the arrow. The fact that the -1 branch of the Lambert function has to be chosen comes from the fact that, when $N_{\text{end}} - N \to \infty$, one must have $x \to \infty$. On the other hand, when $N_{\text{end}} - N \to 0$, $x \to x_{\text{end}} > \beta^{-3/4}$ and this is again consistent with the choice of the -1 branch.

Finally, one can use the CMB normalization to calculate the mass scale M. Without any approximation on top of slow-roll, this leads to the following expression

$$ \left( \frac{M}{M_{\text{Pl}}} \right)^4 = 1280 \pi^2 \alpha^2 x_*^{2/3} e^{-2\beta x_*^{4/3}} \left( 1 - \beta x_*^{4/3} \right)^2 \left( 1 - \alpha x^{4/3} e^{-\beta x_*^{4/3}} \right)^{-2} \frac{Q_{\text{rms}-\text{PS}}^2}{T^2}. \qquad (6.58) $$
<figure><img src="image_64.png" /><figcaption>Fig. 61. Lambert functions $W_0(x)$ (dashed line) and $W_{-1}(x)$ (solid line). During Kähler moduli inflation II, inflation proceeds along the “-1” branch in the direction specified by the arrow.</figcaption></figure>

Making use of the approximated trajectory and of the expression for the scale $M$, one roughly obtains

$$V_s \approx \frac{\Delta N_*}{\pi \sqrt{720}} \frac{1}{(M_{Pl} l_s^3)^3} \left[ \frac{4 B a_n (W_0 l_s^3)^2}{3 \gamma \lambda_n} \right] \ln^{-5/4} \left( \frac{16 \alpha \beta^2}{9} \Delta N_* \right) \frac{T}{Q_{\text{rms}-\text{PS}}}. \quad (6.59)$$

Given that $a_n, B, \gamma, \lambda_n, W_0 l_s^3$ are a priori coefficients of order one, we see that the above expression roughly implies that $V$ is of the order $10^6 l_s$. The reheating consistent slow-roll predictions for the Kähler moduli inflation II models are displayed in Fig. 175, for $V \in [10^5, 10^7]$, and taking $\alpha = V^{5/3}$ and $\beta = V^{2/3}$. One can check that even if one adds $\mathcal{O}(1)$ factors in these relations, the slow-roll predictions do not depend significantly on them. Also, we notice that $\epsilon_1$ is typically extremely small and that $\epsilon_2$ is almost independent of $V$. These effects can be analytically understood. Working out Eq. (6.55) and Eqs. (6.52), (6.53), and (6.54) in the large field limit, one obtains

$$\epsilon_{1*} \approx \frac{1}{324 \beta^{3/2} (\Delta N_*)^2} \ln^{5/2} \left( 16 \sqrt{\frac{9}{8}} \alpha \beta^{1/2} \Delta N_* \right), \quad \epsilon_{2*} \approx \frac{2}{\Delta N_*}, \quad \epsilon_{3*} \approx \frac{1}{\Delta N_*}, \quad (6.60)$$

from which one deduces that

$$n_s \approx 1 - \frac{2}{\Delta N_*}, \quad r \approx \frac{4}{81 \beta^{3/2} (\Delta N_*)^2} \ln^{5/2} \left( 16 \sqrt{\frac{9}{8}} \alpha \beta^{1/2} \Delta N_* \right), \quad \alpha_s \approx -\frac{2}{\Delta N_*^2}. \quad (6.61)$$

Firstly, we see that the slow-roll parameters at Hubble crossing depend on $\alpha$ logarithmically only. This explains the weak dependence in the $\mathcal{O}(1)$ factors mentioned above. Secondly, we also notice that $\epsilon_{2*}$ and $\epsilon_{3*}$ do not depend on $\beta$. In a third place, $\epsilon_1$ is a very small number since it is proportional to the inverse of $\beta^{3/2}$. This also means that, when $V$ increases, $\epsilon_1$ decreases. All these considerations can be checked in Fig. 175 and the amount of gravitational waves predicted by this model is very small. This is in agreement with the rough estimates given in Refs. [384,387,388,390]. However, contrary to what is claimed in Ref. [390], the predicted value for the running of the spectral index is not excluded by observations since, according to the Planck results [187,188], $\alpha_s = 0.0011 \pm 0.0099$ while, for the fiducial value $\Delta N_* \approx 55$, one obtains $\alpha_s \approx -0.0006$.

## 6.4. Logamediate Inflation (LMI)

Logamediate inflation has been discussed in Refs. [566,567] and refers to inflationary scenarios in which the scale factor evolves according to

$$a(t) = a_0 \exp \left[ A \left( \ln \frac{t}{t_0} \right)^{\lambda} \right], \quad (6.62)$$

where $A$ and $\lambda$ are two dimensionless parameters and where $t_0$ has the dimension of a cosmic time. This evolution form for the scale factor is required to occur “at late times”, i.e. when $t \gg t_0$. If $\lambda = 1$, one recovers the power law model (see Section 5.8), and in that case, $t_0$ can be absorbed in a rescaling of the scale factor. Otherwise, these three parameters are relevant and one therefore expects LMI to be a two parameters models according to our classification. Following Ref. [566], from Eq. (6.62), one has

$$H \equiv \frac{\dot{a}}{a} = \frac{A \lambda}{t} \left( \ln \frac{t}{t_0} \right)^{\lambda-1}, \quad (6.63)$$

from which one deduces that $A\lambda > 0$ in order to have expansion ($H > 0$). From Eq. (6.62), one can also establish that

$$\frac{\ddot{a}}{a} = \frac{A\lambda}{t^2} \left( \ln \frac{t}{t_0} \right)^{\lambda-1} \left[ (\lambda-1) \left( \ln \frac{t}{t_0} \right)^{-1} - 1 + A\lambda \left( \ln \frac{t}{t_0} \right)^{\lambda-1} \right], \quad (6.64)$$

from which one deduces that in order to have inflation at late times (when $t \gg t_0$), one must have $\lambda > 1$, or if $\lambda = 1$, $A > 1$. If this inflationary scenario is implemented within a single minimally coupled scalar field $\phi$, one can derive the corresponding potential. From the Friedmann-Lemaître
and Klein–Gordon equations one can show that [566]

$$ \frac{\phi(t)}{M_{Pl}} = \frac{\sqrt{2A\lambda}}{t} \left( \ln \frac{t}{t_0} \right)^{\frac{\lambda-1}{2}}. \qquad (6.65) $$

This equation can easily be integrated into

$$ \frac{\phi(t)}{M_{Pl}} = \frac{\phi_0}{M_{Pl}} + 2 \frac{\sqrt{2A\lambda}}{\lambda+1} \left( \ln \frac{t}{t_0} \right)^{\frac{\lambda+1}{2}}. \qquad (6.66) $$

Combining the Friedmann-Lemaître equation $3M_{Pl}^2H^2 = V(\phi) + \dot{\phi}^2/2$ and the relation $2M_{Pl}^2\dot{H} = -\dot{\phi}^2$, one obtains $V(\phi) = 3M_{Pl}^2H^2 + M_{Pl}^2\dot{H}$, namely

$$ V(\phi) = \frac{3M_{Pl}^2 A^2 \lambda^2}{t^2} \left( \ln \frac{t}{t_0} \right)^{2(\lambda-1)} + \frac{M_{Pl}^2 A \lambda}{t^2} (\lambda-1) \left( \ln \frac{t}{t_0} \right)^{\lambda-2} - \frac{M_{Pl}^2 A \lambda}{t^2} \left( \ln \frac{t}{t_0} \right)^{\lambda-1}. \qquad (6.67) $$

Together with Eq. (6.66), this gives a parametric representation of the field potential in terms of $t$. It can be further simplified since the Logamediate regime occurs in the limit $t \gg t_0$. If $\lambda > 1$, the first term of this expression dominates at late times and one has $V(\phi) = 3M_{Pl}^2 A^2 \lambda^2 (\ln t/t_0)^{2(\lambda-1)}/t^2$. Defining $x \equiv (\phi - \phi_0)/M_{Pl}$, one makes use of Eq. (6.66) to obtain

$$ V(\phi) = M^4 x^\alpha \exp(-\beta x^\gamma), \qquad (6.68) $$

where the new parameters are defined by

$$ \alpha = 4 \frac{\lambda - 1}{\lambda + 1}, \qquad \beta = 2 \left( \frac{\lambda + 1}{2\sqrt{2A\lambda}} \right)^{2/(\lambda+1)}, \qquad \gamma = \frac{2}{\lambda + 1}, \qquad (6.69) $$

and

$$ \frac{M^4}{M_{Pl}^4} = \frac{3A^2 \lambda^2}{M_{Pl}^2 t_0^2} \left( \frac{\lambda + 1}{2\sqrt{2A\lambda}} \right)^{4 \frac{\lambda - 1}{\lambda + 1}}. \qquad (6.70) $$

The same potential has been studied for $\alpha = 2$, $\beta = 1/8$ and $\gamma = 2$ within tachyon inflation models in Ref. [513]. The case $\lambda = 1$ is particular. At late times, the first term and the last term must be kept in Eq. (6.67), such that $V(\phi) = (3A-1)AM_{Pl}^2/t^2$. In that situation, one has $x = \sqrt{2A\ln t/t_0}$, and the derived potential shares the same expressions for $\alpha$, $\beta$ and $\gamma$ as in Eq. (6.69) but evaluated at $\lambda = 1$. There is a difference however because $M^4$ now reads $M^4 = (3A-1)AM_{Pl}^2/t_0^2$. We recover explicitly that $\lambda = 1$ corresponds to power law inflation and has already been treated in Section 5.8.

In the following, we will work only with the derived parameters $\beta$, $\gamma$ and $M^4$, noticing that

$$ \alpha = 4(1 - \gamma). \qquad (6.71) $$

The restrictions $A\lambda > 0$ and $\lambda \ge 1$ translates into the conditions $0 < \gamma \le 1$ and $\beta > 0$. Following Ref. [567], since there is no fundamental reasons preventing it, we will generalize this model to any possible values of these parameters supporting inflation.

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

$$ \epsilon_1 = \frac{(\alpha - \beta\gamma x^\gamma)^2}{2x^2}, \qquad \epsilon_2 = \frac{2}{x^2} [\alpha + \beta(\gamma - 1)\gamma x^\gamma], \qquad (6.72) $$

$$ \epsilon_3 = \frac{\alpha - \beta\gamma x^\gamma}{x^2} \frac{2\alpha - \beta(\gamma - 2)(\gamma - 1)\gamma x^\gamma}{\alpha + \beta(\gamma - 1)\gamma x^\gamma}. \qquad (6.73) $$

The potential and the Hubble flow functions in the slow-roll approximation have been represented in Fig. 62.

Inflation can proceed in two regimes: either at decreasing field values, left to the top of the potential (LMI1), or at increasing field values, right to the top of the potential (LMI2). Notice that from Eq. (6.66), $\phi$ has to increase with time to reproduce the scale factor expansion Eq. (6.62) and this happens only in the regime LMI2 for large values of $x$. As can be seen in Fig. 62, the slow-roll parameter $\epsilon_1$ diverges when $x$ approaches zero, it vanishes at the top of the potential for $x = x_{Vmax}$ and it is maximal at $x = x_{\epsilon_1^{max}}$ with

$$ x_{Vmax} \equiv \left( \frac{\alpha}{\beta\gamma} \right)^{1/\gamma}, \qquad x_{\epsilon_1^{max}} = \left[ \frac{\alpha}{\beta\gamma(1-\gamma)} \right]^{1/\gamma}. \qquad (6.74) $$

Finally it asymptotes to zero for large values of the field. The value of the local maximum of $\epsilon_1$ reads

$$ \epsilon_1^{\max} = \frac{\alpha^2}{2} \left[ \frac{\beta\gamma(1-\gamma)}{\alpha} \right]^{\frac{2}{\gamma}} \left( \frac{\gamma}{1-\gamma} \right)^2. \qquad (6.75) $$

Thus in the regime LMI1, inflation always stops naturally as $\epsilon_1$ becomes larger than unity whereas in the regime LMI2, this may occur only if $\epsilon_1^{\max} > 1$ and if inflation has started from $x_{ini} < x_{\epsilon_1^{max}}$. Otherwise, if inflation starts with $x_{ini} > x_{\epsilon_1^{max}}$, or if $\epsilon_1^{\max} < 1$, one needs to add an extra-parameter $x_{end}$ encoding an unspecified mechanism to end inflation. In that situation, the model becomes a three parameters one. If one makes use of $\alpha = 4(1-\gamma)$, one obtains $\epsilon_1^{\max} = 8\gamma^2(\beta\gamma/4)^{2/\gamma}$. Solving $\epsilon_1^{\max} \ge 1$ for $\beta$ gives

$$ \beta \ge \frac{4}{\gamma (8\gamma^2)^{\gamma/2}}. \qquad (6.76) $$

This condition is therefore required for the model LMI2, if one wants inflation to end naturally. As we will see below, LMI2 inflating in the domain $x_{Vmax} < x < x_{\epsilon_1^{max}}$ is a very fine-tuned situation which is strongly disfavored by the observations. Notice that if one assumes $0 < \gamma \le 1$, this conditions translates into $\beta > \sqrt{2}$.
<figure><img src="image_65.png" /><figcaption>Fig. 62. Log-mediated Inflation (LMI). Upper panels: the potential and its logarithm for $\beta = 2, \gamma = 0.95$. Bottom left panel: Hubble flow function $\epsilon_1$ for a potential with $\beta = 2, \gamma = 0.95$ (blue curve) and $\beta = 2, \gamma = 0.8$ (green curve). The position of the maximum of $\epsilon_1$ with respect to one depends on $\gamma$. The shaded region indicates where inflation stops. Bottom right panel: slow-roll parameters $\epsilon_2$ (solid line) and $\epsilon_3$ (dotted line) for a potential with $\beta = 2, \gamma = 0.7$.</figcaption></figure>

Finally, let us notice that for the value of $\epsilon_2$ at the top of the potential to be smaller than some maximal value $\epsilon_{2,\text{top}}^{\text{max}}$, one needs to impose the condition

$$ \beta < \beta^{\text{max}}(\gamma, \epsilon_{2,\text{top}}^{\text{max}}) = 2^{2-3\gamma/2} \left(\epsilon_{2,\text{top}}^{\text{max}}\right)^{\gamma/2} \frac{(1-\gamma)^{1-\gamma/2}}{\gamma^{1+\gamma/2}}. \qquad (6.77) $$

In the LMI1 model, a slow roll regime of inflation can proceed only if such a condition is verified (with typically $\epsilon_{2,\text{top}}^{\text{max}} \simeq 10^{-1}$).

The slow-roll trajectory can be integrated thanks to the hypergeometric function [281,282] $_2F_1$, leading to

$$ N - N_{\text{end}} = \frac{x_{\text{end}}^2}{2\alpha} {}_2F_1 \left[ 1, \frac{2}{\gamma}, \frac{2}{\gamma} + 1, \left( \frac{x_{\text{end}}}{x_{\text{V-max}}} \right)^{\gamma} \right] - \frac{x^2}{2\alpha} {}_2F_1 \left[ 1, \frac{2}{\gamma}, \frac{2}{\gamma} + 1, \left( \frac{x}{x_{\text{V-max}}} \right)^{\gamma} \right]. \qquad (6.78) $$

One can notice that inserting $\alpha = 4(1 - \gamma)$, as a function of $x/x_{\text{V-max}}$, this trajectory only involves $\gamma$. Plugging $x = x_{\text{V-max}}$ into Eq. (6.78) one gets an infinite number of e-folds. This means that the required number of e-folds to solve the problems of the standard Big-Bang scenario can always be realized, both in the decreasing branch of the potential and the increasing one, provided that inflation starts close enough to $x_{\text{V-max}}$. However, it can numerically be checked that in the case of LMI2 with $\epsilon_1^{\text{max}} > 1$ and inside the $x_{\text{V-max}} < x < x_{\text{e-max}}$ region, one has to fine-tune $x_{\text{ini}}$ and $x_*$ extremely close to $x_{\text{V-max}}$. In that situation $n_s = 1$, with vanishing $r$ and vanishing running of the spectral index, can be considered as generic predictions of the model. For this reason, it is more natural to consider LMI2 in the large field regime, namely $x > \max(x_{\text{V-max}}, x_{\text{e-max}})$, together with the extra-parameter $x_{\text{end}}$.

The trajectory in Eq. (6.78) cannot be inverted analytically. However, one can perform some consistency checks in the limit $x/x_{\text{V-max}} \gg 1$ in which

$$ N - N_{\text{end}} \approx \frac{1}{\beta\gamma(2-\gamma)} (x^{2-\gamma} - x_{\text{end}}^{2-\gamma}), \qquad (6.79) $$

and

$$ x \approx \left[ x_{\text{end}}^{2-\gamma} + \beta\gamma(2-\gamma)(N - N_{\text{end}}) \right]^{\frac{1}{2-\gamma}}. \qquad (6.80) $$

These expressions can be compared to Eq. (6.66)

$$ x = 2 \frac{\sqrt{2A\lambda}}{\lambda + 1} \left( \ln \frac{t}{t_0} \right)^{\frac{\lambda+1}{2}}, \qquad (6.81) $$

where $t$ in terms of the number of e-folds $N$ can be obtained from Eq. (6.62). With $N - N_0 = A (\ln t/t_0)^{\lambda}$, one gets

$$ x = 2 \frac{\sqrt{2A\lambda}}{\lambda + 1} \left( \frac{N - N_0}{A} \right)^{\frac{\lambda+1}{2\lambda}}. \qquad (6.82) $$
<figure><img src="image_66.png" /><figcaption>Fig. 63. Top left panel: Twisted Potential Inflation (TWI) for $\phi_0 = 0.02 M_p$. Top right panel: logarithm of the potential for the same value of $\phi_0$. Bottom left panel: slow-roll parameter $\epsilon_1$ with $\phi_0 = 0.02 M_p$ (solid blue line) and $\phi_0 = 0.05 M_p$ (solid green line). The shaded area indicates the non-inflationary region. Bottom right panel: slow-roll parameters $\epsilon_2$ (solid line) and $\epsilon_3$ (dotted line) with $\phi_0 = 0.02 M_p$.</figcaption></figure>

The previous calculations are consistent since, making use of Eq. (6.69), Eq. (6.80) and Eq. (6.82) are the same when setting the constants $N_0 = N_{ini}$ and $x_0 = x_{ini} = 0$. This means that in the late times limit $x/x_{vmax} \gg 1$, the slow-roll trajectory coincides with the exact one, as expected.

The amplitude of the CMB anisotropies fixes the value of the parameter M according to

$$ \frac{M^4}{M_{Pl}^4} = 720\pi^2 (\alpha - \beta\gamma x_*^\gamma)^2 e^{\beta x_*^\gamma} x_*^{\alpha-2} \frac{Q_{rms-PS}^2}{T^2}, \qquad (6.83) $$

where $x_*$ is the observable field value obtained by solving Eq. (3.48) given some assumptions on the reheating. The reheating consistent slow-roll predictions for the models LMI1 and LMI2 (at $x > x_{vmax}$) are displayed in Figs. 176, 177, and 178 for LMI1, and in Figs. 179, 182, and 185 for LMI2. In the case of LMI2, the turning points in the plots precisely correspond to the case where inflation occurs in the fine-tuned domain $x_{vmax} < x_* < x_{vmax}^1$ and in which the model behaves like a pure de Sitter era.

## 6.5. Twisted Inflation (TWI)

### 6.5.1. Theoretical justifications

This model was introduced in Ref. [568] and is based on higher dimensional supersymmetric gauge theories. The idea is to assume that, in higher dimensions, we have a flat direction $\phi$ in the potential. Since the theory is supersymmetric, this flat direction will not receive corrections because the bosonic and fermionic contributions exactly cancel out. Then, we compactify the theory down to 3 + 1 dimensions but with boundary conditions that break supersymmetry. The typical example given in Ref. [568] is “twisted” circle compactification, hence the name of the model. Since supersymmetry is broken, the “Kaluza-Klein” masses of bosons and fermions will differ. Typically, they can be written as

$$ m_b = \sqrt{\phi^2 + \frac{n^2}{R^2}}, \qquad m_f = \sqrt{\phi^2 + \frac{(n+1/2)^2}{R^2}}, \qquad (6.84) $$

where R is the radius of compactification and n an integer. Since $m_b \neq m_f$, this time, the potential will receive one loop corrections which lift the potential. However, it is clear that, when $\phi R \gg n$, one has approximately $m_b \approx m_f$. Therefore, in this regime, we expect the corrections to vanish and the flat direction to remain flat. This is thus particularly well-suited for inflation. In practice, the higher dimensional model considered to implement the above discussed mechanism is a maximally supersymmetric $4+1$ U(N) Yang-Mills theory compactified on a circle of radius R. A priori, we have therefore two parameters: N and the compactification scale R.

### 6.5.2. Slow-roll analysis

As shown in Ref. [568], the above considerations leads to the following expression for the inflaton potential

$$ V(\phi) = M^4 \left[ 1 - A \left( \frac{\phi}{\phi_0} \right)^2 e^{-\phi/\phi_0} \right], \qquad (6.85) $$
where A is a constant parameter given by

$$A = \frac{32}{93\zeta(5)} \approx 0.33, \qquad (6.86)$$

and where $\phi_0$ is related to the compactification scale R through the following equation

$$\frac{\phi_0}{M_{Pl}} = \frac{1}{2\pi R M_{Pl}}. \qquad (6.87)$$

Since the radius R must be larger than the Planck length, i.e. $RM_{Pl} \gg 1$, this implies that $\phi_0/M_{Pl} \ll 1$. On the other hand, the overall normalization can be expressed as

$$M^4 = \frac{8N}{A\pi^2(2\pi R)^4}. \qquad (6.88)$$

We see that the scale M depends on the compactification radius R but also on the number $N$. In addition, one must have $\phi < \sqrt{3/N} M_{Pl}$ or $\phi \ll M_{Pl}$ which guarantees that the higher order Planck suppressed operators do not alter the potential. The potential (6.85) is the small coupling limit of the model, while the strong coupling limit corresponds to a BI model with $p = 3$, see Section 6.19.

The potential Eq. (6.85), as well as its logarithm, is displayed in Fig. 63. Inflation is supposed to take place for vev's larger than the scale $\phi_0$, i.e. for $\phi > \phi_0$, in the increasing branch of the potential. This means that it proceeds from the right to the left in the direction indicated by the arrow. The minimum of the potential is located at $\phi/\phi_0 = 2$.

Let us now turn to the calculation of the Hubble flow parameters. If one defines $x$ by $x \equiv \phi/\phi_0$, then they are given by

$$\epsilon_1 = \frac{A^2}{2} \left( \frac{M_{Pl}}{\phi_0} \right)^2 e^{-2x} \left[ \frac{x(x-2)}{1 - Ax^2 e^{-x}} \right]^2, \qquad \epsilon_2 = 2A \left( \frac{M_{Pl}}{\phi_0} \right)^2 e^{-2x} \frac{2Ax^2 + e^x (x^2 - 4x + 2)}{(1 - Ax^2 e^{-x})^2}, \qquad (6.89)$$

and

$$\epsilon_3 = A \left(\frac{M_{Pl}}{\phi_0}\right)^2 x(2-x) e^{-2x} \frac{4A^2 x^3 - e^{2x}(x^2 - 6x + 6) - A x e^x (x^3 - 6x^2 + 18x - 12)}{(1 - Ax^2 e^{-x})^2 [2Ax^2 + e^x (x^2 - 4x + 2)]}. \qquad (6.90)$$

They are displayed in Fig. 63. The first slow-roll parameter $\epsilon_1$ vanishes at the minimum of the potential when $x = 2$, then increases with $x$ and reaches a maximum at $x_{e\text{max}}$, and finally decreases to zero when $x$ goes to infinity. The value of $\epsilon_1$ at this local maximum is larger than one if $\phi_0$ is smaller than some value that can only be determined numerically. We find

$$\phi_0 < 0.04228 M_{Pl}. \qquad (6.91)$$

Therefore, a priori, inflation could stop by slow-roll violation. However, by numerically integrating the exact trajectory (i.e. if one does not make use of the slow-roll approximation), one realizes that, in fact, the first Hubble flow function, which is defined by $\epsilon_1^H = -\dot{H}/H^2$, remains smaller than one for all field values, see Fig. 64. This is due to the fact that while the inflaton rolls down its potential and approaches its minimum, the slow-roll parameters continuously increase and the slow-roll approximation is broken before $\epsilon_1$ becomes $\mathcal{O}(1)$. Usually, this leads only to small corrections at the end of inflation. However, in the case of twisted inflation, this leads to a radically different picture because the potential does not vanish at its minimum and, therefore, acts as a cosmological constant. In practice, the numerical calculations indicate that the field oscillates around its minimum but always such that $\epsilon_1^H < 1$ and independently on the value of $\phi_0$, see Fig. 64. In principle, inflation can never stops in this model since the final stage of the evolution corresponds to an inflaton field sitting for ever at the bottom of the potential and, as already mentioned, it acts as a cosmological constant. However, as explained in Ref. [568], the interactions of the inflaton field with the other degrees of freedom of the standard model starts to play a role in this regime. As a consequence, the energy contained in the inflaton field should quickly be transferred to other fields and a phase of reheating starts. The details of this process are complicated and are discussed in Ref. [568]. In order to model the end of inflation, we therefore introduce the extra parameter $x_{\text{end}}$ giving the vev at which inflation stops. As a consequence, TWI is in fact a two parameter model, $\phi_0$ and $\phi_{\text{end}}$.

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

$$N_{\text{end}} - N = \left(\frac{\phi_0}{M_{Pl}}\right)^2 \left\{ \frac{1}{2A} \left[ \text{Ei}(x_{\text{end}}) - \text{Ei}(x) \right] - \frac{e^2}{2A} \left[ \text{Ei}(x_{\text{end}} - 2) - \text{Ei}(x - 2) \right] \right. \\ \left. + x_{\text{end}} - x + 2 \ln \left( \frac{x_{\text{end}} - 2}{x - 2} \right) \right\}, \qquad (6.92)$$

where $N_{\text{end}}$ is the number of e-folds at the end of inflation and $Ei$ is the exponential integral function [281,282]. This expression is the one used in the ASPIC library. However, if one makes the vacuum dominated approximation, $x \gg 1$, then a simpler formula can be derived for the trajectory, namely

$$N_{\text{end}} - N \approx \frac{1}{A} \left(\frac{\phi_0}{M_{Pl}}\right)^2 \left(\frac{e^x}{x^2} - \frac{e^{x_{\text{end}}}}{x_{\text{end}}^2}\right). \qquad (6.93)$$

This allows us to obtain an approximated expression for the vev of the field at Hubble radius crossing which reads

$$x_* \approx \ln \left[ 4A \Delta N_* \left( \frac{M_{Pl}}{\phi_0} \right)^2 \right]. \qquad (6.94)$$
<figure><img src="image_67.png" /><figcaption>Fig. 64. Left panel: slow-roll parameter $\epsilon_1$ as a function of the field $\phi/\phi_0$, for $\phi_0/M_{\text{pl}} = 0.02 < 0.04228$, see Eq. (6.91). The solid black line corresponds to the approximated slow-roll formula of Eq. (6.89), i.e. $\epsilon_1^V = M_{\text{pl}}^2/2V_0^2/V^2$, while the solid blue line represents the exact $\epsilon_1^H = -\dot{H}/H^2$ obtained from a numerical integration starting at $\phi_{\text{uni}}/M_{\text{pl}} = 0.33$ and vanishing initial velocity. We see that the exact $\epsilon_1^H$ remains in fact always smaller than one and that inflation never stops. The inflaton eventually oscillates around the minimum of its potential located at $\phi = 2\phi_0$ (the arrows indicate the direction of the first oscillations). Right panel: Maximum value taken by $\epsilon_1^V$ (solid black line) and $\epsilon_1^H$ (solid blue line) for different values of $\phi_0$. One can see that $\epsilon_1^H$ remains smaller than one for any value of $\phi_0$. When $\phi_0$ increases, the slow-roll parameters, which scale proportional to $M_{\text{pl}}^2/\phi_0^2$, decrease so that the slow-roll approximation becomes more and more efficient and eventually starts matching the numerical exact predictions.</figcaption></figure>

It is valid provided $\phi_0/M_{\text{pl}} \ll 1$, i.e. precisely in the regime for which the TWI potential was derived. Using this formula, one can estimate the value of the three first Hubble flow parameters at Hubble crossing. One arrives at

$$ \epsilon_{1*} \approx \frac{A^2}{2} \left( \frac{M_{\text{Pl}}}{\phi_0} \right)^2 e^{-2x_*} x_*^4 \approx \frac{1}{32\Delta N_*^2} \left( \frac{\phi_0}{M_{\text{Pl}}} \right)^2, \\ \epsilon_{2*} \approx \frac{\epsilon_{3*}}{2} \approx 2A \left( \frac{M_{\text{Pl}}}{\phi_0} \right)^2 e^{-x_*} x_*^2 \approx \frac{1}{2\Delta N_*}. \qquad (6.95) $$

Finally, we can derive an expression for the tensor-to-scalar ratio, the spectral index

$$ r \approx 8A^2 \left( \frac{M_{\text{Pl}}}{\phi_0} \right)^2 e^{-2x_*} x_*^4 \approx \frac{1}{2\Delta N_*^2} \left( \frac{\phi_0}{M_{\text{Pl}}} \right)^2, \quad n_s - 1 \approx -2A \left( \frac{M_{\text{Pl}}}{\phi_0} \right)^2 x_*^2 e^{-x_*} \approx \frac{1}{2\Delta N_*}, \qquad (6.96) $$

and the running

$$ \alpha_s \approx -2A^2 \left( \frac{M_{\text{Pl}}}{\phi_0} \right)^4 x_*^4 e^{-2x_*} \approx -\frac{1}{8\Delta N_*^2}. \qquad (6.97) $$

These estimates are in agreement with the ones of Ref. [568], up to a missing factor 4 in Eq. (6.94). However, we have checked that this does not affect the predictions in a significant way.

It is also interesting to discuss the value of the scale M since this is important from the model building point of view. The CMB normalization gives

$$ \frac{M^4}{M_{\text{Pl}}^4} = 720\pi^2 A^2 \left( \frac{M_{\text{Pl}}}{\phi_0} \right)^2 \frac{[e^{-x_*} x_* (x_* - 2)]^2}{(1 - A x_*^2 e^{-x_*})^3} \frac{Q^2_{\text{rms-PS}}}{T^2}. \qquad (6.98) $$

In the vacuum dominated approximation, the above expression simplifies and gives $M^4/M_{\text{pl}}^4 \approx 45\pi^2/\Delta N_*^2 \phi_0^2/M_{\text{pl}}^2 Q^2_{\text{rms-PS}}/T^2$. This leads to

$$ M_{\text{Pl}} R = \sqrt{\frac{2N}{45A} \frac{\Delta N_*}{\pi^3}} \frac{T}{Q_{\text{rms-PS}}} \approx 1.2 \times 10^5 \sqrt{N}, \qquad (6.99) $$

where we have taken $\Delta N_* \approx 60$. This also implies that

$$ \frac{\phi_0}{M_{\text{Pl}}} \approx \frac{1.35}{\sqrt{N}} \times 10^{-5}. \qquad (6.100) $$

Therefore, we have a rough determination of the compactification radius. The model seems consistent since we obtain that $M_{\text{pl}}R \gg 1$, in agreement with the assumptions made at the beginning of this section.

The predictions for TWI are presented in Fig. 188. The reheating equation of state parameter $\bar{w}_{\text{reh}}$ has been taken to be 0 since the potential is quadratic close to its minimum. However, since the details of reheating depend on the details of the interactions between the inflaton field and the others degrees of freedom in the theory, this parameter is a priori unspecified and could very well take different values. In the ASPIC code, $\bar{w}_{\text{reh}}$ can be freely chosen. Anyway, since the reheating temperature is in fact fully degenerate with the parameter $x_{\text{end}}$, these two parameters cannot be constrained independently. One can check that the rough description provided by Eqs. (6.96) is correct: the model typically predicts a small amount of gravitational waves which increases with $\phi_0$, and a deviation from scale invariance which does not significantly depends on $\phi_0$. When $\phi_0/M_{\text{pl}} = O(1)$, however, one notices a turning point (at fixed values of $\phi_0$). This corresponds to the separation between two regimes, one where $x_* < x_{\text{max}}$ and $\epsilon_1$ is an increasing function of $x$ (hence $\epsilon_{1*}$ increases with $x_{\text{end}}$) and another where $x_* > x_{\text{max}}$ and $\epsilon_1$ is a decreasing function of $x$ (hence $\epsilon_{1*}$ decreases with $x_{\text{end}}$). If a sufficient number of e-folds can be realized in the $2 < x < x_{\text{max}}$ part of the potential, then $\epsilon_{2*}$ can become negative. However, this mostly happens for fine-tuned values of $x_{\text{end}} \approx 2$.
<figure><img src="image_68.png" /><figcaption>Fig. 65. GMSSM Inflation (GMSSMI). Top left panel: GMSSM Inflation potential Eq. (6.102) for $\alpha = 0.1, 0.7, 1.5, 2.5$, as a function of $\phi/\phi_0$. Top right panel: logarithm of the potentials for the same value of $\alpha$. Bottom left panel: slow-roll parameter $\epsilon_1$ for a potential with the same values of $\alpha$. Bottom right panel: slow-roll parameter $\epsilon_2$ for a potential with the same values of $\alpha$. See discussion in the text body.</figcaption></figure>

## 6.6. Generalized MSSM Inflation (GMSSMI)

As for the MSSMI models, see Section 5.17, GMSSMI scenarios are based on the Minimal Supersymmetric Model (MSSM) in which a flat direction is lifted by soft supersymmetry breaking terms and by superpotential corrections. The potential is of the form

$$V(\phi) = \frac{1}{2} m_{\phi}^2 \phi^2 - A \frac{\lambda_n}{n} \frac{\phi^n}{M_{Pl}^{n-3}} + \lambda_n^2 \frac{\phi^{2(n-1)}}{M_{Pl}^{2(n-3)}}. \qquad (6.101)$$

The MSSMI model corresponds to $n = 6$ and $A^2 = 8(n-1)m_{\phi}^2$. This last relation is of crucial importance since it implies an exactly flat inflection point. Following Refs. [486,487,490,569–572], one may wonder whether the model is robust when this relation is not exactly satisfied. In order to investigate this question, we therefore relax the condition $A^2 = 8(n-1)m_{\phi}^2$. In this more general case, the potential can be reparametrized in the form

$$V(\phi) = M^4 \left[ \left( \frac{\phi}{\phi_0} \right)^2 - \frac{2}{3} \alpha \left( \frac{\phi}{\phi_0} \right)^6 + \frac{\alpha}{5} \left( \frac{\phi}{\phi_0} \right)^{10} \right], \qquad (6.102)$$

where $\phi_0 \simeq 10^{14}$ GeV, this value being the same as the one found in Section 5.17. The positive dimensionless parameter $\alpha$ encodes any deviations from the MSSM case for which it equals unity, $\alpha_{MSSM} = 1$.

The potential is displayed in Fig. 65, where four cases can be distinguished. In the following, we define the quantity $x$ by the expression

$$x \equiv \frac{\phi}{\phi_0}. \qquad (6.103)$$

If $\alpha < 9/25$, the second derivative of the potential does not vanish and the potential is convex everywhere. This corresponds to the case $\alpha = 0.1$ case in Fig. 65. If $9/25 < \alpha < 1$, the potential has two inflection points $x_{V''=0}^{\pm}$ and is concave in between. It remains an increasing function of the field since its first derivative never vanishes. This is illustrated with the case $\alpha = 0.7$ in Fig. 65. If $\alpha = 1$, this is the MSSM inflation models (see Section 5.17) where the potential has a flat inflection point. If $1 < \alpha < 9/5$, the potential decreases in between $x_{V'=0}^{\pm}$ but remains positive everywhere. This is exemplified by the case $\alpha = 1.5$ in Fig. 65. Finally, if $\alpha > 9/5$, the potential becomes negative (hence is not properly defined) between the two points $x_{V'=0}^{\pm}$ (see $\alpha = 2.5$ in Fig. 65). The values of the field $\text{vev}$s appearing in this discussion are given by the following formulas:

$$x_{V''=0}^{\pm} = \left[ \frac{5}{9} \left( 1 \pm \sqrt{1 - \frac{9}{25\alpha}} \right) \right]^{1/4}, \qquad x_{V'=0}^{\pm} = \left( 1 \pm \sqrt{1 - \frac{1}{\alpha}} \right)^{1/4}, \qquad (6.104)$$
and

$$x_{V=0}^{\pm} = \left[ \frac{5}{3} \left( 1 \pm \sqrt{1 - \frac{9}{5\alpha}} \right) \right]^{1/4}. \qquad (6.105)$$

Let us now calculate the first three Hubble flow functions in the slow-roll approximation. They are given by

$$\begin{align}
\epsilon_1 &= 450 \left( \frac{M_{\mathrm{Pl}}}{\phi_0} \right)^2 \frac{(1 - 2\alpha x^4 + \alpha x^8)^2}{x^2 (15 - 10\alpha x^4 + 3\alpha x^8)^2}, \nonumber \\
\epsilon_2 &= 60 \left( \frac{M_{\mathrm{Pl}}}{\phi_0} \right)^2 \frac{15 + 40\alpha x^4 + \alpha (20\alpha - 78) x^8 + 3\alpha^2 x^{16}}{x^2 (15 - 10\alpha x^4 + 3\alpha x^8)^2}, \tag{6.106}
\end{align}$$

and

$$\begin{align}
\epsilon_3 = 60 \left( \frac{M_{\mathrm{Pl}}}{\phi_0} \right)^2 & \left[ 225 - 1800\alpha x^4 + 60\alpha (69 + 10\alpha) x^8 - 40 (189 - 100\alpha) \alpha^2 x^{12} \right. \nonumber \\
& \left. \qquad + 10\alpha^2 (243 - 504\alpha + 402\alpha^2) x^{16} + 40\alpha^3 (117 - 20\alpha) x^{20} + 12\alpha^3 (10\alpha - 123) x^{24} \right. \nonumber \\
& \left. \qquad + 72\alpha^4 x^{28} + 9\alpha^4 x^{32} \right] \times \left[ 3375x^2 + 4500\alpha x^6 - 600\alpha (27 + 10\alpha) x^{10} \right. \nonumber \\
& \left. \qquad + 100\alpha^2 (261 - 20\alpha) x^{14} + 10\alpha^2 (200\alpha^2 - 840\alpha - 621) x^{18} + 60\alpha^3 (69 - 20\alpha) x^{22} \right. \nonumber \\
& \left. \qquad + 48\alpha^3 (10\alpha - 9) x^{26} - 180\alpha^4 x^{30} + 27\alpha^4 x^{34} \right]^{-1}. \tag{6.107}
\end{align}$$

The first two slow-roll parameters diverge when $x \to 0$ and vanish asymptotically. In between, their shape depends on $\alpha$ as it is represented in Fig. 65. If $\alpha < 1$, $\epsilon_1$ first decreases, reaches a local non-zero minimum where $\epsilon_2$ vanishes, then increases to reach a local maximum where $\epsilon_2$ vanishes again, and eventually decreases again. Let $x_{\epsilon_2=0}^{\pm}$ be the position of these two local extrema. From Ferrari's solutions for depressed quartic equations one gets

$$x_{\epsilon_2=0}^{\pm} = \left[ \frac{1}{2\alpha} \sqrt{\frac{5}{3}} \left( \sqrt{\Sigma} \pm 2 \sqrt{\frac{39}{5}\alpha - 2\alpha^2 - \frac{\Sigma}{4} - \frac{12}{\sqrt{15\Sigma}}\alpha^2} \right) \right]^{1/4}, \qquad (6.108)$$

where

$$\begin{align}
\delta &= \frac{736\alpha^2}{25} - \frac{208\alpha^3}{15} + \frac{16\alpha^4}{9}, \nonumber \\
\Delta &= -\frac{430336\alpha^4}{625} + \frac{612352\alpha^5}{1125} - \frac{20992\alpha^6}{225} + \frac{256\alpha^8}{243}, \nonumber \\
\sigma &= -\frac{12896\alpha^3}{125} + \frac{2944\alpha^4}{25} - \frac{416\alpha^5}{15} + \frac{64\alpha^6}{27} + \frac{6}{5}\sqrt{15\Delta\alpha}, \tag{6.109} \\
\Sigma &= \frac{52\alpha}{5} - \frac{8\alpha^2}{3} + \frac{\delta}{\sigma^{1/3}} + \sigma^{1/3}, \nonumber
\end{align}$$

are intermediate quantities introduced solely to reduce the size of Eq. (6.108). If $\alpha > 1$, $\epsilon_1$ has two local minimums located at $x_{\epsilon_2=0}^{\pm}$ where it vanishes. In between it reaches a local maximum or may even diverge for $\alpha > 9/5$ (see Fig. 65). The slow-roll parameter $\epsilon_2$ vanishes when $\epsilon_1$ reaches these local maxima, or diverge when $\epsilon_1$ does (for $\alpha > 9/5$). As explained in Section 5.17, inflation is meant to proceed at $\phi \lesssim \phi_0$. Let us assume that inflation can end for $\epsilon_1 > 1$ between $x = 0$ and the position of the first minimum $x_{\epsilon_1}^{\min}$. Following the previous considerations, this latter location is defined as

$$x_{\epsilon_1^{\min}} = \begin{cases} x_{\epsilon_2=0}^{-} & \text{if } \alpha < 1 \\ x_{V'=0}^{-} & \text{if } \alpha > 1 \end{cases}, \qquad (6.110)$$

and provides an upper bound to $x_{\mathrm{end}}$ the solution of $\epsilon_1(x_{\mathrm{end}}) = 1$. This one can only be determined numerically. The values of $x_{\epsilon_2=0}^{\pm}$ and $x_{V'=0}^{\pm}$ in terms of $\alpha$ are displayed in the left panel in Fig. 66 together with $x_{\epsilon_1^{\min}}$. The right panel of Fig. 66 represents the value of the first slow-roll parameter at this minimum, $\epsilon_1^{\min} = \epsilon_1(x_{\epsilon_1^{\min}})$. For $\alpha < 1$, one can see that $\epsilon_1^{\min} < 1$ only if the parameter $\alpha \lesssim 1$. This defines a minimum value for $\alpha$, which depends on $\phi_0$, such that inflation can take place within this domain. When $\alpha \simeq 1$, one can derive an approximated version of Eq. (6.108), namely, $x_{\epsilon_2=0}^{-} \simeq 1 - (1 - \alpha)/32$. Plugging it into the expression for $\epsilon_1$ one obtains

$$\epsilon_1^{\min} \approx \frac{225}{32} (\alpha - 1)^2 \frac{M_{\mathrm{Pl}}^2}{\phi_0^2}, \qquad (6.111)$$

from which one gets

$$\alpha > 1 - \frac{4\sqrt{2}}{15} \frac{\phi_0}{M_{\mathrm{Pl}}}. \qquad (6.112)$$

For the value suggested in Ref. [483], $\phi_0/M_{\mathrm{Pl}} \simeq 10^{-4}$, one obtains $\alpha > 1 - 10^{-5}$, which is in agreement with Ref. [569], and shows that the model needs to be sufficiently fine-tuned (i.e. sufficiently close to regular MSSM inflation) in order to be a viable inflationary model.
<figure><img src="image_69.png" /><figcaption>Fig. 66. GMSSM Inflation (GMSSMI). Left panel: $x_{\epsilon_2=0}^{\pm}$ defined in Eq. (6.108) and $x_{\nu=0}^{\pm}$ defined in Eq. (6.104) together with $x_{\epsilon_{min}}$ [see Eq. (6.110)] as a function of $\alpha$. Right panel: minimal value of the slow-roll parameter $\epsilon_1$ (rescaled by $\phi_0^2/M_n^2$) as a function of $\alpha$. When it is greater than unity, inflation cannot occur.</figcaption></figure>

On top of that, as shall be seen now, the constraints on $\alpha$ are even tighter if one wants a sufficient number of e-folds to be produced. Let us thus turn to the slow-roll trajectory. It can be integrated, and leads to

$$N_{\text{end}} - N = \frac{\phi_0^2}{M_{\text{Pl}}^2} \left\{ -\frac{x_{\text{end}}^2 - x^2}{20} - \frac{b_+}{10\sqrt{a_+}} \left[ \arctan(\sqrt{a_+}x_{\text{end}}^2) - \arctan(\sqrt{a_+}x^2) \right] - \frac{b_-}{10\sqrt{a_-}} \left[ \arctan(\sqrt{a_-}x_{\text{end}}^2) - \arctan(\sqrt{a_-}x^2) \right] \right\}, \quad (6.113)$$

where

$$a_{\pm} = -\alpha \pm \sqrt{\alpha^2 - \alpha}, \quad b_{\pm} = 2 \frac{a_{\pm} + \alpha/3}{a_{\pm} - a_{\mp}}, \quad (6.114)$$

A few remarks are in order. Firstly, even if the terms appearing in the previous expression are complex, their imaginary contributions cancel out and the resulting expression is truly a real quantity. Then, one can check that formally, when $\alpha \to 0$, one has $a_{\pm} \to 0$ and $b_{\pm} \to 1$, hence $N \simeq - (x^2 - x_{\text{ini}}^2)/4$, which is precisely the LFI slow-roll trajectory for $p=2$, see Section 5.2. This is just a formal check since $\alpha$ is meant to be tuned close to 1 in the GMSSMI scenario. Finally, let us notice that, in the case $\alpha < 1$, and contrary to the MSSM models ($\alpha=1$), the number of e-folds never diverges at a given point $x$. Therefore, the total number of e-folds is bounded from above for the field vev's considered here. Working out the limit of Eq. (6.113) when $\alpha \to 1$, one has

$$N_{\text{end}} - N_{\text{ini}} \le \left( \frac{\phi_0}{M_{\text{Pl}}} \right)^2 \frac{\pi}{30} \frac{1}{\sqrt{1-\alpha}}. \quad (6.115)$$

If one require at least $\Delta N = N_{\text{end}} - N_{\text{ini}}$ e-folds during inflation, then $\alpha$ has to be fine-tuned to

$$\alpha > 1 - \left( \frac{\phi_0}{M_{\text{Pl}}} \right)^4 \frac{\pi^2}{900 \Delta N^2}. \quad (6.116)$$

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. (6.112). Taking $\phi_0/M_{\text{Pl}} \simeq 10^{-4}$ and $\Delta N \simeq 50$, one obtains $\alpha > 1 - 10^{-22}$. This is clearly an extreme fine-tuning which can even make the numerical investigation of the model challenging.<fnref n="8" /> As explained below, the same condition $|\alpha - 1| < \phi_0^4/M_{\text{Pl}}^4/\Delta N^2$ also applies to the case $\alpha > 1$ in order to maintain an acceptable deviation from scale invariance. This makes GMSSM inflation a severely fine-tuned scenario. Let us also notice that our parameter $\alpha$ is related to the parameter $\delta$ of Ref. [570] by $\delta = \sqrt{\alpha^{-2} - 1}$. Ref. [570] finds that, in order for the model to be compatible with the data, $\delta \simeq 10^{-20}$. Therefore, although our method slightly differs from that of Ref. [570], our results are in broad agreement.

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

$$\left(\frac{M}{M_{\text{Pl}}}\right)^4 = 2880\pi^2 \frac{M_{\text{Pl}}^2}{\phi_0^2} \frac{(1 - 2\alpha x_*^4 + \alpha x_*^8)^2}{x_*^4 \left(1 - \frac{2}{3}\alpha x_*^4 + \frac{\alpha}{5}x_*^8\right)^3} \frac{Q_{\text{rms-PS}}^2}{T^2}. \quad (6.117)$$

As explained in Section 5.17, this leads to $M/M_{\text{Pl}} \simeq 10^8$ GeV for $\phi_0/M_{\text{Pl}} \simeq 10^{-4}$.

The reheating consistent slow-roll predictions of the GMSSMI models are displayed in Figs. 191, 193, for $\alpha > 1$ and $\alpha < 1$, respectively. 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 MSSM predictions are recovered, see Fig. 150. The amount of gravitational waves $r$ seems to be quite independent on $\alpha$ and, therefore, is similar to its regular MSSM counterpart. On the other hand, the spectral index $n_S$ strongly depends on $\alpha$. In the case $\alpha > 1$, larger values of $\alpha - 1$ worsens the spectral index problem, already present in standard MSSMI. 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. 193, the spectral index varies so quickly with $\alpha$ that one has to fine-tune the power of the fine-tuning to remain inside the two-sigma contours. In Refs. [487,569–572], it is argued that, since

<footnote n="8">⁸ This exceeds the usual 64 bits precision on floating point numbers (FP64).</footnote>
<figure><img src="image_70.png" /><figcaption>Fig. 67. Top left panel: Generalized Renormalizable Point Inflation (GRIPi) potential given by Eq. (6.119) for $\alpha = 0.7, 0.85, 1, 1.094, 1.188$, as a function of $\phi/\phi_0$. Top right panel: logarithm of the potentials for the same values of $\alpha$. Bottom left panel: slow-roll parameter $\epsilon_1$ rescaled by $M_n^2/\phi_0^2$, for GRIPi models with the same values of $\alpha$. Bottom right panel: slow-roll parameter $\epsilon_2$, rescaled by $M_n^2/\phi_0^2$. A description of these various quantities can be found in the text.</figcaption></figure>

the flat saddle point condition is robust against radiative corrections, such a fine-tuning may not be a problem. However, as explained here and in Section 5.17, if the flat saddle point condition is exactly satisfied, the model is disfavored by the observations because the spectral index is too red. The only way out is therefore to detune the condition $\alpha = 1$ at an extremely fine-tuned level.

## 6.7. Generalized Renormalizable Point Inflation (GRIPi)

As for the MSSMI models (see Section 5.17) and for the RIPI models (see Section 5.18), the GRIPi models have a potential of the form

$$V(\phi) = \frac{1}{2} m_{\phi}^2 \phi^2 - A \frac{\lambda_n}{n} \frac{\phi^n}{M_{Pl}^{n-3}} + \lambda_n^2 \frac{\phi^{2(n-1)}}{M_{Pl}^{2(n-3)}}. \qquad (6.118)$$

In Section 5.18, the particular example $n = 3$ is discussed in the case where the potential has a flat inflection point, i.e. when $A^2 = 16m_{\phi}^2$. Then, as studied in Section 6.6 for MSSMI, comes the question of what happens when we relax this condition. To address this issue, it is convenient to reparametrize the potential as

$$V(\phi) = M^4 \left[ \left( \frac{\phi}{\phi_0} \right)^2 - \frac{4}{3} \alpha \left( \frac{\phi}{\phi_0} \right)^3 + \frac{\alpha}{2} \left( \frac{\phi}{\phi_0} \right)^4 \right], \qquad (6.119)$$

where the positive dimensionless parameter $\alpha$ encodes the deviation from the RIPI case (that is to say $\alpha_{RIPI} = 1$). This model was studied in Ref. [573] and in Refs. [574,575]. In the first reference, the mass $m_{\phi}$ is fixed by the soft supersymmetry breaking terms and, in Section 5.18, it was shown that this leads to $\phi_0 \simeq 10^{14}$ GeV. However, in Refs. [574,575], the scale $m_{\phi}$ is no longer controlled by the soft supersymmetry breaking terms but by the right-handed neutrino mass in Type I supersymmetric seesaw and this leads to a different value for $\phi_0$, namely $\phi_0 \simeq 10^{17}$ GeV. Therefore, in what follows, we will use both values.

The potential is displayed in Fig. 67, where four cases can be distinguished. In the following, for convenience, we use the quantity $x$ defined by

$$x \equiv \frac{\phi}{\phi_0}. \qquad (6.120)$$

If $\alpha < 3/4$, the second derivative of the potential does not vanish and the potential is convex everywhere. This corresponds to the case $\alpha = 0.7$ case in Fig. 67. If $3/4 < \alpha < 1$, the potential has two inflection points $x_{V''=0}^{\pm}$ and is concave in between. It remains an increasing function of the field since its first derivative never vanishes. This is illustrated by the case $\alpha = 0.85$ in Fig. 67. If $\alpha = 1$, then this is the RIPI model (see Section 5.18) where the potential has a flat inflection point. If $1 < \alpha < 9/8$, then the potential decreases between the two values of $x$, $x_{V''=0}^{\pm}$, for which the derivative is zero, but remains positive everywhere. Typically, this corresponds to the case $\alpha = 1.094$ in Fig. 67. Finally, if $\alpha > 9/8$, then
the potential becomes negative (hence is not properly defined everywhere) between $x_{V=0}^{\pm}$ (see the case $\alpha = 1.188$ in Fig. 67). The values of the field $v_{\nu}$ in this discussion are given by the following formulas:

$$x_{V''=0}^{\pm} = \frac{2}{3} \left( 1 \pm \sqrt{1 - \frac{3}{4\alpha}} \right), \quad x_{V'=0}^{\pm} = 1 \pm \sqrt{\frac{\alpha - 1}{\alpha}}, \qquad (6.121)$$

and

$$x_{V=0}^{\pm} = \frac{4}{3} \left( 1 \pm \sqrt{1 - \frac{9}{8\alpha}} \right). \qquad (6.122)$$

Let us now calculate the first Hubble flow functions in the slow-roll approximation. They are given by

$$\epsilon_1 = 72 \left( \frac{M_{Pl}}{\phi_0} \right)^2 \frac{(1 - 2\alpha x + \alpha x^2)^2}{x^2 (6 - 8\alpha x + 3\alpha x^2)^2}, \qquad (6.123)$$

$$\epsilon_2 = 24 \left( \frac{M_{Pl}}{\phi_0} \right)^2 \frac{6 - 16\alpha x + (3 + 16\alpha)\alpha x^2 - 12\alpha^2 x^3 + 3\alpha^2 x^4}{x^2 (6 - 8\alpha x + 3\alpha x^2)^2},$$

and

$$\epsilon_3 = 24 \left( \frac{M_{Pl}}{\phi_0} \right)^2 \left[ \begin{aligned} & 36 - 216\alpha x + 30\alpha (3 + 16\alpha) x^2 - 8(45 + 64\alpha) \alpha^2 x^3 \\ & + 2(27 + 276\alpha + 128\alpha^2) \alpha^2 x^4 - 2(208\alpha + 81) \alpha^3 x^5 + 9(1 + 28\alpha) \alpha^3 x^6 \\ & - 72\alpha^4 x^7 + 9\alpha^4 x^8 \end{aligned} \right] \times \left[ x^2 (6 - 8\alpha x + 3\alpha x^2)^2 (6 - 16\alpha x + 3\alpha x^2 + 16\alpha^2 x^2 - 12\alpha^2 x^3 + 3\alpha^2 x^4) \right]^{-1}. \qquad (6.124)$$

The first two slow-roll parameters diverge when $x \to 0$ and asymptotically goes to zero when $x \to \infty$. In between, their behavior depends on $\alpha$ as can be seen in Fig. 67. If $\alpha < \alpha_0$, where

$$\alpha_0 = \frac{3}{16} \left[ 5 - 3^{2/3} (6 - 2\sqrt{3})^{-1/3} - 2^{-2/3} (9 - 3\sqrt{3})^{1/3} \right] \approx 0.4671, \qquad (6.125)$$

$\epsilon_1$ monotonically decreases with $x$. If $\alpha_0 < \alpha < 1$, $\epsilon_1$ first decreases, reaches a local non-vanishing minimum at a value of $x$ for which $\epsilon_2$ vanishes, then increases to reach a local maximum where $\epsilon_2$ vanishes again, and eventually decreases for $x \to \infty$, as already mentioned. Let $x_{\epsilon_2=0}^{\pm}$ be the position of these two local extrema. Similarly to Eq. (6.108) for the generalized MSSM inflation models, analytic expressions can be obtained for these two quantities using Ferrari's solutions for depressed quartic equations. They are implemented in ASPIC but are not displayed here since this does not add much to the discussion. If $\alpha > 1$, $\epsilon_1$ has two local minima located at $x_{V'=0}^{\pm}$ where it vanishes. In between it reaches a local maximum or may even diverge for $\alpha > 9/8$ (see Fig. 67). The slow-roll parameter $\epsilon_2$ vanishes when $\epsilon_1$ reaches these local maxima, or diverge when $\epsilon_1$ itself diverges (for $\alpha > 9/8$).

As explained in Section 5.18, inflation is supposed to proceed at $\phi \le \phi_0$. Let us assume that inflation ends by violation of slow-roll between $x=0$ and the position of the first minimum $x_{\epsilon_1}^{\min}$. Following the previous considerations, this latter value of $x$ is defined by

$$x_{\epsilon_1}^{\min} = \begin{cases} x_{\epsilon_2=0}^{-} & \text{if } \alpha_0 < \alpha < 1 \\ x_{V'=0}^{-} & \text{if } \alpha > 1 \end{cases}, \qquad (6.126)$$

and, moreover, provides an upper bound to determine $x_{\text{end}}$ [i.e. the solution of the equation $\epsilon_1(x_{\text{end}}) = 1$]. Let us emphasize that this one can only be determined numerically. The values of $x_{\epsilon_2=0}^{\pm}$ and $x_{V'=0}^{\pm}$ in terms of $\alpha$ are displayed in the left panel of Fig. 68 together with $x_{\epsilon_1}^{\min}$. The right panel of Fig. 68 represents the value of the first slow-roll parameter at this minimum, $\epsilon_1^{\min} = \epsilon_1(x_{\epsilon_1}^{\min})$. For $\alpha < \alpha_0$, one has $\epsilon_1(x=1) > 1.5 M_{Pl}^2/\phi_0^2$ and, recalling that typically $\phi_0 \simeq 10^{14}$ GeV or $\phi_0 \simeq 10^{17}$ GeV, one sees that inflation cannot proceed in this case. For $\alpha_0 < \alpha < 1$, one has $\epsilon_1^{\min} < 1$ only if the parameter $\alpha \le 1$. This defines a minimum value for $\alpha$, which depends on $\phi_0$, allowing for inflation to take place. When $\alpha \simeq 1$, one can derive an approximated formula for $x_{\epsilon_2=0}^{-}$, namely, $x_{\epsilon_2=0}^{-} \simeq 1 - (1 - \alpha)/2$. Plugging it into the expression for $\epsilon_1$ one obtains

$$\epsilon_1^{\min} \simeq 72(\alpha - 1)^2 \frac{M_{Pl}^2}{\phi_0^2}, \qquad (6.127)$$

from which it follows that

$$\alpha > 1 - \frac{\sqrt{2}}{12} \frac{\phi_0}{M_{Pl}}. \qquad (6.128)$$

With $\phi_0/M_{Pl} \simeq 10^{-1}$, one obtains $\alpha > 0.99$, which shows that the model needs to be sufficiently fine-tuned such that it becomes very similar to the regular RIPI scenario. If, on the other hand, $\phi_0/M_{Pl} \simeq 10^{-4}$, the constraint is much tighter. As discussed in Refs. [574,575], one of the main advantage of the model studied in those references is that a value $\phi_0 \simeq 10^{17}$ GeV leads to a less severe fine tuning problem than $\phi_0 \simeq 10^{14}$ GeV.