Exact rational counterexamples to the triangle inequality for the Komálovics–Molnár distances d_(1/k), 2 ≤ k ≤ 8

Abstract

For p>0 and positive semidefinite matrices X,Y, Komálovics and Molnár set XκₚY=(X^(p/4)Y^(p/2)X^(p/4))^(1/p) and dₚ(X,Y)=(Tr((X+Y)/2-XκₚY))^(1/2), a one-parameter family through the quantum Hellinger (p=1) and Bures (p=2) distances, and asked (their Problem 2) whether dₚ on Mₙ(ℂ)₊, n ≥ 2, is a metric for each 1<p<2 and fails to be one for each 0 ≤ p<1, reporting numerical evidence for both halves. Zhang (arXiv:2602.11922) settled the single value p=1/2 by an explicit triple in M₂(ℂ)₊₊. We settle p=1/3,1/4,1/5,1/6,1/7,1/8, and give an independent, entirely rational witness at p=1/2. The mechanism is that at p=1/k the outer exponent is the natural number k: if X=X₁⁴ᵏ and Y=Y₁⁴ᵏ with X₁,Y₁ rational and positive semidefinite, then Tr(Xκ_(1/k)Y)=Tr((X₁Y₁²X₁)ᵏ) is a rational number produced by ring arithmetic alone, and the triangle inequality becomes a comparison of two rationals. Each triple is built from one rational reflection W of ℝ²: A₁=diag(1,1/10), B₁=WA₁W, and C₁ a polynomial in W with two rational eigenvalues chosen for each k, so that dₚ(A,C)=dₚ(C,B) and the failure reads dₚ(A,B)²-4 dₚ(A,C)²>1/10. The same triples satisfy the triangle inequality at p=1, the failure is by a factor below 2, and a rank-one triple satisfies the inequality at each p=1/k, as Komálovics and Molnár's Proposition 8 predicts. All statements are verified in Lean 4 against Mathlib. The half 1<p<2 of Problem 2 is untouched here; we record, as an unproved reduction, a two-variable inequality which would settle the whole range 0<p<1 at once.

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