7
$\begingroup$

A triangulation of a two-manifold $M$ is three-colorable if all vertices of the triangulation can be colored red, green, or blue without any two adjacent vertices having the same color.

My question: do all two-manifolds admit a three-colorable triangulation?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
24
$\begingroup$

Sure. Start with an arbitrary triangulation $T$, and then takes its barycentric subdivision $T'$. That means that the vertices of $T'$ come in three types: (1) vertices of $T$, (2) midpoints of edges of $T$ and (3) centers of faces of $T$.

Color these three types red, blue and green respectively.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ Indeed. Any superficial resemblance to four-color-type problems is misleading. :) $\endgroup$
    – paul garrett
    Commented Dec 23, 2023 at 0:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .