The traveling salesman problem in the Heisenberg group: upper bounding curvature

We show that if a subset $K$ in the Heisenberg group (endowed with the Carnot-Carath\'{e}odory metric) is contained in a rectifiable curve, then it satisfies a modified analogue of Peter Jones's geometric lemma. This is a quantitative version of the statement that a finite length curve has a tangent at almost every point. This condition complements that of \cite{FFP} except a power 2 is changed to a power 4. Two key tools that we use in the proof are a geometric martingale argument like that of \cite{Schul-TSP} as well as a new curvature inequality in the Heisenberg group.