4.8 Curvature and Characteristic Classes

The excess of the sum of the angles of a triangle formed by shortest lines over two right angles is equal to the total curvature of the triangle.

— Carl Friedrich Gauss, abstract of General Investigations of Curved Surfaces (1827), trans. Morehead & Hiltebeitel

Try to comb the hair on a coconut so that it lies flat everywhere. You will fail: somewhere a tuft will stand up, or the hair will swirl around a bald spot. Now take a lump of clay shaped like a ball and squash it, dent it, pull out a nose. Its curvature changes everywhere — yet, added up over the whole surface, it is always exactly the same number.

Both facts are instances of one theme: local geometry knows global topology. A vector field is a choice of arrow at every point; curvature is a measurement at every point. Neither seems to know anything about the overall shape. Yet the zeros of every vector field, counted properly, add up to the Euler characteristic, and so does the total curvature, divided by $2 π .$ The bookkeeping that makes these facts precise is a family of cohomology classes, the characteristic classes, which measure how a family of vector spaces twists over a space. They also give a famous argument for why electric charge should come in whole units, and they explain why certain electrical measurements are exact integers.

Families of vector spaces

Stand at a point $p$ of a sphere. The directions in which you can walk without leaving the surface form a flat plane touching the sphere at $p :$ the tangent plane $T_{p} S^{2}$ of §2.4 Manifolds and Surfaces. Every point has its own tangent plane, and as $p$ moves the plane tilts with it. Each one is a copy of $R^{2},$ a vector space in which arrows can be added and scaled, but there is no single, natural way to identify the plane at the north pole with the plane at the equator. This collection of planes, one attached at every point and varying continuously, is called the tangent bundle $T S^{2} .$

A simpler example lives over a circle. Attach to every point of a circle a short segment, a copy of the real line $R$ (we draw only a piece of it). If all the segments stand upright, they sweep out a cylinder. If instead they turn slowly as we go round, so that after one full turn the segment has turned upside down, they sweep out a Möbius band. Look at either one near any point and you see the same thing — a little strip, a piece of circle times a piece of line. The difference is only global.

The cylinder $S^{1} \times R$ is the trivial line bundle over the circle. The Möbius band is a line bundle over the circle too, and we are about to prove that it is not trivial. The tangent bundle of a surface is a bundle of rank $2 .$

Local triviality is where the previous chapter comes in. Cover the base by open sets $U_{i}$ over which the bundle is a product, with coordinates chosen on each. On an overlap $U_{i} \cap U_{j}$ a fibre gets two coordinate systems, related by an invertible linear map $g_{ij} (b),$ an invertible $n \times n$ matrix depending continuously on $b .$ These transition functions satisfy $g_{ik} = g_{ij} g_{j k}$ on triple overlaps — a Čech cocycle condition, written multiplicatively. For a line bundle each $g_{ij} (b)$ is a nonzero number. For the Möbius band, covered by two arcs whose overlap has two pieces, we can take $g = + 1$ on one piece and $g = - 1$ on the other; for the cylinder, $+ 1$ on both. Changing coordinates on the pieces changes $g$ by a coboundary, exactly as in §4.7 Sheaves and Čech Cohomology.

Sections: choosing a vector at every point

Here is a test that tells the cylinder from the Möbius band. Take a pencil and draw a line once around the band, crossing every fibre exactly once, without ever touching the core circle that runs down the middle. On the cylinder it is easy: draw a line parallel to the core. On the Möbius band, try it in the figure below. The line you are drawing has a name.

Every bundle has the zero section. The interesting question is the opposite one:

For a trivial bundle the answer is yes: take a constant nonzero vector in $B \times R^{n} .$ So whenever the answer is no, the bundle must be twisted — and the pencil test suggests that the Möbius band is the first example.

Sections of the cylinder and the Möbius band Drag the gold points in the strip · switch bundles · drag the 3D view

1 zero — always an odd number on the Möbius band, so never none.

Drag the gold points in the strip

Figure 4.8.1 Above, the bundle: each blue segment is a fibre, a copy of

R

whose midpoint is

0;

the golden curve is a section. Below, the same band cut open along one fibre and laid flat. For the cylinder the two cut edges are glued directly; for the Möbius band they are glued upside down, so a curve that starts above the middle must arrive below it. Zeros of the section glow red.

For line bundles the question is decisive: a line bundle has a nowhere-zero section exactly when it is trivial. (Given such a section $s,$ every vector in the fibre over $b$ is $t s (b)$ for exactly one number $t,$ and $(b, t)$ are coordinates making the bundle a product.) So the Möbius band really is twisted, and in a moment we will see that it is the only twisted line bundle over the circle.

There is a cohomology class that tells them apart. Only the sign of a transition function matters topologically (a positive number can be slid continuously to $1,$ but never through $0$ to a negative one), so a line bundle gives a Čech cocycle of signs $\pm 1,$ and its class lives in $H ˇ^{1} (B; {\pm 1}) = H^{1} (B; Z /2) .$

For the Möbius band, $w_{1}$ is the nonzero element of $H^{1} (S^{1}; Z /2) ≅ Z /2$ — the very class you met as the orientation sheaf’s obstruction in §4.7 Sheaves and Čech Cohomology. In fact real line bundles over a reasonable space $B$ correspond exactly to elements of $H^{1} (B; Z /2)$ (Hatcher 2017, Prop. 3.10): there are $2$ over a circle (the cylinder and the Möbius band), $4$ over a torus ( $H^{1} (T^{2}; Z /2) ≅ (Z /2)^{2}),$ and only the trivial one over a sphere.

A little history From vector fields to a rented computer

Eduard Stiefel 1909 – 1978

A Zürich painter’s son, he wrote his thesis under Heinz Hopf in 1935 on fields of directions on manifolds: how many vector fields, independent at every point, can a manifold carry? The obstructions he found, generalising the Euler characteristic, were the first characteristic classes, discovered at the same time by Hassler Whitney and now named after both. He served in the Swiss army, rising to colonel, and was so often away on duty that Olga Taussky-Todd, advised by Hopf to talk to him, kept missing him. In 1948 his career turned: he founded an institute for applied mathematics at ETH Zürich, and in 1949, hearing that Konrad Zuse’s Z4 computer was standing in a German Alpine village, he went to see it and rented it. Among the methods his group developed is the conjugate gradient algorithm (O’Connor & Robertson 2008).

Combing a hairy ball

Now take the tangent bundle of the sphere. A nowhere-zero section would be a way to comb hair on a ball with no cowlick and no bald spot. In §3.5 Maps, Invariance, and First Triumphs you proved, using homology, that this is impossible: every continuous tangent vector field on $S^{2}$ vanishes somewhere. On a torus, by contrast, you can comb the hair flat around the hole. Here is the picture. In the next two sections we find out what the zeros of a field must add up to, counted properly, and why.

Combing the sphere and the torus Choose a field · drag to rotate

Zeros: centre (+1), centre (+1).

sum of indices = 1 + 1 = 2 = χ (S^{2})

Figure 4.8.2 Hairs combed along a tangent vector field, with particles drifting along it. Every field on the sphere has zeros — two centres, a source and a sink, or a single double cowlick — and their indices always add up to

2 .

On the torus the hair can lie flat; a steady wind blowing across a doughnut creates four zeros, whose indices add up to

0 .

The index of a zero

A zero of a vector field can look like many things: a source (everything flows out), a sink (everything flows in), a centre (everything circulates), a saddle (flowing in along one line and out along another). To measure a zero by a single number, draw a small loop around it that encloses no other zero, walk once around the loop anticlockwise, and watch the arrow of the field. It comes back to where it started, so it has turned a whole number of times: the winding number of §2.3 Homotopy, now applied to the direction of the field.

Walking around a source, the arrow always points away from the centre, so it turns once with you: index $+ 1 .$ The same happens for a sink (the arrow points inwards and still turns once) and for a centre (the arrow points sideways, and again turns once). At a saddle the arrow turns once the other way: index $- 1 .$ There are zeros of index $+ 2$ (a dipole, like the field lines of a bar magnet seen from far away) and $- 2$ (a monkey saddle), and of every other integer. The index does not depend on the size of the loop: shrinking it changes the winding number continuously, and an integer that changes continuously does not change.

Index explorer Choose a zero and walk around it · in the playground, drag zeros and the loop

Walking once around the loop, the field arrow turns 0.00 times so far — a full walk gives the index,

ind = 1

start once around

Figure 4.8.3 The colour shows the direction of the field (gold points right, violet up-left, teal down-left…), so around a zero of index

k

the colours go round

k

times — backwards for negative

k .

In the playground, the winding number of the dashed loop is always the sum of the indices of the zeros inside it, and a zero at infinity tops the total up to

2 .

The playground reveals the key additivity property. A big loop around several zeros has winding number equal to the sum of their indices: cut the region inside the loop into small pieces, each around one zero, and the turnings along the interior cuts cancel because each cut is walked twice, in opposite directions. It also reveals something stranger. Think of the plane as a sphere with one point missing — stereographic projection from §2.4 Manifolds and Surfaces — and fill that point back in. The field acquires one more zero, at infinity, and its index is $2 - w$ where $w$ is the sum of the finite indices. However you arrange the zeros, the total on the sphere is $2 .$

Poincaré–Hopf: the zeros know the shape

Add up the indices in Figure 4.8.2. On the sphere every field comes to $2 :$ spinning about an axis gives two centres, $1 + 1;$ flowing from pole to pole gives a source and a sink, $1 + 1;$ the dipole has a single zero of index $2 .$ On the torus the total is $0 :$ the field around the hole has no zeros at all, and the wind blowing across the doughnut has a source and a sink on the outside and two saddles on the inside, $1 + 1 - 1 - 1 .$ Two surfaces, two totals, and neither budges however the hair is combed. You have seen both numbers before: $χ (S^{2}) = 2$ and $χ (T^{2}) = 0 .$

(The same statement holds for closed manifolds of any dimension, with the index defined using spheres instead of loops.) Henri Poincaré proved the theorem for surfaces (Poincaré 1885), and Heinz Hopf proved it in every dimension (Hopf 1927); the place to read about it is Milnor (1965, §6), a slim classic.

The theorem makes predictions about surfaces we have not combed. A double torus has $χ = - 2 .$ Stand it on end and let water run down it: there is a source at the top, a sink at the bottom and four saddles, one at the top and bottom of each hole, and indeed $1 + 1 - 4 = - 2 .$

Why it is true: a sketch

A proof has two parts. First, the total index is the same for every vector field on $M .$ Two fields can be deformed into each other; during the deformation the zeros move, and they can collide, merge or split, but the total index never jumps — a loop surrounding all the zeros involved in a collision has a winding number that changes continuously, hence not at all. (Making this airtight takes care, but this is the heart of it.) Second, compute the total for one cleverly chosen field. Triangulate $M$ (§2.5 Simplicial Complexes) and let everything flow away from the vertices, along the edges to the midpoints of the edges, and then into the middles of the triangles.

The field that counts V − E + F

Figure 4.8.4 On a triangulated surface, put a source at every vertex (index

+ 1),

a saddle at the middle of every edge (index

- 1 :

flow arrives from the two ends and leaves towards the two neighbouring faces) and a sink at the centre of every face (index

+ 1).

The total index is

V - E + F = χ (M) .

This field has one zero at each vertex with index $+ 1,$ one at each edge with index $- 1$ and one at each face with index $+ 1 .$ So its total index is $V - E + F,$ which is $χ (M)$ by §2.6 The Euler Characteristic. By the first part, every field has the same total. That is Poincaré–Hopf.

The Euler class

Poincaré–Hopf says that the number “sum of indices of a section” does not depend on the section. Nothing in that idea needs the planes to be tangent planes: any family of planes over a surface has sections, their zeros have indices, and the same count can be made. The total is the first example of a characteristic class, and it turns the question “is there a nowhere-zero section?” into arithmetic. Here it is for oriented bundles of rank $2$ over a closed oriented surface $M$ (the tangent bundle of an orientable surface is one).

Two more descriptions show why $e$ deserves to be a cohomology class. First, it is an obstruction cocycle: try to build a nowhere-zero section one cell at a time. Over the vertices and edges of a triangulation it is easy. Over each triangle, the section is already defined on the boundary loop, and it extends without zeros across the triangle exactly when its winding number around that loop is $0 .$ Those winding numbers form a $2$ -cochain, in fact a $2$ -cocycle, and its class is $e (E) .$ Second, by Poincaré duality (§4.6 Poincaré Duality), the Euler class is the cohomology class dual to the zero set of a generic section, counted with signs. In every dimension, an oriented bundle of rank $n$ has an Euler class in $H^{n} :$ it is the first obstruction to a nowhere-zero section, it is dual to the zeros of a generic section, and for the tangent bundle of a closed oriented manifold it adds up to $χ (M)$ (Milnor & Stasheff 1974, §9, Cor. 11.12, Thm 12.5).

Curvature from scratch

The second half of our theme starts with a more elementary notion. How curved is a curve?

Curves: turning per unit length

Drive along a road. On a straight stretch your steering wheel is centred; on a bend you turn it. The curvature $κ$ of a curve at a point is the rate at which its direction turns, per unit of distance travelled: $κ = \frac{d θ}{d s},$ where $θ$ is the angle of the tangent direction and $s$ is arc length. A straight line has $κ = 0 .$ A circle of radius $r$ turns through $2 π$ radians in a length of $2 π r,$ so $κ = 1/ r :$ small circles are very curved, huge ones nearly straight. For a plane curve we keep a sign: positive for left turns, negative for right.

Add up the curvature all the way around a closed curve and you get the total angle through which its tangent turns, $\oint κ d s .$ Since the direction returns to where it started, the total is a whole number of full turns.

The total turning of a closed curve Drag the gold points · try the presets

The tangent turns through 360° in all: 1 full turn. A simple closed curve, traversed anticlockwise: always exactly one turn.

Drag the gold points

Figure 4.8.5 The comb shows the curvature along the curve: gold teeth where it turns left, teal where it turns right, longer where it turns more sharply. However you drag the points, the dial shows a whole number of turns:

+ 1

for a simple loop traversed anticlockwise,

0

for a figure eight,

2

with an extra loop.

This is a one-dimensional preview of Gauss–Bonnet: a local quantity, $κ,$ adds up to a topological one. For a polygon, all the turning happens at the corners, and the theorem becomes a school fact: the exterior angles of a convex polygon add up to $36 0^{\circ} .$

Surfaces: two curvatures and their product

At a point $p$ of a surface, slice the surface with planes that contain the normal line at $p .$ Each slice is a curve through $p$ with its own curvature, the normal curvature in that direction (counted positive if the curve bends towards the chosen normal). As the slicing plane turns around the normal, the normal curvature varies between a largest value $k_{1}$ and a smallest $k_{2},$ the principal curvatures; Leonhard Euler showed, in a memoir written in 1763, that they occur in perpendicular directions (Euler 1767).

Sphere of radius $r$ : every slice is a great circle, $k_{1} = k_{2} = 1/ r,$ so $K = 1/ r^{2} > 0 .$
Cylinder of radius $r$ : around the cylinder $k = 1/ r,$ along it $k = 0$ (a straight line); so $K = 0 .$ A cylinder is a rolled-up sheet of paper, and curvature in only one direction does not count.
Saddle (a mountain pass, or a Pringle): it curves up in one direction and down in the perpendicular one, so $k_{1} > 0 > k_{2}$ and $K < 0 .$
Torus: the outer half is curved like a ball ( $K > 0),$ the inner half, around the hole, like a saddle ( $K < 0),$ and the top and bottom circles have $K = 0 .$

Flipping the normal changes the sign of both $k_{1}$ and $k_{2},$ so their product $K$ does not care which side is which: it makes sense even on a non-orientable surface. In 1827 Gauss discovered something much deeper about it.

“If a curved surface is developed upon any other surface whatever, the measure of curvature in each point remains unchanged.” — Gauss (1827, art. 12)

“Developed” means bent without stretching, as you can bend paper but not a ping-pong ball. Gauss’s Theorema Egregium (“remarkable theorem”) says that $K,$ although we defined it by looking at the surface from outside, can be measured by an ant who lives in the surface and only measures lengths and angles there. It explains why a slice of pizza stiffens when you fold it across (once bent one way, $K = 0$ forbids bending the other way too), and why no flat map of the Earth can be perfect: the sphere has $K > 0,$ the paper $K = 0,$ and no amount of bending without stretching can change that.

Triangles on curved surfaces

How could the ant measure $K ?$ With triangles. On a surface, the straightest possible paths are called geodesics; on a sphere they are arcs of great circles, like the routes of long-distance flights. A triangle whose sides are geodesics is a geodesic triangle. In the plane its angles add up to $π$ (180°). On a sphere they add up to more.

Angle excess on a sphere Drag the gold corners · drag elsewhere to turn the sphere

α + β + γ = 90. 0^{\circ} + 90. 0^{\circ} + 90. 0^{\circ} = 270. 0^{\circ}

excess

α + β + γ - π = 1.571

area ÷ R²

area / R^{2} = \iint_{T} K d A = 1.571

The triangle covers 12.5% of the sphere. Drag a corner: the two numbers move together, always equal.

Drag a gold corner to move it · drag elsewhere to turn the sphere

Figure 4.8.6 A geodesic triangle on a sphere of radius

R .

Its angles add up to more than

18 0^{\circ},

and the excess

α + β + γ - π

equals the area divided by

R^{2},

which is the curvature

K = 1/ R^{2}

integrated over the triangle. The octant triangle has three right angles: excess

π /2,

and area one eighth of

4 π R^{2} .

The symbol $\iint_{T} K d A$ means: chop $T$ into tiny pieces, multiply the curvature at each piece by its area, and add up — the total curvature of $T .$ Gauss wrote that this theorem, “if we mistake not, ought to be counted among the most elegant in the theory of curved surfaces” (Gauss 1827, art. 20). It is the local seed of everything that follows.

The Gauss–Bonnet theorem

Now cover a whole closed surface with geodesic triangles and add up Gauss’s formula over all of them. The right-hand sides add up to the total curvature of the surface. On the left, the angles that meet at each vertex fill a full turn, $2 π,$ so all the angles together come to $2 π V,$ and the left-hand sides add up to $2 π V - π F .$ A little counting, done in the proof below, turns that into $2 π (V - E + F) .$ The shape has dropped out. Needham (2021) gives four different geometric proofs of the result.

Round sphere of radius $r :$ $K = 1/ r^{2}$ and the area is $4 π r^{2},$ so the total is $4 π = 2 π \cdot 2 .$
Any deformed sphere — a potato, an egg, a dumbbell with a thin neck — also has total curvature exactly $4 π .$ Where you add positive curvature by pulling out a bump, negative curvature appears around its base to compensate.
Torus: the positive curvature of the outer half and the negative curvature of the inner half cancel exactly, $2 π \cdot 0 = 0 .$
Surface of genus $g$ : $2 π (2 - 2 g);$ for the double torus, $- 4 π .$

The polyhedral version, proved

For surfaces made of flat triangles, all the curvature is concentrated at the vertices, and Gauss–Bonnet becomes exact arithmetic. At a vertex of a polyhedron, the triangles meeting there have angles adding up to less than $2 π$ (on a convex corner) or more (on a saddle-like corner).

A cube has three right angles at each corner, defect $2 π - 3 π /2 = π /2,$ and eight corners: total $4 π .$ René Descartes noticed, around 1630, that the total defect of every convex polyhedron is $4 π$ (in his units, eight right angles) (Richeson 2008). Here is the general statement, with its complete proof.

Nothing in the proof used the shape: only that the triangles are flat and fit together into a closed surface. The angle defect at a vertex, divided by the area around the vertex, is a good estimate of the Gaussian curvature there, and as the triangles get finer the sum of defects becomes the integral $\iint K d A$ (Crane 2025, §5.5). Now play.

The Gauss–Bonnet sculptor Play the deformation · click the surface to push or pull it · drag to rotate

total curvature ÷ 2π 2.000 = χ = 2

positive part —

negative part —

V - E + F = 10242 - 30720 + 20480 = 2

round sculpted

Figure 4.8.7 A sphere made of 20,480 flat triangles (or a torus of 28,800), coloured by curvature — angle defect per unit area: gold where the surface curves like a ball, teal where it curves like a saddle, a bright line where the curvature changes sign. Pull out bumps, ripple it, squash it, twist it: the colours change everywhere, the positive and negative parts swell and shrink, and the total stays exactly

2 π χ :

4 π

for the sphere, where

χ = 2,

and

0

for the torus.

Watch the two bars under the figure. Pulling out a bump adds positive curvature at its tip and creates a ring of negative curvature around its base; the two changes cancel exactly, to the last decimal place. Only a change of topology — which no amount of pushing, pulling or twisting can make — could change the total.

Why curvature is a cohomology class

Here is the point of view that generalises. On an oriented surface $M,$ the expression $K d A$ is a $2$ -form (§4.3 Differential Forms). On a surface every $2$ -form is closed, so it has a de Rham cohomology class in $H_{dR}^{2} (M)$ (§4.4 de Rham Cohomology). Changing the shape (the way lengths are measured) changes $K d A$ only by an exact form $d η,$ and by Stokes’ theorem an exact form integrates to zero over a closed surface. So the class of $\frac{1}{2 π} K d A$ does not depend on the shape at all: it is the Euler class of the tangent bundle, and Gauss–Bonnet says $\frac{1}{2 π} \iint_{M} K d A = e (T M) = χ (M) .$

This idea — curvature, a local quantity, represents a characteristic class, a global one — is called Chern–Weil theory. In 1944 Shiing-Shen Chern gave an intrinsic proof of the Gauss–Bonnet formula for manifolds of every even dimension (Chern 1944), and in 1946 he introduced the characteristic classes of complex vector bundles that now bear his name (Chern 1946). The standard short account is the appendix on connections and curvature in Milnor & Stasheff (1974, App. C).

In higher dimensions Chern’s Gauss–Bonnet formula integrates a more complicated expression in the curvature, and it is natural to ask whether the sign of the curvature still decides the sign of $χ,$ as it does for surfaces: a sphere curves positively and has $χ = 2,$ a surface of genus $2$ can be curved negatively everywhere and has $χ = - 2 .$

Conjecture Hopf’s sign conjecture

Open since the 1930s Posed by Heinz Hopf, the 1930s

If a closed manifold $M$ of dimension $2 n$ has a metric whose curvature is negative in every direction (negative sectional curvature), then $(- 1)^{n} χ (M) > 0;$ if the curvature is only nonpositive, then $(- 1)^{n} χ (M) \geq 0 .$

In dimension $2,$ $n = 1$ and the conjecture says $- χ > 0$ for negatively curved surfaces, which Gauss–Bonnet proves in one line: the integral of a negative function is negative. From dimension $6$ on, the expression Chern integrates no longer has a sign that curvature controls, and nobody has found another way.

What is known. For surfaces it follows at once from Gauss–Bonnet, and in dimension $4$ from Chern’s version of it, whose integrand then has the right sign. It is known for several special classes, such as Kähler manifolds of negative curvature (Gromov, 1991), and open from dimension $6$ on. Hopf also asked whether positive curvature forces $χ > 0,$ and whether $S^{2} \times S^{2}$ can be curved positively in every direction; in August 2026 Simon Brendle and Pei-Ken Hung posted a metric of positive curvature on $S^{2} \times S^{2},$ which experts are now checking.

A way in. Check the surfaces: a round sphere (positive curvature, $χ = 2),$ a flat torus ( $χ = 0)$ and a surface of genus $2$ with negative curvature ( $χ = - 2).$ In dimension $4,$ the product of two surfaces of genus $2$ has nonpositive curvature; compute its Euler characteristic with $χ (M \times N) = χ (M) χ (N)$ and compare with the prediction.

(Chern 1944; Brendle & Hung 2026) Status as of October 2026 All open problems

A little history A week through Africa and South America

Shiing-Shen Chern 1911 – 2004

Taught mathematics as a boy by his father, a lawyer with a classical Confucian education, he was one of a class of four mathematics students at Nankai University in Tianjin. In Paris in 1936–37 he worked with Élie Cartan, from whom, he remembered, a letter usually came the day after each meeting: “After you left, I thought more about your questions …”. He came home in 1937 to a country at war with Japan, and taught at a university that had fled to Kunming. Invited to Princeton, he was given a seat on an almost empty US Air Force plane from Calcutta to Miami; the trip took a week, through Africa and South America. Both pieces of work above belong to the two years that followed, 1943–45, in Princeton (O’Connor & Robertson 2014).

Characteristic classes

Follow the sphere through the chapter so far. Its number $2$ turned up as the Euler characteristic $V - E + F,$ as the total index of every combing (Poincaré–Hopf), as the Euler class $e (T S^{2}),$ and as the total curvature divided by $2 π$ (Gauss–Bonnet). A count of cells, a count of cowlicks, a cohomology class and an integral: the theorems of this chapter say they are one invariant, the Euler class of the tangent bundle, which measures how the tangent planes twist as they are carried around the sphere. The Möbius band gave us a second invariant of the same kind, $w_{1} .$ Both belong to a family.

class	for	lives in	measures
Stiefel–Whitney $w_{1}$	real bundles	$H^{1} (B; Z /2)$	orientability (the Möbius band)
Stiefel–Whitney $w_{i}$	real bundles	$H^{i} (B; Z /2)$	finer twisting; an oriented bundle has a spin structure exactly when $w_{2} = 0$
Euler $e$	oriented real bundles of rank $n$	$H^{n} (B; Z)$	nowhere-zero sections; $e (T M) = χ$ (hairy ball)
Chern $c_{1}$	complex line bundles	$H^{2} (B; Z)$	twisting of phases (monopoles, quantum Hall effect)
Chern $c_{i}$	complex bundles of rank $n,$ $i \leq n$	$H^{2 i} (B; Z)$	the top one, $c_{n},$ is the Euler class of the bundle seen as a real one
Pontryagin $p_{i}$	real bundles	$H^{4 i} (B; Z)$	for a closed 4-manifold, $⟨ p_{1} (T M), [M]⟩$ is three times the signature

The higher classes $w_{i},$ $c_{i}$ and $p_{i}$ get no more than their row in this book; Milnor & Stasheff (1974, §4, §9, §14, §15) is the classic guide to all of them. We look more closely at the one that physics needs most, $c_{1} .$

Characteristic classes also forbid structures. A manifold is almost complex if each of its tangent spaces can be made into a complex vector space, smoothly from point to point; then its tangent bundle has Chern classes, and they must satisfy conditions that many manifolds fail. Among the spheres, only $S^{2}$ and $S^{6}$ pass. The 2-sphere is even a genuine complex manifold, the Riemann sphere, with complex coordinates that fit together. Whether $S^{6}$ is one was a question for almost eighty years.

Problem A complex structure on the 6-sphere

Solution claimed in 2026 Posed by Heinz Hopf, the 1940s; claimed by Levent Alpöge, working with the AI model Claude, 2026

Can the 6-sphere be given complex coordinates: is $S^{6}$ a complex manifold of complex dimension $3 ?$

An almost complex structure lives in each tangent space separately; a complex structure needs coordinates. The gap between the two is exactly what made the question hard.

How it was settled. On 23 August 2026 Alpöge, a mathematician at Anthropic, posted a 108-page manuscript, written with Claude, that builds a complex structure on $S^{6}$ out of a family of complex 2-dimensional tori lying over a sphere, completed over three special points. Within weeks other geometers had used it to construct further complex structures on $S^{6} .$ As of October 2026 it has not been refereed.

What is known. Using characteristic classes, Borel and Serre showed in 1953 that among all spheres only $S^{2}$ and $S^{6}$ admit even an almost complex structure, a rotation $J$ of every tangent space with $J^{2} = - 1 .$ The 2-sphere is a complex manifold, the Riemann sphere. The 6-sphere has an almost complex structure coming from the octonions, but it does not come from complex coordinates, and LeBrun showed in 1987 that no complex structure can fit the round shape of the sphere. Michael Atiyah announced a proof that none exists at all in 2016; it was not accepted, and the 2026 construction, if correct, shows the answer is yes.

A way in. Give the 2-sphere complex coordinates: use stereographic projection from the north pole and from the south pole, and check that the change of coordinates between the two charts is $z \mapsto 1/ z,$ which is complex-differentiable where both charts apply. The question was whether anything like this can be done for $S^{6} .$

(Borel & Serre 1953; Alpöge 2026; Viaclovsky 2026) Status as of October 2026 All open problems

Complex line bundles and the first Chern class

A complex line bundle has fibres that are copies of the complex numbers $C .$ Its transition functions are nonzero complex numbers $g_{ij} (b),$ and topologically only their direction matters — a point of the unit circle, $e^{i α},$ a phase. Over the sphere, cover $S^{2}$ by a northern patch $U_{N}$ and a southern patch $U_{S},$ each a disk, overlapping in a band around the equator. Over each disk the bundle is trivial (a disk is contractible), so the whole bundle is described by a single transition function on the band, which up to deformation is a map from the equator to the circle of phases, $g_{N S} (φ) = e^{in φ},$ where $φ$ is the longitude and $n$ is an integer: the number of times the phase winds around as you go once around the equator. It must be an integer because $g_{N S}$ has to return to its starting value after a full turn.

The tangent bundle of the sphere, with each tangent plane regarded as a copy of $C$ (rotating by $9 0^{\circ}$ is multiplying by $i),$ is a complex line bundle with Chern number $2$ — once more, $χ (S^{2}) .$ And Chern–Weil theory says that $c_{1}$ can be computed from a curvature, just as Gauss–Bonnet computes $e :$ a way of comparing phases in nearby fibres (a connection) has a curvature $2$ -form $F,$ and $\frac{1}{2 π} \iint F$ is the Chern number. We have not defined connections in this book; think of $F$ as a curvature density, like $K d A .$ Physicists know it better as a magnetic field.

Monopoles, Berry phases and topological matter

This section is a guided tour rather than a course, with simplifications flagged as we go. All three stories are about complex line bundles over the two surfaces we know best, the sphere and the torus, and in each of them a Chern number becomes something nature measures.

Dirac’s monopole and the quantisation of charge

A magnet always has two poles; cut it in half and you get two smaller magnets. But nothing in the equations of electromagnetism forbids an isolated magnetic pole, a magnetic monopole, whose field points straight out like the electric field of a point charge: $B = g \frac{r ^}{r ^{2}}, flux through any sphere around it = 4 π g .$ Away from the monopole $B$ has zero divergence, so as a $2$ -form it is closed — exactly the non-exact form on $R^{3} ∖ {0}$ that you met in §4.4 de Rham Cohomology.

Quantum mechanics describes magnetism not by $B$ but by a vector potential $A$ with $curl A = B :$ the potential is what shifts the phase of a charged particle’s wavefunction. But no such $A$ can exist on a whole sphere around the monopole. If it did, Stokes’ theorem would give $\iint_{S^{2}} B \cdot d S = \iint_{S^{2}} curl A \cdot d S = 0$ (a closed surface has no boundary), contradicting the flux $4 π g .$ In 1931 Paul Dirac found a way around this (Dirac 1931). In 1975 Tai Tsun Wu and Chen Ning Yang gave it its cleanest form (Wu & Yang 1975), with two patches: use one potential $A_{N}$ on the northern patch and another, $A_{S},$ on the southern patch. On the overlap, the two describe the same field, so they differ by a gradient, $A_{N} - A_{S} = \nabla χ, χ = 2 g φ,$ (here $χ$ is a function, not the Euler characteristic) and the wavefunctions of a particle of charge $q$ in the two descriptions differ by the phase $e^{i q χ /ℏ}$ (in units where the speed of light is $1).$

A monopole in two patches Change the charge · let it take any value · drag to rotate

g_{N S} (φ) = e^{i φ}

winding number = c_{1} = 1

flux 4 π g = 2 π ℏ/ q

The phase comes back to itself after one turn, so the two patches fit together: a legitimate monopole with Chern number 1.

Monopole charge n 1 Let n be any number

Figure 4.8.8 Field lines stream out of the monopole. The northern patch (violet) and southern patch (teal) overlap in a band where the two descriptions are related by the phase

e^{in φ},

drawn as a ribbon that twists

n

times around the equator. For a whole number

n

the ribbon closes up; let

n

take other values and it tears — the phase is no longer single-valued.

A wavefunction must have a single value at each point, so the phase $e^{i q χ /ℏ} = e^{i (2 q g /ℏ) φ}$ must come back to itself after one turn around the equator. That forces $\frac{2 q g}{ℏ} = n \in Z .$ This is Dirac’s quantisation condition, and it is exactly the statement that the transition function is a map of winding number $n :$ the monopole is a complex line bundle over the sphere with Chern number $n .$ Now follow the consequence. If a single magnetic monopole exists anywhere in the universe, every electric charge $q$ must be a whole multiple of $ℏ/2 g :$ electric charge is quantised, as it is observed to be. No monopole has ever been found, so the explanation stays conditional, and physicists still hunt for monopoles in cosmic rays and in particle colliders.

Berry’s phase

Take a quantum system that depends on some external knobs — a spin sitting in a magnetic field whose direction you can turn. Turn the knobs slowly around a loop and back. The system ends where it started, except for a phase, and in 1984 Michael Berry showed that part of this phase is geometric: it depends only on the loop of knob settings, not on how fast you went around (Berry 1984). For a spin- $\frac{1}{2}$ particle it is half the solid angle swept out by the direction of the field. Barry Simon recognised in 1983, having seen Berry’s work before publication, that this phase is the holonomy of a connection on a complex line bundle over the space of knob settings: the phase a fibre picks up when it is carried around a loop (Simon 1983). Its curvature, the Berry curvature, integrates over a closed surface of settings to $2 π$ times a Chern number. For the spin it is the field of a monopole of strength $\frac{1}{2}$ sitting at the setting where the field vanishes: Dirac’s monopole, reappearing in the space of parameters.

The quantum Hall effect and topological matter

In 1980 Klaus von Klitzing measured the Hall resistance of a thin layer of electrons in a strong magnetic field at very low temperature and found plateaus at the values $h / (ν e^{2}),$ with $ν$ a whole number (von Klitzing et al. 1980), so precisely reproducible that the effect now serves as the international standard of electrical resistance. In 1982 David Thouless, Mahito Kohmoto, Peter Nightingale and Marcel den Nijs (TKNN) explained the integers for electrons in a crystal (Thouless et al. 1982), and the following year Simon recognised their formula as a Chern number. The possible momenta of an electron in a crystal form a torus (the Brillouin zone), each filled band of electron states is a complex line bundle over that torus, and the Hall conductance is $σ_{x y} = \frac{e ^{2}}{h} filled bands \sum C_{n},$ where $C_{n}$ is the Chern number of the $n$ -th band (signs depend on conventions). In units of $e^{2} / h,$ a number read off a laboratory instrument is a sum of characteristic numbers.

In 1988 Duncan Haldane showed that a suitably designed crystal could have bands with nonzero Chern number even without an overall magnetic field — a Chern insulator (Haldane 1988). Since about 2005, related invariants (not Chern numbers, but $Z /2$ -valued cousins suited to materials with time-reversal symmetry) have been found and observed in the materials called topological insulators, which insulate in their interior but conduct along their surfaces (Hasan & Kane 2010). In 2016 the Nobel Prize in Physics went to Thouless, Haldane and J. Michael Kosterlitz “for theoretical discoveries of topological phase transitions and topological phases of matter” (Nobel Prize 2016).

Exercises

Exercise Defects of polyhedra

(a) Check that the total angle defect of a regular octahedron and of a regular icosahedron is $4 π .$ (b) A “picture frame” torus is made by cutting a square hole through a thick square slab. Its outer box has 8 corners, each with three right angles; at each of the 8 corners of the hole, two faces of the hole meet the top or bottom face, which has an angle of $27 0^{\circ}$ there. Find the total defect.

Solution

(a) Octahedron: 6 vertices, four equilateral triangles each, defect $2 π - 4 π /3 = 2 π /3,$ total $4 π .$ Icosahedron: 12 vertices, five triangles each, defect $2 π - 5 π /3 = π /3,$ total $4 π .$

(b) Outer corners: defect $2 π - 3 \cdot π /2 = π /2$ each, $8 \cdot π /2 = 4 π .$ Inner corners: angles $π /2 + π /2 + 3 π /2 = 5 π /2,$ defect $- π /2$ each, $8 \cdot (- π /2) = - 4 π .$ Total $0 = 2 π χ (T^{2}) .$ (The faces here are not all triangles, and the top face is not even a disk, but cutting the faces into flat triangles does not change the angle sums at the corners, and any new vertices inside a face or on an edge have angles adding up to exactly $2 π,$ defect $0 .)$

Exercise Dirac’s argument

Suppose a monopole of strength $g$ exists and particles of charges $q_{1}$ and $q_{2}$ both satisfy Dirac’s condition $2 q_{k} g /ℏ \in Z .$ Show that every charge is a whole multiple of a smallest unit, and identify the unit. Why does the argument need only one monopole in the whole universe?

Solution

Dirac’s condition says $q_{k} = n_{k} \cdot \frac{ℏ}{2 g}$ with $n_{k} \in Z .$ So every charge is an integer multiple of $q_{0} = ℏ/ (2 g),$ and the ratio $q_{1} / q_{2} = n_{1} / n_{2}$ of any two charges is a rational number. (The charges that actually occur may all be multiples of a larger unit, but that unit is then itself a multiple of $q_{0} .)$ The condition comes from the requirement that each particle’s wavefunction be single-valued around a sphere enclosing the monopole, and such spheres exist wherever the particle is; so the existence of a single monopole anywhere constrains every charge everywhere.

Exercise Turning numbers

(a) Explain why the exterior angles of a convex polygon add up to $36 0^{\circ} .$ (b) What is the turning number of a figure eight, and why?

Solution

(a) Walking around the polygon anticlockwise, you turn left by the exterior angle at each corner and go straight along each side; after one lap you face your original direction, having turned exactly once: $36 0^{\circ} .$ (b) $0 .$ The two lobes are mirror images of each other in the vertical line through the crossing, and the curve runs through the second lobe as the mirror image of how it ran through the first. A reflection swaps left turns and right turns, so whatever the first lobe turns ( $- 27 0^{\circ}$ for the curve $(sin t, sin t cos t),$ starting at the crossing), the second lobe turns the opposite amount, and the total is $0 .$

Summary

Recap

A vector bundle is a continuously varying family of vector spaces, locally a product; its transition functions form a Čech cocycle. The cylinder is trivial; the Möbius band is twisted.
A section chooses a vector in every fibre. Every section of the Möbius band vanishes somewhere (intermediate value theorem); a line bundle is trivial exactly when it has a nowhere-zero section. Its twist is $w_{1} \in H^{1} (B; Z /2)$ .
Zeros of vector fields have indices (winding numbers): source, sink, centre $+ 1,$ saddle $- 1 .$ Poincaré–Hopf: the indices of any field on a closed surface add up to $χ (M);$ proof idea: a triangulation flow has total $V - E + F$ . The Euler class packages this as a cohomology class.
Curvature: $κ = d θ / d s$ for curves, $K = k_{1} k_{2}$ for surfaces (intrinsic, by the Theorema Egregium). Geodesic triangles have angle excess $\iint_{T} K d A .$
Gauss–Bonnet: $\iint_{M} K d A = 2 π χ (M),$ exactly true for angle defects of any triangulated surface; curvature represents the Euler class (Chern–Weil).
Chern classes: complex line bundles over $S^{2}$ are classified by the winding number $n$ of a transition function $e^{in φ} .$ Physics: Dirac’s monopole (charge quantisation), Berry’s phase, and the quantum Hall effect, where the measured conductance, in units of $e^{2} / h,$ is a sum of Chern numbers.

Curvature and Characteristic Classes

Families of vector spaces

Sections: choosing a vector at every point

Combing a hairy ball

The index of a zero

Poincaré–Hopf: the zeros know the shape

Why it is true: a sketch

The Euler class

Curvature from scratch

Curves: turning per unit length

Surfaces: two curvatures and their product

Triangles on curved surfaces

The Gauss–Bonnet theorem

The polyhedral version, proved

Why curvature is a cohomology class

Characteristic classes

Complex line bundles and the first Chern class

Monopoles, Berry phases and topological matter

Dirac’s monopole and the quantisation of charge

Berry’s phase

The quantum Hall effect and topological matter

Exercises

Summary

Further reading

References