Kac's Walk on Rotation Matrices Mixes in $\boldsymbol{\Theta(n^2)}$ Steps: A Proof Discovered with AI

Abstract

Let $N=\binom n2=\dim\mathrm{SO}(n)$. We prove that the coordinate-plane Kac walk on $\mathrm{SO}(n)$ has total-variation mixing time of order $N$: for every fixed $0<\varepsilon<1$, \[ t_{\mathrm{mix}}^{(n)}(\varepsilon)=\Theta_\varepsilon(n^2). \] The lower bound is the dimensional singularity obstruction before $N$ steps. The upper bound removes the final logarithm from the previously known $O(n^2\log n)$ estimate. The proof combines the discrete Malliavin coupling and low-degree pseudo-mixing inputs with a new log-free analysis of the derivative shells. Its static core is a circuit-anchored, arbitrary-spectrum root/pass identity for the physical five-box prime. Keeping one normalization base per original circuit permits simultaneous scalar regluing without paying for artificial cuts. Its temporal core is an exact chronological calculus: passive singleton runs acquire a coboundary/Riesz gain, while root-interrupted components are allocated by vertex-labelled packets before absolute values are taken. The curvature split into pure-Weyl and Ricci parts is kept at its physical tensor type. All-Weyl packets retain a full $N^{-1}$ resource; mixed packets contain a typed $O(n^{-1/2})$ Ricci debit; and the final packetless Ricci cell is closed by a joint invariant-column estimate on its two root-hit circuits and an exact causal restoration of the marked root time. These estimates yield an $O(n)$ squared first-derivative shell and a summable all-order marked-shell expansion through logarithmic degree. The resulting score energy is $O(n/c^2)$ after $cN$ steps. A weighted submersion integration-by-parts argument and the Haar log-Sobolev inequality then give the uniform total-variation upper bound. No cutoff profile or cutoff window is asserted.