lectures on quadratic forms

8/3/2019 Lectures on Quadratic Forms

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Lectures on Quadratic Forms

By

C.L. Siegel

Tata Institute of Fundamental Research, Bombay

1957

(Reissued 1967)


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Lectures on Quadratic Fomrs

By

C.L. Siegel

Notes by

K. G. Ramanathan

No part of this book may be reproduced in any

form by print, microffilm of any other means with-

out written permission from the Tata Institute ofFundamental Research, Colaba, Bombay 5

Tata Institute of Fundamental Research, Bombay

1955 56

(Reissumed 1967)


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Contents

1 Vector groups and linear inequalities 1

1 Vector groups . . . . . . . . . . . . . . . . . . . . . . . 12 Lattices . . . . . . . . . . . . . . . . . . . . . . . . . . 7

3 Characters . . . . . . . . . . . . . . . . . . . . . . . . . 10

4 Diophantine approximations . . . . . . . . . . . . . . . 13

2 Reduction of positive quadratic forms 21

1 Quadratic forms . . . . . . . . . . . . . . . . . . . . . . 21

2 Minima of definite forms . . . . . . . . . . . . . . . . . 27

3 Half reduced positive forms . . . . . . . . . . . . . . . . 33

4 Two auxiliary regions . . . . . . . . . . . . . . . . . . . 40

5 Space of reduced matrices . . . . . . . . . . . . . . . . 47

6 Binary forms . . . . . . . . . . . . . . . . . . . . . . . 567 Reduction of lattices . . . . . . . . . . . . . . . . . . . 61

3 Indefinite quadratic forms 67

1 Discontinuous groups . . . . . . . . . . . . . . . . . . . 67

2 The H - space of a symmetric matrix . . . . . . . . . . . 72

3 Geometry of the H-space . . . . . . . . . . . . . . . . . 78

4 Reduction of indefinite quadratic forms . . . . . . . . . 80

5 Binary forms . . . . . . . . . . . . . . . . . . . . . . . 85

4 Analytic theory of Indefinite quadratic forms 99

1 The theta series . . . . . . . . . . . . . . . . . . . . . . 992 Proof of a lemma . . . . . . . . . . . . . . . . . . . . . 103

iii


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iv Contents

3 Transformation formulae . . . . . . . . . . . . . . . . . 105

4 Convergence of an integral . . . . . . . . . . . . . . . . 1155 A theorem in integral calculus . . . . . . . . . . . . . . 127

6 Measure of unit group and measure of representation . . 133

7 Integration of the theta series . . . . . . . . . . . . . . . 142

8 Eisenstein series . . . . . . . . . . . . . . . . . . . . . . 148

9 Main Theorem . . . . . . . . . . . . . . . . . . . . . . 157

10 Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . 161


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Chapter 1

Vector groups and linear

inequalities

1 Vector groups1

Let K be the field of real numbers and V a vector space of dimension n

over K. Let us denote element ofV by small Greek letters and elements

of K by small Latin letters. The identity element ofV will be denoted

by 0 and will be called the zero element ofV. We shall also denote by 0

the zero element in K.

Let 1, . . . , n be a base ofV so that for any V =

i

iixi, xi K.

We call x1, . . . , xn the coordinates of . Suppose 1

, . . . , n is anotherbasis ofV, then

i =

j

jaji, i = 1, . . . , n

where aji K and the matrix M = (aji) is non-singular. If in terms of

1, . . . , n

= i

iyi, yi K

1


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2 1. Vector groups and linear inequalities

then it is easy to see that x1

:

:

xn

= My1

:

:

yn

(1)Suppose 1, . . . , m is any finite set of elements ofV. We denote by

L(1, . . . , m) the linear subspace generated in V by 1, . . . , m. This2

means that L(1, . . . , m) is the set of elements of the form

1x1 + + mxm, xi X.

It is clear that L(1, . . . , m) has dimension

Min(n, m).Let Rn denote the Euclidean space of n dimensions, so that every

point P in Rn has coordinates x1, . . . , xn, xi K. Let 1, . . . , n be abasis of V and let x1, . . . , xn be the coordinates of in V with regard

to this basis. Make correspond to , the point in Rn with coordinates

x1, . . . , xn. It is then easily seen that this correspondence is (1, 1). For

any V define the absolute value || by

||2 =n

i=1

x2i

where x1, . . . , xn are coordinates of . Then

| |satisfies the axioms of a

distance function in a metric space. We introduce a topology in V by

prescribing a fundamental system of neighbourhoods of to be the set

of{S d} where S d is the set of in V with

| | < d (2)

S d is called a sphere of radius dand center . The topology above makes

V a locally compact abelian group. The closure S d of S d is a compact

set. From (1), it follows that the topologies defined by different bases of

V are equivalent.

A subgroup G of V is called a vector group. The closure G ofG in3

V is again a vector group. We say that G is discrete if G has no limitpoints in V. Clearly therefore a discrete vector group is closed.


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1. Vector groups 3

Suppose G is discrete, then there is a neighbourhood of zero which

has no elements ofG not equal to zero in it. For, if in every neighbour-hood of zero there exists an element ofG, then zero is a limit point ofG

in V. This contradicts discreteness ofG. Since G is a group, it follows

that all elements ofG are isolated in V. As a consequence we see that

every compact subset ofV has only finitely many elements ofG in it.

We now investigate the structure of discrete vector groups. We shall

omit the completely trivial case when the vector group G consists only

of the zero element.

Let G {0} be a discrete vector group. Let 0 be an element ofG. Consider the intersection

G1 = G

L().

Let d > 0 be a large real number and consider all the y > 0 for

which y is in G1 and y d. If d is large, then this set is not empty.

Because G is discrete, it follows that there are only finitely many y with

this property. Let q > 0 be therefore the smallest real number such that

1 = q G1. Let = x be any element in G1. Put x = hq + kwhereh is an integer and 0 k < q. Then x and 1h are in G1 and so kis in 4

G1. But from definition ofq, it follows that k = 0 or

= 1h, h integer.

This proves that

G1 = {1},the infinite cyclic group generated by 1.

If in G there are no elements other than those in G1, then G = G1.

Otherwise let us assume as induction hypothesis that in G we have found

m( n) elements 1, . . . , m which are linearly independent over K and

such that G L(1, . . . , m) consists precisely of elements of the form1g1 + +mgm where g1, . . . , gm are integers. This means that

Gm = G L(1, . . . , m) = {1} + + {m}is the direct sum ofm infinite cyclic groups. If in G there exist no other

elements than in Gm then G = Gm. Otherwise let

G, Gm. Put

Gm+1 = G L(1, . . . , m, ).


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Consider the elements in Gm+1 G of the form = 1x1 + +mxm +y, xi K.

where y 0 and y dwith d a large positive real number. This set C of

elements is not empty since it contains . Put now xi = gi + ki where

gi is an integer and 0 ki < 1, i = 1, . . . , m. Let = 1g1 + +mgm,then Gm and so

= 1k1 + +mkm +yis an element of Gm+1. Thus for every Gm+1 there exists a =5 G with the property

= 1k1 + +mkm +y0 ki < 1, y d. Thus all those s lie in a closed sphere of radius

(m + d2)12 . Since G is discrete, this point set has to be finite. Thus for

the s in G the y can take only finitely many values.

Therefore let q > 0 be the smallest value of y for which m+1 =

1t1 + +mtm +q is in G. Let = 1x1 + +mxm +y

be in Gm+1. Put y = qh + kwhere h is an integer and 0 k < q. Then

m

+1

h = 1

(x1

t1

h) +

+m

(xm

tm

h) +k

is in Gm+1. By definition of q, k = 0. But in that case by induction

hypothesis xi tih = hi is an integer. Thus = 1h1 + +mhm +m+1h

h1, . . . , h are integers. This proves that

Gm+1 = {1} + + {m+1}is a direct sum ofm + 1 infinite cyclic groups.

We can continue this process now but not indefinitely since 1, . . . ,

m+1, . . . are linearly independent. Thus after r n steps, the processends. We have hence the


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1. Vector groups 5

Theorem 1. Every discrete vector group G {0} in V is a direct sum of6r infinite cyclic groups, 0 < r n.

Conversely the direct sum of cyclic infinite groups is a discrete vec-

tor group. We have thus obtained the structure of all discrete vector

groups.

We shall now study the structure of all closed vector groups.

Let G be a closed vector group. Let S d be a sphere of radius d

with the zero element ofG as centre. Let r(d) be the maximum number

of elements of G which are linearly independent and which lie in S d.

Clearly r(d) satisfies

0 r(d) n.

Also r(d) is an increasing function ofd and since it is integral valued ittends to a limit when d 0. So let

r = limd0

r(d).

This means that there exists a d0 > 0 such that for d d0

r = r(d).

We call r the rankofG.

Clearly 0 r n. Suppose r = 0, then we maintain that G is

discrete; for if not, there exists a sequence 1, . . . , n, . . . of elements of

G with a limit point in V. Then the differences {k 1}, k 1 willform a set of elements of G with zero as a limit point and so in every

neighbourhood of zero there will be elements ofG which will mean that

r > 0.

Conversely if G is discrete there exists a sphere S d, d > 0 which 7

does not contain any point of G not equal to zero and containing zero.

This means r = 0. Hence

r = 0 G is discrete.

Let therefore r > 0 so that G is not discrete. Let d be a real number

0 < d < d0 so that r(d)=

r. Let S d be a sphere around the zero elementofG and of radius d. Let 1, . . . , r be elements ofG in S d which are


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linearly independent. Let t > 0 be any real number and let d1 > 0

be chosen so that d1 < Min(d,t

n). Then r(d1) = r. If 1, . . . , r be

elements ofG which are linearly independent and which are contained

in the sphere S d1 around the zero element of G, then L(1, . . . , r) L(1, . . . , r) since S d1 S d. But since both have dimension r,

L(1, . . . , r) = L(1, . . . , r).

Since 1, . . . , r are in S d1 we have

|i| d1 t

n, i = 1, . . . , r.

Let

L(1

, . . . , r). Then by above

= 1x1 + +rxr.

Put xi = gi + ki where gi is an integer and 0 ki < 1. Put = 1g1 +

+rgr. Since 1, . . . , r G, will also be in G. Now

|| = |1k1 + +rkr| |1k1| + + |rkr| r+ s, rand s being integers determined by theorem 2. If G thenfor V

() = x1(1) + + xn(n).If however G then () is integral. Therefore

(i) =

0 i rinteger r < i r+ sarbitrary real i > r+ s.

Thus for G

() =

r+si=r+1

(i) xi

If G, then because of definition of 1, . . . , n it follows thateither at least one of xr+1, . . . , xr+s is not an integer or at least one of


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xr+s+1, . . . , xn is not zero. Suppose that = i 1xi, xr+1 0(mod 1).Define the linear function on V by 15(i) =

1 ifi = r+ 10 ifi r+ 1.Then is a character ofG and

() = (r+1)xr+1 = xr+1 0(mod 1)

The same thing is true if xr+i 0(mod 1), 1 i s. Suppose now that =

iixi and one of xr+s+1, . . . , xn say xn 0. Define linear on V

by

(i) =

0 ifi n

1

2xnifi = n.

Then is a character of G and () =1

2 0(mod 1). Hence if

G there is a character ofG which is not integral for . We have thus

proved.

Theorem 4. Let V. Then G if and only if for every character of G, () is integral.

Let us fix a basis 1, . . . , n of V so that 1, . . . , r+s is a basis of

G. If G then (i) = ci where

ci =

0 i rinteger r < i r+ sreal i > r+ a

If (c1, . . . , cn) is any set of n real numbers satisfying the above condi-

tions, then the linear function defined on V by

() =

ni=1

cixi


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4. Diophantine approximations 13

where = i ixi, is a character of G. If Rn denotes the space of real16 n-tuples (x1, . . . , xn) then the mapping (c1, . . . , cn)

is seen to be an isomorphism of G into Rn. Thus G is a closed vectorgroup of rankn r s.

It can be proved easily that G the character group ofG is isomor-phic to G.

4 Diophantine approximations

We shall study an application of the considerations in 3 to a problemin linear inequalities.

Let

Li(h) =

mj=1

ai jhj, (i = 1, . . . , n)

be n linear forms in m variables h1, . . . , hm with real coefficient ai j. Let

b1, . . . , bn be n arbitrarily given real numbers. We consider the problem

of ascertaining necessary and sufficient conditions on the ai js so that

given a > 0 there exist integers h1, . . . , hm such that

|Li(h)

bi

|< , (i = 1, . . . , n).

In order to study this problem, let us introduce the vector space V of

all a rowed real columns

=

a1...

an

, ai K.V has then dimension n over K. Let 1, . . . , n be elements ofV defined 17

by

i=

a1i...

ani , i = 1, . . . , m


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and let G be the vector group consisting of all sumsm

i=1 igi where gisare integers. Let be the vector =

b1...

bn

Then our problem on linear forms is seen to be equivalent to that of

obtaining necessary and sufficient conditions that there be elements in

G as close to as one wishes; in other words the condition that be in

G. Theorem 4 now gives the answer, namely that

() 0(mod 1)for every character ofG.

Let us choose a basis 1, . . . , n ofV where

i =

0...

1

0...

0

i = 1, . . . , n

with zero everywhere except at the i th place. Now in terms of this basis

k = 1a1k + + nank, k = 1, . . . , mTherefore if is a character ofG18

(k) =

ni=1

aikci

where (i) = ci, i = 1, . . . , n. Also (k) 0(mod 1). Furthermore ifc1, . . . , cn be any real numbers satisfying

ni=1

ciaik 0(mod 1), k = 1, . . . , m,


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then the linear function defined on V by (i) = ci is a character ofG.

By theorem 4 thereforen

i=1

cibi 0(mod 1)

We have therefore the theorem due to Kronecker.

Theorem 5. A necessary and sufficient condition that for every t > 0,

there exist integers h1, . . . , hm satisfying

|Li(h) bi| < t, i = 1, . . . , n,is that for every set c1, . . . , cn of real numbers satisfying

ni=1

ciaik 0(mod 1), k = 1, . . . , m,

we should haven

i=1

aibi 0(mod 1).

We now consider the special case m > n. Let m = n + q, q 1. Letthe linear forms be

q

j=1 ai jhj + gi, i = 1, . . . , nin the m variables h1, . . . , hq, g1, . . . , gn. Then the vectors 1, . . . , m 19

above are such that

q+i = i, i = 1, . . . , n.

This means that if is a character ofG, ci = (i) is an integer. Thus

Corollary 1. The necessary and sufficient condition that for every t > 0,

there exist integers h1, . . . , hq, g1, . . . , gn satisfying

q

j=1ai jhj + gi bi < t, i = 1, . . . , n


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is that for every set c1, . . . , cn of integers satisfyingi

ciai j 0(mod 1), j = 1, . . . , q

we have i

cibi 0(mod 1).

We now consider another special case q = 1. The linear forms are

of the type

aih + gi bi i = 1, . . . , n

a1, . . . , an, b1, . . . , bn being real numbers. Suppose now we insist thatthe condition on b1, . . . , bn be true whatever b1, . . . , bn are. This will

mean that from above Corollary c1 = c2 = . . . = cn = 0 or, in other

words, that a1, . . . , an have to satisfy the condition thati

ciai 0(mod 1), ci integral

if and only if ci = 0, i = 1, . . . , n. This is equivalent to saying that20

the real numbers 1, a1, . . . , a1 are linearly independent over the field of

rational numbers.

Let us denote by Rn the Euclidean space ofn dimensions and by Fn

the unit cube consisting of points (x1, . . . , xn) with

0 xi < 1 i = 1, . . . , n.

For any real number x, let ((x)) denote the fractional part ofx, i.o. ((x)) =

x [x]. Then

Corollary 2. If1, a1, . . . , an are real numbers linearly independent over

the field of rational numbers, then the points (x1, . . . , xn) where

xi = ((hai)) i = 1, . . . , n

are dense in the unit cube, if h runs through all integers.


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We consider now the homogeneous problem namely of obtaining

integral solutions of the inequalities

|Li(h)| < t, i = 1, . . . , n

t > 0 being arbitrary. Here we have to insist that h1, . . . , hm should not

all be zero.

We study only the case m > n. As before introduce the vector space

V of n-tuples. Let 1, . . . , m and G have the same meaning as before.

If the group G is not discrete, it will mean that the inequalities will have

solutions for any t, however small. If however G is discrete then since 21

m > n the elements 1, . . . , m have to be linearly integrally dependent.

Hence we have integers h1

, . . . , hm

not all zero such that

1h1 + + mhm = 0.

We have hence the

Theorem 6. If m > n, the linear inequalities

|Li(h)| < t, i = 1, . . . , n

have for every t > 0, a non-trivial integral solution.


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Bibliography

[1] O. Perron : Irrationalzahlen Chelsea, 1948.

[2] C. L. Siegel : Lectures on Geometry of Numbers, New York Uni-

versity, New York, 1946.

19


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Chapter 2

Reduction of positive

quadratic forms

1 Quadratic forms22

Let V be a vector space of dimension n over the field K of real numbers.

Define an inner product between vectors , ofV by

i) Kii) =

iii) ( + ) = +

iv) (a) = ()a, a K.Obviously if1, . . . , n is a base ofV and , have the expression =

i

iai, =

iibi then

=

ni,j=1

aibj(i j).

If we denote by S the n-rowed real matrix S = (si j), si j = ij then S is

symmetric and = aS b (1)

21


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22 2. Reduction of positive quadratic forms

where a = a1...an , b =

b1...bn and a denotes the transpose of the column

vector a. (1) is a bilinear form in the 2n quantities a1, . . . , an, b1, . . . , bn.

In particular

2 = aS a

is a quadratic form in a1, . . . , an.

Suppose that 1

, . . . , n is another base ofV. Then23

i =

j

jai j i = 1, . . . , n.

and the matrix A = (ai j) is non-singular. IfS 1 = (i

j) then one sees

easily thatS 1 = S [A] = A

S A.

Thus ifS with regard to one base is non-singular, then the S correspond-

ing to any other base is also non-singular.

Conversely let S by and real n-rowed symmetric matrix and 1, . . . ,

n a base ofV over K. Put

i j = si j (j, i = 1, . . . , n)

and extend it by linearity to any two vectors of V. Then we have in V an

inner product defined.

If = i

ixi is a generic vector ofV over K,

2 = xS x = S [x] =

i,j

xixj si j.

The expression on the right is a quadratic form in the n variables x1, . . . ,

xn and we call S its matrix. The quadratic form is degenerate or non-

degenerate according as its matrix S is or is not singular.

Let xS x =n

k,l=1

sklxkxl be a quadratic form in the n variables x1, . . . ,

xn and let s1 = s11 0. We may write

xS x = s1x21 + 2s12x1x2 + + 2s1nx1xn + Q(x2, . . . , xn)


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1. Quadratic forms 23

so that Q(x2, . . . , xn) is a quadratic form in the n1 variables x2, . . . , xn.24We now write, since s1 0,

xS x = s1

x1 +

s12

s1x2 + +

s1n

s1xn

2 s

212

s1x22 . . .

s1n2

s1x2n + Q(x2, . . . , xn).

We have thus finally

xS x = s1y21 +R(x2, . . . , xn)

where y1 = x1 +s12

s1x2 + +

s1n

s1xn and R(x2, . . . , xn) is a quadratic form

in the n 1 variables x2, . . . , xn. If we make a change of variablesy1 = x1

s12

s1x2 + +

s1n

s1xn

y1 = xi i > 1

(2)then we may write

xS x =

s1 0

0 S 1

y1

...

yn

where S 1 is the matrix of the quadratic form R(x2, . . . , xn). Using matrix

notation we have

S =

s1 qq S 2

= s1 00 S 1

1 s1

1q

0 E

(3)

where E is the unit matrix of order n 1, q is a column ofn 1 rows and 25

S 1 = S 2 s11 qq;which, incidentally gives an expression for the matrix ofR.

More generally suppose S =

S 1 QQ S 2

where S 1 is a k-rowed matrix

and is non-singular. Put x =y

z

where y is a column ofkrows and z has

n

krows. Then

S [x] = S 1[y] +yQz +zQy + S 2[z],


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which can be written in the form

S [x] = S 1[y + S11 Qz] + W[z] (4)

where W = S 2 QS 1Q. In matrix notation we have

S =

S 1 0

0 W

E S 1

1Q

0 E

(5)

the orders of the two unit matrices being evident. In particular, we have

|S | = |S 1| |W|.

Let S be a real, non-singular, n-rowed, symmetric matrix. It is wellknown that there exists an orthogonal matrix V such that

S [V] = VS V = D

where D = [d1, . . . , dn] is a real diagonal matrix. The elements d1, . . . ,

dn ofD are called the eigen-values of S . Let L denote the unit sphere26

L : x x = 1

so that a generic point x on L is an n-tuple x =

x1...

xn

of real numbers.

Let m and N denote the smallest and largest of the eigen values of S .Then for any x on L.

m S [x] M

For, if we put

y1...yn

= y V1x, then yy = 1 andS [x] = D[V1x] = D[y] = d1y21 + + dny2n.

But then

S [x] = (d1 M)y21 + + (dn M)y2n + M M.

The other inequality is obtained by changing S to S .


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1. Quadratic forms 25

More generally we have, for any arbitrary real vector x

mx x S [x] M xx. (6)

If x = 0, the statement is obvious. Let x 0. Then t2 = xx 0. Puty = t1x. Then yy = 1 and so m S [y] M. Multiplying throughoutby t2 we get the result in (6).

We now define a quadratic form xS x to be positive definite (or sim-ply positive) ifS [x] > 0 for all vectors x 0. It is positive semi-definite

if S [x] 0 for real x 0. We shall denote these by S > 0 and S 0respectively. IfS > 0, then obviously |S | 0. For, if |S | = 0, then there 27exists x 0 such that S x = 0. Then

0 = xS x > 0

which is absurd.

IfS > 0 and |A| 0 and A is a real matrix, then T = S [A] is againpositive. For, if x = 0, the Ax y 0 and so

T[x] = S [Ax] = S [y] > 0.

We now prove two lemmas for later use.

Lemma 1. A matrix S is positive definite if and only if |S r| > 0 forr = 1, . . . , n, where S r is the matrix formed by the first r rows and

columns of S .

Proof. We shall use induction on n. Ifn = 1, the lemma is trivial. Let

therefore lemma be proved for matrices of order n 1 instead ofn. Let

S =

S n1 qq a

IfS > 0 then S n1 > 0 and so |S n1 | 0. We can therefore write

S =

S n1 0

0 l

E S 1

n1q0 l

(7)

so that |S | = |S n1|l. Induction hypothesis shows that |S n1| > 0 andl > 0 so that |S | > 0 and |S r| > 0 for all r.


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The converse also follows since by hypothesis |S | > 0 and |S n1| > 280. So 1 > 0. But by induction hypothesis S n1 > 0.Lemma 2. If S > 0 and S = (skl), then

|S | s1 . . . snwhere skk = sk, k = 1, . . . , n.

Proof. We again use induction on n. From the equation (7) we have

|S | = |S n1| l.

But l = sn qS 1n1q > 0 since S 1n1 > 0 and sn > 0. If we assumelemma proved for n

1 instead ofn we get

|S | s1 . . . sn1l s1 . . . sn.

More generally we can prove that ifS > 0 and S =

S 1 S 12S

12S 2

then

|S | |S 1| |S 2| (8)

It is easy to see that equality holds in (8) if and only ifS 12 = 0.

Let S > 0, then s1, . . . , sn are all positive. We can write as in (3)

S =

s1 0

0 W

1 s1

1q

0 E

But since now W > 0, its first diagonal element is different from zero29

and we can write W also in the form (3). In this way we get

S =

d1 0

. . .

0 dn

1, d12, . . . , d1n0, 1, d23, . . . , d2n: . . . . . .

0 . . . . . . 1

= D[V] (9)where D = [d1, . . . , dn] is a diagonal matrix and V = (dkl) is a triangle

matrix with dkk = 1, k = 1, . . . , n, and dkl = 0, k > 1. We can therefore

write

S [x] =

nk=1

dk(xk + dk k+1xk+1 + + dknxn)2


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2. Minima of definite forms 27

The expression S = D[V] is unique. For ifS = D1[V1] where D1 is

a diagonal matrix and V1 is triangular, then

D[W] = D1

where W = VV11

is also a triangular matrix. In this case, it readily

follows that W = E and D = D1.

In general we have the fact that if

S =

S 0

0 S 2

E T

0 E

(10)

where S 1 has order k then S 1, S 2 and T are unique.

We call the decomposition (9) of S the Jacobi transformation of

S .

2 Minima of definite forms30

Let S and T be two real, non-singular n-rowed symmetric matrices.

They are said to be equivalent (denoted S T) if there exists a uni-modular matrix U such that

S [U] = T.

Since the unimodular matrices form a group, the above relation is an

equivalence relation. We can therefore put the n-rowed real symmetric

matrices into classes of equivalent matrices. Evidently, two matrices in

a class have the same determinant.

If S = S is real and t is a real number, we say that S represents tintegrally, if there is an integral vector x such that

S [x] = t.

In case t = 0, we insist that x 0. The representation is said to be

primitive, if x is a primitive vector. Obviously if S T then S and Tboth represent the same set of real numbers.

If S > 0, then all the eigen values of S are positive. Let m > 0 bethe smallest eigen value of S . Let t > 0 be a large real number. Then if


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S [x] < t then mxx < tand so the elements of x are bounded. Therefore

there exist only finitely many integral vectors x satisfying

S [x] < t.

This means that if x runs through all non-zero integral vectors, S [x] has

a minimum. We denote this minimum by (S ). There is therefore an31

integral x such that

S [x] = (S )., x 0.

Moreover x is a primitive vector. For if x is not primitive, then x = qy

where q > 1 is an integer, and y is a primitive vector. Then

(S ) = S [x] = q2S [y] > S [y]

which is impossible. Furthermore ifS T then (S ) = (T). For, letS = T[U] where U is unimodular. If x is a primitive vector such that

(S ) = S [x], then

(S ) = S [x] = T[U x] (T).

Also if(T) = T[y], then

(T) = T[y] = S [U1

y] (S ).This proves the contention.

IfS > 0 and t is a real number, then (tS ) = t(S ). But |tS | = tn|S |so that it seems reasonable to compare (S ) with |S |1/n.

We not prove the following important theorem due to Hermite.

Theorem 1. If(S ) is the minimum of the positive matrix S of n rows,

there exist a constant cn depending only on n, such that

(S ) cn|S |1/n

Proof. We use induction on n.


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If n = 1, then S is a positive real number s. If x 0, and integral,32

then sx2

> s unless x = 1 so thatc1 = 1.

Let us assume theorem proved for n 1 instead ofn. Let x be the primi-tive integral vector such that (S ) = S [x]. Complete x into a unimodular

matrix U. Then T = S [U] has first diagonal element equal to (S ). Also

(S ) = (T) by our remarks above. Furthermore |S | = |T|. Thereforein order to prove the theorem we may assume that the first diagonal

element s1 ofS is equal to (S ).

Let S = s1 q

q S 1 . ThenS =

s1 0

0 W

1 s1

1q

0 E

where W = S 1 qs11 q! Also |S | = s1|W|.

Let x =

x1y

be a vector and let y have n 1 rows, so that

S [x] = s1(x1 + s11 q

y)2 + W[y]. (11)

Since W > 0, we can choose an integral y so that W[y] is minimum. x1can now be chosen integral in such a manner that

12

x1 + s11 qy 1

2(12)

using (11) and (12) and induction hypothesis we get 33

(S ) S [x] (S )4

+ cn1W1/(n1)

Substituting for |W| we get

(S )

4

3cn1

n1

n

|S | 1n

which proves the theorem.


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Using c1 = 1 and computing successively from the recurrence for-

mula cn =4

3cn1 n1n we see that

cn = (4/3)n1

2 (13)

is a possible value ofcn. This estimate is due to Hermite.

The best possible value for cn is unknown except in a few cases. We

shall show that c2 =

4

3and that it is the best possible for n = 2. From

Hermites estimate (13), we see that for a positive binary matrix S ,

(S ) 43 12 |S | 12 .

Consider now the positive quadratic from x2 + xy +y2 whose matrix

S =

1 1

212

1

For integral x, y not both zero, x2 + xy +y2 1 so that (S ) = 1. Also|S | = 3

4. We have

1 = 43 12 |S | 12

which proves that

4

3is the best possible value of c2.34

We shall now obtain a finer estimate for cn due to Minkowski. This

estimate is better than Hermites for large values of n. To this end we

make the following consideration.

Let Rn denote the Euclidean space of n dimensions regarded as a

vector space of ordered n-tuples (x1, . . . , xn). A point set L in Rn is

said to be convex if whenever A and B are two points of it,A + B

2, the

mid point of the line joining A and B, is also a point ofL. It is saidto be symmetric about the origin if whenever x belongs to it, x also


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belongs to it. Obviously ifL is both convex and symmetric, it contains

the origin.IfL is a point set in Rn and h is any point in Rn we denote by Lh

the set of points x such that x Lh if and only if x h is a point ofL.With this notation L = L0.

IfL is an open, bounded symmetric convex set, the L has a mea-

sure (L) in the Jordon sense and for h Rn

(L) = (Lh).

We call a point P = (x1, . . . , xn) in Rn a lattice pointif x1, . . . , xn are

all integers. The lattice points form a lattice in Rn considered as a vector

group. We shall denote points of this lattice by the letters g, g, . . ..The following lemma, due to Minkowski, shows the relationship be-tween convex sets and lattices.

Lemma 3. IfL is an open, bounded, symmetric and convex set of vol- 35

ume > 2n, thenL contains a lattice point other than the origin.

Proof. We shall assume that L has no lattice point in it other than the

origin and then prove that (L) 2n.

So let L have no lattice point in it other than the origin. Define the

point set M by x M if and only if 2x L. Then M is an open,symmetric, bounded and convex set. Also

(L) = 2n(M). (14)

Consider now the translatesMg ofM by the lattice points. Ifg g

thenMg andMg are disjoint sets. For, if x Mg Mg then x g andx g are points ofM. Since M is symmetric and convex.

g g2

=(x g) + (g x)

2

is a point ofM. By definition ofM, g g is a point ofL. But g g.Thus L has a lattice point other than the origin. This contradicts ourassumption. Thus the Mg for all g are distinct.


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Let denote the unit cube, that is the set of points x = (x1, . . . , xn)

with 0 xi < 1, i = 1, . . . , n. By the property ofMgs aboveg

(Mg ) = (

g

Mg) () = 1 (15)

But (Mg ) = (M g) so that by (15)

1

g

( Mg) =

g

(g M) = (

g

g M).

But the g cover Rn completely without gaps or overlapping when g36runs over all lattice points. Hence

(M)

1.

Using (14) our theorem follows.

We can now prove the following theorem due to Minkowski.

Theorem 2. If S > 0 and(S ) is its minimum, then

(S ) 4

n

2+ 1

2/n|S |1/n

Proof. In Rn let us consider the point set L defined by the set of x = x1...

xn

with

S [x] <

It is trivially seen to be open and symmetric. Also since S > 0, L is

bounded. To see that it is convex, write S = AA and put Ax1 = y1,Ax2 = y2

. Then a simple calculation proves that

2

y1 +y2

2

y1

+y2

2

y

1y

1+y

2y

2.

This shows that L is a convex set. The volume ofL is

(L) =n/2n/2

( n2

+ 1)

S 1/2

If we put =

(S ), then L contains no lattice point other than the37origin. Minkowskis lemma then proves theorem 2.


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3. Half reduced positive forms 33

Denote the constants in Hermites and Minkowskis theorems by cn

and cn respectively. If we use stirlings formula for the -function in theform

log (x) x log x.

We get log n = log4

+

2

nlog

n

2+ 1

log n whereas log cn =

n 12

log 4/3 n where is an absolute constant. This shows thatfor large n, Minkowskis estimate is better than Hermites.

3 Half reduced positive forms

We now consider the space Rh, h = n(n + 1)2

of real symmetric n-rowed

matrices and impose on it the topology of the h-dimensional real Eu-

clidean space. Let P denote the subspace of positive matrices. If

S P then all the principal minors of S have positive determinant.This shows that P is the intersection of a finite number of open subsets

ofRh and hence is open.

Let S be a matrix in the frontier ofP in Rh. Let S 1, S 2, . . . be a

sequence of matrices in P converging to S . Let x 0 be any real

column vector. Then S k[x] > 0 and hence by continuity S [x] 0. Fromthe arbitrariness of x, it follows that S 0. On the other hand let S beany positive semi-definite matrix in Rh. Let E denote the unit matrix oforder n. Then for > 0, S + E is a positive matrix, which shows that

in every neighbourhood ofS there are points ofP. This proves that the 38

frontier ofP in Rh consists precisely of positive semi-definite matrices.

Let denote the group of unimodular matrices. We represent in

Rh as a group of transformations S S [U], S Rh. Also U and Uload to the same representation in Rh. It is easy to see that the only

elements in which keep every element of Rh fixed are E. Thus if weidentify in , the matrices U and U then S S [U] gives a faithfulrepresentation of 0 in Rh, 0 = / E. If U runs over all elementsof and S Rh, S [U] runs through all matrices in the class of S . Weshall now find in each class of positive matrices, a matrix having certainnice properties.


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Let T P and let u run over the first columns of all the matrices in. There u are precisely all the primitive vectors. Consider the valuesT[u] as u runs over these first columns. Then T[u] has a minimum,

which is none other than (T). Let this be attained for u = u1. It is

obvious that u1 is not unique for, u1, also satisfies this condition. Inany case, since T > 0, there are only finitely many us with the property

T[u] = T[u1]. Let u1 be fixed and let u run over the second columns of

all unimodular matrices whose first column is u1. The us now are notall

the primitive vectors (for instance u u1). T[u] again has a minimum

say for u = u2 and by our remark above39

T[u1] T[u2]

Also there are only finitely many u with T[u] = T[u2]. Consider now all

unimodular matrices whose first two columns are u1, u2 and determine

a u3 such that T[u3] is minimum. Continuing in this way one finally

obtains a unimodular matrix

U = (u1, . . . , un)

and a positive matrix S = T[U].

S T and by our construction, it is obvious, that S is not unique inthe class ofT. We shall study the matrices S and U more closely.

Suppose we have constructed the columns u1, . . . , uk

1. In order to

construct the k-th column we consider all unimodular matrices V whosefirst k 1 columns are u1, . . . , uk1 in that order. Using the matrix Uabove which has this property,

U1V =Ek1 A

0 B

(16)

where Ek1 is the unit matrix of order k 1 and A and B are integralmatrices. Since U and V are unimodular, B is unimodular. Ifw =

w1...

wn

denotes the first column of the matrix

AB then, since B is unimodular

(wk, wk+1, . . . , wn) = 1. (17)


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The k-th column yk

ofV is Uw. Conversely let w be any integral column40

satisfying (17). Then wk, . . . , wn can be made the first column of a uni-modular matrix B of order n k+ 1. Choosing any integral matrix A ofk 1 rows and n k+ 1 columns, whose first column is w1, . . . , wk1, weget a matrix V whose first k 1 columns are u1, . . . , uk1 (by means ofthe equation (16)). Thus the k-th column of all the unimodular matrices

with first k 1 columns equal to u1, . . . , uk1 is of the form Uw, wherew is an arbitrary integral vector with (wk, . . . , wn) = 1.

Consider the matrix S = T[U]. By the choice ofuk, we have if w

satisfies (17), then

S [w] = T[Uw] T[uk] = sk

where S = (skl). We have thus proved that in each class of T there existsa matrix S satisfying

I) s1 > 0

II) S [w] sk, k = 1, . . . , n

for every integral column w =

w1...

wn

with (wk, . . . , wn) = 1.

Matrices which satisfy (I) and (II) shall be called half reduced and

the subset ofP of matrices S , half reduced, shall be denoted R0.

In the sequel we shall denote by e1, . . . , en, the n columns in order

of the unit matrix of order n and by an admissible k-vector w we shallunderstand an integral vector w of n rows, satisfying (17). ek is clearly 41

an admissible k-vector.

Since ek+1 is an admissible k+ 1-vector, we have

sk+1 = S [ek+1] skwhich shows that

s1 s2 . . . sn. (18)

Let u =

x1...

xn

be an integral vector with xk = 1, xl = 1, xi = 0 for

i k, i l and k < l. Then u is an admissible l-vector and so

sk + 2skl + sl = S [u] sl.


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This means that 2skl sk. Changing the sign of xk we get 2skl sk.Hence sk 2skl sk, 1 k < l n (19)Remark. Suppose S is a real symmetric matrix satisfying (II). Let S 1be the matrix obtained from S by deleting the h1-th, h2-th,...,hl-th rows

and columns from S . Then S 1 also has properties similar to S since we

have only to consider such admissible vectors w for which the h1, . . . , hl-

th elements are zero.

We now prove the

Theorem 3. Let S be a real, symmetric n-rowed matrix with the prop-

erty (II). Then S 0. If, in addition, it satisfies (I), then S > 0.42Proof. Suppose s1 = 0. Then by (19) we have

0 = s1 2s1l s1 = 0

which shows that S has the form

S =

0 0

0 S 1

If s2 = 0, we again have a similar decomposition for S 1, since S 1, by

our remark above, also satisfies II. Thus either S = 0 or else there is a

first diagonal element sk, such that sk 0. Then

S =

0 0

0 S k

S k having sk for its first diagonal element. We shall now show that S k >

0. Observe that S k satisfies both I) and II) and therefore for proving the

theorem it is enough to show that if S satisfies I and II, then S > 0.

Ifn = 1, the theorem is trivially true. Let therefore theorem proved

for n 1 instead ofn. Put

S = S 1 qq sn


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where q is a column of n 1 rows. S 1 satisfies I and II and so by43induction hypothesis S 1 > 0. Also since sn s1, therefore sn > 0.

Let x =yz

be a column ofn rows, y having n 1 rows and let z be

a real number. Then

S [x] = S 1[y + S11 qz] + (sn qS 11 q)z2.

We assert that sn qS 11 q > 0. For let sn qS 11 q. Then for > 0 andevery x 0

S [x] S 1[y + S 11 qz] + z2 (20)Consider the quadratic form on the right side of the inequality above. It

is of order n, positive and has a determinant |S 1|. Therefore we mayfind a column vector x =

yz

, integral, such that the value of the right

side is a minimum and so by Hermites theorem

S 1[y + S11 qz] + z

2 cn|S 1|1/n1/n.

Using (20) and observing that s1 is the minimum ofS [x] we get, for this

x,

0 < s1 S [x] cn|S 1|1/n1/n (21)Since can be chosen arbitrarily small we get a contradiction from (21).

Thus sn

qS 1

1q > 0. This means the S > 0.

We have thus shown that all matrices satisfying (I) and (II) are inP.

We prove now the following important theorem due to Minkowski. 44

Theorem 4. If S is a positive half-reduced matrix, then

1 s1 . . . sn|S | bn

where bn is a constant depending only on n.

Proof. The left hand side inequality has already been proved in lemma

2 even for all matrices in P. In order to prove the right hand sideinequality we use induction.


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Consider now the ratios

sn

sn1,

sn1sn2

, . . . ,s2

s1.

Since S is half-reduced, all these ratios are 1. Let = n(n 1)4

. For

the above ratios, therefore, one of two possibilities can happen. Either

there exists a k, 2 k n such thatsn

sn1< ,

sn1sn2

< , . . . ,sk+1

sk<

sk

sk

1

(22)

or thatsn

sn1, . . . ,

s2

s1< (23)

Note that in the case n = 2, the second possibility cannot occur since

then =1

2and

s2

s1 1.

Consider (23) first. We have45

s1 . . . sn

sn1

< n(n 1)

2

and sinces1 . . . sn

|S | =s1 . . . sn

sn1

sn

1

|S |we get, using Hermites inequality

s1 . . . sn

|S | < cnn

n(n 1)2

which proves theorem.

Suppose now that (22) is true and so k 2. Write

S = S k1 QQ1 R


43/170


where S k1 has k 1 rows. Let x = y

z where y is a column with k 1rows. We have, by completion of squaresS [x] = S k1[y + S 1k1Qz] + (R QS 1k1 Q)[z] (24)

Also |R QS 1k1Q| = |S |/|S k1|. Choose z to be an integral primitive

vector such that (R QS 1k1 Q)[z] is minimum. By Hermites theorem

therefore

(R QS 1k1 Q)[z] cnk+1(|S |/|S k1|)1/nk+1 (25)

Put y + S 1k1

Qz = w so that w = w1..

.wk1 . Choose now y to be an integralvector such that

12

wi 1

2, i = 1, . . . , k 1. (26)

By the choice of z, it follows that x =y

z

is an admissible k-vector. 46

Hence

sk S [x]. (27)Also since S k1 is half-reduced, we get

S k1[w] =k

1p,q=1

spqwpwq k(k 1)8

sk1.

Using (22) we get

S k1[w] sk

2(28)

From (24), (25), (27) and (28) we get

sk 2cnk+1(|S |/|S k1|)1/(nk+1) (29)

Since

s1 . . . sn|S | = s1 . . . sk1|S k1 | |Sk1 ||S | snk+1k sk . . . snsnk+1

k


44/170


we get by induction hypothesis on S k1, that

s1 . . . sn

|S | bk1 (2cnk+1 )nk+1 (n k)(n k+ 1)

2

which proves the theorem completely.

The best possible value ofbn is again unknown except in a few sim-

ple cases. We shall prove that

b2 = 4/3 (30)

and it is the best possible value.

Let ax2 + 2bxy + cy2 be a half-reduced positive form. Then 2b a c. The determinant of the form is d = ac b2. Thus47

ac = ac b2 + b2 d+ a2

4 d+ ac

4

which gives

ac 43

d (31)

Consider the binary quadratic form x2 + xy + y2. It is half-reduced

because if x and y are two integers not both zero, then x2 + xy +y2 1.The determinant of the form is 3/4. Product of diagonal elements is

unity. Hence

1 =4

3d

and this shows that 4/3 is the best possible value.

4 Two auxiliary regions

Let R0 denote the space of half-reduced matrices. Define the point set

Rt for t > bn 1 as the set ofS satisfying0 < sk < tsk+1 k = 1, . . . , n 1

t < sklsk

< t 1 k < 1 n

s1 . . . sn|S | < t

(32)


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4. Two auxiliary regions 41

Because of (18), (19) and theorem 4, it follows that

R0 Rt . (33)

But what is more important is that

limtR

t = P (34)

This is easy to see. For, if S P, let t be chosen larger than the 48maximum of the finite number of ratios

sk

sk+1, k = 1, . . . , n 1; skl

sk, 1

k < 1 n, s1 . . . sn|S | and bn. Then S Rt for this value oft.

Let S Rt and consider the Jacobi transformation of S ; namely

S =

d1 0

. . .

0 dn

1, t12, . . . t1n

0 . . . . . . 1

= D[T] (35)Then

skl = dktkl +

k1h=1

dhthkthl, 1 k l n.

In particular, putting k = 1, and using the fact that d1, . . . , dn are all

positive, we getsk

dk 1. (36)

Also since |S | = d1 . . . dn, we haven

k=1

sk

dk=

s1 . . . sn

|S | < t. Since t > 1,we have

sk

dk< t (k = 1, . . . , n).

Using (32) we get

dk

dk+1=

dk

sk sk

sk+1 sk+1

dk+1< t2. (37)

Now s1l = d1 t1l so that

|t1l| = |s1l|d1

= |s1l|s1

s1d1

< t2


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Let us assume that we have proved that 49

abs tgl < u0, 1 g k 1, g < l n (38)for a constant u0 depending on tand n. Then

abs tkl abs skldk

+

k1h=1

dh

dkabs thk abs thl < u1,

because of (37) and (38), u1 depending only on t and n. It therefore

follows that if u is the maximum of u0, u1, t2, then for the elements of

D and T in (35) we have

0 < dk < udk+1, k = 1, . . . , n 1

abs tkl < u, k < l. (39)We now define Ru to be the set of points S P such that if S =

D[T] where D = [d1, . . . , dn] is a diagonal matrix and T = (tkl) is a

triangle matrix then D and T satisfy (39) for some u. Since the Jacobi

transformation is unique, this point set is well defined.

From what we have seen above, it follows that given Rt , there existsa u = u(t, n) such that

Rt Ru

Conversely one sees easily that given Ru there exists a t = t(u, n) suchthat

Ru Rt .In virtue of (34), it follows that

limuR

u = P. (40)

50

We now prove two lemmas useful later.

Let S P and let t be a real number such that S Rt . Let S 0denote the matrix

S 0 =

s1 0. . .

0 sn

(41)

We prove


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Lemma 4. There exists a constant c = c(t, n) such that whatever be the

vector x,1

cS 0[x] S [x] cS 0[x].

Proof. Let P1 denote the diagonal matrix P1 = [

s1, . . . ,

sn]. Put

W = S [P]. In order to prove the lemma, it is enough to show that if

xx = 1 then1

c W[w] c.

Let W = (wkl). Then wkl = skl/ sksl. Because S Rt we have

abs wkl = absskl

sk

sk

s1< t c1, k l (42)

where c1 depends only on t and n. W being symmetric, it follows that

the elements ofW are in absolute value less than a constant c2 = c2(t, n).

Consider now the characteristic polynomial f() = |E W|. By(42) all the coefficients of the polynomial f() are bounded in absolute

value by a constant c3 = c3(t, n). Also since W > 0, the eigen values 51

ofW are bounded by c4 = c4(t, n). Let 1, . . . , n be these eigen values.

Then1 . . . n = |W| =

|S |s1 . . . sn

> t1

which means that there exists a constant c5 = c5(t, n) such that

i > c5(t, n), i = 1, . . . , n.

(6) then gives the result of lemma 4.

Next we prove

Lemma 5. If S Rt and S = S 1 S 12

S 12

S 2 , then S 1

1S 12 has all its ele-

ments bounded in absolute value by a constant depending only on t andn.


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Proof. By the Jacobi transformation we have S = D[T]. Since Rt Ru for u = u(t, n), the elements of T are u in absolute value. Write

T =

T1 T120 T2

, D =

D1 0

0 D2

where T1 and D1 have the same number of rows and columns as S 1. We

have S 1 = D1[T1] and S 12 = T1D1T12 so that

S 11 S 12 = T11 T12.

Since T1 is a triangle matrix, so is T11

and its elements are u1 inabsolute value, u1 = u1(t, n). The elements ofT12 are already u. Ourlemma is proved.

We are now ready to prove the following important52

Theorem 5. Let S and T be two matrices in Rt . Let G be an integralmatrix such that1) S [G] = T and2) abs |G| < t. Then the elements of Gare less, in absolute value, then a constant c depending only on t and n.

Proof. The constants c1, c2, . . . occurring in the following proof depend

only on t and n. Also bounded shall mean bounded in absolute value

by such constants.

Let G = (gkl) and let g1

, . . . , gn

denote the n columns ofG. We then

have

S [gl] = tl l = 1, . . . , n.

Introducing the positive diagonal matrix of lemma 4, we obtain

S 0[gl] c1S [gl] = c1tl.

But S 0[gl] =k

skg2kl

so that

skg2kl c1t1 k, l = 1, . . . , n (43)

Consider now the matrix G. Since |G| 0, there exists in its expan-sion a non-zero term. That means there is a permutation l1, . . . , ln of 1,

2, 3, . . . , n such thatg1l1 g2l2 . . . gnln 0.


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From (43) therefore we get

sk skg2klk c1tlk k = 1, . . . , n.

Consider now the integers k, k+ 1, . . . and lk, lk+1, . . . , ln. All of the 53

latter cannot be > k. So there is an i k such that li k. Hence

si c1tli .

So, since S and T are in R0

,

sk c2tk, k = 1, . . . , n. (44)

On the other handn

k=1

tk

sk=

t1 . . . tn

|T| |S |

s1 . . . sn|G|2

and all the factors on the right are bounded. Therefore

nk=1

tk

sk< c3.

Using (44), it follows that

tk c4 sk, (k = 1, 2, . . . , n). (45)Combining (43) and (45) we have the inequality

skg2kl < c5 sl k, l = 1, . . . , n. (46)

Let p now be defined to be the largest integer such that

sl c5 sl, k p, l p 1. (47)

If p = 1, this condition does not exist. From the definition of p, it

follows that for every integer g with p + 1 g n, there exists a kg gand an lg < g such that

skg < c5 slg (48)


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This holds for p = 1, but if p = n, it does not exist. 54

Let c6 be a constant such that

sk < c6 sl k l. (49)

This exists since S Rt . Using (48) and (49) and putting c7 = c5c26 wehave

sg < c7 sg1 g p + 1 (50)(49) and (50) give the important inequality

1

c8 1.


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5. Space of reduced matrices 47

In order to prove the theorem we use induction. Ifn = 1, the theorem

is trivially true. Assume theorem therefore proved for n 1 instead ofn. Split S and T in the form

S =

S 1 S 12S

12S 2

T =

T1 T12

T12

T2

where S 1 and T1 are p 1 rowed square matrices. Because S [G] = T,we get

S 1[G1] = T1

G1S 1G12 + G1S 12G2 = T12

(55)

By considerations above G21 = 0 therefore

|G

|=

|G1

| |G2

|. Since G

is integral it follows that abs |G1| < t. Also S 1 and T1 are p 1 rowedsquare matrices which are in R

t,p1, where Rt,p1 is the same as R

t

with p 1 instead ofn. Ey induction hypothesis and (55) we see that G1is bounded.

Using the fact that G1

S 1 = T1G11

we get

G12 = G1T11 T12 S 11 S 12G2.

Using lemma 5, it follows that the elements ofG12 are bounded.

Our theorem is completely proved.

In particular,

Corollary. If S and T are inRt and S [U] = T for a unimodular U, then 56U belongs to a finite set of unimodular matrices determined completely

by t and n.

5 Space of reduced matrices

We have seen that given any matrix T > 0, there exists in the class ofT,

a half-reduced matrix S . Consider now the 2n unimodular matrices of

the form

A=

a1 0. .

.0 an


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where ai = 1. If S is half-reduced, then S [A] also is half-reduced.For, if x = x1...

xn

is an admissible k-vector, then Ax = x1... xn is also anadmissible k-vector. Also, the diagonal elements of S and S [A] are the

same. We shall choose A properly so that S [A] satisfies some further

conditions.

Since S [A] = S [A], there is no loss in generality if we assumea1 = 1. Denote by 1, . . . , n the n columns of the matrix A. Consider

now 1

S 2. This equals a2 s12. If s12 0 choose a2 so that

a2 s12 0.

If s12=

0, a2 may be chosen arbitrarily. Having chosen a1, . . . , ak con-sider k

S k+1

= akak+1 skk+1. Since ak has been chosen, we choose

ak+1 = 1 by the condition

akak+1 skk+1 0,

provided skk+1 0. If skk+1 = 0, ak+1 may be arbitrarily chosen. We57

have thus shown that in each class of equivalent matrices, there is a

matrix S satisfying

) s1 > 0

) skk+1 0, k = 1, . . . , n 1.) S [x] sk 0, k = 1, . . . , n for every admissible k-vector.

We shall call a matrix satisfying the above conditions a reduced ma-

trix, reduced in the sense ofMinkowski. Let Rdenote the set of reduced

matrices, then

R R0. (56)Since the elements of S P are coordinates of the point S , the

conditions ) and ) above show that R is defined by the intersection

of an infinity of closed half spaces ofP. We shall denote the linear

functions in ) and ) by Lr, r=

1, 2, 3, . . .. It is to be noted that weexclude the case when an Lr is identically zero. This happens when in


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), x is the admissible k-vector equal to ek. We may therefore say thatR is defined by

) s1 > 0, ) Lr 0 r = 1, 2, 3, . . . (57)We shall see presently that the infinite system of linear inequalities can

be replaced by a finite number of them.

In order to study some properties of the reduced space R, we first 58

make some definitions.

Definition. i) S is said to be an inner point ofR if s1 > 0 and

Lr(S ) > 0 for all r.

ii) It is said to be a boundary point ofR if s1 > 0 Lr(S )

0 for all r

and Lr(S ) = 0 at least for one r.

iii) It is said to be an outer point ofR if s1 > 0 and Lr(S ) < 0 at least

for one r.

We first show that R has inner points.

Consider the quadratic form

S |x| = x21 + + x2n + (p1x1 + + pnxn)2

where p1, . . . , pn are n real numbers satisfying

0 < p1 < p2 . . . < pn < 1.

The matrix S = (skl) is then given by

sk = 1 + p2k, k = 1, . . . , n

skl = pkpl, k l.

We assert that S is an inner point ofR. In the first place

s1 > 0, skk+1 = pkpk+1 > 0; k = 1, . . . , n 1.Next let x be an admissible k-vector not equal to ek. Then at least oneof xk, . . . , xn has to be different from zero. If at least two of then, say xj,

x1 are different from zero, so that k

1 < j

n, then

S [x] x21 + x2j + 2 > 1 + p2k = sk.


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Consider now the outer points ofR. Let S be one such. Then at least

for one r, Lr(S ) < 0. Since the Lr are linear functions of the coordinatesand hence continuous, we may choose a neighbourhood of S consisting

of points for all of which Lr < 0. This means that the set of outer points

ofR is open. Note that here it is enough to deal with one inequality

alone unlike the previous one where one had to deal with all the Lrs.

Let now S be a boundary point ofR. Let S be an inner point.Consider the points T defined by

T = S

+ (1 )S .

These are points on the line joining S and S and every neighbourhoodofS contains points T with > 0 and points T with < 0.

Consider the points T with 0 < 1. These are the points betweenS and S . Let Lr be one of the linear polynomials defining R. NowLr(S ) 0, and Lr(S ) > 0, for all r. Thus

Lr(T) = Lr(S) + (1 )Lr(S ) > 0.

Hence T is an inner point. 61

Let now T be a point with < 0. Since S is a boundary point, there

is an r such that Lr(S ) = 0. For this r

Lr(T) = Lr(S) < 0

which proves that T is an outer point.

Since linear functions are continuous, the limit of a sequence of

points ofR is again a point ofR. This proves

Theorem 7. R is a closed set in P and the boundary points ofR

constitute the frontier ofR in the topology ofP.

We now prove the following

Theorem 8. Let S and S be two points ofR such that S [U] = S fora unimodular U E. Then S and S are boundary points ofRand Ubelongs to a finite set of unimodular matrices determined completely by

the integer n.


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Proof. The second part of the theorem follows readily from the Corol-

lary to Theorem 5. To prove the first part, we consider two cases: (1) Uis a diagonal matrix, and (2) U is not a diagonal matrix.

Let U be a diagonal matrix, U = (a1, . . . , an), with ai = 1. We mayassume, since S [U] = S [U] that a1 = 1. Let ak+1 be the first element= 1. Then, with usual notation,

skk+1 = skk+1.

But S and S being points ofRwe have62

0 skk+1 = skk+1 0

which means that skk+1 = 0 = skk+1. Hence S and S are both boundarypoints ofR.

Suppose U is not a diagonal matrix and denote its columns by u1,

. . . , un. Let uk be the first column different from the corresponding col-

umn of a diagonal matrix. Hence ui = ei, i = 1, . . . , k 1. (Note that kmay very well be equal to 1). Then

U =

D 0 V

where D is a diagonal matrix which is unimodular. V is a unimodular

matrix. FurthermoreU1 =

D1 0 V1

is unimodular. Let wk be the k-th column ofU

1. Then wk ek. Now

sk = S [uk] skand

sk = S[wk] sk

which proves that S [uk] sk = 0 = S [wk] sk and therefore S and S are boundary points ofR.

Suppose now that S is a boundary point ofR. By Theorem 7, there-fore, there exists a sequence of outer points S 1, S 2, . . . converging to S .63


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If the suffix k is sufficiently large, then all the S ks lie in a neighbour-

hood of S . Therefore they are all contained in an Rt for some t. Foreach k let Uk be a unimodular matrix such that S k[Uk] is in R. Since

R Rt , we have for all sufficiently large k, S k and S k[Uk] are both inRt . It follows therefore by Theorem 5, that Uks belong to a finite set ofmatrices. There exists therefore a subsequence S k1 , S k2 , . . . converging

to S such that one unimodular matrix U, among these finitely many, car-

ries S ki into R. Also Limn

S kn = S and therefore lim S kn [U] = S [U] is a

point ofR. Since S is a point ofR, it follows from the above theorem

that S [U] is also a boundary point ofR. Furthermore U E since S kare all outer points and S k[U] R. Hence

Theorem 9. If S is a boundary point ofR, there exists a unimodularmatrix U E and belonging to the finite set determined by Theorem8, such that S [U] is again a boundary point ofR.

By Theorem 8, there exist finitely many unimodular matrices say

U1, . . . , Ug which occur in the transformation of boundary points into

boundary points. Ifuk is the k-th column of one of these matrices, then

uk is an admissible k-vector. Suppose it is ek. Then for all S R,S [uk] sk 0. Let us denote by L1,L2, . . . ,Lh all the linear forms, notidentically zero, which result from all the uks k = 1, . . . , n occurring in 64

the set U1, . . . , Ug. Let L1, . . . ,Lh also include the linear forms skk+1,

k = 1, . . . , n

1; then from above we see that for a boundary point S of

R, there is an r h such that Lr(S ) = 0 (not identically). Also for allpoints ofR

s1 > 0,L1(S ) 0, . . . ,Lh(S ) 0. (59)But what is more important, we have

Theorem 10. A point S ofP belongs to R if and only if s1 > 0 and

Lr(S ) 0 for r = 1, . . . , h.Proof. The interest in the theorem is in the sufficiency of the conditions

(59).

Let S be a point ofP satisfying (59). Suppose S is not in R. Sinceit is inP, it is an outer point ofR. Therefore Lr(S ) < 0 for some r > h.


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Let S be an inner point ofR. Consider the points T,

T = S + (1 )S

for 0 < < 1, in the open segment joining S and S . Since the set ofinner points ofR is open and S is assumed to be an outer point, there

exists a 0 such that T0 is on the frontier ofR and 0 < 0 < 1. By our

remarks above, there exists for T0 an s h such that Ls(T0 ) = 0. Thismeans that

0 = Ls(T0 ) = 0Ls(S ) + (1 0)Ls(S ).

But (1 0)Ls(S ) > 0 so that Ls(T0 ) > 0. This is a contradiction.Therefore S R.

We have therefore proved that R is bounded by a finite number of65

planes all passing through the origin. R is thus a pyramid.

Let now R denote the closure ofR in the space Rh. At every point

S or R one has, because of continuity of linear functions,

s1 0, Lr(S ) 0, r = 1, 2, 3, . . .

If S R but not in R, then s1 = 0. In virtue of the other inequalities,we see that

S =

0 0

0 S 1.

S 1 again has similar properties. Thus either S = 0 or

S =

0 0

0 S k

where S k is non-singular and is a reduced matrix of order r, 0 < r < n.

We thus see that the points ofR which are not in Rare the semi-positive

reduced matrices.

Consider now the space P and the group . IfU , the mappingS S [U] is topological and takes P onto itself. For U denote byRU the set of matrices S [U] with S R. Because U and U lead to thesame mapping, we have RU

=RU. Since in every class of matrices

there is a reduced matrix we see that


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1) URU = Pwhere in the summation we identify U and U. Thus the RUs66cover P without gaps.

Let Uand Vbe in and U V. Consider the intersection ofRUand RV. Let S RU RV. Then T1 = S [U1] and T2 = S [V1]are both points ofR. Moreover T1 = T2[VU

1] and VU1 Eso that T1 is a boundary point ofR. Since the mapping S S [U]is topological S is a boundary point ofRU and also ofRV. Hence

2) If UV1 E and U and V are unimodular, then RU andRVcan have at most boundary points in common.

In particular, if U E, R and RU can have only boundarypoints in common. If S R RU then S and S [U1] are in Rand by Theorem 9, U belongs to a finite set of matrices depending

only on n. If we call RU a neighbourofR ifRRU is not empty,then we have proved

3) R has only finitely many neighbours.

Let K now be a compact subset ofP. It is therefore bounded in

P and hence there exists a t > 0 such that K Rt . Suppose RU,for a unimodular U, intersects K. Let S RU K. There is thena T R such that T[U] = S . For large t, R Rt . Then T andS are both in Rt and S = T[U]. Therefore U belongs to a finiteset of matrices. Hence there exist a finite number of unimodular

matrices, say U1, . . . , Up such that

K p

i=1

RUi

Hence 67

4) Every compact subset ofP is covered by a finite number of im-

ages RU ofR.

We have thus obtained the fundamental results of Minkowskisreduction theory.


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We now give a simple application.

Suppose S is a positive, reduced, integral matrix. Then since s1 s2 . . .sn bn|S |, s1, . . . , sn are positive and bn depends only on n, it followsthat for a given |S |, there exist only finitely many integer values fors1, . . . , sn. Also

sk 2skl sk, k < lso that skl being integers, there are finitely many values of skl satisfying

the above inequalities. We have therefore the

Theorem 11. There exist only finitely many positive, integral, reduced

matrices with a given determinant and number of rows.

Since all matrices in a class have the same determinant, and in each

class there is at least one reduced matrix, we get the

Theorem 12. There exist only a finite number of classes of positive in-

tegral matrices with given determinant and number of rows.

It has to be noticed, that in virtue of property 3) above, one has, in

general, only one reduced matrix in a class.

6 Binary forms68

We now study the particular case n = 2.

Let S = a bb c be a positive binary matrix and x = xy a vector. Thequadratic form S [x] = ax2 +2bxy +cy2 is positive definite. By the results

of the previous section, we see that, ifS is reduced then

a > 0, 0 2b a c. (60)

We shall now prove that any matrix S satisfying (60) is reduced.

Let x =xy

be an admissible one-vector. Ify = 0, then x = 1. Ify

0, then x and y are coprime integers. Consider the value ax2 + 2bxy + cy2

for admissible one-vectors. We assert that ax2 + 2bxy + cy2 a. In thefirst case S [x] = a. In the second case, because of (60)

ax2 + 2bxy + cy2 a(x2 xy +y2).


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6. Binary forms 57

But x and y are not both zero. Thus x2 xy +y2 1 which means thatS [x] a.

Let now x =xy

be an admissible two-vector. Then y = 1. Ifx = 0,

then S [x] = c. Let x 0, then

S [x] = ax2 2bx + c = c + x(ax 2b).

Because of (60), it follows that x(ax 2b) 0. Thus S satisfies condi-tions I) and II) of half reduction. Also b 0. This proves that S > 0and reduced.

(60) thus gives the necessary and sufficient conditions for a binary 69

quadratic form to be reduced.

In the theory of binary quadratic forms, one discusses some-timesequivalence not under all unimodular matrices, but only with respect to

those unimodular matrices whose determinant is unity. We say that two

binary matrices S and T are properly equivalent if there is a unimodular

matrix U such that

S = T[U], |U| = 1. (61)The properly equivalent matrices constitute a proper class. Note that

the properly unimodular matrices form a group. Two matrices S and T

which are equivalent in the sense of the previous sections, but which do

not satisfy (61) are said to be improperly equivalent. Note that improper

equivalence is notan equivalence relation.

In order to obtain the reduction theory for proper equivalence we

proceed thus: If S 1 =

a1 b1b1 c1

is positive, then there is a unimodular

matrix U such that S = S 1[U] =

a bb c

satisfies (60). If |U| = 1 we call

S a properly reducedmatrix. If |U| = 1, then consider W

W =

1 0

0 1

(62)

Then V = UW has the property |V| = 1. Now S [W] =

a bb c

and we

call this properly reduced. In any case we see that S is properly reduced 70

meansa > 0, 0 |2b| a c. (63)


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If we denote by R the reduced domain, that is the set of reduced

matrices in the old sense and R the properly reduced domain, one seesimmediately that

R

= R+ RW

where W has the meaning in (62).

We shall now give two applications.

Let S =

a bb c

be a positive integral matrix. Because of conditions

(63) and the additional condition (31), it follows that for given |S |, thereexist only finitely many properly reduced integral matrices. Consider

now the case |S | = 1. Then because of (31),

ac 4

3 (64)

and hence the only integers a, b, c, satisfying (63) and (64) are a = c =

1, b = 0. This proves

i) Every binary integral positive quadratic form of determinant unity

is properly equivalent to x2 +y2.

Let now p be a prime number > 2. Let p be representable by the

quadratic form x2 +y2. We assert that then p 1(mod 4). For, if x andy are integers such that

x2 +y2 = p

then x and y cannot be congruent to each other mod 2. So let x be odd71

and y even. Then p = x2 +y2 1(mod 4).We will now prove that conversely if p 1(mod 4), the form x2 +y2

represents p (integrally). For, let be a primitive root mod p. There is

then an integer k, 1 k < p 1 such that

k 1(mod p).

This means that 2k 1(mod p) and by definition of primitive root, weget k = p1/2. But p 1(mod 4) so that kis an even integer. Therefore

1 (k/2)2(mod p).


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6. Binary forms 59

There is thus an integer b, 1 b p 1 such that b2 1(mod p). Putb

2

= 1 + p, 1 an integer.Consider the binary form px2 + 2bxy + y2. Its determinant is p

b2 = 1. By the result obtained in i), this form is equivalent to x2 + y2.

But px2 + 2bxy + y2 represents p, (x = 1,y = 0). Therefore x2 + y2

represents p. Thus

ii) If p is a prime > 2, then x2 +y2 = p has a solution if and only if

p 1(mod 4).

Results i) and ii) are due originally to Lagrange.

Let S [x] = ax2 + 2bxy + cy2 be a real, positive, binary quadratic

form. We can write

S [x] = a(x y)(x y) (65)

where is a root, necessarily complex, of the polynomial az2 + 2bz + c 72

and is its conjugate. Let = + i have positive imaginary part.

Let V =

be a real matrix of unit determinant and consider the

mapping

S S [V].Then S [V x] is given by

S [V x] = a(x

y)(x

y) (66)

where a = a( ) ( ) is necessarily real and positive, and

= V1() = + . (67)

It is easy to see that also has positive imaginary part. Let us also

observe that =b + i |S |

a.

Consider now the relationship between S and . IfS is given, then

(65) determines a with positive imaginary part. Now given , (65)

itself shows that S is determined only upto a real factor. This real factor

can be determined by insisting that the associated quadratic forms havea given determinant. In particular, if |S | = 1 then the is uniquely


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determined by S and conversely. If = + i, > 0, then the S is given

by

S =

1 0

0

1 0 1

(68)

LetP denote the space of positive binary forms of unit determinant73

and G the upper half complex -plane. By what we have seen above the

mapping S in (65) is (1, 1) both ways. Let denote the group ofproper unimodular matrices. It acts on G as a group of mappings

U() = +

+ , U = (69)ofG onto itself. If we define two points 1, 2 in G as equivalent if

there is a U such that 1 = U(2), then the classical problem ofconstructing a fundamental region in G for , is seen to be the same

as selecting from each class of equivalent points one point so that the

resulting point set has nice properties.

By means of the (1, 1) correspondence, we have established in (68)

betweenP and G, we have S 1 = S 2[U] if and only if the corresponding

points 1, 2 respectively satisfy

1 = U1(2).

We define the fundamental region F in G to be the set of points such

that the matrices corresponding to them are properly reduced; in other

words, they satisfy (63). For the S in (68), S [x] =1

(x2 2xy + (2 +

2)y2). Therefore F consists of points = + i for which

|2

| 1

2 + 2 1 (70)


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7. Reduction of lattices 61

This is the familiar modular region in the upper half -plane. That 74

it is a fundamental region follows from the properties of the space of

reduced matrices in P. The points P and Q are the complex numbers1 + i 32

, and so for any point in F,

3

2. This means that for a

positive reduced binary form ax2 + 2bxy + cy2 of determinant d

ad

23

,

which we had already seen in Theorem 1.

7 Reduction of lattices

Let V be the Euclidean space of n dimensions formed by n-rowed real

columns

=

a1...

an

.Let 1, . . . , n be a basis ofV so that

i =

a1i...

ani

, i = 1, . . . , n.

Denote by A the matrix (akl). Obviously |A| 0.


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Let L be a lattice in V and let 1, . . . , n be a basis of this lattice. L

then consists of elements 1g1 + + ngn where g1, . . . , gn are integers.We shall call A the matrix of the lattice.

Conversely if A is any non-singular n-rowed matrix, then the75

columns of A, as elements of V are linearly independent and therefore

determine a lattice.

Let L be the lattice above and let 1, . . . , n be any other base of L

and B its matrix, then

B = AU

where U is a unimodular matrix. Also ifU runs through all unimodular

matrices, then AU runs through all bases of L. We now wish to single

out among these bases one which has some distinguished properties.

Let us introduce in V, the inner product of two vectors and by

= a1b1 + + anbn

where =

a1...

an

, =

b1...bn

. The square of the length of the vector isgiven by

2 = a21 + + a2n.Let A be the matrix of a base 1, . . . , n of L. Consider the positive

matrix S = AA. IfS is given A is determined only upto an orthogonalmatrix P on its left. For, if AA = A

1

A1 then AA11

= P is orthogonal.

But multiplication on the left by an orthogonal matrix implies a rotation

in V about the origin.

We shall call a base B ofL reduced ifS 1 = BB is a reduced matrix.

Obviously in this case

0 < 21 . . . 2nkk+1 0, k = 1, . . . , n 1.

From the way reduced matrices are determined we see that a reduced76

base 1, . . . , n of L may be defined to be a base such that for every set

of integers x1, . . . , xn such that (xk, . . . , xn) = 1 the vector

= 1x1 + +nxn


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7. Reduction of lattices 63

satisfies

2

2k (k = 1, . . . , n.)

Also

k k+1 0(k = 1, . . . , n + 1).If follows therefore that

21 . . . 2n cn|AA| = cn|A|2

cn being a constant depending only on n. Also abs |A| is the volume ofthe parallelopiped formed by the vectors 1, . . . , n.

consider the case n = 2.

We have, because of (30)

21 22 4

3|A|2 (72)

Let now denote the acute angle between the vectors 1 and 2.

Since the area of the parallelogram formed by 1 and 2 on the one hand

equals abs |A| and on the other

212

2 sin , we see that 77

sin2 34

(73)

Since 0

2 , it follows from (73) that

3

2.

Hence for a two dimensional lattice we may choose a basis in such a

manner that the angle (acute) between the basis vectors is between 60

and 90.


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Bibliography

[1] C.F. Gauss Disquisitiones Arithmeticae Ges. Werke 1, (1801).

[2] C. Hermite Oeuvres Vol.1, Paris (1905) P. 94-273.

[3] H. Minkowski Geometrie der Zahlen, Leipzig (1896).

[4] H. Minkowski Discontinuitatsbereich fur arithmetische Aquiv-

alenz. Ges. Werke, Bd.2, (1911), P.53 - 100.

[5] C.L. Siegel Einheiten quadratischer Formen Abh. Math. Sem. Han-

sischen Univ. 13, (1940), P. 209-239.

65


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Chapter 3

Indefinite quadratic forms

1 Discontinuous groups78

In the previous chapter we had met with the situation in which a group

of transformations acts on a topological space and we constructed, by

a certain method, a subset of this space which has some distinguished

properties relative to the group. We shall now study the following gen-

eral situation.

Let be an abstract group and T a Hausdorff topological space on

which has a representation

t t, t T, (1)

carrying T into itself. We say that this representation of is discontin-

uous if for every point t T, the set of points {t} for has nolimit point in T. The problem now is to determine, for a given , all the

spaces T on which has a discontinuous representation. For an arbitrar-

ily given group, this problem can be very difficult. We shall, therefore,

impose certain restrictions on and T. Let us assume that there is a

group , of transformations of T onto itself, which is transitive on T.

This means that ift1 and t2 are any two elements ofT, there exists such that

t1 = t2. (2)

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68 3. Indefinite quadratic forms

Let us further assume that is a subgroup of . Let t0 be a point in T

and consider the subgroup of consisting of such that79t0 = t0. (3)

Ift is any point ofT, we have because of transitivity,

t = t0

for some . Because of (3), we get

t = t0.

Conversely if is such that t = t0, then t0 = t0 or .Thus every point t T determines a coset of\ that is, the space ofright cosets of modulo . Conversely if is any coset, then t = t0

is a point determined by . Hence the mapping

t (4)

of T on \ is (1, 1) both ways. In order to make this correspondencetopological, let us study the following situation.

Let be locally compact topological group and T a Hausdorff topo-

logical space on which has a representation

t t (5)as a transitive group of mappings. Let us assume that this representation

is open and continuous. We recall that (5) is said to be open if for every

open set P in and every t T the set {t}, P is an open set inT. Then it follows that the subgroup of leaving t0 T fixed isnot only a closed subgroup but that the mapping (4) of T on \ is ahomeomorphism.

Let be a subgroup of which has on T a discontinuous represen-80

tation. Then has trivially a representation in \. By the remarksabove, the representation

, (6)


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1. Discontinuous groups 69

is discontinuous in \.On the other hand, let be any closed subgroup of . Then the

representation

of on \ is open and continuous. It is clearly transitive. In order,therefore, to find all spaces on which has a discontinuous representa-

tion, it is enough to consider the spaces of right cosets of with regard

to closed subgroups of.

Suppose is a closed subgroup of and has a discontinuous

representation on \. Let K be a closed subgroup of contained in. Then has a discontinuous representation on K\. For, if K isa coset such that the set of cosets

{K

},

has a limit point in

K\, then the set {}, also has a limit point in \ and so (6)would not be discontinuous. In particular, if we take for K the subgroup

consisting only of the identity element e, then is discontinuous in is

clearly equivalent to is a discrete subgroup of.

Thus if there exists some subgroup of such that is discontin-

uous in \, then necessarily has to be discrete. It can be proved thatif has a countable basis of open sets, then is enumerable.

Suppose now that is a locally compact group with a countable ba- 81

sis of open sets. Let be a discrete subgroup of. If is any compact,

hence closed, subgroup of then it follows that the representation (6)

of in

\ is discontinuous. This can be seen by assuming that for a

certain , the set n, n has limit point and this will lead to acontradiction because of the discreteness of .

In general the fact (6) is discontinuous in \ does not entail that is compact. Let us, therefore, consider the following situation.

Let be a locally compact group possessing a countable basis of

open sets. Then there exists in a right invariant Haar measure d

which is determined uniquely upto a positive multiplicative factor. Let

be a discrete subgroup of. There exists then in a subset Fpossessing

the following properties: 1)

aFa = , 2) the sets {Fa} for a are

mutually disjoint and 3) F is measurable in terms of the Haar measure

d. F is then said to be a fundamental set relative to

. Note that if Fis a fundamental set then so if Fa for any a so that a fundamental


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70 3. Indefinite quadratic forms

set is not unique. 1) and 2) assert that F intersects each coset of \in exactly one point so that F has to be formed in by choosing oneelement from each coset \. The interesting point is that, under theconditions on , this can be done in such a way that the resulting set F

is measurable. Let us now assume thatF

d < . (7)

It can then be shown that the value of the integral in (7) is independent82

of the choice of F. We now state, without proof, the important

Theorem 1. Let be a locally compact topological group with a count-

able basis of open sets. Let be a discrete subgroup of and F a funda-

mental set in relative to . Let F have finite Haar measure in . If

is any closed subgroup of , then has a discontinuous representation

in \ if and only if is compact.

The interest in the theorem lies in the necessity part of it.

Let us assume that is, as will be in the applications, a Lie group.

Let be a discrete subgroup of . For any closed subgroup of, the

dimensions of, \ and are connected by

dim + dim \ = dim .

If F is a fundamental set in with regard to and is of finite measure,

in terms of the invariant measure in , then by Theorem 1, will be

discontinuous in \ if and only if is compact. In order, therefore,to obtain a space T = \ of smallest dimension in which has adiscontinuous representation, one has to consider a which is compact

and maximal with this property.

Let us consider the following example.83

Let be the group of n-rowed real matrices A is a Lie group.Let us determine first all compact subgroups of . Let K be a compact

subgroup of. IfC K, then |C|

lectures on quadratic forms

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