Convex polyhedron and its Gauß-curvature

Question

I am trying to prove:

A convex polyhedron has positive Gauß-Curvature at every vertex.

What we know:

• Gauß-Curvature at every vertex is given by $K(p) = 2\pi - \sum\limits_{\text{angle } \alpha_i \text{ around } p} \alpha_i$.
• Gauß-Bonnet state $\sum\limits_{\text{every vertex }p} K(p) = 2\pi\chi(S)$, where $S$ the convex polyhedron. Therefore we can also write$\sum\limits_{\text{every vertex }p} K(p) = 4\pi$.

My attempt:

I plugged the Gauß-Curvature in the Gauß-Bonnet-Formula and obtained

$$ \sum\limits_{\text{every vertex }p} K(p) = \sum\limits_{\text{every vertex }p} \left(2\pi - \sum\limits_{\text{angle } \alpha_i \text{ around } p} \alpha_i\right) = 4\pi. $$ Assume now that there exists a $\tilde{p}$ such that $2\pi - \sum\limits_{\tilde{p}} \alpha_i < 0$. Then $2\pi < \sum\limits_{\tilde{p}} \alpha_i$...

From here i do not know how to get further on. Does anyone have an alternative of proving this statement or at least can somebody tell me how to complete my proof?

Thx in advance!

Answer 1

This is a near-duplicate answer from MathOverflow, where (on 1Jul2015) the question is [on hold], which means it will likely be closed and eventually deleted.

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Let me try to sketch a proof of "the evident fact that sum of angles around a vertex of a convex polytope is less than $2\pi$," as phrased by Alexandre Eremenko in a comment on MathOverflow.

Let $v$ be a vertex of the convex polyhedron $P$. Intersect $P$ with a small sphere $S$ centered on $v$. The intersection forms a convex spherical polygon $Q$ (blue below) lying within one hemisphere of $S$. (The hemisphere rim (red below) corresponds to a plane supporting $P$ at $v$.) We can take the radius of $S$ to be $1$ by appropriately scaling $P$ prior to intersecting. Note that the angles of the faces of $P$ incident to $v$ become the arcs of $Q$.

Now I would like to claim that the perimeter of a convex spherical polygon $Q$ lying within a hemisphere of a unit sphere $S$ is at most $2\pi$. I will mimic a proof in a Noam Elkies MO posting for a planar analog of the claim. The proof is by induction.

Let $Q_0$ be the hemisphere rim; its length is $2\pi$. $Q_0$ contains $Q$: $Q_0 \supseteq Q$. We will form a succession of more tightly enclosing convex spherical polygons, $Q_0 \supseteq Q_1 \supseteq Q_2 \ldots \supseteq Q$, with decreasing perimeters (more accurately: non-increasing perimeters).

Let $k$ be the number of edges of $Q$ that do not lie along the boundary of $Q_i$. If $k=0$, then $Q_i=Q$ and we are finished. Let $e$ be some edge of $Q$ that does not lie on the boundary of $Q_i$. Extend $e$ to a great-circle arc that cuts $Q_i$ at points $a$ and $b$. Define $Q_{i+1}$ as following $Q_i$ but then the "shortcut" $ab$. Because $ab$ is a geodesic, it is a shortest path between $a$ and $b$, and therefore cannot be longer than the portion of $Q_i$ cut off. So the perimeter of $Q_{i+1}$ is at most the perimeter of $Q_i$. Continuing in this manner produces the nested sequence of enclosing spherical polygons, ending with $k=0$.

If no edge of $Q$ lies along the original $Q_0$, then the first cut will be a semicircle (green below), and the perimeter of $Q_1$ is $\pi+\pi$, i.e., still $2 \pi$.

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So the perimeter of $Q$ is at most $2\pi$, the perimeter of $Q_0$. Therefore the sum of the angles of the faces of $P$ incident to $v$ is at most $2\pi$. Therefore, $P$ has positive angle deficit at $v$, i.e., it has positive Gaussian curvature there.

Note convexity is used crucially, for a nonconvex spherical polygon could have an arbitrarily large perimeter, still fitting inside a hemisphere.

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Added (2Jul2015). The example below is intended to show it is not correct to project the angles incident to $v$ to a plane and argue that the angles only enlarge under projection. Below, one angle is $68^\circ$ incident to $v$ in 3D, but that angle projects to $45^\circ$ in 2D.

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As you can see, some angles grow, some shrink, under projection.

Answer 2

Given any convex polyhedron $P$, consider the Minkowski sum of $P$ with $\bar{B}(\epsilon)$, a closed ball centered at $0$ with small radius $\epsilon$.

$$P_{\epsilon} \stackrel{def}{=} P + \bar{B}(\epsilon) = \{\; \vec{p} + \vec{q} : \vec{p} \in P, |\vec{q}| \le \epsilon\; \}$$

Let $K$ be the Gaussian curvature on the boundary $\partial P_{\epsilon}$. Let $V, E, F$ be the number of vertices, edges and faces of $P$ respectively. The boundary $\partial P$ is composed of $V + E + F$ fragments.

• $F$ planar polygons, one for each face.

  On these planer polygons, the Gaussian curvature $K$ vanishes.

• $E$ cylindrical fragments with radius $\epsilon$, one for each edge.

  For any edge $e$ of $P$, let $\ell_e$ be its length. Let $\psi_e$ be the angle between the two outward pointing normals of the two faces of $P$ attached to $e$. The cylindrical fragment is the "cartesian product" of a line segment of length $\ell_e$ and a circular arc of length $\psi_e\epsilon$. Since one of the principal curvatures vanishes on a cylinder, the Gaussiaon curvature $K$ again vanishes on these cylindrical fragments.

• $V$ spherical fragments with radius $\epsilon$, one for each vertex.

  Let $p$ be a vertex of $P$ and $e_1, e_2, \ldots, e_{d}$ be the edges in contact with $p$. For simplicity of description, we will extend the definition of $e_i$ for other integer $i$ by periodicity. We will assume the edges are ordered so that $e_{i}$ and $e_{i+1}$ are adjacent to each other. Let

  • $\alpha_i$ be the angle between the edge $e_i$ and $e_{i+1}$.
  • $\hat{n}_i$ be the outward pointing normal vector for the face between $e_{i}$ and $e_{i+1}$.

  The spherical fragment associated with $p$ will be a geodesic polygon having vertices at $\vec{v}_i = \vec{p} + \epsilon \hat{n}_i$. The geodesic between $\vec{v}_{i-1}$ and $\vec{v}_i$ has length $\epsilon \psi_{e_i}$.

  If we apply Gauss Bonnet Theorem to this geodesic polygon, we find its area is given by the formula $$\epsilon^2 ( 2\pi - \sum_{i=1}^d \beta_i)$$ where $\beta_i$ is the change of angle of the tangent vectors of the arcs corresponds to $e_{i}$ and $e_{i+1}$ at $\vec{p} + \epsilon\hat{n}_i$. The key is $\beta_i = \alpha_i$. To see this, switch to a new coordinate system where

  • $p$ is the origin.
  • the edge $e_1$ is along the $x$-axis, i.e. the direction $(1,0,0)$.
  • the edge $e_2$ is along the direction $(\cos\alpha_1, \sin\alpha_1, 0)$.
  • $\hat{n}_1$ is along the $z$-axis, i.e the direction $(0,0,1)$.

  It is easy to see

  • the plane determined by $\hat{n}_0$ and $\hat{n}_1$ is the $yz$-plane.
  • the plane determined by $\hat{n}_1$ and $\hat{n}_2$ is the one $-x\sin\alpha_1 + y\cos\alpha_2 = 0$.

  This means the angle between these two planes is equal to $\alpha_1$. This in turn implies, $\beta_1$, the change of angle of tangent vectors at $\hat{n}_1$ is equal to $\alpha_1$.

  As a result, the integral of $K$ over such a spherical segment is equal to to the angular deficit of corresponding vertex $p$. $$\frac{1}{\epsilon^2} \times \epsilon^2 ( 2 \pi - \sum_{i=1}^{d} \beta_i ) = 2\pi - \sum_{i=1}^d \alpha_i$$

These settles two questions:

1. Why the angular deficit can be viewed as a concentration of Gaussian curvature?

   This is because it is equal to the integral of Gaussian curvature of a smoothed version of $P$ around a neighborhood of corresponding vertex.

2. Why the angle deficit of a vertex is positive?

   The answer is close to trivial. The angular deficit is equal to the solid angle of corresponding spherical fragment!