One point in every big enough shape

Take two vectors in the plane and form every whole-number combination of them. The result is a lattice: an infinitely repeating grid of points, generally slanted, and the parallelogram spanned by the two vectors tiles the plane with exactly one lattice point per tile.

That parallelogram’s area is the determinant of the matrix whose columns are the two vectors — the same number that, read as a map, says what the unit square becomes. Read that way the number stops being a property of a map and becomes a property of the point set — and it is the only number the point set needs.

The area does not depend on the basis #

A lattice has many bases. The vectors $(1,0)$ and $(0,1)$ generate the integer grid; so do $(2,1)$ and $(1,1)$, and so do infinitely many other pairs, all describing exactly the same set of points.

Two bases generate the same lattice exactly when the matrix carrying one to the other has whole-number entries both ways round. Such a matrix and its inverse both have integer determinants whose product is one, so each is $\pm 1$; and by the product rule the two bases’ determinants differ by that factor. Their absolute values agree.

So $|\det|$ is an invariant of the lattice rather than of any description of it, and it has a name: the covolume, or the determinant of the lattice. Everything below is about what that one number controls.

The exchange rate #

Cover a large region with translates of the cell. Each holds one lattice point, so a region of area $A$ holds about $A / \det$ points, with an error that comes only from the cells the boundary cuts through.

The figures measure exactly that. A window of area $100$ is drawn, the points inside are counted, and the count is asserted to match $100/\det$ up to a boundary allowance proportional to the perimeter. That is the whole content of “the determinant is the area per point”, stated as something that can fail and does not.

The boundary term is not a technicality to be waved away — it is where the difficulty of every lattice-counting problem lives. Counting lattice points in a circle of radius $r$ gives $\pi r^2$ plus an error, and how small that error can be is an open problem that has resisted a century of work. What is easy is the leading term, and the leading term is the determinant doing its job.

Counting points, and where the counting stops being easy #

The exchange rate is a statement about large regions, and it is worth separating what is exact from what is asymptotic, because the two get conflated constantly.

Exact: the cell tiles the plane with one point per tile. Nothing approximate about it — it is a bijection between lattice points and translates of the cell.

Asymptotic: a region of area $A$ holds $A/d$ points up to the boundary. The error is at most the number of cells the boundary passes through, which is proportional to the perimeter divided by the cell’s diameter. For a growing disc of radius $r$ that error is at most a constant times $r$, against a main term of order $r^2$, so the ratio tends to $1/d$ and the counting is asymptotically exact.

Pick’s theorem is the case where the error term disappears completely: for a polygon with lattice-point corners, the area is exactly the interior count plus half the boundary count minus one, with no approximation anywhere. That exactness is special to polygons whose corners are on the lattice, and it is what makes Pick’s theorem a theorem rather than an estimate. For a circle nothing like it exists, and how well the error can be bounded is the Gauss circle problem, still open.

The reason to be careful here is that Minkowski’s theorem needs none of it. Its proof does not count points at all; it compares two areas and finds an overlap. So the hardest part of lattice counting — the boundary — never appears, which is why the theorem is provable in four sentences while the counting question it sits beside is a century old.

Minkowski’s theorem #

Here is the statement, and it is one of the most useful facts in the subject.

Let $L$ be a lattice of determinant $d$, and let $K$ be a region that is convex, symmetric about the origin, and of area greater than $4d$. Then $K$ contains a lattice point other than the origin.

No hypothesis about the shape beyond convexity and symmetry; no hypothesis about the lattice beyond its determinant. Area alone, past a threshold, forces a point.

The figures compute the sharp version rather than the guarantee: for the drawn lattice they find the smallest ellipse of the given shape containing a non-zero point, and assert that its area is at most $4d$. That is Minkowski’s bound checked at a particular lattice, which is what a figure can do; the theorem is the claim that it holds at every lattice and every convex symmetric shape at once.

The proof is a fold #

The argument is due to Blichfeldt and it is a pigeonhole in disguise, which puts it in the same family as eight socks in seven drawers.

Shrink $K$ by half, giving a region $K/2$ of area greater than $d$. Now chop $K/2$ along the lattice cells and slide every piece back into one single cell by subtracting the appropriate lattice vector. The pieces have total area greater than $d$ and the cell has area exactly $d$, so two pieces must overlap.

An overlap means there are two points $x$ and $y$ in $K/2$ with $x - y$ a non-zero lattice vector. Now use the two hypotheses. Symmetry says $-y$ is in $K/2$ as well; convexity says the midpoint of $x$ and $-y$ is in $K/2$; and that midpoint is $(x-y)/2$. So $(x - y)/2$ lies in $K/2$, which means $x - y$ lies in $K$ — and $x - y$ is a non-zero lattice point.

Every hypothesis was used exactly once, which is the sign of an argument with nothing to spare in it.

Why both hypotheses are needed #

Drop symmetry and the theorem fails immediately: a long thin triangle far out along the diagonal can have any area at all and contain no lattice point, since nothing forces it near the origin.

Drop convexity and it fails too, and more interestingly. Take a cross or a starfish shape, symmetric about the origin, with thin arms reaching out between the lattice lines. Its area can be made as large as wanted while every arm threads the gaps, because the arms can be made thin and long without limit. Convexity is what stops a region from being large and evasive at the same time: a convex region containing two far-apart points must contain the whole segment between them, and that segment is what runs into the lattice.

The bound cannot be improved #

Four times the determinant is exactly right, and the witness is the simplest possible.

Take the integer grid, whose determinant is $1$, and the open square with corners at $(\pm 1, \pm 1)$. It is convex, it is symmetric, its area is $4$ — and the only lattice point it contains is the origin, since every other integer point has a coordinate of absolute value at least one and the square is open. So area $4$ is not enough, and the theorem’s “greater than $4d$” cannot be relaxed to “at least”.

That sharpness is worth noticing because it explains the constant. The four is $2^n$ in dimension $n$ — the shrinking step halves each of $n$ coordinates — and in dimension $n$ the theorem reads: area greater than $2^n d$ forces a point. The same open cube witnesses sharpness there.

What it buys #

Minkowski’s theorem is a machine for turning geometry into arithmetic, and three of its outputs are already in this collection under other names.

Approximation by fractions. How close a fraction can get to an irrational is a pigeonhole argument in its elementary form and a Minkowski argument in its general one: the region $|x| \le N$, $|αx - y| \le 1/N$ is convex, symmetric, and has area $4$, so it holds a lattice point, and that point is a fraction within $1/N^2$ of $α$. The essay that proved it by drawers names this generalisation and does not build it; this is the construction it was pointing at.

Sums of two squares. That a prime leaving remainder one on division by four is a sum of two squares can be proved by choosing a lattice adapted to that prime and applying the theorem to a disc. The essay that proves it by descent takes a different route; the lattice route is shorter and less self-contained, since it needs the square root of minus one modulo the prime before it starts.

Four squares. Every whole number is a sum of four squares — Lagrange proved it in 1770 by an argument in the same family as the descent that shows a square cannot shrink forever — and the standard modern proof is Minkowski’s theorem applied to a four-dimensional lattice and a ball. The lattice is chosen so that every one of its points has a length whose square is divisible by the number in question, the ball is chosen large enough for the theorem to bite, and the point it produces is the representation. Nothing about the argument is four-dimensional in spirit; the dimension is simply where the four squares live. That is where the theorem earns the generality of its statement — there is no picture at all, and the argument is unchanged.

A lattice with no good basis #

One more property the determinant does not see, because it explains why the subject continues past this rung.

Two lattices of the same determinant can be very differently shaped. The integer grid and the lattice generated by $(100, 0)$ and $(0, 1/100)$ both have determinant one, and a disc of area $5$ around the origin contains several points of the first and none of the second — the second’s points are strung out along two nearly-perpendicular directions at wildly different spacings. Minkowski’s theorem still applies, and what it forces is a point of the second lattice inside a region of area past four, which will be a long thin ellipse rather than a disc.

The quantity that measures the difference is the length of the shortest non-zero vector, and the ratio between that and the determinant is where the geometry of numbers begins in earnest — reduction theory, successive minima, and the question of finding a basis whose vectors are as short as they can be. Minkowski’s theorem gives the first bound in that subject: the shortest vector is never longer than a constant times the square root of the determinant, because a disc of that size has area past the threshold. Everything sharper is harder, and in high dimensions finding the shortest vector at all is the problem a good deal of modern cryptography is built on.

Two vectors, and the whole of the plane #

A last observation, because it is the one that makes the covolume feel inevitable rather than defined.

The lattice generated by two vectors is a subgroup of the plane, and the cell is a set of representatives for the plane modulo that subgroup. So the determinant is the area of the quotient — the plane with lattice-equivalent points glued together, which is a torus — and “one point per cell” is the statement that the quotient map is one-to-one on the cell.

That reading makes the basis-independence obvious rather than computed: the quotient does not know which basis was used to describe the subgroup, so any quantity attached to the quotient is automatically an invariant. It also says what the theorem is really about. Minkowski’s argument shrinks the region by half and wraps it onto the torus; the overlap is two points of the region that are the same point of the torus; and the difference of two such points is a lattice vector by construction. Read on the torus, the pigeonhole is the statement that a set of area greater than the torus’s own area cannot inject into it.

Every essay in this collection that glues a plane into a surface is doing the same construction with different intentions, and the determinant is what that surface’s area turns out to be.

What the pictures cannot show #

The figures draw one lattice at a time and one ellipse at a time. The theorem is universally quantified over both, and no drawing quantifies over anything — so what the figures check is the instance: this lattice’s cell has this area, this window holds this many points, this ellipse’s smallest scale holding a point has an area under the bound.

The folding argument has no figure here at all, and it is the part a reader most wants to see. Drawing it honestly would need the region chopped along cell boundaries and the pieces slid back into one cell, overlapping — five or six panels of a construction whose whole content is that the total area does not fit. It is a reasonable figure to want and it is not one this essay has; the argument is four sentences and the sentences are exact.

The other gap is dimension. Every consequence above except the two-dimensional ones happens in higher dimensions, where the theorem is identical and the pictures stop. The four-squares proof lives in four dimensions and is the clearest case: the geometry is doing all the work, and none of the geometry can be looked at.

Where the ladder goes next #

So far the determinant has measured things — an area, a volume, the room a lattice leaves. It also counts: there are matrices whose determinant is the number of spanning trees of a graph, and matrices whose determinant is the number of ways several paths can be drawn without touching. In both cases the minus signs in the permutation sum are what does the counting, by cancelling exactly the configurations that ought not to be counted.