0
$\begingroup$

Suppose $X$ is a non-singular irreducible projective variety over an algebraically closed field of characteristic zero. Let $L$ be an ample and globally generated line bundle on $X$. Then the complete linear system of $L$ gives a finite morphism (because of ampleness) to Projective space.

Can we find ample globally generated line bundles $L$ such that $h^0(X,L) = n+1$ where $n$ is the dimension of $X$?

We know that because $L$ is ample, $h^0(X,L)\geq n+1$. But in what situations/under what conditions does equality hold. In case of curves I can think of examples. Does this happen for higher dimensional varieties.

|cite|improve this question
$\endgroup$
1
$\begingroup$

More often than not, this can not be done. For a definite example (similar examples are plenty) take a smooth hypersurface $X$ of degree $d$ in $\mathbb{P}^4$. It is well known that the Picard group is just $\mathbb{Z}\mathcal{O}(1)$ by Grothendieck-Lefscetz. The ample line bundles are $\mathcal{O}_X(r)$ with $r>0$ and none of them have the property that the dimension of global sections is $\dim X+1=4$, if $d\geq 2$.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ if $X$ is a variety admitting a finite map $f$ to Projective space of the same dimension as $X$, and if $f_*O_X$ is again the structure sheaf. Then it looks like I can find line bundles with the required condition $\endgroup$ – user349424 Apr 15 '17 at 15:07
  • 1
    $\begingroup$ As my answer said, the answer is no. Take the hypersurface as above, take any finite map to projective 3-space, but you will not get an ample bundle with 4 sections. $\endgroup$ – Mohan Apr 15 '17 at 15:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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