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Consider the standard Cremona involution $i:\mathbb{P}^2\dashrightarrow \mathbb{P}^2$, $[x:y:z]\rightarrow [yz:xz:xy]$.

Let $Y$ be the blow-up of $\mathbb{P}^2$ in the three base points of $i$, so that $i$ lifts to an automorphism $\widetilde{\imath}:Y\rightarrow Y$, and $X := Y/\left\langle \widetilde{\imath}\right\rangle$ the quotient of $Y$ by the action of $\widetilde{i}$.

What kind of surface is $X$? Can it be described explicitly?

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    $\begingroup$ $i$ is a birational involution, so what do you mean by the quotient? $\endgroup$
    – abx
    Commented Jun 13 at 14:28
  • $\begingroup$ I made my question more precise. $\endgroup$
    – Robert B
    Commented Jun 13 at 14:48
  • $\begingroup$ There is no such thing as the standard involution; the answer to your question depends very much of the way you write it. For instance, the way you define it, $i$ fixes the curve $y^2=xz$, while the usual form has only 4 fixed points. The quotient surface will be different in each case. $\endgroup$
    – abx
    Commented Jun 13 at 19:25
  • $\begingroup$ What's the usual form form you? $\endgroup$
    – Robert B
    Commented Jun 13 at 20:01
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    $\begingroup$ What do you want to know, precisely? It is a log del Pezzo surface with four $A_1$ singularities. It had an embedding in $\mathbb{P}^5$ as a codimension two subvariety of $\text{Sym}^2(\mathbb{P}^2)$. If you want the defining equations, you can get them from a computer algebra program. $\endgroup$
    – Jason Starr
    Commented Jun 14 at 1:51

1 Answer 1

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The quotient surface is the cubic $$ u_0u_1u_2 + u_0u_1u_3 + u_0u_2u_3 + u_1u_2u_3 = 0, $$ the only cubic with four nodes.

EDIT. To see this one can first observe that the involution acts biregularly on the sextic del Pezzo surface and has exactly 4 fixed points. It follows that the quotient is a suface with ample anticanonical class of degree $6/2 = 3$ with 4 nodes. Finally, there is only one such cubic surface up to isomorphism, and the above equation gives it.

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  • $\begingroup$ Thanks. How did you get this equation? $\endgroup$
    – Robert B
    Commented Jun 14 at 8:18
  • $\begingroup$ I added some explanation to my answer. $\endgroup$
    – Sasha
    Commented Jun 14 at 17:17

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