We know that there are 27 lines lie in a smooth cubic surface. And we can get a cubic surface by blowing up six general points, say $P_1,P_2,P_3,P_4,P_5$ and $P_6$, in $\mathbb{P}^2$.

Those exceptional lines are 6 out of 27.

Let $L_{ij}$ be the line in $\mathbb{P}^2$ through $P_1$ and $P_2$, then its strict transform is a line. And there are $C_2^6=15$ lines out of 27.

Let $C_i$ be the plane conic through those 6 points but $P_i$, the strict transform is a line. And there are $C_1^6=6$ lines out of 27.

My question is How can we see that strict transform of $L_{ij}$ and of $C_i$ are linear subspace in $\mathbb{P}^3$ where the cubic surface is embedded.

The embedging of the blownup surface is given by Which divisor? $|H+\sum(E_i)|$ ?

And also how the strict transform of a conic becomes a line in $\mathbb{P}^3$


  • 1
    $\begingroup$ The linear system is $3H-\sum P_j$ down in $\mathbb P^2$ or $3\pi^*H-\sum E_i$ up on the cubic, where $\pi$ is the projection. The image of $L_{ij}$ meets a hyperplane only once, since it meets a cubic in $\P^2$ in three points, but two of those intersections get "blown apart" in the blowup. Upstairs, your intersection with the hyperplane class is $3\cdot 1-2$, since the strict transform of the line meets two of the exceptional curves. A similar calculation holds for the conics. $\endgroup$ Oct 13 '15 at 5:24

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