% !TeX program = pdflatex % Standalone source; compile twice with pdfLaTeX. No external files are needed. % Extracted and simplified from constant_width_volume_borsuk(1).tex, Section 3. % Main simplifications: one fixed affine lift, a half-reflection, and a direct % partition argument. Gaussian Poincare inequalities are cited in the text. \documentclass[11pt,reqno,letterpaper]{amsart} \usepackage[letterpaper,margin=1in]{geometry} \usepackage[T1]{fontenc} \usepackage{lmodern} \usepackage{amsmath,amssymb,amsthm,mathtools} \usepackage[expansion=false]{microtype} \usepackage{needspace} \usepackage{tikz} \usetikzlibrary{arrows.meta} \usepackage[colorlinks=true,allcolors=blue]{hyperref} \newtheorem{theorem}{Theorem}[section] \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \newcommand{\R}{\mathbb R} \newcommand{\B}{\mathbb B} \newcommand{\Sph}{\mathbb S} \newcommand{\E}{\mathbb E} \newcommand{\Prob}{\mathbb P} \newcommand{\ip}[2]{\langle #1,#2\rangle} \newcommand{\good}[1]{\mathcal H_{#1}} \DeclareMathOperator{\conv}{conv} \DeclareMathOperator{\diam}{diam} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Lip}{Lip} \numberwithin{equation}{section} \allowdisplaybreaks[1] % ----------------------------------------------------------------------------- % FIRST-PAGE PROVENANCE / PUBLICATION TEMPLATE % Edit the fields below when reusing this template for another note. % Before posting this note, complete the AI-tools and permanent-URL fields. % Use the provider's recorded model names/versions; do not infer them. % Add any substantive human mathematical input to \NoteOrigin as appropriate. % Keep the version date fixed: recompiling must not change the cited version. % The verification statement below records the curator's confirmation of full % checking; it is not a claim that the front-matter editor rechecked the proof. % ----------------------------------------------------------------------------- \newcommand{\NoteTitle}{An exponential improvement for Borsuk's problem} \newcommand{\NoteCurator}{Andriy Prymak} \newcommand{\NoteDate}{October 6, 2026} \newcommand{\NoteAITools}{\textit{ChatGPT-6 Astra, Sept.--Oct.~2026}} % Replace the entire value below by \url{https://...} once the URL is fixed. % Prefer a stable landing page that identifies this dated version. % \newcommand{\NotePermanentURL}{\textit{https://prymak.net/aiomn/2610-01v1.pdf}} \newcommand{\NoteOrigin}{% The main proof arguments were primarily generated using \NoteAITools with discussion and prompts by Andrii~Arman, Andriy~Bondarenko, Andriy~Prymak and Danylo~Radchenko. \NoteCurator{} curates this note but does not claim authorship of its mathematical results. The primary context provided to AI was the study of small spherical bodies of constant width by Hall, Prymak and Sujsuntinukul arXiv:2606.30960v2. Sources of the key proof ingredients are credited in the text.} \newcommand{\NoteVerification}{% The complete mathematical argument has been checked by \NoteCurator{}, who believes the statements and proofs to be correct. This does not assert formal verification or independent peer review.} % Expand the title once so amsart retains its usual capitalization. \expandafter\title\expandafter{\NoteTitle} % No author byline: the curator is identified only in the disclosure/citation. \date{\NoteDate} \makeatletter \@namedef{subjclassname@2020}{% \textup{2020} Mathematics Subject Classification} \makeatother \subjclass[2020]{Primary 52A37; Secondary 52A23, 52A55} \keywords{Borsuk's partition problem, asymptotic convex geometry, spherical polarity, self-dual cones, Gaussian measure, Gaussian Poincar\'e inequality} % In amsart, these unnumbered title-page footnotes also work without \author. \thanks{\textbf{Origin and curation.} \NoteOrigin} % \thanks{\textbf{Mathematical context.} \NoteContext} \thanks{\textbf{Verification: fully checked.} \NoteVerification} % \thanks{\textbf{Permanent URL.} \NotePermanentURL} % \thanks{\textbf{License.} To the extent that \NoteCurator{} holds copyright % or related rights in this manuscript and its \LaTeX{} source, those rights % are licensed under % \href{https://creativecommons.org/licenses/by/4.0/}{Creative Commons % Attribution 4.0 International (CC BY 4.0)}. % Third-party material retains its own terms.} % \thanks{\textbf{Cite as.} \emph{\NoteTitle}. AI-originated mathematical notes, % curated by \NoteCurator{}, version dated \NoteDate. % \NotePermanentURL. % Identify the curator's role without assigning mathematical authorship.} \hypersetup{ pdftitle={\NoteTitle}, pdfauthor={}, pdfsubject={AI-generated mathematical note; curated and fully checked by Andriy Prymak}, pdfkeywords={Borsuk's partition problem, asymptotic convex geometry, spherical polarity, self-dual cones, Gaussian measure, Gaussian Poincare inequality} } \begin{document} \begin{abstract} We present a method for improving the classical exponential base $\sqrt{3/2}$ in the upper bound for Borsuk's partition number. The emphasis is on explaining the method rather than optimizing the bound. The problem is reduced to constructing directions which make an acute angle with every member of a prescribed spherical set. These play the same role as the illuminating directions in Schramm's method. The geometric ingredients are an antipodal pairing, a lift by one coordinate, and reflection in a self-dual cone. This reflection folds Euclidean space into the cone. A translation opposite to the horizontal part of its Gaussian mean makes the required supporting inequalities hold with high probability. The Gaussian Poincar\'e inequality controls the mean, and the formula for translating a Gaussian density bounds the measure cost of this construction. \end{abstract} \maketitle \section{The result and the ideas} Let $b(n)$ be the least integer such that every bounded set of positive diameter in $\R^n$ can be partitioned into at most $b(n)$ sets of strictly smaller diameter. Schramm~\cite{Schramm} and Bourgain--Lindenstrauss~\cite{BL} established the classical bound \[ b(n)\le\left(\sqrt{\frac32}+o(1)\right)^n. \] The purpose of this note is to improve its exponential base with simple constants and a fully explained proof\footnote{A sharper bound, which is not yet verified as of October~6, 2026, can be found at \url{https://prymak.net/aiomn/}.}. For further background on Borsuk's problem, see Kalai's 2015 survey~\cite{Kalai}. \begin{theorem}\label{thm:main} As $n\to\infty$, \[ b(n)\le(C+o(1))^n, \qquad C=\sqrt{\frac{9536}{6375}}=1.22304666\ldots. \] In particular, $b(n)\le 1.224^n$ for all sufficiently large $n$, whereas $\sqrt{3/2}=1.22474487\ldots$. \end{theorem} For a compact nonempty $A\subset\Sph^{n-1}$, write \[ A_+=\{v\in\Sph^{n-1}:\ip uv>0\text{ for every }u\in A\}. \] This is the open positive polar: the directions making an acute angle with every member of $A$. Let $\nu_j$ be probability surface measure on $\Sph^j$. The estimate behind the theorem is the following stable form. \begin{proposition}\label{prop:polar} For every $\beta0$, there are $\delta>0$ and $n_0$ such that \begin{equation}\label{eq:stable} \nu_{n-1}(A_+)\ge \beta^n \quad\text{whenever } n\ge n_0, \quad \diam_{\Sph}A\le\frac\pi3+\delta. \end{equation} Here $\diam_{\Sph}$ denotes angular diameter, and the choices are uniform over all compact nonempty $A$. \end{proposition} \noindent\emph{Why this is relevant to partitions.} Fix a small $\varepsilon>0$ and join two points of a diameter-one set whenever their distance is at least $1-\varepsilon$. A direction $v$ selects a point $x$ if $\ip v{x-y}>0$ for every neighbour $y$: viewed along $v$, the point $x$ lies strictly ahead of all its neighbours. Two neighbours cannot both be selected, because this would require both $\ip v{x-y}>0$ and $\ip v{y-x}>0$. The class selected by $v$ therefore has diameter at most $1-\varepsilon$. A collection of directions selecting every point gives a covering by such classes, and assigning each point to one class gives the desired partition. Within a small neighborhood in the set, collect all unit vectors pointing from nearly diametral neighbours towards points of that neighborhood. Their angular diameter is close to $\pi/3$, and any direction in their open positive polar selects the whole neighborhood. Proposition~\ref{prop:polar} gives a uniform lower bound for the probability of this event. Only exponentially many neighborhoods are needed; a union bound then requires an additional factor of order $n$ in the number of sampled directions. This is a direct analogue of Schramm's approach to illuminating bodies of constant width~\cite{Schramm}. At a boundary point, the outward unit normals have angular diameter at most $\pi/3$, and the negative of a direction in their open positive polar illuminates that point. Schramm used spherical cap-containment and an antipodal pairing argument to estimate the measure of admissible directions. Discretization and random sampling then gave the exponential base $C=\sqrt{3/2}$, both for illumination and, as a consequence, for Borsuk's problem. \Needspace{5\baselineskip} \noindent\emph{How the polar estimate improves the exponent.} We first prove the polar estimate for the set $A$ in Proposition~\ref{prop:polar} under the exact assumption $\diam_{\Sph}A\le\pi/3$; the small increase of the diameter is handled in Section~\ref{sec:finish}. Central projection (also known as gnomonic projection) writes $A$ as \[ A=\left\{\frac{p+x}{\sqrt{1+|x|^2}}:x\in L\right\}, \] where $p$ is a suitable pole and $L$ lies in the unit ball of $p^\perp$. If a unit vector $u\in p^\perp$ satisfies $h_L(u):=\sup_{x\in L}\ip xu\le a$, then \[ v_\theta=\cos\theta\,p-\sin\theta\,u\in A_+, \qquad 0\le\theta<\arctan(1/a). \] This sector supplies the exponential measure factor $(1+a^2)^{-n/2}$. The value $a=1/\sqrt2$ corresponds to the classical base $\sqrt{3/2}$, so we seek many directions satisfying the stronger inequality with $a=7/10$. An antipodal pairing first reduces the task to the outer shell $E=\{x\in L:|x|^2\ge3/4\}$, losing only a factor two in measure. The lift \[ x\longmapsto\left(x,\frac76-|x|^2\right) \] makes the shell acute: every pair of lifted vectors has nonnegative scalar product. A self-dual cone $D$ containing the lift has a reflection $R$ which is even, preserves length, and is nonexpansive, meaning $|R(z)-R(w)|\le|z-w|$. Evenness and Gaussian Poincar\'e force $\mu=\E R(G)$ to have length at least $\sqrt{n/2}$. The cone also ensures $R(G)-G\in D$, giving supporting inequalities for every lifted point at once. We translate $G$ opposite to the horizontal part of $\mu$, by a fraction $k=1/16$ of that mean. The supporting inequalities and the mean bound then show that the direction of the translated horizontal Gaussian is good for the entire shell with probability tending to one. Returning to ordinary Gaussian measure costs a factor at worst $\exp(-n/512-o(n))$. Together with the sector factor, this gives the reciprocal exponential base $e^{-k^2/2}/\sqrt{1+a^2}$. Once the support bound is established, these two factors improve the classical exponent provided that \begin{equation}\label{eq:tradeoff} (1+a^2)e^{k^2}<\frac32. \end{equation} For $a=7/10$ and $k=1/16$, the reciprocal base is $e^{-1/512}/\sqrt{149/100}$. Its reciprocal is less than the value $C$ in Theorem~\ref{thm:main}, hence less than $\sqrt{3/2}$. \noindent\emph{Sources of the ideas.} The study of small spherical bodies of constant width by Hall, Prymak and Sujsuntinukul~\cite{HPS}, including Gaussian methods near the central width $\pi/2$, provided a crucial context for this work. The central projection and antipodal pairing follow Schramm's method~\cite{Schramm}. The outer-shell pairing used here is a generalization of that idea. The key geometric fact about self-dual cones comes from Moreau's decomposition~\cite{Moreau}: every vector $z$ is the difference of two orthogonal vectors $P_Dz$ and $P_D(-z)$ in the cone $D$, where $P_D$ is the nearest-point projection onto $D$. Replacing their difference by their sum defines the length-preserving folding $R$ and ensures $R(z)-z\in D$. The additional ingredient in the present argument is the radial lift which makes this cone geometry available for the outer shell. \section{From good directions to smaller-diameter pieces} \label{sec:partition} We first prove Theorem~\ref{thm:main} from Proposition~\ref{prop:polar}. This is Schramm's discretization and random-direction strategy, expressed directly in terms of partition classes. It uses packing, random sampling, and the observation that two points cannot both lie ahead of each other. Fix $0<\beta0\text{ for every }y\in K \text{ with }|x-y|\ge1-\varepsilon\}. \] If $v\in(A_i)_+$, then $F_i\subset B_v$. If $A_i$ is empty, this inclusion holds for every $v$. Consequently the $B_{v_j}$ cover $K$. If $x,y\in B_v$ had $|x-y|\ge1-\varepsilon$, the definition would give both $\ip v{x-y}>0$ and $\ip v{y-x}>0$, an impossibility. Hence \[ \diam B_v\le1-\varepsilon<1. \] Assigning each point to the first covering set containing it gives a partition. By \eqref{eq:neighborhoodcount}, $b(n)\le O_\varepsilon(n)\beta^{-n}$. Letting $\beta\uparrow C^{-1}$ proves the theorem. \section{Central projection and an outer shell} \label{sec:geometry} We now prove Proposition~\ref{prop:polar}, starting with the exact diameter assumption $\diam_{\Sph}A\le\pi/3$, or equivalently $\ip z{z'}\ge1/2$ for all $z,z'\in A$. \subsection{Flattening the spherical set} Let $q$ be the nearest point of $\conv A$ to the origin and set $p=q/|q|$. For any convex combination $q=\sum_i\alpha_i z_i$, \[ |q|^2=\sum_{i,j}\alpha_i\alpha_j\ip{z_i}{z_j}\ge\frac12. \] The nearest-point inequality gives $\ip zq\ge|q|^2$ for $z\in A$, so $\ip zp\ge1/\sqrt2$. Central projection onto the tangent hyperplane $p+p^\perp$ therefore writes \begin{equation}\label{eq:gnomonic} A=\left\{\frac{p+x}{\sqrt{1+|x|^2}}:x\in L\right\}, \qquad L\subset\B^{n-1}\subset p^\perp, \end{equation} where $\B^{n-1}$ is the unit ball in $p^\perp$. The spherical diameter bound $\diam_{\Sph}A\le\pi/3$ says that any two points of $A$ have scalar product at least $1/2$. Substituting their representations from \eqref{eq:gnomonic} therefore gives \begin{equation}\label{eq:constraint} \frac{1+\ip xy}{\sqrt{(1+|x|^2)(1+|y|^2)}}\ge\frac12 \qquad(x,y\in L). \end{equation} For a compact set $F$ put $h_F(u)=\sup_{x\in F}\ip xu$; for the empty set use $h_\varnothing=-\infty$. Fix throughout \begin{equation}\label{eq:constants} a=\frac7{10},\qquad r^2=\frac34,\qquad k=\frac1{16}. \end{equation} Let $E=\{x\in L:|x|\ge r\}$ and $\good F=\{u\in\Sph^{n-2}:h_F(u)\le a\}$. \subsection{Support bounds give a spherical sector} We first relate the measure of $\good L$ to that of the positive polar $A_+$. For every $u\in\good L$ and $0\le\theta<\theta_a:=\arctan(1/a)$, the direction $v_\theta=\cos\theta\,p-\sin\theta\,u$ belongs to $A_+$, because \[ \ip{(p+x)/\sqrt{1+|x|^2}}{v_\theta} =\frac{\cos\theta-\sin\theta\ip xu}{\sqrt{1+|x|^2}}>0 \qquad(x\in L). \] Figure~\ref{fig:sector} illustrates the projection and this sector. Polar integration gives \begin{equation}\label{eq:sector} \nu_{n-1}(A_+) \ge\nu_{n-2}(\good L)\, \frac{\int_0^{\theta_a}\sin^{n-2}\theta\,d\theta} {\int_0^\pi\sin^{n-2}\theta\,d\theta}. \end{equation} The ratio is at least $(\sin\theta_a)^n\exp(-o(n))=(1+a^2)^{-n/2}\exp(-o(n))$. For an elementary verification, bound the denominator by $\pi$ and integrate the numerator only over $[\theta_a-1/n,\theta_a]$; this loses only a polynomial factor and an absolute constant. Thus the remaining task is to find enough support directions with threshold $a<1/\sqrt2$ to compensate for their probability cost, as in \eqref{eq:tradeoff}. \begin{figure}[htbp] \centering \begin{tikzpicture}[scale=1.2,>=Stealth,font=\small] \coordinate (O) at (0,0); \coordinate (p) at (0,2); \coordinate (z) at ({2*sin(25)},{2*cos(25)}); \coordinate (q) at ({2*tan(25)},2); \coordinate (T) at (1.4,2); \fill[blue!8] (O) -- (90:2) arc[start angle=90,end angle=145.008,radius=2] -- cycle; \draw[gray!55] (2,0) arc[start angle=0,end angle=180,radius=2]; \draw[gray!70] (-1.6,2) -- (2.65,2) node[right,text=black!75] {$p+p^\perp$}; \draw[->,thick] (O) -- (p); \draw[->,thick] (O) -- (2,0) node[right] {$u$}; \draw[red!65!black,dashed] (O) -- (q); \draw[gray,densely dotted] (O) -- (T); \draw[blue!65!black,dashed] (O) -- (145.008:2); \draw[->,blue!65!black,thick] (O) -- (125:2) node[above left] {$v_\theta$}; \draw[blue!65!black] (90:0.48) arc[start angle=90,end angle=125,radius=0.48]; \node[blue!65!black] at (107.5:0.73) {$\theta$}; \draw[gray] (0.1147,0.1638) -- (-0.0492,0.2785) -- (-0.1638,0.1147); \fill (O) circle (1.1pt) node[below left] {$0$}; \fill (p) circle (1.1pt) node[above left] {$p$}; \fill[red!65!black] (z) circle (1.3pt) node[below right,fill=white,inner sep=1pt] {$z\in A$}; \fill[red!65!black] (q) circle (1.3pt) node[above left=2pt] {$p+x$}; \fill[gray] (T) circle (1.1pt) node[above right,text=black!75] {$p+au$}; \end{tikzpicture} \caption{A two-dimensional section through $p$ and $u$. Central projection follows the ray through $z$ to $p+x$. When $\ip xu\le a$, directions with $0\le\theta<\theta_a$ lie in the positive polar. The sector's limiting boundary ray is orthogonal to $p+au$.} \label{fig:sector} \end{figure} \subsection{Pairing opposite directions} \begin{lemma}\label{lem:pairing} The outer shell controls the whole set up to a factor two: \begin{equation}\label{eq:pairing} \nu_{n-2}(\good L)\ge\tfrac12\nu_{n-2}(\good E). \end{equation} \end{lemma} \begin{proof} We first exclude $x,y\in L$ and a unit vector $u$ with \[ |x|\le r,\qquad \ip xu\ge a,\qquad \ip yu\le-a. \] Put $z=\sqrt{|x|^2-a^2}$ and $w=\sqrt{|y|^2-a^2}$. Decomposing along $u$ bounds $\ip xy$ above by $-a^2+zw$. By \eqref{eq:constraint}, $1+\ip xy>0$, so this upper bound can be squared after adding one. For the next step, apply Cauchy--Schwarz to the two vectors \[ \left(\frac{1-a^2}{\sqrt{1+a^2}},z\right), \qquad (\sqrt{1+a^2},w). \] Their scalar product is $1-a^2+zw$, so \[ (1-a^2+zw)^2 \le\left(\frac{(1-a^2)^2}{1+a^2}+z^2\right)(1+a^2+w^2). \] Consequently, \begin{align*} \frac{(1+\ip xy)^2}{(1+|x|^2)(1+|y|^2)} &\le\frac{(1-a^2+zw)^2}{(1+a^2+z^2)(1+a^2+w^2)}\\ &\le\frac{(1-a^2)^2+(1+a^2)z^2} {(1+a^2)(1+a^2+z^2)}\\ &\le\frac{1-3a^2+(1+a^2)r^2}{(1+a^2)(1+r^2)}<\frac14. \end{align*} The expression on the second line increases with $z^2$, and $z^2\le r^2-a^2$. The strict last inequality is exactly \begin{equation}\label{eq:pairgap} 13a^2-3-3(1+a^2)r^2=\frac7{400}>0. \end{equation} This contradicts \eqref{eq:constraint}. Now suppose $u\in\good E\setminus\good L$. A point witnessing $h_L(u)>a$ must lie inside the shell. The excluded configuration shows that $h_L(-u)\le a$. Thus $\good E\subset\good L\cup(-\good L)$, and antipodal symmetry proves \eqref{eq:pairing}. \end{proof} The point of this pairing is that we need not control the inner part of $L$ probabilistically. If an inner point ``spoils'' a direction already good for the shell, the pairing argument shows that the opposite direction is good for all of $L$. \subsection{A lift with one formula} Set $g(t)=7/6-t$ and lift each $x\in E$ to \[ \lambda_x=(x,g(|x|^2))\in\R^n. \] Geometrically, these points lie on the graph of the paraboloid $s=7/6-|x|^2$, restricted to $x\in E$. The last coordinate is its height above the horizontal space $p^\perp$. These lifted vectors have positive pairwise scalar products. Indeed, for $t=|x|^2$ and $s=|y|^2$ in $[3/4,1]$, \eqref{eq:constraint} gives \begin{equation}\label{eq:acutelift} \ip{\lambda_x}{\lambda_y} \ge g(t)g(s)+\tfrac12\sqrt{(1+t)(1+s)}-1. \end{equation} The right-hand side is concave in each variable separately, so its minimum occurs at a corner. The three distinct corner values are \[ \frac7{144},\qquad \frac5{72}+\sqrt{\frac78}-1,\qquad\frac1{36}. \] All are positive: for the middle value, $7/8-(67/72)^2=47/5184>0$. The extra coordinate compensates for possible negative products $\ip xy$. Its height decreases towards the outside, where \eqref{eq:constraint} already gives better products. This is why a radius-dependent lift is useful. \section{Reflection and its Gaussian mean} \label{sec:reflection} For a closed convex cone $D\subset\R^n$ let $D^*=\{y:\ip xy\ge0\ (x\in D)\}$. A cone is \emph{acute} when $D\subset D^*$ and \emph{self-dual} when $D=D^*$. We use the classical completion result of Barker and Foran~\cite{BF}, which is an easy consequence of Zorn's lemma. \begin{lemma}[Barker--Foran completion]\label{lem:completion} Every closed acute convex cone is contained in a closed self-dual cone in the same Euclidean space. \end{lemma} By \eqref{eq:acutelift}, the closed convex cone generated by the $\lambda_x$ is acute. Complete it to a self-dual cone $D\subset\R^n$. If $E$ is empty, $\good E$ is the whole sphere and this and the next section may be skipped. Let $P_D$ be nearest-point projection onto $D$ and put $R=2P_D-I$. Moreau's decomposition reads \[ z=P_Dz-P_D(-z),\qquad P_Dz\perp P_D(-z). \] For completeness, if $p_0=P_Dz$, the projection inequality gives $z-p_0\in-D$ and $\ip{z-p_0}{p_0}=0$. Writing $q_0=p_0-z\in D$, the same inequality shows $q_0=P_D(-z)$. Thus $R(z)=p_0+q_0\in D$: the reflection folds the space into the cone. It also follows that \begin{equation}\label{eq:reflection} R(-z)=R(z),\qquad |R(z)|=|z|,\qquad R(z)-z\in D. \end{equation} The folding is illustrated in Figure~\ref{fig:folding}. \begin{figure}[htbp] \centering \begin{tikzpicture}[scale=1.05,>=Stealth,font=\small] \coordinate (O) at (0,0); \coordinate (p0) at (2,0); \coordinate (q0) at (0,1.5); \coordinate (z0) at (2,-1.5); \coordinate (Rz) at (2,1.5); \fill[blue!8] (O) rectangle (2.65,2.05); \draw[->,gray!60] (O) -- (2.8,0); \draw[->,gray!60] (O) -- (0,2.18); \node[blue!65!black] at (0.85,1.75) {$D$}; \draw[gray,dashed] (-36.87:2.5) arc[start angle=-36.87,end angle=36.87,radius=2.5]; \draw[gray,densely dotted] (q0) -- (Rz) -- (z0); \draw[->,thick] (O) -- (p0); \draw[->,thick] (O) -- (q0); \draw[->,red!65!black,thick] (O) -- (z0); \draw[->,blue!65!black,thick] (O) -- (Rz); \draw (0.22,0) -- (0.22,0.22) -- (0,0.22); \fill (O) circle (1.1pt) node[below left] {$0$}; \fill (p0) circle (1.1pt) node[below left] {$p_0$}; \fill (q0) circle (1.1pt) node[left] {$q_0$}; \node[red!65!black,right] at (2.12,-1.5) {$z=p_0-q_0$}; \node[blue!65!black,right] at (2.12,1.5) {$R(z)=p_0+q_0$}; \end{tikzpicture} \caption{Moreau's orthogonal pair $p_0=P_Dz$, $q_0=P_D(-z)$. Replacing their difference by their sum folds $z$ into $D$ without changing its length. The dashed arc records the equal lengths; the shaded quadrant, in the plane spanned by $p_0,q_0$, lies in $D$.} \label{fig:folding} \end{figure} Projection also satisfies \[ |P_Dz-P_Dw|^2\le\ip{P_Dz-P_Dw}{z-w}. \] Expanding a square gives $|R(z)-R(w)|\le|z-w|$. For $D=\R_+^n$, we have $R(z)=(|z_1|,\ldots,|z_n|)$: this folds each coordinate to the positive half-line and preserves length. The general reflection has the same three properties in \eqref{eq:reflection}. Let $G$ be a standard Gaussian vector in $\R^n$ and $\mu=\E R(G)$. The coordinates of $G$ are independent real Gaussians of mean zero and variance one. We use the ordinary Gaussian Poincar\'e inequality and its even version: \begin{equation}\label{eq:poincare} \Var f(G)\le\E|\nabla f(G)|^2, \qquad \Var f(G)\le\tfrac12\E|\nabla f(G)|^2\quad(f(-z)=f(z)). \end{equation} The first inequality is the standard Gaussian Poincar\'e inequality; see~\cite{Janson}. For the second, apply~\cite[Lemma~2.3]{Lehec} to $f-\E f(G)$: an even function is orthogonal to every linear function under Gaussian measure. Both inequalities extend to Lipschitz functions by smooth approximation, preserving evenness when needed. Each coordinate function $R_i$ is even and $1$-Lipschitz. The second inequality therefore gives $\Var R_i(G)\le1/2$. Summing these scalar inequalities yields \[ \E|R(G)-\mu|^2=\sum_{i=1}^n\Var R_i(G)\le\frac n2. \] Since $\E|R(G)|^2=n$, we obtain \begin{equation}\label{eq:mean} \frac n2\le|\mu|^2\le n. \end{equation} This is the source of the gain: an even contraction preserving Gaussian length cannot have a small mean. \section{A translation makes good directions typical} \label{sec:translation} Let $e$ be the added coordinate direction and let $\Pi$ be orthogonal projection onto $e^\perp$. We use the horizontal part and the height of the mean: \[ m=\Pi\mu,\qquad d=\frac{|\ip\mu e|}{\sqrt n}. \] The projection $\Pi$ keeps the original horizontal coordinates and discards the coordinate added in the lift. Thus $m$ is the average horizontal position of the folded Gaussian $R(G)$, while $d$ is the absolute value of its average height, on the scale $\sqrt n$. Now define the shifted horizontal Gaussian and its direction: \[ X=\frac{\Pi G-km}{\sqrt n},\qquad U=\frac X{|X|}. \] The vector $X$ is shifted opposite to $m$ and rescaled to have typical length of order one. Its direction $U$ is the candidate good support direction; $X\ne0$ almost surely. The cone projection supplies the auxiliary vector \[ W=\Pi P_DG-km. \] It is the horizontal part of $P_DG$, shifted by $-km$. Since $P_DG=(G+R(G))/2$, the candidate splits as \begin{equation}\label{eq:split} \sqrt n\,X=W+\tfrac12\Pi(G-R(G)). \end{equation} The full vector $G-R(G)$ belongs to $-D$, so its scalar product with every lifted point is nonpositive. Thus the second term has an upper support bound supplied by the added coordinate. It remains to control the length of $W$. \subsection{Three scalar estimates} Because $R$ is even, $\E\ip{\Pi G}{\Pi R(G)}=0$. Also $\E P_DG=\mu/2$. Consequently \begin{align*} \E|W|^2 &=\frac14\bigl(\E|\Pi G|^2+\E|\Pi R(G)|^2\bigr) -(k-k^2)|m|^2\\ &\le\frac n2-\frac{|\ip\mu e|^2}{4} -(k-k^2)\left(\frac n2-|\ip\mu e|^2\right)\\ &=n(A_0-B_0d^2), \end{align*} where \eqref{eq:mean} was used and \begin{equation}\label{eq:AB} A_0=\frac{1-k+k^2}{2}=\frac{241}{512},\qquad B_0=(1/2-k)^2=\frac{49}{256}. \end{equation} Notice the term linear in $k$ in $A_0$: the geometric improvement starts at first order in the translation, while its measure cost will be quadratic. The ordinary inequality in \eqref{eq:poincare} and Chebyshev's inequality now imply, with probability $1-o(1)$, \begin{equation}\label{eq:typical} |X|\ge1-o(1),\qquad \frac{|W|}{\sqrt n}\le\sqrt{A_0-B_0d^2}+o(1),\qquad \frac{\ip{R(G)-G}{e}}{\sqrt n}\le d+o(1). \end{equation} Here are the details needed for uniformity. The functions $|\Pi G-km|$, $|W|$, and $\ip{R(G)-G}{e}$ have Lipschitz constants at most $1,1,2$, respectively, so their variances are bounded by $1,1,4$. Each differs from its mean by at most $n^{1/4}$ with probability $1-O(n^{-1/2})$. Moreover, \[ \E|\Pi G-km|^2=n-1+k^2|m|^2\ge n-1, \] so $\E|\Pi G-km|\ge\sqrt{n-2}$; and $\E|W|\le\sqrt{\E|W|^2}$. These observations prove \eqref{eq:typical}, with errors independent of $D$. \subsection{Controlling the whole shell at once} For $x\in E$, put $t=|x|^2$. Since $\lambda_x\in D=D^*$ and $R(G)-G\in D$, \[ \ip x{\Pi(G-R(G))}\le g(t)\ip{R(G)-G}{e}. \] Combining this with \eqref{eq:split}, on the single event \eqref{eq:typical} we obtain, simultaneously for all $x\in E$, \begin{align} \ip xU &\le\frac{\sqrt t\,|W|+\tfrac12g(t)\ip{R(G)-G}{e}} {\sqrt n\,|X|}\notag\\ &\le\sqrt{t(A_0-B_0d^2)}+\tfrac12g(t)d+o(1)\notag\\ &\le\sqrt{A_0\left(t+\frac{64}{49}(7/6-t)^2\right)}+o(1). \label{eq:support} \end{align} The last inequality is Cauchy--Schwarz applied to $(\sqrt{A_0-B_0d^2},\sqrt{B_0}d)$ and $(\sqrt t,g(t)/(2\sqrt{B_0}))$. This step handles every possible orientation of the mean. Only a quadratic on $[3/4,1]$ remains. It is convex, and its endpoint values are \[ \left[t+\frac{64}{49}(7/6-t)^2\right]_{t=3/4} =\frac{1723}{1764}<1, \qquad \left[t+\frac{64}{49}(7/6-t)^2\right]_{t=1} =\frac{457}{441}<\frac{26}{25}. \] Therefore the square of the main term in \eqref{eq:support} is less than \begin{equation}\label{eq:supportgap} \frac{241}{512}\frac{26}{25} =\frac{3133}{6400}<\frac{3136}{6400}=a^2. \end{equation} It follows that $h_E(U)\le a$ with probability $1-o(1)$. There is no union bound over $E$: the same three scalar estimates control every point of this possibly infinite set. \subsection{Paying for the translation} Let $v=km$ and define \[ \mathcal E=\{z\in\R^n:\Pi z\ne0, \ h_E(\Pi z/|\Pi z|)\le a\}. \] We have just proved $\Prob(G-v\in\mathcal E)=1-o(1)$, and $|v|^2\le k^2n$ by \eqref{eq:mean}. Write $\gamma_n$ for standard Gaussian measure, whose density is $(2\pi)^{-n/2}e^{-|z|^2/2}$. This density gives the exact change-of-variables identity \begin{equation}\label{eq:density} \gamma_n(\mathcal E) =\E\left[\mathbf1_{\mathcal E}(G-v) \exp\left(\ip vG-\frac{|v|^2}{2}\right)\right]. \end{equation} Since $\Var\ip vG=|v|^2=O(n)$, Chebyshev gives $\Prob(\ip vG<-n^{3/4})=o(1)$. Restricting the expectation to the intersection of the two good events yields \[ \gamma_n(\mathcal E)\ge\exp(-k^2n/2-o(n)). \] The direction of $\Pi G$ is uniform on $\Sph^{n-2}$, so \begin{equation}\label{eq:shellmeasure} \nu_{n-2}(\good E)\ge\exp(-n/512-o(n)). \end{equation} This calculation is the whole measure comparison. In probabilistic terms, we first make the desired event typical by shifting a Gaussian, then read its probability under the original Gaussian from the density. \section{The final bound and stability} \label{sec:finish} Combining \eqref{eq:pairing}, \eqref{eq:shellmeasure}, and \eqref{eq:sector} proves the uniform estimate \begin{equation}\label{eq:exactpolar} \nu_{n-1}(A_+) \ge\left(\frac{e^{-1/512}}{\sqrt{149/100}}\right)^n\exp(-o(n)). \end{equation} If $E$ is empty, \eqref{eq:pairing} gives $\nu_{n-2}(\good L)\ge1/2$, which is stronger than needed. Finally, $e^x<(1-x)^{-1}$ for $0