Degenerate Versus Full Ma–Trudinger–Wang Conditions

Definitions and mathematical statement

Let $(M,g)$ be a Riemannian manifold, and consider the squared-distance cost $c(x,y)=\tfrac12d_g(x,y)^2$. The factor $1/2$ is a positive normalization. Work on the intended domain $\mathcal D\subseteq M\times M$ where $c$ is $C^4$ and its mixed Hessian $(c_{i\bar j})$ is invertible, ordinarily off the cut-locus relation. Bars distinguish derivatives in the second variable. Let $(c^{\bar a b})$ be the inverse matrix.

One conventional sign choice for the MTW biquadratic form is \[ \operatorname{MTW}_{(x,y)}(\xi,\eta)= \left(-c_{ij\bar k\bar l} +c_{ij\bar a}c^{\bar a b}c_{b\bar k\bar l}\right) \xi^i\xi^j\eta^{\bar k}\eta^{\bar l}, \] where $\xi\in T_xM$, $\eta\in T_yM$, and repeated indices are summed. This fixes the sign convention; a positive overall factor does not affect nonnegativity. The cost-orthogonality constraint is \[ c_{i\bar j}\xi^i\eta^{\bar j}=0. \] Because the two vectors lie at different points, this is not an undefined ordinary inner product between $T_xM$ and $T_yM$; an alternative formulation using two vectors at $x$ requires the mixed-Hessian identification.

The degenerate/weak orthogonal condition $\operatorname{MTW}^{\perp}\geq0$ means nonnegativity for every $(x,y)\in\mathcal D$ and every cost-orthogonal pair $(\xi,\eta)$. The stronger full condition $\operatorname{MTW}\geq0$ means the same inequality for all pairs, with no orthogonality restriction. The source asks whether, specifically for Riemannian squared-distance costs, \[ \left[\forall(x,y)\in\mathcal D\ \forall\xi,\eta \text{ with }c_{i\bar j}\xi^i\eta^{\bar j}=0, \ \operatorname{MTW}_{(x,y)}(\xi,\eta)\geq0\right] \] \[ \Longrightarrow\quad \left[\forall(x,y)\in\mathcal D\ \forall\xi,\eta, \ \operatorname{MTW}_{(x,y)}(\xi,\eta)\geq0\right]. \] The full condition plainly includes the orthogonal tests; the requested direction removes the constraint. No implication for arbitrary transport costs, no strict positivity, and no quantitative lower curvature bound is asserted. The source does not specify the manifold class, exact domain, or whether a pointwise variant rather than a domain-wide hypothesis is intended.

The supplied assessment dated 2026-08-21 reports the orthogonal-to-full question as open and says known regularity results use stronger assumptions. It does not provide those hypotheses or justify treating a regularity theorem as proof of this implication.

Short English statement

For squared Riemannian distance, does nonnegative MTW curvature on cost-orthogonal vector pairs force nonnegativity on every pair? The smooth cost domain and tensor sign/identification conventions must be fixed, and no stronger positivity assumption is part of the question.

Sources

• Local UnsolvedMath registry, permanent ID $30000997$, exact source code catalogue entry, proposed year 2008; no individual proposer is supplied. MTW expands as Ma--Trudinger--Wang.
• Calculus of Variations (2009), Oberwolfach Reports, DOI 10.4171/owr/2008/31, pp. 1707--1770. https://doi.org/10.4171/owr/2008/31.
• Supplied reference labelled "Kim--McCann curvature paper": https://math.ubc.ca/~yhkim/yhkim-home/research/counterexample-A3w.pdf.
• The supplied corrected_verified note reports checking mathematical content against the official report. The open label and assessment checked 2026-08-21 are dated supplied triage requiring independent verification of exact cost, curvature conventions, and current status.