Math-4038 · mathematics · Geometry

Math-4038

Scalar Curvature Question [?64]: Are all extremal convex polyhedraP are mean convexly extremal

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$\mathbb N=\{1,2,\ldots\}$, and $\mathbb Z,\mathbb Q,\mathbb R,\mathbb C$ denote the integers, rationals, reals and complex numbers. Any other conventions are specified locally.

For a Riemannian domain, mean curvature $H$ of a smooth boundary face is the trace of its second fundamental form with the convention that the boundary of a Euclidean ball has positive mean curvature. Mean convexity means $H\geq0$. Dihedral angles are interior angles between adjacent faces, measured between their tangent spaces along their common codimension-two faces. Let $P\subset\mathbb R^n$ be a compact convex polyhedron with nonempty interior, with codimension-one faces $Q_i$, $i\in I$, and interior dihedral angles $\alpha_{ij}$ at adjoining faces. Two polyhedra are combinatorially equivalent if their face posets are isomorphic, with the indicated correspondence of faces. Call $P$ extremal if every convex polyhedron $P'$ combinatorially equivalent to $P$ satisfying $\alpha'_{ij}\leq\alpha_{ij}$ at all adjoining face pairs necessarily satisfies equality at every such pair. Call $P$ mean-convexly extremal if there is no Euclidean domain $P'\subset\mathbb R^n$ with smooth faces and corners, diffeomorphic to $P$ with its corresponding faces, satisfying all of the following: every face $Q'_i$ has $H\geq0$; at every point of $Q'_i\cap Q'_j$ the interior dihedral angle is at most $\alpha_{ij}$; and at some point of some such intersection the angle is strictly smaller than $\alpha_{ij}$.

Statement recorded in the supplied source.
Is every extremal convex polyhedron mean-convexly extremal? [S2]

Short English statement

Does extremality against convex polyhedra also prevent any mean-convex domain of the same face type from decreasing all dihedral angles and strictly decreasing at least one?

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