TITLE: When is a triangulation of sphere two-colorable?
QUESTION [8 upvotes]: Let $T$ be a triangulation of sphere. We say that $T$ is $k$-colorable if the triangles of $T$ can be assigned with $k$ colors such that any two triangles with a common edge have different colors.
I am interested in $2$-colorable triangulations. Easy examples are boundaries of bipyramids over even polygons such as the simplicial comlex $\{123,134,145,152, 623,634,645,652\}$
Question: Are there efficient criteria to tell if a triangulation is $2$-colorable? How to construct $2$-colorable triangulations? Do they arise in some contexts?
I am also interested in triangulations of other manifolds, so any such references are welcome.

REPLY [11 votes]: As Fedor Petrov mentions in the comments, a necessary and sufficient condition is that each vertex has even degree.  Here is a proof.
Let $T^*$ be the dual graph of the triangulation.  That is, the vertices of $T$ are the faces of the triangulation, and two faces are adjacent if they share an edge.  Rephrased, your question is asking when $T^*$ is bipartite.  So, it suffices to prove that a planar graph $G$ is bipartite if and only if the dual graph $G^*$ is Eulerian (all vertices have even degree). For one direction, if $G$ is bipartite, then all cycles of $G$ are even.  In particular, all facial cycles are even, and hence $G^*$ is Eulerian.  For the other direction, suppose that the dual graph $G^*$ is Eulerian.  Thus, each face of $G$ is even.  Since the cycle space of $G$ is generated by the facial cycles of $G$ (take the symmetric difference of all faces inside the cycle), this implies that every cycle of $G$ is even.  Hence $G$ is bipartite.