3
$\begingroup$

Let $p_1,\dots,p_r\in \mathbb P^2$ be points in general position, $r\leq 6$. Let $S$ be the blowup of these $r$ points, then the anticanonical divisor $K_S$ is very ample and it gives an embedding, in particular since $K_{\mathbb P^2}=\mathcal O(-3)$ and each point has an exceptional divisor $E_i$, we have that this embedding can be interpreted as the embedding in $\mathbb P^{9-r}$ by the linear system of cubics passing through $p_1,\dots,p_r$. This case is well explained in Beauville or Hartshorne books.

This is a del Pezzo variety in the classical sense (very ample anticanonical divisor). However, in modern texts, people started to allow the anticanonical divisor to be only ample, meaning a multiple of it gives the embedding. This includes $r=7,8$ as del Pezzo varieties. My question is, in such cases can we still understand the embedding as a linear system of cubics? I am sure for $r=8$ this cannot happen because of dimension counting. Is this embedding then a linear system of quartics or some other power?

$\endgroup$
2
  • $\begingroup$ Multiples of the anticanonical class $3H$ are of the form $3kH$, so it would be a linear system of (probably) sextics or degree $9$ curves. $\endgroup$ Commented Feb 28, 2023 at 9:19
  • $\begingroup$ This was my first guess, but I saw comments (without any proof) that it should be done with quartics, and I do not understand. $\endgroup$
    – ett
    Commented Feb 28, 2023 at 11:56

1 Answer 1

2
$\begingroup$

For $r=7,8$ the anticanonical system is only ample and not very ample, in both cases it can be seen from a dimension computation. Indeed $h^0(S,\mathcal{O}_S(-K_S))=10-r$, if the points are in general position, therefore, for $r=7,8$ we get $3,2$. In first case the image of the anticanonical map is contained in $\mathbb{P}^2$ so it cannot be an embedding, you can prove that this map is $2:1$ cover and compute the branching (quartic curve). In the second case the image has dimension $1$.

There are lots of embeddings for $S$, for example the natural ones, as suggested by Tabes, are given by $-2K_S, -3K_S$, they are very ample if (respectively) $r=7,8$. The point is that usually lower degree (and codimension) embeddings are easier to handle, so we seek for a very ample divisor as small as possible.

In your explicit situation, the next candidates are the ones given by quartic curves, in particular you can prove (by similar arguments as for cubics) that quartics through points $p_1,\dots, p_r$ give you an embedding for $r=7,8$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .