THE FOLD / CO-OP / THE PULL REQUEST / THE EULER CHARACTERISTIC
THE EULER CHARACTERISTIC
a number three solids cannot tell apart
1 WHAT IT IS · WHAT IT DOES · FACT OR FICTION
Count the vertices of any convex polyhedron, subtract the edges, add the faces. The answer is 2. Not approximately, not usually — always, for a tetrahedron and for a dodecahedron and for a sphere chopped into five thousand triangles. Euler noticed it in 1750 and could not prove it; the number turns out to depend on nothing about the shape except how many holes it has. Punch one hole through and it becomes 0, permanently, for every possible triangulation.
LIT verified live: all five Platonic solids give χ = 2 — tetrahedron 4−6+4, cube 8−12+6, octahedron 6−12+8, dodecahedron 20−30+12, icosahedron 12−30+20; subdividing a sphere five times takes the face count from 20 to 5120 and the vertex count from 12 to 2562 while χ does not move off 2; and four torus grids — 3×3, 4×5, 6×6, 8×12 — all give χ = 0, confirmed by explicitly enumerating and de-duplicating the edge set rather than applying a formula.
LIT verified live: all five Platonic solids give χ = 2 — tetrahedron 4−6+4, cube 8−12+6, octahedron 6−12+8, dodecahedron 20−30+12, icosahedron 12−30+20; subdividing a sphere five times takes the face count from 20 to 5120 and the vertex count from 12 to 2562 while χ does not move off 2; and four torus grids — 3×3, 4×5, 6×6, 8×12 — all give χ = 0, confirmed by explicitly enumerating and de-duplicating the edge set rather than applying a formula.
2 HOW IT WAS WEAVED · AI + HUMAN
David (human) seated this at THE PULL REQUEST: everyone triangulates differently and the merge produces the same number regardless.
AVAN (AI) built the torus check twice on purpose. Once from the counting formula — an m×n grid on a torus has mn vertices, 3mn edges and 2mn triangles — and once by actually constructing the triangulation, inserting every edge into a set keyed on its endpoint pair, and counting what survived de-duplication. The formula version is the kind of thing that is right until an edge is shared by more or fewer faces than assumed, and the enumerated version cannot make that mistake. They agree. Worth being clear on scope: what is verified here is that χ is constant across these triangulations, not that it is a topological invariant in general — that is a theorem, cited and not proved by any amount of counting.
AVAN (AI) built the torus check twice on purpose. Once from the counting formula — an m×n grid on a torus has mn vertices, 3mn edges and 2mn triangles — and once by actually constructing the triangulation, inserting every edge into a set keyed on its endpoint pair, and counting what survived de-duplication. The formula version is the kind of thing that is right until an edge is shared by more or fewer faces than assumed, and the enumerated version cannot make that mistake. They agree. Worth being clear on scope: what is verified here is that χ is constant across these triangulations, not that it is a topological invariant in general — that is a theorem, cited and not proved by any amount of counting.
3 ONE DIMENSION
V, E and F run away. The alternating sum does not.
4 TWO DIMENSIONS · INTERACTIVE
Subdivide, and watch the only number that refuses to change.
5 THREE DIMENSIONS + AVAN’S INVERSE
The green forward object: a surface being cut finer and finer, holding one number steady.
AVAN’s addition (the inverse-companion): the forward reading is “χ measures the surface.” The inverse is that χ is what remains after you cancel everything that depended on the choice. Split a face and you add a face and an edge; split an edge and you add an edge and a vertex — every local move you can make changes two terms with opposite signs, so the alternating sum is designed to be blind to how you cut. Read backwards, χ is not a fact about the shape that happens to be stable; it is the residue left when the alternating sum has annihilated every arbitrary decision, and its stability is the construction rather than a discovery about it.
LIT all five Platonic solids give chi = 2 - tetrahedron 4-6+4, cube 8-12+6, octahedron 6-12+8, dodecahedron 20-30+12, icosahedron 12-30+20; subdividing a sphere five times takes the face count from 20 to 5120 and the vertex count from 12 to 2562 while chi does not move off 2; and four torus grids (3x3, 4x5, 6x6, 8x12) all give chi = 0, confirmed by explicitly enumerating and de-duplicating the edge set rather than applying a formula
FIG The torus check was built twice on purpose: once from the counting formula (mn vertices, 3mn edges, 2mn triangles) and once by actually constructing the triangulation, inserting every edge into a set keyed on its endpoint pair, and counting what survived de-duplication. The formula version is right until an edge is shared by more or fewer faces than assumed; the enumerated version cannot make that mistake. They agree. Scope: what is verified is that chi is constant across THESE triangulations, not that it is a topological invariant in general - that is a theorem, cited and not proved by any amount of counting.
FIG The torus check was built twice on purpose: once from the counting formula (mn vertices, 3mn edges, 2mn triangles) and once by actually constructing the triangulation, inserting every edge into a set keyed on its endpoint pair, and counting what survived de-duplication. The formula version is right until an edge is shared by more or fewer faces than assumed; the enumerated version cannot make that mistake. They agree. Scope: what is verified is that chi is constant across THESE triangulations, not that it is a topological invariant in general - that is a theorem, cited and not proved by any amount of counting.
◆ sealed .dlw.fold → folded to ROOT_0 · a sphere of THE PULL REQUEST · David Lee Wise (ROOT0), with AVAN