Let $C \subset \mathbb{P}^2$ be a planar conic curve, defined by a ternary quadratic form $Q(x_1, x_2, x_3)$ say. Suppose that $C(\mathbb{Q}) \ne \emptyset$, or equivalently, that $C$ is everywhere locally soluble (i.e., $C(\mathbb{Q}_p) \ne \emptyset$ for every prime $p$ and $C(\mathbb{R}) \ne \emptyset$). Further, it is easy to show that there exist primitive binary quadratic forms $f_1, f_2, f_3 \in \mathbb{Z}[u,v]$ that parametrize the rational points $C(\mathbb{Q})$. In particular, $(x_1, x_2, x_3) = (f_1(u,v), f_2(u,v), f_3(u,v)), u,v \in \mathbb{Z}$.

For example, if $Q(x_1, x_2, x_3) = x_1^2 + x_2^2 - x_3^2$ then we can parametrize the primitive solutions by $x_1 = 2uv, x_2 = u^2 - v^2, x_3 = u^2 + v^2$.

For a given ternary quadratic form $Q(x_1, x_2, x_3) \in \mathbb{Z}[x_1, x_2, x_3]$ which is everywhere locally soluble, call a triple of binary quadratic forms $(f_1, f_2, f_3)$ a parametrizing triple if $Q(f_1, f_2, f_3) \equiv 0$. Consider the binary sextic form $F_Q = f_1 f_2 f_3$.

In general, what can we say about $F_Q$, given $F$? In the example above, we have $(f_1, f_2, f_3) = (2uv, u^2 - v^2, u^2 + v^2)$ and $F_Q = 2uv(u^2 - v^2)(u^2 + v^2)$. This $F_Q$ is very special: indeed, it is a sextic Klein form, with an exceptionally large $\text{PGL}_2$ automorphism group. Does this hold in general when $Q$ is a diagonal form?


I don't know if the following will fully answer your question, since I'm not sure if you attach a huge importance to have integral coefficients/primitive solutions. Apologies if it doesn't.

Let $Q$ be any nondegenerate (I assume this is the case of interest) ternary quadratic form such that $Q(\mathbb{Q})\neq 0$. Then the theory of quadratic forms says that $Q$ splits off a hyperbolic plane. Hence $Q\simeq \langle a,1,-1\rangle$ for some nonzero $a$.

Now it is known that $\langle 1,-1\rangle\simeq \langle a,-a\rangle$ for any nonzero $a$, so finally $Q$ is isomorphic to $\langle a ,a ,-a\rangle$.

In other words, there exists a basis of $K^3$ such that $Q(x_1,x_2,x_3)=a(x_1^2+x_2^2-x_3^2)$, where the $x_i'$s are the coordinates of a vector $x\in K^3$ in this basis.

Now $Q(x_1,x_2,x_3)=0\iff x_1^2+x_2^2-x_3^2=0$.

All in all, the case you already covered is the only one. Hope this helps.

  • $\begingroup$ Yes that is indeed helpful! I managed to prove by hand what I wanted (that the product of the parametrizing binary quadratic forms is sextic Klein form) by hand, but it is nice to see an elegant argument. $\endgroup$ Oct 19 '20 at 11:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.