Transform quadratic ternary form to normal form

Question

Does anyone know of an integral transform which transforms the normal form $Ax^2 + By^2 + Cz^2 + Dxy + Eyz + Fzx = 0$ to the form $ax^2 + by^2 + cz^2 = 0$ ?

Thanks in advance.

Answer 1

It is a two step iterative algorithm:

1. Take the first variable $x$. If $x^2$ appears there with non-zero coefficient, complete the square with $x^2$ and the double product $xy$ with the next variable. Continue with the next variables.
2. If $x^2$ is not there, only the double product with the next variable $xy$, then make the substitution $x=u+v$, $y=u-v$. This makes the term $u^2$ appear. Go to step $1$ with the variable $u$.

This will give you a change of variable with rational coefficients. You will have to multiply the original form by a convenient factor to clear denominators.

Example:

$$xy+y^2+z^2.$$

The order to input the variables to the algorithm could be other, but let us do it with $x$ as the first variable.

We need step $2$ because there is no $x^2$. We get $$(u+v)(u-v)+(u-v)^2+z^2=2u^2-2uv+z^2.$$

Now $u$ is our first variable. We go to step $1$.

$$2(u^2-uv+v^2/4)-v^2/4+z^2=2(u-v/2)^2-v^2/4+z^2.$$

In the new variables $z_1=u-v/2$, $z_2=v$, $z_3=z$ (where $u=(x+y)/2$ and $v=(x-y)/2$) we get $$2z_1^2-z_2^2/4+z_3^2.$$

Notice we can, in this case, multiply the whole form by $16$ and get rid of the denominators in the change of variable.

Answer 2

I also found a two step transformation that will do the required thing. First we make the transformation $x \rightarrow x - (By + Cz)/(2A)$ (if $A = 0$ then we switch the variables). This will result in an equation of the form $A'x^2 + D'y^2 + E'yz + F'z^2 = 0$ after clearing the denominators. Then the transform $y \rightarrow y - E'z/(2D')$ would give us the form $ax^2 + by^2 + cz^2 = 0$.