Six groups are made of glides

Four groups are made of mirrors took located mirror lines and closed them under composition, and found that exactly four plane groups are generated by their mirrors alone: pmm, p4m, p3m1 and p6m. They are the kaleidoscopes, the groups whose mirrors cut the plane into copies of a polygon. Closing the plane from two centres had done the same with rotation centres and reached the rotation groups. One kind of generator was left over, and the mirror essay named it. A glide line is a location, like a mirror line, but a glide also carries a slide along its line, a second piece of data, and two glides on the same line with different slides are different motions. Whether the seventeen fall out of closing located glides, as the kaleidoscopes fall out of closing mirrors, was the question it left.

They do not, and the census that shows it is short. Of the twelve plane groups containing a reflection, six are generated by their glides alone: pg, pgg, p4g, p3m1, p31m and p6m. Only two of those, p3m1 and p6m, are among the four made of mirrors. Together the two kinds generate eleven of the twelve, every group with a reflection except pm. The five groups with no reflection at all are out of reach of any closure of reflections. The difference between the two lists has a one-line cause. A mirror done twice is nothing, and a glide done twice is a translation.

What “generated” has to mean #

The census needs a definition, and the obvious one is wrong in an instructive way. A set of operations generates a group when every operation of the group is a product of them. For a plane group that includes the lattice translations. They are operations of the group like any other, and a set of reflections that cannot produce them has not generated the group, however many cosets its products reach.

It is tempting to work modulo the lattice, keeping track of each operation’s linear part and translation only up to lattice vectors. That is how the cosets of a plane group are usually listed, and it hands every set of generators the whole lattice for free. Done that way, pm would be generated by its mirrors, since its two cosets are the identity and the mirror and the mirrors reach both. But pm’s mirrors are all parallel. Two parallel mirrors compose to a translation across them, twice the distance between them, and nothing composes to a translation along them. The subgroup the mirrors actually generate has translations in one direction only. It is not a plane group at all but a frieze group, sitting inside pm at infinite index, as three reflections, and never four remarked.

So the test here has two halves. Products of the generators, reduced modulo the lattice, must reach every coset. And the translations among products of the generators themselves, computed exactly in the affine group, must span the whole lattice. The products are enumerated as words of length up to six. That is enough, because in every case the translations either span the lattice by length four or visibly never will.

The census #

The picture at the head of this essay is the whole census. Reading down the columns gives the three lists. Mirrors alone make four groups, the kaleidoscopes the earlier essay found by a different route, which is a check on the definition. Glides alone make six. Both together make eleven.

Reading across the rows shows how each group is built. p3m1 and p6m are made either way. p6m has so many reflections that any one kind is enough, and in p3m1 every glide line lies halfway between two mirror lines, as it must, and the glides generate the mirrors back. pmm and p4m are made of mirrors and not of glides: pmm has no glides at all, and p4m’s glides generate a subgroup of index two. pg, pgg, p4g and p31m are the reverse, made of glides and not of mirrors. pg and pgg have no mirrors. p4g’s mirrors generate a subgroup of index two. p31m’s mirrors reach all of its point operations, but the translations their products make form a lattice of index three in its own.

Three groups need both kinds: cm, pmg and cmm. Their mirrors alone and their glides alone each fall short, and the two together close. And pm is not made of anything. Its only reflections are parallel mirrors, and no combination of them produces the translation along them.

A glide squared is a translation #

The reason glides generate more than mirrors can be drawn in one cell.

A glide reflects across its line and slides along it by half a lattice vector. Done twice, the two reflections cancel and the two slides add, so the result is a translation along the line by a whole lattice vector. A mirror done twice is the identity. So a single glide already produces one direction of the lattice, the one along its own line, which no family of parallel mirrors can ever produce.

The product of two parallel glides at different positions supplies the other direction. The two reflections combine into a translation across the lines, twice the distance between them, and the two slides add along the line. In pg that product and the square of one glide span the whole lattice. So pg is generated by two parallel glides, while pm, the same arrangement with mirrors in place of glides, is not generated by any number of its mirrors. The single difference between the two groups, a half-lattice slide on each reflection, is exactly the difference between generating the translations along the lines and not generating them.

Why the slide is always half a lattice vector #

The slide of a glide is not a free second parameter, although the earlier essay was right that it is a second datum. A glide done twice is a translation by twice its slide, and that translation belongs to the group, so it must be a lattice vector. So the slide is half of a lattice vector lying along the glide’s line. It is not itself a lattice vector, or the operation would be a mirror followed by a translation, which the census counts as a mirror. That leaves very little room. The lattice vectors along one line form a one-dimensional lattice, the multiples of a shortest one $u$. The slides a glide on that line can have are $\tfrac12 u$ plus a multiple of $u$, and any two of them differ by a lattice translation along the line.

So a line of the plane carries at most one glide, up to translations along the line, in the same way that a line carries one screw in space. Two glides on the same line with different slides are different motions, as the earlier essay said, but they differ by a translation the group already has, so as generators they are interchangeable. What is left of the slide as a datum is its location: which lines of the lattice carry glides and which carry mirrors. In cm that is the whole of the difference between the two kinds. Mirror lines and glide lines alternate at half-spacings, and each glide’s slide is half the lattice vector along the lines.

This is also why the census had to count individual operations, not cosets. The mirror through one line and the glide through the line halfway to the next have the same linear part, and in cm they lie in the same coset of the lattice. A census that asked of each coset only whether it contains a mirror would have called cm a group with no glides at all.

Two glides that cross #

Two mirrors crossing at an angle $\theta$ compose to a rotation by $2\theta$ about their crossing point, the doubling law that where the product is establishes. Two glides crossing at the same angle compose to a rotation by the same $2\theta$, since the linear parts are the same. The slides then displace the centre of that rotation away from the crossing.

In pgg the two generating glides are perpendicular, so their product is a half-turn. Its centre sits a quarter of the cell away from the crossing in each direction, half of each slide, and so do the centres of all the other products. That is why the two-fold centres of pgg lie on no reflection line, a fact usually read off the group’s diagram. Here it is a consequence of how the group is made: its rotations are products of glides, and a glide product’s centre is where the slides put it, not where the lines cross.

The same displacement explains p4g. Its generating pair is a glide along an axis and a glide along a diagonal, at forty-five degrees. Their product is a quarter-turn about a point displaced by the slides. Because the displacement moves the four-fold centres off the mirror lines, p4g has four-fold centres on no mirror, the property that distinguishes it from p4m.

How few glides are enough #

For five of the six groups two glides are enough. They are parallel for pg, perpendicular for pgg, at forty-five degrees for p4g and at sixty degrees for p31m and p6m. For p3m1 no two of its glides within one cell generate it, and three are needed. Its glides run in three directions, sixty degrees apart, and every pair tried, whether in two directions or in one, falls short. Which subgroup each pair does generate, and whether a pair of glides further apart than one cell could succeed, the search does not say. p6m, with reflections of both kinds in six directions, builds everything from two.

The counts sit alongside the mirror counts of the earlier essay. There pmm needed four mirrors, the sides of a rectangle, and p3m1 three, the sides of an equilateral triangle. The kaleidoscope’s polygon sets the number of mirrors. No polygon sets the number of glides, since glides do not bound a region the way mirrors do. The small numbers come instead from the squares and products doing the work that extra generators would do.

Where both kinds are needed #

The three groups that need both kinds show the division of labour most clearly. In cm every reflection line is parallel to every other, alternating mirror and glide. The mirrors alone make translations across the lines and nothing along them, like pm’s. The glides alone make the squares along the lines and products across them, a lattice, but of index two, missing the centring. A mirror composed with its neighbouring glide gives exactly the missing translation, half a step along and half a step across. That translation is the centring vector centring, and why cm is not pm described, here produced by the two kinds of reflection together and by neither alone.

pmg needs both for a different reason. Its mirrors and glides are perpendicular, and each kind alone generates a subgroup of index two: the mirrors a pm and the glides a pg. The half-turns of pmg are products of one mirror and one glide, so a set lacking either kind cannot make them. cmm combines the two cases, with centring as in cm and perpendicular families as in pmm, and needs both kinds for both reasons.

What the closure answers #

The earlier essay’s question was whether the classification falls out of closing located glides with their slides. The answer is a list, not a yes or no. Closing glides alone produces six groups, closing mirrors alone four, and closing both eleven. Of the five groups without reflections, four, p2, p3, p4 and p6, are reached by closing rotation centres. p1 has no operation but its translations, so there is nothing to close. That leaves pm, which has point operations and still cannot be made from them, because it needs a translation that none of them supplies. Closing located rotation centres, located mirrors and located glides therefore produces fifteen of the seventeen groups. The other two need a translation given outright.

That is not a failure of the method. It is a statement about pm: apart from p1, which has no other operations, it is the one plane group in which a lattice translation cannot be written as a product of the group’s other operations. Every other group can express at least one of its lattice directions through its point operations. In pm, the translation along the mirrors is independent of everything else in the group. The classification proof by case analysis, which the classification proof runs branch by branch, never has to notice this, because it starts from a lattice. The closure starts from located operations and has to produce the lattice, and pm is where that fails.

What these pictures cannot show #

The enumeration itself has a horizon, and it is worth saying why the horizon does not matter. Words of up to six generators are formed, and the translations among them are collected. In every group that its reflections generate, the translations spanning the whole lattice have already appeared by words of length four: a square, or a product of two parallel reflections, and a product of two such products. In the groups they do not generate, the shortfall is structural, not a matter of length. pm’s mirrors are parallel, so every product of them is a translation across them or a mirror, and no word of any length has a component along them. cm’s glides alone produce only the translations of the index-two lattice, because each is a glide by half the lattice vector along the lines, and sums of those never reach the centring vector. Longer words would add nothing to either.

The drawings show generators, lines and slides, and in two cases the operations they produce. They cannot show the claim itself, that a set generates the whole group. That claim rests on the enumeration of words, whose coverage of every coset and whose translations are counted, not drawn. The minimal generating sets were found by searching the glides within one cell of the origin. For p3m1 the statement that three are needed is a statement about that search: no two glides within one cell generate the group. The argument about its three directions of glide makes it plausible that no two anywhere would, but the search does not prove it.

What the closure has to refuse #

The first refused claim is the one the question invited: that closing located glides reaches all seventeen. It reaches six. The second is the definition that would have hidden the answer, that a set generates a group when it reaches every coset modulo the lattice. pm’s mirrors do that and generate a proper subgroup of infinite index. Every count in this essay depends on refusing it, and the four groups made of mirrors, which the earlier essay found with a different method, are the check that the refusal is the right one.

Glides in space #

In three dimensions a glide plane carries a slide along one of the directions within it, and the five glide planes counted the kinds. The same closure question applies. Which of the two hundred and thirty space groups are generated by their glide planes alone, which by their mirrors, which by screws, and which need a translation given outright? Screws square to translations along their axes as glides square to translations in their planes, so a three-dimensional census should find more groups made of their non-symmorphic operations than of their point operations. Seven space groups cannot make their own translations runs that census over all 219 types. It finds 177 made of their screws and glides against 139 of their point operations, and the part of pm played by seven groups, the glide-free plane groups repeated along a line they fix.