Physlib.Particles.BeyondTheStandardModel.RHN.AnomalyCancellation.PlusU1.Basic

For a given natural number $n$ representing the number of fermion generations, this definition constructs the Anomaly Cancellation Condition (ACC) system for the Standard Model extended with right-handed neutrinos and an additional $U(1)$ gauge symmetry. The system is defined over the charge space $V = \mathbb{Q}^{6n}$, where for each generation $i \in \{0, \dots, n-1\}$, the charges are assigned to the fermion representations: the left-handed quark doublet $Q_i$, the right-handed up-type quark $u_i$, the right-handed down-type quark $d_i$, the left-handed lepton doublet $L_i$, the right-handed charged lepton $e_i$, and the right-handed neutrino $\nu_i$. The system consists of the following equations that a charge configuration $S \in V$ must satisfy: 1. **Four Linear ACCs**: - The gravitational anomaly: $\sum_{i=0}^{n-1} (6 Q_i + 3 u_i + 3 d_i + 2 L_i + e_i + \nu_i) = 0$ - The $SU(2)$ gauge anomaly: $\sum_{i=0}^{n-1} (3 Q_i + L_i) = 0$ - The $SU(3)$ gauge anomaly: $\sum_{i=0}^{n-1} (2 Q_i + u_i + d_i) = 0$ - The $Y^2$ mixed anomaly: $\sum_{i=0}^{n-1} (Q_i + 8u_i + 2d_i + 3L_i + 6e_i) = 0$ 2. **One Quadratic ACC**: - The $f(S) = \sum_{i=0}^{n-1} (Q_i^2 - 2u_i^2 + d_i^2 - L_i^2 + e_i^2) = 0$ 3. **One Cubic ACC**: - The cubic $U(1)$ anomaly: $\sum_{i=0}^{n-1} (6 Q_i^3 + 3 u_i^3 + 3 d_i^3 + 2 L_i^3 + e_i^3 + \nu_i^3) = 0$

In the $n$-generation Standard Model with right-handed neutrinos, let $S$ be a configuration of rational charges $(Q_i, u_i, d_i, L_i, e_i, \nu_i)_{i=0, \dots, n-1} \in \mathbb{Q}^{6n}$ that satisfies the linear anomaly cancellation conditions of the system. Then the gravitational anomaly $\text{accGrav}(S)$ is zero, where the anomaly is defined as: $$\text{accGrav}(S) = \sum_{i=0}^{n-1} (6 Q_i + 3 u_i + 3 d_i + 2 L_i + e_i + \nu_i) = 0.$$

For a charge configuration $S$ that satisfies the linear anomaly cancellation conditions of the $n$-generation Standard Model with right-handed neutrinos and an additional $U(1)$ gauge symmetry, the $SU(2)$ gauge anomaly cancellation condition is satisfied: $$\sum_{i=0}^{n-1} (3 Q_i + L_i) = 0$$ where $Q_i$ and $L_i$ represent the rational charges assigned to the left-handed quark doublet and the left-handed lepton doublet of the $i$-th generation, respectively.

In the $n$-generation Standard Model with right-handed neutrinos, let $S$ be a charge configuration that satisfies the linear anomaly cancellation conditions of the $U(1)$ extension. Then the $SU(3)$ gauge anomaly cancellation condition is satisfied for $S$, which means: $$\sum_{i=0}^{n-1} (2 Q_i + u_i + d_i) = 0$$ where $Q_i, u_i,$ and $d_i$ are the rational charges of the $i$-th generation left-handed quark doublet, right-handed up-type quark, and right-handed down-type quark, respectively.

For a given natural number $n$ representing the number of fermion generations, let $S$ be a charge configuration that satisfies the linear anomaly cancellation conditions (ACCs) for the Standard Model extended with right-handed neutrinos and an additional $U(1)$ gauge symmetry. Then, $S$ satisfies the $Y^2$ mixed anomaly cancellation condition, which is given by $$\text{accYY}(S) = \sum_{i=0}^{n-1} (Q_i + 8u_i + 2d_i + 3L_i + 6e_i) = 0$$ where $Q_i, u_i, d_i, L_i,$ and $e_i$ are the rational charges assigned to the left-handed quark doublet, right-handed up-type quark, right-handed down-type quark, left-handed lepton doublet, and right-handed charged lepton of the $i$-th generation, respectively.

For any charge configuration $S$ belonging to the set of quadratic solutions for the $n$-generation Standard Model with right-handed neutrinos and an additional $U(1)$ gauge symmetry, the quadratic anomaly cancellation condition holds: $$\sum_{i=0}^{n-1} (Q_i^2 - 2u_i^2 + d_i^2 - L_i^2 + e_i^2) = 0$$ where $Q_i, u_i, d_i, L_i, e_i$ are the rational charges assigned to the $i$-th generation of fermions.

Consider the anomaly cancellation condition (ACC) system for the $n$-generation Standard Model with right-handed neutrinos. For any solution $S$ of this system, the cubic anomaly cancellation condition $\mathcal{A}_{\text{cube}}$ evaluated at $S$ is zero: $$\mathcal{A}_{\text{cube}}(S) = \sum_{i=0}^{n-1} \left( 6 Q_i(S)^3 + 3 u_i(S)^3 + 3 d_i(S)^3 + 2 L_i(S)^3 + e_i(S)^3 + \nu_i(S)^3 \right) = 0$$ where $Q_i, u_i, d_i, L_i, e_i, \nu_i$ represent the rational charges assigned to the left-handed quark doublet, right-handed up-type quark, right-handed down-type quark, left-handed lepton doublet, right-handed charged lepton, and right-handed neutrino of the $i$-th generation, respectively.

Given a charge configuration $S \in \mathbb{Q}^{6n}$ for the $n$-generation Standard Model with right-handed neutrinos, this definition constructs an element of the linear solution space $\text{LinSols}$ for the anomaly cancellation system. This requires proofs that $S$ satisfies the four linear anomaly cancellation conditions (ACCs): 1. The gravitational anomaly: $\text{accGrav}(S) = \sum_{i=0}^{n-1} (6 Q_i + 3 u_i + 3 d_i + 2 L_i + e_i + \nu_i) = 0$ 2. The $SU(2)$ gauge anomaly: $\text{accSU2}(S) = \sum_{i=0}^{n-1} (3 Q_i + L_i) = 0$ 3. The $SU(3)$ gauge anomaly: $\text{accSU3}(S) = \sum_{i=0}^{n-1} (2 Q_i + u_i + d_i) = 0$ 4. The $Y^2$ mixed anomaly: $\text{accYY}(S) = \sum_{i=0}^{n-1} (Q_i + 8u_i + 2d_i + 3L_i + 6e_i) = 0$ where $Q_i, u_i, d_i, L_i, e_i, \nu_i$ are the rational charges for the $i$-th generation of fermions.

For $n$ generations of the Standard Model with right-handed neutrinos, let $S$ be a configuration of rational charges that satisfies the four linear anomaly cancellation conditions (the gravitational, $SU(2)$, $SU(3)$, and $Y^2$ mixed anomalies). If $S$ additionally satisfies the quadratic anomaly cancellation condition (ACC), defined as $$\sum_{i=0}^{n-1} (Q_i^2 - 2u_i^2 + d_i^2 - L_i^2 + e_i^2) = 0,$$ where $Q_i, u_i, d_i, L_i, e_i$ are the charges for the $i$-th generation, then $S$ is an element of the space of quadratic solutions $\text{QuadSols}$ for the system.

Given a charge configuration $S$ in the space of quadratic solutions $\text{QuadSols}$ for the $n$-generation Standard Model with right-handed neutrinos (meaning $S$ already satisfies the four linear and one quadratic anomaly cancellation conditions), and a proof $hc$ that $S$ also satisfies the cubic anomaly cancellation condition $\mathcal{A}_{\text{cube}}(S) = 0$, this function constructs an element of the set of complete anomaly-free solutions $\text{Sols}$.

For a configuration of rational charges $S \in \mathbb{Q}^{6n}$ in the $n$-generation Standard Model with right-handed neutrinos, this definition constructs an element of the space of quadratic solutions $\text{QuadSols}$. This requires providing proofs that $S$ satisfies the four linear anomaly cancellation conditions (ACCs) and the one quadratic ACC: 1. **The gravitational anomaly**: $\sum_{i=0}^{n-1} (6 Q_i + 3 u_i + 3 d_i + 2 L_i + e_i + \nu_i) = 0$ 2. **The $SU(2)$ gauge anomaly**: $\sum_{i=0}^{n-1} (3 Q_i + L_i) = 0$ 3. **The $SU(3)$ gauge anomaly**: $\sum_{i=0}^{n-1} (2 Q_i + u_i + d_i) = 0$ 4. **The $Y^2$ mixed anomaly**: $\sum_{i=0}^{n-1} (Q_i + 8u_i + 2d_i + 3L_i + 6e_i) = 0$ 5. **The quadratic anomaly**: $\sum_{i=0}^{n-1} (Q_i^2 - 2u_i^2 + d_i^2 - L_i^2 + e_i^2) = 0$ where $Q_i, u_i, d_i, L_i, e_i$, and $\nu_i$ are the rational charges assigned to the $i$-th generation of the left-handed quark doublet, right-handed up-type quark, right-handed down-type quark, left-handed lepton doublet, right-handed charged lepton, and right-handed neutrino, respectively.

For an $n$-generation Standard Model with right-handed neutrinos, let $S \in \mathbb{Q}^{6n}$ be a configuration of rational charges assigned to the fermions $Q_i, u_i, d_i, L_i, e_i$, and $\nu_i$ for each generation $i \in \{0, \dots, n-1\}$. If $S$ satisfies the following six anomaly cancellation conditions (ACCs): 1. **Gravitational anomaly**: $\sum_{i=0}^{n-1} (6 Q_i + 3 u_i + 3 d_i + 2 L_i + e_i + \nu_i) = 0$ 2. **$SU(2)$ gauge anomaly**: $\sum_{i=0}^{n-1} (3 Q_i + L_i) = 0$ 3. **$SU(3)$ gauge anomaly**: $\sum_{i=0}^{n-1} (2 Q_i + u_i + d_i) = 0$ 4. **$Y^2$ mixed anomaly**: $\sum_{i=0}^{n-1} (Q_i + 8u_i + 2d_i + 3L_i + 6e_i) = 0$ 5. **Quadratic anomaly**: $\sum_{i=0}^{n-1} (Q_i^2 - 2u_i^2 + d_i^2 - L_i^2 + e_i^2) = 0$ 6. **Cubic anomaly**: $\sum_{i=0}^{n-1} (6 Q_i^3 + 3 u_i^3 + 3 d_i^3 + 2 L_i^3 + e_i^3 + \nu_i^3) = 0$ then this function constructs an element of the set of complete anomaly-free solutions $\text{Sols}$ for the system.

For an $n$-generation Standard Model with right-handed neutrinos, let $S$ be a configuration of rational charges that satisfies the four linear anomaly cancellation conditions (gravitational, $SU(2)$, $SU(3)$, and $Y^2$ mixed anomalies). If $S$ additionally satisfies the quadratic anomaly cancellation condition $$\sum_{i=0}^{n-1} (Q_i^2 - 2u_i^2 + d_i^2 - L_i^2 + e_i^2) = 0$$ and the cubic anomaly cancellation condition $$\sum_{i=0}^{n-1} (6 Q_i^3 + 3 u_i^3 + 3 d_i^3 + 2 L_i^3 + e_i^3 + \nu_i^3) = 0,$$ then this function constructs an element of the set of complete anomaly-free solutions $\text{Sols}$ for the system.

For a given number of fermion generations $n \in \mathbb{N}$, this definition constructs the group action of the permutation group $(S_n)^6$ on the Anomaly Cancellation Condition (ACC) system for the Standard Model with right-handed neutrinos. The group $(S_n)^6$ acts on the charge space $\mathbb{Q}^{6n}$ via the representation `repCharges`, which permutes the generation indices for each of the six particle species ($Q, u, d, L, e, \nu$) independently. This action is shown to be a symmetry of the ACC system, as it leaves invariant the four linear anomalies (gravitational, $SU(2)$, $SU(3)$, and $Y^2$ mixed), the quadratic anomaly, and the cubic $U(1)$ anomaly.