Physlib.QFT.QED.AnomalyCancellation.Odd.BasisLinear

For any natural number $n$, the following equality holds: $$(1 + n) + n = 2n + 1$$ This identity is used to partition the $2n+1$ charges into groups of size $1$, $n$, and $n$.

For a natural number $n$, this function maps an index $j \in \text{Fin } n$ to the set of $2n+1$ indices $\text{Fin } (2n+1)$. It represents the inclusion of the first $n$ indices into the set of $2n+1$ charges, based on the decomposition $2n+1 = (n+1) + n$.

For a natural number $n$, the function maps an index $j \in \{0, \dots, n-1\}$ to the index $(n+1) + j$ in the set of indices $\{0, \dots, 2n\}$. This represents the inclusion of the second group of $n$ charges into the total $2n+1$ charges used to define the linear solutions for the $U(1)$ anomaly cancellation conditions.

For a system with $2n+1$ charges indexed by $\{0, 1, \dots, 2n\}$, the constant `oddMid` represents the middle index $n$. This index corresponds to the single charge that remains when the total set of charges is partitioned into two groups of $n$ elements.

For any natural number $n$, let $S: \text{Fin}(2n+1) \to \mathbb{Q}$ be a function representing a set of $2n+1$ charges. The total sum of these charges can be decomposed as: $$\sum_{i=0}^{2n} S_i = S_{\text{oddMid}} + \sum_{j=0}^{n-1} (S_{\text{oddFst}(j)} + S_{\text{oddSnd}(j)})$$ where: - $S_{\text{oddMid}}$ is the charge at the middle index $n$. - $S_{\text{oddFst}(j)}$ are the charges at the first $n$ indices (from $0$ to $n-1$). - $S_{\text{oddSnd}(j)}$ are the charges at the last $n$ indices (from $n+1$ to $2n$).

The function maps an index $j \in \{0, \dots, n-1\}$ to the index $j + 1$ within the set of $2n+1$ charges. This corresponds to the inclusion of the first block of $n$ charges in a system partitioned into $1 + n + n$ components, where the first component is a single charge at index 0.

The function maps an index $j \in \{0, 1, \dots, n-1\}$ to an index in the set $\{0, 1, \dots, 2n\}$. Specifically, it maps $j$ to the value $n + 1 + j$. This identifies the second block of $n$ indices within the partition of $2n+1$ indices into $1 + n + n$.

The element $\text{oddShiftZero}$ is the first index in the set of $2n+1$ indices (represented by $\{0, 1, \dots, 2n\}$), identified as the single element in the first group when partitioning the indices into three groups of sizes $1$, $n$, and $n$.

For a sequence of $2n+1$ rational numbers $S_i$ indexed by $i \in \{0, 1, \dots, 2n\}$, the sum of all elements is equal to the value at the first index plus the sum of the two subsequent blocks of $n$ elements: $$\sum_{i=0}^{2n} S_i = S_0 + \sum_{j=0}^{n-1} (S_{j+1} + S_{n+1+j})$$ where the indices are partitioned into the set $\{0\}$ and two blocks $\{1, \dots, n\}$ and $\{n+1, \dots, 2n\}$.

For any natural number $n$, the following identity holds: $((1 + n) + 1) + (n + 1) = 2(n + 1) + 1$, where $n + 1$ denotes the successor of $n$.

In the context of $2(n+1)+1$ charges for a natural number $n$, the index set $\text{Fin}(2n+3)$ is partitioned into four consecutive segments of sizes $1, n, 1,$ and $n+1$. The definition `oddShiftShiftZero` represents the index $0$ in this set, which corresponds to the single element in the first partition of size $1$.

In the context of the "shifted-shifted" split of $2n+3$ charges for the group $U(1)$, this function maps an index $j$ from the set $\{0, \dots, n-1\}$ to the index $j + 1$ within the total index range $\{0, \dots, 2n+2\}$. This mapping identifies the first group of $n$ charges in the partition $((1+n)+1) + (n+1)$.

For a natural number $n$, let $n_{\text{succ}} = n + 1$. This definition identifies the specific index $n+1$ within the set of indices $\{0, 1, \dots, 2n_{\text{succ}}\}$ (totaling $2n+3$ charges). In the context of the "shifted-shifted" partition of the indices into groups of sizes $1$, $n$, $1$, and $n_{\text{succ}}$, this value represents the element in the third group (the second singleton).

For a natural number $n$, this function defines an inclusion mapping from the set of indices $\{0, 1, \dots, n\}$ to the set of indices $\{0, 1, \dots, 2n + 2\}$. It maps each index $j$ to the index $j + n + 2$. This mapping identifies the second group of charges in the "shifted shifted split," which partitions the $2n+3$ total charges into blocks of size $1, n, 1,$ and $n+1$.

For a natural number $n$, consider a system with $2n+3$ charges. The index $\text{oddShiftShiftZero}$—which represents the first element in the $((1+n)+1) + (n+1)$ partition of the charge indices—is equal to the index $\text{oddFst}(0)$, which is the first element of the $(n+1) + (n+1)$ symmetric partition.

For any natural number $n$, in a system with $2n+3$ charges, the index $\text{oddShiftShiftZero}$—representing the first element in the $((1+n)+1) + (n+1)$ partition of charge indices—is equal to the index $\text{oddShiftZero}$, which represents the first element in the $1 + (n+1) + (n+1)$ partition.

In a system of $2n+3$ charges, for any index $j \in \{0, \dots, n-1\}$, the index mapping $\text{oddShiftShiftFst}(j)$ defined in the "shifted-shifted" split is equal to the index mapping $\text{oddFst}(j+1)$ defined in the "symmetric" split. Here, $\text{oddShiftShiftFst}$ identifies the $j$-th element of the first group in the $((1+n)+1) + (n+1)$ partition, while $\text{oddFst}$ identifies elements of the first group in the $(n+1) + (n+1)$ partition.

For any index $j \in \{0, \dots, n-1\}$, the index mapping $\text{oddShiftShiftFst}(j)$, which identifies the $j$-th charge in the first group of $n$ charges within a system of $2n+3$ charges (under the "shifted-shifted" split), is equal to the index mapping $\text{oddShiftFst}(j)$ within the same system of $2n+3$ charges (viewed as a "shifted" split with groups of size $n+1$).

For a system of $2n+3$ charges, the index $\text{oddShiftShiftMid}$ (which identifies the specific index $n+1$ as the second singleton in the "shifted-shifted" partition of sizes $1, n, 1, n+1$) is equal to the middle index $\text{oddMid}$ (the index $n+1$ in the set $\{0, 1, \dots, 2n+2\}$).

In a system of $2n+3$ charges, consider two ways of partitioning the indices $\{0, 1, \dots, 2n+2\}$. In the first partition (the "shifted split" of sizes $1, n+1, n+1$), the mapping $\text{oddShiftFst}$ identifies indices in the second block. In the second partition (the "shifted-shifted split" of sizes $1, n, 1, n+1$), the constant $\text{oddShiftShiftMid}$ identifies the single index in the third block (the index $n+1$). This theorem states that $\text{oddShiftShiftMid}$ is equal to the image of the last index $n$ under the mapping $\text{oddShiftFst}$: $$\text{oddShiftShiftMid} = \text{oddShiftFst}(n)$$

In the context of a system with $2n+3$ charges (corresponding to the $n+1$ case for odd linear solutions), consider two ways to index the second group of $n+1$ charges. The mapping $\text{oddSnd}$ identifies the second group of charges in the standard split, and the mapping $\text{oddShiftShiftSnd}$ identifies the second group of charges in the "shifted shifted split" (which partitions charges into blocks of size $1, n, 1,$ and $n+1$). For any index $j \in \{0, 1, \dots, n\}$, these two mappings produce the same index in the total set of $2n+3$ charges: $$\text{oddShiftShiftSnd}(j) = \text{oddSnd}(j)$$

For a natural number $n$ and an index $j \in \{0, 1, \dots, n\}$, consider a system of $2n+3$ charges. There are two ways to partition these charges: 1. The "shifted split," which partitions the $2n+3$ indices into blocks of sizes $1, n+1,$ and $n+1$. The mapping $\text{oddShiftSnd}$ identifies the indices in the second block of size $n+1$. 2. The "shifted shifted split," which partitions the $2n+3$ indices into blocks of sizes $1, n, 1,$ and $n+1$. The mapping $\text{oddShiftShiftSnd}$ identifies the indices in the final block of size $n+1$. This theorem states that for any index $j$, these two mappings identify the same position in the total set of indices: $$\text{oddShiftShiftSnd}(j) = \text{oddShiftSnd}(j)$$

In a system of $2n+1$ charges, we consider two different ways to partition the set of indices $\{0, 1, \dots, 2n\}$. 1. The standard split partitions the indices into a group of $n+1$ indices and a group of $n$ indices. The mapping $\text{oddSnd}$ identifies the indices in the second group of size $n$. 2. The "shifted split" partitions the indices into groups of size $1, n,$ and $n$. The mapping $\text{oddShiftSnd}$ identifies the indices in the final group of size $n$. For any index $j \in \{0, 1, \dots, n-1\}$, the theorem states that these two mappings yield the same index in the total set: $$\text{oddSnd}(j) = \text{oddShiftSnd}(j)$$

In a system of $2n+1$ charges, let the indices be partitioned into groups. Let $\text{oddShiftZero}$ denote the first index in a $1 + n + n$ partitioning scheme. Let $\text{oddFst}$ be the mapping that embeds the first $n+1$ indices into the total set of $2n+1$ indices (corresponding to an $(n+1) + n$ partition). Then $\text{oddShiftZero}$ is equal to the image of the first index $0$ under the mapping $\text{oddFst}$.

In a system of $2n+1$ charges, let $\text{oddFst}$ be the mapping that embeds the first $n+1$ indices into the set of all indices $\text{Fin}(2n+1)$, following an $(n+1) + n$ partition. Let $\text{oddShiftFst}$ be the mapping that embeds the first block of $n$ indices into the set $\text{Fin}(2n+1)$ for a $1 + n + n$ partition, which maps an index $j \in \{0, \dots, n-1\}$ to the index $j+1$. For any index $j \in \{0, \dots, n-1\}$, the image of $j$ under $\text{oddShiftFst}$ is equal to the image of its successor $j+1$ under $\text{oddFst}$: $$\text{oddShiftFst}(j) = \text{oddFst}(j+1)$$

In a system of $2n+1$ charges, let $\text{oddShiftFst}$ be the mapping that assigns an index $j \in \{0, \dots, n-1\}$ to the index $j + 1$ in the set $\{0, \dots, 2n\}$. The image of the last index of this first block, $\text{Fin.last } n = n-1$, under $\text{oddShiftFst}$ is equal to the middle index $n$, denoted by $\text{oddMid}$.

For any index $j \in \{0, 1, \dots, n-1\}$, the mapping to the second group of $n$ indices in the $1 + n + n$ split of $2n+1$ charges, denoted as $\text{oddShiftSnd}(j)$, is equal to the mapping to the second group of $n$ charges in the symmetric split, denoted as $\text{oddSnd}(j)$. Specifically, both mappings identify the index $n + 1 + j$ in the set of indices $\{0, 1, \dots, 2n\}$.

For a given index $j \in \{0, 1, \dots, n-1\}$, this definition provides a vector of rational charges $Q \in \mathbb{Q}^{2n+1}$ for a pure $U(1)$ theory with $2n+1$ fermions. The components $Q_i$ of this vector (for $i \in \{0, 1, \dots, 2n\}$) are defined as: \[ Q_i = \begin{cases} 1 & \text{if } i = \text{oddFst}(j) \\ -1 & \text{if } i = \text{oddSnd}(j) \\ 0 & \text{otherwise} \end{cases} \] where $\text{oddFst}(j)$ identifies the index of the $j$-th fermion in the first group of $n$ fermions, and $\text{oddSnd}(j)$ identifies the index of the $j$-th fermion in the second group of $n$ fermions. These vectors serve as the basis for the first of two planes that satisfy the linear and cubic anomaly cancellation conditions in the odd $2n+1$ case.

For a natural number $n$ and an index $j \in \{0, 1, \dots, n-1\}$, let $Q^{(j)} \in \mathbb{Q}^{2n+1}$ be the $j$-th basis vector of the first plane of charges (as defined in `basisAsCharges`) for a pure $U(1)$ theory with $2n+1$ fermions. The component of the vector $Q^{(j)}$ at the index $\text{oddFst}(j)$ is equal to $1$.

For a pure $U(1)$ theory with $2n+1$ fermions, let $Q^{(k)} \in \mathbb{Q}^{2n+1}$ be the $k$-th basis vector of the first plane (the "vector-like" plane) for $k \in \{0, \dots, n-1\}$. For any two distinct indices $k, j \in \{0, \dots, n-1\}$, the component of $Q^{(k)}$ at the index $i = \text{oddFst}(j)$ is zero, i.e., $Q^{(k)}_{\text{oddFst}(j)} = 0$. Here, $\text{oddFst}(j)$ denotes the index mapping that embeds the first group of $n$ fermions into the total $2n+1$ charges.

For a pure $U(1)$ gauge theory with $2n + 1$ fermions, let $Q^{(k)} \in \mathbb{Q}^{2n+1}$ be the $k$-th basis vector of the first plane for $k \in \{0, \dots, n-1\}$. For any index $j \in \{0, \dots, 2n\}$, if $j \neq \text{oddFst}(k)$ and $j \neq \text{oddSnd}(k)$, then the $j$-th component of the vector $Q^{(k)}$ is zero, i.e., $Q^{(k)}_j = 0$. Here, $\text{oddFst}(k)$ and $\text{oddSnd}(k)$ are the index mapping functions that embed the $k$-th fermion of the first and second groups into the total $2n+1$ charges.

Consider the basis vectors $Q^{(j)} \in \mathbb{Q}^{2n+1}$ (for $j \in \{0, 1, \dots, n-1\}$) that define the first plane of linear solutions for a pure $U(1)$ theory with $2n+1$ fermions. For any indices $i, j \in \{0, 1, \dots, n-1\}$, the components of these basis vectors satisfy the relation: \[ Q^{(j)}_{\text{oddSnd}(i)} = -Q^{(j)}_{\text{oddFst}(i)} \] where $\text{oddFst}(i)$ and $\text{oddSnd}(i)$ are the index mapping functions that identify the $i$-th fermion in the first and second groups of $n$ fermions, respectively.

For a natural number $n$ and an index $j \in \{0, 1, \dots, n-1\}$, let $Q^{(j)} \in \mathbb{Q}^{2n+1}$ be the $j$-th basis vector of the first plane of charges (the "vector-like" plane) for a pure $U(1)$ theory with $2n+1$ fermions. The component of the vector $Q^{(j)}$ at the index $\text{oddSnd}(j)$ is equal to $-1$, where $\text{oddSnd}(j)$ denotes the index mapping that embeds the second group of $n$ fermions into the total $2n+1$ charges.

For a pure $U(1)$ theory with $2n+1$ fermions, let $Q^{(k)} \in \mathbb{Q}^{2n+1}$ be the $k$-th basis vector of the first plane (the "vector-like" plane) for $k \in \{0, \dots, n-1\}$. For any two distinct indices $k, j \in \{0, \dots, n-1\}$, the component of $Q^{(k)}$ at the index $i = \text{oddSnd}(j)$ is zero, i.e., $Q^{(k)}_{\text{oddSnd}(j)} = 0$. Here, $\text{oddSnd}(j)$ denotes the index mapping that embeds the second group of $n$ fermions into the total $2n+1$ charges.

Consider a system of $2n+1$ fermions with rational charges. Let $Q^{(j)} \in \mathbb{Q}^{2n+1}$ be the $j$-th basis vector for the first plane of linear solutions to the $U(1)$ anomaly cancellation conditions, where $j \in \{0, 1, \dots, n-1\}$. For any such $j$, the component of the vector $Q^{(j)}$ at the middle index $n$ is $0$.

For a pure $U(1)$ gauge theory with $2n+1$ fermions, let $Q^{(j)} \in \mathbb{Q}^{2n+1}$ be the $j$-th basis vector of the first plane, where $j \in \{0, 1, \dots, n-1\}$. The gravitational anomaly of this charge vector, defined as the sum of its components $\sum_{i=0}^{2n} Q^{(j)}_i$, is zero. This confirms that these basis vectors satisfy the linear anomaly cancellation condition.

For a given natural number $n$, this defines the $j$-th basis vector (for $j \in \{0, 1, \dots, n-1\}$) of the first plane of linear solutions for a pure $U(1)$ theory with $2n+1$ fermions. The vector is represented by a charge assignment $Q \in \mathbb{Q}^{2n+1}$, where the component at the $j$-th index of the first group of $n$ fermions is $1$, the component at the $j$-th index of the second group of $n$ fermions is $-1$, and all other components (including the middle fermion) are $0$. This vector $Q$ is shown to satisfy the linear gravitational anomaly cancellation condition $\sum_{i=0}^{2n} Q_i = 0$, thereby identifying it as an element of the space of linear solutions ($\text{LinSols}$) for the system.

For a given vector of rational coefficients $f = (f_0, f_1, \dots, f_{n-1}) \in \mathbb{Q}^n$, this function defines a charge vector in the space $\mathbb{Q}^{2n+1}$ for a pure $U(1)$ theory with $2n+1$ fermions. The resulting charge vector $P(f)$ is defined as the linear combination of the basis vectors $Q^{(i)}$ of the first plane of linear solutions: \[ P(f) = \sum_{i=0}^{n-1} f_i Q^{(i)} \] where each $Q^{(i)}$ corresponds to the $i$-th basis vector `basisAsCharges i`. This vector represents a point in the span of the first part of the basis for the linear solution space.

For a pure $U(1)$ gauge theory with $2n+1$ fermions, let $f \in \mathbb{Q}^n$ be a vector of rational coefficients. Let $P(f) \in \mathbb{Q}^{2n+1}$ be the charge vector representing a point in the first plane of linear solutions, defined as $P(f) = \sum_{i=0}^{n-1} f_i Q^{(i)}$. For any index $j \in \{0, 1, \dots, n-1\}$, the component of $P(f)$ at the index $\text{oddFst}(j)$ is given by: \[ P(f)_{\text{oddFst}(j)} = f_j \] where $\text{oddFst}(j)$ is the index in $\{0, 1, \dots, 2n\}$ corresponding to the $j$-th fermion of the first group.

For a pure $U(1)$ gauge theory with $2n+1$ fermions, let $f \in \mathbb{Q}^n$ be a vector of rational coefficients and $P(f) \in \mathbb{Q}^{2n+1}$ be the charge vector defined by the inclusion of the first plane of linear solutions as $P(f) = \sum_{i=0}^{n-1} f_i Q^{(i)}$. For any index $j \in \{0, 1, \dots, n-1\}$, the component of $P(f)$ at the index $\text{oddSnd}(j)$ is given by: \[ P(f)_{\text{oddSnd}(j)} = -f_j \] where $\text{oddSnd}(j)$ is the index mapping that embeds the second group of $n$ fermions into the total $2n+1$ charges.

For a pure $U(1)$ gauge theory with $2n+1$ fermions, let $P(f) \in \mathbb{Q}^{2n+1}$ be the charge vector defined by the inclusion of the first plane of linear solutions for a given vector of coefficients $f \in \mathbb{Q}^n$. The component of the charge vector $P(f)$ at the middle index $n$ is zero: \[ P(f)_n = 0 \]

For a pure $U(1)$ gauge theory with $2n+1$ fermions, let $f = (f_0, f_1, \dots, f_{n-1}) \in \mathbb{Q}^n$ be a vector of rational coefficients and $P(f) \in \mathbb{Q}^{2n+1}$ be the charge vector defined by the inclusion of the first plane of linear solutions. The gravitational anomaly of $P(f)$, which is defined as the sum of its charge components $\sum_{i=0}^{2n} P(f)_i$, is zero: \[ \text{accGrav}_{2n+1}(P(f)) = 0 \] This confirms that every point in the first plane satisfies the linear anomaly cancellation condition.

For a pure $U(1)$ gauge theory with $2n+1$ fermions, let $f = (f_0, f_1, \dots, f_{n-1}) \in \mathbb{Q}^n$ be a vector of rational coefficients and $P(f) \in \mathbb{Q}^{2n+1}$ be the charge vector defined by the inclusion of the first plane of linear solutions. The cubic anomaly of $P(f)$, defined as the sum of the cubes of its charge components, is zero: \[ \text{accCube}_{2n+1}(P(f)) = \sum_{i=0}^{2n} (P(f)_i)^3 = 0 \] This theorem confirms that every point in the first plane of the linear solution space also satisfies the cubic anomaly cancellation condition (ACC).

For a pure $U(1)$ gauge theory with $2n+1$ fermions, let $f \in \mathbb{Q}^n$ be a vector of rational coefficients and $P(f) \in \mathbb{Q}^{2n+1}$ be the corresponding charge vector in the first plane of linear solutions, defined by $P(f) = \sum_{i=0}^{n-1} f_i Q^{(i)}$. If $P(f) = 0$, then for every $i \in \{0, 1, \dots, n-1\}$, the coefficient $f_i$ is zero.

For a given natural number $n$ and a sequence of rational coefficients $f = (f_0, f_1, \dots, f_{n-1}) \in \mathbb{Q}^n$, the function $P'$ defines a solution to the linear anomaly cancellation conditions for a pure $U(1)$ theory with $2n+1$ fermions. This solution is constructed as the linear combination of the basis vectors $\text{basis}_i$ belonging to the "first plane" of the solution space: \[ P'(f) = \sum_{i=0}^{n-1} f_i \cdot \text{basis}_i \] where each $\text{basis}_i$ is a specific basis vector in the space of linear solutions ($\text{LinSols}$) for the system.

In a pure $U(1)$ gauge theory with $2n+1$ fermions, for any vector of rational coefficients $f = (f_0, f_1, \dots, f_{n-1}) \in \mathbb{Q}^n$, the charge vector associated with the linear anomaly cancellation solution $P'(f)$ is equal to the charge vector $P(f)$. Specifically, if $P'(f) = \sum_{i=0}^{n-1} f_i Q^{(i)}$ is the solution in the first plane of the linear solution space and $P(f)$ is the same linear combination viewed as a vector in the space of charges $\mathbb{Q}^{2n+1}$, then $(P' f).val = P f$.

In a pure $U(1)$ gauge theory with $2n+1$ fermions, the set of basis vectors $\{\text{basis}_0, \text{basis}_1, \dots, \text{basis}_{n-1}\}$ for the first plane of linear anomaly cancellation solutions is linearly independent over the field of rational numbers $\mathbb{Q}$. Each basis vector $\text{basis}_j$ is defined by having a charge of $1$ at the $j$-th index, $-1$ at the $(n+j)$-th index, and $0$ elsewhere.

For each index $j \in \{0, 1, \dots, n-1\}$, this definition specifies a charge assignment (a vector $v_j \in \mathbb{Q}^{2n+1}$) for a pure $U(1)$ gauge theory with $2n+1$ fermions. The components of the vector $v_j$ are given by: \[ (v_j)_i = \begin{cases} 1 & \text{if } i = j+1 \\ -1 & \text{if } i = n+1+j \\ 0 & \text{otherwise} \end{cases} \] where the index $i$ ranges from $0$ to $2n$. These vectors form the basis for the second plane of linear solutions to the anomaly cancellation conditions in the odd-dimensional case.

For any natural number $n$ and any index $j \in \{0, 1, \dots, n-1\}$, the $j$-th basis vector $v_j$ of the second plane of linear solutions for a pure $U(1)$ gauge theory with $2n+1$ fermions satisfies $(v_j)_{j+1} = 1$, where the index $j+1$ corresponds to the mapping `oddShiftFst j`.

For a pure $U(1)$ gauge theory with $2n+1$ fermions, let $v_k \in \mathbb{Q}^{2n+1}$ (for $k \in \{0, \dots, n-1\}$) be the basis vectors for the second plane of linear solutions to the anomaly cancellation conditions. Let $f(j) = j+1$ be the mapping that identifies the first block of $n$ indices within the set of $2n+1$ charges. For any $k, j \in \{0, \dots, n-1\}$, if $k \neq j$, then the $f(j)$-th component of the vector $v_k$ is zero, i.e., $(v_k)_{f(j)} = 0$.