Every corner pays for itself

A cube has $8$ corners, $12$ edges and $6$ faces. Subtract the edges from the corners and add the faces:

$$8 - 12 + 6 = 2.$$

A tetrahedron: $4 - 6 + 4 = 2$. An octahedron: $6 - 12 + 8 = 2$. A pyramid on a square base: $5 - 8 + 5 = 2$. A football, with its twenty hexagons and twelve pentagons: $60 - 90 + 32 = 2$.

What is being measured #

The counts differ wildly and the combination does not, which is the signal that the combination is measuring something the individual counts are not.

The first thing to establish is how robust it is. Take a cube and push a corner in, and the counts do not change. Slice a corner off, and $V$, $E$ and $F$ all change — a cut corner replaces one vertex with three, adds three edges, and adds one face, so the sum shifts by $+2 - 3 + 1 = 0$. Draw a new edge across a face, splitting it in two: one edge added, one face added, and the sum is unmoved because they cancel.

Every operation that keeps the surface a surface leaves the total alone. The quantity does not care about the shape at all — only about the fact that the shape is a sphere in disguise.

That is what makes it a topological invariant, and the word earns itself here: $V - E + F$ is a property of the underlying surface, and the polyhedron is merely a way of writing the surface down. Any two polyhedra that could be deformed into each other without tearing give the same answer, and every convex polyhedron can be deformed into a sphere.

The proof, by flattening #

The cleanest argument removes the third dimension first, which is the move Euler made with the bridges and the same one made here.

Imagine the polyhedron is made of rubber. Puncture one face and stretch the hole open until the whole thing lies flat on the table. Every vertex and edge survives; every face survives too, except that the punctured one has become the infinite region outside the drawing.

That last clause is the whole trick and the place every first reading goes wrong. The flattened cube looks like it has five faces. It has six, because the outside counts — the face that was punctured did not disappear, it became unbounded.

Now the counts are about a flat graph, and the sum can be reduced. Two moves suffice:

• Remove an edge that separates two faces. One edge goes, and the two faces merge into one, so $E$ and $F$ each drop by one and the sum is unchanged.
• Remove an edge with a loose end, along with that end. One edge and one vertex go, and no face changes, so $V$ and $E$ each drop by one and the sum is unchanged.

Alternating those two, any connected planar graph can be stripped down to a single vertex: no edges, one vertex, and one face — the outside. And $1 - 0 + 1 = 2$.

Since nothing along the way changed the total, the total was $2$ to begin with. The proof is complete and it never mentions a solid.

Two conditions are being used and both are worth surfacing, since each is a place the argument fails if dropped. The graph must be connected — two separate components strip down to two vertices rather than one, giving $2 - 0 + 1 = 3$, and indeed two cubes standing side by side have $\chi = 4$. And it must be planar, drawn without crossings, or the regions it cuts the plane into are not well defined. The flattening supplies both, which is why the puncture-and-stretch is doing more than convenience.

The reduction also gives a bound worth having. Each face of a simple planar graph is bounded by at least three edges and each edge borders at most two faces, so $2E \ge 3F$. Substituting into the formula gives

$$E \le 3V - 6,$$

which says a planar graph cannot have many edges — a graph on $5$ vertices can have at most $9$, so the complete graph on five vertices, with its $10$, cannot be drawn flat. That is the fact underneath the four colour theorem’s restriction to planar graphs, and it falls straight out of a counting argument about faces.

What the two is #

The natural next question is why the answer is $2$ rather than some other number, and the answer is that $2$ is a property of the sphere.

Run the same count on a surface with a hole through it — a doughnut, tiled with polygons — and it comes out $0$. Two holes gives $-2$. In general, a surface with $g$ handles has

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

So the alternating sum counts handles, and it does so from data that never mentions them: a list of corners, edges and faces, with no notion of a hole anywhere in it. That is a genuinely startling piece of bookkeeping. A creature living on the surface, able to survey its own triangulation but unable to see the surface from outside, could compute the number of holes by counting and subtracting.

This quantity is called the Euler characteristic, written $\chi$, and it is the first topological invariant anyone meets. It is complete for closed orientable surfaces — two of them are the same shape if and only if their characteristics agree — which is a rare and powerful situation. Most invariants distinguish some things and miss others; this one settles the whole classification with a single integer.

It also has a consequence for maps rather than shapes: because the sphere and the disc have no holes, a continuous map of either into itself must leave a point where it was, and on the annulus — where $\chi$ is $0$ — it need not. The characteristic and the fixed-point guarantee are two readings of the same absence.

It stops being complete the moment orientability is dropped. A Möbius band has $\chi = 0$, and so does a cylinder, and they are not the same surface — so the characteristic alone cannot tell them apart, and the classification needs orientability as a second ingredient.

The five solids, again #

The formula has an immediate payoff, and it is a second route to a result this collection has already reached one way.

Suppose a solid has every face a regular $p$-gon and every vertex meeting $q$ edges. Count the edge-ends two ways: each face contributes $p$ of them and each edge has two, so $pF = 2E$; each vertex contributes $q$ and again each edge has two, so $qV = 2E$. Substituting into $V - E + F = 2$ and dividing by $2E$:

$$\frac1p + \frac1q = \frac12 + \frac1E.$$

Since $E$ is positive, the left side exceeds $\tfrac12$. With $p$ and $q$ at least $3$, the whole-number solutions are $(3,3)$, $(4,3)$, $(3,4)$, $(5,3)$ and $(3,5)$ — five of them, which are the five Platonic solids.

The angle-budget argument in that essay reaches the same list from a completely different direction, and it is worth being precise about what each one supplies that the other does not. The angle argument makes it obvious why the list is finite — there is only so much angle at a corner. This one makes it obvious why the entries pair up — swapping $p$ and $q$ swaps the members of each dual pair, and the swap is visible in the notation.

Neither picture carries both facts comfortably, which is the usual situation when one answer has two proofs.

The count that is not a count #

There is a reframing that explains where the alternating signs come from, and it is worth a paragraph because the minus sign otherwise looks arbitrary.

The pattern $+V - E + F$ continues in higher dimensions as $+ - + -$, and the reason is that the pieces have dimensions $0, 1, 2$ and each is being weighted by $(-1)^{\text{dimension}}$. So the characteristic is a signed count of cells, and the alternation is what makes the total insensitive to how the surface was cut up.

That insensitivity is the point. Chop a face into two and both $E$ and $F$ rise by one, with opposite signs, cancelling. Add a vertex in the middle of an edge and both $V$ and $E$ rise by one, cancelling. The alternating sum is precisely the combination that every legal subdivision leaves alone, and it is essentially the only one — which is why it is the quantity that survives to be a property of the surface rather than of the drawing.

Read that way, the number is not counting anything about the polyhedron. It is what remains after everything that depends on the choice of polyhedron has been cancelled out, and the surprise is that anything remains at all.

The exceptions, and the century they took #

The formula as usually stated is false, and the history of noticing that is a good corrective to how clean the previous sections look.

Take a cube with a square tunnel bored through it. Counting carefully gives $\chi = 0$, not $2$ — because the shape is a doughnut with corners, not a sphere with corners. Take two cubes joined at a single vertex: $\chi = 3$. Take a cube with a square hollow inside it, a shell within a shell: $\chi = 4$.

None of these is a trick. They are ordinary solids, and each was offered during the nineteenth century as a counterexample to a theorem that had been considered proved since 1750. Lakatos’s Proofs and Refutations is built around this episode, and its argument is that the word polyhedron was not stable: each counterexample forced a revision, either by excluding the offending shape or by adjusting the theorem.

The resolution was to be explicit about the hypothesis. Euler’s formula holds for polyhedra whose surface is topologically a sphere — connected, no tunnels, no shells inside shells, every face a disc. Once stated that way it has no exceptions, and every counterexample above is seen to violate it.

That is the ordinary fate of a theorem that arrives before its definitions. The proof was not wrong; it was a proof about a class nobody had yet delimited, and delimiting it took a hundred years and a good deal of argument.

What the picture cannot show #

The flattening figure draws one polyhedron pushed flat, and the argument needs every polyhedron to be flattenable. That the puncture-and-stretch always works is the substance of the proof and it is assumed by the drawing rather than demonstrated — a shape that could not be flattened would look exactly as convincing before the attempt.

The outer face is worse, because the picture cannot draw it. It is the unbounded region, extending past every edge of the page, and a reader counting faces in the figure will count five and get $1$ rather than $2$. The single most error-prone step in the proof is invisible by construction.

Nor can any drawing here show what the answer means. That $\chi = 2 - 2g$ counts handles is the reason the quantity is interesting, and no picture of a sphere-like solid contains any information about doughnuts. The figures establish that the sum is $2$ for the cases drawn, and the content is what happens for the cases not drawn.

The ladder from here #

Rungs above: the flattening proof drawn stage by stage, with the graph reduced to a point. Non-convex polyhedra and the tunnel counterexample, counted. The classification of surfaces, where $\chi$ and orientability together settle everything. The characteristic for higher-dimensional complexes, with the alternating sum continued. Descartes’ angle-defect theorem, which is this formula in geometric clothing and totals $720°$ for the same reason the sum is $2$. Gauss–Bonnet, the smooth version, where curvature integrates to $2\pi\chi$. Homology, which is what the alternating sum is a shadow of. Triangulations, and the theorem that the answer does not depend on which one is used. And the Königsberg graph, where planarity and this formula together give Kuratowski’s criterion.

The shape of the idea #

The habit worth taking from this is the one the whole subject is built on: look for the combination of measurements that does not change.

Every individual count here is fragile. $V$, $E$ and $F$ change under the mildest deformation and carry no information about the surface. The combination $V - E + F$ is immovable, and because it is immovable it belongs to the surface rather than to the description.

That is what an invariant is, and finding them is most of topology. The rest of this collection is full of the same move at smaller scale: the determinant is what survives a change of basis, the degree parities are what survive deleting a city, the width is what survives turning a shape. In each case a great deal of information is thrown away deliberately, and what is left answers a question the full description could not.