Affine Bernstein rigidity: why dimension ten is the first failure

Source and verification

This article explains family 353 of the October 6, 2026 OpenAI collection. Research conclusions are attributed to its manuscripts. The review covers statements, introductions, and selected proof mechanisms; it is not an independent verification of the complete proofs.

The affine Bernstein problem asks whether a complete convex hypersurface that is stationary for affine area must be an elliptic paraboloid. Two manuscripts in this family place a dimension boundary between rigidity and smooth counterexamples. The first proves rigidity for Euclidean-complete graphs in dimensions three through nine; the second constructs a smooth, strictly convex, nonquadratic entire graph in dimension ten. (OpenAI, 2026) (OpenAI, 2026a)

The prerequisites are multivariable calculus, convexity, positive definite matrices, ordinary differential equations, and the meaning of an interior estimate for an elliptic PDE. We introduce the geometric terminology below. The counterexample offers a relatively concrete entry point: its fourth-order PDE reduces to a radial system whose global existence follows from a three-dimensional invariant box.

🏷️ Convex graphs and affine area

Let $\Omega \subset {\mathbb{R}}^n$ be an open convex set, and let $u:\Omega \to \mathbb{R}$ be smooth with positive definite Hessian $D^{2}u$. Its graph is an $n$-dimensional hypersurface in ${\mathbb{R}}^{n+1}$. The phrase locally uniformly convex means that the Hessian has a positive lower bound on each compact subset. It does not require one positive lower bound on all of $\Omega$.

Ordinary graph area uses the integrand $\sqrt{1+|Du{|}^2}$. The affine area functional instead uses

$$\begin{array}{c}\mathcal{A}(u)={\int }_{\Omega }(\det D^{2}u{)}^{\delta }dx,\\ \delta =\frac{1}{n+2}.\end{array}$$

Here stationarity is tested against smooth compactly supported variations. Introduce the cofactor matrix $U$ and a determinant power $w$:

$$\begin{array}{rl}U& =(\det D^{2}u)(D^{2}u{)}^{-1},\\ w& =(\det D^{2}u{)}^{-(n+1)/(n+2)}.\end{array}$$

The cofactor matrix of a Hessian is divergence free: ${\partial }_i{U}^{ij}=0$. This identity follows from the symmetry of third derivatives and the alternating determinant expansion. Differentiating affine area in the direction $\eta$ gives $\delta \int w{U}^{ij}{\eta }_{ij}$, with repeated indices summed. Two integrations by parts therefore give the classical affine maximal equation

$$U^{ij}{w}_{ij}=0.$$

The equation is fourth order in $u$, although it is second order and elliptic in the auxiliary quantity $w$. A positive definite quadratic polynomial has constant Hessian and constant $w$, so every elliptic paraboloid is a solution. The rigidity question is whether completeness forces this elementary example to be the only one.

Two metrics must be distinguished:

$$\begin{array}{rl}{g}_E& =I+Du\otimes Du,\\ g_{\mathrm{aff}}& =(\det D^{2}u{)}^{-1/(n+2)}{D}^{2}u.\end{array}$$

The first is the Euclidean metric induced on the graph. The second is the Berwald–Blaschke, or affine, metric. Completeness refers to the chosen metric. Every smooth entire graph is Euclidean complete: a Cauchy sequence in graph distance projects to a Euclidean Cauchy sequence, and near its limiting base point the graph metric is comparable to the ordinary metric. A graph over a proper domain can also be Euclidean complete when its height diverges sufficiently strongly at the boundary.

🏷️ The two conclusions and their hypotheses

Rigidity in dimensions three through nine

The rigidity manuscript proves the following. Let $3\le n\le 9$ and let $u$ be smooth and locally uniformly convex on a nonempty open convex domain $\Omega \subset {\mathbb{R}}^n$. If $u$ solves the classical affine maximal equation and its graph is complete for $g_E$, then $\Omega ={\mathbb{R}}^n$ and $u$ is a positive definite quadratic polynomial. No growth bound or completeness assumption for $g_{\mathrm{aff}}$ is imposed. (OpenAI, 2026b)

Thus the conclusion includes extension of the domain to all of Euclidean space. It is stronger than a classification stated only for functions already assumed entire.

A smooth dimension-ten counterexample

The construction manuscript produces $u\in C^{\infty }({\mathbb{R}}^{10})$ with $D^{2}u>0$ everywhere, solving the same classical equation, which is not quadratic. Its graph is consequently Euclidean complete. The paper asserts neither a uniform global lower bound on $D^{2}u$ nor completeness for the affine metric. (OpenAI, 2026a)

The word “dimension” counts independent variables: the counterexample is a hypersurface in ${\mathbb{R}}^{11}$. The older singular model recalled by the manuscript is $\sqrt{|x{|}^9+t^2}$, with $x\in {\mathbb{R}}^9$. The new construction replaces that singular profile by one smooth and strictly convex through $x=0$.

The manuscripts recall the earlier Euclidean-complete surface theorem of Trudinger and Wang, which supplies the two-dimensional case. These results make ten the first failure beyond the rigidity range covered here. The dimension-ten construction by itself does not assert examples in every larger dimension. Adding an extra quadratic variable changes the dimension-dependent exponent in the classical equation, so such a propagation would require a separate argument.

🏷️ A separated construction in one extra variable

Write $n=m+1$ and seek a function of the form

$$\begin{array}{c}u(x,t)=\sqrt{F(x{)}^2+t^2},\\ x\in {\mathbb{R}}^m.\end{array}$$

Assume $F>0$ and $H=D_x^{2}F>0$. As an identity of quadratic forms,

$$D^{2}u=\frac{F}{u}H+\frac{F^2}{u^3}{\left(dt-\frac{t}{F}dF\right)}^2.$$

For a nonzero vector with nonzero $x$ component the first term is positive; for a purely vertical vector the second is positive. Hence the full Hessian is positive definite. Using the basis $e_i+(t/F)F_i{e}_t,e_t$, whose determinant is one, also gives

$$\det D^{2}u=F^{m+2}{u}^{-m-3}\det H.$$

Introduce a second positive function $P$ and set

$$\begin{array}{c}k=m+2,\gamma =\frac{m+3}{m+2},\\ \det H=F{P}^{-\gamma },\\ \mathrm{tr}(H^{-1}{D}_x^{2}P)+\frac{kP}{F}=0.\end{array}$$

These two equations imply the affine maximal equation for $u$. To see the cancellation, put $\zeta =t/F$ and $h(\zeta )=(1+{\zeta }^2{)}^{k/2}$. The determinant formula gives $w=Ph(\zeta )$. In the adapted basis, taking the inverse-Hessian trace reduces the equation to the second displayed equation because

$$(1+{\zeta }^2)h^{\prime \prime }-m\zeta h^{\prime }=kh.$$

The choice $k=m+2$ is exactly what makes the right side a constant multiple of $h$. This calculation explains why the auxiliary determinant equation and trace equation were chosen together. (OpenAI, 2026a)

Radial equations and smoothness at the origin

Take $F$ and $P$ radial, put $r=|x|$, and write $y=F^{\prime }$. The eigenvalues of $D_x^{2}F$ are $y^{\prime }$ in the radial direction and $y/r$ in the other $m-1$ directions. The system becomes

$$\begin{array}{rl}{y}^{m-1}{y}^{\prime }& =r^{m-1}F{P}^{-\gamma },\\ (y^{m-1}{P}^{\prime }{)}^{\prime }& =-k{r}^{m-1}{P}^{1-\gamma },\\ F^{\prime }& =y.\end{array}$$

For the counterexample, $m=9$, $k=11$, and $\gamma =12/11$. Starting this system at $r=0$ needs care: a standard ODE theorem away from zero does not ensure smoothness of a radial function at the origin.

The manuscript uses $s=r^2/2$, writes $F=f(s)$ and $P=p(s)$, and prescribes $f(0)=p(0)=1$. With $q=y/r$, integration of the two divergence equations yields

$$\begin{array}{rl}q(s{)}^m& =m{\int }_0^1{v}^{m-1}{F}_s(v)dv,\\ F_s(v)& =f(v^{2}s)p(v^{2}s{)}^{-\gamma },\\ j(s)& ={\int }_0^1{v}^{m-1}p(v^{2}s{)}^{1-\gamma }dv,\\ f^{\prime }(s)& =q(s),\\ p^{\prime }(s)& =-kj(s)q(s{)}^{1-m}.\end{array}$$

Here $j=-y^{m-1}{P}^{\prime }/(k{r}^m)$ for $r>0$. On a small interval around $s=0$, these formulas define a contraction for $(f,p)$ near $(1,1)$: the integrands are Lipschitz while $f,p$ remain in a fixed positive interval, and the final integration in $s$ supplies the small factor. Differentiating the integral formulas bootstraps the fixed point to smoothness. Working on both sides of $s=0$ makes $f(|x{|}^2/2)$ and $p(|x{|}^2/2)$ smooth functions of $x$.

At the origin, $q(0)=1$ and $j(0)=1/m$. Thus $D^{2}F(0)=I$ and, when $m=9$, $D^{2}P(0)=-(11/9)I$. Initially $F,P,y,y^{\prime }$ are positive and $P^{\prime }<0$. The remaining task is to prevent $P$ from vanishing, or the other quantities from becoming singular, at a finite radius.

🏷️ The invariant box and global continuation

The useful variables are logarithmic derivatives:

$$\begin{array}{c}A=\frac{ry}{F},B=\frac{r{y}^{\prime }}{y},\\ C=-\frac{r{P}^{\prime }}{P},D=\frac{C}{B}.\end{array}$$

Let a dot denote differentiation in $\log r$. Direct substitution into the radial equations gives

$$\begin{array}{rl}\dot{A}& =A(1+B-A),\\ \dot{B}& =B\left(9+A+\frac{12}{11}BD-9B\right),\\ \dot{D}& =(11-D)A-8D\\ & +BD\left(1-\frac{D}{11}\right).\end{array}$$

The smooth initial solution approaches $(A,B,D)=(0,1,0)$ as $r\downarrow 0$. The key region is

$$\mathcal{B}=\left[0,\frac{9}{2}\right]\times \left[0,\frac{7}{2}\right]\times \left[0,\frac{33}{7}\right].$$

Its upper corner is stationary. Moreover, each component of the vector field is nondecreasing in the other two coordinates throughout this box. For example, the nonzero cross derivatives of the third component are $11-D$ and $D(1-D/11)$, both nonnegative because $0\le D\le 33/7<11$.

On an upper face, increasing the other coordinates to their upper values can only increase the outward component; at the corner that component is zero. Therefore it is nonpositive on the whole face. On the three lower faces the relevant components are respectively $0$, $0$, and $11A$, so they point inward or tangent to the box. This proves forward invariance. To handle weak inequalities rigorously, the source applies Grönwall’s inequality to squared distance from the box, using the local Lipschitz property of the vector field. A strict inward-pointing condition is unnecessary. (OpenAI, 2026a)

In particular,

$$\begin{array}{c}0<A\le \frac{9}{2},0<B\le \frac{7}{2},\\ 0<C=BD\le \frac{33}{2}.\end{array}$$

Suppose the maximal radial solution ended at some finite $R$. Fix a positive $r_0<R$. Integrating these logarithmic bounds on $[r_0,R)$ gives

$$\begin{array}{rl}1& \le \frac{F(r)}{F(r_0)}\le (R/r_0{)}^{9/2},\\ 1& \le \frac{y(r)}{y(r_0)}\le (R/r_0{)}^{7/2},\\ (r_0/R{)}^{33/2}& \le \frac{P(r)}{P(r_0)}\le 1.\end{array}$$

Also $|P^{\prime }|\le (33/2)P(r_0)/r_0$. Away from $r=0$, the equations form a regular first-order system for $(F,y,P,P^{\prime })$ whenever $F,y,P>0$. The bounds keep this state in a compact subset of that regular region. The ordinary continuation theorem therefore extends the solution past $R$, a contradiction.

We obtain global positive $F,P$ with $D^{2}F>0$. The separated construction then yields the promised entire affine maximal graph. It is visibly nonquadratic on the vertical line:

$$\begin{array}{rl}u(0,t)& =\sqrt{1+t^2},\\ u_{tt}(0,t)& =(1+t^2{)}^{-3/2}.\end{array}$$

This also directly demonstrates why a global positive Hessian lower bound is unavailable. The invariant-box argument proves global existence; it does not require or establish convergence of the trajectory to the stationary corner, nor does it establish completeness of the affine metric.

🏷️ Tangent sections and the rigidity mechanism

We now return to $3\le n\le 9$. A tangent section measures the height above a tangent plane. For a base point $a$, define

$$\begin{array}{rl}{\ell }_a(x)& =u(a)+Du(a)\cdot (x-a),\\ S_u(a,t)& =\{x:u(x)<{\ell }_a(x)+t\},\\ K_a(t)& =\overline{S_u(a,t)}-a.\end{array}$$

The central geometric conclusion is a single balance constant $\rho >0$ working at every base point and every height:

$$-\rho K_a(t)\subset K_a(t).$$

This says that a section cannot stretch arbitrarily far in one direction relative to the opposite direction measured from its tangent point. It is much stronger than positivity of the Hessian at each individual point. The rigidity manuscript derives this global balance from completeness and stationarity. (OpenAI, 2026b)

Compact caps and additional recession directions

First, Euclidean completeness makes tangent sections compact inside $\Omega$. If $u$ stayed bounded along points approaching a finite boundary point, convexity would give a lifted line segment with finite length approaching the missing boundary, contradicting completeness. Local positive definiteness near $a$, followed by convexity along rays, also gives a positive linear lower bound for $u-{\ell }_a$ outside a small ball. This prevents sections from escaping to infinity.

Consider the closed epigraph $\{(x,z):z\ge u(x)\}$ and all full-dimensional local limits of its invertible affine images. Such limits may have flat faces, so differential identities will be applied to smooth approximating hypersurfaces, not directly to a potentially nonsmooth limit.

The source organizes these limits by recession directions. A $k$-direction model has coordinates $(s,y)$ with $s\in {\mathbb{R}}^k$ and $y\in {\mathbb{R}}^{n+1-k}$, lies in $s_i\ge 0$, contains the nonnegative coordinate orthant with $y=0$, and has compact caps cut out by $\sum s_i\le T$. Its transverse fibers are convex bodies $K_s$. Moving $s$ coordinatewise upward enlarges the fiber. Tangent-plane coordinates provide at least one such direction.

Choose the largest possible $k$. A compactness argument shows that fibers of all maximal models are uniformly centered at their distinguished origin. Otherwise normalize a badly centered fiber by its John ellipsoid, pass to a limit where the origin lies on a supporting face, and magnify thin slices next to that face. This produces an additional recession direction, contradicting maximality. John normalization here means using a maximal inscribed ellipsoid to bound the ratio between an inner and outer Euclidean ball after a linear map; the map must preserve the distinguished origin.

There is a necessary compactness detail: normalizing one thin slice must control the entire cap. The source obtains that control by contracting any point in a cap toward the normalized slice and interpolating with a fixed point on the new axis. This bounds all transverse coordinates by a constant times the cap height. Consequently the further affine limit exists with compact caps. The centering constant is uniform over models and their coordinates, which will eventually give one $\rho$ for all tangent sections.

The analytic work is to exclude maximal models with $k\ge 2$.

🏷️ A dimension-dependent energy inequality

For $2\le k\le n$, put $m=n+1-k$. Parametrize a smooth positive fiber by its support function $h(s,e)$, where $e\in S^{m-1}$ is the outward normal direction in the transverse variables. Set

$$\begin{array}{rl}B& =-D_s^{2}h,\\ R& ={\nabla }_S^{2}h+hI,\\ a& =(\det B/\det R{)}^{1/(n+2)},\\ f& =\log (a/h).\end{array}$$

The matrices $B,R$ are positive definite on these smooth tube charts. For $m=1$, the sphere $S^0$ consists of the two endpoints of an interval; the angular matrix is empty and its determinant is one. Thus this endpoint case is included.

Two densities encode the geometry:

$$\begin{array}{rl}d\mathcal{A}& =(\det B{)}^{\delta }(\det R{)}^{1-\delta }dsd\omega ,\\ d\mu & =h^{-n/2}\sqrt{\det B\det R}dsd\omega .\end{array}$$

The second is the volume density of the metric with blocks $B/h$ and $R/h$. Let $p=D_s\log h$, $d=D_{s}f$, and $d_e={\nabla }_{S}f$, with norms and pairings taken using the corresponding inverse metric blocks. Weighted integration by parts in the affine-area equation produces an energy whose integrand is

$$\begin{array}{rl}{\mathcal{F}}_{\lambda }=& k-2\lambda \\ & +(n+1-2\lambda )|p{|}^2\\ & +\frac{n+2}{2}⟨p,d⟩\\ & +\lambda (|d{|}^2+|d_e{|}^2).\end{array}$$

The paired quadratic form in $p,d$ has matrix

$$M_{\lambda }=\left(\begin{array}{cc}n+1-2\lambda & (n+2)/4\\ (n+2)/4& \lambda \end{array}\right).$$

At $\lambda =1$, its determinant is $(n-2)(10-n)/16$. This identifies the critical dimension, but the scalar term $k-2$ vanishes when $k=2$. The source instead chooses $\lambda =15/16$, for which

$$\begin{array}{rl}k-2\lambda & \ge \frac{1}{8},\\ \det M_{\lambda }& =\frac{-8n^2+88n-137}{128}\\ & \ge \frac{7}{128}(3\le n\le 9).\end{array}$$

The diagonal entries are positive as well. Applying Young’s inequality to the cutoff terms absorbs the mixed derivatives and bounds the resulting positive energy by cutoff-gradient energy. Thus the dimension restriction is a specific coercivity calculation within a larger geometric argument. Failure of this calculation at dimension ten would not itself construct a counterexample; that is the role of the separate radial manuscript. (OpenAI, 2026b)

Polynomial mass growth versus exponential mass growth

Pass to logarithmic coordinates ${\alpha }_i=\log s_i$ and define

$$d\sigma =\left(1+\sum _{i=1}^k|D_s\log s_i{|}^2\right)d\mu .$$

The weighted energy estimate implies, for compactly supported cutoffs $v(\alpha )$,

$$\int v^{2}d\sigma \le C_n\int |{\nabla }_{\alpha }v{|}^{2}d\sigma .$$

The gradient on the right is Euclidean in the $\alpha$ variables. This change is useful because multiplying $s_i$ by a positive constant translates ${\alpha }_i$, while the geometric measure is unchanged under the corresponding coordinate normalization.

The source supplies two estimates needed to turn this inequality into a contradiction. First, a fixed inner box has a uniformly positive amount of mass. A small convex quadratic replacement inside a cap, together with concavity of the affine-area integrand and stationarity, gives positive cap area. Stretching a strip next to a face shows that its area tends to zero with its thickness. Therefore the positive mass cannot all disappear onto the coordinate faces. This argument also excludes the endpoint $k=n+1$, where no transverse direction remains.

Second, every unit logarithmic cell has uniformly bounded $\sigma$ mass after normalizing its central fiber. The affine-area first variation against a cap-supported displacement toward an interior point controls inverse-Hessian terms. Bounds on the gradient images of convex support functions control the complementary determinant factors. Hölder’s inequality combines these estimates. Uniform fiber centering is what makes their constants independent of the cell’s location.

Write $Q_L=\{|\alpha {|}_{\infty }\le L\}$. Covering by unit cells gives polynomial growth,

$$\sigma (Q_L)\le C(L+1{)}^k.$$

On the other hand, insert a cutoff equal to one on $Q_{\ell }$, zero outside $Q_{\ell +1}$, with bounded gradient in the intervening shell. The energy inequality gives

$$\begin{array}{rl}\sigma (Q_{\ell })& \le C_{\ast }(\sigma (Q_{\ell +1})-\sigma (Q_{\ell })),\\ \sigma (Q_{\ell +1})& \ge (1+C_{\ast }^{-1})\sigma (Q_{\ell }).\end{array}$$

Starting from the positive inner mass, iteration forces exponential growth, contradicting the polynomial upper bound.

All these estimates are used on smooth approximants. The correct order is to choose a finite $L$ large enough for the two growth laws to conflict, then choose one approximant sufficiently far along for every cell and cutoff in that finite argument. No differentiation of the limiting convex set and no interchange of infinite limits is needed. Thus the maximal recession count is one, and uniform centering gives the desired balance of every tangent section.

🏷️ From section balance to a quadratic polynomial

The final argument explains why the global section estimate is strong enough for rigidity. Restrict $u$ to a line from $a$ and subtract its tangent:

$$\begin{array}{rl}g(r)=& u(a+re)-u(a)\\ & -rDu(a)\cdot e.\end{array}$$

Apply balance to a section based at $a+r_{0}e$ that just reaches $a$. It also reaches the point at parameter $(1+\rho )r_0$, giving

$$g((1+\rho )r_0)\le (1+\rho )r_0{g}^{\prime }(r_0).$$

Convexity bounds $g^{\prime }(r_0)\le g(R)/(R-r_0)$ for $r_0<R$. Choosing $r_0=(2+\rho )R/[2(1+\rho )]$ gives a uniform doubling inequality $g(\beta R)\le Ag(R)$ with

$$\begin{array}{c}\beta =1+\frac{\rho }{2}>1,\\ A=\frac{(2+\rho )(1+\rho )}{\rho }.\end{array}$$

It also proves that $\beta R$ belongs to the line’s domain. Iteration extends every ray indefinitely, establishing $\Omega ={\mathbb{R}}^n$.

Translate an arbitrary base point to zero and subtract its tangent. John normalization and section balance produce invertible linear maps $A_t$ such that

$$\begin{array}{c}{v}_t(y)=t^{-1}u(A_{t}y),\\ B_1\subset S_{v_t}(0,1)\subset B_R,\end{array}$$

with fixed $R$, independent of $t$. Balance bounds both directions of every small section near the center. Iterating the doubling inequality backward then supplies a single positive function $b(r)$ such that $S_{v_t}(x,b(r))\subset x+B_r$ for $x\in B_{1/4}$, uniformly in $t$. This is a quantitative modulus of strict convexity, not merely a Hessian positive at each point.

The manuscript now invokes the Trudinger–Wang interior estimates for the affine maximal equation. Applied to these normalized solutions with their uniform modulus, they give

$$D^2{v}_t(0)\ge cI,|D^3{v}_t(0)|\le C,$$

where the constants do not depend on section height. This analytic estimate is an imported ingredient; the article does not reprove its underlying Monge–Ampère and linearized regularity theory.

Let $H=D^{2}u(0)$, which is fixed. Since $D^2{v}_t(0)=t^{-1}{A}_t^{T}H{A}_t$, the lower bound implies $\Vert A_t^{-1}\Vert \le C_0{t}^{-1/2}$. Transforming third derivatives back gives

$$\begin{array}{rl}|D^{3}u(0)|& \le t|D^3{v}_t(0)|\Vert A_t^{-1}{\Vert }^3\\ & \le C_1{t}^{-1/2}.\end{array}$$

Letting $t\to \infty$ forces $D^{3}u(0)=0$. The base point was arbitrary, so the Hessian is constant throughout the domain, and $u$ is a positive definite quadratic polynomial. This is the final rigidity mechanism: uniform estimates after normalization become vanishing third derivatives on larger and larger original sections. (OpenAI, 2026b)

🏷️ The immersed consequence and verification boundary

The rigidity manuscript also treats connected smooth open (noncompact and without boundary) locally uniformly convex immersed hypersurfaces of dimensions three through nine that are affine complete and classically affine maximal. It imports a theorem of Trudinger and Wang converting affine completeness to Euclidean completeness and a global convex graph representation after an affine change of coordinates. The graph theorem above then identifies the hypersurface as an elliptic paraboloid. That reduction has an additional external dependency beyond the proof for an already given graph. (OpenAI, 2026b)

The two papers therefore address carefully specified versions of the Bernstein question. In low dimensions, Euclidean completeness forces global section control and then quadraticity. In dimension ten, pointwise strict convexity and an entire domain coexist with a nonquadratic solution. The construction’s decaying vertical Hessian illustrates precisely why those hypotheses should not be strengthened silently.

The source review followed both manuscripts’ proof architecture, including the seven recorded TeX files, the radial origin construction and continuation argument, the convex-limit geometry, and the logarithmic mass estimates. Exact symbolic checks reproduced the separated-profile identity, the invariant-box stationary corner, and the displayed coercivity determinants. These elementary checks do not independently verify the full analytic proofs or the external regularity and completeness theorems.

🏷️ Key steps and proof outlines

The radial normalization converts blowup risks into bounded ratios

The regular radial system uses $F^{\prime }=y$, $P^{\prime }=Z$, and

$$y^{\prime }=\frac{r^{8}F{P}^{-12/11}}{y^8},Z^{\prime }=-8\frac{y^{\prime }}{y}Z-\frac{11r^8{P}^{-1/11}}{y^8}.$$

For $r>0$ it is smooth whenever $F,y,P>0$. The issue is therefore not a singular coefficient at an arbitrary positive radius, but possible escape toward infinity or toward the boundary of this positive region. Logarithmic derivatives $A,B,C$ measure exactly those possibilities.

The smooth origin expansion gives $F=1+O(r^2)$, $y=r+O(r^3)$, and $P^{\prime }=-(11/9)r+O(r^3)$. Thus the trajectory enters the stated invariant box near $(0,1,0)$. To justify invariance even at corners, project a point $z$ to the box as $\Pi z$. The face inequalities imply $(z-\Pi z)\cdot V(\Pi z)\le 0$. Local Lipschitz continuity then gives

$$\frac{d}{d\log r}|z-\Pi z{|}^2\le 2L|z-\Pi z{|}^2.$$

Gronwall prevents exit. Integrating the bounded logarithmic derivatives on any finite radial interval keeps $F,y,P$ uniformly bounded above and away from zero. The regular ODE continuation theorem then gives an entire solution. (OpenAI, 2026a)

The rigidity energy needs both a positive matrix and a positive scalar term

At the formal choice $\lambda =1$, the determinant $(n-2)(10-n)/16$ suggests the critical dimension, but the scalar coefficient $k-2$ vanishes for two recession directions. Choosing $\lambda =15/16$ repairs that endpoint while preserving positive definiteness for dimensions three through nine.

Consequently the integrated support-function identity controls a strictly positive density, not just its derivative terms. Young’s inequality absorbs the cutoff cross terms into that coercive quadratic form. Passing to logarithmic coordinates makes the cutoff cost invariant under independent scaling of the recession variables, which is necessary for an estimate uniform over distant cells. (OpenAI, 2026b)

One finite growth contradiction rules out every multidirectional limit

The source first proves positive mass on an interior logarithmic box and a uniform upper bound on each unit logarithmic cell. The first excludes loss of all mass to coordinate faces; the second excludes arbitrarily large mass in a single normalized cell. Covering yields polynomial growth $C(L+1{)}^k$.

The cutoff energy inequality, applied between boxes of radii $\ell$ and $\ell +1$, forces a fixed factor of growth at each step. Starting from positive interior mass gives exponential growth. Choose one finite $L$ where these bounds conflict, and only then choose a smooth approximant accurate enough for all finitely many tests. This avoids differentiating the limiting convex body.

The resulting section balance is used in the affine-complete rigidity theorem. The dimension-ten example instead obtains Euclidean completeness from being an entire graph; it does not assert affine-metric completeness. These different completeness hypotheses explain why the explicit counterexample and the affine-complete consequence do not contradict one another.

🏷️ Manuscripts and verification scope

• A Smooth Nonquadratic Entire Affine Maximal Graph in Dimension Ten — pinned manuscript and companion source.
• The affine Bernstein theorem in dimensions three through nine — pinned manuscript and companion source.

The family scope page describes selected formalization coverage. The selected statement is Euclidean-complete graph rigidity in dimensions three through nine. The immersed consequence is explicitly outside that selection, and the comparator list does not select the smooth dimension-ten counterexample despite the broader page heading. No Lean build, axiom audit, or manuscript-to-formal-statement equivalence check was performed for these notes.

OpenAI collection provides prerequisites and the reading sequence; coverage register records completed and pending families.

🐻  OpenAI 2026a. A Smooth Nonquadratic Entire Affine Maximal Graph in Dimension Ten.

🐻  OpenAI 2026b. The affine Bernstein theorem in dimensions three through nine.