Answers are grounded in this paper and cite their sources.
We construct an infinite family of rectangular \(T\)-meshes whose conformality-vector-space dimension is constant over all structurally similar geometric realizations, although the meshes are not diagonalizable. For every integer \(d≥ 2\), the construction has \(2(d+2)\) \(\)-edges and intersection graph \(Kd+2,d+2\). Every \(\)-edge contains at least \(d+2\) vertices, so no edge is vanishable under the usual vertex-count criterion for bidegree \((d,d)\) and smoothness \((d-1,d-1)\). We prove that the conformality-vector space is one-dimensional throughout the structural similarity class. Its generator is the outer product of two vectors of barycentric interpolation weights, extended by four zero coordinates. Non-diagonalizability follows by examining the last edge in any proposed diagonalizing ordering. Thus, for the geometric structural similarity class specified here, diagonalizability is not a necessary condition for dimensional absolute stability.
No takes yet. Share an insight, caveat, or question.
Mao Shi (2026) studied this question.
Synapse has enriched 8 closely related papers on similar clinical questions. Consider them for comparative context: