One way up and p ways down

How many sublattices counts the ways to thin a lattice, and at a prime index the count needs no arithmetic at all. A sublattice of index $p$ is a line through the origin in a plane over the field of $p$ elements, and that plane has $p + 1$ lines. So every lattice has $p + 1$ sublattices of index $p$, the same number whatever its shape, and the number says nothing about the sublattices themselves.

How many lattices share a determinant counted something else: the lattices whose integer metric has a given discriminant. Those two essays never meet, and they ought to. Each of the $p + 1$ sublattices is again a lattice with an integer metric, so it has a discriminant of its own and a place in that count. Thinning is a map from the list of lattices to itself. A map from a list to itself is a graph, and this graph turns out to have a shape so definite that it has a name in another subject.

This essay draws that graph. The result is one rule with three cases, tested on every lattice in a large census, and the rule explains at a stroke several facts that are usually met separately: why the square lattice has a rotated copy of itself at index five and none at index three, why the hexagonal lattice has its familiar root-seven superlattices and no root-two one, and why the class number grows by the factor it does.

What a sublattice can do to a discriminant #

The convention first, since everything below depends on it. A lattice here is a positive definite binary form $[a, b, c]$ with integer entries and no common factor, meaning the metric $a x^2 + b x y + c y^2$, and it is taken up to a change of basis of determinant one and up to overall scale. Its discriminant is $D = b^2 - 4ac$, a negative integer. Two forms that differ by a reflection are counted as two lattices when they are not also related by a rotation, which is a different convention from the earlier essay on class numbers, where a lattice and its mirror image were one thing. That choice matters here because the graph’s cycles run through mirror pairs, and folding them together would hide the cycle.

Now restrict a form to a sublattice of index $p$. The new basis is two integer combinations of the old, with determinant $p$, and the new form’s discriminant is $p^2 D$ exactly. Its entries may then share a factor. If they do, the factor can only be $1$, $p$ or $p^2$, because of how the discriminant scales, and dividing it out gives a primitive form of discriminant $p^2 D$, $D$ or $D / p^2$. Those are the only three things that can happen.

So each of the $p + 1$ edges from a lattice goes down, to a larger discriminant by a factor of $p^2$; across, to the same discriminant; or up, to a smaller one. Any other outcome is impossible, and the computation counts an edge that does anything else as a failure.

The smallest example makes the three visible at once, and it is the lattice everybody knows.

At index two, one sublattice is a square lattice again, and the other two are rectangles of aspect ratio two. At index three, all four are the wrong shape. At index five, two are square, turned by $\pm 26.6^\circ$, and four are not. The sublattices that stay square gave the reason in terms of sums of two squares: five is $1 + 4$, three is not a sum of two squares at all, and two is $1 + 1$ in a degenerate way. The new observation is what the other edges do. The rectangles of index two have discriminant $-16$, the ones at index three $-36$, the ones at index five $-100$. They are not random lattices. They are the level below.

The shape, for one discriminant #

Take a discriminant with a few lattices in it and follow the edges. At $D = -23$ there are three lattices: $[1, 1, 6]$, $[2, 1, 3]$ and its mirror image $[2, -1, 3]$. At the prime two each has three sublattices of index two. Two of those edges go across, and they join the three lattices in a triangle. The third goes down, to discriminant $-92$. Follow it, and the lattice there has one edge back up and two more going down, to discriminant $-368$. Each of those has one edge up and two down, and so on without end.

The shape has a name. In the theory of elliptic curves, where lattices in the plane correspond to curves and sublattices to the maps between them, the graph is called an isogeny volcano: a crater at the rim, with slopes running down from it. David Kohel described the structure in his 1996 thesis, and Mireille Fouquet and François Morain gave it the name and the algorithms in 2002. It is now routine in the arithmetic of elliptic curves, and it has been studied for reasons with no connection to crystals at all. Nothing about curves is used here. Forms are restricted and reduced, and the edges are read off.

The statement is this. Fix a prime $p$ and a discriminant $D$ with nothing above it, meaning a fundamental discriminant, or at least one not divisible by $p^2$ in the relevant way. Then every lattice at a level below the top has exactly one edge up, exactly $p$ edges down, and none across. And every lattice at the top has no edge up and a number of edges across fixed by the prime, with the rest going down.

Three craters, from one symbol #

The number of edges across at the top is the part that looks like it should depend on the lattice, and it does not. It depends only on $D$ and $p$, through a single value: whether $D$ is a square modulo $p$. That value is the Kronecker symbol $(D/p)$, and it is $1$, $0$ or $-1$. The number of edges across is $1 + (D/p)$.

Number theory has names for the three cases. When $(D/p) = -1$ the prime is inert: there are no edges across, and the top lattices sit isolated, each with $p + 1$ edges down. When $(D/p) = 0$ the prime is ramified: one edge across. When $(D/p) = 1$ the prime splits: two edges across, and these join up into cycles, since each top lattice has one edge in and one edge out.

The reason is short enough to give whole, and it is the surprising part. A plane lattice with an integer metric is not just a set of points. It has a set of similarities: complex numbers $\alpha$, written with the plane as the complex plane, such that $\alpha L \subseteq L$. Every integer is one. For most lattices there are no others, but for a lattice with an integer metric there always are, and they form a ring whose discriminant is exactly $D$. For the square lattice that ring is the Gaussian integers; for the hexagonal lattice, the Eisenstein integers.

So a sublattice can be asked a question that has nothing to do with shape in the first instance: is it carried into itself by the same ring? Those that are, and only those, keep the discriminant. They are the ideals of norm $p$ in that ring, and how many of those there are is what “split”, “ramified” and “inert” mean. A prime of a ring of integers can factor into two different prime ideals, or into one prime ideal squared, or not factor at all. So the edges across are counted by a factorisation in a ring that a crystallographer would have no other reason to mention. The lattice’s own similarities decide which of its sublattices repeat its discriminant.

A sublattice that is not an ideal loses part of the ring: its ring of similarities has index $p$ in the original, and that smaller ring is what a larger discriminant means. Going down, then, is losing similarities. Below the top, a lattice’s ring has already lost something, and exactly one of its $p + 1$ sublattices gets the lost part back, which is the edge up. The other $p$ lose more.

Familiar superlattices, read off the symbol #

The rule immediately sorts the superlattices a crystallographer actually uses.

For the square lattice, $D = -4$. The symbol is one at primes congruent to one modulo four, minus one at those congruent to three, and zero at two. So square superlattices of prime index exist at $2$, $5$, $13$, $17$, $29$ and no others in that range, at two orientations each except at two, where there is one. A surface scientist’s $\sqrt{5} \times \sqrt{5}$ cell turned by 26.6 degrees is an edge across at $p = 5$. There is no $\sqrt{3} \times \sqrt{3}$ on a square lattice, and the crater with no edge across is why.

For the hexagonal lattice, $D = -3$. The symbol is one at primes congruent to one modulo three, minus one at those congruent to two, and zero at three. So there is a $\sqrt{3} \times \sqrt{3}$ cell, turned by 30 degrees, one edge across, and it is the most common superlattice in surface science. There is a pair of $\sqrt{7} \times \sqrt{7}$ cells at $\pm 19.1^\circ$, two edges across, and they are the two cells that are mirror images of one another on which coincidence arithmetic also turns. There is no hexagonal cell of index two or five. The three sublattices of index two are all one rectangular shape, of aspect ratio $\sqrt{3}$, because $p = 2$ is inert there and all three edges go down.

None of this is new as a list. The sublattices that are the same shape arrived at the same list by a similarity test. What the graph adds is that the rest of the $p + 1$ are accounted for, that their shapes are the next level down rather than leftovers, and that the same structure holds at a discriminant with no symmetry in it. $[2, 1, 3]$ is an oblique lattice with no rotation at all, and it has two sublattices of index two that are related to itself by the same ideal arithmetic: not similar to itself, but similar to the other members of its cycle. The symmetric lattices are the cases where the cycle has length one, so that an edge across comes back to where it started and looks like a self-similarity. A cycle of length seven at $D = -71$ is the same fact, with no symmetry to distract from it.

The class number, one level down #

Counting the vertices of a level gives a classical formula with no effort, because the edges count them. Each lattice at the top has $p - (D/p)$ edges down, and each lattice one level below has exactly one edge up, so the level below has $h(D)\,(p - (D/p))$ lattices, where $h(D)$ is the number at the top. For every further level the count multiplies by $p$. Only the discriminants $-3$ and $-4$ need a correction, because their lattices have extra rotations that make several downward edges arrive at the same lattice: divide by three and by two respectively.

This is the class number relation that appears in any text on binary quadratic forms, usually with its proof by counting cosets of units. Here it is simply a statement about a graph’s levels, and both sides are computed: the prediction from the edge rule, the other side by enumerating reduced forms at the larger discriminant. The two are not derived from one another in the calculation, which is what gives the comparison its force.

The census behind the table’s last line runs over all 94 fundamental discriminants down to $-300$, at each of the primes $2$, $3$, $5$ and $7$, to two levels below the top wherever the discriminant stays within reach. That is 18,413 lattices, each restricted to all its sublattices and each edge classified. The rule failed on none. The class number formula was compared at 605 steps and failed at none. That is a test of a known theorem rather than a discovery, and the point of running it is that a form reduction with an error in it, or an incorrect sublattice list, would show up immediately as an edge of a kind the rule forbids. A list of $p + 1$ matrices of determinant $p$ that gives only two different lattices is an easy mistake to make, and it breaks the rule at the first split prime.

Since the prime-by-prime symbol sets the width of the crater and the class group sets how it closes, the cycle lengths carry real information. At $D = -23$ and $p = 2$ the crater has length three because the class group has order three. At $D = -71$ the crater has length seven, the whole class group, because the prime form above two generates it. There is no requirement that a prime generates: the crater at a prime is the cycle of the group generated by that prime’s form, and when that is a proper subgroup the top level breaks into several separate craters of equal length. Lattices that agree at every prime met this class group from the side of congruences, where discriminant $-23$ was the first place the genus and the class came apart. Here the same three lattices form one triangle, joined by thinning.

What thinning does to the shape #

Every level has a place in the space every lattice lives in. A form $[a, b, c]$ is a point $\tau = (-b + i\sqrt{|D|})/2a$ in the modular region, after reduction. So a level is a set of points in one familiar picture, and the question of what repeated thinning does to shape can be looked at directly.

The points spread out. Those at low levels sit in a small number of places; by the fifth level they fill the region, with a visible density that thickens towards the bottom. That density is not a defect of the drawing. The natural measure on this region, the one how far one lattice is from another derived from the requirement that no change of basis can alter a distance, weights height $y$ by $1 / y^2$, so the low part of the region really does hold most of the area.

That convergence is a measurement here, and a theorem elsewhere. In 1988 William Duke proved that the points of discriminant $D$, as $|D|$ grows, become equidistributed for exactly this measure, whatever sequence of discriminants is taken. The levels of a volcano are one such sequence. So a crystallographer with a superlattice of a superlattice of a superlattice, of an arbitrary but prime-power index, can make no prediction about what shape a typical one has except that shapes come up in proportion to hyperbolic area. Almost all very deep sublattices of any integer lattice, including the square one, are the same distribution of shapes. The lattice’s own symmetry is forgotten as soon as one has descended far enough.

The contrast with the top of the graph is the interesting part. At the top, thinning at a split prime can return the lattice exactly. A few levels down, thinning returns it to an ever larger class in which its starting shape is a vanishingly small share. The two statements are about one structure: the crater is finite and the slopes grow by a factor of $p$ at each step.

What the picture leaves out #

Several limits belong next to these figures.

The graphs here are drawn for integer forms, and most lattices are not integer. A generic lattice has no similarities beyond the integers, no discriminant in this sense, and no crater. What it has instead is a tree: each of its $p + 1$ sublattices is a new shape, none of them repeats, and no chain of thinning and thickening comes back to where it began. Jean-Pierre Serre’s book on trees makes this exact: the lattices up to scale at a prime form an infinite tree in which every vertex has $p + 1$ neighbours. The volcano is what that tree looks like when it is folded up by a ring of similarities. It is the tree quotiented by a group of units and class-group elements, and a cycle exists only where a lattice is similar to a sublattice of itself. So the figures show the exceptional lattices rather than the general case, and those happen to be the only lattices with an integer metric, which is why a crystallographer meets them far more often than their measure would suggest.

The drawing of a level as a tree is also an arrangement rather than a measurement. Nothing about the positions of the dots in the volcano figure means anything: the graph has no geometry of its own, and the distances between the dots do not correspond to anything about the lattices. Only the edges carry information. The modular region figure has true positions and shows no edges at all, and no single picture here can hold both.

And mirror images count as distinct, which is the convention chosen at the start. Under the convention of the earlier class-number essay the triangle at $-23$ folds to a line with a loop on it, and the seven-cycle at $-71$ folds to a path of four lattices with a loop at one end. The arithmetic is unchanged; only the drawing is less informative. Which convention a crystallographer prefers depends on whether a handed superlattice, one of a mirror pair of cells on a surface, should count twice. Physically it should, since a surface domain and its mirror image are distinguishable, and that is the reading here.

How the rule was tested #

The census runs on the edges themselves. For every lattice in the list, each of the $p + 1$ bases is applied, the content is divided out, the form is reduced, and the edge is classified by its discriminant. The expected numbers of up, across and down edges are then compared with the rule. The mirror-image convention needs a reduction that keeps them apart, and the one used here is written for that purpose: the standard one elsewhere in these essays folds mirror images together, which would have made the rule look wrong at the first split prime.

The refusal that does the most work is the simplest. A number congruent to two or three modulo four is not the discriminant of any integer form, and the enumeration refuses it, so a mistake in the bookkeeping of $p^2$ cannot quietly produce a list of forms for a discriminant that does not exist. The other checks are the known answers: two square sublattices at five and none at three, a pair of root-seven cells on the hexagonal lattice, a triangle at $-23$. Each is a place where a wrong rule would be visibly wrong rather than subtly so.

Still open: what the graph does in three dimensions #

The whole structure depends on the plane in one respect that is easy to overlook. An integer lattice in the plane has a commutative ring of similarities, because the plane’s rotations commute. In three dimensions there is no such ring. A lattice in space has at most a finite group of isometries and essentially no similarities other than scalar multiples, unless it is one of a few special lattices. The three that stay cubic and the shapes a lattice in space can thin to both found how much harder thinning is there. In three dimensions the sublattices of index $p$ number $p^2 + p + 1$, and so do those of index $p^2$ whose quotient is two copies of the cyclic group, and the two families are not the same kind of object: one is the lines of a projective plane over the field of $p$ elements and the other is its points.

So the natural question has no obvious answer. The analogue of the top of the graph would be sublattices of a lattice that have the same set of integer metric invariants, and the analogue of an ideal would be a sublattice stable under the lattice’s own quaternion order, for those few lattices that have one. Whether any lattice in space has a graph with this one-up, many-down rule, and what replaces the Kronecker symbol if it does, is the first thing to find out. The cubic lattices are the place to begin, since similar sublattices of them are already known to exist. It has not been computed here.