A theorem you can operate

Two From One

You can cut a solid ball into a handful of pieces and reassemble them — moving each one rigidly, no stretching — into two solid balls, each the same size as the first. That is the Banach–Tarski theorem, and it is true. Not in a workshop: in the mathematics. Here is the engine most retellings skip — and the exact place the knife stops.

The doubling doesn't come from “infinity,” and nothing gets stretched. It comes from one group — and you can hold it. Below is the free group on two letters, drawn as an infinite tree where every point is a word. Take the quarter of it that starts with a⁻¹, shift the whole quarter by a single letter, and watch it swell to cover three-quarters. Do that with all four quarters and one tree becomes two, each complete. That swelling is Banach–Tarski, stripped to its skeleton.

Instrument IThe tree that doubles itself

Every reduced word in a, a⁻¹, b, b⁻¹ — a string with no letter next to its own inverse — is one node. From the centre (the empty word) four branches leave; each node after splits into three, forever. Colour every node by the letter it starts with, and the tree falls into four coloured pieces.

The move that matters: the branch that starts with a⁻¹ is one quarter of the tree. Prepend a single a to every word in it — the a·a⁻¹ cancels — and each word loses its first letter. What comes out is every word that doesn't start with a: three quarters of the whole group, plus the centre. A quarter, rigidly shifted by one letter, became three quarters. Nothing was added; the group is just big enough, in the right way, to be its own proper part.

So the whole group is W(a) together with a·W(a⁻¹) — the a-branch, plus the shifted a⁻¹-branch — reassembled by one rigid motion. And independently, it is also W(b) together with b·W(b⁻¹). Four pieces of one group; two copies of it out. Press double it to watch each copy fill in — every node of each new tree wearing exactly two of the four original colours.

The bridgeFrom words to a spinning ball

A group of words is not yet a ball. The link is that this exact free group can be made of rotations. Take two rotations of a sphere, each by the angle arccos(⅓) ≈ 70.5°, about two perpendicular axes:

Compose these two rotations in any order, never letting one immediately undo the other, and you never once return to where you started — no sequence of them is the do-nothing rotation except the empty one. That is exactly the condition for a free group: the rotations obey no law but the unavoidable one. So the abstract tree above is a set of rotations, and its self-doubling becomes a doubling of the sphere's points — then, radius by radius, of the ball. (This step needs the Axiom of Choice to pick one point per orbit, and a little bookkeeping to absorb the rotation axes' fixed points; see the apparatus.) Świerczkowski proved the two rotations free in 1958.

Instrument IIWhy the ball, and never the disk

Here is the whole difference between three dimensions and two. Build a word by tapping the letters. On the flat disk, the two rotations are rotations of the plane — and plane rotations commute: spin then spin, the order never matters. On the sphere, the two arccos(⅓) rotations about different axes do not commute. Watch what that does to the same word.

On the plane, a·b·a⁻¹·b⁻¹ brings the marker exactly home: the rotations undo each other, the word collapses to nothing, and every longer word collapses too. A commuting pair can never spell a free group — so on the plane there is nothing to double. On the sphere the same word lands somewhere new and stays there. In exact arithmetic it sends the point (1,0,0) to (33, −44√2, 40)⁄81 — still dead on the unit sphere (33² + (44√2)² + 40² = 81²), but not home. That refusal to return, compounded forever, is the free group — and the whole reason a ball can be doubled and a coin cannot.

The apparatus is the pointWhere the knife stops

The reason this never becomes a way to counterfeit gold is not a loophole — it is the whole content of the theorem, and the part worth keeping.

The pieces have no volume

If the pieces had a well-defined volume, rigid motion would preserve it, and one ball's worth of volume could not equal two. So at least one piece is non-measurable: not a solid, not a shape with a boundary, but a fog of points with no volume at all. Nothing about volume is violated — volume is simply undefined on these sets, so the addition that would give a contradiction never applies. You could no more machine these pieces than 3D-print a cloud of individually chosen points.

It costs the Axiom of Choice

Picking one representative from each of uncountably many orbits, all at once, is exactly the Axiom of Choice — and the construction genuinely needs it. Drop that axiom (there are consistent mathematics where every set of reals is measurable) and the paradox vanishes. Banach–Tarski is a fact about a particular choice of axioms, not a procedure anyone could run.

The plane is safe — and that is a theorem too

Why can space be doubled but not the plane? Because in 1923 Banach proved that on the line and on the plane you can hand every single set an area — a finitely-additive, motion-invariant area extending the ordinary one — so no rearrangement by rigid motions can ever change a total. In three dimensions no such volume exists. Tarski pinned down precisely why: a shape can be paradoxically doubled by a group of motions exactly when that group is non-amenable, and the rotations of space are non-amenable because they contain the free group you just operated. The plane's motions are amenable; they contain no such free pair (two plane rotations always commute). The free group is not a curiosity in this story — it is the hinge the whole dimension turns on.

$ node research/banach-tarski/verify.mjs

PART A — the free group F2 = <a,b>, doubled by exact word combinatorics
  ok  reduced-word counts follow 1, 4, 4·3^(n−1)
  ok  W(a),W(a⁻¹),W(b),W(b⁻¹) are disjoint and cover F2 ∖ {e}
  ok  copy 1:  F2 = W(a) ⊔ a·W(a⁻¹)   — a whole group from 2 pieces
  ok  copy 2:  F2 = W(b) ⊔ b·W(b⁻¹)   — a second whole group from 2 more
  ok  two disjoint copies of F2 assembled from the 4 pieces of one

PART B — the free group lives in SO(3), not SO(2): exact ℤ[√2]
  ok  SO(2): every two rotations commute ⇒ a·b·a⁻¹·b⁻¹ = e  (no free group)
  ok  SO(3): the same word MOVES (1,0,0) → (33+0√2, 0−44√2, 40+0√2) / 3^4
  ok  no reduced word up to length 8 is the identity rotation (13120 tested)

ALL CHECKS PASS — 8 passed, 0 failed.