Discrete Gauß and geodesic curvature Imagine that you have an n-polygon $S$ and you wanted to calculated the discrete Gaussian or gedoesic curvature. How are they defined? If $p$ is a vertex of $S$ then
Gauß-Bonnet suggests that the full integral $$\sum_{ p \in S} K(p) = 2\pi - \sum_{i=1}^n \theta_i ,$$
where $\theta_i$ are the external angles.
Also, one is tempted to interpret $$\sum_{p \in S} \kappa_g(p) = \sum_{i=1}^n \theta_i$$ as the full geodesic curvature. 
Are these two equations correct and if yes, how is the Gaußian and geodesic curvature of a single point defined in the discrete case?
If anything is unclear, please let me know.
 A: The notion of "discrete geodesic curvature" of a polygon is easy to define; as you suggest, one should take the geodesic curvature $\varepsilon_v$ at a vertex $v$ to be the exterior angle of the polygon at that vertex. As an argument that this is the right definition, we have a Gauss-Bonnet formula; for geodesic polygons on a Riemannian surface $S$, this says exactly that
$$
   \int_S K = 2\pi - \sum_v \varepsilon_v.
$$
A good reference is Lee's book Riemannian Manifolds; this is Theorem 9.3. For more general polygons (where the edges are nice enough smooth curves), an extra term appears on the right that is exactly the negative of the integral of the ordinary geodesic curvatures over the smooth parts of the curve, justifying calling $\varepsilon_v$ a "discrete geodesic curvature".
Now to move to discrete Gaussian curvature. For one thing, "discrete Gaussian curvature" is either meaningless or zero (depending on your philosophy) at the verticies of a polygon in a Riemannian surface. The fact that the edge of the polygon is "bent" is of no consequence to the Gaussian curvature, which is a measure of how much the surface "curves" -- it's a two-dimensional phenomenon, and for Riemannian surfaces, everything is smooth, so there's no interesting discreteness going on.
So instead we imagine the faces of a $3$-polytope; this is a flat surface except at some verticies, and we'd like to have an analog of the Gauss-Bonnet formula (and the Gauss-Bonnet theorem). For such a thing to be true we'll need a notion of Gaussian curvature that applies to the verticies (since elsewhere there is no curvature -- note that edges don't cause curvature, for the same reason that it's easy to fold a piece of paper).
The correct definition of the Gaussian curvature $K_v$ of the $3$-polytope at a vertex $v$ is that $K_v$ is the angle deficit at the vertex. That is, if $\theta_1, \dots, \theta_k$ are the interior angles at the vertex $v$ of the faces touching $v$, then
$$
  K_v = 2\pi - \sum \theta_i
$$
which can be positive or negative. Positively curved points have "too little" angle; negatively curved points have "too much".
As examples, the cube has a discrete Gaussian curvature of $\pi/2$ at each of its verticies, the tetrahedron a curvature of $\pi$, the octahedron a curvature of $2\pi/3$.
A picture I like to have is to imagine gluing $n$ equilateral triangles around a common vertex. When $n = 6$, the triangles will flatten; when $n = 5$, it's easy to see that one gets a cone-like object, which is positively curved (it looks like part of a discrete $2$-sphere). When $n = 7$, well -- try it with paper if you can't picture it -- you get something that looks more like a discrete saddle point -- negatively curved. 
In any case, the point is that there's a Gauss-Bonnet formula for polygons on such polytopes. In particular, let $P$ is a polygon in our polytope that bounds a homeomorphic disc $D$ and passes through no verticies of the polytope. Let $V_D$ be the set of verticies of the polytope inside $D$, and let $V_P$ be the set of verticies of the polygon. Then the expected analog of the Gauss-Bonnet theorem holds:
$$
  \sum_{v \in V_D} K_v = 2\pi - \sum_{v \in V_P} \varepsilon_v.
$$
This is something you can prove with just Euclidean geometry -- in particular I encourage you to think through the case where $V_D$ is a single point. You can also do examples very easily with a piece of paper and some scissors and tape; just cut out or glue in the desired angle deficit to make a vertex, and draw a polygon on the paper, and check that this formula holds "in real life". 
In addition, this implies an analog of the Gauss-Bonnet theorem for $3$-polytopes: The sum of the angle deficits at the verticies is exactly $2\pi$ times the Euler characteristic. 
