Physlib.QFT.QED.AnomalyCancellation.LineInPlaneCond

For a triple of rational numbers $(s_1, s_2, s_3) \in \mathbb{Q}^3$, this proposition is satisfied if $s_1 = s_2$, $s_1 = -s_2$, or $2s_3 + s_1 + s_2 = 0$.

Let $S = (S_0, S_1, \dots, S_{n-1}) \in \mathbb{Q}^n$ be a linear solution. The condition `LineInPlaneCond S` holds if for every triple of pairwise distinct indices $i_1, i_2, i_3 \in \{0, \dots, n-1\}$, the components of $S$ satisfy at least one of the following: - $S_{i_1} = S_{i_2}$ - $S_{i_1} = -S_{i_2}$ - $2S_{i_3} + S_{i_1} + S_{i_2} = 0$

Let $S = (S_0, S_1, \dots, S_{n-1}) \in \mathbb{Q}^n$ be a linear solution to the Pure $U(1)$ anomaly cancellation system. The line-in-plane condition holds for $S$ if for every triple of pairwise distinct indices $i_1, i_2, i_3 \in \{0, \dots, n-1\}$, at least one of the following is true: - $S_{i_1} = S_{i_2}$ - $S_{i_1} = -S_{i_2}$ - $2S_{i_3} + S_{i_1} + S_{i_2} = 0$ If $S$ satisfies the line-in-plane condition, then for any permutation $M$ in the symmetric group $S_n$, the permuted solution $M \cdot S$ also satisfies the line-in-plane condition.

Let $S = (S_0, S_1, \dots, S_{n+1})$ be a linear solution of $n+2$ rational components satisfying the line-in-plane condition. The line-in-plane condition states that for any three pairwise distinct indices $i_1, i_2, i_3 \in \{0, \dots, n+1\}$, at least one of the following holds: $S_{i_1} = S_{i_2}$, $S_{i_1} = -S_{i_2}$, or $2S_{i_3} + S_{i_1} + S_{i_2} = 0$. If the squares of the last two components are not equal, $S_n^2 \neq S_{n+1}^2$, then the components $S_n$ and $S_{n+1}$ satisfy $(2-n)S_{n+1} = -(2-n)S_n$.

Let $S = (S_0, S_1, \dots, S_{n+4})$ be a linear solution with $n+5$ rational components satisfying the line-in-plane condition. The line-in-plane condition holds if for every triple of pairwise distinct indices $i_1, i_2, i_3 \in \{0, \dots, n+4\}$, at least one of the following is satisfied: - $S_{i_1} = S_{i_2}$ - $S_{i_1} = -S_{i_2}$ - $2S_{i_3} + S_{i_1} + S_{i_2} = 0$ Then the last two components of the solution have the same absolute value, i.e., $|S_{n+3}| = |S_{n+4}|$.

Let $S = (S_0, S_1, \dots, S_{n+4})$ be a solution to the linear anomaly cancellation conditions for the Pure $U(1)$ system with $n+5$ fermions. Suppose $S$ satisfies the line-in-plane condition, which states that for any three pairwise distinct indices $i_1, i_2, i_3 \in \{0, \dots, n+4\}$, at least one of the following holds: - $S_{i_1} = S_{i_2}$ - $S_{i_1} = -S_{i_2}$ - $2S_{i_3} + S_{i_1} + S_{i_2} = 0$ Then, for every pair of distinct indices $i, j \in \{0, \dots, n+4\}$, the components of the solution have equal absolute values, i.e., $|S_i| = |S_j|$.

Let $S = (S_0, S_1, \dots, S_{n+4})$ be a linear solution to the pure $U(1)$ anomaly equations with $n+5$ components (where $n$ is a natural number). Suppose $S$ satisfies the line-in-plane condition, which states that for any three pairwise distinct indices $i_1, i_2, i_3 \in \{0, \dots, n+4\}$, at least one of the following holds: - $S_{i_1} = S_{i_2}$ - $S_{i_1} = -S_{i_2}$ - $2S_{i_3} + S_{i_1} + S_{i_2} = 0$ Then the solution $S$ satisfies the constant absolute value condition, meaning $|S_i| = |S_j|$ for all $i, j \in \{0, \dots, n+4\}$.

Let $S = (S_0, S_1, S_2, S_3) \in \mathbb{Q}^4$ be a solution to the pure $U(1)$ anomaly equations for $n=4$. If $S$ satisfies the line-in-plane condition—meaning that for any distinct indices $i_1, i_2, i_3 \in \{0, 1, 2, 3\}$, either $S_{i_1} = S_{i_2}$, $S_{i_1} = -S_{i_2}$, or $2S_{i_3} + S_{i_1} + S_{i_2} = 0$—then the first two components of the solution have equal absolute values, $|S_0| = |S_1|$.

Let $S = (S_0, S_1, S_2, S_3) \in \mathbb{Q}^4$ be a solution to the pure $U(1)$ anomaly equations for $n=4$, such that $\sum_{i=0}^3 S_i = 0$ and $\sum_{i=0}^3 S_i^3 = 0$. Suppose $S$ satisfies the line-in-plane condition: for every triple of pairwise distinct indices $i_1, i_2, i_3 \in \{0, 1, 2, 3\}$, at least one of the following holds: - $S_{i_1} = S_{i_2}$ - $S_{i_1} = -S_{i_2}$ - $2S_{i_3} + S_{i_1} + S_{i_2} = 0$ Then for all pairs of distinct indices $i, j \in \{0, 1, 2, 3\}$, the components of the solution have equal absolute values, $|S_i| = |S_j|$.

Let $S = (S_0, S_1, S_2, S_3)$ be a solution to the $n=4$ pure $U(1)$ anomaly equations, such that $\sum_{i=0}^3 S_i = 0$ and $\sum_{i=0}^3 S_i^3 = 0$. Suppose $S$ satisfies the line-in-plane condition: for every triple of pairwise distinct indices $i_1, i_2, i_3 \in \{0, 1, 2, 3\}$, at least one of the following holds: - $S_{i_1} = S_{i_2}$ - $S_{i_1} = -S_{i_2}$ - $2S_{i_3} + S_{i_1} + S_{i_2} = 0$ Then the solution $S$ has constant absolute value, meaning $|S_i| = |S_j|$ for all $i, j \in \{0, 1, 2, 3\}$.

Let $N = n + 4$ for some natural number $n$ (such that $N \ge 4$). Let $S = (S_0, S_1, \dots, S_{N-1})$ be a solution to the pure $U(1)$ anomaly equations, satisfying $\sum_{i=0}^{N-1} S_i = 0$ and $\sum_{i=0}^{N-1} S_i^3 = 0$. Suppose $S$ satisfies the line-in-plane condition, which states that for every triple of pairwise distinct indices $i_1, i_2, i_3 \in \{0, \dots, N-1\}$, at least one of the following holds: - $S_{i_1} = S_{i_2}$ - $S_{i_1} = -S_{i_2}$ - $2S_{i_3} + S_{i_1} + S_{i_2} = 0$ Then the solution $S$ has constant absolute value, meaning $|S_i| = |S_j|$ for all $i, j \in \{0, \dots, N-1\}$.