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Consider $X$ a smooth cubic surface in $\mathbb{P}^3$, and let $l_1,...,l_6$ be six disjoint lines contained in $X$.

What is the linear system giving the blow-down map $X \to \mathbb{P}^2$, so that the lines $l_k$ are contracted to points ?

The other way round is well-known : if $p_1,\dots,p_6$ are six points in general position, the rational map $\mathbb{P}^2 \to \mathbb{P}^3$ obtained by the linear system of cubic containing the six points has image a cubic surface.

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  • $\begingroup$ In what terms? How do you describe the Picard group of $X$? $\endgroup$
    – abx
    Commented Jan 28, 2023 at 5:15
  • $\begingroup$ For example, one can take the base given by Hartshorne in the book "algebraic geometry", the chapter on cubic surfaces, this is given by the hyperplane section h, and the six exceptional curves. $\endgroup$
    – Xavier49
    Commented Jan 28, 2023 at 16:14
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    $\begingroup$ Then your map is given by the linear system $\lvert h \rvert$… $\endgroup$
    – abx
    Commented Jan 28, 2023 at 17:13
  • $\begingroup$ Thank you very much $\endgroup$
    – Xavier49
    Commented Jan 28, 2023 at 19:09
  • $\begingroup$ Your "hyperplane section", is apparently π*, via the blow down map π, of the class of a line in P^2, Hartshorne p. 401, Notation 4.7.3. Hence it is by definition the linear series defining the blow down map. Alternatively, since the inverse map is given by cubic plane curves with base points, the image in P^3 of a line by this map, must be the divisor you want. The class of this rational cubic curve on X can be represented by a line m on X and two disjoint exceptional lines e1, e2, meeting m, e.g. I.e. then m + e1 + e2 ≈ π* of a line in P^2 meeting 2 blownup pts. $\endgroup$
    – roy smith
    Commented Jan 31, 2023 at 22:46

1 Answer 1

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As you probably know, if a representation of $X$ as blowup is given, $$ K_X = -3h + \sum l_i, $$ where $h$ is the line class. Consequently, the linear system $$ |-K_X + \sum l_i| $$ gives the required contraction.

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  • $\begingroup$ I do not see the consequence, can you explain please ? $\endgroup$
    – Xavier49
    Commented Jan 28, 2023 at 16:12
  • $\begingroup$ Just observe that all divisors in this linear system are pullbacks from the plane. $\endgroup$
    – Sasha
    Commented Jan 28, 2023 at 16:17
  • $\begingroup$ Thank you ; following abx idea it is rather $|-K_X+\sum l_i|$ but divided by $3$. $\endgroup$
    – Xavier49
    Commented Jan 28, 2023 at 19:08
  • $\begingroup$ These two define the same contraction. $\endgroup$
    – Sasha
    Commented Jan 28, 2023 at 19:21
  • $\begingroup$ yes, thank you (using your divisor is the map to $\mathbb{P^2}$ followed by a Veronese embedding) $\endgroup$
    – Xavier49
    Commented Jan 28, 2023 at 19:47

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