Parity is not enough

A necessary condition that is easy to check is a fine thing to have, and it is a dangerous thing to mistake for the answer. This is the essay where the mistake gets made and then measured.

A loop of paper folds flat when it has an even number of creases round it and a Möbius band when it has an odd number. The two statements have the same shape, they are proved the same way, and one of them is a complete account of when the sheet folds. The other is not, and the gap is not a technicality at the edge: no Möbius band creased square across the strip folds flat at any crease count whatever.

The parity is right every second time and the answer is no every time.

The claim, made concrete #

Take a strip of paper, join the ends with a half twist, and put three creases across it perpendicular to the long edges. Three is odd. The panel colouring exists — three minus ones from the creases and one from the seam multiply to plus one — so the condition that decides a loop of paper is satisfied.

Now try to press it flat. It will not go, and it will not go for a reason that has nothing to do with skill or with the paper being stiff.

What the composition sees #

The instrument that sees it is the closure condition. Walk the folded state round the band, composing a reflection at every crease, and the composition has to equal the map that identifies the two ends. On a Möbius band that map is a slide along the band together with a flip across its axis.

Now compose three reflections in parallel lines. Reflection in a line at angle $\phi$ has a linear part determined by $\phi$; composing two of them gives a rotation by twice the difference of their angles, which for parallel lines is no rotation at all. Composing three gives a reflection again, in a line at the angle of the alternating sum $\phi_1 - \phi_2 + \phi_3$ — and if all three are square across the strip, that alternating sum is the same right angle they all sit at.

So the composition is a reflection in a line across the band. The gluing map’s flip is a reflection in a line along it. Those are ninety degrees apart, and no choice of where the creases sit changes either of them.

That is the argument in full. It is three lines long, it needs no computation, and the counting condition cannot express any part of it, because the counting condition never looks at an angle.

At every odd count, not just at three #

It is worth checking that the obstruction is not an accident of the number three, and the check is immediate once the argument is stated in terms of alternating sums.

With $k$ creases all at ninety degrees, the alternating sum is ninety degrees for every odd $k$ — the terms alternate in sign and one is left over. So the composition is always a reflection across the band, the gluing map always wants a reflection along it, and the mismatch is the same mismatch at three, five, seven and beyond.

The size of the refusal #

A bit says which bands refuse. It does not say how badly, and here the difference in size is the difference between two kinds of failure.

The two failures are worth naming separately. A band whose translation is wrong is a band that would close if the strip were a different length or the creases were somewhere else; the positions are unknowns, the condition is linear in them, and a solution either exists or the geometry rules it out. A band whose linear part is wrong cannot be repaired by moving anything at all, because the linear part does not depend on where the creases are.

The counting argument is a statement about the determinant of the linear part and nothing else. It notices whether the composition turns the paper over, which is one bit of the four numbers in the matrix, and it is silent about the other three.

What the condition on the angles is #

If the parity is one bit of the linear part, the rest of the linear part is a condition too, and it is short.

The composition of $k$ reflections at angles $\phi_1, \ldots, \phi_k$ has a linear part determined by the alternating sum $\phi_1 - \phi_2 + \phi_3 - \cdots$. For the composition to match the Möbius band’s gluing map, that alternating sum has to be a multiple of a straight angle — which is a condition on the angles alone, decided before any crease is placed anywhere.

Square-creased bands fail it, at every odd count, because their alternating sum is a right angle. What the admissible angles look like is a separate matter, and the short version is that they form a curve in the space of possible triples rather than a region.

And a band that does fold #

The counter-example to the counter-example, so that the whole thing does not read as an impossibility result.

Give the three creases angles of sixty, a hundred and twenty and sixty degrees. The alternating sum is nought, the linear part matches, and what remains is two linear equations for the positions. Solve them and the band folds — into an equilateral triangle, on a strip a little over one and three-quarter times its own width.

So the parity was necessary after all, and it was doing about as much work as noticing that a jigsaw has the right number of pieces.

Trying it, which takes two minutes #

The claim is unusual enough that it is worth putting a strip of paper against it, and the experiment is quick.

Cut a strip about three centimetres wide and thirty long. Join the ends with a half twist and a small piece of tape, taking care that the tape holds the front of one end against the back of the other rather than front to front — a full turn gives an ordinary cylinder and the experiment then tests nothing.

Mark three lines across the band, square to the long edges, at roughly a third of the way round from each other. Crease them, in whichever directions seem most promising, and press the band down onto the table.

It buckles. Not at one place, and not because the creases were in the wrong spots: pushing the paper flat forces a fourth crease to appear somewhere, and the fourth crease is the sheet’s way of saying that three were not going to do it. Move the three creases anywhere else and repeat, and the same thing happens. Try five and it happens again.

Then do the sixty-degree version. Same strip, same tape, but the three creases at sixty degrees to the edges, slanting alternately. That one goes flat, into a triangle, with almost no persuasion.

The value of doing both is that the second removes the obvious escape route. If only the first experiment is performed, the natural conclusion is that a Möbius band cannot be folded flat at all — which is false, and is the sort of thing a failed attempt is very good at suggesting.

The one place the positions could have helped #

There is a step in the argument that deserves to be poked at, because it is where the result would fail if it were going to.

The claim is that no arrangement of an odd number of square creases works. The composition’s linear part is fixed by the angles, so moving the creases about cannot change it — that is the argument. But the creases could also be reordered, or two of them could be made to coincide, or one could be moved outside the strip.

Reordering does nothing: the alternating sum of $k$ equal angles is the same however they are permuted. Making two coincide is worse, not better — two coincident creases are one crease of doubled thickness and the count drops by two, which keeps the parity and keeps the mismatch. And moving a crease outside the strip removes it, which does the same.

The only remaining freedom is the one the result is about: pointing the creases somewhere other than square across. That is a continuous parameter, it enters the linear part, and it is exactly the thing the parity discards.

What this says about the two-colouring’s status #

The colouring is the subject’s one global condition and it has a long-standing reputation for being the obstruction — the thing that catches what the vertex conditions miss. That reputation is earned on discs and annuli and it is worth qualifying.

On a loop of paper it is the complete answer. On a Möbius band it is one condition of several and the weakest of them. On a general sheet with vertices it is neither: it is implied by the vertex conditions wherever every loop bounds, and it adds something exactly where a loop does not.

The right way to hold it is as a shadow of the closure condition. The composition of the reflections round a loop has a determinant, and the determinant’s sign is the colouring. Everything else about the composition — three of the four numbers in the matrix, and both components of the translation — is invisible to it.

That framing also explains why the colouring is so cheap to compute. Signs multiply; matrices do not commute. The colouring gets its speed by discarding the part of the problem that is hard, and on the sheets where the discarded part is trivial, it loses nothing.

Necessary conditions in this subject, ranked by how much they leave #

It is useful to have the conditions of this subject laid out by what each one throws away, because the answer is different for each and the differences are not usually stated.

Developability throws away nothing continuous — it is an exact statement about the angles at a point, and a vertex either has a full turn of paper or does not.

Kawasaki likewise is an equation among the angles, and it is exact at a point.

Maekawa throws away the angles entirely and keeps a count. It is a parity in disguise and it is the reason this essay’s mistake has a family.

The big-little-big lemma keeps one comparison among the angles — which sector is strictly smallest — and discards the rest of them, which is why it says nothing at all when no sector is strictly smallest.

The two-colouring throws away the geometry completely and keeps one bit per loop.

The closure condition throws away nothing, and costs accordingly: it is six numbers per loop rather than one bit, and it needs the coordinates.

Read down that list and the general rule is visible. The cheap conditions are cheap because they forget something, and each of them has a family of counter-examples built precisely out of what it forgot. A subject with six conditions has six such families, and the square-creased Möbius band is one of them.

How a counting argument fails, in general #

The pattern here is not confined to bands, and it is worth stating in the abstract because this subject is full of counting conditions.

A counting condition is a statement about an invariant that takes finitely many values — a parity, a difference, a residue. It is derived by observing that some quantity must be preserved, and it is necessary because the quantity is preserved. It fails to be sufficient exactly when the object it is about has continuous data as well as discrete data, and the counting throws the continuous data away.

Maekawa’s condition is the same shape: it counts letters at a vertex and says nothing about the angles, so a vertex can satisfy it and fail to fold because Kawasaki’s condition is about the angles. Neither is sufficient alone and the pair is sufficient at a vertex, which is a happy accident of that case rather than a general principle — the pair is not sufficient for a whole sheet.

The band is the same story with the roles rearranged. The parity is the discrete half of the closure condition; the angles and positions are the continuous half; and the discrete half is one bit of a four-number matrix.

Why the loop of paper is different #

That last figure is the reason the mistake is so easy to make.

On an annulus with radial creases, the parity really is the whole answer. Any number of rays from the hole to the rim, evenly spaced or not, folds if and only if the count is even. There is no angle condition left over, because the creases all pass through a common centre and the composition of reflections in concurrent lines is a rotation whose angle is twice the alternating sum — and the identification a loop of paper needs is a rotation too, so the two are asking for the same kind of object and only the amount has to match.

A band’s creases are not concurrent. They are, in the interesting cases, not even parallel. The composition can be a translation or a glide, the identification is a specific one of those, and matching them is a genuine two-part condition where the annulus’s was one part.

So the annulus is the misleading case, and it is the case everybody meets first.

The claim was checked rather than reasoned about #

The result above is short enough to prove on paper and it is asserted in code as well, and the reason for doing both is that a three-line argument is exactly the length at which an error survives review.

Every band drawn here has its composition computed and compared against the map its gluing demands, and the comparison is a distance rather than a test. A band claimed to fold is a band whose distance is at rounding. A band claimed not to is a band whose distance was measured and reported, and the assertion that no square-creased Möbius band closes is checked at every odd count the figures use rather than at the one the argument was written about.

It refuses to draw if that ever stops being true. That is the useful half: an assertion which has never rejected anything is a comment, and this one has a case it must reject — the one-crease band, whose parity is right, whose colouring exists, and whose composition is a single reflection in a line that would have to run along the strip and does not.

The general lesson, stated once #

Two conditions are being confused throughout, and separating them is the whole content of this essay.

The first is does the sheet permit an assignment at all — a question about counts and signs, answered by arithmetic, and cheap.

The second is do the pieces of paper actually reach the positions the fold requires — a question about isometries of the plane, answered by composing matrices, and not cheap.

On a loop of paper the second question has no content, because the creases are concurrent and any rotation the composition produces is a rotation the identification can absorb. That case is the one everyone learns from, and it teaches that the first question is the whole problem.

On a band it is not. The creases point somewhere, where they point matters, and a condition that never looks at an angle cannot see it. Half of all crease counts are refused correctly by the parity, and of the half that survive, every single one that is creased square across the strip is wrong.

Where the other conditions sit #

For a reader assembling the whole picture, the band’s conditions in order of what they refuse.

The parity refuses half the crease counts, and it is the same parity a loop of paper obeys.

The alternating sum refuses all but a curve of the angle triples.

The fitting refuses all but a half-line of the lengths.

And not one of the subject’s vertex conditions refuses anything at all here, because a band has no vertices — which is the situation a sheet with a hole is also in.

What is safe to take away #

The parity condition is exact on a loop of paper, necessary and far from sufficient on a band, and its scope is decided by whether the creases have any freedom in their directions.

Stated in the form that survives: the composition of the reflections has to equal the sheet’s own identification map, and the parity is the sign of that equation’s determinant. Checking the sign is cheap and worth doing first, because it refuses half of all crease counts in one arithmetic step and refuses them correctly.

Checking only the sign is how one concludes that a strip with three square creases folds into a Möbius band, which it does not, and which the paper in one’s hands will decline to do in about five seconds.