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The converse Alexandrov-to-distributional sectional curvature conjecture for $C^1$ metrics
Converse curvature conjecture. The distributional sectional curvature of $(M,g)$ is bounded from below (respectively, above) by $k$ .
Conjectureopenalexandrov-spaces low-regularity-metrics riemannian-geometryarXiv posted 2 days ago

The converse Alexandrov-to-distributional sectional curvature conjecture for C1C^1C1 metrics