Let $e = \lbrace u,v\rbrace$, $e' = \lbrace v,u'\rbrace$ be edges of an undirected graph $G$ and $ee'$ be the path from $u$ through $v$ to $u'$. The following defintions make sense for *every* graph and thus are purely combinatorial:

Definition 1: $ee'$ is apre-geodesicwhen it is the unique shortest path between $u$ and $u'$

Definition 2: A pre-geodesic $ee'$ is ageodesicwhen there is no other edge $f$ through $v$ such that $ef$ or $e'f$ is a pre-geodesic.

The first condition is motivated by the fact that geodesics in differential geometry are locally *shortest* paths and as such locally *unique*. It prevents the graph

from having geodesics.

The second condition reflects the fact, that geodesics usually do not “split”. It prevents the graph

from having geodesics.

But now, let's restrict to **polyhedral graphs**. I wonder if it's true for polyhedral graphs that only a small number of patterns of geodesics going through a given vertex $v$ are possible:

Which patterns?

How to prove this?

Have a look at this gallery of vertices $v$ (grey, drawn with *all* their neighbours) with degree at most 6 and at most three geodesics (red, blue, green) passing through them (all the other vertices may or must have *further* neighbours):

Is this gallery essentially complete?

If so: how to prove it?

Otherwise:

Is there a polyhedral graph with a vertex of degree 7 with a geodesic passing through it?

Is there a polyhedral graph with a vertex with four geodesics passing through it?

isa geodesic according to your definition. $\endgroup$ – Joseph O'Rourke Feb 6 '13 at 1:15nota pre-geodesic because they are not auniqueshortest path. $\endgroup$ – Hans-Peter Stricker Feb 6 '13 at 17:56