\documentclass[11pt]{amsart} \usepackage[a4paper,margin=25mm]{geometry} \usepackage[T1]{fontenc} \usepackage{lmodern,amsmath,amssymb,amsthm,mathtools,booktabs,array,mathrsfs} \usepackage{xcolor,microtype,hyperref,fancyhdr,enumitem} \hypersetup{colorlinks=true,linkcolor=blue!40!black,citecolor=blue!40!black, urlcolor=blue!40!black,pdftitle={A two-thirds-power deficit for complete arcs}, pdfauthor={}} \setlength{\parindent}{0pt} \setlength{\parskip}{5pt} \setlength{\headheight}{14pt} \setlist{nosep,leftmargin=*} \newtheorem{theorem}{Theorem}[section] \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \numberwithin{equation}{section} \newcommand{\F}{\mathbf F} \newcommand{\PP}{\mathbf P} \newcommand{\ord}{\operatorname{ord}} \newcommand{\mult}{\operatorname{mult}} \newcommand{\Sing}{\operatorname{Sing}} \newcommand{\Fr}{\operatorname{Fr}} \newcommand{\pa}{p_a} \newcommand{\ceil}[1]{\left\lceil#1\right\rceil} \newcommand{\floor}[1]{\left\lfloor#1\right\rfloor} \pagestyle{fancy}\fancyhf{} \fancyhead[L]{\small Complete arcs over nonsquare binary fields} \fancyhead[R]{\small Duality and degree bounds} \fancyfoot[C]{\thepage} \title{Complete planar arcs over nonsquare binary fields} \author{Jos\'e Felipe Voloch} \address{School of Mathematics and Statistics, University of Canterbury, Christchurch 8140, New Zealand} \email{felipe.voloch@canterbury.ac.nz} \urladdr{https://www.math.canterbury.ac.nz/~f.voloch/} %\date{24 October 2026} \begin{document} \maketitle\thispagestyle{plain} \begin{abstract} Let $m'_2(q)$ be the largest size of a complete arc in $\operatorname{PG}(2,q)$ other than a hyperoval. We prove \[ m'_2(q)\le q-\floor{q^{2/3}-q^{1/3}} \qquad(q=2^h\ge512,\ h\text{ odd}). \] The geometric statement underlying the proof is a degree obstruction for arbitrary geometrically irreducible Frobenius nonclassical plane curves; an origin from an arc is unnecessary. The argument uses the curve and a Frobenius twist of its dual, exact weighted point counts, and a local conductor inequality that allows arbitrary singularities. We give a sharper bound obtained by optimizing over powers of two, determine its leading constant according to $h\bmod3$, and prove further refinements using a characteristic-two normal form. Exact finite computations give, for example, $m'_2(128)\le105$, $m'_2(512)\le452$, and $m'_2(2048)\le1875$. \end{abstract} \section{Introduction} \subsection{AI use} \label{ai} Apart from this subsection (and the first paragraph in the next one), this paper is AI generated. I had some notes making the modest improvement of my old bound $m'_2(q)\le q -\sqrt{2q}+2$ to $m'_2(q)\le q -\sqrt{2q}$, (for $q$ even and nonsquare) following the same strategy as the old result, which I uploaded to ChatGPT 5.6 to check the proof and suggest possible improvements. After some back and forth, ChatGPT generated the argument below using the dual curve, substantially improving the bound. I also had Claude Opus 5 run an audit of the manuscript and included corrections from the output of this audit. \subsection{Statements and conventions} Throughout $PG(2,q)$ denotes the projective plane over the finite field of $q$ elements. An arc is a set of points in $PG(2,q)$ with no three collinear; it is complete if it cannot be enlarged. The quantity $m'_2(q)$ is defined to be the largest cardinality of a complete arc other than the $(q+2)$-arcs, or hyperovals. Segre's classical bound is $m'_2(q)\le q -\sqrt{q}+1$, $q$ even which is tight when $q$ is a square (see, e.g. \cite{BL18} for the full historical development). I improved it to $m'_2(q)\le q -\sqrt{2q}+2$ in \cite{V91} for $q$ nonsquare that had seen no improvement until the present work. Write \begin{equation}\label{eq:kt} k=q+2-t \end{equation} for the size of an arc. Thus an upper bound on $k$ is a lower bound on $t$. All plane curves in the geometric assertions are geometrically irreducible. Their genus and their number $N$ of rational points refer to the smooth projective normalization. We separately write $M_{\rm sm}$ for the number of smooth rational points on a plane model. A plane curve is $q$-Frobenius nonclassical if the tangent at a general point $P$ contains $\Fr_q(P)$; its nontrivial Frobenius order then equals its generic tangent contact order. In characteristic two this order is a power of two. \begin{theorem}\label{thm:main} Suppose $q=2^h\ge512$ and $h$ is odd. Then \begin{equation}\label{eq:main} \boxed{m'_2(q)\le q-\floor{q^{2/3}-q^{1/3}}.} \end{equation} More precisely, put \begin{align} A_\nu(q)&=\frac{q+3\nu+2}{\nu+2},\label{eq:A}\\ B_\nu(q)&=\frac{1-\nu+\sqrt{(\nu-1)^2+4\nu(q+2)}}2,\label{eq:B}\\ T(q)&=\min_{\substack{\nu=2^e\\1\le eq^{2/3}-q^{1/3}+1. \end{equation} If $d\le q/8$ and $\nu$ is its Frobenius order, then the more informative necessary conditions are \begin{equation}\label{eq:scalar} \boxed{(\nu+2)d\ge q+3\nu+2,\qquad d(d+\nu-1)\ge\nu(q+2).} \end{equation} \end{theorem} The curve theorem is useful because it isolates the main geometric work. The passage to arcs uses only the envelope construction and the St\"ohr--Voloch bound. No assumption of ordinary singularities, rational tangent directions at singular points, or irreducibility of the entire envelope is made. \begin{corollary}\label{cor:asymptotic} As $h$ tends to infinity through odd integers, \begin{equation}\label{eq:asymptotic} m'_2(2^h)\le 2^h-c_h2^{2h/3}+O(2^{h/3}), \end{equation} where \[ c_h=\begin{cases} 1,&h\equiv0\pmod3,\\ 2^{1/3},&h\equiv1\pmod3,\\ 2^{1/6},&h\equiv2\pmod3. \end{cases} \] \end{corollary} Section~\ref{sec:refine} proves the additional exact bound \begin{equation}\label{eq:cubeintro} \boxed{m'_2(u^3)\le u^3-u^2+\frac u2, \qquad u=2^{2r+1}\ge8.} \end{equation} Section~\ref{sec:numerical} records finite bounds, including $q=32,128$, where the convenient restriction $t\le q/8$ does not cover the critical range. The comparison with $q-\sqrt{2q}$ in that section refers to the bound that started this investigation. The classical nonsquare bound is $q-\sqrt{2q}+2$ from \cite{V91}, see subsection \ref{ai}. \section{From a complete arc to a curve}\label{sec:envelope} \begin{lemma}\label{lem:envelope} Let $q$ be even and let $\mathcal A$ be a complete $k$-arc other than a hyperoval, with $k=q+2-t$ and $k>t+2$. Its Segre envelope has degree $t$. Every geometric irreducible component $C$ is defined over $\F_q$, has degree $2\le d\le t$, and contains at least $kd$ distinct smooth $\F_q$-points. In particular, $N\ge kd$ on its normalization. \end{lemma} \begin{proof} Use Segre's envelope theorem in the form of \cite[Theorem 11]{BL18}. For each $a\in\mathcal A$, let $L_a$ be its line in the dual plane. The restriction of the degree-$t$ envelope polynomial to $L_a$ has exactly $t$ distinct zeros, the points representing the unisecants through $a$. All these zeros are rational, and all intersections are simple. Consequently the envelope is reduced, and each intersection with any $L_a$ is smooth on the entire envelope. A geometric component of degree $d$ meets each $L_a$ in $d$ such points. The resulting $kd$ points are distinct: a unisecant cannot pass through two points of the arc. At any one of these rational smooth points there is a unique geometric component, so Frobenius fixes that component. It is therefore defined over $\F_q$. A line component would correspond to a rational point $R$ in the original plane such that every line $Ra$, $a\in\mathcal A$, is a unisecant. The point $R$ cannot belong to the arc, since an $L_a$ is not an envelope component. Thus $R$ could be added to the arc, contradicting completeness. Hence $d\ge2$. \end{proof} Let $\lambda$ be the nontrivial Frobenius order of the line series on $C$, using $\lambda=1$ in the Frobenius classical case. St\"ohr--Voloch \cite{SV86}, together with $2g-2\le d(d-3)$, gives \begin{equation}\label{eq:SV} N\le\frac{\lambda(2g-2)+(q+2)d}{2} \le\frac{d\{\lambda(d-3)+q+2\}}2. \end{equation} The normalization convention in this formula is also stated explicitly in \cite[Theorem 1.1]{BH17}. For an envelope component, therefore, \begin{equation}\label{eq:arcSV} \lambda(d-3)\ge q+2-2t. \end{equation} If $3t 1$ and $d(x^{\mu})=0$, gives $da=x^\mu db$. Here $db\ne0$, since otherwise $da=db=0$ would contradict $K=\F_q(a,b)$. In coordinates $(B,A)$ on $C^\dagger$, the tangent line at $(b,a)$ is \[ A+x^\mu B+y^\mu=0. \] Its Gauss field is $\F_q(x^\mu,y^\mu)=K^\mu$, and its image is a coefficient Frobenius twist of $C$. The tangent therefore has generic contact order $\mu$, and its image has degree $d$. The second identity of \eqref{eq:rootFrob} says precisely that $(b^q,a^q)$ lies on this tangent, proving Frobenius nonclassicality. \end{proof} It is the curve $C^\dagger$, with its $\F_q$-isomorphic function field, that is used below. This avoids identifying the original curve with the normalization of its ordinary dual without accounting for the twist. \begin{lemma}[Weighted point identity]\label{lem:weighted} With the preceding notation, \begin{equation}\label{eq:weighted} \sum_{P\in C(\F_q)_{\rm plane}}\mult_P(C) =(q+1)d-\nu D. \end{equation} Consequently \begin{equation}\label{eq:upperN} \boxed{N\le(q+1)d-\nu D,\qquad N\le(q+1)D-\mu d.} \end{equation} \end{lemma} \begin{proof} On the normalization, the sections $X_iX_j^q-X_jX_i^q$ have total degree $(q+1)d$. After removal of their common divisor they define the Gauss map: generically the line joining $P$ and $P^{(q)}$ is the tangent at $P$. The resulting base-point-free system has degree $\nu D$. Their common zeros are exactly the branches centered at rational plane points. At such a center take rational affine coordinates $(u,v)$ vanishing there. The minimum valuation of the cross-product sections is the minimum of the valuations of $u^q-u$ and $v^q-v$, namely $m_\beta=\min\{\ord_\beta u,\ord_\beta v\}$. Summing over the branches at $P$ gives $\mult_P(C)$, proving \eqref{eq:weighted}. A rational point of the normalization contributes at least one to this weighted sum; nonrational branches at rational centers only increase the sum. Hence the first inequality in \eqref{eq:upperN}; apply the same argument to $C^\dagger$ for the second. \end{proof} \section{The conductor and two symmetric degree inequalities}\label{sec:local} Work over an algebraic closure for local computations. If $C:F=0$ is Frobenius non-classical of degree $d$, let $B$ be the common base divisor of its first partial derivatives on the normalization, and set \[ E=\deg B,\qquad \Delta=\pa(C)-g,\qquad \pa(C)=\frac{(d-1)(d-2)}2. \] At each geometric branch $\beta$, choose an affine chart, a uniformizer $z$, and a generic separating linear coordinate $x$, and put \[ \rho_\beta=\ord_z(dx/dz) =\min\{\ord_z(dx/dz),\ord_z(dy/dz)\},\qquad R=\sum_\beta\rho_\beta. \] The plane conductor/different identity is \begin{equation}\label{eq:conductor} E=2\Delta+R. \end{equation} Locally, if $c_\beta$ is the conductor exponent, it says that the polar base valuation is $c_\beta+\rho_\beta$, and $\sum c_\beta=2\Delta$. For an individual branch, choose a separating coordinate \(x\), and let \(f(x,Y)\in k[[x]][Y]\) be the monic distinguished polynomial defining its completed local ring, obtained by formal Weierstrass preparation. Then $\ord(f_y)=2\delta_\beta+\ord(dx/dz)$; intersections with the other branches supply the remaining conductor contributions. These are identities of differents and conductors, valid also for wild projections. See \cite[Theorem 1.3 and Section 3]{NU} for the arbitrary-characteristic conductor/Jacobian formulation. (Suggested correction from Claude: Theorem 1.3 of \cite{NU} is ``$X$ is a local complete intersection iff $\widehat{\mathscr{I}}(\mathscr{O}_X)=\mathfrak{C}_X$'' , an lci characterization, not a conductor/Jacobian formula. What this section needs is $\operatorname{Jac}X\cdot\mathcal{O}{X'}=\mathfrak{C}_X\cdot\operatorname{Jac}(X'/X)$, which appears there only as the intermediate equality (3.8.1). Citing the theorem as stated implies you need the hard direction; you need the trivial one (a plane curve is a hypersurface, hence lci), and \cite{NU} dispatch that direction in one line, by appeal to Remark 2.6(1) with the parenthetical "which is still true in our case" plus Lemma 3.11. An alternative reference is \cite{K}.) The polar system and \eqref{eq:conductor} give \begin{equation}\label{eq:class} \nu D=d(d-1)-E=2g-2+2d-R. \end{equation} Use $E^\dagger,R^\dagger,\Delta^\dagger$ for $C^\dagger$; then \begin{equation}\label{eq:dualclass} \mu d=D(D-1)-E^\dagger=2g-2+2D-R^\dagger. \end{equation} In particular, \begin{equation}\label{eq:polar} \nu D\le d(d-1),\qquad \mu d\le D(D-1). \end{equation} \begin{lemma}[Local genus loss]\label{lem:local} Every individual geometric branch satisfies \[ 2\delta_\beta\ge(\nu-1)\rho_\beta. \] Consequently \begin{equation}\label{eq:local} (\nu-1)R\le2\Delta. \end{equation} This includes $\nu=2$. \end{lemma} \begin{proof} The inseparability of the Gauss map gives, locally, $y=A(z)^\nu+xB(z)^\nu$. By the choice of $x$, $B^\nu=y'/x'$ is regular, and then $A$ is regular as well. Center $x,y$ at the branch point. For every nonidentity embedding $\sigma$ of the separable extension $\overline{\F}_q((z))/\overline{\F}_q((x))$ into its normal closure, \[ y-\sigma y=(A-\sigma A)^\nu+x(B-\sigma B)^\nu. \] Differences of regular power series are divisible by $z-\sigma z$ in the integral closure. Thus \[ v(y-\sigma y)\ge\nu\,v(z-\sigma z). \] Sum over the embeddings, with the valuation normalized by $v(z)=1$. The product of the conjugate differences of $y$ is $f_y$, up to a unit, for a Weierstrass equation of the branch. The corresponding product for $z$ generates the different of $\overline{\F}_q[[z]]$ over $\overline{\F}_q[[x]]$, of valuation $\rho_\beta$. These rings are monogenic for this purpose by Weierstrass preparation. Therefore \[ 2\delta_\beta+\rho_\beta=v(f_y)\ge\nu\rho_\beta. \] Finally $\sum_\beta\delta_\beta\le\Delta$, since intersections of distinct branches make additional nonnegative contributions to $\Delta$. \end{proof} \begin{proposition}\label{prop:symmetric} The degrees of $C$ and $C^\dagger$ satisfy \begin{align} d(q+2\nu+1-d)&\le\nu(\nu+1)D,\label{eq:localC}\\ D(q+2\mu+1-D)&\le\mu(\mu+1)d.\label{eq:localD} \end{align} \end{proposition} \begin{proof} Substitute $R=2g-2+2d-\nu D$ into \eqref{eq:local}. Rearrangement gives \[ \nu(2g-2)\le d(d-2\nu-1)+\nu(\nu-1)D. \] By $R^\dagger\ge0$ and \eqref{eq:dualclass}, \[ \nu(2g-2)\ge qd-2\nu D. \] Combining proves \eqref{eq:localC}; interchange the two curves for \eqref{eq:localD}. \end{proof} \begin{lemma}[Smooth rational points]\label{lem:lower} For $C$, \begin{equation}\label{eq:lower} N\ge M_{\rm sm}\ge d(q-d+2). \end{equation} The same statement holds for $C^\dagger$, with $D$ in place of $d$. \end{lemma} \begin{proof} The weighted point identity and the polar degree formula give the exact identity \begin{align*} M_{\rm sm} &= (q+1)d-\nu D -\sum_{P\in\Sing(C)(\F_q)}\mult_P(C)\\ &= d(q-d+2)+2\Delta+R -\sum_{P\in\Sing(C)(\F_q)}\mult_P(C). \end{align*} At every singular point of multiplicity $m\ge2$, $2\delta_P\ge m(m-1)\ge m$. Thus the correction is nonnegative. Every smooth rational plane point lifts to a rational normalization point. \end{proof} The lower bound is the one supplied by Borges--Homma \cite[Theorem 1.3]{BH17}; the proof above records it directly from the identities already needed here. The distinction between smooth plane points, weighted plane points, and normalization points is essential. Combining \eqref{eq:SV} with \eqref{eq:lower} gives \begin{equation}\label{eq:scalarfirst} \nu(d-3)\ge q+2-2d, \quad\text{equivalently}\quad (\nu+2)d\ge q+3\nu+2. \end{equation} Applying \eqref{eq:lower} to $C^\dagger$ and \eqref{eq:upperN} to $C$ also gives \begin{equation}\label{eq:H} H(D)\le0,\qquad H(X)=X(q+\nu+2-X)-(q+1)d. \end{equation} \section{The large dual and the second scalar inequality}\label{sec:large} \begin{lemma}[Large dual]\label{lem:large} Let $q=2^h$ with $h$ odd. If $C$ is Frobenius nonclassical and $d\le q/8$, then \begin{equation}\label{eq:large} \boxed{D>q+\nu+2-2d.} \end{equation} \end{lemma} \begin{proof} We have \[ H(2d)=d(q+2\nu+3-4d)>0. \] The quadratic $H$ is symmetric about $(q+\nu+2)/2$, so \eqref{eq:H} implies \begin{equation}\label{eq:twoalternatives} D<2d\quad\text{or}\quad D>q+\nu+2-2d. \end{equation} Suppose $D<2d$, and set $r=D/d$. From the lower bound for $C$ and the upper bound for $C^\dagger$, \[ r\ge\frac{q-d+2+\mu}{q+1}>\frac78. \] Since $D<2d\le q/4$, \eqref{eq:H} gives \[ r\le\frac{q+1}{q+\nu+2-D}<\frac43. \] Because $\nu\mu=q=2^h$ and $h$ is odd, at least one of $\nu^2\le q/2$ and $\mu^2\le q/2$ holds. If $\nu^2\le q/2$, then \eqref{eq:localC} implies \[ q+1-d\le r\nu^2+(r-2)\nu<\frac{2q}{3}, \] whereas $q+1-d\ge7q/8+1$. If $\mu^2\le q/2$, then \eqref{eq:localD} implies \[ r(q+1-D)\le\mu^2+(1-2r)\mu\le\frac q2. \] But its left side is greater than $(7/8)(3q/4)=21q/32$. Both alternatives are impossible. This excludes the first case of \eqref{eq:twoalternatives}. \end{proof} This is precisely where the nonsquare hypothesis enters. For square $q$, the balanced case $\nu=\mu=\sqrt q$ is available, and the preceding dichotomy fails. \begin{proposition}\label{prop:secondscalar} Under the hypotheses of Lemma~\ref{lem:large}, \begin{equation}\label{eq:secondscalar} \boxed{d(d+\nu-1)\ge\nu(q+2).} \end{equation} \end{proposition} \begin{proof} Set $U=d(d-1)/\nu$. By \eqref{eq:polar}, $U\ge D$. Lemma~\ref{lem:large} puts $D$ strictly to the right of the vertex of $H$, since $d\le q/8$. The quadratic is decreasing there, so $H(U)\le H(D)\le0$. Direct factorization gives \begin{equation}\label{eq:factor} H(U)=\frac{d(d-1-\nu)}{\nu^2} \bigl\{\nu(q+2)-d(d+\nu-1)\bigr\}. \end{equation} The first factor is positive. Indeed, a generic tangent gives $d\ge\nu+1$; equality would imply $D\le U=d$, contradicting Lemma~\ref{lem:large}. The expression in braces is therefore nonpositive, as required. \end{proof} Equations \eqref{eq:scalarfirst} and \eqref{eq:secondscalar} prove the scalar assertions in Theorem~\ref{thm:curve}. Their dominant terms are $d\gtrsim q/\nu$ and $d\gtrsim\sqrt{q\nu}$. Balancing them gives $\nu\asymp q^{1/3}$ and $d\asymp q^{2/3}$. \section{Uniform bound, optimization, and the arc conclusion}\label{sec:optimization} \subsection{Eliminating the Frobenius order} \label{frob} Assume $d\le q/8$. The two scalar inequalities give \[ \frac{q+2-2d}{d-3}\le\nu \le\frac{d(d-1)}{q+2-d}. \] The denominators are positive: \eqref{eq:scalarfirst} forces $d>3$. Multiplication and simplification yield \begin{equation}\label{eq:cubic} P_q(d):=d^3-6d^2+3(q+3)d-(q+2)^2\ge0. \end{equation} For real $x$, \[ P'_q(x)=3\bigl((x-2)^2+q-1\bigr)>0. \] Put $Q=q^{1/3}$. An exact substitution gives \begin{equation}\label{eq:cubiceval} P_q(Q^2-Q+1)=-Q^2(3Q^2-4Q+3)<0. \end{equation} Thus \begin{equation}\label{eq:strictd} d>Q^2-Q+1. \end{equation} For $q\ge512$, we have $Q\ge8$ and $q/8\ge Q^2>Q^2-Q+1$. Therefore \eqref{eq:strictd} also holds automatically if $d>q/8$. This proves the uniform curve bound globally. \subsection{The exact optimization over powers of two} For a fixed $\nu$, the two scalar inequalities are exactly $d\ge A_\nu(q)$ and $d\ge B_\nu(q)$. Taking ceilings and minimizing gives $d\ge T(q)$ whenever $d\le q/8$. To extend this conclusion to every $d$, observe that \begin{equation}\label{eq:Tsmall} T(q)\le q/8\qquad(q\ge512). \end{equation} Indeed, choose $\nu=8$ and $d=q/8$. The first scalar inequality is $q/4\ge26$. The second, after multiplication by $64$, is $q^2-456q-1024\ge0$, which holds for $q\ge512$. Here $q/8$ is an integer. The conclusion is then immediate also for $d>q/8$. This completes Theorem~\ref{thm:curve}. \subsection{Deduction for complete arcs} Let $\mathcal A$ be a complete arc other than a hyperoval, and put $t=q+2-|\mathcal A|$. If $t>q/8$, then both conclusions of Theorem~\ref{thm:main} follow from \eqref{eq:Tsmall} and $q/8>Q^2-Q+1$. If $t\le q/8$, Lemma~\ref{lem:envelope} applies, and \eqref{eq:arcSV} excludes Frobenius classical components. Choose any component. It is Frobenius nonclassical and $d\le t$, so Theorem~\ref{thm:curve} gives \[ t\ge T(q),\qquad t>Q^2-Q+1. \] Since $t$ is an integer, the second inequality is $t\ge\floor{Q^2-Q}+2$. Substitution into $k=q+2-t$ proves \eqref{eq:main} and \eqref{eq:optimized}. \subsection{Leading constants} An adjacent power of two gives $T(q)=O(Q^2)$, where $Q=q^{1/3}$. The two scalar inequalities then force every minimizing $\nu$ to be bounded above and below by fixed positive multiples of $Q$; this follows from the two displayed bounds on $\nu$ at the beginning of subsection \ref{frob}. Uniformly for such $\nu$, \[ \max\{A_\nu(q),B_\nu(q)\} =\max\{q/\nu,\sqrt{q\nu}\}+O(Q). \] The expression on the right decreases up to $\nu=Q$ and increases afterward. Thus its minimum on the powers of two is attained at a power adjacent to $Q$. Powers outside these adjacent choices cannot lower the leading term. Write $h=3a+j$. \[ \begin{array}{c|c|c} j&\text{relevant values of }\nu& \min\max\{q/\nu,\sqrt{q\nu}\}/q^{2/3}\\\hline 0&2^a&1\\ 1&2^a,\ 2^{a+1}&2^{1/3}\\ 2&2^{a+1}&2^{1/6} \end{array} \] This proves Corollary~\ref{cor:asymptotic}. The two values in the middle row tie at leading order; retaining the exact formulas resolves the rounding and lower-order terms. \section{Normal forms and sharper exact families}\label{sec:refine} The following criterion is independent of finite-field Frobenius nonclassicality. It sharpens the sufficient condition $Eq-2d$. If $\nu\le u/2$, then \[ (\nu+2)d<(u/2+2)u^2\le3u^3/42u(q-2d)\ge\tfrac32u^4>d^2, \] contradicting \eqref{eq:polar}. Hence $\nu=u$. The second scalar inequality excludes $d\le u^2-u/2$, since at $d_0=u^2-u/2$ its left side is \[ d_0(d_0+u-1)=u^4-\tfrac54u^2+\tfrac12u0. \] Since $D$ is also on the decreasing side and $H(D)\le0$, it follows that $D>L$, hence $D\ge q-d+2$. Consequently \[ E=d(d-1)-uD\le\tfrac34u^2-\tfrac32uq-2d$ and $\nu D\le d(d-1)4u^4. \] For $d\le2u^2-2u$ the left side is at most $4u^4-4u^2$. Therefore $\nu=u$, and the first scalar inequality forces $d=T_u$. Let $\sigma=(u+2)d-(q+3u+2)$. It equals $2$ when $u=16$ and $20$ when $u\ge64$, so $\sigma