Slid, but never turned

Every dissection on this ladder so far turns something. The triangle-to-rectangle step turns two pieces a half turn each; the staircase slides one piece and slides it only because the two pieces were cut to match; the Pythagorean rearrangement turns four of its five.

Take the turn away. The pieces may be cut anywhere and slid anywhere, and that is all — no rotation, no reflection, nothing but translation. The operation set is strictly smaller, so the reach can only shrink, and this time it does.

A triangle and a square of the same area cannot be cut into pieces that slide from one to the other, however many pieces are allowed. The obstruction is the quantity in that figure, it was found by Hadwiger and Glur in 1951, and it is a second worked example of the manoeuvre the rung below used on the Dehn invariant.

The quantity #

Take a polygon and walk round its boundary anticlockwise. Each edge has an outward normal — a direction pointing away from the interior — and a length.

For a direction $u$, define

$$\Phi_P(u) = \sum_{\text{edges with normal } u}\!\!|e| \;-\; \sum_{\text{edges with normal } -u}\!\!|e|.$$

Only edges pointing exactly along $u$ or exactly against it contribute, so the quantity is nought at almost every direction and has a spike at each of the polygon’s edge normals. It is a finite list of numbers rather than a function in any useful sense, and the figure draws it as spikes for that reason.

For a rectangle the four normals are $\pm\hat x$ and $\pm\hat y$, and the two edges facing $\hat x$ and $-\hat x$ have the same length. So the difference is nought, and the same holds vertically: a rectangle’s quantity vanishes in every direction. A parallelogram’s does too, for the same reason — opposite sides are parallel and equal.

For a triangle the three normals are three different directions with no opposite pair among them, so each spike is the whole length of its edge and nothing cancels it. A triangle’s quantity does not vanish, and the three spikes are the three side lengths.

Why cutting cannot change it #

The whole of the argument is one observation about a cut.

Cut a polygon along a straight segment. The segment becomes an edge of each of the two resulting pieces — and it is the same segment, so the two new edges have the same length and opposite outward normals, because what is outward for one piece is inward for the other.

So when the two pieces’ quantities are added, the two new contributions cancel exactly. Every other edge is inherited unchanged from the original, with its length and its direction intact. Cutting preserves the total.

Sliding preserves it too, and this is where the restriction earns its keep. A translation moves an edge to a new place and leaves its direction alone, so every normal survives. A rotation does not: it turns every normal, so the spikes move round the dial, and a quantity whose spikes have moved is not the quantity it was.

That is the entire proof. The quantity is preserved by both allowed moves and destroyed by the forbidden one, so two shapes that differ in it cannot be connected by the allowed moves — and a triangle differs from a rectangle in it.

The shape of every invariant on this site #

This is the fourth invariant in the collection, and it is worth putting the four side by side, because the recipe is identical each time and only the moves change.

The linking number is a signed count that a deformation cannot change, because a deformation creates and destroys crossings in pairs. The sign of a permutation is a parity that a redrawing cannot change, because a redrawing adds and removes crossings in pairs. Dehn’s quantity is a sum over edges that a cut cannot change, because a cut splits an angle into two adding back to it. And this one is a signed sum over edges that a cut cannot change, because a cut adds two edges with opposite normals.

All four are of the form: find something the move alters in a way that cancels, and quotient out exactly what does not cancel — the same instinct that produces the degree of a number over the rationals as the obstruction to a straightedge-and-compass construction. Here the quotient is the subtraction — the quantity is a difference between opposite directions rather than a total over all of them — and the subtraction is what makes a cut invisible.

The test of whether a quotient is right is the same each time. Quotient too little and the moves change the quantity; quotient too much and it is nought for everything. A quantity that summed edge lengths without the sign would be the perimeter, which every cut changes; the signed version is invisible to a cut and still distinguishes a triangle from a rectangle, which is exactly the balance wanted.

The theorem, in both directions #

Hadwiger and Glur proved in 1951 that this quantity is not merely an obstruction but the only one.

Two polygons are translation-equidecomposable exactly when they have equal area and $\Phi_P(u) = \Phi_Q(u)$ for every direction $u$.

That is a complete classification, and the shape of it is exactly the shape of Sydler’s theorem one dimension up: area is not enough, one more quantity is added, and the pair together is sufficient. The rung below recorded that in space the answer was that “one number was not enough and two are”; here the plane needs one number and a function-shaped thing, and it is the restriction of the moves rather than the change of dimension that made the extra quantity appear.

Two consequences are worth reading off.

Any polygon with a centre of symmetry slides into a rectangle. A centrally symmetric polygon has its edges in opposite pairs of equal length, so every spike cancels, so the quantity vanishes, so it is translation-equidecomposable with any rectangle of the same area. Parallelograms, regular hexagons and centrally symmetric octagons all qualify.

The converse is false, and the default figure carries the counterexample. The L-shaped piece has no centre of symmetry at all, and its quantity vanishes anyway: it has two edges facing right, of lengths $0.25$ and $0.5$, against one facing left of length $0.75$, and the totals cancel even though no individual pair does. So central symmetry is sufficient for the quantity to vanish and is not necessary, and the criterion the theorem actually uses is the weaker one.

That distinction cost a version of the figure. The first implementation compared each edge’s length against the total facing the other way, rather than comparing total against total, and reported the L as obstructed — a shape that in fact slides into a rectangle without difficulty. The assertion that caught it was not about the L; it was the requirement that a centrally symmetric shape’s quantity vanish, which the corrected grouping satisfies and the original did not.

A triangle slides into nothing but a triangle-shaped family. Its quantity has three spikes at three specific directions with three specific values, and any polygon sharing them is reachable. That is a genuine family and not a single shape — a hexagon built by cutting corners appropriately can match a triangle’s spikes — but it is a small family and no rectangle is in it.

The one dissection on this ladder that needs no turn #

There is a construction on the rung below that survives the restriction untouched, and it is worth pointing at because it is the cheapest one there.

Two rectangles have vanishing quantities, so the theorem says a translation-only dissection between them exists, and the staircase is one. The previous rung noted that this was the awkward dissection to hinge, precisely because its motion is a translation rather than a rotation.

So the same property makes it easy here and hard there, and the two restrictions are not nested. Hinging demands rotations and forbids lifting; translation-only demands slides and forbids rotations; and a dissection can satisfy either without satisfying the other. The staircase satisfies the second and not the first; the triangle’s three pieces satisfy the first and not the second.

That is worth registering because “a smaller operation set” sounds like it should produce a chain of ever-weaker reaches, and it does not. Narrowing in two different directions gives two incomparable classes, and the intersection — dissections that are both hinged and translation-only — is very small, since a hinge’s motion is a rotation and a rotation is what is forbidden.

What one turn buys #

The natural question is how much has to be added back to recover the full theorem, and the answer is clean and surprising.

Allow half turns as well as translations and equal area is enough again. A half turn is a rotation by $180°$, so it reverses every normal — it sends the spike at $u$ to a spike at $-u$ and negates the difference. That changes the quantity’s sign rather than its structure, and with half turns available a triangle can be brought to a rectangle in three pieces, which is the construction the whole ladder rests on.

So the obstruction is destroyed by the single most restricted rotation there is. The operation set that is genuinely weaker than the plane theorem’s is translations alone, and adding any half turn at all recovers the whole thing.

That is the sharpest available answer to “how much does the restriction cost”. It costs a $180°$ rotation, and no less than one.

What the restriction is worth knowing about #

It would be reasonable to ask why anybody would forbid the turn, since nothing physical does. The answer is that the restricted question is the one that generalises, and the generalisation is where the subject went.

Replace “the plane with translations” by “a space with a group of motions acting on it”, and equidecomposability becomes a question about that group rather than about the shapes. Every group has its own family of invariants; the larger the group, the fewer survive, because more has to be respected. Translations are a small group and have many invariants; all rigid motions form a larger one and leave only area; and the sequence continues in both directions.

Two ends of it are worth naming. At the small end, the group with only the identity in it makes equidecomposability into congruence-of-the-pieces-in-place, and every polygon is equidecomposable with itself and nothing else — the invariants are everything. At the large end, allow all measure-preserving bijections and every pair of equal-area sets is equidecomposable trivially, with the pieces being arbitrary and unusable.

The interesting groups are in the middle, and the plane’s rigid motions turn out to sit at exactly the place where area alone suffices. That is the content of the Bolyai–Gerwien theorem restated in these terms, and it is a much less obvious statement in that form: the group is precisely large enough to kill every invariant but one, and one step smaller it is not.

The Banach–Tarski construction is the far end of the same axis. Its group is the rigid motions of space, which is not large enough to make equal volume sufficient — Dehn’s invariant survives — and what makes the construction work is dropping the requirement that the pieces be measurable, which removes the volume invariant itself rather than enlarging the group. Two different ways of breaking a classification, and only one of them is about the moves.

Where it needs care #

The quantity is a function on directions and not a number. It cannot be written as a single value, which is the same difficulty Dehn’s quantity has for a different reason. What is checkable is that it is non-zero somewhere, which is all an impossibility proof needs.

Orientation of the boundary has to be fixed. The outward normal is outward only if the polygon is traversed in a chosen sense, and the figures reverse any shape given the other way round before computing anything. A sign error here would report a rectangle as having a non-vanishing quantity, which is the kind of mistake that looks like a discovery.

Degenerate edges break it. A polygon with a zero-length edge, or with two collinear edges meeting at a straight angle, has normals that are undefined or duplicated, and the sums are then wrong in a way no drawing shows. The generator refuses a zero-length edge outright.

And the classification is for polygons. Regions with curved boundary have a normal at every point and the sums become integrals; the corresponding statement exists and is not what is drawn here.

Where it came from, and what it belongs to #

Hadwiger’s work on this is part of a much larger programme, and knowing which one explains why the answer has the shape it has.

He was classifying the valuations on polygons: functions assigning a number to each polygon such that the value of a union is the sum of the values minus the value of the overlap. Area is one. Perimeter is one. The quantity here is one for each direction. Hadwiger’s theorem — the famous one, from 1957 — says that in $n$ dimensions the valuations invariant under all rigid motions and continuous in the right sense form a space of dimension $n+1$, spanned by volume, surface area, and the intermediate quantities. The Euler characteristic is the last of them, at the opposite end from volume, and it is a valuation for exactly the same reason: cutting a shape in two adds the piece counts and subtracts the cut.

Restrict the group of motions and the space of invariant valuations grows, because fewer things have to be respected. Translation-only dissection has more invariants than rigid-motion dissection for exactly that reason, and this quantity is one of the extra ones. The 1951 result is a special case of a general principle: the smaller the group, the finer the classification, and the more obstructions there are.

That reading also predicts the previous section’s answer. Adding half turns enlarges the group from translations to translations-with-half-turns, and the extra invariants that do not survive the enlargement are exactly the ones that change sign under a half turn — which is this one, and, as it turns out, all of them.

What the pictures cannot show #

The dials draw spikes at a handful of directions and the quantity is defined at every direction of the circle. Between the spikes it is nought, so nothing is being hidden — but a reader looking at a dial with three marks on it is looking at a function that is zero on a set of full measure, and the drawing gives no sense of that.

The figure shows four shapes and the theorem is about all of them. That two of these four have a vanishing quantity is a computation; that a vanishing quantity is sufficient for the dissection to exist is Hadwiger and Glur’s theorem, and no picture here contains any part of its proof.

And no figure on this page shows a translation-only dissection succeeding. The parallelogram’s quantity vanishes, so it can be slid into a rectangle, and the construction that does it — cut off a triangle from one end and slide it to the other — is not drawn, because the essay is about the shapes for which no such construction exists. The impossibility is the subject and impossibility has no picture, so what is drawn is the quantity that establishes it.

The ladder from here #

Rungs above: the piece count, which asks how expensive a dissection that does exist has to be. Hadwiger’s classification of valuations, of which this quantity is one member and area another. Translation-equidecomposability in three dimensions, where the invariant becomes a function on the sphere of directions. Equidecomposability under a general group of motions, where the invariants are the group’s own cohomology and this becomes a special case with a name. And Sydler’s theorem, which completes the space classification the rung below left open in the same way this one completes the plane’s.

Removing a move to see what it was doing #

The habit is the one the rung below named and this rung is its cleanest instance: an invariant is built against the moves, not against the shapes.

Nothing in the definition of the quantity mentions triangles or rectangles. It mentions edges, lengths and directions, and it was designed so that the two allowed operations leave it alone. That it distinguishes a triangle from a rectangle is a consequence discovered afterwards, and if it had distinguished nothing the quantity would have been correct and useless.

The corollary is a practical way to look for one. Take the operation that is allowed, ask what it changes, and look for the part of the change that cancels. A cut adds two edges — the cancellation is that their normals are opposite. A deformation adds two crossings — the cancellation is that their signs are opposite. A move in a congestion game changes one resource’s load by one — and the potential’s change is exactly the mover’s gain, one field away, from the same instinct applied to a quantity that decreases rather than one that is preserved.