For sure answers to my questions are well known - but I never saw them anywhere.

Let $X$ be a smooth projective (or just proper) variety over an algebraically closed field $k$. Let $A_i$ be the subset of $\text{Pic }X$ of all line bundles $L$ with nonzero $H^i(X, L)$.

General question. What does $(\text{Pic }X, A_0, \ldots, A_d)$ ($d=\dim X$) look like when seen from far far away?

Here are some more specific questions:

Question 1. Does the property $L\in A_i$ depend only on the numerical class of $L$, for $L$ ,,large enough''? Precisely: does there exist a bounded region $C$ in $NS(X)$ such that for all $L\in \text{Pic }X$, $M\in \text{Pic}^\tau X$ we have $\dim H^i(X, L) = \dim H^i (X, L\otimes M)$ when $L\notin C$?

Let $B_i$ be the image of $A_i$ in $NS(X)$.

Question 2. Does $B_i$ look like a union of finitely many ,,translated strictly convex cones''?

For example, when $X=G/B$ then $B_i=A_i$ is a union of the interiors of the Weyl chambers corresponding to the length $i$ elements of the Weyl group, shifted by half the canonical class.

Question 3. What can one say about the intersections of $B_i$ (again far away from zero)?

E.g. in the above example of $G/B$, there is at most one non-vanishing cohomology group. This seems to hold for many varieties as soon as we are ,,far away from zero''. So in addition to ,,ample directions'' and ,,anti-ample directions'' (Serre duality) there seem to be ,,$H^i$-directions'' as well... As far as I remember, something similar holds for abelian varieties.

Motivation. The only examples I know pretty well are curves, abelian varieties and $G/B$ and in a sense they look similar.

Note. I'm sure MO users will quickly post counterexamples or comment on how I could make the questions more precise or reasonable. If that is okay with MO policy, I plan to edit the question to make it more complete and less silly.

  • $\begingroup$ Yes, that is certainly in accordance with MO policy. $\endgroup$ – Will Sawin Aug 28 '12 at 4:08
  • $\begingroup$ Is $f$ assumed to be linear? Nonlinear but convex? $\endgroup$ – Will Sawin Aug 28 '12 at 5:08

Here's an example where $B_1$ is not a finite union of cones: Let $X$ be a K3 surface of Picard number 3, such that the cone of effective divisors, $Eff(X)=Nef(X)$ is one of the components of the $[D\in NS(X) | D^2\ge 0]$ (for example, a K3 surface without $(-2)$-curves, such surfaces can be constructed as in this paper). In this case $Eff(X)$ is non-rational polyhedral. Then it is easy to see that $$ B_0=Eff(X), B_2=-Eff(X) \mbox{ and } B_1= (B_0\cup B_2)^c $$In particular, $B_1$ is not a finite union of cones.

As for the the 'General question', I think the shapes of the $B_i$ are related to Alex Kuronya's asymptotic cohomological functions. These are basically higher cohomology versions of the volume function of a big line bundle and measure the asymptotic growth of cohomology. The definition is $$ \hat{h}^{i}(X,D) = \limsup_{m}\frac{h^i(X,O_X(mD))}{m^n/n!} . $$One of the main theorems in his paper is that the $\hat{h}^i$ define continuous functions on the Neron-Severi space $NS(X)=A^1(C)\otimes \mathbb{R}/\equiv$. The vanishing of these functions should be related to your question.

See his paper for a lot of examples of asymptotic cohomology vanishing (flag varieties, abelian varieites,..). In most of these examples it is clear that the regions of vanishing cohomology are unions of convex cones.

These functions have been used to study certain positivity conditions of line bundles. For example, in the paper "Higher cohomology of divisors on a projective variety" by T. de Fernex, A. Kuronya, R. Lazarsfeld, the authors show that a divisor $D$ is ample if and only if the higher asymptotic cohomological functions vanish in a neighbourhood of $D$ in $NS(X)$.

There is also the concept of $q-$ampleness, introduced by Demailly-Peternell-Schneider, Arapura, and Totaro among others. This is a generalization of the notion of an ample line bundle in the sense that high tensor powers of a line bundle are required to kill cohomology of coherent sheaves in degrees $>q$ (so $0$-ampleness coincides with ordinary ampleness.). This is related to $\hat{h}^i(X,D)$ in the sense that it is expected that the local vanishing of the $\hat{h}^i(X,D)$ in degrees $>q$ is equivalent to $q$-ampleness of $D$. In general it is known (and easy to prove) that the cones of $q$-ample line bundles are star-shaped.

  • $\begingroup$ wouldn't that expectation imply that the trivial bundle of $\mathbb P^1$ is $0$-ample? $\endgroup$ – Will Sawin Aug 28 '12 at 15:08
  • $\begingroup$ What do you mean? $\hat{h}ˆ1(\mathbb{P}^1,D)$ is certainly not vanishing in a neighbourhood of $O_\mathbb{P^1}$? $\endgroup$ – J.C. Ottem Aug 28 '12 at 15:30
  • $\begingroup$ What is a neighborhood of $\mathcal O_{\mathbb P^1}$? It seems to me it should be just $\mathcal O_{\mathbb P^1}$, which has no cohomology so every tensor power of it has no cohomology. $\endgroup$ – Will Sawin Aug 28 '12 at 15:31
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    $\begingroup$ I mean a neighbourhood of the class of $O_{P^1}$ in $NS^1(P^1)$, so $\mathbb{R}$-divisors with negative coefficients are allowed. $\endgroup$ – J.C. Ottem Aug 28 '12 at 15:35
  • $\begingroup$ Sorry I misunderstood. $\endgroup$ – Will Sawin Aug 28 '12 at 16:13

There are many examples for which Question 1 fails.

Let $Z$ be such that $\mathrm{Pic}^\circ Z\neq 0$. Then if $\mathscr N\in \mathrm{Pic}^\circ Z$ and $\mathscr N\not\simeq \mathscr O_Z$, then $H^0(Z,\mathscr N)=0$, so if $X=Z\times Y$ for any (say) projective $Y$, then for any ample line bundle $\mathscr L$ on $Y$, the line bundles $p_2^*\mathscr L$ and $p_2^*\mathscr L\otimes p_1^*\mathscr N$ are numerically equivalent, but the first may have a non-zero $H^0$, while the second cannot.

By Serre duality this gives a counter-example for $H^{\dim X}$ as well. I think that with a little more work one can prove that the same idea gives counter-examples for any $H^i$. For that it might be useful to know how these $A_i$ behave on $\mathrm{Pic}^\circ X$ in general.

In fact, Green and Lazarsfeld started the study of these sets in this paper. They use the notation $$ S^i(X)=\{ L\in \mathrm{Pic}^\circ X | H^i(X,L)\neq 0 \} $$ and prove among other things that

Theorem (Green-Lazarsfeld, 1987) $$ \mathrm{codim}( S^i(X), \mathrm{Pic}^\circ X )\geq a(X)-i $$ where $a(x)=\dim (\mathrm{im}[ X\to \mathrm{Alb}(X)])$ is the dimension of the image of the Albanese morphism of $X$.

In particular, this implies that $H^i(X,L)=0$ for $i<a(X)$ and for a general $L\in \mathrm{Pic}^\circ X$.

So if in the above example you take $Z$ to have maximal Albanese dimension (say an Abelian variety), $Y$ to be the same dimension as $Z$ and $\mathscr N\in \mathrm{Pic}^\circ Z$ general, then the above example should work for all $i\leq \dim Z$ and give the other half of the spectrum by Serre duality.

For Question 2, I don't think you can do this with one $f$ for all $i$'s simultaneously. For individual ones, I think you run into the same problem as in Question 1.

Question 3 seems hard, but it is also pretty vague.

I think the main issue with what I am guessing you want is that "far away" from zero in $NS(X)$ depends a lot on what direction you are going. Going in the direction of something ample might give you reasonable results for these kind of questions, but going in other directions is highly unpredictable.

  • $\begingroup$ Thank you! Regarding Question 2: my $f$ was supposed to depend on $i$. I will edit the question to clarify this. $\endgroup$ – Piotr Achinger Aug 28 '12 at 5:00
  • $\begingroup$ You can get different $H^i$s using the Kunneth formula. Take an $i$-dimensional projective variety times a $n−i$ -dimensional projective variety, and tensor a very minus ample line bundle on the first (with only $H^i$ nontrivial) with a nontrivial numerically-equivalent-to-0 line bundle on the second. By the Kunneth formula, $H^i$ is $0$, but if you replace the second with the trivial line bundle you get nontrivial $H^i$. $\endgroup$ – Will Sawin Aug 28 '12 at 5:07
  • $\begingroup$ @Will: yeah, that works, too. I like the phrase "very minus ample". :) For some reason I was trying to avoid the Künneth formula. In any case, it seems that this is very unlikely to hold in any non-trivial case. $\endgroup$ – Sándor Kovács Aug 28 '12 at 6:34

The answer to question $1$ is no. For a product of two curves $C_1 \times C_2$ of positive genus, there are infinitely many numerical equivalence classes where $H^0(X,L)$ is not determined by the numerical equivalence class, since for any very ample line bundle $L_1$ on $C_1$ and degree $0$ nontrivial line bundle $L_2$ on $C_2$, $H^0(X,L_1^{\otimes n} \otimes L_2)=0<H^0(X,L_1^{\otimes n} \otimes \mathcal O_X)$, but the two bundles are numerically equivalent, so there are infinitely many numerical equivalence classes without this property.

I would expect that instead you can make the statement: for any numerical equivalence class $D_1$ and ample numerical equivalence class $D_2$, for sufficiently large $n$, $D_1+nD_2$ and $D_1-nD_2$ have this property, by a vanishing of cohomology argument.

Question 2: I think $B_0$ and $B_d$ will look, at least much more like a cone than the middle $B_i$. Consider $\mathbb P^1 \times \mathbb P^1$. $B_0$ and $B_2$ are cones, but $B_1$ is not.

For $B_0$ and $B_d$, choose any numerical equivalence class of curves $C$ that contains a curve missing each codimension $1$ subset. The for $\mathcal O(D)$ to have a nonzero global section, it must certainly have a nonzero global section restricted to some curve in $C$, so $C \cdot D\geq 0$. By Serre duality, for $\mathcal O(D)$ to lie in $B_d$, we must have $C \cdot D \leq C \cdot K$. So we can choose $f= C \cdot D$ and $-C \cdot D$ and $c=0$ and $- C \cdot K$ respectively.

To construct such a $C$, if the variety is projective, we can take the intersection with a linear subspace of the appropriate dimension and get a curve we can move around anywhere. If it is not then I think a birational transformation to a projective variety will work, but I'm not sure.

For an example where $B_i$ is not a union of cones, consider the self-product of an elliptic curve with complex multiplication by a ring of integers $\mathcal O_K$. The Neron-Severi group is the group of Hermitian matrices over $\mathcal O_K$, and the Euler characteristic is the determinant. So when $\det(M)<0$, we know $H^1>0$. However, I thin when $\det(M)>0$, since the bundle is either ample or minus ample, $H^1$ vanishes or vanishes except within some bounded distance of the boundary $\det(M)=0$. The boundary is a smooth bicone $xy=z^2+w^2$ so the negative side cannot be covered by finitely many convex cones, similar to how the complement of a circle in the plane cannot be covered by finitely many convex bodies.

No idea about question $3$.

  • $\begingroup$ Thank you! Concerning Question 2: thanks for the easy counterexample. Looking at the more complicated $G/B$, we see that $B_i$ is a union of shifted cones indexed by words of length $i$ in the Weyl group. It's this kind of interesting behaviour that I would like to see generalized. I will edit Question 2 to mention this. $\endgroup$ – Piotr Achinger Aug 28 '12 at 5:08
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    $\begingroup$ Note that my example is just the flag variet of $sl_2 \times sl_2$. The decomposition into cones seems related to the function $\chi$ on the Neron-Severi group, which we can extend onto a $\mathbb Q$-vector space. If the vanishing set of $\chi$ is not a union of hyperplanes, I'm guessing that you will find the cone decomposition will not occur. $\endgroup$ – Will Sawin Aug 28 '12 at 5:38

A friend told me that the paper

Nathan Broomhead, Artie Prendergast-Smith Partially ample line bundles on toric varieties

seems to address something similar to Question 2 for toric varieties.

  • $\begingroup$ That's true. More precisely, the paper talks about the notion of $q$-ampleness, as mentioned by J.C Ottem in his answer. To learn about that notion, though, one should read the paper by Burt Totaro, "Line bundles with partially vanishing cohomology", arXiv:1007.3955. $\endgroup$ – user5117 Aug 28 '12 at 21:45

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