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I'm trying to understand the relationship between the different models of del Pezzo surfaces of degree $2$.

Let $k$ be a field of characteristic not equal to $2$. Usually, del Pezzo surfaces of degree $2$ are considered as hypersurfaces of degree $4$ in $\mathbb{P}(1,1,1,2)$: $$S: \quad w^2 = f(x,y,z)$$ with $\deg f = 4$.

There are however other natural models, such as hypersurfaces of bidegree $(2,2)$ in $\mathbb{P}^1 \times \mathbb{P}^2$: $$q(s,t,x,y,z) = 0,$$ where $q$ is bihomogeneous of degree (2,2) in $(s,t)$ and $(x,y,z)$, respectively.

Call these models of type $1$ and $2$, respectively. I'm trying to understand how one goes between these different models.

Given a surface of type $2$, how does one construct a model of type $1$ (i.e. a model in weighted projective space)?

I know how one should go about doing this. The anticanonial class $-K_S$ is the fibre of the projection onto $\mathbb{P}^2$. So a basis for the anticanonical sections is $x,y,z$. Now the space $H^0(S, -2K_S)$ is $7$-dimensional, so there should be a section $w$ in here which is linearly indenpendent from the obvious sections $x^2,xy,xz,y^2,yz,z^2$. With this in hand, we obtain the required map $$S \to \mathbb{P}(1,1,1,2), \quad (s,t) \times (x,y,z) \mapsto (x,y,z,w).$$ But what is $w$? I don't see how to determine $w$ explicitly.

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    $\begingroup$ Is it just the following: write $q(s,t,x,y,z) = A(x,y,z)s^2 + B(x,y,z)st + C(x,y,z)t^2$ and let $f(x,y,z)$ be $B^2-4AC$? $\endgroup$
    – Jason Starr
    Commented Jun 9, 2016 at 15:04
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    $\begingroup$ I guess you can find $f$ up to constants because it cuts out the ramification locus of the anticanonical map. View $q$ as a quadratic form in $s,t$ with coefficients involving $x,y,z$; then its discriminant is a polynomial of degree $4$ in $x,y,z$, which is your $f$. $\endgroup$
    – Martin Bright
    Commented Jun 9, 2016 at 15:07
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    $\begingroup$ @MartinBright. That is what I was thinking as well. $\endgroup$
    – Jason Starr
    Commented Jun 9, 2016 at 15:07
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    $\begingroup$ Beaten to it! Yes, what Jason Starr said. $\endgroup$
    – Martin Bright
    Commented Jun 9, 2016 at 15:08
  • $\begingroup$ This looks great! But there is still something I'm confused about: how does one view $\sqrt{B^2 - 4AC}$ (in Jason's notation) as a global section of $-2K_S$? It looks horribly undefined as written. $\endgroup$
    – Daniel Loughran
    Commented Jun 9, 2016 at 20:05

1 Answer 1

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One can transform birationally $S \subset \mathbb{P}^1 \times \mathbb{P}^2$ to $S' \subset\mathbb{P}(1,1,1,2)$ as follows. Choose a point $P \in \mathbb{P}^1$ such that the conic $$ C := (P \times \mathbb{P}^2) \cap S $$ is smooth. Blow up $P \times C \subset \mathbb{P}^1 \times \mathbb{P}^2$. Then blow down the proper preimage of $\mathbb{P}^1 \times C$. Then you will get $\mathbb{P}_{\mathbb{P}^2}(O \oplus O(-2))$. The strict transform of $P \times \mathbb{P}^2$ is the exceptional section and it does not intersect the strict transform of $S$. If you contract it, you get $\mathbb{P}(1,1,1,2)$ with an embedded image of $S$.

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