A rule with no mirror hands every strip a glide

Seven friezes classified the ways a motif can repeat along a strip, and the friezes inside the seventeen found them again as what survives of a wallpaper group along one of its rows. In both, the group comes first. Somebody chooses the operations, and the pattern is whatever those operations make of a motif.

There is another way for a frieze to arise, and it reverses the order. Choose a rule instead of a group, a local rule that fixes each entry of a strip from its neighbours, and let the rule fill the strip. Then ask what symmetry the result has. The rule may promise translations, since it is applied identically at every position. It promises nothing else, and in particular nothing that turns the strip over.

The rule here is one H. S. M. Coxeter introduced in 1971, and the answer he and John Conway found two years later is the surprise this essay is about. Every strip the rule fills, provided it closes, has a glide reflection, turning it upside down and moving it half a period. No step of the rule mentions a reflection. The glide comes out of the arithmetic, and it decides which frieze groups can occur. Four of the seven can, and three never do.

A strip filled by one rule #

A frieze pattern of numbers has a row of ones along the top. Below it is a row of positive integers, then further rows, each offset by half a step from the one above, so that every four neighbouring entries form a diamond: $b$ above, $a$ and $d$ to either side, $c$ below. The rule is

$$ad - bc = 1$$

for every diamond. Given the row of ones and the second row, it fixes the third row entry by entry, $c = (ad - 1)/b$, then the fourth from the second and third, and so on downward. Nothing guarantees that the divisions come out whole, or that the entries stay positive, or that the process ever ends. A pattern closes when a row of ones appears again, with every entry between positive. The next row would then be zeros, and the strip stops there.

The figure runs the rule on a second row of seven numbers repeated: $1, 3, 2, 1, 4, 1, 3$. Three rows follow and then the ones return, so this strip has six rows counting both rows of ones. The rule is symmetric under translation along the strip, because it is the same at every place. It is also symmetric under reading the diamond left to right or right to left, since $ad - bc$ does not change when $a$ and $d$ are swapped. It is not symmetric under turning the strip upside down in any obvious sense, because the computation runs downward from a chosen second row, and the second-to-last row is an output, not an input.

The order of a closed frieze is $n$ when the second row repeats with period $n$; the example has order seven, and a frieze of order $n$ has $n - 1$ rows from ones to ones.

What Conway and Coxeter found #

Coxeter noticed in his 1971 paper that every closed frieze he computed was invariant under a glide reflection, and proved it. The second-to-last row is the second row again, shifted half a period. The third-to-last is the third, and so on. In coordinates: index the entries by pairs $i < j$, with $m(i, i + 1) = 1$ for the top row of ones and $m(i, j)$ in row $j - i - 1$. Then

$$m(i, j) = m(j, i + n)$$

for every entry. Reading $(i, j)$ as a position, the right-hand side is the entry in the mirror-image row, $n - 2 - (j - i - 1)$ rows down, moved along the strip by $n/2$. That is a glide.

Conway and Coxeter’s 1973 papers then found what the closed friezes are. The second rows that close are exactly the sequences counting triangles at the vertices of a triangulated polygon. Cut an $n$-gon into triangles by non-crossing diagonals, write at each vertex the number of triangles meeting there, and go round: that is a second row that closes after $n - 1$ rows, and every second row that closes arises this way from exactly one triangulation. The example comes from the heptagon drawn beside it in the next figure. The numbers of closed friezes of each order are therefore the numbers of triangulations, the Catalan numbers 2, 5, 14, 42, 132 for orders four to eight.

Why the glide is forced #

The proof is short enough to give whole, and it shows exactly where the glide comes from.

Attach to each position $k$ along the strip a vector $V_k$ in the plane, with integer coordinates, such that consecutive vectors have determinant one, $\det(V_k, V_{k+1}) = 1$, and such that $V_{k+1} = a_k V_k - V_{k-1}$, where $a_k$ is the second-row entry at $k$. Start from $V_0 = (1, 0)$ and $V_1 = (0, 1)$ and the recurrence determines the rest. Then set

$$m(i, j) = \det(V_i, V_j).$$

These numbers satisfy everything the frieze does. $m(i, i) = 0$, $m(i, i + 1) = 1$, and for fixed $i$ they obey the same recurrence in $j$ as the vectors, which is the diamond rule read one diagonal at a time. The diamond rule itself is the identity among four $2 \times 2$ determinants that algebraists call a Plücker relation. So the frieze is a table of determinants of pairs of vectors.

Now closing. The frieze closes after $n$ steps exactly when $m(i, i + n) = 0$ and $m(i, i + n - 1) = 1$ for all $i$. In terms of vectors: $V_{i+n}$ is parallel to $V_i$, and the determinant conditions fix the factor. Closing means $V_{k+n} = -V_k$ for every $k$: after $n$ steps the vectors come back pointing the other way. The glide follows in one line:

$$m(j, i + n) = \det(V_j, V_{i+n}) = \det(V_j, -V_i) = \det(V_i, V_j) = m(i, j).$$

The minus sign from closing and the minus sign from swapping the arguments of a determinant cancel. The glide is a consequence of the strip closing with a sign, and the sign is not optional. If $V_{i+n} = c\,V_i$, then $1 = \det(V_{i+n-1}, V_{i+n}) = c \det(V_{i+n-1}, V_i) = -c\, m(i, i + n - 1) = -c$, so the returning row of ones forces $c = -1$ and never $+1$.

So the rule alone does not make the glide. A second row that does not close gives a strip with no glide, and the next section shows one. What makes it is the rule together with the requirement that the ones come back.

When the strip does not close #

Change one entry of a second row that closes and run the rule again. The hexagon’s second row $1, 3, 1, 3, 1, 3$ closes; with its first entry raised to $2$ the entries still come out whole numbers, but where the row of ones should return, the rule produces $1, 1, 4, 7, 7, 4$.

That strip, cut off where a frieze of order six would end, has mirrors across it, because the changed row $2, 3, 1, 3, 1, 3$ reads the same backwards from its first entry, and nothing else. Its group is p1m1, one of the three groups no closed frieze ever has. The rule was applied with the same care in both cases. The difference is only whether the ones came back, which is what the vectors coming back reversed amounts to.

Four of the seven #

Every closed frieze has a glide or a mirror along its length, so its group must contain an operation that turns the strip over in place. Of the seven frieze groups, three have no such operation: p1 has translations only, p1m1 adds mirrors across the strip, and p2 adds half-turns. No closed frieze has any of these three groups. The other four each contain a glide or a mirror along the strip, and all four occur.

The figure below finds all four among friezes of order eight, by testing every isometry of the strip against every entry. The test is the same filter why sixteen become seven runs on operations, run here on a pattern instead: each candidate is a reflection or not along the strip, a reflection or not across it, and a shift by a multiple of half a step, and it is kept when it carries every entry of one period to an equal entry.

The four are not equally easy to reach. The plain glide, p11g, needs nothing. Each of the others needs the polygon to have a symmetry, and the triangulation must keep it.

The polygon decides which #

The polygon’s symmetries act on the frieze directly. Rotating a triangulation by one vertex shifts its second row by one place, and so shifts the whole frieze by one step. A triangulation fixed by a rotation through $t$ vertices therefore gives a frieze with translation $t$, shorter than the period $n$. Reflecting the triangulation reverses its second row, and reverses the frieze left to right. A triangulation fixed by a reflection gives a frieze with mirrors across the strip.

The rest follows by composing with the glide. The glide moves the strip by $n/2$ and turns it over. If the frieze also has a translation by $n/2$, the two together turn the strip over and move it nowhere: a mirror along the strip, and the group contains p11m. So a mirror along the strip appears exactly when the triangulation has a half-turn, which requires $n$ to be even. A triangulation’s rotational symmetry can only be a half-turn or a third of a turn, since its centre lies on a diagonal or inside a triangle. A third of a turn shortens the period to $n/3$ but leaves the glide a glide, since $n/2$ is then an odd multiple of half the new period.

The full p2mm needs a half-turn and a reflection together, and that needs more than an even order. With a half-turn, the triangulation’s central diagonal is a diameter of the polygon, and the two triangles on either side of it are exchanged by the half-turn. A reflection keeping the triangulation must keep that diameter, so its axis lies along it or across it. Along it, the reflection must exchange the two central triangles the same way the half-turn does, and that puts their apexes a quarter of the way round from each other, which needs $n/2$ even. Across it, the axis must pass through each central triangle’s apex, and an axis across a diameter between two vertices passes through vertices only when $n/2$ is even. Either way p2mm needs the order to be a multiple of four.

The census confirms all of it. Of the 23,712 closed friezes of order four to twelve, every one has the glide, verified entry by entry rather than by citing the theorem. Counted up to translation and reflection, there are 1, 1, 3, 4, 12, 27, 82, 228 and 733 distinct patterns of each order, the numbers of triangulations up to the polygon’s own symmetry. Every one of them has the group the polygon predicts. p11m never appears at an odd order, and p2mm appears only at orders four, eight and twelve, once, once and twice.

The distribution says something the theorem does not. At order four the only frieze is p2mm; at order twelve, 671 of 733 are plain p11g. Most triangulations of a large polygon have no symmetry, so most friezes have exactly what the arithmetic forces and nothing more. The glide is the floor, and the floor is where almost everything ends up.

The glide, rolled up #

A closed frieze of order $n$ repeats every $n$ steps, so it can be wrapped round a cylinder of circumference $n$ with no seam. Seven friezes round a cylinder did this to the seven groups and found the seven families of axial point groups. Here the wrapping says something definite about every frieze of numbers at once. On the cylinder the translation by $n$ is the identity, and the glide, applied twice, is that translation. So the glide becomes an operation of order two that turns the cylinder upside down and carries it half way round: the composition of a reflection in the cylinder’s waist with a half-turn about its axis, which is an inversion through the cylinder’s centre. Every frieze of numbers, rolled up at its own period, is centrosymmetric. A triangulation with no symmetry at all gives a cylinder with exactly that symmetry and nothing else, the point group $\bar{1}$.

In orbifold notation the four groups the friezes allow are the four whose symbols carry a cross-cap or a mirror along the strip’s length, $\infty\times$, $\infty*$, $2{*}\infty$ and $*22\infty$. The three that never occur are $\infty\infty$, $*\infty\infty$ and $22\infty$, whose orbifolds keep the strip’s two edges apart. The diamond rule, by forcing the glide, forces the strip’s two edges to be the same edge, and that is the whole of what the census found, said in Conway’s own notation.

What the counts rest on #

Every count here is an enumeration, and the enumeration was checked against things known independently. The second rows were generated by cutting ears: every triangulation of an $n$-gon has a vertex in only one triangle, removing it leaves a triangulation of an $(n - 1)$-gon, and reversing the step from every smaller triangulation in every position, with duplicates discarded, generates every second row of order $n$. The counts came out as the Catalan numbers at every order, which a missed or repeated triangulation would have broken. Every frieze was then computed from its second row by the rule, in exact integers, and checked to close.

The groups were found from the patterns, not from the polygons. For each frieze, every candidate isometry of the strip, keeping or reversing the rows, keeping or reversing left and right, and shifting by any multiple of half a step within two periods, was tested against every entry of a period. That the resulting group always matches the polygon’s symmetry is a second, independent check, made on all 1,091 distinct patterns to order twelve.

The refused claim, that an odd polygon can produce a mirror along the strip, is tempting because odd friezes do have reflections along the strip. They are glides, though, and the census shows the distinction is sharp: none of the 260 patterns of odd order up to eleven has a mirror along its length.

What the strip cannot show #

The figures show a few periods of an infinite strip and a handful of polygons, and the claims are about all of them. That every closed frieze has the glide is the determinant argument, which no drawing contains; the drawings show the glide holding where they are drawn. That three groups never occur is a statement about an absence. No figure can show a strip that does not exist, only the reason it cannot.

Nor do the figures show what the numbers mean. The entries count things in the triangulation: Broline, Crowe and Isaacs showed in 1974 that each is the number of ways of matching certain vertices to triangles under a fixed rule. The same friezes reappear in the theory of cluster algebras, where the diamond rule is an exchange relation. None of that is needed for the symmetry, which is why none of it is drawn.

There is a convention in the census that changes its numbers. A frieze and its mirror image are counted once among the patterns, and a frieze and its translate once. Counted without identifying mirror images, the numbers rise and the groups’ shares change, but which groups occur does not.

A local rule and a global symmetry #

One tension runs through much of crystallography: a symmetry is a global property, and the things that produce patterns are usually local. What forces a lattice found that local conditions can force periodicity, and a tiling of the whole plane, decided on one tile’s edge found a global symmetry read off a boundary. The frieze is the cleanest case of the kind, because the local rule and the global symmetry are both completely explicit and the gap between them is a single sign.

It also changes how the seven groups look. In the four motions, the glide is the awkward one, the motion that is neither a reflection nor a translation and that the frieze classification makes unavoidable. Here it is the inevitable one. A rule that knows nothing of turning over produces a pattern that turns over, and the group it lands in is the plainest group that does.

Who found it #

Frieze patterns of numbers are Coxeter’s, from a paper in Acta Arithmetica in 1971, which proved the glide. Their correspondence with triangulated polygons is due to Conway and Coxeter, in two problems published in the Mathematical Gazette in 1973, with the solutions. The determinant description of the entries is standard in the later literature, including the survey by Morier-Genoud of 2015. The census by frieze group, and the condition that p2mm needs a multiple of four, are computed and argued here.

Still open: friezes that do not close, and friezes that are not of numbers #

The census stops at closed friezes, and the essay’s point rests on closing. Two kinds of strip lie beyond it. An infinite frieze, whose ones never return, is known to correspond to a triangulation of a polygon with infinitely many vertices or of an annulus. It has no glide, since the vectors never come back reversed. Whether it can have mirrors across it, and which of the three groups without a reflection along the strip each kind allows, is the mirror image of the question answered here. The other direction is a frieze with entries in another ring, such as the integers modulo a prime, where closing is easier and positivity means nothing. Such friezes close with a sign as well, so they should keep the glide. Whether the four groups are still the only ones, or whether the modular rule allows strips whose polygon has no meaning, has not been computed here. Counting what a group cannot tell apart counts patterns up to symmetry in exactly the way that half of the question needs.