Uniformity of ampleness

Question

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 $\widetilde X$.

Answer 1

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.

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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 $p\in X$.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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.