Homology in Brief

The fundamental group is a powerful invariant but a hard one to handle: groups given by presentations can be impossible to compare (7A.5 Computing π₁). Homology replaces it with something much easier: abelian groups, computed by linear algebra on a triangulation, one in each dimension. $H_0$ counts components, $H_1$ counts independent loops that don't bound, $H_2$ counts closed surfaces that don't bound, and so on. For many purposes this is enough. In dimension one, $H_1$ is exactly $\pi_1$ made abelian.

Poincaré invented both invariants, and in 1900 he conjectured that homology alone recognises the 3-sphere. In 1904 he found a counterexample: the Poincaré homology sphere, which has the homology of $S^3$ but a fundamental group of order $120$. That is why the Poincaré conjecture is about $\pi_1$. This chapter explains homology just far enough to make that story precise, and to supply two facts the proof uses later: Poincaré duality for three-manifolds, and the Hurewicz theorem.

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

• compute the simplicial homology of simple complexes, and know the homology of spheres, surfaces and projective spaces;
• relate the Betti numbers to the Euler characteristic;
• state the Hurewicz theorem in degree one, $H_1 = \pi_1^{\mathrm{ab}}$, and use it;
• state Poincaré duality for closed orientable three-manifolds and deduce what it says about their homology;
• explain why the Poincaré homology sphere shows homology cannot detect $S^3$;
• describe persistent homology and how it detects holes in data.

8.1Holes in a sensor network

8.2Chains, boundaries and homology

A simplicial complex is a space built from vertices, edges, triangles, tetrahedra and their higher analogues ($k$-simplices), glued along faces. Orient each simplex by an ordering of its vertices, up to even permutations. The $k$-chains $C_k$ are formal integer combinations of oriented $k$-simplices (with $-\sigma$ the simplex with the opposite orientation). The boundary of a simplex is the alternating sum of its faces,

where the hat means "omit": $\partial[a, b] = b - a$, and $\partial[a, b, c] = [b, c] - [a, c] + [a, b]$, the three edges of the triangle oriented around it. Extended linearly, $\partial : C_k \to C_{k-1}$ satisfies

(Exercise 8.4): the boundary of a boundary is zero. A chain with zero boundary is a cycle, and a chain that is the boundary of something is a boundary; every boundary is a cycle.

$H_k$ measures the $k$-dimensional cycles that are not boundaries: holes. Homology doesn't depend on the triangulation, and it is defined for all spaces (as singular homology, using continuous maps of simplices) and is a homotopy invariant: homotopy equivalent spaces have isomorphic homology groups (Lee, chapter 13).

Examples.

• A circle, triangulated as a triangle's boundary with vertices $a$, $b$, $c$. $C_1$ has basis $[a, b]$, $[b, c]$, $[c, a]$; the cycle $z = [a, b] + [b, c] + [c, a]$ has $\partial z = 0$, and there are no 2-simplices, so $H_1 = \mathbb{Z}$, generated by $z$. $H_0 = \mathbb{Z}$ (one component).
• The sphere $S^2$, as the boundary of a tetrahedron: $H_0 = \mathbb{Z}$, $H_1 = 0$, $H_2 = \mathbb{Z}$, generated by the sum of the four faces, suitably oriented (Exercise 8.5).
• The torus: $H_0 = \mathbb{Z}$, $H_1 = \mathbb{Z}^2$ (the two circles $a$, $b$), $H_2 = \mathbb{Z}$.
• $\mathbb{R}P^2$: $H_0 = \mathbb{Z}$, $H_1 = \mathbb{Z}/2$, $H_2 = 0$. With the cell structure of 7A.5 Computing π₁, the 2-cell's boundary is $2a$, so $a$ is a cycle with $2a$ a boundary: torsion.
• $S^n$: $H_0 = H_n = \mathbb{Z}$, all others $0$ (for $n \geq 1$).

For cell complexes there is a shortcut, cellular homology, with one generator for each cell and boundary maps computed from degrees of attaching maps (7A.7 Smooth Topology); it gives the same groups.

Euler characteristic. For a finite complex,

the second equality by rank–nullity (1A.2 Linear Maps and Matrices) applied to each $\partial$. Since the Betti numbers are topological invariants, so is $\chi$: the independence from the decomposition promised in 7A.3 Manifolds and Surfaces.

8.3Homology and the fundamental group

So $H_1$ is $\pi_1$ with the order of loops forgotten. Comparing with 7A.5 Computing π₁: $H_1(\Sigma_g) = \mathbb{Z}^{2g}$, $H_1(\mathbb{R}P^n) = \mathbb{Z}/2$, $H_1$ of a knot complement is $\mathbb{Z}$, and $H_1(L(p, q)) = \mathbb{Z}/p$. When $\pi_1$ is abelian, $H_1$ and $\pi_1$ agree; when it is not, $H_1$ loses information, and in the extreme case of a perfect group ($G = [G, G]$), $H_1 = 0$ although $\pi_1 \neq 1$.

Witold Hurewicz (1935) proved the higher version: if $X$ is $(n - 1)$-connected ($\pi_k(X) = 0$ for $k < n$) with $n \geq 2$, then $\pi_n(X) \cong H_n(X)$. This is how $\pi_3$ of a simply connected three-manifold is computed (Exercise 8.9).

8.4Poincaré duality

In general Poincaré duality says that for a closed orientable $n$-manifold, $H_k \cong H^{n-k}$, the cohomology in the complementary dimension (Hatcher, section 3.3); for $n = 3$ it gives the statements above. Geometrically, a surface in a three-manifold and a loop crossing it pair off: each non-trivial 2-cycle is detected by a 1-cycle meeting it a non-zero number of times. Every closed odd-dimensional manifold has $\chi = 0$.

8.5Homology spheres and the conjecture

A closed three-manifold $M$ is a homology sphere if $H_*(M) \cong H_*(S^3)$: by Poincaré duality, for orientable $M$ this just means $H_1(M) = 0$, which by Hurewicz means $\pi_1(M)$ is perfect.

In 1900 Poincaré asked, in effect, whether every closed three-manifold with the homology of $S^3$ is homeomorphic to $S^3$. In the fifth supplement to Analysis Situs (1904) he constructed a counterexample and computed its fundamental group, showing it is non-trivial: the Poincaré homology sphere $P = S^3/I^*$ (7A.6 Covering Spaces). Its fundamental group, the binary icosahedral group, has the presentation

of order $120$, and it is perfect: abelianising the relations gives $2s + 2t = 3s = 5t$, which force $s = t = 0$ (Exercise 8.7). So $H_1(P) = 0$ and $P$ is a homology sphere, but $\pi_1(P) \neq 1$, so $P$ is not $S^3$. Poincaré then posed the question correctly: is every closed three-manifold with trivial fundamental group homeomorphic to $S^3$? That is the Poincaré conjecture (7A.9 The Poincaré Conjecture, Precisely).

There are infinitely many homology spheres (Dehn constructed many by surgery in 1910, 10A.2 Building Three-Manifolds), and the Poincaré sphere is the only one, besides $S^3$, with finite fundamental group (a consequence of the classification of spherical space forms together with geometrization, 10A.5 Thurston’s Eight Geometries). Homology spheres remain central in three- and four-dimensional topology.

8.6Persistent homology

Given points $x_1, \dots, x_N$ and a scale $\varepsilon$, the Vietoris–Rips complex has a simplex for every set of points at pairwise distance at most $\varepsilon$. As $\varepsilon$ increases the complexes grow, and the inclusions induce maps on homology. A homology class born at one scale may die at a later one, when it becomes a boundary. Herbert Edelsbrunner, David Letscher and Afra Zomorodian (2002), and Zomorodian and Gunnar Carlsson (2005), showed how to compute all the births and deaths at once by reducing one boundary matrix, and that the result is a list of intervals, the barcode, which is stable under small perturbations of the data. Long bars are features; short bars are noise.

8.7History

Poincaré introduced Betti numbers (named after Enrico Betti) and torsion in Analysis Situs (1895) and its first supplements (1899–1900), proved his duality theorem there, and posed and then refuted the homology version of his conjecture in 1900 and 1904. The treatment of homology as groups rather than numbers is due to Emmy Noether and others in the mid-1920s. Witold Hurewicz's theorem dates from 1935. Persistent homology was introduced by Edelsbrunner, Letscher and Zomorodian in 2002, and de Silva and Ghrist's coverage criterion appeared in 2007.

8.8Exercises

Solution

$H_0 = \mathbb{Z}$, $H_1 = \mathbb{Z}/p$, $H_2 = 0$, $H_3 = \mathbb{Z}$. Lens spaces with the same $p$ but different $q$ have the same homology and the same $\pi_1$, but need not be homeomorphic (Reidemeister, 1935).

Solution

(a) $H_1 = \pi_1^{\mathrm{ab}} = 0$; Poincaré duality gives $b_2 = b_1 = 0$ and $H_2$ free, so $H_2 = 0$; $H_3 = \mathbb{Z}$. (b) Hurewicz with $n = 2$ needs $\pi_1 = 0$, which holds. (c) With $\pi_1 = \pi_2 = 0$, Hurewicz with $n = 3$ applies: the generator of $\pi_3(M)$ is a map $f : S^3 \to M$ with $f_*$ sending the generator of $H_3(S^3)$ to the generator of $H_3(M)$, that is, a map of degree $1$.