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Consider a convex polyhedron $\mathcal{P}$ in $\mathbb{E}^3$ with $n$ edges. By convex polyhedron, I mean the convex hull of a finite set of points in $\mathbb{E}^3$ whose affine hull is all of $\mathbb{E}^3$. And by edge, I mean a line segment along which precisely 2 distinct faces meet at an angle less than $\pi$. Let $\theta_\mathrm{max}(\mathcal{P})$ be the maximal dihedral angle of $\mathcal{P}$. It seems intuitive to me that, as $n$ grows, $\theta_\mathrm{max}(\mathcal{P})$ must approach $\pi$, regardless of which polyhedron I choose. To make this more precise, I am suggesting that $$\inf\{\theta_\mathrm{max}(\mathcal{P})~|~\mathcal{P}\text{ a convex polyhedron with }n\text{ edges}\}\longrightarrow\pi\quad\text{as}\quad n\to\infty.$$

Is this statement true? If so, how does one prove it? Please reference any literature you might have used in an answer.

Edit 25/06/20 – I believe this problem is interesting because if we ask the question about vertices and their conical angles instead, we can show that the maximum always approaches $2\pi$ by just using Gauss-Bonnet. This is in some way an analogous statement question one dimension higher.

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  • $\begingroup$ What do you mean, I don't need $n\to\infty$? I don't just want to come up with a sequence of polyhedra whose maximal dihedral angle tends to $\pi$. I want to show that for any sequence of polyhedra with increasing number of edges, the maximal dihedral angle tends to $\pi$. Another way of looking at this is to say that if you bound the maximal dihedral angle strictly away from $\pi$, then the number of edges is bounded. $\endgroup$
    – Tom Sharpe
    Commented Jun 24, 2020 at 14:19
  • $\begingroup$ I'm sorry. I overlooked the $\inf$. $\endgroup$
    – saulspatz
    Commented Jun 24, 2020 at 14:25
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    $\begingroup$ The Gauss-Bonnet theorem, applied to a small sphere centred on a vertex, gives the solid angle in terms of the dihedral angles: $$\Omega_v=2\pi-\sum_{e\supset v}(\pi-\theta_e)$$ $$=2\pi-\pi\sum_{e\supset v}1+\sum_{e\supset v}\theta_e$$ so, summing over vertices (and noting that each edge gets counted twice), $$\sum_v\Omega_v=2\pi\sum_v1-\pi\cdot2\sum_e1+2\sum_e\theta_e.$$ Thus the average dihedral angle is $$\frac{\sum_e\theta_e}{\sum_e1}=\pi-\frac12\frac{\sum_v(2\pi-\Omega_v)}{\sum_e1}.$$ I don't know where to go from here.... $\endgroup$
    – mr_e_man
    Commented Jun 24, 2020 at 15:36
  • $\begingroup$ Well, each vertex has at least three edges, so our denominator is $$2\sum_e1=\sum_v\sum_{e\supset v}1\geq\sum_v3;$$ $$\frac{\sum_e\theta_e}{\sum_e1}\geq\pi-\frac13\frac{\sum_v(2\pi-\Omega_v)}{\sum_v1}.$$ Therefore, if the average solid angle approaches $2\pi$, then the average dihedral angle approaches $\pi$. $\endgroup$
    – mr_e_man
    Commented Jun 24, 2020 at 16:03
  • $\begingroup$ But you want the maximum, not the average. And, for example, a pyramid with increasing number of edges doesn't have the average approach $\pi$; it has $n$ rising edges with angles approaching $\pi$, but also $n$ base edges with angles below $\pi/2$, so the average stays below $3\pi/4$. $\endgroup$
    – mr_e_man
    Commented Jun 24, 2020 at 16:18

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No, it's not true.

Take a regular $n$-gon prism, and cut off the vertices, with planes sloped at some angle $\beta$, so that the original edges are reduced to points. (You can ensure that the horizontal and vertical edges disappear at the same time by adjusting the height of the prism. But we don't really need such precision; we only need to get rid of the vertical edges, which are the ones with a large dihedral angle.) The result is a rectified prism.

The square (rather, rhombus) faces have normal vectors

$$a_k=\left(\cos\frac{(2k+1)\pi}{n},\;\sin\frac{(2k+1)\pi}{n},\;0\right),$$

the triangular faces have normal vectors

$$b^\pm_k=\left(\sin\beta\cos\frac{2k\pi}{n},\;\sin\beta\sin\frac{2k\pi}{n},\;\pm\cos\beta\right),$$

and the $n$-gon faces have normal vectors

$$c^\pm=(0,0,\pm1).$$

Consider these as points on the unit sphere (see Gauss map). The exterior (or supplementary) dihedral angle at an edge is just the spherical distance between the two faces' normal vectors. The angle deficit at a vertex is the area of the spherical polygon formed by the surrounding faces' normal vectors. In fact, the Gauss map gives another realization of the dual polyhedron, as a tiling of the sphere. So your question is whether a spherical tiling with a large number of vertices (or edges, or tiles) must have a short edge.

This particular polyhedron has only two types of edge. The exterior dihedral angles are given by

$$\cos\theta_{ab}=a_k\cdot b_k^\pm=a_k\cdot b_{k+1}^\pm=\sin\beta\cos\frac\pi n$$

(between a square and a triangle), and

$$\cos\theta_{bc}=b_k^\pm\cdot c^\pm=\cos\beta$$

(between a triangle and an $n$-gon). Clearly, as $n\to\infty$, neither $\theta_{ab}$ nor $\theta_{bc}\to0$.

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  • $\begingroup$ Yep, you're absolutely right! Well done and thanks. I'll have to think of a modified statement involving something more subtle than just dihedral angle. $\endgroup$
    – Tom Sharpe
    Commented Jun 26, 2020 at 14:09

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