Physlib.Particles.BeyondTheStandardModel.RHN.AnomalyCancellation.Basic

For a natural number $n$ representing the number of fermion generations, `SMνCharges n` defines the charge space for the Standard Model with right-handed neutrinos. It constructs an anomaly cancellation condition (ACC) system with $6n$ total charges, corresponding to the six fermion species per generation (the left-handed quark doublet $Q$, the right-handed up-type quark $u$, the right-handed down-type quark $d$, the left-handed lepton doublet $L$, the right-handed charged lepton $e$, and the right-handed neutrino $\nu$).

For a given natural number $n$ representing the number of fermion generations, `SMνSpecies n` defines the anomaly cancellation condition (ACC) charge system for a single species of fermion. It represents an $n$-dimensional vector space where each dimension corresponds to the charge of that specific fermion species (such as the up-quark or the electron) across the $n$ generations.

For a given natural number $n$ representing the number of fermion generations, this equivalence identifies the space of charges for the Standard Model with right-handed neutrinos—which consists of $6n$ total rational charges—with the space of functions from $\{0, 1, \dots, 5\} \times \{0, 1, \dots, n-1\}$ to $\mathbb{Q}$. This map splits the total charge vector into a species-generation grid, where the six species indices correspond to the left-handed quark doublet $Q$, the right-handed up-type quark $u$, the right-handed down-type quark $d$, the left-handed lepton doublet $L$, the right-handed charged lepton $e$, and the right-handed neutrino $\nu$.

For a given natural number $n$ representing the number of fermion generations and an index $i \in \{0, 1, \dots, 5\}$, the map `toSpecies i` is a $\mathbb{Q}$-linear map from the total charge space of the $n$-generation Standard Model with right-handed neutrinos, $(\text{SM}\nu\text{Charges } n).\text{Charges} \cong \mathbb{Q}^{6n}$, to the charge space of a single fermion species across $n$ generations, $(\text{SM}\nu\text{Species } n).\text{Charges} \cong \mathbb{Q}^n$. This map projects the total charge configuration $S$ onto the charges associated with the $i$-th species, where the indices correspond to the fermion types $Q, u, d, L, e,$ and $\nu$.

For any two charge configurations $S$ and $T$ in the space of charges for the $n$-generation Standard Model with right-handed neutrinos, $S$ is equal to $T$ if and only if their projections onto each of the six fermion species (representing $Q, u, d, L, e,$ and $\nu$) are identical, i.e., $\text{toSpecies}_i(S) = \text{toSpecies}_i(T)$ for all $i \in \{0, 1, \dots, 5\}$.

For a given natural number $n$ representing the number of fermion generations, let $f : \{0, 1, \dots, 5\} \to \{0, 1, \dots, n-1\} \to \mathbb{Q}$ be a function that assigns rational charges to each of the six fermion species ($Q, u, d, L, e, \nu$) across $n$ generations. Let $S = \text{toSpeciesEquiv}^{-1}(f)$ be the total charge configuration in $(\text{SM}\nu\text{Charges } n).\text{Charges} \cong \mathbb{Q}^{6n}$ corresponding to $f$. For any species index $i \in \{0, 1, \dots, 5\}$, the projection of $S$ onto that species, denoted as $\text{toSpecies}_i(S)$, is equal to the original vector of charges $f(i)$.

For a charge configuration $S$ in the 1-generation Standard Model with right-handed neutrinos, the projection of $S$ onto the $j$-th fermion species (where $j \in \{0, 1, \dots, 5\}$ corresponds to species $Q, u, d, L, e, \nu$) evaluated at the unique generation index $0$ is equal to the $j$-th component of $S$. That is, $\text{toSpecies}_j(S)_0 = S_j$.

For a given natural number $n$ representing the number of fermion generations, `SMνCharges.Q` is a $\mathbb{Q}$-linear map that projects the total charge configuration of the Standard Model with right-handed neutrinos (an element of the charge space $\mathbb{Q}^{6n}$) onto the charges of the left-handed quark doublet $Q$. The resulting value is a vector in $\mathbb{Q}^n$, where the $j$-th component represents the rational charge assigned to the $Q$ quark in the $j$-th generation.

For an $n$-generation Standard Model with right-handed neutrinos, this defines the $\mathbb{Q}$-linear map that projects the total charge configuration $S \in \mathbb{Q}^{6n}$ onto the charges of the right-handed up-type quark species $u$ across all $n$ generations. The result is an element of $\mathbb{Q}^n$, representing the vector of charges $(u_1, u_2, \dots, u_n)$ assigned to the $u$ quarks.

For an $n$-generation Standard Model with right-handed neutrinos, $D$ is a $\mathbb{Q}$-linear map that projects the total charge configuration of the system onto the charges associated with the right-handed down-type quarks. It maps an element from the total charge space $\mathbb{Q}^{6n}$ to the species charge space $\mathbb{Q}^n$, representing the charges $d_j$ for each generation $j \in \{0, \dots, n-1\}$.

For an $n$-generation Standard Model with right-handed neutrinos, `SMνCharges.L` is the $\mathbb{Q}$-linear map from the total charge space $(\text{SM}\nu\text{Charges } n).\text{Charges} \cong \mathbb{Q}^{6n}$ to the charge space of a single species $(\text{SM}\nu\text{Species } n).\text{Charges} \cong \mathbb{Q}^n$. It extracts the vector of rational charges $(L_1, L_2, \dots, L_n) \in \mathbb{Q}^n$ corresponding to the left-handed lepton doublets $L$ from the total charge configuration.

For a natural number $n$ representing the number of fermion generations, `SMνCharges.E` is the $\mathbb{Q}$-linear map that projects the total charge configuration of the Standard Model with right-handed neutrinos, $(\text{SM}\nu\text{Charges } n).\text{Charges} \cong \mathbb{Q}^{6n}$, onto the charges associated with the right-handed charged leptons (the $e$ species) across all $n$ generations. The result is a vector in $(\text{SM}\nu\text{Species } n).\text{Charges} \cong \mathbb{Q}^n$, where each component represents the rational charge of the right-handed charged lepton for a specific generation.

For a given natural number $n$ representing the number of fermion generations, the map $N$ is a $\mathbb{Q}$-linear projection from the total charge space of the $n$-generation Standard Model with right-handed neutrinos, $(\text{SM}\nu\text{Charges } n).\text{Charges} \cong \mathbb{Q}^{6n}$, to the charge space of the right-handed neutrinos across $n$ generations, $(\text{SM}\nu\text{Species } n).\text{Charges} \cong \mathbb{Q}^n$. This map extracts the rational charges assigned to the right-handed neutrino species from a total charge configuration $S$.

For a given natural number $n$ representing the number of fermion generations, the gravitational anomaly map is a $\mathbb{Q}$-linear map $\text{accGrav} : \mathbb{Q}^{6n} \to \mathbb{Q}$. For a total charge configuration $S$ in the Standard Model with right-handed neutrinos, the map is defined by the weighted sum of charges across all generations $i \in \{0, \dots, n-1\}$: $$\text{accGrav}(S) = \sum_{i=0}^{n-1} (6 Q_i + 3 u_i + 3 d_i + 2 L_i + e_i + \nu_i)$$ 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 $S$ be a total charge configuration. The gravitational anomaly $\text{accGrav}(S)$ can be decomposed into the weighted sums of charges for each fermion species across all generations: $$\text{accGrav}(S) = 6 \sum_{i=0}^{n-1} Q_i + 3 \sum_{i=0}^{n-1} u_i + 3 \sum_{i=0}^{n-1} d_i + 2 \sum_{i=0}^{n-1} L_i + \sum_{i=0}^{n-1} e_i + \sum_{i=0}^{n-1} \nu_i$$ where $Q_i, u_i, d_i, L_i, e_i$, and $\nu_i$ represent 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 any two charge configurations $S$ and $T$ in the $n$-generation Standard Model with right-handed neutrinos, if for every fermion species $j \in \{0, \dots, 5\}$ (corresponding to $Q, u, d, L, e, \nu$) the sum of charges over all generations $i \in \{0, \dots, n-1\}$ is equal, such that $$\sum_{i=0}^{n-1} (\text{toSpecies } j)(S)_i = \sum_{i=0}^{n-1} (\text{toSpecies } j)(T)_i$$ then their gravitational anomalies are equal: $\text{accGrav}(S) = \text{accGrav}(T)$.

For a Standard Model with $n$ generations and right-handed neutrinos, the $SU(2)$ anomaly cancellation condition is defined as a $\mathbb{Q}$-linear map from the space of rational charges to $\mathbb{Q}$. For a given charge configuration $S$, the map calculates the sum over all generations $i$: \[ \text{accSU2}(S) = \sum_{i=1}^{n} (3 Q_i + L_i) \] where $Q_i$ and $L_i$ represent the rational charges of the left-handed quark doublet and the left-handed lepton doublet of the $i$-th generation, respectively.

For a charge configuration $S$ in the $n$-generation Standard Model with right-handed neutrinos, the $SU(2)$ anomaly cancellation condition is equal to three times the sum of the left-handed quark doublet charges plus the sum of the left-handed lepton doublet charges: $$\text{accSU2}(S) = 3 \sum_{i=1}^n Q_i + \sum_{i=1}^n L_i$$ where $Q_i$ and $L_i$ are the rational charges of the left-handed quark doublet and the left-handed lepton doublet of the $i$-th generation, respectively.

Let $S$ and $T$ be two charge configurations for the $n$-generation Standard Model with right-handed neutrinos. If for every fermion species $j \in \{0, 1, \dots, 5\}$ (corresponding to $Q, u, d, L, e, \nu$), the sum of charges across all $n$ generations is equal for $S$ and $T$, such that \[ \sum_{i=1}^{n} (\text{toSpecies } j)(S)_i = \sum_{i=1}^{n} (\text{toSpecies } j)(T)_i \] then the $SU(2)$ anomaly cancellation condition is the same for both configurations, i.e., $\text{accSU2}(S) = \text{accSU2}(T)$.

This $\mathbb{Q}$-linear map represents the $SU(3)$ anomaly cancellation condition for the Standard Model with $n$ generations of fermions and right-handed neutrinos. It maps a total charge configuration $S$ to the sum $$\sum_{i=0}^{n-1} (2 Q_i + U_i + D_i)$$ where $Q_i$, $U_i$, and $D_i$ are the rational 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 charge configuration $S$ in the $n$-generation Standard Model with right-handed neutrinos, the $SU(3)$ anomaly cancellation condition $\text{accSU3}(S)$ can be decomposed into the sums of the charges of the quark species as follows: $$\text{accSU3}(S) = 2 \sum_{i=0}^{n-1} Q(S)_i + \sum_{i=0}^{n-1} U(S)_i + \sum_{i=0}^{n-1} D(S)_i$$ where $Q(S)_i$, $U(S)_i$, and $D(S)_i$ represent the rational 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.

Consider two charge configurations $S$ and $T$ in the Standard Model with $n$ generations and right-handed neutrinos. If for every fermion species $j \in \{0, \dots, 5\}$ (corresponding to the species $Q, u, d, L, e,$ and $\nu$), the sum of the charges across all $n$ generations is the same for both $S$ and $T$, i.e., $$\sum_{i=0}^{n-1} (\text{toSpecies } j)(S)_i = \sum_{i=0}^{n-1} (\text{toSpecies } j)(T)_i,$$ then the $SU(3)$ anomaly cancellation condition values for $S$ and $T$ are equal: $$\text{accSU3}(S) = \text{accSU3}(T).$$

For a given natural number $n$ of fermion generations, the $\mathbb{Q}$-linear map $\text{accYY} : \mathbb{Q}^{6n} \to \mathbb{Q}$ represents the $Y^2$ anomaly cancellation condition for the Standard Model with right-handed neutrinos. For a charge configuration $S$, it is defined by the sum $$\text{accYY}(S) = \sum_{i=0}^{n-1} \left( Q(S)_i + 8U(S)_i + 2D(S)_i + 3L(S)_i + 6E(S)_i \right)$$ where $Q(S)_i$, $U(S)_i$, $D(S)_i$, $L(S)_i$, and $E(S)_i$ are the rational charges assigned to the left-handed quark doublet, the right-handed up-type quark, the right-handed down-type quark, the left-handed lepton doublet, and the right-handed charged lepton for the $i$-th generation, respectively.

For a charge configuration $S$ in the $n$-generation Standard Model with right-handed neutrinos, the $Y^2$ anomaly cancellation condition $\text{accYY}(S)$ can be decomposed as: $$\text{accYY}(S) = \sum_{i=0}^{n-1} Q_i + 8 \sum_{i=0}^{n-1} u_i + 2 \sum_{i=0}^{n-1} d_i + 3 \sum_{i=0}^{n-1} L_i + 6 \sum_{i=0}^{n-1} e_i$$ where $Q_i, u_i, d_i, L_i,$ and $e_i$ are the rational charges assigned to the left-handed quark doublet, the right-handed up-type quark, the right-handed down-type quark, the left-handed lepton doublet, and the right-handed charged lepton for the $i$-th generation, respectively.

For any two charge configurations $S$ and $T$ in the $n$-generation Standard Model with right-handed neutrinos, if the sum of rational charges across all generations $i \in \{0, \dots, n-1\}$ for each of the six fermion species $j$ (where $j$ represents the left-handed quark doublet $Q$, right-handed up-type quark $u$, right-handed down-type quark $d$, left-handed lepton doublet $L$, right-handed charged lepton $e$, and right-handed neutrino $\nu$) is equal for both $S$ and $T$, such that $$\sum_{i=0}^{n-1} (\text{toSpecies}_j(S))_i = \sum_{i=0}^{n-1} (\text{toSpecies}_j(T))_i$$ for all $j \in \{0, \dots, 5\}$, then the $Y^2$ anomaly cancellation condition $\text{accYY}$ evaluates to the same value for both configurations: $$\text{accYY}(S) = \text{accYY}(T)$$

For an $n$-generation Standard Model with right-handed neutrinos, the symmetric bilinear map $B: \mathbb{Q}^{6n} \times \mathbb{Q}^{6n} \to \mathbb{Q}$ is defined as the sum over all generations $i \in \{0, \dots, n-1\}$ of the product of charges: $$B(S, T) = \sum_{i} \left( Q_i(S)Q_i(T) - 2U_i(S)U_i(T) + D_i(S)D_i(T) - L_i(S)L_i(T) + E_i(S)E_i(T) \right)$$ where $S$ and $T$ are total charge configurations, and $Q_i, U_i, D_i, L_i, E_i$ 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 for the $i$-th generation, respectively.

For the $n$-generation Standard Model with right-handed neutrinos, given two charge configurations $S$ and $T$ in the charge space $\mathbb{Q}^{6n}$, the symmetric bilinear form $B(S, T)$ (represented by `quadBiLin`) can be decomposed as: $$B(S, T) = \sum_{i=0}^{n-1} Q_i(S)Q_i(T) - 2\sum_{i=0}^{n-1} U_i(S)U_i(T) + \sum_{i=0}^{n-1} D_i(S)D_i(T) - \sum_{i=0}^{n-1} L_i(S)L_i(T) + \sum_{i=0}^{n-1} E_i(S)E_i(T)$$ where $Q_i, U_i, D_i, L_i$, and $E_i$ are the rational charges for 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 an $n$-generation Standard Model with right-handed neutrinos, the quadratic anomaly cancellation condition is a homogeneous quadratic map $f: \mathbb{Q}^{6n} \to \mathbb{Q}$. For a charge configuration $S$, it is defined as the quadratic form $f(S) = B(S, S)$ associated with the symmetric bilinear form $B$, explicitly given by: $$f(S) = \sum_{i=0}^{n-1} \left( Q_i(S)^2 - 2U_i(S)^2 + D_i(S)^2 - L_i(S)^2 + E_i(S)^2 \right)$$ where $Q_i(S), U_i(S), D_i(S), L_i(S)$, and $E_i(S)$ 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 for the $i$-th generation, respectively.

For an $n$-generation Standard Model with right-handed neutrinos, the quadratic anomaly cancellation condition $\text{accQuad}(S)$ for a given charge configuration $S \in \mathbb{Q}^{6n}$ is decomposed as the following sum over generations $i \in \{0, \dots, n-1\}$: $$\text{accQuad}(S) = \sum_{i=0}^{n-1} Q_i(S)^2 - 2\sum_{i=0}^{n-1} U_i(S)^2 + \sum_{i=0}^{n-1} D_i(S)^2 - \sum_{i=0}^{n-1} L_i(S)^2 + \sum_{i=0}^{n-1} E_i(S)^2$$ where $Q_i(S), U_i(S), D_i(S), L_i(S)$, and $E_i(S)$ denote 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.

Let $S, T \in \mathbb{Q}^{6n}$ be two charge configurations for the $n$-generation Standard Model with right-handed neutrinos. Let $X_{j,i}(S)$ denote the rational charge of the $j$-th fermion species in the $i$-th generation for configuration $S$, where the species $j \in \{0, \dots, 5\}$ correspond to the left-handed quark doublet $Q$, right-handed up-type quark $u$, right-handed down-type quark $d$, left-handed lepton doublet $L$, right-handed charged lepton $e$, and right-handed neutrino $\nu$. If for every species $j$, the sum of the squares of the charges across all generations is equal for $S$ and $T$, i.e., $$\sum_{i=0}^{n-1} X_{j,i}(S)^2 = \sum_{i=0}^{n-1} X_{j,i}(T)^2$$ then the quadratic anomaly cancellation condition $\text{accQuad}$ yields the same value for both configurations: $$\text{accQuad}(S) = \text{accQuad}(T)$$

The symmetric trilinear form $f: V \times V \times V \to \mathbb{Q}$ associated with the cubic anomaly cancellation condition (ACC) for the $n$-generation Standard Model with right-handed neutrinos, where $V$ is the charge space $\mathbb{Q}^{6n}$. For three charge configurations $S, T, R \in V$, the form is defined as the sum over all $n$ generations of the weighted products of the charges of the six fermion species: \[ f(S, T, R) = \sum_{i=0}^{n-1} \left( 6 Q_i(S) Q_i(T) Q_i(R) + 3 U_i(S) U_i(T) U_i(R) + 3 D_i(S) D_i(T) D_i(R) + 2 L_i(S) L_i(T) L_i(R) + E_i(S) E_i(T) E_i(R) + N_i(S) N_i(T) N_i(R) \right) \] where $Q_i, U_i, D_i, L_i, E_i, N_i$ denote the rational charges for 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. The coefficients (6, 3, 3, 2, 1, 1) correspond to the dimensions of the representations of these species under the Standard Model gauge group $SU(3)_C \times SU(2)_L$.

For any three charge configurations $S, T, R$ in the $n$-generation Standard Model with right-handed neutrinos, where the charge space is $\mathbb{Q}^{6n}$, the symmetric trilinear form $f(S, T, R)$ associated with the cubic anomaly cancellation condition is decomposed as: \[ f(S, T, R) = 6 \sum_{i=0}^{n-1} (Q_i(S) Q_i(T) Q_i(R)) + 3 \sum_{i=0}^{n-1} (U_i(S) U_i(T) U_i(R)) + 3 \sum_{i=0}^{n-1} (D_i(S) D_i(T) D_i(R)) + 2 \sum_{i=0}^{n-1} (L_i(S) L_i(T) L_i(R)) + \sum_{i=0}^{n-1} (E_i(S) E_i(T) E_i(R)) + \sum_{i=0}^{n-1} (N_i(S) N_i(T) N_i(R)) \] where $Q_i, U_i, D_i, L_i, E_i, N_i$ denote the rational charges for 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 $V$ be the space of rational charges $\mathbb{Q}^{6n}$. The cubic anomaly cancellation condition (ACC) is the homogeneous cubic map $\mathcal{A}_{\text{cube}}: V \to \mathbb{Q}$ defined by evaluating the symmetric trilinear form of the system on the diagonal. For a charge configuration $S \in V$, the map is given by the sum over all $n$ generations: \[ \mathcal{A}_{\text{cube}}(S) = \sum_{i=0}^{n-1} \left( 6 Q_i(S)^3 + 3 U_i(S)^3 + 3 D_i(S)^3 + 2 L_i(S)^3 + E_i(S)^3 + N_i(S)^3 \right) \] where $Q_i, U_i, D_i, L_i, E_i, N_i$ 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, right-handed charged lepton, and right-handed neutrino of the $i$-th generation, respectively. The integer coefficients $(6, 3, 3, 2, 1, 1)$ correspond to the dimensions of the representations of these fermion species under the gauge group $SU(3)_C \times SU(2)_L$.

For any charge configuration $S$ in the $n$-generation Standard Model with right-handed neutrinos, the cubic anomaly cancellation condition $\mathcal{A}_{\text{cube}}(S)$ is decomposed as the weighted sum of the cubes of the rational charges for each fermion species across all generations: \[ \mathcal{A}_{\text{cube}}(S) = 6 \sum_{i=0}^{n-1} Q_i(S)^3 + 3 \sum_{i=0}^{n-1} U_i(S)^3 + 3 \sum_{i=0}^{n-1} D_i(S)^3 + 2 \sum_{i=0}^{n-1} L_i(S)^3 + \sum_{i=0}^{n-1} E_i(S)^3 + \sum_{i=0}^{n-1} N_i(S)^3 \] where $Q_i, U_i, D_i, L_i, E_i, N_i$ represent 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 any two charge configurations $S$ and $T$ in the $n$-generation Standard Model with right-handed neutrinos, if for every fermion species $j \in \{Q, u, d, L, e, \nu\}$, the sum of the cubes of the charges across all generations is the same for $S$ and $T$, such that: \[ \sum_{i=0}^{n-1} q_{j, i}(S)^3 = \sum_{i=0}^{n-1} q_{j, i}(T)^3 \] where $q_{j, i}$ denotes the rational charge assigned to the $j$-th fermion species in the $i$-th generation, then the cubic anomaly cancellation condition $\mathcal{A}_{\text{cube}}$ evaluates to the same value for both configurations: $\mathcal{A}_{\text{cube}}(S) = \mathcal{A}_{\text{cube}}(T)$.