Take $\mathbb{P}^2$ and blow up three points on the same line, let the resulting surfaces be $S$. Use $H$ to denote the hyperplane and $E_i,i=1,2,3$ to denote the exceptional divisors. Consider the complete linear system $|3H-E_1-E_2-E_3|$, gives a morphism $f:S\to \mathbb{P}^6$ where the image $f(S)$ is a degree $6$ singular del Pezzo obtained by contracting the $-2$ curve $H-E_1-E_2-E_3$ on $S$.

My question: if we consider the projection of $\mathbb{P}^6$ from the double point on $f(S)$, what would be a minimal compactification of the image of $f(S)$ in $\mathbb{P}^5$?

Clearly the three exceptional curves $E_i$ will be contracted by the projection since they are straight lines through the double point.



First you have to blowup the singular point of the sextic del Pezzo; then what you get is your surface $S$; the exceptional divisor of this blowup is the unique curve $E$ in the linear system $H - E_1 - E_2 - E_3$. Then the projection is given by the linear system $$ (3H - E_1 - E_2 - E_3) - E = (3H - E_1 - E_2 - E_3) - (H - E_1 - E_2 - E_3) = 2H. $$ The result, of course, is just the Veronese surface in $\mathbb{P}^5$.

  • $\begingroup$ Thank you very much for the answer. Just to make sure I understand this fully, in a similar situation if $S$ is blow up $\mathbb{P}^2$ at two infinitesimally near points, let $f(S)$ be the image of the $|3H-E_1-E_2|$ ($E_1,E_2$ exceptional and $E_1-E_2$ is the $-2$ curve.), then (compatification) of the image of projection from the double point of $f(S)$ would be the same as the image in $|3H-E_1-E_2-(E_1-E_2)|=|3H-2E_1|$? $\endgroup$ – Rust Q Feb 26 at 20:30

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