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Let $X$ be a complex projective manifold of complex dimension $n$ and $A\to X$ an ample line bundle. Let $L\to X$ be a line bundle such that $$ c_1(L)^k\cdot c_1(A)^{n-k}>0,\quad k=1,\dots,n. $$ Is it true that then $L$ is big?

The answer is yes if $n\le 2$: for $n=1$ there is nothing to prove, and for $n=2$ the positivity of the top self intersection $c_1(L)^2>0$ says that $L$ or its dual is big. But then $c_1(L)\cdot c_1(A)>0$ implies that in fact $L$ is big.

The answer is again yes in all dimensions if $X$ is an abelian variety: in this case $L$ is moreover ample. This is because one can represent $c_1(L)$ and $c_1(A)$ by "constant" hermitian forms, the second being positive definite, and thus the intersection conditions simply tell that the elementary symmetric polynomials in the eigenvalues of the hermitian form representing $c_1(L)$ are all positive. Thus, $L$ is positively curved and hence ample.

I strongly suspect anyway that the result is false in general.

Could you give for instance a counterexample in dimension three?

Thanks in advance.

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  • $\begingroup$ Maybe I'm missing something, but isn't 'big' usually defined as 'ample plus effective'? (Or equivalently, in the interior of the effective cone). Your definition seems to give that many non effective line bundles on $\mathbb{P}^1\times \mathbb{P}^1$ are big.. $\endgroup$
    – J.C. Ottem
    Commented Mar 24, 2011 at 18:39
  • $\begingroup$ You can define big as you wish, for example what your are saying is Kodaira's lemma: a line bundle $L$ is big if and only if for every ample line bundle $A$, some power of $L$ twisted by the dual $A^*$ is effective (that is $L$ is $\mathbb Q$-linearly equivalent to the sum of an ample plus an effective). In my mind big means that the growth rate of sections of its multiples is the same as it was ample (which is equivalent to the former). My definition of what? $\endgroup$
    – diverietti
    Commented Mar 24, 2011 at 18:43
  • $\begingroup$ In any case, of course you can have a non effective line bundle being big. Even more, you can have a non effective line bundle which is ample. Bigness as well as ampleness are asymptotic condition on multiples of the line bundle. $\endgroup$
    – diverietti
    Commented Mar 24, 2011 at 18:45
  • $\begingroup$ @JC: the divisor class 2H1−H2 on P1×P1 has negative self-intersection, if this is what you had in mind. @divierietti: It is relatively common to call a line bundle effective if its class in Pic lies in the effective cone, regardless of whether it has a section. $\endgroup$
    – Jack Huizenga
    Commented Mar 24, 2011 at 19:22
  • $\begingroup$ I guess it is relatively common when one talks about $\mathbb Q$ divisors... Otherwise the common terminology is pseudoeffective. In any case, I didn't give any definition in my question, so I don't understand what you are talking about... $\endgroup$
    – diverietti
    Commented Mar 24, 2011 at 19:30

1 Answer 1

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Here is a very straightforward contre-example. Let $X=\mathbb CP^2\times \mathbb CP^1$ blown up in one point. Denote by $E$ the exceptional divisor, and denote by $\pi$ the projection of $X$ to $\mathbb CP^2$, and take the following bundle:

$$L_n=\pi^*(O(n))\otimes O(E),$$ where $n$ satisfies two conditions: $$c_1(O(n))^2\cdot c_1(A)>-c_1(O(E))^2\cdot c_1(A),\;\;\;\; c_1(O(n))\cdot c_1(A)^2>-c_1(O(E))\cdot c_1(A)^2,$$ it is obvious that such $n$ exists.

To that this bundle is what you want we just need the following two simple facts: $c_1(\pi^*(O(n)))\cdot c_1(O(E))=0$ and $c_1(O(E))^3=1$. $L_{n}$ is not big because the $H^0(kL_n)$ grows quadratically with $k$.

Idea of this example works in dimensions $2m+1$. We take a semi-ample line bundle $L_{sa}$ with Itaka dimension $2m$ http://en.wikipedia.org/wiki/Iitaka_dimension (in particular it is not big), and tensor it with a line bundle corresponding to an exceptional divisor $E$. We chose them so that $c_1(L_{sa})\cdot c_1(O(E))=0$, i.e., these bundles "don't interact". For large $n$ the class $(c_1(nL_{sa}))^k$ (provided $k\le 2m$) is represented by a cycle of a high degree (with respect to $A$), so it "beats" $(c_1(O(E)))^k$. Finally $E^{2m+1}=1$.

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  • $\begingroup$ Thanks for the answer and for the link... Fortunately for me, I did know what the Iitaka dimension is... :) I'll check your answer in the afternoon... $\endgroup$
    – diverietti
    Commented Mar 25, 2011 at 10:07
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    $\begingroup$ Diviertti, I understand of course that you know this :), just decided to put the link, for convenience of any potential reader who might be not so familiar with this Big line bundles... $\endgroup$
    – Dmitri Panov
    Commented Mar 25, 2011 at 10:14
  • $\begingroup$ Did you mean $c_1(\mathcal O(n))^2\cdot c_1(A)>-c_1(\mathcal O(E))^2\cdot c_1(A)$ and $c_1(\mathcal O(n))\cdot c_1(A)^2>-c_1(\mathcal O(E))\cdot c_1(A)^2$? $\endgroup$
    – diverietti
    Commented Mar 25, 2011 at 10:33
  • $\begingroup$ And in the last sentence, did you mean that it beats $c_1(E)^k$? $\endgroup$
    – diverietti
    Commented Mar 25, 2011 at 10:39
  • $\begingroup$ Yes, you are completely right, I'll fix it $\endgroup$
    – Dmitri Panov
    Commented Mar 25, 2011 at 10:47

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