由单位根和互反对扭曲的舒尔多项式：挠率滤子、融合商与奇数阶全单模性
Schur polynomials twisted by roots of unity and reciprocal pairs: torsion filters, fusion quotients, and total unimodularity at odd order
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中文总结 AI 辅助
本文研究由单位根和互反对扭曲的舒尔多项式，解决其求值的零化问题，证明奇数阶对应全单模性，涉及挠率滤子、融合商等，部分结论依赖猜想。
中文摘要 AI 辅助
记μ_t为t次单位根，z^{±1}为r个自由互反对。我们研究Φ_{t,r}(β)=s_λ(μ_t,z^{±1})，其中β=λ+δ，以及r=1时配套论文留下的未解决问题：它何时为零？我们将该求值分解为经典分支再经挠率滤子，其形式取决于t的奇偶性：该点属于行列式为(-1)^{t+1}的正交群。当t为奇数时，它属于恒等分支：即普通限制SO_{2R'+1}↓SO_{2m'+1}×SO_{2r}，滤子为阶h+1的主元处的奇正交特征；当t为偶数时，它属于另一分支：即扭化、虚拟展开以及正则但非主元的挠率元素，我们在此证明带符号的滤子。两种情形可统一表述为：仅当移位点正则半单时，滤子非零。二者均为极小级融合投影：偶情形为C型，奇情形为B型的张量扇区。仿射折叠对应0、±1；其未覆盖部分以猜想形式存在。最高存活权重为分子牛顿多面体的主导顶点减去分母的顶点——后者已被证明，前者依赖于单轨道性质——且该类猜想为本原的，即±秩1商的生成元。对于奇数t和一个Λ，由等秩特征公式，分子为{0,±1}中的带符号横截计数，仅余一次除法。我们沿等差数列将其以闭形式求逆；商为±ε_t det M，其中M为显式的0/±1矩阵——即带符号的区间矩阵，故全单模，由此解决(L1)。纤维计数为永久值，仅当为1时为奇数，因此在主导索引处，多重命中纤维和为零。仍有两个极值命题待证。未证明部分在两种奇偶性下均有度量。
英文摘要
Write $μ_t$ for the $t$-th roots of unity and $z^{\pm1}$ for $r$ free reciprocal pairs. We study $Φ_{t,r}(β)=s_λ(μ_t,z^{\pm1})$, $β=λ+δ$, and the question the companion paper left open after $r=1$: when does it vanish? We factor the evaluation into classical branching followed by a torsion filter, and the shape depends on the parity of $t$: the point lies in the orthogonal group with determinant $(-1)^{t+1}$. For odd $t$ it sits in the identity component: an ordinary restriction $SO_{2R'+1}\downarrow SO_{2m'+1}\times SO_{2r}$, the filter an odd orthogonal character at a principal element of order $h+1$. For even $t$ in the other: a twining, a virtual expansion, and a torsion element regular but not principal; there we prove the filter, with its sign. One description covers both: the filter is nonzero exactly when the shifted point is regular semisimple. Both are minimal-level fusion projections: the even of type $C$, the odd's tensor sector of type $B$. Affine folding accounts for $0,\pm1$; what it does not survives as conjectures. The highest surviving weight is the dominant vertex of the numerator's Newton polytope minus the denominator's --- the latter proved, the former conditional on a single-orbit property --- and the class there is conjecturally primitive, $\pm$ the generator of the rank-one quotient. For odd $t$ and one $Λ$ that numerator is a signed transversal count in $\{0,\pm1\}$ by the equal-rank character formula, leaving one division. We invert it in closed form, along an arithmetic progression; the quotient is $\pmε_t\det M$ for an explicit $0/{\pm}1$ matrix --- an interval matrix up to signs, hence totally unimodular, which settles (L1). The fibre count is a permanent, odd only when $1$, so at a dominant index a multi-hit fibre sums to zero. Two extremal statements remain. What is unproved is measured, in both parities.
