Embedding degree 1 Del Pezzo surfaces in $\mathbb{P}(1,1,2,3)$

Question

In the projective bundle $\mathbb{P}(\mathcal{O}(-1)\oplus \mathcal{O}(-1)\oplus \mathcal{O})\rightarrow\mathbb{P}^1$ consider the hypersruface $$ X := \{a_{00}y_0^2+a_{01}y_0y_1+a_{02}y_0y_2+a_{11}y_1^2+a_{12}y_1y_2+a_{22}y_2^2=0\} $$ where the $a_{ij}$ are homogeneous polynomials in the coordinates $s,t$ of the base $\mathbb{P}^1$, and $y_0,y_1,y_2$ are coordinates of the fiber. Assume that the $a_{ij}$ have degree $3,3,2,3,2,1$. So $X$ is a conic bundle and also a Del Pezzo surface of degree $1$. As such $X$ must be embeddable in $\mathbb{P}(1,1,2,3)$.

Following the standard description of the embedding one should take the two sections of $-K_X$ which are $y_0,y_1$ then look at the sections of $-2K_X$ and of $-3K_X$. The problems is that, since $-K_X$ spans an extremal ray of the effective cone of $X$, the only way I see of producing sections of powers of $-K_X$ is taking monomials of type $y_0^ay_1^b$. Clearly, in this way one can only obtain a map of $X$ onto a curve.

How can one write down the additional sections of $-2K_X$ and $-3K_X$, giving the embedding in $\mathbb{P}(1,1,2,3)$, in terms of the variables $s,t,y_0,y_1,y_2$?

Answer 1

Assume for simplicity $a_{22}=t$. Let $a_{ij,k}$ be the coefficients of $s^{\deg a_{ij}-k} t^k$ in $a_{ij}$. Then the missing section of $-2 K_X$ is

$$ \frac{ s a_{00,0} y_0^2 + s a_{01,0} y_0 y_1 + s a_{11,0} y_1^2 + a_{02,0} y_0 y_2 + a_{12,0} y_1 y_2}{t}$$

This is a meromorphic section of $-2K_X$ since the bidegree of each term matches the bidegree of $y_1^2$. It can only have a pole at $t=0$, but we chose the coefficients in the numerator to cancel the defining equation of $X$ when $t=0$ and thus vanish when $t=0$, so there is no pole where $t=0$.

For the missing section of $-3 K_X$ write $$q_1 = y_0^2 a_{00,1} +y_0 y_1 a_{01,1} + y_1^2 a_{11,1} $$ $$q_0 = y_0^2 a_{00,0} +y_0 y_1 a_{01,0} + y_1^2 a_{11,0} $$ $$l_0 = a_{02,0} y_0 + a_{12,0} y_1$$ $$l_1 = a_{02,1} y_0 + a_{12,1} y_1$$

so that the defining equation of $X$, mod $t^2$, specialized to $s=1$ is $ q_0 + t q_1 + (l_0 + t l_1) y_2 + t y_2^2 $. Multiplying this defining equation by $l_0 - t y_2$ we obtain, modulo $t^2

$ l_0 q_0 + l_0 t q_1 + (l_0^2 + t l_0 l_1 - t q_0) y_2 $ and rehomegenizing in $s$ and $t$ we obtain $ s l_0 (s q_0 + tq_1) + (s l_0^2 + t l_0 l_1 -t q_0) y_2$.

Then as long as $l_0 \nmid q_0$ we can take the missing section of $-3K_X$ to be $$ \frac{ s l_0 (s q_0 + tq_1) + (s l_0^2 + t l_0 l_1 -t q_0) y_2}{ t^2} $$ since the numerator vanishes to order $2$ when $t=0$.

To check linear independence, observe that none of the formulas for these sections involve $y_2^2$ so because the equation does involve $y_2^2$, any linear relation between these sections cannot involve the equation. So the section of $-2K_X$ is independent of $y_0^2, y_0y_1, y_1^2$ since its coefficient of $y_2$ is $l_0 / t \neq 0 $ and the section of $-3 K_X$ is independent of the previous ones because its coefficient of $y_2$ is $(sl_0^2+ tl_0 l_1 - tq_0 ) /t^2$ which is not a multiple of $l_0$.

Answer 2

Me and my graduate student have a method, which this margin is too narrow to contain. Please get in touch if this is of further interest.