# Gaussian curvature of a planar graph (with edges of unit length)

Let's take a graph (with vertices and edges), which is in some sense 2D. In the simplest case consider a flat square lattice (or triangular, or hexagonal). But as well it can be a graph on a sphere (e.g. icosahedron) or a tessellation, e.g. uniform tilling in hyperbolic plane.

Is there a way to assign its Gaussian curvature? (Especially if it is constant.)

For a simple case, inspired by a game HyperRogue (a screenshot is below):

• each vertex (tile) has 6 or 7 edges (neighbouring tiles),
• length of each edge is 1 (it takes one turn to get to every neighbouring site).

It that case, how does suffice to know ratio of heptagons to hexagons?

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The standard way to assign a curvature would be to build a piecewise linear surface, where each face of the graph is a regular $n$-gon. Then the curvature would be sitting on the vertices, and the curvature at a vertex $v$ would be $K(v)= 1 - \frac{\alpha(v)}{2\pi}$, where $\alpha(v)$ is the sum of the angles of the faces at $v$.

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Sure (for n-gons it's $K = 1-n/6$). However, a graph does not provide unique angles. Is there a way to choose the 'right' angles or to argue that regardless of the choice, an averaged $K(v)$ is the same? – Piotr Migdal Dec 2 '12 at 15:36
If you have a closed surface, the averaged $K$ will be the same, no matter how you prescribe the angles. For non-closed surfaces you will get some kind of boundary term, or you have to normalize on the boundary. For infinite graphs, if you have uniformly bounded degree, you should also be able to show that for any sensible choice of angles (you don't want polygons degenerating in the limit) any sensible averaging mechanism will get you the same average curvature in the limit. – Lukas Geyer Dec 3 '12 at 17:22
For closed surfaces I see it. For infinite - I suspect that some restrictions on angles are needed as well (not to vary too much, I guess). – Piotr Migdal Dec 3 '12 at 18:40