## Minkowski Theory

Since I’ve only gotten one reply to my further algebraic number theory thread – which is about what I expected – I’m going to go with the standard treatment of the subject and talk about Minkowski theory. Intuitively, Minkowski proved theorems that relate to lattices, like Z^n, so that we can view the ring of integers of a number field (which is additively identical to Z^n) as a geometric object.

First, the case of interest is the vector space R^n. Although R is usually more important to calculus and analysis than algebra and number theory, it has one key property: you can define length on it. You can define length, area, volume, etc. on many fields and vector spaces, but they won’t have the familiar properties unless they’re of the form R^n.

Second, a lattice in R^n is the additive group generated by a set of elements that are linearly independent over R. Since R^n has dimension n, the additive group is generated by at most n elements, so it’s of the form Z^r for some r <= n. Minkowski theory is interested mostly in the case when r = n.

Third, an easy way to characterize a lattice is by writing the elements that generate it in matrix form. For example, in R^2, the lattice generated by (1, 0) and (2, 5) can be written as [1, 2; 0, 5]. The matrix will have nonzero determinant, because the vectors are linearly independent. Of course, the matrix is not unique. The lattice generated by (1, 0) and (2, 5) is the same as this generated by (1, 0) and (0, 5), so we can also write it as [1, 0; 0, 5]. But the absolute value of the determinant of the matrix is always the same, in this case 5.

Fourth, one way to visualize the absolute value of the determinant is by volumes (or, in the case of R^2, areas). Take any region in R^2 of area 1, say the unit square, consisting of all points (x, y) with 0 <= x < 1 and 0 <= y < 1. Now, look at its image under a matrix A, say [1, 0; 0, 5]. The point (x, y) will get mapped to (x, 5y), so the image of the unit square is the set of all points of the form (x, 5y) with 0 <= x, y < 1. That’s a rectangle of length 1 and height 5, whose area is 5 = |A|.

This in fact applies to all matrices: for proof, note that switching rows and adding rows to other rows changes neither the absolute value of the determinant nor the volume of the image of a region of volume 1, and that these two operations can turn every matrix into a diagonal matrix, for which the result follows by stretching or compressing.
Fifth, every lattice L = Z^r is an additive subgroup of V = R^n, so we can form a quotient, V/L. If you think about it right, it’s fairly clear that V/L has the structure of (R/Z)^r * R^(nr). Although R/Z is not a vector space over any field – at most it’s an additive group – we can still define a length on it, inherited from R. All we have to do is look at it as the interval [0, 1] with 0 = 1.

Sixth and finally for this post, if L = Z^n has determinant m, then the total volume of V/L is equal to m. To see why, let v1, v2, …, v(n) be an integral basis for L, which will also be a basis for V over R. We can write every vector in V in exactly one way as a1v1 + a2v2 + … + a(n)v(n) where the a(i)’s are in R. Let S be the region of all vectors whose a(i)’s are all between 0 and 1, including 0 but not 1.

Since S is the image of the unit (hyper-)cube in R^n under the transformation induced by the matrix of L, the volume of S is m. In V/L, no two points of S collapse to one point. Otherwise, a1v1 + … + a(n)v(n) = b1v1 + … + b(n)v(n) in V/L, so (a1-b1)v1 + … + (a(n)-b(n))v(n) is in L; since 0 <= a(i), b(i) < 1, we have -1 < a(i)-b(i) < 1, and by the definition of an integral basis, a(i)-b(i) is an integer, so that it’s 0. Then the two points are equal, and we’re done.

Also, every point of V is equivalent in V/L to a point in S, since we can write a(i) as c(i) + b(i) where c(i) is an integer and 0 <= b(i) < 1. So we have a bijective function from S to V/L. Without getting too much into topology, the quotient function in this case preserves volumes, so V/L has the same volume as S.