[Math] How to show that two quadratic forms are equivalent

linear algebraquadratic-forms

To show to quadratic forms are not equivalent, we can find rank, or discriminant or some element which is represented by either one only etc. But Is there a general criterion to show that two binary(right now I am only concerned for binary) quadratic forms are equivalent.
Like here is an example which uses variable transformation, but every time find what transformation will apply is tough.

$\textbf{Example-}$ $X^2-Y^2 $ and $4XY$. replacing $X,Y$ by $X+Y,X-Y$ do the job here. Nnow there are criterion which gives an equivalent condition to check whether two binary forms are equivalent or not , like

$\textbf{Criterion 1-}$ On algebraically closed fields, same dimension and same rank suffice.

$\textbf{Criterion 2-}$ On Reals, same rank and same signature suffices.

$\textbf{Criterion 3-}$ On finite fields,same discriminant is enough.

But in general if we are given field as some arbitrary $K$, then what to do? I am very new at quadratic forms. Any help will be appreciated.

I know that we can check similarity of their corresponding matrices, but is there any other way, beside checking matrices are similar or not. Like how would you proceed on these couple of question below. If there is no other method, can somebody please explain me the method to check whether two matrices are similar or not, I am really weak with matrices.

$\textbf{Question 1-}$ Prove $X^2-4XY+3Y^2, X^2-Y^2,aX^2-aY^2$ are equivalent over $K$.

$\textbf{Question 2-}$ Show

i) $X^2-Y^2 \sim XY$ over $K$, ii) $3X^2-2Y^2 \sim X^2-6Y^2$ (over $Q$)

Best Answer

Q1)

a) $X^2-Y^2$ is already in diagonal from. It's matrix is $\bigl( \begin{smallmatrix} 1& 0 \\ 0 & -1 \end{smallmatrix} \bigr)$.
b) $aX^2-aY^2$ is already in diagonal from. It's matrix is $\bigl( \begin{smallmatrix} a& 0 \\ 0 & -a \end{smallmatrix} \bigr)$. We have $\bigl( \begin{smallmatrix} a& 0 \\ 0 & -a \end{smallmatrix} \bigr) \sim \bigl( \begin{smallmatrix} 1& 0 \\ 0 & -1 \end{smallmatrix} \bigr)$
c) $X^2-4XY+3Y^2$ can also be written as $(X-2Y)^2-Y^2$
Result a) and b): if $K=\Bbb{R}$ or $\Bbb{C}$ they are always equivalent. If $\sqrt(a)$ is rational and $K=\Bbb{Q}$ then also, and when $K$ is a finite field then if the characteristic is $2$ then always otherwise only when $a=\pm 1$.
Result (b) and (c) always equivalent (same discriminant, rank and signature).

Q2)

i) if in the second we replace $X=U+V$ and $Y=U-V$ then we get $XY=(U+V)(U-V)=U^2-V^2$, so equivalence.
ii) $3X^2-2Y^2 \sim 3(3X^2-2Y^2)=(3X)^2-6Y^2=U^2-6V^2$

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[Math] Equivalence of Quadratic Forms that represent the same values

As you mentioned on Chat, Kap and I wrote a note on forms of different discriminants (but positive definite forms, meaning negative discriminants). This was corrected and extended by John Voight, now at Dartmouth; also published.

The best known examples are the pair $x^2 + xy + y^2$ and $x^2 + 3 y^2.$ The proof that these represent the same numbers is some 2 by 2 matrices, some things mod 2. Same for the indefinite pair $x^2 + xy - y^2$ and $x^2 - 5 y^2.$

Probably worth pointing out that the forms $x^2 + xy + 2ky^2$ and $x^2 + (8k-1)y^2$ represent all the same odd numbers, including any odd primes. The latter form does not represent $2$ or $-2,$ if you can say the same about the former form they agree on primes. We called these "Trivial Pairs." Um; as with Gauss, we discard these if the discriminant is square, meaning we demand $8k - 1 \neq -w^2,$ or $k \neq \frac{1 - w^2}{8}.$

The question changes if you allow square discriminants.

There may be infinitely many other indefinite pairs, we did not check.

If the discriminant is not a square, two forms of the same discriminant that share even a single prime are $GL_2 \mathbb Z$ equivalent. In traditional terms, they are either equivalent or opposite.

Forms with square discriminant, such as $xy$ or $x^2 - y^2,$ are unusual in representing entire arithmetic progressions. Primes do not control things.

For self study, I recommend Buell, Binary Quadratic Forms. I find it easier reading than Buchmann and Vollmer. I also recommend L. E. Dickson Introduction to the Theory of Numbers. For just the first section, I also like Cox, Primes of the Form $x^2 + n y^2.$ Cox does a good job on positive forms, genera, composition. No indefinite forms, though, no Pell. As you can see from my answers, I like the first chapter in Conway, The Sensual Quadratic Form. The wonderful thing there is the "Topograph" construction. I have written a bunch of software to tell me how to avoid arithmetic mistakes in drawing those. These give the best way for talking about a fixed indefinite form $A x^2 + B x y + C y^2$ with $B^2 - 4 AC > 0 $ but not a square. The "cycle" method of Lagrange does not do well when $|n|$ is too large, in finding all solutions to $A x^2 + B x y + C y^2 = n.$ Lagrange's method gives all answers when $|n| < \frac{1}{2} \sqrt{B^2 - 4 AC};$ this result is Theorem 85 in Dickson. Oh, both Lagrange and Conway are talking about primitive representations, $\gcd(x,y) = 1.$

The relation between binary quadratic forms and quadratic orders

A binary quadratic form, with integer coefficients, is some $$ f(x,y) = A x^2 + B xy + C y^2. $$ The discriminant is $$ \Delta = B^2 - 4 A C. $$ We will abbreviate this by $$ \langle A,B,C \rangle. $$ It is primitive if $\gcd(A,B,C)=1. $ Standard fact, hard to discover but easy to check: $$ (A x^2 + B x y + C D y^2 ) (C z^2 + B z w + A D w^2 ) = A C X^2 + B X Y + D Y^2,$$ where $ X = x z - D yw, \; Y = A xw + C yz + B yw. $ This gives us Dirichlet's definition of "composition" of quadratic forms of the same discriminant, $$ \langle A,B,CD \rangle \circ \langle C,B,AD \rangle = \langle AC,B,D \rangle. $$ In particular, if this $D=1,$ the result represents $1$ and is ($SL_2 \mathbf Z$) equivalent to the "principal" form for this discriminant. Oh, duplication or squaring in the group; if $\gcd(A,B)=1,$ $$ \langle A,B,AD \rangle^2 = \langle A^2,B,D \rangle. $$ This comes up with positive forms: $ \langle A,B,C \rangle \circ \langle A,-B,C \rangle = \langle 1,B,AC \rangle $ is principal, the group identity. Probably should display some $SL_2 \mathbf Z$ equivalence rules, these are how we calculate when things are not quite right for Dirichlet's rule: $$ \langle A,B,C \rangle \cong \langle C,-B,A \rangle, $$ $$ \langle A,B,C \rangle \cong \langle A, B + 2 A, A + B +C \rangle, $$ $$ \langle A,B,C \rangle \cong \langle A, B - 2 A, A - B +C \rangle. $$

Imaginary first. Suppose we want to know about $\mathbf Q(\sqrt {-47}).$ Reduced positive forms $ \langle A,B,C \rangle $ obey $|B| \leq A \leq C$ and $B \neq -A,$ also whenever $A=C$ we have $B \geq 0.$ Our group of binary forms is

-47
  class  number  5
 all  
( 1, 1, 12)
( 2, -1, 6)
( 2, 1, 6)
( 3, -1, 4)
( 3, 1, 4)

This is an abelian group in any case, so it is cyclic of order 5. These are also the five elements in the ring of integers of $\mathbf Q(\sqrt {-47}).$ Here is the mapping from forms to ideals: given $ \langle A,B,C \rangle, $ drop the letter $C.$ That's it. $$ \langle A,B,C \rangle \mapsto \left[ A, \frac{B + \sqrt \Delta}{2} \right]. $$ Oh, why is this an ideal, rather than just some $\mathbf Z$-lattice? Because, given $\alpha,\beta$ rational integers, $$ \left[ \alpha, \frac{\beta + \sqrt \Delta}{2} \right] $$ is an ideal if and only if $$ 4 \alpha | ( \Delta - \beta^2 ). $$

Group: we already see how to do $$ \langle 2,1,6 \rangle^2 \cong \langle 4,1,3 \rangle \cong \langle 3,-1,4 \rangle; $$ $$ \langle 2,1,6 \rangle \circ \langle 3,-1,4 \rangle \cong \langle 2,5,9 \rangle \circ \langle 3,5,6 \rangle \cong \langle 6,5,3 \rangle \cong \langle 3,-5,6 \rangle \cong \langle 3,1,4 \rangle; $$ $$ \langle 2,1,6 \rangle \circ \langle 3,1,4 \rangle \cong \langle 6,1,2 \rangle \cong \langle 2,-1,6 \rangle. $$ $$ \langle 2,1,6 \rangle \circ \langle 2,-1,6 \rangle \cong \langle 1,1,12 \rangle $$ in any case.

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[Math] Equivalence of Quadratic Forms that represent the same values

The relation between binary quadratic forms and quadratic orders

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