# Surface of type $(2,2)$ on the Segre cubic scroll $\mathbb{P}^1 \times \mathbb{P}^2 \subset \mathbb{P}^5$

Let $$S=\mathbb{P}^1 \times \mathbb{P}^2 \subset \mathbb{P}^5$$ embedded with the Segre embedding given by $$\mathcal{O}_S(1,1)$$. If we intersect $$S$$ with a general smooth quadric $$Q \subset \mathbb{P}^5$$ we get a smooth surface $$X \subset S$$ of type $$(2,2)$$. Since $$deg(S)=3$$ by Bertini we have that $$deg(X)=6$$.

By construction we have that (up to identifying every $$\mathbb{P}^2$$ in the ruling by projecting from the $$\mathbb{P}^1$$ factor) for every point $$p \in \mathbb{P}^2$$ there exist only $$2$$ conics $$C_1,C_2$$ passing through $$p$$, so this is not a surface of the type $$Y=\mathbb{P}^1 \times \mathbb{P}^1$$ embedded with bidegree $$(2,1)$$. In fact we have that $$Y \subset \mathbb{P}^5$$ but $$deg(Y)=4 \neq deg(X)$$.

My question now is how to prove that $$X$$ is rational? (I've the feeling that it must be so but I'm currently not able to figure it out).

Also I had the feeling that $$X$$ could be viewed as the total space of a conic $$\mathcal{C}$$ in the $$\mathbb{P}^5=\mathbb{P}(\mathbb{C}[x^2,y^2,z^2,xy,xz,yz])$$ where $$[x,y,z]$$ are homogeneous coordinates of $$\mathbb{P}^2$$, where by total space I mean the space consisting of the union of the conics $$C_x$$ corresponding to the point $$x \in \mathcal{C}$$. This would imply that there are $$6$$ singular conics among the conics in the $$\mathbb{P}^2$$ fibers of $$S$$, since $$deg(\mathcal{C} \cap \mathbb{S}ec_2(v_2(\mathbb{P}^2)))=2 \cdot 3=6$$.

Anyone knows if $$6$$ is the correct number of singular conics in a smooth quadric section of the Segre cubic scroll in $$\mathbb{P}^5$$? The fact that $$6$$ is also the degree is a coincidence?

Here is another approach. Take a smooth surface $$X$$ of type $$(d,2)$$ in $$S$$. The general fiber of the projection $$\pi:X\rightarrow \mathbb{P}^1$$ is a smooth conic. So $$X$$ is rationally connected and since $$X$$ has dimension $$2$$ it is rational. This does not depend on $$d$$.

The bi-homogeneous polynomial cutting out $$X$$ in $$S$$ of the the following form:

$$a_{0,0}(u,v)x^2+a_{0,1}(u,v)xy+a_{0,2}(u,v)xz+a_{1,1}(u,v)y^2+a_{1,2}(u,v)yz+a_{2,2}(u,v)z^2$$

where $$[u:v]$$ and $$[x:y:z]$$ are homogeneous coordinates on $$\mathbb{P}^1$$ and $$\mathbb{P}^2$$ respectively, and the $$a_{i,j}(u,v)$$ are homogeneous polynomials of degree $$d$$. The matrix of the conic $$C_{u,v}$$ over the point $$[u:v]\in\mathbb{P}^1$$ is a $$3\times 3$$ matrix whose entries are homogeneous polynomials of degree $$d$$. So its determinant is a homogeneous polynomial of degree $$3d$$. The singular conics in the conic bundle correspond to the zeros of the determinat. So you have $$3d$$ singular conics.

This surface $$X$$ satisfies $$p_g(X)=q(X)=0$$. Moreover, if $$H_X$$ is the hyperplane section, by adjunction we find $$K_X^2=2, \quad K_X H_X=-4.$$

Therefore no multiple of the canonical divisor can be effective, in particular $$P_2(X)=q(X)=0$$ and $$X$$ is rational by Castelnuovo criterion.

In fact, $$X$$ is a conic bundle with six singular fibres, that turns out to be the blow-up at six points of the quadric $$\mathbb{F}_0=\mathbb{P}^1 \times \mathbb{P}^1$$. This is Example 1.9, p. 236 in

S. Di Rocco, K. Ranestad: On surfaces in $$\mathbb{P}^{6}$$ with no trisecant lines (http://dx.doi.org/10.1007/BF02384319), Ark. Mat. 38, No. 2, 231-261 (2000). ZBL1035.14011.

• thank you very much!
– gigi
Nov 27 '20 at 15:10