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Use the scalar components of the three chiral superfields, and set $a=|{\varphi }_0{|}^2$, $b=|{\varphi }_{+}{|}^2$, $c=|{\varphi }_{-}{|}^2$ and $\Sigma =a+b+c$. We assume the physical sign $g>0$ and a nonzero coupling $\lambda$. The coefficient $g$ below is exactly the coefficient in the supplied D-term scalar potential; it should not be silently replaced by a differently normalized squared gauge coupling. The canonical Kähler potential has identity Kähler metric, and the three Kähler covariant derivatives of a superpotential areSubstitution in the supergravity F-term potential givesIn particular, the subtraction cancels only half of the six ${\kappa }^{2}abc$ terms contributed by the three squares. The full scalar potential isIts nonnegativity follows term by term from $a,b,c\ge 0$ and $g>0$. This is a special property of this cubic superpotential, not of a general supergravity F-term potential.

The D-term vacuum branches of a cubic superpotential must be described as branches: there are not two isolated field-space minima. First suppose $\zeta >0$. Vanishing of $V_F$ requires $ab=ac=bc=0$, and vanishing of the D-term scalar potential requires $b-c=\zeta$. Together these conditions give the vacuum expectation valuesThe phase of ${\varphi }_{+}$ lies on a gauge orbit, and one may choose it positive and real. At this global minimum, all $D_{i}W$ and the real auxiliary field of the vector multiplet vanish. It is a supersymmetric vacuum, even though the internal gauge symmetry is spontaneously broken. For $\zeta <0$, interchange ${\varphi }_{+}$ and ${\varphi }_{-}$: the nonzero vacuum expectation value has magnitude $\sqrt{-\zeta }$. For $\zeta =0$, the supersymmetric vacuum is the entire family ${\varphi }_{+}={\varphi }_{-}=0$, with ${\varphi }_0$ arbitrary.

The other branch has ${\varphi }_{+}={\varphi }_{-}=0$ and arbitrary ${\varphi }_0$. Here $W=D_{i}W=0$ but the real auxiliary field is nonzero for $\zeta \ne 0$, andThus this branch has pure D-term supersymmetry breaking and a flat direction of a scalar potential in the neutral scalar. It is not stable at every neutral-field value. At a fixed $a=|{\varphi }_0{|}^2$, expansion in the charged scalars yieldsHence the two charged squared masses and the stability condition areAbove this threshold, both charged directions have positive mass squared. Continuity makes them uniformly positive in a sufficiently small neighborhood of the chosen neutral-field value, so the stationary branch is a genuine, non-strict local minimum valley. Its neutral direction stays flat. Below the threshold, a charged mass squared is negative and the branch is a saddle point. At equality it is a boundary of the stable valley, not a local minimum in the full field space: changing $a$ slightly downward and switching on the unstable charged field lowers $V$. For example, for $\zeta >0$, put $a=a_c-ϵ$, $b={ϵ}^2$, $c=0$. Since $M(a)=|\lambda {|}^{2}a{e}^{{\kappa }^{2}a}$ has $M^{\prime }(a_c)>0$, one finds $V-g{\zeta }^2=-M^{\prime }(a_c){ϵ}^3+O({ϵ}^4)<0$.

There are no further local-minimum branches under these generic assumptions. If $b+c>0$, the expression for $V_F$ is strictly increasing in $a$, so a stationary configuration with a charged condensate must have $a=0$. At $a=0$, an interior critical point with $b,c>0$ is impossible, becauseOn the boundary $c=0$, the relevant scalar potential is $g(b-\zeta {)}^2$; on $b=0$ it is $g(c+\zeta {)}^2$. Their nonnegative-domain minima give exactly the supersymmetric vacuum described above, or the origin of the charged-field branch. Thus, for nonzero $\zeta$, the two types are the zero-energy supersymmetric gauge orbit and the positive-energy D-breaking minimum valley above the threshold. If $\zeta =0$, the two types coalesce into the flat supersymmetric vacuum; the literal two-minimum assertion needs this parameter qualification.

In the global supersymmetry limit $\kappa \to 0$, the F-term scalar potential becomes $|\lambda {|}^2(ab+ac+bc)$ and the D-term scalar potential retains its form. The same two types of branches occur, with the threshold $|\lambda {|}^{2}a>2g|\zeta |$. Supergravity instead introduces the exponential and additional positive terms. The goldstino of global supersymmetry breaking is a physical fermion; in local supersymmetry it participates in the super-Higgs mechanism. A positive vacuum energy also gravitates in supergravity. In particular, the positive-energy branch here has $W=0$ and hence vanishing gravitino mass from a superpotential; it is not the tuned zero-energy broken vacuum of Question 3.

There is a further gauge-consistency qualification. In conventional two-derivative matter-coupled supergravity, a constant Fayet–Iliopoulos term gauges an R-symmetry and requires a corresponding nontrivial transformation of the superpotential. The apparent charges $0,+1,-1$ make the supplied cubic superpotential neutral, so a nonzero independent constant $\zeta$ does not by itself specify a consistent conventional supergravity action. The computed branches are those of the supplied scalar potential; a full embedding needs compatible charges or additional fields. This restriction is explained in Van Proeyen's derivation of the superpotential gauge-covariance condition.

More generally, the supergravity moment-map constraint on D-term breaking follows directly from gauge covariance. If $k^i{\partial }_{i}W=-{\kappa }^{2}rW$ and the moment map is $\mathcal{P}=i(k^i{K}_i-r)$, then at a point with $W\ne 0$,Consequently all chiral auxiliary fields vanishing implies the D-term also vanishes there. This relation has no analogue forcing that implication in an arbitrary global supersymmetry model. Its $W\ne 0$ hypothesis matters: the formal D-breaking branch above has $W=0$, so it does not contradict the identity. These conventional gauge-completion restrictions are additional to the elementary minimization calculation.