Topological triangulations of a sphere and disk

Question

The euler characteristic V-E+F=1 for the triangulation of a disk and V-E+F=2 for the triangulation of a sphere are necessary conditions for such triangulations to exist. What are sufficient conditions for a triangulation to be of a disk or sphere?

I'm ideally looking for something that can tell me if given the valence of the vertices, does there exist a triangulation that has those valences. For example, if the valences were four valence 3 vertices, then you could say there does exist a triangulation of a sphere (tetrahedron) with those valences. If the valences were five valence 10 vertices, then you could say no there doesn't exist a triangulation of a sphere with those valences.

I'd appreciate any sufficient conditions though, whether or not they're expressed in terms of the valence of the vertices. I don't care about the embedding of the triangulations. Just their connectivity.

Answer 1

It seems like you are asking whether a graph can be extended to a triangulation of a disk or a sphere. This about whether the graph is planar. The Wikipedia page has many criteria.

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If instead you want a way to check if a triangulation is of a disk or a sphere, here is an algorithm.

Let's assume the triangulation has the property that every face is defined by its three vertices.

Step 1, check if it is connected.

Step 2, check if it is a manifold. In 2D, you need to make sure

• each edge has one or two incident triangles.
• for each vertex, the incident triangles (the star) form a disk. In other words, the triangles can be listed in an order where each triangle is incident to the previous one. The last triangle might or might not be incident to the first depending on whether the vertex is on the boundary.

Step 3, count the number of boundary components. Make sure there's not more than one.

Step 4, calculate $V-E+F$. If it is $1$, it is a disk, and if it is $2$, it is a sphere.