Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Good Morning,

I've been trying to brush up a bit on linear systems lately, and I've ran into the following (seemingly) contradictory statements. Hopefully someone can tell me where I'm going wrong here (and say how to go right).

Consider $H^0(\mathbb{P}^1, \mathcal{O}(3))$ which is the $k$ vector space spanned by degree 3 monomials in 2 variables, say $x^3, x^2y, xy^2, y^3$. Let $V$ be the sub-vector space spanned by the first 3 symbols; and consider the map induced by the corresponding two-dimensional linear system $L_V$:

$ \phi_V : \mathbb{P}^1 \to \mathbb{P}^2 $

$(x,y) \mapsto (x^3, x^2y, xy^2)=(x^2, xy, x^2)$

This map is easily seen to be an embedding, and hence we say that $L_V$ is very ample. However, $L_V$ has a base point; namely the point $(0,1)$ which is a common zero of every element of $V$.

Now, I'm under the impression that every very ample linear system should be base point free (since being very ample means the associated map is an embedding, which means in particular that it's a morphism, and hence the linear system is base point free.

Cheers, Robert

edit: I seem to have several different accounts - could someone please point me in the right direction as to how I can merge them all?

share|improve this question
    
@Robert Garbary when you write $(x^3,x^2y,x^y^2)=(x^2,xy,x^2)$ do you mean $(x^3,x^2y,x^y^2)=(x^2,xy,y^2)$(brutaly calnceling a $x$)? If it is so you have the Veronese embedding and actually you are looking at the complete $H^0(\mathbb{P}^1,\mathcal{O}(2))$ –  unknown Mar 17 '11 at 18:56

2 Answers 2

up vote 2 down vote accepted

The sections $s_1=x^3,s_2=x^2y,s_3=xy^2$ and $t_1=x^2,t_2=xy,t_3=y^2$ are not the same sections, in fact they correspond to different line bundles. The $s_i$ do have a base-point, and as you remark, they give the same map as the $t_i$ outside the base-locus.

And see this meta thread for how to merge your used acounts: http://tea.mathoverflow.net/discussion/605/merge-two-user-ids/#Item_0.

share|improve this answer

I will also point out something minor with regards to the statement: ``every very ample linear system should be base point free''. You need a complete linear system -- in particular, leaving out the $y^3$ term means the linear system you are using is not complete.

EDIT: To clarify the confusion, a correct statement would be: ``every very ample complete linear system is base point free''.

Of course, essentially every line bundle has non-complete linear subsystems. (Choose the empty system, or in most cases choose a system with only one element in it).

share|improve this answer
    
Ok, so I'm still a bit confused - is my linear system from above very ample or not? –  Robert Garbary Mar 17 '11 at 14:12
    
It is not very ample by the reason you gave: It doesn't even give a morphism. Karl points out that your linear system is not complete. –  J.C. Ottem Mar 17 '11 at 14:21
    
Indeed, that's right. Your linear system comes from a very ample, but it doesn't have everything it could from the very ample, and so it is not complete. –  Karl Schwede Mar 17 '11 at 14:25

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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