4.6 Poincaré Duality · Homology & Cohomology

…geometry is the art of reasoning well from badly drawn figures; however, these figures, if they are not to deceive us, must satisfy certain conditions; the proportions may be grossly altered, but the relative positions of the different parts must not be upset.

— Henri Poincaré, Analysis Situs (1895), translated by John Stillwell

Write down the Betti numbers of the torus: $1, 2, 1 .$ Of the sphere: $1, 0, 1 .$ Of the surface with two handles: $1, 4, 1 .$ They read the same forwards and backwards. Henri Poincaré noticed in the 1890s that this is no accident: for every closed, orientable manifold of dimension $n,$ the $k$ -th Betti number equals the $(n - k)$ -th (Poincaré 1895).

Behind the numerical palindrome hides a structural one: a second way of cutting a manifold into cells, in which every $k$ -dimensional piece is traded for an $(n - k)$ -dimensional one — and in which homology and cohomology trade places. This chapter builds that mirror.

Betti numbers that read the same backwards

Recall from §3.3 Homology Groups that the Betti number $b_{k}$ counts the independent $k$ -dimensional holes of a space: $b_{0}$ counts pieces, $b_{1}$ independent loops, $b_{2}$ enclosed voids, and so on. Make a list of them for a few familiar spaces and something odd jumps out. Try it in the figure.

The Betti mirror Pick a space · reflect it in the mirror

reads the same backwards

T^{2} : (b_{0}, b_{1}, b_{2}) = (1, 2, 1)

χ = 1 - 2 + 1 = 0

The middle number is matched with itself: the two loops cross once, so the fence of each one detects the other.

Figure 4.6.1 Bars show the Betti numbers

b_{0}, \dots, b_{n};

dashed outlines show the same list read backwards. For closed orientable manifolds the outlines fit exactly. The Klein bottle and

RP^{2}

fit only over

Z /2;

the last two spaces (dashed buttons) are not closed manifolds at all, and the mirror breaks.

Every closed surface you can build with handles — the sphere, the torus, $Σ_{g}$ — has Betti numbers $1, 2 g, 1 .$ The 3-sphere $S^{3}$ has $1, 0, 0, 1;$ the 3-dimensional torus $T^{3}$ has $1, 3, 3, 1;$ the complex projective plane $CP^{2},$ a four-dimensional space, has $1, 0, 1, 0, 1 .$ All palindromes.

The failures are just as instructive. The wedge $S^{2} \lor S^{2}$ of two spheres touching at a point has $1, 0, 2$ — but it is not a manifold: near the touching point it does not look like a flat plane. The disk has $1, 0, 0$ — it is a manifold, but it has a boundary, a rim where it stops. And the Klein bottle has rational Betti numbers $1, 1, 0 :$ it is a closed manifold, but it is not orientable — an ant walking around it can come back mirror-reversed (§2.4 Manifolds and Surfaces). Yet the Klein bottle’s Betti numbers mod 2 are $1, 2, 1,$ a palindrome again.

The fundamental class

A mirror needs a frame. Before we can match $k$ -dimensional holes with $(n - k)$ -dimensional ones, we need the whole manifold as a single object of algebra, and the natural candidate lives in the top dimension. Take a closed manifold $M$ of dimension $n,$ cut into $n$ -simplices (triangles for a surface, tetrahedra for a 3-manifold). Because $M$ is a manifold without boundary, every $(n - 1)$ -dimensional face — every edge of a triangulated surface — is shared by exactly two top simplices, one on each side.

Now try to orient all the top simplices compatibly, as in §3.4 Computing Homology: neighbouring simplices should induce opposite orientations on the face they share, so that in the boundary of the sum, every face appears once with a plus sign and once with a minus sign. If this can be done, the sum

[M] = σ top-dimensional \sum \pm σ

has boundary zero: it is an $n$ -cycle, and it uses every top simplex exactly once. Being able to choose the signs consistently is exactly what it means for the triangulated manifold to be orientable.

Mod 2 the sign problem disappears. With $Z /2$ coefficients every face is counted twice, and $1 + 1 = 0,$ so the plain sum of all top simplices is always a cycle. Every closed manifold, orientable or not, has a mod-2 fundamental class $[M]_{2} \in H_{n} (M; Z /2) .$

The fundamental class is the “whole manifold” as a cycle, and measuring a top-dimensional cochain on it means integrating over all of $M .$ That is how, in the previous chapter, we turned a 2-cochain on the torus into a number: $H^{n} (M; Z) ≅ Z$ by $ω \mapsto ⟨ ω, [M]⟩$ for closed, connected, orientable $M .$

Cutting a manifold the other way

The key idea begins with a marriage far older than topology. In his Harmonices Mundi of 1619, Johannes Kepler saw “two noteworthy weddings” among the Platonic solids, the cube with the octahedron and the dodecahedron with the icosahedron, and one solid left over, the tetrahedron, “as it were bachelor or hermaphrodite … because it is inscribed in itself” (Kepler 1619). What marks the partners, he wrote, is “angles opposite planes”. Put a dot in the middle of each face of a cube and join the dots on neighbouring faces: you have built an octahedron, whose corners sit where the cube had faces and whose faces sit where the cube had corners.

Poincaré’s insight, in effect, was that Kepler’s wedding can be performed on any triangulated manifold. It is the picture to keep in mind for the rest of the chapter. Take a triangulated surface and build a second decomposition of it into cells, as follows.

In the middle of each triangle, put a dual vertex — the triangle’s centre of mass, its barycentre.
Across each edge, draw a dual edge: it runs from the centre of the triangle on one side, through the midpoint of the edge, to the centre of the triangle on the other side. It crosses the original edge exactly once.
Around each vertex, the dual edges of the edges at that vertex close up into a polygon: the dual face of the vertex. It has one side for each edge at the vertex and one corner for each triangle at the vertex.

Dual cells Point at any cell · play or scrub between the two worlds

original dual

vertex (0-cell) ⟷ dual face (2-cell)

The dual face around this vertex is a 6-sided region: one side for each edge at the vertex, one corner for each triangle.

7 interior vertices ⟷ 7 complete hexagons

Figure 4.6.2 A triangulation (ivory) and its dual (violet). Point at a vertex, an edge or a triangle and its partner lights up: a dual face, a dual edge crossing it, a dual vertex inside it. The timeline fades one decomposition into the other, and you point at whichever is in front. In the plane the dual of the triangular grid is the honeycomb; on the torus every dual face is a (slanted) hexagon.

The dual cells again cut the surface into vertices, edges and faces — just not triangles any more: the faces are polygons. And the counts are swapped:

V^{*} = F, E^{*} = E, F^{*} = V .

For the $3 \times 3$ torus that is $18$ dual vertices, $27$ dual edges and $9$ hexagonal dual faces. For an octahedron ( $6$ vertices, $12$ edges, $8$ faces) it is $8, 12, 6 :$ the dual of the octahedron is the cube, and Kepler’s “angles opposite planes” is exactly the swap $V^{*} = F,$ $F^{*} = V .$ The seven-vertex torus has a dual with seven hexagons, known in geometry as the Szilassi polyhedron. Since the alternating sum $V - E + F$ on a surface reads the same in both directions, the Euler characteristic does not notice the swap.

A little history Weddings, barrels and a trial

Johannes Kepler 1571 – 1630

Tycho Brahe’s successor as Imperial Mathematician, Kepler is remembered for three laws of planetary motion; the third appeared in the same book as the polyhedral weddings, Harmonices Mundi. He wrote it while his mother, Katharina, was on trial for witchcraft. He enlisted the law faculty at Tübingen in her defence, and on the coach journey to Württemberg he read a treatise on music by Galileo’s father, which the book cites again and again. Its second book holds the first systematic treatment of tilings of the plane. Even his second wedding produced mathematics: at the celebrations he watched the wine barrels being measured with a rod pushed in through the bung-hole, and began the study of volumes that became his New Stereometry of Wine Barrels (1615) (Field 1999).

(The barycentric subdivision was met in §2.5 Simplicial Complexes; on a surface, it cuts each triangle into six small triangles, and the dual face of a vertex is the union of the small triangles touching it — exactly the shaded region in the figure.)

The most important property of the construction is that it reverses the face relation:

The duality theorem

Now watch what the reversal does to the algebra. Recall from §4.2 Cohomology Groups that a $k$ -cochain is an assignment of a number to each $k$ -simplex, and that the coboundary of the cochain that is $1$ on a single simplex $σ$ and $0$ elsewhere is a signed sum of the $(k + 1)$ -simplices that have $σ$ as a face:

δ (1_{σ}) = τ \supset σ \sum \pm 1_{τ} .

On the dual side, the cells $τ^{*}$ of those bigger simplices are exactly the faces of $σ^{*} .$ So the boundary of the dual cell $σ^{*}$ is a signed sum of the same things:

\partial (σ^{*}) = τ \supset σ \sum \pm τ^{*} .

Match each $k$ -cochain $φ$ with the dual $(n - k)$ -chain $\sum_{σ} φ (σ) σ^{*} .$ The two displays say that, under this matching, the coboundary map $δ$ of the original triangulation becomes the boundary map $\partial$ of the dual decomposition — as long as the signs agree. Cocycles become cycles, coboundaries become boundaries, and therefore

H^{k} (K) ≅ H_{n - k} (K^{*}) ≅ H_{n - k} (M),

where the last step holds because homology does not depend on how the manifold is cut into cells (§3.6 Exact Sequences and Mayer–Vietoris).

Where do the signs come from? Each dual cell needs an orientation, and it has to be chosen in a way that fits with the orientation of $σ$ and the orientation of the manifold. On an oriented surface there is a natural recipe: the dual edge $e^{*}$ is the edge $e$ turned a quarter-turn counterclockwise, and each dual face is oriented counterclockwise. A direct computation on the $3 \times 3$ torus confirms that this works: with these choices the boundary matrices of the dual complex are exactly the transposes of the original ones (one of them up to an overall minus sign), that is, the coboundary matrices. On a non-orientable surface “counterclockwise” cannot be chosen consistently, and the signs cannot be made to match — except mod 2, where signs do not matter.

On a surface the isomorphism has a face you can draw. A degree-1 cohomology class can be drawn as a fence, as in §4.5 The Cup Product. Poincaré duality says that the fence, read as a loop, is the dual homology class.

Loop and fence Drag γ to slide it · drag elsewhere to rotate · choose the curves

signed crossings

γ \cdot C = + 1

intersection number

det (1001) = 1 \cdot 1 - 0 \cdot 0 = 1

class of C

class of γ

Figure 4.6.3 The same curve

C,

seen as a loop (gold, with beads showing its direction) or as a fence (teal, with chevrons pointing from its left to its right). The fence measures the violet test loop

γ

by counting signed crossings. Drag

γ

across the surface: the crossings move, but the count is always the intersection number

γ \cdot C,

which depends only on the classes of the two curves.

Three consequences

The cap product: duality as a machine

Dual cells explain why duality holds. To compute with it, there is a formula that turns a cochain into a chain directly, without building the dual decomposition. It is the sibling of the cup product.

Compare with the cup product: there both factors were measurements, read on the front and the back face. Here one factor is a measurement and the other a place; the measurement is spent on the front face, and the back face survives as a smaller place. The two are tied together by the identity $⟨ ψ, c ⌢ φ ⟩ = ⟨ φ ⌣ ψ, c ⟩$ (check it on a single triangle: both sides are $φ ([v_{0}, v_{1}]) ψ ([v_{1}, v_{2}])).$

The formula $\partial (σ ⌢ φ) = (- 1)^{p} (\partial σ ⌢ φ - σ ⌢ δ φ)$ (Hatcher 2002, p. 240) shows that since $\partial [M] = 0,$ cocycles go to cycles and coboundaries go to boundaries. So $D$ is defined on classes, $D : H^{p} (M) \to H_{n - p} (M) .$ The modern statement of Poincaré duality is that this map is an isomorphism.

Let us run it on the torus. Every lower triangle $L$ of our grid hands its back edge (vertical) the value of $φ$ on its front edge (horizontal); every upper triangle $U,$ counted with a minus sign, hands its back edge (horizontal) the value of $φ$ on its front edge (vertical). Take $φ = α,$ the vertical fence: it is $1$ exactly on the horizontal edges of its column (and on the diagonals, which are never front edges here). So $D (α)$ is the sum of the vertical back edges of the lower triangles in that column — a vertical loop running right beside the fence. Similarly $D (β)$ is the horizontal loop just above the fence $β,$ traversed backwards: $D (β) = - a .$

The cap product at work Choose a class · tap vertices to add coboundaries

the cocycle

φ = α

its dual, a 1-chain (rose)

D (φ) = [T^{2}] ⌢ φ = \sum_{σ} \pm φ (front) \cdot back

is it a cycle?

yes — its boundary is zero

its homology class

[D (φ)] = b \in H_{1} (T^{2})

Tap vertices to add $δ f,$ where $f$ is $1$ on the tapped vertices. The rose cycle changes, yet its class does not: duality is well defined on classes.

Figure 4.6.4

D (φ) = [T^{2}] ⌢ φ

on the

3 \times 3

torus: the rose 1-chain. For the fence

α

it is the vertical loop beside the fence, of class

b;

for

β

it is

- a .

Tap vertices to change

φ

by a coboundary: the rose chain changes — sometimes into something that hardly looks like a loop — but it is always a cycle, and its homology class never moves.

Notice how this matches the fence picture. In §4.5 The Cup Product, $α$ was the fence of the vertical loop $b$ running upwards, so $α = PD [b];$ and $β$ was the fence of the horizontal loop running leftwards, $β = PD [- a] .$ The cap product undoes exactly this: $D (α) = b$ and $D (β) = - a .$ The fence and the loop really are two faces of one object.

Intersection numbers

We have been counting crossings with signs all along; let us make it official. Take two oriented closed curves $C_{1}, C_{2}$ on an oriented surface, in general position — they cross at finitely many points, each crossing a clean “×”, never a tangency.

The key fact is that the intersection number depends only on the homology classes of the two curves. Slide one curve across the other and crossings are created or destroyed in pairs with opposite signs — you saw this in the bent fence of Figure 4.5.4, which crossed three times with signs $+ 1, - 1, + 1 .$ Two consequences are immediate: $C_{2} \cdot C_{1} = - C_{1} \cdot C_{2}$ (swapping the order reverses every turn), and $C \cdot C = 0$ on an orientable surface (push $C$ off itself to a parallel copy that never meets it).

On the torus, write $(p, q)$ for the class $p a + q b$ of a curve that winds $p$ times around horizontally and $q$ times vertically. Straight curves of two classes cross in exactly $∣ p_{1} q_{2} - p_{2} q_{1} ∣$ points, all with the same sign, and in general

(p_{1}, q_{1}) \cdot (p_{2}, q_{2}) = p_{1} q_{2} - p_{2} q_{1} = det (p_{1} p_{2} q_{1} q_{2}) .

The intersection game Change the classes · drag the teal curve

p₁ 1

q₁ 2

p₂ 2

q₂ 1

Gold curve $C_{1}$ goes $p_{1}$ times around horizontally and $q_{1}$ times vertically; teal $C_{2}$ likewise with $p_{2}, q_{2} .$ Drag the teal curve to slide it around the torus.

crossings

3 in all: 0 positive, 3 negative → signed sum −3

intersection number

C_{1} \cdot C_{2} = det (1221) = 1 \cdot 1 - 2 \cdot 2 = - 3

cup product of the duals (7 × 7 triangulation)

⟨ PD [C_{1}] ⌣ PD [C_{2}], [T^{2}]⟩ = - 3

Figure 4.6.5 Two straight closed curves on the torus, their crossings marked

+

(filled) or

-

(hollow). Wherever you drag the teal curve, the signed count is the determinant

p_{1} q_{2} - p_{2} q_{1} .

The last line computes, by front face × back face on a hidden

7 \times 7

triangulation, the cup product of the two fences — the same number.

This is the precise form of the slogan at the end of the previous chapter, with our sign conventions (other books use others; the content is the same). It holds in every dimension: on a closed oriented $n$ -manifold, the Poincaré dual of a $p$ -dimensional submanifold is a class of degree $n - p,$ and the cup product of two such classes is dual to the intersection of the submanifolds. The last line of Figure 4.6.5 is the surface version at work: it computes the cup product by front face × back face, without looking at a single crossing, and always lands on the same number as the count.

Poincaré duality can also be phrased entirely in terms of intersections. The intersection pairing between $H_{k} (M)$ and $H_{n - k} (M)$ counts the signed intersections of a $k$ -cycle with an $(n - k)$ -cycle, and it is perfect (with coefficients in a field, or over $Z$ once torsion is ignored): for every class that is not zero (or torsion) there is a class of complementary dimension that it meets a nonzero number of times (Hatcher 2002, Proposition 3.38, phrased with cup products). In other words, a hole in one dimension can always be detected by a hole in the complementary dimension that passes through it.

The intersection form and the signature

When the manifold has even dimension $n = 2 m,$ the middle dimension pairs with itself, and the cup product gives a square table of integers.

For a surface ( $m = 1)$ we have met it already: the torus gives $(0 - 1 10),$ and the genus- $g$ surface gives $g$ copies of that block — the multiplication table of Figure 4.5.6. For four-dimensional manifolds ( $m = 2)$ the form is symmetric, and it is where the modern study of those spaces begins.

$CP^{2} :$ $H^{2} = Z,$ and $Q = (1) .$ The class is dual to a complex projective line inside $CP^{2},$ and two such lines always meet in exactly one point — so a line meets a pushed-off copy of itself once.
$S^{2} \times S^{2} :$ $H^{2} = Z^{2},$ dual to the two spheres $S^{2} \times {pt}$ and ${pt} \times S^{2} .$ They meet once, and each can be pushed off itself, so $Q = (0110) .$
Reversing the orientation of $CP^{2}$ gives a manifold, usually written $CP^{2},$ with $Q = (- 1) .$

The signature of a symmetric form is the number of positive eigenvalues minus the number of negative ones; for $CP^{2}$ it is $1,$ for $S^{2} \times S^{2}$ it is $0 .$ The signature, the rank, and whether $Q (x, x)$ is always even are invariants that already distinguish many 4-manifolds with the same Betti numbers (see the exercises).

Remark Why 4-manifolds care

Two results of the early 1980s, each of which won its author a Fields Medal, put the intersection form at the centre of four-dimensional topology. Michael Freedman showed in 1982 that a closed simply connected topological 4-manifold is determined, up to homeomorphism, by its intersection form together with one extra mod-2 invariant, and that every unimodular symmetric form occurs (Freedman 1982). Simon Donaldson showed in 1983 that for smooth such manifolds with a definite form (all eigenvalues of one sign), the form must be equivalent to $\pm$ the identity matrix (Donaldson 1983). Together they imply, for example, that the topological 4-manifold whose form is the eight-dimensional “ $E_{8}$ form” admits no smooth structure at all. We only glimpse this world; it lies far beyond this book.

Which forms do occur for smooth 4-manifolds? For forms that are not definite, the answer is still unknown, and the open case concerns even forms, those with $Q (x, x)$ even for every $x,$ like that of $S^{2} \times S^{2} .$

Conjecture The 11/8 conjecture

Open since 1982 Posed by Yukio Matsumoto, 1982

Every closed smooth spin 4-manifold satisfies $b_{2} \geq \frac{11}{8} ∣ σ ∣,$ where $b_{2}$ is the second Betti number and $σ$ the signature of the intersection form. (For a simply connected manifold, spin means that $Q (x, x)$ is even for every $x .)$

The two numbers in the conjecture are the ones just defined: $b_{2}$ is the size of the matrix $Q,$ and $σ$ its number of positive eigenvalues minus its number of negative ones. A smooth manifold, the conjecture says, cannot have a large signature without paying for it in extra homology.

What is known. The bound would be sharp: the K3 surface, the complex surface $x^{4} + y^{4} + z^{4} + w^{4} = 0$ in $CP^{3},$ has $b_{2} = 22$ and $σ = - 16,$ and $22 = \frac{11}{8} \cdot 16 .$ Donaldson’s theorem rules out definite forms. In 2001 Mikio Furuta proved $b_{2} \geq \frac{10}{8} ∣ σ ∣ + 2$ when $σ \neq = 0,$ using the Seiberg–Witten equations, and in 2022 Hopkins, Lin, Shi and Xu reached $\frac{10}{8} ∣ σ ∣ + 4$ for simply connected manifolds other than $S^{4},$ $S^{2} \times S^{2}$ and K3, showing also that this is as far as that method can go.

A way in. Plot the pairs $(∣ σ ∣, b_{2})$ of the manifolds you know: $S^{2} \times S^{2}$ gives $(0, 2),$ K3 gives $(16, 22),$ and connected sums add both numbers. Which points can you reach from these two, and where do they lie relative to the line $b_{2} = \frac{11}{8} ∣ σ ∣ ?$

(Matsumoto 1982; Furuta 2001; Hopkins, Lin, Shi & Xu 2022) Status as of October 2026 All open problems

A little history

Poincaré stated the symmetry of Betti numbers in his 1895 memoir Analysis Situs and tried to prove it there by intersecting cycles with one another (Poincaré 1895). Heegaard’s counterexample sent him back to the problem (§3.2 Chains and the Boundary Operator), and in 1899 he explained duality through the “reciprocal” subdivision, the dual cells of this chapter (Poincaré 1899). Gaston Darboux recalled that Poincaré drafted each memoir “all in one go, with only a few erasures, and did not return to what he had written”; John Stillwell, who translated the memoir and its supplements, advises reading them the same way, taking in “their general sweep without lingering too long over gaps and errors” (Poincaré 1895). The formula $D (φ) = [M] ⌢ φ$ had to wait for the cup and cap products of the late 1930s; it was Hassler Whitney who named the cap (Whitney 1938).

Boundaries and complements: Lefschetz and Alexander

Poincaré duality needs a closed manifold. Two classical variations relax this: one lets the manifold have a rim, the other looks at the space left over around a shape sitting inside a sphere.

Manifolds with boundary

For a compact manifold $M$ with boundary $\partial M,$ duality survives if one side of the mirror is taken relative to the boundary. Relative homology $H_{k} (M, \partial M)$ was introduced in §3.6 Exact Sequences and Mayer–Vietoris: its cycles are chains whose boundary lies in $\partial M$ — paths allowed to end on the rim. Its cohomological twin uses cochains that vanish on every simplex of $\partial M :$ measurements that ignore the rim. They form the relative cohomology groups $H^{k} (M, \partial M) .$

The annulus $A$ (a ring) shows both halves at once, and the rule is the one we already know: a curve is matched with the fence along it. Its absolute $H_{1}$ is generated by the core circle $C .$ The fence along $C$ stays away from the rim, so it is a measurement that ignores the rim — a class in $H^{1} (A, \partial A)$ — and that class is the dual of $C .$ A path crossing the ring from rim to rim meets it once. In the other direction, take a rung $R,$ a radial segment from the inner rim to the outer one. Read as a relative cycle, it generates $H_{1} (A, \partial A),$ and its dual is the fence along it: the measurement that counts how many times a loop goes around the ring, a class in $H^{1} (A) .$

Lefschetz duality on the annulus Switch between the two pairings

The fence along C (teal chevrons) never touches the rim, so it is a class in $H^{1} (A, \partial A)$ . Duality matches it with C itself. A path from rim to rim crosses it once.

Figure 4.6.6 Each curve is matched with the fence along it. The core circle goes with its own fence, which never touches the rim; the rung, read as a relative cycle, goes with the fence along the rung. Each fence is tested by the other curve (dashed violet), which crosses it exactly once, at the rose point.

Complements: knots in space

Now take a shape $K$ sitting inside a sphere $S^{n}$ (think of $S^{3}$ as ordinary space $R^{3}$ with one point at infinity added) and ask about the space around it, the complement $S^{n} ∖ K .$ Its homology is determined by the cohomology of $K$ alone — no matter how $K$ is tangled up inside.

Two examples show its reach.

Curves in the plane. A circle $C$ drawn in $S^{2}$ (the plane with a point at infinity), however wiggly, has $H ~_{0} (S^{2} ∖ C) ≅ H ~^{1} (C) = Z :$ the complement has exactly two pieces, an inside and an outside. This is the Jordan curve theorem, met in §3.5 Maps, Invariance, and First Triumphs.
Knots in space. A knot $K$ is a circle embedded in $S^{3} .$ Then $H_{1} (S^{3} ∖ K) ≅ H^{1} (K) = H^{1} (S^{1}) = Z :$ every knot complement has the first homology of a circle, generated by a small loop around the knot’s strand, its meridian. §3.7 Knots and Links proved this with Mayer–Vietoris; Alexander duality says why it had to be so. The homology of the complement depends only on the cohomology of the knot, and every knot is a circle.

A loop $γ$ in the complement is therefore some multiple $n [μ]$ of the meridian class, and the number $n$ is the linking number of $γ$ with $K :$ how many times $γ$ winds around the knot. Gauss wrote down an integral formula for it in a brief diary note dated 22 January 1833, which stayed unknown until it was printed in his collected works in 1867 (Ricca & Nipoti 2011); the figure computes that integral numerically. Dually, $H^{1} (S^{3} ∖ K) ≅ Z$ is generated by the measurement “linking number with $K ”,$ and that measurement has a fence — now a surface, since we are in three dimensions: a Seifert surface, an oriented surface whose boundary is the knot, which Seifert’s algorithm (§3.7 Knots and Links) builds for every knot (Seifert 1935). For the unknot it is a disk, and the linking number of a loop is the signed number of times it pierces the disk.

Loops around a knot Drag γ along the knot · drag elsewhere to rotate · pick a knot and a loop

Gauss linking integral

lk (γ, K) = 1.000

class of the loop

[γ] = [μ] \in H_{1} (R^{3} ∖ K) ≅ Z

Figure 4.6.7 The complement of any knot has

H_{1} ≅ Z,

generated by a meridian

μ .

The class of a loop is its linking number with the knot, computed here by the Gauss linking integral:

1

for a meridian,

2

for a loop going twice around the strand,

0

for a loop that does not link. Drag the loop along the knot and its class stays put. For the unknot, the glassy disk is a fence for the class “linking number”: the signed piercings agree with the integral.

Exercises

Exercise Forms that tell 4-manifolds apart

The manifolds $S^{2} \times S^{2}$ and $CP^{2} # CP^{2}$ (the connected sum of $CP^{2}$ and its mirror image) both have Betti numbers $1, 0, 2, 0, 1 .$ Their intersection forms are $(0110)$ and $(10 0 - 1) .$ Show that no change of basis turns one form into the other, and conclude that the two manifolds are not homotopy equivalent.

Hint

Look at the numbers $Q (x, x)$ for integer vectors $x .$

Solution

For the first form, $Q (x, x) = 2 x_{1} x_{2}$ is always even. For the second, $Q ((1, 0), (1, 0)) = 1$ is odd. A change of basis over $Z$ replaces $x$ by another integer vector but does not change the set of values $Q (x, x),$ so an “even” form cannot become an “odd” one. A homotopy equivalence would induce an isomorphism of cohomology rings preserving the fundamental class up to sign, hence an equivalence of the forms up to sign — and $- Q$ is even exactly when $Q$ is. So the manifolds are not homotopy equivalent. (Both forms have signature $0;$ evenness is what tells them apart.)

Summary

Recap

A closed orientable $n$ -manifold has a fundamental class $[M] \in H_{n} (M; Z) ≅ Z :$ all top simplices, coherently oriented. Every closed manifold has one mod 2.
The dual cell decomposition trades each $k$ -simplex for an $(n - k)$ -cell that crosses it once; it reverses the face relation and turns coboundary maps into boundary maps (up to sign).
Poincaré duality: $H^{k} (M; Z) ≅ H_{n - k} (M; Z)$ for closed orientable $M,$ and $H^{k} (M; Z /2) ≅ H_{n - k} (M; Z /2)$ for every closed $M .$ The map is the cap product with the fundamental class, $D (φ) = [M] ⌢ φ .$
Consequences: $b_{k} = b_{n - k};$ $H_{n} ≅ Z;$ $χ (M) = 0$ in odd dimensions; torsion pairs up in complementary degrees.
On a surface, the Poincaré dual of a loop is its fence; intersection numbers count signed crossings, and $⟨ PD [C_{1}] ⌣ PD [C_{2}], [M]⟩ = C_{1} \cdot C_{2} .$ On the torus, $(p_{1}, q_{1}) \cdot (p_{2}, q_{2}) = p_{1} q_{2} - p_{2} q_{1} .$
In dimension $4,$ the cup product on $H^{2}$ is the intersection form; its rank, signature and parity are powerful invariants of 4-manifolds.
Lefschetz duality handles boundaries ( $H^{k} (M, \partial M) ≅ H_{n - k} (M));$ Alexander duality handles complements ( $H ~_{i} (S^{n} ∖ K) ≅ H ~^{n - i - 1} (K))$ — every knot complement has $H_{1} ≅ Z,$ generated by a meridian, and the class of a loop is its linking number.

Poincaré Duality