Crystallography & Symmetry
patterns derived from their groups
Generator

The shells of the hexagonal lattice

The shells of the hexagonal lattice
The shells of the hexagonal lattice. Every point of the hexagonal lattice within a squared distance of 24, with a circle drawn at each length that occurs. The form is x² + xy + y², and the number of points on each circle is a coefficient of the lattice's theta series: 6 at 1, 0 at 2, 6 at 3, 6 at 4, 0 at 5, 0 at 6, 12 at 7, 0 at 8. The gaps matter as much as the counts — a circle with no points on it is a length the lattice does not have, and which lengths those are is a question in number theory rather than in geometry.

Every point of the hexagonal lattice within a squared distance of 24, with a circle drawn at each length that occurs. The form is x² + xy + y², and the number of points on each circle is a coefficient of the lattice's theta series: 6 at 1, 0 at 2, 6 at 3, 6 at 4, 0 at 5, 0 at 6, 12 at 7, 0 at 8. The gaps matter as much as the counts — a circle with no points on it is a length the lattice does not have, and which lengths those are is a question in number theory rather than in geometry.

12 essays call theta-series. The drawing above is what it returns with no arguments at all; every call below passes it something, because a placement that passes nothing draws whichever member of the family the generator happens to default to rather than the one its essay argues about. Every one of this site's 522 essays names its parameters at the call site, which the standard pass of 2026-08-09 established and param-floor holds.

Where it is called

Changing this generator changes every one of these figures.

The shells of the hexagonal lattice. Every point of the hexagonal lattice within a squared distance of 24, with a circle drawn at each length that occurs. The form is x² + xy + y², and the number of points on each circle is a coefficient of the lattice's theta series: 6 at 1, 0 at 2, 6 at 3, 6 at 4, 0 at 5, 0 at 6, 12 at 7, 0 at 8. The gaps matter as much as the counts — a circle with no points on it is a length the lattice does not have, and which lengths those are is a question in number theory rather than in geometry. Lattices

How many vectors of each length

Counting the lattice points at each distance from the origin turns out to be a question about divisors, and the answer explains something a crystallographer meets every day: why a cubic powder pattern has no line at seven.

 In the plane, the lengths do name the lattice. Every reduced binary form with coefficients up to 20 — 1750 lattices — with its theta series computed to 120 terms. No two of them agree. That is Schiemann's theorem for binary forms, which says the theta series determines the lattice in two dimensions and in three, confirmed here as far as the search reaches rather than proved. The closest pair is worth the space: two lattices whose shortest vectors both have squared length twenty agree for 38 terms — because neither has any vector before then — and part at the next one. Lattices

The lengths do not name the lattice

Seventeen hundred plane lattices, every one with a theta series shared with no other — the lengths determine the lattice, and an exhaustive search says so. In sixteen dimensions two different lattices have identical counts at every distance, and the example is sixty years old.

 a lattice triangle: 1 inside, 6 on the edge, area 3. a lattice triangle on its lattice, with the 1 points strictly inside it in the first colour and the 6 points on its boundary in the measured colour. Pick's theorem says the area is the interior count plus half the boundary count less one, which is 1 + 6/2 − 1 = 3; the shoelace formula on the same integer coordinates gives twice the area as 6. The two agree, and both sides are integers, so the check has no tolerance in it. The theorem holds for a non-convex polygon and a polygon with no interior point alike, neither of which the usual triangle-and-square picture makes obvious. Lattices

How many points a shape holds

Draw a polygon on a lattice, count the points inside, then double the polygon and count again. The counts are not approximately a polynomial in the scale — they are one, exactly, with the area as its leading coefficient and a constant term of one for every polygon there is.

 How many different lattices share a determinant. One bar per determinant: the number of inequivalent integral lattices whose metric has that determinant, which is the class number of the corresponding discriminant. Area does not decide shape — at determinant 1 and 2 there is one lattice each, and by 11 there are four — and the count does not grow steadily either. Each bar is computed twice: once by enumerating the reduced forms directly, and once by reducing every form in a box and collecting the distinct results, which is a search followed by an algorithm rather than a search over answers. The two agree at every bar. Lattices

How many lattices share a determinant

Area does not decide shape. The number of inequivalent lattices whose metric has a given determinant is a class number, computed by enumerating reduced forms — and checked by reducing every form in a box and counting what comes back distinct.

 The whole space of plane lattices, and its corner. Every plane lattice appears exactly once in this picture. Scaling changes no density, so the leading coefficient is fixed at one; reduction then confines the other two to 0 ≤ b ≤ 1 ≤ c, and every lattice has exactly one reduced form. The curves are the levels of constant density, which are parabolas — a density d needs 4c − b² to equal (π/2d)². They crowd toward the corner b = c = 1, which is the hexagonal lattice at π/√12 ≈ 0.9069; the square lattice sits on the left edge at π/4 ≈ 0.7854. The picture is a search over a region rather than over a list, which is what makes the answer a decision: there is nowhere else for a lattice to be. Symmetry at work

The densest lattice in the plane

Which arrangement of equal discs covers the most floor is a question about infinitely many lattices, and reduction turns it into a question about a two-parameter region with a corner. The answer is at the corner, and the argument finishes.

 Both sides of the transformation, on five lattices. A Gaussian of width set by t on every point of a lattice, summed; and the same sum over the dual lattice with the width inverted and the covolume divided out. The two agree to the last bit a double carries, at every t and on lattices with no symmetry in them, so nothing here is a coincidence of parameters. The identity is exact and the reason to have it is that the two sides do not cost the same. Lattices

The sum that turns a lattice into its dual

Put a Gaussian on every point of a lattice and add them up. The answer equals the same sum over the dual lattice with the width inverted and the covolume divided out — exactly, to the last bit a double carries, on lattices with no symmetry in them. The identity is free and the reason to have it is that the two sides do not cost the same: at one end of the range the direct sum needs forty thousand terms and the dual sum needs a hundred and twenty-five.

 The same terms, added in two shapes. Partial sums of the alternating 1/r sum over the simple cubic lattice, taken over expanding cubes and over expanding spheres. The terms are identical and only the order differs. The cubes creep towards 1.747565 — 1.7258 by the last point drawn — and the spheres do not settle at all, landing at -3.527 after passing through values on both sides of it. A conditionally convergent sum has no value until the order is named. Lattices

The sum whose answer depends on the shape

Give the points of a cubic lattice alternating signs and add up one over the distance. Added over expanding cubes the total creeps towards 1.747565; added over expanding spheres it does not converge at all, landing on both sides of that number and never settling. The terms are identical and only the order differs. Splitting the sum in two with the theta transformation gives it a value — ten decimal places from a few thousand terms.

 Three lattices no congruence can separate. The three reduced forms of discriminant minus twenty-three, with the integers each represents. The principal form represents one and the others do not; the others represent two and it does not. So they are genuinely different lattices — and they represent exactly the same residues modulo twenty-three, so they are in one genus and no congruence condition of any kind distinguishes them. Lattices

Lattices that agree at every prime

Counting the plane lattices with a given metric determinant is a class number. Above it sits a coarser count — the genus, which is what congruences can see — and for most small determinants the two agree. At discriminant minus twenty-three they part: three lattices representing exactly the same residues modulo everything, and different integers. No argument modulo any number can tell them apart, and they are not the same lattice.

 Face- and body-centred cubic change places at t = 1. The Gaussian sum of the face-centred cubic lattice minus that of the body-centred cubic lattice, both at unit volume, across widths from 0.35 to 2.8. Below t = 1 the difference is positive, so the body-centred lattice has the smaller sum; above, it is negative and the face-centred lattice does. At t = 1 the two are equal to the last digit a computer carries, and not by accident: at unit volume each lattice's sum at width t is the other's at width 1/t times a known factor, and at t = 1 the factor is one. Neither lattice is best at every width, and no other lattice can be either. Lattices

No lattice in space wins at every width

Spread a Gaussian over every point of a lattice and add them up. In the plane one lattice gives the smallest sum at every width — the hexagonal one — and that is a theorem. In space no lattice does, and the proof is two lines: at large widths the winner must be the densest lattice, at small widths it must be the densest lattice's dual, and in space those are face- and body-centred cubic. They change places at exactly the width where each is the other's dual.

 E₈ has the smallest sum at every width. For four eight-dimensional lattices of unit covolume — D₈, its dual D₈, the cubic lattice Z⁸ and two copies of D₄ — the ratio of their Gaussian sums, less the constant term, to E₈'s, at widths from 0.2 to 5, logarithmic. Every ratio exceeds one everywhere: E₈ wins at every width against every rival. The closest approach is at t = 1, where D₈ and its dual are equal, a factor of 1.31 above E₈. D₈ is the better of the pair at wide widths and D₈ at narrow ones, and neither comes close to E₈ at either end. Lattices

In eight dimensions one lattice wins at every width

In space no lattice has the smallest Gaussian sum at every width, because the densest lattice wins when the Gaussian is narrow and its dual wins when it is wide, and fcc is not bcc. The argument fails exactly when the densest lattice is its own dual. In eight dimensions it is: E₈, with 240 shortest vectors, beats D₈, its dual, the cubic lattice and a sum of two four-dimensional lattices at every width tried, by a factor of at least 1.31, and its sum is mirror-symmetric in the logarithm of the width.

 Almost every descent ends at D₄. At five widths of the Gaussian, twelve local minimisations of the lattice sum over all four-dimensional lattices of unit covolume, each from a random lattice, each marked by where it ends and labelled with the number of shortest vectors of the lattice it ends at. Fifty-nine of the sixty end at D₄, with twenty-four shortest vectors. One, at the widest Gaussian, ends at A₄*, with ten, a lattice whose sum is higher than D₄'s but which no small change lowers. Lattices

The two lattices where a four-dimensional descent stops short

In four dimensions the argument that rules out a single best lattice in space has nothing to act on, so whether one lattice wins at every width is left to be measured. A search over every four-dimensional lattice finds D₄ from almost anywhere. The exceptions are the finding: two lattices, A₄ and its dual, where a descent can stop short, each at one end of the widths and neither in the middle, for the same reason Voronoi found four dimensions has two perfect forms.

 The window in which A₄ and its dual are both minima. The exact second derivative of the Gaussian sum at A₄ and at A₄ along their five-dimensional family of changes of shape, for widths within six hundred-thousandths of one. A₄'s turns positive at t = 0.99998110 and A₄'s turns negative at 1.00001890, the reciprocal. Between them both are positive, so both lattices are local minima; at t = 1 itself both equal 8.59e-6. Lattices

The self-dual width is not flat

A search over four-dimensional lattices found that A₄ and its dual stop being traps for the Gaussian sum almost exactly at the self-dual width, and read that as a flat of the sum sitting on the width itself. Computed exactly rather than by finite differences, the curvature there is not zero. It is six parts in a million of the other curvature, positive, and the same for both lattices, because duality makes their Hessians one matrix at that width. Both are minima, in a window four parts in a hundred thousand wide, with basins a few millionths across.

The whole library · All essays