Physlib.QFT.QED.AnomalyCancellation.ConstAbs

For a pair of rational numbers $(q_1, q_2) \in \mathbb{Q} \times \mathbb{Q}$, this proposition represents the condition that their squares are equal, $q_1^2 = q_2^2$. This is equivalent to the condition that the two rational numbers have the same absolute value, $|q_1| = |q_2|$.

For a charge assignment $S \in \mathbb{Q}^n$ within a pure $U(1)$ gauge theory with $n$ fermions, this proposition states that all charges have the same absolute value. Formally, it is defined as the condition that for all indices $i, j \in \{0, 1, \dots, n-1\}$, the charges satisfy $(S_i)^2 = (S_j)^2$, which is equivalent to $|S_i| = |S_j|$.

In a pure $U(1)$ gauge theory with $n$ fermions, let $S = (S_0, S_1, \dots, S_{n-1}) \in \mathbb{Q}^n$ be a charge configuration. For any permutation $\sigma$ in the symmetric group $S_n$, the permuted charge configuration $\sigma(S) = (S_{\sigma(0)}, S_{\sigma(1)}, \dots, S_{\sigma(n-1)})$ has constant absolute values (i.e., $|S_{\sigma(i)}| = |S_{\sigma(j)}|$ for all $i, j$) if and only if the original configuration $S$ has constant absolute values (i.e., $|S_i| = |S_j|$ for all $i, j$).

In a pure $U(1)$ gauge theory with $n$ fermions, let $S = (S_0, S_1, \dots, S_{n-1}) \in \mathbb{Q}^n$ be a charge configuration. If all charges in $S$ have the same absolute value, such that $|S_i| = |S_j|$ for all $i, j \in \{0, 1, \dots, n-1\}$, then the configuration obtained by sorting $S$ also satisfies the condition that all its components have the same absolute value.

For a charge assignment $S = (S_1, S_2, \dots, S_n) \in \mathbb{Q}^n$ within a pure $U(1)$ gauge theory with $n$ fermions, the property `ConstAbsSorted` is satisfied if the charges are sorted in non-decreasing order, $S_1 \le S_2 \le \dots \le S_n$, and all charges have the same absolute value, such that $|S_i| = |S_j|$ for all $i, j \in \{1, \dots, n\}$.

Let $S = (S_0, S_1, \dots, S_n)$ be a charge assignment for a pure $U(1)$ gauge theory with $n+1$ fermions such that the charges are sorted in non-decreasing order ($S_0 \le S_1 \le \dots \le S_n$) and all charges have the same absolute value ($|S_i| = |S_j|$ for all $i, j$). For any indices $i, k \in \{0, 1, \dots, n\}$, if $i \le k$ and the charge $S_k$ is non-positive ($S_k \le 0$), then $S_i = S_k$.

Let $S = (S_0, S_1, \dots, S_n)$ be a charge assignment for a pure $U(1)$ gauge theory with $n+1$ fermions such that the charges are sorted in non-decreasing order ($S_0 \le S_1 \le \dots \le S_n$) and all charges have the same absolute value ($|S_i| = |S_j|$ for all $i, j \in \{0, 1, \dots, n\}$). If for some index $i$, the charge $S_i$ is non-positive ($S_i \le 0$), then $S_i$ must be equal to the first charge $S_0$.

Consider a pure $U(1)$ gauge theory with $n+1$ fermions and a charge assignment $S = (S_0, S_1, \dots, S_n) \in \mathbb{Q}^{n+1}$. Suppose the charges are sorted in non-decreasing order ($S_0 \le S_1 \le \dots \le S_n$) and all charges have the same absolute value ($|S_i| = |S_j|$ for all indices $i, j$). If for some index $k$, the charge $S_k$ is non-negative ($0 \le S_k$), then for any index $i$ such that $k \le i$, it holds that $S_i = S_k$.

Let $S = (S_0, S_1, \dots, S_n) \in \mathbb{Q}^{n+1}$ be a charge assignment for a pure $U(1)$ gauge theory with $n+1$ fermions. Suppose the charges are sorted in non-decreasing order ($S_0 \le S_1 \le \dots \le S_n$) and all charges have the same absolute value ($|S_i| = |S_j|$ for all $i, j \in \{0, \dots, n\}$). If the first charge is non-negative ($0 \le S_0$), then for any index $i \in \{0, \dots, n\}$, it holds that $S_i = S_0$.

Let $S = (S_0, S_1, \dots, S_n) \in \mathbb{Q}^{n+1}$ be a charge assignment for a pure $U(1)$ gauge theory with $n+1$ fermions that is sorted ($S_0 \le S_1 \le \dots \le S_n$) and where all charges have the same absolute value ($|S_i| = |S_j|$ for all $i, j$). For any indices $i, j \in \{0, 1, \dots, n\}$, if $S_i \le 0$ and $S_j \ge 0$, then $S_i = -S_j$.

Let $S = (S_0, S_1, \dots, S_n)$ be a vector of rational charges for a pure $U(1)$ gauge theory with $n+1$ fermions. Suppose that $S$ satisfies the condition that its charges are sorted in non-decreasing order ($S_0 \le S_1 \le \dots \le S_n$) and all charges have the same absolute value ($|S_i| = |S_j|$ for all $i, j$). If the first charge is zero ($S_0 = 0$), then all charges in the vector are zero ($S = 0$).

Let $S = (S_0, S_1, \dots, S_n) \in \mathbb{Q}^{n+1}$ be a vector of charges for an anomaly cancellation system (ACC) of a pure $U(1)$ gauge theory with $n+1$ fermions. For an index $k \in \{0, 1, \dots, n-1\}$, $k$ is defined as a **boundary** if the $k$-th charge is strictly negative ($S_k < 0$) and the $(k+1)$-th charge is strictly positive ($S_{k+1} > 0$).

Let $S = (S_0, S_1, \dots, S_n)$ be a charge assignment for a pure $U(1)$ gauge theory with $n+1$ fermions, such that the charges are sorted in non-decreasing order ($S_0 \le S_1 \le \dots \le S_n$) and all charges have the same absolute value ($|S_i| = |S_j|$ for all $i, j \in \{0, \dots, n\}$). If an index $k \in \{0, \dots, n-1\}$ is a boundary, defined by the condition $S_k < 0 < S_{k+1}$, then the $k$-th charge is equal to the first charge ($S_k = S_0$).

Let $S = (S_0, S_1, \dots, S_n) \in \mathbb{Q}^{n+1}$ be a charge assignment for a pure $U(1)$ gauge theory with $n+1$ fermions, such that the charges are sorted in non-decreasing order ($S_0 \le S_1 \le \dots \le S_n$) and all charges have the same absolute value ($|S_i| = |S_j|$ for all $i, j \in \{0, \dots, n\}$). If an index $k \in \{0, \dots, n-1\}$ is a boundary, defined by the condition $S_k < 0 < S_{k+1}$, then the $(k+1)$-th charge is the negative of the first charge, $S_{k+1} = -S_0$.

For any natural number $n$ and any index $k \in \{0, \dots, n-1\}$, the sum of the numerical value of the successor of $k$ (denoted $k+1$) and the difference between the total number of elements $n+1$ and $k+1$ is equal to $n+1$.

For a pure $U(1)$ gauge theory with $n+1$ Weyl fermions and a charge vector $S \in \mathbb{Q}^{n+1}$, the gravitational anomaly, defined as the sum of all charges $\sum S_i$, is equal to the sum of the charges $S_i$ where the index $i$ ranges over a set of size $(k+1) + (n+1 - (k+1))$ for any $k \in \{0, \dots, n-1\}$.

Let $S = (S_0, S_1, \dots, S_n) \in \mathbb{Q}^{n+1}$ be a charge assignment for a pure $U(1)$ gauge theory with $n+1$ fermions. Suppose the charges are sorted in non-decreasing order ($S_0 \leq S_1 \leq \dots \leq S_n$) and all charges have the same absolute value ($|S_i| = |S_j|$ for all $i, j$). If an index $k \in \{0, \dots, n-1\}$ is a boundary, defined by $S_k < 0 < S_{k+1}$, then the gravitational anomaly (the sum of the charges) is given by: $$\sum_{i=0}^n S_i = (2k + 1 - n) S_0$$ where $k$ is the index of the last negative charge and $S_0$ is the value of the first charge.

Let $S = (S_0, S_1, \dots, S_n) \in \mathbb{Q}^{n+1}$ be a vector of charges for a pure $U(1)$ gauge theory with $n+1$ fermions. The charge vector $S$ is said to **have a boundary** if there exists an index $k \in \{0, 1, \dots, n-1\}$ that satisfies the boundary condition, namely that $S_k < 0$ and $S_{k+1} > 0$.

Let $S = (S_0, S_1, \dots, S_n) \in \mathbb{Q}^{n+1}$ be a charge assignment for a pure $U(1)$ gauge theory with $n+1$ fermions such that the charges are sorted in non-decreasing order ($S_0 \le S_1 \le \dots \le S_n$) and all charges have the same absolute value ($|S_i| = |S_j|$ for all $i, j \in \{0, 1, \dots, n\}$). If the charge vector $S$ does not have a boundary (meaning there is no index $k$ such that $S_k < 0 < S_{k+1}$) and the first charge is negative ($S_0 < 0$), then all charges in the vector are equal to the first charge, i.e., $S_i = S_0$ for all $i \in \{0, 1, \dots, n\}$.

Let $S = (S_0, S_1, \dots, S_n) \in \mathbb{Q}^{n+1}$ be a charge assignment for a pure $U(1)$ gauge theory with $n+1$ fermions. Suppose the charges are sorted in non-decreasing order ($S_0 \le S_1 \le \dots \le S_n$) and all charges have the same absolute value ($|S_i| = |S_j|$ for all $i, j \in \{0, \dots, n\}$). If the charge vector $S$ does not have a boundary (meaning there is no index $k$ such that $S_k < 0 < S_{k+1}$), then all charges in the vector are equal to the first charge, i.e., $S_i = S_0$ for all $i \in \{0, 1, \dots, n\}$.

Let $S = (S_0, S_1, \dots, S_n) \in \mathbb{Q}^{n+1}$ be a charge assignment for a pure $U(1)$ gauge theory with $n+1$ fermions such that the charges are sorted in non-decreasing order ($S_0 \le S_1 \le \dots \le S_n$) and all charges have the same absolute value ($|S_i| = |S_j|$ for all $i, j \in \{0, \dots, n\}$). If the charge vector $S$ does not have a boundary (meaning there is no index $k$ such that $S_k < 0 < S_{k+1}$), then the gravitational anomaly, defined as the sum of the charges $\sum_{i=0}^n S_i$, is equal to $(n+1) S_0$.

Consider a pure $U(1)$ gauge theory with $n+1$ fermions. Let $A = (A_0, A_1, \dots, A_n) \in \mathbb{Q}^{n+1}$ be a vector of rational charges that satisfies the linear anomaly cancellation condition $\sum_{i=0}^n A_i = 0$. Suppose the charges are sorted in non-decreasing order ($A_0 \le A_1 \le \dots \le A_n$) and all charges have the same absolute value ($|A_i| = |A_j|$ for all $i, j \in \{0, \dots, n\}$). If the first charge is non-zero ($A_0 \neq 0$), then the charge vector $A$ has a boundary, meaning there exists an index $k \in \{0, \dots, n-1\}$ such that $A_k < 0$ and $A_{k+1} > 0$.

Consider a pure $U(1)$ gauge theory with an odd number of fermions $2n + 1$. Let $A = (A_0, A_1, \dots, A_{2n}) \in \mathbb{Q}^{2n+1}$ be a charge assignment that satisfies the linear anomaly cancellation condition $\sum_{i=0}^{2n} A_i = 0$. Suppose the charges are sorted such that $A_0 \le A_1 \le \dots \le A_{2n}$ and all charges have the same absolute value $|A_i| = |A_j|$ for all $i, j$. If the first charge is non-zero ($A_0 \neq 0$), then the charge vector $A$ has no boundary, meaning there is no index $k \in \{0, \dots, 2n-1\}$ such that $A_k < 0 < A_{k+1}$.

Consider a pure $U(1)$ gauge theory with an odd number of fermions $2n + 1$. Let $A = (A_0, A_1, \dots, A_{2n}) \in \mathbb{Q}^{2n+1}$ be a vector of rational charges that satisfies the linear anomaly cancellation condition $\sum_{i=0}^{2n} A_i = 0$. Suppose the charges are sorted in non-decreasing order ($A_0 \le A_1 \le \dots \le A_{2n}$) and all charges have the same absolute value ($|A_i| = |A_j|$ for all $i, j \in \{0, \dots, 2n\}$). Then the first charge $A_0$ must be equal to zero.

Consider a pure $U(1)$ gauge theory with an odd number of fermions $2n + 1$. Let $A = (A_0, A_1, \dots, A_{2n}) \in \mathbb{Q}^{2n+1}$ be a charge assignment that satisfies the linear anomaly cancellation condition $\sum_{i=0}^{2n} A_i = 0$. If the charges are sorted in non-decreasing order ($A_0 \le A_1 \le \dots \le A_{2n}$) and all charges have the same absolute value ($|A_i| = |A_j|$ for all $i, j$), then $A$ must be the zero vector, i.e., $A = 0$.

Consider a pure $U(1)$ gauge theory with $2n+2$ fermions having rational charges $x = (x_0, x_1, \dots, x_{2n+1})$. Suppose the charges satisfy the linear anomaly cancellation condition $\sum_{i=0}^{2n+1} x_i = 0$, are non-zero ($x_0 \neq 0$), and are sorted such that they all have the same absolute value ($x_0 \le x_1 \le \dots \le x_{2n+1}$ and $|x_i| = |x_j|$ for all $i, j$). If an index $k \in \{0, \dots, 2n\}$ is a boundary such that $x_k < 0 < x_{k+1}$, then $k = n$.

Consider a pure $U(1)$ gauge theory with $2n+2$ fermions. Let $x = (x_0, x_1, \dots, x_{2n+1})$ be a vector of rational charges that satisfies the linear anomaly cancellation condition $\sum_{j=0}^{2n+1} x_j = 0$. Suppose the charges are sorted in non-decreasing order ($x_0 \le x_1 \le \dots \le x_{2n+1}$) and all have the same absolute value ($|x_i| = |x_j|$ for all $i, j \in \{0, \dots, 2n+1\}$). If the first charge is non-zero ($x_0 \neq 0$), then for any index $i \in \{0, 1, \dots, n\}$, the $i$-th charge is equal to the first charge ($x_i = x_0$).

Consider a pure $U(1)$ gauge theory with $2n+2$ fermions. Let $x = (x_0, x_1, \dots, x_{2n+1})$ be a vector of rational charges that satisfies the linear anomaly cancellation condition $\sum_{j=0}^{2n+1} x_j = 0$. Suppose the charges are sorted in non-decreasing order ($x_0 \le x_1 \le \dots \le x_{2n+1}$) and all have the same absolute value ($|x_j| = |x_k|$ for all $j, k \in \{0, \dots, 2n+1\}$). Then for any index $i \in \{0, 1, \dots, n\}$, the $i$-th charge is equal to the first charge, $x_i = x_0$.

Consider a pure $U(1)$ gauge theory with $2n+2$ fermions having rational charges $x = (x_0, x_1, \dots, x_{2n+1})$. Suppose the charges satisfy the linear anomaly cancellation condition $\sum_{j=0}^{2n+1} x_j = 0$, are non-zero ($x_0 \neq 0$), and are sorted such that they all have the same absolute value ($x_0 \le x_1 \le \dots \le x_{2n+1}$ and $|x_j| = |x_k|$ for all $j, k \in \{0, \dots, 2n+1\}$). Then for any index $i \in \{0, 1, \dots, n\}$, the charge at index $n+1+i$ is equal to the negative of the first charge, $x_{n+1+i} = -x_0$.

Consider a pure $U(1)$ gauge theory with $2n+2$ fermions and rational charges $x = (x_0, x_1, \dots, x_{2n+1})$. If the charges satisfy the linear anomaly cancellation condition $\sum_{j=0}^{2n+1} x_j = 0$, are sorted in non-decreasing order ($x_0 \le x_1 \le \dots \le x_{2n+1}$), and all have the same absolute value ($|x_j| = |x_k|$ for all $j, k$), then for any index $i \in \{0, 1, \dots, n\}$, the charge at index $n+1+i$ is equal to the negative of the first charge, $x_{n+1+i} = -x_0$.

Consider a pure $U(1)$ gauge theory with an odd number of fermions $2n + 1$. Let $S = (S_0, S_1, \dots, S_{2n}) \in \mathbb{Q}^{2n+1}$ be a charge assignment that satisfies the linear anomaly cancellation condition $\sum_{i=0}^{2n} S_i = 0$. If all charges in $S$ have the same absolute value, such that $|S_i| = |S_j|$ for all $i, j \in \{0, 1, \dots, 2n\}$, then $S$ must be the zero vector, i.e., $S = 0$.

Consider a pure $U(1)$ gauge theory with an even number of fermions $2n + 2$ (where $n \in \mathbb{N}$). Let $S = (x_0, x_1, \dots, x_{2n+1})$ be a vector of rational charges that satisfies the linear anomaly cancellation condition $\sum_{i=0}^{2n+1} x_i = 0$. If all charges in $S$ have the same absolute value, $|x_i| = |x_j|$ for all $i, j$, then the charge assignment $S$ is vector-like. That is, if the charges are sorted in non-decreasing order $x'_0 \le x'_1 \le \dots \le x'_{2n+1}$, they satisfy $x'_i = -x'_{n+1+i}$ for all $i \in \{0, 1, \dots, n\}$.