Harmonic dimension beyond the Euclidean count: changing links and averaged growth

Source and verification

This article explains family 361 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.

🏷️ Counting harmonic functions by their growth

A harmonic function satisfies ${\Delta }_{g}u=0$. On Euclidean space, a harmonic function with polynomial growth is a harmonic polynomial. Family 361 asks whether nonnegative Ricci curvature forces the same upper bound on the number of such functions. Its two manuscripts give independent counterexample constructions: one in an even dimension, with a concrete finite selection in dimension sixteen, and another in dimension three.

For a complete manifold and a base point $o$, define

$$\begin{array}{c}{\mathcal{H}}_d(M,g)=\{u:{\Delta }_{g}u=0,\\ |u(x)|\le C_u(1+d_g(o,x){)}^d\forall x\},\\ h_d(M,g)={\dim }_{\mathbb{R}}{\mathcal{H}}_d(M,g).\end{array}$$

The constant can depend on the function. Changing the base point changes only this constant. The bound is pointwise at every radius; estimates on a sparse sequence of spheres are insufficient.

Ricci curvature is the trace of sectional curvature over planes containing a given direction. Nonnegative Ricci curvature therefore does not imply that every sectional curvature is nonnegative. This distinction matters especially for the three-dimensional construction.

For an integer $k\ge 1$, the Euclidean comparison value is

$$\begin{array}{c}{h}_k({\mathbb{R}}^n)\\ =\left(\genfrac{}{}{0ex}{}{n+k-1}{k}\right)+\left(\genfrac{}{}{0ex}{}{n+k-2}{k-1}\right).\end{array}$$

Here is a useful reason that no additional Euclidean functions enter this count. A smooth radial averaging kernel of radius $R$ reproduces a harmonic function by the mean-value property. Moving $k+1$ derivatives onto the kernel bounds those derivatives at a fixed point by $O(R^k{R}^{-k-1})$. Letting $R\to \infty$ makes them zero, so the function is a polynomial of degree at most $k$.

In dimension three, the space of homogeneous harmonic polynomials of degree $l$ has dimension $2l+1$. Thus

$$h_k({\mathbb{R}}^3)=\sum _{l=0}^k(2l+1)=(k+1{)}^2.$$

The question concerns this exact integer-degree count. Finite dimensionality, bounds of the correct asymptotic order, and counterexamples at noninteger degrees are different statements. (OpenAI, 2026) (OpenAI, 2026b)

🏷️ Two counterexamples and their quantifiers

The even-dimensional construction stated in the first paper

There is a smooth complete metric on an even-dimensional Euclidean space with nonnegative Ricci curvature and more degree-$k$ harmonic functions than the Euclidean comparator. It is Euclidean near the origin and has asymptotic volume ratio strictly between zero and one. The finite selection supplied in the appendix permits ambient dimension $16$ and integer degree $50,000$. (OpenAI, 2026a)

The asymptotic volume ratio compares large metric balls with Euclidean balls:

$$\mathrm{AVR}(g)=\underset{R\to \infty }{\lim }\frac{{\mathrm{vol}}_g{B}_g(o,R)}{{\omega }_n{R}^n},$$

where ${\omega }_n$ is the Euclidean unit-ball volume.

The independent three-dimensional construction

For each $4/9<v<1$ and $1<c<9v/4$, every sufficiently large integer $k$ admits a smooth complete metric $g_k$ on ${\mathbb{R}}^3$, Euclidean near the origin, with nonnegative Ricci curvature, $\mathrm{AVR}(g_k)=v$, and $h_k({\mathbb{R}}^3,g_k)\ge c(k+1{)}^2$. Its Ricci curvature is positive outside a compact set. The metric is chosen after $k$ and may depend on it. (OpenAI, 2026b)

For any $1<c<9/4$, these examples can have arbitrarily small global bi-Lipschitz distortion from Euclidean space, after choosing the volume ratio sufficiently close to one and then taking $k$ large. This is a comparison of metric lengths, not a claim of uniform control of all derivatives of the metric. No optimality or endpoint assertion at $c=9v/4$ is made.

Feature	Even-dimensional construction	Three-dimensional construction
Finite spectral input	A strict mean-rate selection; concrete data at dimension 16	A long cycle of low sphere eigenlines
Angular spaces	Fixed finite-dimensional spaces preserved exactly	Moving low spaces coupled to infinitely many high modes
Exact continuation	Tune each full period to a signed cycle	Match all crossing controls simultaneously
Metric quantifier	One metric for the chosen degree	A metric for each sufficiently large degree

Neither paper disproves finite dimensionality. The three-dimensional theorem also does not assert a fixed-factor asymptotic excess for one metric at all large degrees.

🏷️ Cones turn eigenvalues into growth rates

Consider a metric cone

$$g=d{r}^2+r^{2}h$$

over an $m$-dimensional compact link. If $-{\Delta }_h\varphi =\lambda \varphi$, substitution of $u=r^d\varphi$ into the cone Laplacian gives

$$d(d+m-1)=\lambda .$$

The nonnegative root is the growing harmonic rate. On a fixed cone, each angular eigenfunction has its own fixed rate. In both counterexamples, the angular metric changes slowly with logarithmic radius $t=\log r$. A selected harmonic function is arranged to spend successive intervals in different angular directions, so its long-term growth depends on an average of rates.

An elementary model illustrates the idea. Suppose two directions grow at rates $d_1$ and $d_2$. Exact exchanges after equal intervals make each selected solution experience both rates, giving mean $(d_1+d_2)/2$. That mean can be below $k$ even when $d_2>k$.

The word exact matters. A small unwanted component in a faster direction can overwhelm the intended solution during a long interval. Slowly varying coefficients or approximate exchanges alone do not establish the claimed growth bound. The main analytic work in both manuscripts constructs genuine harmonic functions whose transfers realize the desired averages. (OpenAI, 2026a) (OpenAI, 2026b)

🏷️ The first construction: fixed angular blocks

The first paper uses volume-normalized Berger metrics on odd-dimensional spheres. These stretch a distinguished circle direction relative to its orthogonal directions while retaining enough symmetry that every round harmonic space $V_l$ remains invariant under the angular Laplacian.

Write $m=n-1$. The dimension of the real degree-$l$ block is

$$N_l=\left(\genfrac{}{}{0ex}{}{m+l}{m}\right)-\left(\genfrac{}{}{0ex}{}{m+l-2}{m}\right),$$

with the second term zero for $l<2$. For one fixed nonround link, order the corresponding growing rates as $d_{l,1}\le \cdots \le d_{l,N_l}$. The spectral selection seeks integers $M_l$ satisfying

$$\begin{array}{c}\sum _{l=2}^L{M}_l>h_k({\mathbb{R}}^{m+1}),\\ {\overline{d}}_l:=\frac{1}{M_l}\sum _{i=1}^{M_l}{d}_{l,i}<k(M_l>0).\end{array}$$

The selected span need not remain invariant during the construction; the full block $V_l$ does. Within that block, geometric pulses supply enough finite-dimensional control to realize a signed cyclic permutation of the selected axes. This simultaneous control across the finitely many blocks is a source theorem used here without developing its algebraic proof.

The spectral argument uses the multiplicities of circle-direction charges inside $V_l$. Their normalized squares approach a beta distribution. Selecting by the mean of a prefix, rather than requiring every selected rate to be below $k$, produces the surplus. The paper proves an analytic selection theorem and separately supplies an exact finite arithmetic instance. (OpenAI, 2026a)

🏷️ What the finite calculation certifies

The supplied check uses $n=16$, $k=50,000$ and rational link parameters. It computes angular multiplicities exactly and uses strict rational upper bounds on the growth roots. If $h=m-1$ and $\lambda =N/D$, choose $Q=10^8$ and an integer $S$ with

$$S^{2}D>Q^2(h^{2}D+4N).$$

Then the positive root of $d(d+h)=\lambda$ satisfies

$$d<\frac{S-hQ}{2Q}.$$

The program obtains $S$ using an integer square root plus one. It accepts a selected group only when the sum of these strict upper bounds is less than $k$ times the group’s dimension. Partial selection within a repeated eigenvalue is allowed: it amounts to choosing that many independent real eigenvectors. Degrees zero and one are omitted from the selected total but remain in the Euclidean comparator.

The local rerun completed successfully, matched the repository’s reference output, and established a strict integer surplus. The selected-to-Euclidean ratio is approximately $1.000126031$; this decimal is informational, while the decision uses exact integers. The positive link-curvature margin and the accompanying radial-scale parameter inequalities were also checked exactly.

This verifies the finite spectral input. It does not construct the metric, tune its controls, or certify the entire counterexample. Those remain analytic arguments in the manuscript. The pinned verification script and reference output record that limited task. (OpenAI, 2026a)

🏷️ Tuning the entire period

For the first metric, an angular coefficient vector satisfies a matrix equation in logarithmic radius:

$$Y^{\prime \prime }+p(t)Y^{\prime }-A_l(t)Y=0.$$

Smoothness at the Euclidean origin selects its regular solution space. A symmetric matrix $P_l$ describes that space by $Y^{\prime }=P_{l}Y$ and satisfies the Riccati equation

$$P_l^{\prime }=A_l-p{P}_l-P_l^2.$$

Positive lower and upper matrix barriers keep $P_l$ bounded, and any value vector at one time determines a unique regular solution. Long round intervals reset $P_l$ close to a scalar matrix.

Period $j$ consists of a pulse, an upward ramp, a nonround dwell, a downward ramp, and a round rest. Each short piece has duration $T_j=j^6$, while the dwell has duration $L_j=j^7$. The accumulated logarithmic radius is $t_j\sim j^8/8$.

A pulse controls both values and radial derivatives. Even when its value map is close to the desired cycle, a derivative error can feed a fast mode during the following dwell. The paper therefore factors out the whole subsequent diagonal leg. If $U,V$ are its propagation matrices for initial value and derivative, respectively, and ${\theta }_l$ is the reference round rate, set

$$\begin{array}{c}D=U+{\theta }_{l}V,K=D^{-1}V,\\ 0\le K\le {\theta }_l^{-1}\mathrm{Id}.\end{array}$$

This bound does not deteriorate with the leg’s length. If $\mathcal{B}$ is the pulse value map and $P_e$ its exiting regular derivative matrix, the full transfer obeys the exact identity

$$D^{-1}\mathcal{T}=[\mathrm{Id}+K(P_e-{\theta }_l\mathrm{Id})]\mathcal{B}.$$

After removing a scalar determinant factor, this map and its first parameter derivative approach the ideal control map. Its derivative is invertible at the target cycle. A contraction argument chooses parameters attaining that cycle exactly, uniformly over all earlier admitted controls. Inductively, every full period has the form

$$\begin{array}{c}{\mathcal{T}}_{l,j}=s_{l,j}{D}_{l,j}{\Pi }_l,\\ |\log s_{l,j}|=O(j^6),\end{array}$$

where ${\Pi }_l$ is a signed cycle and $\log (D_{l,j}{)}_{ii}=j^7(d_{l,i}+o(1))$.

Over a fixed-length cycle, consecutive weights $j^7$ differ by $O(j^6)$. Summing gives total main growth ${\overline{d}}_l{t}_J$ with errors $o(t_J)$. A whole period has length $o(t_j)$, and the bounded Riccati matrices control growth inside it. Hence the strict exponent bound extends to every radius, not just period endpoints. Finite-dimensional angular norm bounds then give the required pointwise estimate. (OpenAI, 2026a)

🏷️ Keeping Ricci curvature nonnegative

Slow angular motion does not automatically preserve curvature. Both papers introduce a concave radial factor that provides a positive radial Ricci contribution large enough to pay for the angular variation.

For the Berger construction, the mixed radial-angular Ricci block vanishes by the symmetry of its distinguished direction. The positive radial contribution in period $j$ is of order $j^{-10}$ after multiplication by $r^2$, while the adverse squared angular-speed term is $O(j^{-12})$. A strict intrinsic link-curvature margin handles the tangential block. The estimates hold for every admissible control sequence before the sequence is chosen.

The three-dimensional metric has the form

$$\begin{array}{c}g=d{r}^2+f(r{)}^2{H}_{\ast }(\log r),\\ f(r)/r\to a=\sqrt{v}.\end{array}$$

Its sphere metrics have one fixed area form and move with phase speed $\eta (t)=t^{-3/4}$. Here the mixed Ricci block is generally nonzero. In the tensor $r^2\mathrm{Ric}$, the tangential block has a fixed positive lower bound, the mixed block is $O(\eta )$, and the radial contribution is a chosen large multiple of ${\eta }^2$ minus controlled errors.

For a symmetric block matrix, positivity requires the radial entry to dominate the mixed block times the inverse tangential block times its transpose. This Schur complement loss is $O({\eta }^2)$, so the radial coefficient can absorb it. The coefficient is chosen before moving the start of the angular program farther out. Discarding the mixed block would miss a necessary part of this argument.

The Euclidean core ensures smoothness at the origin. Radial distance equals $r$, making centered closed balls compact and the metric complete. The fixed area form gives $\mathrm{AVR}=a^2=v$ directly by integrating $4\pi f(r{)}^2$. (OpenAI, 2026a) (OpenAI, 2026b)

🏷️ Why the three-dimensional count can exceed the square

The independent construction moves the spectrum of near-round metrics on $S^2$. At an isolated double eigenvalue, two smooth eigenlines can pass through each other: following a line differs from renaming it according to increasing eigenvalue. A sequence of adjacent exchanges cycles a long band of ordered positions while protecting a gap above the selected low spectrum.

Put $a=\sqrt{v}$ and choose

$$\sqrt{c}<\vartheta <\frac{3a}{2}.$$

For the chosen large integer $k$, take $M=\lfloor \vartheta k\rfloor$, leave the positions through degree $L=\lfloor \sqrt{k}\rfloor$ uncycled, and select all $p=(M+1{)}^2$ positions through degree $M$. Thus $p/(k+1{)}^2\to {\vartheta }^2>c$.

Each band label visits every ordered position at each phase over a full label cycle. The relevant round-sphere average satisfies

$$\frac{{\sum }_{l=L+1}^M(2l+1)(l+1)}{(M+1{)}^2-(L+1{)}^2}\sim \frac{2M}{3}.$$

Radial scaling multiplies the leading frequency by $a^{-1}$. Since $2\vartheta /(3a)<1$, a sufficiently small angular distortion leaves every selected average strictly below $k$. The lower, uncycled modes have rates only $O(\sqrt{k})$.

This explains the range $c<9v/4$: it allows both the count inequality and the mean-frequency inequality. It is not a claim that this range is optimal. Constructing the prescribed isolated crossings, while excluding unwanted multiplicities and protecting the outer gap, is a substantial source theorem rather than a consequence of this counting calculation alone. (OpenAI, 2026b)

🏷️ Exact inward maps control the high modes

The low eigenspaces now move, so harmonic functions can acquire arbitrarily high angular components. The companion handles them through exact whole-ball Dirichlet maps. Given data on a sphere of radius $R$, solve the harmonic equation throughout the enclosed ball and restrict to radius $r<R$.

These maps preserve constants and the common spherical mean. Positivity gives the pointwise inequality $(Tv{)}^2\le T(v^2)$, and integrating proves $\Vert Tv{\Vert }_2\le \Vert v{\Vert }_2$. This contraction is valid at every intermediate radius and already incorporates regularity at the origin.

At integer logarithmic radii, the map $T_j$ is approximated by the inward transfer of a frozen link. Two estimates have different jobs: an $o(1)$ operator-norm estimate separates the low spectrum from the tail, while a sharper $o({\delta }_j)$ estimate on the finite smooth reference frame detects crossing controls, where ${\delta }_j=j^{-9/16}$. The sharper bound is not claimed on the whole Hilbert space.

The protected outer gap yields invariant spaces that are graphs over the first $p$ reference modes. Their high-mode components have norm $O({\delta }_j)$. Restricting the exact maps to these graphs and inverting gives finite outward matrices

$$A_j=\mathrm{diag}(e^{d_i(j)})+O({\delta }_j).$$

The high-mode graph depends on future controls as well as earlier geometry. The proof consequently chooses the controls simultaneously. A crossing control changes a selected off-diagonal entry to first order with a fixed nonzero sign, uniformly over all the other controls. Calculations use the continued reference frame, since perturbed eigenvectors can rotate substantially near a double eigenvalue. (OpenAI, 2026b)

🏷️ Matching the lines and obtaining entire functions

Seek columns $v_i(j)$ close to the coordinate vectors, with

$$\begin{array}{c}{A}_j{v}_i(j)={\sigma }_i(j)v_i(j+1),\\ {\sigma }_i(j)>0.\end{array}$$

For each pair of coordinates, this becomes a scalar recurrence. Solve it forward where it contracts and backward where the reverse recurrence contracts. At a crossing, the two propagated values can disagree. The source expresses the mismatch as a weighted sum and proves that the two extreme choices of its control give opposite signs, independently of all other choices.

A self-map updates the column coordinates by these stable propagations and updates each control by its mismatch, clipped to $[-1,1]$. Finite-dimensional fixed points and a coordinatewise compact limit give one simultaneous fixed point. The strict endpoint signs keep every control in the interior, forcing every mismatch to vanish. Thus all selected lines satisfy exact recurrences for one common metric.

The invariant graphs lift these columns to full boundary traces. Multiplying by the accumulated ${\sigma }_i$ makes traces on successive spheres compatible under the inward maps. Harmonic extension to nested whole balls and uniqueness then give $p$ independent smooth harmonic functions on all of ${\mathbb{R}}^3$.

Their logarithmic growth has the desired phase mean. If ${\Theta }_i(s)$ is the periodic reference frequency with mean $m_i<k$, a bounded primitive of ${\Theta }_i-m_i$ and $s^{\prime }(t)=t^{-3/4}$ give

$${\int }_J^T({\Theta }_i(s(t))-m_i)dt=O(T^{3/4}).$$

The mesh error is $O(T^{1/4})$, and the accumulated transfer error is $O(T^{7/16})$; the remaining radial-frequency error has vanishing average. All are sublinear in $T=\log r$. Every selected spherical $L^2$ norm therefore has exponent $m_i<k$.

The all-radius contraction fills the gaps between integer logarithmic radii. Uniform elliptic estimates on fixed-width logarithmic cylinders turn these $L^2$ estimates into pointwise bounds. This last step is what proves membership in ${\mathcal{H}}_k$ as originally defined. (OpenAI, 2026b)

🏷️ Geometry at infinity and review limits

A tangent cone at infinity is a limit seen after rescaling larger and larger regions of the manifold to unit size. Because angular motion becomes slower but continues indefinitely, different subsequences freeze different links. The first construction has nonunique cone limits; the three-dimensional one has uncountably many pairwise nonisometric pointed cone limits. A long-term averaged growth exponent need not equal one frozen cone’s homogeneous degree.

The three-dimensional examples also have negative radial sectional planes arbitrarily far out, despite positive Ricci curvature there. They therefore do not meet stronger comparison hypotheses requiring nonnegative sectional curvature.

The nodal-bound family concerns the size of eigenfunction zero sets on compact manifolds. The present family counts polynomial-growth harmonic functions on noncompact manifolds; the shared name does not make these the same comparison problem.

The review checked the finite arithmetic certificate, selected spectral and transfer statements, the radial curvature constructions, exact-matching mechanisms, and the final every-radius growth arguments. It did not independently verify the full simultaneous-control theorem, angular crossing construction, pulse estimates, freezing estimates, invariant-graph proof, or crossing-mismatch bounds. Both geometric counterexamples remain conclusions attributed to their manuscripts.

🏷️ Key steps and proof outlines

Long pulse estimates need an integrated error bound

A reset error of size $j^{-2}$ cannot simply be carried through a pulse of length $T_j=j^6$: that estimate would permit an error of order $j^4$. The first manuscript instead keeps the common scalar growth exactly. Let $v_l$ solve the scalar Riccati equation for the round angular eigenvalue, and approximate the regular matrix by

$$Q_l(t)=v_l(t)\mathrm{Id}+T_j^{-1}{R}_l((t-t_j)/T_j).$$

Here $R_l$ is the bounded ideal control profile. Substitution into the matrix Riccati equation cancels the scalar terms and leaves a residual ${\rho }_l$ of size $O({\eta }_j/T_j)$, where ${\eta }_j=j^{-2}+T_j^{-1}$. For $E_l=P_l-Q_l$, the exact, noncommutative difference equation is

$$E_l^{\prime }={\rho }_l-(p\mathrm{Id}+P_l)E_l-E_l{Q}_l.$$

Both coefficient matrices have positive lower eigenvalue bounds. Taking the Frobenius inner product with $E_l$ gives

$$\frac{d}{dt}\Vert E_l\Vert \le \Vert {\rho }_l\Vert -\gamma \Vert E_l\Vert .$$

After integration, the initial discrepancy contributes its size divided by $\gamma$, and the forcing contributes its time integral divided by $\gamma$. Thus $\int \Vert E_l\Vert dt=O({\eta }_j)$ over the entire long pulse. Removing the exact scalar factor $\exp (\int v_l)$ now leaves a transfer map close to the ideal profile, including one control derivative. This is the analytic input required by the subsequent exact tuning argument. (OpenAI, 2026a“The response of one pulse”)

A simple crossing creates a Gaussian summation scale

In the three-dimensional construction, consider a scalar off-diagonal recursion $u(j+1)=r(j)u(j)+F(j)$ near a crossing. The eigenvalue gap vanishes linearly in angular phase, and phase advances at speed $\eta$. A stable product over $m$ steps therefore satisfies

$$\prod r_{\mathrm{stable}}\le C\exp (-c\eta m^2),\sum _{m\ge 0}C{e}^{-c\eta m^2}\le C^{\prime }{\eta }^{-1/2}.$$

The square in the exponent comes from summing the increasing gap, $\eta (1+\cdots +m)$. A forcing of size ${\eta }^{3/4}$ consequently produces a trial solution of size ${\eta }^{1/4}$, strictly smaller than its permitted box ${\eta }^{1/8}$.

At a crossing where the forward and backward stable solutions arrive from opposite sides, their mismatch is a sum of the forces with positive weights. On the central $O({\eta }^{-1/2})$ steps, these weights stay bounded below. Setting the crossing control to either endpoint $\pm 1$ fixes the sign of the leading forcing there and produces mismatch of magnitude at least $c{\eta }^{1/4}$. Outside the fixed phase plateau the weights are exponentially small in $1/\eta$. Thus other controls cannot reverse that sign.

Finite-dimensional fixed points solve all countably many matches

For each mismatch $D_{\nu }$, update its control by projecting $z_{\nu }-D_{\nu }$ onto $[-1,1]$. Update the column coordinates by the stable trial formulas. This maps the full product of boxes into itself. Freeze all but finitely many coordinates, apply Brouwer, and extract a coordinatewise convergent subsequence. Uniform summability of the backward tails makes each output coordinate continuous, so the limit solves every fixed-point equation.

The endpoint signs exclude fixed controls at $\pm 1$. Every control is therefore interior, where the projected equation forces $D_{\nu }=0$. This closes the exact matching conditions simultaneously; merely tuning one crossing after another could disturb previous matches. (OpenAI, 2026b“One scalar obstruction at each crossing” and “Simultaneous matching”)

🏷️ Manuscripts and verification scope

• A counterexample to integer-degree harmonic dimension comparison — pinned manuscript and companion source.
• A Three-Dimensional Counterexample to Integer-Degree Harmonic Dimension Comparison — pinned manuscript and companion source.

The family scope page describes selected formalization coverage. The documented selection is the even-dimensional counterexample with dimension 16 and degree 50000, including its Euclidean core and positive asymptotic volume ratio below one. Separate tangent-cone and cone-spectrum conclusions are excluded. The independent three-dimensional companion is not listed in this scope page. 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 counterexample to integer-degree harmonic dimension comparison.

🐻  OpenAI 2026b. A Three-Dimensional Counterexample to Integer-Degree Harmonic Dimension Comparison.