# blowing up general k points on the plane

Del Pezzo surfaces are obtained by blowing up $1 \leq k \leq 8$ points on general position in $\mathbb{P}^2$. What does it happen when the number of points is larger than nine? In this sense, Beauville's book in surfaces presents the topic in the context of linear system of cubics: Nine points in the plane determine a cubic curve, and del Pezzo surfaces $S_{9-k}$, with $k\leq 6$, are embedded into $\mathbb{P}^{9-k}$ by the linear system of cubics through the $k$ points. Is there a nice interpretation of the surfaces obtained by linear system of plane curves of degree $d$?

I suppose this is well known but I cannot find a reference. Thanks!!

• I'm having some trouble understanding the question. Are you asking whether the blowup of P^2 in more than 9 points is embedded in projective space by a linear system of curves of degree higher than 3? (By the way, your assertion that "del Pezzo surfaces S^9−k are embedded into P^9−k by the linear system of cubics through k≤9 points" is only true for k <= 6; after that the linear system of cubics through the points fails to be very ample.)
– user5117
Sep 16, 2012 at 21:17
• Another small correction: the first sentence should say "1<=k<=8".
– user5117
Sep 16, 2012 at 21:21

The canonical divisor of the blow up $\pi: X\to \mathbb P^2$ at $k$ ordinary points is $$K_X = -3\pi^*L +\sum_{i=1}^k E_i,$$ where $L\subset \mathbb P^2$ is a hyperplane and $E_i$ is an exceptional curve of the first kind. Choosing the representatives right and an easy computation shows that $$K_X^2 = 9 - k,$$ so $-K_X$ could possibly be ample only if $0\leq k\leq 8$.

As Artie points out, embedding into $\mathbb P^{9-k}$ only works for $0\leq k\leq 6$. It is easy to see that this cannot be true for $k>6$ as then $9-k\leq 2$ and there is no way you can embed a surface different from $\mathbb P^2$ into $\mathbb P^2$, $\mathbb P^1$, or $\mathbb P^0$.

So, the interesting question is what you get if $k=7$ or $8$.

It is relatively easy to see that $-K_X$ is not very ample (unlike in the $0\leq k\leq 6$ case): By looking at the short exact sequences, $$0\to \pi^*\omega_{\mathbb P^2}^{-1}(-\sum_{i=1}^r E_i) \to \pi^*\omega_{\mathbb P^2}^{-1}(-\sum_{i=1}^{r-1} E_i)\to \mathscr O_{E_r}\to 0$$ one can see easily that $$\dim H^0(X, \omega_X^{-1}) = 10-k.$$ In fact the $\mathbb P^{9-k}$ above is just the projectivization of this linear space.

Remark Since the OP is talking about embedding a blow-up of $\mathbb P^2$ at $k$ points into $\mathbb P^{9-k}$, I assume they mean the classical definition of Del Pezzos, although the fact why $k$ can't be bigger than $8$ works for any definition, in particular for the now commonly used one asking only for that $-K_X$ is ample.

• Sándor: are you saying that the blowup of P^2 in 7 or 8 points is not del Pezzo?
– user5117
Sep 16, 2012 at 21:45
• Artie, it depends on what definition of Del Pezzo you use. The classical one demanded that $-K_X$ is very ample, not just ample, but perhaps I should remove that statement anyway. Sep 16, 2012 at 21:47
• I must admit I've never seen the definition that demands -K very ample, but I am an ignorant youth. Anyway, as long as it's clear how the terminology is being used, I guess it's just a matter of taste.
– user5117
Sep 16, 2012 at 21:53
• Hartshorne, Remark V.4.7.1, page 401. Sep 16, 2012 at 21:56
• (And sorry if my first comment was annoying. I just wanted to make sure I understood your answer.)
– user5117
Sep 16, 2012 at 22:03

A minor correction: the blowup of $\mathbb{P}^{2}$ at 9 points cannot be del Pezzo, since its anticanonical class has self-intersection equal to 0.

Much of the recent interest in the blowup $X_{k}$ of $\mathbb{P}^{2}$ at $k \geq 9$ points in general position centers around the ample cone of $X_{k},$ rather than a specific embedding of $X_{k}$ in projective space.

Let $H$ be the pullback of the hyperplane class via the blowup ${\pi}: X_{k} \rightarrow \mathbb{P}^{2},$ and let $E=\sum_{i=1}^{k}E_{i}$ be the sum of the $k$ exceptional divisors on $X_{k}.$ The anticanonical class $3H-E$ fails to be ample for $k \geq 9,$ but we can instead ask the following: for which positive integers $d,r$ is the divisor $dH-rE$ ample?

Since $H$ spans a boundary ray of the ample cone of $X_{k},$ we know that $H-tE$ is ample for $0 < t << 1,$ e.g. that $dH-rE$ is ample for $d >> r.$ So what we are really interested in is $$t_{k}:=\sup \{ t > 0 : H-tE \hskip5pt {\rm ample} \}$$ An upper bound for $t_{k}$ may be obtained from the positive value of $t$ for which $(H-tE)^{2}=0,$ i.e. ${1}/{\sqrt{k}}.$

$\textbf{Nagata's conjecture:} \hskip10pt t_{k}=1/{\sqrt{k}}.$

This statement holds when $k=m^{2}$ is a perfect square which is at least 9; this can be seen by looking at the ample cone of the blowup of $\mathbb{P}^{2}$ at a general complete intersection of two degree-$m$ plane curves and noting that the ample cone of a surface can only shrink upon specialization.

There is a large body of work on Nagata's conjecture and its generalizations. A nice overview can be found in "Remarks on the Nagata Conjecture" by B. Strycharz-Szemberg and T. Szemberg, available at

www.uni-due.de/~mat903/preprints/nagata1.pdf

• Nice answer! Let me mention that for k=9 a bit more than Nagata's conjecture is known: actually the structure of whole ample cone is known, by a result of Borcea (a paper in Crelle, around 1991, whose title I forget). In recent years, de Fernex has given a partial description of the K-positive part of the ample cone for k=10,11 (I think), but a description of the whole cone is not known.
– user5117
Sep 16, 2012 at 21:57
• Thanks! I wasn't aware of either of these--I'll have to take a look. Sep 16, 2012 at 22:30