Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For a variety, if it has a base point free line bundle, then one can define a morphism from the variety to a $\mathbb{P}^n$. And if a variety is a projective variety(in the sense of Hartshorne), it is equivalent to have a very ample line bundle.

In the same vein, I was wondering if there are properties of line bundle corresponding to birational map. To be precisely:

Let $V$ be a variety, $L$ be a line bundle on $V$. Which conditions are needed to ensure the $map$ defined by $L$ (i.e. using all the basis of $H^0(V,L)$)is a birational map to its image in $\mathbb{P}^n$? Here $n=dim\ H^0(V,L)-1$.

share|improve this question
3  
Note that $L$ doesn't define a unique morphism to a projective space. You need to specify finitely many global sections which generate $L$. –  user18119 Jan 1 '13 at 22:02
    
@QiL But I guess $L$ will define a morphism up to a PGL(n) action. –  Li Zhan Jan 2 '13 at 1:15
1  
Yes, but in general you don't need to take a basis of $H^0(V, L)$. Any subset of sections generating the sheaf $L$ induces a morphism to some projective space. –  user18119 Jan 2 '13 at 10:46
    
@Should I require the set defined the morphism is exactly the set of $H^0(V,L)$? –  Li Zhan Jan 2 '13 at 19:16
    
This is a winy detail, but i think you meant $n = dim H^0(V,L) - 1$ instead of $n = dim H^0(V,L)$.. I though i'd point it out, but its not really important of course.. –  Joachim Jan 3 '13 at 12:41

1 Answer 1

up vote 4 down vote accepted

I don't have a satisfying answer. Anyway here is one.

Let $L$ be a basepoint-free line bundle on a projective variety $V$ over a field $k$. Let $$\pi : V \to \mathbb P(H^0(V, L))\simeq \mathbb P^n$$ be the morphism associated to the complete linear system $|L|$. If we choose a basis $s_0, \dots, s_n$ of $H^0(V, L)$, then set-theoretically $$\pi(v)=[s_0(v):\dots : s_n(v)].$$ If $F(X_0,\dots, X_n)$ is a homogeneous polynomial of degree $m$, then $\pi^{-1}(D_+(F))$ is $V_f$ where $f=F(s_0,\dots, s_n)\in H^0(V, L)^{\otimes m}$ and $$ V_f:=\{ v\in V \mid f(v)\ne 0\}.$$ The morphism $V_f\to D_+(F)$ is given by $$ k[T_0, \dots, T_n]_{(F)} \to O_V(V_f), \quad P/F^r \mapsto P(s)/F(s)^r.$$

We know that if $L$ is very ample, then for any section $f\in H^0(V, L)$, $V_f$ is an affine open subset and the above homomorphism is surjective. The same is true if we take $f$ in $H^0(V, L)^{\otimes m}$ for any $m\ge 1$.

In general $\pi$ is birational if and only if there exists a dense open subset $U$ of $\mathbb P^n$ such that $\pi: \pi^{-1}(U) \to U$ is a closed immersion (an isomorphism from $\pi^{-1}(U)$ onto $U\cap \pi(X)$). The complement of $U$ is contained in a hypersurface $V_+(F)$ of deree $m$. Shrinking $U$ if necessary, we can suppose $U=D_+(F)$. Then $\pi^{-1}(U)=V_f$ where $f=F(s)$ as above. The condition for $V_f\to U$ to be a closed immersion is: $V_f$ is affine and $O_V(V_f)$ is generated, as $k$-algebra, by $H(s)/f(s)$ where $H$ runs through monomials of degree $m$.

Summarizing the above discussion, we can say that $\pi$ induces a birational morphism from $V$ to its image if and only if there exists $m\ge 1$ and $f\in H^0(V, L)^{\otimes m}$ such that:

  1. $V_f$ is affine,
  2. $O_V(V_f)$ is generated by $H(s_0,\dots, s_n)/f$ where $H$ are the monomials of degree $m$.

Everything remains true if we replace $H^0(V, L)$ with a vector subspace generating $L$.

share|improve this answer
    
Dear QiL, Thank you so much! I am very happy to have your answer,it clarified a lot thing about birational morphisms. I did not get you in two places in your answer, could you please explain a little bit more? (1)you wrote: $\pi:\pi^−1(U) \to U$ is a closed immersion. Is this because we assume X is proper, and thus $\pi(X)$ is closed? (2) you wrote: The complement of $U$ is contained in a hypersurface $V_+(F)$ of degree $m$. I cannot see why this is true, and moreover, I cannot see why this is used in the following argument? –  Li Zhan Jan 4 '13 at 1:52
    
Dear @LiZhan: (1) yes because $\pi(X)$ is closed. (2) The complement is defined by a non-zero ideal $I$. Let $F\in I$ be homogeneous and non-zero, then $V_+(F)$ contains this complement. Subsequently the complement is replaced by $V_+(F)$ so $U$ is affine. This implies that $V_f=\pi^{-1}(U)$ is affine. –  user18119 Jan 4 '13 at 13:56

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.