# Asymptotic nonvanishing of syzygies
of algebraic varieties

Jinhyung Park  Department of Mathematical Sciences, KAIST, 291 Daehak-ro,
Yuseong-gu, Daejeon 34141, Republic of Korea<EMAIL_ADDRESS>
###### Abstract.

We establish precise nonvanishing results for asymptotic syzygies of smooth
projective varieties. This refines Ein–Lazarsfeld’s asymptotic nonvanishing
theorem. Combining with the author’s previous asymptotic vanishing result, we
completely determine the asymptotic shapes of the minimal free resolutions of
the graded section modules of a line bundle on a smooth projective variety as
the positivity of the embedding line bundle grows.

###### Key words and phrases:

asymptotic syzygies, Koszul cohomology, algebraic varieties, line bundles

###### 1991 Mathematics Subject Classification:

14C20, 14J60, 13D02

J. Park was partially supported by the National Research Foundation (NRF)
funded by the Korea government (MSIT) (NRF-2019R1A6A1A10073887 and
NRF-2022M3C1C8094326).

##  1\. Introduction

The purpose of this paper is to address the main theme of [8] and [19]: _the
asymptotic behavior of syzygies of algebraic varieties is surprisingly
uniform_. After the pioneering work of Green [16], there has been a
considerable amount of research to understand syzygies of algebraic varieties.
It is an interesting problem to describe the overall asymptotic behavior of
syzygies of graded section modules of a line bundle on a smooth projective
variety as the positivity of the embedding line bundle grows (see [16, Problem
5.13] and [7, Problem 4.4]). The influential paper [8] of Ein–Lazarsfeld opens
the door to asymptotic syzygies of algebraic varieties, and the asymptotic
nonvanishing theorem was proved there. In the present paper, we provides a new
approach to nonvanishing of asymptotic syzygies, and together with the
author’s asymptotic vanishing theorem [19], we exhibit the uniform behavior of
all asymptotic syzygies.

Throughout the paper, we work over an algebraically closed field $\mathbf{k}$
of arbitrary characteristic. Let $X$ be a smooth projective variety of
dimension $n$, and $B$ be a line bundle on $X$. For an integer $d\geq 1$, set

$L_{d}:=\mathscr{O}_{X}(dA+P)~{}~{}\text{ and }~{}~{}r_{d}:=h^{0}(X,L_{d})-1,$

where $A$ is an ample divisor and $P$ is an arbitrary divisor on $X$. We
assume that $d$ is sufficiently large so that $L_{d}$ is very ample and
$r_{d}=\Theta(d^{n})$. Here, for a nonnegative function $f(d)$ defined for
positive integers $d$, we define:

$\begin{array}[]{rcl}f(d)\geq\Theta(d^{k})&\Longleftrightarrow&\begin{array}[]{l}\text{there
is a constant $C_{1}>0$ such that}\\\ \text{$f(d)\geq C_{1}d^{k}$ for any
sufficiently large positive integer $d$};\end{array}\\\\[12.0pt]
f(d)\leq\Theta(d^{k})&\Longleftrightarrow&\begin{array}[]{l}\text{there is a
constant $C_{2}>0$ such that}\\\ \text{$f(d)\leq C_{2}d^{k}$ for any
sufficiently large positive integer $d$};\end{array}\\\\[12.0pt]
f(d)=\Theta(d^{k})&\Longleftrightarrow&\begin{array}[]{l}\text{there are
constants $C_{1},C_{2}>0$ such that }\\\ \text{$C_{1}d^{k}\leq f(d)\leq
C_{2}d^{k}$ for any sufficiently large positive integer
$d$.}\end{array}\end{array}$

For simplicity, we write $f(d)=\Theta(1)$ if $f(d)$ is a constant including
$0$ for any sufficiently large positive integer $d$. Let
$S_{d}:=\bigoplus_{m\geq 0}S^{m}H^{0}(X,L_{d})$. By Hilbert syzygy theorem,
the finitely generated graded section $S_{d}$-module

$R_{d}=R(X,B;L_{d}):=\bigoplus_{m\in\mathbf{Z}}H^{0}(X,B\otimes L_{d}^{m})$

admits a minimal free resolution

$\textstyle{0}$$\textstyle{R_{d}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\ignorespaces\ignorespaces\ignorespaces\ignorespaces\cdots}$$\textstyle{\ignorespaces\ignorespaces\ignorespaces\ignorespaces
E_{r}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\ignorespaces\ignorespaces\ignorespaces\ignorespaces
0,}$

where

$E_{p}=\bigoplus_{q}K_{p,q}(X,B;L_{d})\otimes_{\mathbf{k}}S_{d}(-p-q).$

Here the _Koszul cohomology group_ $K_{p,q}(X,B;L_{d})$ can be regarded as the
space of $p$-th syzygies of weight $q$. When
$X\subseteq\mathbf{P}H^{0}(X,L_{d})=\mathbf{P}^{r_{d}}$ is projectively
normal, the minimal free resolution of $R_{d}$ with $B=\mathscr{O}_{X}$
contains all information about the defining equations of $X$ in
$\mathbf{P}^{r_{d}}$ and their syzygies. Based on the experience of the case
of curves, it was widely believed that the minimal free resolutions of $R_{d}$
become simpler as $d$ increases. However, Ein–Lazarsfeld [8] showed that this
had been misleading, and they instead proposed that there would be a uniform
asymptotic vanishing and nonvanishing behavior of $K_{p,q}(X,B;L_{d})$ when
$d$ is sufficiently large.

It is elementary to see that

$K_{p,q}(X,B;L_{d})=0~{}~{}\text{ for $q\geq n+2$}.$

The cases $q=0$ and $q=n+1$ are well understood due to Green, Schreyer,
Ottaviani–Paoletti, and Ein–Lazarsfeld: [8, Proposition 5.1 and Corollary 5.2]
state that

$\begin{array}[]{rcl}K_{p,0}(X,B;L_{d})\neq 0&\Longleftrightarrow&0\leq p\leq
h^{0}(B)-1;\\\\[3.0pt] K_{p,n+1}(X,B;L_{d})\neq
0&\Longleftrightarrow&r_{d}-n-h^{0}(X,\omega_{X}\otimes B^{-1})+1\leq p\leq
r_{d}-n.\end{array}$

For $1\leq q\leq n$, let $c_{q}(d)$ be the number such that

$K_{c_{q}(d),q}(X,B;L_{d})\neq 0~{}~{}\text{ and
}~{}~{}K_{p,q}(X,B;L_{d})=0~{}~{}\text{ for $0\leq p\leq c_{q}(d)-1$},$

and $c_{q}^{\prime}(d)$ be the number such that

$K_{r_{d}-c_{q}^{\prime}(d),q}(X,B;L_{d})\neq 0~{}~{}\text{ and
}~{}~{}K_{p,q}(X,B;L_{d})=0~{}~{}\text{ for $r_{d}-c_{q}^{\prime}(d)+1\leq
p\leq r_{d}$}.$

After interesting nonvanishing results of Ottaviani–Paoletti [18] and
Eisenbud–Green–Hulek–Popescu [12], Ein–Lazarsfeld proved the asymptotic
nonvanishing theorem ([8, Theorem 4.1]): For each $1\leq q\leq n$, if $d$ is
sufficiently large, then

$K_{p,q}(X,B;L_{d})\neq 0~{}~{}\text{ for $\Theta(d^{q-1})\leq p\leq
r_{d}-\Theta(d^{n-1})$}.$

In particular, $c_{q}(d)\geq\Theta(d^{q-1})$ and
$c_{q}^{\prime}(d)\leq\Theta(d^{n-1})$. In [8, Conjecture 7.1], Ein–Lazarsfeld
conjectured that

$K_{p,q}(X,B;L_{d})=0~{}~{}\text{ for $0\leq p\leq\Theta(d^{q-1})$},$

and this was confirmed by the present author [19, Theorem 1.1] using Raicu’s
result in the appendix of [20]. In particular, $c_{q}(d)=\Theta(d^{q-1})$.

Despite the aforementioned results on asymptotic syzygies of algebraic
varieties, at least two problems remain. First, it is unclear whether
vanishing and nonvanishing of $K_{p,q}(X,B;L_{d})$ can alternate in a few
steps after $c_{q}(d)$ or before $r_{d}-c_{q}^{\prime}(d)$. Second, the
previous results do not say anything about $K_{p,q}(X,B;L_{d})$ for
$r_{d}-\Theta(d^{n-1})\leq p\leq r_{d}$. In this paper, we completely resolve
these two issues: We show that vanishing and nonvanishing of
$K_{p,q}(X,B;L_{d})$ can alternate only at $c_{q}(d)$ and
$r_{d}-c_{q}^{\prime}(d)$, and we give estimations for $c_{q}(d)$ and
$c_{q}^{\prime}(d)$. Consequently, we could determine the precise vanishing
and nonvanishing range of $p$ for $K_{p,q}(X,B;L_{d})$. It is worth noting
that $c_{q}^{\prime}(d)$ heavily depends on $H^{q-1}(X,B)$ while $c_{q}(d)$
depends only on $d$ asymptotically.

######  Theorem 1.1.

Let $X$ be a smooth projective variety of dimension $n\geq 1$, and $B$ be a
line bundle on $X$. For an integer $d\geq 1$, set

$L_{d}:=\mathscr{O}_{X}(dA+P)~{}~{}\text{ and }~{}~{}r_{d}:=h^{0}(X,L_{d})-1,$

where $A$ is an ample divisor and $P$ is an arbitrary divisor on $X$. Fix an
index $1\leq q\leq n$. Then there exist functions $c_{q}(d)$ and
$c_{q}^{\prime}(d)$ with

(1.1) $c_{q}(d)=\Theta(d^{q-1})~{}~{}\text{ and
}~{}~{}c_{q}^{\prime}(d)=\begin{cases}\Theta(d^{n-q})&\text{if
$H^{q-1}(X,B)=0$ or $q=1$}\\\ q-1&\text{if $H^{q-1}(X,B)\neq 0$ and $q\geq
2$}\end{cases}$

such that if $d$ is sufficiently large, then

(1.2) $K_{p,q}(X,B;L_{d})\neq 0~{}~{}\Longleftrightarrow~{}~{}c_{q}(d)\leq
p\leq r_{d}-c_{q}^{\prime}(d).$

To prove Theorem 1.1, we do not use Ein–Lazarsfeld’s asymptotic nonvanishing
theorem [8, Theroem 4.1], but we adopt their strategy in [8]. Let $H$ be a
suitably positive very ample line bundle on $X$, and choose a general member
$\overline{X}\in|H|$. Put

$\overline{L}_{d}:=L_{d}|_{\overline{X}},~{}~{}\overline{B}:=B|_{\overline{X}},~{}~{}\overline{H}:=H|_{\overline{X}},~{}~{}V_{d}^{\prime}:=H^{0}(X,L_{d}\otimes
H^{-1}).$

We have noncanonical splitting

$K_{p,q}(X,\overline{B};L_{d})=\bigoplus_{j=0}^{p}\wedge^{j}V_{d}^{\prime}\otimes
K_{p-j,q}(\overline{X},\overline{B};\overline{L}_{d})~{}~{}\text{ and
}~{}~{}K_{p,q}(X,\overline{B}\otimes\overline{H};L_{d})=\bigoplus_{j=0}^{p}\wedge^{j}V_{d}^{\prime}\otimes
K_{p-j,q}(\overline{X},\overline{B}\otimes\overline{H};\overline{L}_{d}).$

There are natural maps

$\theta_{p,q}\colon
K_{p+1,q-1}(X,\overline{B}\otimes\overline{H};L_{d})\longrightarrow
K_{p,q}(X,B;L)~{}~{}\text{and }~{}~{}\theta_{p,q}^{\prime}\colon
K_{p,q}(X,\overline{B};L_{d})\longrightarrow K_{p,q}(X,B;L_{d}).$

When $q=1$ or $2$, the map $\theta_{p,q}$ should be modified (see Section 3).
In [8, Sections 3 and 4], the secant constructions were introduced to carry
nonzero syzygies by highly secant planes. This essentially shows that the map
$\theta_{p,q}$ is nonzero for $\Theta(d^{q-1})\leq p\leq
r_{d}-\Theta(d^{n-1})$. Instead of utilizing the secant constructions, in this
paper, we apply the asymptotic vanishing theorem [19, Theorem 1.1] to get the
estimations of $c_{q}(d)$ and $c_{q}^{\prime}(d)$ (Proposition 4.4) and to see
that the maps $\theta_{c_{q}(d),q}$ and
$\theta_{r_{d}-c_{q}^{\prime}(d),q}^{\prime}$ are nonzero (Proposition 4.6).
The latter statement means that there are syzygies $\alpha$ and $\beta$ in

($\star$)
$K_{c_{q}(d)+1-j_{0},q-1}(\overline{X},\overline{B}\otimes\overline{H};\overline{L}_{d})~{}~{}\text{
and
}~{}~{}K_{r_{d}-c_{q}^{\prime}(d)-j_{0}^{\prime},q}(\overline{X},\overline{B};\overline{L}_{d})$

for some $0\leq j_{0}\leq c_{q}(d)+1$ and $\dim
V_{d}^{\prime}-c_{q}^{\prime}(d)\leq j_{0}^{\prime}\leq\dim V_{d}^{\prime}$
that are lifted to syzygies in $K_{c_{q}(d),q}(X,B;L_{d})$ and
$K_{r_{d}-c_{q}^{\prime}(d),q}(X,B;L_{d})$ via the maps $\theta_{c_{q}(d),q}$
and $\theta_{r_{d}-c_{q}^{\prime}(d),q}^{\prime}$, respectively. Since
($\star$ ‣ 1) survives in

$\text{$K_{p,q}(X,\overline{B}\otimes\overline{H};L_{d})$ for $c_{q}(d)\leq
p\leq r_{d}-\Theta(d^{n-1})$ and $K_{p,q}(X,\overline{B};L_{d})$ for
$\Theta(d^{n-1})\leq p\leq r_{d}-c_{q}^{\prime}(d)$},$

we can argue that the syzygies $\alpha$ and $\beta$ in ($\star$ ‣ 1) are also
lifted to syzygies in $K_{p,q}(X,B;L_{d})$ for $c_{q}(d)\leq p\leq
r_{d}-c_{q}^{\prime}(d)$ via the maps $\theta_{p,q}$ and
$\theta_{p,q}^{\prime}$, respectively (Theorem 3.1).

The paper is organized as follows. After recalling preliminary results on
syzygies of algebraic varieties in Section 2, we show how to lift syzygies
from hypersurfaces (Theorem 3.1) in Section 3. Section 4 is devoted to the
proof of Theorem 1.1. Finally, in Section 5, we present complementary results
and open problems on asymptotic syzygies of algebraic varieties.

### Acknowledgements

The author is very grateful to Lawrence Ein, Yeongrak Kim, and Wenbo Niu for
inspiring discussions.

##  2\. Preliminaries

In this section, we collect relevant basic facts on Koszul cohomology and
Castelnuovo–Mumford regularity.

###  2.1. Koszul Cohomology

Let $V$ be an $r$-dimensional vector space over an algebraically closed field
$\mathbf{k}$, and $S:=\bigoplus_{m\geq 0}S^{m}V$. Consider a finitely
generated graded $S$-module $M$. The _Koszul cohomology group_ $K_{p,q}(M,V)$
is the cohomology of the Koszul-type complex

$\wedge^{p+1}V\otimes M_{q-1}\xrightarrow{~{}\delta~{}}\wedge^{p}V\otimes
M_{q}\xrightarrow{~{}\delta~{}}\wedge^{p-1}V\otimes M_{q+1},$

where the Koszul differential $\delta$ is given by

$\delta(s_{1}\wedge\cdots\wedge s_{p}\otimes
t)\longmapsto\sum_{i=1}^{p}(-1)^{i}s_{1}\wedge\cdots\wedge\widehat{s}_{i}\wedge\cdots\wedge
s_{p}\otimes s_{i}t.$

Then $M$ has a minimal free resolution

$\textstyle{0}$$\textstyle{M\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\ignorespaces\ignorespaces\ignorespaces\ignorespaces\cdots}$$\textstyle{\ignorespaces\ignorespaces\ignorespaces\ignorespaces
E_{r}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\ignorespaces\ignorespaces\ignorespaces\ignorespaces
0,}$

where

$E_{p}=\bigoplus_{q}K_{p,q}(M,V)\otimes_{\mathbf{k}}S(-p-q).$

We may regard $K_{p,q}(M,V)$ as the vector space of $p$-th syzygies of weight
$q$. Let

$0\longrightarrow M^{\prime}\longrightarrow M\longrightarrow
M^{\prime\prime}\longrightarrow 0$

be a short exact sequence of finitely generated graded $S$-modules. By [16,
Corollary (1.d.4)] (see also [2, Lemma 1.24]), this induces a long exact
sequence

$\cdots\longrightarrow K_{p+1,q-1}(M,V)\longrightarrow
K_{p+1,q-1}(M^{\prime\prime},V)\longrightarrow
K_{p,q}(M^{\prime},V)\longrightarrow K_{p,q}(M,V)\longrightarrow\cdots.$

Consider an injective map

$\iota^{\prime}\colon\wedge^{p+1}V\longrightarrow
V\otimes\wedge^{p}V,~{}~{}s_{1}\wedge\cdots\wedge
s_{p+1}\longmapsto\sum_{i=1}^{p+1}(-1)^{i}s_{i}\otimes
s_{1}\wedge\cdots\wedge\widehat{s}_{i}\wedge\cdots\wedge s_{p+1}.$

It is straightforward to check that the following diagram commutes:

$\textstyle{\wedge^{p+1}V\otimes
M_{q}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\delta}$$\scriptstyle{\iota^{\prime}\otimes\operatorname{id}_{M_{q}}}$$\textstyle{\wedge^{p}V\otimes
M_{q+1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\iota^{\prime}\otimes\operatorname{id}_{M_{q+1}}}$$\textstyle{V\otimes\wedge^{p}V\otimes
M_{q}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{-\operatorname{id}_{V}\otimes\delta}$$\textstyle{V\otimes\wedge^{p-1}V\otimes
M_{q+1}.}$

Then the map $\iota^{\prime}$ induces a map

$\iota\colon K_{p+1,q}(M,V)\longrightarrow V\otimes K_{p,q}(M,V).$

This map is the glueing of the evaluation maps $\operatorname{ev}_{x}\colon
K_{p+1,q}(M,V)\longrightarrow K_{p,q}(M,V)$ for $x\in V^{\vee}$ in [2,
Subsection 2.2.1].

We now turn to the geometric setting. Let $X$ be a projective variety, $B$ be
a coherent sheaf on $X$, and $L$ be a very ample line bundle on $X$. Put
$V:=H^{0}(X,L)$ and $S:=\bigoplus_{m\geq 0}S^{m}V$. Then the section module

$R(X,B;L):=\bigoplus_{m\in\mathbf{Z}}H^{0}(X,B\otimes L^{m}),$

is finitely generated graded $S$-module. We define the _Koszul cohomology
group_ as

$K_{p,q}(X,B;L):=K_{p,q}(R(X,B;L),V).$

In this paper, $L$ is always assumed to be sufficiently positive, so we have

$H^{0}(X,B\otimes L^{-m})=0~{}~{}\text{ for $m>0$}.$

Then $K_{p,q}(X,B;L)=0$ for $q<0$. It is clear that if $K_{p_{0},q}(X,B;L)=0$
for $0\leq q\leq q_{0}$, then $K_{p,q}(X,B;L)=0$ for $p\geq p_{0}$ and $0\leq
q\leq q_{0}$. Let $M_{L}$ be the kernel bundle of the evaluation map
$\operatorname{ev}\colon H^{0}(X,L)\otimes\mathscr{O}_{X}\rightarrow L$. We
have a short exact sequence

$0\longrightarrow M_{L}\longrightarrow
H^{0}(X,L)\otimes\mathscr{O}_{X}\longrightarrow L\longrightarrow 0.$

We frequently use the following well-known facts.

######  Proposition 2.1 (cf. [8, Proposition 3.2, Corollary 3.3, Remark 3.4],
[19, Proposition 2.1]).

Assume that

$H^{i}(X,B\otimes L^{m})=0~{}~{}\text{ for $i>0$ and $m>0$.}$

For any $p\geq 0$, the following hold:
$(1)$ If $q\geq 2$, then

$\begin{array}[]{rcl}K_{p,q}(X,B;L)&=&H^{1}(X,\wedge^{p+1}M_{L}\otimes
B\otimes L^{q-1})\\\ &=&H^{2}(X,\wedge^{p+2}M_{L}\otimes B\otimes L^{q-2})\\\
&&~{}\text{ }~{}\text{ }~{}\text{ }~{}\text{ }~{}\text{ }~{}\text{ }~{}\text{
}~{}\vdots\\\ &=&H^{q-1}(X,\wedge^{p+q-1}M_{L}\otimes B\otimes L).\end{array}$

Consequently, $K_{p,q}(X,B;L_{d})=0$ for $p\geq r_{d}-q$, and
$K_{p,q}(X,B;L_{d})=0$ for $q\geq\dim X+2$.
$(2)$ If $q\geq 2$ and $H^{q-1}(X,B)=H^{q}(X,B)=0$, then
$K_{p,q}(X,B;L)=H^{q}(X,\wedge^{p+q}M_{L}\otimes B)$.
$(3)$ If $q=1$, then

$K_{p,1}(X,B;L)=\operatorname{coker}\big{(}\wedge^{p+1}H^{0}(X,L)\otimes
H^{0}(X,B)\longrightarrow H^{0}(X,\wedge^{p}M_{L}\otimes B\otimes L)\big{)}.$

If $H^{1}(X,B)=0$, then $K_{p,1}(X,B;L)=H^{1}(X,\wedge^{p+1}M_{L}\otimes B)$.
$(4)$ If $q=0$ and $H^{0}(X,B\otimes L^{-m})=0$ for $m>0$, then
$K_{p,0}(X,B;L)=H^{0}(X,\wedge^{p}M_{L}\otimes B)$.

###### Proof.

We have a short exact sequence

(2.1)
$0\longrightarrow\wedge^{p+1}M_{L}\longrightarrow\wedge^{p+1}H^{0}(X,L)\otimes\mathscr{O}_{X}\longrightarrow\wedge^{p}M_{L}\otimes
L\longrightarrow 0.$

Using the Koszul-type complex and chasing through the diagram, we see that

$K_{p,q}(X,B;L)=\operatorname{coker}\big{(}\wedge^{p+1}H^{0}(X,L)\otimes
H^{0}(X,B\otimes L^{q-1})\longrightarrow H^{0}(X,\wedge^{p}M_{L}\otimes
B\otimes L^{q})\big{)}.$

Then the proposition easily follows. See [2, Section 2.1], [7, Section 1], [8,
Section 3]. ∎

######  Proposition 2.2 (cf. [1, Proposition 2.4], [8, Proposition 3.5]).

Put $n:=\dim X$. Assume that $X$ is smooth, $B$ is a line bundle, and

$H^{i}(X,B\otimes L^{m})=0~{}~{}\text{ for $i>0$ and $m>0$ or $i<n$ and
$m<0$.}$

For any $p\geq 0$, the following hold:
$(1)$ If $q=n+1$, then $K_{p,n+1}(X,B;L)=K_{r-p-n,0}(X,\omega_{X}\otimes
B^{-1};L)^{\vee}$.
$(2)$ If $q=n$, then there is an exact sequence

$\wedge^{p+n}H^{0}(X,L)\otimes H^{n-1}(X,B)\longrightarrow
K_{p,n}(X,B;L)\longrightarrow K_{r-p-n,1}(X,\omega_{X}\otimes
B^{-1};L_{d})^{\vee}\longrightarrow 0.$

If $H^{n-1}(X,B)=0$, then $K_{p,n}(X,B;L)=K_{r-p-n,1}(X,\omega_{X}\otimes
B^{-1};L)^{\vee}$.
$(3)$ If $2\leq q\leq n-1$, then there is an exact sequence

$\begin{array}[]{l}\wedge^{p+q}H^{0}(X,L)\otimes H^{q-1}(X,B)\longrightarrow
K_{p,q}(X,B;L)\\\\[3.0pt] \text{ }~{}\text{ }~{}\text{ }\longrightarrow
K_{r-p-n,n+1-q}(X,\omega_{X}\otimes
B^{-1};L_{d})^{\vee}\longrightarrow\wedge^{p+q}H^{0}(X,L)\otimes
H^{q}(X,B).\end{array}$

If $H^{q-1}(X,B)=H^{q}(X,B)=0$, then
$K_{p,q}(X,B;L)=K_{r-p-n,n+1-q}(X,\omega_{X}\otimes B^{-1};L_{d})^{\vee}$.
$(4)$ If $q=1$, then there is an exact sequence

$0\longrightarrow K_{p,1}(X,B;L)\longrightarrow
K_{r-p-n,n}(X,\omega_{X}\otimes
B^{-1};L_{d})^{\vee}\longrightarrow\wedge^{p+1}H^{0}(X,L)\otimes H^{1}(X,B)$

If $H^{1}(X,B)=0$, then $K_{p,1}(X,B;L)=K_{r-p-n,n}(X,\omega_{X}\otimes
B^{-1};L_{d})^{\vee}$.
$(5)$ If $q=0$, then $K_{p,0}(X,B;L)=K_{r-p-n,n+1}(X,\omega_{X}\otimes
B^{-1};L)^{\vee}$.

###### Proof.

Let $V:=H^{0}(X,L)$. From (2.1), we get an exact sequence

$\wedge^{p+q}V\otimes H^{q-1}(X,B)\longrightarrow
H^{q-1}(X,\wedge^{p+q-1}M_{L}\otimes B\otimes L)\longrightarrow
H^{q}(X,\wedge^{p+q}M_{L}\otimes B)\longrightarrow\wedge^{p+q}V\otimes
H^{q}(X,B).$

By Proposition 2.1, $H^{q-1}(X,\wedge^{p+q-1}M_{L}\otimes B\otimes
L)=K_{p,q}(X,B;L)$ when $q\geq 2$. By Serre duality,

$H^{q}(X,\wedge^{p+q}M_{L}\otimes
B)=H^{n-q}(X,\wedge^{p+q}M_{L}^{\vee}\otimes\omega_{X}\otimes B^{-1})^{\vee}.$

Since $\operatorname{rank}M_{L}=r$ and $\det M_{L}=L^{-1}$, it follows that
$\wedge^{p+q}M_{L}^{\vee}=\wedge^{r-p-q}M_{L}\otimes L$. Thus

$H^{q}(X,\wedge^{p+q}M_{L}\otimes
B)=H^{n-q}(X,\wedge^{r-p-q}M_{L}\otimes\omega_{X}\otimes B^{-1}\otimes
L)^{\vee}.$

Using Proposition 2.1, the assertions easily follow. See [2, Section 2.3], [8,
Section 3]. ∎

In the situation of Proposition 2.2, if we further assume $H^{i}(X,B)=0$ for
$1\leq i\leq n-1$, i.e., $R(X,B;L)$ and $R(X,\omega_{X}\otimes B^{-1};L)$ are
Cohen–Macaulay, then

$K_{p,q}(X,B;L)=K_{r-p-n,n+1-q}(X,\omega_{X}\otimes B^{-1};L)^{\vee}.$

######  Lemma 2.3.

If $H^{q}(X,M_{L}\otimes\wedge^{p}M_{L}\otimes B)=0$ and $H^{q}(X,B)=0$, then
$H^{q}(X,\wedge^{p+1}M_{L}\otimes B)=0$.

###### Proof.

Let $V:=H^{0}(X,L)$. Consider the commutative diagram with exact rows

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{M_{L}\otimes\wedge^{p}M_{L}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{V\otimes\wedge^{p}M_{L}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\wedge^{p}M_{L}\otimes
L\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\wedge^{p+1}M_{L}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\wedge^{p+1}V\otimes\mathscr{O}_{X}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\wedge^{p}M_{L}\otimes
L\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0,}$

which gives rise to the following commutative diagram with exact rows

$\textstyle{H^{q-1}(X,\wedge^{p}M_{L}\otimes B\otimes
L)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{H^{q}(X,M_{L}\otimes\wedge^{p}M_{L}\otimes
B)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{V\otimes
H^{q}(X,\wedge^{p}M_{L}\otimes
B)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{H^{q-1}(X,\wedge^{p}M_{L}\otimes
B\otimes
L)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{H^{q}(X,\wedge^{p+1}M_{L}\otimes
B)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\wedge^{p+1}V\otimes
H^{q}(X,B).}$

Since $H^{q}(X,B)=0$, the middle vertical map is surjective. Thus the lemma
follows. ∎

###  2.2. Castelnuovo–Mumford Regularity

Let $X$ be a projective variety, and $L$ be a very ample line bundle on $X$. A
coherent sheaf $\mathscr{F}$ on $X$ is said to be _$m$ -regular_ with respect
to $L$ if

$H^{i}(X,\mathscr{F}\otimes\mathscr{O}_{X}(m-i))=0~{}~{}\text{ for $i>0$}.$

By Mumford’s theorem ([17, Theorem 1.8.5]), if $\mathscr{F}$ is $m$-regular
with respect to $L$, then $\mathscr{F}\otimes L^{m+\ell}$ is globally
generated, the multiplication map

$H^{0}(X,\mathscr{F}\otimes L^{m})\otimes H^{0}(X,L^{\ell})\longrightarrow
H^{0}(X,\mathscr{F}\otimes L^{m+\ell})$

is surjective, and $\mathscr{F}$ is $(m+\ell)$-regular with respect to $L$ for
every $\ell\geq 0$.

######  Lemma 2.4 (cf. [3, Corollary 3.2]).

If $\mathscr{O}_{X}$ is $k$-regular with $k\geq 1$ and $\mathscr{F}$ is
$m$-regular with respect to $L$, then there are finite dimensional vector
spaces $W_{N},\ldots,W_{1},W_{0}$ over $\mathbf{k}$ and a resolution of
$\mathscr{F}$ of the form

$W_{N}\otimes L^{-m-Nk}\longrightarrow\cdots\longrightarrow W_{1}\otimes
L^{-m-k}\longrightarrow W_{0}\otimes
L^{-m}\longrightarrow\mathscr{F}\longrightarrow 0.$

###### Proof.

By [17, Theorem 1.8.5], $\mathscr{F}\otimes L^{m}$ is globally generated.
Letting $W_{0}:=H^{0}(X,\mathscr{F}\otimes L^{m})$, we have a short exact
sequence

$0\longrightarrow M_{0}\longrightarrow W_{0}\otimes
L^{-m}\longrightarrow\mathscr{F}\longrightarrow 0.$

By [17, Theorem 1.8.5], the map

$W_{0}\otimes H^{0}(X,L^{(m+k)-m-1})\longrightarrow H^{0}(X,\mathscr{F}\otimes
L^{(m+k)-1})$

is surjective. Note that

$H^{i}(X,L^{(m+k)-m-i})=0~{}~{}\text{ for $i\geq
1$}~{}~{}and~{}~{}H^{i-1}(X,\mathscr{F}\otimes L^{(m+k)-i})=0~{}~{}\text{ for
$i\geq 2$}.$

Thus $M_{0}$ is $(m+k)$-regular with respect to $L$. Replacing $\mathscr{F}$
by $M_{0}$ and continuing the arguments, we obtain the lemma. ∎

##  3\. Lifting Syzygies from Hypersurfaces

The aim of this section is showing how to lift syzygies from hypersurfaces
(see Theorem 3.1). This is the main ingredient of the proof of Theorem 1.1. We
start by setting notations. Let $X$ be a smooth projective variety, $B$ be a
line bundle on $X$, and $L$ be a very ample line bundle on $X$. Assume that
$n:=\dim X\geq 2$. Take a very ample line bundle $H$ on $X$, and suppose that

(3.1) $\begin{array}[]{l}H^{i}(X,B\otimes L^{m})=0~{}~{}\text{ for $i>0$ and
$m>0$ or $i<n$ and $m<0$};\\\\[3.0pt] H^{i}(X,B\otimes H\otimes
L^{m})=H^{i}(X,B\otimes H^{-1}\otimes L^{m})=0~{}~{}\text{ for $1\leq i\leq
n-1$ and $m\in\mathbf{Z}$};\\\\[3.0pt] H^{0}(X,B\otimes H\otimes
L^{m})=H^{0}(X,B\otimes H^{-1}\otimes L^{m})=0~{}~{}\text{ for
$m<0$};\\\\[3.0pt] H^{n}(X,B\otimes H\otimes L^{m})=H^{n}(X,B\otimes
H^{-1}\otimes L^{m})=0~{}~{}\text{ for $m>0$}.\end{array}$

In particular, $R(X,B\otimes H;L)$ and $R(X,B\otimes H^{-1};L)$ are
Cohen–Macaulay. Choose a general member $\overline{X}\in|H|$, and put

$\begin{array}[]{l}\overline{L}:=L|_{\overline{X}},~{}~{}\overline{B}:=B|_{\overline{X}},~{}~{}\overline{H}:=H|_{\overline{X}};\\\\[3.0pt]
V:=H^{0}(X,L),~{}~{}V^{\prime}:=H^{0}(X,L\otimes
H^{-1}),~{}~{}\overline{V}:=H^{0}(X,\overline{L});\\\\[3.0pt] r:=\dim
V-1,v^{\prime}:=\dim
V^{\prime},~{}~{}\overline{r}:=\dim\overline{V}-1\end{array}$

so that $r=v^{\prime}+\overline{r}$. Fix a splitting
$V=V^{\prime}\oplus\overline{V}$. As in [8, Lemma 3.12], we get

(3.2)
$\displaystyle\wedge^{p+1}M_{L}|_{\overline{X}}=\bigoplus_{j=0}^{p+1}\wedge^{j}V^{\prime}\otimes\wedge^{p+1-j}M_{\overline{L}}~{}~{}\text{
for $p\geq 0$}.$

By Proposition 2.1, we obtain

(3.3) $\begin{array}[]{l}\displaystyle
K_{p,q}(X,\overline{B};L)=\bigoplus_{j=0}^{p}\wedge^{j}V^{\prime}\otimes
K_{p-j,q}(\overline{X},\overline{B};\overline{L})~{}~{}\text{ for $p,q\geq
0$};\\\\[3.0pt] \displaystyle
K_{p,q}(X,\overline{B}\otimes\overline{H};L)=\bigoplus_{j=0}^{p}\wedge^{j}V^{\prime}\otimes
K_{p-j,q}(\overline{X},\overline{B}\otimes\overline{H};\overline{L})~{}~{}\text{
for $p,q\geq 0$}.\end{array}$

Now, put $S:=\bigoplus_{m\geq 0}S^{m}V$ and $\overline{S}:=\bigoplus_{m\geq
0}S^{m}\overline{V}$. Consider the following short exact sequence

(3.4) $0\longrightarrow B\longrightarrow B\otimes
H\longrightarrow\overline{B}\otimes\overline{H}\longrightarrow 0.$

By (3.1), we have an exact sequence of finitely generated graded $S$-modules

$0\longrightarrow R(X,B;L)\longrightarrow R(X,B\otimes H;L)\longrightarrow
R(X,\overline{B}\otimes\overline{H};L)\longrightarrow
H^{1}(X,B)\longrightarrow 0.$

Let $\overline{R}(X,\overline{B}\otimes\overline{H};L)$ be the kernel of the
map $R(X,\overline{B}\otimes\overline{H};L)\longrightarrow H^{1}(X,B)$, which
is a finitely generated graded $\overline{S}$-module. Then we get a short
exact sequence of finitely generated graded $S$-modules

$0\longrightarrow R(X,B;L)\longrightarrow R(X,B\otimes
H;L)\longrightarrow\overline{R}(X,\overline{B}\otimes\overline{H};L)\longrightarrow
0.$

By [16, Corollary (1.d.4)], this induces a connecting map

(3.5)
$\theta_{p,q}\colon\underbrace{\overline{K}_{p+1,q-1}(X,\overline{B}\otimes\overline{H};L)}_{:=K_{p+1,q-1}(\overline{R}(X,\overline{B}\otimes\overline{H};L),V)}\longrightarrow
K_{p,q}(X,B;L).$

Notice that

(3.6)
$\overline{K}_{p+1,q-1}(X,\overline{B}\otimes\overline{H};L)=\bigoplus_{j=0}^{p+1}\wedge^{j}V^{\prime}\otimes\underbrace{\overline{K}_{p+1-j,q-1}(\overline{X},\overline{B};\overline{L})}_{\mathclap{:=K_{p+1-j,q-1}(\overline{R}(X,\overline{B}\otimes\overline{H};L),\overline{V})}}.$

On the other hand, we also have a short exact sequence of finitely generated
graded $S$-modules

$0\longrightarrow\overline{R}(X,\overline{B}\otimes\overline{H};L)\longrightarrow
R(X,\overline{B}\otimes\overline{H};L)\longrightarrow
H^{1}(X,B)\longrightarrow 0.$

Since

$K_{p,q}(H^{1}(X,B),V)=\begin{cases}\wedge^{p}V\otimes H^{1}(X,B)&\text{if
$q=0$}\\\ 0&\text{if $q\geq 1$},\end{cases}$

we get an exact sequence

(3.7)
$\begin{array}[]{l}0\longrightarrow\overline{K}_{p+1,0}(X,\overline{B}\otimes\overline{H};L)\xrightarrow{~{}\psi_{0}~{}}K_{p+1,0}(X,\overline{B}\otimes\overline{H};L)\\\\[3.0pt]
~{}\text{ }~{}\text{ }~{}\text{ }~{}\text{ }~{}\text{
}~{}\longrightarrow\wedge^{p+1}V\otimes
H^{1}(X,B)\xrightarrow{~{}\varphi~{}}\overline{K}_{p,1}(X,\overline{B}\otimes\overline{H};L)\xrightarrow{~{}\psi_{1}~{}}K_{p,1}(X,\overline{B}\otimes\overline{H};L)\longrightarrow
0\end{array}$

and an isomorphism

(3.8)
$\psi_{q-1}\colon\overline{K}_{p+1,q-1}(X,\overline{B}\otimes\overline{H};L)\longrightarrow
K_{p+1,q-1}(X,\overline{B}\otimes\overline{H};L)~{}~{}\text{ for $q\geq 3$}.$

In view of (3.6), there is a map

$\overline{\psi}_{p+1-j,q-1}\colon\overline{K}_{p+1-j,q-1}(\overline{X},\overline{B}\otimes\overline{H};\overline{L})\longrightarrow
K_{p+1-j,q-1}(\overline{X},\overline{B}\otimes\overline{H};\overline{L})$

such that

$\psi_{q-1}=\bigoplus_{j=0}^{p+1}\operatorname{id}_{\wedge^{j}V^{\prime}}\otimes\overline{\psi}_{p+1-j,q-1}.$

Next, consider the following short exact sequence

(3.9) $0\longrightarrow B\otimes H^{-1}\longrightarrow
B\longrightarrow\overline{B}\longrightarrow 0.$

By (3.1), we have a short exact sequence of finitely generated graded
$S$-modules

$0\longrightarrow R(X,B\otimes H^{-1};L)\longrightarrow
R(X,B;L)\longrightarrow R(X,\overline{B};L)\longrightarrow 0.$

By [16, Corollary (1.d.4)], this induces a restriction map

(3.10) $\theta_{p,q}^{\prime}\colon K_{p,q}(X,B;L)\longrightarrow
K_{p,q}(X,\overline{B};L).$

######  Theorem 3.1.

Fix an index $q\geq 1$. Then we have the following:
$(1)$ Suppose that the map $\theta_{p,q}$ in (3.5) is a nonzero map for $p=c$
with $0\leq c\leq r-\overline{r}-1$. Then the map $\theta_{p,q}$ is a nonzero
map for $c\leq p\leq r-\overline{r}-1$, and consequently,

$K_{p,q}(X,B;L)\neq 0~{}~{}\text{ for $c\leq p\leq r-\overline{r}-1$}.$

$(2)$ Suppose that the map $\theta_{p,q}^{\prime}$ in (3.10) is a nonzero map
for $p=r-c^{\prime}$ with $0\leq c^{\prime}\leq\overline{r}$. Then
$\theta_{p,q}^{\prime}$ is a nonzero map for $\overline{r}\leq p\leq
r-c^{\prime}$, and consequently,

$K_{p,q}(X,B;L)\neq 0~{}~{}\text{ for $\overline{r}\leq p\leq r-c^{\prime}$}.$

###### Proof.

$(1)$ There is
$\alpha_{c+1}\in\overline{K}_{c+1,q-1}(X,\overline{B}\otimes\overline{H};L)$
such that $\theta_{c,q}(\alpha_{c+1})\neq 0$. We may assume that

$\alpha_{c+1}=s_{1}^{\prime}\wedge\cdots\wedge
s_{j_{0}}^{\prime}\otimes\alpha^{\prime}\in\wedge^{j_{0}}V^{\prime}\otimes\overline{K}_{c+1-j_{0},q-1}(\overline{X},\overline{B};\overline{L})\subseteq\overline{K}_{c+1,q-1}(X,\overline{B}\otimes\overline{H};L)$

for some $0\leq j_{0}\leq c+1$. We proceed by induction on $p$. For $c\leq
p-1\leq r-\overline{r}-2$, we may assume that there is

$\alpha_{p}=s_{1}^{\prime}\wedge\cdots\wedge
s_{j}^{\prime}\otimes\alpha^{\prime}\in\wedge^{j}V^{\prime}\otimes\overline{K}_{c+1-j_{0},q-1}(\overline{X},\overline{B};\overline{L})\subseteq\overline{K}_{p,q-1}(X,\overline{B}\otimes\overline{H};L),$

where $j=p-(c+1-j_{0})$, such that $\theta_{p-1,q}(\alpha_{p})\neq 0$.
Consider the commutative diagram

$\textstyle{\overline{K}_{p+1,q-1}(X,\overline{B}\otimes\overline{H};L)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\theta_{p,q}}$$\scriptstyle{\iota}$$\textstyle{K_{p,q}(X,B;L)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\iota}$$\textstyle{V\otimes\overline{K}_{p,q-1}(X,\overline{B}\otimes\overline{H};L)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\operatorname{id}_{V}\otimes\theta_{p-1,q}}$$\textstyle{V\otimes
K_{p-1,q}(X,B;L).}$

Take any $s_{j+1}^{\prime}\in V^{\prime}$ with
$s_{1}^{\prime}\wedge\cdots\wedge s_{j}^{\prime}\wedge s_{j+1}^{\prime}\neq
0$, and let

$\alpha_{p+1}:=s_{1}^{\prime}\wedge\cdots\wedge
s_{j+1}^{\prime}\otimes\alpha^{\prime}\in\wedge^{j+1}V^{\prime}\otimes\overline{K}_{c+1-j_{0},q-1}(\overline{X},\overline{B};\overline{L})\subseteq\overline{K}_{p+1,q-1}(X,\overline{B}\otimes\overline{H};L).$

Then

$\iota(\alpha_{p+1})=\underbrace{\textstyle\sum_{i=1}^{j+1}(-1)^{i}s_{i}^{\prime}\otimes
s_{1}^{\prime}\wedge\cdots\wedge\widehat{s_{i}^{\prime}}\wedge\cdots\wedge
s_{j+1}^{\prime}\otimes\alpha^{\prime}}_{\text{\tiny$\in
V^{\prime}\otimes\overline{K}_{p,q-1}(X,\overline{B}\otimes\overline{H};L)$}}+\underbrace{s_{1}^{\prime}\wedge\cdots\wedge
s_{j+1}^{\prime}\otimes\iota(\alpha^{\prime})}_{\text{\tiny$\in\overline{V}\otimes\overline{K}_{p,q-1}(X,\overline{B}\otimes\overline{H};L)$}}.$

Notice that

$\operatorname{id}_{V}\otimes\theta_{p-1,q}\big{(}(-1)^{j+1}s_{j+1}^{\prime}\otimes\alpha_{p}\big{)}=(-1)^{j+1}s_{j+1}^{\prime}\otimes\theta_{p-1,q}(\alpha_{p})\neq
0$

in $\langle s_{j+1}^{\prime}\rangle\otimes K_{p-1,q}(X,B;L)$. Observe then
that all other terms in $\iota(\alpha_{p+1})$ go into complements of $\langle
s_{j+1}^{\prime}\rangle\otimes K_{p-1,q}(X,B;L)$ via the map
$\operatorname{id}_{V}\otimes\theta_{p-1,q}$. Thus

$(\iota\circ\theta_{p,q})(\alpha_{p+1})=((\operatorname{id}_{V}\otimes\theta_{p-1,q})\circ\iota)(\alpha_{p+1})\neq
0,$

and hence, $\theta_{p,q}(\alpha_{p+1})\neq 0$.

$(2)$ There is $\beta_{r-c^{\prime}}\in K_{r-c^{\prime},q}(X,B;L)$ such that
$\theta_{r-c^{\prime},q}^{\prime}(\beta_{r-c^{\prime}})$ has a nonzero term

$s_{1}^{\prime}\wedge\cdots\wedge
s_{j_{0}}^{\prime}\otimes\beta^{\prime}\in\wedge^{j_{0}}V^{\prime}\otimes
K_{r-c^{\prime}-j_{0},q}(\overline{X},\overline{B};\overline{L})\subseteq
K_{r-c^{\prime},q}(X,\overline{B};L)$

for some $r-\overline{r}-c^{\prime}\leq j_{0}\leq r-\overline{r}$. We proceed
by reverse induction on $p$. For $\overline{r}+1\leq p+1\leq r-c^{\prime}$, we
may assume that there is $\beta_{p+1}\in K_{p+1,q}(X,B;L)$ such that
$\theta_{p+1,q}^{\prime}(\beta_{p+1})$ has a nonzero term

$s_{1}^{\prime}\wedge\cdots\wedge
s_{j+1}^{\prime}\otimes\beta^{\prime}\in\wedge^{j+1}V^{\prime}\otimes
K_{r-c^{\prime}-j_{0},q}(\overline{X},\overline{B};\overline{L})\subseteq
K_{p+1,q}(X,\overline{B};L),$

where $j+1=p+1-(r-c^{\prime}-j_{0})$. Consider the commutative diagram

$\textstyle{K_{p+1,q}(X,B;L)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\theta_{p+1,q}^{\prime}}$$\scriptstyle{\iota}$$\textstyle{K_{p+1,q}(X,\overline{B};L)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\iota}$$\textstyle{V\otimes
K_{p,q}(X,B;L)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\operatorname{id}_{V}\otimes\theta_{p,q}^{\prime}}$$\textstyle{V\otimes
K_{p,q}(X,\overline{B};L).}$

We have

$\iota(s_{1}^{\prime}\wedge\cdots\wedge
s_{j+1}^{\prime}\otimes\beta^{\prime})=\underbrace{\textstyle\sum_{i=1}^{j+1}(-1)^{i}s_{i}^{\prime}\otimes
s_{1}^{\prime}\wedge\cdots\wedge\widehat{s_{i}^{\prime}}\wedge\cdots\wedge
s_{j+1}^{\prime}\otimes\beta^{\prime}}_{\text{\tiny$\in
V^{\prime}\otimes\wedge^{j}V^{\prime}\otimes
K_{r-c^{\prime}-j_{0},q}(\overline{X},\overline{B};\overline{L})\subseteq
V^{\prime}\otimes
K_{p,q}(X,\overline{B};L)$}}+\underbrace{s_{1}^{\prime}\wedge\cdots\wedge
s_{j+1}^{\prime}\otimes\iota(\beta^{\prime})}_{\text{\tiny$\in\overline{V}\otimes
K_{p,q}(X,\overline{B};L)$}}.$

Note that all terms of $\theta_{p+1,q}^{\prime}(\beta_{p+1})$ not in
$\wedge^{j+1}V^{\prime}\otimes\langle\beta^{\prime}\rangle$ go into
complements of
$V^{\prime}\otimes\wedge^{j}V^{\prime}\otimes\langle\beta^{\prime}\rangle$ via
the map $\iota$. Observe then that the term

$(-1)^{j+1}s_{j+1}^{\prime}\otimes s_{1}^{\prime}\wedge\cdots\wedge
s_{j}^{\prime}\otimes\beta^{\prime}\in
V^{\prime}\otimes\wedge^{j}V^{\prime}\otimes
K_{r-c^{\prime}-j_{0},q}(\overline{X},\overline{B};\overline{L})\subseteq
V\otimes K_{p,q}(X,\overline{B};L)$

of $(\iota\circ\theta_{p+1,q}^{\prime})(\beta_{p+1})$ cannot be cancelled in
$V^{\prime}\otimes\wedge^{j}V^{\prime}\otimes\langle\beta^{\prime}\rangle$.
Thus there is $\beta_{p}\in K_{p,q}(X,B;L)$ such that
$(\operatorname{id}_{V}\otimes\theta_{p,q}^{\prime})((-1)^{j+1}s_{j+1}^{\prime}\otimes\beta_{p})$
has the above term, so $\theta_{p,q}^{\prime}(\beta_{p})$ has a nonzero term

$s_{1}^{\prime}\wedge\cdots\wedge
s_{j}^{\prime}\otimes\beta^{\prime}\in\wedge^{j}V^{\prime}\otimes
K_{r-c^{\prime}-j_{0},q}(\overline{X},\overline{B};\overline{L})\subseteq
K_{p,q}(X,\overline{B};L).$

We complete the proof. ∎

##  4\. Precise Asymptotic Nonvanishing Theorem

After establishing key steps as propositions, we finish the proof of Theorem
1.1 at the end of this section. We start by setting notations. Let $X$ be a
smooth projective variety of dimension $n\geq 1$, and $B$ be a line bundle on
$X$. For an integer $d\geq 1$, set

$L_{d}:=\mathscr{O}_{X}(dA+P)~{}~{}\text{and }~{}~{}r_{d}:=h^{0}(X,L_{d})-1,$

where $A$ is an ample divisor and $P$ is an arbitrary divisor on $X$. We
assume throughout that $d$ is sufficiently large so that $L_{d}$ is a
sufficiently positive very ample line bundle and $r_{d}=\Theta(d^{n})$.
Furthermore, we have

$H^{i}(X,B\otimes L_{d}^{m})=0~{}~{}\text{ for $i>0$ and $m>0$ or $i<n$ and
$m<0$}.$

For $1\leq q\leq n$, let $c_{q}(d)$ be the number such that

$K_{c_{q}(d),q}(X,B;L_{d})\neq 0~{}~{}\text{ and
}~{}~{}K_{p,q}(X,B;L_{d})=0~{}~{}\text{ for $0\leq p\leq c_{q}(d)-1$},$

and $c_{q}^{\prime}(d)$ be the number such that

$K_{r_{d}-c_{q}^{\prime}(d),q}(X,B;L_{d})\neq 0~{}~{}\text{ and
}~{}~{}K_{p,q}(X,B;L_{d})=0~{}~{}\text{ for $r_{d}-c_{q}^{\prime}(d)+1\leq
p\leq r_{d}$}.$

If $K_{p,q}(X,B;L_{d})=0$ for all $p$, then we set $c_{q}(d):=r_{d}+1$ and
$c_{q}^{\prime}(d):=r_{d}+1$. We will see in Proposition 4.4 that this cannot
happen. Recall from [19, Theorem 1.1] that $c_{q}(x)\geq\Theta(d^{q-1})$.

######  Lemma 4.1.

Assume that $H^{i}(X,B)=0$ for $1\leq i\leq n-1$, i.e., $R(X,B;L_{d})$ is
Cohen–Macaulay. Fix an index $1\leq q\leq n$. Then we have the following:
$(1)$ If $K_{p_{0},q}(X,B;L_{d})=0$ for some $p_{0}\leq\Theta(d^{q-1})$, then
$K_{p,q}(X,B;L_{d})=0$ for $p\leq p_{0}$.
$(2)$ If $K_{p_{0},q}(X,B;L_{d})=0$ for some $p_{0}\geq
r_{d}-\Theta(d^{n-q})$, then $K_{p,q}(X,B;L_{d})=0$ for $p\geq p_{0}$.

###### Proof.

By Proposition 2.2, [8, Proposition 5.1], and [19, Theorem 1.1], if $0\leq
q^{\prime}\leq q-1$, then

$K_{p,q^{\prime}}(X,B;L_{d})=K_{r_{d}-p-n,n+1-q^{\prime}}(X,\omega_{X}\otimes
B^{-1};L_{d})^{\vee}=0~{}~{}\text{ for $p\geq
r_{d}-\Theta(d^{n-q^{\prime}})$}.$

Thus $K_{p_{0},q^{\prime}}(X,B;L_{d})=0$ for $0\leq q^{\prime}\leq q$, so the
assertion $(2)$ follows. Now, by Proposition 2.2, the assertion $(2)$ implies
the assertion $(1)$. ∎

######  Remark 4.2.

If $R(X,B;L_{d})$ is Cohen–Macaulay, then [8, Theorem 4.1], [19, Theorem 1.1],
and Lemma 4.1 imply Theorem 1.1. In general, we know from [8, Theorem 4.1] and
[19, Theorem 1.1] that $c_{q}(d)=\Theta(d^{q-1})$. Using Boij–Söderberg theory
[4], [13], one can show that vanishing and nonvanishing of
$K_{p,q}(X,B;L_{d})$ do not alternate for a while after $c_{q}(d)$. This means
that

$K_{p,q}(X,B;L_{d})\neq 0~{}~{}\text{ for $c_{q}(d)\leq p\leq
r_{d}-\Theta(d^{n-1})$}.$

However, we will not use this remark in our proof of Theorem 1.1.

######  Lemma 4.3.

Theorem 1.1 holds when $n=1$.

###### Proof.

When $n=1$, [8, Proposition 5.1 and Corollary 5.2] imply

$K_{p,1}(X,B;L_{d})\neq 0~{}~{}\text{ for $h^{0}(B)\leq p\leq
r_{d}-h^{0}(X,\omega_{X}\otimes B^{-1})-1$}.$

This shows that $c_{1}(d)=\Theta(1)$ and $c_{1}^{\prime}(d)=\Theta(1)$. Then
Lemma 4.1 implies the lemma. ∎

Assume henceforth that $n=\dim X\geq 2$. Let $H$ be a very ample line bundle
(independent of $d$) on $X$ such that

$H^{i}(X,B\otimes H)=H^{i}(X,B\otimes H^{-1})=0~{}~{}\text{ for $1\leq i\leq
n-1$}.$

As $d$ is sufficiently large, we have

(4.1) $H^{i}(X,B\otimes H\otimes L_{d}^{m})=H^{i}(X,B\otimes H^{-1}\otimes
L_{d}^{m})=0~{}~{}\text{ for $1\leq i\leq n-1$ and $m\in\mathbf{Z}$}.$

This means that $R(X,B\otimes H;L_{d})$ and $R(X,B\otimes H^{-1};L_{d})$ are
Cohen–Macaulay. Clearly,

(4.2) $\begin{array}[]{l}H^{0}(X,B\otimes H\otimes L_{d}^{m})=H^{0}(X,B\otimes
H^{-1}\otimes L_{d}^{m})=0~{}~{}\text{ for $m<0$};\\\\[3.0pt] H^{n}(X,B\otimes
H\otimes L_{d}^{m})=H^{n}(X,B\otimes H^{-1}\otimes L_{d}^{m})=0~{}~{}\text{
for $m>0$};\end{array}$

Choose a general member $\overline{X}\in|H|$, and put

$\begin{array}[]{l}\overline{L}_{d}:=L_{d}|_{\overline{X}},~{}~{}\overline{B}:=B|_{\overline{X}},~{}~{}\overline{H}:=H|_{\overline{X}};\\\\[3.0pt]
V_{d}:=H^{0}(X,L_{d}),~{}~{}V_{d}^{\prime}:=H^{0}(X,L_{d}\otimes
H^{-1}),~{}~{}\overline{V}_{d}:=H^{0}(X,\overline{L}_{d});\\\\[3.0pt]
r_{d}:=\dim V_{d}-1,~{}~{}v_{d}^{\prime}:=\dim
V_{d}^{\prime},~{}~{}\overline{r}_{d}:=\dim\overline{V}_{d}-1\end{array}$

so that $r_{d}=v_{d}^{\prime}+\overline{r}_{d}$ and
$\overline{r}_{d}=\Theta(d^{n-1})$. As in the previous section, fix a
splitting $V_{d}=V_{d}^{\prime}\oplus\overline{V}_{d}$. Then (3.2) and (3.3)
hold. Furthermore, the short exact sequences (3.4) and (3.9) induce the map
$\theta_{p,q}$ in (3.5) and $\theta_{p,q}^{\prime}$ in (3.10), respectively.
In view of [16, Corollary (1.d.4)], they fit into the following exact
sequences

(4.3a) $\displaystyle K_{p+1,q-1}(X,B\otimes
H;L_{d})\longrightarrow\overline{K}_{p+1,q-1}(X,\overline{B}\otimes\overline{H};L_{d})\xrightarrow{\theta_{p,q}}K_{p,q}(X,B;L_{d})\longrightarrow
K_{p,q}(X,B\otimes H;L_{d});$ (4.3b) $\displaystyle K_{p,q}(X,B\otimes
H^{-1};L_{d})\longrightarrow
K_{p,q}(X,B;L_{d})\xrightarrow{\theta_{p,q}^{\prime}}K_{p,q}(X,\overline{B};L_{d})\longrightarrow
K_{p-1,q+1}(X,B\otimes H^{-1};L_{d}).$

######  Proposition 4.4.

For each $1\leq q\leq n$, we have

$c_{q}(d)=\Theta(d^{q-1})~{}~{}\text{ and
}~{}~{}c_{q}^{\prime}(d)=\begin{cases}\Theta(d^{n-q})&\text{if
$H^{q-1}(X,B)=0$ or $q=1$}\\\ q-1&\text{if $H^{q-1}(X,B)\neq 0$ and $q\geq
2$}.\end{cases}$

###### Proof.

We proceed by induction on $n$. As the assertion holds for $n=1$ by Lemma 4.3,
we assume that $n\geq 2$ and the assertions of the lemma hold for
$\overline{X}$.

First, we consider $c_{q}^{\prime}(d)$. Suppose that $H^{q-1}(X,B)\neq 0$ and
$q\geq 2$. Then

$K_{r_{d}-q+1,q}(X,B;L_{d})=H^{q-1}(X,\wedge^{r_{d}}M_{L_{d}}\otimes B\otimes
L_{d})=H^{q-1}(X,B)\neq 0.$

Since $K_{p,q}(X,B;L_{d})=0$ for $p\geq r_{d}-q$, it follows that
$c_{q}^{\prime}(d)=q-1$. Suppose that $H^{q-1}(X,B)=0$ when $q\geq 2$ or
$q=1$. By Proposition 2.2 and [19, Theorem 1.1],

$K_{p,q}(X,B;L)\subseteq K_{r_{d}-p-n,n+1-q}(X,\omega_{X}\otimes
B^{-1};L_{d})^{\vee}=0~{}~{}\text{ for $0\leq r_{d}-p-n\leq\Theta(d^{n-q})$},$

so $c_{q}^{\prime}(d)\geq\Theta(d^{n-q})$. For $2\leq q\leq n$, let
$\overline{c}_{q-1}^{\prime}(d)$ be the number such that

$K_{\overline{r}_{d}-\overline{c}_{q-1}^{\prime}(d),q-1}(\overline{X},\overline{B}\otimes\overline{H};\overline{L}_{d})\neq
0~{}~{}\text{ and
}~{}~{}K_{p,q-1}(\overline{X},\overline{B}\otimes\overline{H};\overline{L}_{d})=0~{}~{}\text{
for $p\geq\overline{r}_{d}-\overline{c}_{q-1}^{\prime}(d)+1$}.$

By induction, $\overline{c}_{q-1}^{\prime}(d)=\Theta(d^{n-q})$. Recall that
$H^{1}(X,B)=0$ when $q=2$. By considering (3.3), (3.7), (3.8), we get

$\overline{K}_{r_{d}-\overline{c}_{q-1}^{\prime}(d),q-1}(X,\overline{B}\otimes\overline{H};L_{d})=K_{r_{d}-\overline{c}_{q-1}^{\prime}(d),q-1}(X,\overline{B}\otimes\overline{H};L_{d})\neq
0~{}~{}\text{ for $2\leq q\leq n$}.$

Possibly replacing $H$ by more positive $H$ (still independent of $d$), we may
assume that

(4.4) $h^{0}(X,B\otimes H)>h^{0}(X,B).$

Then
$\overline{K}_{0,0}(\overline{X},\overline{B}\otimes\overline{H};\overline{L}_{d})\neq
0$. Putting $\overline{c}^{\prime}_{0}(d):=\overline{r}_{d}=\Theta(d^{n-1})$,
we get from (3.6) that

$\overline{K}_{r_{d}-\overline{c}_{0}^{\prime}(d),0}(X,\overline{B}\otimes\overline{H};L_{d})\neq
0.$

For $1\leq q\leq n$, thanks to (4.1), Proposition 2.2 and [19, Theorem 1.1]
yield that

$K_{r_{d}-\overline{c}_{q-1}^{\prime}(d),q-1}(X,B\otimes
H;L_{d})=K_{\overline{c}_{q-1}^{\prime}(d)-n,n+2-q}(X,\omega_{X}\otimes
B^{-1}\otimes H^{-1};L_{d})^{\vee}=0,$

since $\overline{c}_{q-1}^{\prime}(d)-n=\Theta(d^{n-q})<\Theta(d^{n+1-q})$.
Then the map

$\theta_{r_{d}-\overline{c}_{q-1}^{\prime}(d)-1,q}\colon\overline{K}_{r_{d}-\overline{c}_{q-1}^{\prime}(d),q-1}(X,\overline{B}\otimes\overline{H};L_{d})\longrightarrow
K_{r_{d}-\overline{c}_{q-1}^{\prime}(d)-1,q}(X,B;L_{d})$

in (4.3a) is a nonzero injective map. Thus
$c_{q}^{\prime}(d)\leq\overline{c}_{q-1}^{\prime}(d)+1=\Theta(d^{n-q})$, so
$c_{q}^{\prime}(d)=\Theta(d^{n-q})$.

Next, we consider $c_{q}(d)$. We know from [19, Theorem 1.1] that
$c_{q}(d)\geq\Theta(d^{q-1})$. When $q=n$, Proposition 2.2 says that there is
a surjective map

$K_{p,n}(X,B;L_{d})\longrightarrow K_{r_{d}-p-n,1}(X,\omega_{X}\otimes
B^{-1};L_{d})^{\vee}.$

As we have seen in the previous paragraph that
$K_{r_{d}-p-n,1}(X,\omega_{X}\otimes B^{-1};L_{d})^{\vee}\neq 0$ for some
$p=\Theta(d^{n-1})$, we have $c_{n}(d)\leq\Theta(d^{n-1})$. Hence
$c_{n}(d)=\Theta(d^{n-1})$. Assume that $1\leq q\leq n-1$. Let
$\overline{c}_{q}(d)$ be the number such that

$K_{\overline{c}_{q}(d),q}(\overline{X},\overline{B};\overline{L}_{d})\neq
0~{}~{}\text{ and
}~{}~{}K_{p,q}(\overline{X},\overline{B};\overline{L}_{d})=0~{}~{}\text{ for
$p\leq\overline{c}_{q}(d)-1$}.$

By induction, $\overline{c}_{q}(d)=\Theta(d^{q-1})$. Note that

$K_{\overline{c}_{q}(d),q}(X,\overline{B};L_{d})\neq 0$

thanks to (3.3). By [19, Theorem 1.1],

$K_{\overline{c}_{q}(d)-1,q+1}(X,B\otimes H^{-1};L_{d})=0$

since $\overline{c}_{q}(d)-1=\Theta(d^{q-1})<\Theta(d^{q})$. Then the map

$\theta_{\overline{c}_{q}(d),q}^{\prime}\colon
K_{\overline{c}_{q}(d),q}(X,B;L_{d})\longrightarrow
K_{\overline{c}_{q}(d),q}(X,\overline{B};L_{d})$

in (4.3b) is a nonzero surjective map. Thus
$c_{q}(d)\leq\overline{c}_{q}(d)=\Theta(d^{q-1})$, so
$c_{q}(d)=\Theta(d^{q-1})$. ∎

Next, we prove the following technical lemma.

######  Lemma 4.5.

Let $B^{\prime}$ be a line bundle on $X$ (independent of $d$), and $L$ be a
very ample line bundle on $X$ (independent of $d$) such that

$\displaystyle H^{i}(X,L^{m})=0~{}~{}\text{ for $i>0$ and $m>0$ or $i<n$ and
$m<0$};$ $\displaystyle H^{i}(X,B^{\prime}\otimes L^{m})=0~{}~{}\text{ for
$i>0$ and $m>0$ or $i<n$ and $m<0$};$ $\displaystyle H^{i}(X,M_{L_{d}}\otimes
L^{m})=0~{}~{}\text{ for $i>0$ and $m>0$}.$

Put $H:=L^{n+1}$. For $1\leq p\leq\Theta(d^{q-1})$ and $1\leq q\leq n+1$, we
have the following:
$(1)$ If $q\geq 2$ and $H^{q-1}(X,\wedge^{p}M_{L_{d}}\otimes B^{\prime}\otimes
L_{d})=0$, then $H^{q-1}(X,\wedge^{p+1}M_{L_{d}}\otimes B^{\prime}\otimes
H\otimes L_{d})=0$.
$(2)$ If $q\leq n$ and $H^{q}(X,\wedge^{p}M_{L_{d}}\otimes B^{\prime})=0$,
then $H^{q}(X,\wedge^{p+1}M_{L_{d}}\otimes B^{\prime}\otimes H)=0$.

###### Proof.

Notice that $\mathscr{O}_{X}$ and $M_{L_{d}}$ are $(n+1)$-regular with respect
to $L$. By Lemma 2.4, there are finitely dimensional vector spaces
$\ldots,W_{1},W_{0}$ over $\mathbf{k}$ and an exact sequence

(4.5) $\cdots\longrightarrow W_{1}\otimes H^{-2}\longrightarrow W_{0}\otimes
H^{-1}\longrightarrow M_{L_{d}}\longrightarrow 0.$

$(1)$ By Lemma 2.3, it is sufficient to prove that

$H^{q-1}(X,M_{L_{d}}\otimes\wedge^{p}M_{L_{d}}\otimes B^{\prime}\otimes
H\otimes L_{d})=0.$

For this purpose, consider the exact sequence from (4.5):

$\begin{array}[]{l}\cdots\longrightarrow
W_{1}\otimes\wedge^{p}M_{L_{d}}\otimes B^{\prime}\otimes H^{-1}\otimes
L_{d}\longrightarrow W_{0}\otimes\wedge^{p}M_{L_{d}}\otimes B^{\prime}\otimes
L_{d}\\\\[3.0pt] ~{}\text{ }~{}\text{ }~{}\text{ }~{}\text{ }~{}\text{
}~{}\text{ }~{}\text{ }~{}\text{ }~{}\text{ }~{}\text{ }~{}\text{ }~{}\text{
}~{}\text{ }~{}\text{ }~{}\text{ }~{}\text{ }~{}\text{ }~{}\text{ }~{}\text{
}~{}\text{ }~{}\text{ }~{}\text{ }~{}\text{ }~{}\text{ }~{}\text{ }~{}\text{
}~{}\text{ }\longrightarrow M_{L_{d}}\otimes\wedge^{p}M_{L_{d}}\otimes
B^{\prime}\otimes H\otimes L_{d}\longrightarrow 0.\end{array}$

In view of [17, Proposition B.1.2], it suffices to show that

$H^{q-1+i}(X,\wedge^{p}M_{L_{d}}\otimes B^{\prime}\otimes H^{-i}\otimes
L_{d})=K_{p-q-i+1,q+i}(X,B^{\prime}\otimes H^{-i};L_{d}).=0~{}~{}\text{ for
$i\geq 0$}.$

When $i=0$, this is the given condition. When $i\geq 1$, this follows from
[19, Theorem 1.1] since $p-q-i+1\leq\Theta(d^{q-1})<\Theta(d^{q+i-1})$.

$(2)$ By Lemma 2.3, it is sufficient to prove that

(4.6) $H^{q}(X,M_{L_{d}}\otimes\wedge^{p}M_{L_{d}}\otimes B^{\prime}\otimes
H)=0.$

Let $M_{0}$ be the kernel of the map $W_{0}\otimes H^{-1}\to M_{L_{d}}$ in
(4.5). We have a short exact sequence

(4.7) $0\longrightarrow M_{0}\otimes B^{\prime}\otimes H\longrightarrow
W_{0}\otimes B^{\prime}\longrightarrow M_{L_{d}}\otimes B^{\prime}\otimes
H\longrightarrow 0.$

As $H^{q}(X,\wedge^{p}M_{L_{d}}\otimes B^{\prime})=0$, the claim (4.6) is
implied by the injectivity of the map

$\rho\colon H^{q+1}(X,\wedge^{p}M_{L_{d}}\otimes B^{\prime}\otimes
H)\longrightarrow W_{0}\otimes H^{q+1}(X,\wedge^{p}M_{L_{d}}\otimes
B^{\prime}).$

This map fits into the following commutative diagram

$\textstyle{H^{q}(X,\wedge^{p}M_{L_{d}}\otimes M_{0}\otimes B^{\prime}\otimes
H\otimes
L_{d})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{W_{0}\otimes
H^{q}(X,\wedge^{p}M_{L_{d}}\otimes B^{\prime}\otimes
L_{d})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{H^{q+1}(X,\wedge^{p}M_{L_{d}}\otimes
M_{0}\otimes B^{\prime}\otimes
H)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\rho}$$\scriptstyle{\psi}$$\textstyle{W_{0}\otimes
H^{q+1}(X,\wedge^{p}M_{L_{d}}\otimes
B^{\prime})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\wedge^{p}V_{d}\otimes
H^{q+1}(X,M_{0}\otimes B^{\prime}\otimes
H)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\varphi}$$\textstyle{W_{0}\otimes\wedge^{p}V_{d}\otimes
H^{q+1}(X,B^{\prime}).}$

For the injectivity of the map $\rho$, it is enough to check that $\psi$ and
$\varphi$ are injective. From (4.5), we have an exact sequence

$\cdots\longrightarrow W_{2}\otimes H^{-3}\longrightarrow W_{1}\otimes
H^{-2}\longrightarrow M_{0}\longrightarrow 0.$

By [19, Theorem 1.1],

$H^{q+i}(X,\wedge^{p}M_{L_{d}}\otimes B^{\prime}\otimes H^{-i-1}\otimes
L_{d})=K_{p-q-i,q+i+1}(X,B^{\prime}\otimes H^{-i-1};L_{d})=0~{}~{}\text{ for
$i\geq 0$}$

since $p-q-i\leq\Theta(d^{q-1})<\Theta(d^{q+i})$. By [17, Proposition B.1.2],

$H^{q}(X,\wedge^{p}M_{L_{d}}\otimes M_{0}\otimes B^{\prime}\otimes H\otimes
L_{d})=0,$

so $\psi$ is injective. On the other hand, we get from (4.7) that

$H^{n+2-q}(X,M_{0}\otimes B^{\prime}\otimes H)=W_{0}\otimes
H^{n+2-q}(X,B^{\prime}),$

so $\varphi$ is an isomorphism. ∎

Now, take a very ample line bundle $L$ on $X$ (independent of $d$) such that

(4.8a) $\displaystyle H^{i}(X,L^{m})=0~{}~{}\text{ for $i>0$ and $m>0$ or
$i<n$ and $m<0$};$ (4.8b) $\displaystyle H^{i}(X,B\otimes L^{m})=0~{}~{}\text{
for $i>0$ and $m>0$ or $i<n$ and $m<0$};$ (4.8c) $\displaystyle
H^{i}(X,M_{L_{d}}\otimes L^{m})=0~{}~{}\text{ for $i>0$ and $m>0$};$ (4.8d)
$\displaystyle H^{i}(X,M_{L_{d}}\otimes\omega_{X}\otimes B^{-1}\otimes
L^{m})=0~{}~{}\text{ for $i>0$ and $m>0$};$

By Proposition 4.4, we can take an integer $c\geq
c_{1}(d)+1,c_{n}^{\prime}(d)-n+1$ independent of $d$. Successively applying
Lemma 4.5 and possibly replacing $L$ by a higher power of $L$ (still
independent of $d$), we may assume that

$\begin{array}[]{l}K_{c-1,1}(X,B\otimes
L^{n+1};L_{d})=H^{1}(X,\wedge^{c}M_{L_{d}}\otimes B\otimes
L^{n+1})=0;\\\\[3.0pt] K_{c-1,1}(X,\omega_{X}\otimes B^{-1}\otimes
L^{n+1};L_{d})=H^{1}(X,\wedge^{c}M_{L_{d}}\otimes\omega_{X}\otimes
B^{-1}\otimes L^{n+1})=0.\end{array}$

By Lemma 4.1, we have

(4.9a) $\displaystyle K_{c_{1}(d),1}(X,B\otimes
L^{n+1};L_{d})=H^{1}(X,\wedge^{c_{1}(d)+1}M_{L_{d}}\otimes B\otimes
L^{n+1})=0;$ (4.9b) $\displaystyle
K_{c_{n}^{\prime}(d)-n,1}(X,\omega_{X}\otimes B^{-1}\otimes
L^{n+1};L_{d})=H^{1}(X,\wedge^{c_{n}^{\prime}(d)-n+1}M_{L_{d}}\otimes\omega_{X}\otimes
B^{-1}\otimes L^{n+1})=0.$

From now on, replace $H$ by $H:=L^{n+1}$. Then (4.1) holds thanks to (4.8b),
so $R(X,B\otimes H;L_{d})$ and $R(X,B\otimes H^{-1};L_{d})$ are
Cohen–Macaulay. Clearly, (4.2) is satisfied.

######  Proposition 4.6.

Assume that $n\geq 2$. For $1\leq q\leq n$, we have the following:
$(1)$ If $K_{p,q}(X,B;L_{d})=0$ for $p<c_{q}(d)$, then $K_{p+1,q}(X,B\otimes
H;L_{d})=0$. Consequently, the map

$\theta_{c_{q}(d),q}\colon\overline{K}_{c_{q}(d)+1,q-1}(X,\overline{B}\otimes\overline{H};L_{d})\longrightarrow
K_{c_{q}(d),q}(X,B;L_{d})$

in (4.3a) is a nonzero surjective map.
$(2)$ If $K_{p,q}(X,B;L_{d})=0$ for $p>r_{d}-c_{q}^{\prime}(d)$, then
$K_{p-1,q}(X,B\otimes H^{-1};L_{d})=0$. Consequently, the map

$\theta_{r_{d}-c_{q}^{\prime}(d),q}^{\prime}\colon
K_{r_{d}-c_{q}^{\prime}(d),q}(X,B;L_{d})\longrightarrow
K_{r_{d}-c_{q}^{\prime}(d),q}(X,\overline{B};L_{d})$

in (4.3b) is a nonzero injective map.

###### Proof.

$(1)$ Recall that $R(X,B\otimes H;L_{d})$ is Cohen–Macaulay. Then Lemma 4.1
says that

$K_{c_{q}(d),q}(X,B\otimes
H;L_{d})=0~{}~{}\Longrightarrow~{}~{}K_{p+1,q}(X,B\otimes
H;L_{d})=0~{}~{}\text{ for $p\leq c_{q}(d)-1$}.$

Thus it suffices to show that $K_{c_{q}(d),q}(X,B\otimes H;L_{d})=0$. The case
$q=1$ is nothing but (4.9a). Assume that $2\leq q\leq n$. Note that the given
condition is

$H^{q-1}(X,\wedge^{c_{q}(d)+q-2}M_{L_{d}}\otimes B\otimes
L_{d})=K_{c_{q}(d)-1,q}(X,B;L_{d})=0.$

Then Lemma 4.5 $(1)$ yields

$K_{c_{q}(d),q}(X,B\otimes
H;L_{d})=H^{q-1}(X,\wedge^{c_{q}(d)+q-1}M_{L_{d}}\otimes B\otimes H\otimes
L_{d})=0.$

$(2)$ Recall that $R(X,B\otimes H^{-1};L_{d})$ is Cohen–Macaulay. Then Lemma
4.1 says that

$K_{r_{d}-c_{q}^{\prime}(d),q}(X,B\otimes
H^{-1};L_{d})=0~{}~{}\Longrightarrow~{}~{}K_{p-1,q}(X,B\otimes
H^{-1};L_{d})=0~{}~{}\text{ for $p\geq r_{d}-c_{q}^{\prime}(d)+1$}.$

Thus it suffices to show that $K_{r_{d}-c_{q}^{\prime}(d),q}(X,B\otimes
H^{-1};L_{d})=0$. By Proposition 2.2,

$K_{r_{d}-c_{q}^{\prime}(d),q}(X,B\otimes
H^{-1};L_{d})=K_{c_{q}^{\prime}(d)-n,n+1-q}(X,\omega_{X}\otimes B^{-1}\otimes
H;L_{d})^{\vee}.$

We need to show that

(4.10)
$H^{n+1-q}(X,\wedge^{c_{q}^{\prime}(d)-q+1}M_{L_{d}}\otimes\omega_{X}\otimes
B^{-1}\otimes H)=0.$

When $q=n$, (4.10) is the same to (4.9b). Assume that $1\leq q\leq n-1$. If
$q\geq 2$ and $H^{q-1}(X,B)\neq 0$, then $c_{q}^{\prime}(d)=q-1$ so that
(4.10) holds by (4.8b). Assume $H^{q-1}(X,B)=0$ when $q\geq 2$. Then
$c_{q}^{\prime}(d)=\Theta(d^{n-q})$. The given condition and Serre duality
yield

$\begin{array}[]{rcl}H^{n+1-q}(X,\wedge^{c_{q}^{\prime}(d)-q}M_{L_{d}}\otimes\omega_{X}\otimes
B^{-1})&=&H^{n+1-q}(X,\wedge^{r_{d}-c_{q}^{\prime}(d)+q}M_{L_{d}}^{\vee}\otimes\omega_{X}\otimes
B^{-1}\otimes L_{d}^{-1})\\\\[3.0pt]
&=&H^{q-1}(X,\wedge^{r_{d}-c_{q}^{\prime}(d)+q}M_{L_{d}}\otimes B\otimes
L_{d})^{\vee}\\\\[3.0pt]
&=&K_{r_{d}-c_{q}^{\prime}(d)+1,q}(X,B;L_{d})^{\vee}~{}=~{}0.\end{array}$

Then the claim (4.10) follows from Lemma 4.5 $(2)$. ∎

Theorem 1.1 now follows at once from the previous propositions and Theorem
3.1.

###### Proof of Theorem 1.1.

By Lemma 4.3, we may assume that $n\geq 2$. Note that (1.1) is proved in
Proposition 4.4. For $1\leq q\leq n$, by Theorem 3.1 $(1)$ and Proposition 4.6
$(1)$,

$K_{p,q}(X,B;L_{d})\neq 0~{}~{}\text{ for $c_{q}(d)\leq p\leq
r_{d}-\overline{r}_{d}-1$}.$

On the other hand, [8, Proposition 5.1] says $K_{p,0}(X,B;L_{d})=0$ for
$p>\Theta(1)$. Thus we obtain (1.2) for $q=1$. For $2\leq q\leq n$, by Theorem
3.1 $(2)$ and Proposition 4.6 $(2)$,

$K_{p,q}(X,B;L_{d})\neq 0~{}~{}\text{ for $\overline{r}_{d}\leq p\leq
r_{d}-c_{q}^{\prime}(d)$}.$

As $r_{d}-\overline{r}_{d}-1=\Theta(d^{n})>\Theta(d^{n-1})=\overline{r}_{d}$,
we obtain (1.2) for $2\leq q\leq n$. ∎

##  5\. Complements and Problems

In this section, we show some additional results, and discuss some open
problems. Recall that the asymptotic vanishing theorem ([19, Theorem 1.1])
holds for singular varieties with coherent sheaves. Precisely, let $X$ be a
projective variety of dimension $n$, and $B$ be a coherent sheaf on $X$. For
an integer $d\geq 1$, let $L_{d}:=\mathscr{O}_{X}(dA+P)$, where $A$ is an
ample divisor and $P$ is an arbitrary divisor on $X$. For each $1\leq q\leq
n+1$, if $d$ is sufficiently large, then

$K_{p,q}(X,B;L_{d})=0~{}~{}\text{ for $0\leq p\leq\Theta(d^{q-1})$}.$

We expect that Theorem 1.1 also holds in this setting.

######  Conjecture 5.1.

Theorem 1.1 still holds when $X$ is a projective variety and $B$ is a coherent
sheaf on $X$ with $\operatorname{Supp}B=X$.

Note that the expected nonvanishing of $K_{p,q}(X,B;L_{d})$ for
$q>\dim\operatorname{Supp}B+1$ may not hold.

######  Remark 5.2.

In the proof of Theorem 1.1, we use the assumption that $X$ is smooth and $B$
is a line bundle only when we apply Serre duality. Thus Theorem 1.1 holds when
$X$ is Cohen–Macaulay and $B$ is a vector bundle.

From now on, we assume that $X$ is smooth and $B$ is a line bundle as in
Theorem 1.1. In the remaining, fix an index $1\leq q\leq n$. It is very
natural to study the asymptotic growth of $c_{q}(d)$ and $c_{q}^{\prime}(d)$
as $d\to\infty$. In the spirit of [23], we give an effective upper bound for
each of $c_{q}(d)$ and $c_{q}^{\prime}(d)$. For this purpose, we introduce
some notations. Choose suitably positive very ample divisors
$H_{1},\ldots,H_{n-1}$ on $X$ such that

$\overline{X}_{i}:=H_{1}\cap\cdots\cap H_{i}$

is a smooth projective variety for every $0\leq i\leq n-1$. Note that
$\overline{X}_{0}=X$. For each $0\leq i\leq n-1$, put

$\begin{array}[]{l}\overline{H}_{i}:=\mathscr{O}_{X}(H_{i+1})|_{\overline{X}_{i}},~{}\overline{B}_{i}:=B|_{\overline{X}_{i}},~{}\overline{B}_{i}^{\prime}:=B(H_{1}+\cdots+H_{i})|_{\overline{X}_{i}};\\\\[3.0pt]
\overline{L}_{d}:=L_{d}|_{\overline{X}_{i}},~{}\overline{r}_{i}(d):=h^{0}(\overline{X}_{i},\overline{L}_{d})-1=\Theta(d^{n-i}),~{}\overline{r}_{n}(d):=0,\end{array}$

and assume that (4.1), (4.2), (4.4) hold for

$X=\overline{X}_{i},~{}B=\overline{B}_{i},\overline{B}_{i}^{\prime},\omega_{X_{i}}\otimes\overline{B}_{i}^{-1},~{}H=\overline{H}_{i},~{}L_{d}=\overline{L}_{d}.$

######  Proposition 5.3.

$c_{q}(d)\leq\overline{r}_{n+1-q}(d)-q+1$ and
$c_{q}^{\prime}(d)\leq\overline{r}_{q}(d)+q$.

###### Proof.

We proceed by induction on $n=\dim X$. When $n=1$, the assertion is trivial.
Assume that $n\geq 2$. For $c_{q}^{\prime}(d)$, we may assume that
$H^{q-1}(X,B)=0$ or $q=1$. In the proof of Proposition 4.4, we proved that

$\theta_{r_{d}-\overline{r}_{q}(d)-q,q}\colon\overline{K}_{r_{d}-\overline{r}_{q}(d)-(q-1),q-1}(X,\overline{B}\otimes\overline{H};L_{d})\longrightarrow
K_{r_{d}-\overline{r}_{q}(d)-q,q}(X,B;L_{d})$

is nonzero, so $c_{q}^{\prime}(d)\leq\overline{r}_{q}(d)+q$. We also have
$K_{r_{d}-\overline{r}_{1}(d)-1,1}(X,\omega_{X}\otimes B^{-1};L_{d})\neq 0$.
By Proposition 2.2, $K_{r_{1}(d)-n+1,1}(X,B;L_{d})\neq 0$. Thus we obtain
$c_{n}(d)\leq\overline{r}_{1}(d)-n+1$. For $1\leq q\leq n-1$, in the proof of
Proposition 4.4, we proved that

$\theta_{\overline{r}_{n+1-q}(d)-q+1,q}^{\prime}\colon
K_{\overline{r}_{n+1-q}(d)-q+1,q}(X,B;L_{d})\longrightarrow
K_{\overline{r}_{n+1-q}(d)-q+1,q}(X,\overline{B};L_{d})$

is nonzero, so $c_{q}(d)\leq\overline{r}_{n+1-q}(d)-q+1$. ∎

######  Remark 5.4.

In Proposition 5.3, we do not assume that $R(X,B;L_{d})$ is Cohen–Macaulay.
However, when $R(X,B;L_{d})$ is Cohen–Macaulay, by a more careful analysis,
one can improve bounds for $c_{q}(d)$ and $c_{q}^{\prime}(d)$ as in [23]. In
particular, one can recover [8, Theorem 6.1]: If
$X=\mathbf{P}^{n},B=\mathscr{O}_{\mathbf{P}^{n}}(b),L_{d}=\mathscr{O}_{\mathbf{P}^{n}}(d)$
and $b\geq 0,d\gg 0$, then

(5.1) $c_{q}(d)\leq{d+q\choose q}-{d-b-1\choose q}-q~{}~{}\text{ and
}~{}~{}c_{q}^{\prime}(d)\leq{d+n-q\choose n-q}-{n+b\choose q+b}+q.$

We leave the details to interested readers.

In characteristic zero, David Yang [22, Theorem 1] confirmed that $c_{1}(d)$
is a constant. This gives an answer to [8, Problem 7.2]. On the other hand,
for Veronese syzygies, Ein–Lazarsfeld [10, Conjecture 2.3] conjectured that
equalities hold in (5.1) whenever $d\geq b+q+1$. In particular, $c_{q}(d)$ and
$c_{q}^{\prime}(d)$ are polynomials. One may hope that the same is true in
general.

######  Question 5.5 (cf. [6, Remark 3.2]).

$(1)$ Does the limit

$\lim_{d\to\infty}\frac{c_{q}(d)}{d^{q-1}}$

exist? If so, is the function $c_{q}(d)$ a polynomial of degree $q-1$ for
sufficiently large $d$? What can one say about the leading coefficient
$a_{q-1}:=\lim_{d\to\infty}c_{q}(d)/d^{q-1}$ of $c_{q}(d)$?
$(2)$ Suppose that $H^{q-1}(X,B)=0$ if $q\geq 2$. Does the limit

$\lim_{d\to\infty}\frac{c_{q}^{\prime}(d)}{d^{n-q}}$

exist? If so, is the function $c_{q}^{\prime}(d)$ a polynomial of degree $n-q$
for sufficiently large $d$? What can one say about the leading coefficient
$a_{n-q}^{\prime}:=\lim_{d\to\infty}c_{q}^{\prime}(d)/d^{n-q}$ of
$c_{q}^{\prime}(d)$?

When $\operatorname{char}(\mathbf{k})=0$, a geometric meaning of the constant
$c_{1}(d)$ was explored as follows. Ein–Lazarsfeld–Yang [11, Theorem A] proved
that if $B$ is $p$-jet very ample, then $c_{1}(d)\geq p+1$. Agostini [1,
Theorem A] proved that if $c_{1}(d)\geq p+1$, then $B$ is $p$-very ample.
These results are higher dimensional generalizations of the gonality
conjecture on syzygies of algebraic curves, which was established by
Ein–Lazarsfeld [9] and Rathmann [21]. On the other hand,
Eisenbud–Green–Hulek–Popescu [12] related the nonvanishing of
$K_{p,q}(X,L_{d})$ to the existence of special secant planes. In particular,
if the property $N_{k}$ holds for $L_{d}$, then $L_{d}$ is $(k+1)$-very ample.
It would be exceedingly interesting know whether the nonexistence of special
secant planes implies the vanishing of certain $K_{p,q}(X,L_{d})$.

When $X$ is a smooth projective curve and $\operatorname{char}(\mathbf{k})=0$,
Rathmann [21, Theorem 1.2] showed that if $H^{1}(X,L_{d})=0$ and
$H^{1}(X,B^{-1}\otimes L_{d})=0$, then $K_{p,1}(X,B;L_{d})=0$. It is natural
to extend this effective result to higher dimensions. When $B=\mathscr{O}_{X}$
and $L_{d}=\mathscr{O}_{X}(K_{X}+dA)$, the following problem is closely
related to Mukai’s conjecture [7, Conjecture 4.2].

######  Problem 5.6.

$(1)$ Suppose that $c_{q}(d)$ is a polynomial of degree $q-1$ for sufficiently
large $d$. Find an effective bound for $d_{0}$ such that $c_{q}(d)$ becomes a
polynomial for $d\geq d_{0}$.
$(2)$ Suppose that $H^{q-1}(X,B)=0$ if $q\geq 2$ and $c_{q}^{\prime}(d)$ is a
polynomial of degree $n-q$ for sufficiently large $d$. Find an effective bound
for $d_{0}^{\prime}$ such that $c_{q}^{\prime}(d)$ becomes a polynomial for
$d\geq d_{0}$.

Now, we turn to the asymptotic behaviors of the _Betti numbers_

$\kappa_{p,q}(X,B;L_{d}):=\dim K_{p,q}(X,B;L_{d}).$

Ein–Erman–Lazarsfeld conjectured that the Betti numbers
$\kappa_{p,q}(X,B;L_{d})$ are normally distributed [5, Conjecture B], and they
verified the conjecture for curves [5, Proposition A]. The normal distribution
conjecture suggests the following unimodality conjecture.

######  Conjecture 5.7.

The Betti numbers $\kappa_{p,q}(X,B;L_{d})$ form a unimodal sequence.

As the cases of very small $p$ and very large $p$ for
$\kappa_{p,q}(X,B;L_{d})$ are negligible in the normal distribution
conjecture, the unimodality conjecture is not a consequence of the normal
distribution conjecture. Finally, we verify the unimodality conjecture for
curves following the strategy of Erman [15] based on Boij–Söderberg theory.
Eisenbud–Schreyer [13] and Boij–Söderberg [4] showed that the Betti table of a
graded module over a polynomial ring is a positive rational sum of pure
diagrams (see [14, Theorem 2.2] for the precise statement). Let $C$ be a
smooth projective curve, $B$ be a line bundle, and $L$ be a very ample line
bundle of sufficiently large degree $d$. Put $c:=h^{0}(X,B)$ and
$c^{\prime}:=h^{1}(X,B)$. By [8, Proposition 5.1 and Corollary 5.2],
$\kappa_{p,0}(C,B;L)=0$ for $p\geq c$ and $\kappa_{p,2}(C,B;L)=0$ for $p\leq
r-1-c^{\prime}$, where $r:=h^{0}(X,L)-1$. By Riemann–Roch theorem,
$r=d-g\approx d$ since $d$ is sufficiently large. Let $\pi$ be the Betti table
of $R(C,B;L)$. Then [14, Theorem 2.2] says that

(5.2) $\pi=\sum_{i=0}^{c}\sum_{j=0}^{c^{\prime}}a_{i,j}\pi_{i,j}~{}~{}\text{
for some rational numbers $a_{i,j}\geq 0$
with}~{}~{}\sum_{i=0}^{c}\sum_{j=0}^{c^{\prime}}a_{i,j}=d,$

where each $\pi_{i,j}$ is the pure diagram of the form:

| $0$ | $\cdots$ | $i-1$ | $i$ | $\cdots$ | $r-j-1$ | $r-j$ | $\cdots$ | $r-1$
---|---|---|---|---|---|---|---|---|---
$0$ | * | $\cdots$ | * | - | $\cdots$ | - | - | $\cdots$ | -
$1$ | - | $\cdots$ | - | * | $\cdots$ | * | - | $\cdots$ | -
$2$ | - | $\cdots$ | - | - | $\cdots$ | - | * | $\cdots$ | *

Here “*” indicates a nonzero entry, and “-” indicates a zero entry. We have

$\begin{array}[]{l}\displaystyle\kappa_{p,0}(\pi_{i,j})=\frac{(r-1)!(i-p)(r-j+1-p)}{(r+1-p)!p!}~{}~{}\text{
for $0\leq p\leq i-1$};\\\\[3.0pt]
\displaystyle\kappa_{p,1}(\pi_{i,j})=\frac{(r-1)!(p+1-i)(r-j-p)}{(r-p)!(p+1)!}~{}~{}\text{
for $i\leq p\leq r-j-1$};\\\\[3.0pt]
\displaystyle\kappa_{p,2}(\pi_{i,j})=\frac{(r-1)!(p+2-i)(p-r+j+1)}{(r-p-1)!(p+2)!}~{}~{}\text{
for $r-j\leq p\leq r-1$}.\end{array}$

######  Proposition 5.8.

$(1)$ The Betti table of $R(C,B;L)$ is asymptotically pure:

$\frac{a_{i,j}}{d}\rightarrow\begin{cases}1&\text{if $i=c$ and
$j=c^{\prime}$}\\\ 0&\text{otherwise}\end{cases}~{}~{}\text{ as
$d\to\infty$}.$

$(2)$ $\kappa_{p,0}(C,B;L),\ldots,\kappa_{c-1,0}(C,B;L)$ is increasing, and
$\kappa_{r-c^{\prime},2}(C,B;L),\ldots,\kappa_{r-1,2}(C,B;L)$ is decreasing.
$(3)$ The Betti numbers $\kappa_{p,1}(C,B;L)$ form a unimodal sequence.

###### Proof.

$(1)$ Let $\overline{a}_{i,j}:=a_{i,j}/d$. Then (5.2) says

$\sum_{i=0}^{c}\sum_{j=0}^{c^{\prime}}\overline{a}_{i,j}=1.$

Notice that $\kappa_{0,0}(\pi_{i,j})\to i/d$ and $\kappa_{r-1,2}(\pi_{i,j})\to
j/d$ as $d\to\infty$. Since $\kappa_{0,0}(C,B;L)=c$ and
$\kappa_{r-1,2}(C,B;L)=c^{\prime}$, it follows from (5.2) that

$\sum_{i=0}^{c}\sum_{j=0}^{c^{\prime}}i\overline{a}_{i,j}\to c~{}~{}\text{ and
}~{}~{}\sum_{i=0}^{c}\sum_{j=0}^{c^{\prime}}j\overline{a}_{i,j}\to
c^{\prime}~{}~{}\text{ as $d\to\infty$}.$

Then we have

$\sum_{i=0}^{c}\sum_{j=0}^{c^{\prime}}(c-i)\overline{a}_{i,j}\to 0~{}~{}\text{
and
}~{}~{}\sum_{i=0}^{c}\sum_{j=0}^{c^{\prime}}(c^{\prime}-j)\overline{a}_{i,j}\to
0~{}~{}\text{ as $d\to\infty$}.$

Thus $\overline{a}_{i,j}\to 0$ as $d\to\infty$ unless $i=c$ and
$j=c^{\prime}$, and hence, $\overline{a}_{c,c^{\prime}}\to 1$ as $d\to\infty$.

$(2)$ As $d\gg 0$, we have $r\approx d$ and

$\kappa_{p,0}(\pi_{i,j})\approx\frac{i-p}{p!}d^{p-1}~{}~{}\text{ for $0\leq
p\leq i-1$}.$

Then $(1)$ implies that

$\kappa_{p,0}(C,B;L)\approx\frac{c-p}{p!}d^{p}~{}~{}\text{ for $0\leq p\leq
c-1$}.$

Thus the first assertion holds, and the second assertion follows from
Proposition 2.2.

$(3)$ For $(r-2+c)/2\leq p\leq r-j-1$, we find

$\frac{\kappa_{p+1,1}(\pi_{i,j})}{\kappa_{p,1}(\pi_{i,j})}=\frac{(r-p)(p+2-i)(r-j-p-1)}{(p+2)(p+1-i)(r-j-p)}\leq
1$

since $r-p\leq p+2$ and

$(p+2-i)(r-j-p-1)-(p+1-i)(r-j-p)=(r-j-p-1)-(p+1-i)\leq 0.$

For $(r-2+c)/2\leq p\leq r-2$, we get

$\kappa_{p+1,1}(C,B;L)=\sum_{i=0}^{c}\sum_{j=0}^{\min\\{c^{\prime},r-p-2\\}}a_{i,j}\kappa_{p+1,1}(\pi_{i,j})\leq\sum_{i=0}^{c}\sum_{j=0}^{\min\\{c^{\prime},r-p-1\\}}a_{i,j}\kappa_{p,1}(\pi_{i,j})=\kappa_{p,1}(C,B;L),$

so the Betti numbers $\kappa_{p,q}(C,B;L)$ with $(r-2+c)/2\leq p\leq r-1$ form
a decreasing sequence. Now, as in [5, Proof of Proposition A], we compute

(5.3) $\kappa_{p,1}(C,B;L)=\chi(C,\wedge^{p}M_{L}\otimes B\otimes
L)-{r+1\choose p+1}c={r\choose
p}\left(b-\frac{pd}{r}-\frac{(r+1)c}{p+1}\right)$

for $c-1\leq p\leq r-c^{\prime}$, where $b:=r+\deg B+1$. For $(r-1)/2\leq
p\leq(r-2+c)/2$, we have

$\frac{\kappa_{p+1,1}(C,B;L)}{\kappa_{p,1}(C,B;L)}=\frac{(r-p)(b-(p+1)d/r-(r+1)c/(p+2))}{(p+1)(b-pd/r-(r+1)c/(p+1))}\leq
1$

since $r-p\leq p+1$ and

$\left(b-\frac{(p+1)d}{r}-\frac{(r+1)c}{p+2}\right)-\left(b-\frac{pd}{r}-\frac{(r+1)c}{p+1}\right)=-\frac{d}{r}+\frac{(r+1)c}{(p+1)(p+2)}\approx-1+\frac{dc}{d/2\cdot
d/2}<0.$

Thus $\kappa_{p+1,1}(C,B;L)\leq\kappa_{p,1}(C,B;L)$. We have shown that the
Betti numbers $\kappa_{p,q}(C,B;L)$ with $(r-1)/2\leq p\leq r-1$ form a
decreasing sequence. By Proposition 2.2, the Betti numbers
$\kappa_{p,q}(C,B;L)$ with $0\leq p\leq(r-1)/2$ form an increasing sequence. ∎

######  Remark 5.9.

When $B=\mathscr{O}_{C}$, Proposition 5.8 $(1)$ is the main theorem of [15].
In view of Proposition 5.8 $(2)$, one may expect that nonzero entries of the
top row ($q=0$) of the Betti table form an increasing sequence and nonzero
entries of the bottom row ($q=n+1$) of the Betti table form a decreasing
sequence in higher dimensions.

######  Example 5.10.

Recall that a log-concave sequence of positive terms is unimodal. It is
tempting to expect that $\kappa_{p,1}(C,B;L)$ form a log-concave sequence.
Unfortunately, this may fail when $p$ is small. For instance, let $C$ be a
general smooth projective complex curve of genus $3$, and $B:=\omega_{C}(x)$
for a point $x\in C$. Note that $B$ is not base point free, $\deg B=5$, and
$h^{0}(C,B)=3$. If $L$ is a very ample line bundle on $C$ of degree $d\gg 0$,
then [9, Theorem C] says that $\kappa_{1,1}(C,B;L)$ is a polynomial in $d$ of
degree

$\gamma_{1}(B):=\dim\\{\xi\in C_{2}\mid\underbrace{\text{$H^{0}(C,B)\to
H^{0}(\xi,B|_{\xi})$ is not
surjective}}_{\Longleftrightarrow~{}h^{1}(C,B(-\xi))=h^{0}(C,\mathscr{O}_{C}(\xi-x))=1~{}\Longleftrightarrow~{}x\in\xi}\\}=1$

On the other hand, from (5.3), we find

$\begin{array}[]{l}\kappa_{2,1}(C,B;L)={d-3\choose
2}\left(d+3-\frac{2d}{d-3}-\frac{(d-2)3}{3}\right)\approx\frac{3}{2}d^{2};\\\\[3.0pt]
\kappa_{3,1}(C,B;L)={d-3\choose
3}\left(d+3-\frac{3d}{d-3}-\frac{(d-2)3}{4}\right)\approx\frac{1}{24}d^{4}.\end{array}$

Thus $\kappa_{2,1}(C,B;L)^{2}<\kappa_{1,1}(C,B;L)\cdot\kappa_{3,1}(C,B;L)$.

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