Physlib.QFT.QED.AnomalyCancellation.Even.LineInCubic

For a vector of charges $S \in \mathbb{Q}^{2(n+1)}$ satisfying the linear anomaly cancellation condition $\sum x_i = 0$, the property `LineInCubic` holds if for every decomposition of $S$ into the form $P(g) + P'(f)$ (where $P(g)$ and $P'(f)$ are vectors defined by the basis of the linear solution space), any linear combination $a P(g) + b P'(f)$ with $a, b \in \mathbb{Q}$ also satisfies the cubic anomaly cancellation condition: \[ \sum_{i=1}^{2(n+1)} (a P(g) + b P'(f))_i^3 = 0 \] In other words, the entire line spanned by the basis components of $S$ lies within the cubic hypersurface defined by the gauge anomaly.

In a pure $U(1)$ gauge theory with $2(n+1)$ Weyl fermions, let $S$ be a vector of rational charges satisfying the linear anomaly cancellation condition $\sum S_i = 0$. Suppose $S$ satisfies the `LineInCubic` property, which means that for any decomposition of $S$ into the form $P(g) + P^\prime(f)$ (where $g: \{0, \dots, n\} \to \mathbb{Q}$ and $f: \{0, \dots, n-1\} \to \mathbb{Q}$ are vectors from the basis of linear solutions), any linear combination $a P(g) + b P^\prime(f)$ for $a, b \in \mathbb{Q}$ also satisfies the cubic anomaly cancellation condition. Then, for any such decomposition and any $a, b \in \mathbb{Q}$, the following identity holds: \[ 3ab \left( a \mathcal{A}(P(g), P(g), P^\prime(f)) + b \mathcal{A}(P^\prime(f), P^\prime(f), P(g)) \right) = 0 \] where $\mathcal{A}(x, y, z) = \sum_{i=1}^{2(n+1)} x_i y_i z_i$ is the symmetric trilinear form associated with the cubic anomaly.

In a pure $U(1)$ gauge theory with $2(n+1)$ Weyl fermions, let $S$ be a vector of rational charges that satisfies the linear anomaly cancellation condition $\sum S_i = 0$. Suppose $S$ satisfies the `LineInCubic` property, which implies that for any decomposition of $S$ into basis vectors $P(g)$ and $P'(f)$ (where $g: \{0, \dots, n\} \to \mathbb{Q}$ and $f: \{0, \dots, n-1\} \to \mathbb{Q}$), every point on the line spanned by these vectors satisfies the cubic anomaly cancellation condition. Then, for any such decomposition $S = P(g) + P'(f)$, the symmetric trilinear form $\mathcal{A}$ evaluated at $(P(g), P(g), P'(f))$ is zero: \[ \mathcal{A}(P(g), P(g), P'(f)) = 0 \] where $\mathcal{A}(x, y, z) = \sum_{i=1}^{2(n+1)} x_i y_i z_i$.

For a vector of rational charges $S \in \mathbb{Q}^{2(n+1)}$ satisfying the linear anomaly cancellation condition $\sum S_i = 0$ in a pure $U(1)$ theory with $2(n+1)$ fermions, the property `LineInCubicPerm` holds if, for every permutation $M \in S_{2(n+1)}$ of the charges, the permuted vector $M(S)$ satisfies the `LineInCubic` property. The `LineInCubic` property signifies that the entire line spanned by the basis decomposition of the vector (into components $P(g)$ and $P'(f)$) lies within the cubic hypersurface defined by the gauge anomaly cancellation condition $\sum x_i^3 = 0$.

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

Consider a pure $U(1)$ gauge theory with $2(n+1)$ fermions, where $n \in \mathbb{N}$. Let $S = (S_1, \dots, S_{2(n+1)})$ be a vector of rational charges satisfying the linear anomaly cancellation condition $\sum_{i=1}^{2(n+1)} S_i = 0$. Suppose $S$ satisfies the property `LineInCubicPerm`, which means that for every permutation $M$ in the symmetric group $S_{2(n+1)}$, the permuted vector $M(S)$ satisfies the `LineInCubic` property (where the entire line spanned by the basis decomposition of the vector lies within the cubic hypersurface $\sum x_i^3 = 0$). Then, for any permutation $M' \in S_{2(n+1)}$, the permuted vector $M'(S)$ also satisfies the `LineInCubicPerm` property.

Consider a pure $U(1)$ gauge theory with $2(n+1)$ fermions, where $n \in \mathbb{N}$. Let $S = (S_1, \dots, S_{2(n+1)})$ be a vector of rational charges satisfying the linear anomaly cancellation condition $\sum S_i = 0$. Suppose $S$ satisfies the property `LineInCubicPerm`, which means that for every permutation $M$ in the symmetric group $S_{2(n+1)}$, the permuted vector $M(S)$ satisfies the `LineInCubic` property (where the entire line spanned by the basis decomposition of the vector lies within the cubic hypersurface $\sum x_i^3 = 0$). For any $j \in \{0, \dots, n-1\}$ and any decomposition of $S$ into the basis components $S = P(g) + P'(f)$, where $g: \{0, \dots, n\} \to \mathbb{Q}$ and $f: \{0, \dots, n-1\} \to \mathbb{Q}$, the following identity holds: $$(S_{i_{2j+1}} - S_{i_{2j}}) \cdot \mathcal{A}(P(g), P(g), B'_j) = 0$$ where $S_{i_{2j}}$ and $S_{i_{2j+1}}$ are the charges at the indices `evenShiftFst j` and `evenShiftSnd j`, $B'_j$ is the $j$-th basis charge vector `basis!AsCharges j`, and $\mathcal{A}(x, y, z) = \sum_{i=1}^{2(n+1)} x_i y_i z_i$ is the symmetric trilinear form associated with the cubic anomaly.

Consider a pure $U(1)$ gauge theory with $2n+4$ fermions (where $n \in \mathbb{N}$). Let $S = (S_1, \dots, S_{2n+4})$ be a vector of rational charges that satisfies the linear anomaly cancellation condition $\sum S_i = 0$. Suppose $S$ is represented as $Pa(f, g)$ for some functions $f$ and $g$. Let $\mathcal{A}$ be the symmetric trilinear form defining the cubic anomaly, given by $\mathcal{A}(x, y, z) = \sum x_i y_i z_i$. For the basis charge vector $B_n$ associated with the last index, the following identity holds: $$\mathcal{A}(P(f), P(f), B_n) = -(S_{i_1} + S_{i_2})(2S_{i_3} + S_{i_1} + S_{i_2})$$ where $S_{i_1}$ and $S_{i_2}$ are the charges at the indices `evenShiftFst` and `evenShiftSnd` for the last basis index, and $S_{i_3}$ is the charge at the index `evenShiftLast`.

Consider a pure $U(1)$ gauge theory with $2n+4$ Weyl fermions ($n \in \mathbb{N}$). Let $S = (S_1, \dots, S_{2n+4}) \in \mathbb{Q}^{2n+4}$ be a vector of rational charges that satisfies the linear anomaly cancellation condition $\sum_{i=1}^{2n+4} S_i = 0$. If $S$ satisfies the `LineInCubicPerm` property—meaning that for every permutation $\sigma$ of the charges, the permuted vector $\sigma(S)$ satisfies the property that the line through it and the two different planes formed by the basis of linear solutions lies within the cubic hypersurface $\sum x_i^3 = 0$—then the charges at the specific indices $i_1 = \text{evenShiftFst}(\text{last } n)$, $i_2 = \text{evenShiftSnd}(\text{last } n)$, and $i_3 = \text{evenShiftLast}$ satisfy the line-in-plane condition: $$S_{i_2} = S_{i_1}, \quad S_{i_2} = -S_{i_1}, \quad \text{or} \quad 2S_{i_3} + S_{i_2} + S_{i_1} = 0$$

Consider a pure $U(1)$ gauge theory with $2n+4$ fermions (where $n \in \mathbb{N}$). Let $S = (S_0, S_1, \dots, S_{2n+3}) \in \mathbb{Q}^{2n+4}$ be a vector of rational charges that satisfies the linear anomaly cancellation condition $\sum_{i=0}^{2n+3} S_i = 0$. Suppose $S$ satisfies the `LineInCubicPerm` property, meaning that for every permutation $\sigma \in S_{2n+4}$, the permuted vector $\sigma(S)$ has the property that the line passing through it and the planes formed by the basis of linear solutions lies entirely within the cubic hypersurface $\sum x_i^3 = 0$. Then $S$ satisfies the `LineInPlaneCond` property: for every triple of pairwise distinct indices $i, j, k \in \{0, \dots, 2n+3\}$, the components of $S$ satisfy at least one of the following: $$S_i = S_j, \quad S_i = -S_j, \quad \text{or} \quad 2S_k + S_i + S_j = 0.$$

Consider a pure $U(1)$ gauge theory with $N = 2n+4$ fermions (where $n \in \mathbb{N}$). Let $S = (S_0, S_1, \dots, S_{N-1}) \in \mathbb{Q}^N$ be a solution to the anomaly cancellation conditions, such that $\sum_{i=0}^{N-1} S_i = 0$ and $\sum_{i=0}^{N-1} S_i^3 = 0$. If $S$ satisfies the `LineInCubicPerm` property—meaning that for every permutation $\sigma$ of the indices, the permuted vector $\sigma(S)$ has the property that the line passing through it and the planes spanned by the basis of linear solutions lies entirely within the cubic hypersurface defined by $\sum x_i^3 = 0$—then the solution $S$ has constant absolute value, i.e., $|S_i| = |S_j|$ for all $i, j \in \{0, \dots, N-1\}$.

Consider a pure $U(1)$ gauge theory with $N = 2n+4$ fermions (where $n \in \mathbb{N}$). Let $S = (S_0, S_1, \dots, S_{N-1}) \in \mathbb{Q}^N$ be a solution to the anomaly cancellation conditions, satisfying $\sum_{i=0}^{N-1} S_i = 0$ and $\sum_{i=0}^{N-1} S_i^3 = 0$. If $S$ satisfies the `LineInCubicPerm` property—meaning that for every permutation $\sigma$ of the indices, the permuted vector $\sigma(S)$ has the property that the line through it and the planes spanned by the basis of linear solutions lies entirely within the cubic hypersurface defined by the gauge anomaly—then the solution $S$ is vector-like. Specifically, if the charges are sorted in non-decreasing order as $x_0 \le x_1 \le \dots \le x_{N-1}$, they satisfy $x_i = -x_{(N/2)+i}$ for all $i \in \{0, 1, \dots, (N/2)-1\}$.

Consider a pure $U(1)$ gauge theory with $N = 2n + 4$ fermions (where $n \in \mathbb{N}$). Let $S$ be a solution to the anomaly cancellation conditions, comprising a vector of rational charges $(S_0, S_1, \dots, S_{N-1}) \in \mathbb{Q}^N$ that satisfies the gravitational anomaly $\sum S_i = 0$ and the gauge anomaly $\sum S_i^3 = 0$. If $S$ satisfies the `LineInCubicPerm` property—meaning that for every permutation $\sigma \in S_N$, the line passing through the permuted solution $\sigma(S)$ and the planes formed by the basis of linear solutions lies entirely within the cubic hypersurface $\sum x_i^3 = 0$—then there exists a permutation $M \in S_N$ such that the permuted charge vector $M(S)$ lies in the subspace spanned by the first part of the basis of linear solutions.