THEORY-189: Nested angular cones in the Poincaré ball, symmetric about the outward ray with an aperture depending only on the apex's radius, are transitive only if the aperture is at most π/2 and r sin ψ(r)/(1 − r²) does not increase outward; so no such cone can be defined near the origin, the widest admissible family is sin ψ = K(1 − r²)/r, and every point a cone contains lies farther from the origin than its apex

Source

Ganea, Bécigneul and Hofmann (2018), LIT-866: §3, Lemma 2, Theorems 3–5 and Appendices E–G, as read in NOTE-671.

What was actually shown

The setting. A cone at x in the Poincaré ball Dⁿ is the exponential map of the tangent vectors within angle ψ(x) of the outward direction through x. The source asks four things of a family of such cones:

1. symmetry about the outward ray;
2. an aperture depending only on ‖x‖;
3. a continuous aperture;
4. transitivity: if y lies in x’s cone, y’s cone lies inside x’s.

Under the fourth, “y in S_x” is a transitive relation, and with the radial point below it, a partial order.

Necessity, proved.

• Lemma 2: the aperture is at most π/2. If it were wider, the points on the cone’s boundary would need apertures of at most π/2. Continuity along the boundary back to the apex then gives a contradiction.
• Theorem 3: h(r) = r sin ψ(r)/(1 − r²) cannot increase with r. The proof applies the hyperbolic law of sines in the triangle formed by the origin, an apex and a point on its cone’s boundary.
• Consequences. h tends to 0 at the origin, so no family with non-zero apertures covers a neighbourhood of it. The order has no top element in the geometry, and the family lives outside a ball B(O, ε).
• The widest family. For a given K = h(ε), the pointwise widest apertures are sin ψ = K(1 − r²)/r, with K ≤ ε/(1 − ε²), which is what sin ψ(ε) ≤ 1 requires. Cones narrow toward the boundary at that rate.

Sufficiency, proved here. The source states that this family is transitive (Theorem 4) and gives no proof. The proof is in the note of 2026-10-19 below, for every dimension n ≥ 2. Numerical checks found no violation, in 3,539 sampled chains in two dimensions and in 35,862 in two, three and five.

Radial monotonicity, derived here. Membership is the angle test π − ∠Oxy ≤ ψ(x) ≤ π/2 (the source’s Theorem 5), so ∠Oxy ≥ π/2. In a hyperbolic triangle the side opposite an angle of at least π/2 is the longest: cosh d(O, y) ≥ cosh d(O, x) cosh d(x, y) > cosh d(O, x). So every point of x’s cone is strictly farther from the origin than x. Along any chain of the order, distance from the origin increases, and the relation is antisymmetric.

What could have come out otherwise. The necessity results could have left the aperture free near the origin, or allowed cones wider than a half-space. They do not.

What this does not say

• It does not say which partial orders embed. Cones of incomparable points can overlap, so a point may lie below two incomparable apexes, and a DAG with shared descendants is expressible. Nothing here bounds the dimension a given poset needs, or says that every finite poset embeds at all.
• It does not give a lattice. The intersection of two cones is not in general a cone, so meets and joins have no geometric counterpart. A concept lattice can at best be embedded as an order, not as a lattice (CLAIM-119, QUESTION-025).
• The constraints are conditional on the four requirements. Drop rotation invariance or axial symmetry, and other families, including ones defined at the origin, are not excluded by these proofs. “Optimal” in the source means widest pointwise for a given K, not unique or best by any other measure.
• It says nothing about learned embeddings. That trained cones realise the order, or that the hyperbolic version outperforms because of curvature, are separate empirical claims. The source’s own experiment leaves both open: it has single runs, a failure of every method when only the transitive reduction is trained on, and Euclidean cones that match hyperbolic ones when initialised alike.
• It is not a statement about linear directions. It neither supports nor contradicts THEORY-185. It describes a different encoding of implication, by region containment, beside the distance encoding of THEORY-196. The precision cost that THEORY describes applies to the narrow cones near the boundary, but that connection is the record’s, not the source’s.

Note of 2026-10-19: Theorem 4 proved, in every dimension

The promotion audit (batch 043, ADR-047 v2) proved the source’s Theorem 4: the family sin ψ(x) = K(1 − ‖x‖²)/‖x‖ on Dⁿ \ B(O, ε), with K ≤ ε/(1 − ε²), is transitive in every dimension n ≥ 2. The source asserts this and gives no proof.

Setting. Write a = d(O, x). Then sinh a = 2r/(1 − r²), so the family is sinh(d(O, x)) sin ψ(x) = 2K =: c, a constant. For a point w, θ_x(w) is the angle at x between the outward ray and the geodesic to w. S_x is the image under exp_x of the closed solid cone θ ≤ ψ(x). Since exp_x is a diffeomorphism, S_x is closed, and ∂S_x \ {x} is a smooth hypersurface, the image of the cone’s lateral surface.

1. Law of sines. In the triangle O, x, w, the angle at x is π − θ and the angle at w is α. Then sin α = sinh(d(O, x)) sin θ / sinh(d(O, w)). If θ ≤ π/2, the angle at x is at least π/2, so α < π/2.
2. Continuation. Let z ∈ S_y with z ≠ y, and let δ be the direction at z that continues the geodesic from y. The angle between δ and z’s outward ray is the triangle’s angle α at z. By step 1 and z ∈ S_y, sin α ≤ sinh(d(O, y)) sin ψ(y) / sinh(d(O, z)) = c / sinh(d(O, z)) = sin ψ(z). Both angles are acute, so α ≤ ψ(z): δ lies in z’s cone of directions.
3. Tangency on the boundary. Let z ≠ x lie on ∂S_x, on the generator from x, and let g be the direction continuing it. By rotational symmetry about the axis through O and x, the tangent hyperplane of ∂S_x at z is spanned by g and the directions orthogonal to the plane of O, x and z. z’s outward ray lies in that plane at angle α from g, and by step 1 sin α = c / sinh(d(O, z)) = sin ψ(z), so α = ψ(z). So z’s cone of directions lies in the closed half-space on S_x’s side and touches the hyperplane only along g.
4. Transitivity. Let y ∈ S_x and w ∈ S_y, and suppose w ∉ S_x. Every point of the segment from y to w after y lies in S_y, so it is farther from O than y (radial monotonicity above) and is not x. Let z be the segment’s last point in S_x. Then z ∈ ∂S_x and z ≠ x, and by step 2 the segment’s direction δ at z lies in z’s cone. By step 3, either δ points strictly into S_x, so the segment is inside S_x just after z, or δ = g, so the segment runs along the generator from x, which lies in S_x. Either way z is not the last point, a contradiction. So w ∈ S_x.

With Theorem 4 proved, the family is admissible. Theorem 3 bounds every transitive family with the same K = h(ε) pointwise by this one, so it is the widest. A numerical check in D², D³ and D⁵ (K = 0.02, 0.1 and 0.3; 35,862 random chains, 30% with points on cone edges) found no violation. The largest excess θ − ψ was 5 × 10⁻¹², the tangency of step 3. A control in which h increases with r gave 2,249 violations.

The same review corrected the bound on K. Eq. 25 of the source reads K ≤ ε/(1 − ε²), equivalently ε ≥ 2K/(1 + √(1 + 4K²)). The earlier 2ε/(1 − ε²) was a misreading.