# Curvature of Regular Tilings of the Hyperbolic Plane

Given a regular tiling of the hyperbolic plane, with p-sided polygons and q polygons meeting at each vertex, is the Gaussian curvature of the plane determined by (p,q)?

Intuitively it seems like having five pentagons meeting at each point would need greater (negative) curvature than four pentagons. But I'm just guessing? (And have never formally studied hyperbolic geometry.)

• Isn't the curvature of the hyperbolic plane $-1$ independent of what you draw on it? – arctic tern Jun 3 '17 at 0:50
• @articterm the curvature of the MODELS of hyperbolic geometry is almost always $-1$ but for a real hyperbolic plane it depends on the unit of measurement in Yrast question I suppose the side lengths (in his idea) stay the same while in yours the side lengths grow and the angles get smaller – Willemien Jun 3 '17 at 12:22
• Yeah I assumed that a regular tiling (e.g. using just one regular polygon to tile the entire plane) would mean the polygons are each made of straight lines with equal lengths and would have equal areas. And if that is true it seems like using more tiles around each point would require greater curvature. – Yrast Jun 4 '17 at 20:28

Usually the convention is that you use a hyperbolic plane with curvature $-1$ unless indicated otherwise. In such a plane, you can observe that small figures look almost Euclidean, but the larger a figure is, the more pronounced the difference in geometry becomes. Actually the angle deficit can be directly used as a measure of area. So for five pentagons meeting at each corner, you need a smaller interior angle, which means a larger angle deficit and thus a larger area than for for pentagons meeting at each corner.
The formula $\text{surplus angle}=\text{curvature}\times\text{area}$ holds in general. For hyperbolic geometry you have negative curvature and an angle deficit which is a negative surplus. For elliptic geometry surplus and curvature are positive, and for Euclidean both are zero. To fit more pentagons around a corner, you need more deficit and therefore have to increase either the absolute value of the curvature or the area of each pentagon. If you have a $p$-sided polygon, the Euclidean interior angle is $\frac{p-2}p\pi$. If you fit $q$ polygons around each corner the hyperbolic interior angle is $\frac2q\pi$. So the surplus angle is $\left(\frac2q-\frac{p-2}p\right)\pi$. For four pentagons this is $-0.1\pi$, for five pentagons it is $-0.2\pi$ so the value of the curvature would have to be twice as high in the latter case if you fix the area, or the area would have to be twice as large (corresponding to a $\sqrt2$ increase in edge length) if you fix the curvature.