Physlib.Particles.StandardModel.AnomalyCancellation.NoGrav.One.Lemmas

For a set of 1-family Standard Model rational charges $S = (Q, U, D, L, E)$ satisfying the anomaly cancellation conditions (ACCs) without gravity—specifically the $SU(2)$ anomaly $3Q + L = 0$, the $SU(3)$ anomaly $2Q + U + D = 0$, and the cubic anomaly $6Q^3 + 3U^3 + 3D^3 + 2L^3 + E^3 = 0$—the charge of the left-handed quark doublet $Q$ is zero if and only if the charge of the right-handed charged lepton $E$ is zero.

For a set of 1-family Standard Model rational charges $S = (Q, U, D, L, E)$ satisfying the anomaly cancellation conditions (ACCs) without gravity—specifically the $SU(2)$ anomaly $3Q + L = 0$, the $SU(3)$ anomaly $2Q + U + D = 0$, and the cubic anomaly $6Q^3 + 3U^3 + 3D^3 + 2L^3 + E^3 = 0$—if the charge of the left-handed quark doublet $Q$ is zero, then the gravitational anomaly equation $6Q + 3U + 3D + 2L + E = 0$ is satisfied.

For a set of rational charges $S = (Q, U, D, L, E)$ in the 1-family Standard Model that satisfies the gauge anomaly cancellation conditions—specifically the $SU(2)$ anomaly $3Q + L = 0$, the $SU(3)$ anomaly $2Q + U + D = 0$, and the cubic anomaly $6Q^3 + 3U^3 + 3D^3 + 2L^3 + E^3 = 0$—if the charge of the left-handed quark doublet $Q$ is non-zero, then the gravitational anomaly cancellation condition is satisfied: $$6Q + 3U + 3D + 2L + E = 0$$

For a single generation of fermions in the Standard Model, let the rational charges be denoted by $Q$ (left-handed quark doublet), $U$ (right-handed up-type quark), $D$ (right-handed down-type quark), $L$ (left-handed lepton doublet), and $E$ (right-handed charged lepton). If these charges satisfy the gauge anomaly cancellation conditions (ACCs) without gravity: 1. $3Q + L = 0$ 2. $2Q + U + D = 0$ 3. $6Q^3 + 3U^3 + 3D^3 + 2L^3 + E^3 = 0$ then they necessarily satisfy the gravitational anomaly cancellation condition: $$6Q + 3U + 3D + 2L + E = 0$$