Physlib.QFT.QED.AnomalyCancellation.Odd.LineInCubic

For a linear solution $S$ to the anomaly cancellation conditions of a pure $U(1)$ theory with $2n+1$ fermions, the property `LineInCubic S` holds if, for any decomposition of $S$ into vectors $P(g)$ and $P!(f)$ (where $g, f \in \mathbb{Q}^n$ parametrize the basis planes), every linear combination $a P(g) + b P!(f)$ for $a, b \in \mathbb{Q}$ satisfies the cubic anomaly condition. That is, \[ \sum_{i=1}^{2n+1} (a P(g)_i + b P!(f)_i)^3 = 0. \] In geometric terms, this means the plane spanned by the vectors $P(g)$ and $P!(f)$ lies entirely within the cubic surface defined by the gauge anomaly.

In a pure $U(1)$ gauge theory with $2n+1$ fermions, let $S$ be a charge vector satisfying the linear anomaly cancellation condition $\sum_{i=1}^{2n+1} S_i = 0$. Suppose $S$ satisfies the `LineInCubic` property, which means that for any decomposition $S = P(g) + P'(f)$ (where $P(g)$ and $P'(f)$ are vectors in the charge space), any linear combination $a P(g) + b P'(f)$ for $a, b \in \mathbb{Q}$ satisfies the cubic anomaly condition $\sum x_i^3 = 0$. Under these conditions, for any such decomposition $S = P(g) + P'(f)$ and for any rational numbers $a$ and $b$, the symmetric trilinear form $\mathcal{A}(x, y, z) = \sum_{i=1}^{2n+1} x_i y_i z_i$ satisfies the following identity: \[ 3ab \left( a \mathcal{A}(P(g), P(g), P'(f)) + b \mathcal{A}(P'(f), P'(f), P(g)) \right) = 0 \]

In a pure $U(1)$ gauge theory with $2n+1$ fermions, let $S$ be a charge vector satisfying the linear anomaly cancellation condition $\sum_{i=1}^{2n+1} S_i = 0$. Suppose $S$ satisfies the `LineInCubic` property, which means that for any decomposition $S = P(g) + P'(f)$ (where $g, f \in \mathbb{Q}^n$ parametrize the basis planes), every linear combination $a P(g) + b P'(f)$ for $a, b \in \mathbb{Q}$ satisfies the cubic anomaly condition. Under these conditions, for any such decomposition $S = P(g) + P'(f)$, the symmetric trilinear form $\mathcal{A}(x, y, z) = \sum_{i=1}^{2n+1} x_i y_i z_i$ satisfies the identity: \[ \mathcal{A}(P(g), P(g), P'(f)) = 0 \]

For a rational charge vector $S = (S_1, \dots, S_{2n+1})$ satisfying the linear anomaly cancellation condition $\sum_{i=1}^{2n+1} S_i = 0$ in a pure $U(1)$ gauge theory with $2n+1$ fermions, the property `LineInCubicPerm S` holds if, for every permutation $\sigma \in S_{2n+1}$, the permuted vector $\sigma(S)$ satisfies the `LineInCubic` property. The `LineInCubic` property signifies that the plane spanned by the vectors $P(g)$ and $P'(f)$ in the decomposition of the charge vector lies within the cubic surface defined by the gauge anomaly $\sum_{i=1}^{2n+1} x_i^3 = 0$.

In a pure $U(1)$ gauge theory with $2n+1$ fermions, let $S = (S_1, \dots, S_{2n+1})$ be a vector of rational charges satisfying the linear anomaly cancellation condition $\sum_{i=1}^{2n+1} S_i = 0$. If $S$ satisfies the `LineInCubicPerm` property—meaning that for every permutation $\sigma$ of its components, the permuted vector $\sigma(S)$ satisfies the `LineInCubic` property—then the original vector $S$ itself satisfies the `LineInCubic` property.

In a pure $U(1)$ gauge theory with an odd number of fermions $2n+1$, let $S = (S_1, \dots, S_{2n+1})$ be a rational charge vector satisfying the linear anomaly cancellation condition $\sum_{i=1}^{2n+1} S_i = 0$. If $S$ satisfies the property `LineInCubicPerm` (meaning that every permutation $\sigma$ of the vector $S$ satisfies the `LineInCubic` property, where the plane associated with the charge vector lies within the cubic anomaly surface), then for any permutation $M$ in the symmetric group $S_{2n+1}$, the permuted charge vector $M(S)$ also satisfies the `LineInCubicPerm` property.

Consider a pure $U(1)$ gauge theory with an odd number of fermions $2n+3$, where $n \in \mathbb{N}$. Let $S = (S_1, \dots, S_{2n+3})$ be a vector of rational charges satisfying the linear anomaly cancellation condition $\sum_{i=1}^{2n+3} S_i = 0$. Suppose $S$ satisfies the `LineInCubicPerm` property, which means that for every permutation $\sigma$ of the components of $S$, the permuted vector $\sigma(S)$ satisfies the `LineInCubic` property. For any decomposition of $S$ into the form $S = Pa(g, f)$, where $g, f: \{0, \dots, n\} \to \mathbb{Q}$ are functions parameterizing the solution space, and for any index $j \in \{0, \dots, n\}$, the following identity holds: \[ (S_{\text{snd}, j} - S_{\text{fst}, j}) \cdot \mathcal{A}(P(g), P(g), e'_j) = 0 \] where $S_{\text{fst}, j}$ and $S_{\text{snd}, j}$ denote the components of the charge vector $S$ at the indices `oddShiftFst j` and `oddShiftSnd j` respectively, $\mathcal{A}(x, y, z) = \sum_{i=1}^{2n+3} x_i y_i z_i$ is the symmetric trilinear form associated with the cubic anomaly cancellation condition, $P(g)$ is a charge vector derived from the function $g$, and $e'_j$ is the basis charge vector `basis!AsCharges j`.

Consider a pure $U(1)$ gauge theory with an odd number of fermions $2(n+2)+1$. Let $S$ be a vector of rational charges that satisfies the linear anomaly cancellation condition $\sum S_i = 0$. Suppose $S$ is represented via the decomposition $Pa(f, g)$ for functions $f, g: \{0, 1, \dots, n+1\} \to \mathbb{Q}$. Let $\mathcal{A}(x, y, z) = \sum x_i y_i z_i$ be the symmetric trilinear form associated with the cubic anomaly condition. The value of this form evaluated at $(P(f), P(f), e_0)$, where $e_0$ is the basis charge vector corresponding to the first index, is given by: \[ \mathcal{A}(P(f), P(f), e_0) = (S_{\text{fst}, 0} + S_{\text{snd}, 0}) (2 S_{\text{zero}} + S_{\text{fst}, 0} + S_{\text{snd}, 0}) \] Here, $S_{\text{zero}}$, $S_{\text{fst}, 0}$, and $S_{\text{snd}, 0}$ denote the specific components of the charge vector $S$ at the indices `oddShiftZero`, `oddShiftFst 0`, and `oddShiftSnd 0` respectively, and $P(f)$ is a charge vector derived from the component function $f$.

Consider a pure $U(1)$ gauge theory with an odd number of fermions $2(n+2)+1$, where $n \in \mathbb{N}$. Let $S = (S_1, \dots, S_{2n+5})$ be a vector of rational charges satisfying the linear anomaly cancellation condition $\sum_{i=1}^{2n+5} S_i = 0$. Suppose $S$ satisfies the `LineInCubicPerm` property, meaning that for every permutation $\sigma$ of the components of $S$, the permuted vector $\sigma(S)$ satisfies the property that the line through that point and the planes formed by the basis of linear solutions lies within the cubic surface $\sum x_i^3 = 0$. Then, the specific components of the charge vector $S_{\text{snd}, 0}$, $S_{\text{fst}, 0}$, and $S_{\text{zero}}$ (corresponding to the indices `oddShiftSnd 0`, `oddShiftFst 0`, and `oddShiftZero`) satisfy the condition: \[ S_{\text{snd}, 0} = S_{\text{fst}, 0}, \quad S_{\text{snd}, 0} = -S_{\text{fst}, 0}, \quad \text{or} \quad 2S_{\text{zero}} + S_{\text{snd}, 0} + S_{\text{fst}, 0} = 0. \]

Consider a pure $U(1)$ gauge theory with an odd number of fermions $2(n+2)+1$, where $n$ is a natural number. Let $S = (S_0, S_1, \dots, S_{2n+4}) \in \mathbb{Q}^{2n+5}$ be a vector of rational charges satisfying the linear anomaly cancellation condition $\sum_{i=0}^{2n+4} S_i = 0$. If $S$ satisfies the `LineInCubicPerm` property (meaning that for every permutation $\sigma$ of the components of $S$, the permuted vector $\sigma(S)$ satisfies the property that the line through that point and the planes formed by the basis of linear solutions lies within the cubic surface $\sum x_i^3 = 0$), then $S$ satisfies the `LineInPlaneCond` property. Specifically, for any three distinct indices $i, j, k \in \{0, 1, \dots, 2n+4\}$, the components of $S$ must satisfy at least one of the following conditions: - $S_i = S_j$ - $S_i = -S_j$ - $2S_k + S_i + S_j = 0$

Consider a pure $U(1)$ gauge theory with an odd number of fermions $N = 2(n+2)+1$, where $n \in \mathbb{N}$ (so $N \ge 5$). Let $S = (S_1, \dots, S_N) \in \mathbb{Q}^N$ be a vector of rational charges satisfying the linear anomaly cancellation condition $\sum_{i=1}^N S_i = 0$. Suppose $S$ satisfies the `LineInCubicPerm` property, meaning that for every permutation $\sigma$ of the indices $\{1, \dots, N\}$, the permuted vector $\sigma(S)$ satisfies the property that the line through that point and the planes formed by the basis of linear solutions lies within the cubic surface defined by the gauge anomaly $\sum_{i=1}^N x_i^3 = 0$. Then, the vector $S$ must satisfy the constant absolute value condition, meaning $|S_i| = |S_j|$ for all $i, j \in \{1, \dots, N\}$.

Consider a pure $U(1)$ gauge theory with an odd number of fermions $N = 2(n+2)+1$, where $n \in \mathbb{N}$ (so $N \ge 5$). Let $S = (S_1, \dots, S_N) \in \mathbb{Q}^N$ be a vector of rational charges satisfying the linear anomaly cancellation condition $\sum_{i=1}^N S_i = 0$. Suppose $S$ satisfies the `LineInCubicPerm` property, meaning that for every permutation $\sigma$ of the indices $\{1, \dots, N\}$, the permuted vector $\sigma(S)$ satisfies the property that the line through that point and the planes formed by the basis of linear solutions lies within the cubic surface defined by the gauge anomaly $\sum_{i=1}^N x_i^3 = 0$. Then $S$ is the zero vector, i.e., $S = (0, \dots, 0)$.