Physlib.QFT.QED.AnomalyCancellation.Sorts

For a pure $U(1)$ gauge theory with $n$ fermions, let $S = (S_0, S_1, \dots, S_{n-1}) \in \mathbb{Q}^n$ represent the vector of rational charges. The property that $S$ is sorted is defined by the condition that for all indices $i, j \in \{0, 1, \dots, n-1\}$, if $i \le j$, then $S_i \le S_j$.

Given a vector of rational charges $S = (S_0, S_1, \dots, S_{n-1}) \in \mathbb{Q}^n$ for a pure $U(1)$ gauge theory with $n$ fermions, this function returns the sorted charge assignment. The resulting vector $\text{sort}(S)$ is a permutation of $S$ such that its components are in non-decreasing order, i.e., for any indices $i, j \in \{0, 1, \dots, n-1\}$, if $i \le j$, then $(\text{sort } S)_i \le (\text{sort } S)_j$.

For any natural number $n$ and any vector of rational charges $S = (S_0, S_1, \dots, S_{n-1}) \in \mathbb{Q}^n$ for a pure $U(1)$ gauge theory with $n$ fermions, the vector produced by the sorting function, $\text{sort}(S)$, is sorted. That is, for any indices $i, j \in \{0, 1, \dots, n-1\}$, if $i \le j$, then $(\text{sort } S)_i \le (\text{sort } S)_j$.

For any natural number $n$, let $S = (S_0, S_1, \dots, S_{n-1}) \in \mathbb{Q}^n$ be the vector of rational charges for a pure $U(1)$ gauge theory with $n$ fermions. Let $M$ be an element of the permutation group $S_n$ acting on these charges. Then, sorting the permuted charge vector $\text{rep}(M, S)$ into non-decreasing order results in the same vector as sorting the original vector $S$: $$\text{sort}(\text{rep}(M, S)) = \text{sort}(S)$$

For a natural number $n$, let $S = (S_0, S_1, \dots, S_{n-1}) \in \mathbb{Q}^n$ be a vector of rational charges for a pure $U(1)$ gauge theory. Let $\text{sort}(S)$ be the vector containing the components of $S$ rearranged in non-decreasing order. Then, for any index $j \in \{0, 1, \dots, n-1\}$, the $j$-th component of the sorted vector is given by $(\text{sort } S)_j = S_{\sigma(j)}$, where $\sigma$ is the permutation of indices (denoted as `Tuple.sort S`) that sorts the vector $S$.

For any natural number $n$, let $S \in \mathbb{Q}^n$ be a vector of rational charges for a pure $U(1)$ gauge theory with $n$ fermions. If the sorted charge vector $\text{sort}(S)$ is equal to the zero vector $0$, then the original charge vector $S$ is also equal to $0$.

For any natural number $n$ and any vector of rational charges $S = (S_0, S_1, \dots, S_{n-1}) \in \mathbb{Q}^n$ in a pure $U(1)$ gauge theory with $n$ fermions, the sorting operation is idempotent. That is, sorting an already sorted vector of charges yields the same vector: $$\text{sort}(\text{sort } S) = \text{sort } S$$ where $\text{sort}(S)$ denotes the vector $S$ with its components rearranged in non-decreasing order.

For a pure $U(1)$ gauge theory with $n$ fermions, let $\text{LinSols} \subset \mathbb{Q}^n$ be the space of charge vectors $S = (x_1, \dots, x_n)$ satisfying the linear anomaly cancellation condition $\sum_{i=1}^n x_i = 0$. The function `sortAFL` maps a solution $S \in \text{LinSols}$ to another solution in $\text{LinSols}$ by permuting its components into a sorted order. This is achieved by applying the representation of the permutation group $S_n$ on the space of linear solutions, using the specific permutation that sorts the components of the vector $S$.

In a pure $U(1)$ gauge theory with $n$ fermions, let $S$ be a vector of rational charges $(x_1, \dots, x_n) \in \mathbb{Q}^n$ that satisfies the linear anomaly cancellation condition $\sum_{i=1}^n x_i = 0$. Let $\text{sortAFL}(S)$ denote the sorted version of $S$ within the space of linear solutions, and let $\text{sort}(S.val)$ denote the vector $S$ with its components rearranged in non-decreasing order. The theorem states that the underlying vector of charges for the sorted linear solution is equal to the sorted vector of the original charges: $$(\text{sortAFL } S).val = \text{sort}(S.val)$$

For a pure $U(1)$ gauge theory with $n$ fermions, let $S$ be a vector of rational charges $(x_1, \dots, x_n) \in \mathbb{Q}^n$ that satisfies the linear anomaly cancellation condition $\sum_{i=1}^n x_i = 0$. If the sorted version of this linear solution, $\text{sortAFL}(S)$, is equal to the zero vector, then the original solution $S$ is also equal to the zero vector.