# Space of rational conics

Let $$K$$ be a field of characteristic different from two. Conics over $$K$$ (that is curves of degree two in $$\mathbb{P}^2_K$$) are parametrized by $$\mathbb{P}(k[x,y,z]_2) = \mathbb{P}^5_K$$.

Conisider the set $$R = \{C \in \mathbb{P}(k[x,y,z]_2) \: | \: C \text{ has a K-point}\}\subseteq\mathbb{P}^5$$.

If $$K$$ is algebraically closed then $$R = \mathbb{P}^5$$ but if not there are conics without points.

When $$K$$ is not algebraically closed does $$R$$ have any meaningful geometric structure?

• It is an orbit of $\operatorname{PGL}(3,K)$ acting on $\mathbb{P}^5_K$.
– abx
Jan 12 at 19:53
• If I understand correctly you take a conic with a point and look at its orbit under $PGL(3,K)$. But what is the dimension of this orbit?
– Arty
Jan 12 at 20:03
• It is a set; it has no geometric structure. In any case the orbit is Zariski dense. Jan 12 at 20:16
• Since the OP does not exclude singular conics, it is not true that $\mathrm{PGL}(3,K)$ acts transitively on $R$. There is one orbit consisting of smooth conics (rank 3), one for double lines (rank 1), and a whole family for rank 2, parametrized by $K^\times/K^{\times2}$, with $d\in K^\times$ coding for the conic $x^2-dy^2$. Jan 13 at 12:41
• @DanielLoughran: the orbit is not Zariski-dense if (and only if) $K$ is finite. But in this case, of course, $R=\mathbb{P}^5(K)$. And also, what we mean by Zariski-dense is debatable. Jan 15 at 9:37

Let $$X: \quad a_{0,0}x^2 + a_{1,1}y^2 + a_{2,2}z^2 + a_{0,1}xy + a_{0,2}xz + a_{1,2}yz = 0 \quad \subset \mathbb{P}^2 \times \mathbb{P}^5$$ be the total space of the family of all plane conics. Then your set $$R$$ is the image $$\pi_2(X(K)) \subset \mathbb{P}^5(K)$$ of the $$K$$-rational points on $$X$$ with respect to the second projection.