Topology

In the last chapter rigour reached its limit: a proof became a chain of symbols that a machine can check, and it turned out that even such a machine is left with unanswerable questions. This chapter is built the other way round: lots of handwork and few symbols, like a session of a maths circle. You’ll need modelling clay, strips of paper, scissors, sticky tape, a marker pen, a piece of string and an old map.

A topologist, they say, is someone who can’t tell a coffee mug from a doughnut. Mould a mug with a handle out of clay and, without tearing off or sticking on a single bit, push in the bottom until the mug becomes a flat cake, then draw the cake into the handle. You get a doughnut. But a ball can’t be moulded into a doughnut that way: the hole would have to be punched through. Why are a mug and a doughnut “the same”, but a ball and a doughnut not? And what does “the same” even mean if the shapes have different sizes and bends?

Modelling clay: what “the same” means

The rules of the circle are these. A shape may be stretched, squeezed and bent as much as you like. You may not tear it or glue together points that were separate. You may cut it, provided you glue the cut back exactly as it was. The properties that survive all this are studied by topology; the word was coined by the German mathematician Johann Listing in 1847, from the Greek “topos”, place.

“No tearing” is the continuity of the chapter on limits, only with the distance between numbers replaced by the distance between points: a map $f$ is continuous at a point $x$ if for every $\varepsilon > 0$ there’s a $\delta > 0$ such that points closer than $\delta$ to $x$ go to points closer than $\varepsilon$ to $f(x)$.

Being one-to-one (a bijection) forbids gluing, continuity forbids tearing, and continuity of the inverse map prevents cheating at a seam. Wind the half-open interval $[0, 2\pi)$ round a circle: $t \mapsto (\cos t, \sin t)$. This is one-to-one and continuous, but the inverse map tears the circle at the point $(1, 0)$: points just above it go back near zero, points just below near $2\pi$.

Homeomorphisms can be surprising. A circle is homeomorphic to the outline of a square: project the outline onto the circle by rays from the centre. The interval $(-1, 1)$ is homeomorphic to the whole line: the map $x \mapsto \frac{x}{1 - |x|}$ takes $0.5$ to $1$, $0.9$ to $9$, $0.99$ to $99$, and the inverse, $y \mapsto \frac{y}{1 + |y|}$, pulls the line back. Length, straightness, even finiteness don’t interest a topologist. So how do you prove that shapes are not homeomorphic? Going through all the maps is impossible.

This device already came to our rescue in the chapter on Gödel, where the remainder of the number of letters I on division by $3$ proved that the word MU can’t be obtained. The simplest topological invariant is what happens to a shape if a point is removed from it.

The trick is to remove a point from each: the circle stays in one piece, while the segment falls into two.

Suppose a homeomorphism $f$ of the circle onto the segment $[0, 1]$ exists. It takes exactly one point of the circle to the midpoint $m = \frac12$ of the segment; call that point $p$. Remove $p$ from the circle. What’s left is an arc: from any of its points to any other you can travel without leaving it. Remove $m$ from the segment. It falls into two halves, $[0, \frac12)$ and $(\frac12, 1]$. Since $f$ is one-to-one, the arc without $p$ goes exactly onto the segment without $m$. Take a point $a$ of the arc that goes to the left half and a point $b$ that goes to the right half, and travel along the arc from $a$ to $b$. The coordinate of the image point changes continuously; at the start it’s less than $\frac12$, at the end greater. By the intermediate value theorem (chapter 25) at some moment it equals $\frac12$: some point of the arc went to $m$. But only $p$ goes to $m$, and it isn’t on the arc. A contradiction: the homeomorphism doesn’t exist. We used only that $f$ is continuous and one-to-one, so a circle can’t even be unrolled continuously into a segment without gluing.

Draw the letters as plain sticks, without serifs. Which of them is not homeomorphic to the letter T?

Y E F X

Remove a point from the letter and count the pieces. T can’t be split into more than three pieces: the worst point is where three sticks meet. Y, E and F also have one point where three sticks meet (for E and F, where the middle bar joins the stem) and three free ends, so they can be straightened into T. But in X four sticks meet at the centre, and without the centre it falls into four pieces.

A marker pen: a number that can’t be kneaded

Draw a grid on a ball with a marker pen: dots as vertices and lines as edges, so that the surface is divided into pieces, faces, with no holes inside. Whatever you draw, the number of vertices minus the number of edges plus the number of faces will be $2$. It’s Euler’s formula from the chapter on graphs: a grid on a sphere can be punctured in one face and stretched out over the plane.

Now draw $6$ rings along an inflatable ring and $4$ rings around it. They cross at $24$ points and cut the surface into $24$ quadrilaterals. Four edges leave each vertex, and each edge joins two vertices, so there are $24 \cdot 4 / 2 = 48$ edges. The result: $24 - 48 + 24 = 0$. Not $2$.

The number of vertices of the grid. The number of edges, the lines between vertices. The number of faces, the pieces into which the edges cut the surface. Each piece must be homeomorphic to a disc: an annulus doesn’t count as a face. Example: a square picture frame with a square cross-section is a torus. It has $16$ vertices, $32$ edges and $16$ faces, $\chi = 16 - 32 + 16 = 0$. A cube, that is, a sphere, has $8 - 12 + 6 = 2$.

The definition makes sense only if the number doesn’t depend on the grid. The widget below covers a surface with a real grid of thousands of triangles and honestly counts them. Change the fineness of the grid and knead the mug: $V$, $E$ and $F$ change by thousands, but their combination doesn’t.

The proof splits into an easy part and a hard part. The easy part: refining a grid doesn’t change the number.

Every refinement is made up of three small moves; let’s check each one.

Here’s a piece of a grid: two faces, five vertices, six edges. For a piece homeomorphic to a disc $V - E + F = 5 - 6 + 2 = 1$: it’s a sphere without one face. Move one: a new vertex $X$ on the edge $CD$. The edge splits in two: $V$ goes up by $1$, $E$ also by $1$, and $V - E + F$ doesn’t change. Move two: a new edge $XB$ across a face, between two vertices of its boundary. It cuts the face in two: $E$ and $F$ go up by $1$, and the sum stays the same. Move three: a new vertex $Y$ inside a face, joined by an edge to the vertex $A$ of its boundary. $V$ and $E$ go up by $1$. The face with such a slit is still one face and still homeomorphic to a disc: the slit can be opened out, like a slit in cloth. Each new vertex lies on an old edge or inside a face, and each new edge runs inside some face, so any refinement is a chain of such moves, and $V - E + F$ is preserved at each one.

The hard part is finding a common refinement of any two grids. If the edges of the two grids cross at finitely many points, their union is a common refinement. But an edge of one grid may cross an edge of the other infinitely many times, as the graph of $x \sin\frac{1}{x}$ crosses the axis near zero, and then the grid has to be carefully wiggled first. Today invariance is usually derived from homology, a tool of algebraic topology that lies beyond the course.

Since the number doesn’t depend on the grid, it doesn’t change under a homeomorphism either: a grid can be carried over to another surface along with the map. The sphere ($\chi = 2$) isn’t homeomorphic to the torus ($\chi = 0$), nor the torus to the pretzel ($\chi = -2$). The number of handles, that is, through-holes, as in a doughnut, is called the genus of the surface.

The genus: the number of handles. The sphere has $g = 0$, the torus $g = 1$. Example: the pretzel has $\chi = 2 - 2 \cdot 2 = -2$, the surface with three holes $-4$. Each handle takes away two; we’ll prove this once we learn to attach handles.

A square frame $3 \times 3 \times 1$ without the central cube is built from eight cubes. The surface of the frame is divided into unit squares: $32$ vertices and $32$ squares. How many edges are there?

The frame is a torus, so $V - E + F = 0$ and $E = V + F = 64$. Let’s check by counting directly. On the top face the grid points form a $4 \times 4$ lattice and there are $4 \cdot 3 + 4 \cdot 3 = 24$ edges there: the edges of the central hole are edges too. The same on the bottom, and $16$ more vertical edges join the top points to the bottom ones: $24 + 24 + 16 = 64$.

A strip of paper: one side

In 1858 this band was found independently by August Möbius and by Listing, whom we’ve already met. Listing published his find first, but the name went to Möbius.

“One side” describes how the band sits in space. A topologist prefers a property that an inhabitant of the surface itself would notice. Let a flat clock, drawn on the paper itself, crawl along the band. Having gone round the Möbius band along its middle line, it comes back mirror-reversed: the hand goes anticlockwise.

A Möbius band is cut with scissors along its middle line. What do you get?

Two Möbius bands One band twice as long, and it has two sides Two separate bands One Möbius band of the same length

A cut along the middle line doesn’t separate the upper half of the band from the lower: the end of one half is glued to the start of the other. The result is one band of double length. The proof is below, and you can check it with scissors.

Lay the band out on the table: it’s a rectangle whose left side is glued to its right side upside down.

A rectangle $ABCD$ of length $4$ and width $2$. The arrows show the gluing: the single arrow on the left matches the single arrow on the right, the double arrow the double one. So $A$ is glued to $C$, and $D$ to $B$. Cut the rectangle along the middle line. The gluing of the short sides remains in force, only each short side is now split in half. The upper half of the right side (the double arrow at $C$) is glued to the lower half of the left side (the double arrow at $A$). So the right end of the upper strip is glued to the left end of the lower one. Likewise the right end of the lower strip is glued to the left end of the upper one. Turn the lower strip over and lay its left end against the right end of the upper one so that the double arrows lie on top of each other. We have one strip of length $8$ and width $1$. What remains glued are its own ends: the left end of the upper half and the right end of the flipped lower half. The single arrows on them now point the same way, that is, the ends are glued without a twist. A strip glued without a twist is a cylinder: it has two sides and two edges. The length has doubled, the width halved.

In space the new band has two full twists. The flat drawing doesn’t see this: it remembers what’s glued to what, but not how the band lies in space. Use the same method to work out the cut into thirds: the middle third goes to itself and remains a thin Möbius band, while the outer thirds join into one long band threaded through the middle one.

The Euler characteristic can be computed for a surface with boundary too. The rectangle glued into a Möbius band has $2$ vertices (the corners are glued in pairs), $3$ edges and $1$ face: $\chi = 0$. The cylinder has the same. The number $\chi$ alone isn’t enough to tell surfaces apart: what distinguishes the cylinder from the Möbius band is orientability.

A sheet of paper: gluing a square

A torus is a square with its opposite sides glued. The gluing is written as a word: go round the square clockwise and write down the sides; identical letters are glued together, and the power $-1$ means we traverse the arrow on that side backwards. The torus has the word $aba^{-1}b^{-1}$. If the left side is glued to the right upside down, you get a Möbius band, and if then the top is also glued to the bottom, a Klein bottle.

Neither can be glued together in our space: the surface would have to pass through itself. But you can live in them. Imagine a flat creature for whom the square is the whole universe: leaving through the right edge, it appears on the left.

The Klein bottle is drawn as a bottle whose neck has entered its own side and opened out from inside into the bottom. The place where the neck passes through the wall is a flaw of the picture, not of the surface: in four-dimensional space the neck would bypass the wall, as a flyover bypasses a junction.

The idea of the proof, and why the whole of it won’t fit here

A closed surface without self-intersections divides space into an inside and an outside. This is the Jordan–Brouwer theorem, the three-dimensional relative of the fact that a closed curve divides the plane; we relied on it in the chapter on graphs. Then at each point of the surface we can choose an “outward” direction that changes continuously. A clock that on any path keeps its face turned outwards can’t come back reversed. The hard part is the Jordan–Brouwer theorem itself: its proof relies on homology and is beyond the scope of the course.

The Euler characteristic can be found directly from the gluing. There’s one face, the polygon itself; there are as many edges as distinct letters; and as many vertices as the separate points into which the gluing brings the corners. For the torus all four corners of the square come together at one point (in the widget you can see four corners of different copies meeting at a corner), so $\chi = 1 - 2 + 1 = 0$.

The number of vertices after gluing: the number of separate points into which the corners of the polygon come together. The number of edges: a $2n$-gon has its sides glued in pairs, leaving $n$ edges, one per letter. There’s one face, the polygon itself. Example: for the projective plane $abab$ the corners come together at two points, $\chi = 2 - 2 + 1 = 1$. For the Klein bottle at one, $\chi = 0$, like the torus. For the octagon $aba^{-1}b^{-1}cdc^{-1}d^{-1}$ all eight corners come together at one point, $\chi = 1 - 4 + 1 = -2$: it’s the pretzel.

The sides of a hexagon are glued by the word $abca^{-1}b^{-1}c^{-1}$: each side is glued to the opposite one, and the arrows on opposite sides point the same way. Find the Euler characteristic of the resulting surface.

There are $3$ edges and one face. Number the corners round the hexagon $1, \dots, 6$ so that side $a$ goes from corner $1$ to corner $2$. Gluing $a$ to $a^{-1}$ identifies the starts and ends of the arrows: corner $1$ with corner $5$, corner $2$ with corner $4$. Likewise $b$ glues $2$ to $6$ and $3$ to $5$, and $c$ glues $3$ to $1$ and $4$ to $6$. We get two groups, $\{1, 3, 5\}$ and $\{2, 4, 6\}$, that is, two vertices, and $\chi = 2 - 3 + 1 = 0$. Each letter appears with the powers $1$ and $-1$, so the surface is orientable: it’s a torus. No wonder hexagons tile the plane like a honeycomb: a shift to the opposite side takes a hexagon to the neighbouring cell.

All surfaces on one shelf

Two surfaces can be made into one: cut a small disc out of each and glue the edges of the holes together. Two tori give a pretzel this way.

The characteristic of the first surface. The characteristic of the second. Each surface loses a face, where the disc was cut out. Example: $\chi(T \# T) = 0 + 0 - 2 = -2$, the pretzel. Two projective planes: $1 + 1 - 2 = 0$, and that’s the Klein bottle; it really can be cut into two Möbius bands.

We’ll cut out not a disc but a triangular face of a grid: a homeomorphism can’t tell a triangle from a disc anyway.

Draw grids on $M$ and on $N$, each with a triangular face. If there isn’t one, it can be obtained by refinement, which, as we’ve proved, doesn’t change $\chi$. Cut out these triangles with their insides. Each surface has one face fewer: we get $\chi(M) - 1$ and $\chi(N) - 1$. The edges and vertices of the triangles remain: they’re the edges of the holes. Glue the edges of the holes, vertex to vertex and edge to edge. Three pairs of vertices become three vertices, three pairs of edges three edges: $V$ and $E$ both go down by $3$, and $V - E + F$ doesn’t change. We get $\chi(M \# N) = (\chi(M) - 1) + (\chi(N) - 1) = \chi(M) + \chi(N) - 2$.

The sum of $g$ tori is a sphere with $g$ handles, so the formula $\chi = 2 - 2g$ is proved. The sum of $k$ projective planes can be pictured as a sphere with $k$ holes cut out and a Möbius band glued into each (its edge is a circle too). There are no other closed surfaces.

Connected means in one piece. The proof took half a century to assemble: orientable surfaces were described by Möbius (1863) and Camille Jordan (1866), non-orientable ones by Walther von Dyck (1888); the first complete proof was given by Henry Brahana in 1921, and in 1925 Tibor Radó showed that any closed surface can be divided into triangles. The idea: cut the surface along a grid, lay the triangles out as a single polygon, write down its gluing word, and with scissors and glue bring the word to the form $aba^{-1}b^{-1}\,cdc^{-1}d^{-1}\dots$ or $aabbcc\dots$. The most vivid account is John Conway’s “ZIP proof”, published by George Francis and Jeffrey Weeks in 1999; here it would take several pages of case analysis.

A projective plane is attached to a torus. What surface do you get?

A Klein bottle A pretzel The sum of three projective planes A torus with a hole

The surface is non-orientable: the Möbius band from the projective plane hasn’t gone anywhere. Its characteristic is $0 + 1 - 2 = -1$, and the sum of $k$ projective planes has $\chi = 2 - k$, so $k = 3$. A handle next to a Möbius band turns into two bands, as von Dyck already noticed.

A piece of string: knots

As a shape every knot is homeomorphic to a circle, so what matters isn’t what the curve is but how it sits in space. A knot is drawn as a shadow on the plane, and at each crossing the lower strand is drawn with a break. Such a drawing is a knot diagram, and the pieces of strand from break to break are its arcs.

It was proved by Kurt Reidemeister in 1927 and independently, a year earlier, by James Alexander with Garland Briggs. The idea: a deformation of the knot can be slightly adjusted so that the shadow changes its structure only at isolated moments, and each such moment turns out to be one of the three moves. The case analysis takes several pages, and we’ll skip it. But the theorem suggests how to prove that a knot is knotted: find a property of diagrams that no move changes.

Call the colours by the numbers $0$, $1$ and $2$. “All the same or all different” means that the sum of the numbers of the three arcs is divisible by $3$: three different ones give $0 + 1 + 2 = 3$, three the same give $3a$, and two the same and one different give a sum not divisible by $3$.

The colour number of the upper arc at the crossing. The colour number of the lower arc on one side of the crossing. The colour number of the lower arc on the other side. Congruences are from chapter 41. The rule is equivalent to $a + b + c$ being divisible by $3$: the numbers $2a$ and $-a$ differ by $3a$. Example: $0$ on top, $1$ and $2$ below: $2 \cdot 0 = 0$ and $1 + 2 = 3 \equiv 0$, so it’s fine. $0$ on top, $0$ and $1$ below: $0 \not\equiv 1$, not fine.

A move changes the diagram only inside a small disc. We’ll show that the colouring inside the disc is uniquely determined by the colours of the ends on its boundary, and that after the move the ends get the same colours.

The first move. The curl has two pieces: the incoming one, of colour $a$, and the arc of the curl, of colour $x$. At the crossing $x$ meets on top, and $a$ and $x$ below. Two colours coincide, so the third does too: $x = a$. The curl is coloured uniquely and in the same colour as the strand. The second move. A strand of colour $a$ lies over a strand of colour $b$, which breaks into three pieces: $b$, the middle one $c$, and $b'$. The first crossing requires $c \equiv 2a - b$, the second $b' \equiv 2a - c = b$. The outgoing end gets its old colour. If $a \ne b$, the middle piece gets the third colour; if $a = b$, everything is one colour, as before. The third move. A strand $t$ passes over the crossing of strands $m$ and $b$. Before the move, strand $b$ passes first under $m$ and becomes $2m - b$, then under $t$: $2t - (2m - b) = 2t - 2m + b$. After the move it passes first under $t$ and becomes $2t - b$, then under strand $m$, which by then has taken the colour $2t - m$: $2(2t - m) - (2t - b) = 2t - 2m + b$. In both cases strand $m$ leaves with colour $2t - m$. In the drawing $t = 0$, $m = 1$, $b = 2$. In all three moves the colours on the boundary of the disc are preserved, and inside the colouring extends uniquely. If all the arcs outside the disc were one colour, everything inside would be one colour too. So in a colouring with at least two colours, different colours appear outside as well, and after the move they stay there. Tricolourability survives each move, and by Reidemeister’s theorem any deformation of the knot.

Compare the trefoil with the circle by tricolourability.

The usual diagram of the trefoil has three crossings and three arcs. Colour the three arcs in three different colours. All three arcs meet at each crossing, and the colours are different: the rule holds. The trefoil is tricolourable. A circle has one arc, and it can be coloured with only one colour: the unknot isn’t tricolourable. Tricolourability is a property of the knot, so the trefoil and the circle are different knots.

With the figure-eight it won’t work: the widget runs through all $3^4 = 81$ colourings of its four arcs, and only the one-colour ones are valid. Does that mean the figure-eight can be untied? No: a colouring proves knottedness when one is found, but its absence proves nothing. The “unknot in disguise” isn’t tricolourable either, and it really can be untied: it’s a trefoil with one crossing flipped. To catch the figure-eight, take five colours and the same rule modulo $5$; the proof of invariance goes through word for word, since in it we only added and subtracted remainders. The figure-eight is five-colourable, but the circle isn’t, so the figure-eight is knotted too. Such colourings were invented by the American topologist Ralph Fox.

A knot $K$ is tricolourable, and a knot $L$ isn’t. What follows?

$K$ and $L$ are different knots $L$ is the unknot $K$ is the unknot Nothing follows

Tricolourability is a property of the knot: the same knots have the same tricolourability. Since $K$ and $L$ differ in it, they’re different knots. Whether $L$ is knotted can’t be seen from this.

An old map: a fixed point

In the chapter on functional analysis we laid on the table a map of the city we were sitting in and found the point lying exactly over the place it depicts. There the map was a neat reduced copy, and Banach’s contraction theorem found such a fixed point by iterations. Now crumple the map and drop it on the table so that the ball of paper lies entirely over the city. A point lying over its own image exists even now, though there’s no contraction any more.

The crumpled map defines a continuous map of the city to itself: to each point of the city assign the point over which its image lies. Let’s start with the one-dimensional case, with a segment instead of a map.

How far and in which direction the point $x$ has moved. A zero of $g$ is a fixed point of $f$. Where the map takes $x$. Where the point was. Example: $f(x) = 1 - x^2$ maps $[0, 1]$ to itself. Here $g(0) = 1$, $g(1) = -1$, and the zero of $g$ is the root of the equation $x^2 + x - 1 = 0$, the number $\frac{\sqrt5 - 1}{2} \approx 0.618$. Iterations don’t find it: from $0.5$ we get $0.75$, $0.44$, $0.81$, $0.35$, …, and the values drift apart towards $0$ and $1$. It isn’t a contraction, but there’s still a fixed point.

Draw the graph of $f$ in the unit square. A fixed point is where the graph crosses the diagonal $y = x$.

The map $f$ is continuous and takes $[0, 1]$ to itself, so its graph lies entirely in the square. The function $g(x) = f(x) - x$ is continuous too. On the left side of the square $f(0) \ge 0$, since the values of $f$ lie in $[0, 1]$. So $g(0) = f(0) \ge 0$. On the right side $f(1) \le 1$, so $g(1) = f(1) - 1 \le 0$. If $g(0) = 0$ or $g(1) = 0$, the fixed point is an end of the segment. Otherwise $g(0) > 0 > g(1)$, and by the intermediate value theorem (chapter 25) there’s a $c$ with $g(c) = 0$. At that point $f(c) = c$. A continuous graph that starts no lower than the diagonal and ends no higher can’t avoid it.

It was proved by the Dutch mathematician Luitzen Brouwer in 1912; special cases had been obtained shortly before him by Piers Bohl and Jacques Hadamard. A square map will do: a square is homeomorphic to a disc.

The idea of the proof for a disc

Suppose a continuous $f$ has no fixed points. For each point $x$ of the disc draw a ray from $f(x)$ through $x$ to the boundary of the disc and call the exit point $r(x)$. The map $r$ is continuous, takes the disc to the boundary circle, and leaves the points of the boundary itself where they are. Look at how many times the image of the circle of radius $s$ with the same centre winds round the centre. At $s = 1$ the image is the boundary itself: one turn. At $s = 0$ the image is a point: zero turns. But the number of turns of a closed curve round a point it doesn’t pass through doesn’t change under continuous deformation: it’s an integer and can’t jump. We used this number in the chapter on complex analysis. A contradiction. Defining the winding number rigorously and proving that it doesn’t change is technically long work, though the idea is the same.

Brouwer says a fixed point exists, but not where it is or whether it’s unique: a badly crumpled map in the widget sometimes gives three pins. This theorem underlies, for example, John Nash’s theorem on equilibria in games with many players (1950–1951), for which he received the Nobel Memorial Prize in Economics in 1994. And coffee stirred without splashing is a continuous map of the liquid in the cup to itself, and the cup is homeomorphic to a ball: some drop, after stirring, has ended up exactly where it was.

A hairy ball that can’t be combed

For the sphere it was proved by Henri Poincaré in 1885, for spheres of any even dimension by Brouwer in 1912. The wind at the Earth’s surface is a tangent field; if we regard it as continuous, then at every moment there’s somewhere on Earth where the horizontal wind is zero, for example in the eye of a cyclone.

The idea of the proof

Suppose the field $v$ is nowhere zero; divide each vector by its length. Move each point $x$ of the sphere along a great circle in the direction $v(x)$: after time $t$ it’ll be at the point $\cos t \cdot x + \sin t \cdot v(x)$, which lies on the sphere, because $v(x)$ is perpendicular to $x$. At $t = 0$ this is the identity map, at $t = \pi$ the central symmetry $x \mapsto -x$. So the identity map of the sphere deforms continuously into the central symmetry. But maps of the sphere to itself have an integer invariant under deformation, the degree: the number of times the image covers the sphere, with sign. The identity map has degree $1$, while the central symmetry reverses the orientation of the sphere and has degree $-1$. A contradiction. On a circle the central symmetry is a rotation by $180^\circ$, which doesn’t reverse orientation, and a circle can be combed.

Let’s count the crowns. Let the sphere be a planet with mountains and hollows, and let each hair lie in the direction water would roll down fastest. The bald spots will be where there’s no slope: at peaks, at the bottoms of hollows, and at passes, the saddle points from the chapter on functions of several variables.

Points higher than all the points nearby. Saddle points: in two directions the ground rises, in two it falls. Points lower than all the points nearby. Example: on a planet with three peaks and three hollows there must be four passes: $3 - 4 + 3 = 2$. On a torus with three peaks and three hollows there are six. Occasionally the widget shows a “monkey saddle”, where three ridges and three valleys meet; it counts as two passes.

The idea of the proof: a flood

Slowly flood the planet and follow the Euler characteristic of the flooded part. While the water level passes between special heights, the flooded part only expands without changing its structure; this is the hardest point of the proof, and it’s ensured by the lines of steepest descent. When the water reaches the bottom of a hollow, a new lake appears, a disc, and $\chi$ goes up by $1$. When the water spills over a pass, a bridge glued at both ends is added to the flooded part, and $\chi$ goes down by $1$. When a peak goes under, the last little island in a lake is covered over, and $\chi$ goes up by $1$. At the end the whole planet is under water.

For the sphere this relation was noticed by James Clerk Maxwell in his note “On Hills and Dales” (1870). The general Poincaré–Hopf theorem (Heinz Hopf, 1926) is about any tangent field: the sum of the “indices” of its zeros equals $\chi$; peaks and hollows have index $+1$, passes $-1$. Since the sphere has $\chi = 2 \ne 0$, a field without zeros is impossible on it: another explanation of why a hairy ball can’t be combed. The torus has $\chi = 0$, and it can be combed.

A globe: antipodal twins

Plot the air temperature on a globe at some moment and walk along the equator. There will be two diametrically opposite points of the equator with the same temperature, and once again it’s the intermediate value theorem.

How much warmer it is at the point of the equator with longitude $\theta$ than at the opposite point. The temperature at the point $\theta$. The temperature at the diametrically opposite point. Example: if it’s $+20^\circ$ at longitude zero and $+12^\circ$ at longitude $180$, then $g(0) = 8$, and $g(\pi) = 12 - 20 = -8$.

Compare the temperature at each point with the temperature at its antipode.

A point of the equator is given by the angle $\theta$, its antipode by $\theta + \pi$. The curve in the drawing is the temperature: the further out from the circle, the warmer. The function $g(\theta) = T(\theta) - T(\theta + \pi)$ is continuous. Take any point $\theta_0$ and its antipode and find the difference $g(\theta_0)$. From the antipode the picture is the same, but with the opposite sign: $g(\theta_0 + \pi) = T(\theta_0 + \pi) - T(\theta_0 + 2\pi) = -g(\theta_0)$, since the angle $\theta_0 + 2\pi$ gives the same point as $\theta_0$. If $g(\theta_0) = 0$, the pair has been found. Otherwise at the ends of the half-circle from $\theta_0$ to $\theta_0 + \pi$ the function $g$ takes values of different signs, and by the intermediate value theorem somewhere on it $g(\theta^*) = 0$. At the point $\theta^*$ and its antipode the temperatures are equal. The argument works for any continuous quantity on any circle.

The question was asked by Stanisław Ulam and proved by Karol Borsuk in 1933. If temperature and pressure at the Earth’s surface vary continuously, then at every moment there are two antipodes on Earth with the same temperature and the same pressure.

The idea of the proof

Suppose there are no such points. Then the vector of differences $F(x) = \bigl(T(x) - T(-x),\ P(x) - P(-x)\bigr)$ is nowhere zero and $F(-x) = -F(x)$. Let’s see how many times it turns round zero as $x$ goes round the equator. On the way from a point to its antipode the vector turns into its opposite, that is, it turns through an odd number of half-turns. The second half of the way is a mirror copy of the first and gives the same number. Over the whole circuit that’s an odd number of full turns. But the equator can be shrunk continuously over the northern hemisphere to the pole, where the number of turns is zero, and under continuous deformation it doesn’t change. A contradiction.

The Borsuk–Ulam theorem implies the ham sandwich theorem: a sandwich of bread, ham and cheese, however unevenly they lie, can be cut by one straight cut so that each ingredient is divided exactly in half. It was published by Hugo Steinhaus in 1938, who credited the idea to Stefan Banach; the general case was proved by Arthur Stone and John Tukey in 1942.

The universe: the Poincaré conjecture

How can a flat inhabitant find out that their world is a sphere and not a torus, if leaving it is impossible? Throw a lasso. On a sphere any loop can be pulled tight to a point without leaving the surface, while on a torus a loop round the hole can’t be. The classification of surfaces implies that the lasso can be pulled tight only on the sphere.

Our universe looks like ordinary space on a small scale, that is, it’s a three-dimensional manifold; what its overall shape is, nobody knows. In 1904 Henri Poincaré asked: if a three-dimensional world is closed and every lasso in it can be pulled tight, must it be a three-dimensional sphere? The three-dimensional sphere is the set of points of four-dimensional space at distance $1$ from the origin; it can be pictured as two balls glued along their boundary spheres.

In higher dimensions the analogue turned out to be easier: for dimensions five and up it was proved by Stephen Smale in 1961, for dimension four by Michael Freedman in 1982. In the same year, 1982, Richard Hamilton proposed the “Ricci flow”, a way of smoothing out the shape of a manifold, as heat evens out temperature. In 2002–2003 Grigori Perelman posted three papers publicly in which he carried Hamilton’s programme through to the end. Checking them took several years. In 2006 Perelman declined the Fields Medal, and in 2010 the Clay Mathematics Institute’s million-dollar prize. It’s the only one of the seven Millennium Problems whose solution has been officially recognised; the others are in the last chapter. The proof runs to hundreds of pages and goes far beyond the scope of the course.

Where next

Let’s bring back distances and angles. A torus is glued from a square, and nothing obstructs the gluing: the four corners of the square meet at one point, $4 \cdot 90^\circ = 360^\circ$, exactly one full turn. That’s why on a glued torus you can do geometry as on the plane, with straight lines and triangles whose angles add up to $180^\circ$. A pretzel is glued from the octagon $aba^{-1}b^{-1}cdc^{-1}d^{-1}$, and all eight of its corners meet at one point. For exactly one full turn to come out around it, each angle must be $360^\circ : 8 = 45^\circ$. But any octagon in the plane has angles adding up to $1080^\circ$, and it can’t have eight angles of $45^\circ$. Where can you get an octagon with angles of $45^\circ$? In a geometry where many lines parallel to a given line pass through a point off it. It was discovered in attempts to prove Euclid’s fifth postulate: the next chapter.