发表机构
Department of Mathematics, Hacettepe University（哈塞特佩大学数学系）
机构由 AI 辅助整理，请以论文原文为准。AI 中文总结
本文研究三维协动递归过程中转置对尾部结构的影响，发现尾部指数不变但尾部水平可能不对称，并揭示算子手性等价于Kalman秩条件，通过极-扭机制和数值模拟验证了可探测性障碍。
AI 中文摘要
我们探讨在d=3维中，转置是否改变协动递归过程$x_{t+1}=Q_tBQ_t^\top x_t+Q_t\eta_t$的平稳重尾。该模型包含独立的Haar旋转、协方差为$\Sigma$的高斯体框架噪声，以及一个显式的正度量$\mathsf g$。尾部指数$\alpha_\star$仅依赖于奇异值，因此$\alpha_\star(B)=\alpha_\star(B^\top)$。结构定理和极-扭机制是无条件的。在$\mathsf g=I$坐标下，设$B=S+\widehat{\boldsymbol\omega}$：$\Delta_6(B)=-16\det[\boldsymbol\omega,S\boldsymbol\omega,S^2\boldsymbol\omega]$。则$B\not\sim_{O(3)}B^\top$、$\Delta_6(B)\ne0$、$\operatorname{rank}[\boldsymbol\omega,S\boldsymbol\omega,S^2\boldsymbol\omega]=3$以及$(S,\boldsymbol\omega)$的可控性等价；内禀算子手性是Kalman秩条件。与指数不同，尾部水平不必是转置不变的。精确的极-扭恒等式在镜像下旋转各向异性噪声框架，展示了水平不对称如何进入。非零不对称仅条件性和数值性地建立，而非通过闭式示例：在SPD图的假设R下，当$A\to0$时，$\Delta\log C=\langle A,G\rangle+o(\\|A\\|)$，其中$G=2H\circ[\Sigma,\partial_\Sigma\log C]$；有限阈值模拟与此一阶定律一致。严格Haar平均使每个滞后二阶交叉矩转置盲；第四径向矩保留一个奇通道。没有有限标量加权径向矩组合能统一消除偶二次型而保留奇协向量。
英文摘要
We ask whether transposition changes the stationary heavy tail of the co-moving recursion $\mathbf x_{t+1}=Q_tBQ_t^\top\mathbf x_t+Q_t\boldsymbolη_t$ in $d=3$. The model has independent Haar rotations, Gaussian body-frame noise with covariance $Σ$, and an explicit positive metric $\mathsf g$. The tail index $α_\star$ depends only on singular values, so $α_\star(B)=α_\star(B^\top)$. The structure theorem and polar-twist mechanism are unconditional. In the $\mathsf g=I$ chart, write $B=S+\widehat{\boldsymbolω}$: $Δ_6(B)=-16\det[\boldsymbolω,S\boldsymbolω,S^2\boldsymbolω]$. Then $B\not\sim_{O(3)}B^\top$, $Δ_6(B)\ne0$, $\operatorname{rank}[\boldsymbolω,S\boldsymbolω,S^2\boldsymbolω]=3$, and controllability of $(S,\boldsymbolω)$ are equivalent; intrinsic operator chirality is a Kalman rank condition. Unlike the exponent, the tail level need not be transpose-invariant. The exact polar-twist identity rotates the anisotropic noise frame under mirroring, showing how level asymmetry can enter. A nonzero asymmetry is established only conditionally and numerically, not by a closed-form example: under Assumption R on the SPD chart, $Δ\log C=\langle A,G\rangle+o(\|A\|)$ as $A\to0$, with $G=2H\circ[Σ,\partial_Σ\log C]$; finite-threshold simulations are consistent with this first-order law. Strict-Haar averaging makes every lagged second-order cross-moment transpose-blind; a fourth radial moment retains an odd channel. No finite scalar-weighted radial-moment combination uniformly cancels the even quadratic form while retaining the odd covector.
Comments29 pages, 1 figure; 47-page Supplement included as an ancillary file
