Physlib.QFT.QED.AnomalyCancellation.Even.BasisLinear

Given a system of $2(n+1)$ charges, this function defines the inclusion of the index set $\text{Fin}(n+1)$ into the larger index set $\text{Fin}(2(n+1))$. Specifically, it maps an index $j \in \{0, \dots, n\}$ to the same value $j$ considered as an element of the first half of the set $\{0, \dots, 2n+1\}$.

This function maps an index $j$ from the set of indices $\{0, 1, \dots, n\}$ to the index $j + (n + 1)$ in the set of indices $\{0, 1, \dots, 2n + 1\}$. It represents the inclusion of the second block of $n+1$ charges into the total set of $2(n+1)$ charges.

Let $n$ be a natural number. For any two charge assignments $S, T : \text{Fin}(2(n+1)) \to \mathbb{Q}$, if $S$ and $T$ agree on the first $n+1$ indices (the range of `evenFst`) and on the second $n+1$ indices (the range of `evenSnd`), then $S = T$.

For a sequence of $2(n+1)$ rational numbers $S: \text{Fin}(2(n+1)) \to \mathbb{Q}$, the total sum of its elements can be decomposed into a sum over the first and second halves of the index set: $$\sum_{i \in \text{Fin}(2(n+1))} S(i) = \sum_{j \in \text{Fin}(n+1)} (S(\text{evenFst}(j)) + S(\text{evenSnd}(j)))$$ where $\text{evenFst}(j) = j$ and $\text{evenSnd}(j) = j + n + 1$ represent the indices for the first and second blocks of $n+1$ charges, respectively.

For any natural number $n$, the identity $1 + ((n + n) + 1) = 2(n + 1)$ holds.

Given a natural number $n$, this function maps an index $j \in \{0, \dots, n-1\}$ to the index $j + 1$ in the set $\{0, \dots, 2n+1\}$. This corresponds to the inclusion of the first group of $n$ indices within the "shifted even split" of $2(n+1)$ charges, which partitions the indices into segments of size $1, n, n,$ and $1$.

For any natural number $n$, this function maps an index $j \in \{0, \dots, n-1\}$ to the index $n + j + 1$ in the set of $2(n+1)$ indices. It represents the inclusion of the second block of $n$ indices in the "shifted even split" of charges, which partitions the $2(n+1)$ total charges into groups of sizes $1$, $n$, $n$, and $1$.

This definition identifies the first index (index $0$) in the set of indices $\text{Fin}(2(n+1))$ for $2(n+1)$ charges. It corresponds to the first element in a "shifted even split" partition, which divides the $2(n+1)$ charges into groups of sizes $1, n, n,$ and $1$.

The definition `evenShiftLast` identifies the last index in the set of $2(n+1)$ charges. Specifically, within the shifted even splitting of charges into groups of sizes $1$, $n$, $n$, and $1$, this element corresponds to the value $2n + 1$ as an element of the finite set $\text{Fin}(2(n+1))$.

For any natural number $n$ and any function $S: \text{Fin}(2n+2) \to \mathbb{Q}$ representing a collection of charges, the total sum of these charges can be decomposed according to the "shifted even split" ($1 + n + n + 1$) as follows: $$\sum_{i=0}^{2n+1} S_i = S_{\text{evenShiftZero}} + S_{\text{evenShiftLast}} + \sum_{j=0}^{n-1} (S_{\text{evenShiftFst}(j)} + S_{\text{evenShiftSnd}(j)})$$ where the indices are defined as: - $\text{evenShiftZero} = 0$ - $\text{evenShiftLast} = 2n + 1$ - $\text{evenShiftFst}(j) = j + 1$ for $j \in \{0, \dots, n-1\}$ - $\text{evenShiftSnd}(j) = n + j + 1$ for $j \in \{0, \dots, n-1\}$

For a system with $2(n+1)$ charges, the index designated as the first element of the shifted even split, denoted as $\text{evenShiftZero}$, is equal to the first index of the first group in the standard even split, denoted as $\text{evenFst}(0)$.

For any natural number $n$, the index representing the last charge in the shifted splitting of $2(n+1)$ charges, denoted as $\text{evenShiftLast}$, is equal to the image of the last index of the second block under the inclusion map $\text{evenSnd}: \text{Fin}(n+1) \to \text{Fin}(2n+2)$. That is, $\text{evenShiftLast} = \text{evenSnd}(n)$.

For any natural number $n$ and any index $j \in \{0, \dots, n-1\}$, let $\text{evenShiftFst} : \text{Fin}(n) \to \text{Fin}(2(n+1))$ be the mapping that identifies indices in the first group of the shifted split, and $\text{evenFst} : \text{Fin}(n+1) \to \text{Fin}(2(n+1))$ be the mapping for the first group of the standard even split. Then the identity $\text{evenShiftFst}(j) = \text{evenFst}(j+1)$ holds.

For any natural number $n$ and any index $j \in \{0, \dots, n-1\}$, the second shifted even index $\text{evenShiftSnd}(j)$ is equal to the index $\text{evenSnd}(j)$ from the second block of the standard even split of $2(n+1)$ charges.

For a pure $U(1)$ gauge theory with $2(n+1)$ fermions, the $j$-th basis charge vector $v_j \in \mathbb{Q}^{2(n+1)}$ (where $j \in \{0, \dots, n\}$) is defined by assigning a charge of $1$ to the $j$-th fermion and a charge of $-1$ to the $(j+n+1)$-th fermion, with all other charges being $0$. Formally, for an index $i \in \{0, \dots, 2n+1\}$, the $i$-th component of the vector $v_j$ is: $$(v_j)_i = \begin{cases} 1 & \text{if } i = j \\ -1 & \text{if } i = j + n + 1 \\ 0 & \text{otherwise} \end{cases}$$ These vectors represent the first part of a basis used to span the linear solutions of the anomaly cancellation conditions in the even-dimensional case.

For a pure $U(1)$ gauge theory with $2(n+1)$ fermions, let $v_j \in \mathbb{Q}^{2(n+1)}$ be the $j$-th basis charge vector for the vector-like plane, where $j \in \{0, \dots, n\}$. Let $\text{evenFst}(j)$ denote the inclusion of the index $j$ into the first block of the $2(n+1)$ charges. The theorem states that the $(\text{evenFst}(j))$-th component of the vector $v_j$ is equal to $1$.

For a pure $U(1)$ gauge theory with $2(n+1)$ fermions, let $v_k$ be the $k$-th basis charge vector for the vector-like plane, where $k \in \{0, \dots, n\}$. For any index $j \in \{0, \dots, n\}$ such that $k \neq j$, the component of $v_k$ at the index $evenFst(j)$ (which represents the $j$-th position in the first half of the $2(n+1)$ charges) is $0$.

For a pure $U(1)$ gauge theory with $2(n+1)$ fermions, let $v_k \in \mathbb{Q}^{2(n+1)}$ be the $k$-th basis charge vector for the vector-like plane, where $k \in \{0, \dots, n\}$. For any index $j \in \{0, \dots, 2n+1\}$, if $j \neq k$ and $j \neq k + n + 1$, then the $j$-th component of the basis vector is zero, i.e., $(v_k)_j = 0$.

For a pure $U(1)$ gauge theory with $2(n+1)$ fermions, let $v_j \in \mathbb{Q}^{2(n+1)}$ denote the $j$-th basis charge vector for $j \in \{0, \dots, n\}$. Let the set of $2(n+1)$ indices be divided into two halves, where $\text{evenFst}(i) = i$ represents the $i$-th index in the first half and $\text{evenSnd}(i) = i + n + 1$ represents the $i$-th index in the second half, for $i \in \{0, \dots, n\}$. The theorem states that for any $j, i \in \{0, \dots, n\}$, the component of the basis vector $v_j$ at the index in the second half is the negative of its component at the corresponding index in the first half: $$(v_j)_{\text{evenSnd}(i)} = -(v_j)_{\text{evenFst}(i)}$$

For a pure $U(1)$ gauge theory with $2(n+1)$ fermions, let $v_j \in \mathbb{Q}^{2(n+1)}$ be the $j$-th basis charge vector for the vector-like plane, where $j \in \{0, \dots, n\}$. Let $\text{evenSnd}(j)$ denote the inclusion of the index $j$ into the second block of the $2(n+1)$ charges, which corresponds to the index $j + n + 1$. The theorem states that the $(\text{evenSnd}(j))$-th component of the vector $v_j$ is equal to $-1$.

For a pure $U(1)$ gauge theory with $2(n+1)$ fermions, let $v_k \in \mathbb{Q}^{2(n+1)}$ denote the $k$-th basis charge vector for the vector-like plane, where $k \in \{0, \dots, n\}$. Let $\text{evenSnd}(j)$ represent the index of the $j$-th fermion in the second half of the index set, specifically $\text{evenSnd}(j) = j + n + 1$ for $j \in \{0, \dots, n\}$. If $k \neq j$, then the component of the basis vector $v_k$ at the index $\text{evenSnd}(j)$ is zero: $$(v_k)_{\text{evenSnd}(j)} = 0$$

In a pure $U(1)$ gauge theory with $2(n+1)$ fermions, the $j$-th basis charge vector $v_j \in \mathbb{Q}^{2(n+1)}$ of the vector-like plane (where $j \in \{0, \dots, n\}$) satisfies the gravitational anomaly cancellation condition. That is, the sum of its components is zero: $$\sum_{i=0}^{2n+1} (v_j)_i = 0$$ where $(v_j)_i$ is the $i$-th component of the charge vector $v_j$.

In a pure $U(1)$ gauge theory with $2(n+1)$ Weyl fermions, let $v_j \in \mathbb{Q}^{2(n+1)}$ be the $j$-th basis charge vector for the vector-like plane, where $j \in \{0, \dots, n\}$. This vector is defined by $(v_j)_i = 1$ if $i=j$, $(v_j)_i = -1$ if $i=j+n+1$, and $(v_j)_i = 0$ otherwise. The cubic anomaly cancellation condition (ACC) evaluated at $v_j$ is zero: \[ \sum_{i=0}^{2n+1} (v_j)_i^3 = 0 \]

For a pure $U(1)$ gauge theory with $2(n+1)$ fermions, the $j$-th basis charge vector $v_j \in \mathbb{Q}^{2(n+1)}$ (where $j \in \{0, \dots, n\}$) is defined as a member of the space of linear solutions to the anomaly cancellation conditions. The vector $v_j$ is given by: $$(v_j)_i = \begin{cases} 1 & \text{if } i = j \\ -1 & \text{if } i = j + n + 1 \\ 0 & \text{otherwise} \end{cases}$$ This definition confirms that $v_j$ satisfies the linear gravitational anomaly condition $\sum_{i=0}^{2n+1} (v_j)_i = 0$, thereby treating it as a formal linear solution.

Given a sequence of rational coefficients $f \in \mathbb{Q}^{n+1}$, this function defines a charge vector $P(f) \in \mathbb{Q}^{2(n+1)}$ for a pure $U(1)$ gauge theory with $2(n+1)$ fermions. The charge vector is constructed as the linear combination of the basis vectors $v_i$ (defined as `basisAsCharges i` for $i \in \{0, \dots, n\}$) spanning the first (vector-like) plane: $$P(f) = \sum_{i=0}^{n} f_i v_i$$ where each $v_i$ is a vector in $\mathbb{Q}^{2(n+1)}$ with components $(v_i)_j = 1$ if $j=i$, $(v_i)_j = -1$ if $j=i+n+1$, and $0$ otherwise.

In a pure $U(1)$ gauge theory with $2(n+1)$ fermions, let $P(f) = \sum_{i=0}^{n} f_i v_i$ be a charge vector in the vector-like plane, where $f = (f_0, f_1, \dots, f_n) \in \mathbb{Q}^{n+1}$ is a sequence of rational coefficients and $v_i$ are the basis vectors. For any index $j \in \{0, \dots, n\}$, the component of the charge vector $P(f)$ corresponding to the $j$-th fermion in the first half of the system (denoted as $\text{evenFst}(j)$) is equal to the coefficient $f_j$.

In a pure $U(1)$ gauge theory with $2(n+1)$ fermions, let $P(f) = \sum_{i=0}^{n} f_i v_i$ be a charge vector in the vector-like plane, where $f = (f_0, f_1, \dots, f_n) \in \mathbb{Q}^{n+1}$ is a sequence of rational coefficients and $v_i$ are the basis vectors. For any index $j \in \{0, \dots, n\}$, the component of the charge vector $P(f)$ at the index corresponding to the second half of the system, defined by the mapping $\text{evenSnd}(j) = j + n + 1$, is equal to the negative of the coefficient $f_j$: $$(P(f))_{\text{evenSnd}(j)} = -f_j$$

In a pure $U(1)$ gauge theory with $2(n+1)$ fermions, let $P(f) \in \mathbb{Q}^{2(n+1)}$ be a charge vector in the vector-like plane defined by a sequence of rational coefficients $f = (f_0, \dots, f_n) \in \mathbb{Q}^{n+1}$. Let $\text{evenFst}: \{0, \dots, n\} \to \{0, \dots, 2n+1\}$ and $\text{evenSnd}: \{0, \dots, n\} \to \{0, \dots, 2n+1\}$ be the inclusion maps for the first and second blocks of $n+1$ charges, respectively. Then the components of the charge vector $P(f)$ in the second block are the negatives of its components in the first block, satisfying: $$P(f) \circ \text{evenSnd} = -P(f) \circ \text{evenFst}$$

In a pure $U(1)$ gauge theory with $2(n+1)$ Weyl fermions, let $P(f) = \sum_{i=0}^{n} f_i v_i$ be a charge vector in the vector-like plane, where $f = (f_0, f_1, \dots, f_n) \in \mathbb{Q}^{n+1}$ is a sequence of rational coefficients. The gravitational anomaly of this charge vector, which is the sum of its components $\sum_{i=0}^{2n+1} (P f)_i$, is equal to $0$.

In a pure $U(1)$ gauge theory with $2(n+1)$ Weyl fermions, let $f = (f_0, f_1, \dots, f_n) \in \mathbb{Q}^{n+1}$ be a sequence of rational coefficients. Let $P(f) \in \mathbb{Q}^{2(n+1)}$ be the charge vector in the vector-like plane defined as the linear combination $P(f) = \sum_{i=0}^{n} f_i v_i$, where the basis vectors $v_i$ result in components $(P(f))_j = f_j$ and $(P(f))_{j+n+1} = -f_j$ for $j \in \{0, \dots, n\}$. The cubic anomaly cancellation condition (ACC) evaluated at $P(f)$ is zero: \[ \sum_{i=1}^{2(n+1)} (P(f))_i^3 = 0 \]

In a pure $U(1)$ gauge theory with $2(n+1)$ fermions, let $P(f) = \sum_{i=0}^{n} f_i v_i$ be a charge vector in the vector-like plane, where $f = (f_0, f_1, \dots, f_n) \in \mathbb{Q}^{n+1}$ is a sequence of rational coefficients and $v_i$ are the basis vectors of the plane. If $P(f) = 0$ in the space of charges, then $f_i = 0$ for all $i \in \{0, \dots, n\}$.

Given a sequence of rational coefficients $f = (f_0, f_1, \dots, f_n) \in \mathbb{Q}^{n+1}$, this function defines a linear solution to the anomaly cancellation conditions for a pure $U(1)$ gauge theory with $2(n+1)$ fermions. The solution is constructed as the linear combination of the basis vectors $v_i$ of the vector-like plane: $$P'(f) = \sum_{i=0}^{n} f_i v_i$$ where each $v_i$ is a basis vector as defined for the vector-like even plane.

For a pure $U(1)$ gauge theory with $2(n+1)$ fermions, let $f = (f_0, f_1, \dots, f_n) \in \mathbb{Q}^{n+1}$ be a sequence of rational coefficients. The charge vector corresponding to the linear anomaly solution $P'(f)$ is equal to the charge vector $P(f)$ defined in the space of all charges $\mathbb{Q}^{2(n+1)}$.

In a pure $U(1)$ gauge theory with $2(n+1)$ fermions, the set of basis vectors $\{v_j\}_{j=0}^n$ spanning the first (vector-like) plane of linear solutions to the anomaly cancellation conditions is linearly independent over the field of rational numbers $\mathbb{Q}$. Each basis vector $v_j \in \mathbb{Q}^{2(n+1)}$ is defined by its components: $$(v_j)_i = \begin{cases} 1 & \text{if } i = j \\ -1 & \text{if } i = j + n + 1 \\ 0 & \text{otherwise} \end{cases}$$ for $j \in \{0, \dots, n\}$.

For a pure $U(1)$ gauge theory with $2(n+1)$ fermions, let $S$ be a charge vector in the space of linear solutions to the anomaly cancellation conditions (satisfying $\sum S_i = 0$). If $S$ is vector-like (meaning that when its components are sorted as $x_0 \le x_1 \le \dots \le x_{2n+1}$, they satisfy $x_i = -x_{i+n+1}$ for $i \in \{0, \dots, n\}$), then there exists a permutation $M$ of the $2(n+1)$ indices such that the permuted solution $M(S)$ lies in the $\mathbb{Q}$-linear span of the basis vectors $\{v_j\}_{j=0}^n$. Here, each basis vector $v_j$ is defined such that its $j$-th component is $1$, its $(j+n+1)$-th component is $-1$, and all other components are $0$.

For a given index $j \in \{0, \dots, n-1\}$, this definition specifies a charge vector $q \in \mathbb{Q}^{2n+2}$ belonging to the second plane of the linear solution space. The components $q_i$ of the vector are defined as: \[ q_i = \begin{cases} 1 & \text{if } i = \text{evenShiftFst}(j) \\ -1 & \text{if } i = \text{evenShiftSnd}(j) \\ 0 & \text{otherwise} \end{cases} \] where $\text{evenShiftFst}(j) = j+1$ and $\text{evenShiftSnd}(j) = n+j+1$ for $i \in \{0, \dots, 2n+1\}$.

Consider a pure $U(1)$ theory with $2(n+1)$ fermions. Let $q^{(j)}$ be the $j$-th basis charge vector of the second plane for $j \in \{0, \dots, n-1\}$. Then the component of $q^{(j)}$ at the index $\text{evenShiftFst}(j) = j + 1$ is equal to $1$.

In a pure $U(1)$ gauge theory with $2(n+1)$ charges, let $q^{(k)}$ denote the $k$-th basis charge vector for the second plane, where $k \in \{0, \dots, n-1\}$. For any index $j \in \{0, \dots, 2n+1\}$, if $j \neq \text{evenShiftFst}(k)$ and $j \neq \text{evenShiftSnd}(k)$, then the $j$-th component of $q^{(k)}$ is $0$.

Let $n \in \mathbb{N}$ and consider the basis charge vectors $q^{(k)} \in \mathbb{Q}^{2n+2}$ for the second plane of the pure $U(1)$ anomaly cancellation system with $2(n+1)$ fermions, where $k \in \{0, \dots, n-1\}$. For any two distinct indices $k, j \in \{0, \dots, n-1\}$, the component of the $k$-th basis vector at the index defined by the first shifted split $\text{evenShiftFst}(j)$ is zero, i.e., $q^{(k)}_{\text{evenShiftFst}(j)} = 0$.

For a pure $U(1)$ anomaly cancellation system with $2(n+1)$ fermions, let $q^{(j)} \in \mathbb{Q}^{2n+2}$ be the $j$-th basis charge vector of the second plane, where $j \in \{0, \dots, n-1\}$. For any index $i \in \{0, \dots, n-1\}$, the components of $q^{(j)}$ at the shifted indices $\text{evenShiftFst}(i) = i+1$ and $\text{evenShiftSnd}(i) = n+i+1$ satisfy: \[ q^{(j)}_{\text{evenShiftSnd}(i)} = -q^{(j)}_{\text{evenShiftFst}(i)} \]

Consider a pure $U(1)$ anomaly cancellation system with $2(n+1)$ fermions. Let $q^{(j)} \in \mathbb{Q}^{2n+2}$ be the $j$-th basis charge vector of the second plane for $j \in \{0, \dots, n-1\}$. The component of $q^{(j)}$ at the index $\text{evenShiftSnd}(j) = n + j + 1$ is equal to $-1$.

Let $n \in \mathbb{N}$. In a pure $U(1)$ anomaly cancellation system with $2(n+1)$ fermions, let $q^{(k)} \in \mathbb{Q}^{2n+2}$ be the $k$-th basis charge vector of the second plane, where $k \in \{0, \dots, n-1\}$. For any $j \in \{0, \dots, n-1\}$ such that $k \neq j$, the component of $q^{(k)}$ at the index $\text{evenShiftSnd}(j) = n + j + 1$ is zero, i.e., $q^{(k)}_{\text{evenShiftSnd}(j)} = 0$.

In a pure $U(1)$ gauge theory with $2(n+1)$ fermions, for any index $j \in \{0, \dots, n-1\}$, the $j$-th basis charge vector $q^{(j)}$ for the second plane of linear solutions is zero at the first index of the shifted even split ($i = 0$). That is, $q^{(j)}_0 = 0$.

In a pure $U(1)$ gauge theory with $2(n+1)$ fermions, let $q^{(j)} \in \mathbb{Q}^{2n+2}$ denote the $j$-th basis charge vector for the second plane of the linear solution space, where $j \in \{0, \dots, n-1\}$. The component of $q^{(j)}$ at the index $2n+1$ (the last index in the shifted even split) is equal to $0$.

In a pure $U(1)$ gauge theory with $2(n+1)$ Weyl fermions, let $q^{(j)} \in \mathbb{Q}^{2n+2}$ be the $j$-th basis charge vector of the second plane for an index $j \in \{0, \dots, n-1\}$. This vector satisfies the gravitational anomaly cancellation condition (ACC), meaning the sum of its components is zero: $$\sum_{i=0}^{2n+1} q^{(j)}_i = 0$$

For any index $j \in \{0, \dots, n-1\}$, the $j$-th basis charge vector $q^{(j)}$ of the second plane for a pure $U(1)$ gauge theory with $2(n+1)$ fermions satisfies the cubic anomaly cancellation condition (ACC): \[ \text{accCube}_{2n+2}(q^{(j)}) = \sum_{i=1}^{2n+2} (q^{(j)}_i)^3 = 0 \] where the components of $q^{(j)}$ are $1$ at index $j+1$, $-1$ at index $n+j+1$, and $0$ otherwise.

For a pure $U(1)$ gauge theory with $2n+2$ fermions, this definition provides the $j$-th basis vector for the second plane within the space of linear solutions. For an index $j \in \{0, \dots, n-1\}$, the vector is defined as a charge vector $x \in \mathbb{Q}^{2n+2}$ where the components are $x_{j+1} = 1$, $x_{n+j+1} = -1$, and $x_k = 0$ for all other $k$. This vector satisfies the linear gravitational anomaly cancellation condition $\sum_{i=1}^{2n+2} x_i = 0$.