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Dimensionally absolutely stable but non-diagonalizable $T$-meshes
arXiv cs.NA0 views
arXiv:2611.05923v1 Announce Type: new Abstract: 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 \(d\ge 2\), the construction has \(2(d+2)\) \(\Tl\)-edges and intersection graph \(K_{d+2,d+2}\). Every \(\Tl\)-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.
Source:https://arxiv.org/abs/2611.05923PreviousLearning Coarse Spaces for Two-Grid Methods from Smoothed Errors with Applications to High-Contrast Porous Media FlowNextMagnetic Bloch eigenvalue problems in nonuniform periodic magnetic fields: a Fourier-finite difference approach
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