The flow of a Killing vector field consists of local isometries. Isometries preserve both the metric and its unique torsion-free metric-compatible Levi-Civita connection, hence ${\mathcal{L}}_K\nabla =0$. Set $X=K$ and $Y=Z=U$ in part (b). Since ${\nabla }_{U}U=0$,and antisymmetry of the first two curvature arguments givesThis is the geodesic deviation equation with connecting field $K$: applying the isometry to the original geodesic produces a neighboring geodesic, and curvature determines their relative acceleration.

Write ${\mathcal{L}}_{X}Y=[X,Y]$. Expanding the definition on $fZ$ gives terms proportional to derivatives of $f$ with coefficientAll remaining terms carry an overall factor $f$, soThe same argument gives linearity in $Y$, confirming that the Lie derivative of an affine connection is tensorial.

Using the torsion-free identities $[X,Y]={\nabla }_{X}Y-{\nabla }_{Y}X$ and $[X,Z]={\nabla }_{X}Z-{\nabla }_{Z}X$, expandThis is the required formula.

A Killing vector field $K$ obeys ${\mathcal{L}}_{K}g=0$, or equivalently ${\nabla }_{(a}{K}_{b)}=0$. Along an affinely parametrized geodesic with tangent $U$,The second term vanishes by the geodesic equation, while the first contracts the symmetric tensor $U^a{U}^b$ with the antisymmetric part selected by the Killing equation. Thus $g(K,U)$ is a geodesic conserved quantity from a Killing vector. In Minkowski spacetime, translational Killing fields give conserved energy-momentum and rotational or boost Killing fields give the corresponding angular-momentum and boost charges; the same construction works for any spacetime isometry.

The component interactions include the Yukawa vertexand the quartic scalar vertex $-|g{|}^2|\varphi {|}^4$, with cubic scalar interactions fixed by the same $m$ and $g$. At one loop, the scalar propagator has a bosonic tadpole from the quartic vertex and a closed fermion loop with two Yukawa vertices. The loop's fermionic minus sign makes their quadratic cutoff dependence equal and opposite; diagrams involving the dimensionful cubic scalar vertex are at most logarithmically divergent. This is the supersymmetric cancellation of quadratic divergences, and it relies on the coupling relations imposed by supersymmetry.

Eliminating the auxiliary field givesExplicitly,The two supersymmetric vacua for $g\ne 0$ are $\varphi =0$ and $\varphi =-m/g$.

Weak-coupling matching givesbecauseThus $W_{\mathrm{eff}}=W_{\mathrm{tree}}$. The holomorphic superpotential receives neither perturbative nor nonperturbative corrections compatible with the limits and symmetries. The Kähler potential and wavefunction normalization can still be renormalized, so canonically normalized couplings can run through anomalous dimensions.

The ratiois neutral under both $U(1)$ symmetries. The unique monomial with mass dimension three and charges $(0,2)$ isHolomorphy and the spurion symmetries therefore requireExpanding a term givesRegularity as $g\to 0$ requires $n\ge 2$, while regularity as $m\to 0$ requires $n\le 3$. Hence only $n=2,3$ can occur.

Take an ordinary $U(1)$ and an $R$-symmetry, assigning $q(\Phi )=R(\Phi )=1$. The superpotential has chargesbecause $d^2\theta$ is neutral under the first symmetry and has $R$-charge $-2$. Formal invariance of each term then assigns

At $x_0=\sqrt{3}-1$ and $\beta =2-\sqrt{3}$,The stationary point is therefore a local minimum along the real direction. In fact $V(x)$ is positive away from this point on the real axis, tends to $+\infty$ as $x\to \pm \infty$, decreases to the zero-valued minimum at $x_0\approx 0.732$, and then rises again. Its sketch is a single well tangent to the $V=0$ axis at $x_0$.

At a real Minkowski vacuum, $V=0$ giveswhere the branch relevant to the sub-Planckian solution has been chosen. Since the bracket multiplying $e^{x^2}$ vanishes, stationarity givesCombining the equations yields $2x+\beta =\sqrt{3}$. Solving givesThus $X=3$, $Y=2$, $U=-1$, and $Z=3$.

For $K=|\varphi {|}^2$,and the supergravity F-term potential isThe potential is invariant under complex conjugation. Writing $\varphi =x+iy$, its derivative with respect to $y$ vanishes at $y=0$, so the extremisation equations consistently admit a real solution. On that slice,

A vacuum preserves supersymmetry exactly whenA nonzero expectation value alone therefore does not decide the issue. On the real slice this condition is $1+{\varphi }^2+\beta \varphi =0$, which has real solutions only when $\beta \ge 2$. The sub-Planckian Minkowski solution found below has $\beta =2-\sqrt{3}<2$, so $D_{\varphi }W\ne 0$ there and supersymmetry is spontaneously broken.

The three Weyl components form one fundamental representation of the global $SU(3)$, so coupling $G$ to background gauge fields reveals a nonzero $SU(3{)}^3$ 't Hooft anomaly. This does not destroy the global symmetry in flat space or make the theory inconsistent; it constrains any infrared description through 't Hooft anomaly matching. If $G$ were instead intended as a dynamical gauge symmetry, the same uncancelled gauge anomaly would make the model inconsistent and additional chiral matter would be required. Ordinary quantum corrections cannot generate the forbidden superpotential, and the non-renormalization theorem protects its absence.

For one $SU(3)$ triplet and no gauge multiplet, the only nonzero $G$-invariant soft term is the universal scalar mass,An invariant bilinear does not exist, and the formal trilinear $A{ϵ}^{ijk}{\varphi }_i{\varphi }_j{\varphi }_k$ vanishes because the scalar fields commute. Thus the phrase “general soft terms” does not generate further interactions for the field content stated in the question.

Soft supersymmetry breaking means adding supersymmetry-breaking operators with positive-mass-dimension coefficients that do not create new ultraviolet quadratic divergences. Standard examples are scalar masses, holomorphic bilinear and trilinear scalar couplings, gaugino masses, and singlet tadpoles when the field content and symmetries allow them.

The F-term scalar potential isSince the invariant superpotential is constant, all derivatives vanish andThere are no gauge fields in the stated model, so there is no D-term potential.

The canonical Kähler potential isUnder $\Phi \mapsto U\Phi$ with $U\in SU(3)$, it becomes ${\Phi }^{†}{U}^{†}U\Phi =K$, so it respects $G$.

The eight Gell-Mann matrices generate $G=SU(3)$, and the three superfields form one fundamental triplet. A renormalizable superpotential may have degree at most three. There is no invariant vector or symmetric rank-two tensor in the fundamental. The only apparent cubic singlet isbut chiral superfields are Grassmann-even and commute, so this contraction of a symmetric product with ${ϵ}^{ijk}$ vanishes identically. Consequently the most general invariant superpotential as the question is written is only a constant,which has no effect in global supersymmetry. In particular, $W=\lambda {\Phi }_1{\Phi }_2{\Phi }_3$ is not invariant under a general $SU(3)$ rotation.