5
$\begingroup$

Let $X$ be a smooth projective surface not contained in $\mathbb{P}^3$. Is there any known condition on $X$ under which I can embed it into $\mathbb{P}^3$ such that the its image contains at most simple surface singularities. Any idea or references on this topic will be very helpful.

$\endgroup$
5
  • $\begingroup$ I am not an expert on this, but at least over C, cant you always embed into P^3? Just by taking lines through a point which is not on any chord or tangent reduces dimension by by one. Dimension the variety of chords and tangents shows you can keep finding such a point until you get down to P^3. $\endgroup$
    – Steve
    May 10, 2013 at 17:45
  • 2
    $\begingroup$ Related: mathoverflow.net/questions/32938/… $\endgroup$
    – user5117
    May 10, 2013 at 18:41
  • $\begingroup$ @Steve, definitely not every surface embeds in $\mathbb{P}^3$ (for example, Abelian surfaces). With regards to your argument, you can't find lines which intersect your variety only once (without tangency) in general. $\endgroup$ May 10, 2013 at 18:52
  • 1
    $\begingroup$ What is your definition of simple surface singularity? $\endgroup$
    – rita
    May 10, 2013 at 20:52
  • $\begingroup$ oops. I was thinking geometric surface, but algebraic curve. $\endgroup$
    – Steve
    May 10, 2013 at 21:49

1 Answer 1

9
$\begingroup$

Let $S' \subset \mathbb{P}^3$ be the birational projection of a smooth surface $S \subset \mathbb{P}^4$. The generic projection theorem of Gruson-Peskine (http://arxiv.org/abs/1010.2399v2) tells you that either $S'$ is smooth or has a curve of double points.

For instance, if $S$ is the Veronese surface in $\mathbb{P}^4$, then its projection in $\mathbb{P}^3$ (the Steiner surface) has a curve of double points.

So the answer to your question is "never", unless your surface embedds in $\mathbb{P}^3$ (which you don't want).

Edit : If by "embed into $\mathbb{P}^3$", you mean project down to $\mathbb{P}^3$ from another embedding, my answer is correct, but there is a much simpler proof : you can always embed your surface in $\mathbb{P}^5$ because the secant variety of a surface in any projective space has at most dimension $5$. Then a projection to $\mathbb{P}^3$ factors as a projection to $\mathbb{P}^4$ and a projection to $\mathbb{P}^3$. If you choose your projection generically from $\mathbb{P}^5$ to $\mathbb{P}^4$ then it has only simple singularities in $\mathbb{P}^4$. But a simple count of dimension shows that for any point $p \in \mathbb{P}^4$, there is a $1$-dimensional family of bisecants to $S$ passing through $p$. This gives you a curve of double points in $\mathbb{P}^3$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.