3
$\begingroup$

Let $k$ be a fixed algebraically closed field and $X/k$ an irreducible scheme smooth and proper over $k$. Can there exist a line bundles $\mathcal{L}, \mathcal{M}$ and an integer $m > 0$ so that

1.) $\dim_k \Gamma(\mathcal{L}) = 0$

2.) $\dim_k \Gamma(\mathcal{M}) > 0$

With $\mathcal{L}^m \cong \mathcal{M}^m$. If not, does an example exist if you drop smoothness (then replace $\mathcal{M}$ with the line bundle of an effective Weil divisor) or the condition that $k$ is algebraically closed?

$\endgroup$

1 Answer 1

7
$\begingroup$

Yes. Take $\mathcal L$ the trivial line bundle, with a one-dimensional space of global sections, and $\mathcal M$ a nontrivial torsion line bundle, so $\mathcal M^k=\mathcal L$.

Then $\Gamma(\mathcal M)$ is certainly zero-dimensional, since otherwise $\mathcal M$ would have a nonvanishing section and be trivial or a somewhere vanishing section and then $\mathcal L$ would have a somewhere vanishing section and be nontrivial.

This provides a counterexample. Such examples exist on any smooth proper variety with a nontrivial Picard variety, meaning $\operatorname{dim}_kH^1(X,\mathcal O_X)>0$, such as curves of positive genus.

$\endgroup$
14
  • 1
    $\begingroup$ Yes. Take a curve of genus $2$ or higher, then most degree $1$ line bundles have a zero-dimensional space of global sections. Take the $n$th tensor power of this line bundle for $n\geq g$. Now it has degree $n$ and so a positive-dimensional space, by Riemann-Roch. $\endgroup$
    – Will Sawin
    Oct 27, 2012 at 18:37
  • 3
    $\begingroup$ Yes. Take a curve of genus $g\geq 2$. Then the space of line bundles of degree $1$ is a complex torus of dimension $g$. The space of line bundles which have a nontrivial global section is the image of the curve under the map sending a point to the line bundle corresponding to its divisor, so has dimension $1$. Fix one such line bundle. Now consider all the line bundles that are it tensor a torsion line bundle. These correspond to translations of the torus by torsion points on its Jacobian, so these points are dense, so at least one of them does not have a nontrivial global section. $\endgroup$
    – Will Sawin
    Oct 27, 2012 at 20:04
  • 1
    $\begingroup$ It is not too hard to see by counting that $m=2$ suffices on a hyperelliptic curve. If $f$ is the hyperelliptic projection to $\mathbb P^1$ there are $2g+2$ points such that $\mathcal O(P)^{\otimes 2}=f^*\mathcal O(1)$, but there $2^{2g}$ degree $1$ divisors such that $\mathcal L^{\otimes 2} =f^* \mathcal O(1)$, so there must be one line bundle of that form with a nontrivial global section and one line bundle of that form without a nontrivial global section. $\endgroup$
    – Will Sawin
    Oct 27, 2012 at 20:13
  • 1
    $\begingroup$ If there were any more global sections you would get a degree $2$ map to $\mathbb P^2$, which makes your curve a conic in $\mathbb P^2$, which makes it rational. I guess this is a special case of Clifford's Theorem (Hartshorne 4.5.4) + Riemann-Roch: By Clifford's theorem, if $f^* \mathcal O(1)$ is special then the dimension of the space of global sections is at most $2/2+1=2$. On the other hand if it's general then the dimension of the space of global sections is $2+1-g=3-g\leq 1$. So the dimension of the space of global sections is exactly $2$, so all of them come from the pullback. $\endgroup$
    – Will Sawin
    Oct 28, 2012 at 3:21
  • 1
    $\begingroup$ Another thing to generalize it from is Lemma 4.5.1, which says the canonical linear system has no base points, plus Riemann-Roch, to show the dimension of the space of global sections of divisor of degree $d>0$ on a curve of genus $g\geq 2$ is at most $d$. $\endgroup$
    – Will Sawin
    Oct 28, 2012 at 3:23

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.