Let $X$ be a smooth projective variety over $\mathbb C$ and $A\to X$ an ample line bundle.

Is there an integer $k_0$ such that for all line bundle $L\to X$, the tensor product $A^{\otimes k}\otimes L$ is ample for every $k\ge k_0$?

Of course the answer is yes if we let $k_0$ depend on $L$, but the point here is that $k_0$ must be an universal constant depending only on $X$ and $A$.

Thanks in advance.

EDIT: In fact the answer is NO in general: take for example $X=\mathbb P^n$ and $A=\mathcal O(1)$. Then for each $k$ there exists a line bundle $L$ on $\mathbb P^n$, namely $L=\mathcal O(-k-1)$ such that $A^{\otimes k}\otimes L$ is not ample. So, is there any reasonable condition on $X$ and $A$, or on the family in which we let the bundles $L$ vary, such that the statement holds true?

My question was motivated by the following on positivity:

Let $X$ be a compact complex manifold endowed with a positive (in the sense that it carries a smooth hermitian metric whose Chern curvature tensor is positive definite) line bundle $A\to X$. For every pair of points $p,q\in X$ consider the blow-up $\sigma\colon\widetilde X\to X$ at this two point and let $E_p$, $E_q$ be the two exceptional divisors (if $p=q$ then consider $2E_p$). Is there a uniform $k_0>0$ (independent of $p$ and $q$) such that $$ \sigma^* A^{\otimes k}\otimes\mathcal O_{\widetilde X}(-E_p-E_q) $$ is positive whenever $k\ge k_0$?

EDIT bis. Here is a beginning of a possible proof. Le us take the case $p=q$ and suppose the contrary. Then, there is a sequence of points $(p_k)\subset X$ such that $$ \sigma^* A^{\otimes k}\otimes\mathcal O_{\widetilde X}(-2E_{p_k}) $$ is not positive definite. Since $X$ is compact, after extracting a subsequence we may suppose that $p_k\to p\in X$. Let $k_0$ be a integer such that $$ \sigma^* A^{\otimes k}\otimes\mathcal O_{\widetilde X}(-2E_{p}) $$ is positive definite for all $k>k_0$. This integer certainly exists, since $\mathcal O_{E_p}(-E_p)$ is positive.

If we are able to construct for all points $q$ nearby $p$ a smooth hermitian metric such that $$ \sigma^* A^{\otimes k}\otimes\mathcal O_{\widetilde X}(-2E_{q}) $$ is positive definite whenever $k\ge k_0$ we then get a contradiction and we are done.

For the moment I am not still able to do this last part. This should be achieved by a somehow clever use of a partition of unity on $\widetilde X$.

  • $\begingroup$ Not in general. Take $X=\mathbb{P}^1$, $A=\mathcal{O}(1)$. $\endgroup$ – Francesco Polizzi Dec 5 '10 at 10:41
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    $\begingroup$ And probably never. Assume that $k_0$ exists, and take $L=A^{-k_0-1}$. Then $A^{k_0} \otimes L=A^{-1}$, which is not ample... $\endgroup$ – Francesco Polizzi Dec 5 '10 at 10:45
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    $\begingroup$ I think you must require that the degree of $L$ (with respect to a fixed polarization, for instance with respect to A) is bounded form below. Otherwise, for any fixed $k$, you can always find tensor products $A^{k} \otimes L$ of negative degree, which are not ample... $\endgroup$ – Francesco Polizzi Dec 5 '10 at 10:53
  • $\begingroup$ You are definitely right. In any case, this condition on the degree seems to be enough in order to control the intersection products in the Nakai-Moishezon criterion, right? $\endgroup$ – diverietti Dec 5 '10 at 11:18
  • $\begingroup$ mmm... maybe not... $\endgroup$ – diverietti Dec 5 '10 at 11:19

Remark: If $p=q$, then one can just work with $E_p$ instead of $2E_p$ and then at the end take $2k_0$ instead of $k_0$, so we may assume that $p\neq q$ and in particular that the exceptional divisor to be subtracted is reduced.


Lemma 1. Let $D$ be a semi-ample Cartier divisor on a smooth projective variety $Y$ and assume that for any proper curve $C\subseteq Y$, the intersection number $D\cdot C>0$. Then $D$ is ample.

Proof. Consider the morphism induced by a large enough multiple of $D$. If it had a positive dimensional fiber, then it would contain a proper curve which would have intersection $0$ with $L$ which contradicts the assumption. Q.E.D.

Lemma 2. Let $X$ be a a smooth projective variety, $\sigma: Y\to X$ the blow up of the points $p_1,\dots,p_r\in X$ with (reduced) exceptional divisor $E\subset Y$ and finally let $A$ be an ample Cartier divisor on $X$. Assume that $k_0,k_1\in\mathbb N$ are such that $k_0A-K_X$ is ample and $k_1(\sigma^*A)-E$ is nef. Then $k(\sigma^*A)-E$ is ample for any $k\geq k_0+\dim X \cdot k_1+1$.

Proof. First observe that $K_Y\sim \sigma^* K_X + (\dim X−1) E$ and hence with $m=k-k_0-\dim X\cdot k_1> 0$, $$ (k(\sigma^*A)-E)-K_Y\sim m(\sigma^*A) + \sigma^*(k_0 A-K_X) + \dim X \cdot(k_1(\sigma^*A)-E) $$ is nef and big. Then by the Basepoint-free theorem $k(\sigma^*A)-E$ is semi-ample. It also follows that for any proper curve $C\subseteq Y$, the intersection number $(k(\sigma^*A)-E)\cdot C>0$. Indeed, if $C\subseteq E$, then $\sigma^*A\cdot C=0$ and $-E\cdot C>0$ and if $C\not\subseteq E$, then $(\sigma^*A)\cdot C>0$ and $(k_0(\sigma^*A)-E)\cdot C\geq 0$. This is enough as $k>k_0$. Finally, then $k(\sigma^*A)-E$ is ample by Lemma 1. Q.E.D.

Claim The motivating positivity statement of the question is true.

Proof. Let $k_0$ be such that $k_0A-K_X$ is ample and let $k_1$ be an integer that is larger than $1/\varepsilon_p+1/\varepsilon_q$ for the Seshadri constants $\varepsilon_p=\varepsilon(A, p)$ and $\varepsilon_q=\varepsilon(\sigma_p^*A-E_p, q)$, where $\sigma_p$ is the blow up of $p$ alone. Then Lemma 2 implies the desired statement.

Note that for a fixed $X$ and fixed $A$ there is a lower bound for the Seshadri constants that works for all $p\in X$.

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  • $\begingroup$ Just a typos, I guess that $K_Y\sim\sigma^*K_X+(\dim X-1)E$. But this change just the numerical constants. $\endgroup$ – diverietti Dec 6 '10 at 7:10
  • $\begingroup$ Thanks very much Sándor, do you think that if I find a more elementary proof, which does not rely on deep theorems such as the Basepoint-free one, could be interesting? $\endgroup$ – diverietti Dec 6 '10 at 7:12
  • $\begingroup$ I think that finding a direct way to construct a positive metric would be interesting. It is always interesting to see algebraic and analytic methods produce the same result in completely different manners. You are right about the coefficient. I will fix that. $\endgroup$ – Sándor Kovács Dec 6 '10 at 7:20
  • $\begingroup$ Also, one might need to work out the right bound for these Seshadri constants. $\endgroup$ – Sándor Kovács Dec 6 '10 at 7:24
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    $\begingroup$ Dear Sandor, In the proof of the claim, "Let $k_1$ be ..." should read "Let $k_0$ be ...". Best wishes, Matt $\endgroup$ – Emerton Dec 7 '10 at 6:09

I see that Sándor already gave an answer to the question, but I wanted to expand on my comment, giving an answer that does not rely on the Basepoint-free Theorem. It exploits the fact that ampleness is open both in the cone of divisors and in families, and that the family of divisors that we consider is parameterized by a quasi-compact scheme (namely $X \times X$, or more precisely $(X \times X \setminus \Delta) \sqcup \Delta$, where $\Delta $ denote the diagonal of $X \times X$).

Let $X$ be a smooth projective variety and let $A$ be an ample divisor on $X$. For $p \in X$ denote by $E_p$ the exceptional divisor of the blow up of $X$ at $p$; to simplify the notation, denote also by $A$ the pull-back of $A$ to the blow up of $X$ at $p$.

Lemma. Let $p$ be a point of $X$ and let $A$ be an ample divisor on $X$. For large enough $k \in \mathbb{R}$ the divisor $kA-E_p$ is ample on $X_p$.

Proof. Let $B$ be an ample divisor on $X_p$. Write $B=B'-tE_p$, where $B'$ is the pull-back of a divisor on $X$ and $t \in \mathbb{R}$; note that the ampleness of $B$ implies that $t>0$. Choose $h$ large enough so that $hA-B'$ is ample on $X$; we deduce that the divisor \[ hA-tE_p = (hA-B') + (B'-tE_p) = (hA-B') + B \] is ample on $X_p$, and the lemma follows.

Observe that the Lemma implies that for $p,q \in X$ the divisor $kA-E_p-E_q$ is ample on $(X_p)_q$ for sufficiently large $k$, by similar arguments.

For $k \in \mathbb{R}$ define $A_k := \{(p,q) \in X \times X \mid kA-E_p-E_q {\textrm{ is ample}}\}$. Since the property of being ample is an open property (see Lazarsfeld's Positivity I, Theorem 1.2.17), it follows that $A_k$ is (the set of closed points of) an open set. Moreover, applying the Lemma, we deduce that $\cup _k A_k = X \times X$. Finally, by quasi-compactness of $X \times X$ and the inclusions $A_k \subset A_{k'}$ for $k < k'$, we conclude that a fixed $k$ suffices for all pairs of points.

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    $\begingroup$ This is a really nice proof! $\endgroup$ – J.C. Ottem Dec 6 '10 at 15:40
  • $\begingroup$ Very very nice! $\endgroup$ – diverietti Dec 7 '10 at 1:04

You are essentially asking about a bound on the Seshadri constant of $A$ (on $X$). In fact, you are asking for something a little stronger. You can see the definition and basic properties of Seshadri constants in Lazarsfeld's Positivity in algebraic geometry.

The answer to your first question is obviously no as already pointed out by Francesco and JC.

In the second question, do you want something independent of $X$? This is not going to happen for special points of $X$. Miranda gave examples (see Lazarsfeld's book) of surfaces with arbitrarily small Seshadri constants which implies that you cannot have a bound that works for all $X$.

If your bound may depend on $X$, then something along the lines of what JC suggests should work.

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  • $\begingroup$ Dear Sándor, I am not essentially asking about Seshadri constant. I don't want to measure locally the positivity of A and I do know the example of Miranda. My question (apart from the first, which is obviously false, as I do have also pointed out), after a moment of reflection, was about smooth hermitian metrics construction on the (line bundle associated to) exceptional divisor of blow-ups nearby the point already blown-up. Of course the bound on the multiple of $A$ may depend on $X$: my question was not universal w.r.t the line bundle $A$ but w.r.t the point on $X$ I am blowing-up. $\endgroup$ – diverietti Dec 5 '10 at 22:50

I don't think it would be possible to put restrictions on $X$ and $A$ such that $A^k\otimes L$ is ample for any line bundle $L$. If such a condition holds, then in particular every line bundle is ample (which is impossible unless $X$ is a point).

Perhaps by restricting the set of line bundles $L$ would give you better results, but even in this case it might require specific conditions on $X$, e.g., letting $L$ vary over the effective line bundles, would give that the closures of the ample cone and the effective cone are equal, which is a strong requirement.

Comment to the 2nd question: How about something like this: Let $I=I_{p,q}$ be the ideal sheaf of $p\cup q$ and let $p:X'\to X$ be the blow-up of $X$ with center $I$ with exceptional divisor $E$. Let $A=O(1)$ and let $F$ be a coherent sheaf on $X'$. We want to show that $L=p^*O(n) \otimes O(-E)$ is ample, or equivalently that large powers of $L$ kill the higher cohomology of $F$. By the Leray spectral sequence,

$$ H^i(X',L^{\otimes k}\otimes F)=H^i(X, I^k(nk) \otimes p_* F). $$ This latter group vanishes for $i>0$ and $k/n$ large, since $A=O(1)$ is ample $(p_*O(−kE)=I^k$, all the higher direct images $R^i O(-kE)$ vanish, and $p_*F$ is coherent). So if we can show that this $n$ can be chosen independently of $p,q$ we should be done. Note that when $F=O_X$, $n$ can be chosen independently of $p,q$ since all $I$ have the same Hilbert polynomial.

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  • $\begingroup$ Thanks, I start thinking that my question is not well-posed... Or, in other words, that I haven't thought enough to this question before asking it... $\endgroup$ – diverietti Dec 5 '10 at 12:21
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    $\begingroup$ Yes, your argument looks correct (modulo a few minor typos). But you'll need a stronger vanishing statement to get ampleness of $A-E_p-E_q$. $\endgroup$ – Donu Arapura Dec 5 '10 at 17:34
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    $\begingroup$ I mean $nA-E_p-E_q$. $\endgroup$ – Donu Arapura Dec 5 '10 at 17:35
  • $\begingroup$ I mean that there a number of cohomological tests for ampleness (e.g. Serre, $-1$-regularity,...) which you could try to use here. However, $H^i(X,L^n)=0$ for $i>0$, $n\gg 0$ is insufficient. I'm sorry if this is a bit cryptic, but I don't have time to write a longer explanation. $\endgroup$ – Donu Arapura Dec 5 '10 at 18:11
  • $\begingroup$ Yes, of course you are right. I'll try to fix the problem. $\endgroup$ – J.C. Ottem Dec 5 '10 at 18:41

1. Here is an elementary and constructive proof from a hermitian point of view.

I will reproduce it only in the case of curves and blow-up equal to the identity, the general case being just more complicated to write.

Let $C$ be a compact curve and $A\to C$ a positive line bundle. If the uniformity is not true, then there exists a sequence $(p_k)\subset C$ such that $$ A^{\otimes k}\otimes\mathcal O_C(-p_k) $$ is not positive. Up to subsequence we can suppose that $p_k\to p\in C$. Let $k_0$ be an integer such that $$ A^{\otimes k_0}\otimes\mathcal O_C(-p) $$ is positive when $\mathcal O_C(-p)$ is endowed with the following metric:

Take a trivialization of $\mathcal O_C(-p)$ consisting in a coordinate chart $U$ centered in $p$ and in the open set $V_p=C\setminus\{p\}$ complementary of the point $p$. Put the trivial metrics $h_{U}$ and $h_{V_p}$ on this two trivialization and glue them together with the very simple partition of unity given by a smooth function $\vartheta\colon C\to\mathbb[0,1]$ which has compact support contained in $U$ and is identically equal to $1$ in an open neighborhood $\Omega$ of $p$; the metric on $\mathcal O_C(-p)$ is thus given by $h_{\mathcal O_C(-p)}=\vartheta\,h_U+(1-\vartheta)\,h_{V_p}$.

Then, for every $q$ in the aforesaid open neighborhood we can put the metric $h_{\mathcal O_C(-q)}$ on $\mathcal O_C(-q)$ given by $\vartheta\,h_U+(1-\vartheta)\,h_{V_q}$ where $h_{V_q}$ is the constant metric on the trivialization $V_q=C\setminus\{q\}$ of $\mathcal O_C(-q)$.

These two metrics are in fact equal Therefore, the two curvatures coincide and $$ A^{\otimes k_0}\otimes\mathcal O_C(-q) $$ is positive, contradiction.

2. If you prefer a less constructive and more global approach, here it is:

Let $p$ and $q$ two distinct points on a compact curve $C$, the $\mathcal O_C(p-q)$ has zero first Chern class, and thus this line bundle admits a metric $h$ of identically zero curvature $i\,\Theta(\mathcal O_C(p-q),h)$ (this is just the $\partial\bar\partial$-lemma). Now put any smooth hermitian metric on $h_q$ on $\mathcal O_C(q)$. Then $h_p=h\otimes h_q$ is a metric on $\mathcal O_C(p)$.

Therefore $$ \begin{aligned} i\,\Theta(\mathcal O_C(p),h_p)&=i\,\Theta(\mathcal O_C(p),h_p)-i\,\Theta(\mathcal O_C(q),h_q)+i\,\Theta(\mathcal O_C(q),h_q) \\ &=i\,\Theta(\mathcal O_C(p-q),\underbrace{h_p\otimes h_q^{-1}}_{=h})+i\,\Theta(\mathcal O_C(q),h_q)\\ &=i\,\Theta(\mathcal O_C(q),h_q), \end{aligned} $$ so that the same constant $k_0$ can be taken for all prime divisor $p$ in order to have that $$ A^{\otimes k}\otimes\mathcal O_C(-p) $$ has positive definite Chern curvature.

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  • $\begingroup$ The curve case is perhaps a bit degenerate in this question. $\endgroup$ – J.C. Ottem Dec 7 '10 at 0:56
  • $\begingroup$ Not at all!! You can reproduce word-by-word the first argument just replacing the points with the exceptional divisors... $\endgroup$ – diverietti Dec 7 '10 at 1:02
  • $\begingroup$ I mean, you have to arrange it a little bit, but nothing mysterious... Just use a tubular neighborhood of the exceptional divisor E on the blown-up manifold $\tilde X$ and extend the natural metric of $\mathcal O_E(−E)$ in an arbitrary way to a metric on $\mathcal_{\tilde X} O(-E)$. $\endgroup$ – diverietti Dec 7 '10 at 1:13
  • $\begingroup$ On the other hand, of course the "global approach" two doesn't work in higher dimension, since the difference of two exceptional divisors coming form the blow-up of two different points is never zero in cohomology. But it was so simple, that I wanted to post it anyway. :) $\endgroup$ – diverietti Dec 7 '10 at 12:42
  • $\begingroup$ Ottem, would you prefer a complete proof in the general case? $\endgroup$ – diverietti Dec 7 '10 at 15:04

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