Lattices and determinants

2 JAI,

Theorem 5.4 Let n > 2, and let

6.1 Lattices and determinants

6

Geometry of numbers

But of [Minkowskil it might be said as of Saul that he went out to look after his father's asses and found a kingdom.

H. Weyl [1271

168 6. Geometry of numbers

e2

- (0, 1,0,0,...,0)

e"

- (0,0,0.0,1).

The integer lattice Z" is the lattice with basis {e1... e" I. The elements of Z" are

the vectors of the form u - (uI_., where u, E Z for i - 1, ..., n.

The closure of the set X in R" will be denoted X. For X E R", let B(x, e) denote the open ball with center x and radius s > 0. A group G in R" is discrete if there exists e > 0 such that B(u, s) n G - {u} for every u E G.

Theorem 6.1 Let A C R". Then A is a lattice if and only if A is a discrete

subgroup that contains n linearly independent vectors.

Theorem 6.2 Let A be a l a t t i c e in R", and letb 1 . . .. . b,, be n linearly independent vectors contained in A. Then there exists a basis (a1, ... , an }for A such that each of the vectors bi is of the form

bi - i

vi.iai,

wherevi,i E Z f o r j - 1,...,n andi - 1,..., j.

Proof We shall prove both theorems at the same time.

Let A be a lattice in R". Then A contains n linearly independent vectors. The integer lattice Z" is discrete, since B (g, 1)rZ" - {g} for all g E Z". Let (al, ..., a,,) be a basis for the lattice A, and let T : R" -> R" be the linear transformation defined by T (ai) - e; fori - 1, ... , n. Then T is an isomorphism, and T (A) - V.

Let U - T-'(B(0, 1)). Since T is continuous, U is an open set in R", and so there exists s > 0 such that

0EB(O,s)cU.

If u B(O, e) n A, then

T(u) E B(0, 1) n Z",

which implies that T (u) - 0; hence u - 0. Therefore, B(0, e) n A - (0).

Let u,u'EA.Then u-u'EA.If u'EB(u,e),then In-u'l <a, and so u-u'EB(O,e)nA-(0)

and so u - u' - 0. Thus,

B(u, e) n A - (u)

for all u E A. This proves that the group A is discrete.

Conversely, let A (0) be a discrete subgroup of R" and let (bi, ... , b,} be a maximal set of linearly independent vectors contained in A. Then I < r < n. Fix k E (l, r}, and let Ak be the set of all vectors u E A of the form

u-xibi+-+xkbk,

where xk > 0 and 0 < x, < I for i - I , ... , k - 1. Note that bk E Ak,

and so At f 0. Let Ck be the set of all of the kth coordinates xk of vectors in Ak, and let Ck,k - inf Ck. Then there exists a sequence of vectors

u, - x1.5b1 +... +xk-Isbk-I +xk.,bk E Ak

such that lim,.,,. xk., - ck.k. Since 0 < xi., < I for i - 1, ... , k - I and

s - 1, 2, ..., there exists a subsequence {u,) g At that converges to a vector ak - CI.kb1 +... +Ck-l.kbk-I+ q.jbk E R".

Since A is a discrete subset of R", the sequence {u,, } is eventually constant, and so at E Ak and Ck.k > 0. Since the vectors b1, ..., b, are linearly independent, it follows that the vectors a1, ... , a, are also linearly independent.

We shall show that every element of A is an integral linear combination of a1... a,. Let U E A. The vectors a1, ..., a, span the vector subspace generated by A, and so there exist real numbers u 1, ... , u, such that

Suppose that uj 1 Z for some j E [1, r]. Let k be the greatest integer such that

Uk ' Z. and let Uk - gk +Xk, where gk E Z and 0 < xk < 1. The vector gkak + Uk+Iak+l + " + Urar

belongs to the group A since it is an integral linear combination of a t.. . .. a,.

Then

U

-

U - (gkak + Uk+lak+l + - - + Urar)

-

also belongs to A. Since aj - E; _I c;. b; for j = 1, ..., k, it follows that u' can be written as a linear combination of the vectors b1, ... , bk in the form

u' -ujbl+...+u'

k_Ibk_I+XkcL.kbk

For i - I... k - 1, let u; - g' + xi, where g' E Z and 0 < xi < 1. Then

g',b1 E A, and so

u"

-

u' - (gibe +... +gR-1bk-1)

-

x1b1 +... +xk-lbk-1 +XkCk kbk

E A.

Since 0 < xi < I and xkck.k > 0, it follows that u" E At, which is impossible, because the inequality

0 < XkCk.k < Ck.k

170 6. Geometry of numbers

contradicts the minimality of ck.k. Thus, every vector u E A must be an integral linear combination of the linearly independent vectors a1, ..., a,. If A contains n linearly independent vectors, then r - n and A is a lattice. This proves Theorem 6.1.

To prove Theorem 6.2, let bl, ... , bn be n linearly independent vectors in the lattice A. Since A is discrete, the preceding argument shows that there exists a basis (a,, ... , an) for A such that each vector aj is of the form

aj - Lci.jbi, i

i-1

where c1,1 E R for j - 1... n , i - 1... j, and cj.j > 0. Solving these

equations for b1... b, we obtain real numbers vi.j such that

bj -

vi,ja;

(-1

f o r j - 1 , ... , n. Since (al,.. , an } is a basis for A, it follows that vi,j E Z for

j - 1,...,n and i - 1,..., j. Also, vj_j - 1/cj,j > I for j - I,...,n. This

completes the proof.

The basis of a lattice is not uniquely determined by the lattice. For example, let A be the lattice in Z2 generated by the vectors a, - (7, 5) and a2 - (4, 3). Since a1, a2 E Z2, it follows that A c Z2. Conversely, since

e,-3a,-5a2EA

and

e2 - -4a1 + 7a2 E A

it follows that Z2 c A. Thus, Z2 - A, and the sets ((1, 0), (0, 1)} and ((7, 5), (4, 3)}

are distinct bases for Z2. Observe that a, - 7e, + 5e2 , a2 - 4e, + 3e2, and the determinant

7 4

5 3

This example can be generalized. Let U - (uij) be an n x n unimodular matrix, that is, a matrix with integer entries and determinant det(U) - ± 1. Then the inverse matrix U-1 - V - (vij) also has integer entries and det(V) - det(U) - ±1. Since UV - V U - I, where I is then x n identity matrix,

n n

I

ifi-j

E uik Vkj - E vikukj

- dij -

^{ ₀

if i f j.

k-1 k-1

Let a, .... , an be a basis for the lattice A in R". We shall use the matrix U to construct another basis for A. For j - 1, ... , n, let

a' -

^{uijai E A.}

i-I

Let A' be the group in R" generated by then vectors (a,, ..., a;, }. Since a'. E A for j - 1, ... , n, it follows that A' C A. Since V is the inverse of the matrix U,

n n n

uk;ak

- E vki

^ujkaj

k-I k-I j-I

n

(E)u

^{jk Vki}

^aj

j-1 k-I

n

E Sjiaj

j-I ai E A',

and so A C A'. Thus, A - A', and the vectors a,_., an and a' 1, ... , a',, are both bases for the lattice A.

Conversely, let (a,, ... , an } and (a, , ... , a,,) be two bases for the lattice A. For i, j - 1, ..., n, there exist integers u; j and v; j such that

/I

aj - Eu;jai

i-I and

aj n

n

E

_k-I

n n

n vectors a1.... , an are linearly independent, it follows that

E uikukj -aij

k-I f o r i, j - 1, ..., n.Similarly,

n

E vikukj - sij

k-1

for i, j - 1, ... , n. Let U and V be the matrices U - (u,1) and V - (v1). Then V - U-I and, since the matrix elements u,1 and v;1 are integers, det(U) - det(V) - f 1.

Thus, any two bases for a lattice A in R" are related by a unimodular matrix.

172 6. Geometry of numbers

Leta,,...,a,, E R", and let

it

a, = E a j e;

for j = 1, ...

, n, where aid are the coordinates of at with respect to the standard

basis vectors e,, ..., e,,. The n x n matrix A = (aid) is called the matrix of the

vectors a,, ..., a,,. The vectors a,, ..., a,, are linearly independent if and only if det(A) O.

The determinant of the lattice A, denoted det(A), plays a fundamental role in the geometry of numbers. This determinant is defined by

det(A) _ I det(A)I,

where A is the matrix of a basis Jai, ..., a,,) for A. Then det(A) ,' 0, since a basis for a lattice is a set of n linearly independent vectors in R". We shall prove that det(A) is independent of the choice of basis for the lattice A.

Let (a,, ..., a } and {a'1 , ... , a',) be two bases for the lattice A, and let U - (u,1) be the unimodular matrix such that

a' = ui i-i

Let

and

aj _

wife;

rr

aj =

aid e; ,

;-t

where a,f and a' f are the coordinates of a; and a1 with respect to the standard basis

(e,, Then det(U) - f 1. Let A = (a,3) and A' _ (a'J) be the matrices of

the bases Jai,^. a } and {a'1 , ... , a;,), respectively. We shall show that A' = A U.

Observe that

aifei = of

rr

uk f ak k-1

,r rr

ukj a;ke;

k-1 ;-1

n

(Ea'u)

^{k f e,,}

i-I k-I

and so

n

E aikukj

k-1

f o r i, j - 1, ... , n. This is equivalent to the matrix equation A' - AU.

It follows that

I det(A') I - I det(AU)I - I det(A) I I det(U) I - I det(A)I. (6.1) This proves that det(A) is well defined.

, ... ,

The fundamental parallelepiped of the lattice A with respect to the basis (a an) is the set

F(A) -

F(A; al, ... , an)