2.5 Simplicial Complexes · Homology & Cohomology

In these days the angel of topology and the devil of abstract algebra fight for the soul of each individual mathematical domain.

— Hermann Weyl, “Invariants”, Duke Mathematical Journal (1939)

A sphere has infinitely many points. So do a doughnut, a coffee cup and the surface of the Earth. Yet if we want to count things about a shape — how many pieces it has, how many loops, how many hollows — or ask a computer to do the counting for us, infinitely many points is far too many. The way out is to build every shape from a few of the simplest possible pieces — points, line segments, triangles and their higher-dimensional cousins — glued together by two strict rules. The whole shape then becomes a finite list you could write on a napkin, and questions about holes become questions about the list.

From shapes to lists

Look at any animated film or video game and you are looking at triangles. A dragon, a face, a waving flag: each is stored in the computer as a mesh — a long list of corner points, and a list saying which three corners make up each little triangle. The computer never stores “a smooth surface”. It stores finitely many numbers, and the smooth-looking surface is something your eye assembles from them.

Topology can do the same thing, and for a better reason than convenience. Remember that topology does not care about lengths, angles or curvature: a shape may be stretched and bent as much as you like, so long as nothing is torn and nothing new is glued (a homeomorphism, in the language of §2.1 Spaces and Continuity). So when we chop a shape into triangles, the exact sizes and angles of the triangles cannot matter. What survives every stretch is only which pieces touch which. Poincaré said it in the sentence that opens §0.1 The Shape of a Question: a drawing may distort the proportions as much as it likes, as long as the relative positions of the parts are right.

This suggests a plan. If we glue simple pieces together under rules strict enough that the gluing is completely described by a list — “these three corners form a triangle, these two corners form an edge” — then the list is the shape, as far as topology is concerned. Everything we want to know about holes must be readable from the list. In §2.2 Gluing Spaces Together you already built a torus by gluing the sides of a square. This chapter makes that kind of gluing systematic and finite.

Here is a puzzle to carry through the chapter: how few pieces does it take to build a torus? By the end you will have three answers — $42,$ $6$ and $4$ — depending on how strict the gluing rules are, and the strictest answer hides a small gem: a torus on which every corner is joined to every other. First the pieces, then the rules for gluing them. Then comes the surprise that makes everything else possible: once the rules are in force, the geometry can be thrown away, and what remains is a list.

Simplices: points, edges, triangles and beyond

Let us meet the pieces one dimension at a time. In each dimension we want the simplest possible solid shape: the one with the fewest corners.

In dimension 0, the simplest shape is a single point. It has one corner: itself.
In dimension 1, it is a line segment, an edge, with two corners (its endpoints). One corner is not enough to make anything one-dimensional.
In dimension 2, it is a filled-in triangle, with three corners. Two corners only span a segment; three corners, if they are not in a line, span a flat region.
In dimension 3, it is a solid tetrahedron — a pyramid with a triangular base — with four corners.

The pattern is: a simplest shape of dimension $n$ has $n + 1$ corners. Each such shape is called a simplex (plural simplices; the word is Latin for “simple”). A point is a 0-simplex, an edge a 1-simplex, a triangle a 2-simplex, a tetrahedron a 3-simplex. The pattern does not stop at three. A 4-simplex has five corners and lives in four-dimensional space; we cannot see it, but we can describe it perfectly well, and we can draw its shadow.

There is one catch. Three points in a row do not span a triangle — they span only a segment — and four points lying in one plane do not span a tetrahedron. So we insist that the corners are in general position (mathematicians also say affinely independent): no corner lies on the line, plane or flat space passing through the others. With that proviso, the corners determine the simplex completely: it is everything you can reach by “filling in” between them.

Do not worry about the formula with the $t_{i}$ yet; the next section is devoted to it. For now, notice the off-by-one that trips everybody up at first: an $n$ -simplex has $n + 1$ vertices. A triangle is two-dimensional but has three corners.

Faces, and Pascal’s triangle

Look at a tetrahedron. Its boundary is made of four triangles; the triangles meet along six edges; the edges meet at four corners. Each of these smaller pieces is itself a simplex, spanned by some of the tetrahedron’s vertices. That is the general pattern.

So the faces of a triangle $[v_{0}, v_{1}, v_{2}]$ are its three vertices $[v_{0}],$ $[v_{1}],$ $[v_{2}];$ its three edges $[v_{0}, v_{1}],$ $[v_{0}, v_{2}],$ $[v_{1}, v_{2}];$ and the triangle itself. How many $k$ -faces does an $n$ -simplex have? A $k$ -face is determined by choosing which $k + 1$ of the $n + 1$ vertices it uses, and every choice works (any vertices of a simplex are in general position). So the number of $k$ -faces is the number of ways to choose $k + 1$ things from $n + 1 :$

# {k -faces of an n -simplex} = (k + 1 n + 1),

read “ $n + 1$ choose $k + 1 ”.$ For the tetrahedron ( $n = 3)$ this gives $(1 4) = 4$ vertices, $(2 4) = 6$ edges, $(3 4) = 4$ triangles and $(4 4) = 1$ tetrahedron. These binomial numbers are exactly the entries of Pascal’s triangle, in which every number is the sum of the two above it. Counting the single 1 at the top as row 0, row $n + 1$ of Pascal’s triangle lists the faces of the $n$ -simplex, dimension by dimension. Try it in the figure.

The simplices and their faces Drag to rotate · tap a count or an entry

Δ^{3}

is a tetrahedron

(2 4) = 6

counted 0 of 6

Row 4 of Pascal’s triangle lists the faces of $Δ^{3}$ . Tap an entry.

Figure 2.5.1 Choose a simplex

Δ^{0}, \dots, Δ^{4}

below, then a kind of face; the faces light up one at a time while they are counted, and the count appears in Pascal’s triangle. (

Δ^{4}

does not fit in space: what you see is its shadow, and the edge through the middle does not really meet the triangle it passes.)

Adding up a whole row gives the total number of faces. Each vertex is either used or not used — two choices per vertex, $2^{n + 1}$ choices in all — but the choice “use nothing” is not a face, so an $n$ -simplex has $2^{n + 1} - 1$ faces altogether. The tetrahedron has $4 + 6 + 4 + 1 = 15 = 2^{4} - 1 .$

Barycentric coordinates: an address for every point

Now the formula from the definition. Pick a triangle with corners $v_{0}, v_{1}, v_{2},$ and imagine hanging weights at the corners: a weight $t_{0}$ at $v_{0},$ $t_{1}$ at $v_{1},$ $t_{2}$ at $v_{2},$ with the weights adding up to $1 .$ The triangle would balance on a pin placed at one particular point — the centre of mass of the weights — and that point is $p = t_{0} v_{0} + t_{1} v_{1} + t_{2} v_{2} .$ (To add points and multiply them by numbers, read each point as its list of coordinates and do the arithmetic coordinate by coordinate.) Heavy weight at $v_{0}$ pulls the balance point towards $v_{0};$ equal weights put it in the middle.

Every point of the triangle arises this way, from exactly one choice of weights. The three numbers $(t_{0}, t_{1}, t_{2})$ are called the barycentric coordinates of $p$ (from the Greek barys, heavy: they describe a centre of weight). There is a second, purely geometric way to see them: join $p$ to the three corners, cutting the triangle into three smaller triangles. The small triangle opposite $v_{0}$ takes up exactly the fraction $t_{0}$ of the whole area, and likewise for the others.

Barycentric coordinates Drag the point · use arrow keys

p = 0.25 v_{0} + 0.35 v_{1} + 0.40 v_{2}

t_{0} + t_{1} + t_{2} = 1.00

The point is inside the triangle: all three coordinates are positive.

Show areas

Figure 2.5.2 Drag

p

around. The discs at the corners show the weights

t_{0}, t_{1}, t_{2};

the shaded pieces show the same numbers as shares of the area. Slide

p

onto an edge and the coordinate of the opposite corner becomes

0;

push it outside and a coordinate turns negative.

Barycentric coordinates tell you at a glance where a point is.

All three coordinates positive: the point is strictly inside the triangle.
Exactly one coordinate zero, say $t_{2} = 0 :$ the point lies on the edge $[v_{0}, v_{1}]$ opposite $v_{2} .$ The coordinates that are not zero name the face the point lives in.
Two coordinates zero: the point is a corner.
A negative coordinate: the point is outside the triangle altogether — it is not in the simplex.

The same works in any dimension: a point of an $n$ -simplex has $n + 1$ barycentric coordinates, all at least $0$ and adding up to $1,$ and the coordinates that are not zero name the smallest face containing it. The special point where all coordinates are equal, $t_{0} = t_{1} = \dots = t_{n} = \frac{1}{n + 1},$ is the barycentre of the simplex: the midpoint of an edge, the centre of a triangle. It will be the star of the section on subdivision.

Barycentric coordinates also show that all $n$ -simplices are really the same. Take the $n + 1$ points of $R^{n + 1}$ with a single coordinate equal to $1$ and the rest $0 :$ for $n = 2,$ these are $(1, 0, 0),$ $(0, 1, 0)$ and $(0, 0, 1) .$ The simplex they span is the standard $n$ -simplex $Δ^{n} = {(t_{0}, \dots, t_{n}) \in R^{n + 1} all t_{i} \geq 0, \sum_{i} t_{i} = 1},$ read “the set of all lists $(t_{0}, \dots, t_{n})$ of real numbers such that every $t_{i}$ is at least zero and they add up to one”. A point of $Δ^{n}$ is its own list of barycentric coordinates. Any other $n$ -simplex $[v_{0}, \dots, v_{n}]$ is a copy of it: send $(t_{0}, \dots, t_{n})$ to $t_{0} v_{0} + \dots + t_{n} v_{n} .$ That is why the gallery above could call its shapes $Δ^{0}$ to $Δ^{4} .$

A little history A geometer’s weights

August Ferdinand Möbius 1790 – 1868

Barycentric coordinates were introduced by August Ferdinand Möbius in his book Der barycentrische Calcul (“The barycentric calculus”) of 1827 (Möbius 1827). Möbius, the son of a dancing teacher, was by profession an astronomer: he spent his whole career at Leipzig, where he supervised the rebuilding of the observatory. He was, it seems, a poor lecturer, and had to advertise his courses as free of charge before students thought them worth taking. His idea was the one we just used: describe a point by the weights that, hung at the corners of a fixed triangle, would make it the balance point. It is the same Möbius whose one-sided band you met in §2.2 Gluing Spaces Together, and, as we shall see in a moment, the first person known to have written down the smallest triangulated torus.

Intuition Why simplices, and not balls?

Up to homeomorphism a simplex is nothing new: it is a ball. Stand at the barycentre and slide every point along its ray from there, stretching or shrinking each ray so that the boundary lands on a round sphere; the filled triangle becomes a disk, the solid tetrahedron a solid ball. So why build with simplices at all? Look at the boundaries. The boundary of a ball is a sphere, which is not a ball, and nothing about it says how to cut it into pieces. The boundary of an $n$ -simplex comes already cut: it is $n + 1$ simplices of dimension $n - 1,$ one opposite each vertex, and their boundaries are cut in the same way, all the way down to points. Simplices stay simplices when you take boundaries, and that is what will let a whole shape, boundaries and all, be written down as a list. (The third part of Boarbarktree’s video series You Could Have Invented Homology asks the same question, and answers it the same way.)

Gluing simplices: the two rules

A single simplex is a rather boring shape: every simplex can be squashed continuously down to a point, so as far as holes are concerned it has none. Interesting shapes come from gluing many simplices together — but carefully. We want the gluing to be so tidy that knowing which simplices we used tells us everything. Two rules achieve this.

Rule 1 says the collection is complete: if a triangle is in it, so are its three edges and three vertices. It would be strange to have a triangle but not its edges, and the rule forbids it. Rule 2 says that simplices may only touch “corner to corner, edge to edge”: two triangles may share a whole edge, or a single vertex, but they may not overlap, cross, or have the corner of one resting in the middle of an edge of the other. Here are the possibilities side by side.

Which gluings are allowed?

Glued along an edge

They meet in a whole common edge.

Glued at a corner

They meet in a single common vertex.

A missing edge

Rule 1 fails: a face of the triangle is not in the list.

A corner on an edge

Rule 2 fails: they meet in a point that is a vertex of one but not of the other.

Overlapping triangles

Rule 2 fails: they share a whole region that is not a face of either.

Edges that cross

Rule 2 fails: two edges meet at a point that is not a vertex.

Figure 2.5.3 The two pictures on top obey both rules. The others each break one: a missing face breaks rule 1; a corner resting on an edge, an overlap or a crossing breaks rule 2, because the two pieces meet in something that is not a face of both.

A little vocabulary, which we will use constantly. The dimension of a complex is the largest dimension of its simplices: a complex made of edges and vertices is 1-dimensional (a graph, the main character of §3.1 Cycles and Boundaries); a surface made of triangles is 2-dimensional. The number of $k$ -simplices of $K$ is written $n_{k},$ and the list $(n_{0}, n_{1}, n_{2}, \dots)$ is the f-vector of $K :$ the number of vertices, edges, triangles, and so on. The hollow tetrahedron has f-vector $(4, 6, 4);$ keep an eye on f-vectors, because the next chapter will find something hiding in them. The union of all the simplices — the actual set of points in space — is the underlying space of $K,$ written $∣ K ∣ .$ A sub-collection that obeys the rules by itself is a subcomplex; the most useful ones are the skeletons: the $k$ -skeleton $K^{(k)}$ consists of all simplices of dimension at most $k .$ The 1-skeleton of a hollow tetrahedron is a graph with 4 vertices and 6 edges; its 0-skeleton is just its 4 vertices.

Now build some complexes yourself. The figure checks both rules as you go.

Build a simplicial complex Pick a tool · tap to add · drag to move

vertices6

edges8

triangles2

(n_{0}, n_{1}, n_{2}) = (6, 8, 2)

This is a simplicial complex.

Drag a vertex to move it.

Fill in missing faces (rule 1)

Start from:

Figure 2.5.4 Add vertices, edges and triangles, or drag vertices around. With “fill in missing faces” switched on, rule 1 is enforced for you (new edges flash teal); switch it off and missing faces show up as dashed amber lines. Anything that breaks rule 2 is circled in rose. Try to make a crossing, then drag a vertex until it disappears.

Abstract simplicial complexes: shapes as lists

Here is the payoff of the two rules. Because simplices only meet in common faces, and because every simplex is determined by its vertices, a simplicial complex is completely described by saying which sets of vertices span simplices. The positions of the vertices in space add nothing that topology cares about. So we can throw the positions away.

Rule 1 has become the second bullet (“subsets of simplices are simplices”), and rule 2 has disappeared — there is nothing left that could overlap or cross, because there are no positions any more. For example, with vertices $0, 1, 2,$ the hollow triangle is the list $K = {{0}, {1}, {2}, {0, 1}, {0, 2}, {1, 2}},$ and the filled triangle is the same list with ${0, 1, 2}$ added. Since rule 1 fills in all the faces automatically, it is enough to list the facets — the simplices that are not faces of any bigger simplex. The hollow triangle has facets ${0, 1}, {0, 2}, {1, 2};$ the filled triangle has the single facet ${0, 1, 2} .$

This is the moment when, in Hermann Weyl’s image at the top of the chapter (Weyl 1939, p. 500), the devil of algebra gets its foot in the door of topology. A list can be counted, sorted, fed to a computer and turned into matrices — and that is exactly what Part III will do. The angel need not worry: we will keep checking that every count means something you can see.

Going back from a list to a shape in space is called geometric realisation: choose a point for each vertex, and draw the simplex spanned by the points of each set in the list. Can that always be done without breaking rule 2? Yes — and there is a simple recipe.

Why does this recipe never create a bad crossing? Points of the drawn simplices can be described by barycentric coordinates: a point of the simplex on ${v_{1}, v_{2}, v_{5}}$ is $t_{1} e_{1} + t_{2} e_{2} + t_{5} e_{5},$ whose coordinates in $R^{m}$ are just $(t_{1}, t_{2}, 0, 0, t_{5}, 0, \dots) .$ So each point of the drawing remembers, in its coordinates, exactly which vertices it uses; two simplices can only share points that use vertices common to both, which is to say points of a common face. The price is a lot of dimensions — one for every vertex. Far fewer are enough: a complex of dimension $d$ can always be realised in $R^{2 d + 1}$ (Edelsbrunner & Harer 2010, ch. III). For instance every graph ( $d = 1)$ can be drawn in ordinary space without crossings, even though some graphs cannot be drawn in the plane without crossings — a fact we will prove in §2.6 The Euler Characteristic.

Finally, relabelling the vertices does not change the shape: two abstract complexes are isomorphic if there is a one-to-one correspondence between their vertices that matches simplices with simplices. The hollow triangle on $0, 1, 2$ and the one on $a, b, c$ are isomorphic.

Triangulations of familiar shapes

A simplicial complex is only useful to a topologist if it is the shape we care about, up to homeomorphism.

Let us triangulate the spine examples of this book, starting small.

The circle. The hollow triangle — three vertices, three edges — is a circle: inflate it like a balloon, and its three straight sides round out into a circle without tearing. Could two vertices do? With two vertices there is only one possible edge (an edge is determined by its two endpoints), and one edge is a segment, not a circle. So three vertices is the minimum. This “hollow triangle” is the first spine example of the book, and it will reappear in almost every chapter as the simplest shape with a hole.

The disk and the sphere. The filled triangle is a disk. The hollow tetrahedron — 4 vertices, 6 edges, 4 triangles, with no solid interior — is a sphere: imagine a light at its centre projecting each point outward onto a surrounding globe. The hollow octahedron (6 vertices, 12 edges, 8 triangles) is another triangulation of the same sphere. One shape, many triangulations: that observation will drive the whole of the next chapter.

The torus, and why a 2 × 2 grid is not enough

In §2.2 Gluing Spaces Together you made a torus from a square by gluing the left side to the right and the bottom to the top (the gluing word $ab a^{- 1} b^{- 1}).$ To triangulate it, cut the square into a grid of little squares and cut each little square along a diagonal. After gluing, points on opposite sides of the big square are the same point of the torus, so they get the same label.

The square torus, cut into triangles Hover or tap a label · switch grids

$9$ vertices, $27$ edges, $18$ triangles. Every edge has its own pair of endpoints and every triangle its own three corners

Figure 2.5.5 The square with its sides glued as the arrows show; a label appears twice (or four times, at the corners) when the gluing makes those points one. In 3D: the same triangles drawn on a torus, where every label appears once. With a

3 \times 3

grid this is a simplicial complex. Switch to

2 \times 2 :

two different edges (rose and violet) join the same two vertices.

With a $3 \times 3$ grid we get $9$ vertices, $27$ edges (each little square owns its bottom edge, its left edge and its diagonal: $3 \times 9)$ and $18$ triangles (two per square). Every edge joins its own pair of vertices, and every triangle has its own three corners, so the list of vertex sets describes the torus faithfully. This is the second spine example of the book.

A $2 \times 2$ grid seems simpler, but it fails, and the reason is a counting argument. After gluing it has $4$ vertices, yet it needs $12$ edges and $8$ triangles. Four vertices only have $(2 4) = 6$ pairs and $(3 4) = 4$ triples. Twelve edges on six pairs means some pairs of vertices are joined by two different edges — and in a simplicial complex an edge is determined by its endpoints, so the list cannot tell those edges apart. Likewise each triple of vertices would have to carry two different triangles. The $2 \times 2$ grid is a perfectly good way to glue a torus, just not a simplicial complex. (We will meet a framework that allows it — Δ-complexes — at the end of the chapter.)

The smallest torus: seven vertices

Nine vertices is not the record. The torus can be triangulated with just seven vertices, $21$ edges and $14$ triangles, using the triangles ${i, i + 1, i + 3} and {i, i + 2, i + 3} for i = 0, 1, \dots, 6,$ where the labels are read modulo 7 (on a 7-hour clock: $5 + 3 = 1,$ as in §1.2 Equivalence and Quotients). It is found in the notes of August Möbius, published after his death (Lutz 2008). Now count its edges against its vertices. Seven vertices can be paired up in $(2 7) = 21$ ways, and there are $21$ edges, so no pair is left out: every vertex is joined to every other. And seven is the minimum — you will prove that with the next chapter’s tools.

The easiest way to see it is to tile the whole plane with equilateral triangles and label the lattice point $a$ steps along one direction and $b$ steps along the other with $(a + 3 b) mod 7 .$ The labels repeat, and rolling the plane up so that equal labels coincide wraps it exactly once around a torus. Around any point, the six neighbours carry the six other labels — which is why every pair of vertices ends up joined.

Seven vertices, every pair joined Tap a vertex · drag to rotate

Vertex 0 is joined to 1, 2, 3, 4, 5, 6 — all six others. Pairs checked: 6 of 21

Figure 2.5.6 The labelled triangular tiling; the 14 shaded triangles form one complete copy of the torus. Tap a vertex: each of its copies is joined to the six others. In 3D: the same seven vertices on a torus in space — or as Császár’s polyhedron, built from 14 flat triangles with no self-crossings. The counter shows that all 21 pairs are edges.

Can the seven-vertex torus be built in space from flat triangles and straight edges, without crossing itself? Yes: in 1949 the Hungarian mathematician Ákos Császár found coordinates for its seven vertices that do it (Császár 1949), and the result is called Császár’s polyhedron — a polyhedron with a hole through it and no diagonals at all, since every pair of corners is already joined by an edge. (The coordinates in our figure are those listed by Frank Lutz (Lutz 2008), squashed vertically to fit the screen — a stretch or squash cannot create crossings — and we checked by computer that no two triangles cross.) Its dual, found by Lajos Szilassi in 1977, has seven hexagonal faces, each touching all six others. Colour its faces so that neighbours differ and you need seven colours: a map drawn on a torus can need seven colours, not the four that always suffice on a sphere.

The projective plane in six vertices

The real projective plane $RP^{2}$ of §2.2 Gluing Spaces Together — a disk whose opposite boundary points are glued — has a famous small triangulation too: $6$ vertices, $15$ edges and $10$ triangles. Again $(2 6) = 15,$ so every pair of vertices is an edge. One way to build it is to take a regular icosahedron (20 triangles) and glue each point to the diametrically opposite point; that halves everything, so it is also called the hemi-icosahedron. Here it is drawn as a disk: one vertex in the middle, and the other five each appearing twice on the boundary circle, at opposite points, because opposite points are glued.

ℝP² with six vertices Tap a vertex

Opposite points of the circle are glued (matching colours and arrows). Tap a vertex.

Figure 2.5.7 The 6-vertex projective plane. Boundary arcs of the same colour, with matching arrows, are glued. Tap a vertex to light up all its copies and edges.

For the Klein bottle, seven vertices are not enough. A seven-vertex triangulation of it would also have to join every pair of vertices (the next chapter’s counting shows why), and Philip Franklin proved in 1934 that seven points cannot all be joined to each other on a Klein bottle without crossings (Franklin 1934). Eight vertices is the minimum (Lutz 2008), and the $3 \times 3$ grid with one pair of sides glued with a twist gives a simple nine-vertex triangulation.

Can everything be triangulated? Every surface can, as Tibor Radó proved in 1925 (Radó 1925), and so can every shape in this book. Whether every manifold can was one of the long-standing questions of topology.

Conjecture The triangulation conjecture

Disproved in 2013 Asked since the 1920s; disproved by Ciprian Manolescu, 2013

Every manifold can be cut into simplices: it is homeomorphic to a simplicial complex.

The answer took nearly a century and, in the end, a new kind of homology. Remarkably, the obstruction lives in dimension three: whether high-dimensional manifolds can be cut into simplices was decided by a question about 3-manifolds.

How it was settled. Galewski and Stern, and independently Matumoto, had reduced the question in dimensions $5$ and higher to a problem about 3-manifolds with the homology of the 3-sphere. Manolescu solved that problem in 2013 with a new version of Floer homology, a homology theory built from solutions of differential equations, and the answer was no: in every dimension from $5$ on there are manifolds that cannot be triangulated.

Before and after. Every surface can be triangulated (Radó, 1925), every 3-manifold (Moise, 1952), and every smooth manifold of any dimension. In dimension $4,$ manifolds that cannot be triangulated were found in the 1980s, from Freedman’s $E_{8}$ manifold and Casson’s invariant.

A way in. Triangulate the torus with as few vertices as possible (seven suffice, and fewer do not), and compute its homology from your triangulation.

(Radó 1925; Galewski & Stern 1980; Manolescu 2016) Status as of October 2026 All open problems

This is one reason later chapters develop homology for spaces that are not given to us in pieces.

Orientation: giving simplices a direction

In §3.2 Chains and the Boundary Operator we will add and subtract simplices like numbers. To subtract, you need a sense of direction: walking along an edge from $v_{0}$ to $v_{1}$ should be the opposite of walking from $v_{1}$ to $v_{0} .$ An orientation is that sense of direction.

For an edge it is just an arrow. Writing $[v_{0}, v_{1}]$ means the edge traversed from $v_{0}$ to $v_{1};$ writing $[v_{1}, v_{0}]$ means the same edge traversed backwards, and we will write $[v_{1}, v_{0}] = - [v_{0}, v_{1}] .$

For a triangle, an orientation is a direction of travel around it. The order $[v_{0}, v_{1}, v_{2}]$ means “go from $v_{0}$ to $v_{1}$ to $v_{2}$ and back to $v_{0} ”.$ Starting the same tour at a different corner — $[v_{1}, v_{2}, v_{0}]$ or $[v_{2}, v_{0}, v_{1}]$ — goes round in the same direction, so these three orders give the same orientation. The other three orders, such as $[v_{1}, v_{0}, v_{2}],$ go round the other way. In a picture, one orientation is anticlockwise and the other clockwise.

You can tell the two kinds of order apart without drawing anything. Starting from $[v_{0}, v_{1}, v_{2}],$ count how many swaps of two neighbouring entries it takes to reach the order you have. Rotating, $[v_{0}, v_{1}, v_{2}] \to [v_{1}, v_{0}, v_{2}] \to [v_{1}, v_{2}, v_{0}],$ takes two swaps; reversing a pair takes one. An even number of swaps keeps the orientation, an odd number reverses it. Different routes to the same order may use different numbers of swaps, but the parity (even or odd) always comes out the same — the hint to the exercise “Same orientation or opposite?” below explains why.

Orientations Press the buttons

Edge:

[v_{0}, v_{1}]

points from

v_{0}

v_{1}

Triangle:

[v_{0}, v_{1}, v_{2}]

is an even rearrangement of

[v_{0}, v_{1}, v_{2}]

— the same orientation (anticlockwise here). Swaps so far: 0.

Figure 2.5.8 Rearrange the vertices of the edge and the triangle. The arrows show the direction each order describes: an even number of swaps (gold) keeps the triangle anticlockwise, an odd number (violet) turns it clockwise. Notice that a rotation is two swaps.

For a tetrahedron $[v_{0}, v_{1}, v_{2}, v_{3}],$ the two orientations are the two “handednesses” of a screw: curl the fingers of your right hand around the face $[v_{1}, v_{2}, v_{3}]$ in the direction $v_{1} \to v_{2} \to v_{3};$ your thumb then points either towards $v_{0}$ or away from it, and swapping any two vertices switches which. You will not need to picture higher orientations; the swap-counting rule handles every dimension at once.

An oriented simplex hands a direction down to its edges. Travelling around $[v_{0}, v_{1}, v_{2}]$ runs along $v_{0} \to v_{1},$ then $v_{1} \to v_{2},$ then $v_{2} \to v_{0} .$ Two of these agree with the default directions $[v_{0}, v_{1}]$ and $[v_{1}, v_{2}],$ but the third goes against $[v_{0}, v_{2}] .$ Keeping track of exactly this kind of agreement and disagreement will give the formula $\partial [v_{0}, v_{1}, v_{2}] = [v_{1}, v_{2}] - [v_{0}, v_{2}] + [v_{0}, v_{1}]$ for the boundary of a triangle in §3.2 Chains and the Boundary Operator, the minus sign marking the edge travelled backwards.

Finally, orientations connect to orientability from §2.4 Manifolds and Surfaces. Orient every triangle of a triangulated surface. The orientations are coherent if, wherever two triangles share an edge, they run along it in opposite directions. Draw two triangles side by side on paper and orient both anticlockwise: along their shared edge one goes up and the other comes down, automatically. Coherence asks for the same thing all over a surface that need not lie flat. A surface can be coherently oriented exactly when it is orientable: the torus triangulations above can be, the projective plane cannot. On $RP^{2},$ however you orient the ten triangles, somewhere two neighbours will run along their shared edge in the same direction.

Simplicial maps and subdivision

So far each complex has stood alone. But topology is as much about maps as about spaces: a circle winding twice around another, a disk squashed flat onto a segment. Since a simplicial complex is “vertices plus which sets span simplices”, the natural maps between complexes send vertices to vertices and respect those sets.

A simplicial map is really a continuous map in disguise: extend it to every point using barycentric coordinates, $f (t_{0} v_{0} + \dots + t_{k} v_{k}) = t_{0} f (v_{0}) + \dots + t_{k} f (v_{k}),$ so each simplex is carried straight onto its image, stretching or squashing as needed. Nearby points go to nearby points, so the result is continuous.

Simplicial maps Pick an example · play or scrub the map

f (0) = f (3) = w_{0},

f (1) = f (4) = w_{1},

f (2) = f (5) = w_{2}

Every edge

[i, i + 1]

lands on an edge of the triangle, so the hexagon goes round the triangle twice.

source image

Figure 2.5.9 Carry each vertex to its image and the edges follow. A hexagon wraps twice around a triangle (its second lap is dashed); a triangle collapses onto an edge.

The “wrap twice” map is a good example to keep in mind: it is the combinatorial version of the map that winds a circle twice around another, and in §3.5 Maps, Invariance, and First Triumphs such maps will acquire a degree (here, 2). Simplicial maps are the stand-ins for continuous maps throughout the simplicial theory of homology.

Barycentric subdivision

Often we want a finer triangulation of the same shape — smaller pieces, more of them. The standard way to get one uses barycentres. Put a new vertex at the barycentre of every simplex of $K :$ the old vertices stay, every edge gets its midpoint, every triangle its centre, and so on. Then join them up: a set of new vertices spans a new simplex exactly when the old simplices they came from are nested, each a face of the next. A new triangle, for example, has corners at the barycentres of a vertex, an edge containing that vertex, and a triangle containing that edge — a flag $v \subset e \subset t .$

Barycentric subdivision Step forward · hover a small triangle

V - E + F = 4 - 4 + 1 = 1

A triangle with an edge attached. Step forward to subdivide.

old vertex centre of an edge centre of a triangle

The complex K

Figure 2.5.10 A triangle with an edge attached, subdivided once and twice. Each triangle becomes six and each edge two. Hover a small triangle to see the flag — vertex, edge, triangle — it was built from. The quantity

V - E + F

never changes.

The result is the barycentric subdivision $sd K :$ a new simplicial complex with the same underlying space, so it triangulates the same shape. Counting flags tells you how many pieces you get. In a triangle a flag is a choice of a vertex (3 ways), then an edge containing it (2 ways), so a triangle splits into $3 \times 2 = 6$ small triangles; a tetrahedron splits into $4 \times 3 \times 2 = 24,$ and an $n$ -simplex into $(n + 1)! .$ The subdivided triangle has $7$ vertices, $12$ edges and $6$ triangles.

Subdivision earns its keep in three ways.

It makes pieces as small as we like: each subdivision shrinks every simplex by a definite factor.
It repairs looser gluings: subdividing a Δ-complex (next section) twice always produces an honest simplicial complex (Hatcher 2002, §2.1, Exercise 23).
It lets simplicial maps imitate any continuous map. The simplicial approximation theorem says that any continuous map between triangulated spaces can be deformed into a simplicial map, provided the domain is subdivided finely enough (Hatcher 2002, Theorem 2C.1).

Did you notice the readout in the figure? The f-vector changes from $(4, 4, 1)$ to $(9, 14, 6)$ to $(29, 64, 36),$ yet $V - E + F$ stays at $1 .$ Hold that thought for one chapter.

Looser gluings: Δ-complexes and CW complexes

Simplicial complexes are concrete, but they are expensive. The smallest simplicial torus needs $7$ vertices, $21$ edges and $14$ triangles — $42$ simplices — although the gluing square of §2.2 Gluing Spaces Together used a single square. Relax the rules a little and the numbers collapse; topologists use two such relaxations all the time.

Δ-complexes

A Δ-complex (“delta complex”) is built from simplices too, but they may be glued more freely: the faces of a single simplex may be glued to each other, and several simplices may share exactly the same vertices. The only requirement is that faces are glued to faces matching the order of their vertices, so the gluing is still recorded by a finite list of which face goes where. The $2 \times 2$ grid torus, illegal before, is a perfectly good Δ-complex — and so is something much smaller.

Three Δ-complexes

A circle

1 vertex, 1 edge

A sphere

3 vertices, 3 edges, and 2 triangles with the same three corners

A torus

1 vertex, 3 edges, 2 triangles

Figure 2.5.11 Not simplicial complexes, but legitimate Δ-complexes. A circle from one edge whose two ends are the same vertex. A sphere from two triangles glued along their whole boundaries, so they have the same three corners. A torus from two triangles

U

and

L :

the square cut by one diagonal

c,

with all four corners glued into one vertex.

The two-triangle torus has $1$ vertex, $3$ edges and $2$ triangles. Hatcher’s standard textbook computes the homology of the torus with exactly these six pieces (Hatcher 2002, §2.1, Example 2.3), and everything in Part III works for Δ-complexes as well as for simplicial complexes.

CW complexes

The second relaxation goes further: drop the triangles altogether and build with cells. An $n$ -cell is an $n$ -dimensional disk: a 0-cell is a point, a 1-cell an arc (an open interval), a 2-cell an open disk, a 3-cell an open ball. We build a space in stages, one dimension at a time.

Start with some points: the 0-cells.
Attach 1-cells: take arcs, and glue the two ends of each arc to points already there.
Attach 2-cells: take disks, and glue the boundary circle of each disk to the part already built, by a continuous map called its attaching map. The inside of the disk stays as it was.
And so on in higher dimensions, gluing the boundary sphere of each $n$ -ball to what has been built so far.

The result is a CW complex. (The letters stand for two technical conditions, “closure-finite” and “weak topology”, from J. H. C. Whitehead’s original definition of 1949 (Whitehead 1949); Hatcher explains them in an appendix (Hatcher 2002, Appendix, p. 520). For the finite complexes in this book both hold automatically.) The space built after the $k$ -th stage is the $k$ -skeleton, just as for simplicial complexes. Let us see how few cells our favourite shapes need.

Circle: one point and one arc, with both ends of the arc glued to the point.
Sphere: one point and one disk, with the whole boundary circle of the disk glued to the point — like pulling the drawstring of a bag tight. No edges at all.
Torus: one point $v;$ two loops $a$ and $b$ attached at $v;$ and one disk, attached along the loop that runs around $a,$ then $b,$ then $a$ backwards, then $b$ backwards — the gluing word $ab a^{- 1} b^{- 1}$ of the square, read as an instruction for the disk’s edge. That is $1 + 2 + 1 = 4$ cells instead of $42$ simplices.
Projective plane: one point, one loop $a,$ and one disk attached along $aa$ (the edge goes round the loop twice). Klein bottle: one point, two loops, and a disk attached along $aba b^{- 1} .$

Building a torus from cells Step through · drag to rotate

1 point 0 loops 0 disks

c_{0} - c_{1} + c_{2} = 1 - 0 + 0 = 1

One point: a 0-cell

Figure 2.5.12 Step through the construction. A single point; a gold loop

a

and a teal loop

b,

both attached at it; then a disk whose edge is glued along

a,

b,

a^{- 1},

b^{- 1}

— watch it grow from the far side until it closes up on the two loops, then follow the glowing point around its rim. Switch to the sphere to see a disk closing up like a drawstring bag.

Here is something to wonder about before the next chapter. Count the pieces of each torus we have built, alternating signs by dimension: $9 - 27 + 18 = 7 - 21 + 14 = 1 - 3 + 2 = 1 - 2 + 1 = 0.$ Four very different ways of cutting up a torus, and the same answer every time. The sphere gives $4 - 6 + 4 = 6 - 12 + 8 = 1 - 0 + 1 = 2 .$ Coincidence? §2.6 The Euler Characteristic explains.

Exercises

Exercise Grids on the torus

(a) In the $2 \times 2$ grid on the square torus, find two different edges with the same two endpoints. (b) What goes wrong with a $1 \times 1$ grid (the square cut by one diagonal)? (c) Show that an $n \times n$ grid has $n^{2}$ vertices, $3 n^{2}$ edges and $2 n^{2}$ triangles after gluing.

Hint

Use the labels of the figure: the vertex in column $i,$ row $j$ is $i + nj,$ reading $i$ and $j$ modulo $n .$

Solution

(a) With labels $0, 1$ on the bottom row (and $2, 3$ above), the bottom edge of the left square runs from $0$ to $1,$ and the bottom edge of the right square runs from $1$ to the corner, which is glued to $0 .$ Two different edges, both joining $0$ and $1 .$ (b) With a $1 \times 1$ grid all four corners are the same vertex, so every edge — the sides $a,$ $b$ and the diagonal — starts and ends at that one vertex, which no simplex can do, and the two triangles have the “same” three corners. (It is the two-triangle Δ-complex torus.) (c) Each of the $n^{2}$ small squares contributes its bottom-left corner, three edges (bottom, left, diagonal) and two triangles, and after gluing every vertex, edge and triangle is counted exactly once.

Exercise Same orientation or opposite?

Which of these orders give the same orientation as $[0, 1, 2] :$ $[2, 0, 1],$ $[1, 0, 2],$ $[2, 1, 0] ?$ Which give the same orientation as the tetrahedron $[0, 1, 2, 3] :$ $[1, 0, 3, 2],$ $[1, 2, 3, 0],$ $[3, 2, 1, 0] ?$

Hint

Count the pairs that are “out of order” (a larger label before a smaller one). Each swap of neighbours changes that count by one, so its parity is the parity of the number of swaps.

Solution

For the triangle: $[2, 0, 1]$ has 2 out-of-order pairs (2 before 0, 2 before 1): same orientation. $[1, 0, 2]$ has 1: opposite. $[2, 1, 0]$ has 3: opposite. For the tetrahedron: $[1, 0, 3, 2]$ has 2: same. $[1, 2, 3, 0]$ has 3 (each of 1, 2, 3 before 0): opposite. $[3, 2, 1, 0]$ has all 6 pairs out of order: same. (Note that a “rotation” of four labels is three swaps, so unlike for a triangle it reverses the orientation.)

Exercise Subdividing a tetrahedron

Find the f-vector $(n_{0}, n_{1}, n_{2}, n_{3})$ of the barycentric subdivision of a solid tetrahedron, and compute $n_{0} - n_{1} + n_{2} - n_{3} .$

Hint

New vertices are the 15 faces of the tetrahedron. A new edge is a pair “face inside a bigger face”; a new tetrahedron is a full flag $v \subset e \subset t \subset T .$

Solution

Vertices: one per face, $4 + 6 + 4 + 1 = 15 .$ Edges: one per pair $σ ⊊ τ$ of faces; a face with $m$ vertices has $2^{m} - 2$ proper faces, so the count is $6 \cdot 2 + 4 \cdot 6 + 1 \cdot 14 = 50 .$ Tetrahedra: one per full flag $v \subset e \subset t \subset T,$ so $4 \times 3 \times 2 = 24 .$ Triangles: one per flag with one step missing. Missing $T :$ $v \subset e \subset t,$ $4 \cdot 3 \cdot 2 = 24 .$ Missing $t :$ $v \subset e \subset T,$ $4 \cdot 3 = 12 .$ Missing $e :$ $v \subset t \subset T,$ $4 \cdot 3 = 12 .$ Missing $v :$ $e \subset t \subset T,$ $6 \cdot 2 = 12 .$ That is $60$ in all. So the f-vector is $(15, 50, 60, 24)$ and $15 - 50 + 60 - 24 = 1$ — the same as for the tetrahedron itself, $4 - 6 + 4 - 1 = 1 .$

Exercise How few vertices?

Show that a simplicial complex homeomorphic to a circle needs at least 3 vertices, and one homeomorphic to a sphere needs at least 4.

Hint

With 3 vertices, how many different triangles can there be? Can a single triangle, possibly with extra edges, be a sphere?

Solution

Circle. With 1 vertex there are no edges at all; with 2 vertices there is at most one edge (an edge is determined by its endpoints). So the complex is a point, two points, or a segment — never a circle. Three vertices suffice: the hollow triangle.

Sphere. A sphere is two-dimensional, so the complex must contain a triangle. (That a one-dimensional complex — a graph — cannot be homeomorphic to a sphere is believable, and it is proved properly in §3.5 Maps, Invariance, and First Triumphs.) With only 3 vertices there is at most one triangle, ${0, 1, 2},$ and rule 1 forces in its edges and vertices: the complex is a filled triangle, which is a disk. But a disk has a rim — the points in the middle of its three edges sit on the edge of the shape, with surface on one side only — while a sphere has no rim anywhere. So at least 4 vertices are needed, and the hollow tetrahedron shows that 4 suffice.

Summary

Recap

An $n$ -simplex is the filled-in shape spanned by $n + 1$ points in general position: point, edge, triangle, tetrahedron, …. It has $(k + 1 n + 1)$ faces of dimension $k$ — row $n + 1$ of Pascal’s triangle.
Barycentric coordinates $(t_{0}, \dots, t_{n}),$ all $\geq 0$ and adding to $1,$ locate every point of a simplex; the non-zero ones name the face it lies in. The standard simplex $Δ^{n}$ is the set of all such lists.
A simplicial complex is a collection of simplices closed under taking faces (rule 1) in which any two meet in a common face or not at all (rule 2). Its f-vector $(n_{0}, n_{1}, n_{2}, \dots)$ counts simplices by dimension.
An abstract simplicial complex is just a list of vertex sets closed under taking subsets; every one can be realised in space. For topology, the list is the shape.
Triangulations: the hollow triangle (circle), hollow tetrahedron (sphere), the $3 \times 3$ grid torus (the $2 \times 2$ grid fails), Möbius’s 7-vertex torus and Császár’s polyhedron, the 6-vertex projective plane.
An orientation is an ordering of the vertices up to an even number of swaps; each simplex has two. Default: increasing labels.
Simplicial maps send simplices to simplices; barycentric subdivision refines a complex without changing its shape.
Δ-complexes and CW complexes glue more freely and need far fewer pieces: the torus is 1 point, 2 loops and 1 disk attached along $ab a^{- 1} b^{- 1} .$

Simplicial Complexes

From shapes to lists

Simplices: points, edges, triangles and beyond

Faces, and Pascal’s triangle

Barycentric coordinates: an address for every point

Gluing simplices: the two rules

Abstract simplicial complexes: shapes as lists

Triangulations of familiar shapes

The torus, and why a 2 × 2 grid is not enough

The smallest torus: seven vertices

The projective plane in six vertices

Orientation: giving simplices a direction

Simplicial maps and subdivision

Barycentric subdivision

Looser gluings: Δ-complexes and CW complexes

Δ-complexes

CW complexes

Exercises

Summary

Further reading

References