Ampleness equivalence and dominance for vector bundles

TL;DR

This paper extends Hartshorne's ampleness criterion from vector bundles to flag manifolds, establishing equivalences for ampleness of associated line bundles and vector bundle dominance, with a new proof of the Ampleness Dominance theorem.

Contribution

It generalizes ampleness equivalence to flag manifolds and provides a novel proof of the Ampleness Dominance theorem independent of Littlewood-Richardson saturation.

Findings

Ampleness of line bundles on flag manifolds corresponds to ampleness of associated vector bundles.

The determinant line bundle on Grassmannians is ample if and only if the wedge power of the bundle is ample.

A new proof of the Ampleness Dominance theorem that does not rely on saturation properties.

Abstract

Hartshorne in "Ample vector bundles" proved that $E$ is ample if and only if $\OOO_{P(E)}(1)$ is ample. Here we generalize this result to flag manifolds associated to a vector bundle $E$ on a complex manifold $X$: For a partition $a$ we show that the line bundle $\it Q_a^s$ on the corresponding flag manifold $\mathcal{F}l_s(E)$ is ample if and only if $ \SSS_aE $ is ample. In particular $\det Q$ on $\it{G}_r(E)$ is ample if and only if $\wedge ^rE$ is ample.\\ We give also a proof of the Ampleness Dominance theorem that does not depend on the saturation property of the Littlewood-Richardson semigroup.

Equations

$$(V=V_{s_0}\supset V_{s_1}\supset \dots \supset V_{s_m}=\{0\})=F,$$

$$a_1$$

$$a_{s_1+1}$$

$$a_{s_2+1}$$

$$\dots$$

$$Q_a^s=Q_{s_1}^{a_{s_1}}\otimes Q_{s_2}^{a_{s_2}}\otimes \dots \otimes Q_{s_m}^{a_{s_m}}$$

$$Q_{s_j}{|}_F=\det (V_{s_{j-1}}/V_{s_j}),1\le j\le m.$$

$${\pi }_{\ast }((Q_a^s{)}^m)\simeq {\mathcal{S}}_{ma}E$$

$$R^q{\pi }_{\ast }((Q_a^s{)}^m)\simeq 0,\text{mbox}ifq>0$$

$$\pi =p_{|C}:C⟶p(C)=C_1$$

$$i^{\ast }{p}^{\ast }{p}_{\ast }G⟶i^{\ast }G⟶0.$$

$$\begin{array}[]{rcl}C&\stackrel{{\scriptstyle i}}{{\hookrightarrow}}&Y\\ {\pi}{\downarrow}&&\downarrow{p}\\ C_{1}&\stackrel{{\scriptstyle i_{1}}}{{\hookrightarrow}}&X\end{array}$$

$${\mathcal{S}}^m(i^{\ast }{p}^{\ast }(p_{\ast }G))={\pi }^{\ast }{i}_1^{\ast }({\mathcal{S}}^m(p_{\ast }G)).$$

$${\pi }_{\ast }((Q_a^s{)}^r)\simeq {\mathcal{S}}_{ra}E.$$

$$ra\simeq a$$

$$a_1$$

$$a_1+a_2$$

$$\dots$$

$$a_1+\dots +a_{d-1}$$

$$a_1+\dots +a_{d-1}+a_d$$

$$a⪯b\text{mbox}if{n}_{1}a⪯n_{2}b\text{mbox}forsome{n}_1,n_2\in \mathbb{N},$$

$$a⪯b\text{mbox}if|b|a⪯|a|b.$$

$$b\vee c=(b_1,\dots ,b_s,c_1,\dots c_t).$$

$$\mathcal{L}(d)=\{(l_1,\dots ,l_r)|r,l_i\in \mathbb{N},l_1+l_2+\dots +l_r=d\}.$$

$$1_i=(\underset{itimes}{\underbrace{1,1,\dots ,1}}),$$

$$0_{\mathbf{i}}=(\underset{itimes}{\underbrace{0,\grave{a},\dots ,0}}).$$

$$C(a)=\{b=(b,\dots ,b_d)\in {\mathbb{Q}}_{\ge 0}^d|b_1\ge b_2\ge \dots \ge b_d\ge 0\text{mbox}andb⪯a\}.$$

$$v(L,a)=\frac{|a(L,1)|}{l_1}{1}_{l_1}\vee \dots \vee \frac{|a(L,r)|}{l_r}{1}_{l_r}.$$

$$\Sigma (n)=\{b\in {\mathbb{Q}}_{\ge 0}^d|\sum _{k=1}^n{b}_k=\sum _{k=1}^n{a}_k\},$$

$$D(n)=\{b\in {\mathbb{Q}}_{\ge 0}^d|b_n=b_{n+1}\},$$

$$\Sigma (n)\cap C(a)\simeq P((a_1,\dots ,a_n))\times P((a_{n+1},\dots ,a_d)),$$

$$b=c+\sum _{L\in \mathcal{L}(d)}{m}_{L}v(L,\mu (l)a),c\in \sigma (a)$$

$$b=\sum _{L\in \mathcal{L}(d)}{\tau }_{L}v(L,\mu (d)a)$$

$$v(L,\mu (d)a)\in a^{\otimes \mu (d)}.$$

Full text

We give also a proof of the Ampleness Dominance theorem that does not depend on the saturation property of the Littlewood-Richardson semigroup.

1. Introduction

Let $E$ be a complex vector bundle of rank $d$ on a compact complex manifold $X$ and $s$ be a sequence of integers $s=(s_{0},s_{1},\ldots s_{m})$ such that

$0=s_{0}<s_{1}<\ldots<s_{m}=d.$ For $x\in X$ and the corresponding fiber $V=E_{x},$ consider the manifold $\mathcal{F}l_{s}(V)$ of incomplete flags:

[TABLE]

where $V_{s_{i}}$ is a vector subspace of $V,\ \operatorname{codim}V_{s_{j}}=s_{j}.$ When $x$ varies, the $\mathcal{F}l_{s}(E_{x})$ together form a manifold $\mathcal{F}l_{s}(E).$

For a partition $a=(a_{1},a_{2},\ldots)$ of length $l\leq d$ such that

[TABLE]

there is a corresponding line bundle $\it Q_{a}^{s}$ over $\mathcal{F}l_{s}(E).$ When restricted to $\mathcal{F}l_{s}(V)$ it has the form

[TABLE]

where

[TABLE]

By Bott’s formula [1], for any $m\geq 0$

[TABLE]

[TABLE]

where $\pi:\mathcal{F}l_{s}(E)\longrightarrow X$ is the natural projection and ${\mathcal{S}}_{a}$ is the Schur functor corresponding to $a$.

Our Main Theorem is

Theorem 1.1**.**

The line bundle $\it Q_{a}^{s}$ on $\mathcal{F}l_{s}(E)$ is ample if and only if the vector bundle ${\mathcal{S}}_{a}E$ on $X$ is ample.

A useful special case is

Corollary 1.2**.**

The line bundle $\det Q$ on the grassmannian bundle $\it{G}_{r}(E)$ is ample if and only if the vector bundle $\wedge^{r}E$ is ample on $X.$

Note that by Theorem 1.1, the vanishing theorem of Demailly [2] is valid under the minimal hypothesis ${\mathcal{S}}_{a}E$ ample.

Moreover we give a new and simple proof of the Ampleness Dominance result:

Theorem 1.3**.**

Let $E$ be a vector bundle of rank $d$ and $a,b$ be partitions. If $b\preceq a,$ then ${\mathcal{S}}_{a}E$ ample implies ${\mathcal{S}}_{b}E$ ample.

Corollary 1.4**.**

${\mathcal{S}}_{a}E\otimes{\mathcal{S}}_{b}E$* is ample if and only if ${\mathcal{S}}_{a+b}E$ is ample.

For any positive integer $m$, ${\mathcal{S}}_{a}E$ is ample if and only if ${\mathcal{S}}_{ma}E$ is ample.

If $\Lambda^{k}E$ is ample and $s>0$ then $\Lambda^{s+k}E$ is ample.*

2. Proof of Theorem 1.1.

For the “if” direction of Theorem(1.1) we will use the following lemma which was first pointed out to us by L. Gruson and which we have already used in ([8]. lemma 2.6). To our knowlege there is no proof in the literature, so we prove it here.

Lemma 2.1**.**

*Let $G$ be a vector bundle on $Y$ and $p:Y\longrightarrow X$ be a proper morphism satisfying:

1. $p_{*}G$ is an ample vector bundle,

2. $G$ is ample along the fibers and

3. the map $p^{*}p_{*}G\longrightarrow G$ is surjective.

Then $G$ is ample.*

Proof.

We will use the following result of Gieseker (Lemma 2.1.page 101, [4]):

Suppose $Y$ is proper over a field $k$ and $E$ is a bundle over $Y$ generated by its global sections. Then $E$ is ample if and only if every quotient line bundle of $E_{|C}$ is ample for every curve $C$ in $Y.$

Now we will prove Lemma(2.1):

Since $p_{*}G$ is ample, for large $m,\ {\mathcal{S}}^{m}(p_{*}G)$ is generated by global sections. By $3),$ ${\mathcal{S}}^{m}(p^{*}p_{*}G)\longrightarrow{\mathcal{S}}^{m}G$ is surjective. Hence ${\mathcal{S}}^{m}G$ is generated by sections.

Let $C$ be any curve in $Y.$

If $C\subset p^{-1}(x)$ for some $x\in X,$ then ${\mathcal{S}}^{m}G|_{C}$ is ample by 2). This implies $G|_{C}$ ample.

If $C$ is not contained in any fiber, then

[TABLE]

is a finite morphism. By $3)$ we have a surjective map

[TABLE]

The commutative diagram

[TABLE]

gives

[TABLE]

Now $\pi^{*}i_{1}^{*}({\mathcal{S}}^{m}(p_{*}G))$ is ample by assupmption 1). Hence by (2.1) $i^{*}G=G|_{C}$ is ample. This shows that $G$ is ample by Gieseker’s result stated at the beginning of the proof. ∎

I. Proof of the “if” direction of Theorem 1.1.

We apply Lemma(2.1) with $Y=\mathcal{F}l_{s}(E),\ \ G=\it Q_{a}^{s}$ and $p=\pi.$ By using equation(1.1), it is clear that all assumptions in Lemma(2.1) are satisfied. Hence $\it Q_{a}^{s}$ is ample.

II. Proof of the ’only if’ direction of Theorem 1.1.

Some preparations:

Theorem 2.2**.**

Let $\pi:Y\longrightarrow X$ be a submersion between two complex manifolds and $L$ be an ample line bundle on $Y.$ Then there exists $r_{0}$ such that $\forall r\geq r_{0},\ \ \pi_{*}(L^{r})$ is ample.

Proof.

We use the following result due to Mourougane (Theorem 1, [9]):

If $\pi:Y\longrightarrow X$ is a submersion between two complex manifolds and $L$ is an ample line bundle on $Y,$ then $\pi_{*}(L\otimes K_{Y/X})$ is ample or zero.

On the other hand, ampleness of $L$ implies that there exists $r_{0},$ so that $\forall\ r\geq r_{0},\ L^{r}\otimes K_{Y/X}^{-1}$ is ample.

Thus $\pi_{*}(K_{Y/X}\otimes(L^{r}\otimes K_{Y/X}^{-1}))=\pi_{*}(L^{r})$ is ample by Mourougane’s result. ∎

Now we come to the proof of the ’only if’ direction of Theorem(1.1).

If $\it Q_{a}^{s}$ is ample by Theorem(2.2) there exists $r_{0}$ such that

$\forall r\geq r_{0},\ \ \pi_{*}((Q_{a}^{s})^{r})$ is ample. By equation (1.1)

[TABLE]

Ampleness of ${\mathcal{S}}_{a}E$ is deduced by the Ampleness Dominance

Theorem(1.3), since

[TABLE]

in the dominance partial order.

Remark 2.3**.**

The proof of the Ampleness Dominance result (Theorem 3.7 p.175 in [6]) uses the saturation property of the Littlewood-Richardson semigroup for $r$ summands, $r$ arbitrary (p. 172 in [6]). This property had been proven by Knutson and Tao [5] for $r=2$. We assumed that the extension to $r$ arbitrary is immediate, but this appears not to be the case. A new and simpler proof that does not use saturation at all will be given in the next section.

3. Dominance partial order on partitions and ampleness

Definition 3.1**.**

A partition $a=(a_{1},a_{2},\ldots a_{d})$ of length $d$ is a non-increasing sequence of non-negative integers $a_{i}.$ The weight $\sum_{i=1}^{d}a_{i}$ is denoted by $|a|.$

The dominance partial ordering of partitions is defined in [3] by:

Let $a=(a_{1},a_{2},\ldots a_{d})$ and $b=(b_{1},b_{2},\ldots b_{d})$ be two partitions of the same weight.

Then   $a\preceq b$   if

[TABLE]

We extend this definition to non-increasing sequences of non-negative rational numbers:

[TABLE]

where  $n_{1}a$  and  $n_{2}b$ are partitions of the same weight.

When $a$ and $b$ are non-zero partitions of not necessarily equal weight an equivalent formulation is:

[TABLE]

If $a\preceq b\ \ {\mbox{and}}\ \ b\preceq a,$ then we say $a$ is equivalent to $b$ and write $a\simeq b.$

Notation 3.2**.**

For finite sequences $b=(b_{1},\ldots,b_{s})$ and $c=(c_{1},\ldots c_{t})$ we set

[TABLE]

For a given positive integer $d$ let

[TABLE]

If $L=(l_{1},\ldots,l_{r})\in{\mathcal{L}}(d)$ and $a\in{\mathbb{Q}}_{\geq 0}^{d},$ then for $1\leq i\leq r,$ $a(L,i)$ is a sequence of length $l_{i},$ such that $a=a(L,1)\vee a(L,2)\ldots\vee a(L,r).$

Notation 3.3**.**

Let $a,b$ be partitions of length $d$. If $V$ is a vector space of dimension $d$ we denote by $a\otimes b$    the set of all partitions appearing in the decomposition of the tensor product $\ \ {\mathcal{S}}_{a}V\otimes{\mathcal{S}}_{b}V$ as a direct sum of irreducible representations of $GL(V)$, and analogously for tensor products with more than two factors.

For ease of understanding we write the partition $\bf{1_{d}}$ instead of $\det,$ where

[TABLE]

and ${\bf{1_{i}}}\vee{\bf{0_{d-i}}}$ instead of $\wedge^{i},$ where

[TABLE]

The only properties of the Littlewood-Richardson rules for the tensor product of irreducible representions we need to use in the sequel are the following.

Proposition 3.4**.**

*Let $a,b,c$ be partitions of length $d$.

If $c\in a\otimes b$, then $c\preceq(a+b).$*

Proposition 3.5**.**

*Let $a,b,c,d,e,f$ be partitions of length $d$. If

$c\in a\otimes b$ and $f\in d\otimes e,$ then   $(c+f)\in(a+d)\otimes(b+e).$*

Proposition 3.6**.**

*Let $a,b,c$ be partitions of length $d$ and

$L=(l_{1},l_{2},\ldots,l_{r})\in{\mathcal{L}}(d).$ If

$c(L,i)\in a(L,i)\otimes b(L,i)$ for $i=1,\ldots,r,$ then

$c\in a\otimes b$.*

These propositions follow immediately from the rules of Littlewood-Richardson.For some background see Zelevinsky [10].

Remark 3.7**.**

Since $d\in a\otimes b\otimes c$ is equivalent to the existence of a partition $e$ with $e\in a\otimes b$ and $d\in e\otimes c$ the three properties generalize immediately to tensor products with more than two factors.

The partitions $c$ such that $c\preceq a,$ are the integral points of a rational cone $C(a)$. More precisely:

Definition 3.8**.**

Let $a$ be a partition. Let

[TABLE]

Lemma 3.9**.**

Let $a=(a_{1},\ldots,a_{d})$ be a partition. The cone $C(a)$ is generated by the set of sequences   $\{v(L,a)\ |\ L\in{\mathcal{L}}(d)\},$ where

[TABLE]

Proof.

Consider the hyperplanes

[TABLE]

[TABLE]

$n=1,\ldots,d$, where $b_{d+1}=0$.

$P(a)=C(a)\cap\Sigma(d)$ is a convex polytope in $\Sigma(d)$. Its vertices are given by an intersection of $d$ of the $2d$ hyperplanes just introduced, with $\Sigma(d)$ included among them.

We will prove the lemma by induction on $d$. For $d=1$ the claim is obvious. For a given vertex $v,$ consider the case where $v\in\Sigma(n)$ for some $n\in\{1,\ldots,d-1\}$. One has

[TABLE]

since $(b_{1},\ldots,b_{n})\in P((a_{1},\ldots,a_{n}))$ implies   $b_{n}\geq a_{n}$ and

$(b_{n+1},\ldots,b_{d})\in P((a_{n+1},\ldots,a_{d}))$ implies $a_{n+1}\geq b_{n+1}$. The vertices of $P((a_{1},\ldots,a_{n}))$ and $P((a_{n+1},\ldots,a_{d}))$ are known by induction. If $v\in D(d),$ then $v\in\Sigma(d-1)$, too. The only remaining possibility is $v=\Sigma(d)\cap D(1)\cap D(2)\cap\ldots\cap D(d-1)$, in which case the vertex is $v(L,a)$ for $L=(d)$. ∎

Notation 3.10**.**

For $l\in{\mathbb{N}}$   let $\mu(l)={\it lcm}\ \{1,2,\ldots,l\}$.

Lemma 3.11**.**

Let $a$ be a partition. There is a finite set $\sigma(a)$ of partitions such that if $b\in C(a)\cap{\mathbb{Z}}_{\geq 0}^{d},$ then $b$ can be written as

[TABLE]

and $m_{L}\in{\mathbb{Z}}_{\geq 0}$  for all $L\in{\mathcal{L}}(d)$.

Proof.

Note that each $v(L,\mu(d)a)$ is a partition. According to Lemma 3.9 one has

[TABLE]

with $\tau_{L}\in{\mathbb{Q}}_{\geq 0}$ for all $L\in{\mathcal{L}}(d)$. The set of partitions $b$ with $0\leq\tau_{L}<1$ for all $L\in{\mathcal{L}}(d)$ is a bounded subset of ${\mathbb{Z}}_{\geq 0}^{d},$ thus finite. ∎

Lemma 3.12**.**

Let $a$ be a partition. For any $L\in{\mathcal{L}}(d)$

[TABLE]

The proof of this lemma will be subdivided into three elementary steps.

Lemma 3.13**.**

Let $k,m,q$ be non-negative integers with $k\leq d$ and let $mk=dq+s$ with $0\leq s<k.$ Then

[TABLE]

Proof.

By induction on $m,$ with $m=1$ as trivial case. One has

[TABLE]

and

[TABLE]

for $s+k<d$,

[TABLE]

for $s+k\geq d$. ∎

Lemma 3.14**.**

For any partition $a$ of weight ${|a|}$ and length $d$,

[TABLE]

Proof.

Let $\tilde{a}=(t_{1},t_{2},\ldots)$ be the transpose of $a$. Then $a=\wedge^{t_{1}}+\wedge^{t_{2}}+\ldots.$ By Lemma(3.13) the result is true for $\wedge^{k}$, $k=1,2,\ldots.$ By Proposition(3.5) it is true for $a$. ∎

The proof of Lemma(3.12) follows immediately from Proposition(3.6) and Lemma(3.14).

4. Proof of Theorem 1.3

For the proof we need to recall a particular case of Lemma(3.3) in [7].

Lemma 4.1**.**

A vector bundle $E$ on $X$ is ample if and only if the following condition is true: Given any coherent sheaf $\mathcal{F}$ on $X$, there exists $N(\mathcal{F})\in{\mathbb{N}}$ such that for any $n\geq N(\mathcal{F})$ one has

[TABLE]

Assuming ${\mathcal{S}}_{a}E$ ample and   $b\preceq a$  we want to prove that ${\mathcal{S}}_{b}E$ is ample.

According to Lemma 4.1 we have a map $N_{a}$ from the coherent sheaves on $X$ to ${\mathbb{N}}$ such that

[TABLE]

We want to prove the analogous property for ${\mathcal{S}}_{b}E$.

For any partition $c$ contained in $b^{\otimes m}$, according to Lemma 3.4 and 3.11 one has a decomposition $c=f+g$, where $f\in\sigma(a)$ and $g=\displaystyle{\sum_{L\in{\mathcal{L}}(d)}\tau_{L}\ v(L,\mu(d)\ a)},$ with $\tau_{L}$ a non-negative integer for all $L\in{\mathcal{L}}(d).$

According to Proposition 3.5 and Lemma 3.12 we have $g\in a^{\otimes\mu(d)s}$, where $s=\displaystyle{\sum_{L\in{\mathcal{L}}(d)}\tau_{L}}$. Define a map $N_{b}$ from the coherent sheaves on $X$ to ${\mathbb{N}}$ by

[TABLE]

where the symbol $[\ ]$ is the integral part.

By Proposition(3.5) $H^{q}(X,({\mathcal{S}}_{b}E)^{\otimes m}\otimes\mathcal{F})$ is a direct sum of subgroups of the cohomology groups $H^{q}(X,({\mathcal{S}}_{a}E)^{\otimes\mu(d)s}\otimes({\mathcal{S}}_{f}E\otimes\mathcal{F}))$ with $f\in\sigma(a)$, $m|b|=|f|+\mu(d)|a|s$. The latter groups vanish for $m\geq N_{b}(\mathcal{F})$. Thus ${\mathcal{S}}_{b}E$ is ample. This finishes the proof of Theorem 1.3.

Acknowledgement: The first author would like to thank Dublin Institute for Advanced Studies for its hospitality. We would also like to thank Nagaraj D.S. for useful discussion.