2.6 The Euler Characteristic · Homology & Cohomology

Without much exaggeration, it can be said that only one important topological concept came to light before Poincaré. This was the Euler characteristic of surfaces…

— John Stillwell, introduction to Poincaré’s Papers on Topology (2010)

Take any box, die, pyramid or gemstone. Count its corners, its edges and its flat faces. Subtract the second number from the first, and add the third. A cube gives $8 - 12 + 6 = 2 .$ A pyramid gives $5 - 8 + 5 = 2 .$ A football, stitched from 32 panels, gives $60 - 90 + 32 = 2 .$ The answer is always 2. This chapter is about why — and about what happens when it is not 2, which turns out to be even more interesting.

A curious count

A polyhedron is a solid whose surface is made of flat polygons (with no holes in them), the faces, joined along straight edges, with the edges meeting at vertices (corners). The cube is one; so is a pyramid, a prism, a cut diamond. For each polyhedron we count three numbers: $V,$ the number of vertices; $E,$ the number of edges; and $F,$ the number of faces. Then we compute $V - E + F .$ Try it in the gallery below; the figure does the counting with you.

A gallery of polyhedra Choose a solid · tap a count · drag to rotate

Cube

V - E + F = 8 - 12 + 6 = 2

Angle defects: every corner falls 90° short of a flat 360°, and 8 × 90° = 720°.

Tap a count to watch it being counted.

Figure 2.6.1 Pick a solid and tap

V,

E

F

to watch the vertices, edges or faces counted one by one. However the numbers change,

V - E + F

is always

2 .

(The note about angle defects will make sense later in the chapter.)

The five Platonic solids — the convex polyhedra whose faces are identical regular polygons, the same number meeting at every corner — make a good first test.

Solid	Faces	V	E	F	V − E + F
Tetrahedron	4 triangles	4	6	4	2
Cube	6 squares	8	12	6	2
Octahedron	8 triangles	6	12	8	2
Dodecahedron	12 pentagons	20	30	12	2
Icosahedron	20 triangles	12	30	20	2

The individual numbers vary wildly; the combination $V - E + F$ does not. It is not just the Platonic solids. Take a prism over a polygon with $n$ sides — two copies of the polygon, top and bottom, joined by $n$ rectangles. Drag the number to change it: with $n =$ 6 there are $2 n$ vertices, $3 n$ edges ( $n$ on top, $n$ on the bottom, $n$ upright) and $n + 2$ faces, and $12 - 18 + 8 = 2$ . A pyramid over the same polygon has $n + 1$ vertices, $2 n$ edges and $n + 1$ faces: $7 - 12 + 7 = 2$ . A double pyramid has $n + 2$ vertices, $3 n$ edges and $2 n$ faces: $8 - 18 + 12 = 2$ . In each case the algebra shows why: the $n ’$ s cancel, and 2 is left over.

Euler’s polyhedron formula

“Convex” means the solid has no dents: the straight segment between any two of its points stays inside it. Every polyhedron in the gallery is convex. The more general phrasing says the same thing topologically: the surface must be a sphere in disguise — homeomorphic to a sphere, in the language of §2.1 Spaces and Continuity. That condition is not decoration. We will see that the whole content of the formula is a fact about the sphere, and that other surfaces give other numbers.

A little history A lost notebook and a letter

Leonhard Euler 1707 – 1783

The formula was announced in a letter to Christian Goldbach in November 1750 and published, with an attempted proof, in 1758 (Euler 1758; Euler 1758). Euler’s insight was to ignore angles and lengths and count only corners, edges and faces. Euler went almost entirely blind soon after he returned to St Petersburg in 1766; working from a remarkable memory, with his sons’ help, he went on to produce almost half of all his work. René Descartes had come close around 1630, in a notebook about polyhedra. After his death in Stockholm the notebook was shipped back to Paris, the boat sank in the Seine, and the pages spent three days under water. Leibniz copied them in 1676; then both the original and the copy vanished, until the copy turned up in Hanover around 1860 (Federico 1982). The whole story is told in Richeson (2008).

Descartes’s notebook contained a related theorem that you can already check in the gallery. At each corner of a convex polyhedron, the angles of the faces meeting there add up to less than a full turn of $360°$ — otherwise the corner would be flat. The shortfall is the angle defect at that corner. A cube has three right angles at each corner, so each defect is $360° - 270° = 90°,$ and the eight corners together fall short by $8 \times 90° = 720° .$ Descartes found that the total defect of every convex polyhedron is exactly $720°,$ two full turns (Federico 1982) — and two is the same 2 as in Euler’s formula. That should seem odd. Adding up angles is measuring; counting corners, edges and faces is not. Why should one know about the other? Keep the question in mind: the last section of the chapter answers it in three lines.

Euler’s formula is one relation among the numbers of vertices, edges and faces. Finding all such relations, for polytopes of every dimension, is still an active subject, and some of its simplest questions are open.

Conjecture Kalai’s 3ᵈ conjecture

Open since 1989 Posed by Gil Kalai, 1989

A convex polytope in $d$ dimensions that is symmetric about its centre has at least $3^{d}$ faces, counting the vertices, the edges, and so on up to the polytope itself.

Count everything: a cube has $8$ vertices, $12$ edges, $6$ faces and $1$ solid, $27 = 3^{3}$ in all, and the octahedron, with $6 + 12 + 8 + 1,$ also has $27 .$ Among solids that are symmetric about their centre, these seem to be the most economical.

What is known. The cube shows the bound is right: a face of $[- 1, 1]^{d}$ is chosen by deciding, for each coordinate, whether it is $- 1,$ $+ 1$ or free, so there are exactly $3^{d}$ of them. The conjecture is proved in dimensions up to $4$ (Sanyal, Werner and Ziegler, 2009) and for several special families of polytopes; from dimension $5$ on it is open.

A way in. Count the faces of the square ( $4 + 4 + 1 = 9),$ the regular hexagon, the cube and the octahedron, and check your counts against Euler’s formula $V - E + F = 2 .$ Then look for a centrally symmetric solid with more than $27$ faces in all.

(Kalai 1989; Sanyal, Werner & Ziegler 2009) Status as of October 2026 All open problems

Why it is true

A good proof should explain the 2. Here are two. The first is Cauchy’s, which changes the polyhedron step by step without ever changing $V - E + F;$ the second counts the edges with two trees.

Cauchy’s proof: shrink the problem without changing the answer

The idea is to keep simplifying the picture by moves that never change $V - E + F,$ until we reach something trivial.

Remove one face and flatten. Take out one face, say the front face of the cube, and imagine the rest is made of stretchy rubber. Pull the hole wide open and lay the surface flat on a table. You get a flat network of points and lines — for the cube, a small square inside a big one, joined at the corners. Vertices and edges are untouched; only one face is gone. So for the flat network, $V - E + F$ is exactly one less than for the polyhedron: Euler’s formula is the same as saying the flat network has $V - E + F = 1 .$ (The removed face has not really disappeared: it has become the region outside the network.)
Cut every face into triangles. Draw diagonals across each polygon until only triangles are left. Each diagonal adds one edge, and splits one face into two, so it adds one face too. $E$ and $F$ both go up by 1, and $V - E + F$ is unchanged.
Remove triangles from the outside, one at a time. A triangle on the outside of the network touches the outside along either one edge or two edges.
- One edge: removing the triangle removes that edge and one face. $E$ and $F$ both drop by 1; no change.
- Two edges: removing the triangle removes both edges, the corner between them, and one face. $V$ drops by 1, $E$ by 2 and $F$ by 1; the change in $V - E + F$ is $- 1 + 2 - 1 = 0 .$
Stop when one triangle is left. It has $3 - 3 + 1 = 1 .$ Since nothing ever changed, the flat network had $V - E + F = 1$ all along, and the polyhedron, with its face put back, has $V - E + F = 1 + 1 = 2 .$

Cauchy’s proof, on a cube Step through, or press play

V - E + F = 8 - 12 + 6 = 2

A cube has 8 vertices, 12 edges and 6 faces. We will show that V − E + F = 2 by changing the picture in steps that never change V − E + F.

A cube

Figure 2.6.2 Each step changes

V,

E

and

F,

but never

V - E + F .

The dashed rose outline shows what the last step removed. The final count, one triangle, is

3 - 3 + 1 = 1;

adding back the face we took out gives

2 .

Look at where the proof used the shape of the polyhedron. Only in step 1: removing a face had to leave something that can be stretched flat. That is true for anything shaped like a sphere — a sphere with a hole punched in it is a disk — and it is exactly what will fail for a polyhedron with a tunnel through it.

A second proof: two trees

Flatten the polyhedron as in step 1 (or simply draw its edges on a sphere). Now choose a spanning tree: a set of edges that connects all $V$ vertices without forming any loop. A tree on $V$ vertices always has exactly $V - 1$ edges, because building it up one edge at a time, each new edge brings in exactly one new vertex. Colour these edges teal.

Now look at the edges you did not use. Put a dot in the middle of every face (including one for the outside face), and for each unused edge, draw a short path between the two faces on either side of it, crossing that edge. These paths join the $F$ face dots into a network of their own, the dual graph. Two facts make the proof work. The paths join up all the faces — if a group of faces were cut off from the rest, the teal edges around it would have to form a loop, and a tree has none. And the paths contain no loop — a loop of paths would fence off some vertices from the others, but the teal tree connects everything. So the unused edges form a tree on the $F$ faces, with $F - 1$ edges. Every edge is in exactly one of the two trees, so $E = (V - 1) + (F - 1), that is, V - E + F = 2.$ The 2 is $1 + 1 :$ one for each tree. This proof goes back to Karl von Staudt’s Geometrie der Lage of 1847 (von Staudt 1847), and it is the favourite of David Eppstein, who has collected twenty-one different proofs of the formula (see the reading list).

Two trees Hover a grey edge · try another tree

7 tree edges + 5 dual-tree edges = 12 edges

E = (V - 1) + (F - 1)

⟹ V - E + F = 2

Teal: a spanning tree — it reaches all 8 corners, with no loops, using 7 edges. Violet: the edges it leaves out, crossed by paths joining the 6 faces (the outside counts) into a tree with 5 edges. Hover a grey edge.

Figure 2.6.3 The flattened cube. Teal: a spanning tree of the 8 corners (7 edges). Violet: paths across the other 5 edges, joining the 6 faces — counting the outside — into a tree. Every one of the 12 edges is used by exactly one of the two trees. Each grey edge, added to the teal tree, would close a gold loop.

Cutting finer never changes it

Cauchy’s proof hides a more general idea, worth pulling into the light. Take any surface cut into polygons — a polyhedron, or a triangulated torus from §2.5 Simplicial Complexes — and refine it by one of three moves.

Split an edge by putting a new vertex in its middle: $V$ goes up by 1, and the edge becomes two, so $E$ goes up by 1. Change in $V - E + F :$ $+ 1 - 1 = 0 .$
Cut a face with a new edge between two of its corners: $E$ goes up by 1 and $F$ goes up by 1. Change: $- 1 + 1 = 0 .$
Put a vertex inside a face with $k$ corners and join it to all of them: $V$ goes up by 1, $E$ by $k,$ and the face becomes $k$ triangles, so $F$ goes up by $k - 1 .$ Change: $1 - k + (k - 1) = 0 .$

Try to break it.

Subdivide it yourself Pick a move · click an edge or face · drag to turn

8vertices

12edges

6faces

V − E + F 2

Choose a tool, then click an edge or a face of the cube. Drag to turn it.

Figure 2.6.4 Every click changes

V,

E

and

F;

the readout shows by how much. The glowing number

V - E + F

never moves from 2. “Ten random moves” lets the computer try.

Now we can define the number in general. A finite simplicial complex has an f-vector $(n_{0}, n_{1}, n_{2}, \dots)$ — the numbers of vertices, edges, triangles, tetrahedra and so on — and a CW complex has its numbers of cells $c_{0}, c_{1}, c_{2}, \dots .$

This is what makes $χ$ a topological invariant: a number attached to a space that homeomorphisms cannot change. If two spaces have different Euler characteristics, they are definitely not homeomorphic — no amount of stretching will turn one into the other.

Some quick checks with complexes from the previous chapter. A single point: $χ = 1 .$ A solid triangle: $3 - 3 + 1 = 1 .$ A solid tetrahedron: $4 - 6 + 4 - 1 = 1 .$ All of these can be squashed down to a point, and all have $χ = 1 .$ The circle, as a hollow triangle: $3 - 3 = 0 .$ The barycentric subdivisions of the last chapter changed the f-vector of a triangle with an edge attached from $(4, 4, 1)$ to $(9, 14, 6)$ to $(29, 64, 36),$ and every time $χ = 1 .$ Hatcher’s textbook opens with a stranger example, the “house with two rooms”: a box divided by a floor into two rooms, each reached only by a tunnel through the other. Cut into the obvious pieces it has $29$ vertices, $51$ edges and $23$ faces, and $29 - 51 + 23 = 1$ — a hint that, against all appearances, this house too can be squashed to a point (Hatcher 2002, Example 0.2).

Polyhedra with tunnels

In 1813 Simon Lhuilier pointed out polyhedra for which Euler’s formula fails: polyhedra with tunnels through them (Lhuilier 1813). Take a square picture frame, with a square hole through the middle. Its top and bottom are each made of four trapezoids, the outside of four rectangles, and the inside of the tunnel of four more: $16$ faces. There are $16$ corners (eight around the top, eight around the bottom) and $32$ edges. So $V - E + F = 16 - 32 + 16 = 0.$ Every face is a flat polygon, every edge is shared by two faces, nothing is weird — except that the surface is not a sphere in disguise. It is a torus.

Polyhedra with tunnels Switch shapes · change the number of tunnels · drag to rotate

Lhuilier’s picture frame

16vertices

32edges

16faces

V - E + F = 16 - 32 + 16 = 0

2 - 2 g = 2 - 2 \cdot 1 = 0

Not 2! Each face is a flat polygon and every edge joins two faces — but there is a tunnel through the solid.

Figure 2.6.5 Lhuilier’s picture frame has

V - E + F = 0 .

Switch to the slab of cubes and add tunnels: each one lowers

V - E + F

by 2, so a slab with

g

tunnels has

V - E + F = 2 - 2 g .

Why exactly 2 per tunnel? Cauchy’s proof gives a clue. A surface with a tunnel cannot be laid flat after removing just one face: you must also cut the tube of the tunnel, and cut around the ring of the frame. Here is a direct count instead. Drilling a tunnel removes a small square face from the top and one from the bottom, and puts in a tube of four rectangular walls. The 8 corners of the two little squares stay, as corners where the walls meet the top and bottom; their 8 edges stay too, as the top and bottom rims of the tube. The tube adds 4 new upright edges and 4 new faces, and we lost 2 faces. Total change: $Δ V - Δ E + Δ F = 0 - 4 + (4 - 2) = - 2.$ (This assumes the two little squares were already faces of the surface; you can always arrange that by subdividing first, which does not change $χ .)$ Lhuilier had already found the general rule: a polyhedron pierced by $g$ separate tunnels has $V - E + F = 2 - 2 g .$

Lhuilier had another “monster”, as Lakatos would later call such examples: a cube with a cube-shaped cavity hidden inside it, like a hollow box with thick walls. Its surface is made of two separate pieces — the outside of the box and the inside of the cavity — each a cube’s surface. Counting everything gives $V = 8 + 8 = 16,$ $E = 12 + 12 = 24,$ $F = 6 + 6 = 12,$ and $V - E + F = 16 - 24 + 12 = 4 = 2 + 2.$ This one is easy to explain: the count adds up over separate pieces, and each piece is a sphere in disguise. The Euler characteristic of several separate pieces is the sum of their Euler characteristics.

Careful Faces must be disks

Lhuilier found a third kind of exception, subtler than the other two: put a small cube on top of a big one, in the middle of its top face. The big cube’s top face is now a square with a square hole in it — a ring. (Lhuilier expected such solids to be common in nature, and had seen clusters of crystals shaped like this in a friend’s mineral collection.) Counting naively, $V = 16,$ $E = 24,$ $F = 11,$ and $V - E + F = 3 .$ Nothing is wrong with the surface (it is still a sphere in disguise); the problem is the ring-shaped face. Add one edge across the ring, from an inner corner to an outer corner, and the ring becomes a single disk-shaped face: now $E = 25,$ $F$ is still $11,$ and $V - E + F = 2$ again. The Euler characteristic counts faces that are disks — or, in a CW complex, cells that are open disks — which is why our definition of a polyhedron asked for polygons with no holes.

χ for every closed surface

We can now compute the Euler characteristic of every surface we met in §2.2 Gluing Spaces Together and §2.4 Manifolds and Surfaces, using any decomposition we like.

Surface	Decomposition	Count	χ
Sphere	hollow tetrahedron hollow octahedron point + disk	4 − 6 + 4 6 − 12 + 8 1 − 0 + 1	2
Torus	3 × 3 grid 7-vertex torus point, 2 loops, disk	9 − 27 + 18 7 − 21 + 14 1 − 2 + 1	0
Projective plane	6-vertex point, 1 loop, disk	6 − 15 + 10 1 − 1 + 1	1
Klein bottle	twisted 3 × 3 grid point, 2 loops, disk	9 − 27 + 18 1 − 2 + 1	0
Genus-2 surface	two 7-vertex tori, glued point, 4 loops, disk	11 − 39 + 26 1 − 4 + 1	−2

The cell decompositions (the last line of each row) come straight from the gluing words of §2.2 Gluing Spaces Together: a polygon whose sides are glued in pairs so that all its corners become a single point gives one 0-cell, one 1-cell for each letter, and one 2-cell. The surface of genus $g$ — a sphere with $g$ handles, glued from a $4 g$ -gon with the word $a_{1} b_{1} a_{1}^{- 1} b_{1}^{- 1} \dots a_{g} b_{g} a_{g}^{- 1} b_{g}^{- 1}$ — has $2 g$ letters, so $χ (Σ_{g}) = 1 - 2 g + 1 = 2 - 2 g .$ For $g =$ 3 handles that is $2 - 2 \cdot 3 = - 4$ . Gluing $k$ projective planes together (the word $a_{1} a_{1} a_{2} a_{2} \dots a_{k} a_{k})$ gives a non-orientable surface with $χ (N_{k}) = 1 - k + 1 = 2 - k,$ so for $k =$ 3 it is $2 - 3 = - 1$ .

The same formulas fall out of the connected sum of §2.4 Manifolds and Surfaces, which cuts a small disk out of each of two surfaces and glues them along the two boundary circles. Triangulate both surfaces and cut out one triangle from each: each $χ$ drops by 1 (one face fewer). Then glue the two triangular holes together: the 3 vertices and 3 edges of one hole are matched with those of the other, so we lose $3$ vertices and $3$ edges, which changes nothing ( $- 3 + 3 = 0).$ Hence $χ (A # B) = χ (A) + χ (B) - 2.$ Adding a handle — a connected sum with a torus — lowers $χ$ by $2;$ adding a projective plane lowers it by $1 .$

Counting by cutting and pasting

The connected-sum argument is one instance of a rule that deserves a name of its own. Suppose a complex $X$ is the union of two subcomplexes $A$ and $B,$ which overlap in the subcomplex $A \cap B .$ Count the cells of $X$ by counting those of $A$ and those of $B :$ the cells in the overlap get counted twice, so we subtract them once. Since this works for vertices, edges and faces separately, it works for the alternating sum: $χ (A \cup B) = χ (A) + χ (B) - χ (A \cap B) .$ This is the same inclusion–exclusion you would use to count the people in two overlapping clubs. Combined with two facts we already know — a disk has $χ = 1,$ and a circle, a cylinder and a Möbius band all have $χ = 0$ — it computes Euler characteristics by cutting shapes into simple pieces.

Sphere = northern hemisphere ∪ southern hemisphere, two disks meeting along the equator, a circle: $χ = 1 + 1 - 0 = 2 .$
Torus = two cylinders (slice the doughnut like a bagel into an upper and a lower half; each half is a cylinder), meeting along two circles, the inner and the outer rim: $χ = 0 + 0 - (0 + 0) = 0 .$
Projective plane = a Möbius band with a disk sewn onto its single boundary circle: $χ = 0 + 1 - 0 = 1 .$
Klein bottle = two Möbius bands sewn together along their boundary circles: $χ = 0 + 0 - 0 = 0 .$

Each answer agrees with the table. In §3.6 Exact Sequences and Mayer–Vietoris this cut-and-paste reasoning grows into the Mayer–Vietoris sequence, which does the same thing for the holes themselves rather than only for their alternating count.

Two numbers decide everything

In §2.4 Manifolds and Surfaces you met the classification of closed surfaces: every closed connected surface is a sphere, a sphere with handles, or a connected sum of projective planes (Armstrong 1983, ch. 7). Combine that with the formulas we just found.

Indeed, among orientable surfaces $χ = 2 - 2 g$ determines the number of handles $g,$ and among non-orientable ones $χ = 2 - k$ determines the number of projective planes $k .$ Neither test is enough alone: the torus and the Klein bottle both have $χ = 0,$ but one is orientable and the other is not; the sphere and the torus are both orientable, but have $χ = 2$ and $χ = 0 .$

Every closed surface has its own cell Hover or tap a surface

T^{2}

, the torus

Built from one point, 2 loops and one disk glued along

ab a^{- 1} b^{- 1}

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

Orientable, with 1 handle:

χ = 2 - 2 g = 2 - 2 \cdot 1 = 0

Figure 2.6.6 One direction: the Euler characteristic. The other: orientable or not. Every closed connected surface occupies exactly one cell, and no cell holds two. Orientable surfaces only have even

χ,

since

2 - 2 g

is always even.

Now we can keep a promise from the last chapter: why can no torus be triangulated with fewer than seven vertices? In a triangulated closed surface, $E = 3 (V - χ)$ (see the exercise “Counting a triangulated surface”). So a triangulated torus has $E = 3 V,$ and since at most $(2 V)$ edges are possible, $3 V \leq V (V - 1) /2,$ which forces $V \geq 7 .$ The seven-vertex torus meets this bound exactly — which is why every pair of its vertices had to be joined. The same argument gives $V \geq 6$ for the projective plane and $V \geq 4$ for the sphere.

χ for graphs: pieces minus loops

A graph — vertices joined by edges, with no faces — is a one-dimensional complex, and its Euler characteristic is simply $χ = V - E .$ What does it measure? Build a graph one step at a time. Adding a lone vertex creates a new piece, and raises $χ$ by 1. Adding an edge lowers $χ$ by 1, and it does one of two things: either it joins two pieces into one, or both its ends are already in the same piece and it closes a new loop. So every step either changes the number of pieces or the number of loops, and keeping track gives $V - E = (number of pieces) - (number of independent loops) .$ A tree has one piece and no loops, so $V - E = 1;$ a triangle has one piece and one loop, so $3 - 3 = 0 .$

Pieces and loops Tap to add vertices and edges · switch modes · hover a gold edge

χ = V - E = 5 - 6 = - 1

1pieces

b_{0}

2independent loops

b_{1}

b_{0} - b_{1} = 1 - 2 = - 1

= χ

Each gold edge closes a loop. Hover one to see its loop.

Figure 2.6.7 Each colour is one piece. Gold dashed edges are the “extra” edges: each closes one loop (hover to see it). However you edit the graph,

V - E

equals the number of pieces minus the number of independent loops.

For a connected graph this says that the number of independent loops is $E - V + 1,$ a formula found by Gustav Kirchhoff in 1847 while studying electrical circuits (Kirchhoff 1847) (it counts how many independent loop currents a circuit can carry). The words “independent loop” need care — the graph $K_{4}$ has seven different loops but only $3$ independent ones — and §3.1 Cycles and Boundaries will make them precise. In that chapter the number of pieces is called $b_{0}$ and the number of independent loops $b_{1} .$

A shadow of something deeper

Look at what just happened with graphs. On the left of $V - E = b_{0} - b_{1}$ is a count of the pieces of a decomposition: vertices and edges, which depend on how we drew the graph. On the right is a count of features of the shape itself: how many components it has and how many independent loops. The equation says the alternating count of pieces equals the alternating count of holes.

The same is true one dimension up. For a closed surface, let $b_{0}$ be the number of components, $b_{1}$ the number of independent loops that do not fence off a piece of the surface, and $b_{2}$ the number of enclosed hollows. For the sphere these are $1, 0, 1 :$ one piece; no such loop, since every loop on a sphere cuts it in two; one hollow inside. For the torus they are $1, 2, 1 :$ the loop around the hole and the loop around the tube, neither of which cuts the torus apart, and the hollow inside the tube. And indeed $χ (S^{2}) = 1 - 0 + 1 = 2, χ (T^{2}) = 1 - 2 + 1 = 0, χ (Σ_{g}) = 1 - 2 g + 1 = 2 - 2 g .$ This is the Euler–Poincaré formula, $χ = b_{0} - b_{1} + b_{2} - b_{3} + \dots,$ where the numbers $b_{k},$ the Betti numbers, count “ $k$ -dimensional holes”. Making “hole” precise, defining the $b_{k}$ properly, and proving the formula is the business of Part III; it happens in §3.3 Homology Groups. The proof is the alternating ledger you saw with graphs, in every dimension: each new cell either creates a new hole or fills an old one, and the alternating signs keep the books balanced.

The Euler characteristic turns up far beyond counting. Here is the promised answer to Descartes’s puzzle. A face with $k$ corners has angles adding up to $(k - 2) \times 180° .$ Add this up over all the faces: since every edge borders two faces, the $k ’$ s add up to $2 E,$ and the angles of all the faces together come to $2 E \times 180° - F \times 360° = 360° \times (E - F) .$ The total defect is a full turn at every vertex minus all those angles: $360° \times V - 360° \times (E - F) = 360° \times χ .$ That holds for any closed polyhedral surface — $720°$ for a sphere, and $0°$ for the picture frame, where each of the eight outer corners falls $90°$ short of flat while each of the eight corners around the tunnel overshoots by $90° .$ For smooth surfaces the defects become curvature, and the Gauss–Bonnet theorem says the total curvature of a closed surface is $2 π χ,$ whatever its shape. A vector field on a closed surface — a breeze blowing along it — must have calm points whose “indices” add up to $χ;$ since $χ (S^{2}) = 2 \neq = 0,$ every breeze on the Earth has a calm spot somewhere (the hairy ball theorem), while a torus, with $χ = 0,$ can be combed smooth. These stories are told in §4.8 Curvature and Characteristic Classes.

And the alternating sum makes sense in every dimension. The 3-dimensional sphere, triangulated as the boundary of a 4-simplex, has $5 - 10 + 10 - 5 = 0;$ in fact every closed manifold of odd dimension has $χ = 0,$ a consequence of the Poincaré duality of §4.6 Poincaré Duality. Not bad for a count Euler made with polyhedra.

Still, a single number is a blunt instrument, and the minus signs are to blame. A circle has $χ = 0;$ so do two separate circles, because the extra piece and the extra loop cancel. A sphere and a pair of points both have $χ = 2 .$ The alternating sum balances the books and throws away the entries. To get the entries back — $b_{0},$ $b_{1},$ $b_{2},$ one at a time — we have to say exactly what a hole is, which, you may have noticed, nobody has yet done, and then count holes straight from a list of simplices. That is the machine Part III builds, starting in §3.1 Cycles and Boundaries with the pieces and loops of a graph.

Exercises

Exercise Descartes’s defects

(a) A regular pentagon has angles of $108° .$ Find the angle defect at each corner of a regular tetrahedron, octahedron and dodecahedron, and check that in each case the defects add up to $720° .$ (b) A corner can also have a negative defect. In Lhuilier’s picture frame each outer corner meets two trapezoids, with angles of $45°$ there, and two rectangles; each corner around the tunnel meets two trapezoids, with angles of $135°,$ and two rectangles. Find the total defect, and compare it with $360° \times χ .$

Solution

(a) Tetrahedron: three $60°$ angles at each corner, defect $360° - 180° = 180°,$ and $4 \times 180° = 720° .$ Octahedron: four $60°$ angles, defect $120°,$ and $6 \times 120° = 720° .$ Dodecahedron: three $108°$ angles, defect $36°,$ and $20 \times 36° = 720° .$ (b) An outer corner has $45° + 45° + 90° + 90° = 270°,$ a defect of $90°;$ an inner corner has $135° + 135° + 90° + 90° = 450°,$ a defect of $- 90° .$ The total is $8 \times 90° - 8 \times 90° = 0°,$ which is $360° \times 0,$ as it should be for a surface with $χ = 0 .$

Exercise Counting a triangulated surface

A closed surface is triangulated with $V$ vertices, $E$ edges and $F$ triangles. Show that $E = 3 (V - χ)$ and $F = 2 (V - χ) .$ Deduce that a triangulated torus needs at least 7 vertices, a projective plane at least 6, and a sphere at least 4.

Hint

Every triangle has 3 edges and every edge lies in exactly 2 triangles, so $3 F = 2 E .$ And there are at most $(2 V)$ edges.

Solution

From $3 F = 2 E$ we get $F = \frac{2}{3} E,$ and $χ = V - E + \frac{2}{3} E = V - \frac{1}{3} E,$ so $E = 3 (V - χ)$ and $F = 2 (V - χ) .$ Since an edge is determined by its two endpoints, $E \leq (2 V) = \frac{V ( V - 1 )}{2} .$ For the torus ( $χ = 0):$ $3 V \leq V (V - 1) /2$ gives $V - 1 \geq 6,$ so $V \geq 7 .$ For the projective plane ( $χ = 1):$ $3 (V - 1) \leq V (V - 1) /2$ gives $V \geq 6 .$ For the sphere ( $χ = 2):$ $3 V - 6 \leq V (V - 1) /2,$ i.e. $V^{2} - 7 V + 12 = (V - 3) (V - 4) \geq 0,$ so $V \leq 3$ or $V \geq 4;$ and $V = 3$ is impossible (it would need $E = 3$ and $F = 2 :$ two triangles with the same three corners). So $V \geq 4 .$ All three bounds are attained, by the 7-vertex torus, the 6-vertex projective plane and the hollow tetrahedron.

Exercise Five towns, ten roads

Five towns are each to be joined to each of the others by a road, on flat ground, with no crossings and no bridges. Show that this is impossible. (In graph language: the complete graph $K_{5}$ cannot be drawn in the plane without crossings.)

Hint

A connected drawing in the plane has $V - E + F = 2$ (counting the outside region as a face). Each face is bounded by at least 3 edges.

Solution

Suppose it could be drawn. Then $V = 5,$ $E = 10,$ and Euler’s formula gives $F = 2 - V + E = 7 .$ Each face is surrounded by at least 3 edges and each edge borders at most 2 faces, so $3 F \leq 2 E,$ that is $21 \leq 20$ — a contradiction. (In general this argument shows $E \leq 3 V - 6$ for any crossing-free drawing with $V \geq 3,$ and $K_{5}$ has $10 > 9 .)$ On a torus the corresponding bound is $E \leq 3 V,$ and the seven-vertex torus shows that even $K_{7},$ with $21 = 3 \cdot 7$ edges, fits.

Summary

Recap

Euler’s polyhedron formula: $V - E + F = 2$ for every polyhedron whose surface is a sphere in disguise. Descartes’s angle defects always add up to $720° .$
Cauchy’s proof: remove a face and flatten ( $- 1),$ triangulate (no change), remove triangles from the outside (no change), end with one triangle ( $1).$ The two-trees proof: $E = (V - 1) + (F - 1) .$
Splitting an edge, cutting a face, or putting a vertex in a face never changes $V - E + F .$ In general $χ = n_{0} - n_{1} + n_{2} - \dots,$ and it depends only on the space: it is a topological invariant.
Polyhedra with tunnels break Euler’s formula: each tunnel lowers $χ$ by 2, and the picture frame has $χ = 0 .$ Faces must be disks.
$χ (S^{2}) = 2,$ $χ (T^{2}) = 0,$ $χ (RP^{2}) = 1,$ $χ (Klein bottle) = 0;$ $χ (Σ_{g}) = 2 - 2 g,$ $χ (N_{k}) = 2 - k;$ $χ (A # B) = χ (A) + χ (B) - 2 .$
Closed surfaces are classified by two facts: orientable or not, and $χ .$
For a graph, $χ = V - E$ = (number of pieces) − (number of independent loops).
Coming in §3.3 Homology Groups: $χ = b_{0} - b_{1} + b_{2} - \dots,$ the alternating count of holes.

The Euler Characteristic

A curious count

Euler’s polyhedron formula

Why it is true

Cauchy’s proof: shrink the problem without changing the answer

A second proof: two trees

Cutting finer never changes it

Polyhedra with tunnels

χ for every closed surface

Counting by cutting and pasting

Two numbers decide everything

χ for graphs: pieces minus loops

A shadow of something deeper

Exercises

Summary

Further reading

References