No-scale supergravity

In a no-scale sector, the Kähler metric satisfies $K_i{K}^{i\bar{j}}{K}_{\bar{j}}=3$. If the superpotential is independent of that sector, its covariant derivatives contribute $3e^K|W{|}^2$, cancelling the universal $-3e^K|W{|}^2$ in the supergravity F-term potential. Spectator sectors can still contribute to the potential. This cancellation permits zero-energy minima with broken supersymmetry and exact flat scalar directions.

For $s=S+\bar{S}>0$, $t=T+\bar{T}-|C{|}^2>0$, $K=-\log s-3\log t$ and a holomorphic superpotential independent of $T$, the exact F-term potential is $V=e^K[s^2|D_{S}W{|}^2+(t/3)|W_C{|}^2]$. Inverting the mixed $T,C$ metric gives $K^{T\bar{T}}=t(t+|C{|}^2)/3$, $K^{T\bar{C}}=t\bar{C}/3$, $K^{C\bar{T}}=tC/3$, and $K^{C\bar{C}}=t/3$. Substitution of $D_{T}W=-3W/t$ and $D_{C}W=W_C+3\bar{C}W/t$ cancels all mixed terms, leaving $3|W{|}^2+(t/3)|W_C{|}^2$; the first term cancels the supergravity negative term.

If a no-scale supergravity scalar potential has the displayed form with $y>0$ and nonnegative $A,B$ independent of the volume coordinate $y$, every point with $A+B>0$ has a direction of strictly decreasing potential. Thus a finite stationary vacuum requires $A=B=0$. If that common zero set is empty, the potential has infimum zero along $y\to \infty$ but no finite minimum. If the common zero set is nonempty, varying $y$ on it is a flat direction of a scalar potential. This distinguishes an attained zero-energy vacuum from a decompactification runaway.

For $W=C^3+A+b$, $A=a{e}^{-\alpha S}$, the potential is $V=|C^3+b+(1+\alpha s)A{|}^2/(s{t}^3)+3|C{|}^4/(s{t}^2)$. A finite zero-energy vacuum requires $C=0$ and $b+(1+\alpha s)A=0$. With $a,b\ne 0$, such a solution exists exactly when $0<|b/a|\le 2e^{-1/2}$: the magnitude equation is $|b/a|=(1+x)e^{-x/2}$, $x=\alpha s>0$, whose maximum occurs at $x=1$. At that vacuum $F^S=F^C=0$ but $F^T=e^{K/2}t\bar{W}\ne 0$. Both components of $T$ are flat, while $C$ is quartically lifted. If no root exists, the positive potential falls toward zero as $t\to \infty$ and has no finite minimum. If $a=b=0$, the $C=0$ family is supersymmetric instead.

For the cubic-matter no-scale vacuum with an exponential dilaton, put $x=S+\bar{S}$ and $A=a{e}^{-\alpha S}$. At $C=0$, zero energy requires $G(S)=b+(1+\alpha x)A=0$. On a fixed-imaginary-part slice, $s=\mathrm{Re}S$ gives $dG/ds=\alpha (1-2\alpha s)A$. At the double root $s_0=1/(2\alpha )$, the first derivative vanishes but the second is $-2{\alpha }^{2}A$, so $G=-{\alpha }^{2}A(s-s_0{)}^2+O((s-s_0{)}^3)$. Therefore $V=|G{|}^2/(x{y}^3)$ begins at fourth order. This real scalar has zero quadratic mass but is not an exact flat direction of a scalar potential. The imaginary part of $S$ is still lifted quadratically, and the matter field $C$ also has a quartic leading potential. Exact flat directions instead come from the complex volume modulus $T$ along the zero-energy locus.

The no-scale supergravity identity cancels the negative gravitino contribution, leaving $V=e^{|C{|}^2}|3C^2+\bar{C}(C^3+B){|}^2/(T+\bar{T}{)}^3$. Zero-energy minima occur at $C=0$ and, for $B\ne 0$, three phase-related nonzero solutions with $|B|=|C|(3+|C{|}^2)$. Their $T$ supergravity auxiliary field is nonzero if $B\ne 0$, while both real components of $T$ are flat on each vacuum branch. For $B=0$, the $C=0$ branch is instead supersymmetric and has zero gravitino mass from a superpotential.

For $K=-\log (S+\bar{S})-3\log (T+\bar{T})$ and $W=a{e}^{-\alpha S}+b$ with $\alpha >0$, the potential is $|b+[1+\alpha (S+\bar{S})]a{e}^{-\alpha S}{|}^2/[(S+\bar{S})(T+\bar{T}{)}^3]$. With nonzero $a,b$, a finite zero-energy vacuum exists exactly when $0<|b/a|\le 2e^{-1/2}$. To prove this, set $x=\alpha \mathrm{Re}S>0$; the magnitude equation is $|b/a|=(1+2x)e^{-x}$, whose derivative changes sign at $x=1/2$ and whose maximum is $2e^{-1/2}$. The phase fixes the imaginary part of $S$. At a zero-energy solution the $T$ auxiliary field remains nonzero, so supersymmetry is broken and both real components of $T$ are flat. If $a=b=0$, the family is instead supersymmetric.

Let $K=-3\log \Gamma$ where $\Gamma >0$ is twice differentiable and homogeneous of degree one. If the Kähler Hessian is invertible, the Euler theorem for homogeneous functions identities give ${\tau }_i{K}_{ij}=3{\Gamma }_j/\Gamma$, then $K_{ij}^{-1}{\Gamma }_j/\Gamma ={\tau }_i/3$ and ${\Gamma }_i{K}_{ij}^{-1}{\Gamma }_j/{\Gamma }^2=1/3$. Multiplication by nine gives the no-scale identity. A positive degree-one $\Gamma$ need not induce an invertible or positive Kähler metric: $\Gamma ={\tau }_1+{\tau }_2$ gives a rank-one metric for $K$. The Hessian matrix $({\Gamma }_{ij})$ itself always has the radial null vector $\tau$, whereas the Hessian matrix of $K$ can be invertible. In a larger theory the relevant full inverse metric, or a decoupled block, must obey the identity.