For every metric-measure space (X,d,mathfrak m), every K∈ R, and every 1<N<∞, is BCD(K,N) equivalent to RCD(K,N)? Equivalently, does the barycenter curvature-dimension condition characterize Riemannian curvature-dimension spaces?

Ambrosio's barycenter curvature-dimension problem

The problem asks whether barycenter entropy inequalities characterize Riemannian curvature-dimension spaces, through the relationship between $\mathrm{BCD}(K,N)$ and $\mathrm{RCD}(K,N)$ . The recent claim also addresses an almost- $\mathrm{BCD}$ condition and non-branching geometry.

Known results

Han and Liu, 2025: introduced $\mathrm{BCD}(K,\infty)$ and its finite-dimensional version, with entropy inequalities for Wasserstein barycenters.
Han and Liu, 2025: showed that on geodesic spaces, $\mathrm{BCD}$ implies the Lott–Sturm–Villani curvature-dimension condition and proved stability under measured Gromov–Hausdorff convergence.

September 2026 claimed solution

A September 2026 preprint by Bang-Xian Han and Deng-Yu Liu claims the equivalence of $\mathrm{BCD}(K,N)$ and $\mathrm{RCD}(K,N)$ , introduces an almost- $\mathrm{BCD}$ condition, and derives essential non-branching consequences. This is a claimed resolution, not an independently verified one.

Current status (as of September 2026): Han and Liu claim the $\mathrm{BCD}(K,N)$ – $\mathrm{RCD}(K,N)$ equivalence, but the new preprint is unrefereed and independent verification is not recorded.

Ambrosio's barycenter curvature-dimension problem

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Additional references

Progress summary

Solutions 0