# Conics and triples of binary quadratic forms

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.

• 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. Oct 19 '20 at 11:07