12 pentagons!

I’ve been reading Sternberg’s “Group Theory and Physics”, Cambridge University, reprinted 1999. On pages 43 to 44 he says, “… every fullerene has exactly 12 pentagons. This is not an accident.”

The stable structure of carbon which has 60 carbon atoms arranged like the vertices of a soccer ball is called a buckyball. It turns out that, in similar structures, we can have any even number, greater than 18, of carbon atoms except for 22. This is equivalent to polyhedra having 12 pentagons and any number of hexagons except 1.

This family of structures consists of polyhedra whose faces are either pentagons or hexagons. In chemistry they are labeled by the number of carbon atoms, so they talk about

I find it unforgettable and marvelous that the number of pentagons is always exactly 12. And I can prove it.

Sternberg’s proof is somewhat different from mine, because I choose to use two general equations which we have seen before. One equation depends on the fact that every edge separates exactly 2 faces; the other depends on the fact that every edge joins exactly 2 vertices.

Imagine that we have a polyhedron. Choose one face, and count the number of edges it has. Do this for all the faces.

We will have counted each edge twice. Let be the number of faces with k edges; then, for example, is the number of pentagons. We have the general formula

.

(I think I first showed you that here.)

For example, the term says that we counted 3 edges for each triangular face we touched. But the 2e on the LHS says that we will have counted each of those edges as part of another face, too. And yes, an edge can join two different kinds of polygons.

Similarly, if we count the number of edges by counting the number at each vertex, we will have counted each edge twice. Let be the number of vertices with k edges; we have the formula

(I believe I have not shown you that equation, except in the very special case 2 e = 3 v, for triangulations.)

We also have Euler’s formula in general

and

in particular.

Let’s see what the smallest case looks like: 12 pentagons.

This is the dodecahedron, one of the five platonic solids. (Now is a good time to remark that people used to think of them as 3D solids, but for quite some time mathematicians have treated them as 2D surfaces bounding 3D volumes. Similarly, the sphere is the surface which bounds the ball.) We see that there are three edges at every vertex.

It has .

I note that it has 20 vertices. It corresponds to . That’s why the list starts at 20.

What about a soccer ball (buckyball)? Mathematica® knows this as the TruncatedIcosahedron.

It has . This corresponds to carbon 60. Or to the most common soccer ball. (Not all soccer balls have 32 faces, apparently.)

Note also that every vertex has 3 edges.

Not every polyhedron — not even every regular polyhedron — has 3 edges at every vertex. Here is the regular 20-sided Platonic solid, the icosahedron:

Incidentally, it has 5 edges at every vertex. We are going to assume 3 edges at every vertex.

Here is the question. If

• every face of a polyhedron is either a pentagon or a hexagon,
• and every vertex has three edges,

what are the possible polyhedra?

The answer has two parts:

• any such polyhedron must have 12 pentagons;
• it can have any number of hexagons other than 1.

What I will actually prove is that no number of pentagons is possible, except 12.

Our general formula for edges and faces

becomes

,

because for every other k.

Our general formula for edges and vertices becomes

because for every k ≠ 3.

The total number of faces f is

,

and the total number of vertices v is

.

Thus, we wish to investigate solutions of the following 4 equations:

.

It turns out that Mathematica® is extraordinarily cooperative if I write separate equations for

and consider a set of 5 equations:

When I tell it to eliminate , Mathematica® tells me that :

It’s not too difficult to work this out by hand; you can do it without Mathematica.

This algebra places no restrictions on the number of hexagons. An it’s easy enough to exhibit solutions for any number of hexagons. Even for just one hexagon! Ruling out that case is a challenge; the proof dates only from 1963.

We know a couple of cases. We know that (no hexagons) is a dodecahedron. And (32 faces total) is a soccer ball or buckyball.

Whether or not we can actually build a polygon with any value of is another question. It turns out that there is only one nonnegative value of which does not work: we can’t have just one hexagon.

We can have a polyhedron with any number except n=1; that is, only cannot exist. But i haven’t proved either that cannot exist, or that anything does exist, other than and , which I have drawn.

I am not about to try to prove either that cannot be constructed, or that every other value can be. Maybe someday, but not today. (Among other things, I have no idea if all the hexagonal faces can be chosen the same size, nor even if they are all regular hexagons. My geometric intuition is nil — but the algebra is clear. Well, maybe it isn’t: the algebra says that simply by counting vertices, edges, and faces we can rule out anything that doesn’t have 12 pentagons. It says northing about the shape or even about the existence.)

Exactly 12 pentagons. I think that is amazing. And the equations I used are nicely general, and can be used for other things – such as showing that there can be no regular polyhedra other than the 5 Platonic solids.

For more about buckyballs, you can try this Wikipedia article.

Here is a drawing that suggests why we can’t add just one hexagon to a dodecahedron.

Within my personal library, the best reference on polyhedra is Hartshorne’s “Geometry: Euclid and Beyond”. (See my bibliographies page.)

Ah, a book newly acquired since I drafted this is: Richeson, “Euler’s Gem”, Princeton University 2008; most of it should be accessible to a high school student.

My reference for the two “2 e” equations is Firby & Gardiner, also on my bibliographies page.

A proof that cannot exist can be found in Branko Grunbaum’s “Convex Polytopes”, Springer, 2nd Edition, 2003. (Also newly acquired. It’s a graduate text.)

Simplicial curvature & simplicial Gauss-Bonnet

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 times its Euler characteristic .

(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 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 which touch v, add up those angles, and subtract the sum from . (That’s why it’s also called the angle defect: it measures how far from , 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 , 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 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 , 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 be a vertex. If is a face (polygon) containing v, let denote the angle at v in . The curvature of K at v (or the angle defect at v) is defined to be the number d(v) given by

where the are the faces of K containing v.

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

.

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

so the angle defect at one vertex is

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

By direct computation, the Euler characteristic of a tetrahedron is .

Therefore, we do indeed have

.

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 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 ,

so the angle defect is

.

(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 . That’s where the 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 and two angles from the other two faces. At each of these 3 faces, the sum of angles is

,

so the angle defect is

.

If we look at either the yellow or blue vertices, the portion that is not faces is certainly larger than , and in fact we can see that it’s , which is, indeed, .

Now we add up all the angle defects

.

Once again, that’s .

What about a cube?

Looking at any of the vertices shown as blue dots, we should guess that the angle defect is at each of the 8 vertices, so the total is, once again,

.

Alternatively, 3 rectangles meet at each vertex, so the sum of the angles is

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

.

OTOH, the Euler characteristic of a cube is

,

so .

(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 (), so we know the total simplicial curvature should be , but let’s work it out.

Here’s a regular hexagonal cylinder, or prism. There are 12 vertices; each is touched by 2 rectangles and a hexagon .

(Oh, cut the hexagon into 6 equilateral triangles; each angle at a vertex of the hexagon is made up of two angles.)

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

,

so the angle defect is

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

.

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

,

But that triangulation has 18 faces (that’s the easy number), 27 edges, and 9 vertices. .

Good. We’ve finally seen a case that didn’t add up to .

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

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 , that will be zero. And zero integrated over 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 , so cos u = -1 and the curvature < 0.
The green curve is , so cos u = 0 = curvature.

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

The Euler Characteristic: Teasers

These are things i came across when I first started looking at the Euler characteristic, in fact, when I was looking at triangulations in particular.

n-manifolds

The Euler characteristic generalizes to dimensions other than 2, and there are at least three noteworthy theorems involving the Euler characteristic. I’m not going to say much about them, because they, like so much else, are still outside my comfort zone. I’ll just barely tell you what they are, and leave you to chase them down if they interest you.

As we’ve seen, the Euler characteristic of a polyhedron is given by

,

where v, e, f are the numbers of vertices, edges, and faces. Homeomorphic polyhedra have the same Euler characteristic, and that means we can define the Euler characteristic of a topological surface as the Euler characteristic of any polyhedron which is homeomorphic to it.

This alternating-sign sum of the numbers of 0-, 1-, and 2- simplices generalizes in the obvious way: for an n-simplex, we take the sum, with alternating signs, of the numbers of k-simplices, for k <= n. As for surfaces, so for n-manifolds: this is a topological invariant, and we want to define the Euler characteristic of an n-manifold as the Euler characteristic of any k-simplex homeomorphic to it.

But we’ve been told that topological, simplicial, and differential structures on manifolds do not always coincide for n > 3. We can define the Euler characteristic for k-simplices and for manifolds with simplicial structure, but I am going to look skeptically at the Euler characteristic of topological and differentiable manifolds until I learn more about the different structures.

BTW, we can go down a dimension, too: we can define the Euler characteristic of a graph as v – e.

Betti numbers

The Euler characteristic of an n-manifold has an extremely interesting equivalence: not only does the formula generalize from 2 to n, but it can also be generalized to a topological space X: we can write

where is the kth Betti number of X, and we assume that eventually (i.e. for k large enough) . (Lee, “Introduction to Topological Manifolds”; or Rotman, “An Introduction to Algebraic Topology”.)

But hold on just a minute. The Euler characteristic was defined for things with simplicial structure, and the Betti numbers are defined (not by me!) for topological spaces. But we know that simplicial & topological do not coincide for higher dimensions than 3, so we probably have to be rather careful here.

When all else fails, look carefully at the book. What we actually have is two definitions, a theorem, and an extension of the definition. We define the Euler characteristic of a simplicial complex, and the Betti numbers of a topological space; the theorem says that for a finite simplicial complex, the Euler characteristic is equal to the alternating sum of the Betti numbers:

Having the theorem, we then extend the definition of the Euler characteristic to any topological space: we define the Euler characteristic as the alternating sum of the Betti numbers; the theorem assures us that the two definitions of the Euler characteristic coincide on simplicial complexes.

We’re still not out of the woods. Looking in Rotman, I am reminded that there are different homology theories. Do they all give the same Betti numbers? I have no idea. (Well, I suppose they do, but I’m ready for anything.) Gee, no wonder algebraic topology is still on my list of things to study.

As an aside, in Massey’s combined book, “A Basic Course in Algebraic Topology”, I find a note which says that until about 1930, mathematicians focused on the Betti numbers (and the torsion coefficients) of homology groups, rather than on the groups themselves.

Index of a surface

Next, on this detour, we have the Poincare-Hopf index theorem, which says that the Euler characteristic of a surface is equal to the index of the surface.

Hey what?

As the Betti numbers came out of left field, the study of vector fields on a surface is coming out of right field. If you’ve ever been told that you can’t comb the hair on a ball, you’ve encountered one of the key theorems of the subject.

There are two ways to look at the index theorem. In the special setting of dynamical systems (i.e. differential equations), we are looking at points where the vector field vanishes (critical points, which are equilibrium points, but not necessarily stable equilibria). More specifically, we are interested in the flow pattern in the vicinity of critical points.

In this context, we break flow patterns into sectors each of which is called parabolic, hyperbolic, or elliptic; and we have a recipe (due to Bendixson) for planar systems which says that the index of a critical point (wrt the vector field) is

I = 1 + (e-h)/2,

where e and h are the numbers of elliptic and hyperbolic sectors. (The parabolic sectors don’t contribute to the index.)

The index of the surface wrt the vector field is the sum of the indices of the critical points. The Poincare-Hopf theorem says that the

index of the surface (in fact, of an n-manifold) wrt the vector field is equal to the Euler characteristic of the manifold.

(I believe that Poincare proved it for surfaces and Hopf extended it to higher dimensions. I am not sure that the definition of sectors applies in higher dimensions; that is, I am not sure how the index is defined in higher dimensions.)

This imposes significant constraints on the possible vector fields.

That’s one way to do it. My references for this approach are Perko, “Differential Equations and Dynamical Systems” and Firby & Gardiner, “Surface Topology”.

I was hoping to at least construct illustrations of the three kinds of sectors, but I’m finding it difficult and very annoying to update my version 5 drawings to version 6 of Mathematica®. (Backwards compatibility? What’s that?)

From Firby & Gardiner, I extract the following description. A sector is parabolic if all paths lead to the critical point, or all paths lead away. A sector is elliptic if all paths begin and end at the critical point. A sector is hyperbolic if all paths sweep past the critical point.

But there is another way to define and compute the index of a critical point. In fact, we can compute the index of any point wrt a curve, but if it’s not a critical point, then its index is zero; therefore we can confine our attention to critical points. This approach starts with winding numbers. Fulton, “Algebraic Topology: A First Course”, is excellent for surfaces (with multiple chapters on winding numbers and vector fields), and even suggests how to generalize to higher dimensions. Also, Sieradski, “An Introduction to Topology and Homotopy”, has a few sections, rather than chapters, on winding numbers and vector fields.

I’m beginning to look forward to tackling Fulton.

Gauss-Bonnet

The Gauss-Bonnet theorem, as O’Neill states it (“Elementary Differential Geometry”) says that the gaussian curvature, integrated over all of a compact orientable surface, is equal to , where is the Euler characteristic of the surface.

Oh, but doesn’t that require differentiable structure?

Yes and no.

Bloch shows us the Gauss-Bonnet theorem for simplicial surfaces! It turns out we can associate curvature with the vertices of a simplicial surface, in such a way that the total curvature (the sum of the curvatures at each vertex) is equal to .

I’ll talk about this when I discuss chapter 3 of Bloch. I finished the chapter eons ago, but I haven’t tallied it up yet.

surfaces: visualizing the gluing of them

Quite some time ago, a friend asked me what would happen if we tried to construct a torus by gluing all 4 sides of a sheet of paper together, instead of first one pair then the other. Didn’t the math have to specify first one pair then the other?

One reason I’ve been hesitating over this post is that it doesn’t seem to be “real” mathematics – though any number of people might howl that PCA / FA isn’t “real” mathematics either. This is just a small drawing that I cobbled together to show that the homeomorphism between a circle and a line segment with endpoints identified… well, it doesn’t have to correspond to a physical process. (Why don’t I refer to “the glued line”?)

“… algebra provides rigor while geometry provides intuition.”
from the preface to “A Singular Introduction to Commutative Algebra” by Greuel & Pfister

It helped me to go back and read my original comment when I acklowledged Jim’s question to this post. I see that I did not understand that what matters to the formalism, the algebra, is before and after; what matters to the geometry is between or during. We reconcile them by permitting some things in the geometric visualization that we would not permit in the formal algebra, if the algebra even formalized the process: points passing thru points; or even some tearing, if when we reglue it we restore it rather than take the opportunity to change it.

That’s what bloch says, on p.57, discussing the physical process of getting from the knotted torus

to the regular torus.

In contrast, the sphere movie says going thru is ok, but tearing is not. (That’s the same link cited in my original comment.)

Keep it simple. Jim’s comment was that deforming and gluing a flat sheet to form a torus requires another dimension, namely time. To be specific, Jim considered 3 processes:

1. glue the left and right sides together, then glue the two circles.
2. glue the top and bottom sides together, then the two circles.
3. glue both pairs of sides together simultaneously.

He cannot see that the third process generates a torus. Well, neither can I.

I’d like to think that it does, but it doesn’t have to.

The formal mathematics doesn’t actually deal with the deformation process we’re imagining; rather it shows “before” and “after”. Don’t get me wrong: the formal math is motivated by the deformation process, and if the formal math hadn’t led to results which were by-and-large plausible, we would have changed the math.

The mathematics requires that a function describing “before” and “after” be a homeomorphism: 1-1 and both the function and its inverse are continuous. But this isn’t a description of “during”. We describe the result of gluing, not the process of gluing.

Let me show you a much simpler drawing than a torus, simpler even than a cylinder. I start with a line segment of the x-axis, and I want to transform it into a circle. If I imagine that I am wrapping the line onto the circle – wrapping a strip of paper mache’ onto a frame – then I could come up with the following mapping.

It starts at the origin and moves CWW (counter-clockwise); t runs from 0 to .

Now, draw lines from “before” points on the line to “after” points on the circle. Let’s do it first for just the leftmost quarter of the line segment, :

add the next quarter, :

We see a problem as we move toward the top. As one example of many, the line L to the top of the circle, mapping the point on the line to the point (0,2) on the circle, passes through the circle. That is, it passes thru points that have already been mapped to the circle. If we imagine that these lines had been traced out at constant speed, but stopping when they reach their destination on the circle, then the line L hits and goes thru a point on the circle.

The problem is even clearer when we add mappings from the last half of the line segment, since all the lines which are mapped to the left of the circle must pass thru the right of the circle, and all of those points have already been “installed”, as it were. Let’s see this with the 3rd quarter of the line segment :

This process was a far cry from “wrapping” the line segment onto the circle. But such processes have nothing to do with the homeomorphism between the glued line and the circle.

The homeomorphism is a relationship between the glued line and the circle; not a relationship from the glued line to the circle. There are many processes from the glued line to the circle, but not all of them are nice. And this is one that isn’t nice.

The Euler characteristic (triangulations 2)

I want to talk about some things I saw when I was looking at triangulations. This also continues my reading in Bloch.

The major thing I saw was the Euler characteristic. We usually define it for polyhedra, as the alternating sum / difference of the number of vertices, edges, and faces (0-, 1- and 2- simplices)…

Then we would define it for a surface by taking the Euler characteristic of any polyhedron which is homeomorphic to that surface.

For that to be well-defined, of course, requires that all polyhedra which are homeomorphic to a surface S have the same Euler characteristic (as they do).

By the same token, we could define the Euler characteristic of a surface from any triangulation of the surface, after we show that all triangulations of a surface have the same Euler characteristic. Oh, and we’d better actually prove that every topological surface (every topological 2-manifold) can be triangulated.

Yes, they can be. Not true for topological 4-manifolds, and I think it’s still wide open for higher dimensions. In contrast, I believe that every differentiable n-manifold supports a unique piecewise linear (PL) structure (which is the generalization, they tell me, of triangulations).

Details.

So we have the Euler characteristic in one hand. With the other hand, let’s grab the classification theorem. One form said that every compact connected surface is a connected sum of tori T or of projective planes P:

gT, g = 0,1,2,3…

or

gP, g = 1,2,3,…

(where the sphere is 0T, the “torus” with no holes; 2T = T#T, etc., and the integer g is called the genus of the surface.)

Is there any chance we can easily compute the Euler characteristic of the connected sum A#B from the Euler characteristics of A and B? If we could, then we could compute the Euler characteristic of nT and nP, hence of all connected compact surfaces.

Of course we can, and here’s the equation:

Oh, now go get the Euler characteristics of triangulations of the sphere, torus, and projective plane.

Ok, we can do that.

But it would be simpler to get the Euler characteristic from the polygonal disks from which we constructed the basic surfaces. We would have to know that counting the v, e, f (after identifications) of the polygonal disk gives us the same answer, but it does.

Yes, having specifically studied Bloch because he distinguished topological surfaces from simplicial surfaces, I am now effectively going to compute the Euler characteristics of topological surfaces instead of simplicial surfaces.

I never said I didn’t want to mix it all together; I just wanted to see the separate ingredients that went into the stew.

In principal we only need 3 surfaces: sphere, torus, and projective plane. Nevertheless, since the Klein bottle has an equally simple polygonal disk, I’ll show it too. One set of representations is

Edges which are to be identified (“glued”) have the same symbol; an arrow shows the directions which are used. Points which are to be identified have the same color. I considered using colors for the edges, but we can actually do algebra with the symbols: the sphere can be called , the torus , etc.) If it helps, the colors of the vertices are a consequence of the identifications of the edges.

From them, we compute that the Euler characteristics are

Note that the largest Euler characteristic is 2, and it corresponds to a sphere. Note that all four are non-negative. Note that an Euler characteristic of 1 corresponds to the (non-orientable) projective plane. Then for Euler characteristic of 0, we have two surfaces, one orientable, one not.

Then we compute that the Euler characteristic of gT is 2 – 2g, i.e. the set of even numbers less than or equal to 2. And the Euler characteristic of gP is 2 – g, i.e. the set of integers less than or equal to 1. (Maybe I should restrict g > 0 for nP, but I’m not sure it’s worth worrying about.) So if we have an odd negative Euler characteristic, the surface is unique and non-orientable; if the Euler characteristic is even and negative, there are two surfaces, one orientable, one not.

Note that all the new surfaces we construct have Euler characteristic less than zero.

Note that the connected sum of a sphere and any surface A has the same Euler characteristic as A. in fact, the connected sum is homeomorphic to A; gluing spheres to surfaces does not change them.

We know, however, another form of the classification theorem. The non-orientable surfaces gP can be written as a connected sum of tori plus either one projective plane or one Klein bottle:

mT#P

or

mT#K

for some m (and I am paying no attention to the relationships between these m’s and the genus g). To prove that, we need to know that

P # P = K

and

P # P # P = P # T.

We can now compute that the Euler characteristic of mT#P is (2 – 2m) + 1 – 2 = 1 – 2m, i.e. the set of odd integers less than or equal to 1.

The Euler characteristic of mT#K is (2-2m) + 0 – 2 = -2m, i.e. the set of even integers less than or equal to 0.

So, in fact, if the Euler characteristic is an odd negative, there is a unique surface of the form mT#P; if the Euler characteristic is even negative, there is an orientable surface of the form nT and a non-orientable surface of the form mT#K.

References for this material are

• Bloch, “A First Course in Geometric Topology & Differential Geometry”.
• Lee, “Introduction to Topological Manifolds”.
• Firby & Gardiner, “Surface Topology”.
• Massey, “Algebraic Topology: An Introduction” or “A Basic Course in Algbraic Topology”

See the bibliography.

Triangulations of Surfaces: minimum number of triangles

Edited 4 Sep. search on “edit”.

Take a cube. If you cut it along a few edges, you could lay it out flat. To restore the cube, we identify certain edges with each other. Similarly for a tetrehedron, or a theoretical soccer ball (with flat faces), or any polyhedron. For studying surfaces by looking at polyhedra (i.e. picewise linear structure), it is convenient to use only triangular faces (2D simplices) rather than arbitrary polygonal faces. The analog of our cut and flattened cube is called a triangulation. As before, we want to identify certain edges with each other.

In particular, the following

is offered as a triangulation of a torus (with the top & bottom edges identified, and the left & right, as we’ve seen before).

Why are there so many triangles? I have wondered that since the first time I saw it.

“Surely fewer would suffice,” I thought. Compared to the simple square which we use to generate a torus, that triangulation looks unnecessarily complicated.

Yes, it is, but not by much. It is true that fewer would suffice, but not a whole lot fewer.

I could hand you the answer and be done with it. But I had fun chasing this down and saw many things along the way. It was a fine little gambol through the park.

On second thought, this post is getting way too long, so I’d better hand you the answer and save the cultural remarks for another post.

Back up a moment. What exactly is the question?

I’m going to answer several questions.

• Why are there so many triangles in a triangulation of a torus?
• Is there a minimum number of triangles required in a triangulation of a torus?
• If so, what is it?
• Is there a minimum number of triangles required in a triangulation of any given surface?
• If so, is there a formula?
• If so, how do we derive it?

As I said quite a while ago, I wanted to get more comfortable with simplicial complexes; this took me into a bunch of algebraic topology books. While I was wandering and wondering, I came across this, in Rotman (Intro Algebraic Topology) p. 133, speaking of our picture: “This triangulation of the torus has 18 triangles…. It is known that the minimum number of triangles in a triangulation of the torus is 14 (see [Massey (1967), p. 34, Exercise 2] for an inequality implying this result.”

As I said, that triangulation is not much larger (18 triangles) than it needs to be (14 triangles), but it is way larger than I would have expected.

Come, Watson, the hounds are afoot.

Massey (Algebraic Topology, Introduction, p. 34 or A Basic Course… p. 31), exercise 2 is: “For any triangulation of a compact surface, show that”

1) 3 f = 2 e

2)

3)

where f, e, v are the numbers of triangles (= number of faces), edges, and vertices. (That appears to say there is a minimum number of triangles required for any triangulation of a given surface, including the torus, and we have a formula. Unfortunately, it does not quite stand by itself.)

You might want to scribble these down, with the numbers 1-3. For example, I’m generally going to refer to “Massey’s (3)” rather than display the inequality over and over again.

Ok. What’s the minimum number of triangles in a triangulation of the torus? We need the definition of the Euler characteristic

and we need to know that the Euler characteristic of the torus (T) is 0: .

While we’re at it, the Euler characteristic of the sphere is 2, of the projective plane is 1, and of the Klein bottle is 0. Yes, the Klein bottle and the torus have the same Euler characteristic. More on that in another post.

Now, Massey’s (3) implies that

.

For the minimum v = 7 we would have

Then

3 f = 2 e = 42, so f = 14.

That’s what Rotman told us. So there must be at least 14 triangles in a triangulation of the torus.

Now let’s start getting Massey’s formulae.

The first thing I did was observe that once we have the definition of , we can combine it with (1) to get (2).

So what about (1)?

Massey gives us the crucial property of a triangulation. Since a triangulation is homeomorphic to a surface, every point on it – including points on the “outside” edges, must be homeomorphic to a 2D disk. That means that the “outside” edges must be identified with other edges so that each of the “outside” edges is an edge of exactly two triangles (or two faces in general). Of course this also holds for the “interior” edges, which by construction are edges of exactly two triangles.

To put it another way, there cannot actually be any “outside” edges; they must all be interior.

That is, a triangulation is characterized by the fact that every edge is an edge of exactly two triangles. The formulae not withstanding, that may be the most important fact in this post. There is, however, another fact almost as important (later).

I like the way Firby & Gardiner do the next step. Suppose that what we have is a 2D complex – which may contain general polygons rather than just triangles – homeomorphic to a surface. That is, it may have faces with more than 3 sides. Instead of saying that every edge is the edge of exactly two triangles, we say that every edge is an edge of exactly two polygons.

We try to count the edges by counting faces. Let denote the number of faces with i edges. ( is the number of triangles, is the number of quadrilaterals, etc.) Then the weighted sum of faces

counts the number of edges associtaed with each face, but it counts them exactly twice because each edge is the edge of exactly two faces – that’s crucial. We have, then

If, now, there are only triangles , we get

i.e. Massey’s first equation

One down. and as I said, this equation and gives us the second equation (just eliminate f):

.

Two down, one to go.

Firby & Gardiner eventually state a theorem and a conjecture due to Heawood. The theorem is that if M is a compact surface with euler characteristic , then any map on M can be colored in no more than

colors. ( denotes the floor function, the largest integer less than or equal to x.) Note that the hypothesis of the theorem excludes the sphere ; and note that the number of colors required for a map on the sphere is the same for a finite rectangle in the plane. That is, Heawood did not prove the 4-color theorem.

The number h is called the Heawood number of the surface; the number of colors which suffice to color all maps on a surface is called the chromatic number of the surface.

And the number of vertices is bounded by the expression whose floor is the Heawood number. Hmm. This could be promising.

Heawood’s theorem says that if the Euler characteristic , then the chromatic number of a compact surface is its Heawood number.

Heawood’s conjecture was that the theorem was true for any : that the Heawood number is equal to the chromatic number for any compact surface. It is false, but just barely.

It turns out that there is exactly one exception to the conjecture: for the Klein bottle, the chromatic number is 6 while the Heawood number is 7. (The Heawood number for the sphere is 4; for the projective plane 6; and for the torus it’s the same as for the Klein bottle, 7, because they have the same , namely 0.)

edit:
The previous paragraph is correct. But my distinction between the Heawood conjecture and the Heawood theorem makes no sense. How can the theorem include when the counterexample is for ? The simplest correction might seem to be that the theorem is for , but even that isn’t right. (I think that correct would be the theorem as stated, but for orientable surfaces only.)

I based my summary on Firby & Gardiner, but I misread them. Upon review, I believe – they didn’t put it this way – that they said the theorem was that the chromatic number is less than or equal to the Heawood number for strictly less than zero; the conjecture, as I read them, is two generalizations: remove the restriction on , and replace the inequality by equality (chromatic number equal to Heawood number).

Unfortunately, I think their distinction is wrong, too. Now, I’m arguing about the history of mathematics, not the mathematics. I believe (Biggs et al.) that the theorem which Heawood stated in 1890 was: chromatic number equals Heawood number for orientable surfaces with . I’m not sure he proved it all, but I think that’s what he thought he was proving.

The Heawood theorem, as I see it, asserts that the chromatic number is equal to the Heawood number for orientable surfaces with . The conjecture would generalize that to all and to non-orientable surfaces. The Klein bottle, being non-orientable, is an exception – the only exception – to the conjecture, not to the theorem.
end edit

So the Heawood conjecture implied that the 4-color theorem was true (for the sphere, hence for the plane).

Back to Massey. The Heawood number occurs in map coloring…. but how do I get from it to vertices? Turns out I don’t. (Although I’m sure there’s a connection; I probably needed to read a couple of proofs.)

Fulton (Algebraic Topology First Course) gave me the last clue, on p. 115, problem 8.9. “Show that, for any triangulation of a sphere with g handles:”

1) 2 e = 3 f

2)

3)

I needed to have seen that

for a sphere with g handles, but I had. We’re looking at a special case of Massey’s (3).

Even though Fulton has asked for a special case, can we do the general case? No problem: the key is Fulton’s inequality (2).

Can we get Massey’s (3) from Fulton’s (2)? (You may very well have recognized (2) and said, “Oh, yeah!” Not me. My focus was on whether I could use Fulton’s (2) to get Massey’s (3)).

Piece of cake. Really. Consider it done.

Famous last words. It’s a piece of cake up to a point, and then it begins to look like cardboard instead of food. We’ll come to this. For now, assume we can get Massey’s (3) from Fulton’s (2).

Only then did I ask: can I prove Fulton’s (2)? I don’t know about you, but once I looked at (2) in its own right, it was crystal clear. This is the number of distinct pairs of v objects without repetition. Given a pile of vertices, v in number, choose one of them (v possible choices), choose a second one (v-1 choices), and divide by two because order is irrelevant.

That’s the maximum number of edges I can draw using v vertices, when loops and multiple edges are not allowed (the ends of an edge must be distinct, and no more than one edge may join two given vertices). I need not, however, draw all possible edges: the actual number e of edges is bounded above by the maximum number of edges, hence , so we have Fulton’s (2). QED.

So Massey’s inequality is far easier to obtain than any of the arguments concerning the Heawood number and map colorings. The Heawood number per se was interesting but apparently irrelevant.

So I thought. Then I figured I really ought to write out the derivation.

I learn so much when I look at the gory details.

From and , we get

.

From that and , we get

.

NOTE that the quadratic is always a parabola opening upward. It is negative between its two real zeroes, if they exist. So what we’ve actually shown is that the number of vertices must lie outside the two roots of that quadratic.

For the torus, , so we get . That quadratic has zeroes at v = 0 and v = 7, and we’re not going to allow negative numbers of vertices, so we assert that v must be 7 or more for a triangulation of a torus. That, we saw, required 14 triangles.

What about a sphere? , so we get . That quadratic has zeroes at 3 and 4…. Both are perfectly fine positive numbers, and in fact no positive integer is excluded by the quadratic bound.

As the dealer says in poker, “No help.” That quadratic inequality would not let us rule out 3 vertices or 2 or even 1. Something else must do that. My piece of cake is beginning to taste like cardboard.

Let’s recall Massey’s (1)

1) 3 f = 2 e

For both e and f to be positive integers, that equation requires a minimum of f = 2 and e = 3. OK, we can rule out 1 or 2 edges.

Is there something wrong with 3 edges?

Yes. Turns out there’s another fact we need.

Well, when we define a simplicial complex – which I know haven’t done – or when we imagine gluing triangles together, there are some obvious cases which we want to exclude. We don’t want one triangle partly inside another, and we also don’t want to share edges among too many triangles.

(I don’t know about your neighborhood, but in mine, houses are offset: i share a back fence with two neighbors, not one. We don’t want to allow that with triangulations.)

What your book says – whatever book you have – if it says this at all, is that if two triangles intersect at all, they intersect in either a vertex or an edge. The example they invariably draw shows 2 triangles having part of an edge in common. That’s a no-no.

But it’s also a no-no to have more than one edge in common. At least, I think so. I have not seen anyone emphasize that “an edge” excludes multiple edges, but it had better.

We know how to make a square into something homeomorphic to a sphere: glue the right edge to the top edge, and the left edge to the bottom edge. Fine, now imagine that we first put one diagonal inside that square, and then glue it that same way. What we get is still homeomorphic to a sphere; it looks like two triangles glued together on all three edges. It’s just not a triangulation because the two triangles do not intersect in an edge, but in 3 edges.

In short, the quadratic bound merely says that any integer number of vertices would work for a sphere, but 3 is too few to make a triangulation at all. Then the quadratic bound assures us that 4 should be allowed.

I should probably exhibit one such for both the sphere, but I’ll pass. I hope 4 vertices works, and i’ll leave it at that.

In summary,

• Why are there so many triangles in a triangulation of a torus? Because there can’t be any “outside” edges.
• Is there a minimum number of triangles required in a triangulation of a torus? Yes.
• If so, what is it? 14.
• Is there a minimum number of triangles required in a triangulation of any given compact surface? Yes.
• If so, is there a formula? Sort of: no less than 4, and no less than the larger root of a particular quadratic.
• If so, how do we derive it? Done.

Topological Surfaces: Bloch Ch 2.

Let me talk about the second chapter, “Topological Surfaces” in Bloch’s “A First Course in Geometric Topology and Differential Geometry”. I finished it quite a while ago, but I’ve had trouble deciding how to talk about it. I don’t want to just summarize it. Instead, I think I’ll try asking and answering some leading questions.

First of all, why did I choose to read Bloch? A few years ago, I was seriously shaken up by the following, from Freed & Uhlenbeck’s “Instantons and Four-Manifolds”, p. 1: “A basic problem is to ascertain when a topological manifold admits a PL [piecewise linear] structure and, if it does, whether there is also a compatible smooth [differential] structure. By the early 1950’s it was known that every topological manifold of dimension less than or equal to three admits a unique smooth structure.” They were setting us up for the fact that it isn’t generally true in dimensions greater than 3.

To be explicit about the challenge we face: topological, simplicial (which Bloch tells me generalizes to PL), and differential structures on a manifold coincide for low dimensions but not for high, and “high” means “greater than 3″.

This was all news to me. I knew, for a small value of knew, differential geometry, and I’d seen some simplicial stuff – triangulating and cutting up surfaces – and I was acquainted with topology, possibly for a large value of acquainted, but their “basic problem” was nothing I’d ever heard before.

Bloch is presenting separately the various structures for 2D manifolds – surfaces – even though they coincide for 2D. We can still keep the concepts separate. We will need to know what came from which structure when we move to 4D. I was ready for Bloch when I found him. (“When the student is ready, the teacher will appear.”)

What’s the big result in the chapter? The “classification theorem for compact connected surfaces (2D manifolds) in R^n”: any compact connected surface in R^n is homeomorphic to the sphere, a connected sum of tori, or a connected sum of projective planes.” Further, “the surfaces in this list are all distinct”. We could write

S
nT = T # T # … # T, a connected sum of tori;
nP = P # P # … # P, a connected sum of projective planes.

What’s the biggest disappointment in the chapter? That he doesn’t prove the classification theorem in this chapter. He does it at the end of the next chapter, after he’s introduced simplexes. Damn. There is a chance – skimming the proof – that he needs simplexes in order to prove that “the surfaces in this list are all distinct”. Either way, damn again. I was hoping for a proof that didn’t invoke simplicial structure.

What happened to the Klein bottle K? I don’t see it on that list. It’s there, as the connected sum of two projective planes: K = P # P = 2P.

What about the Möbius strip? It’s not a surface – because it has a boundary. Well, what’s a surface? it’s a subset of R^n every point of which has an open neighborhood that is homeomorphic to the open unit disk. No boundaries allowed.

What’s the big deal with boundaries? I’ve seen “manifolds with boundary”. Well, if we restrict ourselves to surfaces as being 2D manifolds without boundary, then, for example, it is true that every closed surface in R^3 (not in R^n) is orientable. Since the Möbius strip is a closed subset of R^3 but not orientable, better to not call it a surface. (We see something similar when we talk about unique factorization into primes: if we say that 1 is a prime, then 4 = 2×2 cannot be written uniquely as a product of primes, because we can put any number of 1s in the product.)

What’s the most fundamental theorem in the chapter? I have to name two: “schönflies theorem” and “invariance of domain”. Schönflies is just a tad more technical than I want to state here – but note that it is for R^2 not R^n; but invariance of domain is trivial to state: “if is homeomorphic to R^n, then U is open in R^n.”

What’s the big deal? A corollary is: if m and n are distinct integers, then R^m is not homeomorphic to R^n. (It’s the corollary which sounds more like “invariance of domain”.) We know that R^m and R^n are distinct vector spaces; now we know they are distinct topological spaces. There is another big deal to “invariance of domain”. In the differential geometry section of the book, I found: “Many differential geometry texts avoid using Invariance of Domain … by restricting to the use of coordinate patches that have continuous inverses.” Yes, I’ve seen that, they’re called “proper patches”. Now I know that they’re just a technical decision, not a fundamental one.

What’s the most satisfying theorem in the chapter? It’s the one that says a gluing scheme gives us a surface, and every surface has a gluing scheme. Hey what? A gluing scheme formalizes what we do when we take, for example, a square and imagine gluing its four sides by lining up the arrows and symbols. Here’s how we show gluing the sides of a square to get a torus:

The goal of describing surfaces as topological manifolds would have failed without this theorem, specifically, without a proof of this theorem that does not use simplicial structure. I can find plenty that do use it!

What was the cutest part of the chapter? (How’s that for an adjective applied to mathematics?) The connected sum # of compact connected surfaces is associative, commutative, and it has an identity (the sphere); that’s lovely, but what it doesn’t have is inverses! In fact, we can show that if A # B = S, then A = S and B = S (and I am using = to denote “homeomorphic to”); i.e. only the sphere S has an inverse, namely itself. (It must; after all, it is the identity.)

What did i like best about the chapter? The formalization of gluing schemes using identification spaces and quotient maps. Everyone shows me pictures of gluing schemes, but how do we describe them formally? OK, done.

What did I get out of other books? Having finished this chapter of Bloch, i picked up Firby & Gardiner’s “Surface Topology”, and Lee’s “Introduction to Topological Manifolds” (which has a chapter on surfaces).

From Firby & Gardiner, I got a more customary approach: a more casual approach to surfaces, and no distinction between topological and simplicial concepts. They had plenty of caveats and counterexamples, but they just didn’t prove a whole lot.

OTOH, they gave me a different statement of the classification theorem (they write the sphere as 0T, i.e. a torus with zero holes) :
an orientable compact surface is homeomorphic to nT, for some n >= 0.
a non-orientable compact surface is homeomorphic to
either nT # K
or nT # P, for some n >=0.

To show that this is equivalent to Bloch, we need to know that K # P = T # P, in addition to knowing K = P # P.

In Lee, I got to see the major simplification in Bloch. Where Bloch defines a surface as a subset of R^m which for which every point has an open neighborhood homeomorphic to the open unit disk, Lee gives us the more abstract definition. He defines an n-dimensional topological manifold as a Hausdorff second countable topological space which is locally euclidean of dimension n. Bloch gets the properties “Hausdorff second countable” by restricting to a subset of R^m; and, of course, he restricted to surfaces, n =2.

Lee describes the purpose of the two conditions as: Hausdorff tells us there are enough open sets; second countable tells us there aren’t too many of them.

I remind us that every abstract manifold, a la Lee, can be viewed as a manifold embedded in R^m for some m. While the abstraction may be useful – I’m sure it is – it gets us no new manifolds, none that we couldn’t have found by starting with subsets of R^m, a la Bloch.

Let me also remind us that the Klein bottle is a surface (a 2D manifold) but it sits in R^4, not R^3. That’s why I used both m and n in the previous paragraphs: for the Klein bottle, n = 2, m = 4.

Both Firby & Gardiner and Lee introduced the algebra of polygonal disks, typified by that square with arrows which I claim embodies a gluing scheme for the torus. Instead of the drawing, we could write a presentation

by starting at a, and moving CCW. We could relate operations on such “words” to geometric operations. Bloch did none of that.

We could also get the sphere, the projective plane P, and the Klein bottle in similar fashion:

Lee states the classification theorem as Bloch did, but he describes it using presentations; he also uses simplicial structure to prove it.

So, I’m disappointed that the proof of the classification theorem is in the chapter on simplicial structure, but for all I know, we wouldn’t have the classification theorem if we didn’t have a unique simplicial structure for any topological surface. Maybe there can be no proof without simplicial structure.

Still, it was good fun and I’m looking forward to chapter 3.