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.


1 Answer 1


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 $d\geq 2$.

  • $\begingroup$ 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, 2017 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, 2017 at 15:44

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