Friezes of numbers find six groups and never p2

A frieze of numbers is built by one rule. A row of ones runs along the top, a row of positive integers sits under it, and every further entry is fixed by the requirement that each diamond of four neighbours, $b$ above, $a$ and $d$ beside, $c$ below, satisfies $ad - bc = 1$. A rule with no mirror hands every strip a glide took the friezes in which a row of ones comes back further down, closing the strip, and found that every one of them is carried onto itself by a glide reflection. The rule says nothing about reflections, and the closing alone supplies one. Of the seven frieze groups, only the four that contain a glide or a mirror along the strip occur, and which of the four is decided by the symmetry of a triangulated polygon.

That census stopped at friezes that close. Most periodic second rows do not close, and among those some still produce a strip of positive integers going down for ever, a frieze with a top edge and no bottom one. They turn out to supply exactly the groups the closed friezes cannot, apart from one. Between the two kinds, six of the seven frieze groups occur, and the seventh, p2, the group of half-turns and nothing else, cannot occur in any frieze of numbers at all.

A second row that never comes back #

Take the second row $1, 2, 6$ and repeat it. The rule gives a third row of 11, 5 and 1, a fourth of 9, 4 and 5, and the entries go on growing: 19, 33, 26, 51 and beyond, with no row of ones returning and no entry ever reaching zero. The strip is infinitely deep.

What decides whether a second row closes, grows for ever or breaks down is a single matrix. The entries of a frieze are determinants of pairs of vectors in the plane, each vector obtained from the one before by the matrix $\begin{pmatrix} c & -1 \\ 1 & 0 \end{pmatrix}$ for the next entry $c$ of the second row. Moving through one full period of the second row applies the product of $n$ such matrices, the period matrix $M$, which has determinant one. A frieze closes, with a row of ones returning in row $n - 2$ and zeros under it, exactly when $M = -I$. The question of what the other second rows do is a question about $M$, and since its determinant is one, its trace $s$ says almost everything.

The trace sorts the strips #

A matrix of determinant one with trace between $-2$ and 2 is a rotation in disguise. Its powers turn the frieze’s vectors round and round, and two vectors that are turned far enough have a determinant of the opposite sign, so somewhere down the strip an entry is zero or negative. A trace of $-2$ or less, other than $-I$ itself, flips the vectors’ sense every period and makes the signs alternate. And $M = I$ puts a zero in row $n - 1$. That leaves two possibilities for a strip of positive integers: $M = -I$, which closes, and a trace of 2 or more, which grows.

The census confirms the sorting over every second row of period up to six with entries up to seven. Every closed frieze has trace $-2$, every frieze that stays positive without closing has trace 2 or more, and the thousand second rows with traces $-1$, 0 or 1 all break down. The converse does not hold. A trace of 2 or more is necessary and not sufficient. Of the second rows of period up to six with trace at least 2, 145 still hit zero, all of them containing a run of ones. Three ones in a row are fatal whatever follows, since the diamond rule turns them into a zero two rows down.

Knowing that a frieze stays positive “for ever” needs one more step, because a computation sees finitely many rows. Here the period matrix supplies it. A matrix of determinant one and trace $s$ satisfies $M^2 = sM - I$, so down any diagonal of the frieze an entry two periods on is $s$ times the entry one period on, less the entry now. With $s \ge 2$, once a diagonal’s entries are positive and no smaller than those a period earlier, the next period’s are no smaller again, and induction carries positivity on for ever. Every infinite frieze in the census reaches that state within twelve periods, which turns a finite check into a proof for each frieze counted.

Growing by a step or by a factor #

The trace also sets how fast the frieze grows. At trace exactly 2 the period matrix is a shear: down a diagonal, each period adds the same amount, and the entries grow in a straight line. The simplest case is the second row $2, 2, 2, \ldots$, whose diagonals read 1, 2, 3, 4 and so on, the integers. Above trace 2 the matrix has a real eigenvalue $\lambda > 1$ with $\lambda + 1/\lambda = s$, and every period multiplies the entries by about $\lambda$.

The second row $3, 3, 3, \ldots$ is the cleanest example of growth: its diagonals read 1, 3, 8, 21, 55, 144, every other term of the Fibonacci sequence, and its factor is $(3 + \sqrt5)/2$, the square of the golden ratio. The same factor turns up in the Fibonacci chain as its inflation, and in the frieze with second row $1, 2, 6$, which has the same trace and grows by the same factor once every three steps. Trace 2 is the boundary between closing and growing: a closed frieze’s period matrix is $-I$, a growing one’s has an eigenvalue above one, and trace 2 is the shear between them, with growth by a fixed step rather than a fixed factor.

The boundary can be approached from the closed side. Triangulate an $n$-gon as a fan, every diagonal drawn from one vertex. That vertex meets $n - 2$ triangles, its two neighbours one each, and every other vertex two, so the second row is $n - 2, 1, 2, 2, \ldots, 2, 1$. Its frieze closes, as every triangulation’s must, and in the middle of the strip its diagonals read 1, 2, 3, 4 and onward, exactly the frieze of twos, until the far side of the polygon turns them back to a row of ones. For the twelve-gon the run reaches ten before it turns. As the polygon grows, the closed fan agrees with the infinite frieze of twos over a wider and wider stretch. The infinite frieze of trace 2 is what a fan looks like from its middle when its far edge has gone to infinity, and that is the sense in which trace 2 sits on the border of the closed friezes.

Binomial coefficients where closing gave Catalan numbers #

Closed friezes are counted by a famous number. Conway and Coxeter proved in 1973 that the closed friezes of order $n$ correspond one for one to the triangulations of an $n$-gon, so there are as many as the Catalan number $C_{n-2}$. Infinite friezes are endless in number at any period, since the trace can be as large as one likes, but at a fixed trace there are finitely many, and the counts are again recognisable.

At trace 2 the counts for periods one to six are 1, 3, 10, 35, 126 and 462, the binomial coefficients $C(2n - 1, n)$. At trace 3 they are 1, 2, 6, 20, 70 and 252, the central binomials $C(2n - 2, n - 1)$. At trace 4 they are 1, 4, 12, 40, 140 and 504, twice the trace-3 counts from period two on. The search tried every second row with entries up to the trace plus the period plus three, and the largest entry any counted frieze needed was one less than the trace plus the period, so the bound never cut a count short. Set against the closed friezes, the open ones of a single trace are already the majority. Closed friezes of order four, five and six number 2, 5 and 14. The infinite friezes of trace 2 alone number 35, 126 and 462 at the same periods, seventeen and a half, twenty-five and thirty-three times as many, the ratio rising by about eight with each step in the period. Closing is the exception among positive strips, not the rule, and the census of a rule with no mirror was a census of the exception.

That the counts are binomials is a measurement on six periods and three traces, not a theorem proved here. Baur, Parsons and Tschabold showed in 2016 that periodic infinite friezes come from triangulations of a polygon with a puncture or a hole, the way closed ones come from triangulations of a polygon, and a count of those triangulations is where a proof would begin.

No bottom edge, so nothing turns the strip over #

The symmetry of an infinite frieze is constrained before any number is looked at. Six of the seven frieze groups contain an operation that exchanges the strip’s two edges: a glide along the strip, a mirror along it, or a half-turn. An infinite frieze has a top edge of ones and no bottom edge, so nothing can exchange them. Only translations and mirrors across the strip are left, and that is p1 or p1m1, the two groups the closed friezes never have. Which of the two a given infinite frieze has is read off its second row: a mirror across the strip exists exactly when the second row reads the same backwards from some starting point, because the entries of a frieze are built from the second row by an expression that is unchanged when its letters are reversed. Forty friezes of period four, tested entry by entry, agree with the reading of their second rows.

The p1 friezes have a property the closed ones never have. A glide reflection reverses orientation, so every closed frieze, carrying its glide, is carried onto its own mirror image: closed friezes are never one-handed. An infinite frieze in p1 has no operation that reverses orientation, and its mirror image is a different frieze. The second rows $1, 2, 6$ and $1, 6, 2$ give such a pair, a left hand and a right hand. Up to period six and trace four there are 278 infinite friezes in p1 and 29 in p1m1, and the 278 fall into 139 mirror pairs. The local rule has no hand, and the closed strips cannot have one; the open strips nearly always do. The share grows with the period. Every second row of period one or two reads the same backwards, so the seven friezes of those periods are all p1m1. At period three, 8 of 9 are p1; at period six, 188 of 201. A random long second row is almost never a palindrome, so almost every infinite frieze is one-handed, in the same way that almost every triangulation of a large polygon has no symmetry and almost every closed frieze is plain p11g.

Chirality in a crystal is the same story at a larger scale: the groups one hand may sit in are exactly those with no operation that reverses orientation, and the infinite friezes in p1 are the strip’s version of a crystal built from one hand.

Six of the seven, and why never p2 #

Put the two censuses side by side and every frieze group is accounted for except one. Closed friezes, to order twelve, give p11g 975 times, p2mg 82, p11m 30 and p2mm 4. Infinite friezes, to period six and trace four, give p1 278 times and p1m1 29. Nobody gives p2.

The census only shows that p2 is absent up to those sizes. The reason it is absent at every size is three steps long. A half-turn about a point of the strip carries the top edge onto the bottom edge, so a frieze with a half-turn has a bottom edge of ones, and its strip closes. Every closed frieze has the glide, by the theorem the earlier essay computed through. And a half-turn about a point of the strip’s midline, followed by a glide along that midline, is a reflection in a line across the strip, which is how the seven groups multiply, as why sixteen become seven worked out case by case. So a frieze of numbers with a half-turn has a mirror across it as well, and its group is p2mg or p2mm, never p2. The one group whose only non-trivial point operation is a half-turn is the one group no frieze of numbers can have.

It is worth noticing what that argument does not use. It never asks which numbers are in the frieze, only that the frieze closes and that closing forces the glide. Any rule that forces a glide on every closed strip and leaves open strips without a bottom edge excludes p2 in the same way. It is a fact about strips made by such a rule and not about strips in general. A lattice row of the plane group p2 keeps the half-turns centred on it and nothing else, and so carries the frieze group p2, which is the kind of fact the friezes inside the seventeen reads off row by row. A pattern of motifs can have any of the seven groups; a pattern of numbers obeying the diamond rule can have six.

What the account has to refuse #

The refused claim is the tempting shortcut that a period matrix of trace 2 always gives a frieze. The second row of six ones has a period matrix equal to the identity, trace 2, and a zero in its third row, and a further 144 second rows of trace 2 or more break down the same way, through runs of ones. The other tests guard the counts and the symmetry separately: the binomials against an exhaustive search whose bound is shown not to bind, the mirrors against the second rows, and the mirror pairs by checking that the reverse of every p1 second row is another p1 second row and not a rotation of the same one.

What the figures cannot show #

Every figure shows finitely many rows of something infinitely deep, and a strip that grows by a factor of 2.6 a period quickly holds numbers too long to print. The claim that a frieze stays positive for ever rests on the recurrence from the period matrix, checked at the twelfth period; the drawings are evidence of the shape and not of the infinity. The counts are complete only within their search: second rows up to period six, traces up to four, entries up to a stated bound. The binomial formulae are a pattern in those counts, consistent with the correspondence to punctured and holed polygons, and not proved here.

The conventions matter too. A frieze group here is a group of isometries of a strip of the plane, the seven named as in the symbols the frieze groups take, and the friezes are counted up to translation; a frieze and its mirror image are counted twice, which is what makes a mirror pair a pair. Friezes whose entries are not positive integers, with zeros allowed or entries in another number system, are not considered, and for those the argument against p2 needs restating, because closing no longer means a returning row of ones.

Who worked it out #

Coxeter introduced frieze patterns of numbers in 1971, and Conway and Coxeter proved the correspondence with triangulated polygons in 1973. Infinite friezes were studied by Tschabold, and by Baur, Parsons and Tschabold in 2016, who showed that the periodic ones come from triangulations of punctured discs and annuli, and introduced the growth coefficient, which is the trace of the period matrix used here. The census by trace and its binomial counts, the finite check made final by the recurrence, the mirror pairs and the argument that no frieze of numbers has group p2 are made here.

Still open: friezes that close with the other sign #

Over the integers a positive frieze closes only with the period matrix equal to $-I$. Allow the entries to be integers modulo a prime, where positivity means nothing, and a second row can also close with the period matrix equal to $+I$. The row that returns is then a row of minus ones, and the identity that gave every closed frieze its glide becomes a glide that carries each entry to its negative. In the language of two colours and a swap, that is an antisymmetry, a glide that exchanges two colours, with the sign as the colour. Which of the frieze groups with a colour-reversing operation such friezes realise, and whether p2 is still excluded once a half-turn may reverse the colour rather than keep it, are finite questions over each prime, and counting what a group cannot tell apart has the counting they need. None of it has been computed here.