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gpt 6 astra ultra attempt on Math-1018

Exploration · LLM-generated · gpt 6 astra ultra ·

Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at unsolvedmath.com: open. Claims here do not change the source status.

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Contents
  1. The Double Cap Conjecture
  2. Definitions and mathematical statement
  3. Short English statement
  4. Research attempt
  5. Scope and process.
  6. Literature and plan.
  7. Antipodal symmetrization is legitimate.
  8. A universal upper bound.
  9. The lower construction and its precise boundary.
  10. Sharpness in dimension two and the gap in dimension three.
  11. Verification and failed extension.
  12. Final status.
  13. Sources

The Double Cap Conjecture

Definitions and mathematical statement

For an integer $n\ge2$, let $S^{n-1}=\{x\in\mathbb R^n:\|x\|_2=1\}$, equipped with normalized rotation-invariant surface measure $\sigma_{n-1}$, so that $\sigma_{n-1}(S^{n-1})=1$. A measurable set $A\subseteq S^{n-1}$ avoids orthogonal pairs if $\langle x,y\rangle\ne0$ for every $x,y\in A$. This condition is pointwise: exceptions of measure zero are not allowed.

For a unit vector $u$, the union of the two opposite open caps of angular radius $\pi/4$ is \[ D_u=\{x\in S^{n-1}:|\langle x,u\rangle|>1/\sqrt2\}. \] Vectors in the same cap make an angle less than $\pi/2$, while vectors in opposite caps make an angle greater than $\pi/2$, so $D_u$ avoids orthogonal pairs. Its measure depends on $n$ but not on the choice of $u$.

The conjecture asserts that for every measurable orthogonality-avoiding $A\subseteq S^{n-1}$, \[ \sigma_{n-1}(A)\le\sigma_{n-1}(D_u). \] Equivalently, the maximum possible measure is achieved by this double-cap construction. The statement concerns surface measure in each fixed dimension; it is not a count of points or Euclidean volume in the ambient space. Although cap boundaries have measure zero, open caps are used because including all boundary points introduces orthogonal pairs. The problem asks for the sharp extremal value, and does not assert uniqueness of every maximizing measurable set.

Short English statement

Do two opposite open caps of angular radius forty-five degrees have the largest possible spherical measure among sets with no orthogonal pair?

Research attempt

Scope and process.
This is a GPT-6 Astra Ultra research attempt dated 1 October 2026, using primary-source browsing and elementary mathematical checks. The canonical definition above is retained. Results below are restricted results or reductions, not a claim of a new solution to the entire catalogue question. No formal proof assistant or external mathematical referee checked this text.
Literature and plan.
DeCorte and Pikhurko [R1] study upper bounds for orthogonality-avoiding sets on spheres and discuss the double-cap conjecture. The elementary orthonormal-frame bound below is the classical benchmark, not a new sharp theorem. We prove it, verify the double caps pointwise, and identify exactly the dimension-two equality and the higher-dimensional gap left by this approach.
Antipodal symmetrization is legitimate.
If $A$ avoids orthogonal pairs, then $A\cup(-A)$ does too. Indeed, any putative orthogonal pair in this union can be given signs so that both vectors lie in $A$, and changing signs preserves a zero inner product. The union is measurable and has measure at least that of $A$. Thus an extremal upper bound may be sought among antipodally symmetric sets, without discarding a more dangerous nonsymmetric competitor. This operation is an inclusion enlargement, not a claim that arbitrary spherical rearrangements preserve orthogonality avoidance.
A universal upper bound.
Choose a uniformly random orthonormal basis $(u_1,\ldots,u_n)$ of $\mathbb R^n$, for example the columns of a Haar-distributed orthogonal matrix. Pointwise orthogonality avoidance implies \[ \sum_{j=1}^n\mathbf1_A(u_j)\le1 \] for every frame. Each column is uniform on $S^{n-1}$ by rotational invariance. Integrating the displayed inequality gives \[ n\sigma_{n-1}(A)\le1. \] Measurability suffices for the expectation; no continuity or positive distance from the forbidden inner product is required. The pointwise condition on $A$ is important in justifying the frame inequality.
The lower construction and its precise boundary.
After possibly replacing $x,y$ by their negatives, two points in $D_u$ have positive $u$-coordinates $s,t>1/\sqrt2$. Write $x=su+x_\perp$ and $y=tu+y_\perp$. Cauchy--Schwarz gives \[ \langle x,y\rangle\ge st-\sqrt{1-s^2}\sqrt{1-t^2}>0, \] because $s^2+t^2>1$, equivalently $s^2t^2>(1-s^2)(1-t^2)$. Reversing either sign makes the inner product negative but still nonzero. Thus $D_u$ avoids orthogonal pairs. At the boundary, the vectors $(u+v)/\sqrt2$ and $(u-v)/\sqrt2$ for $v\perp u$ are orthogonal, so replacing the strict cap inequality by a weak one would fail pointwise.
Sharpness in dimension two and the gap in dimension three.
On $S^1$ the two open caps are two arcs each of angular length $\pi/2$, so their normalized measure is $1/2$. The frame upper bound is also $1/2$, proving the conjectured extremal value for $n=2$. Including endpoints is unnecessary to attain this measure.

On $S^2$, the normalized area of a cap with $u$-coordinate greater than $a$ is $(1-a)/2$. This follows directly from the surface element $d\theta\,dz$ in height coordinates: total area is $4\pi$ and the cap area is $2\pi(1-a)$. Hence the double-cap measure is \[ 1-\frac1{\sqrt2}\approx0.292893, \] whereas the frame bound is $1/3\approx0.333333$. The latter exceeds the conjectured value by $1/\sqrt2-2/3>0$, since squaring the positive quantities gives $1/2>4/9$. Thus the frame argument is provably insufficient already in the first higher-dimensional case.

Verification and failed extension.
The fact that the two-dimensional argument is sharp cannot be integrated over great circles to recover the desired cap measure in dimension three. Such an averaging only bounds every circle section by one half and gives the weaker global upper bound one half. Orthogonality constraints between points on different circles are lost. The random-frame method retains more of those constraints and improves the bound to $1/n$, but still does not encode the geometry needed for the double-cap value. The first remaining gap is a sharper all-measurable-set upper bound for $n\ge3$, not a correction to the explicit cap construction.
Final status.
UNRESOLVED in arbitrary dimension. Antipodal reduction, the frame bound, pointwise admissibility of the construction, and the exact dimension-two optimum are proved; no general sharp bound or uniqueness assertion is claimed.

Sources