Physlib.Particles.BeyondTheStandardModel.RHN.AnomalyCancellation.FamilyMaps

For natural numbers $n$ and $m$ representing the number of fermion generations, let $f: \mathbb{Q}^n \to \mathbb{Q}^m$ be a $\mathbb{Q}$-linear map that transforms the charges of a single fermion species across generations. The function `chargesMapOfSpeciesMap` constructs a $\mathbb{Q}$-linear map between the total charge spaces of the Standard Model with right-handed neutrinos, mapping from the $n$-generation space to the $m$-generation space. This total map is defined by applying $f$ independently to the charge vectors of each of the six fermion species (the quark doublet $Q$, up-type quark $u$, down-type quark $d$, lepton doublet $L$, charged lepton $e$, and right-handed neutrino $\nu$). Specifically, if $S$ is a total charge configuration, the resulting configuration's $i$-th species component is given by $f(S_i)$, where $S_i \in \mathbb{Q}^n$ is the charge vector for species $i \in \{0, \dots, 5\}$.

For any natural numbers $n$ and $m$ representing the number of fermion generations, let $f: \mathbb{Q}^n \to \mathbb{Q}^m$ be a $\mathbb{Q}$-linear map that transforms the charges of a single fermion species across generations. Let $S$ be a total charge configuration for the $n$-generation Standard Model with right-handed neutrinos (an element of the charge space $\mathbb{Q}^{6n}$). For any species index $j \in \{0, \dots, 5\}$ corresponding to the fermion species ($Q, u, d, L, e, \nu$), the projection of the transformed total charges onto the $j$-th species is equal to applying $f$ to the $j$-th species component of the original configuration $S$. Mathematically, $$\text{toSpecies}_j (\text{chargesMapOfSpeciesMap}(f)(S)) = f(\text{toSpecies}_j(S)),$$ where $\text{toSpecies}_j$ is the $\mathbb{Q}$-linear projection onto the $j$-th species and $\text{chargesMapOfSpeciesMap}(f)$ is the linear map that applies $f$ independently to each of the six fermion species.

For any natural numbers $n$ and $m$ such that $n \leq m$, this is the $\mathbb{Q}$-linear map that projects the charges of a single fermion species across $m$ generations onto the space of charges for $n$ generations. Specifically, given a charge vector $S \in \mathbb{Q}^m$ (represented as an element of $(\text{SM}\nu\text{Species } m).\text{Charges}$), the map produces a vector in $\mathbb{Q}^n$ by retaining only the first $n$ components, defined by the composition $S \circ \iota$ where $\iota: \{0, \dots, n-1\} \hookrightarrow \{0, \dots, m-1\}$ is the natural inclusion.

For natural numbers $n$ and $m$ such that $n \leq m$, this $\mathbb{Q}$-linear map projects the total charge configuration of the $m$-generation Standard Model with right-handed neutrinos onto the charge space of the $n$-generation model. The map is defined by applying a truncation to each of the six fermion species (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 $\nu$). Specifically, for each species $i \in \{0, \dots, 5\}$, the charge vector $S_i \in \mathbb{Q}^m$ is mapped to $S_i \circ \iota \in \mathbb{Q}^n$, where $\iota: \{0, \dots, n-1\} \hookrightarrow \{0, \dots, m-1\}$ is the natural inclusion, thereby retaining only the first $n$ family components.

For natural numbers $m$ and $n$ representing the number of fermion generations, this definition provides a $\mathbb{Q}$-linear map from the space of charges for $m$ generations $\mathbb{Q}^m$ to the space of charges for $n$ generations $\mathbb{Q}^n$. For a given vector of charges $S = (q_0, q_1, \dots, q_{m-1}) \in \mathbb{Q}^m$, the resulting vector in $\mathbb{Q}^n$ has its $i$-th component equal to $q_i$ if $i < m$, and equal to $0$ if $i \geq m$.

For natural numbers $m$ and $n$ representing the number of fermion generations, this map defines a $\mathbb{Q}$-linear embedding from the charge space of an $m$-generation Standard Model with right-handed neutrinos to that of an $n$-generation model. The map is constructed by applying a species-wise embedding to each of the six fermion species (the quark doublet $Q$, up-type quark $u$, down-type quark $d$, lepton doublet $L$, charged lepton $e$, and right-handed neutrino $\nu$). Specifically, for each species, a vector of charges $(q_0, q_1, \dots, q_{m-1}) \in \mathbb{Q}^m$ is mapped to a vector in $\mathbb{Q}^n$ where the $i$-th component remains $q_i$ for $i < m$, and any additional components for $i \geq m$ are set to $0$.

For a given number of generations $n \in \mathbb{N}$, this defines a $\mathbb{Q}$-linear map from the space of charges for a single fermion species in a 1-generation model to the corresponding space in an $n$-generation model. Specifically, it maps a rational charge $q \in (SM\nu Species\ 1).Charges \cong \mathbb{Q}$ to an $n$-dimensional vector $(q, q, \dots, q) \in (SM\nu Species\ n).Charges \cong \mathbb{Q}^n$, assigning the same charge $q$ to every generation $i \in \{0, \dots, n-1\}$.

For a given natural number $n$ representing the number of fermion generations, the map `SMRHN.familyUniversal` is a $\mathbb{Q}$-linear map from the charge space of a 1-generation Standard Model (with right-handed neutrinos) to the charge space of an $n$-generation model. This map performs a universal embedding by taking the charges $(q_Q, q_u, q_d, q_L, q_e, q_\nu) \in \mathbb{Q}^6$ of the six fermion species in the 1-generation case and duplicating them across all $n$ generations. Specifically, for each species $j \in \{Q, u, d, L, e, \nu\}$, the $n$-dimensional charge vector in the target space is $(q_j, q_j, \dots, q_j) \in \mathbb{Q}^n$, meaning every generation is assigned the same charge as the single generation in the source.

For a given natural number $n$ and a fermion species index $j \in \{0, \dots, 5\}$, let $S$ be a charge configuration for the 1-generation Standard Model with right-handed neutrinos. Let $\text{familyUniversal}(n, S)$ be the $n$-generation charge configuration obtained by duplicating the charges of $S$ across $n$ generations. For any generation index $i \in \{0, \dots, n-1\}$, the charge of the $j$-th fermion species in the $i$-th generation of $\text{familyUniversal}(n, S)$ is equal to the charge of the $j$-th species in the original 1-generation configuration $S$.

For any natural number $n$ representing the number of generations and any exponent $m \in \mathbb{N}$, let $S$ be a charge configuration for the 1-generation Standard Model with right-handed neutrinos, and let $j \in \{0, \dots, 5\}$ be the index for a fermion species. If we construct an $n$-generation charge configuration by duplicating the charges of $S$ across all $n$ generations (via the universal family embedding), then the sum over all generations $i \in \{0, \dots, n-1\}$ of the $m$-th power of the charges of the $j$-th species is equal to $n$ times the $m$-th power of the charge of the $j$-th species in the original 1-generation configuration $S$. That is, $$\sum_{i=0}^{n-1} (q_{j,i})^m = n \cdot (q_j)^m$$ where $q_j$ is the charge of species $j$ in $S$, and $q_{j,i}$ is the charge of species $j$ in the $i$-th generation of the $n$-generation model.

For any natural number $n$ representing the number of fermion generations and any fermion species index $j \in \{0, \dots, 5\}$, let $S$ be a charge configuration for the 1-generation Standard Model with right-handed neutrinos. If we construct an $n$-generation charge configuration by duplicating the charges of $S$ across all $n$ generations (via the universal family embedding), then the sum over all generations $i \in \{0, \dots, n-1\}$ of the charges of the $j$-th species is equal to $n$ times the charge of the $j$-th species in the original 1-generation configuration $S$. That is, $$\sum_{i=0}^{n-1} q_{j,i} = n \cdot q_j$$ where $q_j$ is the charge of species $j$ in the 1-generation configuration $S$, and $q_{j,i}$ is the charge of species $j$ in the $i$-th generation of the $n$-generation model.

For a natural number $n$, let $S$ be a charge configuration for the 1-generation Standard Model with right-handed neutrinos, and $T$ be a charge configuration for the $n$-generation model. Let $j \in \{0, \dots, 5\}$ be an index representing one of the six fermion species. Let $\text{familyUniversal}(n, S)$ be the $n$-generation charge configuration obtained by replicating the charges of $S$ across all generations. The sum over generations $i \in \{0, \dots, n-1\}$ of the product of the charge of species $j$ in the universal embedding of $S$ and the charge of species $j$ in configuration $T$ is equal to the charge of species $j$ in the 1-generation configuration $S$ multiplied by the sum of charges of species $j$ in $T$ over all generations: $$\sum_{i=0}^{n-1} \left( \text{toSpecies}_j (\text{familyUniversal}(n, S))_i \cdot \text{toSpecies}_j(T)_i \right) = \text{toSpecies}_j(S)_0 \cdot \sum_{i=0}^{n-1} \text{toSpecies}_j(T)_i$$ where $\text{toSpecies}_j(\cdot)_i$ denotes the charge of the $j$-th fermion species in the $i$-th generation.

For any natural number $n$, let $S$ be a charge configuration for the 1-generation Standard Model with right-handed neutrinos, and let $T$ and $L$ be charge configurations for the $n$-generation model. For any fermion species $j \in \{0, \dots, 5\}$, the sum over all generations $i$ of the product of the $i$-th generation charges of species $j$ from the universal embedding of $S$, $T$, and $L$ satisfies: $$\sum_{i=0}^{n-1} (q_{S,j,i} \cdot q_{T,j,i} \cdot q_{L,j,i}) = q_{S,j,0} \cdot \sum_{i=0}^{n-1} (q_{T,j,i} \cdot q_{L,j,i})$$ where $q_{S,j,i}$ is the charge of species $j$ in the $i$-th generation of the universal embedding of $S$ (which is equal to the single-generation charge $q_{S,j,0}$ for all $i$), and $q_{T,j,i}$ and $q_{L,j,i}$ are the charges of species $j$ in the $i$-th generation of configurations $T$ and $L$ respectively.

For any natural number $n$ and any charge configuration $S$ in a 1-generation Standard Model with right-handed neutrinos, let $\text{familyUniversal}(n, S)$ be the $n$-generation charge configuration obtained by replicating the charges of $S$ across all $n$ generations. The gravitational anomaly $\text{accGrav}$ of this $n$-generation configuration is equal to $n$ times the gravitational anomaly of the original 1-generation configuration $S$: $$\text{accGrav}(\text{familyUniversal}(n, S)) = n \cdot \text{accGrav}(S)$$

For a given natural number $n$ representing the number of fermion generations and a charge configuration $S$ for the 1-generation Standard Model with right-handed neutrinos, let the $n$-generation configuration be formed by duplicating $S$ across all $n$ generations (the universal embedding). Then the $SU(2)$ anomaly cancellation condition (ACC) for this $n$-generation model is equal to $n$ times the $SU(2)$ ACC for the original 1-generation model. That is, $$\text{accSU2}(\text{familyUniversal}(n, S)) = n \cdot \text{accSU2}(S)$$ where $\text{accSU2}$ denotes the linear combination of charges $\sum_{i=1}^n (3Q_i + L_i)$.

For any natural number $n$ representing the number of fermion generations and any charge configuration $S$ in the 1-generation Standard Model with right-handed neutrinos, the $SU(3)$ anomaly cancellation condition evaluated on the universal family embedding of $S$ into $n$ generations is equal to $n$ times the $SU(3)$ anomaly cancellation condition of the original configuration $S$. Mathematically, this is expressed as: $$\text{accSU3}(\text{familyUniversal}(n, S)) = n \cdot \text{accSU3}(S)$$ where $\text{familyUniversal}(n, S)$ is the $n$-generation configuration where each generation is assigned the exact same charges as $S$.

For any natural number $n$ representing the number of fermion generations, let $S$ be a charge configuration for the 1-generation Standard Model with right-handed neutrinos. If we construct an $n$-generation charge configuration by duplicating the charges of $S$ across all $n$ generations (using the universal family embedding), then the $Y^2$ anomaly cancellation condition ($\text{accYY}$) of the resulting $n$-generation configuration is equal to $n$ times the $\text{accYY}$ value of the original 1-generation configuration: $$\text{accYY}(\text{familyUniversal}(n, S)) = n \cdot \text{accYY}(S)$$ where for $n$ generations, $\text{accYY}(S) = \sum_{i=0}^{n-1} (Q_i + 8u_i + 2d_i + 3L_i + 6e_i)$.

For a natural number $n$, let $S$ be a charge configuration for the 1-generation Standard Model with right-handed neutrinos, and $T$ be a charge configuration for the $n$-generation model. Let $B$ denote the symmetric bilinear form `quadBiLin`, and let $\text{familyUniversal}(n, S)$ be the $n$-generation charge configuration obtained by replicating the charges of $S$ across all $n$ generations. Then, the bilinear form evaluates as: $$B(\text{familyUniversal}(n, S), T) = Q(S) \sum_{i=0}^{n-1} Q_i(T) - 2 U(S) \sum_{i=0}^{n-1} U_i(T) + D(S) \sum_{i=0}^{n-1} D_i(T) - L(S) \sum_{i=0}^{n-1} L_i(T) + E(S) \sum_{i=0}^{n-1} E_i(T)$$ where $Q(S), U(S), D(S), L(S), E(S)$ are the rational charges of the five fermion species (left-handed quark doublet, right-handed up-type quark, right-handed down-type quark, left-handed lepton doublet, and right-handed charged lepton) in the 1-generation configuration $S$, and $Q_i(T), U_i(T), D_i(T), L_i(T), E_i(T)$ are the charges of the corresponding species in the $i$-th generation of the configuration $T$.

For any natural number $n$ representing the number of generations, let $S$ be a charge configuration for the 1-generation Standard Model with right-handed neutrinos. Let $\text{familyUniversal}(n, S)$ be the $n$-generation charge configuration constructed by duplicating the charges of $S$ across all $n$ generations. Then the quadratic anomaly cancellation condition $\text{accQuad}$ of the resulting $n$-generation configuration is equal to $n$ times the $\text{accQuad}$ of the original 1-generation configuration: $$\text{accQuad}(\text{familyUniversal}(n, S)) = n \cdot \text{accQuad}(S)$$ where $\text{accQuad}$ is the quadratic form defined by the sum over generations $i$ of the charges $Q_i^2 - 2U_i^2 + D_i^2 - L_i^2 + E_i^2$.

For any natural number $n$, let $S$ be a charge configuration for the 1-generation Standard Model with right-handed neutrinos, and let $T$ and $R$ be charge configurations for the $n$-generation model. The symmetric trilinear form $f$ associated with the cubic anomaly cancellation condition (ACC), evaluated on the universal embedding of $S$ into $n$ generations and the configurations $T$ and $R$, is given by: $$f(\text{familyUniversal}(n, S), T, R) = 6 S_Q \sum_{i=0}^{n-1} Q_{T,i} Q_{R,i} + 3 S_u \sum_{i=0}^{n-1} u_{T,i} u_{R,i} + 3 S_d \sum_{i=0}^{n-1} d_{T,i} d_{R,i} + 2 S_L \sum_{i=0}^{n-1} L_{T,i} L_{R,i} + S_e \sum_{i=0}^{n-1} e_{T,i} e_{R,i} + S_\nu \sum_{i=0}^{n-1} \nu_{T,i} \nu_{R,i}$$ where $S_j$ denotes the rational charge of fermion species $j$ in the single-generation configuration $S$, and $j_{T,i}$ (resp. $j_{R,i}$) denotes the rational charge of fermion species $j$ in the $i$-th generation of configuration $T$ (resp. $R$). The species indices $j$ 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 any natural number $n$, let $S$ and $T$ be charge configurations for the 1-generation Standard Model with right-handed neutrinos, and let $R$ be a charge configuration for the $n$-generation model. The symmetric trilinear form $f$ associated with the cubic anomaly cancellation condition (ACC), evaluated on the universal embeddings of $S$ and $T$ into $n$ generations and the configuration $R$, is given by: $$f(\text{familyUniversal}(n, S), \text{familyUniversal}(n, T), R) = 6 S_Q T_Q \sum_{i=0}^{n-1} Q_{R,i} + 3 S_u T_u \sum_{i=0}^{n-1} u_{R,i} + 3 S_d T_d \sum_{i=0}^{n-1} d_{R,i} + 2 S_L T_L \sum_{i=0}^{n-1} L_{R,i} + S_e T_e \sum_{i=0}^{n-1} e_{R,i} + S_\nu T_\nu \sum_{i=0}^{n-1} \nu_{R,i}$$ where $S_j$ and $T_j$ denote the rational charges of fermion species $j$ in the 1-generation configurations $S$ and $T$ respectively, and $j_{R,i}$ denotes the rational charge of fermion species $j$ in the $i$-th generation of configuration $R$. The fermion species indices $j$ 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 natural number $n$ representing the number of fermion generations, let $S$ be a charge configuration for the 1-generation Standard Model with right-handed neutrinos. Let $\text{familyUniversal}_n(S)$ be the $n$-generation charge configuration obtained by duplicating the charges of $S$ across all $n$ generations. The cubic anomaly cancellation condition $\mathcal{A}_{\text{cube}}$ for this $n$-generation configuration is equal to $n$ times the cubic anomaly cancellation condition for the 1-generation configuration $S$: $$\mathcal{A}_{\text{cube}}(\text{familyUniversal}_n(S)) = n \cdot \mathcal{A}_{\text{cube}}(S)$$