An octagon only the hyperbolic plane can hold

Every surface is a sphere with handles built the two-handled surface from an octagon. Label its sides around the boundary $a, b, a^{-1}, b^{-1}, c, d, c^{-1}, d^{-1}$ and glue each side to the one with the same letter, matching the arrows; the result is a closed surface with two handles, and every word driven to a normal form showed that every orientable surface of that kind can be reached this way. All of that is topology. The octagon is a rubber sheet, its sides are stretched and bent to meet each other, and nothing about lengths or angles has been said. The fewest corners a surface needs closed the counting questions about such surfaces and named the next one: what shape can the surface actually have?

The question has a definite answer, and it is surprising the first time one meets it. The two-handled surface can be built from an octagon of rigid, uniform material — every point looking like every other, the way every point of a flat plane or a round sphere does — but not from a flat octagon or a spherical one. It needs the third uniform geometry, the hyperbolic plane. This essay shows why, builds the octagon that works, measures its angles and its area, and computes the four motions that glue it together, finding that they satisfy exactly the one relation the surface’s loops satisfy.

The angle the gluing demands #

The decisive fact is about the corners. When the octagon’s sides are glued by the word above, its eight corners do not become eight points on the surface, or four, or two. They become one. Follow a corner round: the end of side $a$ is glued to the start of $a^{-1}$, which sits next to the end of $b^{-1}$, which is glued to the start of $b$, and so on through all eight, before the chain returns to where it began. The Euler characteristic of the result is one vertex, minus four edges, plus one face: $1 - 4 + 1 = -2$, which is $2 - 2g$ for two handles.

A uniform surface has no special points, so the single point where all eight corners meet must look like every other point: a full turn of angle around it, no more and no less. The eight corner angles of the octagon therefore have to add to $360°$, which makes each one $45°$ if the octagon is regular.

A flat octagon cannot do that. The angles of any flat polygon with eight sides add to $(8 - 2) \times 180° = 1{,}080°$, three full turns, and that sum is fixed by flatness, whatever the octagon’s shape. Glued up, a flat octagon would have three turns of angle bunched at one point, a cone with too much material, which is not what a uniform surface looks like. The same count for a square, glued by $a\,b\,a^{-1}\,b^{-1}$ into a torus, gives four corners becoming one point and angles needing to add to $360°$ — exactly what a flat square has. That is why the torus can be flat, and it is the whole of the difference between one handle and two.

The flat octagon is not useless; it describes the same surface with its geometry arranged differently. Glued flat, the octagon gives a surface that is perfectly flat everywhere except at the single corner point, where three full turns of angle are packed into a space that should hold one. That point is a cone point with an angle excess of $4\pi$, and by the polyhedral rule that seven hundred and twenty degrees of gap measured — curvature at a corner is the full turn minus the angle actually there — its curvature is $2\pi - 6\pi = -4\pi$. All of the surface’s curvature is gathered at that one point. The question of a uniform geometry is whether the same total, $-4\pi$, can be spread evenly over the whole surface instead, so that no point is special. The answer is yes, and spreading it evenly means curvature $-1$ over an area of $4\pi$.

Shrinking angles by growing the polygon #

The geometry that can supply small angles is the one where triangles are thin. On a sphere a triangle’s angles add to more than $180°$ by an amount proportional to its area, which is what the triangle that a globe gets wrong measured. In the hyperbolic plane they add to less, by an amount proportional to the area too. A small hyperbolic polygon is nearly flat and has nearly flat angles; a large one has much smaller angles; and as it grows towards infinity its angles shrink towards nought.

For a regular polygon with $n$ sides and corners at hyperbolic distance $R$ from its centre, splitting it into $2n$ right triangles from the centre gives the relation

$$\cosh R = \cot\frac{\pi}{n}\,\cot\frac{\alpha}{2}$$

between the size $R$ and the corner angle $\alpha$. As $R$ goes from nought to infinity, $\alpha$ falls continuously from the flat value to nought, passing every angle in between exactly once. For the octagon the angle $45°$ is reached when $\cosh R = \cot^2(\pi/8) = 3 + 2\sqrt2$, at $R = 2.4485$. The figure draws the curve for four, eight and twelve sides, and for the octagon and the twelve-sided polygon it also measures the angle directly from the drawing — the angle between the two arcs that meet at a corner in the Poincaré disc, computed from their tangent lines — and checks that it agrees with the formula to a millionth of a degree.

The square is the exception that proves the rule. It needs $90°$, which is the flat square’s own angle, and the hyperbolic square has that angle only in the limit of zero size. There is no hyperbolic torus of this kind; the torus’s natural geometry is flat.

Drawing the plane in a disc #

The hero figure draws the hyperbolic plane by the method of the Poincaré disc: the whole infinite plane is squeezed into the inside of a circle, straight lines become arcs of circles that meet the rim at right angles, and distances are stretched by the factor $2/(1 - |z|^2)$ so that the rim is infinitely far from everything. Angles in the drawing are the true hyperbolic angles, which is why the $45°$ corners can be read straight off the page; lengths and areas are not, which is why the octagon looks large and its neighbours look small.

The octagon that works has its corners at $84\%$ of the way to the rim. Its eight sides are arcs bulging inward, and each corner is visibly sharp, as $45°$ should be. Around it are drawn its neighbours: the copies of the octagon across each side, and the copies across those. In the drawing they crowd towards the rim, ever smaller, but in hyperbolic terms every copy is exactly the same size as the first, and exactly eight of them meet at every corner — which is the gluing condition seen from the outside. Those copies tile the whole hyperbolic plane, and that tiling is the universal cover of the surface: walking across a side of one tile into the next is the same as walking across the glued seam on the surface itself.

The area is fixed by the topology #

The hyperbolic octagon’s area is not a free parameter. For a hyperbolic polygon the area is the angle deficit: the flat polygon’s angle sum minus the actual one. The octagon’s flat sum is $6\pi$ and its actual sum is $2\pi$, so its area is $4\pi$. For the $4g$-sided polygon of the surface with $g$ handles, the same subtraction gives $(4g - 2)\pi - 2\pi = 4\pi(g - 1)$.

The figure checks this by measurement rather than formula. In the disc the hyperbolic area element is $4/(1 - |z|^2)^2$ times the ordinary one, and integrating it along 16,000 rays per side from the centre out to each polygon’s arcs gives $12.566370$ for the octagon, against $4\pi = 12.566371$; the polygons for three to six handles agree just as closely with $8\pi$, $12\pi$, $16\pi$ and $20\pi$.

Written in terms of the Euler characteristic $\chi = 2 - 2g$, the area is $-2\pi\chi$. That is the Gauss–Bonnet theorem for a surface of constant curvature $-1$: the total curvature of a closed surface is $2\pi\chi$, whatever its shape, and on a surface whose curvature is $-1$ everywhere the total curvature is minus its area. Seven hundred and twenty degrees of gap met the polyhedral version, where the angle defects at the corners of a convex solid always add to $4\pi = 2\pi \times 2$. Here the defect is spread evenly over the whole surface instead of concentrated at corners, and it has the opposite sign. Every hyperbolic surface with two handles, of whatever shape, has area exactly $4\pi$.

Four motions and one relation #

The gluing has a second description, which turns the topology into algebra. Each pair of sides is matched by a motion of the hyperbolic plane — an isometry that carries one side of the octagon onto its partner and carries the octagon onto its neighbour across that side. In the disc, the orientation-preserving isometries are exactly the maps

$$z \mapsto \frac{\alpha z + \beta}{\gamma z + \delta}$$

that preserve the unit disc, the Möbius transformations the sphere that complex numbers live on studied as rotations of the Riemann sphere. Each is recorded as a two-by-two complex matrix of determinant one.

Each pairing is computed here as two reflections in succession: first the octagon’s own mirror symmetry that swaps a side with its partner, then the reflection in the partner’s arc, which carries the octagon across that side. The composition is orientation-preserving, and the figure checks that it maps every point of one side onto the other side’s circle. All four matrices have trace $3.414214$, which is $2 + \sqrt2$. A trace above two means the motion is a hyperbolic translation, sliding the plane along an axis without fixing any point inside it, and the distance it slides is $2\,\operatorname{arcosh}(|\mathrm{trace}|/2) = 2.2568$.

Then the decisive check. The fundamental group of the two-handled surface, its group of loops that cannot be pulled tight, is generated by four loops $a, b, c, d$ subject to exactly one relation, which comes from walking once round the octagon’s boundary:

$$a\,b\,a^{-1}\,b^{-1}\,c\,d\,c^{-1}\,d^{-1} = 1.$$

Multiplying the four computed matrices in that pattern — the commutator of the first two times the commutator of the last two — gives the identity matrix to within $4 \times 10^{-14}$, rounding error. The relation that topology derived by counting corners is satisfied by four concrete motions of the hyperbolic plane. The group they generate is the fundamental group, now acting by isometries; the octagon is a tile for that action; and the surface is the plane divided by the group.

Why the copies never overlap #

That the four motions satisfy the relation is necessary for the octagon’s copies to fit together, but it is not obviously enough. A family of motions could map the octagon to copies that overlap one another, so that the plane is covered several times over, or covered with gaps. The guarantee that this does not happen is a theorem of Poincaré’s from the 1880s about polygons and their side pairings. If every side is paired with a side of the same length by an isometry, and the corners fall into cycles whose angles add to a full turn, then the copies of the polygon under the group the pairings generate tile the plane exactly: they cover it, they meet only along sides and corners, and the group acts without any element other than the identity fixing a point.

For the octagon, every side has the same length, the eight corners form one cycle, and their angles add to $8 \times 45° = 360°$. Poincaré’s theorem therefore applies, and it delivers everything the figures observe: the eight copies at each corner, the tiling of the whole disc, and the fact that the octagon is a faithful picture of the surface. Had the angles been $40°$, the copies would have left a gap of $40°$ at every corner; had they been $50°$, the copies would have overlapped; only the angle the gluing demands makes the tiles fit.

Copies that grow like the plane does #

The last figure counts the tiles. Applying every word of up to six of the gluing moves and their inverses to the octagon, and discarding repeats, gives 155,577 distinct copies; the figure counts how many have centres within hyperbolic distance $R$ of the first.

There is one copy until $R$ reaches $3.06$, the distance across a side to the eight neighbours; then nine, then twenty-five, and by $R = 7$ there are 265, against the 274 that the area of a hyperbolic disc of radius 7, $2\pi(\cosh 7 - 1)$, divided by $4\pi$, predicts. A hyperbolic disc’s area grows like $e^R$, and the count follows it. In the flat plane, the copies of a torus’s square within distance $R$ grow like $R^2$. That contrast is a theorem with a name: John Milnor proved in 1968 that the fundamental group of a compact manifold of negative curvature grows exponentially, so the number of distinct words of length $L$ in $a, b, c, d$ grows exponentially with $L$ — a fact about algebra that the geometry makes visible. A torus’s group, by contrast, grows only polynomially.

Why every surface has a geometry #

The octagon is one example of a general theorem. Every closed orientable surface can be given exactly one of three uniform geometries: the sphere’s, if it has no handles; the flat plane’s, if it has one; and the hyperbolic plane’s, if it has two or more. The sign of the curvature is the sign of the Euler characteristic, which is the polyhedral angle-defect count turned into geometry. The theorem behind it, uniformisation, was proved by Henri Poincaré and Paul Koebe independently in 1907, after Poincaré had developed the hyperbolic tilings and their groups in the 1880s and noticed that the transformations of his tilings were the ones he had already met in non-Euclidean geometry.

The octagon used here is not the only hyperbolic shape the two-handled surface can take. Any octagon with angles adding to $2\pi$ and paired sides of equal length works, and so do many other polygons; the pants decomposition records the shapes by three lengths and three twists, so the two-handled surface has a six-dimensional family of hyperbolic shapes, its Teichmüller space. Gluing the same regular octagon’s opposite sides instead, by the word $a\,b\,c\,d\,a^{-1}\,b^{-1}\,c^{-1}\,d^{-1}$, gives a different point of that family, the Bolza surface, which has more symmetries than any other surface with two handles.

Still open: the shortest loop #

Every hyperbolic surface has a shortest closed geodesic that cannot be shrunk to a point, its systole, and the natural question is how long it can be. On the octagon surface here, the side-pairing translations slide by $2.2568$, so loops of that length exist; the Bolza surface does better. For two handles the answer is known, and it is the Bolza surface: no hyperbolic surface with two handles has a longer shortest loop. For most numbers of handles beyond two, the surface with the longest systole is not known.

The general asymptotic question is open too. The systole of a hyperbolic surface with $g$ handles can be at most about $2\log g$, by an area argument — a disc of radius half the systole must fit on the surface, and the area is only $4\pi(g - 1)$ — and surfaces built from congruence subgroups reach about $\tfrac43 \log g$, a 1994 construction by Peter Buser and Peter Sarnak. Whether the true maximum grows like $2\log g$, like $\tfrac43\log g$, or like something in between, is not known.

A topology that chooses its geometry #

The octagon began as a rubber sheet with a word written round its edge. The word decides that the eight corners become one point, the single point decides that the angles must add to one turn, and one turn of angle in eight corners decides that the octagon must be hyperbolic, with corners of $45°$, a corner $2.4485$ from its centre, and area $4\pi$. The same word, read as a relation among loops, is satisfied by four computed motions of the plane to fourteen decimal places, and the copies of the octagon those motions produce fill the plane at the exponential rate its area dictates. Nothing in the topology mentioned curvature; the curvature was forced, and so was its amount.