Supergravity F-term potential

In Planck units, chiral multiplets in four-dimensional $\mathcal{N}=1$ supergravity have

Let $k^i$ generate a holomorphic gauge transformation, with $k^i{\partial }_{i}W=-{\kappa }^{2}rW$ and moment map $\mathcal{P}=i(k^i{K}_i-r)$. Substitution of the Kähler covariant derivative of a superpotential gives the displayed identity. Consequently, a finite point with $W\ne 0$ and all $D_{i}W=0$ also has $\mathcal{P}=0$. Pure D-term breaking with vanishing chiral auxiliary fields can therefore evade this argument only where $W=0$ or outside its hypotheses. In conventional two-derivative matter-coupled supergravity, a constant Fayet–Iliopoulos term gauges an R-symmetry and constrains the transformation of $W$; a neutral superpotential and an independently assigned nonzero constant shift are not generally consistent. The gauge-covariance restriction is derived in Van Proeyen's discussion of Fayet–Iliopoulos terms and R-symmetry.

The gravitino mass parameter in four-dimensional supergravity is the displayed expression. In a zero-energy supersymmetry breaking vacuum with only F-term breaking, $K_{i\bar{j}}{F}^i\overline{F^j}=3m_{3/2}^2/{\kappa }^2$. Writing this norm as ${\Lambda }_{\mathrm{SUSY}}^4$ gives $m_{3/2}={\Lambda }_{\mathrm{SUSY}}^2/(\sqrt{3}{m}_p)$. It is small relative to ${\Lambda }_{\mathrm{SUSY}}$ when the breaking scale is below the Planck mass. The super-Higgs mechanism supplies the longitudinal gravitino states.

The Kähler covariant derivative of a superpotential transforms by the same holomorphic factor as the superpotential under a Kähler transformation. This follows by differentiating $e^{-f}W$ and cancelling the extra derivative of $f$ against the shift of $K_i$.

The Polonyi model uses a canonical Kähler potential and a superpotential linear in one chiral superfield to obtain spontaneous supersymmetry breaking.

For $m\ne 0$ and positive $\beta$, the Polonyi model has a unique zero-energy global minimum at $\beta =2-\sqrt{3}$, $⟨z⟩=\sqrt{3}-1$. Writing $u=\mathrm{Re}z-(\sqrt{3}-1)$ and $y=\mathrm{Im}z$, its potential bracket is $(u^2+y^2+\sqrt{3}u{)}^2+(2\sqrt{3}-3)u^2+(4-2\sqrt{3})y^2$, a sum of nonnegative terms. The real and imaginary curvatures are $4\sqrt{3}$ and $8-4\sqrt{3}$. The other zero-energy stationary point with $\beta =2+\sqrt{3}$ and $z=-\sqrt{3}-1$ is a saddle point. The stable vacuum has nonzero supergravity auxiliary field, so it has supersymmetry breaking even with zero cosmological constant. The $m=0$ branch does not fix these values.

For the Polonyi model with canonical Kähler potential and $W=m^2(z+\beta )$, $\beta >0$, a constant supersymmetric vacuum with $m\ne 0$ requires real $z=(-\beta \pm \sqrt{{\beta }^2-4})/2$ and exists exactly for $\beta \ge 2$. Its supergravity F-term potential is negative because $W\ne 0$. For $0<\beta <2$, the supergravity auxiliary field cannot vanish. If $m=0$, the model is flat with unbroken supersymmetry. This parameter-dependent statement is sharper than asserting that every linear superpotential breaks supersymmetry in supergravity.