16. Homology¶
The fundamental group sees one-dimensional holes: a loop caught around a hole. But the sphere \(S^2\) has a hole too, the hollow inside, and every loop on the sphere shrinks to a point, so \(\pi_1\) cannot see it. Higher homotopy groups can, but they are notoriously hard to compute. Homology is the other answer: count holes of every dimension with linear algebra. In this chapter you will build spaces from triangles and their higher-dimensional cousins, define the boundary of a chain, prove that the boundary of a boundary is zero, and compute the homology of the circle, the sphere, the torus and the projective plane by hand and by computer. You will then prove that the Euler characteristic of Chapter 2 is the alternating sum of these hole counts, and see how homology and the fundamental group are related.
In this chapter
You will learn:
- why a second invariant is needed, and what "a hole of dimension \(n\)" should mean
- simplices and simplicial complexes, and how to triangulate the circle, the sphere, the torus and \(\mathbb{R}P^2\)
- chains, the boundary map, cycles and boundaries, and the proof that \(\partial \circ \partial = 0\)
- how to compute homology groups and Betti numbers, by hand and with a short program
- the proof that the Euler characteristic equals \(b_0 - b_1 + b_2 - \cdots\)
- the Hurewicz theorem: \(H_1\) is the abelianised fundamental group, and where homology and homotopy part ways
You need: Chapter 13 and Chapter 15 (fundamental groups, abelianisation), Chapter 2 (Euler characteristic) and Chapter 10 (surfaces from polygons). From linear algebra: matrices, rank, kernel and image, and the rank-nullity theorem, as in Linear Algebra Chapter 7. Quotient groups from Algebra Chapter 15.
Levels: 16.1 is intermediate; 16.2 to 16.4 are advanced; 16.5 is advanced; 16.6 is expert (statements with proof sketches).
16.1 Why another invariant¶
Level: intermediate.
Line up four spaces: a point, the circle \(S^1\), the sphere \(S^2\) and the torus \(T^2\). Their fundamental groups are \(\{1\}\), \(\mathbb{Z}\), \(\{1\}\) and \(\mathbb{Z}^2\). So \(\pi_1\) cannot tell the point from the sphere. Yet they are clearly different: the sphere encloses a cavity, and the point does not. (They are not even homotopy equivalent, but we cannot yet prove it.)
The fix suggested by the fundamental group is to use spheres instead of loops: maps \(S^2 \to X\) up to homotopy, giving the second homotopy group \(\pi_2(X)\), and in general \(\pi_n(X)\) from maps \(S^n \to X\). These exist, and \(\pi_2(S^2) \cong \mathbb{Z}\) does detect the cavity. But higher homotopy groups are extraordinarily hard to compute. Even for the sphere \(S^2\), the groups \(\pi_n(S^2)\) are not all known: Heinz Hopf showed in 1931 that \(\pi_3(S^2) \cong \mathbb{Z}\), which means there are maps from the three-sphere onto the two-sphere that cannot be deformed to a constant, even though the three-sphere is "bigger" (Chapter 20).
Homology takes a different route. Instead of maps of spheres, it uses pieces of the space itself, added and subtracted like vectors:
- a cycle is a combination of pieces with no boundary, like a closed curve or a closed surface;
- a boundary is a cycle that is the edge of something one dimension higher, like the equator, which bounds a hemisphere;
- a hole of dimension \(n\) is an \(n\)-cycle that is not a boundary.
On the torus, the circle around the tube is a 1-cycle that bounds nothing: a 1-dimensional hole. The whole torus surface is a 2-cycle (it has no edge) and bounds nothing in the torus: a 2-dimensional hole, the cavity. On the sphere, every closed curve bounds a piece of the sphere, so there are no 1-dimensional holes, but the sphere itself is a 2-cycle that bounds nothing.
Turning this into mathematics requires a way to cut a space into pieces (simplices) and an algebra of boundaries. The reward is that everything becomes linear algebra: matrices, ranks and kernels. Homology groups are abelian and computable, by a finite calculation, for any space built from finitely many simplices. The price is that homology forgets some information that homotopy keeps (16.6).
16.2 Simplicial complexes¶
Level: advanced.
Simplices¶
The building blocks are the simplest convex shapes in each dimension:
- a 0-simplex is a point;
- a 1-simplex is a segment;
- a 2-simplex is a triangle (filled in);
- a 3-simplex is a tetrahedron (solid);
- in general an \(n\)-simplex is the convex hull of \(n + 1\) points \(v_0, \dots, v_n\) in some \(\mathbb{R}^N\) that are affinely independent, meaning the vectors \(v_1 - v_0, \dots, v_n - v_0\) are linearly independent.
The \(v_i\) are the vertices. A face of a simplex is the simplex spanned by some nonempty subset of its vertices. A triangle has seven faces: itself, three edges and three vertices. An \(n\)-simplex has \(\binom{n+1}{k+1}\) faces of dimension \(k\), and \(2^{n+1} - 1\) faces in all.
Complexes¶
Definition. A (finite) simplicial complex \(K\) is a finite collection of simplices in some \(\mathbb{R}^N\) such that
- every face of a simplex of \(K\) is also in \(K\), and
- any two simplices of \(K\) meet in a common face, or not at all.
The union of its simplices, with the subspace topology, is a space \(|K|\), called the polyhedron of \(K\). A triangulation of a space \(X\) is a simplicial complex \(K\) with a homeomorphism \(|K| \to X\).
Condition 2 rules out triangles that overlap in their middles, or that meet edge to vertex in a mismatched way. In practice one rarely draws the complex in \(\mathbb{R}^N\). All the information is combinatorial: which sets of vertices span simplices. An abstract simplicial complex is a finite set of vertices and a collection of nonempty subsets (the simplices) that contains every one-element subset and is closed under taking nonempty subsets. Every abstract complex can be realised geometrically (put the vertices at the standard basis vectors of a big \(\mathbb{R}^N\)), so the two notions carry the same information.
Examples.
- The circle. Three vertices \(a, b, c\) and the three edges \(ab\), \(ac\), \(bc\), without the triangle: the boundary of a triangle, homeomorphic to \(S^1\). Two vertices are not enough, because two edges \(ab\) would be the same simplex.
- The disc. Add the triangle \(abc\).
- The sphere. The four triangles of a tetrahedron's surface, with their six edges and four vertices: the boundary of a 3-simplex, homeomorphic to \(S^2\) by radial projection from the centre.
- The torus. Cut the square of Chapter 6 into a \(3 \times 3\) grid and each small square into two triangles, then glue opposite sides. There are 9 vertices (the grid points, with the edges glued), 27 edges and 18 triangles. The labels in the picture show which grid points become the same vertex.
Why not use the square with a single diagonal, cut into two triangles? After gluing it has one vertex, three edges and two triangles, and both triangles have the same three vertices (all equal to the one vertex), which condition 2 forbids. That coarse picture is a perfectly good \(\Delta\)-complex (a looser structure allowed in Hatcher's book), but for simplicial complexes we need finer cuts. A \(2 \times 2\) grid is still too coarse; the \(3 \times 3\) grid works. The smallest triangulation of the torus has 7 vertices, 21 edges and 14 triangles.
The projective plane. Take the icosahedron, which has 12 vertices, 30 edges and 20 triangles, and glue each point to its opposite. Opposite faces of the icosahedron are disjoint, so the result is a simplicial complex with 6 vertices, 15 edges and 10 triangles, homeomorphic to \(\mathbb{R}P^2\) (the sphere with antipodes glued). With vertices \(0, \dots, 5\) its triangles are
Check: every pair of the six vertices is an edge (\(\binom{6}{2} = 15\)), and every edge lies in exactly two triangles, as it must in a surface.
Not every space can be triangulated, and a space can have many triangulations. Homology will be defined from a triangulation, so we will need to know that the answer does not depend on which triangulation is used. That is a theorem (16.5), and it is the reason homology is a topological invariant.
Orientation¶
To add and subtract simplices like vectors we need signs, as with directed edges in a circuit. An orientation of a simplex is an ordering of its vertices, with two orderings counted as the same orientation if they differ by an even permutation. We write \([v_0, v_1, \dots, v_n]\) for the oriented simplex, and agree that swapping two vertices changes the sign:
For an edge, \([a, b]\) is the edge directed from \(a\) to \(b\), and \([b, a] = -[a, b]\) is the same edge directed the other way. For a triangle, an orientation is a direction of travel around its edge, anticlockwise or clockwise. In computations we order the vertices by number and use \([v_0, \dots, v_n]\) with \(v_0 < \dots < v_n\) as the standard orientation.
16.3 Chains, boundaries and cycles¶
Level: advanced.
Chains¶
Definition. The group of \(n\)-chains \(C_n(K)\) is the free abelian group with one generator for each \(n\)-simplex of \(K\) (with its standard orientation). Its elements are formal sums
where the \(\sigma_i\) are the \(n\)-simplices. If \(K\) has \(c_n\) simplices of dimension \(n\), then \(C_n(K) \cong \mathbb{Z}^{c_n}\). For \(n\) below \(0\) or above the dimension of \(K\), \(C_n(K) = 0\).
Think of an \(n\)-chain as a bookkeeping of \(n\)-dimensional pieces with multiplicities: a 1-chain \(2[a, b] - [b, c]\) means "edge \(ab\) twice forwards, edge \(bc\) once backwards". The opposite orientation of a simplex is identified with the negative of the generator.
The boundary map¶
Definition. The boundary of an oriented \(n\)-simplex is the alternating sum of its \((n-1)\)-dimensional faces:
where the hat means "leave this vertex out". Extending linearly gives a homomorphism \(\partial_n : C_n(K) \to C_{n-1}(K)\). (For \(n = 0\) the boundary is \(0\).)
Worked examples.
- \(\partial [a, b] = b - a\): the end minus the start, as for a path.
- \(\partial [a, b, c] = [b, c] - [a, c] + [a, b] = [a, b] + [b, c] + [c, a]\): the three edges, travelled once around the triangle in the direction \(a \to b \to c \to a\). The signs make the edges join head to tail.
- \(\partial [0, 1, 2, 3] = [1, 2, 3] - [0, 2, 3] + [0, 1, 3] - [0, 1, 2]\): the four faces of the tetrahedron, each oriented so that, seen from outside, they all turn the same way.
Before the definition makes sense one must check it respects the sign convention: swapping two vertices of \(\sigma\) changes the sign of \(\partial\sigma\). For adjacent vertices \(v_i\) and \(v_{i+1}\) this is a direct check on the formula (the two terms that omit \(v_i\) or \(v_{i+1}\) trade places with a sign change, and every other term has the two vertices swapped); every permutation is a product of adjacent swaps.
The boundary of a boundary is zero¶
Take the boundary of a triangle: the three edges, going around. Its boundary is \(\partial([b, c] - [a, c] + [a, b]) = (c - b) - (c - a) + (b - a) = 0\). A closed loop of edges has no ends. This is the central fact of the subject.
Theorem 16.1. \(\partial_{n-1} \circ \partial_n = 0\) for every \(n\).
Proof. It is enough to check it on one oriented simplex \(\sigma = [v_0, \dots, v_n]\), since both maps are homomorphisms. Apply \(\partial\) twice. The first time, leave out \(v_i\), with sign \((-1)^i\). The second time, leave out another vertex \(v_j\). Its position in the shortened list is \(j\) if \(j < i\), and \(j - 1\) if \(j > i\), because removing \(v_i\) moves the later vertices one place to the left. So
Each \((n-2)\)-face, with two vertices \(v_p\) and \(v_q\) removed (\(p < q\)), appears exactly twice: once in the first sum, with \(i = q\), \(j = p\) and sign \((-1)^{q+p}\), and once in the second sum, with \(i = p\), \(j = q\) and sign \((-1)^{p+q-1}\). These signs are opposite, so the two terms cancel, and the whole sum is \(0\). \(\square\)
Check on the tetrahedron:
and each of the six edges appears once with each sign. The total is \(0\).
Cycles, boundaries and homology¶
Two subgroups of \(C_n(K)\) matter:
- the \(n\)-cycles \(Z_n = \ker \partial_n\): chains with zero boundary, the closed things;
- the \(n\)-boundaries \(B_n = \operatorname{im} \partial_{n+1}\): chains that are the boundary of an \((n+1)\)-chain.
By Theorem 16.1 every boundary is a cycle: if \(c = \partial d\), then \(\partial c = \partial\partial d = 0\). So \(B_n \subseteq Z_n\), and we can form the quotient.
Definition. The \(n\)-th homology group of \(K\) is
Its elements are cycles, where two cycles count as the same (are homologous) when their difference is a boundary. The \(n\)-th Betti number \(b_n\) is the rank of \(H_n(K)\): the number of \(\mathbb{Z}\) summands, ignoring any finite part.
So \(H_n\) is "closed \(n\)-dimensional things, modulo those that bound", the precise version of the holes of 16.1.
Computing with ranks. A finitely generated abelian group like \(H_n(K)\) is \(\mathbb{Z}^{b_n}\) plus a finite part called the torsion. The Betti numbers can be computed with ordinary linear algebra over the rational numbers: treat each \(\partial_n\) as a matrix with \(c_{n-1}\) rows and \(c_n\) columns, and let \(r_n\) be its rank. By rank-nullity, \(\dim \ker \partial_n = c_n - r_n\), and \(\dim \operatorname{im} \partial_{n+1} = r_{n+1}\). So
(That the rank of the integer group \(H_n\) equals the dimension of the rational one is a short argument: tensoring with \(\mathbb{Q}\) kills torsion and keeps ranks. The torsion part needs integer methods: the Smith normal form of the matrix of \(\partial_{n+1}\), a diagonal form reached by integer row and column operations, has diagonal entries \(d_1, d_2, \dots\), and the torsion of \(H_n\) is the product of the groups \(\mathbb{Z}/d_i\) for those \(d_i > 1\).)
\(H_0\) counts pieces¶
Theorem 16.2. \(H_0(K) \cong \mathbb{Z}^k\), where \(k\) is the number of path components of \(|K|\).
Proof. Every 0-chain is a cycle, since \(\partial_0 = 0\). A boundary of an edge is \(b - a\), so two vertices are homologous when they are joined by an edge, and hence (adding up along a chain of edges) when they are joined by an edge path, that is, when they lie in the same component. Conversely, the homomorphism sending a 0-chain to the list of its coefficient sums on each component, in \(\mathbb{Z}^k\), sends every \(b - a\) to \(0\) (both ends lie in one component), so it kills \(B_0\); and its kernel is exactly \(B_0\), since a chain with coefficient sum \(0\) on a component is a sum of differences of vertices in that component, each a boundary. It is onto. By the first isomorphism theorem, \(H_0 = C_0/B_0 \cong \mathbb{Z}^k\). \(\square\)
16.4 Computing homology: circle, sphere, torus¶
Level: advanced.
The circle¶
Take the hollow triangle with vertices \(a, b, c\) and edges \(e_1 = [a, b]\), \(e_2 = [a, c]\), \(e_3 = [b, c]\). There are no 2-simplices. The boundary map \(\partial_1\) has matrix (columns \(e_1, e_2, e_3\), rows \(a, b, c\))
Its first column is \(\partial e_1 = b - a\), and so on.
\(H_1\). A 1-chain \(x e_1 + y e_2 + z e_3\) is a cycle when \(\partial_1\) sends it to \(0\): from the rows, \(-x - y = 0\), \(x - z = 0\) and \(y + z = 0\). So \(z = x\) and \(y = -x\), and
That generator is the loop once around. There are no 2-simplices, so \(B_1 = 0\) and \(H_1 = \mathbb{Z}\).
\(H_0\). The space is connected, so \(H_0 \cong \mathbb{Z}\) by Theorem 16.2. Check with ranks: \(\partial_1\) has rank \(2\) (the rows add to zero, and the first two columns are independent), so \(b_0 = c_0 - r_1 = 3 - 2 = 1\).
So \(H_0(S^1) \cong \mathbb{Z}\), \(H_1(S^1) \cong \mathbb{Z}\), and \(H_n(S^1) = 0\) for \(n \ge 2\).
Fill it in. Now add the triangle \([a, b, c]\), making a disc. Then \(\partial_2[a, b, c] = [b, c] - [a, c] + [a, b] = e_1 - e_2 + e_3\), which is exactly the generator of \(Z_1\). Now \(B_1 = Z_1\) and \(H_1 = 0\): the loop bounds, the hole is filled. Also \(H_2 = \ker \partial_2 = 0\), since \(\partial_2\) of a nonzero multiple of the triangle is not zero. The disc has the homology of a point.
The sphere¶
Take the boundary of the tetrahedron, with vertices \(0, 1, 2, 3\), six edges and four triangles.
\(H_2\). There are no 3-simplices, so \(H_2 = Z_2 = \ker \partial_2\). Which combinations of the four triangles have zero boundary? Each edge lies in exactly two triangles, so the coefficient of the edge in \(\partial\) of the chain involves only those two coefficients; for the boundary to vanish, the two triangles at each edge must have coefficients that cancel there. Because you can walk from any triangle to any other across edges, one coefficient determines all the others, and the cycle condition is solved by
and its multiples. (This \(\Sigma\) is \(\partial[0, 1, 2, 3]\), so \(\partial\Sigma = 0\) by Theorem 16.1; but \([0,1,2,3]\) is not in the complex, so \(\Sigma\) is not a boundary here.) So \(H_2 \cong \mathbb{Z}\), generated by the whole sphere with coherent orientation. This is the cavity, finally detected.
\(H_1\). By ranks: \(c_1 = 6\), and \(\partial_1\) has rank \(3\) (the graph of edges is connected with 4 vertices, so \(r_1 = 4 - 1\)). \(\partial_2\) has rank \(4 - 1 = 3\) (its kernel has dimension 1). So \(b_1 = (6 - 3) - 3 = 0\). Every closed edge path on the sphere is a boundary; for example the triangle \([0,1] + [1,2] + [2,0]\) is \(\partial[0,1,2]\), and the square \([0,1] + [1,2] + [2,3] + [3,0]\) is \(\partial([0,1,2] + [0,2,3])\). The Smith normal form has no entries bigger than \(1\), so there is no torsion: \(H_1 = 0\).
So \(H_0(S^2) \cong \mathbb{Z}\), \(H_1(S^2) = 0\), \(H_2(S^2) \cong \mathbb{Z}\). Compare \(\pi_1(S^2) = 1\): homology sees the cavity that the fundamental group missed. In general, triangulating \(S^n\) as the boundary of an \((n+1)\)-simplex, one finds \(H_0(S^n) \cong H_n(S^n) \cong \mathbb{Z}\) and all other groups zero (for \(n \ge 1\)). This fact powers Chapter 17.
The torus¶
The \(3 \times 3\) grid triangulation has \(c_0 = 9\), \(c_1 = 27\), \(c_2 = 18\). By hand, the ranks are found as follows.
- \(r_1 = 9 - 1 = 8\), since the edge graph is connected (for a connected graph with \(V\) vertices the boundary map has rank \(V - 1\): its image is the chains with coefficient sum \(0\), as in Theorem 16.2).
- \(\ker \partial_2\): orient every triangle anticlockwise in the square picture. Each interior edge of the square is then traversed in opposite directions by its two triangles, and the gluing of the sides of the square matches the edges on opposite sides with the same direction, so they also cancel. So the sum \(T\) of all 18 triangles is a 2-cycle. As for the sphere, any 2-cycle is a multiple of \(T\), because neighbouring triangles must have equal coefficients. So \(\dim \ker \partial_2 = 1\) and \(r_2 = 18 - 1 = 17\).
Then
The two 1-dimensional holes are the loops around the two directions of the torus: in the picture, the bottom row of edges \([0, 1] + [1, 2] + [2, 0]\) and the left column \([0, 3] + [3, 6] + [6, 0]\). Neither bounds, and no combination \(m \cdot (\text{row}) + n \cdot (\text{column})\) with \((m, n) \ne (0, 0)\) bounds. A computer check of the Smith normal form shows there is no torsion, so
The projective plane: torsion¶
For the 6-vertex \(\mathbb{R}P^2\), the same computation gives Betti numbers \(b_0 = 1\), \(b_1 = 0\), \(b_2 = 0\). So rationally it looks like a point. But the Smith normal form of \(\partial_2\) has a diagonal entry \(2\), and
There is a loop (the image of a half great circle) that does not bound, but twice it does: that is torsion. And \(H_2 = 0\) because there is no way to orient all the triangles coherently, since the surface is non-orientable: whatever orientations you choose, adding up all the triangles leaves at least one edge with coefficient \(\pm 2\) instead of \(0\). In general a closed connected surface has \(H_2 \cong \mathbb{Z}\) if it is orientable and \(H_2 = 0\) if not. Homology detects orientability.
Try it in Python¶
This program (it uses sympy) builds the boundary matrices from a list of triangles, checks \(\partial\partial = 0\), and reports the numbers of simplices, the Betti numbers and the torsion of \(H_1\).
from itertools import combinations
from sympy import Matrix, ZZ
from sympy.matrices.normalforms import smith_normal_form
def boundary(higher, lower):
"""Matrix of the boundary map from chains on `higher` to chains on `lower`."""
index = {s: i for i, s in enumerate(lower)}
M = [[0] * len(higher) for _ in lower]
for j, s in enumerate(higher):
for k in range(len(s)):
face = s[:k] + s[k + 1:] # leave out vertex number k
M[index[face]][j] += (-1) ** k
return Matrix(M)
def homology_of_surface(triangles):
tris = sorted(tuple(sorted(t)) for t in triangles)
edges = sorted({e for t in tris for e in combinations(t, 2)})
verts = sorted({(v,) for t in tris for v in t})
d1, d2 = boundary(edges, verts), boundary(tris, edges)
assert (d1 * d2).is_zero_matrix # the boundary of a boundary is zero
r1, r2 = d1.rank(), d2.rank()
betti = (len(verts) - r1, len(edges) - r1 - r2, len(tris) - r2)
snf = smith_normal_form(d2, domain=ZZ)
torsion = [abs(snf[i, i]) for i in range(min(snf.shape)) if abs(snf[i, i]) > 1]
return (len(verts), len(edges), len(tris)), betti, torsion
V = lambda i, j: 3 * (i % 3) + (j % 3) # the 3 x 3 grid torus
torus = [t for i in range(3) for j in range(3)
for t in ((V(i, j), V(i + 1, j), V(i + 1, j + 1)), (V(i, j), V(i, j + 1), V(i + 1, j + 1)))]
sphere = [(0, 1, 2), (0, 1, 3), (0, 2, 3), (1, 2, 3)]
rp2 = [(0, 1, 2), (0, 2, 3), (0, 3, 4), (0, 4, 5), (0, 5, 1),
(1, 2, 4), (2, 3, 5), (3, 4, 1), (4, 5, 2), (5, 1, 3)]
for name, K in (("sphere", sphere), ("torus", torus), ("RP2", rp2)):
print(name, *homology_of_surface(K))
# sphere (4, 6, 4) (1, 0, 1) []
# torus (9, 27, 18) (1, 2, 1) []
# RP2 (6, 15, 10) (1, 0, 0) [2]
16.5 Euler characteristic and Betti numbers¶
Level: advanced.
In Chapter 2 the Euler characteristic of a surface was \(V - E + F\), and it gave \(2\) for every polyhedron shaped like a sphere and \(0\) for every torus. We could check it on many examples but could not explain why it never depends on how the surface is cut up. Homology explains it.
Definition. The Euler characteristic of a finite simplicial complex is
where \(c_n\) is the number of \(n\)-simplices. For a surface this is \(V - E + F\).
Theorem 16.3 (Euler-Poincaré). \(\chi(K) = \sum_{n \ge 0} (-1)^n b_n = b_0 - b_1 + b_2 - \cdots\).
Proof. Work over \(\mathbb{Q}\), so that chain groups are vector spaces of dimension \(c_n\) and \(b_n = \dim Z_n - \dim B_n\). Rank-nullity for \(\partial_n : C_n \to C_{n-1}\) says
Take the alternating sum over all \(n\) from \(0\) to the top dimension \(d\), with \(B_{-1} = 0\) and \(B_d = 0\):
where the second sum was reindexed with \(m = n - 1\), which changes \((-1)^n\) into \(-(-1)^m\). Combining, \(\sum_n (-1)^n (\dim Z_n - \dim B_n) = \sum_n (-1)^n b_n\). \(\square\)
Checks.
| Space | \(c_0, c_1, c_2\) | \(c_0 - c_1 + c_2\) | \(b_0, b_1, b_2\) | \(b_0 - b_1 + b_2\) |
|---|---|---|---|---|
| Sphere (tetrahedron) | 4, 6, 4 | 2 | 1, 0, 1 | 2 |
| Torus (\(3 \times 3\) grid) | 9, 27, 18 | 0 | 1, 2, 1 | 0 |
| Torus (7 vertices) | 7, 21, 14 | 0 | 1, 2, 1 | 0 |
| \(\mathbb{R}P^2\) (6 vertices) | 6, 15, 10 | 1 | 1, 0, 0 | 1 |
| Circle (triangle) | 3, 3, 0 | 0 | 1, 1, 0 | 0 |
The left side depends on the triangulation; the right side depends only on the homology groups. So the Euler characteristic is a topological invariant, provided homology is. That is the following theorem.
Theorem 16.4 (topological invariance, statement). If \(|K|\) and \(|L|\) are homeomorphic, or even homotopy equivalent, then \(H_n(K) \cong H_n(L)\) for all \(n\).
The proof goes through singular homology, which uses all continuous maps of simplices into a space instead of a fixed triangulation. Singular homology is defined for every space, is visibly a topological invariant, is unchanged by homotopy equivalence, and agrees with simplicial homology on simplicial complexes. Each of these steps takes real work; see Hatcher, Chapter 2 (Section 2.1 for the comparison theorem). We use the result.
Consequences.
- Euler's formula. Any triangulation of the sphere has \(V - E + F = 2\), since \(b_0 - b_1 + b_2 = 1 - 0 + 1\). This holds even for wildly irregular triangulations, which the edge-removal proof of Chapter 2 had to handle case by case.
- Surfaces. The orientable surface \(M_g\) of genus \(g\) has \(b_0 = 1\), \(b_2 = 1\) (orientable) and \(\chi = 2 - 2g\), so \(b_1 = 2 - \chi = 2g\): the \(2g\) independent loops, two for each handle. The non-orientable \(N_k\) has \(b_2 = 0\) and \(\chi = 2 - k\), so \(b_1 = 1 - \chi = k - 1\), plus torsion \(\mathbb{Z}/2\) in \(H_1\).
- Graphs. For a connected graph, \(b_0 - b_1 = V - E\), so \(b_1 = E - V + 1\), the rank of the free fundamental group found in Chapter 15.
- Dimension. For a point, \(\chi = 1\). A space with \(\chi \ne 1\) is not contractible. So \(S^2\) (\(\chi = 2\)) is not homotopy equivalent to a point: the question left open in 16.1.
16.6 Homology and homotopy¶
Level: expert. Statements, with proof sketches and references.
We now have two families of invariants: the homotopy groups \(\pi_n\), defined by maps of spheres, and the homology groups \(H_n\), defined by chains. How are they related?
The first homology group¶
There is a natural map from loops to 1-cycles. In singular homology a loop \(f : I \to X\) is itself a singular 1-simplex, and since \(f(0) = f(1)\) its boundary \(f(1) - f(0)\) is \(0\): it is a cycle. In a simplicial complex, an edge loop \(v_0 \to v_1 \to \cdots \to v_k = v_0\) gives the cycle \([v_0, v_1] + [v_1, v_2] + \cdots + [v_{k-1}, v_k]\) (each edge written with the right sign).
Theorem 16.5 (Hurewicz, dimension 1). For a path-connected space \(X\), the map \(h : \pi_1(X, x_0) \to H_1(X)\) sending a loop to its cycle is a surjective homomorphism whose kernel is the commutator subgroup. So
Sketch of proof. Homotopic loops give homologous cycles: a path homotopy is a map of a square, which can be cut into two triangles, and its boundary relation says the two loops differ by a boundary. Joining loops adds cycles, since a map of a triangle with edges \(f\), \(g\) and \(f \cdot g\) shows \(f + g - (f \cdot g)\) is a boundary. So \(h\) is a homomorphism to an abelian group, and it kills all commutators. Surjectivity: in a 1-cycle, the edges can be joined up into closed loops, each made into a loop at \(x_0\) by a path to the base point and back. The kernel is no bigger than the commutator subgroup: this is the delicate part, and Hatcher (Theorem 2A.1) gives a careful two-page proof. \(\square\)
Checks against Chapter 15.
| Space | \(\pi_1\) | \(H_1\) |
|---|---|---|
| Circle | \(\mathbb{Z}\) | \(\mathbb{Z}\) |
| Figure eight | \(F_2\) (non-abelian) | \(\mathbb{Z}^2\) |
| Torus | \(\mathbb{Z}^2\) | \(\mathbb{Z}^2\) |
| Genus \(g\) surface \(M_g\) | one relator, \(2g\) generators | \(\mathbb{Z}^{2g}\) |
| \(\mathbb{R}P^2\) | \(\mathbb{Z}/2\) | \(\mathbb{Z}/2\) |
| Klein bottle | \(\langle a, b \mid abab^{-1} \rangle\) | \(\mathbb{Z} \oplus \mathbb{Z}/2\) |
Each \(H_1\) is the abelianisation computed in Chapter 15, and the Betti numbers \(b_1\) agree with 16.5.
What homology forgets¶
The figure eight and the torus have the same \(H_1\) but different fundamental groups, so \(H_1\) forgets the non-abelian structure (though here \(H_2\) still tells them apart). A sharper example: in 1904 Poincaré constructed a closed three-dimensional manifold, now called the Poincaré homology sphere, with exactly the homology of the three-sphere \(S^3\), but whose fundamental group has 120 elements (the binary icosahedral group). That group equals its own commutator subgroup, so its abelianisation is trivial, consistent with Hurewicz. Homology cannot tell this space from \(S^3\); the fundamental group can. This example is why the Poincaré conjecture (Chapter 19) is phrased with the fundamental group and not with homology.
Higher dimensions¶
Theorem 16.6 (Hurewicz, statement). Let \(n \ge 2\) and let \(X\) be a path-connected space with \(\pi_k(X) = 0\) for all \(1 \le k \le n - 1\). Then \(H_k(X) = 0\) for \(1 \le k \le n - 1\), and \(\pi_n(X) \cong H_n(X)\).
Witold Hurewicz introduced the higher homotopy groups and proved this in 1935. Applied to \(S^n\) (with \(n \ge 2\), which is simply connected and, by a cellular approximation argument, has \(\pi_k(S^n) = 0\) for \(k < n\)), it gives \(\pi_n(S^n) \cong H_n(S^n) \cong \mathbb{Z}\): the degree of a map of spheres, which Chapter 17 uses to prove Brouwer's theorem in every dimension.
Beyond the first nonzero group, homology and homotopy separate. For the two-sphere, \(H_n(S^2) = 0\) for all \(n \ge 3\), but \(\pi_3(S^2) \cong \mathbb{Z}\), generated by the Hopf map, and Jean-Pierre Serre proved in the early 1950s that infinitely many of the groups \(\pi_n(S^2)\) are nonzero. Homology is the computable shadow; homotopy is the full, and mostly unknown, object.
What homology does well¶
Homology has three properties that make it the standard tool, each stated here without proof (all in Hatcher, Chapter 2):
- Functoriality. A continuous map \(f : X \to Y\) induces homomorphisms \(f_* : H_n(X) \to H_n(Y)\), with \((g \circ f)_* = g_* \circ f_*\), exactly as for \(\pi_1\) (Theorem 13.13).
- Homotopy invariance. Homotopic maps induce the same homomorphisms, so homotopy equivalent spaces have isomorphic homology.
- Mayer-Vietoris. For \(X = U \cup V\) (open sets) there is a long exact sequence \(\cdots \to H_n(U \cap V) \to H_n(U) \oplus H_n(V) \to H_n(X) \to H_{n-1}(U \cap V) \to \cdots\), the homology analogue of van Kampen's theorem, which computes the homology of a glued space from its pieces.
A short history¶
Poincaré's 1895 Analysis Situs introduced Betti numbers (named after Enrico Betti, who had studied similar counts of cuts in 1871) and torsion coefficients, defined by counting, and proved a first version of the Euler-Poincaré formula. The idea that these numbers are ranks of groups (cycles modulo boundaries) took hold in the mid-1920s; Pavel Alexandroff and Heinz Hopf credited Emmy Noether with urging it in her lectures in Göttingen, and Leopold Vietoris and Walther Mayer developed the algebra. The invariance of homology under homeomorphism, the hard part of Theorem 16.4, was settled by James Alexander and others in the 1910s and 1920s for complexes, and the singular homology that makes it routine was put on a firm footing by Samuel Eilenberg in 1944.
Check yourself¶
1. Compute \(\partial[0, 1, 2, 3]\) and check directly that \(\partial\partial[0, 1, 2, 3] = 0\).
\(\partial[0, 1, 2, 3] = [1, 2, 3] - [0, 2, 3] + [0, 1, 3] - [0, 1, 2]\). Applying \(\partial\) to each term gives the four rows displayed in 16.3; each of the six edges \([0,1], [0,2], [0,3], [1,2], [1,3], [2,3]\) appears once with \(+\) and once with \(-\). The total is \(0\).
2. Show that swapping the first two vertices changes the sign of the boundary of a triangle: compare \(\partial[b, a, c]\) with \(-\partial[a, b, c]\).
\(\partial[b, a, c] = [a, c] - [b, c] + [b, a] = [a, c] - [b, c] - [a, b]\), and \(-\partial[a, b, c] = -[b, c] + [a, c] - [a, b]\). They are equal.
3. A figure eight is triangulated as two hollow triangles sharing a vertex: vertices \(0, 1, 2\) and \(0, 3, 4\), six edges. Find \(b_0\) and \(b_1\).
\(c_0 = 5\), \(c_1 = 6\), no triangles. The graph is connected, so \(r_1 = 5 - 1 = 4\) and \(b_0 = 1\). Then \(b_1 = (6 - 4) - 0 = 2\), generated by the two triangles' edge loops. This matches \(H_1 = \pi_1^{\mathrm{ab}} = F_2^{\mathrm{ab}} = \mathbb{Z}^2\).
4. Triangulate the surface of a square pyramid (split the square base into two triangles). Count \(V\), \(E\), \(F\) and say what the Betti numbers must be.
\(V = 5\) (four base corners and the apex). Edges: four base sides, one base diagonal, four slant edges: \(E = 9\). Faces: four sides and two base triangles: \(F = 6\). \(\chi = 5 - 9 + 6 = 2\). The surface is a sphere, so \(b_0 = 1\), \(b_1 = 0\), \(b_2 = 1\), and indeed \(1 - 0 + 1 = 2\).
5. Show that in a tree (a connected graph without cycles) every 1-cycle is zero, and deduce \(H_1 = 0\).
A nonzero 1-cycle has, at every vertex, incoming and outgoing coefficients that balance. Take an edge with nonzero coefficient and leave its end vertex along another edge with nonzero coefficient (one must exist, or the boundary is not zero there), and continue. In a finite graph you must eventually revisit a vertex, which gives a cycle in the graph, impossible in a tree. So \(Z_1 = 0\) and \(H_1 = 0\). Check with ranks: \(b_1 = E - V + 1 = (V - 1) - V + 1 = 0\).
6. The orientable surface of genus 3 has \(\chi = -4\). Find all its Betti numbers.
It is connected and orientable, so \(b_0 = b_2 = 1\). Then \(1 - b_1 + 1 = -4\) gives \(b_1 = 6 = 2g\).
7. The Klein bottle has \(H_1 \cong \mathbb{Z} \oplus \mathbb{Z}/2\) and \(H_2 = 0\). Check the Euler characteristic, and explain why \(H_2 = 0\).
\(b_0 = 1\), \(b_1 = 1\) (the torsion does not count), \(b_2 = 0\), so \(\chi = 1 - 1 + 0 = 0\), matching \(V - E + F = 0\) for any triangulation (as for \(N_2\), \(\chi = 2 - 2 = 0\)). \(H_2 = 0\) because the Klein bottle is non-orientable: the triangles cannot be oriented so that every edge cancels, so there is no nonzero 2-cycle.
8. On the sphere, the equator is a 1-cycle. Why does that not give a nonzero element of \(H_1(S^2)\)? On the torus, why does the loop around the tube give one?
The equator is the boundary of the upper hemisphere (the sum of its triangles, oriented coherently), so it lies in \(B_1\) and is zero in \(H_1\). On the torus, the loop around the tube does not bound any 2-chain: cutting along it leaves a cylinder, not two pieces, and any 2-chain with that loop as boundary would have to be a region with that one edge, which does not exist. Its class is one of the two generators of \(H_1(T^2) \cong \mathbb{Z}^2\).
9. Give two spaces with isomorphic \(H_1\) but non-isomorphic \(\pi_1\), and two spaces with isomorphic \(\pi_1\) but different homology.
The figure eight (\(\pi_1 = F_2\)) and the torus (\(\pi_1 = \mathbb{Z}^2\)) both have \(H_1 \cong \mathbb{Z}^2\). A point and the sphere \(S^2\) both have trivial \(\pi_1\), but \(H_2(S^2) \cong \mathbb{Z}\) while \(H_2(\text{point}) = 0\).
10. Use the Euler-Poincaré formula to prove that no triangulation of the torus has \(V - E + F \ne 0\), and that any triangulated closed surface with \(V - E + F = 2\) has \(b_1 = 0\).
For any triangulation, \(V - E + F = b_0 - b_1 + b_2\), and the Betti numbers of the torus are \(1, 2, 1\) whatever the triangulation (Theorem 16.4), so the sum is \(0\). For a closed connected surface \(b_0 = 1\) and \(b_2 \le 1\), so \(2 = 1 - b_1 + b_2 \le 2 - b_1\), which forces \(b_1 = 0\) and \(b_2 = 1\).
11. In a simplicial complex every edge lies in exactly two triangles and \(V - E + F = 1\). Can it be an orientable closed surface?
For an orientable closed connected surface, \(b_0 = b_2 = 1\), so \(\chi = 2 - b_1\), and \(b_1 = 2g\) is even; \(\chi\) is even. But \(\chi = 1\) is odd. So it is not orientable; if it is a closed connected surface, it is \(N_1 = \mathbb{R}P^2\), since \(\chi(N_k) = 2 - k\).
12. Using Hurewicz, explain why the Poincaré homology sphere \(P\) has \(H_1(P) = 0\) although \(\pi_1(P) \ne 1\).
\(H_1(P) \cong \pi_1(P)^{\mathrm{ab}}\). The group \(\pi_1(P)\), of order 120, equals its own commutator subgroup (it is a perfect group), so its abelianisation is trivial. Homology sees only the abelian part of \(\pi_1\), so it misses \(P\)'s one-dimensional complexity entirely.
What you learned¶
- A simplicial complex builds a space from simplices glued along faces; the circle, sphere, torus and projective plane have small triangulations.
- Chains are formal sums of oriented simplices, and the boundary map takes the alternating sum of faces; the boundary of a boundary is always zero, because each face of a face appears twice with opposite signs.
- Homology \(H_n = Z_n/B_n\) is "closed \(n\)-dimensional pieces modulo those that bound"; \(H_0\) counts components and the Betti numbers come from ranks of boundary matrices.
- \(H_*(S^1) = \mathbb{Z}, \mathbb{Z}\); \(H_*(S^2) = \mathbb{Z}, 0, \mathbb{Z}\); \(H_*(T^2) = \mathbb{Z}, \mathbb{Z}^2, \mathbb{Z}\); \(H_*(\mathbb{R}P^2) = \mathbb{Z}, \mathbb{Z}/2, 0\), where torsion and missing \(H_2\) record non-orientability.
- The Euler characteristic is the alternating sum of Betti numbers, by rank-nullity, which makes it a topological invariant and proves Euler's formula for every triangulated sphere.
- \(H_1\) is the abelianised fundamental group (Hurewicz), and the first nonzero higher homotopy group equals homology; beyond that the two theories separate.
Next: Chapter 17 puts homology to work: Brouwer's theorem in every dimension, the hairy ball theorem, degree, fixed points, the ham sandwich theorem and persistent homology in data.
Further reading¶
- Allen Hatcher, Algebraic Topology, Cambridge University Press, 2002, Chapter 2 and Section 2.A.
- James R. Munkres, Elements of Algebraic Topology, Addison-Wesley, 1984, Chapter 1 (simplicial homology, with full proofs and many computations).
- P. J. Giblin, Graphs, Surfaces and Homology, 3rd edition, Cambridge University Press, 2010: homology of surfaces at a gentle pace.
- Herbert Edelsbrunner and John Harer, Computational Topology: An Introduction, American Mathematical Society, 2010: computing homology with matrices.
- Henri Poincaré, Papers on Topology: Analysis Situs and Its Five Supplements, translated by John Stillwell, American Mathematical Society, 2010.
- On this site: Linear Algebra Chapter 7, The four subspaces (kernels, images and rank-nullity), Algebra Chapter 15, Homomorphisms and quotients, Combinatorics Chapter 12, Planar graphs and colouring for Euler's formula on graphs, and the U of T guide MAT240 Unit 5, Rank-nullity and matrices.