We’ve received a couple questions on homework problem 2.2 regarding whether a constructive proof is necessary, i.e., do you need to actually come up with examples of triangulations that achieve the desired bounds? The answer is ** no**:

*you simply need to show (algebraically) that these bounds hold.*

But if you *were* looking for a constructive proof, a nice tool to know about is something called the *fundamental polygon*. The basic idea is that a torus of any genus can be cut into a single disk, which can be visualized as a polygon in the plane. Visualizing a triangulation (or other data) on the fundamental polygon is often much simpler than visualizing it on the embedded torus — consider this example, for instance:

Here the arrows mark identifications of edges — for instance, the top and bottom edges get “glued” together along the positive \(x\)-axis. (Try convincing yourself that these gluings really do produce a torus — note that at some point you’ll need to make a \(180^\circ\) “twist.”) One way to visualize the identifications is to imagine that the fundamental polygon tiles the plane:

The torus is interesting because it actually admits two *different* funamental polygons: the hexagon and the square (corresponding to two different tilings of the Euclidean plane). So we could also visualize the torus on the square, leading to an even simpler triangulation:

(By the way, is this really what you’d call a triangulation? Each region certainly has three sides, but each “triangle”‘ has only one vertex! What about the first example? There all triangles have three vertices, but they share the *same* three vertices. So combinatorially these triangles are not distinct — another way of saying this is that they do not form a valid *simplicial complex*. What’s the smallest simplicial decomposition you can come up with for the torus?)

In general the fundamental polygon for a torus of genus \(g\) has \(4g\) sides with identifications \(a_1 b_1 a_1^{-1} b_1^{-1} \cdots a_n b_n a^{-1}_n b_n\), where two edges with the same letter get identified and inverse just means that the edge direction is reversed. For instance, the fundamental polygon for the double torus looks like this:

(Note that all the polygon vertices ultimately get identified with a single point on the surface.) From here it becomes easy to start playing around with tessellations — for instance, here’s how you decompose a surface of any genus into *quadrilaterals*, using only two irregular vertices (can you do better?):

Tiling the Euclidean plane with the fundamental polygon is impossible for a surface of genus \(g \geq 2\), since the interior angles of the fundamental polygon don’t evenly divide \(2\pi\) (proof!). Fortunately, we can still tile the *hyperbolic plane*, i.e., the plane with constant negative curvature. For instance, here’s a tiling of the hyperbolic plane by octagons:

From here there are all sorts of fascinating things to say about covering spaces, uniformization, and especially the *fundamental group* of a surface — if you’re interested I highly recommend you take a look at Allen Hatcher’s (free!) book on algebraic topology.

There is a distinction called “regular triangulation.” In that case identified edges and/or vertices are not allowed. For purely topological arguments more general triangulations (which allow such identifications) are often fine. Once one wants to have a smooth structure and talk about differentiability and such it is often more convenient or even necessary to make sure the triangulation is regular. (One has to be careful with nomenclature here as a regular triangulation in the context of computational geometry is a Delaunay triangulation with weights at the sites. A whole different concept.)