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Suppose $p_1,\ldots,p_n$ are points in $\mathbb{R}^3$, and suppose $\delta_1,\ldots,\delta_n$ are positive real numbers, each less than $2\pi$, whose sum is $4\pi$. Is there a polyhedron $\mathcal P$ which has a vertex at $p_i$ of angular defect $\delta_i$ (meaning that the sum of the interior angles of the faces meeting at $p_i$ is $2\pi-\delta_i$) for each $i=1,\ldots, n$, and possibly some extra vertices all of whose angular defect $0$? That is, I am specifying some of the vertices and angular defects, but allowing the addition of extra vertices on the condition that those vertices have angular defect zero.

If so, is there a constructive proof? If not, are there known constraints on the points and/or defects?

The question can be extended more broadly: the condition that the defects are positive could be dropped and we could ask that $\delta_1+\cdots+\delta_n=2\pi \chi$.

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  • $\begingroup$ Related: Determining a surface in $\mathbb{R}^3$ by its Gaussian curvature. $\endgroup$ Commented Feb 1, 2021 at 2:35
  • $\begingroup$ I think (not certain) you are seeking Alexandrov's Theorem? In more detail: Bobenko, Alexander I., and Ivan Izmestiev. "Alexandrov’s theorem, weighted Delaunay triangulations, and mixed volumes." In Annales de l'institut Fourier, vol. 58, no. 2, pp. 447-505. 2008. $\endgroup$ Commented Feb 1, 2021 at 2:45
  • $\begingroup$ 1. Since $0<\delta_j<1$ your polyherdon must be convex therefore you must have $\chi=2$. 2. You cannot prescribe $\delta_j$ and $p_j$ simultaneously. The $\delta_j$ aready define the polyhedron up to scaling and rotation. The $p_j$ define the polyhedron up to finitely many choices. $\endgroup$ Commented Feb 1, 2021 at 11:49
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    $\begingroup$ @stepnp21: the changes you made do not help. You cannot prescribe both angular defects AND vertices. For example, if you have 4 angular defects each $=\pi$ then the polyhedron MUST be a regular tetrahedron, so you cannot prescribe vertices. $\endgroup$ Commented Feb 1, 2021 at 22:00
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    $\begingroup$ @AlexandreEremenko If you take a cube and attach to four of its faces square-based pyramids whose triangular faces have angles $\pi/4$, $3\pi/8$, $3\pi/8$, then that gives another polyhedron with four vertices, each with angular defect $\pi$; I've added 8 vertices with angular defect 0 to accomplish this. $\endgroup$
    – stepanp21
    Commented Feb 2, 2021 at 19:09

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