Etapas de la resolución de problemas matemáticos

To understand an equivalence relation, a common strategy is to pick one element out of each equivalence class to use as a representative. (Such a representative is often called a “canonical form”.) We now do this for positive definite quadratic forms:

• we define “reduced” forms, and give a simple test for recognising them;

• we give an algorithm for finding the unique reduced form equivalent to a given form;

• we show that there are only a finite number of equivalence classes of posi- tive definite forms with given discriminant.

This gives us a fairly satisfactory classification of positive definite forms.

Definition Let f(x,y) =ax2+bxy+cy2be a positive definite quadratic form (so thatb2−4ac<0,a>0,c>0). We say that f isreducedif either

• c>a,−a<b≤a; or

• c=a, 0≤b≤a.

We are going to show that any positive definite quadratic form is equivalent to a unique reduced form. This requires two steps: first we show that there is a reduced form equivalent to any given form; then we show that if two reduced forms are equivalent then they are equal.

We begin with a simple observation.

Proposition 10.6 let f =ax2+· · ·be a reduced positive definite form. Then a is the smallest positive integer represented by f .

Proof The form f =ax2+bxy+cy2does represent the integera(just put(x,y) = (1,0)). So it suffices to assume that f is reduced (so that, in particular,|b| ≤a≤c), and that ax2+bxy+cy2 <a for some integers x,y(not both zero), and derive a contradiction.

10.3. POSITIVE DEFINITE FORMS 107

• Ify=0, thena>ax2≥a(sincex2≥1).

• Ifx=0, thena>cy2≥c≥a(sincey2≥1).

• Ifx,y6=0 and|x| ≤ |y|, then|bxy| ≤cy2(since|b| ≤c), anda>ax2+bxy+ cy2≥ax2≥a.

• Ifx,y6=0 and|y| ≤ |x|, then|bxy| ≤ax2, anda>ax2+bxy+cy2≥cy2≥

c≥a.

First we look at a particular kind of equivalence. The matrix P=

0 1

−1 k

is unimodular. Our calculations above show that if f(x,y) =ax2+bxy+cy2 is represented byM, then the equivalent form represented byP>MPis

f(y,ky−x) =cx2−(b+2k)xy+ (a+bk+ck2)y2.

We call this theright neighbour of f viak. This is a formula to which we shall return very often! Note that the right neighbour of f via 0 is cx2−bxy+ay2. Any right neighbour of a form · · ·+cy2 looks like cx2+· · ·. Note also that the coefficient ofy2in the right neighbour of f bykis f(1,k).

Theorem 10.7 Any positive definite quadratic form is equivalent to a reduced form.

Proof We will see that the method of proof gives us a construction for finding the reduced form equivalent to a given positive definite form.

Let our form be

f₀=a₀x2+b₀xy+a₁y2. (You will see in a minute why we write it this way.)

Define q and b₁ by b₀ =2a1q−b1, with −a1<b1 ≤a1. (In other words,

divideb0by 2a1, so that the remainder is between−a1anda1.)

Now take the right neighbouring form by−q. This form is

f₁=a₁x2−(b₀−2a1q)xy+ (a0−b0q+a1q2)y2=a1x2+b1xy+a2y2,

wherea₂= f₀(1,−q) =a₀−b₀q+a₁q2.

Ifa₂≥a₁, then stop; otherwise repeat to obtain f2=a2x2+b2xy+a3y2,

108 CHAPTER 10. QUADRATIC FORMS with−a₂<b₂≤a₂. Continue the process, obtaining formsa_nx2+b_nxy+an+1y2

forn=1,2, . . .; stop whenan+1≥an.

We havea1>a2>a3>· · ·, and this sequence cannot continue for ever since

a_i>0. When it terminates we have a form

f_n=a_nx2+b_nxy+a_n+1y2,

with−a_n<b_n≤a_nanda_n≤a_n+1.

This form is reduced unlessa_n+1=a_nand b_n<0; in this case take the right neighbour by 0 to change the sign ofb_n.

Example Find a reduced form equivalent to the positive definite form 31x2+ 22xy+4y2.

First we put 22=8q−b1with−4<b1≤4. The solution isq=3,b1=2, and

f₁=42+2xy+a₂y2, wherea₂=31−22·3+4·9=1, that is, f₁=4x2+2xy+y2. Now we put 2=2q−b₂ with−1<b₂≤1, with solutionq=1,b₂=0. We get f₂=x2+ (4−2+1)y2=x2+3y2, which is reduced.

Theorem 10.8 If the two reduced positive definite forms f =ax2+bxy+cy2and g=a0x2+b0xy+c0y2are equivalent, then they are equal: a=a0, b=b0, c=c0.

The proof of this theorem is a calculation which we omit. Note that the fact thata=a0follows from Proposition 10.6.

Now the theorem gives us an important conclusion. We have seen that equiv- alent forms have the same discriminant but that the converse is false. But the following holds:

Theorem 10.9 There are only finitely many equivalence classes of positive defi- nite forms with given discriminant.

Proof Since there is a unique reduced form in any given equivalence class, it is enough to show that there are only finitely many reduced forms with any given discriminant.

Let f =ax2+bxy+cy2be a reduced form, so thatc≥a≥ |b|, and

−d=−b2+4ac≥3ac≥3c. So we have

−a≤b≤a≤c≤ |d|/3,