A centre at every other ring

A gap the sphere does not have counted every trivalent net of pentagons and hexagons on the projective plane out to six hexagons, by finding the closed cages that halve onto it, and found two counts with no net at all. It then said what a census cannot do. The row is infinite and the enumeration is not, the cages more than double in number with every two atoms, and the question of whether any further count is empty is not one that counting will reach.

What it proposed instead was a construction:

A proof would not count. It would build, for every number of hexagons from three upwards, one cage with a centre — and the natural candidates are tubes capped at both ends by halves of one symmetric cap, lengthened a ring at a time.

The candidates are natural for a good reason. A tube of hexagons is neutral — it charges nothing, so the whole of the twelve pentagons can sit in the two caps — and a tube with the same cap at each end, one turned through half a turn, looks exactly like an object with a centre. Lengthening it ought to give a cage with a centre at every size, which is what the row needs.

It does not. The construction is carried out below, and what it produces is a centre at every other ring and none between. Two families exist and there is no third, so what the construction reaches is two arithmetic progressions rather than a row, and two-thirds of the row is untouched. Along the way it explains one of the census’s two gaps — the cage of twenty-four atoms is the first member of a family whose parity is wrong — and it proves outright something the census could only check: that every odd hexagon count on the sphere is empty of centres before any cage is built.

A cap is a pole and a ring, and that is why there are two families #

Nothing here has coordinates. A cage is the adjacency of its faces — which face touches which — and everything below is a question about that pattern of contacts, as it was when the distances were thrown away.

A tube of circumference n is a stack of rings of n faces each. Every face of a ring touches its two neighbours round the ring, two faces of the ring above and two of the ring below: six sides in all, so a ring away from the ends is a ring of hexagons and charges nothing — which is the fact about a vertex that makes three bonds an atom the interesting case at all. That is the whole reason a tube is the right shape to try. The curvature has nowhere to be except the caps, and the caps are short.

The simplest cap closes the tube with a single pole face and one ring under it. The pole has n sides, since it meets the n faces of that ring. A face of the ring touches its two ring neighbours, the pole above and two faces below — five sides, a pentagon — so the ring is n pentagons, and the cap carries n of them, plus the pole itself when the pole happens to have five sides.

Two caps have to carry twelve pentagons between them, and that is an equation with very few solutions.

At circumference five the pole is a pentagon, the ring is five pentagons, and the cap has six. At circumference six the pole is a hexagon, the ring is six pentagons, and the cap has six again. At every other circumference the sum is not twelve and no cage of this kind exists. The construction has exactly two families because the arithmetic of a cap permits exactly two, and this is settled before a single cage is built.

The five-family starts at the dodecahedron, which is the cap placed directly on its own image with no hexagons between. The six-family starts at a cage of twenty-four atoms with two hexagons, the two poles. Both are lengthened by inserting rings of hexagons, one at a time, exactly as proposed.

The tube of forty atoms halves and the tube of thirty does not #

Take the five-family at two rings. The cage has forty atoms, twelve pentagons and ten hexagons, and twenty symmetries.

The drawing is the placement nobody chose — no coordinate is picked, each atom simply sits where its three neighbours average — and the numbering is read off the cage’s own symmetry. Ten of its twenty symmetries reverse orientation and six of those are of order two; one of the six fixes no face, no bond and no atom, and that one is the centre. It carries every face to a different face, so the faces fall into eleven pairs, and identifying each pair gives a net on a surface of characteristic one.

Now take the same family at one ring, which is thirty atoms.

The two cages are the same construction at successive lengths, they have the same number of symmetries, the same number of reversing symmetries, and the same number of reversing involutions among them. Nothing about the size of the group separates them. What separates them is which kind the reversing involutions are, which is the distinction between a mirror and a centre arriving from the combinatorial side: an operation of order two that reverses orientation either fixes a great circle of the sphere, and so crosses faces and bonds and passes through atoms, or fixes nothing at all.

The order never moves and the kind alternates #

Taking both families to eight rings and sorting their reversing involutions by what each fixes gives the shape of the result.

The five-family has twenty symmetries at every length past the dodecahedron, ten of them reversing, and six reversing involutions. At even numbers of rings, five of the six are mirrors and one is free. At odd numbers, all six are mirrors. The six-family has twenty-four symmetries at every length and twelve reversing; at odd numbers of rings it has eight reversing involutions of which one is free, and at even numbers it has six of which none is.

So the quantity that decides whether the cage halves is not the amount of symmetry. The five-family’s dodecahedron has a hundred and twenty symmetries and one free reversal; its cage of forty atoms has twenty symmetries and one free reversal; its cage of thirty has twenty symmetries and none. A count of symmetries would put the second and third together and separate them from the first, which is the wrong grouping.

The reason for the alternation is worth stating in the language of the tube rather than of the group. Each ring is set half a step round from the ring beneath it, so a tube with an even number of hexagon rings ends with its two caps in the relation a centre needs — every face opposite a face — and a tube with an odd number ends with them in the relation a mirror needs, face opposite face across a plane through the middle ring. Inserting one ring exchanges the two, which is why the construction alternates rather than accumulating. The six-family begins in the opposite phase because its poles are hexagons rather than pentagons, and that single difference is what puts its centres at the odd lengths.

This is the same arithmetic that decides whether a rolled honeycomb comes back on itself, met in a different place. There the rolling vector’s residues decide whether a tube has a repeat at all; here the ring count’s parity decides which of two operations the two ends are related by. Both are the fact that a tube is a stack with a half-step in it.

The first gap the census found is the first member of a family #

The six-family at no rings is a cage of twenty-four atoms with two hexagons, and it is the only cage of twenty-four atoms there is. The census reached it, found six reversing involutions and every one of them fixing four faces, four bonds and four atoms, and recorded a gap at one hexagon on the projective plane.

That reading is correct and it is incomplete. The cage is not an isolated small case that happens to lack a centre; it is the first term of an alternation, and the reason it lacks one is the reason the whole six-family lacks one at even lengths. Its successor at one ring — thirty-six atoms, eight hexagons — has a centre, and halves to a projective net with four hexagons. So the census’s dash at one hexagon and its net at four hexagons are consecutive entries of one family seen from outside it.

The census’s other gap, at two hexagons, has no such explanation. Two hexagons on the projective plane lift to a cage of twenty-eight atoms with four hexagons, and neither family contains a cage of twenty-eight atoms: the five-family goes 20, 30, 40 and the six-family 24, 36, 48. That gap stays a fact about two particular cages and is not a fact about a family, which is a difference the census could not see and the construction makes visible.

A centre pairs the faces, so half the row is empty before anything is built #

One statement about the whole infinite row does come out of the construction, and it comes out of what a centre is rather than out of any cage.

A centre reverses orientation and fixes nothing. Fixing nothing means it carries every face to a different face, and being of order two means it carries that face back, so the faces fall into pairs. A pair is two faces with the same number of sides, since a symmetry preserves the number of sides. So a cage with a centre has an even number of faces of every size at once — an even number of pentagons, which twelve satisfies, and an even number of hexagons, which is a condition on the cage.

The census confirms it — 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 3, 0, 3 carriers at the hexagon counts from none to twelve, with every entry at an odd count a zero — but the census is not where it comes from. It is an argument about an involution acting on a finite set, and it holds at every size. It is also the small end of what a group cannot tell apart: an orbit of size two is what a free involution has everywhere, and an orbit of size one is what it is forbidden.

That halves the question immediately. The projective plane’s row asks about a net with h hexagons, which lifts to a cage with 2h, and 2h is even whatever h is; so the pairing rules out nothing there directly. What it rules out is the other row: the sphere’s cages with an odd number of hexagons — twenty-two atoms, twenty-six, thirty, thirty-four, and so on for ever — can never halve, and half of the sphere’s row is closed with three sentences where the census had to open forty cages of thirty-eight atoms to say nothing about them.

The argument also explains the shape of the construction’s own output. Both families are lengthened by whole rings, so the five-family’s hexagon count moves in fives and the six-family’s in sixes; the centres sit at alternate members, so the hexagon counts with a centre move in tens and twelves. Every one of those is even, as it has to be.

Two progressions do not make a row #

Putting the construction’s reach beside the census’s gives the honest state of the question.

The five-family reaches the projective counts divisible by five and the six-family the counts four more than a multiple of six. Between them that is one entry in five plus one in six less one in thirty, which is one third of the row. The census settles the first seven entries, two of them negatively. Everything else is open, and the first open entry is at seven hexagons — a projective net of twenty-four atoms, lifting to a cage of forty-eight.

Forty-eight atoms is four atoms past where the census stops and it is not close to reachable by extending it: the number of face spirals to wind up more than doubles with every two atoms, and the cages that have a centre stay a handful while the cages that do not run into hundreds. The six-family’s second member has exactly forty-eight atoms and fourteen hexagons, and it is at an even number of rings, so it has no centre. The construction produces a cage at the first open count and the cage is the wrong one, which is as sharp a statement of the failure as the row allows.

What would close the row is not more rings. It is more caps. A cap that carries six pentagons and some hexagons — a longer cap rather than a pole and one ring — would start a family at a different hexagon count while keeping the step, and enough caps with enough different hexagon counts would cover the residues each step leaves. Whether such caps exist for every residue, and whether the parity that alternates here alternates for them too, is not computed anywhere on this page.

What the construction settles and what it does not #

Settled, for every size. A cage with a centre has an even number of faces of each size, so no cage with an odd number of hexagons halves. That is a proof rather than a check, and the census is where it is tested rather than where it comes from.

Settled, for two progressions. There is a net on the projective plane at every hexagon count divisible by five and at every count four more than a multiple of six, for ever. Each is exhibited by a cage that is constructed rather than found, and the alternation is verified at every length to eight rings rather than argued.

Not settled: the rest of the row. Ten of the first twenty-one counts are open and the first is seven. Nothing here says whether any of them is empty, and the two gaps the census found are the only two known.

And the surfaces below characteristic one are untouched. A cage of pentagons and hexagons cannot pay a negative charge, so a heptagon has to pay for a pentagon there, and neither family above has a heptagon in it anywhere.

Not settled: whether the alternation is a theorem. It is measured at nine lengths in each family, and the account given for it — that a ring sets the next one half a step round, so the parity of the ring count decides whether the two caps face each other across a point or across a plane — is a description of what the computation finds rather than a proof that it must.

And the caps are of one kind. Every cap here is a pole with one ring beneath it. That is the only kind whose pentagon count is forced to be n or n + 1, which is what made the two-family result available; it is not the only kind of cap.

The second refusal is the useful one. The tube of thirty atoms has everything a centre would need except the centre: the right number of atoms for its family, the same group as its neighbours, six reversing involutions. Offering it as a cage that halves is the mistake the alternation makes easy, and what rejects it is not a count but a test on each involution for a fixed face, a fixed bond and a fixed atom.

Who built the tubes, and who counted the cages #

The capped tube is older than the question it is used on here. Hugh Christopher Longuet-Higgins and colleagues, and independently several others, described closed carbon cages as tubes with caps in the years after 1985, and the systematic construction of a tube from a cap and a rolling vector is the subject a rolled honeycomb is about. What the chemistry wanted from those constructions was a structure; what is wanted here is a symmetry, and the two questions turn out to have different answers at the same lengths.

The alternation of point groups along a capped tube — the five-family running between the two groups of order twenty and the six-family between the two of order twenty-four — is standard in that literature and is usually stated as a fact about which of two symbols a tube of given length carries. Read as a statement about a centre it says something the symbol does not: one of the two groups in each pair contains an operation that fixes nothing, and the other does not, and that is what decides whether the cage halves.

Branko Grünbaum and Theodore Motzkin’s theorem of 1963, which settles the sphere’s row for every hexagon count but one, has no analogue here and the difference is instructive. Their result is a construction that adds one hexagon at a time to a cage, and one at a time is what a row needs. Every construction on this page adds five or six at a time, because a ring is the smallest thing a tube can be lengthened by, and a step of five leaves four residues untouched however far it is taken.

Still open: a cap for every residue #

The construction fails in a way that says what would fix it. Each family’s step is its circumference, and its starting point is its cap’s own hexagon count, which for a pole-and-ring cap is nought or one. A family whose cap carried two extra hexagons would reach a different residue class with the same step, and five caps of circumference five with hexagon counts nought to four would cover every residue between them.

So the question the row turns on is no longer about tubes at all. It is: for a tube of circumference five, which numbers of hexagons can a cap of six pentagons hold? A cap is a patch with a boundary rather than a closed net, so neither the charge nor the pairing argument applies to it directly, and the enumeration that would answer it is an enumeration of patches, which is a different computation from the enumeration of cages. What is known here is only that the two smallest caps exist and that their families alternate.

The other direction the row could be closed from is the surface itself. Everything above works on the sphere and reads the projective plane off it, which is the method the census established and the reason both gaps were found. A construction working directly on the projective plane — adding one hexagon at a time to a net rather than two to its cover — would have a step of one where every step here is five or six, and whether such a move exists is not a question anybody here has asked.