Physlib.Particles.StandardModel.AnomalyCancellation.NoGrav.Basic

For a natural number $n$ representing the number of fermion generations, the anomaly cancellation system (ACC) for the Standard Model without right-handed neutrinos and gravitational anomalies is defined over the charge space $\mathbb{Q}^{5n}$. A configuration of charges assigns rational values $(Q_i, U_i, D_i, L_i, E_i)$ to the five fermion species for each generation $i \in \{0, \dots, n-1\}$. The system consists of the following equations: 1. Two linear conditions: - The $SU(2)$ anomaly: $\sum_{i=0}^{n-1} (3 Q_i + L_i) = 0$ - The $SU(3)$ anomaly: $\sum_{i=0}^{n-1} (2 Q_i + U_i + D_i) = 0$ 2. No quadratic conditions. 3. One cubic condition: $$\sum_{i=0}^{n-1} (6 Q_i^3 + 3 U_i^3 + 3 D_i^3 + 2 L_i^3 + E_i^3) = 0$$

In the $n$-family Standard Model (without gravity or right-handed neutrinos), let $S$ be a configuration of rational charges $(Q_i, U_i, D_i, L_i, E_i)$ for each generation $i \in \{0, \dots, n-1\}$ that satisfies the linear anomaly cancellation conditions. Then the $SU(2)$ anomaly equation is satisfied: $$\sum_{i=0}^{n-1} (3 Q_i + L_i) = 0$$

In the $n$-family Standard Model without gravity or right-handed neutrinos, let $S$ be a configuration of rational charges $(Q_i, U_i, D_i, L_i, E_i)$ for $i \in \{0, \dots, n-1\}$ that satisfies the linear anomaly cancellation conditions. Then $S$ satisfies the $SU(3)$ anomaly equation: $$\sum_{i=0}^{n-1} (2 Q_i + U_i + D_i) = 0$$ where $Q_i, U_i, D_i$ denote the charges of the left-handed quark doublet, the right-handed up-type quark, and the right-handed down-type quark for the $i$-th generation, respectively.

For any solution $S$ to the anomaly cancellation conditions of the $n$-family Standard Model without gravity, the charges $(Q_i, U_i, D_i, L_i, E_i)$ for each generation $i \in \{0, \dots, n-1\}$ satisfy the cubic anomaly cancellation equation: $$\sum_{i=0}^{n-1} (6 Q_i^3 + 3 U_i^3 + 3 D_i^3 + 2 L_i^3 + E_i^3) = 0$$

For a configuration of rational charges $S \in \mathbb{Q}^{5n}$ in the $n$-family Standard Model without right-handed neutrinos and gravitational anomalies, this definition constructs an element of the space of linear solutions $(SMNoGrav\ n).LinSols$, provided that $S$ satisfies the $SU(2)$ anomaly cancellation condition $\text{accSU2}(S) = 0$ and the $SU(3)$ anomaly cancellation condition $\text{accSU3}(S) = 0$. The charge vector $S$ assigns rational values $(Q_i, U_i, D_i, L_i, E_i)$ to the five fermion species for each generation $i \in \{0, \dots, n-1\}$.

In the $n$-family Standard Model without gravity or right-handed neutrinos, the system of anomaly cancellation equations contains no quadratic conditions. Consequently, any configuration of charges $S$ that satisfies the linear anomaly equations (specifically the $SU(2)$ and $SU(3)$ anomalies) is automatically a solution to the quadratic equations. This function maps a linear solution $S \in \text{LinSols}$ to a quadratic solution $S \in \text{QuadSols}$ by asserting that the empty set of quadratic constraints is vacuously satisfied.

Let $n$ be the number of fermion generations. In the Standard Model without gravity or right-handed neutrinos, consider a configuration of charges $S$ that satisfies the linear anomaly cancellation conditions (specifically the $SU(2)$ and $SU(3)$ anomalies). If $S$ additionally satisfies the cubic anomaly cancellation condition: $$\sum_{i=0}^{n-1} (6 Q_i^3 + 3 U_i^3 + 3 D_i^3 + 2 L_i^3 + E_i^3) = 0$$ then this configuration $S$ is a full anomaly-free solution for the system.

For a configuration of rational charges $S \in \mathbb{Q}^{5n}$ in the $n$-family Standard Model without right-handed neutrinos and gravitational anomalies, this definition constructs an element of the space of quadratic solutions $(SMNoGrav \ n).QuadSols$. It requires that the charge configuration $S$ satisfies the $SU(2)$ anomaly cancellation condition, $\text{accSU2}(S) = 0$, and the $SU(3)$ anomaly cancellation condition, $\text{accSU3}(S) = 0$. Since the system of anomaly equations for this specific model does not include any quadratic constraints, satisfying these linear conditions is sufficient to produce a quadratic solution.

For a configuration of rational charges $S \in \mathbb{Q}^{5n}$ in the $n$-family Standard Model without right-handed neutrinos and gravitational anomalies, if $S$ satisfies the $SU(2)$ anomaly condition, the $SU(3)$ anomaly condition, and the cubic anomaly condition, then $S$ is an anomaly-free solution. Specifically, given: 1. $\text{accSU2}(S) = \sum_{i=0}^{n-1} (3 Q_i + L_i) = 0$ 2. $\text{accSU3}(S) = \sum_{i=0}^{n-1} (2 Q_i + U_i + D_i) = 0$ 3. $\text{accCube}(S) = \sum_{i=0}^{n-1} (6 Q_i^3 + 3 U_i^3 + 3 D_i^3 + 2 L_i^3 + E_i^3) = 0$ this definition constructs an element of the space of anomaly-free solutions $(\text{SMNoGrav } n).\text{Sols}$. Because this specific model does not impose any quadratic or gravitational anomaly cancellation conditions, satisfying these three equations is sufficient for the configuration to be considered anomaly-free.

In the $n$-family Standard Model without gravity or right-handed neutrinos, let $S$ be a configuration of charges that satisfies the linear anomaly cancellation conditions (specifically the $SU(2)$ and $SU(3)$ anomalies). If $S$ additionally satisfies the cubic anomaly cancellation condition: $$\sum_{i=0}^{n-1} (6 Q_i^3 + 3 U_i^3 + 3 D_i^3 + 2 L_i^3 + E_i^3) = 0$$ then $S$ is a full anomaly-free solution for the system. Since this specific model does not impose any quadratic anomaly cancellation conditions, a configuration that satisfies both the linear and cubic constraints is sufficient to be considered anomaly-free.