2-Dimensional Tilings and Curvature
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In Euclidean geometry, if we want to tile the plane with regular polygons of the same size, we can ask which pairs $(p,q)$ are admissible where $p$ is the number of sides of the regular polygon and $q$ is the number of them that would share the same vertex. The sum of the interior angles of a polygon with $p$ sides is $(p-2)\pi$ and if it’s regular, then each angle is $(p-2)\pi/p$. On the other hand, if we have $q$ of them sharing a vertex, to get total coverage, we need $\dfrac{q(p-2)\pi}{p} = 2\pi$ which is equivalent to $q(p-2)=2p$. Rearranging and adding $4$ to both sides gives $pq -2q-2p+4=4$ which then lets us factor the left: $(p-2)(q-2)=4$. The only solutions where $p,q>0$ to this are $(3,6),(4,4),(6,3)$. We can ask the same question but for other 2D geometries such as spherical or hyperbolic geometry. On the sphere, the sum of angles of a triangle is greater than $\pi$ and so the relationship is $(p-2)(q-2)<4$ whereas on the Poincaré hyperbolic disk, it is $(p-2)(q-2)>4$.
