Download The Geometry of Numbers and Minkowski`s Theorem

Number Theory (part
n):
The Geometry of Numbers
(by Evan Dummit, 2014, v. 1.00)
Contents
1.1
1.1
•
The Geometry of Numbers and Lagrange's Four-Squares Theorem
. . . . . . . . . . . . . . . . . . .
1
The Geometry of Numbers and Lagrange's Four-Squares Theorem
In this section, we discuss several results from the geometry of numbers. The underlying idea is (obviously
enough) to use geometric ideas to study number theory.
◦
We have already encountered some of these ideas when discussing the Gaussian integers: they form a
lattice inside the complex plane.
•
First, we recall some terminology:
◦
A set
in
◦
◦
•
B
in
Rn
is convex if, for any
B
in
Rn
is symmetric about the origin if, for any
x
and
y
B,
in
x
all points on the line segment joining
and
y
are also
B.
A set
We denote the set of all points in
R
n
x
B,
in
the point
−x
all of whose coordinates are integers by
is also in
n
Z
B.
.
Our starting point is the following theorem, which says that if a convex set is suciently nice and large
enough, it must contain an integer point.
•
Theorem (Minkowski's Convex Body Theorem): Let
the origin and whose volume is
◦
n
>2
. Then
B
B
be a convex open set in
that is symmetric about
contains a nonzero point all of whose coordinates are integers.
We rst show the following (sometimes called Blichfeldt's principle): if
x and y
volume is greater than 1, then there exist two points
∗
∗
Rn
in
S
S
is a bounded set in
such that
x−y
Rn
whose
has integer coordinates.
Proof: The idea is essentially to use the pigeonhole principle.
a = (a1 , · · · , an ), let R(a) be the box consisting of the points (x1 , · · · , xn )
ai ≤ xi < ai+1 .
X
S(a) = S ∩ R(a), we have
vol(S(a)) = vol(S), because each point of S lies in
For each lattice point
whose coordinates satisfy
∗
If we then set
a∈Zn
exactly one of the boxes
∗
Now imagine translating the set
land inside
∗
R(a).
Then
X
S(0).
vol(S
∗
S(a)
−a:
S ∗ (a).
by the vector
Denote this translated set by
it will preserve volume, but move
S(a)
to
(a)) = vol(S) > 1.
a∈Zn
∗
Now notice that each of the sets
S ∗ (a)
lies inside
S(0),
which has volume 1, so there must be some
overlap.
∗
Hence, there exists some distinct
x − y = a1 − a2
◦
Proof (of Minkowski's Theorem): Suppose
>2
◦
x, y ∈ S
n
, and let
1
B=
2
and
a1 , a2 ∈ Zn
such that
x − a1 = y − a2 :
but then
is a nonzero lattice point.
1
x : x∈B
2
Notice that since vol(B)
> 2n ,
B
is a convex open set symmetric about 0 whose volume is
.
we have
1
vol( B) > 1.
2
Apply Blichfeldt's principle to
1
distinct points x, y ∈
B such that x − y has integer coordinates.
2
∗ Then 2x ∈ B and 2y ∈ B .
∗ Since B is symmetric about the origin, −2y ∈ B .
1
1
B:
2
we obtain
∗
∗
Since
B
is convex, the midpoint of the line segment joining
But this point is simply
x − y,
2x
and
−2y
lies in
B.
which is a nonzero point all of whose coordinates are integers, as
desired.
•
The result of Minkowski's theorem does not apply merely to the lattice of points having integer coordinates.
◦
If
v1 , ... , vn are (R-)linearly independent vectors in Rn , the set Λ of vectors of the form c1 v1 + · · · + cn vn ,
ci ∈ Z, is called a lattice.
where each
◦
A fundamental domain for this lattice can be obtained by drawing all of the vectors
v1 ,
... ,
vn
outward
from the origin, and then lling them in to create a skew box.
◦
A basic fact from linear algebra says: the volume of the fundamental domain is equal to the determinant
of the matrix whose columns are the vectors
•
v1 ,
... ,
vn
Theorem (Minkowski's Theorem for general lattices): Let
has volume
V.
If
B
(expressed in terms of the standard basis of
Λ
be any lattice in
is any open convex centrally-symmetric region in
contains a nonzero point of
R
n
Rn
Rn ).
whose fundamental domain
whose volume is
> 2n · V ,
then
B
Λ.
◦
Proof: Apply a linear transformation
◦
Linear transformations preserve open sets, convex sets, and central symmetry, so the image of
T
sending the basis vectors of
Λ
to the standard basis of
Rn .
B
under
this map is still open, convex, and centrally symmetric.
◦
The volume of
T (B)
is equal to
1/V
open convex centrally-symmetric set
◦
has volume
Applying the previous version of Minkowski's theorem to
all of whose coordinates are integers: then
•
B
> 2n .
times the volume of
T (B)
B
(by standard linear algebra), so this new
T (B) yields that T (B) contains a nonzero point
Λ.
contains a nonzero point of
There are many applications of Minkowski's theorem in number theory.
◦
One very important one, which we do not quite possess the tools to discuss at this stage, is to provide
an eective way to determine whether the ring
◦
√
Z[ D]
possesses unique factorization.
Instead, we will discuss a simpler application: proving that every positive integer can be written as the
sum of four squares.
•
Theorem (Lagrange): Every positive integer
◦
We rst show that if
n
can be expressed as the sum of four squares of integers.
a, b are the sum of four squares,
then so is
ab.
We then show every prime is the sum
of four squares. By applying these results, we immediately see that every positive integer is the sum of
four squares.
◦
Lemma 1: If
∗
a
and
b
are the sum of four squares, then so is
ab.
Proof: This follows from the following magical-seeming identity:
(x21 + x22 + x23 + x24 )(y12 + y22 + y32 + y4 )2
=
(x1 y1 + x2 y2 + x3 y3 + x4 y4 )2 + (x1 y2 − x2 y1 + x3 y4 − x4 y3 )2
+(x1 y3 − x2 y4 − x3 y1 + x4 y2 )2 + (x1 y4 + x2 y3 − x3 y2 − x4 y1 )2
which can be veried (if not understood) simply by multiplying out and verifying that all of the
cross-terms cancel.
∗
Like the corresponding identity for sums of two squares
(a2 + b2 )(c2 + d2 ) = (ac + bd)2 + (ad − bc)2
which arises from the fact that the norm map on
Z[i] is multiplicative, the four-squares identity does
H of quaternions, which
not come from nowhere: it in fact arises from a norm map on the ring
is a noncommutative ring.
(The letter
H
is used because the quaternions were rst described by
Hamilton.)
∗
Explicitly,
H
a + bi + cj + dk , where a, b, c, d are real numbers,
i2 = j 2 = k 2 = ijk = −1. (From these relations one can deduce
ij = −ji = k , jk = −kj = i, and ki = −ik = j .)
is the set of elements of the form
subject to the multiplication rules
explicitly that
2
∗
The conjugation map is
a + bi + cj + dk = a − bi − cj − dk , and the norm map is N (q) = qq :
N (a + bi + cj + dk) = a2 + b2 + c2 + d2 , and the fact that this map
explicitly, one can compute that
is multiplicative amounts to the four-squares identity.
∗
In fact, since the norm of a nonzero quaternion is nonzero, we in fact see that every nonzero quaternion has a multiplicative inverse.
∗
Multiplication in this noncommutative manner using the letters
i, j ,
k might be familiar from
i = h1, 0, 0i, j = h0, 1, 0i,
i × j = k.
and
the algebra of the cross product of vectors in 3-space: often the notation
k = h0, 0, 1i
∗
is used for the basis vectors, and then for example one has
Due to their connection with geometry in 3 dimensions, the quaternions are often used in computer
graphics, applied physics, and engineering, since they cam be used to represent spatial rotations in
3-dimensional space far more eciently than matrices.
◦
Lemma 2: Every prime
∗
can be written as the sum of four squares.
Proof: As shown on a homework assignment,
p:
say,
·
−1 ≡ r2 + s2
(mod
The argument was: if
T
does the set
∗
p
−1
is the sum of two squares modulo
p
for any prime
p).
p
S of squares modulo p contains (p + 1)/2 elements,
−1 − s2 , so they must have a nontrivial intersection.
is odd, the set
of elements of the form
as
Λ be the lattice spanned by the four vectors hp, 0, 0, 0i, h0, p, 0, 0i, hr, s, 1, 0i, and hs, −r, 0, 1i.
p2 , so the volume of
2
the fundamental domain is p .
∗ Let B be the convex, centrally-symmetric open set in R4 dened by x21 + x22 + x23 + x24 < 2p. The
Now let
It is a simple computation to see that the determinant of these four vectors is
volume of this ball can be computed (either with cleverness or a brute-force quadruple integration)
to be
∗
2π 2 p2 .
Since the volume of
2 2
2
2π p > 16p
B
is larger than
24
times the volume of the fundamental domain of
Λ
(since
), we conclude that there is a nonzero element
hx1 , x2 , x3 , x4 i = a hp, 0, 0, 0i + b h0, p, 0, 0i + c hr, s, 1, 0i + d hs, −r, 0, 1i
of
∗
Λ
in
B.
But then
x21 + x22 + x23 + x24
hx1 , x2 , x3 , x4 i
+ x23 + x24 = p.
and since
x21
∗
+
x22
Thus,
p
=
(ap + cr + ds)2 + (bp + cs − dr)2 + c2 + d2
≡
(c2 + d2 )(1 + r2 + s2 ) (mod p)
≡
0 (mod p)
is nonzero and has
x21 + x22 + x23 + x24 < 2p,
the only possibility is that
is the sum of four squares, and we are done.
Well, you're at the end of my handout. Hope it was helpful.
Copyright notice: This material is copyright Evan Dummit, 2014. You may not reproduce or distribute this material
without my express permission.
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