Source
Sarkar, Mukherjee and Sato (2026), LIT-1820, §2 (pp. 2–3) and App. B (p. 7) of the arXiv preprint 2610.37656v1; read in NOTE-1331. The premise is from the companion paper, LIT-1803, read in part in NOTE-1326.
What was actually shown
The setting. Each edge of a graph carries Tᵢⱼ ∈ O(d) with Tⱼᵢ = Tᵢⱼ⁻¹. A gauge transformation is an independent gᵢ ∈ O(d) at every entity, acting by Tᵢⱼ ↦ gᵢTᵢⱼgⱼ⁻¹ (p. 2).
No cycle, no interaction. On an edge set with no cycle, gauge fixing along a spanning tree sends every transport to I. So any invariant function of the transports is constant, and an equivariant update of an edge commutes, at the identity, with the residual diagonal O(d). By Schur’s lemma it is ±I, so U = ±Tᵢⱼ (p. 2). This holds for the transports alone. With entity vectors, xᵢxⱼᵀ is an equivariant edge term with no cycle, as the paper notes.
Triangle invariants. For a triangle, gᵢ = I, gⱼ = Tᵢⱼ, gₖ = TᵢⱼTⱼₖ send (Tᵢⱼ, Tⱼₖ, Tₖᵢ) to (I, I, H), leaving only H ↦ gHg⁻¹. The orbits are the conjugacy classes of H, and the continuous invariants are the class functions, generated by tr Hʳ, r ≤ d (Procesi). For b cycles the invariants are functions of (H₁, …, H_b) under simultaneous conjugation (p. 2). The same move is THEORY-1082’s spanning-tree step, with U(1) replaced by O(d).
The forced edge update (Theorem 1). In the gauge (I, I, H) define Ψ(H) = U(I, I, H). Equivariance under the residual diagonal gives Ψ(gHg⁻¹) = gΨ(H)g⁻¹, and undoing the gauge gives U = Ψ(H)Tᵢⱼ (p. 7). The claim that Ψ lies in the closure of Σᵣ cᵣ(H)Hʳ rests on Procesi’s polynomial classification and Stone–Weierstrass, and is sketched, not proved in full. Ψ = H⁻¹ gives composition through the third vertex, U = TᵢₖTₖⱼ.
Checked here. For random O(3) frames on six entities, the two-hop aggregate Π(Σₖ αₖTᵢₖTₖⱼ), with αₖ from the trace scores of the squares that avoid (i, j), was gauge-covariant and its scores invariant, to machine precision.
What could have failed. An invariant could have depended on a tree edge, or an equivariant update could have used an edge-adjacent transport outside every cycle. The gauge-fixing argument rules out both.
What this does not say
- That one mechanism is forced. Ψ is any conjugation-equivariant map, and scores are any class functions. The paper’s “the unique form the symmetry permits” is true of the family. Its weights, softmax, mean and polar projection are choices within it.
- That a task has the symmetry. The companion paper calls the premise “a chosen inductive bias, not a consequence of fresh names” (LIT-1803, p. 13). Its CLUTRR control, where orthogonal transports cost 64.6 points, is a task where the O(d) form is the wrong one.
- Anything about relations that are not invertible. Every step uses Tⱼᵢ = Tᵢⱼ⁻¹ and Hᵀ = H⁻¹.
- Where the transports come from. They are given. The result constrains how supplied relations may interact. It does not say which relations a system has (QUESTION-130, QUESTION-282).
- That this is new mathematics. The paper’s contribution is the statement on general relational graphs and the reading across three literatures, which it says itself (p. 3).