# I Triangulation of circle and disk in R2

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1. Apr 8, 2016

### pierce15

I am studying topology right now and am a bit confused about the idea of triangulation. The definition is: if a topological space X is homeomorphic to a polyhedron K (union of simplexes) then X is triangulable and K is a (not necessarily unique) triangulation.

Apparently $K_0 \equiv {1} \cup {2} \cup {3} \cup {1,2} \cup {1,3} \cup {2,3}$ is a triangulation of $S^1$, the unit circle in R2, and $K \equiv K_0 \cup {1,2,3}$ is a triangulation of the unit disk in R2. The notation is that ${i}$ is a node, ${i,j}$ is a 1-simplex (edge), and ${i,j,k}$ is a 2-simplex (face).

I am having trouble seeing why these are homeomorphic, can anyone explain?

2. Apr 8, 2016

### andrewkirk

Identify $S^1$ with the unit circle in $\mathbb{R}^2$ and $K_0$ with any equilateral triangle with vertices on that circle ('inscribed triangle').

Define a function $f:K_0\to S^1$ whereby, for $Q\in K_0$, $f(Q)$ is the point where the ray from the origin through $Q$ intersects the circle.

It is straightforward to show that $f$ is a homeomorphism.

Now define $g:K\to D^2$ (the unit disk, which is $S^1$ plus its interior) such that, for $Q\in K$:

$g(Q)$ is the point obtained by following the ray from the origin $O$ through $Q$ for distance $\bar{OQ}\frac{\bar{OB}}{\bar{OA}}$ where $A$ is the point where the ray intersects $K_0$ and $B=f(A)$ is the point where the ray intersects $S^1$.

It is straightforward to show that $g$ is a homeomorphism.

3. Apr 9, 2016

### pierce15

Those make sense. However, there are a few more examples in my book (e.g. torus, klein bottle, etc.) which have more complicated polyhedra that I definitely would not have been able to come up with. Is there a general strategy for finding a homeomorphic polyhedron?

4. Apr 9, 2016

### andrewkirk

My general strategy is the 'rubber sheet geometry' approach that characterises topology. I look for a smooth deformation than can convert one shape into the other - eg a triangle into a circle. If I can imagine such a deformation then it will usually be a bijection and it's just a matter of working out how to formalise it as an aexplicit function.

5. Apr 10, 2016

### lavinia

Every surface such as a torus or the Klein bottle is a polygon with pairs of edges identified. Because of this, it is easy to picture a surface as a polygon and to triangulate it while keeping track of the identified edges.

The only exception is the sphere which is just a tetrahedron.

For higher dimensional manifolds there there is no general method for triangulation although the three dimensional torus is similar to the two dimensional and there are other 3 manifolds that are cubes with pairs of faces identified, What is the triangulation of the 3 dimensional sphere? Can you generalize this to the sphere in n dimensions?

Last edited: Apr 12, 2016
6. Apr 12, 2016

### zinq

Just some additional comments: All manifolds* of dimension 1, 2, 3 can be triangulated. These triangulations are necessarily what is called "piecewise linear". A weaker notion of triangulability is when a manifold is homeomorphic to some d-dimensional polyhedron, but not in a piecewise linear manner. There are some manifolds like that. And perhaps most surprising of all is that, according to a 2013 paper, some manifolds are not even homeomorphic to any polyhedron.

See https://ldtopology.wordpress.com/2013/03/16/manolescu-refutes-the-triangulation-conjecture/ for a brief survey of these results.

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By "manifold" we mean a topological space that for some d is a) locally homeomorphic to an open set of Euclidean space Rd; b) Hausdorff; c) paracompact.

7. Apr 12, 2016

### lavinia

You might note that all smooth manifolds can be triangulated.

8. Apr 12, 2016

### zinq

Excellent idea, L. Every smooth manifold has a piecewise-linear structure, and in dimensions 1, 2, 3 all manifolds possess a smooth structure. In higher dimensions this is not true.

It is also quite surprising to learn that, typically, a smooth manifold has several smooth structures that are not diffeomorphic (the appropriate equivalence relation for smooth manifolds. For instance, the 7-dimensional sphere possesses 15 mutually non-diffeomorphic smooth structures. The 14 that are not the standard structure are called "exotic".

(Likewise, a piecewise-linear manifold typically has several inequivalent piecewise-linear structures, sometimes infinitely many.)

9. Apr 13, 2016

### lavinia

Can you explain the difference between a manifold that does not admit a PL structure and a manifold that is not homeomorphic to a polyhedron?

10. Apr 13, 2016

### zinq

It might be easier to explain what a PL manifold is, and what a manifold that is merely homeomorphic to a polyhedron is without being PL (piecewise linear):

In either case, the manifold is the union of n-dimensional "triangles" — that is, simplices*: An n-dimensional simplex can be defined as the convex hull of n+1 distinct points in (n+1)-dimensional Euclidean space Rn+1. E.g., a 1-simplex is just a line segment; a 2-simplex is just a triangle; a 3-simplex is just a tetrahedron.

And in both cases, if two simplices of the manifold intersect, the intersection must be equal to a common lower-dimensional face. E.g., a 2-simplex and a 3-simplex must intersect in a common 0-simplex (point), 1-simplex, or 2-simplex — or else they do not intersect. Such a union of simplices is called a polyhedron.

We will assume that we have a manifold that is homeomorphic to a polyhedron K.

One way to define a PL manifold is to do so inductively, by dimension. To proceed, we need to define the link of a vertex (0-simplex). For any vertex v of K, its link Lk(v) is defined as the boundary of the union of all simplices having v as a vertex:

Lk(v) = ∂( {σ | σ is a simplex of K and v is a vertex of σ}).​

Then K is a PL manifold if the link of any vertex is a PL sphere.

This is an inductive definition, because if K is 1-dimensional (a union of 1-simplices, i.e., line segments, that intersect along common endpoints if they intersect at all), then for any vertex v of K, we always have Lk(v) = two vertices, which is by definition a PL 0-sphere. Thus every polyhedron K that is a 1-manifold is automatically a PL 1-manifold.

Then if K is a polyhedron homeomorphic to an n-dimensional manifold, it is PL if and only if the link Lk(v) of every vertex v of K is a PL (n-1)-sphere S. This sphere is PL if and only if the link in S of any of its vertices is a PL (n-2) sphere, etc.

It was shown in 1976 by Bob Edwards that the double-suspension of a triangulated homology 3-sphere is topologically a 5-sphere S5. It is an easy consequence of this fact that this S5 is not PL. However, the topological 5-sphere — like a sphere of any dimension — can be given a PL structure just by taking the boundary of a simplex of one higher dimension.

* * *
It has been proved that some topological manifolds are homeomorphic to a polyhedron but have no PL structure at all. And some topological manifolds are not even homeomorphic to a polyhedron.
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* The word "simplices" is the plural of the word "simplex" and is pronounced SIM-pluh-seez.

11. Apr 13, 2016

### lavinia

So a manifold could be homeomorphic to a polyhedron but not to a PL manifold?

12. Apr 13, 2016

### zinq

"So a manifold could be homeomorphic to a polyhedron but not to a PL manifold?"

Yes, that is correct.

There are no examples of this phenomenon for any dimension ≤ 4. But for every dimension n ≥ 5 there exists an n-manifold that has no PL structure but is homeomorphic to a polyhedron (i.e., "triangulable").

So there exist PL manifolds that have no smooth structure; triangulable manifolds that have no PL structure; and topological manifolds that are not triangulable.

Notes:
* Before1956, it was widely believed that all manifolds were uniquely smoothable, which would have implied that all of these complications did not exist. Then in 1956 Milnor discovered exotic smooth structures; within a few years it was discovered by Kervaire that some manifolds admit no smooth structure at all. Then in 1969 Kirby and Siebenmann found that some manifolds admitted no PL structure at all, and among those that did there were often more than one inequivalent PL structures. Etc., etc. So the actual situation is vastly more complicated than anyone ever imagined in 1955.

* Prior to about 1970, the word "triangulable" was often used as a synonym for PL. But in modern usage, "triangulable" just means homeomorphic to a polyhedron.

* All of the above applies only to "metrizable" manifolds —those whose topology is given by a metric space. The situation for non-metrizable manifolds — such as the surface called the "Prüfer manifold" — is almost entirely unexplored.

Last edited: Apr 13, 2016
13. Apr 13, 2016

### lavinia

Can you give an example of a triangulable manifold with no PL structure?
What goes wrong with the link of a vertex? Is it even a topological sphere?

14. Apr 13, 2016

### andrewkirk

If I'm following correctly, that means that the polyhedron itself is not piecewise linear (PL). That is, it is a weird polyhedron in some sense. So the weirdness of the manifold is already there in the polyhedron. It's not that the homeomorphism is somehow blind to some weirdness that is there in the manifold but not in the polyhedron (a difference in weirdness that might prevent the existence of a diffeomorphism, for instance).

In which case the theorem says that, in dimensions greater than four, it's possible to construct some really weird (non-PL) polyhedrons.

15. Apr 13, 2016

### zinq

Can you give an example of a triangulable manifold with no PL structure?
What goes wrong with the link of a vertex? Is it even a topological sphere?

Maybe the simplest (though not simple!) is the 5-manifold

M = E8 x S1

where E8 is the 4-manifold that's called "Freedman's E8 manifold".

The "E8 manifold" is constructed by

a) obtaining a 4-manifold W with boundary by plumbing* together 8 copies of the tangent 2-disk bundle over S2 according to the Coxeter-Dynkin diagram of the exceptional Lie algebra E8 (7 dots connected in a straight line by 6 lines, plus an 8th dot connected to the 5th dot by a 7th line,

b) noting that the boundary ∂W of W is in fact the Poincaré dodecahedral homology 3-sphere, which we will call ∑,

and

c) carefully glueing onto W a certain contractible 4-manifold V whose boundary is also ∑.

Michael Freedman proved that such a V exists, and also that the resulting "E8 manifold" is non-smoothable.

Finally taking the cartesian product of E8 with a circle results in a 5-manifold M that is homeomoprhic to a polyhedron but not homeomorphic to any PL manifold.

I don't know enough about it to tell you just what goes wrong. But if the link of some vertex is not a PL 4-sphere, that by itself is not sufficient to prove that M is not homeomorphic to any PL manifold. For example, the double suspension Susp2( ) of the Poincaré dodecahedral homology 3-sphere ∑ is known (Edwards, 1976) to be homeomorphic to S5. And of course S5 admits a PL structure — despite the fact that Susp2( ∑ ) has vertices whose links are not a PL 4-sphere S4.
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* Each dot in the E8 Dynkin diagram corresponds to a closed tangent 2-disk bundle over S2, and when two dots are connected by a line, they are to be "plumbed" together as follows: In each disk bundle, pick a *local* trivialization D2 x D2 of this bundle, where the first D2 is a closed 2-disk in S2 and the second one is the fibre. Now identify the two copies of D2 x D2 with each other by an identification that switches the base disk with the fibre disk. Finally, smoothing the corners makes this (if you like) into a smooth 4-manifold in a unique way.

16. Apr 14, 2016

### zinq

Correction: I have learned that I may have omitted a relevant condition for a polyhedral n-manifold to be piecewise linear (PL).

I had stated that, inductively on dimension, it was sufficient that the link of each vertex be a PL (n-1)-sphere Sn.

But I am now unsure if that is in fact sufficient.

What I *do* know is sufficient is the following inductive definition:

A polyhedral n-manifold is PL if the link of each k-simplex is a PL (n-k-1)-sphere.

(As far as I know, what I wrote originally — which is the indented condition but only for the case k = 0 — *might* be sufficient, but I am definitely not sure of this.)

I apologize for any confusion that this may have caused.
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Note: The "link of a k-simplex" σ is defined as the boundary of (the union of all simplices that contain σ as a sub-simplex).

For example, in the 4-dimensional regular polytope that is called the 5-cell* — which is the boundary of a 4-simplex — any 1-simplex σ is a sub-simplex of each of three 2-simplices and three 3-simplices. Their union is a double triangular pyramid, and its boundary is the union of six 2-simplices, each of which shares only one 0-simplex with σ.

Conversely, another example is that a 3-simplex — which is a tetrahedron — contains as a sub-simplex each of four 2-simplices, six 1-simplices, and four 0-simplices. The numbers 4, 6, 4 should look familiar.

* The 5-cell is the 4-dimensional analogue of the tetrahedron after you remove the tetrahedron's interior. It contains five 0-simplics, ten 1-simplices, ten 2-simplices, and five 3-simplices. The numbers 5, 10, 10, 5 should also look familiar.

Last edited: Apr 14, 2016
17. Apr 14, 2016

### lavinia

18. Apr 14, 2016

### zinq

In which Prop. 10 confirms that a polyhedral n-manifold is PL if and only if the link of every vertex is an (n-1)-sphere, after all.

Thanks, Lavinia!