\documentclass[11pt]{amsart} \pdfoutput=1 \usepackage[usenames,dvipsnames]{xcolor} % \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} \usepackage[letterpaper,margin=1in]{geometry} \usepackage{lmodern} \usepackage{microtype} \usepackage{etoolbox} % \usepackage{lettrine} \usepackage{mathtools} \usepackage{amsmath,amssymb,amsfonts,amsxtra,bm} \usepackage{amsthm} \usepackage{mathrsfs} \usepackage{latexsym} \usepackage{nicefrac} % \usepackage{booktabs} \usepackage{makecell,multirow} % \usepackage{colortbl} \usepackage{graphicx} \usepackage{epstopdf} \usepackage{float} \floatstyle{plaintop} \restylefloat{table} \usepackage{footnote, tablefootnote} \usepackage{enumitem} \usepackage{algorithm} \usepackage{algorithmic} \usepackage{comment} \usepackage{xspace} \usepackage{url} \usepackage[most]{tcolorbox} \usepackage[round]{natbib} \usepackage{hyperref} \numberwithin{equation}{section} \makeatletter \definecolor{darkblue}{rgb}{0.0,0.0,0.65} \definecolor{darkred}{rgb}{0.68,0.05,0.0} \definecolor{darkgreen}{rgb}{0.0,0.29,0.29} \definecolor{darkpurple}{rgb}{0.47,0.09,0.29} \renewcommand{\topfraction}{0.85} \renewcommand{\bottomfraction}{0.4} \renewcommand{\textfraction}{0.1} \renewcommand{\floatpagefraction}{0.9} \patchcmd{\section}{\scshape}{\fontfamily{ppl}\bfseries\scshape\selectfont}{}{% \PackageWarning{aiproofs}{Could not bold the section font; amsart may have changed}} \newcommand{\aiauthor}{% \author[Suvrit Sra]{Suvrit Sra\textsuperscript{\normalfont*}}% \thanks{*\,TU Munich, Dept.~of Mathematics, School of CIT; Garching, Germany. 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Secondary 15B52, 52A40} \keywords{matrix Gaussian series, random matrices, log-concave functions, small-ball probabilities} \hypersetup{pdftitle={Beyond Small Balls: A Determinant Inequality for Functions of Matrix Gaussian Series}, pdfauthor={Suvrit Sra}} \makeatletter \newcommand{\preprintslink}[2]{% \thanks{First public release: #2, as v1 in the author's preprint repository, \url{https://github.com/suvrit/preprints/tree/main/preprints/#1}; every version is kept there with its timestamp.}} \makeatother \begin{document} \title[Determinant inequality for matrix Gaussian series]{Beyond Small Balls: A Determinant Inequality for Functions of Matrix Gaussian Series} \aiauthor \begin{abstract} We prove a functional extension of Gaussian operator-norm small-ball inequalities. Let $X=\sum_{i=1}^d g_iB_i$, where $g\sim N(0,I_d)$, the matrices $B_i\in\R^{d\times d}$ are symmetric, and $\sum_iB_i^2\preceq I_d$. For every even log-concave $f\colon\R\to[0,\infty)$, we obtain $\E\det f(X)\ge(\E f(\kappa G))^d$ at dilation $\kappa=13/5$, where $G$ is standard Gaussian. The proof reduces each function to one small-ball estimate, obtained from matrix-weighted Poincar\'e inequalities, and controls the rest by averages over the spectrum. We give an equivalent entropy formulation. An explicit counterexample shows that no universal dilation below $\sqrt{\pi/2}$ can work; we conjecture that this value is optimal. % \end{abstract} \preprintslink{gsd}{19 October 2026} \maketitle \addtocontents{toc}{\protect\setcounter{tocdepth}{-1}} \vspace*{-10pt} \section{Introduction}\label{sec:introduction} When working with random vectors, we often need to know how likely their norms are to be small. Small-ball inequalities answer precisely such a question; they estimate the probability that a random vector lies in a small norm ball around the origin. While this topic is classical for Gaussian vectors~\citep{li2001gaussian}, our interest stems from matrix Gaussian series, where small-ball inequalities were instrumental in resolving the matrix Spencer conjecture~\citep{akbas2026spencer}. Concretely, say $X =\sum_{i=1}^d g_iB_i$ is a real symmetric matrix Gaussian series with variance $V_X=\sum_iB_i^2\preceq I_d$, then a small-ball inequality is simply $\Pp(\|X\|\le C)\ge e^{-cd}$ for absolute constants $C$ and~$c$. We extend these small-ball inequalities from the operator-norm ball to functions.% Let $f\colon\R\to[0,\infty)$. Define the expectation $\E\det f(X) :=\E\prod_j f(\lambda_j(X))$; it equals the small-ball probability $\Pp(\|X\|\le C)$ when $f=\ind_{[-C,C]}$, the \emph{indicator} of the interval $[-C,C]$. More generally, we assume $f$ is even and log-concave, and compare the expectation $\E\det f(X)$ with the ``diagonal matrix'' benchmark $(\E f(G))^d$, where $G$ is a standard Gaussian. When the coefficient matrices $B_i$ commute, this comparison succeeds: the eigenvalues of $X$ are linear functions of the Gaussian coefficients, and \v{S}id\'ak's inequality~\citep{sidak1967rectangular} immediately gives $\E\det f(X)\ge(\E f(G))^d$ (Theorem~\ref{thm:commuting}). \vskip4pt \noindent\textbf{Challenges.} Without commutativity, the eigenvalues of $X$ need not be Gaussian, and the comparison with the benchmark can fail (already for a $2\times2$ series built from two Pauli matrices). We therefore ask whether $X$ satisfies a weaker \emph{determinant inequality} against a \emph{dilated Gaussian} $\kappa G$, with $\kappa$ independent of $d$ and $f$. Because $f$ peaks at zero, this inequality says that the spectrum of $X$ is no more spread out than $d$ independent $N(0,\kappa^2)$ variables, so we seek the smallest such $\kappa$. To address this determinant inequality, we must first overcome a more basic obstruction: determinants couple all eigenvalues through products, while moment and trace estimates control only averages. Jensen's inequality can help us by turning bounds on the expected trace of $-\log f$ into bounds on the determinant, but this expectation is infinite whenever $f$ has zeros. We therefore use one small-ball estimate to bound the probability that the spectrum lies in an interval where $-\log f$ is finite, an \emph{averaged comparison} with the scalar Gaussian to control $-\log f$ there, and Gaussian correlation to combine the two. This strategy, with explicit constants, gives our main result. \begin{theorem*}[informal; Theorem~\ref{thm:main}] Let $X$ be as above. For every even log-concave $f\colon\R\to[0,\infty)$, \[ \E\det f(X)\ge\bigl(\E f(\kappa G)\bigr)^d,\qquad\text{with}\ \kappa=13/5. \] \end{theorem*} The choice of $\kappa$ in the theorem is (up to minor fractions) the currently best known value. Matrices with anticommuting coefficients show that $\kappa=13/5$ cannot be lowered below $\cstar=\sqrt{\pi/2}$. This lower bound is approached only as the number of anticommuting coefficients grows; with two of them, the two-Pauli series above needs only $2/\sqrt\pi$ for all even log-concave functions and $\sqrt{2/\pi}$ for interval indicators. We conjecture that dilation $\cstar$ is the best possible in general (Conjecture~\ref{conj:optimal-gsd}). \vskip4pt \noindent\textbf{Main contributions of the paper}. \begin{enumerate}[label=(\arabic*),leftmargin=2.4em,labelsep=0.5em,itemsep=1pt,topsep=2pt] \item The determinant inequality at dilation $\kappa=13/5$. Its indicator case gives a hereditary small-ball bound (Corollary~\ref{cor:hereditary}) for $m\le d$ contractions at radius of order $\sqrt{m\log(4d/m)}$, improving the radius $\sqrt m\,(1+\log(2d/m))^2$ of \citet[Theorem~1.3]{akbas2026spencer}. % \item The comparison inequality $\E\tr\psi(|X|)\le d\E\psi(\cstar|G|)$ for convex nondecreasing $\psi$ with $\psi(0)=0$ (Theorem~\ref{thm:sharp}). We also prove Conjecture~\ref{conj:optimal-gsd} for some special cases in Theorems~\ref{thm:endpoint} and \ref{thm:ellipsoid-endpoint}. \item An entropy characterization of the determinant inequality, via the Gibbs variational principle (Theorem~\ref{prop:entropy-gsd}). It reduces to the averaged comparison at zero entropy, so the entropy quantifies the gap between averages and determinants. \end{enumerate} \subsection{Related work}\label{sec:related} For $d=1$, the determinant inequality at dilation $\kappa$ says that $X$ is more peaked than $\kappa G$ in the sense of \citet{birnbaum1948peakedness}. The Gaussian correlation inequality \citep{royen2014simple} enters the proof of Theorem~\ref{thm:main}. We prove the small-ball estimate with the matrix-weighted Poincar\'e criterion of \citet[Lemma~2.3]{akbas2026spencer}, a convex-domain form of the Brascamp--Lieb variance inequality \citep{brascamp1976extensions} for matrix weights \citep{cordero2019matrix}. With the shifted partial-coloring theorem of \citet{reis2023vector} \citep[in the form of][Theorem~3.1]{dadush2022framework}, the hereditary bound of Corollary~\ref{cor:hereditary} produces signs with small matrix discrepancy, and hence another derivation of matrix Spencer \citep{akbas2026spencer}. Gaussian-measure bounds for slices of the discrepancy body had earlier given matrix Spencer up to polylogarithmic rank \citep{bansal2023matrix}. For random signs in place of Gaussian coefficients, the Boolean small-ball inequality of \citet{akbas2026bsb} bounds determinant-weighted sign counts. Theorem~\ref{prop:entropy-gsd} parallels the entropic dual of forward--reverse Brascamp--Lieb (FRBL) inequalities \citep[Theorem~1]{liu2018frbl}, with a comparison constraint on the average spectrum in place of prescribed marginals. The Gaussian saturation principle of \citet[Theorem~1.3]{courtade2021frbl} reduces linear FRBL inequalities to Gaussian functions. It does not extend to the eigenvalue maps here, since the two-Pauli series satisfies the determinant inequality for every $f(x)=e^{-bx^2}$ at dilation one but fails it for $f(x)=e^{-a|x|}$ with $a$ large (Theorem~\ref{thm:pauli} and \eqref{eq:pauliasymptotics}). \vspace*{-3pt} \section{The determinant inequality for functions}\label{sec:universal}\label{sec:setup} \vspace*{-3pt} Let $B_1,\ldots,B_m\in\Sym_d(\R)$, and define \[ X=\nlsum_{i=1}^m g_iB_i,\qquad V_X :=\nlsum_iB_i^2,\qquad r_X:=\dim\operatorname{span}\{B_i\}, \] where $g_i$ are independent standard Gaussians. We call $V_X$ the (matrix) \emph{variance} and $r_X$ the (coefficient) \emph{rank} of $X$. Throughout $G$ denotes a standard scalar Gaussian. For a real symmetric $A$, the matrix $f(A)$ is defined by spectral functional calculus, so that $\det f(A)=\prod_{j=1}^d f(\lambda_j(A))$. We say that $X$ satisfies the \emph{determinant inequality at dilation $\kappa$} if $\E\det f(X)\ge(\E f(\kappa G))^d$ for all even, log-concave $f\colon\R\to[0,\infty)$. Observe that smaller dilations mean stronger inequalities, since $f$ is maximal at $0$ and nonincreasing on $[0,\infty)$. We are now ready to state our main result. \begin{dashedbox} \begin{theorem}\label{thm:main} Let $X$ be a real symmetric matrix Gaussian series of order $d$ with $V_X\preceq I_d$ and $r_X\le d$. For every even log-concave $f\colon\R\to[0,\infty)$, we have $\E\det f(X)\ge\bigl(\E f(\kappa G)\bigr)^d$ with $\kappa=\tfrac{13}{5}$. \end{theorem} \end{dashedbox} \noindent For $f\not\equiv0$, normalize $f(0)=1$ without loss and define $\psi(t):=-\log f(t)$ for $t\ge0$. Thus $\psi\colon[0,\infty)\to[0,\infty]$ is convex and nondecreasing, with $\psi(0)=0$; we call such a function a \emph{penalty}. Applying the determinant inequality to $x\mapsto f(x/\kappa)$ gives its equivalent \emph{penalty form} \begin{equation}\label{eq:penalty-form} \mathcal Z_\psi(X;\kappa):=\E e^{-\tr\psi(|X|/\kappa)}\ge p_\psi^d, \qquad p_\psi:=\E e^{-\psi(|G|)}. \end{equation} With~\eqref{eq:penalty-form}, our proof of Theorem~\ref{thm:main} then rests on the following penalty based comparison: \begin{dashedbox}[supporting] \begin{theorem}[Averaged comparison]\label{thm:sharp} If $V_X\preceq I_d$, with $r_X$ arbitrary, then \begin{equation}\label{eq:sharp} \E\tr\psi(|X|) \le d\E\psi(\cstar|G|),\qquad \cstar:=\sqrt{\pi/2}, \end{equation} for every convex nondecreasing $\psi\colon[0,\infty)\to[0,\infty]$ with $\psi(0)=0$. The constant $\cstar$ is optimal. % \end{theorem} \end{dashedbox} To understand this special value of the constant $\cstar$ consider $\psi(t)=t$, so that Theorem~\ref{thm:sharp} claims $d^{-1}\E\tr|X|\le \cstar\E|G|=1$. It can be shown that there is an \emph{anticommuting family} (i.e., a matrix Gaussian series with symmetric anticommuting coefficients) which fulfills Theorem~\ref{thm:sharp} in the limit (Section~\ref{sec:sharp-proof}). The same family bounds the dilation in Theorem~\ref{thm:main} from below, because the determinant inequality also controls first moments. Indeed, for $f_\epsilon(x)=e^{-\epsilon|x|}$, both sides of the determinant inequality equal $1$ at $\epsilon=0$, so comparing their derivatives at $\epsilon=0$ gives $\kappa\ge\cstar\,d^{-1}\E\tr|X|$, which tends to $\cstar$ along the family. Thus, if $\kappa_{\mathrm{univ}}$ denotes the infimum of dilations valid for every matrix series $X$ in Theorem~\ref{thm:main} and every even log-concave function, then \begin{equation}\label{eq:interval} \sqrt{\pi/2}\le\kappa_{\mathrm{univ}}\le13/5. \end{equation} \vskip4pt We close the section with the consequence that motivated us: a proof of the hereditary bound needed by partial coloring for matrix Spencer discrepancy~\citep{akbas2026spencer}. \begin{corr}[Hereditary small-ball bound]\label{cor:hereditary} Let $A_1,\ldots,A_n\in\Sym_d(\R)$ be contractions, let $\varnothing\ne S\subseteq\{1,\ldots,n\}$ have size $m$, put $D=\max\{d,m\}$, and let $0a)\le e^{-a^2/2}=\tfrac{cm}{2D}\le\tfrac12. \] Since $\log(1-u)\ge-2u$ for $0\le u\le1/2$, $(1-u)^D\ge e^{-2uD}\ge e^{-cm}$; also $\|\sum_{i\in S}g_iA_i\|=\sqrt m\,\|Y\|$. \end{proof} For $1\le m\le d$ and fixed $c$, Corollary~\ref{cor:hereditary} improves the dependence of the radius on $d/m$ from $(1+\log(2d/m))^2$ to $\sqrt{\log(2d/(cm))}$, compared with \citet[Theorem~1.3]{akbas2026spencer}, whose multiplicative constant $C(c)$ is unspecified; the corollary also covers $m>d$. \section{Proof of the determinant inequality}\label{sec:proofs} The proof of Theorem~\ref{thm:main} rests on the averaged comparison, so we prove Theorem~\ref{thm:sharp} first. The proofs use closed (lower semicontinuous) convex penalties, which by Lemma~\ref{lem:weight-closure} loses nothing. \subsection{The averaged comparison}\label{sec:sharp-proof} A convex function is an integral of hinges, so we first compare their expectations. Moment and tail estimates give the comparison; anticommuting coefficients then establish its optimality. Throughout, $V_X\preceq I_d$. We use the scalar Gaussian functions \begin{gather*} c_0=\sqrt{2/\pi},\qquad q(r)=\Pp(|G|\le r),\qquad T(r)=1-q(r),\\ H(r)=\E(|G|-r)_+=c_0e^{-r^2/2}-rT(r),\qquad M(r)=\frac{T(r)}{c_0e^{-r^2/2}}. \end{gather*} \vskip4pt \noindent\emph{\textbf{Even moments.}} The bound $V_X\preceq I_d$ on the matrix variance controls every even moment: \begin{equation}\label{eq:moments} d^{-1}\E\tr X^{2p}\le(2p-1)!!\qquad(p\ge1). \end{equation} For $p\ge1$, Gaussian integration by parts gives \[ \E\tr X^{2p}=\sum_i\sum_{k=0}^{2p-2} \E\tr(B_iX^kB_iX^{2p-2-k}). \] In a spectral basis of $X$, take absolute values of the scalar monomials and use the weighted arithmetic--geometric mean inequality \[ |x|^k|y|^{2p-2-k} \le\frac{k}{2p-2}|x|^{2p-2} +\frac{2p-2-k}{2p-2}|y|^{2p-2} \] when $p>1$. Summing the squared matrix entries of $B_i$ and using $V_X\preceq I_d$ bounds the term of each $k$ by $\E\tr X^{2p-2}$, so $\E\tr X^{2p}\le(2p-1)\E\tr X^{2p-2}$. The case $p=1$ is $\tr V_X\le d$, and induction proves \eqref{eq:moments}. \vskip4pt \noindent\emph{\textbf{Comparison for hinge functions.}} Let $U$ be the absolute value of a uniformly chosen eigenvalue of $X$, so that $\E U^{2p}=d^{-1}\E\tr X^{2p}$; then $\E U^2\le1$, $\E U^4\le3$, and $\E U^6\le15$. The even-moment bounds \eqref{eq:moments} give, for $t>0$, \begin{equation}\label{eq:stop} \E(U-t)_+\le\min\left\{ \frac1{4t},\frac{81}{256t^3},\frac{15625}{15552t^5}, \frac{2T(t)}{c_0}\right\}. \end{equation} For $0\le t\le1/2$, the stronger bound $1-t$ is available. These bounds imply \begin{equation}\label{eq:sharpstop} \E(U-t)_+\le\cstar H(t/\cstar). \end{equation} \begin{proof}[Proof of \eqref{eq:stop}, \eqref{eq:sharpstop}, and Theorem~\ref{thm:sharp}] The bound $\E(U-t)_+\le1-t$ for $t\le1/2$ follows from the pointwise inequality $(u-t)_+\le u^2/2+1/2-t$ for $u\ge0$. For any $k>1$, \[ (u-t)_+\le\frac{(k-1)^{k-1}}{k^kt^{k-1}}u^k. \] Take $k=2,4,6$ and use \eqref{eq:moments}. The even moments \eqref{eq:moments} also give $\E e^{sU}\le2\E\cosh(sU)\le2e^{s^2/2}$ for $s\ge0$, and Markov's inequality with $s=t$ gives $\Pp(U>t)\le2e^{-t^2/2}$, whose integral over $(t,\infty)$ is the fourth term of \eqref{eq:stop}. On $0\le t\le1/2$, $\cstar H(t/\cstar)\ge\cstar H(0)-t=1-t$. On $[1/2,16/5]$, the interval-arithmetic computation of Appendix~\ref{sec:arithmetic} partitions the interval into 1,350 pieces of width $1/500$. The upper bounds in \eqref{eq:stop} decrease, so the computation evaluates their minimum at the left endpoint of each piece and $\cstar H(t/\cstar)$ at the right endpoint. The smallest certified gap exceeds $0.00125$. For $t\ge16/5$, the Mills inequalities \[ T(t)\le c_0e^{-t^2/2}/t, \qquad H(y)\ge c_0e^{-y^2/2}/(y^2+3) \] reduce the claim to \[ \frac{\cstar c_0t}{2(t^2/\cstar^2+3)} e^{(1-\cstar^{-2})t^2/2}\ge1. \] This inequality holds strictly at $t=16/5$, and its logarithmic derivative is positive thereafter: the rational factor contributes at least $-1/t$, while $(1-\cstar^{-2})t>1/t$. For the lower bound on $H$, the rational upper bound $M(y)\le\bar M(y):=(y^2+2)/(y^3+3y)$ follows by differentiation. Indeed, $\bar M'-y\bar M+1=-6/(y^2(y^2+3)^2)<0$, and integration toward infinity compares $\bar M$ with $M$, which solves $M'=yM-1$. The identity $H(y)=c_0e^{-y^2/2}(1-yM(y))$ finishes the argument; at $y=0$ the bound is immediate. Hence \eqref{eq:sharpstop} holds at every threshold. A finite $\psi$ is a mixture of hinges, $\psi(u)=\int(u-t)_+\,d\nu(t)$ with $\nu\ge0$, and $\E(\cstar|G|-t)_+=\cstar H(t/\cstar)$. Integrating \eqref{eq:sharpstop} against $\nu$ therefore proves Theorem~\ref{thm:sharp}. I% \end{proof} \vskip4pt \noindent\emph{\textbf{Sharpness and the universal lower bound.}} Take real symmetric anticommuting involutions $C_1,\ldots,C_j$ and $X=j^{-1/2}\sum_i g_iC_i$. Then $|X|=(\chi_j/\sqrt j)I_d$ and $V_X=I_d$, where $\chi_j=|g|$ is the Euclidean norm of $(g_1,\ldots,g_j)$. Such matrices exist for every $j$: in dimension $d=2^j$, put \[ C_i=Z^{\otimes(i-1)}\otimes H_2\otimes I_2^{\otimes(j-i)},\quad Z=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\quad H_2=\begin{pmatrix}0&1\\1&0\end{pmatrix}. \] The Pauli relation $ZH_2=-H_2Z$ gives $C_iC_k=-C_kC_i$ for $i\ne k$, and $C_i^2=I_d$. To realize $X=d^{-1/2}\sum_{i=1}^d g_iA_i$ with contractions $A_i$, choose $j$ along powers of two and repeat each $C_i$ exactly $2^j/j$ times. Gaussian aggregation gives the same matrix series in distribution. Since $\E\chi_j/\sqrt j\to1$, these matrices complete the first-moment argument of Section~\ref{sec:universal} and prove the lower bound in \eqref{eq:interval}. \subsection{Localization and convex penalties}\label{sec:localization} Theorem~\ref{thm:sharp} bounds an expected spectral sum, whereas \eqref{eq:penalty-form} concerns the expectation of its exponential. Jensen's inequality connects the two, but the resulting bound is empty for penalties that take the value $+\infty$ and lossy otherwise. On the norm event $\{\|X\|\le R\}$, however, only the values of $\psi$ on a bounded interval matter, and beyond that interval $\psi$ may be replaced by a function of linear growth. The price of restricting to the norm event is its probability, measured per dimension by \[ N_X(R)=-d^{-1}\log\Pp(\|X\|\le R). \] Fix $\kappa>0$. Let $\psi$ be finite and nonzero, and choose $r>0$ with $\psi(r)=1$; this radius sets the scale of the argument. For $\alpha\in(0,1)$, let $\widehat\psi$ agree with $\psi$ up to $\alpha r$ and continue along its right supporting tangent thereafter. On the norm event $E=\{\|X\|\le\kappa\alpha r\}$ the trace penalties $\tr\psi(|X|/\kappa)$ and $\tr\widehat\psi(|X|/\kappa)$ agree, so $\mathcal Z_\psi(X;\kappa)\ge\E[\ind_E e^{-\tr\widehat\psi(|X|/\kappa)}]$. The function $A\mapsto\tr\widehat\psi(|A|/\kappa)$ is convex and even, so the superlevel sets of its negative exponential are symmetric and convex, as is $E$. The Gaussian correlation inequality, integrated over these sets, gives \[ \mathcal Z_\psi(X;\kappa) \ge \Pp(E)\, \E e^{-\tr\widehat\psi(|X|/\kappa)}. \] Jensen's inequality and Theorem~\ref{thm:sharp}, applied to the convex function $\widehat\psi$ of linear growth, give \begin{equation}\label{eq:localization-main} \mathcal Z_\psi(X;\kappa)^{1/d} \ge \exp\{-N_X(\kappa\alpha r)-\E\widehat\psi(b_f|G|)\}, \qquad b_f=\cstar/\kappa. \end{equation} It remains to compare the right side of \eqref{eq:localization-main} with $p_\psi$ for every convex penalty. A convex penalty is a mixture of hinges, $\psi(t)=\int(t-h)_+\,d\nu(h)$, as in Section~\ref{sec:sharp-proof}; its \emph{hinge measure} $\nu$ has the initial slope as an atom at zero. A determinant inequality for each hinge separately would not combine, because $p_\psi$ is not a product over hinges, so we compare the two sides hinge by hinge against one measure. Since $\psi(r)=1$, the measure $d\omega(h)=(r-h)\,d\nu(h)$ on $[0,r)$ is a probability measure. Define \[ \psi_{r,h}(t)=\frac{(t-h)_+}{r-h},\qquad p_{r,h}=\E e^{-\psi_{r,h}(|G|)},\qquad \psi_0=\int_{h\alpha r$, the bound $p_{r,h}\le q(r)$ and \eqref{eq:unpriced} give $\log p_{r,h}\le-n$. For $h\le\alpha r$, use \eqref{eq:priced} instead. Integrating both inequalities against $\omega$ and applying the H\"older bound \eqref{eq:holder} proves the result. Finite measures and finite mixtures suffice by approximation; a finite penalty has bounded slope on every compact subinterval of $(0,\infty)$. \end{proof} \subsection{A small-ball bound from matrix-weighted Poincar\'e inequalities}\label{sec:smallball} The scalar inequalities depend on the matrix only through $N_X(R)$, so what remains is a small-ball bound for the operator norm of a matrix Gaussian series. We obtain it by tilting: multiply the Gaussian measure by $Z^{-1}e^{-V(X/R)}$, where $V$ is nonnegative and infinite outside the operator-norm unit ball, so that $Z\le\Pp(\|X\|\le R)$ and a lower bound on $Z$ is an upper bound on $N_X(R)$. Scaling from zero to $X/R$ writes $-\log Z$ as an integral of expectations of spectral quantities under the tilted measure, and matrix-weighted Poincar\'e estimates bound those expectations. The choice of $V$ sets the order of the bound: a determinant weight gives bounds polynomial in $1/R$, while a logarithmic barrier that vanishes on an interior interval gives exponential decay. Large radii require exponential decay, since the benchmark $-\log\Pp(|G|\le r)$ itself decays like $e^{-r^2/2}$. We prove the determinant-weight bound here; Appendix~\ref{app:engine} proves the barrier bounds. The bounds on $N_X$ rest on one external criterion, a matrix-weighted Poincar\'e inequality for Gaussian measures tilted by a determinant or resolvent weight. If $p$ is a positive probability density on an open convex domain $\Omega\subset\R^m$, $K$ is a positive definite matrix function on $\Omega$, and $W=pK$ is smooth and integrable, define \[ \Theta_{ij}(W)=(\partial_iW)W^{-1}(\partial_jW)-\partial_{ij}W. \] Its \emph{normalized curvature} is the block matrix \[ p^{-1}[K^{-1/2}\Theta_{ij}(W)K^{-1/2}]_{ij}. \] If $[\Theta_{ij}(W)]\succeq\gamma I_m\otimes W$ for $\gamma>0$, then \begin{equation}\label{eq:WP} \E F^{\mathsf T}KF-(\E KF)^{\mathsf T}(\E K)^{-1}(\E KF) \le\gamma^{-1}\E\sum_i(\partial_iF)^{\mathsf T}K(\partial_iF). \end{equation} Expectations are under $p$, and $F$ is any $C^1$ vector field with finite weighted square integral. The right side may be infinite. The scalar case $K\equiv1$ is the Poincar\'e inequality for strongly log-concave densities, a consequence of the Brascamp--Lieb variance inequality \citep{brascamp1976extensions}; for matrix weights on $\R^m$, the criterion follows from the Brascamp--Lieb variance inequality of \citet[Theorem~4]{cordero2019matrix}. The convex-domain form used here imposes no zero boundary value on $F$; \citet[Lemma~2.3]{akbas2026spencer} prove it by Neumann estimates on compact convex subdomains and exhaustion, following the route to Brascamp--Lieb-type inequalities through Reilly's formula \citep{reilly1977applications,kolesnikov2017brascamp}. Approximation extends the criterion to the Lipschitz tests used below. Appendix~\ref{sec:regularity} verifies the limiting hypotheses. The simplest barrier is $V(Y)=-\log\det(I-Y^2)$ for $\|Y\|<1$. Its normalizer gives the first bound on $N_X$; Lemma~\ref{lem:small} and the verification of the scalar inequalities (Proposition~\ref{prop:closure}) both use it. Retaining the trace of the matrix variance in the same proof gives Proposition~\ref{prop:gaussian-trace-smallball}. Write $\ell(z)=2/(1+\sqrt{1-4z})$ for $0\le z<1/4$. \begin{prop}\label{prop:det} If $V_X\preceq I_d$ and $R>2$, set $\eta=R^{-2}$. Then \begin{equation}\label{eq:detentropy} N_X(R)\le n_0(R):=2\log\ell(\eta)-\ell(\eta)+1. \end{equation} In particular, \begin{equation*} N_X(R)\le2/R^2\quad(R\ge3). \end{equation*} \end{prop} The proof of Proposition~\ref{prop:det} controls the determinant weight through $\tr V_X$, which can be much smaller than $d\|V_X\|$, and keeping that trace gives a small-ball bound with no condition on the coefficient rank. For $R>0$ and $\beta>0$, let \[ W_{\beta,R}(A)= \begin{cases} \det(I-A^2/R^2)^\beta,&\|A\|0$ and every $R>0$ with $R\ge3\sqrt v$, \begin{equation}\label{eq:gaussian-trace-smallball} \E W_{\beta,R}(X) \ge\exp\bigl(-2\max\{1,\beta\}\tr V_X/R^2\bigr). \end{equation} There is no restriction on the coefficient rank. In particular, $\Pp(\|X\|2$. For $0\le s\le1$, put $Y_s=sX/R$ and tilt the Gaussian measure by $\det(I-Y_s^2)\ind_{\{\|Y_s\|<1\}}$; call the normalizer $Z_s$ and the resulting probability measure $\mu_s$, with density $p_s$. Let $K_s=(I-Y_s)^{-1}$ and $S_s=\E_{\mu_s}K_s$. The derivative of $\log Z_s$ is $-(2/s)\tr(S_s-I)$, as shown below, so the task is an upper bound on the mean resolvent $S_s$, and the weighted Poincar\'e inequality with weight $K_s$ supplies one. With $D_i=sB_i/R$ and $E_i=K_s^{1/2}D_iK_s^{1/2}$, $W_s=p_sK_s$ has normalized curvature \[ \delta_{ij}I+\Gamma^-_{ij}I+\Gamma^+_{ij}I-E_jE_i, \] where $\Gamma^-_{ij}=\tr[(I-Y_s)^{-1}D_i(I-Y_s)^{-1}D_j]$ and $\Gamma^+_{ij}=\tr[(I+Y_s)^{-1}D_i(I+Y_s)^{-1}D_j]$. The block inequality \begin{equation}\label{eq:block} [\tr(E_iE_j)I-E_jE_i]_{ij}\succeq0 \end{equation} proves curvature at least one. For block vectors $u_i$ and $L_b=\sum_i u_i(E_ie_b)^{\mathsf T}$, the quadratic form of the left side of \eqref{eq:block} is $\sum_b(\|L_b\|_F^2-\tr L_b^2) =\tfrac12\sum_b\|L_b-L_b^{\mathsf T}\|_F^2\ge0$. Integrability follows from the identity $\det(I-Y_s^2)(I-Y_s)^{-1}=\det(I+Y_s)\operatorname{adj}(I-Y_s)$ on $\{\|Y_s\|<1\}$. Test~\eqref{eq:WP} with $F=(I-Y_s)z$ for a fixed vector $z$. Evenness gives \[ I-S_s^{-1}\preceq\sum_iD_iS_sD_i\preceq s^2\eta\|S_s\|I. \] Writing $\lambda_s=\|S_s\|$, we obtain $\lambda_s-1\le s^2\eta\lambda_s^2$. Continuity from $\lambda_0=1$ excludes the large root, so $S_s\preceq\ell(s^2\eta)I$. The numerator defining $S_s$ is bounded by a Gaussian-integrable function, and its moving boundary has Gaussian measure zero; these two facts justify continuity, including at $s=0$. To retain $\tr V_X$, conjugate the inequality $I-S_s^{-1}\preceq\sum_iD_iS_sD_i$ by $S_s^{1/2}$ and take traces: \begin{align*} \tr(S_s-I) &\le\sum_i\tr(S_sD_iS_sD_i)\\ &=\sum_i\|S_s^{1/2}D_iS_s^{1/2}\|_F^2 \le\|S_s\|^2\sum_i\tr D_i^2 \le s^2\eta\,\tr V_X\,\ell(s^2\eta)^2. \end{align*} Furthermore, $S_s-I=\E_{\mu_s}[Y_s^2(I-Y_s^2)^{-1}]$ and \[ -\frac{d}{ds}\log Z_s =\frac2s\tr(S_s-I)\le2s\eta\,\tr V_X\,\ell(s^2\eta)^2. \] The determinant vanishes on the boundary, so differentiation creates no boundary term. All integrals may be taken on the coefficient support: removing a kernel preserves $V_X$, and for $s>0$ the boundary lies in the zero set of the nonzero polynomial $\det(I-Y_s^2)$. Since $\ell$ satisfies $(\ell(u)-1)/u=\ell(u)^2$, substituting $u=s^2\eta$ and then $u=(\ell-1)/\ell^2$ gives \begin{equation}\label{eq:trace-det-integral} -\log Z_1\le\tr V_X\int_0^\eta\ell(u)^2\,du =\tr V_X\,n_0(R). \end{equation} Since $\tr V_X\le d$ and $Z_1\le\Pp(\|X\|\le R)$, \eqref{eq:detentropy} follows. For $u\le1/9$, $\ell(u)\le4/3$, so $n_0(R)<2\eta=2/R^2$ when $R\ge3$. For general $v=\|V_X\|>0$, apply \eqref{eq:trace-det-integral} to $X/\sqrt v$ at radius $R/\sqrt v$. This rescaling yields \[ \E W_{1,R}(X)\ge\exp\Bigl(-\frac{\tr V_X}{v}\,n_0(R/\sqrt v)\Bigr) \qquad(R>2\sqrt v). \] The determinant weight $W_{1,R}$ takes values in $[0,1]$. For $\beta\ge1$, Jensen gives $\E W_{1,R}^{\beta}\ge(\E W_{1,R})^\beta$; for $0<\beta\le1$, one has $W_{1,R}^\beta\ge W_{1,R}$. The bound $n_0(R/\sqrt v)\le2v/R^2$ for $R\ge3\sqrt v$ proves \eqref{eq:gaussian-trace-smallball}. The probability bound uses $W_{1,R}\le\ind_{\{\|X\|2$. \subsection{Small radii}\label{sec:small} At small $r$ the norm event $\{\|X\|\le\kappa\alpha r\}$ is itself a small ball, and the tilt estimates, which need $R>2$, no longer apply. The proof instead rescales the coefficient vector by $g\mapsto sg$, with $s\le1$, and retains the Gaussian density ratio. The Jacobian contributes $s^{r_X}\ge s^d$, which is where the hypothesis $r_X\le d$ on the coefficient rank enters. Lemma~\ref{lem:small} gives the resulting determinant bound. Let $R\ge11/4$ and set \[ \lambda=\ell(R^{-2}),\qquad v_R=(R^2+2)^{-1},\qquad s=\kappa r/R\le1. \] Here $n_0(R)=2\log\lambda-\lambda+1$ is the bound \eqref{eq:detentropy}. \begin{lemma}[Localization by Gaussian rescaling]\label{lem:small} If $V_X\preceq I_d$ and $r_X\le d$, then every finite closed penalty $\psi$ with $\psi(r)=1$ satisfies \begin{equation}\label{eq:low} \mathcal Z_\psi(X;\kappa)^{1/d}\ge s\exp\{(1-s^2)(3/2-\lambda)-n_0(R)-\sqrt{v_R}-\log(1-v_R)\}. \end{equation} \end{lemma} The positive term $(1-s^2)(3/2-\lambda)$ comes from the Radon--Nikodym factor of the rescaling. \begin{proof}[Proof of Lemma~\ref{lem:small}] Remove the kernel of the coefficient map, so $m=r_X\le d$. Let $Y=X/R$ and \[ d\mu=Z_R^{-1}\det(I-Y^2)\ind_{\{\|Y\|<1\}}d\gamma_m, \qquad V=-\log\det(I-Y^2). \] Proposition~\ref{prop:det} and its proof give $Z_R\ge e^{-dn_0(R)}$ and \[ A:=\E_\mu\tr[Y^2(I-Y^2)^{-1}]\le d(\lambda-1). \] Scalar Gaussian integration by parts gives the exact identity \begin{equation}\label{eq:coefficientenergy} \E_\mu|g|^2=m-2A. \end{equation} If $Q=[\tr B_iB_j]$, the total potential of $\mu$ has Hessian at least $I_m+2Q/R^2$. The Brascamp--Lieb variance inequality \citep{brascamp1976extensions}, applied to linear functions, yields \[ \Cov_\mu(g)\preceq(I_m+2Q/R^2)^{-1}, \] and evenness makes the mean zero. The eigenvalues of $Q$ sum to at most $d$ and there are at most $d$ of them. Concavity of $u/(1+2u/R^2)$ gives \[ v:=d^{-1}\E_\mu\tr Y^2 \le\frac1{R^2+2}=v_R. \] Spectral Jensen and Cauchy--Schwarz also give \[ d^{-1}\E_\mu V\ge-\log(1-v),\qquad d^{-1}\E_\mu\tr|Y|\le\sqrt v. \] With $s=\kappa r/R$, the change of variables $g\mapsto sg$ gives exactly \[ \mathcal Z_\psi(X;\kappa) =s^m\E_\gamma\exp\{(1-s^2)|g|^2/2-\tr\psi(s|X|/\kappa)\}. \] Restricting this integral to the support of $\mu$, replacing $d\gamma$ by $Z_Re^V\,d\mu$, and applying Jensen's inequality bound it from below. Because $\psi(r)=1$ and $\psi(0)=0$, convexity gives $\psi(rt)\le t$ for $0\le t\le1$, so $\tr\psi(s|X|/\kappa)=\tr\psi(r|Y|)\le\tr|Y|$ on this support. Combining these facts with \eqref{eq:coefficientenergy} gives \begin{align*} d^{-1}\log\mathcal Z_\psi &\ge\frac md\left(\log s+\frac{1-s^2}{2}\right)-n_0(R) -(1-s^2)(\lambda-1)-\log(1-v)-\sqrt v\\ &\ge\log s+(1-s^2)(3/2-\lambda)-n_0(R) -\log(1-v_R)-\sqrt{v_R}. \end{align*} For the second line, $\log s+(1-s^2)/2\le0$ and $m/d\le1$. Moreover, $\sqrt v+\log(1-v)$ increases for $0\le v\le(\sqrt2-1)^2$, and $v_R$ is in this range for $R\ge11/4$. Lemma~\ref{lem:small} follows. Its right side increases as a function of $r$ when $s\le1$: its logarithmic derivative is $r^{-1}[1-2(3/2-\lambda)s^2]>0$. This monotonicity validates the interval comparisons in the proof of Proposition~\ref{prop:closure}. \end{proof} \subsection{Completion of the proof}\label{sec:completion} Together these estimates cover all $r>0$: Lemma~\ref{lem:small} handles small radii, Lemma~\ref{lem:hinge} handles a compact interval, and an analytic tail estimate handles large radii. Proposition~\ref{prop:closure} verifies the scalar inequalities by exact interval arithmetic on the compact interval and by monotonicity at both ends; Appendix~\ref{app:completion} proves it, together with the large-radius tangent bound \eqref{eq:tailfunctional}. \begin{proof}[Proof of Theorem~\ref{thm:main}] Fix $\kappa=13/5$. The zero function is immediate. Otherwise normalize $f(0)=1$; by Lemma~\ref{lem:weight-closure} its penalty $\psi$ may be taken closed, and it suffices to prove the penalty form \eqref{eq:penalty-form}. The zero penalty gives equality. Let $\psi$ be finite and nonzero. Lemmas~\ref{lem:hinge} and~\ref{lem:small} are stated at the radius where the penalty reaches one, so choose $r>0$ with $\psi(r)=1$; by \eqref{eq:profilemass}, $p_\psi\le q(r)$ and $T(r)\le1-p_\psi$. For $r\le24/25$, Lemma~\ref{lem:small} and part~(1) of Proposition~\ref{prop:closure} give $\mathcal Z_\psi(X;\kappa)^{1/d}\ge q(r)\ge p_\psi$. For $24/25\le r\le16/5$, part~(2), with $n=n(\kappa\alpha r)\ge N_X(\kappa\alpha r)$, supplies the hypotheses of Lemma~\ref{lem:hinge}, which gives \eqref{eq:penalty-form}. For $r\ge16/5$, the tangent bound \eqref{eq:tailfunctional} and part~(3) give $-d^{-1}\log\mathcal Z_\psi(X;\kappa)\le(9/50+77/100)(1-p_\psi) <1-p_\psi\le-\log p_\psi$. An extended-valued closed penalty is the increasing limit of the finite penalties $\psi_k(t)=\inf_{0\le u\le t}\{\psi(u)+k(t-u)\}$, and bounded convergence on both sides of \eqref{eq:penalty-form} completes the proof. \end{proof} \section{The determinant inequality at the conjectured dilation}\label{sec:endpoint} Theorem~\ref{thm:main} leaves the gap \eqref{eq:interval} between the dilation $13/5$ and the lower bound $\cstar$ forced by the anticommuting family. The localization proof of Theorem~\ref{thm:main} cannot close this gap: its small-ball estimate decays too slowly to reach any dilation at most $\sqrt2$. The lower bound $\cstar$ suggests the following universal statement; we call $\cstar$ the \emph{conjectured dilation}. \begin{dashedbox} \begin{conj}\label{conj:optimal-gsd} For every integer $d\ge1$, let $X=\sum_{i=1}^m g_iB_i\in\Sym_d(\R)$, where the $g_i$ are independent standard Gaussians and the $B_i$ are real symmetric matrices satisfying $\sum_iB_i^2\preceq I_d$ and $\dim\operatorname{span}\{B_1,\ldots,B_m\}\le d$. Then, for every even log-concave $f\colon\R\to[0,\infty)$, \begin{equation*} \E\det f(X)\ge\bigl(\E f(\cstar G)\bigr)^d, \qquad \cstar=\sqrt{\pi/2}. \end{equation*} \end{conj} \end{dashedbox} \subsection{An entropy route to the conjectured dilation}\label{sec:entropy-route} The averaged comparison alone does not control a determinant. An exact characterization becomes possible when we allow a change of the coefficient law and bound its relative entropy. To state it, regard $X$ as a function of its coefficients, $X(x)=\sum_ix_iB_i$ for $x\in\R^m$, so that $X=X(g)$ with $g\sim\gamma_m$ standard Gaussian on $\R^m$, and let $\gamma_+$ be the law of $|G|$. For probability measures $\nu,\mu$, write $D(\nu\Vert\mu)=\int\log(d\nu/d\mu)\,d\nu$ when $\nu\ll\mu$, and $D(\nu\Vert\mu)=+\infty$ otherwise. For a coefficient law $\nu$ on $\R^m$, define its average absolute spectral law by \[ \int h\,d\sigma_{\nu,\kappa} =\frac1d\int\tr h(|X(x)|/\kappa)\,d\nu(x). \] For laws on $[0,\infty)$ with finite first moments, write $\sigma\preceq_{\mathrm{icx}}\eta$ for \emph{increasing convex order}: $\int h\,d\sigma\le\int h\,d\eta$ for every finite convex nondecreasing $h$, allowing infinite integrals. To enforce this order, minimize the relative entropy of the coefficient law: \begin{equation}\label{eq:entropy-cost} \mathcal I_{X,\kappa}(\eta) :=\inf_{\nu:\,\sigma_{\nu,\kappa}\preceq_{\mathrm{icx}}\eta} D(\nu\Vert\gamma_m). \end{equation} \begin{dashedbox}[supporting] \begin{theorem}[Entropy characterization of the determinant inequality]\label{prop:entropy-gsd} Let $\kappa>0$. A matrix Gaussian series $X$ satisfies the determinant inequality at dilation $\kappa$ if and only if \begin{equation}\label{eq:entropy-gsd} \mathcal I_{X,\kappa}(\eta)\le dD(\eta\Vert\gamma_+) \qquad\text{whenever }D(\eta\Vert\gamma_+)<\infty. \end{equation} For such $\eta$, the infimum in \eqref{eq:entropy-cost} is finite and attained. In particular, \eqref{eq:entropy-gsd} holds at $\kappa=13/5$ under the hypotheses of Theorem~\ref{thm:main}. \end{theorem} \end{dashedbox} \begin{proof}[Proof that \eqref{eq:entropy-gsd} implies the determinant inequality] The Gibbs variational identity connects entropy with the determinant: \begin{equation}\label{eq:gibbs-connections} -\log\int e^{-W}\,d\mu =\inf_\xi\Bigl\{D(\xi\Vert\mu)+\int W\,d\xi\Bigr\} \qquad(W\ge0). \end{equation} Fix a finite closed convex penalty $\psi$ and tilt the scalar law to $d\eta=p_\psi^{-1}e^{-\psi}\,d\gamma_+$. This law has finite entropy. By \eqref{eq:entropy-gsd} and attainment, choose a coefficient law $\nu$ with $D(\nu\Vert\gamma_m)\le dD(\eta\Vert\gamma_+)$ and $\sigma_{\nu,\kappa}\preceq_{\mathrm{icx}}\eta$. Then \[ \begin{aligned} -\log\mathcal Z_\psi(X;\kappa) &\le D(\nu\Vert\gamma_m)+d\int\psi\,d\sigma_{\nu,\kappa}\ \le\ d\Bigl(D(\eta\Vert\gamma_+)+\int\psi\,d\eta\Bigr) =-d\log p_\psi. \end{aligned} \] Increasing finite convex approximations and Lemma~\ref{lem:weight-closure} give all functions. Appendix~\ref{app:connections} proves attainment and the converse: convex separation enforces finite collections of hinge constraints, and entropy compactness passes to all hinges. \end{proof} At $\eta=\gamma_+$, the entropy bound \eqref{eq:entropy-gsd} forces $\nu=\gamma_m$ and reduces to the averaged comparison $\sigma_{\gamma_m,\kappa}\preceq_{\mathrm{icx}}\gamma_+$. Theorem~\ref{thm:sharp} establishes this case at $\kappa=\cstar$; the determinant inequality requires the entropy bound for every scalar law $\eta$. Conjecture~\ref{conj:optimal-gsd} is therefore equivalent to \eqref{eq:entropy-gsd} at $\kappa=\cstar$ for every matrix Gaussian series it covers. \subsection{Two special cases} The localization proof uses Jensen's inequality in \eqref{eq:localization-main}. At $\kappa=\cstar$, even a linear penalty $\psi(t)=at$ leaves the strict gap $e^{-a\E|G|}<\E e^{-a|G|}$ for $a>0$, so \eqref{eq:localization-main} cannot reach $\cstar$ for this penalty, even if the norm event has probability one. The entropy route of Section~\ref{sec:entropy-route}, with a Gaussian trial law for the coefficients, avoids this loss for linear--quadratic penalties and reaches $\cstar$. The result depends on the ratio $r_X/d$ of coefficient rank to order, through $\theta_*:=1/(\pi-2)=0.875969\ldots$. \begin{theorem}[Linear--quadratic penalties at the conjectured dilation] \label{thm:endpoint} Suppose the matrix variance and coefficient rank satisfy $\tr V_X\le d$ and $r_X\le\theta_*d$; since $\theta_*>7/8$, the condition $r_X\le7d/8$ suffices. For every $a,b\ge0$, the function $f(x)=e^{-a|x|-bx^2}$ satisfies \begin{equation}\label{eq:endpoint} \E e^{-a\tr|X|/\cstar-b\tr X^2/\cstar^2} \ge\bigl(\E e^{-a|G|-bG^2}\bigr)^d. \end{equation} The constant $\cstar$ cannot be decreased uniformly over this class, even when $b=0$. \end{theorem} With the trial law $N(0,s^2I)$ in \eqref{eq:gibbs-connections}, the penalty term is proportional to $d$ and the entropy to $r_X$. Optimizing over $s$ reduces the proof to a scalar comparison with $-\log\E e^{-a|G|}$; Appendix~\ref{app:endpoint-calculus} proves it and identifies $\theta_*$ as the largest value of $r_X/d$ allowed by this estimate. To treat every even log-concave function, we compare the trace penalty with a sum of scalar penalties of $d$ Gaussian projections of the coefficients. This comparison keeps the product inside the expectation, where the Gaussian correlation inequality applies. An ellipsoidal bound $\|X(x)\|\le\sqrt{x^{\mathsf T}Qx}$ with $\tr Q\le1$ on the operator norm makes it possible even when the absolute eigenvalues differ, and gives the determinant inequality at $\cstar$ once $d$ exceeds a threshold depending on $Q$ (Theorem~\ref{thm:ellipsoid-endpoint}). The isotropic case $Q=I_r/r$ covers all sufficiently large members of the anticommuting family that forces $\cstar$ (Proposition~\ref{prop:radial-endpoint}). Orthogonal sums and averages of orthogonal conjugates extend the determinant inequality to further matrix Gaussian series (Appendix~\ref{sec:orthogonal-sums}). \begin{aiusage} % The results in this paper were developed through extensive use of AI assistants: OpenAI's GPT~Pro for the main line of argument, including Theorem~\ref{thm:main}, the verification of its scalar inequalities, and Theorem~\ref{thm:endpoint}; and Anthropic's Claude for the results under an ellipsoidal norm bound in Appendix~\ref{sec:ellipsoid-endpoint}, Proposition~\ref{prop:gaussian-trace-smallball}, Appendix~\ref{app:connections}, and the writing, positioning, and literature search. 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Remove the kernel of the coefficient map $x\mapsto X(x)$, so that it is injective on $\R^r$ and $r=r_X$. \begin{prop}[Small balls on the coefficient support]\label{prop:rank-smallball} Suppose $x\mapsto X(x)$ is injective on $\R^r$ and $r\ge1$. Define \[ K=\{x:\|X(x)\|\le1\},\qquad v_K=(2\pi)^{-r/2}\operatorname{vol}_r(K),\qquad R_K=\max_{x\in K}|x|. \] For every $t>0$, \begin{equation}\label{eq:rank-smallball} v_K t^r e^{-R_K^2t^2/2} \le\Pp(\|X\|\le t)\le v_K t^r. \end{equation} Consequently $\Pp(\|X\|\le t)\sim v_K t^r$ as $t\downarrow0$. If $X$ satisfies the determinant inequality at a finite dilation $\kappa$, then $r\le d$; when $r=d$, necessarily $v_K\kappa^d\ge(2/\pi)^{d/2}$. \end{prop} \begin{proof} Injectivity makes $\|X(\cdot)\|$ a norm, so $K$ is a compact convex body with positive volume. Changing variables $x=ty$ gives \[ \Pp(\|X\|\le t) =(2\pi)^{-r/2}t^r\int_{K}e^{-t^2|y|^2/2}\,dy. \] Bounding the integrand proves \eqref{eq:rank-smallball} and its asymptotic. The interval-indicator instance of the determinant inequality gives $\Pp(\|X\|\le\kappa t)\ge\Pp(|G|\le t)^d$. Since $\Pp(|G|\le t)\sim\sqrt{2/\pi}\,t$, comparison of powers and then coefficients proves the final assertions. \end{proof} \subsection{Support endpoints of log-concave functions}\label{app:closure} The proofs use closed penalties (Section~\ref{sec:proofs}); the following lemma justifies that convention. \begin{lemma}[Support-endpoint closure]\label{lem:weight-closure} Let $f$ be a normalized even log-concave function. There is an upper-semicontinuous normalized even log-concave function $\bar f$ that agrees with $f$ except possibly at the two endpoints of a bounded positive support. For every real symmetric matrix Gaussian series $X$ and every $\kappa>0$, the scalar and determinant expectations agree: \[ \E f(G)=\E\bar f(G),\qquad \E\det f(X/\kappa)=\E\det\bar f(X/\kappa). \] \end{lemma} \begin{proof} The positive support of $f$ is a symmetric interval containing zero. If it is $\{0\}$, then $f$ is already upper semicontinuous. Otherwise $\log f$ is finite and concave in the interior of its support, and hence continuous there. If the support has finite radius $a>0$, extend $f$ at $\pm a$ by its interior one-sided limit and keep it zero outside. The endpoint value of $f$ is at most this limit, by log-concavity along the segment from zero. Extending $\log f$ by its one-sided limits preserves concavity, so the resulting $\bar f$ is even, log-concave, and upper semicontinuous. An unbounded symmetric support is all of $\R$, where no extension is needed. The scalar Gaussian assigns zero mass to $\pm a$. For the matrix expectation, write $X$ on its Gaussian coefficient support. Each polynomial $\det(X\mp\kappa aI_d)$ is nonzero at the zero coefficient vector and therefore vanishes on a Lebesgue-null set. Consequently $X/\kappa$ almost surely has no eigenvalue equal to $\pm a$. The products defining the determinants then agree almost surely, proving both identities. \end{proof} \section{Matrix-weighted Poincar\'e estimates for small-ball probabilities}\label{app:engine} This appendix proves the sharper bounds on the negative log probability $N_X(R)$ stated in Section~\ref{sec:smallball}. Throughout, $V_X\preceq I_d$, the coefficient rank $r_X$ is arbitrary, and $\ell$ is the function of Section~\ref{sec:smallball}. \subsection{Convex tilts} The tilts used below are all even and log-concave, so the following comparison applies to them. \begin{lemma}[Even convex observables under an even convex tilt] \label{lem:tilt-domination} Let $d\mu=Z^{-1}e^{-V}d\gamma_m$, where $V\colon\R^m\to(-\infty,\infty]$ is even and convex, and $0u)\le\gamma(F>u)$, and integration of the tails proves the claim. Truncation handles an infinite expectation. Adding a constant to $V$ does not affect the argument. \end{proof} For $Y=\sum_i g_iD_i$ with $\sum_iD_i^2\preceq\xi I$, change the Gaussian density by a factor proportional to $e^{-V}$, where $V$ is even and convex. Under the resulting measure $\mu$, Lemma~\ref{lem:tilt-domination}, applied to $\tr Y^{2p}$, gives \begin{equation}\label{eq:tiltmoments} \E_\mu\tr Y^{2p}\le d(2p-1)!!\xi^p. \end{equation} The barrier argument of Section~\ref{app:flat} uses the following tail bound: \begin{equation}\label{eq:momenttail} \E_\mu\tr\ind_{\{|Y|>u\}} \le\frac32d\exp\left(-\frac{u^2}{2\xi}\right) \quad\text{if }u/\sqrt\xi\ge4. \end{equation} Its proof uses the factorial estimate $(2p-1)!!\le\sqrt2(2p/e)^p$: if $a_p=(2p-1)!!/(2p/e)^p$ and $t=(2p+1)^{-1}$, then \[ \log(a_{p+1}/a_p) =1-\frac{\operatorname{arctanh}t}{t}-\frac12\log(1-t^2) =\nlsum_{k\ge1}\left(\frac1{2k}-\frac1{2k+1}\right)t^{2k}>0. \] Stirling's formula gives $a_p\to\sqrt2$, establishing the bound. For $x=u/\sqrt\xi\ge4$, put $A=x^2/2\ge8$ and take the integer $p$ nearest to $A$. Taylor's theorem for $v\log(v/A)-v+A$ gives \[ p\log(p/A)-p+A\le\frac1{8(A-1/2)}\le\frac1{60}. \] Markov's inequality and \eqref{eq:tiltmoments} now yield $\sqrt2e^{1/60}de^{-x^2/2}<\frac32de^{-x^2/2}$. Indeed, $e^{1/60}\le60/59$ and $2(60/59)^2<9/4$. \subsection{Block Gram identities}\label{sec:block} Besides the block inequality \eqref{eq:block}, the curvature computations below use two further block inequalities for symmetric matrices $D_i$ and $E_i$. The calculation behind \eqref{eq:block}, followed by Frobenius Cauchy--Schwarz, gives \begin{equation}\label{eq:mixed} [D_jE_i+E_jD_i]_{ij} \preceq b[\tr(D_iD_j)]_{ij}\otimes I +b^{-1}[\tr(E_iE_j)]_{ij}\otimes I\quad(b>0). \end{equation} Also, if $\sum_iD_i^2\preceq tI$, then $[D_jD_i]_{ij}\preceq tI$: use $\sum_{ij}\langle D_ju_i,D_iu_j\rangle \le\sum_{ij}\|D_ju_i\|^2$. \subsection{Removing the quadratic term from the barrier}\label{sec:cubic} The bound of Proposition~\ref{prop:det} is of order $R^{-2}$, the order of the quadratic part of $-\log\det(I-Y^2)$. Removing that part from the barrier, and bounding $\tr Y^2$ by the convex-tilt domination of Lemma~\ref{lem:tilt-domination} instead, improves the order to $R^{-3}$. \begin{theorem}\label{thm:cubic} For $\sum_iB_i^2\preceq I_d$, $R>\sqrt5$, define \[ \eta=R^{-2},\quad \lambda=\frac2{1+\sqrt{1-4\eta/(1-\eta)}},\quad D=\frac{\lambda^2(\lambda+4)}{(1-\eta)^2}. \] Then \begin{equation}\label{eq:cubic} N_X(R)\le n_1(R):=\frac{\sqrt D}{R^3}+\frac{D}{2R^4}. \end{equation} The leading coefficient in~\eqref{eq:cubic} is $\sqrt5$ as $R\to\infty$. \end{theorem} \begin{proof} Let $s(x)=-\log(1-x^2)-x^2$ for $|x|<1$. For $b>0$, put $\alpha=1+b^{-1}$ and use the tilt \[ V(Y)=\frac b2\tr Y^2+\alpha\tr s(Y),\qquad d\mu_t=Z_t^{-1}e^{-V(Y_t)}\mathbf1_{\{\|Y_t\|<1\}}d\gamma_m, \quad Y_t=tX/R. \] The scalar function $s$ is convex and nonnegative, so $\mu_t$ is an even log-concave tilt of Gaussian measure. Lemma~\ref{lem:tilt-domination}, applied to the even convex function $\tr Y_t^2$, gives \begin{equation}\label{eq:secondmoment} \E_{\mu_t}\tr Y_t^2\le d\eta t^2. \end{equation} We next verify curvature for the resolvent weight $K=(I-Y_t)^{-1}$. Write $D_i=tB_i/R$, $E_i=K^{1/2}D_iK^{1/2}=D_i+\Delta_i$, and $\Gamma_D=[\tr D_iD_j]$, $\Gamma_\Delta=[\tr\Delta_i\Delta_j]$. If $\mathcal H$ is the coefficient Hessian of $\tr s(Y_t)$, then \begin{equation}\label{eq:tailcurvature} \Gamma_\Delta\preceq\mathcal H. \end{equation} To check \eqref{eq:tailcurvature}, diagonalize $Y_t$ and set $r_a=(1-x_a)^{-1}$, $p_a=(1+x_a)^{-1}$. The respective coefficients in these Gram forms are $(\sqrt{r_ar_b}-1)^2$ and $r_ar_b+p_ap_b-2$. The second dominates the first because $p_ap_b+2\sqrt{r_ar_b}\ge3$, by AM--GM and $p_ap_br_ar_b\ge1$. By~\eqref{eq:block} and~\eqref{eq:mixed}, \[ [E_jE_i]\preceq\eta t^2I+b\Gamma_D\otimes I+ (1+b^{-1})\mathcal H\otimes I. \] The last two terms are exactly the Hessian of $V$. Thus $p_tK$ has curvature at least $c_t=1-\eta t^2$. All matrix weights are integrable: $\alpha>1$ and the remaining factor $e^{(\alpha-b/2)\tr Y_t^2}$ is bounded on the spectral domain. The inverse-weight test and continuation argument from Proposition~\ref{prop:det} give \[ \|S_t\|\le\lambda_t:=\frac2{1+\sqrt{1-4\zeta_t}},\qquad S_t=\E_{\mu_t}K,\qquad \zeta_t=\frac{\eta t^2}{1-\eta t^2}<\frac14. \] Let $M_t=\tr(S_t-I)$ and $Q_t=(I-Y_t^2)^{-1}$. We claim \begin{equation}\label{eq:fourth} \E_{\mu_t}\tr[Y_t^4Q_t] \le\zeta_t(\lambda_t+4)M_t \le dD\eta^2t^4. \end{equation} Apply~\eqref{eq:WP} to the columns of $F=Y_t^2$ and sum. The weighted-mean term is $\tr[(S_t-I)S_t^{-1}(S_t-I)] \le(\lambda_t-1)M_t/\lambda_t$. For the derivative term, evenness permits replacing $K$ by $Q_t$. In an eigenbasis of $Y_t$, write $q_a=(1-x_a^2)^{-1}$. The elementary inequality \[ (x_a+x_b)^2(q_a+q_b) \le4(x_a^2q_a+x_b^2q_b) \] follows from $(x_a+x_b)^2\le2(x_a^2+x_b^2)$ and $(x_a^2-x_b^2)(q_a-q_b)\ge0$. Symmetrizing the derivative energy over $a,b$ therefore gives \[ \E_{\mu_t}\nlsum_i\tr[(D_iY_t+Y_tD_i)Q_t(D_iY_t+Y_tD_i)] \le4\eta t^2M_t. \] Since $(\lambda_t-1)/\lambda_t=\zeta_t\lambda_t$, $M_t\le d\zeta_t\lambda_t^2$, and $\lambda_t\le\lambda$, the claim follows. Finally, differentiating the partition function yields \[ -\frac d{dt}\log Z_t =\frac1t\E_{\mu_t}\{b\tr Y_t^2+2\alpha\tr[Y_t^4Q_t]\}. \] Integrating~\eqref{eq:secondmoment} and~\eqref{eq:fourth} gives \[ -d^{-1}\log Z_1\le \frac{b\eta}{2} +\frac{(1+b^{-1})D\eta^2}{2}. \] Choose $b=\sqrt{D\eta}$ and use $Z_1\le\Pp(\|X\|\le R)$. Hence \eqref{eq:cubic} holds. \end{proof} \subsection{Bounds from higher moments}\label{app:moment} Keeping more moments of the logarithmic barrier improves the leading coefficient in \eqref{eq:cubic}. For fixed $P\ge4$, the resulting leading coefficient is $\sqrt3$. \begin{prop}[Higher-moment bound]\label{prop:completion} For $R>\sqrt5$, set $\eta=R^{-2}$, $z=\eta/(1-\eta)$, and $\delta=\ell(z)-1$. For an integer $P\ge3$ with $d_P=1-\delta-P^2z>0$, define \begin{equation*} B_P(R)=\nlsum_{j=2}^{P-2}\frac{(2j-1)!!}{j}\eta^j +\frac{(2P-3)!!}{(P-1)d_P}\eta^{P-1}. \end{equation*} The sum is empty for $P=3$. Then \begin{equation}\label{eq:momententropy} N_X(R)\le n_P(R):=\sqrt{2\eta B_P(R)}+B_P(R). \end{equation} \end{prop} The verification of the scalar inequalities (Proposition~\ref{prop:closure}) uses the best of \eqref{eq:detentropy}, \eqref{eq:cubic}, and \eqref{eq:momententropy} with $3\le P\le8$. \begin{proof} Fix $\beta>0$ and use the tilt of Theorem~\ref{thm:cubic} with $b=\beta$, $Y_t=tX/R$, variance $\xi_t=t^2/R^2$, curvature $1-\xi_t$, and $z_t=\xi_t/(1-\xi_t)$. Put \[ K=(I-Y_t)^{-1},\quad Q=(I-Y_t^2)^{-1},\quad S=\E_\mu K=\E_\mu Q, \quad\delta_t=\ell(z_t)-1, \] and $J_p=\E_\mu\tr Y_t^{2p}Q$. Apply the weighted Poincar\'e inequality to the columns of $F=Y_t^p$. Its mean term is at most $\|\E K Y_t^p\|_F^2$ because $S\succeq I$. If $p$ is even, $\E KY_t^p=\E QY_t^p$. Factoring the integrand as $(Y_tQ^{1/2})(Y_t^{p-1}Q^{1/2})^T$ and using $\E Y_t^2Q=S-I\preceq\delta_t I$ bounds the squared Frobenius norm by $\delta_t J_{p-1}$. If $p$ is odd, evenness gives $\E KY_t^p=\E QY_t^{p+1}$; use the factors $Y_t^2Q^{1/2}$ and $Y_t^{p-1}Q^{1/2}$ instead. The operator bound $\E Y_t^4Q\preceq S-I$ gives the same conclusion. These operator bounds are weighted Cauchy--Schwarz inequalities; no independence is used. For the derivative term, put $D_p(x,y)=\sum_{k=0}^{p-1}x^ky^{p-1-k}$ and $q_x=(1-x^2)^{-1}$. The integral formula for the divided difference, followed by convexity of $|\cdot|^{2p-2}$, gives \[ |D_p(x,y)|^2\le\frac{p^2}{2}(|x|^{2p-2}+|y|^{2p-2}). \] Monotonicity of $q_x$ as a function of $|x|$ then implies \[ |D_p(x,y)|^2(q_x+q_y) \le p^2(x^{2p-2}q_x+y^{2p-2}q_y). \] Spectral summation and $\sum_i(tB_i/R)^2\preceq\xi_tI$ bound the Dirichlet energy by $p^2\xi_tJ_{p-1}$. Thus \[ J_p\le(\delta_t+p^2z_t)J_{p-1}. \] All tests have finite weighted energy by Section~\ref{sec:regularity}. Write $M_j=\E_\mu\tr Y_t^{2j}$, so $J_j=M_j+J_{j+1}$. If $d_P=1-\delta_1-P^2z_1>0$, then uniformly in $t\le1$, \[ J_{P-1}\le M_{P-1}/d_P,\qquad J_2\le\nlsum_{j=2}^{P-2}M_j+M_{P-1}/d_P. \] Use the tilted moment bound \eqref{eq:tiltmoments} to bound each $M_j$. The normalizer satisfies \[ -\frac1d\frac d{dt}\log Z_t =\frac1{dt}\E_\mu\{\beta\tr Y_t^2+2(1+\beta^{-1})\tr Y_t^4Q\}. \] Integrating each moment term yields $-d^{-1}\log Z_1\le\beta\eta/2+(1+\beta^{-1})B_P(R)$. Since the tilt stays fixed during the integration, minimizing afterwards in $\beta>0$ gives \eqref{eq:momententropy}. \end{proof} \subsection{A logarithmic barrier with a flat interior}\label{sec:wall} The large-radius bound tilts Gaussian measure by a logarithmic barrier that vanishes on $[-a,a]$ and agrees with $-\log(1-x)$ up to a linear term near $x=1$. Vanishing on the interior is the point: the derivative energy in Section~\ref{app:flat} is then supported on eigenvalues beyond $b0$, the Gaussian tilt \begin{equation}\label{eq:flatpotential} V(Y)=\frac{\beta L^2}{2}\tr Y^2+ (1+\beta^{-1})\tr v_a(Y) \end{equation} has resolvent matrix weight $pK$ with curvature at least $1-L^2\xi$ in~\eqref{eq:WP}. \end{lemma} \begin{proof} In a spectral basis, the multiplier of $D_{ab}$ in $U_i$ is \[ u(x,y)=\sqrt{r(x)r(y)}-u_a(x)u_a(y). \] It suffices to prove $u(x,y)^2\le h_a'[x,y]$, with the divided difference interpreted as a derivative on the diagonal. If $x,y\le a$, the left side is zero. If $x,y>a$, then \[ u(x,y)^2=(\sqrt{r(x)r(y)}-L)^2\le r(x)r(y)=h_a'[x,y]. \] If $x>a\ge y$, division of the desired inequality by its positive right side leaves \[ r(y)(x-y)\frac{\sqrt{r(x)}-\sqrt L} {\sqrt{r(x)}+\sqrt L}\le1. \] Both factors on the left are in $[0,1]$: in particular $r(y)(x-y)=(x-y)/(1-y)\le1$. This factor bound proves the first Gram inequality by the spectral Hessian formula. The second follows because every entry multiplier in $C_i$ is at most $L$. Finally, \[ \nlsum_iC_i^2=u_a(Y)\nlsum_iD_iu_a(Y)^2D_iu_a(Y) \preceq L u_a(Y)\Bigl(\nlsum_iD_i^2\Bigr)u_a(Y) \preceq L^2\xi I. \] Expand $E_i=C_i+U_i$. The block inequalities~\eqref{eq:block} and \eqref{eq:mixed}, and the last block inequality in Section~\ref{sec:block} give \[ [E_jE_i]_{ij}\preceq L^2\xi I+ \beta\Gamma_C\otimes I+(1+\beta^{-1})\Gamma_U\otimes I \preceq L^2\xi I+ \beta L^2[\tr D_iD_j]\otimes I+ (1+\beta^{-1})\mathcal H_a\otimes I. \] For the weight $pK$, the normalized block curvature is \[ [\delta_{ij}I+\partial_{ij}V\,I-E_jE_i]_{ij}. \] The Hessian of~\eqref{eq:flatpotential} absorbs the last two terms in the preceding bound. The Hessian of $\tr h_a(-Y)$ is additional and nonnegative. The remaining curvature is at least $1-L^2\xi$. \end{proof} \subsection{An exponential small-ball bound}\label{app:flat} The preceding determinant estimates give polynomial decay in $R$. For the barrier of Section~\ref{sec:wall}, the derivative energy can instead be bounded by the expected number of eigenvalues beyond $b$. The Gaussian tail estimate then gives exponential decay. Fix $0L^2+4$, set \[ z=\frac1{R^2-L^2},\quad\delta=\ell(z)-1,\quad C_{a,b}(R)= \frac{3+2z/((1-a^2)w^2)}{1-3\delta/2-2z/(1-b)^2}. \] The expression is used only when its denominator is positive. \begin{prop}[Exponential bound for the negative log probability]\label{prop:flat} If $bR\ge4$ and the preceding conditions hold, define \[ A_R=\frac{L^2}{2R^2},\qquad B_R=\frac{3C_{a,b}(R)}{b^2R^2}e^{-b^2R^2/2}. \] Then \begin{equation}\label{eq:flatnew} N_X(R)\le2\sqrt{A_RB_R}+B_R. \end{equation} \end{prop} Behind Proposition~\ref{prop:flat} is a bound on the derivative energy of a spectral ramp, obtained by treating the eigenvalues beyond $a$ and those in the transition band $(b,a)$ separately. Let $Y=\sum_ig_iD_i$ with $\sum_iD_i^2\preceq\xi I$, let $\mu$ be the Gaussian tilt~\eqref{eq:flatpotential} of Lemma~\ref{lem:clipped}, and let $F=\min\{I,(|Y|-bI)_+/w\}$ and $Q=(I-Y^2)^{-1}$. Then \begin{equation}\label{eq:newderivative} \E_\mu\nlsum_i\tr[(\partial_iF)(I-Y)^{-1}(\partial_iF)] \le\frac{2\xi}{(1-b)^2}\E_\mu\tr(FQF) +\frac{2\xi}{(1-a^2)w^2}\E_\mu\tr\ind_{\{|Y|>b\}}. \end{equation} The coefficient of the first term depends on $1-b$, not the much smaller transition width $a-b$. \begin{proof}[Proof of \eqref{eq:newderivative} and Proposition~\ref{prop:flat}] Fix $a,b,L,w$ as in Proposition~\ref{prop:flat}, and use the tilt \eqref{eq:flatpotential} at $Y_t=tX/R$. Its curvature is $1-L^2\xi_t$, where $\xi_t=t^2/R^2$. As in Proposition~\ref{prop:det}, the inverse-weight test and continuation from $S_0=I$ give \[ I\preceq S_t=\E_{\mu_t}(I-Y_t)^{-1}=\E_{\mu_t}Q_t \preceq\ell(z_t)I,\quad z_t=\frac{\xi_t}{1-L^2\xi_t}. \] The condition $R^2>L^2+4$ ensures $z_t<1/4$. Let $s(x)=\min\{1,(|x|-b)_+/w\}$, $F=s(Y_t)$, and \[ \mathcal A_t=\E_{\mu_t}\tr(FQ_tF),\qquad P_t=\E_{\mu_t}\tr\ind_{\{|Y_t|>b\}},\qquad \delta_t=\ell(z_t)-1. \] The weighted mean term is at most $3P_t+(3/2)\delta_t\mathcal A_t$. Indeed, split $\E K_tF=\E F+\E(K_t-I)F$ and use $\|u+v\|^2\le3\|u\|^2+(3/2)\|v\|^2$. The first squared norm is at most $P_t$, and weighted Cauchy--Schwarz with the exact identity \[ \E(K_t-I)K_t^{-1}(K_t-I)=S_t-I \] bounds the second by $\delta_t\mathcal A_t$. To prove \eqref{eq:newderivative}, symmetrize the derivative energy using evenness and work in a spectral basis. For an unordered pair of distinct indices its coefficient is $s[x,y]^2(q_x+q_y)$. If both indices have absolute eigenvalue at least $a$, this coefficient is zero. If precisely one index is high, put $v=|x|\ge a$. The explicit ramp gives $|s[x,y]|\le(v-b)^{-1}$ and $q_y\le q_x$. The resolvent term splits as \begin{equation}\label{eq:remaindersplit} \frac{q_v}{(v-b)^2} =\frac{q_v}{(1-b)^2} +\frac{1+v-2b}{(1+v)(v-b)^2(1-b)^2}. \end{equation} The second term decreases in $v$. At $v=a$, it is at most $M=((1-a^2)w^2)^{-1}$, since its ratio to $M$ equals $1-w^2/(1-b)^2\le1$. Consequently the high-index contribution is at most \[ 2\xi_t\left\{(1-b)^{-2}\tr_{\{|Y_t|\ge a\}}Q_t +M\,\tr\ind_{\{|Y_t|\ge a\}}\right\}. \] For pairs, including diagonal terms, with both eigenvalues in $(-a,a)$, the resolvent is at most $(1-a^2)^{-1}$ and the ramp Lipschitz constant is $1/w$. Its divided difference vanishes unless at least one index lies in the transition band $b<|x|b\}$, so their expected traces sum to $P_t$. Also, $\tr_{\{|Y_t|\ge a\}}Q_t\le\tr(FQ_tF)$. Hence \eqref{eq:newderivative} holds. To verify monotonicity of the remainder in \eqref{eq:remaindersplit}, set $u=v-b$. Its logarithmic derivative is \[ \frac{2b}{(1-b+u)(1+b+u)}-\frac2u<0. \] Weighted Poincar\'e therefore yields \[ \left(1-\frac32\delta_t-\frac{2z_t}{(1-b)^2}\right)\mathcal A_t \le\left(3+\frac{2z_t}{(1-a^2)w^2}\right)P_t. \] The denominator stays positive by hypothesis; replacing $z_t,\delta_t$ by their values at one gives $\mathcal A_t\le C_{a,b}(R)P_t$. This bound holds for each $\beta>0$ and uniformly in $t\le1$. By the tilted tail bound \eqref{eq:momenttail} and $bR/t\ge4$, $P_t\le(3/2)d e^{-b^2R^2/(2t^2)}$. Also $\E_{\mu_t}\tr Y_t^2\le d\xi_t$ and $xv_a'(x)\le2(1-x^2)^{-1}\ind_{\{|x|>a\}}$. Differentiation of the normalizer gives \[ -\frac1d\frac d{dt}\log Z_t \le\frac{\beta L^2t}{R^2} +\frac{3(1+\beta^{-1})C_{a,b}(R)}{t} e^{-b^2R^2/(2t^2)}. \] For $c>0$, $\int_0^1e^{-c/t^2}\,dt/t\le e^{-c}/(2c)$. Integration hence gives $\beta A_R+(1+\beta^{-1})B_R$. Optimize at $\beta=\sqrt{B_R/A_R}$ to prove \eqref{eq:flatnew}. Section~\ref{sec:regularity} justifies the regularity and limiting steps. \end{proof} Specializing the parameters makes the large-radius decay explicit. \begin{theorem}[Exponential small-ball bound]\label{thm:large-norm} If $V_X\preceq I_d$, then for every $R\ge83/10$, \begin{equation}\label{eq:cleantail} N_X(R)\le \frac{160}{R^2}e^{-5329R^2/40000} +\frac{507}{R^2}e^{-5329R^2/20000}. \end{equation} There is no restriction on $r_X$. \end{theorem} For the unbounded range, take $a=4/5$, $b=73/100$: exact arithmetic gives $C_{a,b}(R)\le90$ for $R\ge83/10$, and then rounds the two coefficients to 160 and 507. For the compact range we use $a=77/100$, $b=7/10$ and retain the unrounded formula. \subsection{Regularity and boundary limits}\label{sec:regularity} For barriers, choose a smooth nonnegative probability density $\rho_\epsilon$ supported in $(a-\epsilon,a)$ and set \[ h_{a,\epsilon}(x)=\int h_z(x)\rho_\epsilon(z)\,dz. \] For $za-\epsilon$. Use the ramp with upper threshold $a-\epsilon$ and lower threshold $b$ in the proof of Proposition~\ref{prop:flat}. The constants tend to the stated ones, and the denominator stays positive for sufficiently small $\epsilon$. Passing to the limit proves the exact formula with $C_{a,b}(R)$; strict slack in its rounded specializations is retained. Lipschitz spectral ramps are admissible by smooth approximation on compact convex subdomains and exhaustion. Their divided-difference formulas hold almost everywhere. The derived energy inequalities provide a finite integrable majorant. For each fixed $R$ and positive tilt parameter, the barrier power is $1+\beta^{-1}>1$, while the weights $(I-Y)^{-1}$ and $(I-Y^2)^{-1}$ have only first-order poles. Density times weight is bounded on the spectral domain, with Gaussian decay in coefficient kernel directions. The same observations justify continuity of the resolvent mean from $S_0=I$ and differentiation of the normalizer on compact subintervals of $t>0$ with no boundary contribution. Integrate first from $\tau>0$ and then let $\tau\downarrow0$. Bounded convergence gives $Z_\tau\to1$. For the determinant weight, the product with its first resolvent is the adjugate expression displayed above and is integrable even though it need not vanish on the boundary. The curvature criterion imposes no zero-boundary test condition. Scalar integration by parts below uses the density itself, which does vanish. Its logarithmic derivative has an integrable first-order pole, and $-\log\det(I-Y^2)$ is integrable under the same density. These facts justify the rescaling proof and all its expectations by exhaustion. \section{The scalar inequalities at dilation $13/5$}\label{app:completion} This appendix proves the large-radius tangent bound \eqref{eq:tailfunctional} and Proposition~\ref{prop:closure}. \subsection{The tangent bound at large radii}\label{app:tail} We bound the trace penalty in \eqref{eq:localization-main} by $1-p_\psi$ and its norm cost by $T(r)$. Let $\psi$ be a finite closed penalty with $\psi(r)=1$ and $r\ge16/5$. Set $\alpha=999/1000$, $\rho=(13/5)\alpha$, $b_f=\cstar/(13/5)$, $a_0(r)=1-H(r)/H(0)$, and $\delta_0=1-e^{-1}$. For $0\le h\le\alpha r$, \begin{equation}\label{eq:kernel} H(h)-H(r)\ge a_0(r)H(h/b_f). \end{equation} The difference vanishes at zero and its derivative has the sign of $-1+(a_0(r)/b_f)T(h/b_f)/T(h)$. The ratio $T(h/b_f)/T(h)$ decreases: its logarithmic derivative is $1/M(h)-1/(b_fM(h/b_f))<0$, since $M$ decreases and $b_f<1$. Thus the minimum of the difference is at an endpoint. At $h=\alpha r$, put $u=\alpha r/b_f$ and use \[ H(\alpha r)-H(r)\ge(1-\alpha)rT(r),\qquad H(u)\le T(u)/u\le T(r)e^{-(u^2-r^2)/2}/u. \] The sufficient inequality \[ e^{-((\alpha/b_f)^2-1)r^2/2} \le(1-\alpha)(\alpha/b_f)r^2 \] holds at $r=16/5$ by the interval arithmetic of Section~\ref{app:certificate}, and persists by monotonicity. Hence \eqref{eq:kernel} holds. For the positive hinge measure $\nu$ of the tangent-truncated $\widehat{\psi}$, \[ \E\widehat{\psi}(b_f|G|)=\int b_fH(h/b_f)\,d\nu(h),\quad \E\min\{\psi(|G|),1\}\ge\int(H(h)-H(r))\,d\nu(h). \] Together with $1-e^{-x}\ge\delta_0\min\{x,1\}$, this inequality gives $\E\widehat{\psi}(b_f|G|)\le b_f(1-p_\psi)/(\delta_0a_0(r))$. Substituting this estimate in the localization bound \eqref{eq:localization-main} gives \begin{equation}\label{eq:tailfunctional} -d^{-1}\log\mathcal Z_\psi(X;13/5) \le N_X(\rho r)+\frac{b_f}{\delta_0a_0(r)}(1-p_\psi). \end{equation} To extend the norm bound to the unbounded range, divide the exponential small-ball bound \eqref{eq:cleantail}, at $R=\rho r$, by $T_-(r)=c_0r(1+r^2)^{-1}e^{-r^2/2}\le T(r)$. Each term is a positive constant times $(1+r^2)r^{-3}e^{-cr^2}$ with $c>0$, because $(73/100)^2\rho^2>2$. These terms decrease. Therefore the single endpoint comparison at $r=16/5$ covers the full unbounded range. \subsection{Verification of the scalar inequalities}\label{app:certificate} \begin{prop}[Scalar inequalities at dilation $13/5$]\label{prop:closure} Fix $\kappa=13/5$ and $b_f=\cstar/\kappa=5\sqrt{\pi/2}/13$. \begin{enumerate} \item For $0\frac45>c_0. \] Since $3/2-\lambda>0$, the ratio $L_{11/4}(r)/r$ decreases. Together with $q(r)\le c_0r$, this monotonicity proves the first assertion for $r\le1/2$. On $[1/2,24/25]$, the check $\max_{R\in\{11/4,3\}}L_R(r_l)>q(r_u)$ on 920 intervals of width $1/2000$ completes the first assertion. For the middle range, 2,240 radius intervals and 8,762 terminal hinge rectangles exhaust the domain. Section~\ref{sec:arithmetic} specifies the outward-rounded integer arithmetic and transcendental remainders. For the unbounded range, Appendix~\ref{app:tail} reduces the comparisons to certified endpoints. \end{proof} Table~\ref{tab:certificate} lists the tangent fractions used on the middle range. \begin{table}[ht] \centering \begin{tabular}{@{}ccc@{}}\toprule Radius interval & Tangent fraction $\alpha$ & Terminal rectangles\\\midrule $[24/25,6/5]$ & $9/10$ & 1366\\ $[6/5,3/2]$ & $93/100$ & 1162\\ $[3/2,9/5]$ & $97/100$ & 900\\ $[9/5,21/10]$ & $197/200$ & 900\\ $[21/10,23/10]$ & $199/200$ & 785\\ $[23/10,5/2]$ & $499/500$ & 800\\ $[5/2,16/5]$ & $999/1000$ & 2849\\\bottomrule \end{tabular} \caption{Complete compact partition for the normalized-hinge comparison.} \label{tab:certificate} \end{table} \subsection{Monotonicity behind the enclosures} The interval arithmetic of Section~\ref{app:certificate} bounds each quantity on a rectangle of radii and hinge positions by its value at a corner; the following monotonicity properties justify these corners. For the normalized hinges write $h=yr$, $0\le y<1$. The scalar partition $p_{r,yr}$ increases in both $r$ and $y$. As $r$ increases, the penalty decreases pointwise. For $y$, fix $r$ and put $s=r-h$. The function $e^{-(t-h)_+/s}$ has Lebesgue integral $r$; its cumulative integral for $t\ge h$ is \[ h+s(1-e^{-(t-h)/s}). \] Its derivative in $h$ is $e^{-(t-h)/s}(t-h)/s\ge0$. Thus its integral against any decreasing density, including the half-normal density, increases in $h$. The hinge term $P(r,y)=b_fH(ry/b_f)/(r(1-y))$ decreases in $r$. On any $y$-interval, its maximum is at an endpoint: under $z=(1-y)^{-1}$, \[ P(r,y)=\E\bigl(1+z(b_f|G|/r-1)\bigr)_+, \] a convex function of $z$. All available norm bounds decrease in $R$ on their valid domains. For the moment bounds their denominators increase with $R$, and for the exponential bounds $C_{a,b}(R)$ decreases, so this assertion follows directly from their formulas. Consequently, for a radius interval $[r_l,r_u]$ and hinge interval $[y_l,y_u]$, an upper bound on the left side of \eqref{eq:priced} is \[ p_{r_u,y_ur_u}\exp\{\max(P(r_l,y_l),P(r_l,y_u))\}, \] while $\exp(-n(\kappa\alpha r_l))$ is a lower bound for its target. The condition \eqref{eq:unpriced} is covered by $q(r_u)$. The computation starts with $[0,\alpha]$ and bisects this hinge interval until the strict enclosure passes. All middle-range radius intervals have width $1/1000$. The final partition is exhaustive, with a maximum hinge depth of six. \subsection{Arithmetic and explicit remainders}\label{sec:arithmetic} The computation uses only integer arithmetic. Every interval has outward-rounded endpoints on the fixed grid $10^{-55}\mathbb Z$. Products, quotients and square roots use integer floor, ceiling, and integer square roots. Machin's identity $\pi=16\arctan(1/5)-4\arctan(1/239)$ is enclosed by alternating Taylor remainders. Exponentials are reduced to arguments in $[0,1]$, summed through degree 60 with a geometric upper remainder, then squared back; negative arguments use reciprocals. The logarithms used are in $[1,2]$ and use the positive $\operatorname{arctanh}$ series with its geometric remainder. Gaussian integrals up to argument eight use the integrated alternating series for $e^{-t^2/2}$ through degree index 200 and the next-term remainder. For Mills ratios above six, the integral representation \[ M(z)=\int_0^\infty e^{-zu-u^2/2}\,du \] and the signed Taylor remainder for $e^{-u^2/2}$ give \[ \nlsum_{j=0}^{17}\frac{(-1)^j(2j-1)!!}{z^{2j+1}} \le M(z)\le \nlsum_{j=0}^{16}\frac{(-1)^j(2j-1)!!}{z^{2j+1}}. \] These bounds are integrated Taylor remainder inequalities, not an assumption that the asymptotic series converges. All intermediate roundoff is outward and included in the final bounds. The complete computation has 20 groups of checks: four for the sharp averaged comparison and moment prefactor, two for small penalties, seven for normalized hinges, and seven for the analytic unbounded tail. The middle range uses the determinant bound on 480 radius intervals, the cubic bound on 354, moment orders three through six on 986 in total, and the exponential bound on 420. \section{Spectral comparisons and entropy}\label{app:spectral-functions} The determinant inequality controls symmetric polynomials and Schatten norms through different scalar comparisons. We derive them and their limitations, then prove the entropy characterization. \subsection{Symmetric polynomials} For a symmetric polynomial $P$ in $d$ variables, write $P(M)=P(\lambda_1(M),\ldots,\lambda_d(M))$ when $M\succeq0$, and put $\mathbf1=(1,\ldots,1)$. To apply the determinant inequality, we bound $P(M)$ below by a power of $\det M$. \begin{prop}[Symmetric polynomial comparison]\label{prop:symmetric-gsd} Suppose $X$ satisfies the determinant inequality at dilation $\kappa$. Let $P$ be a nonzero symmetric homogeneous polynomial of degree $m\ge1$ with nonnegative coefficients. For every even log-concave function $f\colon\R\to[0,\infty)$, \begin{equation}\label{eq:polynomial-gsd} \frac{\E P(f(X/\kappa))}{P(\mathbf1)} \ge\bigl(\E f(G)^{m/d}\bigr)^d. \end{equation} If $m\ge d$, the right side is at least $(\E f(G))^m$. \end{prop} \begin{proof} Write $P(y)=\sum_\alpha c_\alpha y^\alpha$, where $c_\alpha\ge0$ and $\sum_i\alpha_i=m$. Normalize the coefficients by $w_\alpha=c_\alpha/P(\mathbf1)$. Symmetry gives $\sum_\alpha w_\alpha\alpha_i=m/d$ for every $i$. Weighted arithmetic--geometric mean therefore yields, for $y_i>0$, \begin{equation}\label{eq:polynomial-geometric} \frac{P(y)}{P(\mathbf1)} =\nlsum_\alpha w_\alpha y^\alpha \ge\prod_\alpha(y^\alpha)^{w_\alpha} =\left(\prod_{i=1}^d y_i\right)^{m/d}. \end{equation} Continuity gives the same inequality when some coordinates vanish. Taking $y_i=f(\lambda_i(X)/\kappa)$ in \eqref{eq:polynomial-geometric} and applying the determinant inequality to the even log-concave function $f^{m/d}$ proves \eqref{eq:polynomial-gsd}. For $m\ge d$, Jensen's inequality gives $\E f(G)^{m/d}\ge(\E f(G))^{m/d}$. \end{proof} The choice $P=e_k$ gives $\E e_k(f(X/\kappa))/\binom dk\ge(\E f(G)^{k/d})^d$ for $1\le k\le d$. The proposition also includes the complete homogeneous symmetric polynomials and products of elementary symmetric polynomials. For a general $P$, the quantity $P(\mathbf1)(\E f(G))^m=P(\E f(G),\ldots,\E f(G))$ need not equal $\E P(f(G_1),\ldots,f(G_d))$ for independent $G_i$: repeated powers in a monomial involve higher scalar moments. For elementary symmetric polynomials, the independent scalar benchmark is $\binom dk(\E f(G))^k$. If the coefficients commute and $V_X\preceq I_d$, applying the commuting case (Theorem~\ref{thm:commuting}) to each $k\times k$ diagonal subseries in a simultaneous eigenbasis and summing gives this benchmark at dilation one. For general coefficients it fails: the two-Pauli series of Theorem~\ref{thm:pauli} has both absolute eigenvalues equal to $\sqrt{g_1^2+g_2^2}/\sqrt2$. Thus, for $f_a=\ind_{[-a,a]}$ and every fixed finite $\kappa>0$, \[ \frac12\E e_1(f_a(X/\kappa)) =1-e^{-\kappa^2a^2}\sim\kappa^2a^2, \qquad \E f_a(G)\sim\sqrt{2/\pi}\,a. \] The different powers of $a$ rule out that benchmark as $a\downarrow0$. \subsection{Norms and their Laplace transforms} Power penalties control the Schatten norms $\|A\|_{S_p}=(\tr|A|^p)^{1/p}$, for $1\le p<\infty$. \begin{prop}[Schatten norm comparison]\label{prop:schatten-gsd} If $X$ satisfies the determinant inequality at dilation $\kappa$, then for $1\le p<\infty$ and $t\ge0$, \begin{equation}\label{eq:schatten-laplace} \E e^{-t\|X/\kappa\|_{S_p}^p} \ge\bigl(\E e^{-t|G|^p}\bigr)^d =\E e^{-t\|(G_1,\ldots,G_d)\|_p^p}, \end{equation} where the $G_i$ are independent standard Gaussians. In particular, for $a>0$, \begin{equation*} \Pp(\|X\|_{S_p}\le\kappa a) \ge\sup_{t>0} \frac{\bigl[(\E e^{-t|G|^p})^d-e^{-ta^p}\bigr]_+} {1-e^{-ta^p}}. \end{equation*} \end{prop} \begin{proof} Apply the determinant inequality to $f(x)=e^{-t|x|^p}$. For nonnegative $T$ and $u,t>0$, \[ \E e^{-tT}\le\Pp(T\le u)+e^{-tu}\Pp(T>u). \] Apply this bound with $T=\|X/\kappa\|_{S_p}^p$ and $u=a^p$, use \eqref{eq:schatten-laplace}, and optimize over $t$. \end{proof} The case $p=1$ concerns the nuclear norm; $p=2$ concerns the Frobenius norm. Theorem~\ref{thm:quadratic} gives the latter comparison at $\kappa=1$ when $V_X\preceq I_d$ and $r_X\le d$. The Laplace-transform bound does not imply small-ball domination by the corresponding scalar norm at every radius. For example, $X=gI_d$ satisfies the determinant inequality at dilation one, but $\|X\|_{S_2}=\sqrt d\,|g|$. Gaussian tail asymptotics give \[ \lim_{a\to\infty}\frac1{a^2} \log\Pp(\|X/\kappa\|_{S_2}>a)=-\frac{\kappa^2}{2d}, \qquad \lim_{a\to\infty}\frac1{a^2} \log\Pp(\|(G_1,\ldots,G_d)\|_2>a)=-\frac12. \] Hence a comparison of these small-ball probabilities at every radius would require $\kappa\ge\sqrt d$. A dilation independent of $d$ cannot give that comparison for all unitarily invariant norms. A unitarily invariant norm $N$ depends only on singular values and satisfies $N(A)\le N(I_d)\|A\|$. Thus, if $X$ satisfies the determinant inequality at dilation $\kappa$, then \begin{equation*} \Pp\bigl(N(X)\le\kappa aN(I_d)\bigr) \ge\Pp(|G|\le a)^d,\qquad a>0. \end{equation*} \subsection{Proof of the entropy characterization}\label{app:connections} To complete Theorem~\ref{prop:entropy-gsd}, we construct a coefficient law that satisfies all hinge constraints at the required entropy cost. Compactness will enforce these constraints simultaneously and give attainment of the infimum. \begin{proof}[Completion of the proof of Theorem~\ref{prop:entropy-gsd}] First, the Gibbs variational identity \eqref{eq:gibbs-connections} follows by subtracting its left side from the expression in braces: when the integral is positive, the difference is the relative entropy to the normalized tilt $e^{-W}\mu$. The tilt attains the infimum, since $We^{-W}\le1/e$. If the integral is zero, both sides are infinite. We will use compactness of entropy sublevel sets. To see tightness, the entropy inequality gives \[ a\int |x|^2\,d\nu \le D(\nu\Vert\gamma_m)+\log\int e^{a|x|^2}\,d\gamma_m \qquad(0a_0C+\nlsum_j a_jb_j. \] For $a_0>0$, divide by $a_0$ and apply \eqref{eq:gibbs-connections} and the determinant inequality with $\psi=\sum_j(a_j/a_0)h_{t_j}$. They bound the infimum by \[ -d\log\int e^{-\psi}\,d\gamma_+ \le dD(\eta\Vert\gamma_+)+d\int\psi\,d\eta, \] contradicting separation. If $a_0=0$, replace it by a sufficiently small positive number. Nonnegativity of relative entropy preserves the strict gap, and the same argument applies. On the compact entropy sublevel set $\{\nu:D(\nu\Vert\gamma_m)\le C\}$, the closed constraints $\int S_j\,d\nu\le b_j$ have the finite intersection property by the preceding argument. Hence one law satisfies every hinge constraint with rational $t\ge0$. The inequality $|h_t-h_s|\le|t-s|$ extends the constraints to all $t\ge0$, giving a feasible law of entropy at most $C$. The moment bound above ensures that both spectral and scalar first moments are finite. \end{proof} For interval indicators, take $\eta$ to be $\gamma_+$ conditioned on $[0,a]$, where $a>0$. Its entropy is $-\log q(a)$, and the hinge $h_a$ forces every feasible $\nu$ to be supported on $E_a=\{x:\|X(x)\|\le\kappa a\}$. The least relative entropy among all laws supported on $E_a$ is $-\log\gamma_m(E_a)$, attained by conditioning. Hence \[ -\log\gamma_m(E_a)\le\mathcal I_{X,\kappa}(\eta) \le-d\log q(a), \] which recovers the small-ball bound. The support constraint alone may be weaker than the order constraint $\sigma_{\nu,\kappa}\preceq_{\mathrm{icx}}\eta$ on the average absolute spectral law. \section{The conjectured dilation}\label{app:endpoint} This appendix proves Theorem~\ref{thm:endpoint} and the conjectured dilation under an ellipsoidal norm bound (Theorem~\ref{thm:ellipsoid-endpoint}), applies the latter to the anticommuting family at large order, and collects the exact cases that support Conjecture~\ref{conj:optimal-gsd}: commuting coefficients, quadratic penalties, and the two-Pauli matrix Gaussian series. \subsection{Proof of Theorem~\ref{thm:endpoint}}\label{app:endpoint-calculus} A Gaussian change of covariance reduces the linear--quadratic comparison to the following scalar inequality. For $s>0$, define \[ D(s)=\frac{s^2-1-2\log s}{2},\qquad h(a)=-\log\E e^{-a|G|},\qquad c_0=\sqrt{2/\pi}. \] The number $D(s)$ is the relative entropy per coordinate of $N(0,s^2I)$ with respect to standard Gaussian measure. \begin{lemma}[A scalar Gaussian variational comparison]\label{lem:variational} For $0<\theta\le\theta_*=1/(\pi-2)$ and $a\ge0$, \begin{equation}\label{eq:scalar-variational} H_\theta(a):=\inf_{s>0}\{\theta D(s)+a c_0s\}\le h(a). \end{equation} The constant $\theta_*$ is the largest value of $\theta$ for which this scalar inequality holds for every $a\ge0$. \end{lemma} Section~\ref{sec:scalar-endpoint} proves the lemma by comparing the derivatives of $h$ and $H_{\theta_*}$ through a Riccati equation; the optimality of $\theta_*$ is a second-order condition at $a=0$. \begin{proof}[Proof of Theorem~\ref{thm:endpoint}] Remove the kernel of the coefficient map, and put $m=r_X$ and $\theta=m/d$. The case $X=0$ is immediate. Since $\tr V_X\le d$, \[ \E\tr|X|\le\sqrt{d\,\E\tr X^2}\le d, \qquad \E\tr X^2\le d. \] Use $N(0,s^2I_m)$ as the trial law $\xi$ in the Gibbs variational identity \eqref{eq:gibbs-connections}. As $\cstar^{-1}=c_0$, it gives \begin{equation*} -d^{-1}\log\mathcal Z_{at+bt^2}(X;\cstar) \le\inf_{s>0}\{\theta D(s)+a c_0s+b c_0^2s^2\}. \end{equation*} Set $w=1+2bc_0^2/\theta$ and $v=1+2b$. Substituting $u=s\sqrt w$ shows that the right-hand side is \[ \frac\theta2\log w+H_\theta(a/\sqrt w) \le\frac\theta2\log w+h(a/\sqrt w). \] On the scalar side, Gaussian rescaling gives the exact identity \begin{equation}\label{eq:scalar-ab} -\log\E e^{-a|G|-bG^2} =\frac12\log v+h(a/\sqrt v). \end{equation} Two elementary facts compare these expressions. First, $xh'(x)\le1$ for $x\ge0$: under the half-normal tilt by $e^{-x|G|}$, integration by parts gives $1=\E_xG^2+x\E_x|G|$. Therefore \[ h(c x)-h(x)\le\log c\qquad(c\ge1). \] Second, $t\log(1+z/t)$ increases with $t>0$ for $z\ge0$. If $w\ge v$, monotonicity of $h$ and $\theta\le1$ imply \[ \theta\log w\le\log(1+2bc_0^2)\le\log v, \qquad h(a/\sqrt w)\le h(a/\sqrt v). \] If $w0$, $m(0)=c_0$, and $m'(a)=-\Var_a(|G|)\le0$. Integrating a derivative gives $M'(a)=aM(a)-1$, so \begin{equation}\label{eq:mean-riccati} m'(a)=a m(a)+m(a)^2-1. \end{equation} The objective defining $H_{\theta_*}$ is strictly convex in $s>0$ and has a unique minimizer. By its first-order equation and the envelope identity, $\ell(a):=H_{\theta_*}'(a)$ is the positive root of \begin{equation}\label{eq:root-ode} \ell(a)^2+\beta a\ell(a)-\sigma=0, \qquad \ell(0)=c_0, \qquad \ell'(a)=-\frac{\beta\ell(a)^2}{\ell(a)^2+\sigma}. \end{equation} Since $m$ solves the Riccati equation \eqref{eq:mean-riccati}, we compare $\ell$ with $m$ through that equation: the residual below measures the failure of $\ell$ to solve it, and its sign pattern, with an integrating factor, gives $m\ge\ell$. Put $w=\ell(a)^2/\sigma$. As $a$ increases from zero to infinity, $w$ decreases continuously from one to zero. The Riccati residual of $\ell$ is \begin{align}\label{eq:residual-factor} \mathcal R(a) &:=\ell'(a)-a\ell(a)-\ell(a)^2+1\notag\\ &=\frac{(w-1)\{\sigma(2\sigma-1)w+3\sigma-2\}} {2(1-\sigma)(1+w)}. \end{align} To obtain it, substitute $a\ell=(\sigma-\ell^2)/\beta$ into \eqref{eq:root-ode} and clear the positive denominator. The elementary bounds $3<\pi<22/7$ imply $7/11<\sigma<2/3$. Therefore $\sigma(2\sigma-1)>0$, $3\sigma-2<0$, and $\sigma(2\sigma-1)+3\sigma-2=2(\sigma^2+\sigma-1)>0$. There is exactly one $a_0>0$ at which the braces in \eqref{eq:residual-factor} vanish. Consequently \[ \mathcal R(a)<0\quad(00\quad(a>a_0). \] Let $\Delta=m-\ell$ and define \[ J(a)=\exp\left\{-\int_0^a[t+m(t)+\ell(t)]\,dt\right\}. \] Equations \eqref{eq:mean-riccati}--\eqref{eq:root-ode} give $(J\Delta)'=-J\mathcal R$ and $\Delta(0)=0$. On $[0,a_0]$, integration from zero shows $J\Delta\ge0$. On $[a_0,\infty)$, integrate from infinity instead: \[ J(a)\Delta(a)=\int_a^\infty J(t)\mathcal R(t)\,dt\ge0. \] The endpoint is justified because $00. \] If $d>16\delta_Q^{-2}$, then $X$ satisfies the determinant inequality at dilation $\cstar$: for every even log-concave function $f$, \[ \E\det f(X)\ge(\E f(\cstar G))^d. \] \end{theorem} The ellipsoidal bound \eqref{eq:ellipsoid-envelope} already implies $V_X\preceq I_d$: since $B_i=X(e_i)$, one has $B_i^2\preceq Q_{ii}I_d$, hence $V_X\preceq(\tr Q)I_d\preceq I_d$. The proof starts by choosing unit vectors $u_1,\ldots,u_d$ such that \begin{equation}\label{eq:ellipsoid-projections} \frac1d\nlsum_{j=1}^d|\langle u_j,x\rangle| \ge \frac{\sqrt{x^{\mathsf T}Qx}}{\cstar} \qquad(\text{every }x). \end{equation} The strict inequality $m_Q<1$ provides the margin needed to approximate a spherical average by $d$ projections. The following sampling estimate does so for any distribution of directions; it uses only that each direction has unit norm. \begin{lemma}\label{lem:projection-sampling} Let $\rho$ be a probability measure on the unit sphere in $\R^r$. For every integer $d\ge1$, there are unit vectors $u_1,\ldots,u_d$ such that \[ \sup_{\|x\|_2\le1} \left|\frac1d\nlsum_{j=1}^d|\langle u_j,x\rangle| -\int|\langle u,x\rangle|\,d\rho(u)\right| \le\frac4{\sqrt d}. \] \end{lemma} \begin{proof} Let $U_1,\ldots,U_d$ be independent with law $\rho$, and let $\varepsilon_1,\ldots,\varepsilon_d$ be independent uniform signs. Write $h_\rho(x)=\int|\langle u,x\rangle|\,d\rho(u)$. Symmetrization and contraction give \begin{align*} &\E\sup_{\|x\|_2\le1} \left|\frac1d\nlsum_j|\langle U_j,x\rangle|-h_\rho(x)\right|\\ &\qquad\le\frac2d\E\sup_{\|x\|_2\le1} \left|\nlsum_j\varepsilon_j|\langle U_j,x\rangle|\right| \le\frac4d\E\left\|\nlsum_j\varepsilon_jU_j\right\|_2 \le\frac4{\sqrt d}. \end{align*} The first inequality follows by introducing an independent copy of the $U_j$, applying Jensen's inequality, and then using the symmetry of each difference and the triangle inequality. To verify contraction, for any finite family $(A_t,b_t)$ and any $1$-Lipschitz function $\phi$ one has \[ \frac12\sup_t(A_t+\phi(b_t))+ \frac12\sup_t(A_t-\phi(b_t)) \le\frac12\sup_t(A_t+b_t)+\frac12\sup_t(A_t-b_t). \] Combine the two suprema on the left into a supremum over $(s,t)$ and use $\phi(b_s)-\phi(b_t)\le|b_s-b_t|$; interchanging $s$ and $t$ removes the absolute value on the right. Apply this inequality successively to the signs, with $\phi(t)=|t|$. The absolute supremum costs at most a factor of two, since the index set contains $x=0$ and the two signs of the sum have the same law. Finite subsets dense in the unit ball give the stated bound by continuity. Finally, $\E\|\sum_j\varepsilon_jU_j\|_2^2=d$. Some realization has error at most its expectation, proving the claim. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm:ellipsoid-endpoint}] We first construct unit vectors satisfying \eqref{eq:ellipsoid-projections}. Let $v=Q^{1/2}z$ and define a probability law $\rho$ on the unit sphere by \[ \int h(u)\,d\rho(u) =\frac1{m_Q}\E\left[\|v\|_2 h\!\left(\frac v{\|v\|_2}\right)\right]. \] Since $\E|G|=1/\cstar$, this law satisfies \[ \int|\langle u,x\rangle|\,d\rho(u) =\frac{\sqrt{x^{\mathsf T}Qx}}{\cstar m_Q} \ge\frac{\sqrt{x^{\mathsf T}Qx}}{\cstar}+\delta_Q\|x\|_2. \] Strict Jensen gives $m_Q<\sqrt{\tr Q}\le1$. Lemma~\ref{lem:projection-sampling} approximates this average uniformly on the unit ball with error at most $4/\sqrt d<\delta_Q$, since $d>16\delta_Q^{-2}$; homogeneity gives \eqref{eq:ellipsoid-projections}. Now normalize $f(0)=1$ and write $f(t)=e^{-\psi(|t|)}$; the zero function is immediate. The norm bound, \eqref{eq:ellipsoid-projections}, and convexity give \begin{equation*} \tr\psi(|X|/\cstar) \le d\psi\!\left(\frac{\sqrt{g^{\mathsf T}Qg}}{\cstar}\right) \le\nlsum_{j=1}^d\psi(|\langle u_j,g\rangle|). \end{equation*} Each scalar projection is standard Gaussian. Their interval events are symmetric and convex, so the Gaussian correlation inequality and layer cake give \[ \E\det f(X/\cstar) \ge\E\prod_{j=1}^d f(\langle u_j,g\rangle) \ge(\E f(G))^d. \] The inequalities remain valid for extended-valued convex penalties; Lemma~\ref{lem:weight-closure} handles support-endpoint values. \end{proof} \subsection{Radial matrix Gaussian series and compressions}\label{sec:radial-endpoint} The isotropic case of Theorem~\ref{thm:ellipsoid-endpoint} applies to the anticommuting family that forces the lower bound $\cstar$. \begin{prop}[Radial matrix Gaussian series]\label{prop:radial-endpoint} For an integer $r\ge1$, put \[ \delta_r=\frac{\Gamma(r/2)}{\sqrt\pi\,\Gamma((r+1)/2)} -\sqrt{\frac2{\pi r}}>0. \] Suppose a real symmetric matrix Gaussian series $X$ of order $d$ satisfies $|X|=(\chi_r/\sqrt r)I_d$, where $\chi_r$ is the Euclidean norm of its standard Gaussian coefficient vector in $\R^r$. If $d>16\delta_r^{-2}$, then the determinant inequality holds at $\cstar$ for every even log-concave function. In particular, it holds for all sufficiently large members of the anticommuting family of order $d=2^r$ in Section~\ref{sec:sharp-proof}, and $\cstar$ is optimal uniformly over those members. \end{prop} \begin{proof} The Gaussian law has full support, so continuity extends the identity everywhere. Apply Theorem~\ref{thm:ellipsoid-endpoint} with $Q=I_r/r$. The radial integral gives $\E\chi_r=\sqrt2\,\Gamma((r+1)/2)/\Gamma(r/2)$, hence \[ \delta_Q=\frac1{\cstar\E\chi_r}-\frac1{\cstar\sqrt r} =\delta_r=\frac{\sqrt{2/\pi}}{4r^{3/2}}+O(r^{-5/2}), \] where the expansion follows from Stirling's formula. Thus $16\delta_r^{-2}=O(r^3)<2^r$ for all sufficiently large $r$. The anticommuting lower bound in Section~\ref{sec:sharp-proof} proves uniform optimality of $\cstar$. \end{proof} Compressions give nonradial examples. Let $C_1,\ldots,C_r$ be symmetric anticommuting involutions of order $N$, and let $T\colon\R^d\to\R^N$ be an isometry. The matrix Gaussian series with \[ X(x)=T^{\mathsf T}\left(\nlsum_{i=1}^r (Q^{1/2}x)_iC_i\right)T \] satisfies \eqref{eq:ellipsoid-envelope}, because the square of the matrix in parentheses is $(x^{\mathsf T}Qx)I_N$. Compression need not preserve its equal absolute eigenvalues. Theorem~\ref{thm:ellipsoid-endpoint} therefore applies to every such compression whose order $d$ exceeds $16\delta_Q^{-2}$. \subsection{Orthogonal sums and averages of conjugates}\label{sec:orthogonal-sums} Fixed orthogonal blocks give another way to combine the proved cases. If matrix Gaussian series $X_\ell$ with common Gaussian coefficients each satisfy the determinant inequality at dilation $\kappa$, then so does their orthogonal sum $\bigoplus_\ell X_\ell$, even when the blocks are dependent: determinants factor over orthogonal blocks, each factor $\det f(X_\ell/\kappa)$ has symmetric convex superlevel sets in the Gaussian coefficients, and the Gaussian correlation inequality bounds the expectation of the product below by the product of the expectations. Averaging orthogonal conjugates then produces further matrix Gaussian series without requiring an ellipsoidal bound for their operator norm. \begin{prop}[Averages of orthogonal conjugates]\label{prop:block-gsd} If $X$ satisfies the determinant inequality at dilation $\kappa$ and \[ \Phi(M)=\nlsum_{j=1}^N p_jO_jMO_j^{\mathsf T},\qquad p_j\ge0,\quad \nlsum_jp_j=1, \] where the $O_j$ are fixed orthogonal matrices, then $\Phi(X)$ satisfies the determinant inequality at the same dilation. This operation preserves $V_X\preceq I_d$ and cannot increase coefficient rank. \end{prop} \begin{proof} Fix a normalized function $f$ and its penalty $\psi$. Convexity and orthogonal invariance of $M\mapsto\tr\psi(|M|)$ give \[ \tr\psi(|\Phi(M)|/\kappa) \le\nlsum_jp_j\tr\psi(|O_jMO_j^{\mathsf T}|/\kappa) =\tr\psi(|M|/\kappa). \] Thus $\det f(\Phi(M)/\kappa)\ge\det f(M/\kappa)$ pointwise. For a symmetric $B$, writing $M_j=O_jBO_j^{\mathsf T}$ gives \[ \Phi(B^2)-\Phi(B)^2 =\nlsum_jp_j(M_j-\Phi(B))^2\succeq0. \] Summing over coefficients yields $V_{\Phi(X)}\preceq\Phi(V_X)\preceq I_d$ whenever $V_X\preceq I_d$. Linearity of $\Phi$ proves the rank assertion. \end{proof} \subsection{Commuting coefficients and special penalties}\label{sec:special} The shape of the function and the coefficient geometry permit smaller dilations than $13/5$. \begin{theorem}[Commuting coefficients]\label{thm:commuting} Suppose the $B_i$ commute and $V_X\preceq I_d$. Then $\E\det f(X)\ge(\E f(G))^d$ for every even log-concave function $f$. More generally, any even nonnegative Borel function that is nonincreasing on $[0,\infty)$ suffices. The constant one is optimal uniformly over matrix Gaussian series with commuting coefficients. \end{theorem} \begin{proof} Simultaneous diagonalization gives coordinates $Z_j=\langle b_j,g\rangle$ with $\|b_j\|_2\le1$. The preimages of the interval superlevel sets of $f$ are symmetric and convex, including when $b_j=0$. Gaussian correlation and layer cake give \[ \E\prod_{j=1}^d f(Z_j) \ge\prod_{j=1}^d\E f(Z_j) \ge(\E f(G))^d, \] where the last step uses $Z_j\overset{d}=\|b_j\|_2G$ and monotonicity. The diagonal series $\operatorname{diag}(g_1,\ldots,g_d)$ with $f(t)=e^{-t^2}$ rules out any dilation below one. \end{proof} \begin{theorem}[Exact quadratic scale]\label{thm:quadratic} Let $r_X\le d$ and put $\tau=d^{-1}\tr V_X$. If $\tau>0$, then \[ \E e^{-t\tr X^2/\kappa^2}\ge(\E e^{-tG^2})^d \quad\text{for every }t\ge0 \] holds with $\kappa=\sqrt\tau$. This value is the smallest dilation for which the displayed inequality holds for every $t\ge0$ for the given matrix Gaussian series. In particular, $V_X\preceq I_d$ permits $\kappa=1$. \end{theorem} \begin{proof} The Gram matrix $Q=[\tr(B_iB_j)]_{ij}$ has rank $r_X\le d$. Pad its nonzero eigenvalues with zeros to obtain $\lambda_1,\ldots,\lambda_d\ge0$, with $\sum_j\lambda_j=d\tau$. The Gaussian quadratic-form identity and arithmetic--geometric mean give \[ \E e^{-t\tr X^2/\kappa^2} =\prod_{j=1}^d(1+2t\lambda_j/\kappa^2)^{-1/2} \ge(1+2t\tau/\kappa^2)^{-d/2}. \] At $\kappa^2=\tau$ this lower bound is the required scalar benchmark. Differentiating the proposed inequality at $t=0$ shows that any valid dilation must satisfy $\kappa^2\ge\tau$. If $\tau=0$, then $X=0$ and the displayed inequality is immediate for every positive dilation. \end{proof} \subsection{The real two-Pauli matrix Gaussian series} The two-Pauli example distinguishes the optimal dilation for even log-concave functions from the optimal dilation for interval indicators. Comparing a diamond or square with a disk of the same area gives the upper bounds. \begin{theorem}[Optimal constants for the real two-Pauli matrix Gaussian series] \label{thm:pauli} Let \[ Z=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\qquad H=\begin{pmatrix}0&1\\1&0\end{pmatrix},\qquad X=\frac{g_1Z+g_2H}{\sqrt2}. \] The smallest dilation valid simultaneously for all even log-concave functions is $2/\sqrt\pi$. The smallest dilation for all interval indicators is $\sqrt{2/\pi}$. \end{theorem} \begin{proof} Since $ZH=-HZ$ and $Z^2=H^2=I_2$, we have $|X|=RI_2$, where $R=\chi_2/\sqrt2$ and $\chi_2=(g_1^2+g_2^2)^{1/2}$. For the upper bounds, the zero function is immediate; otherwise normalize $f(0)=1$ and apply Lemma~\ref{lem:weight-closure} to obtain a closed penalty $\psi$ with $\E\det f(X/\kappa)=\E e^{-2\psi(R/\kappa)}$. Among planar sets of a fixed finite area, the centered disk has largest standard Gaussian measure. Indeed, if $A$ and the centered disk $B$ have equal area, the Gaussian density on $A\setminus B$ is at most its boundary value on $B$, whereas on $B\setminus A$ it is at least that value. The two differences have equal area, so $\gamma_2(A)\le\gamma_2(B)$. To apply this comparison to all functions, put $S=(|g_1|+|g_2|)/2$ and $c_2=2/\sqrt\pi$. The diamond $\{|g_1|+|g_2|\le2t\}$ has area $8t^2$, equal to that of the disk of radius $2\sqrt{2/\pi}\,t=\sqrt2\,c_2t$. Thus $\Pp(R/c_2\le t)\ge\Pp(S\le t)$ for every $t\ge0$. Since $e^{-2\psi}$ is nonincreasing, this probability comparison and convexity of $\psi$ give \[ \E e^{-2\psi(R/c_2)} \ge\E e^{-2\psi(S)} \ge\E e^{-\psi(|g_1|)-\psi(|g_2|)} =(\E e^{-\psi(|G|)})^2. \] Extended-valued penalties are covered too: their exponentials are bounded and nonincreasing. For interval indicators, the square $[-t,t]^2$ has area $4t^2$, equal to that of the disk of radius $2t/\sqrt\pi$. The disk comparison gives $\Pp(\chi_2\le2t/\sqrt\pi)\ge q(t)^2$, proving the bound at $\kappa=\sqrt{2/\pi}$. For necessity, the density of $R$ is $2r e^{-r^2}$ on $r>0$. For the linear penalty $\psi(t)=at$, differentiation at $a=0$ forces only $\kappa\ge\E R/\E|G|=\pi/(2\sqrt2)$. The stronger restriction comes from large $a$: \begin{equation}\label{eq:pauliasymptotics} \E e^{-2aR/\kappa}\sim\frac{\kappa^2}{2a^2},\qquad (\E e^{-a|G|})^2\sim\frac{2}{\pi a^2} \quad(a\to\infty). \end{equation} Thus $\kappa\ge2/\sqrt\pi$. For an interval indicator of radius $r\downarrow0$, \[ \Pp(R\le\kappa r)=1-e^{-\kappa^2r^2}\sim\kappa^2r^2, \qquad q(r)^2\sim(2/\pi)r^2, \] so $\kappa\ge\sqrt{2/\pi}$. Each asymptotic follows from rescaling and dominated convergence. \end{proof} \end{document}