Transverse rigidity is prestress stability (Q2081510)

Let \(G\) be a graph with \(n\) vertices spanning \(\mathbb{R}^d\). Assume also that it is isostatic: it has \(d(n-\frac{d+1}{2})\) edges, the ``right number'' for rigidity. Of course, this simple criterion does not guarantee rigidity. \textit{R. Connelly} and \textit{W. Whiteley} [SIAM J. Discrete Math. 9, No. 3, 453--491 (1996; Zbl 0855.52006)] gave a condition which they named ``prestress stability'' that is sufficient to guarantee rigidity of such a framework. \textit{V. Alexandrov} [Beitr. Algebra Geom. 61, No. 2, 355--368 (2020; Zbl 1439.52022)] gave another condition (here referred to as ``transverse rigidity'') that is also sufficient. In this short note, the authors show that the two conditions are in fact equivalent.