Cohomology and Poincaré Duality

Homology (7A.10 Singular Homology and Mayer–Vietoris) measures cycles. Cohomology measures functions on cycles: a $k$-dimensional cohomology class assigns a number to every $k$-cycle, consistently. At first sight this is just the dual of homology, and over a field it nearly is. Three things make it indispensable.

• Cohomology has a product. Cup product makes $H^*(X)$ a ring, and the ring can distinguish spaces whose homology groups agree: $\mathbb{C}P^2$ and $S^2\vee S^4$ have the same homology but different cup products.
• Manifolds are self-dual. For a closed oriented $n$-manifold, Poincaré duality identifies $H^k$ with $H_{n-k}$. It is why a closed oriented 3-manifold's homology is determined by $H_1$, why odd-dimensional closed manifolds have Euler characteristic zero, and why a closed simply connected 3-manifold has the homology of $S^3$, which with Hurewicz and Whitehead makes it a homotopy sphere: the starting point of the Poincaré conjecture.
• Duality produces pairings that are invariants: the linking form of a 3-manifold, which distinguishes lens spaces that homology cannot, and the intersection form of a 4-manifold, which by theorems of Freedman and Donaldson nearly classifies simply connected 4-manifolds and reveals the strangeness of dimension four.

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

• define cochains and cohomology, compute $H^0$ and $H^1$, and use cohomology to state Kirchhoff's laws;
• state the universal coefficient theorem and compute $H^*(\mathbb{R}P^n; \mathbb{Z})$ and $H^*(L(p, q); \mathbb{Z})$;
• define the cup product, compute the cohomology rings of the torus and of surfaces, and use cup products to distinguish spaces with equal homology;
• define orientations and fundamental classes, state Poincaré duality and outline its proof by dual cells;
• deduce the structure of the homology of closed 3-manifolds and that closed simply connected 3-manifolds are homotopy spheres;
• compute linking forms of lens spaces and use them to separate $L(5, 1)$ from $L(5, 2)$; describe intersection forms and the theorems of Freedman and Donaldson.

11.1Potentials and currents

11.2Cochains and cohomology

Let $G$ be an abelian group ($\mathbb{Z}$, $\mathbb{Z}/2$, $\mathbb{Q}$, $\mathbb{R}$). The cochains $C^n(X; G) = \operatorname{Hom}(C_n(X), G)$ assign an element of $G$ to each singular $n$-simplex; the coboundary $\delta : C^n \to C^{n+1}$ is $(\delta\varphi)(\sigma) = \varphi(\partial\sigma)$. Then $\delta\delta = 0$, and the cohomology is $H^n(X; G) = \ker\delta/\operatorname{im}\delta$: cocycles (cochains vanishing on boundaries) modulo coboundaries. Cohomology is contravariant: $f : X \to Y$ induces $f^* : H^n(Y) \to H^n(X)$, by precomposition. It is homotopy invariant and has long exact sequences, excision, Mayer–Vietoris and cellular versions, all obtained by dualising 7A.10 Singular Homology and Mayer–Vietoris.

Low degrees. A 0-cocycle is a function on points that agrees at the two ends of every path: $H^0(X; G)$ is the group of $G$-valued functions constant on path components. A 1-cocycle assigns to each path a value that is additive under concatenation and zero on boundaries of 2-simplices, so it depends only on the homotopy class rel endpoints; on loops it defines a homomorphism $\pi_1(X) \to G$. In fact

for path-connected $X$ (the second because $G$ is abelian and $H_1$ is the abelianisation of $\pi_1$, 7A.9 Homology in Brief). So $H^1(X; \mathbb{Z}) = 0$ when $\pi_1(X)$ is finite, and a closed 3-manifold with finite $\pi_1$, such as a lens space, has $H^1 = 0$ even though $H_1 \neq 0$.

Universal coefficients. Cohomology is not simply the dual of homology; torsion shifts up one degree.

The torsion shift is visible in a single cell: attaching a 2-cell to a circle by a map of degree $m$ gives cellular chains $\mathbb{Z} \xrightarrow{m} \mathbb{Z}$ in degrees $2 \to 1$, homology $H_1 = \mathbb{Z}/m$, $H_2 = 0$; the dual cochains are $\mathbb{Z} \xrightarrow{m} \mathbb{Z}$ in degrees $1 \to 2$, cohomology $H^1 = 0$, $H^2 = \mathbb{Z}/m$.

11.3The cup product

For $\varphi \in C^k(X; R)$ and $\psi \in C^l(X; R)$, $R$ a commutative ring, define $\varphi\smile\psi \in C^{k+l}$ by evaluating on the front $k$-face and the back $l$-face:

Then $\delta(\varphi\smile\psi) = \delta\varphi\smile\psi + (-1)^k\varphi\smile\delta\psi$ (Exercise 11.7), so the product of cocycles is a cocycle and of a cocycle with a coboundary a coboundary: there is an induced product $H^k\times H^l \to H^{k+l}$, associative, with unit $1 \in H^0$, natural ($f^*(a\smile b) = f^*a\smile f^*b$), and graded commutative: $a\smile b = (-1)^{kl}b\smile a$ (Hatcher, Theorem 3.11; the proof is not obvious, since the formula is not symmetric on cochains). With real coefficients on a smooth manifold, the cup product corresponds to the wedge product of differential forms under de Rham's theorem (8A.8 Differential Forms and Stokes’ Theorem): $\int_\sigma\alpha\wedge\beta$ for closed forms.

Cup products distinguish spaces. $\mathbb{C}P^2$ and $S^2\vee S^4$ both have homology $\mathbb{Z}$ in degrees $0$, $2$, $4$. In $\mathbb{C}P^2$, $x\smile x$ generates $H^4$. In $S^2\vee S^4$, the square of the degree-two class is zero, by naturality: it comes from $S^2$ (via the retraction collapsing $S^4$), where $H^4(S^2) = 0$. So they are not homotopy equivalent. Equivalently, the attaching map $S^3 \to S^2$ of the 4-cell of $\mathbb{C}P^2$, the Hopf map, is not null-homotopic: $\pi_3(S^2) \neq 0$ (10A.4 Homotopy Groups, Degree and the Hurewicz Theorem), detected by a cup product (its Hopf invariant is $1$).

Borsuk–Ulam from the ring. If $f : S^n \to \mathbb{R}^n$ had $f(x) \neq f(-x)$ for all $x$, then $g(x) = \frac{f(x) - f(-x)}{|f(x) - f(-x)|}$ would be an odd map $S^n \to S^{n-1}$, inducing a map $\mathbb{R}P^n \to \mathbb{R}P^{n-1}$ that is nonzero on $H^1(\cdot; \mathbb{Z}/2)$ (it lifts a nontrivial loop to a path between antipodes); but then $a^n \neq 0$ in $H^n(\mathbb{R}P^n; \mathbb{Z}/2)$ would be pulled back from $H^n(\mathbb{R}P^{n-1}; \mathbb{Z}/2) = 0$. This is the cohomological proof of the Borsuk–Ulam theorem of 7A.8 Smooth Topology.

11.4Orientations and fundamental classes

For an $n$-manifold $M$ and $x \in M$, excision gives $H_n(M, M\setminus x) \cong H_n(\mathbb{R}^n, \mathbb{R}^n\setminus0) \cong \tilde H_{n-1}(S^{n-1}) = \mathbb{Z}$. A local orientation at $x$ is a choice of generator $\mu_x$; an orientation is a choice varying continuously (each $x$ has a ball $B$ and a class in $H_n(M, M\setminus B)$ restricting to $\mu_y$ for all $y \in B$). For smooth manifolds this agrees with the usual notion via ordered bases (8A.1 Smooth Structures).

For a triangulated closed oriented manifold, $[M]$ is the sum of all the $n$-simplices, oriented compatibly: their boundaries cancel in pairs, which is possible exactly when $M$ is orientable. A simply connected manifold is orientable (7A.12 The Poincaré Conjecture, Precisely): the orientation double cover is trivial.

11.5Poincaré duality

The cap product $H_n(M)\times H^k(M) \to H_{n-k}(M)$, $\sigma\frown\varphi = \varphi(\sigma|_{[e_0, \dots, e_k]})\,\sigma|_{[e_k, \dots, e_n]}$ on chains, pairs a chain with a cochain by evaluating the cochain on a front face and keeping the back face.

The idea of the proof, by dual cells. Triangulate $M$, and form the dual cell decomposition: for each $k$-simplex $\sigma$, the dual cell $\sigma^*$ is the union of the simplices of the barycentric subdivision that meet $\sigma$ only at its barycentre, an $(n - k)$-dimensional cell crossing $\sigma$ transversally at one point (in a surface, a vertex is dual to the polygon around it, an edge to the edge crossing it, a triangle to its centre). The correspondence $\sigma \leftrightarrow \sigma^*$ is a bijection between $k$-cells of the triangulation and $(n - k)$-cells of the dual decomposition, under which the coboundary of the triangulation's cochains becomes the boundary of the dual cells' chains: $\tau$ is a face of $\sigma$ iff $\sigma^*$ is a face of $\tau^*$, with matching signs when $M$ is oriented. So the cellular cochain complex of the triangulation is isomorphic to the cellular chain complex of the dual decomposition, reversed in degree, and both compute (co)homology. This was Poincaré's own argument (1895–1899); the modern proof by induction over open sets with Mayer–Vietoris is in Hatcher, §3.3.

Consequences.

1. Odd-dimensional closed manifolds have $\chi = 0$. With coefficients in a field, $b_k = b_{n-k}$ (duality plus universal coefficients over a field), and for odd $n$ the terms of $\sum(-1)^kb_k$ cancel in pairs. By Poincaré–Hopf (7A.8 Smooth Topology), every closed odd-dimensional manifold has a nowhere vanishing vector field.
2. Closed oriented 3-manifolds. $H_0 = H_3 = \mathbb{Z}$, $H_1 = \pi_1^{\mathrm{ab}}$, and $H_2 \cong H^1 = \operatorname{Hom}(H_1, \mathbb{Z})$, which is free of rank $b_1$; the torsion of $H_1$ reappears in $H^2 \cong H_1$. The whole homology is determined by $\pi_1$.
3. Simply connected 3-manifolds. If $M$ is closed and simply connected, it is orientable, $H_1 = 0$, hence $H_2 = H^1 = 0$: $H_*(M) \cong H_*(S^3)$. By the Hurewicz theorem (10A.4 Homotopy Groups, Degree and the Hurewicz Theorem), $\pi_2(M) = H_2(\tilde M) = 0$ and $\pi_3(M) \cong H_3(M) = \mathbb{Z}$; a generator $S^3 \to M$ has degree one, induces isomorphisms on homology, and by Whitehead's theorem (7A.6 CW Complexes and Cellular Approximation) is a homotopy equivalence. Every closed simply connected 3-manifold is homotopy equivalent to $S^3$. The Poincaré conjecture asks whether it is homeomorphic. (The degree-one map $S^3 \to M$ is the topological seed of the sweepouts used in finite extinction, 12C.3 Finite Extinction.)
4. Homology spheres. A closed 3-manifold has the homology of $S^3$ iff $H_1 = 0$, iff $\pi_1$ is perfect (equal to its commutator subgroup). The Poincaré homology sphere has $\pi_1$ the binary icosahedral group of order $120$, which is perfect (7A.9 Homology in Brief): it has the homology, but not the homotopy type, of $S^3$.

11.6The linking form of a 3-manifold

In a closed oriented 3-manifold $M$, let $a$, $b$ be disjoint loops representing torsion classes in $H_1(M)$. Since $ma = 0$ for some $m$, $ma$ bounds a 2-chain $S$; the linking number is

the intersection number of $S$ with $b$, divided by $m$. It is well defined in $\mathbb{Q}/\mathbb{Z}$, symmetric, and depends only on the homology classes; it is the pairing $\operatorname{Tors}H_1\times\operatorname{Tors}H_1 \to \mathbb{Q}/\mathbb{Z}$ obtained from Poincaré duality $\operatorname{Tors}H_1 \cong \operatorname{Tors}H^2$ and the Bockstein identification $\operatorname{Tors}H^2 \cong \operatorname{Hom}(\operatorname{Tors}H_1, \mathbb{Q}/\mathbb{Z})$, and duality makes it nondegenerate. It is a homotopy invariant of $M$. In $S^3$ (where every loop bounds) it is the classical linking number of two knots, an integer, modulo $1$; in a rational homology sphere it takes fractional values.

Lens spaces. For $L(p, q)$, $H_1 = \mathbb{Z}/p$ with generator $g$ (the core of one of the two solid tori of the genus-one Heegaard splitting, 7A.5 Computing π₁). The meridian disc $D$ of the second solid torus has boundary a curve on the Heegaard torus which, in the first solid torus, is homologous to $p\,g$; so $p\,g$ bounds a 2-chain $S$, made of $D$ and an annulus in the first solid torus, and a parallel copy of $g$ meets $S$ algebraically $q$ times. So

with the sign and the choice between $q$ and its inverse modulo $p$ depending on conventions (Hatcher, Example 3.44 and §3.E). Changing the generator to $ng$ multiplies the value by $n^2$. So the linking forms of $L(p, q)$ and $L(p, q')$ are isomorphic iff $qq' \equiv \pm n^2 \pmod p$ for some $n$, and since the linking form is a homotopy invariant, this is necessary for a homotopy equivalence (J. H. C. Whitehead showed in 1941 that it is also sufficient).

For $p = 5$: the squares mod $5$ are $\pm1$, and $2 \not\equiv \pm n^2$. So $L(5, 1)$ and $L(5, 2)$ are not homotopy equivalent, although they have the same homology and fundamental group $\mathbb{Z}/5$. For $p = 7$: $2 \equiv 3^2 \pmod 7$, so $L(7, 1)$ and $L(7, 2)$ are homotopy equivalent; Kurt Reidemeister proved in 1935, with his torsion invariant, that they are not homeomorphic. The homeomorphism classification ($L(p, q) \cong L(p, q')$ iff $q' \equiv \pm q^{\pm1} \pmod p$) was completed by Brody in 1960. Lens spaces are the first place where homotopy type and homeomorphism type of 3-manifolds part company; by the geometrization theorem, every closed 3-manifold with finite fundamental group is a spherical space form, and among these the lens spaces are the only ones where this happens. For $S^3$, the Poincaré conjecture says it does not.

11.7The intersection form of a 4-manifold

For a closed oriented 4-manifold $M$, the cup product pairing on $H^2(M; \mathbb{Z})$ modulo torsion,

is a symmetric bilinear form, and Poincaré duality makes it unimodular (its matrix has determinant $\pm1$). Geometrically, $Q_M$ counts signed intersection points of surfaces representing the dual classes. Examples: $Q_{\mathbb{C}P^2} = (1)$; $Q_{S^2\times S^2} = H = \begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}$; the K3 surface has $Q = 2(-E_8)\oplus3H$, with $E_8$ the even positive definite form of rank $8$ from the $E_8$ root lattice.

Two theorems of the early 1980s show how much $Q_M$ knows and how strange dimension four is. Michael Freedman (1982): closed simply connected topological 4-manifolds are classified by their intersection form together with one $\mathbb{Z}/2$ invariant (the Kirby–Siebenmann invariant), and every unimodular form occurs. In particular, a topological 4-manifold homotopy equivalent to $S^4$ is homeomorphic to it: the topological 4-dimensional Poincaré conjecture. Simon Donaldson (1983): if a closed simply connected smooth 4-manifold has definite intersection form, the form is diagonal ($\pm(1)\oplus\dots\oplus\pm(1)$). So Freedman's topological manifold with form $E_8$ has no smooth structure. Combined, they produce exotic smooth structures on $\mathbb{R}^4$. Whether $S^4$ has an exotic smooth structure, the smooth 4-dimensional Poincaré conjecture, is open (7A.12 The Poincaré Conjecture, Precisely).

11.8History

Cohomology emerged in the 1930s from work of Alexander, Čech, Kolmogorov and Whitney, and the cup product from Alexander and Whitney (1935–38). Poincaré stated his duality in Analysis Situs (1895) and gave the dual cell proof in its supplements (1899); torsion and the linking form were studied by Seifert and by de Rham in the 1930s. Reidemeister's torsion appeared in 1935, Whitehead's homotopy classification of lens spaces in 1941, and Brody's homeomorphism classification in 1960. Freedman's classification appeared in 1982 and Donaldson's diagonalisation theorem in 1983; both were recognised with Fields Medals in 1986.

11.9Exercises

Solution

(a) If $V$ vanishes on cycles, define $\phi(x) = V(\text{any path from a base vertex to } x)$, well defined because the difference of two paths is a cycle; then $\delta\phi = V$. Conversely $\delta\phi(\text{cycle}) = \phi(\partial\,\text{cycle}) = 0$. Constants are the kernel of $\delta$ on a connected graph. (b) Cycles are $\ker\partial_1$; $\partial_1 : \mathbb{R}^E \to \mathbb{R}^V$ has image the sum-zero vectors (connected graph), of dimension $V - 1$, so the kernel has dimension $E - V + 1$. (c) Coboundaries have dimension $V - 1$ (the image of $\delta$ on $\mathbb{R}^V$ modulo constants), and $(V - 1) + (E - V + 1) = E$. In fact, with the standard inner products, coboundaries and cycles are orthogonal complements: $\langle\delta\phi, I\rangle = \langle\phi, \partial I\rangle$. This orthogonal splitting is the discrete Hodge decomposition, and solving a resistor network is projecting onto it.

Solution

$(\delta\varphi\smile\psi)(\sigma) = \sum_{i=0}^{k+1}(-1)^i\varphi(\sigma|_{[e_0..\hat e_i..e_{k+1}]})\psi(\sigma|_{[e_{k+1}..e_{k+l+1}]})$ and $(-1)^k(\varphi\smile\delta\psi)(\sigma) = \sum_{i=k}^{k+l+1}(-1)^i\varphi(\sigma|_{[e_0..e_k]})\psi(\sigma|_{[e_k..\hat e_i..e_{k+l+1}]})$. The last term of the first sum ($i = k + 1$) and the first term of the second ($i = k$) cancel: both are $\pm\varphi(\sigma|_{[e_0..e_k]})\psi(\sigma|_{[e_{k+1}..e_{k+l+1}]})$ with signs $(-1)^{k+1}$ and $(-1)^k$. The remaining terms are exactly $\sum_i(-1)^i(\varphi\smile\psi)(\text{face}_i\sigma) = (\varphi\smile\psi)(\partial\sigma)$.

Solution

$H^0 = \mathbb{Z}$; $H^k = \mathbb{Z}/2$ for $k$ even, $0 < k \leq n$; $H^k = 0$ for $k$ odd, $k < n$; and $H^n = \mathbb{Z}$ if $n$ is odd. From the cochain complex: the coboundary $C^{k-1} \to C^k$ is the transpose of $d_k$, which is $2$ for $k$ even and $0$ for $k$ odd; so $H^k = \mathbb{Z}/2$ for even $k > 0$ (cokernel of $2$), and the kernel in odd degree $k < n$ is $0$ (the next map is $2$, injective). For odd $n$, $C^n$ has no outgoing map, and the incoming one is $0$: $H^n = \mathbb{Z}$.

Solution

Label the square $[v_0, v_1, v_2, v_3]$ with horizontal edges $a$ and vertical edges $b$ and the diagonal $c = [v_0, v_2]$; the triangles are $T_1 = [v_0, v_1, v_2]$ and $T_2 = [v_0, v_3, v_2]$, with $[T^2] = T_1 - T_2$. Take $\alpha(a) = 1$, $\alpha(b) = 0$, $\alpha(c) = 1$ and $\beta(a) = 0$, $\beta(b) = 1$, $\beta(c) = 1$ (cocycles: on each triangle, the values on the two short edges sum to the value on the diagonal). Then $(\alpha\smile\beta)(T_1) = \alpha([v_0v_1])\beta([v_1v_2]) = 1\cdot1 = 1$ and $(\alpha\smile\beta)(T_2) = \alpha([v_0v_3])\beta([v_3v_2]) = 0\cdot0 = 0$, so $(\alpha\smile\beta)[T^2] = 1$. Similarly $(\beta\smile\alpha)(T_1) = 0$ and $(\beta\smile\alpha)(T_2) = 1$, so $(\beta\smile\alpha)[T^2] = -1 = -(\alpha\smile\beta)[T^2]$, as graded commutativity requires for degree-one classes.

Solution

(a) $b_k = \dim H_k(M; \mathbb{Q}) = \dim H^{n-k}(M; \mathbb{Q}) = \dim H_{n-k}(M; \mathbb{Q}) = b_{n-k}$, using universal coefficients over a field. For odd $n$, $(-1)^k$ and $(-1)^{n-k}$ are opposite, so the terms cancel in pairs. (b) The same computation with $\dim_{\mathbb{Z}/2}$: $\chi$ equals the alternating sum of the ranks of the cellular chain groups, which equals the alternating sum of homology dimensions over any field (the rank–nullity argument of 7A.10 Singular Homology and Mayer–Vietoris).

Solution

(a) Squares mod $5$ are $\{1, 4\}$, so $\pm n^2 \in \{1, 4\}$. $L(5, 1)$ vs $L(5, 4)$: $4 \in\{1, 4\}$, equivalent. $L(5, 2)$ vs $L(5, 3)$: $6 \equiv 1$, equivalent. $L(5, 1)$ vs $L(5, 2)$: $2 \notin\{1, 4\}$, not. Two types: $\{1, 4\}$ and $\{2, 3\}$. (b) Squares mod $7$ are $\{1, 2, 4\}$ and negatives $\{6, 5, 3\}$: every nonzero residue is $\pm n^2$. So all $L(7, q)$ are homotopy equivalent. (c) $q^{-1}$ mod $7$: $1 \to 1$, $2 \to 4$, $3 \to 5$. Classes under $q \mapsto \pm q^{\pm1}$: $\{1, 6\}$, $\{2, 5, 4, 3\}$. Two homeomorphism types: $L(7, 1)$ and $L(7, 2)$.

Solution

(a) Each sphere has trivial normal bundle (push it off itself, $S^2\times\{p'\}$, disjointly), so self-intersection $0$; the two spheres meet transversally in one point, $+1$. (b) $1$, $-1$, $0$ ($Q = (1)\oplus(-1)$), $0$ ($H$ has eigenvalues $\pm1$). (c) $H$ is even ($Q(x, x) = 2x_1x_2$), while $(1)\oplus(-1)$ is odd ($Q(1, 0) = 1$). Isomorphic forms have the same parity, and the intersection form is a homotopy invariant (cup product and fundamental class), so the manifolds are not homotopy equivalent.