The Gaussian propeller conjecture in every dimension

Three 120-degree wedges are the best way to cut a bell-curve cloud, in every dimension.

1 paper September 24, 2026 Explained by AI

Cut space into pieces, and give each piece an arrow: the average position of a random bell-curve point, counted only inside it. Add up the squared arrow lengths. This paper proves that the total never exceeds 9/(8π), about 0.3581 (Lean proof listed). Three equal 120-degree wedges, like a propeller, reach that value. The question had been settled only for up to four pieces.

First step

2009

Khot and Naor, approximate kernel clustering

Significance

Major

If it holds, it settles a conjecture open since 2009. Earlier work covered only up to four pieces.

Trust

Formal proof listed

Field

Convex and metric geometry

1 paper

The explanation

Scatter random points on a flat plane. Most land near a centre spot, and fewer land far away, in a bell-curve pattern. This is the standard Gaussian distribution. Now cut the plane into pieces. For each piece, average every point's position, but count any point outside that piece as sitting at the centre. The result is an arrow from the centre, which the paper calls the piece's centroid. A piece that leans far to one side gets a long arrow. The score of a cut is the sum of its squared arrow lengths.

Try a few cuts. With no cut at all, the arrow has length 0, because the cloud balances at the centre. One straight line through the centre gives two halves. Each arrow has length about 0.3989, so the score is about 0.3183. Three equal wedges of 120 degrees give arrows of about 0.3455 each. Their score is about 0.3581, exactly 9/(8π). Four quarters drop back to about 0.3183, and five equal wedges score about 0.2749. Too few pieces give arrows that cancel out. Too many give short arrows. Three wedges look like a three-bladed propeller.

The propeller conjecture says no cut does better. That includes curved or scattered pieces, pieces of any size, any number of pieces, and spaces of any dimension. Some pieces may even be empty. The question arose in a 2009 paper by Khot and Naor on approximate kernel clustering, a data-grouping problem. They showed that a best cut can be taken to be made of cones. A cone here is a region of straight rays fanning out from the centre. They also found the best score for three pieces.

Progress then came piece by piece. In 2013, Heilman, Jagannath and Naor proved the bound in three-dimensional space. Their proof was computer-assisted. Because the problem places k pieces in a space of k − 1 dimensions, this settled up to four pieces. In 2022, Heilman gave a route for three or four pieces in any dimension. That route assumed an extra stability property, and it still stopped at four pieces. Five or more pieces stayed open.

The obstacle is the sheer freedom. Each piece can have any shape and any probability, and an argument must handle all of them at once. The paper first shrinks the problem. It takes a best cut with the fewest pieces of positive probabilitySource 1. Merging two of its pieces shows that every pair of arrows points more than 90 degrees apartSource 1. The 2013 three-dimensional theorem then rules out four or fewer piecesSource 2.

For five or more pieces, the paper uses two consequences of being the best cutSource 3. Removing one piece must cost a definite amount. Competing against two other pieces limits how the arrows can sit. These facts force lower limits on the pieces' probabilities. Together these limits cannot fit inside a total probability of 1, so no such cut beats the propellerSource 4. Every numerical comparison in this step is checked with exact fractionsSource 5. The paper's argument also takes two earlier theorems as given. One is the three-dimensional resultSource 6. The other is Ehrhard's inequality, a known fact about bell-curve probabilities.

So the paper proves that no cut, in any dimension, scores above about 0.3581. The plane propeller wins everywhere. In more dimensions, stretch its three wedges straight along the extra directions. Those directions add nothing to any arrow, so the score stays 9/(8π)Source 7. In a straight line, a space of one dimension, the bound still holds but nothing reaches it. Equality needs at least two dimensions and at least three pieces.

The result feeds back into kernel clustering. There, a computer splits N items into k groups to maximize total similarity inside the groups. A loss factor of 2 means a method always gets at least half the best possible total. Khot and Naor's 2009 method has loss factor (8π/9)(1 − 1/k). For three groups, that is about 1.862. The paper claims that any fixed smaller factor is NP-hardSource 8. NP-hard means no fast method exists unless every problem in a large class called NP has one. That claim also needs a companion paper in the same release, The Unique Games TheoremSource 9.

How sure is this? A computer proof checker called Lean has checked a formal version of the main bound. It covers every dimension and every number of pieces, plus the propeller reaching the bound. The kernel-clustering hardness claim and a related result on Gaussian maxima are outside that check. No specialists have published a review yet.

What it doesn't show. This does not give a better kernel-clustering method. It claims that beating the known loss factor is NP-hard.

Where it came from

This result comes from an unreleased OpenAI model, published in the openai/math release on 6 October 2026.

Hype check

Major only if the proof holds. Lean has checked the main bound, but the proof takes the 2013 three-dimensional theorem and Ehrhard's inequality as inputs. The kernel-clustering claim is not machine-checked, and no specialist review has been published.

Words used

Standard Gaussian distribution: The bell-curve rule for random points: most land near the centre, fewer far away.
Centroid (Gaussian first moment): The average position of a random point, counting points outside the piece as sitting at the centre.
Propeller: A cut of the plane into three equal 120-degree wedges meeting at the centre.
Cone: A region made of straight rays fanning out from the centre.
Kernel clustering: Splitting items into a fixed number of groups to maximize total similarity inside the groups.
Loss factor: How far below the best possible total a method may fall; 2 means at least half.
NP-hard: No fast method exists unless every problem in the large class NP has one.
Lean: A computer program that checks each step of a formal proof.

The original paper

The Gaussian propeller bound in every dimension (PDF, opens in a new tab)
September 24, 2026 Formal proof listed

Lean notes

All quoted lines (9)

Among extremizers, choose one with the fewest cells of positive measure. Merging two cells shows that its nonzero centroids $z_1,\ldots,z_m$ have strictly negative pairwise inner products.
The Gaussian propeller bound in every dimension Back to text
The theorem of Heilman, Jagannath, and Naor therefore excludes $m\le4$ from any counterexample to the desired bound.
The Gaussian propeller bound in every dimension Back to text
We first replace this infinite-dimensional optimization by a finite list of Gaussian linear scores. Two different consequences of optimality then constrain that list: deleting one score costs a definite amount of expected maximum, and competing against two scores constrains a covariance determinant.
The Gaussian propeller bound in every dimension Back to text
Finally, Section … combines the scalar probability bounds with the identity $\sum_iP_i=1$ . The available probability mass forces the third-largest squared length to be too large relative to the smallest length, contradicting the determinant bound.
The Gaussian propeller bound in every dimension Back to text
Appendix … gives exact rational enclosures for every finite numerical comparison, including the complete interval cover needed for the scalar constraint.
The Gaussian propeller bound in every dimension Back to text
The geometric proof uses the three-dimensional propeller bound and Ehrhard's inequality as its external theorem inputs; the remaining geometric estimates are proved here.
The Gaussian propeller bound in every dimension Back to text
An orthogonal Gaussian factor contributes zero to every centroid, so the construction has the same value in every dimension at least two.
The Gaussian propeller bound in every dimension Back to text
Together with their low-influence theorem and the companion paper The Unique Games Theorem, it gives NP-hardness of every deterministic approximation with a fixed loss factor below $\alpha_k$ .
The Gaussian propeller bound in every dimension Back to text
The separate Unique Games input is needed only for the hardness statement.
The Gaussian propeller bound in every dimension Back to text