Shift Symmetry and Inflation in Supergravity

TL;DR

The paper investigates inflation within supergravity using a shift symmetry to protect the inflaton flat direction while stabilizing moduli via KKLT. By mildly breaking the shift symmetry with an inflaton-dependent superpotential, it achieves a controlled slope and alleviates the eta-problem, enabling viable chaotic-like inflation. The authors introduce mutated chaotic inflation, where a moduli field acts as a curved waterfall along a nontrivial trajectory, and they show how KKLT stabilization can lift AdS vacua to dS, yielding robust inflationary dynamics. They demonstrate, through both analytical structure and numerical simulations, that inflation ends via slow-roll violation rather than instability, and they discuss implications for cosmological perturbations and observational signatures, including two-field effects and possible isocurvature contributions.

Abstract

We consider models of inflation in supergravity with a shift symmetry. We focus on models with one moduli and one inflaton field. The presence of this symmetry guarantees the existence of a flat direction for the inflaton field. Mildly breaking the shift symmetry using a superpotential which depends not only on the moduli but also on the inflaton field allows one to lift the inflaton flat direction. Along the inflaton direction, the eta-problem is alleviated. Combining the KKLT mechanism for moduli stabilization and a shift symmetry breaking superpotential of the chaotic inflation type, we find models reminiscent of ``mutated hybrid inflation'' where the inflationary trajectory is curved in the moduli--inflaton plane. We analyze the phenomenology of these models and stress their differences with both chaotic and hybrid inflation.

Figures (13)

• Figure 1: Potential $V(\bar{\rho}, \bar{\phi} )$ given by Eq. (\ref{['potmod']}) represented in terms of the normalized field $\bar{\phi }$ and $\bar{\rho }$. The potential possesses a minimum in the $\bar{\phi }$ direction but, as is obvious from the figure, not in the $\bar{\rho }$ direction. Hence, the moduli is not stabilized in this model.
• Figure 2: The function $f\left(y_{{\rm min},\tilde{\cal U}}\right)$ defined in Eq. (\ref{['deff']}). The maximum of this function is located at $y_{{\rm min},\tilde{\cal U}}=\sqrt{2}$ and is equal to $1.17387$. The intersection of $f\left(y_{{\rm min},\tilde{\cal U}}\right)$ with the horizontal line (solid line) $3\alpha m/(2\kappa A)$ determines the value(s) of $\rho$ corresponding to the extrema of the function $\tilde{\cal U}$. According to the values of the parameter $m$ and $A$, the line moves up or downwards while the function $f\left(y_{{\rm min},\tilde{\cal U}}\right)$ remains the same. As explained in the text, if the line is above the maximum of $f(y_{{\rm min},\tilde{\cal U}})$ then there are no extrema. If the line is below the maximum, there are always two extrema. The first intersection between $f\left(y_{{\rm min},\tilde{\cal U}}\right)$ and the horizontal line [in the increasing part of the function $f\left(y_{{\rm min},\tilde{\cal U}}\right)$] corresponds to a minimum while the second intersection corresponds to a maximum. However, if the line is below the value $3/{\rm e}\simeq 1.10364$, the corresponding value of $\tilde{\cal U}$ is negative. The effective squared mass $\tilde{\cal U}$ is positive at the minimum if the horizontal line is in the dashed area which, therefore, represents the allowed region.
• Figure 3: Left panel: function $\tilde{\cal U}(\rho )$ for three different values of the ratio $\kappa A/(\alpha m)$. The dashed line corresponds to a situation where there is no minimum, i.e. the line $3\alpha m/(2\kappa A)$ is above the function $f\left(y_{{\rm min},\tilde{\cal U}}\right)$, see Fig. \ref{['studyu']}. The dotted line represents the case where there is a positive minimum, i.e. the line $3\alpha m/(2\kappa A)$ is within the dashed area in Fig. \ref{['studyu']}. Finally, the solid line corresponds to a function $\tilde{\cal U}$ with a negative minimum. Right panel: function $\tilde{\cal V}(\rho )$ for three different values of the ratio $W_0/A$. One can check on the figure that, if $W_0/A<1$ then there is a negative minimum and $\lim _{y\rightarrow 0}\tilde{\cal V}=+\infty$ while, if $W_0/A> 1$ there is no minimum but $\lim _{y\rightarrow 0}\tilde{\cal V}=-\infty$.
• Figure 4: Potential $V(\bar{\rho}, \bar{\phi })$ [what is actually plotted is $10^{11}V(\bar{\rho}, \bar{\phi })/m_{_{\mathrm Pl}}^4$] for the following choice of parameters: $\alpha =\sqrt{2}$, $\beta =1$, $m=10^{-6}m_{_{\mathrm Pl}}$, $\kappa A/(\alpha m)=1.35135$ and $W_0/A=0.41111$. This gives an absolute minimum of the potential located at $\bar{\phi }=0$, $\bar{\rho }_{\rm min, \tilde{\cal V}}\simeq 0.1181\times m_{_{\mathrm Pl}}$. These parameters satisfy the constraints derived in the text. The moduli is stabilized at the value $y=y_{\rm min, \tilde{\cal U}}\simeq 1.016$ or $\bar{\rho }_{\rm min, \tilde{\cal U}}\simeq 0.0040\times m_{_{\mathrm Pl}}$ (for $\bar{\phi }\gg m_{_{\mathrm Pl}}$) as can be checked in the figure where the inflationary valley is clearly seen.
• Figure 5: Trajectory of the valley of stability in the plan $(\bar{\rho} ,\bar{\phi} )$ (the valley is seen from above). For large values of $\bar{\phi}$, $\bar{\phi }\gg m_{_{\mathrm Pl}}$, the valley coincides with the minimum of the function $\tilde{\cal U}$, i.e. is just given by the equation $\bar{\rho} =\bar{\rho }_{\rm min, \tilde{\cal U}}$. For smaller values of the inflaton, the trajectory bends and quickly joins the absolute minimum of the potential, namely $y=y_{\rm min, \tilde{\cal V}}$, $\phi=0$. The parameters used here are: $\alpha =\sqrt{2}$, $\beta =1$, $m=10^{-6}m_{_{\mathrm Pl}}$, $\kappa A/(\alpha m)=1.35135$ and $W_0/A=0.41111$. This gives $y=y_{\rm min, \tilde{\cal U}}\simeq 1.016$ or $\bar{\rho }_{\rm min, \tilde{\cal U}}\simeq 0.0040\times m_{_{\mathrm Pl}}$ and the global minimum of the potential is located at $\bar{\phi }=0$, $\bar{\rho }_{\rm min, \tilde{\cal V}}\simeq 0.1181\times m_{_{\mathrm Pl}}$.