Introduction and text

The last section of Bloch’s chapter 3 (simplicial surfaces) is a long (and to my mind at this time, uninteresting) proof of the 2D Brouwer fixed point theorem: any continuous map from the disk to the disk has a fixed point. Bloch also proves a corollary, the no-retraction theorem, that there is no continuous map r from the disk to the circle such that r(x) = x for all x on the circle.

That one sounds interesting. We’ve seen in before, with the commentary that you can’t map the surface of a drum onto its rim without tearing it. I still don’t see it that way. But it is rather shocking that the map r cannot preserve all the points on the rim.

Anyway, we’re not going to fight with those. For me, the climax of chapter 3 is the simplicial Gauss-Bonnet theorem. It shows that there is a definition of curvature for simplicial surfaces (in fact, for polyhedra in general) such that the total curvature of a surface is equal to $2\ \pi$ times its Euler characteristic $\chi\$.

(A simplicial surface is a polyhedron all of whose faces are triangles. I expect we’ll see this again in another post.)

That the total Gaussian curvature of a surface is equal to $2\ \pi\ \chi$ is called the Gauss-Bonnet theorem. It is a reasonable culmination of a first course in differential geometry. The simplicial version means that we have a definition of curvature for simplicial surfaces and polyhedra which gives us a form of the Gauss-Bonnet theorem. That says it’s a reasonable definition of curvature.

So what is this marvelous definition of simplicial curvature? It’s also called the angle defect, and goes back to Descartes.

First off, it turns out that all the curvature of a polyhedron (or simplicial surface) is concentrated at the vertices. There is no contribution from the edges.

(Either recall or trust me that the Gaussian curvature of a cylinder is zero…. The simplicial curvature of an infinitely long polygonal cylinder ought to be zero. That is, if instead of a circular cross-section, our cylinder has a polygonal cross-section, we would like the total curvature to still be zero. I take it to be infinitely long so that it has no vertices, only edges and faces. Such a thing is not a compact surface, but it may help justify the idea that edges do not contribute to simplicial curvature.)

Imagine some polyhedron. There are at least two ways to describe the simplcial curvature. One is to look at each vertex. For a fixed vertex v, consider the angle at v for each of the faces $\eta$ which touch v, add up those angles, and subtract the sum from $2\ \pi\$. (That’s why it’s also called the angle defect: it measures how far from $2\ \pi\$, i.e. how far from planar, the polygons are when they’re touching. The pictures that follow may help.)

Note that we are looking at one angle from each polygon that touches the vertex v; we are not measuring an angle in space at the vertex. Anyway, add up all the angles at each vertex, and subtract that sum from $2\ \pi\$, and call it the angle defect at v.

Then add up all the angle defects: that sum is the total simplicial curvature, and it’s equal to $2\ \pi\ \chi\$ for that surface.

The other way to do it is to count all the angles first. Add up all the angles in all the faces. Subtract that sum from $2\ \pi\ V\$, where V is the number of vertices. (Each angle in a polygon is associated with one vertex.)

I believe I could write that compactly as:

Let K be a polyhedron, and let $v \in K$ be a vertex. If $\sigma \in K$ is a face (polygon) containing v, let $\angle(v,\ \sigma)$ denote the angle at v in $\sigma\$. The curvature of K at v (or the angle defect at v) is defined to be the number d(v) given by

$d(v) = 2\ \pi\ - \sum_{\eta}\angle(v,\ \eta)$

where the $\eta$ are the faces of K containing v.

The total simplicial curvature of the polyhedron is the sum of the angle defects,

$\sum_{v} d(v)\$.

Let’s look at some examples.

regular tetrahedron

The simplest possible case is a regular tetrahedron: put 4 equilateral triangles together.

There are 4 vertices and every vertex is the same. At each vertex, 3 equilateral triangles meet, so the sum of the angles at one vertex is

$3\ \frac{\pi}{3} = \pi$

so the angle defect at one vertex is

$2\ \pi - \pi = \pi$

and there are 4 vertices, so the total simplicial curvature is $4\ \pi\$.

By direct computation, the Euler characteristic of a tetrahedron is $\chi = v - e + f = 4 - 6 + 4 = 2\$.

Therefore, we do indeed have

$2\ \pi\ \chi = 4 \pi = \text{total simplicial curvature}\$.

slighly irregular tetrahedron

What about an irregular tetrahedron? Suppose we put the vertices at the origin and on each axis at a distance of 1. There are 3 faces on each of the coordinate planes: these faces are $1 - 1 - \sqrt{2}$ right triangles. The 4th face is an equilateral triangle.

The 4th face does not touch the vertex at the origin (shown orange); we have 3 right angles there. The sum of the angles is $3\ \frac{\pi}{2}\$,

so the angle defect is

$2\ \pi - 3\ \frac{\pi}{2} = \frac{\pi}{2}\$.

(Oh, of course! Look at the picture. More importantly, look at the open space at the orange vertex. That open space has an angle of $\frac{\pi}{2}\$. That’s where the $2 \pi$ comes from, and why we subtract from it.

At the other three vertices (shown yellow, blue, black; the black vertex is in 3 pieces), we have one equilateral face $\left(60{}^{\circ} \right)$ and two $45{}^{\circ}$ angles from the other two faces. At each of these 3 faces, the sum of angles is

$\frac{\pi}{3} + 2\ \frac{\pi}{4} = \frac{5\ \pi}{6}\$,

so the angle defect is

$2\ \pi - \frac{5\ \pi}{6} = \frac{7\ \pi}{6}\$.

If we look at either the yellow or blue vertices, the portion that is not faces is certainly larger than $180{}^{\circ} \$, and in fact we can see that it’s $210{}^{\circ} \$, which is, indeed, $\frac{7\ \pi}{6}\$.

Now we add up all the angle defects

$\frac{\pi}{2} + 3\ \frac{7\ \pi}{6} = 4\ \pi\$.

Once again, that’s $2\ \pi\ \chi = 2\ \pi\ 2 = 4\ \pi\$.

Looking at any of the vertices shown as blue dots, we should guess that the angle defect is $\frac{\pi}{2}\$ at each of the 8 vertices, so the total is, once again,

$8\ \frac{\pi}{2} = 4\ \pi\$.

Alternatively, 3 rectangles meet at each vertex, so the sum of the angles is $3\ \frac{\pi}{2}$

and the angle defect at each vertex is directly computed to be

$2\ \pi - 3\ \frac{\pi}{2} = \frac{\pi}{2}\$.

OTOH, the Euler characteristic of a cube is

$\chi = v - e + f = 8 - 12 + 6 = 2\$,

so $2 \pi\ \chi = 4 \pi\$.

(I know perfectly well that the cube is homeomorphic to the sphere, and therefore has the same Euler characteristic, but come on: we compute the Euler characteristics of polyhedra and simplicial surfaces, and then define the Euler characteristic of the sphere from them. I will admit, however, that if don’t get 2 by direct computation (for polyhedra without holes), I know I made a mistake!)

hexagonal cylinder (prism)

How about a finite polygonal cylinder? We know it’s homeomorphic to a sphere ($\chi = 2$), so we know the total simplicial curvature should be $4 \pi\$, but let’s work it out.

Here’s a regular hexagonal cylinder, or prism. There are 12 vertices; each is touched by 2 rectangles $\left(90{}^{\circ} \right)$ and a hexagon $\left(120{}^{\circ} \right)$.

(Oh, cut the hexagon into 6 equilateral triangles; each angle at a vertex of the hexagon is made up of two $60{}^{\circ} \$ angles.)

At each vertex, then, the sum of the angles is

$2\ \frac{\pi}{2} + \frac{2\ \pi}{3} = \frac{5\ \pi}{3}\$,

so the angle defect is $\frac{\pi}{3}$

and then the sum of those (i.e. 12 times the individual defect) is the simplicial curvature,

$12\ \frac{\pi}{3} = 4\ \pi\$.

torus

How about a torus? First off, let’s grab a triangulation from here.

Just look at any interior vertex: there is no gap, no space where a face isn’t. The angle defect should be zero at every interior vertex.

But every vertex is interior. The ones that seem to be on the boundary are identified with those on another boundary.

Each of the angle defects is zero, the sum of the angle defects is zero, and we need

$2\ \pi\ \chi = 0\$,

But that triangulation has 18 faces (that’s the easy number), 27 edges, and 9 vertices. $\chi = 9 - 27 + 18 = 0\$.

Good. We’ve finally seen a case that didn’t add up to $4\ \pi\$.

Gaussian curvature of the smooth torus

One last thing. The smooth torus, like the cylinder, has total gaussian curvature equal to zero. Unlike the cylinder, however, the torus does not have constant curvature equal to zero everywhere.

(When we take a finite cylinder and bend it to make a torus, we have to stretch the material that goes to the outside and compress the material that goes to the inside.)

In case you’re curious, if we parameterize a torus as

$\{r \sin (u),(R+r \cos (u)) \cos(v),(R+r \cos (u)) \sin (v)\}$

then I believe the Gaussian curvature (which I have not shwn you how to compute) is cos(u). (Really? Independent of the radii r and R? Yes! They show up in both the curvature and the element of area dA, and cancel out.) Integrated over $u \in [0,\ 2\pi]\$, that will be zero. And zero integrated over $v \in [0,\ 2\pi]\$ will still be zero.

What did I actually get from that parameterization? (Take r = 1, R = 3.)

The red curve (outer rim) is u = 0, so cos u = 1 and the curvature > 0.
The black curve (inner rim) is $u = \pi\$, so cos u = -1 and the curvature < 0.
The green curve is $u = \frac{\pi}{2}\$, so cos u = 0 = curvature.

So we have seen some examples of computing the curvature of simplicial surfaces and polyhedra.