21. Where geometry goes¶

This book began with points, lines and Eratosthenes measuring the Earth with shadows. It has travelled through Euclid, circles and triangles, the parallel postulate, spheres and hyperbolic planes, projective space, curved surfaces and the spacetime of general relativity. Where does geometry go from here? Everywhere. This last chapter is a guided tour of five directions: Klein's idea that a geometry is defined by its symmetries, which organises everything before it; Thurston's eight geometries of three-dimensional space and Perelman's proof of the Poincaré conjecture; algebraic geometry, where shapes are solution sets of equations; discrete and computational geometry, from convex hulls to the densest packings of spheres in 8 and 24 dimensions; and some easily stated problems that nobody has solved. Each section states results honestly, proves what fits on the page, and says where the rest is.

In this chapter

You will learn:

Klein's Erlangen programme: a geometry is a space with a group of transformations, and its theorems are the invariants
how Euclidean, affine, projective and hyperbolic geometry fit into one hierarchy, with the hyperbolic distance as a cross-ratio
Thurston's eight geometries, Ricci flow, and what Perelman proved
a first look at algebraic geometry: varieties, Bézout's theorem, genus, elliptic curves
convex hulls, Voronoi diagrams and Delaunay triangulations, and how computers do geometry
sphere packing, from Kepler's conjecture (Hales) to dimensions 8 and 24 (Viazovska)
open problems: the moving sofa, the chromatic number of the plane, the inscribed square

You need: most of this book, especially Chapter 10, Chapter 17, Chapter 18 and Chapter 20; some group theory from Algebra, Chapter 13. Section 21.4 needs very little: it can be read after Part II.

Levels: the content is expert, written as a survey. 21.1 and 21.4 can be read by anyone who has reached Part III; 21.2 and 21.3 build on Part IV and Chapter 20; 21.5 states problems a beginner can understand and experts cannot solve.

21.1 Klein's Erlangen programme: geometry as symmetry¶

Level: advanced. By the 1870s mathematicians had a crowd of geometries: Euclidean, spherical, hyperbolic, projective, affine, inversive. Each had its own theorems. Were they rivals, or parts of one subject? In 1872, aged 23 and newly appointed professor at Erlangen, Felix Klein published a programme (Vergleichende Betrachtungen über neuere geometrische Forschungen, "A comparative review of recent researches in geometry") with a simple answer:

A geometry is a space together with a group of transformations of it. The geometry studies exactly the properties of figures that are unchanged (invariant) under every transformation in the group.

Two figures are "the same" in a geometry if some transformation of its group carries one to the other. The group decides what counts as a fact.

The hierarchy of plane geometries. Take the plane and four groups, each containing the last:

Geometry	Group	What it preserves	A typical theorem
Euclidean	isometries (rotations, reflections, translations)	lengths, angles, areas	Pythagoras
similarity	isometries and scalings	angles, ratios of lengths, shape	the angle sum is \(180°\)
affine	\(x \mapsto Ax + b\), \(A\) invertible	lines, parallelism, ratios along a line, ratios of areas	medians are concurrent
projective	\(PGL(3)\), invertible matrices up to scale	lines, incidence, cross-ratio	Desargues, Pappus, Pascal

The bigger the group, the fewer properties survive, and the more general (but fewer) the theorems. In Euclidean geometry a 3-4-5 triangle differs from a 6-8-10 triangle; in similarity geometry they are the same; in affine geometry all triangles are the same, because an affine map can send any three non-collinear points to any three others; and in projective geometry all quadrangles are the same (Chapter 18, Theorem 18.2).

Using the group to prove things. Klein's view gives a proof strategy: to prove an affine theorem, it is enough to prove it for one convenient figure.

Claim. Affine maps preserve ratios of lengths along a line. Proof. If \(Q = P + t(R - P)\) then \(f(Q) = A(P + t(R - P)) + b = f(P) + t(f(R) - f(P))\), so \(f(Q)\) divides \(f(P)f(R)\) in the same ratio \(t\). \(\square\)

Theorem. The three medians of any triangle meet in one point, which divides each in the ratio \(2 : 1\). Proof. Medians (lines from a vertex to the midpoint of the opposite side) and the ratio \(2 : 1\) are affine notions: by the claim, an affine map sends midpoints to midpoints and preserves the ratio. Send the triangle by an affine map to an equilateral triangle. There the medians are also the axes of symmetry, so they meet at the centre, which is at \(\tfrac23\) of the height from each vertex. Map back. \(\square\)

Non-Euclidean geometry inside projective geometry. Klein's most striking example, building on Cayley's work of 1859, puts hyperbolic geometry inside projective geometry. Take the projective plane and a circle (a conic) in it. The projective transformations that map the inside of the circle to itself form a group. Make that the group of the geometry, with the inside of the circle as the space, and chords as lines: this is the Beltrami–Klein model of the hyperbolic plane. Distance comes from the one projective invariant, the cross-ratio: for points \(P, Q\) inside, let the line \(PQ\) meet the circle at \(A\) and \(B\), and define

\[ d(P, Q) = \frac12 \left| \ln (A, B; P, Q) \right| . \]

Because projective transformations preserve the cross-ratio, they preserve this distance: they are the isometries.

Worked example. In the unit disc take \(P = (0, 0)\) and \(Q = (x, 0)\), \(0 < x < 1\). The line meets the circle at \(A = -1\) and \(B = 1\), and

\[ (A, B; P, Q) = \frac{(0 - (-1))(x - 1)}{(0 - 1)(x - (-1))} = \frac{1 - x}{1 + x}, \qquad d = \frac12 \ln\frac{1 + x}{1 - x} = \operatorname{artanh} x . \]

For \(x = 0.6\): \(d = \tfrac12 \ln 4 \approx 0.693\). As \(x \to 1\), \(d \to \infty\): the boundary circle is infinitely far away. (This is the distance formula of the Klein model; the Poincaré disc of Chapter 17 is the same geometry drawn differently.) Use an imaginary conic instead, such as \(x^2 + y^2 + z^2 = 0\), and the same recipe gives elliptic geometry; let the conic degenerate and Euclidean geometry appears as a limiting case. Cayley summed it up in 1859: "descriptive geometry is all geometry", where by "descriptive" he meant what we now call projective.

Further examples of Klein's principle.

Inversive (Möbius) geometry: the group of Möbius transformations (Chapter 13); circles and lines are one kind of object, and the complex cross-ratio is the invariant.
Topology: the group of all homeomorphisms; the invariants are things like connectedness, number of holes, Euler characteristic.
Symmetry in physics: special relativity is the geometry of the Lorentz group, and conservation laws come from symmetries (Noether's theorem, 1918).

Where the programme stops. A general Riemannian manifold (Chapter 20) usually has no symmetries at all, so it is not a Klein geometry. Élie Cartan reconciled the two views in the 1920s: a Cartan geometry looks infinitesimally like a Klein geometry at each point, with curvature measuring how far it is from being one. A surface is locally like the Euclidean plane, with Gaussian curvature measuring the difference. The models of constant curvature, sphere, plane and hyperbolic plane, are exactly the Klein geometries among surfaces, and that is why they keep appearing.

21.2 Thurston's eight geometries and the Poincaré conjecture¶

Level: expert (survey). Chapter 20 ended with uniformization: every closed surface carries one of three geometries, spherical, Euclidean or hyperbolic. What about three-dimensional spaces? A closed 3-manifold is a compact 3-manifold without boundary, a possible shape for a finite universe with no edge. Examples: the 3-sphere \(S^3\) (the unit sphere in \(\mathbb{R}^4\)), the 3-torus (a cube with opposite faces glued), \(S^2 \times S^1\), and many more made by gluing the faces of polyhedra.

The Poincaré conjecture. In 1904 Henri Poincaré asked: if a closed 3-manifold is simply connected (every loop in it can be shrunk to a point, as on a sphere and unlike a torus), must it be the 3-sphere? For surfaces the answer is yes: a simply connected closed surface is a sphere, by the classification of surfaces. In three dimensions it remained open for almost a century, and it became one of the seven Millennium Prize Problems of the Clay Mathematics Institute in 2000.

Thurston's geometrization conjecture. In the late 1970s William Thurston proposed a far more ambitious picture. A model geometry is a simply connected 3-manifold with a Riemannian metric whose group of isometries is as large as possible in a suitable sense, and which covers at least one closed manifold. Thurston showed there are exactly eight:

Geometry	Description
\(S^3\)	the round 3-sphere, constant curvature \(+1\)
\(\mathbb{E}^3\)	Euclidean space, curvature 0
\(\mathbb{H}^3\)	hyperbolic space, constant curvature \(-1\)
\(S^2 \times \mathbb{R}\)	product of a sphere and a line
\(\mathbb{H}^2 \times \mathbb{R}\)	product of a hyperbolic plane and a line
\(\widetilde{SL_2\mathbb{R}}\)	the universal cover of the group of \(2 \times 2\) real matrices of determinant 1
Nil	the Heisenberg group of upper triangular \(3 \times 3\) matrices with ones on the diagonal
Sol	a group built from the matrices \(\operatorname{diag}(e^t, e^{-t})\)

The geometrization conjecture states: every closed orientable 3-manifold can be cut, along a finite collection of embedded spheres and tori in a canonical way, into pieces each of which carries one of the eight geometries. Thurston proved it for a large class (Haken manifolds) and showed that "most" 3-manifolds are hyperbolic. The Poincaré conjecture is a special case: a simply connected closed 3-manifold can only carry the \(S^3\) geometry, and the only such manifold is \(S^3\) itself.

Hyperbolic 3-manifolds are remarkably rigid. By Mostow's rigidity theorem (1968), a closed (or finite-volume) hyperbolic 3-manifold has only one hyperbolic metric, so geometric numbers such as its volume are topological invariants. The space around a figure-eight knot, for example, carries a complete hyperbolic metric of volume about \(2.0299\), made from two regular ideal tetrahedra; no other knot complement has smaller volume.

Ricci flow. In 1982 Richard Hamilton proposed a way to find the geometry: let the metric evolve by the Ricci flow

\[ \frac{\partial g}{\partial t} = -2\, \mathrm{Ric}(g), \]

a heat equation for the metric. Where Ricci curvature is positive the metric shrinks, where it is negative it expands, and one hopes the flow irons out irregularities, as heat flow evens out temperature.

Worked example: the round sphere. On the round \(n\)-sphere of radius \(r\), \(\mathrm{Ric} = \frac{n - 1}{r^2}\, g\), and \(g = r^2 g_1\) where \(g_1\) is the unit sphere's metric. Trying \(g(t) = r(t)^2 g_1\) in the flow: \(\frac{d}{dt}(r^2)\, g_1 = -2 \cdot \frac{n-1}{r^2} \cdot r^2 g_1\), so

\[ \frac{d}{dt}\, r^2 = -2(n - 1), \qquad r(t)^2 = r_0^2 - 2(n - 1)t . \]

The sphere stays round and shrinks to a point at time \(t = r_0^2/(2(n - 1))\). Check: a unit 2-sphere vanishes at \(t = \tfrac12\); a 3-sphere of radius 3 at \(t = 9/4\).

Hamilton proved in 1982 that a closed 3-manifold with positive Ricci curvature flows, after rescaling, to a round metric of constant positive curvature, which settled that case. In general the flow develops singularities: a thin neck in a dumbbell-shaped sphere pinches off before the rest shrinks.

Perelman. In three preprints posted on the arXiv in 2002 and 2003, Grigori Perelman showed how to control every singularity that can form in three dimensions, cut the manifold just before it (Ricci flow with surgery, an idea of Hamilton's), cap off the pieces, and continue. Following the flow to the end decomposes the manifold exactly as Thurston predicted. This proved the geometrization conjecture and with it the Poincaré conjecture. Several teams spent the next years writing out the details (Kleiner and Lott; Morgan and Tian; Cao and Zhu), and by 2006 the experts accepted the proof. Perelman was awarded the Fields Medal in 2006 and the Clay Millennium Prize in 2010, and declined both.

In other dimensions the story is different. The analogue of the Poincaré conjecture (a closed manifold that is homotopy equivalent to a sphere is homeomorphic to it) was proved for dimensions 5 and up by Stephen Smale in 1961 and for dimension 4 by Michael Freedman in 1982. The smooth version in dimension 4 (is every smooth homotopy 4-sphere diffeomorphic to \(S^4\)?) is still open. In dimension 2, Ricci flow on any closed surface converges, after rescaling, to a metric of constant curvature (Hamilton 1988 and Chow 1991), which gives a new proof of uniformization.

21.3 Algebraic geometry: a glimpse¶

Level: expert (survey). Descartes turned curves into equations (Chapter 8). Algebraic geometry turns that around: it studies the shapes defined by polynomial equations, using algebra to see geometry and geometry to see algebra.

An affine variety is the set of common zeros of some polynomials, for example a circle \(x^2 + y^2 - 1 = 0\), a twisted cubic in space, or the set of \(2 \times 2\) matrices of determinant 0. A projective variety is the same in projective space, with homogeneous polynomials (Chapter 18). Working over the complex numbers and in projective space makes the theory clean, for the same reason that complex numbers make every polynomial split (Algebra, Chapter 12): solutions do not disappear off to infinity or into the imaginary.

Bézout's theorem. How many points do two curves have in common? Two lines: one point. A line and a conic: up to two.

Theorem 21.1 (Bézout). Two curves in the complex projective plane, of degrees \(m\) and \(n\), with no common component, meet in exactly \(mn\) points, counted with multiplicity.

(Proved with resultants, by eliminating one variable; Fulton's Algebraic Curves has a complete proof.)

Worked example: a circle and a parabola. \(x^2 + y^2 = 4\) and \(y = x^2 - 2.5\) have degrees 2 and 2, so Bézout predicts 4 points. Substituting, \(x^2 + (x^2 - 2.5)^2 = 4\), i.e. \(x^4 - 4x^2 + 2.25 = 0\), so \(x^2 = 2 \pm \sqrt{1.75}\), which gives four real points \(x \approx \pm 1.823\) and \(\pm 0.823\). Check: at \(x = 0.823\), \(y = 0.677 - 2.5 = -1.823\), and \(0.677 + 3.323 = 4\).

Worked example: concentric circles. \(x^2 + y^2 = 1\) and \(x^2 + y^2 = 4\) have no common real points, yet Bézout promises 4. Homogenise: \(x^2 + y^2 = z^2\) and \(x^2 + y^2 = 4z^2\). Subtracting, \(3z^2 = 0\), so \(z = 0\) (a double root), and then \(x^2 + y^2 = 0\), so \(y = \pm i x\): the two points \([1 : i : 0]\) and \([1 : -i : 0]\), each with multiplicity 2. Every circle passes through these two circular points at infinity, which is why two circles meet in at most 2 ordinary points: the other 2 of their 4 are always these.

Genus. Over the complex numbers, a smooth projective curve is a compact Riemann surface (Chapter 20), and so it has a genus \(g\), the number of holes. For a smooth plane curve of degree \(d\),

\[ g = \frac{(d - 1)(d - 2)}{2} . \]

Lines and conics have genus 0 (they are spheres); smooth cubics have genus 1 (tori); smooth quartics genus 3. The uniformization theorem then gives each curve a geometry: spherical for \(g = 0\), flat for \(g = 1\), hyperbolic for \(g \ge 2\), and remarkably this geometric trichotomy governs arithmetic. Faltings's theorem (1983, conjectured by Mordell in 1922) says a curve of genus at least 2 defined over the rationals has only finitely many rational points.

Elliptic curves. A smooth cubic curve \(y^2 = x^3 + ax + b\) (with a point at infinity added) has genus 1, and its points form a group: to add \(P\) and \(Q\), draw the line through them, find the third intersection point (Bézout: a line meets a cubic in 3 points), and reflect it in the \(x\)-axis. For example on \(y^2 = x^3 - 2\) the point \(P = (3, 5)\) is rational. To double it, take the tangent at \(P\), of slope \(\lambda = 3x^2/(2y) = 27/10\). Its third intersection has \(x = \lambda^2 - 2 \cdot 3 = 129/100\), and the reflection gives

\[ 2P = \left( \frac{129}{100}, -\frac{383}{1000} \right) . \]

Check: \((383/1000)^2 = 0.146689\) and \((1.29)^3 - 2 = 2.146689 - 2 = 0.146689\). This chord-and-tangent construction goes back to Fermat and Newton; it is the heart of Wiles's proof of Fermat's Last Theorem and of elliptic-curve cryptography (see the cryptography book).

Two classical gems. Every smooth cubic surface in complex projective 3-space contains exactly 27 straight lines (Cayley and Salmon, 1849). And the rigour of the subject was rebuilt twice in the twentieth century: by Zariski and Weil in the 1930s and 1940s, and by Grothendieck's theory of schemes in the 1960s, which lets the same geometric language describe equations over the integers and finite fields. Algebra, Chapter 22 describes the algebra side (Hilbert's Nullstellensatz, Gröbner bases).

21.4 Discrete and computational geometry¶

Level: intermediate to expert. Much of today's geometry is about finite sets of points, and about algorithms: how fast can a computer answer a geometric question? Computer graphics, robot motion planning, geographic information systems, mesh generation for engineering simulations and the analysis of spacecraft images all depend on it.

Convex hulls. The convex hull of a finite set of points is the smallest convex polygon containing them, the shape a rubber band takes around nails at the points.

How fast can it be found? The monotone chain algorithm (a variant of Graham's scan of 1972) sorts the points by \(x\), then sweeps left to right building the lower boundary, keeping only left turns, and right to left for the upper boundary. A turn is detected by the sign of a cross product: for points \(o, a, b\), the quantity \((a_x - o_x)(b_y - o_y) - (a_y - o_y)(b_x - o_x)\) is positive for a left turn.

Try it in Python

def hull(points):
    P = sorted(points)
    cross = lambda o, a, b: (a[0]-o[0])*(b[1]-o[1]) - (a[1]-o[1])*(b[0]-o[0])
    lower, upper = [], []
    for p in P:
        while len(lower) >= 2 and cross(lower[-2], lower[-1], p) <= 0:
            lower.pop()
        lower.append(p)
    for p in reversed(P):
        while len(upper) >= 2 and cross(upper[-2], upper[-1], p) <= 0:
            upper.pop()
        upper.append(p)
    return lower[:-1] + upper[:-1]

print(hull([(0, 0), (2, 0), (1, 1), (2, 2), (0, 2), (1, 3)]))
# [(0, 0), (2, 0), (2, 2), (1, 3), (0, 2)]

Each point is pushed and popped at most once per sweep, so after sorting the work is linear: the total time is \(O(n \log n)\). That is optimal in the usual model of computation: if you could find hulls faster you could sort faster, because the hull of the points \((x_i, x_i^2)\) on a parabola lists them in sorted order, and sorting \(n\) numbers by comparisons needs about \(n \log_2 n\) steps.

Voronoi diagrams and Delaunay triangulations. Given \(n\) sites in the plane, the Voronoi cell of a site is the set of points closer to it than to any other site. The cells are convex polygons (intersections of half-planes bounded by perpendicular bisectors) and they tile the plane: the Voronoi diagram. Joining two sites whenever their cells share an edge gives the Delaunay triangulation (when no four sites lie on a circle).

They answer "which is nearest?" questions: the nearest hospital, the nearest cell tower, the region of a map closest to each weather station. Their main properties:

Empty circle property. A triangle of sites is a Delaunay triangle exactly when its circumcircle contains no other site. (The centre of that circle is a Voronoi vertex, a point equidistant from three sites and farther from all others.)
Best angles. Among all triangulations of the sites, the Delaunay triangulation maximises the smallest angle, avoiding thin triangles; this is why it is used to build meshes for engineering simulations.
Few edges. By Euler's formula (Chapter 11), a triangulation of \(n\) points has at most \(3n - 6\) edges. Check: the figure has 12 sites and 23 Delaunay edges, and \(23 \le 30\). So a Voronoi diagram has size proportional to \(n\), and Fortune's sweep-line algorithm (1986) builds it in \(O(n \log n)\) time.

The two structures are dual, like the points and lines of Chapter 18: Voronoi vertices correspond to Delaunay triangles, Voronoi edges to Delaunay edges. There is also a beautiful lift: put each site \((x, y)\) on the paraboloid \(z = x^2 + y^2\); the lower convex hull of the lifted points, projected back down, is the Delaunay triangulation.

Sphere packing. How densely can equal balls fill space? In the plane, circles in the hexagonal arrangement touch six neighbours each and cover a fraction \(\pi/(2\sqrt3) \approx 0.9069\) of the plane, and no arrangement does better (Thue; a complete proof by László Fejes Tóth in 1940).

Check of the density: each circle of radius \(r\) sits in a hexagonal cell of area \(2\sqrt3\, r^2\) (a regular hexagon with inradius \(r\)), and \(\pi r^2 / (2\sqrt3\, r^2) = \pi/(2\sqrt3)\).

In three dimensions, the familiar greengrocer's stacking of oranges (the face-centred cubic packing) fills \(\pi/\sqrt{18} \approx 0.7405\) of space. Kepler asserted in 1611 that nothing denser is possible. This Kepler conjecture resisted proof for almost four centuries. Thomas Hales, partly with his student Samuel Ferguson, announced a proof in 1998 that reduced the problem to a very large number of inequalities checked by computer; it was published in the Annals of Mathematics in 2005, with referees saying they were "99 percent certain" of the computer part. To remove the doubt Hales led the Flyspeck project, which produced a complete formal proof checked by computer proof assistants, finished in 2014 and published in 2017.

Higher dimensions are stranger. In 2016 Maryna Viazovska proved that the \(E_8\) lattice packing is the densest in dimension 8, with density \(\pi^4/384 \approx 0.2537\), using a remarkable "magic function" built from modular forms that certifies a linear programming bound of Cohn and Elkies. Within a week, Viazovska with Henry Cohn, Abhinav Kumar, Stephen D. Miller and Danylo Radchenko extended the method to the Leech lattice in dimension 24, density \(\pi^{12}/12! \approx 0.00193\). Viazovska received the Fields Medal in 2022. Dimensions 8 and 24 are special because these two lattices are extraordinarily symmetric; in dimension 4, and in every dimension other than 1, 2, 3, 8 and 24, the densest packing is unknown.

A related question is the kissing number: how many unit balls can touch one central unit ball? In the plane, 6. In three dimensions, 12 (a question argued over by Newton and David Gregory in 1694, settled by Schütte and van der Waerden in 1953). In dimension 4 it is 24 (Musin, 2003), in dimension 8 it is 240 and in dimension 24 it is 196,560 (Odlyzko and Sloane, and independently Levenshtein, 1979). Packings like these are also the mathematics of error-correcting codes: codewords are the centres of balls that must not overlap if noise is to be corrected.

21.5 Open problems¶

Level: all levels. Geometry is full of questions that a child can understand and that the best mathematicians cannot yet answer. Here are three, with what is known.

The moving sofa problem¶

You are moving house and must carry a sofa (a rigid flat shape) around a right-angled corner in a corridor of width 1. What is the largest area a sofa can have and still get round? Leo Moser posed the question in 1966.

A unit half-disc (area \(\pi/2 \approx 1.571\)) works: slide it to the corner and rotate it about the corner.
Hammersley (1968) did better: cut the half-disc in two, push the quarters apart by \(4/\pi\) and fill between with a rectangle from which a half-disc of radius \(2/\pi\) has been removed (so it can rotate around the inner corner). Its area is \(\pi/2 + 2/\pi \approx 2.2074\).

Gerver (1992) found a sofa with a boundary made of 18 curve pieces and area about \(2.2195\), and conjectured it is optimal.
Upper bounds came much more slowly: Kallus and Romik (2018) proved by a computer-assisted argument that no sofa has area more than about \(2.37\).

In late 2024 Jineon Baek posted a long paper claiming a proof that Gerver's sofa is optimal. Proofs of this kind take experts a long time to check; at the time of writing it was being studied by the community, and anyone reading this should look up its current status before quoting the problem as open or solved.

The chromatic number of the plane (Hadwiger–Nelson problem)¶

Colour every point of the plane so that any two points at distance exactly 1 get different colours. What is the smallest number of colours needed? The question goes back to Edward Nelson in 1950.

At least 4: the Moser spindle. Leo and William Moser (1961) found seven points with eleven pairs at distance 1.

Proof that it needs 4 colours. The spindle is two rhombi, each made of two equilateral triangles of side 1, sharing the bottom vertex \(A\), turned so that their top vertices are at distance 1. Suppose three colours were enough. In each equilateral triangle the three vertices need three different colours. In the first rhombus \(A, B_1, C_1\) use all three colours, and the top vertex \(T_1\), which forms a triangle with \(B_1, C_1\), must then have the colour of \(A\). The same argument gives \(T_2\) the colour of \(A\). But \(T_1\) and \(T_2\) are at distance 1. Contradiction. \(\square\) (The figure was built and checked numerically: all eleven segments have length 1 to within \(10^{-9}\).)

At most 7: hexagons. Tile the plane by regular hexagons of side \(a\) and colour them in seven colours in the repeating pattern in which each hexagon's six neighbours all have different colours from it and each other. Same-coloured hexagons then have centres \(\sqrt{21}\,a\) apart. Two points in one hexagon are less than \(2a\) apart, and two points in different hexagons of the same colour are at least \(\sqrt{21}\,a - 2a \approx 2.58a\) apart. With \(a = 0.45\): within a hexagon distances are below \(0.9 < 1\), and between same-coloured hexagons they are at least \(1.16 > 1\). So no two points at distance 1 share a colour (after a careful choice of colour for boundary points). \(\square\)

For 57 years the answer was known only to be 4, 5, 6 or 7. In 2018 Aubrey de Grey, an amateur mathematician, found a unit-distance graph with 1581 vertices that cannot be coloured with 4 colours, so at least 5 colours are needed; a collaborative Polymath project then found much smaller examples. The answer is now known to be 5, 6 or 7.

The inscribed square problem¶

Otto Toeplitz asked in 1911: does every closed curve in the plane that does not cross itself (a Jordan curve) contain four points that are the corners of a square?

It is known for many classes: convex curves, smooth curves, and curves that are piecewise analytic, among others. In 2020 Joshua Greene and Andrew Lobb proved that every smooth Jordan curve contains the four corners of a rectangle of every aspect ratio, using symplectic geometry. For general continuous curves, which can wiggle infinitely, the square problem is still open: the proofs for smooth curves use a limit argument that can let the square shrink to a point.

A case you can prove. Suppose the curve is star-shaped about a centre \(O\) and symmetric under a half turn about \(O\), written in polar form as \(r = r(\theta)\) with \(r(\theta + \pi) = r(\theta)\). Let \(g(\theta) = r(\theta) - r(\theta + \pi/2)\). Then \(g(\theta + \pi/2) = r(\theta + \pi/2) - r(\theta + \pi) = -g(\theta)\), so \(g\) takes opposite signs (or is zero) at \(\theta\) and \(\theta + \pi/2\), and by the intermediate value theorem \(g(\theta_0) = 0\) for some \(\theta_0\). The four curve points in the directions \(\theta_0\), \(\theta_0 + \pi/2\), \(\theta_0 + \pi\), \(\theta_0 + 3\pi/2\) are all at distance \(r(\theta_0)\) from \(O\) along two perpendicular lines: a quadrilateral whose diagonals are equal, perpendicular and bisect each other. That is a square. \(\square\)

More open problems, briefly.

Kakeya: a set in \(\mathbb{R}^n\) containing a unit segment in every direction must have full (Hausdorff) dimension \(n\). Proved in the plane in 1971 (Davies); for \(n = 3\) a proof was announced by Hong Wang and Joshua Zahl in 2025; open in higher dimensions.
Erdős's unit distance problem: how many pairs among \(n\) points in the plane can be at distance exactly 1? The best upper bound, of order \(n^{4/3}\), has not been improved since 1984, while the best constructions give only slightly more than linear growth.
Smooth 4-dimensional Poincaré conjecture (21.2), and sphere packing in all dimensions except 1, 2, 3, 8 and 24 (21.4).

Geometry began with measuring fields and stars. Its questions are now shared with physics, computer science, number theory and topology, and they are far from finished.

A short history¶

Klein's Erlangen programme (1872) and Lie's theory of continuous groups (from the 1870s) made symmetry the organising principle of geometry. Hilbert's Foundations of Geometry (1899) settled the axioms, and Poincaré's work (1880s to 1900s) created topology and the geometric theory of Riemann surfaces. The twentieth century brought Cartan's geometries, the global Riemannian geometry of Chapter 20, Grothendieck's algebraic geometry, and Thurston's revolution in 3-manifolds (late 1970s), completed by Perelman (2002 to 2003). Computational geometry became a field in the 1970s (Shamos's thesis of 1978 named it), and the twenty-first century has seen computer-verified proofs (Flyspeck, 2014) alongside new ideas such as Viazovska's magic functions (2016). Problems as easy to state as the moving sofa and the coloured plane remind us that elementary geometry is not finished either.

Check yourself¶

1. Which of these are preserved by every affine map: length, angle, midpoint, parallelism, ratio of areas, being a circle?

Preserved: midpoints (ratios along a line), parallelism (parallel lines are sent to parallel lines since \(Ax + b\) sends the common direction \(d\) to \(Ad\)), and ratios of areas (all areas are multiplied by \(|\det A|\)). Not preserved: length and angle (a stretch changes both), and being a circle (a circle can become an ellipse).

2. In the Beltrami–Klein model (the unit disc, distance \(\frac12|\ln(A, B; P, Q)|\)), find the distance from the centre to \((0.6, 0)\) and to \((0.8, 0)\). Is the distance from \((0.6, 0)\) to \((0.8, 0)\) the difference?

From the centre: \(\operatorname{artanh} 0.6 = \tfrac12 \ln 4 \approx 0.693\) and \(\operatorname{artanh} 0.8 = \tfrac12 \ln 9 \approx 1.099\). All three points lie on one line (a diameter), and along a line the distance is additive, so the distance from \((0.6, 0)\) to \((0.8, 0)\) is \(\tfrac12 \ln 9 - \tfrac12 \ln 4 = \tfrac12 \ln(9/4) \approx 0.405\). Check with the cross-ratio: \(A = -1\), \(B = 1\), \(P = 0.6\), \(Q = 0.8\): \((A, B; P, Q) = \dfrac{(1.6)(-0.2)}{(-0.4)(1.8)} = \dfrac{-0.32}{-0.72} = \dfrac49\), and \(\tfrac12 |\ln(4/9)| \approx 0.405\).

3. Under the Ricci flow, when does a round 2-sphere of radius 3 shrink to a point? A round 3-sphere of radius 3?

\(r^2 = r_0^2 - 2(n - 1)t\). For \(n = 2\): \(9 - 2t = 0\) at \(t = 4.5\). For \(n = 3\): \(9 - 4t = 0\) at \(t = 2.25\).

4. Using Bézout, how many points do the parabola \(y = x^2\) and the line \(y = 2x - 1\) have in common? Find them with multiplicity.

Degrees 2 and 1: 2 points with multiplicity. \(x^2 = 2x - 1\) gives \((x - 1)^2 = 0\): the point \((1, 1)\) counted twice. The line is the tangent to the parabola there.

5. Find the genus of a smooth plane curve of degree 4, and of degree 5. Which of them could have infinitely many rational points?

\((d - 1)(d - 2)/2\): degree 4 gives \(3 \cdot 2/2 = 3\), degree 5 gives \(4 \cdot 3/2 = 6\). Both have genus at least 2, so by Faltings's theorem neither has infinitely many rational points.

6. On the elliptic curve \(y^2 = x^3 - 2\), check that \(P = (3, 5)\) lies on the curve and that \(2P = (129/100, -383/1000)\).

\(5^2 = 25 = 27 - 2\). Tangent slope \(\lambda = 3x^2/(2y) = 27/10\). Then \(x_3 = \lambda^2 - 2x = 729/100 - 600/100 = 129/100\), and \(y_3 = \lambda(x - x_3) - y = \tfrac{27}{10} \cdot \tfrac{171}{100} - 5 = \tfrac{4617}{1000} - \tfrac{5000}{1000} = -\tfrac{383}{1000}\). Check: \((383/1000)^2 = 146689/10^6\) and \((129/100)^3 - 2 = 2146689/10^6 - 2 = 146689/10^6\).

7. Use Euler's formula to show that a triangulation of \(n\) points in the plane, whose outer boundary (the convex hull) has \(h\) points, has exactly \(3n - 3 - h\) edges and \(2n - 2 - h\) triangles.

Count faces including the outer one: \(V - E + F = 2\) with \(V = n\). Each inner face is a triangle and the outer face has \(h\) sides, and each edge borders two faces: \(3(F - 1) + h = 2E\). So \(F = (2E - h)/3 + 1\), and \(n - E + (2E - h)/3 + 1 = 2\), giving \(E = 3n - 3 - h\). Triangles: \(F - 1 = (2E - h)/3 = (6n - 6 - 3h)/3 = 2n - 2 - h\). With \(h \ge 3\), \(E \le 3n - 6\).

8. Compute the density of the square packing of circles (each circle touching four others) and compare with the hexagonal packing.

Each circle of radius \(r\) sits in a square of side \(2r\): density \(\pi r^2/(4r^2) = \pi/4 \approx 0.785\). The hexagonal packing gives \(\pi/(2\sqrt3) \approx 0.907\), about 15 percent denser.

9. Show that the points of the Moser spindle really can be placed with all eleven distances equal to 1: find the angle between the two rhombi.

Each rhombus (two equilateral triangles of side 1) has its far vertex at distance \(2 \cdot \tfrac{\sqrt3}{2} = \sqrt3\) from \(A\). The two far vertices must be 1 apart, both at distance \(\sqrt3\) from \(A\): an isosceles triangle with sides \(\sqrt3, \sqrt3, 1\), so the angle \(\theta\) at \(A\) has \(\sin(\theta/2) = \tfrac{1/2}{\sqrt3}\), \(\theta = 2\arcsin\bigl(1/(2\sqrt3)\bigr) \approx 33.56°\).

10. In the 7-colouring of the plane by hexagons of side \(a\), find the range of \(a\) that works.

Points in one hexagon must be less than 1 apart: the diameter \(2a < 1\), so \(a < 0.5\). Same-coloured hexagons must be more than 1 apart: \(\sqrt{21}\,a - 2a > 1\), so \(a > 1/(\sqrt{21} - 2) \approx 0.387\). Any \(a\) with \(0.387 < a < 0.5\) works.

11. Find a square inscribed in the ellipse \(x^2/4 + y^2 = 1\).

The ellipse is symmetric under a half turn, and also under reflections in both axes. Try the corners \((\pm t, \pm t)\): \(t^2/4 + t^2 = 1\) gives \(t^2 = 4/5\), \(t = 2/\sqrt5 \approx 0.894\). The square has side \(2t = 4/\sqrt5 \approx 1.789\). (Here \(r(\theta) = r(\theta + \pi/2)\) at \(\theta = 45°\), as in the proof in 21.5.)

12. Verify that Hammersley's sofa has area \(\pi/2 + 2/\pi\).

Two quarter discs of radius 1: area \(\pi/2\). A rectangle \(4/\pi\) by 1: area \(4/\pi\). Minus a half disc of radius \(2/\pi\): area \(\tfrac12 \pi (2/\pi)^2 = 2/\pi\). Total \(\pi/2 + 4/\pi - 2/\pi = \pi/2 + 2/\pi \approx 1.5708 + 0.6366 = 2.2074\).

What you learned¶

Klein: a geometry is a space with a group, and its theorems are the invariants; Euclidean, similarity, affine and projective geometry form a hierarchy, and hyperbolic distance is a cross-ratio.
Thurston's eight model geometries describe all closed 3-manifolds; Perelman proved this with Ricci flow with surgery, and with it the Poincaré conjecture.
Algebraic geometry studies solution sets of polynomials; Bézout counts intersections, genus measures complexity, and elliptic curves carry a geometric group law.
Convex hulls, Voronoi diagrams and Delaunay triangulations are computed in \(O(n \log n)\) time and underlie much of computing.
Densest sphere packings are known only in dimensions 1, 2, 3, 8 and 24; the proofs range from computer verification to modular forms.
The moving sofa, the chromatic number of the plane and the inscribed square problem show how much elementary geometry is still open.

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