= Solution
Only the Weyl fermions in the charged chiral multiplets contribute to the Abelian triangle <gauge anomaly>. Their cubic coefficient is
$$
\mathcal A_{U(1)^3}=2^3+0^3+(-2)^3=0,
$$
and the mixed gauge-gravitational coefficient is $2+0-2=0$. Diagrammatically, the oppositely charged $X_{2+}$ and $X_{2-}$ fermion triangles have equal magnitudes and opposite signs. The neutral gaugino also contributes nothing, so the theory is anomaly free.
Gauge invariance and renormalizability permit the most general <superpotential>
$$
\boxed{W=W_0+fX_0+\frac{m_0}{2}X_0^2
+mX_{2+}X_{2-}+\frac\lambda3X_0^3
+yX_0X_{2+}X_{2-}}.
$$
After setting the linear and quadratic parameters to zero, the supersymmetric <non-renormalization theorem> ensures that perturbative quantum corrections do not regenerate them in the Wilsonian superpotential.
Write the scalar components as $x_+,x_0,x_-$. For generic nonzero $\lambda,y$, the <F-term scalar potential> and <D-term scalar potential>, including a <Fayet–Iliopoulos term>, are
$$
V_F=|\lambda x_0^2+yx_+x_-|^2
+|yx_0x_-|^2+|yx_0x_+|^2,
$$
$$
V_D=\frac12\left[2g(|x_+|^2-|x_-|^2)+\xi\right]^2,
\qquad V=V_F+V_D.
$$
Every term is nonnegative. The F-flat equations force $x_0=0$ and $x_+x_-=0$. If $\xi>0$, the vacuum
$$
x_+=x_0=0,
\qquad |x_-|^2=\frac{\xi}{2g}
$$
also makes $D=0$; if $\xi<0$, interchange $x_+$ and $x_-$ and use $|x_+|^2=-\xi/(2g)$. Thus for nonzero $\xi$ the charged vacuum expectation value spontaneously breaks the gauged $U(1)$ through the <Higgs mechanism>, but $F_i=D=0$ means supersymmetry remains unbroken. For $\xi=0$, the origin preserves both the gauge symmetry and supersymmetry.
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