The theorem that has no version in space

Pick’s theorem is one of the most satisfying results in elementary geometry, and part of what makes it satisfying is that it seems to be about nothing at all. Take a polygon whose corners are at whole-number coordinates. Count the lattice points strictly inside it, call that $I$, and the lattice points on its boundary, call that $B$. Then the area is

$$A = I + \frac{B}{2} - 1,$$

exactly, with no error term and no conditions beyond the corners being lattice points.

The theorem says that a continuous quantity — area — is completely determined by two discrete counts. That is not how area usually behaves: the ordinary route to it is a limit of finer and finer approximations, and a limit is precisely what a formula in two whole numbers is not. That is a strong claim, and the natural next question is whether the same holds one dimension up. It does not, and the failure is more instructive than the theorem.

The refutation, in one family #

The tetrahedron with corners $(0,0,0)$, $(1,0,0)$, $(0,1,0)$ and $(1,1,r)$ is the Reeve tetrahedron, named for John Reeve, who introduced it in 1957.

Its volume is $r/6$: the three edges from the origin span a parallelepiped of determinant $r$, and a tetrahedron is a sixth of that.

Its lattice points are four, and no more. A point $(x,y,z)$ inside or on it must have $z$ between $0$ and $r$, and the cross-section at height $z$ is a triangle that shrinks as $z$ rises — at height $z$ it has vertices $(z/r, z/r, z)$, $(1, z/r, z)$ and $(z/r, 1, z)$, and unless $z/r$ is a whole number that triangle contains no lattice point at all, while at $z = 0$ it is the unit right triangle with three lattice points and at $z = r$ it has collapsed to the single point $(1,1,r)$.

So $I = 0$ and $B = 4$, for every $r$, while the volume runs $1/6, 2/6, 3/6, \ldots$ without limit.

Any formula that computes volume from $I$ and $B$ would have to give the same answer for all of them. That is not a formula that fails on a hard case; it is a formula that cannot exist. And the counterexample is a tetrahedron — the simplest solid there is, with four vertices, and one of the two simplest anybody would try.

Why the two-dimensional proof does not survive #

It is worth locating precisely which step breaks, because “the analogue is false” is much less useful than “this step is what was two-dimensional”.

The standard proof of Pick’s theorem goes by triangulation. Cut the polygon into triangles whose corners are lattice points and which contain no other lattice point at all — primitive triangles. Two facts then finish it. Every primitive triangle has area exactly $\tfrac12$; and both sides of Pick’s formula are additive when two polygons are glued along an edge. So counting the primitive triangles gives the area, and counting them again by Pick’s expression gives the same number.

The first fact is the two-dimensional one, and its proof is short: a primitive triangle with vertices at lattice points has an edge vector pair whose determinant is $\pm 1$, because a determinant of larger absolute value would leave a lattice point inside or on the triangle. Every primitive triangle is therefore a unimodular copy of the standard one.

In three dimensions the corresponding statement is false, and the Reeve tetrahedron is the witness. It has no lattice point other than its corners — it is primitive by the same definition — and its volume is $r/6$, which is as large as one likes. There is no fixed volume for a primitive simplex in space, so the counting argument has nothing to count.

What a formula would have had to look like #

Before accepting that no formula exists it is worth being precise about what was being ruled out, because “no formula” is a large claim and the argument above rules out something narrower and more interesting.

What the Reeve tetrahedra refute is any function $f(I, B)$ with $\text{vol} = f(I,B)$ for every lattice polytope. They do not refute a formula using more data. Counting the lattice points on each face separately, or on each edge, gives more information, and one might hope that would be enough.

It is not, and the same family says so with one adjustment. The Reeve tetrahedron’s edges each carry two lattice points — their endpoints — except that the edge from the origin to $(1,1,r)$ carries $\gcd(1,1,r) + 1 = 2$ as well. Its four faces are triangles with three lattice points each. So the whole face lattice with its point counts is identical for every $r$: same number of vertices, edges and faces, same counts on each. Every piece of combinatorial data anybody would collect is the same, and the volumes differ by any factor.

That is what makes the counterexample decisive rather than merely awkward. It is not that one particular formula fails; it is that the map from combinatorial data to volume is not a function at all.

What does generalise #

Something does, and it is more informative than the theorem it replaces.

Take a lattice polytope $P$ in any dimension and count the lattice points in the dilated copy $kP$ for each whole number $k$. Ehrhart proved in 1962 that this count is a polynomial in $k$, of degree the dimension of $P$, whose leading coefficient is the volume.

In two dimensions the polynomial is

$$L(k) = A k^2 + \frac{B}{2}k + 1,$$

and every coefficient is meaningful: the area, half the boundary count, and one, which is the Euler characteristic of a convex region. Pick’s theorem is that polynomial’s coefficients, named. Setting $k = 1$ recovers $I + B = A + B/2 + 1$, which rearranges to Pick’s formula.

So the two-dimensional case is not special because a formula exists there; it is special because the coefficients of the polynomial happen to be counts.

The coefficient that has no meaning #

In three dimensions the Ehrhart polynomial of a solid is a cubic

$$L(k) = V k^3 + c_2 k^2 + c_1 k + 1,$$

with $V$ the volume, as it must be, and $c_2$ half the surface area measured in the right normalisation. The remaining coefficient $c_1$ is where the trouble is.

For the Reeve tetrahedron the polynomial is

$$L(k) = \frac{r}{6}k^3 + k^2 + \left(2 - \frac{r}{6}\right)k + 1,$$

and the coefficient of $k$ is negative as soon as $r > 12$.

A negative coefficient cannot be a count of anything, so there is no combinatorial reading of $c_1$ to be found, however cleverly one looks. This is the sharpest statement of the situation: the correct generalisation of Pick’s theorem exists, is a theorem, and has a coefficient that is not a count. The two-dimensional case was not a hint about what the general shape would be. It was a small case whose coefficients happened to be interpretable, and the interpretability was the accident.

What $c_1$ is has been understood since Pommersheim’s work in the 1990s: it is a sum of Dedekind sums over the edges of the polytope, which are number-theoretic objects arising from lattice geometry and having nothing to do with counting anything visible. Getting an explicit formula for it in three dimensions was a substantial piece of work, and it involves the theory of toric varieties. Counting lattice points under a line is an exercise a rectangle settles; counting them under a plane in space is where the Dedekind sums come from, and they were studied for their own reasons long before anybody asked this question of them.

Reading that figure against the previous one is the quickest way to see what Ehrhart’s theorem asserts and what it does not. Two different polygons give two different quadratics, and nothing about either quadratic could be guessed from the shape — what is asserted is only that a quadratic exists, that its degree is the dimension, and that its top coefficient is the area. Everything below the top coefficient is where dimensions start to differ.

What is true instead, and what it costs #

Three consolations, in decreasing order of usefulness.

Volume is still computable, and cheaply. Nothing about the failure of Pick’s theorem makes finding a lattice polytope’s volume hard: triangulate and add determinants. What is lost is a formula in terms of counts, not the ability to compute.

Bounds survive where equalities do not. A lattice polytope in $d$ dimensions with exactly one interior lattice point has volume bounded above — by a very large bound, but a finite one, which is a theorem of Hensley and Lagarias–Ziegler. So the counts do constrain the volume; they simply do not determine it. Constraint without determination is the ordinary situation and equality is the surprise.

The reciprocity is a real consolation and an odd one. Evaluating the Ehrhart polynomial at $-k$ gives, up to a sign depending on the dimension, the number of lattice points in the interior of $kP$. That is Ehrhart–Macdonald reciprocity, it holds in every dimension, and it is genuinely useful — it turns a question about interiors into a question about closures with no extra work. It is also completely unmotivated by anything in the two-dimensional picture: nobody looking at Pick’s theorem would think to substitute a negative number into a count.

And the two-dimensional theorem has genuine higher-dimensional cousins, just not the naive one. Any solid admitting a triangulation into primitive tetrahedra of volume $1/6$ — a unimodular triangulation — does obey a Pick-like formula, and many solids of interest have one. The theorem is not false about a class; it is false about the class of all lattice polytopes, and that class is larger than intuition suggests.

The two-dimensional accident, examined #

It is worth asking what exactly is true in two dimensions that fails above, since the previous sections say primitive triangles have area one half and leave the reason implicit.

A triangle with lattice vertices $O$, $P$, $Q$ has twice its area equal to $|\det(P, Q)|$, a whole number. If that determinant is $\pm 1$ the triangle has area $\tfrac12$; if it is larger, the parallelogram spanned by $P$ and $Q$ has area at least two and therefore, by a short argument, contains a lattice point other than its corners — which lands in the triangle or its mirror image. So primitive forces the determinant to be one.

The three-dimensional statement one would write down is: a tetrahedron with lattice vertices and no other lattice points has $|\det| = 1$. The Reeve tetrahedron has $\det = r$ and no other lattice points, so it is false, and the reason the two-dimensional argument does not carry over is that the pigeonhole step does not. In the plane, a fundamental domain of area two is small enough that a lattice point has nowhere to hide. In space, a fundamental domain of volume $r$ can be long and thin — the Reeve tetrahedron is exactly that, a sliver reaching to height $r$ over a unit base — and a sliver can have large volume and contain nothing.

Thinness is the whole of it. A convex body can be as large as one likes in volume and still miss the lattice entirely if it is thin in some direction, and in the plane there is not enough room to be thin in a way that helps. That is the same phenomenon that makes the geometry of numbers interesting: a bound relating volume to lattice points always has a convexity or symmetry hypothesis, and the hypothesis is there to forbid slivers.

Why the guess was reasonable #

It is worth defending the wrong guess, since the situation is a good example of a pattern that misleads for a structural reason.

The two-dimensional result feels like it is about the relationship between a region and the lattice, which is a relationship that exists in every dimension. What it is actually about is the classification of primitive simplices, and that classification is trivial in two dimensions and unbounded in three. Nothing in the statement of Pick’s theorem mentions the classification, so nothing in the statement signals that the proof used a fact that would not survive.

Small cases lie in a specific way: they lie by having coincidences that look like structure. In two dimensions “primitive” implies “unimodular” and both imply “area one half”. Those three notions come apart in three dimensions, and the theorem is a statement about whichever one the proof used — which is discoverable only by reading the proof.

That is the same shape as the failure of the minimum spanning tree’s cheapness in space, where the containment survives and the linear bound on the triangulation’s size does not; and the same shape again as regular solids, where the two-dimensional case has infinitely many members and the three-dimensional case has five.

What Reeve was actually doing #

The tetrahedra were not built to be a counterexample, which is worth knowing because it is the usual way such things arrive.

Reeve’s 1957 paper is about defining a volume-like invariant for lattice polytopes in three dimensions that does work, and the tetrahedra appear in it as the obstruction his construction has to accommodate. His answer is to count lattice points not only of the original lattice but of the finer lattices obtained by dividing the unit cell — the counts at several dilations, in effect — and to show that finitely many of those counts do determine the volume.

That is Ehrhart’s polynomial arriving five years early, in a special case, invented as a repair rather than as a theory. The general statement, that the counts are a polynomial in the dilation, is the observation that Reeve’s family of counts has structure; and once it is a polynomial, everything about it follows from having enough values.

The order of events is a good illustration of how an impossibility result is usually productive rather than terminal. The impossibility here — no formula in $I$ and $B$ — did not close the subject; it identified exactly what extra data was needed, and the extra data turned out to be a polynomial with a life of its own.

What the pictures cannot show #

A tetrahedron of height twenty-four cannot be drawn beside one of height one. The figures scale each solid to fit its own panel, so a reader sees four similar-looking wireframes and has to read the labels for the heights. The whole content of the family is that the shapes are not the same, and the drawing normalises exactly that away.

The lattice is not drawn in the three-dimensional panels. Drawing the whole lattice around a tall thin tetrahedron produces an unreadable cloud, so only the four points that belong to the solid are marked. The claim that there are no others is the assertion, made by exhaustive search over a box, and the picture cannot display an absence.

And the Ehrhart polynomial is drawn as a table of coefficients. Its content is a statement about a function of a whole number, and its coefficients are the output of a fit. The figure checks the fit against the measured counts at every dilation it computes, which is a real check on four values, and it is not a proof that the count is a polynomial at all — that is Ehrhart’s theorem, and it has no picture.

Where the ladder goes next #

Named here as debts, both raised above: Ehrhart reciprocity, which says that evaluating the polynomial at negative arguments counts the interior points of the dilations, up to a sign — a statement with no obvious meaning that is nevertheless true and is the reason the polynomial is the right object; and the classification of lattice polytopes with few interior points, where the Reeve tetrahedra reappear as the extreme cases.

Sideways, the lattice this all takes place in is the one whose circles count representations as sums of two squares, the packing of solids against it is the sphere-packing question, and the trap of reading a two-dimensional coincidence as a general law is the one the flip graph falls into as well.

What is worth carrying away #

A theorem that determines a continuous quantity from discrete data is remarkable, and the right response is to ask which step of its proof is doing that.

Pick’s theorem determines area from two counts because every primitive triangle has the same area. That single fact is the whole of the miracle, it is a statement about the classification of small lattice triangles, and it is false in three dimensions in the strongest possible way — the primitive tetrahedra have unbounded volume. Everything else in the theorem generalises perfectly.

The habit worth taking is to find the sentence in a proof that is about classification. A step of the form every object of this small kind is one of these is exactly the step that dimension breaks, and a theorem resting on one is a theorem to be suspicious of the moment it is generalised.