Book 7A

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Course 7Book 7A: Topology and the Fundamental GroupChapter 3

Manifolds and Surfaces

Topological manifolds, the classification of surfaces and the Euler characteristic.

16 min read · Updated Oct 3, 2026

Read with Lee, Introduction to Topological Manifolds, chapter 2 (the definition of manifolds), chapter 5 (cell complexes) and chapter 6 (compact surfaces: the classification theorem and its cut-and-paste proof). Hatcher, chapter 0, is a lighter alternative for cell complexes.

In this chapter · 6 sections
  1. 3.1Twelve pentagons
  2. 3.2Manifolds
  3. 3.3Building surfaces from polygons
  4. 3.4The classification of surfaces
  5. 3.5History
  6. 3.6Exercises

A manifold is a space that looks, near each of its points, like ordinary Euclidean space: a surface is locally a piece of the plane, a three-manifold locally a piece of three-dimensional space. The universe, if it is finite, is presumably a three-manifold; the configuration space of a rigid body is the three-manifold of rotations; the Poincaré conjecture is about closed three-manifolds. This chapter defines manifolds precisely and then solves completely the problem that the Poincaré conjecture poses one dimension up: which closed surfaces are there?

The answer, the classification of surfaces, is a list: the sphere, the torus, the surfaces with more holes, and their non-orientable cousins. Each is determined by two pieces of information, orientability and one integer, the Euler characteristic. The integer is the same one that forces a football to have exactly twelve pentagons. In the list, only the sphere has every loop shrinkable to a point: the two-dimensional Poincaré theorem, proved by classification.

By the end of this chapter you will be able to:

  • define topological manifolds, with and without boundary, and say why each condition in the definition is there;
  • describe surfaces by gluing polygons, and compute their Euler characteristics;
  • form connected sums, and recognise orientable and non-orientable surfaces;
  • state the classification of compact surfaces and sketch its proof;
  • prove that a trivalent polyhedron made of pentagons and hexagons has exactly twelve pentagons.

Twelve pentagons

In the world Data Footballs and fullerenes

The classic 32-panel football is a truncated icosahedron: 12 pentagons and 20 hexagons, with three panels meeting at each of its 60 vertices. In 1985 Harold Kroto, Robert Curl and Richard Smalley and their colleagues found that carbon forms a stable molecule of 60 atoms with exactly this structure, buckminsterfullerene, C60\mathrm{C}_{60}, with an atom at each vertex; they received the 1996 Nobel Prize in Chemistry. Larger fullerenes (C70\mathrm{C}_{70}, C84\mathrm{C}_{84}, carbon nanotubes capped at both ends) are also cages of pentagons and hexagons with three bonds at each atom, and every one of them has exactly twelve pentagons, however many hexagons it has.

This is not chemistry but topology. For any polyhedron whose surface is a topological sphere, Euler's formula says V−E+F=2V - E + F = 2. If every vertex has three edges and every face is a pentagon or a hexagon, with pp pentagons and hh hexagons, then counting edge-ends gives 3V=2E3V = 2E and counting face sides gives 5p+6h=2E5p + 6h = 2E. Substituting into Euler's formula forces p=12p = 12, and leaves hh free (Exercise 3.4, Figure 3.1). A cage of hexagons alone cannot close up into a sphere; it must have exactly twelve "defects". On a torus, where V−E+F=0V - E + F = 0, the same count gives p=0p = 0: a closed cage of hexagons alone is possible there, but not on a sphere.

Figure 3.1. The truncated icosahedron (computed from its vertex coordinates; visible faces only), with pentagons shaded. V−E+F=60−90+32=2V - E + F = 60 - 90 + 32 = 2. Every trivalent pentagon–hexagon sphere has p=12p = 12.

Manifolds

Definition 3.1 Topological manifold

A topological manifold of dimension nn is a topological space MM that is

  1. locally Euclidean: every point has a neighbourhood homeomorphic to an open subset of Rn\mathbb{R}^n;
  2. Hausdorff (7A.1 Topological Spaces and Quotients);
  3. second countable (7A.1 Topological Spaces and Quotients).

A homeomorphism from an open set U⊆MU \subseteq M onto an open subset of Rn\mathbb{R}^n is a chart. The dimension is well defined: an open subset of Rm\mathbb{R}^m is never homeomorphic to an open subset of Rn\mathbb{R}^n for m≠nm \neq n (Brouwer's invariance of domain, 1912; proved with homology or degree theory, 7A.7 Smooth Topology, 7A.8 Homology in Brief). The two extra conditions exclude the line with two origins (not Hausdorff) and spaces too big to carry a metric (not second countable).

Examples. Rn\mathbb{R}^n and its open subsets. The sphere SnS^n: the stereographic projections from the north and south poles are two charts that cover it (7A.2 Compactness and Compactification). The torus Tn=Rn/ZnT^n = \mathbb{R}^n/\mathbb{Z}^n and the projective space RPn=Sn/{±1}\mathbb{R}P^n = S^n/\{\pm1\}, where small balls in the covering space map homeomorphically to the quotient (7A.6 Covering Spaces). Products of manifolds, such as S2×S1S^2\times S^1. Graphs of continuous functions Rn→Rk\mathbb{R}^n \to \mathbb{R}^k. The group SO(3)SO(3) of rotations, a three-manifold (7A.6 Covering Spaces).

A manifold with boundary is defined the same way with charts to open subsets of the closed half-space {xn≥0}\{x_n \geq 0\}; its boundary points are those sent to {xn=0}\{x_n = 0\}, and the boundary ∂M\partial M is a manifold of dimension n−1n - 1 without boundary. The closed disc DnD^n has boundary Sn−1S^{n-1}; the Möbius band has boundary a single circle. A closed manifold is one that is compact and has no boundary. The Poincaré conjecture is about closed three-manifolds.

Building surfaces from polygons

A surface can be described by gluing the edges of polygons in pairs, as in 7A.1 Topological Spaces and Quotients. Label the edges of a polygon with letters, going around the boundary; a letter appearing twice means those edges are glued, matching their directions, and a−1a^{-1} means the edge aa traversed backwards. Then:

word surface
aa−1aa^{-1} sphere S2S^2
aba−1b−1aba^{-1}b^{-1} torus T2T^2
aaaa projective plane RP2\mathbb{R}P^2
abab−1abab^{-1} Klein bottle
a1b1a1−1b1−1a2b2a2−1b2−1a_1b_1a_1^{-1}b_1^{-1}a_2b_2a_2^{-1}b_2^{-1} genus-2 surface

The genus-2 octagon was drawn in the hyperbolic disc in 5A.5 Uniformization and the Two-Dimensional Ricci Flow.

More generally, a cell complex is built by attaching discs of increasing dimension along their boundaries: points (0-cells), then edges (1-cells) attached at their endpoints, then polygons (2-cells) attached along loops of edges, and so on (Lee, chapter 5). A triangulation of a surface is a decomposition into triangles meeting edge to edge. Tibor Radó proved in 1925 that every surface can be triangulated, and Edwin Moise proved the same for three-manifolds in 1952 (8A.1 Smooth Structures).

Definition 3.2 Euler characteristic

For a surface decomposed into VV vertices, EE edges and FF faces (polygons), the Euler characteristic is

χ=V−E+F.\chi = V - E + F.

It does not depend on the decomposition, a fact best proved with homology (7A.8 Homology in Brief); here it can be checked on examples. For the sphere as a polygon with word aa−1aa^{-1}: after gluing there are 22 vertices, 11 edge and 11 face, so χ=2\chi = 2; the surface of a cube gives 8−12+6=28 - 12 + 6 = 2 too. For the torus from the square aba−1b−1aba^{-1}b^{-1}: all four corners become 11 vertex, the edges become 22, and there is 11 face, so χ=0\chi = 0. For RP2\mathbb{R}P^2 from aaaa: χ=1−1+1=1\chi = 1 - 1 + 1 = 1.

Connected sum. Given two connected surfaces, remove a small open disc from each and glue the two boundary circles together: the result is the connected sum Σ1#Σ2\Sigma_1\#\Sigma_2 (Figure 3.2). Since removing a disc lowers χ\chi by 11 (one face less) and gluing along a circle doesn't change it (a circle has χ=0\chi = 0),

χ(Σ1#Σ2)=χ(Σ1)+χ(Σ2)−2.\chi(\Sigma_1\#\Sigma_2) = \chi(\Sigma_1) + \chi(\Sigma_2) - 2.
Figure 3.2. The connected sum of two tori is the genus-2 surface (schematic). The tube that joins them is a neck: in Ricci flow, a three-dimensional neck S2×IS^2\times I can pinch, and Perelman's surgery undoes a connected sum by cutting along such necks (11B.4 Singularities, 12B.4 Surgery).

Orientability. A surface is orientable if it contains no Möbius band, equivalently if a consistent choice of "clockwise" can be made over the whole surface. The sphere and torus are orientable; RP2\mathbb{R}P^2 and the Klein bottle contain Möbius bands and are not.

The classification of surfaces

Theorem 3.3 Classification of compact surfaces

Every compact connected surface without boundary is homeomorphic to exactly one of:

  • the sphere S2S^2, with χ=2\chi = 2;
  • the connected sum of g≥1g \geq 1 tori, Σg=T2#…#T2\Sigma_g = T^2\#\dots\#T^2, the orientable surface of genus gg, with χ=2−2g\chi = 2 - 2g;
  • the connected sum of k≥1k \geq 1 projective planes, Nk=RP2#…#RP2N_k = \mathbb{R}P^2\#\dots\#\mathbb{R}P^2, non-orientable, with χ=2−k\chi = 2 - k.

Two compact connected surfaces are homeomorphic if and only if they are both orientable or both non-orientable and have the same Euler characteristic.

Proof. (Sketch; Lee, chapter 6.) Triangulate the surface (Radó). Glue the triangles together one at a time along shared edges, always keeping the result a single polygon; at the end the surface is one polygon with its edges identified in pairs, described by a word. Then simplify the word by cut-and-paste moves, each of which cuts the polygon along a diagonal and reglues it along a pair of identified edges, so the surface is unchanged: cancel adjacent pairs aa−1aa^{-1}, gather all corners into a single vertex, bring each pair …a…a…\dots a\dots a\dots (same direction) together as aaaa, and each interlocked pair …a…b…a−1…b−1…\dots a\dots b\dots a^{-1}\dots b^{-1}\dots together as aba−1b−1aba^{-1}b^{-1}. One more move converts aa bcb−1c−1aa\,bcb^{-1}c^{-1} into aa bb ccaa\,bb\,cc (a torus plus a projective plane is three projective planes). The result is one of the normal forms aa−1aa^{-1}, a1b1a1−1b1−1⋯agbgag−1bg−1a_1b_1a_1^{-1}b_1^{-1}\cdots a_gb_ga_g^{-1}b_g^{-1}, or a1a1⋯akaka_1a_1\cdots a_ka_k. That the surfaces in the list are different follows from orientability and the Euler characteristic, once both are known to be topological invariants (7A.8 Homology in Brief).

The relation RP2#T2≅RP2#RP2#RP2\mathbb{R}P^2\#T^2 \cong \mathbb{R}P^2\#\mathbb{R}P^2\#\mathbb{R}P^2 shows that the connected sum does not have unique "cancellation" when non-orientable surfaces are involved; for orientable surfaces the genus is all there is. The Klein bottle is N2=RP2#RP2N_2 = \mathbb{R}P^2\#\mathbb{R}P^2.

The two-dimensional Poincaré theorem. In 7A.5 Computing π₁ the fundamental group of each surface in the list is computed: it is trivial for S2S^2 and non-trivial for every other surface (for instance Z2\mathbb{Z}^2 for the torus). So a closed surface in which every loop can be shrunk to a point is a sphere. The proof is by classification: list everything, then check. In three dimensions no such list was available, and the Poincaré conjecture asked whether the same conclusion holds anyway. Perelman's proof does produce a kind of classification, through geometrization (10A.5 Thurston’s Eight Geometries, 12C.4 Geometrization).

Where this goes Euler characteristic and curvature

The Euler characteristic is a combinatorial count, but it is also an integral of curvature: for any Riemannian metric on a closed surface, ∫ΣK dA=2πχ(Σ)\int_\Sigma K\,dA = 2\pi\chi(\Sigma) (Gauss–Bonnet, 8A.9 The Curvature of Surfaces). That is why the sign of χ\chi decides which constant-curvature geometry a surface carries (5A.5 Uniformization and the Two-Dimensional Ricci Flow), and why the normalised Ricci flow on a surface converges to curvature of the sign of χ\chi (11A.7 Ricci Flow on Surfaces). The rehearsal below proves a discrete version for polyhedra, where curvature sits at the vertices.

History

René Descartes found the angle-defect form of the polyhedron formula around 1630, in notes published only in 1860; Leonhard Euler stated V−E+F=2V - E + F = 2 in a letter of 1750 and published it in the following years. The classification of orientable surfaces was obtained by August Möbius (1863) and Camille Jordan (1866), and the non-orientable case was added later; the first complete rigorous proof, based on triangulations, was given by Max Dehn and Poul Heegaard in 1907 and in a polished combinatorial form by Henry Brahana in 1921. Radó's triangulation theorem dates from 1925 and Brouwer's invariance of domain from 1912. Buckminsterfullerene was discovered in 1985.

Recall Where we stand

A topological nn-manifold is locally Euclidean, Hausdorff and second countable; closed manifolds are compact without boundary. Surfaces can be built by gluing polygons along words, and their Euler characteristic χ=V−E+F\chi = V - E + F is a topological invariant: 22 for the sphere, 00 for the torus, 11 for the projective plane, and χ(Σ1#Σ2)=χ(Σ1)+χ(Σ2)−2\chi(\Sigma_1\#\Sigma_2) = \chi(\Sigma_1) + \chi(\Sigma_2) - 2. Every closed surface is a sphere, a connected sum of gg tori, or a connected sum of kk projective planes, determined by orientability and χ\chi. Euler's formula forces twelve pentagons in any trivalent pentagon–hexagon sphere. 7A.4 The Fundamental Group introduces the fundamental group, the invariant in the Poincaré conjecture.

Exercises

Exercise 3.4 Twelve pentagons

For a polyhedral sphere with every vertex of degree 33 and every face a pentagon or hexagon (pp and hh of them), show from 3V=2E3V = 2E, 5p+6h=2E5p + 6h = 2E and V−E+F=2V - E + F = 2 that p=12p = 12. What happens if heptagons are allowed too?

Solution

V=23EV = \frac23E, F=p+hF = p + h, and E=5p+6h2E = \frac{5p + 6h}{2}. Then V−E+F=−13E+p+h=−5p+6h6+p+h=p6=2V - E + F = -\frac13E + p + h = -\frac{5p + 6h}{6} + p + h = \frac p6 = 2, so p=12p = 12. With ss heptagons, the same computation gives p6−s6=2\frac p6 - \frac s6 = 2, so p=12+sp = 12 + s: each heptagon needs an extra pentagon (as in some curved carbon structures).

Exercise 3.5 Euler characteristic of the genus-gg surface

Using the 4g4g-gon with word a1b1a1−1b1−1⋯agbgag−1bg−1a_1b_1a_1^{-1}b_1^{-1}\cdots a_gb_ga_g^{-1}b_g^{-1}, show that after gluing there is 11 vertex, 2g2g edges and 11 face, so χ(Σg)=2−2g\chi(\Sigma_g) = 2 - 2g. Check with the connected sum formula.

Exercise 3.6 The Klein bottle

Compute χ\chi of the Klein bottle from the word abab−1abab^{-1}, and check that it equals χ(RP2#RP2)\chi(\mathbb{R}P^2\#\mathbb{R}P^2). Is the Klein bottle homeomorphic to the torus? Why do they have the same Euler characteristic?

Solution

All four corners are identified (11 vertex), 22 edges, 11 face: χ=0=1+1−2\chi = 0 = 1 + 1 - 2. Not homeomorphic: the Klein bottle is non-orientable. Euler characteristic alone does not distinguish orientable from non-orientable surfaces; the theorem needs both invariants.

Exercise 3.7 Spaces that are not manifolds

Explain why each of the following is not a manifold: (a) the union of the two coordinate axes in R2\mathbb{R}^2; (b) a cone {z2=x2+y2}\{z^2 = x^2 + y^2\} (both nappes); (c) the one-point compactification of the open annulus (7A.2 Compactness and Compactification). (Remove the bad point: how many components does a small punctured neighbourhood have, compared with a punctured disc?)

Exercise 3.8 Many decompositions, one number

Compute V−E+FV - E + F for the cube, the octahedron, the icosahedron, and a "picture frame" polyhedron (a square ring with a square hole, made of four rectangular blocks). Which surface is the picture frame's surface?

Solution

Cube 8−12+6=28 - 12 + 6 = 2; octahedron 6−12+8=26 - 12 + 8 = 2; icosahedron 12−30+20=212 - 30 + 20 = 2. The picture frame's surface is a torus, and any decomposition of it gives 00 (for example, with quadrilateral faces: 16−32+16=016 - 32 + 16 = 0).

Exercise 3.9 Rehearsal: curvature at the corners

At a vertex of a convex polyhedron, the angle defect is 2π2\pi minus the sum of the face angles there: a discrete curvature concentrated at the vertex. (a) For the truncated icosahedron, each vertex has one pentagon angle (108°108°) and two hexagon angles (120°120°). Show that the total defect over all 6060 vertices is 720°=4π720° = 4\pi. (b) Prove Descartes' theorem in general: the total defect of a polyhedral sphere is 2πχ=4π2\pi\chi = 4\pi. (Sum the face angles: a face with kk sides contributes (k−2)π(k - 2)\pi, and ∑facesk=2E\sum_{\text{faces}}k = 2E.) This is the discrete Gauss–Bonnet theorem (8A.9 The Curvature of Surfaces); the angle defect is exactly the discrete curvature that discrete surface Ricci flow evolves (5A.5 Uniformization and the Two-Dimensional Ricci Flow).

Solution

(a) Defect per vertex 360°−108°−240°=12°360° - 108° - 240° = 12°, total 720°720°. (b) The total angle is ∑faces(kf−2)π=(2E−2F)π\sum_{\text{faces}}(k_f - 2)\pi = (2E - 2F)\pi. The total defect is 2πV−(2E−2F)π=2π(V−E+F)=2πχ2\pi V - (2E - 2F)\pi = 2\pi(V - E + F) = 2\pi\chi.

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