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 $K=-\log (S+\bar{S})-3\log (T+\bar{T})$ 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.