\documentclass[11pt]{article} \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} \usepackage{amsmath,amssymb,amsthm} \usepackage[margin=2.6cm]{geometry} \usepackage{lmodern} \usepackage[hidelinks]{hyperref} \newtheorem{theorem}{Theorem} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newcommand{\Dval}{6.92} \newcommand{\Dnum}{6.9173} \title{A certified upper bound for the distortion of the trefoil knot} \author{David Niedbala Giraudin\thanks{ORCID 0009-0009-1526-1178. Technical assistance: Claude (Anthropic).}} \date{31 October 2026; this version 1 November 2026} \begin{document} \maketitle \begin{abstract} The distortion of a rectifiable closed curve $\gamma\subset\mathbb R^3$ is the supremum, over pairs of distinct points, of the ratio of the length of the shorter arc of $\gamma$ between them to their distance; the distortion $\delta(K)$ of a knot type $K$ is the infimum over all curves of type $K$. For the trefoil, the published lower bound is $\delta\ge5\pi/3$, valid for every non-trivial knot (a recent preprint claims $\delta>6.5321$), and the only upper bound we found in the literature is the value $7.16$ obtained by numerical simulation. We exhibit an explicit polygon with $192$ vertices, of knot type trefoil, whose distortion is at most $\Dval$. The proof is computer-assisted: the knot type is certified by an explicit piecewise-linear path of embedded polygons starting from a polygon on a torus, and the distortion by exact rational arithmetic on every pair of edges. Two separately written verifiers are supplied. \end{abstract} \section{Introduction} For a rectifiable closed curve $\gamma$ let $d_\gamma(p,q)$ be the length of the shorter of the two arcs of $\gamma$ joining $p$ and $q$, and \[ \delta(\gamma)=\sup_{p\ne q\in\gamma}\frac{d_\gamma(p,q)}{|p-q|},\qquad \delta(K)=\inf\{\delta(\gamma):\gamma\text{ of knot type }K\}. \] Gromov \cite{Gro78,Gro} showed $\delta(\gamma)\ge\pi/2$, with equality only for the round circle, and asked whether $\delta(K)$ is bounded over all knots; Pardon \cite{Par} proved that it is not. Denne and Sullivan \cite{DS} proved $\delta(K)\ge5\pi/3$ for every non-trivial tame knot; a recent computer-assisted preprint \cite{Sch} claims $\delta(K)>6.5321$. For the trefoil, Denne and Sullivan mention that numerical simulations \cite{Mul} produced a trefoil with distortion less than $7.16$; to our knowledge this is the only upper bound recorded, and it is not accompanied by a certificate. \begin{theorem}\label{thm:main} There is a closed polygon $F$ with $192$ vertices, of knot type trefoil, with $\delta(F)\le\Dval$. Consequently $5\pi/3\le\delta(\text{trefoil})\le\Dval$ (and $6.5321<\delta(\text{trefoil})$ if the bound of \cite{Sch} is confirmed). \end{theorem} Since $\delta$ is invariant under all isometries of $\mathbb R^3$, reflections included, the bound holds for both the left- and the right-handed trefoil. The numerical value of $\delta(F)$ is about $\Dnum$; we do not claim that $F$ is close to optimal. \section{Knot type}\label{sec:type} \paragraph{The starting polygon.} Let $\Sigma$ be the polyhedral torus with vertices $V(a,b)=\big((R+r c_b)c_a,\,(R+r c_b)s_a,\,r s_b\big)$, $a,b\in\mathbb Z/4$, where $(c_k,s_k)=(\cos\frac{k\pi}{2},\sin\frac{k\pi}{2})$, $R=8$, $r=5$. Since $R>r>0$, its faces $V(a,b)V(a+1,b)V(a+1,b+1)V(a,b+1)$ are planar convex trapezoids and $\Sigma$ is an embedded torus bounding an unknotted solid torus. On the chart $(\alpha,\beta)\in(\mathbb R/\mathbb Z)^2$ of $\Sigma$ (bilinear on each face, mapping the boundary of each grid square homeomorphically onto the boundary of the face) consider the line $t\mapsto(2t,3t)$. It meets the grid corners only at $t\in\frac14\mathbb Z$, at four distinct corners. Cutting it at the grid lines $\alpha,\beta\in\frac14\mathbb Z$ gives points on the boundaries of the faces; consecutive cut points lie on the boundary of a common face, and $P_0$ is formed by the straight segments joining them. Distinct passes of the line through a grid square are parallel chords, whose endpoints do not interleave on the boundary of the square, hence not on the boundary of the face; two straight chords of a convex polygon with non-interleaving endpoints are disjoint. So $P_0$ is a simple closed curve on $\Sigma$, isotopic on $\Sigma$ (face by face) to the image of the line, a curve of class $(2,3)$, hence a trefoil \cite[\S3.C]{Rolfsen}. Each of its $16$ edges is subdivided into $6$ equal parts; the $96$ vertices are rational and become integers in the unit $1/U$, $U=36\cdot2^{14}$. \begin{lemma}[linear steps]\label{lem:step} Let $C,C'$ be polygons with the same number of vertices, $\delta_0\ge\max_v|C'_v-C_v|$, and for each $s\in[0,1]$ let $C_s=(1-s)C+sC'$. Suppose (i) every two non-adjacent edges of $C$ are at distance $>2\delta_0$, and (ii) at each vertex, with adjacent edges $e_0,e$ and turning angle $\theta$, $2\delta_0<\min(|e_0|,|e|)\cos(\theta/2)$. Then every $C_s$ is an embedded polygon, and $C$ and $C'$ have the same knot type. \end{lemma} \begin{proof} Points of an edge move by at most $\delta_0$, so non-adjacent edges of $C_s$ stay at positive distance. An edge vector moves by at most $2\delta_0<|e|$, so its direction turns by at most $\arcsin(2\delta_0/|e|)$, and by (ii) the turning angle between adjacent edges stays below $\theta+2\arcsin(\cos(\theta/2))=\pi$; adjacent edges meet only at their common vertex. So every $C_s$ is embedded. By compactness there is $\eta>0$ such that, for all $s$, non-adjacent edges of $C_s$ are at distance $>\eta$, edges are longer than $\eta$ and turning angles are below $\pi-\eta$. Cut $[0,1]$ into intervals on which every vertex moves by less than $\varepsilon$, and on each of them move the vertices one at a time. For $\varepsilon$ small in terms of $\eta$, the two triangles swept by one such move meet the polygon only along the two edges being moved, so each move is an elementary ($\Delta$-) move; hence $C$ and $C'$ are equivalent \cite[Ch.~1]{BZ}. \end{proof} \paragraph{The path.} The certificate lists integer polygons $X_0=P_0,X_1,\dots,X_{75}$ (unit $1/U$); then every edge of $X_{75}$ is split at its midpoint, which leaves the polygon unchanged and is exact in the unit $1/(2U)$, and $44$ further polygons follow, the last being $F$. Each step $A\to B$ is subdivided by the integer points $A+\lfloor j(B-A)/m\rfloor$, and every sub-step satisfies Lemma~\ref{lem:step}. In verifier (i) both conditions are checked exactly, in integers. Write $D$ for the largest squared displacement of a vertex (so $\delta_0^2=D$), $d_U=\lfloor\sqrt D\rfloor+1>\delta_0$ and $\ell_i=\lfloor\sqrt{|e_i|^2}\rfloor+1>|e_i|$. Condition (i) follows from $\mathrm{dist}(e_i,e_j)\ge|m_i-m_j|-\frac12(|e_i|+|e_j|)$ ($m$ the midpoints) when $4|m_i-m_j|^2>(4d_U+\ell_i+\ell_j)^2$, and otherwise from the exact rational distance between the two segments; pairs whose midpoints are farther apart are excluded by an exact integer grid. For (ii), with $n_0=|e_0|^2$, $n=|e|^2$, $m=\min(n_0,n)$ and $P=\lfloor\sqrt{n_0n}\rfloor$, the condition $4\delta_0^2