The solid where the answer is not two

Here is a solid. Its faces are flat, its edges are straight, its corners are ordinary corners, and it is a single connected lump of material with a well-defined inside and outside.

Zero, not two. Every count in that caption is made from the object drawn rather than quoted, and the theorem that every corner pays for itself says the answer should have been two.

This is the most useful kind of object in mathematics: one that satisfies every word of a theorem’s statement and none of its conclusion. It does not refute the theorem. It locates the sentence the theorem forgot to say.

What the proof was actually using #

The flattening proof begins by removing one face and pressing the rest of the surface flat into the plane. Try that on the frame. Remove one face and press: the surface will not go flat without tearing, because a loop of it goes round the hole and there is nowhere for that loop to be pressed to.

The two-tree proof fails in a way that is easier to measure. Grow a spanning tree through the sixteen corners: fifteen edges. Seventeen are left over — and they do not form a tree of the faces. They contain two loops, and $17 = (16 - 1) + 2$. The excess is exactly $2g$ for $g = 1$ hole, which is the whole content of

$$V - E + F = 2 - 2g.$$

So the hypothesis being spent, in both proofs and under two different disguises, is that the surface is sphere-like: every closed curve drawn on it separates it into two pieces. On the frame the loop through the hole separates nothing, and that single fact is what breaks both arguments in their own vocabulary.

The same zero, from a square with its edges glued #

The frame is a torus made of flat panels, and there is a completely different way to build a torus that reaches the same zero without any solid at all.

That count is worth doing slowly, because the whole of it happens in the gluing. All four corners of the square are identified to a single point, so $V = 1$. The left and right edges become one edge and the top and bottom become another, so $E = 2$. The square itself is the only face, so $F = 1$. The alternating sum is zero.

Sixteen, thirty-two and sixteen on the frame; one, two and one on the glued square. Two objects with nothing in common as drawings, built by unrelated methods, agreeing on a number — and that agreement is the reason to believe the number belongs to the surface rather than to either construction.

It also shows how little the characteristic cares about size. The frame has sixty-four pieces and the glued square has four, and neither count is more correct. Subdividing any face of the frame into two by drawing a line across it adds one edge and one face, which cancel; adding a vertex in the middle of an edge adds one vertex and one edge, which cancel. Every legal way of cutting the surface up more finely leaves the alternating sum where it was, which is what makes it worth computing from whichever decomposition is convenient.

Which is the counterexample and which is the theorem #

Two responses to the frame are available, and the history of this formula is the argument between them.

The first is to declare it not a polyhedron. Definitions can always be tightened until the awkward object is outside them — require the surface to be convex, or require that every face be simply connected and the surface be homeomorphic to a sphere, and the frame is excluded by fiat. The theorem is saved and nothing is learnt.

The second is to keep the object and ask what it is measuring. That gives $\chi = 2 - 2g$, a formula that covers the sphere, the frame, the two-holed frame and everything else, and turns the awkward case into an instrument: the alternating sum now counts holes.

Imre Lakatos wrote Proofs and Refutations about exactly this choice, staged as a classroom argument in which each proposed counterexample is met by a redefinition, and each redefinition turns out to have a cost. His term for the first response is monster-barring, and his point is not that it is illegitimate — sometimes an object really is outside the intended subject — but that it is a decision, made silently, that determines what the theorem ends up being about.

The second response is the one that produced modern topology. The characteristic became interesting exactly when it stopped always being two.

More monsters, and what each one shows #

The frame is not the only object that satisfies the words. Two others are worth drawing, because each breaks a different unstated assumption, and knowing which is which is the whole exercise.

The nested cube has two boundary surfaces — an outer one and an inner one — and each is an ordinary sphere-like surface with characteristic $2$. Four is $2 + 2$. What this monster shows is that solid and surface are different things, and the theorem is about surfaces: the object has one interior and two boundaries, and the count adds the boundaries up without being told to.

The barring response here is honest and immediate — require the surface to be connected — and it is the right response, because the object is genuinely two surfaces that happen to be nested.

The twin cubes are the subtler case. The surface is connected — the shared corner joins the two — and it is not a sphere with handles either, because the point where they meet is not a point any surface can have: a small disc around it is two discs pinched together, not one disc. The count comes out at $3$, which is $2 + 2$ minus the one vertex shared, and it is not $2 - 2g$ for any $g$.

What this one shows is that the formula assumes a surface is a manifold: every point has a neighbourhood that is an ordinary flat patch. The pinch point is the cheapest possible violation, it is easy to overlook in a definition written in terms of faces and edges, and until it is excluded the theorem is false.

Three objects, three different missing hypotheses: sphere-like, connected, and locally flat. None of them is a pathology dredged up to be difficult. Each is something a person might build out of card.

The count that survives #

What replaces the theorem is better than the theorem was.

Every closed, connected, orientable surface is a sphere with some number of handles, and for a surface with $g$ handles, any subdivision at all into vertices, edges and faces gives

$$V - E + F = 2 - 2g.$$

The subdivision is arbitrary and the answer is not. That is the sentence that makes the characteristic worth having, and the frame is one line of evidence for it: two entirely different ways of cutting the frame into faces give zero, because zero belongs to the frame and not to the cutting.

Read that way, the frame is not a counterexample at all. It is the second entry of a sequence, and the original theorem was the first entry mistaken for the whole.

Where the flattening gets stuck #

It is worth watching the original proof fail rather than being told that it does.

Now attempt it on the frame. Removing one of the sixteen faces leaves a surface that still has a loop running through the hole, and no amount of pressing will lay that loop flat: the material on the inside of the hole would have to pass through the material on the outside. Removing a second face does not help, and neither does choosing more cleverly, because the obstruction is not local. It is the hole.

What can be done instead is to cut the frame along two closed curves — one round the hole and one through it — after which the surface does flatten, into a rectangle. That is the glued square above, arrived at from the other direction, and the two cuts are the two edges of the identified square. Each cut adds material to the count in a way that is exactly bookkeepable, and the bookkeeping is where the $2g$ comes from.

What it costs to keep the monsters #

The general formula is not free, and its cost is that the hypotheses are now long.

The statement that used to be for any polyhedron, $V - E + F = 2$ becomes for any finite cell decomposition of a closed, connected, orientable surface of genus $g$, the alternating sum of the cell counts is $2-2g$ — with cell decomposition carrying conditions of its own, since faces must be discs and not annuli. Drop the disc condition and the frame’s front face can be a single annular face rather than four quads, giving $16 - 32 + 10 = -6$ for the same object, which is nonsense produced by an illegal subdivision.

That is the honest trade. The old statement was memorable and wrong; the new one is correct and has to be read twice. Every condition in it was bought with an object somebody built, and the objects are the reason the conditions are not arbitrary.

What a hole costs, and why it is exactly two #

The number $2g$ has a reading that makes it inevitable rather than empirical.

Cutting a surface along a closed curve that does not separate it — a loop through the hole of the frame, say — leaves a surface that is still connected and has two new boundary circles where the cut was. Capping each of those circles with a disc adds two faces, and the result is a surface with one fewer handle. Following the counts through the operation: the cut adds a copy of the curve’s vertices and edges, which cancel against each other, and the two caps add two faces. So removing a handle raises the characteristic by exactly two, and a surface with $g$ handles is $2g$ below the sphere’s $2$.

That also explains why holes cannot be worth anything else. There is no arrangement of a handle that costs one, or three, because the operation that removes it always adds precisely two discs. The characteristic of a closed orientable surface is therefore always even, and the odd values belong to the surfaces that are not orientable — the projective plane at $1$, and its relatives.

The frame has one handle in exactly this sense, and it is worth noticing that the hole in it — the rectangular void a finger goes through — is not a feature of the surface at all. The surface is the boundary, and what the boundary knows is that it is a torus. A doughnut and a coffee cup are the standard example of the same thing; the frame is the version made of flat panels, and it counts to zero for the same reason.

Where it needs a condition, still #

Orientable is doing work in the general formula, and dropping it changes the arithmetic rather than breaking it. A Klein bottle has $\chi = 0$, like the torus, and is not the torus; the projective plane has $\chi = 1$, which is odd — a value no orientable closed surface can take. So the characteristic alone does not identify a surface: it takes $\chi$ and orientability together, and that pair does identify it completely, which is the classification theorem.

Closed is doing work too. A surface with a boundary — a disc, a cylinder, a Möbius band — has its own characteristic, and the values are $1$, $0$ and $0$ respectively. The disc and the frame’s front face are both $1$, which is a reminder that the characteristic is not counting holes in any everyday sense. It counts a specific algebraic quantity that happens to equal $2 - 2g$ on the closed orientable surfaces.

What the picture cannot show #

The frame is drawn as a wireframe, which is the only way to show all sixteen of its vertices at once, and a wireframe is exactly the picture in which a hole and a dent look the same. Nothing in the figure distinguishes the drawn object from a slab with a rectangular depression in its front face — and that object has $\chi = 2$, not $0$, because a depression is a dent in a sphere.

The counts in the caption are computed from the face list the generator builds, not read off the drawing, so the arithmetic is safe. But a reader checking the claim by counting the dots and lines in the picture is doing something the picture cannot support: some of those edges are behind others, and which is which is the entire difference between the two objects.

That is the standing difficulty with drawing three-dimensional counterexamples, and it is why Lakatos’s book contains no illustrations of its monsters that settle anything. The dispute in it is about what the objects are, and a drawing of a disputed object is a drawing of one side’s opinion.

The ladder from here #

Below: the formula, the two-tree proof, and the angle defect, whose total on the frame is $0°$ rather than $720°$ — the inner corners of the hole have negative defect and cancel the outer ones exactly.

Above and sideways: the classification of surfaces, which the characteristic is one of the two ingredients of; the Heawood bound, where the number of colours a map needs on a surface is computed from its characteristic and the plane’s four is the case the argument cannot reach; and homology, where $\chi$ becomes an alternating sum of ranks and stops being about surfaces at all.

A theorem is what its counterexamples leave #

The lasting point is about how a statement gets its hypotheses.

Nobody wrote sphere-like into Euler’s formula in 1750, because nobody had the concept, and nobody had the concept because nothing had yet forced it. The frame forced it. The list of conditions on the modern statement is not a list of precautions taken in advance by a careful author — it is a list of objects that somebody built, each of which cost the theorem a sentence.

That is the ordinary way a subject becomes precise, and it is worth expecting rather than regretting. The definition of a function was rebuilt around pathological examples; the definition of a set was rebuilt around a list that cannot contain itself; the definition of an integral was rebuilt around functions nobody had thought would be integrated. In every case the monster arrived first and the definition was written afterwards, to say exactly what had been meant all along and had never had to be said.