Theorem (Lagrange)

Every positive integer is the sum of four squares.

To establish this theorem, we shall require 3 lemmata.

Lemma 1

Let be an odd positive integer, then there exist s.t.

.

Proof
We split into 3 cases.
(i) , an odd prime.

Let

Clearly and the elements of (resp. ) are pairwise incogruent modulo . To see this,
assume satisfy . Clearly . Assume
. Then . By Lagrange’s theorem,

has only 2 solutions, and these are and . If , then , which is a contradiction to how was defined.

So , and hence . Similarly for .

So by the pigeonhole priniciple, the 2 sets can’t be distinct, and it follows that there exist integers such that .

(ii) , odd prime.

We proceed by induction. We have the case from before. For some , assume there exist integers such that

.

Then there exists some integer such that

.

Clearly cant divide both and . Assume . Then we have , and so there exists some integer such that . This can be found by solving the equation.

Let .

Then we have

Hence this case follows by induction.

(iii) is an odd positive integer

The case is trivial. So let

where the are odd primes, and are integers.

Then for each , we have integers such that

We can then use the Chinese remainder theorem to find integers such that

for each . It follows that

Lemma 2
If every odd positive integer is the sum of four squares, then every positive integer is the sum of four squares.

Proof
If some integer is the sum of four squares, say

Then we have

We can continue like this to show that is the sum of four squares for any .
As any positive integer is of the form , for some , and some odd integer , the lemma follows.

Lemma 3
Let be the ball of radius in . Then

Proof
It is not hard to see that the volume of this ball is equal to

Solve this by using the substitution .

We are now ready to prove the main theorem
Proof
By lemma 2, it suffices to prove for odd positive integers. Let be an odd positive integer. By lemma 1, there exist integers such that

Let be the lattice with basis vectors

We have , and can be written as

with .
So

for all .

Let be the ball of radius in . It is clear this is a symmetric convex body, amd has volume

.

Hence we can apply Minkowski’s first theorem, and we have contains some non-trivial lattice point

for integers .
Now we have

and we have

since So it follows that and therefore

This proves the theorem.

Let be prime.  If then is the sum of two squares.

Proof  We’ll show that if and there is such that then there exist such that .  In particular if then -1 is a quadratic residue mod p so p is the sum of two squares.

Let

We can assume that .  Since there is a such that .  For a basis of we take

That these form a basis is left as an exercise.  The determinant of the lattice is then:

Now let

This is convex and symmetric and

So by Minkowski I, there exists .  Since we know

and

so

Thence

and so

The Local-Global Principle for Diagonal Ternary Quadratic Forms

Let and set

If for all p then .

Proof
Wlog we may assume , , square-free, and whenever .

Plan:

• Define a lattice with and such that for all .
• Let
• D is convex and symmetric and (exercise).
• By Minkowski I there is , same reasoning as last theorem gives .

So all we need to do is construct .  Write , where the are distinct and or . There are three parts.

(i) Let p be one of the , so .  Wlog say .  So if

then

By hypothesis there exists a p-adic solution to our equation, say such that

We may assume that and that $p$ doesn’t divide all of them.  Since

we must have , otherwise p will divide all three.  Impose the condition on ,

If satisfies this condition then

(ii) Suppose , so are all odd.  We know there exist such that

We also know .  Exactly one of must be even, wlog say .  Then

Impose the conditions

Then

(iii) Suppose , so one of is even, say .  We know there exist such that

In particular

so that .  Thus either are both odd or both even.  If they’re both even then

is divisible by 4 so will be even, a contradiction.  So are both odd, hence

since for any odd c.  Impose the conditions

Then

These conditions give us for any as we wanted.  Now we just have to check the determinant of .

We know that a lattice is a subgroup of , and in fact the determinant of the lattice is equal to the index of this subgroup:

We’ve defined our lattice in terms of a bunch of congruence conditions.  It is a relatively straightforward lemma in group theory that if one has a subgroup I defined by congruences

then the index of I is at most .  In our case that means

As required.

Minkowski I’. Let K be a non-empty symmetric convex body in , be a full rank lattice in and define

Then

It can easily be shown that this corollary is actually equivalent to Minkowski’s first theorem; its advantage is that it is more amenable to generalisation.

Proof that Minkowski I’ implies Minkowski I. Let K be a symmetric convex volume whose volume exceeds We want to show that K contains a point from By I’ and definition of we must have Since is the infimum of the set

there must be some satisfying

Since it follows that K contains a non-zero point in

For a subset S of we denote its linear span by

Definition. Let K and be as above. For we set

is called the k-th successive minima of K with respect to

Notice that has dimension at least k if and only if contains k linearly independent lattice points of Clearly and this new definition of agrees with that given previously.

Minkowski’s Second Theorem. Let K and be as above. Then

(1)

This inequality is best possible.

Example. Consider (so ) and the rectangle centred at the origin where Then and . We have , so .

Proof of Minkowski II. [WORK IN PROGRESS!] By definition of the successive minima, for each there exists with , and linearly independent. We claim that form a basis for Suppose otherwise, then it can easily be checked that there must exist with where each and at least one is non-zero (take a point of our lattice which is not an integer combination of ; this point must be a linear combination of the ; keep subtracting/adding the until this linear combination has the required form).

Let denote the invertible linear map which takes to the standard basis vector . Then is a lattice in of full rank and contains Also is a symmetric convex body with successive minima (with respect to the lattice Moreover, for each we have

Some notation:

• 
• For let which is an n-dim box.

Let V be a vector space over or . A subset is

• convex if for any two points a, b in K the line segment between a and b is contained in K.
• symmetric if implies .

We will be working in Here we can define a body in to be a bounded open set. Clearly a body is Lebesgue measurable (it is open) and this measure, or ‘volume’, is finite (because the body is bounded). Minkowski’s Theorems concern symmetric convex bodies. Notice that a non-empty symmetric convex body contains some point and so also contains the convex combination

As K is also open, we have for some (a fact we will use later).

Blichfeldt’s Lemma

Lemma (Blichfeldt). Let be a full rank lattice in and let be a body (not necessarily symmetric or convex) with Then there exists a shift of K, v+K, containing at least two points from

Define Then the usual conclusion of Blichfeldt (easily seen to be equivalent) is

There exists with

Equivalentlty

There exists b in K with

Proof of Blichfeldt: In lecture 1 we saw that every lattice is the image of under some map. In particular, is countable. We can therefore enumerate without repetitions:

Let denote the fundamental parallelepiped of with respect to some fixed basis. For each i set Last week (lecture 2) Lee showed (a disjoint union). It follows that , another disjoint union. Define Then and each is measurable. Moreover, by translation invariance of Lebesgue measure, I claim that the are not disjoint. Suppose otherwise. Lee showed last week that By elementary properties of measures we therefore have

a contradiction. Thus for some It follows that there exists and with But then which completes the proof.

Although the proof doesn’t look very intuitive, the idea is simple: look at the body K ‘mod ‘; the resulting set is a subset of the fundamental parallelepiped and so must have volume smaller than ; but then not all residue classes ‘mod ‘ can contain 1 or less elements of K, because the volume of K exceeds

Minkowski I

We’ve now all the tools in place to prove Minkowski’s first theorem.

Theorem (Minkowski I). Let be a full rank lattice in and let be a symmetric convex body of volume greater than . Then K contains a non-zero element of

Proof. A symmetric convex body has a very convenient property. First, shrink it by a factor of 1/2 along each axis to obtain the dilation

Then any shift of by an element of is also a subset of K. Re-phrasing this in a more convenient form, it can easily be checked (using convexity and symmetry) that for any we have

Hence, to prove Minkowski I, it suffices to show there exists with But this is the conclusion of Blichfeldt’s lemma applied to the body It therfore suffices to show Since , the proof is complete.

Minkowski’s first theorem yields the following corollary, which is a baby version of Minkowski’s second theorem.

Corollary. Let be a full rank lattice in and K a non-empty symmetric convex body. Define the set

[Notice that M(K) is non-empty, because for some ]

If then

(1)

Proof. Suppose (1) doesn’t hold. Then by Minkowski I, and so contains some non-zero element Since K is open, so is the dilation of K by a non-zero factor It follows that we can find some with Therefore But then is an element of M(K) smaller than the infimum of M(K), a contradiction. This completes the proof.

]]> http://numbertheoryreadinggroup.wordpress.com/2008/04/30/lecture-3-convex-bodies-blichfeldt-and-minkowski-i-sean/feed/ 0 Sean http://numbertheoryreadinggroup.wordpress.com/2008/04/24/geometry-of-numbers-lecture-2-determinant-of-the-lattice-and-the-fundamental-parallelepiped-lee/ http://numbertheoryreadinggroup.wordpress.com/2008/04/24/geometry-of-numbers-lecture-2-determinant-of-the-lattice-and-the-fundamental-parallelepiped-lee/#comments Thu, 24 Apr 2008 13:39:33 +0000 Lee http://numbertheoryreadinggroup.wordpress.com/?p=7 ]]> Sean showed us last time that lattices are additive subgroups of and that any lattice is of the form

for linearly independent vectors in . The number is called the rank of . If then we say that has full rank. The vectors are called a basis for .

Example Take and . These are linearly independent in so form a lattice of full rank in .

This lattice looks like (and is) . We can also think of as vectors in :

They’re still linearly independent so form the basis of a lattice, but this lattice won’t have full rank.

In the above example we noticed that the lattice with basis looked just like . And clearly any vector can be written as

so . And in fact infinitely many pairs of basis vectors give the same lattice, so it would be helpful if these lattices had some kind of invariant which didn’t depend on the choice of basis, and indeed they do.

Given a set of vectors and a basis in we can write the vectors in a matrix by making the columns the vectors with respect to the basis. For example the vectors and using the standard basis in can be put in the matrix

Changing the order of our vectors or the order of our basis will change the matrix, but not the matrix’s determinant. Moreover this determinant is always nonzero provided our vectors are linearly independent. So perhaps the determinant of this matrix is the invariant we’re looking for, but first we need to show that different bases for our lattice give the same determinant.

Let’s let and be two bases for the same lattice . Since they proffer the same lattice any vector from one of the sets can be written in terms of the other set, that is to say for each there exist integers and such that

and

Then we can write the vectors in terms of themselves as follows

The vectors are linearly independent so the sum inside the brackets must be zero whenever , and one when . Similarly, using the other set of vectors, we have

If we let and be the matrices and respectively then the above tells us that . And since they have integer entries the fact that tells us that . So these two bases are related by a unimodular matrix (a matrix with integer entries and determinant ). Specifically we get the matrix for one basis by right-multiplying the matrix of the other basis by a certain unimodular matrix.

Conversely if we have a basis for a lattice and take a unimodular matrix then the lattice with basis where

is again . To see this let be the lattice with basis and note that each , so . Let , then this is also a unimodular matrix and we have

So that , so the two lattices are in fact the same.

So we’ve shown that two bases give the same lattice if and only if their matrices are related by a unimodular matrix. In the simple example where we had as a basis for we can now note that

Recall we proposed that the determinant of the matrix given by the basis vectors might be an invariant of the lattice. Well the above shows this is the case. Given two bases of a lattice we’ve seen their matrices and are related by a unimodular matrix by . And so

So the determinant – up to sign – does not depend on the basis chosen. We call the absolute value of this the determinant of the lattice, denoted .

Another very important feature of a lattice is its fundamental parallelepiped. This depends on the basis chosen, and given a basis it is the set:

In this is the parallelogram cut out by the points and , and in higher dimensions we get generalisations of this, hence the name. It clearly depends on the basis chosen.

The parallelepiped itself may not be unique, but its volume is. If then we have

And so the volume is independent of the basis used.

The fundamental parallelepiped is extra useful as every point in can be written uniquely as the sum of a point on the lattice and a point in the fundamental parallelepiped. That is,

This is intuitively clear and is an extension of the idea of writing any real number as its integer part plus its fractional part, except now we have the lattice replacing integers and the parallelepiped replacing the fractional part. The proof uses this basic principle and simply applies it to dimensions.

Let be an abelian group whose operation we will denote additively. Given a d-tuple of elements in and a d-tuple of integers we define their dot product by the usual formula

The map is then a homomorphism and its image is precisely the subgroup of generated by .

Lattices.

We will study a special type of abelian group, namely the lattices in . To define these groups we need the notion of an isolated point in a topological space: given a subset of a topological space , we say is isolated in if there exists a neighbourhood of with . is said to be discrete if all its points are isolated.

Definition. A lattice in is any additive subgroup of which is discrete. The rank of is the dimension of the linear subspace of generated by . Clearly . If then we say has full rank. If are linearly independent and generate as a group, then we call a basis for .

The integer lattice is our archetypal example of a lattice (clearly this is a discrete additive subgroup of ). The definition excludes subgroups such as . Generalising , a wealth of examples of lattices of rank k can be obtained by choosing a set of k linearly independent vectors , then forming the dot product

Proposition 1. is a lattice in .

Proof. Clearly is an additive subgroup of . It remains to show is discrete. Nathanson does this by noting that by the linear independence of , the function

is a well-defined linear map from the subspace of generated by to . It can be shown that this map is continuous and maps to . Because of this continuity and the discreteness of , must also be discrete. I find the continuity of intuitive, but not obvious (it is an obvious consequence of the continuity of linear maps between finite dimensional normed spaces, but this fact doesn’t seem concrete enough in this setting for my taste). I will therefore present an argument which shows a little more directly why is discrete. We’ll start by showing that there exists a constant such that for any

where the above absolute values represent Euclidean norms in and , respectively. Let

, (2)

and for each i let By the linear independence of the , there exists (at least one of which is non-zero) such that if and satisfy (2), then

Applying Cauchy-Schwarz we have

Taking we obtain (1). Since two distinct points of are at a distance of at least 1, it follows from (1) that two distinct point of are at a distance of at least . Hence is discrete. This completes the proof.

In fact lattices of the form are the only type of lattices, which follows from the next theorem whose proof can be found in Nathanson (Theorem 6.1):

Theorem 1. is a lattice in of rank k if and only if there exists a set of k linearly independent vectors with