Lower and upper bound of discrete gaussian curvature. In a triangle 2-manifold mesh, or symplicial complex (but I'll stick with the former terminology) the discrete gaussian curvature is usually defined
$$
K(v_i) = \frac{1}{A(i)}\left(2\pi - \sum_{(v_i,v_j,v_k)} \theta_i^{jk}\right)
$$
Where $v_i$ is a vertex of the triangle mesh, $(v_i,v_j,v_k)$ is a triangle of the mesh (identified with the triple of vertices) and $\theta_i^{jk}$ is the angle at $v_i$ of the triangle $(v_i,v_j,v_k)$. Moreover $A(i)$ is the voronoi area around $v_i$, namely one third of sum of all triangles area, incident to $v_i$.
However for example in this C++ library (Geometry Central) it seems that the definition is simply
$$
K(v_i) = 2\pi - \sum_{(v_i,v_j,v_k)} \theta_i^{jk}
$$
For this last version (which I don't know whether or not its correct but it doesn't matter too much I guess) the question I have is what is the lower bound and upper bound of such formulas?
I would've guessed that
$$0 \leq K(v_i) \leq 2\pi$$
But because I see sometimes negative values as well maybe I am wrong? I was able to construct by hand few examples of negative $K$ but the ones I managed are only if the vertex $v_i$ is a boundary vertex. A general proof would be useful, or a reference I can look up.
To summarize then

*

*What are the bounds of $K(v_i)$?

*What is the proof of such bounds?

 A: Assuming triangles are Euclidean triangles, the second "angular defect" definition of curvature at a vertex satisfies
$$
K(v) < 2\pi.
$$
The upper bound is strict because every interior angle of a triangle is strictly positive, but a triangle can have an arbitrarily small positive interior angle so the angular defect can be as close to $2\pi$ as we like. There is no lower bound because an arbitrarily large number of congruent triangles can share one vertex. (Think of a polyhedral Elizabethan collar with no neck hole. For example, take four right isosceles triangles whose right angles meet at a point $v$, and which together form a square. Slit this square along a side of two adjacent triangles, and accordion-fold the four triangles along their successive common sides. Now take as many of these units as desired and join them along their free (slit) edges successively and cyclically. The incident angle at the vertex is $2\pi k$ for some arbitrary positive integer $k$.)

The angular defect is (properly analogous to) the total curvature at the vertex $v$. In other words, a polyhedron vertex with total incident angle unequal to $2\pi$ has infinite curvature, but the integral over every neighborhood of the vertex is the angular defect. Three related ways to see/interpret this claim are:

*

*Consider the image of the Gauss map of a cone with its vertex rounded off and take a suitable limit as the radius of rounding decreases to $0$.

*Calculate the holonomy around a closed path enclosing the vertex, cf. the local Gauss-Bonnet theorem.

*Show that for a closed polyhedron, the sum over all vertices $v$ of "$2\pi$ minus the sum of the angles incident at $v$" is $2\pi$ times the Euler characteristic, cf. the global Gauss-Bonnet theorem.

The last two points are explored in Paper Surface Geometry: Surveying a locally Euclidean universe, Amer. Math. Monthly, 123 (2013), 487–499.
The first definition of discrete Gaussian curvature is consistent with $K(v_{i}) A(i)$ being the integral of the angular-defect (second definition) curvature over a vertex neighborhood comprising one-third of each face touching $v_{i}$.
