\section{Introduction} \label{section:introduction}
\subsection{Motivation and Results}
Since Bridgeland \cite{bri07} defined the notion of
stability conditions on derived categories,
its construction on a given threefold has beeen
an important open problem.
It turned out that, to solve this problem,
we need a Bogomolov-Gieseker (BG) type inequality,
involving the third Chern character,
for certain stable objects in the derived category
\cite{bms16, bmt14a, bmsz17}.
There are several classes of threefolds on which
we know the existence of Bridgeland stability conditions
\cite{bms16, bmt14a, bmsz17, kos17, kos18, li19b, li19a,
liu19, mp16a, mp16b, mac14, piy17, sch14, sun19}.
For $K$-trivial threefolds, the only known cases are
the quintic threefolds \cite{li19b},
Abelian threefolds,
and their {\' e}tale quotients \cite{bms16, mp16a, mp16b}.

Among them, Li \cite{li19b} recently treated quintic threefolds,
which is one of the most important cases
for Mirror Symmetry.
The crucial step in his arguments is
to establish the improvement of
the classical BG inequality
for torsion free slope stable sheaves.
Recall that a version of
the classical BG inequality
is the inequality
\begin{equation} \label{eq:BGintro}
\frac{H\ch_2(E)}{H^3\ch_0(E)}
\leq \frac{1}{2}\left( \frac{H^2\ch_1(E)}{H^3\ch_0(E)} \right)^2,
\end{equation}
where $E$ is a slope stable sheaf
with respect to an ample divisor $H$.
For $K$-trivial surfaces,
we can easily get the inequality
stronger than (\ref{eq:BGintro}),
simply by using the Serre duality.
In contrast, such an improvement
of the BG inequality on Calabi-Yau threefolds
is highly non-trivial.

So far, the arguments in \cite{li19b}
has been applied only for quintic threefolds.
The goal of the present paper is
to extend it to two other examples of
Calabi-Yau threefolds,
namely, general weighted hypersurfaces
in the weighted projective spaces
$\mathbb{P}(1, 1, 1, 1, 2)$ and
$\mathbb{P}(1, 1, 1, 1, 4)$.
We call them as
{\it triple/double cover CY3},
since they have finite morphisms
to $\mathbb{P}^3$ of degree $3$, $2$, respectively.
The following is our main result:

\begin{thm}[Theorems \ref{thm:strongBG}, \ref{thm:BGonT2}] \label{thm:mainintro}
Let $X$ be a double or triple cover CY3,
$H$ the primitive ample divisor,
and $E$ a slope stable sheaf
with slope $\mu \in [-1, 1]$.
Then the inequality
\begin{equation} \label{eq:strongBGintro}
\frac{H\ch_2(E)}{H^3\ch_0(E)}
\leq \Xi\left(\left|
	\frac{H^2\ch_1(E)}{H^3\ch_0(E)}
	\right|\right)
\end{equation}
holds. Here the function
$\Xi$ is defined as follows.
\[
\Xi(t):=\left\{ \begin{array}{ll}
t^2-t & (t \in [0, 1/4]) \\
3t/4-3/8 & (t \in [1/4, 1/2]) \\
t/4-1/8 & (t \in [1/2, 3/4]) \\
t^2-1/2 & (t \in [3/4, 1]).
\end{array} \right.
\]
\end{thm}

\begin{figure}[htb]
\begin{center}
\begin{tikzpicture}[scale=4]
\draw[->] (-1, 0) -- (1, 0) node[right]{$\frac{H^2\ch_1}{H^3\ch_0}$};
\draw[->] (0, -1/2) -- (0, 1) node[right]{$\frac{H\ch_2}{H^3\ch_0}$};
\draw[red, thick, domain=-1:-3/4] plot(\x, \x*\x-1/2);
\draw[red, thick, domain=-3/4:-1/2] plot(\x, -1/4*\x-1/8);
\draw[red, thick, domain=-1/2:-1/4] plot(\x, -3/4*\x-3/8);
\draw[red, thick, domain=-1/4:0] plot(\x, \x*\x+\x);
\draw[red, thick, domain=0:1/4] plot(\x, \x*\x-\x);
\draw[red, thick, domain=1/4:1/2] plot(\x, 3/4*\x-3/8);
\draw[red, thick, domain=1/2:3/4] plot(\x, 1/4*\x-1/8);
\draw[red, thick, domain=3/4:1] plot(\x, \x*\x-1/2);
\draw[domain=-1:1] plot(\x, 1/2*\x*\x);
\end{tikzpicture}
\end{center}
\caption{strong BG inequality on double/triple cover CY3s.} \label{fig:BGonXintro}
\end{figure}

Using this stronger BG inequality,
we prove the following BG type inequality
(involving $\ch_3$)
for $\nu_{\beta, \alpha}$-stable objects,
which are certain two term complexes
in the derived category.
For the precise definition of
$\nu_{\beta, \alpha}$-stability,
see Section \ref{section:preliminaries}.

\begin{thm}[Theorem \ref{thm:reduction}, Corollary \ref{cor:bgconj}] \label{thm:main2intro}
Let $X$ be a double or triple cover CY3,
take real numbers $\alpha, \beta \in \mathbb{R}$ with
$\alpha >
\frac{1}{2}\beta^2
+\frac{1}{2}(\beta-\lfloor \beta \rfloor)
(\lfloor \beta \rfloor+1-\beta)$.
Let $E$ be a $\nu_{\beta, \alpha}$-semistable object.
Then the inequality
\[
Q^\Gamma_{\alpha, \beta}(E) \geq 0
\]
holds.
Here, we put $\Gamma:=2/9H^2$ (resp. $1/3H^2$)
when $X$ is a triple (resp. double) cover CY3,
and the quadratic form
$Q^\Gamma_{\alpha, \beta}$
is defined as follows:
\begin{align*}
Q^\Gamma_{\alpha, \beta}(E)&:=
(2\alpha-\beta^2)\left(
\overline{\Delta}_{H}(E)+3\frac{\Gamma.H}{H^3}
	\left(H^3\ch^\beta_0(E)
	\right)^2
	\right) \\
&\quad +2\left(H\ch^\beta_2(E) \right)\left(
	2H\ch^\beta_2(E)-3\Gamma.H \ch^\beta_0(E)
	\right) \\
&\quad -6\left(H^2\ch^\beta_1(E) \right) \left(
	\ch^\beta_3(E)-\Gamma\ch_1^\beta(E)
	\right).
\end{align*}

\end{thm}

The above theorem enables us
to construct an open subset
in the space of Bridgeland stability conditions
\cite{bms16, bmt14a, bmsz17}.
For real numbers $\alpha, \beta, a, b$,
we define a group homomorphism
$Z^{a, b}_{\beta, \alpha} \colon K(X) \to \mathbb{C}$ as
\[
Z^{a, b}_{\beta, \alpha}:=
-\ch_3^{\beta}+bH\ch_2^\beta+aH^2\ch_1^\beta
+i\left(
	H\ch_2^\beta-\frac{1}{2}\alpha^2H^3\ch_0^\beta
	\right).
\]
We denote by $\mathcal{A}^{\beta, \alpha}$
the double-tilted heart
defined in \cite{bmt14a}.

\begin{thm}[Theorem \ref{thm:familystab}] \label{thm:main3intro}
We have a continuous family
$\left(Z^{a, b}_{\beta, \alpha}, \mathcal{A}^{\beta, \alpha} \right)$
of stability conditions
parametrized by real numbers
$\alpha, \beta, a, b$ satisfying
\[
\alpha> 0, \quad
\alpha^2+\left(\beta-\lfloor\beta \rfloor-\frac{1}{2} \right)^2 > \frac{1}{4},
\quad a > \frac{1}{6}\alpha^2+\frac{1}{2}|b|\alpha+\gamma,
\]
where we put $\gamma:=2/9$ (resp. $1/3$)
when $X$ is a triple (resp. double) cover CY3.
Acting by the group $\widetilde{\GL^+}(2; \mathbb{R})$,
it forms an open subset in the space of stability conditions.
\end{thm}

\subsection{Strategy of the proof}
In this subsection, we briefly explain
how to prove Theorem \ref{thm:mainintro}.
Let us first recall the arguments in \cite{li19b}
for a quintic threefold $X_5 \subset \mathbb{P}^4$.
We consider $(2, 2, 5)$, $(2, 5)$, $(2, 2)$
complete intersections
\[
C_{2, 2, 5} \subset T_{2, 5} \subset X_5,
\quad C_{2, 2, 5} \subset S_{2, 2}.
\]
The stronger BG inequality on $X_5$
is proved in the following way:
\begin{enumerate}
\item First we reduce the problem
to proving the same inequality for stable sheaves
on the surface $T_{2, 5} \subset X_5$,
by using the restriction technique.

\item Again using the restriction,
the problem is further reduced to
establishing a stronger Clifford type bounds
on global sections for stable vector bundles
on the curve $C_{2, 2, 5} \subset T_{2, 5}$.

\item Regard the stable vector bundle on $C_{2, 2, 5}$
as a torsion sheaf on the surface $S_{2, 2}$
via the inclusion $C_{2, 2, 5} \subset S_{2, 2}$.
Then a wall-crossing argument
in the space of Bridgeland stability conditions
on the surface $S_{2, 2}$
gives the desired Clifford type bounds.
The argument in this step
first appeared in \cite{fey19}.
\end{enumerate}
In step (3), the crucial fact is
that the surface $S_{2, 2}$ is del Pezzo,
on which a stronger BG inequality holds.

For double/triple cover CY3s,
the situation is quite similar.
In fact, we have smooth complete intersection varieties
\begin{align*}
&C_{2, 2, 6} \subset T_{2, 6} \subset X_6, \quad
C_{2, 2, 6} \subset S_{2, 2} \quad \mbox{ in }
\mathbb{P}(1, 1, 1, 1, 2), \\
&C_{2, 4, 8} \subset T_{2, 8} \subset X_8, \quad
C_{2, 4, 8} \subset S_{2, 4} \quad \mbox{ in }
\mathbb{P}(1, 1, 1, 1, 4),
\end{align*}
where both of the surfaces
$S_{2, 2} \subset \mathbb{P}(1, 1, 1, 1, 2)$ and
$S_{2, 4} \subset \mathbb{P}(1, 1, 1, 1, 4)$
are isomorphic to the quadric surface $\mathbb{P}^1 \times \mathbb{P}^1$,
which is del Pezzo.
Note that we consider $(2, 4)$ complete intersection
in $\mathbb{P}(1, 1, 1, 1, 4)$ instead of $(2, 2)$,
to avoid the singularity.

Hence we are able to
apply the methods in \cite{li19b}
to our cases.
At this moment, we do not know the way
to treat these examples uniformly,
so the author believes
it is still worth writing down the complete proofs.
In fact, it turns out that, in our cases,
we need the modified term $\Gamma$
in Theorem \ref{thm:main2intro},
unlike the quintic case.

\subsection{Open problems}
\begin{enumerate}
\item In Theorem \ref{thm:main2intro}
we expect we can take $\Gamma=0$.
For this, we need a further improvement
of Theorem \ref{thm:mainintro}.

\item Stability conditions we construct in this paper
are said to be `near the large volume limit' in Physics.
For weighted hypersurfaces,
we expect the existence of
the another kind of stability conditions,
called {\it Gepner type}.
Mathematically, it is the stability condition
invariant under the certain autoequivalence
of the derived category.
See \cite{tod14b, tod17} for discussions
on the construction of Gepner type stability conditions.
To construct the heart corresponding to
the Gepner type stability condition,
the first task is to prove a stronger form
of the BG inequality for stable sheaves
with a specific slope equal $-1/2$.
Unfortunately, Theorem \ref{thm:mainintro}
is not enough for this purpose.

\item One might ask whether we can treat
other Calabi-Yau weighted hypersurfaces inside
$\mathbb{P}(a_1, a_2, a_3, a_4, a_5)$
with more general weights $(a_i)$.
Unfortunately, quintic and double/triple cover CY3s
are the only cases where
$\mathbb{P}(a_1, a_2, a_3, a_4, a_5)$
contains a smooth Calabi-Yau hypersurface
and a smooth del Pezzo (or K3) complete intersection surface
at the same time.
Indeed, it happens precisely when
the weighted $\mathbb{P}^4$ has only isolated singularities
and its canonical line bundle can be written
as $L^{\otimes m}$,
where $L$ is a free line bundle
and $m \geq 2$.
These conditions are equivalent to
the following numerical conditions.
\begin{itemize}
\item for any $i$ with $a_i>1$ and for any $j \neq i$,
$a_i$ does not divide $a_j$,
\item $\sum a_i=m \cdot \lcm(a_i)$.
\end{itemize}

An easy but lengthy calculation show that
there are only three solutions.
If we allow smooth Deligne-Mumford stacks,
i.e., if we allow
the weighted $\mathbb{P}^4$ to have
non-isolated singularities,
there are several other solutions.
\end{enumerate}

\subsection{Plan of the paper}
The paper is organized as follows.
In Section \ref{section:preliminaries},
we recall about the notion of tilt stability
in the derived category,
and about the BG type inequality conjecture.
Sections \ref{section:clifford} and \ref{section:strongBG}
are devoted to proving
Theorem \ref{thm:mainintro}
for a triple cover CY3.
The key ingredient is the stronger Clifford type bound
proved in Section \ref{section:clifford}.
In Section \ref{section:2to1}, we treat
the case of a double cover CY3.
In Section \ref{section:BGconj},
we prove Theorem \ref{thm:main2intro}.
Finally, in Section \ref{section:conststab},
we prove Theorem \ref{thm:main3intro}.

\begin{ACK}
The author would like to thank Professor Arend Bayer
for giving him various advices and comments.
The author would also like to thank
Chunyi Li, Masaru Nagaoka, Genki Ouchi
and Professor Yukinobu Toda
for related discussions.
The author was supported by
ERC Consolidator grant WallCrossAG, no.819864.
\end{ACK}

\begin{NaC}
In this paper, we always work over the complex number field $\mathbb{C}$.
We will use the following notations.
\begin{itemize}
\item For an ample divisor $H$ and a real number $\beta \in \mathbb{R}$,
we denote by
$\ch^\beta=(\ch^\beta_0, \cdots, \ch^\beta_n):=e^{-\beta H}\ch$,
the $\beta$-twisted Chern character.

\item $\hom(E, F):=\dim \Hom(E, F)$, and
$\ext^i(E, F):=\dim\Ext^i(E, F)$
for objects $E, F$ in the derived category, and an integer $i$.
\end{itemize}
\end{NaC}

\section{Preliminaries} \label{section:preliminaries}
\subsection{BG type inequality conjecture}
In this subsection, we recall the notion of tilt stability,
and the BG type inequality conjecture.
We mainly follow the notations in the paper \cite{li19b}.
Let $X$ be a smooth projective variety
of dimension $n \geq 2$,
$H$ an ample divisor.
We take real numbers $\alpha, \beta \in \mathbb{R}$
with $\alpha > \frac{1}{2} \beta^2$.
We define a slope function $\mu_H$ as follows:
\[
\mu_H:=\frac{H^{n-1}\ch_1}{H^n\ch_0} \colon
\Coh(X) \to \mathbb{R} \cup \{+\infty\}.
\]
We have the notion of $\mu_H$-stability on $\Coh(X)$,
and the corresponding torsion pair on $\Coh(X)$ :
\begin{align*}
&\mathcal{T}_\beta:=\left\langle
T \in \Coh(X) :
T \mbox{ is } \mu_H \mbox{-stable with } \mu_H(T) >\beta
\right\rangle, \\
&\mathcal{F}_\beta:=\left\langle
F \in \Coh(X) :
F \mbox{ is } \mu_H \mbox{-stable with } \mu_H(F) \leq \beta
\right\rangle.
\end{align*}

Here, $\langle S \rangle$
denotes the extension closure
of a set $S \subset \Coh(X)$
of objects in the category $\Coh(X)$.
By the general theory of torsion pairs \cite{hrs96},
we obtain the new abelian category
\[
\Coh^\beta(X):=\left\langle
\mathcal{F}_\beta[1], \mathcal{T}_\beta
\right\rangle
\subset D^b(X),
\]
which is the heart of a bounded t-structure on $D^b(X)$.
On the heart $\Coh^\beta(X)$,
we define the following slope function :
\[
\nu_{\beta, \alpha}:=
\frac{H^{n-2}\ch_2-\alpha H^n\ch_0}{H^{n-1}\ch_1-\beta H^n\ch_0}
\colon \Coh^\beta(X) \to \mathbb{R} \cup \{+\infty\}.
\]
Then as similar to the $\mu_H$-stability on $\Coh(X)$,
we can define the notion of $\nu_{\beta, \alpha}$-stability on $\Coh^\beta(X)$.
We also call $\nu_{\alpha, \beta}$-stability as {\it tilt-stability}.

\begin{defin} \label{def:bn}
Let $E \in \Coh^0(X)$ be an object.
\begin{enumerate}
\item We define the {\it Brill-Noether (BN)} slope of $E$ as
\[
\nu_{BN}(E):=
\frac{H^{n-2}\ch_2(E)}{H^{n-1}\ch_1(E)}
\in \mathbb{R} \cup \{+\infty\}.
\]

\item We say the object $E$ is
{\it Brill-Noether (BN) (semi)stable}
if it is $\nu_{0, \alpha}$-(semi)stable
for every sufficiently small real number
$0 < \alpha \ll 1$.
\end{enumerate}
\end{defin}

We refer \cite[Section 2]{li19b} for the basic properties
of tilt stability and BN stability.
Let us define the discriminant
of an object $E \in D^b(X)$ as
\[
\overline{\Delta}_H(E):=
(H^{n-1}\ch_1(E))^2-2H^n\ch_0(E)H^{n-2}\ch_2(E).
\]
The following is the main question we investigate in this paper.

\begin{ques}[\cite{bms16, bmt14a, bmsz17}] \label{conj:BG}
Assume that $n=\dim X=3$.
Find a $1$-cycle $\Gamma \in A_1(X)_{\mathbb{R}}$
satisfying $\Gamma.H \geq 0$, and the following property:
Let $E$ be a $\nu_{\beta, \alpha}$-semistable object.
Then the inequality
\[
Q^\Gamma_{\alpha, \beta}(E) \geq 0
\]
holds. Here, the quadratic form $Q^\Gamma_{\alpha, \beta}$ is defined as follows:
\begin{align*}
Q^\Gamma_{\alpha, \beta}(E)&:=
(2\alpha-\beta^2)\left(
\overline{\Delta}_{H}(E)+3\frac{\Gamma.H}{H^3}
	\left(H^3\ch^\beta_0(E)
	\right)^2
	\right) \\
&\quad +2\left(H\ch^\beta_2(E) \right)\left(
	2H\ch^\beta_2(E)-3\Gamma.H \ch^\beta_0(E)
	\right) \\
&\quad -6\left(H^2\ch^\beta_1(E) \right) \left(
	\ch^\beta_3(E)-\Gamma\ch_1^\beta(E)
	\right).
\end{align*}
\end{ques}

The conjectural inequality above
is called {\it the Bogomolov-Gieseker(BG) type inequality conjecture},
proposed in \cite{bms16, bmt14a} with $\Gamma=0$.
It is known that the BG type inequality conjecture with $\Gamma=0$
fails for some classes of threefolds, such as
the blow-up of $\mathbb{P}^3$ at a point
(cf. \cite{kos17, ms19, sch17}).
The question with the modified term $\Gamma$
appeared in \cite{bmsz17}
and proved affirmatively for all Fano threefolds.

The following reduction of Question \ref{conj:BG}
plays an important role in this paper.
\begin{thm}[cf. {\cite[Theorem 3.2]{li19b}}] \label{thm:reduction}
Assume that $n=\dim X=3$.
Let $\Gamma$ be a $1$-cycle with $\Gamma.H \geq 0$.
Suppose that for every BN stable object with
$\nu_{BN}(E) \in [0, 1/2]$, the inequality
$Q^\Gamma_{0, 0}(E) \geq 0$ holds.

Then the inequality in Question \ref{conj:BG} holds
for any choice of real numbers
$\alpha, \beta \in \mathbb{R}$ with
$\alpha >
\frac{1}{2}\beta^2
+\frac{1}{2}(\beta-\lfloor \beta \rfloor)
(\lfloor \beta \rfloor+1-\beta)$.
\end{thm}
\begin{proof}
Exactly the same arguments
as in \cite[Theorem 3.2]{li19b} work
since the following statements are true.
\begin{itemize}
\item Let $(\beta', \alpha') \in \mathbb{R}^2$ be a point
on the line through $p_H(E)$ and $(\beta, \alpha)$
with $\alpha'>\frac{1}{2}\beta'^2$.
Then $Q^{\Gamma}_{\alpha, \beta}(E) <0$ if and only if
 $Q^{\Gamma}_{\alpha', \beta'}(E) <0$.
Here we define a point $p_H(E) \in \mathbb{R}^2$ as
\[
p_H(E):=\left(
	\frac{H^2\ch_1(E)}{H^3\ch_0(E)},
	\frac{H\ch_2(E)}{H^3\ch_0(E)}
	\right).
\]

 \item The quadratic form
 $Q^{\Gamma}_{\alpha, \beta}$
 is semi-negative definite on the kernel of
 $\overline{Z}_{\alpha, \beta}:=H^2\ch_1^\beta+i(H\ch_2-\alpha H^3\ch_0)$.
\end{itemize}
\end{proof}

\subsection{Star-shaped functions and the BG type inequalities}
In this subsection, we explain
the wall-crossing technique
used to obtain the (stronger) BG inequality
for tilt-stable objects.
This idea will also appear in the proof of
the BG type inequality conjecture involving $\ch_3$.
As in the previous subsection,
we denote by $X$ a smooth projective variety of dimension $n$,
and $H$ an ample divisor on $X$.
We use the following notion.
\begin{defin}
A function $f \colon \mathbb{R} \to \mathbb{R}$
is called {\it star-shaped}
if the following condition hold:
For all real numbers
$\alpha, \beta \in \mathbb{R}$ with $\alpha >0$,
the line segment connecting the points
$(\beta, f(\beta))$ and $(0, \alpha)$ is above
the graph of $f$.
\end{defin}

Recall that for an object $E \in D^b(X)$
with $\ch_0(E) \neq 0$,
we define
\[
p_H(E):=\left(
	\frac{H^{n-1}\ch_1(E)}{H^n\ch_0(E)},
	\frac{H^{n-2}\ch_2(E)}{H^n\ch_0(E)}
	\right).
\]
We also define
$\mathcal{D}$
to be a set of objects $E \in \Coh^0(X)$
satisfying one of the following conditions:
\begin{enumerate}
\item $E \in \Coh(X)$ and it is $\mu_H$-semistable with $\ch_0(E)>0$.

\item $\mathcal{H}^{-1}(E)$ is $\mu_H$-semistable,
and $\dim \Supp \mathcal{H}^0(E) \leq n-2$.
\end{enumerate}

We have the following result:
\begin{prop}[cf. \cite{bms16, li19a}]
\label{prop:bg-tilt*}
Let $f \colon \mathbb{R} \to \mathbb{R}$ be a star-shaped function.
Assume that for every object $E \in \mathcal{D}$,
the inequality
\[
\frac{H^{n-2}\ch_2(E)}{H^n\ch_0(E)} \leq
f\left(
\frac{H^{n-1}\ch_1(E)}{H^n\ch_0(E)}
\right)
\]
holds.
Then for every $\alpha>0$ and
a $\nu_{0, \alpha}$-semistable object $E$
with $\ch_0(E) \neq 0$,
its Chern character satisfies the same inequality.
\end{prop}
\begin{proof}
Assume for a contradiction that
there exists a tilt-semistable object $E$
violating the required inequality.
By \cite[Theorem 3.5]{bms16}, the object $E$
satisfies the usual BG inequality
$\overline{\Delta}_H(E) \geq 0$.
Hence we may assume that it has
the minimum discriminant $\overline{\Delta}_H(E)$
among all tilt-semistable objects violating the inequality.

Assume that $E$ becomes strictly $\nu_{0, \alpha_0}$-semistable
for some $\alpha_0>0$.
Then there exists a Jordan-H{\"o}lder factor $F$ of $E$
such that $p_H(F)$ is on the line segment connecting
$p_H(E)$ and $(0, \alpha_0)$.
Since the function $f$ is star-shaped,
the object $F$ also violates the required inequality.
Moreover, by \cite[Corollary 3.10]{bms16} we have
$\overline{\Delta}_H(F) < \overline{\Delta}_H(E)$,
which contradicts the minimality assumption on the discriminant.

Now we can assume that $E$ is
$\nu_{0, \alpha}$-semistable
for all $\alpha \gg 0$.
Hence by \cite[Lemma 2.7]{bms16},
we have $E \in \mathcal{D}$,
which contradicts to our assumption.
\end{proof}

\subsection{Triple cover CY3}
Let us consider a general hypersurface
\[
X:=X_{6} \subset P:=\mathbb{P}(1, 1, 1, 1, 2)
\]
of degree $6$ inside the weighted projective space.
Then $X$ is a smooth projective Calabi-Yau threefold,
which we call {\it triple cover CY3}.
We will use
general $(2, 2, 6)$, $(2, 6)$, $(2, 2)$-complete intersections
\[
C_{2, 2, 6} \subset T_{2, 6} \subset X_6,
\quad C_{2, 2, 6} \subset S_{2, 2}
\]
in $P$.
Since the line bundle $\mathcal{O}_{P}(2)$ is free, they are smooth.
The following are some of
the numerical invariants of
$C:=C_{2, 2, 6}$, $T:=T_{2, 6}$, $S:=S_{2, 2}$, and $X$.

\begin{itemize}
\item $-K_{P}=6H_{P}$, $H_{P}^4=\frac{1}{2}$.
\item $g(C)=25$,
\item $-K_{S}=2H_{S}$, ($-K_{S})^2=8$.
In particular, $S \cong \mathbb{P}^1 \times \mathbb{P}^1$.
\item $C=3(-K_{S})$ as divisors in $S$.
\item $\td_{S}=(1, H, 1)$.
\item $K_{T}=2H_{T}$, $H_{T}^2=6$,
$\td_{T}=(1, -H, 11)$.
\item $\td_{X}=(1, 0, \frac{7}{6}H_{X}^2, 0)$, $H_{X}^3=3$.
\end{itemize}

All the computations are straightforward.
For example, to compute $\td_{X, 2}$,
it is enough to compute $\chi(\mathcal{O}_X(1))$,
which can be calculated using the exact sequence
\[
0 \to \mathcal{O}_P(-5) \to \mathcal{O}_P(1) \to \mathcal{O}_X(1) \to 0.
\]

\section{Clifford type theorem} \label{section:clifford}
Recall from the last subsection that
we denote by
\[
C=C_{2, 2, 6} \subset S=S_{2, 2} \subset P=\mathbb{P}(1, 1, 1, 1, 2)
\]
the weighted complete intersections.
We have $S \cong \mathbb{P}^1 \times \mathbb{P}^1$
and $C \in \left|\mathcal{O}_S(6, 6) \right|$.
In this section, we will prove the following proposition:

\begin{prop} \label{prop:clifford}
Let $F$ be a slope stable vector bundle on $C$
of rank $r$, slope $\mu$.
Put $t:=\mu/12$.
Assume that $t \in [0, 1/2] \cup [3/2, 2]$.
The following inequalities hold:
\begin{enumerate}
\item When $t \in [0, 1/6)$,
we have
$h^0(F)/r \leq
\frac{12t+24}{25}$.

\item When $t \in [1/6, 1/4)$,
we have
$h^0(F)/r \leq \max\left\{
\frac{8t+8}{9},
\frac{10}{19}t+\frac{145}{152}
\right\}$.

\item When $t \in [1/4, 1/2]$,
we have
$h^0(F)/r \leq \max\left\{
4t, \frac{33}{38}t+\frac{69}{76}
\right\}$.

\item When $t \in [3/2, 11/6]$,
we have
$h^0(F)/r \leq \max \left\{
4t,
\frac{231}{32}t-\frac{375}{64}
\right\}$.

\item When $t \in (11/6, \sqrt{14}/2]$,
we have
$h^0(F)/r \leq
\frac{233t-191}{32}$.

\item When $t \in [\sqrt{14}/2, 23/12]$,
we have
$h^0(F)/r \leq
\frac{192t-168}{25}$.

\item When $t \in [23/12, 2]$,
we have
$h^0(F)/r \leq
12t-15$.
\end{enumerate}
\end{prop}

\begin{figure}[htbp]
\begin{tabular}{c}
\begin{minipage}{0.5\hsize}
\begin{center}
\begin{tikzpicture}[xscale=10, yscale=2]
\draw[thick, ->] (0, 0.5)node[below]{$0$}
-- (0.05+1/2, 0.5) node[above]{$t$};
\draw[thick, ->] (0, 0.5) -- (0, 2.5) node[right]{$h^0(F)/r$};
\node[left] at (0, 24/25) {$\frac{24}{25}$};
\draw[dashed] (1/6, 0.5)node[below]{$\frac{1}{6}$} -- (1/6, 12/25*1/6+24/25);
\draw[domain=0:1/6] plot(\x, 12/25*\x+24/25);
\draw[dashed] (1/4, 0.5)node[below]{$\frac{1}{4}$} -- (1/4, 33/38*1/4+69/76);
\draw[domain=1/6:1/4] plot(\x, {max(8/9*\x+8/9, 10/19*\x+145/152)});
\draw[domain=1/4:1/2] plot(\x, {max(4*\x, 33/38*\x+69/76)});
\draw[dashed] (1/2, 0.5)node[below]{$\frac{1}{2}$} -- (1/2, 2);
\draw[dashed] (0, 2)node[left]{$2$} -- (1/2, 2);
\end{tikzpicture}
\end{center}
\end{minipage}

\begin{minipage}{0.5\hsize}
\begin{center}
\begin{tikzpicture}[xscale=10, yscale=1]
\draw[thick, ->] (3/2, 5.5)node[below]{$\frac{3}{2}$}
-- (2.05, 5.5) node[right]{$t$};
\draw[thick, ->] (3/2, 5.5) -- (3/2, 9.5) node[right]{$h^0(F)/r$};
\node[left] at (3/2, 6) {$6$};
\draw[dashed] (11/6, 5.5)node[above left]{$\frac{11}{6}$}
-- (11/6, 231/32*11/6-375/64);
\draw[domain=3/2:11/6] plot(\x, {max(4*\x, 231/32*\x-375/64)});
\draw[domain=11/6:1.87] plot(\x, 233/32*\x-191/32);
\draw[dashed] (1.87, 5.5)node[below]{$\frac{\sqrt{14}}{2}$}
-- (1.87, 1.87*192/25-168/25);
\draw[domain=1.87:23/12] plot(\x, 192/25*\x-168/25);
\draw[dashed] (23/12, 5.5)node[below]{$\frac{23}{12}$}
-- (23/12, 12*23/12-15);
\draw[domain=23/12:2] plot(\x, 12*\x-15);
\draw[dashed] (2, 5.5)node[below]{$2$} -- (2, 9);
\draw[dashed] (3/2, 9)node[left]{$9$} -- (2, 9);
\end{tikzpicture}
\end{center}
\end{minipage}

\end{tabular}
\caption{The strong Clifford type bounds on $C$.}
\label{fig:Cliff}
\end{figure}

\begin{rmk}
The parameter $t=\mu/12$
naturally appears as the BN slope
of the sheaf $\iota_*F$,
where $\iota \colon C \hookrightarrow S$
is an embedding.
Indeed, we have
$\nu_{BN}(\iota_*F)=t-3$,
as we will see
in the proof of Lemma \ref{lem:bnslope} below.

It is also compatible with
the slope function on $T$
in the following sense.
For a vector bundle $F$ on $T$,
we have $t(F|_C)=\mu_{H_T}(F)$.
\end{rmk}

Our strategy of the proof of Proposition \ref{prop:clifford}
is to use Bridgeland stability conditions on
the surface $S$, with the following three steps.
\begin{enumerate}
\item Regard $F$ as a torsion sheaf $\iota_{*}F \in \Coh(S)$,
which is $\nu_{0, \alpha}$-stable for $\alpha \gg 0$.
\item Estimate the first possible wall
for $\iota_{*}F$ on the line $\beta=0$
in $(\alpha, \beta)$ plane,
using the stronger form of the BG inequality on $S$.
\item Bound global sections of BN-stable objects on $S$.
\end{enumerate}

We define a function $\Upsilon$ on $\mathbb{R}$ as
\[
\Upsilon(x):=
\begin{cases}
\frac{1}{2}x^2-\frac{1}{2}(1-\{x\})^2
	& (\{x\} \in (0, 1/2]) \\
\frac{1}{2}x^2-\frac{1}{2}\{x\}^2
	& (\{x\} \in [1/2, 1)) \\
\frac{1}{2}x^2 & (\{x\}=0).
\end{cases}
\]
Here, $\{x\}$ denotes the fractional part of $x \in \mathbb{R}$.
See Figure \ref{fig:bgonS} below
for the shape of $\Upsilon$.
The following stronger BG inequality
on the quadric surface
$S \cong \mathbb{P}^1 \times \mathbb{P}^1$ is well-known.
We include a proof here,
since it demonstrates the technique which
we will frequently use in this section.

\begin{lem} \label{lem:BGdP}
Let $F$ be a slope semistable tosion free sheaf on $S$.
Then we have an inequality
\begin{equation} \label{eq:bgonS}
\frac{\ch_{2}(F)}{H^2\ch_{0}(F)} \leq
\Upsilon\left(\mu_H(F) \right).
\end{equation}
\end{lem}
\begin{proof}
Since we have
$\Upsilon(x+1)=\Upsilon(x)+x+1/2$,
the claim is invariant under
tensoring with the line bundle $\mathcal{O}_S(H)$.
Hence we may assume $\mu_H(F) \in (0, 1)$.
By the stability of $F$ and the Serre duality,
we have
\[
\hom(\mathcal{O}(1), F)=0, \quad
\ext^2(\mathcal{O}(1), F)=\hom(F, \mathcal{O}(-1))=0
\]
and hence
$0 \geq -\ext^1\left(\mathcal{O}(1), F \right)
=\chi\left(\mathcal{O}(1), F \right)$.
By computing the RHS using the Riemann-Roch theorem,
we get the inequality
\[
\frac{\ch_{2}(F)}{H^2\ch_{0}(F)} \leq 0.
\]

On the other hand,
again by the stability of $F$
and the Serre duality,
we also have
\[
\hom(F, \mathcal{O})=0, \quad
\ext^2(F, \mathcal{O})=\hom(\mathcal{O}, F(-2))=0,
\]
which imply the inequality
$0 \geq -\ext^1(F, \mathcal{O})=\chi(F, \mathcal{O})$.
Hence we obtain
\[
\frac{\ch_2(F)}{H^2\ch_0(F)} \leq \mu_H(F)-\frac{1}{2}.
\]

Taking the minimum,
the inequality (\ref{eq:bgonS}) holds.
\end{proof}

\begin{rmk}
In \cite{rud94}, Rudakov proved
an inequality stronger
than (\ref{eq:bgonS}).
However, our inequality is already optimal
at $\mu_H=1/2$
(consider $F=\mathcal{O}_S(1, 0)$).
Because of this fact,
we cannot improve our inequality
in Theorem \ref{thm:mainintro}
at $\mu_H=1/2$,
even if we use the result in \cite{rud94}.
\end{rmk}

We define a function
$\widetilde{\Upsilon} \colon \mathbb{R} \to \mathbb{R}$ as
\[
\widetilde{\Upsilon}(x):=
\begin{cases}
\Upsilon(x) & (|x| \in [0, 1]) \\
\max\left\{
	\Upsilon(x),
	\frac{1}{2} \lfloor |x| \rfloor x
	\right\}
& (|x| \geq 1).
\end{cases}
\]
Here, $\lfloor |x| \rfloor$
denotes the integral part of
the absolute value of
a real number $x \in \mathbb{R}$.
Note that the function
$\widetilde{\Upsilon}$
is star-shaped
and we have $\Upsilon(x) \leq \widetilde{\Upsilon}(x)$
for all $x \in \mathbb{R}$,
see Figure \ref{fig:bgonS}.

\begin{figure}[htbp]
\begin{center}
\begin{tikzpicture}[xscale=1.8]
\draw [->] (-3, 0) -- (3, 0) node[right]{$\frac{H\ch_1}{H^2\ch_0}$};
\draw[->] (0, -1) -- (0, 9/2) node[right]{$\frac{\ch_2}{H^2\ch_0}$};
\draw[red, thick, domain=0:1/2] plot(\x,\x-1/2);
\draw[red, thick] (1/2, 0) -- (1, 0);
\draw[red, thick, domain=1:3/2] plot(\x,2*\x-2);
\draw[red, thick, domain=3/2:2] plot(\x, \x-1/2);
\draw[red, thick, domain=2:5/2] plot(\x, 3*\x-9/2);
\draw[red, thick, domain=5/2:3]
plot(\x, 2*\x-2) node[right]{$\Upsilon$};
\draw[red, thick, domain=-3:-5/2] plot(\x, -2*\x-2);
\draw[red, thick, domain=-5/2:-2] plot(\x, -3*\x-9/2);
\draw[red, thick, domain=-2:-3/2] plot(\x, -\x-1/2);
\draw[red, thick, domain=-3/2:-1] plot(\x,-2*\x-2);
\draw[red, thick] (-1, 0) -- (-1/2, 0);
\draw[red, thick, domain=-1/2:0] plot(\x,-\x-1/2);
\draw[domain=-3:3] plot(\x, 1/2*\x*\x);
\draw[blue, thick, domain=1:4/3] plot(\x, {abs(1/2*\x)});
\draw[blue, dashed, domain=0:4/3] plot(\x, {abs(1/2*\x)});
\draw[blue, thick, domain=-4/3:-1] plot(\x, {abs(1/2*\x)});
\draw[blue, dashed, domain=-4/3:0] plot(\x, {abs(1/2*\x)});
\draw[blue, thick, domain=2:9/4] plot(\x, {abs(\x)});
\draw[blue, dashed, domain=0:9/4] plot(\x, {abs(\x)});
\draw[blue, thick, domain=-9/4:-2] plot(\x, {abs(\x)});
\draw[blue, dashed, domain=-9/4:0] plot(\x, {abs(\x)});
\draw[dashed] (1, 0) node[below]{$1$} -- (1, 1/2);
\draw[dashed] (-1, 0) node[below]{$-1$} -- (-1, 1/2);
\draw[dashed] (2, 0) node[below]{$2$} -- (2, 2);
\draw[dashed] (-2, 0) node[below]{$-2$} -- (-2, 2);
\draw[dashed] (3, 0) node[below]{$3$} -- (3, 9/2);
\draw[dashed] (-3, 0) node[below]{$-3$} -- (-3, 9/2);
\end{tikzpicture}
\end{center}
\caption{The strong BG inequality $\Upsilon$ (red curve) on the quadric surface.
Blue lines show the modified curve $\widetilde{\Upsilon}$.}
\label{fig:bgonS}
\end{figure}

We have the following consequences of
Lemma \ref{lem:BGdP}.

\begin{lem} \label{lem:geomstabS}
The following statements hold:
\begin{enumerate}
\item Fix a positive real number $\alpha >0$.
Let $F \in \Coh^0(S)$
be a $\nu_{0, \alpha}$-semistable object
with $\ch_0(F) \neq 0$.
Then the Chern character of $F$ satisfies
the inequality
\begin{equation} \label{eq:bgonS-tilt}
\frac{\ch_{2}(F)}{H^2\ch_{0}(F)} \leq
\widetilde{\Upsilon}\left(\mu_H(F) \right).
\end{equation}

\item For all real numbers
$\beta, \alpha \in \mathbb{R}$ with
$\alpha > \Upsilon(\beta)$,
the pair
$\left(Z_{\beta, \alpha}, \Coh^\beta(S) \right)$
defines a stability condition on $D^b(S)$.
Here the group homomorphism
$Z_{\beta, \alpha} \colon K(S) \to \mathbb{C}$ is defined as
\[
Z_{\beta, \alpha}:=-\ch_2+\alpha H^2\ch_0
	+i\left(H\ch_1-\beta H^2\ch_0 \right).
\]
\end{enumerate}
\end{lem}
\begin{proof}
(1) The first assertion follows from
Lemma \ref{lem:BGdP} and Proposition \ref{prop:bg-tilt*}.

(2) For the second assertion,
we can apply the arguments
in \cite{ab13, bri08}
by replacing the classical BG inequality with
the stronger one (\ref{eq:bgonS}).
\end{proof}

In the next two lemmas,
we control the position of the first possible wall
for $\iota_*F$, where $F$ is a stable bundle on $C$,
and then bound the slopes of the HN factors of $\iota_*F$
with respect to BN stability.

\begin{lem} \label{lem:1stwall}
Let $F$ be a slope stable vector bundle on $C$
with rank $r$, slope $\mu$.
Let $(\beta_1, \alpha_1), (\beta_2, \alpha_2)$,
$\beta_1 < 0 < \beta_2$,
be the end points of a wall for $\iota_*F$
with respect to $\nu_{\beta, \alpha}$-stability.
Then we have $\beta_2-\beta_1 \leq 6$.
\end{lem}
\begin{proof}
By the Grothendieck-Riemann-Roch theorem, we have
\begin{equation} \label{eq:grr}
\ch(\iota_{*}F)=(0, 6rH, r(\mu-36)).
\end{equation}
Suppose that there exists
a positive integer $\alpha$ and
a destabilizing sequence
\[
0 \to F_{2} \to \iota_{*}F \to F_{1} \to 0
\]
in $\Coh^0(S)$ for $\nu_{0, \alpha}$-stability.
Denote by $W$ the corresponding wall.
Note that $F_2$ is a coherent sheaf.
Let $T \subset F_2$ be a torsion part and put
$Q:=F_2/T$.
We have the following diagram
in the tilted category $\Coh^0(S)$:
\[
\xymatrix{
& &0 \ar[d] &0 \ar[d] & & \\
& &T \ar[d] \ar@{=}[r] &T \ar[d] & & \\
&0 \ar[r] &F_2 \ar[d] \ar[r] &\iota_*F \ar[d] \ar[r] &F_1 \ar[r] \ar@{=}[d] &0 \\
&0 \ar[r] &Q \ar[d] \ar[r] &\iota_*F/T \ar[d] \ar[r] &F_1 \ar[r] &0 \\
& &0 &0 & &
}
\]

By taking the $\Coh(S)$-cohomology
of the bottom row in the above diagram,
we get the exact sequence
\[
0 \to Q/\mathcal{H}^{-1}(F_1) \to \iota_*F/T \to F_1 \to 0
\]
in $\Coh(S)$.
In particular, the sheaf
$Q/\mathcal{H}^{-1}(F_1)$
is scheme-theoretically supported on the curve $C$.
Hence we have a surjection
$Q|_C \twoheadrightarrow Q/\mathcal{H}^{-1}(F_1)$
and so get an inequality
\[
6H^2\ch_0(Q)=H\ch_1(Q|_C) \geq
H\ch_1(Q/\mathcal{H}^{-1}(F_1)).
\]
Note also that we have
$\ch_0(Q)=\ch_0(\mathcal{H}^{-1}(F_1))$.
Now we have
\begin{equation} \label{eq:mudiff}
\begin{aligned}
\mu_H(Q)-\mu_H(\mathcal{H}^{-1}(F_1))
=\frac{H\ch_1(Q/\mathcal{H}^{-1}(F_1))}{H^2\ch_0(Q)} \leq 6.
\end{aligned}
\end{equation}

Now let $(\beta_1, \alpha_1), (\beta_2, \alpha_2)$
be the end points of the wall $W$
with $\beta_1 < 0 < \beta_2$.
By Bertram's nested wall theorem
(see e.g. \cite[Lemma 2.9]{li19b},  \cite{maci14}),
we know that for $0 < \epsilon \ll 1$,
we have
\[
Q \in \Coh^{\beta_{2}-\epsilon}(S), \quad
\mathcal{H}^{-1}(F_{1})[1] \in \Coh^{\beta_{1}+\epsilon}(S),
\]
which in particular imply
\[
\mu_{H}(Q) > \beta_{2}-\epsilon, \quad
\mu_{H}(\mathcal{H}^{-1}(F_{1})) \leq \beta_{1}+\epsilon.
\]
Combining with the inequality (\ref{eq:mudiff}),
we have the desired inequality
\[
\beta_{2}-\beta_{1} \leq 6.
\]
\end{proof}

\begin{lem} \label{lem:bnslope}
Let $F$ be a slope stable vector bundle on $C$
with rank $r$, slope $\mu \in (0, 24)$.
Let $t:=\mu/12$.
The following statements hold:
\begin{enumerate}
\item If $t \in \left(0, 2-\frac{\sqrt{14}}{2} \right]$,
then the sheaf $\iota_*F$ is BN-stable.

\item We have
\[
\nu^+_{BN}(\iota_*F) \leq
\begin{cases}
1-1/2t
	& (t \in (2-\sqrt{14}/2, 1/2] \cup [3/2, \sqrt{14}/2]) \\
\frac{-9t+11}{-8t+7}
	& (t \in [\sqrt{14}/2, 23/12]) \\
3t-5
	& (t \in [23/12, 2]).
\end{cases}
\]

\item We have
\[
\nu^-_{BN}(\iota_*F) \geq
\begin{cases}
\frac{-5(2t-7)}{2(t-6)}
	& (t \in (2-\sqrt{14}/2, 1/2] \cup [3/2, 11/6] ) \\
-2
	& (t \in [11/6, 2]).
\end{cases}
\]
\end{enumerate}
\end{lem}

\begin{proof}
Let $W$ be a wall for $\iota_*F$,
and let $(\beta_1, \alpha_1), (\beta_2, \alpha_2)$
be the end points of the wall $W$
with $\beta_1 < 0 < \beta_2$.
Recall that the wall $W$ is a line segment
with slope
$\nu_{BN}(\iota_*F)=\mu/12-3=t-3$
(see (\ref{eq:grr}) for the second equality).
Since the curve $\Upsilon$ is not continuous
when $\beta \in \mathbb{Z}$,
the points $(\beta_i, \alpha_i)$ are
either on the graph of $\Upsilon$,
or on the vertical lines
\[
L_n:=\left\{
(n, y) : \frac{n^2-1}{2} < y < \frac{n^2}{2}
\right\},
\quad n \in \mathbb{Z}.
\]
When both of the end points
$(\beta_i, \alpha_i)$ are
on the curve $\Upsilon$,
we say that the wall $W$ is
of {\it Type A},
otherwise, we say it is of {\it Type B}.

First assume that $W$ is of Type A.
By Lemma \ref{lem:1stwall},
we know that $\beta_2-\beta_1 \leq 6$.
Hence the slope of the line through
$(\beta_{2}, \Upsilon(\beta_{2}))$ and
$(\beta_{2}-6, \Upsilon(\beta_{2}-6))$
is smaller than or equal to that of $W$, i.e.,
$\beta_{2}-3 \leq t-3$.
We conclude that every Type A wall is
below the line
$y=(t-3)(x-t)+\Upsilon(t)$.

On the other hand,
for a given point $p=(\beta, \alpha) \in L_n$,
let $W_p$ be the line passing through
the point $p$ with slope $t-3$.
It is easy to compute the intersection points
of $W_p$ and $\Upsilon \cup \bigcup_{n \in \mathbb{Z}}L_n$.
Together with the constraint $\beta_2-\beta_1 \leq 6$,
we can find the first possible wall of Type B.

Using these observations,
we can list up the equation of
the first possible wall:
\begin{itemize}
\item When $t \in [0, 1/2]$,
the following is the first possible wall
\[
y=(t-3)(x-t)+t-1/2,
\]
Note that if $t \in [0, 2-\sqrt{14}/2]$,
it is negative at $x=0$,
hence $\iota_*F$ is BN stable.

\item When $t \in [3/2, 2]$,
one of the following is the first possible wall
\[
y=(t-3)(x-t)+t-1/2, \quad
y=(t-3)(x+4)+8.
\]
The first one is the line
passing through the points
$(t, \Upsilon(t))$ and
$(t-6, \Upsilon(t-6))$.
The second one is the line
with slope $t-3$, passing through the point
$(-4, \Upsilon(-4))$.
See Figure \ref{fig:BGonT} below.
\end{itemize}

Let $L$ be the first possible wall
described above,
and let $(\beta_{\max}, \alpha_{\max})$,
$(\beta_{\min}, \alpha_{\min})$
be the intersection points of $L$
with the curve $\widetilde{\Upsilon}$
with $\beta_{\min}<\beta_{\max}$.
Then any wall $W$ should be below the line $L$.
Now consider the maximal destabilizing subobject
$E_{1} \subset \iota_{*}F$
with respect to the BN stability.
We have three numerical constraints on $E_{1}$:
\begin{itemize}
\item Since $\iota_{*}F$ is $\nu_{\alpha, 0}$-stable
for $\alpha$ sufficiently large,
we have $\ch_{0}(E_{1}) >0$.

\item $E_{1}$ satisfies
the BG type inequality (\ref{eq:bgonS-tilt}).

\item The point $p_{H}(E_{1})$ is below the line $L$.
\end{itemize}

Among all points satisfying the above three conditions,
its slope become maximum at the point
$(\alpha_{\max}, \beta_{\max})$,
hence we get the bound
\[
\nu_{BN}(E_{1}) \leq \frac{\alpha_{\max}}{\beta_{\max}}.
\]

Now the straightforward computation shows the result.
Similarly we can get the bound
$\nu^-_{BN}(\iota_{*}F) \geq \alpha_{\min}/\beta_{\min}$.

\end{proof}

\begin{figure}[htbp]
\begin{tabular}{c}
\begin{minipage}{0.5\hsize}
\begin{center}
\begin{tikzpicture}[scale=0.5]
\draw [->] (-5, 0) -- (3, 0) node[right]{$\beta, \frac{H\ch_1}{H^2\ch_0}$};
\draw[->] (0, -1) -- (0, 13) node[right]{$\alpha, \frac{\ch_2}{H^2\ch_0}$};
\draw[red, thick, domain=0:1/2] plot(\x,\x-1/2);
\draw[red, thick] (1/2, 0) -- (1, 0);
\draw[red, thick, domain=1:3/2] plot(\x,2*\x-2);
\draw[red, thick, domain=3/2:2] plot(\x, \x-1/2);
\draw[red, thick, domain=2:5/2] plot(\x, 3*\x-9/2);
\draw[red, thick, domain=5/2:3] plot(\x, 2*\x-2)
node[right]{$\alpha=\Upsilon(\beta)$};
\draw[red, thick, domain=-5:-9/2] plot(\x, -4*\x-8);
\draw[red, thick, domain=-9/2:-4] plot(\x, -5*\x-25/2);
\draw[red, thick, domain=-4:-7/2] plot(\x, -3*\x-9/2);
\draw[red, thick, domain=-7/2:-3] plot(\x, -4*\x-8);
\draw[red, thick, domain=-3:-5/2] plot(\x, -2*\x-2);
\draw[red, thick, domain=-5/2:-2] plot(\x, -3*\x-9/2);
\draw[red, thick, domain=-2:-3/2] plot(\x, -\x-1/2);
\draw[red, thick, domain=-3/2:-1] plot(\x,-2*\x-2);
\draw[red, thick] (-1, 0) -- (-1/2, 0);
\draw[red, thick, domain=-1/2:0] plot(\x,-\x-1/2);
\draw[domain=-5:3] plot(\x, 1/2*\x*\x);
\draw[green, domain=-6:3] plot(\x, -3/2*\x+13/4) node[left]{$L$};
\draw[red, thick] (-4, 15/2) -- (-4, 8);
\draw[red, thick] (2, 2) -- (2, 3/2);
\end{tikzpicture}
\end{center}
\end{minipage}

\begin{minipage}{0.5\hsize}
\begin{center}
\begin{tikzpicture}[scale=0.5]
\draw [->] (-5, 0) -- (3, 0) node[right]{$\beta, \frac{H\ch_1}{H^2\ch_0}$};
\draw[->] (0, -1) -- (0, 13) node[right]{$\alpha, \frac{\ch_2}{H^2\ch_0}$};
\draw[red, thick, domain=0:1/2] plot(\x,\x-1/2);
\draw[red, thick] (1/2, 0) -- (1, 0);
\draw[red, thick, domain=1:3/2] plot(\x,2*\x-2);
\draw[red, thick, domain=3/2:2] plot(\x, \x-1/2);
\draw[red, thick, domain=2:5/2] plot(\x, 3*\x-9/2);
\draw[red, thick, domain=5/2:3] plot(\x, 2*\x-2)
node[right]{$\alpha=\Upsilon(\beta)$};
\draw[red, thick, domain=-5:-9/2] plot(\x, -4*\x-8);
\draw[red, thick, domain=-9/2:-4] plot(\x, -5*\x-25/2);
\draw[red, thick, domain=-4:-7/2] plot(\x, -3*\x-9/2);
\draw[red, thick, domain=-7/2:-3] plot(\x, -4*\x-8);
\draw[red, thick, domain=-3:-5/2] plot(\x, -2*\x-2);
\draw[red, thick, domain=-5/2:-2] plot(\x, -3*\x-9/2);
\draw[red, thick, domain=-2:-3/2] plot(\x, -\x-1/2);
\draw[red, thick, domain=-3/2:-1] plot(\x,-2*\x-2);
\draw[red, thick] (-1, 0) -- (-1/2, 0);
\draw[red, thick, domain=-1/2:0] plot(\x,-\x-1/2);
\draw[domain=-5:3] plot(\x, 1/2*\x*\x);
\draw[green, domain=-6:3] plot(\x, -13/12*\x+11/3) node[left]{$L$};
\draw[red, thick] (-4, 15/2) -- (-4, 8);
\draw[red, thick] (2, 2) -- (2, 3/2);
\end{tikzpicture}
\end{center}
\end{minipage}

\end{tabular}
\caption{The first possible wall $L$
when $t=3/2$ (left) and
$t=23/12$ (right).} \label{fig:BGonT}
\end{figure}

The following lemma gives the upper bound
on the number of global sections for
BN stable objects.
\begin{lem} \label{lem:sectionbn}
Let $F \in \Coh^0(S)$ be a BN stable object.
Then the following inequalities hold:
\begin{itemize}
\item When $-1 < \nu_{BN}(F) < +\infty$, we have
\[
\hom(\mathcal{O}_{S}, F) = \ch_{0}(F) + H\ch_{1}(F) + \ch_{2}(F).
\]

\item When $\nu_{BN}(E) \in (-n-1, -n)$, $n \in \mathbb{Z}_{>0}$,
we have
\[
\hom(\mathcal{O}_{S}, F)
	\leq \ch_{0}(F) + \frac{1}{2n+1}H\ch_{1}(F)
		+ \frac{1}{(2n+1)^2}\ch_{2}(F).
\]

\item When $\nu_{BN}(E)=-n$,
$n \in \mathbb{Z}_{>0}$, we have
\[
\hom(\mathcal{O}_S, F) \leq \ch_0(F)+\frac{1}{4n}H\ch_1(F).
\]
\end{itemize}
\end{lem}
\begin{proof}
First assume that $\nu_{BN}(F) > -1$.
Noting
$\nu_{BN}(\mathcal{O}_S[1])=+\infty$ and
$\nu_{BN}(\mathcal{O}_{S}(-2)[1])=-1$,
we have the following vanishings for any $i \geq 0$:
\begin{align*}
&\hom(\mathcal{O}_{S}, F[1+i])=\hom(F, \mathcal{O}_{S}(-2)[1-i])=0, \\
&\hom(\mathcal{O}_{S}, F[-1-i])=\hom(\mathcal{O}_{S}[1+i], F)=0.
\end{align*}
Hence by the Riemann-Roch, we get
\begin{align*}
\hom(\mathcal{O}_{S}, F)=\chi(F)
&=\int_{S}\ch(F). (1, H, 1) \\
&=\ch_{0}(F) + H\ch_{1}(F) +\ch_{2}(F).
\end{align*}

Next consider the case
$\nu_{BN}(E) \in (-n-1, -n)$, $n \in \mathbb{Z}_{>0}$.
Let
\[
(x_{F}, y_{F}):=p_{H}(F)
=\left(
\frac{H\ch_{1}(F)}{H^2\ch_{0}(F)},
\frac{\ch_{2}(F)}{H^2\ch_{0}(F)}
\right).
\]
Since we assumed
$y_{F}/x_{F}=\nu_{BN}(F) \leq -1<0$,
the line through $(0, 0)$ and $(x_{F}, y_{F})$
intersects with the region
$y \geq 1/2x^2, x<0$.

Take such a point $(\beta, \alpha)$.
Then we know that the objects
$F, \mathcal{O}_S \in \Coh^\beta(S)$ are
$\nu_{\beta, \alpha}$-semistable with
\[
\nu_{\beta, \alpha}(F)
=\nu_{\beta, \alpha}(\mathcal{O}_{S})
=\alpha/\beta=\nu_{BN}(F).
\]
Let us consider the exact triangle
\[
\Hom(\mathcal{O}_{S}, F) \otimes \mathcal{O}_{S} \xrightarrow{ev}
F \to \widetilde{F}:=\Cone(ev).
\]
Since the only Jordan-H\"{o}lder factor
of $\Hom(\mathcal{O}_S, F) \otimes \mathcal{O}_S$
with respect to $\nu_{\beta, \alpha}$-stability
is $\mathcal{O}_S$, the evaluation map
$ev$ must be injective in the category $\Coh^{\beta}(S)$.
Hence it follows that
$\widetilde{F} \in \Coh^{\beta}(S)$
and it is  $\nu_{\beta, \alpha}$-semistable with
$\nu_{\beta, \alpha}(\widetilde{F})
=\nu_{\beta, \alpha}(F)=\nu_{BN}(F)$.
Now choose $\beta$ sufficiently close to zero so that
$\mathcal{O}_{S}(-2n)[1] \in \Coh^{\beta}(S)$.
As before, we have the vanishing statements
\begin{align*}
&\hom(\mathcal{O}_{S}(-2n), \widetilde{F}[1+i])
=\hom(\widetilde{F}, \mathcal{O}_{S}(-(2n+2))[1-i])=0, \\
&\hom(\mathcal{O}_{S}(-2n), \widetilde{F}[-1-i])
=\hom(\mathcal{O}_{S}(-2n)[1+i], \widetilde{F})=0
\end{align*}
for $i \geq 0$.
Hence we have
\begin{align*}
0 &\leq \hom(\mathcal{O}_{S}(-2n), \widetilde{F}) \\
	&=\chi(\mathcal{O}_{S}(-2n), \widetilde{F}) \\
	&=\ch_{2}(F)+(2n+1)H\ch_{1}(F)
		+(2n+1)^2(\ch_{0}(F)-\hom(\mathcal{O}_{S}, F)),
\end{align*}
and so
\[
\hom(\mathcal{O}_{S}, F) \leq
	\ch_{0}(F)+\frac{H\ch_{1}(F)}{2n+1}
	+\frac{\ch_{2}(F)}{(2n+1)^2}
\]
as required.

Finally, assume that
$\nu_{BN}(F)=-n$,
$n \in \mathbb{Z}_{>0}$.
Then the same argument shows that
$\chi(\mathcal{O}_S(-2n+1), \widetilde{F}) \geq 0$,
and we get the inequality
\[
\hom(\mathcal{O}_S, F) \leq
\ch_0(F)+\frac{H\ch_1(F)}{2n}+\frac{\ch_2(F)}{(2n)^2}
=\ch_0(F)+\frac{1}{4n}H\ch_{1}(F).
\]
\end{proof}

Let us define a function
$\Omega \colon \mathbb{R} \times \mathbb{R}_{>0} \to \mathbb{R}_{>0}$ as
\[
\Omega(x, y):=
\begin{cases}
y+x & (x/y > -1) \\
\frac{y}{2n+1}+\frac{x}{(2n+1)^2} & (x/y \in (-n-1, -n), n \in \mathbb{Z}_{>0}) \\
\frac{1}{4n}y & (x/y=-n, n \in \mathbb{Z}_{>0}).
\end{cases}
\]

Now we can prove Proposition \ref{prop:clifford}.
\begin{proof}[Proof of Proposition \ref{prop:clifford}]
Let $F$ be a slope stable vector bundle on $C$
of rank $r$ and slope $\mu$.
Let
\[
0=E_{0} \subset E_{1} \subset \cdots \subset E_{m}=\iota_{*}F
\]
be the HN filtration with respect to the BN stability,
and define $P_{i}:=(\ch_{2}(E_{i}), H\ch_{1}(E_{i}))$.
We have an inequality
\begin{equation} \label{eq:Omegaineq}
h^0(F) \leq
\sum_{i=1}^m \Omega(\overrightarrow{P_{i-1}P_{i}})
\end{equation}
by Lemma \ref{lem:sectionbn}
(cf. \cite[Equation (21)]{li19b}).
We will bound the RHS in the above inequality.
Let us put
\[
P=(x_p, y_p):=(\ch_2(\iota_{*}F), H\ch_{1}(\iota_{*}F))
=(r(\mu-36), 12r),
\]
and $Q=(x_q, y_q)$ to be a point
such that $x_q/y_q$ is the upper bound for
$\nu^+_{BN}(\iota_{*}F)$,
and $(x_p-x_q)/(y_p-y_q)$ is the lower bound for
$\nu^-_{BN}(\iota_{*}F)$, given in Lemma \ref{lem:bnslope}.
We know that the HN polygon of $\iota_{*}F$ with respect to the BN stability
is inside the triangle $OPQ$.
Then as in \cite[Lemma 4.11]{li19b},
we may assume $m=2$,
and the point $P_1$ satisfies one of the following conditions:
\begin{itemize}
\item $P_1=Q$,
\item $P_1$ is on the line segment $OQ$ (resp. $PQ$)
such that the slope of $P_1P$ (resp. $OP_1$)
is $-1/n$ for some $n \in \mathbb{Z}_{>0}$.
\item the lines $OP_1$ and $P_1P$
have slopes $-1/m, -1/n$
for some integers $m, n \in \mathbb{Z}_{>0}$.
\end{itemize}

We now argue case by case.

(0) When $t \in (0, 2-\sqrt{14}/2]$,
the sheaf $\iota_*F$ is BN stable.
The BN slope is
$\nu_{BN}(\iota_*F)=t-3 \in (-3, -2]$,
hence by Lemma \ref{lem:sectionbn},
we have
\[
h^0(F)/r \leq
\frac{12}{5}+\frac{12(t-3)}{25}
=\frac{12t+24}{25}.
\]

(1) Assume $t \in (2-\sqrt{14}/2, 1/6)$.
By Lemma \ref{lem:bnslope}, we have
\begin{itemize}
\item The slope of $\overrightarrow{OP}$ is
$\frac{1}{t-3} \in (-1/2, -1/3)$,

\item The slope of $\overrightarrow{OQ}$ is
$\frac{2t}{2t-1} \in (-1/2, -1/3)$.

\item The slope of $\overrightarrow{QP}$ is
$\frac{2(t-6)}{-5(2t-7)} \in (-1/2, -1/3)$.
\end{itemize}
Hence we may assume $P_1=Q=(x_q, y_q)$.
We get
\begin{align*}
h^0(F) &\leq
\Omega(\overrightarrow{OQ})+\Omega(\overrightarrow{QP}) \\
&=\frac{y_q}{5}+\frac{x_q}{25}
+\frac{y_p-y_q}{5}+\frac{x_p-x_q}{25}
=\frac{12t+24}{25}r.
\end{align*}

(2) Assume $t \in [1/6, 1/4)$.
Then the slopes of the triangle $OPQ$ are
same as the case (2).
The only difference is that
the slope of $\overrightarrow{OQ}$
sits inside the interval $(-1, -1/2]$,
instead of $(-1/2, -1/3)$.
Hence we may take $P_1$ as $Q$
or the point $A$ on the line segment $QP$
with slope $-1/2$.
The coordinates are given as
\[
Q=((2t-1)r, 2tr), \quad
A=\left(
	\frac{24(2t^2-8t+1)}{-6t+11},
	\frac{12(2t^2-8t+1)}{-6t+11}
	\right).
\]
First consider the case of $P_1=Q$.
We have
\begin{align*}
\Omega(\overrightarrow{OQ})+\Omega(\overrightarrow{QP})
&=\frac{y_q}{3}+\frac{x_q}{9}
+\frac{y_p-y_q}{5}+\frac{x_p-x_q}{25} \\
&=\frac{8t+8}{9}r.
\end{align*}

When $P_1=A$, we have
\begin{align*}
\Omega(\overrightarrow{OA})+\Omega(\overrightarrow{AP})
&=\frac{y_a}{8}
+\frac{y_p-y_a}{5}+\frac{x_p-x_a}{25} \\
&=\frac{1}{200}y_a+\frac{12t+24}{25}.
\end{align*}

As a function on $t \in [1/6, 1/2]$,
we have an inequality
\[
y_a(t) \leq
\frac{176}{19}t
-\frac{23}{19},
\]
and hence
\begin{align*}
\Omega(\overrightarrow{OA})+\Omega(\overrightarrow{AP})
&\leq \frac{1}{200}\left(
	\frac{176}{19}t-\frac{23}{19}
	\right)
+\frac{12t+24}{25} \\
&=\frac{10}{19}t+\frac{145}{152}.
\end{align*}

We conclude that
\[
h^0(F)/r \leq \max\left\{
\frac{8t+8}{9},
\frac{10}{19}t+\frac{145}{152}
\right\}.
\]

(3) Assume $t \in [1/4, 1/2]$.
Again the only difference with the cases (1), (2)
is that the slope of $OQ$ is
smaller than or equal to $-1$
(or $+\infty$ when $t=1/2$)
in the present case.
Hence we may choose $P_1$ to be $Q$,
or the points $A, B$ on the line segment $QP$
with slope $-1/2, -1$, respectively.
The coordinate of $Q, A$
are same as in (2),
and we have
\[
B=(x_b, y_b)=\left(
	-\frac{12(2t^2-8t+1)}{8t-23}r,
	\frac{12(2t^2-8t+1)}{8t-23}r
	\right).
\]
We get
\begin{align*}
&\Omega(\overrightarrow{OQ})+\Omega(\overrightarrow{QP})
=y_q+x_q
+\frac{y_p-y_q}{5}+\frac{x_p-x_q}{25}
=4tr, \\
&\Omega(\overrightarrow{OB})+\Omega(\overrightarrow{BP})
=\frac{1}{4}y_b
+\frac{y_p-y_b}{5}+\frac{x_p-x_b}{25}
=\frac{9}{100}y_b+\frac{12t+24}{25}r.
\end{align*}

As a function of $t \in [1/4, 1/2]$,
we have the inequality
\[
y_b(t) \leq
\left(\frac{82}{19}t-\frac{11}{19} \right)r
\]
and hence
\[
\Omega(\overrightarrow{OB})+\Omega(\overrightarrow{BP})
\leq \frac{33}{38}tr+\frac{69}{76}r.
\]
We can directly compute that
\[
\Omega(\overrightarrow{OA})+\Omega(\overrightarrow{AP})
=\frac{1}{200}y_a+\frac{12t+24}{25}
\leq \frac{33}{38}tr+\frac{69}{76}r.
\]
We conclude that
\[
h^0(F)/r \leq \max\left\{
4t, \frac{33}{38}t+\frac{69}{76}
\right\}.
\]

(4) Assume $t \in [3/2, 11/6]$.
In this case, we have
\begin{itemize}
\item The slope of $\overrightarrow{OP}$ is
$\frac{1}{t-3} \in (-1, -1/2)$,

\item The slope of $\overrightarrow{OQ}$ is
$\frac{2t}{2t-1} >0$,

\item The slope of $\overrightarrow{QP}$ is
$\frac{2(t-6)}{-5(2t-7)} \in [-1/2, -1/3)$.
\end{itemize}

There are four choices of the point $P_1$,
say $Q, A, B, C$,
where
$A$ is the point on the line $PQ$
with slope $-1$,
$B$ is the point on the line $OQ$
such that the slope of $BP$ is $-1/2$,
and $C$ is the intersection point of
two lines $OA$ and $BP$.
Explicitly, we have
\begin{align*}
&Q=\left((2t-1)r, 2tr \right), \quad
A=\left(
	-\frac{12(2t^2-8t+1)}{8t-23}r,
	\frac{12(2t^2-8t+1)}{8t-23}
	\right), \\
&B=\left(
	\frac{12(2t-1)(t-1)}{6t-1}r,
	\frac{24t(t-1)}{6t-1}r
	\right), \quad
C=\left(
-12(t-1)r, 12(t-1)r
\right).
\end{align*}

Hence we get
\begin{align*}
&\Omega(\overrightarrow{OQ})+\Omega(\overrightarrow{QP})
=\frac{4}{5}y_q+\frac{24}{25}x_q+\frac{12t+24}{25}r
=4tr, \\
&\Omega(\overrightarrow{OA})+\Omega(\overrightarrow{AP})
=\frac{9}{100}y_a+\frac{12t+24}{25}r, \\
&\Omega(\overrightarrow{OB})+\Omega(\overrightarrow{BP})
=y_b+x_b+\frac{y_p-y_b}{8}
=\frac{7}{8}y_b+x_b+\frac{3}{2}r, \\
&\Omega(\overrightarrow{OC})+\Omega(\overrightarrow{CP})
=\frac{y_c}{4}+\frac{y_p-y_c}{8}
=\frac{3}{2}tr.
\end{align*}
Firstly, we can show that
\[
\frac{9}{100}y_a <
\frac{4}{5}y_q+\frac{24}{25}x_q
\]
for $t \in [3/2, 11/6]$.
On the other hand,
It is easy to see
\[
\Omega(\overrightarrow{OC})+\Omega(\overrightarrow{CP}) \leq
\Omega(\overrightarrow{OB})+\Omega(\overrightarrow{BP})
=\frac{90t^2-96t+21}{2(6t-1)}r
\leq \left(\frac{231}{32}t-\frac{375}{64} \right)r.
\]

Hence we can conclude that
\[
h^0(F)/r \leq \max \left\{
4t,
\frac{231}{32}t-\frac{375}{64}
\right\}.
\]

(5) Assume $t \in (11/6, \sqrt{14}/2]$.
Then we have
\begin{itemize}
\item The slope of $\overrightarrow{OP}$ is
$\frac{1}{t-3} \in (-1, -1/2)$,

\item The slope of $\overrightarrow{OQ}$ is
$\frac{2t}{2t-1} >0$,

\item The slope of $\overrightarrow{QP}$ is
$-1/2$.
\end{itemize}

Hence we can choose the point $P_1$ to be
$B$ or $C$ appeared in the case (4) above.
For $t \in [11/6, \sqrt{14}/2]$, we have
\[
\Omega(\overrightarrow{OC})+\Omega(\overrightarrow{CP}) \leq
\Omega(\overrightarrow{OB})+\Omega(\overrightarrow{BP})
=\frac{90t^2-96t+21}{2(6t-1)}r \leq
\left(\frac{233}{32}t-\frac{191}{32} \right)r.
\]

(6) Assume $t \in [\sqrt{14}/2, 23/12]$.
In this case, we have
\begin{itemize}
\item The slope of $\overrightarrow{OP}$ is
$\frac{1}{t-3} \in (-1, -1/2)$,

\item The slope of $\overrightarrow{OQ}$ is
$\frac{8t-7}{9t-11} >0$,

\item The slope of $\overrightarrow{QP}$ is
$-1/2$.
\end{itemize}

We may choose the point $P_1$
as $Q$ or $A$,
where
$A$ is the point on the line $PQ$
with slope $-1$.
Explicitly, we have
\begin{align*}
&Q=\left(
	\frac{12}{25}(9t-11)r,
	\frac{12}{25}(8t-7)r
	\right), \quad
A=\left(
	-12(t-1)r,
	12(t-1)r
	\right),
\end{align*}

and hence
\begin{align*}
&\Omega(\overrightarrow{OQ})+\Omega(\overrightarrow{QP})
=y_q+x_q+\frac{y_p-y_q}{8}
=\frac{192t-168}{25}r, \\
&\Omega(\overrightarrow{OA})+\Omega(\overrightarrow{AP})
=\frac{y_a}{4}+\frac{y_p-y_a}{8}
=\frac{3}{2}tr. \\
\end{align*}
We can see that
$\Omega(\overrightarrow{OQ})+\Omega(\overrightarrow{QP})
\geq \Omega(\overrightarrow{OA})+\Omega(\overrightarrow{AP})$,
hence we conclude that
\[
h^0(F)/r \leq
\frac{192t-168}{25}.
\]

(7) Assume $t \in [23/12, 2)$. Then we have
\begin{itemize}
\item The slope of $\overrightarrow{OP}$ is
$\frac{1}{t-3} \in (-1, -1/2)$,

\item The slope of $\overrightarrow{OQ}$ is
$\frac{1}{3t-5} >0$,

\item The slope of $\overrightarrow{QP}$ is
$-1/2$.
\end{itemize}
Hence $P_{1}=Q$ or $A$,
where $A$ is
the point on the line segment $PQ$ with slope $-1$, i.e.,
\[
Q=((12t-20)r, 4r), \quad
A=((-12t+12)r, (12t-12)r).
\]
We can calculate as
\begin{align*}
&\Omega(\overrightarrow{OQ})+\Omega(\overrightarrow{QP})
=\left(12t-15 \right)r, \\
&\Omega(\overrightarrow{OA})+\Omega(\overrightarrow{AP})
=\frac{3}{2}tr
\end{align*}
hence taking the maximum, we conclude that
\[
h^0(F)/r \leq
12t-15.
\]
\end{proof}

\section{Stronger BG inequality} \label{section:strongBG}
Using the Clifford type bound obtained
in Proposition \ref{prop:clifford},
we prove the following stronger version of
the (classical) BG inequality on a triple cover CY3
$X:=X_6 \subset \mathbb{P}(1, 1, 1, 1, 2)$.
\begin{thm} \label{thm:strongBG}
Let $X$ be a triple cover CY3.
Let $F \in D^b(X)$
be a $\nu_{\alpha, 0}$-semistable object
for some $\alpha > 0$,
with $\mu_H(F) \in [-1, 1]$.
Then we have the following inequality
\begin{equation} \label{eq:bgonT}
\frac{H\ch_2(F)}{H^3\ch_0(F)}
\leq \Xi \left(\left|\mu_H(F) \right|\right),
\end{equation}
where
\[
\Xi(t):=\left\{ \begin{array}{ll}
t^2-t & (t \in [0, 1/4]) \\
3t/4-3/8 & (t \in [1/4, 1/2]) \\
t/4-1/8 & (t \in [1/2, 3/4]) \\
t^2-1/2 & (t \in [3/4, 1]).
\end{array} \right.
\]
\end{thm}
\begin{proof}
Assume for contradiction that
there is a tilt semistable object $F$
violating the inequality (\ref{eq:bgonT}).
We may assume $\mu_H(F) \geq 0$
by replacing $F$ with $F^\vee$ if necessary.
First observe that the following conditions hold:
\begin{itemize}
\item Let $p=(a, b)$ be an arbitrary point
with $a \in [0, 1], b>\Xi(a)$,
and take a real number
$\alpha>0$
(resp. $\alpha'>1/2$).
Then the line segment connecting the points
$p$ and $(0, \alpha)$
(resp. $(1, \alpha')$)
is above
the graph of $\Xi$.

\item Let $L$ be the line through $p_H(F)$ and $p_H(F(-2H)[1])$.
Then $L$ passes through points $(0, \alpha_0), (-1, \alpha_0')$
with $\alpha_0 >0, \alpha_0' > 1/2$.
Putting $(a, b):=p_H(F)$,
the conditions are equivalent to the inequalities
\[
b > a^2-a, \quad
b > a^2-1/2.
\]
\end{itemize}

Under these conditions,
we can apply the arguments in
\cite[Proposition 5.2, Corollary 5.4]{li19b}.
As a result, by restricting to
the surface $T=T_{2, 6} \subset X_{6}$,
we obtain a tilt-stable object $F$ on $T$
with $\mu_H(F) \in (0, 1)$ and
\begin{equation} \label{eq:violate}
\frac{\ch_2(F)}{H^2\ch_0(F)}
> \Xi \left(\frac{H\ch_1(F)}{H^2\ch_0(F)} \right).
\end{equation}

Furthermore, by the first paragraph
in the proof of \cite[Proposition 5.2]{li19b},
we may assume that
\begin{itemize}
\item $\mu_{H_{T}}(F) \in (0, 1/2]$,
\item $F$ is $\mu_{H_{T}}$-stable {\it coherent sheaf},
\item $F|_{C}, F^{\vee}(2H_{T})|_{C}$ are slope stable.
\end{itemize}

Using the Riemann-Roch and the vanishings
\[
\hom(\mathcal{O}_{T}, F(-2H_{T}))=0=\hom(\mathcal{O}_{T}, F^{\vee})
\]
(both follows from slope stability of $F$
and the assumption on its slope),
we have
\begin{equation} \label{eq:rr}
\begin{aligned}
\ch_{2}(F)-H_{T}\ch_{1}(F)+11\ch_{0}(F)
&=\chi(F) \\
&\leq h^0(F|_C)+h^0(F^{\vee}(2H_{T})|_{C}).
\end{aligned}
\end{equation}
Note that we have
\begin{align*}
&\ch(F |_C)=(\ch_0(F), 2H\ch_1(F)), \\
&\ch(F^{\vee}(2H_{T})|_C)=(\ch_0(F), 4H^2\ch_0(F)-2H\ch_1(F)).
\end{align*}
Applying Proposition \ref{prop:clifford}
to the RHS of (\ref{eq:rr}),
we get
\begin{equation} \label{eq:corcliff}
\frac{\ch_{2}(F)}{H^2\ch_{0}(F)}
\leq \begin{cases}
-\frac{23}{25}\mu_H(F)-\frac{13}{75}
& (\mu_H(F) \in (0, 1/12]) \\
-\frac{1}{5}\mu_H(F)-\frac{7}{30}
& (\mu_{H}(F) \in [1/12, 2-\sqrt{14}/2]) \\
-\frac{641}{4800}\mu_H(F)-\frac{1157}{4800}
& (\mu_{H}(F) \in [2-\sqrt{14}/2, 1/6)) \\
-\frac{421}{3648}\mu_H(F)-\frac{595}{2432}
& (\mu_{H}(F) \in [1/6, 89/496]) \\
-\frac{95}{1728}\mu_H(F)-\frac{883}{3456}
& (\mu_{H}(F) \in [89/496, 37/206]) \\
\frac{13}{27}\mu_H(F)-\frac{19}{54}
& (\mu_{H}(F) \in [37/206, 1/4)) \\
\frac{109}{228}\mu_H(F)-\frac{53}{152}
& (\mu_{H}(F) \in [1/4, 69/238]) \\
\mu_H(F)-\frac{1}{2}
& (\mu_{H}(F) \in [69/238, 1/2])
\end{cases}
\end{equation}
In all cases, the inequalities (\ref{eq:corcliff})
contradict to the inequality (\ref{eq:violate}).
\end{proof}

\section{The case of double cover} \label{section:2to1}
In this section,
we consider the double cover $X$ of $\mathbb{P}^3$
branched along a smooth hypersurface of degree $8$;
$X$ is an another example of Calabi-Yau threefolds.
As in the previous sections,
we treat $X$ as a weighted hypersurface
in $P=\mathbb{P}(1, 1, 1, 1, 4)$ of degree $8$.
Let
\[
C_{2, 4, 8} \subset T_{2, 8} \subset X_8, \quad
C_{2, 4, 8} \subset S_{2, 4}
\]
be smooth $(2, 4, 8)$-, $(2, 8)$-, $(2, 4)$-complete intersections
in $P$.
The following is the list of their numerical invariants we need.

\begin{itemize}
\item $-K_P=8H$, $H_P^4=1/4$,
\item $g(C)=49$,
\item $S \cong \mathbb{P}^1 \times \mathbb{P}^1$,
\item $K_T=2H_T$, $H_T^2=4$, $\td_T=(1, -H_T, 10)$.
\item $\td_X=(1, 0, \frac{11}{6}H_X^2, 0)$, $H_X^3=2$.
\end{itemize}

\begin{rmk}
To make the surface $S$ smooth,
we take a $(2, 4)$-complete intersection
instead of a $(2, 2)$-complete intersection.
\end{rmk}

\subsection{Clifford type bound}
In this subsection, we will prove the Clifford type theorem
for the curve $C=C_{2, 4, 8}$,
using the embedding $\iota$
into the quadric surface $S \cong \mathbb{P}^1 \times \mathbb{P}^1$.
\begin{lem} \label{lem:bnslope2}
Let $F$ be a slope stable vector bundle on $C$
with rank $r$ slope $\mu$.
Put $t:=\frac{\mu}{16}$ and assume
$t \in [0, 1/2] \cup [3/2, 2]$.
Then we have the following
statements hold:
\begin{enumerate}
\item When
$t \in \left[0, \frac{5-\sqrt{23}}{2} \right]$,
the sheaf $\iota_*F$ is BN stable.

\item We have
\[
\nu^+_{BN}(\iota_{*}F) \leq
\begin{cases}
1-\frac{1}{2t} & (t \in [\frac{5-\sqrt{23}}{2}, 1/2] \cup [3/2, \frac{\sqrt{23}-1}{2}]) \\
\frac{-13t+16}{-12t+11} & (t \in [\frac{\sqrt{23}-1}{2}, 31/16]) \\
4t-7  & (t \in [31/16, 2)).
\end{cases}
\]

\item We have
\[
\nu^-_{BN}(\iota_{*}F) \geq
\begin{cases}
\frac{-7(2t-9)}{2(t-8)} & (t \in (\frac{5-\sqrt{23}}{2}, 1/2] \cup [3/2, 15/8]) \\
-3 & (t \in [15/8, 2]).
\end{cases}
\]
\end{enumerate}
\end{lem}
\begin{proof}
The proof is almost identical to that of Lemma \ref{lem:bnslope}.
Hence we just give an outline of the proof.
Let us consider the embedding
$\iota \colon C \hookrightarrow S$.
For $F \in \Coh(C)$, we have
$\ch(\iota_*F)=(0, 8rH, r(\mu-64))$.
Let $W$ be a wall for $\iota_*F$
with respect to the $\nu_{\alpha, 0}$-stability,
and let $\beta_1, \beta_2$ be the $\beta$-coordinates
of the end points of the wall  $W$
with $\beta_1 < 0 < \beta_2$.
Then we can show that $\beta_2-\beta_1 \leq 8$.
We have $\nu_{BN}(\iota_*F)=t-4$,
and we can get the bounds of the first possible wall
as follows:
\begin{itemize}
\item When $t \in [0, 1/2]$,
the equation of the first possible wall is
\[
y=(t-4)(x-t)+t-1/2,
\]
which is the line passing through the points
$(t, \Upsilon(t)), (t-8, \Upsilon(t-8))$.
We can see that $y(0) \leq 0$
for $t \in [0, \frac{5-\sqrt{23}}{2}]$,
hence the sheaf $\iota_*F$
is BN stable.

\item When $t \in [3/2, 2]$,
we have two possibilities of the first wall:
\[
y=(t-4)(x-t)+t-1/2, \quad
y=(t-4)(x+6)+18.
\]
The first equation is
the line passing through the points
$(t, \Upsilon(t)), (t-8, \Upsilon(t-8))$,
and the second one is the line
passing through the point $(-6, \Upsilon(-6))$
with slope $t-4$.
\end{itemize}

As similar to Lemma \ref{lem:bnslope},
we get the bound on
$\nu_{BN}^{\pm}(\iota_*F)$
by computing the end points
of the first possible walls listed above.
\end{proof}

We get the following Clifford type bound:
\begin{prop} \label{prop:clifford2}
Let $F$ be a slope stable vector bundle on $C$
of rank $r$, slope $\mu \in (0, 8] \cup [24, 32)$.
Put $t:=\mu/16$.
The following inequalities hold:
\begin{enumerate}
\item When $t \in (0, 1/8)$, we have
$h^0(F)/r \leq \frac{16(t+3)}{49}$.

\item When $t \in [1/8, 1/6)$, we have
$h^0(F)/r \leq \max\left\{
	\frac{12t+24}{25},
	 \frac{85t}{246}+\frac{481}{492}
	\right\}$.

\item When $t \in [1/6, 1/4)$, we have
$h^0(F)/r \leq \max\left\{
	\frac{8t+8}{9},
	 \frac{17t}{38}+\frac{147}{152}
	\right\}$.

\item When $t \in [1/4, 1/2]$, we have
$h^0(F)/r \leq \max\left\{
	4t,
	 \frac{63t}{82}+\frac{153}{164}
	\right\}$.

\item When $t \in [3/2, 15/8]$, we have
$h^0(F)/r \leq 4t$.

\item When $t \in (15/8, \frac{\sqrt{23}-1}{2}]$,
we have
$h^0(F)/r \leq
\frac{133t-114}{18}$.

\item When $t \in [\frac{\sqrt{23}-1}{2}, 31/16]$,
we have
$h^0(F)/r \leq
\frac{236}{49}t-\frac{148}{21}$.

\item When $t \in [31/16, 2]$,
we have
$h^0(F)/r \leq 16t-23$.
\end{enumerate}
\end{prop}
\begin{proof}
As in the proof of Proposition \ref{prop:clifford},
the problem is reduced to computing
$\Omega(\overrightarrow{OP_1})+\Omega(\overrightarrow{P_1P})$
for appropriate candidate points $P_1$
in the triangle $OPQ$.
Here the points $P, Q$ are defined as before, namely,
$P:=(\ch_2(\iota_*F),H\ch_1(\iota_*F))=(16(t-4)r, 16r)$,
and $Q=(x_q, y_q)$ is the point
such that $x_q/y_q$ is the upper bound for
$\nu^+_{BN}(\iota_{*}F)$,
and $(x_p-x_q)/(y_p-y_q)$ is the lower bound for
$\nu^-_{BN}(\iota_{*}F)$, given in Lemma \ref{lem:bnslope2}.

(0) First assume that
$t \in \left[0, \frac{5-\sqrt{23}}{2} \right]$.
In this case, the sheaf $\iota_*F$ is BN stable
with BN slope
$\nu_{BN}(\iota_*F)=t-4 \in (-4, -3)$.
Hence we have
\[
h^0(F)/r \leq
\frac{16}{7}+\frac{16(t-4)}{49}
=\frac{16(t+3)}{49}
\]
by Lemma \ref{lem:sectionbn}.

(1) Assume $t \in \left[\frac{5-\sqrt{23}}{2}, \frac{1}{8} \right)$.
By Lemma \ref{lem:bnslope2}, we have
\begin{itemize}
\item The slope of $\overrightarrow{OP}$
is $\frac{1}{t-4} \in (-1/3, -1/4)$,

\item The slope of $\overrightarrow{OQ}$
is $\frac{2t}{2t-1} \in (-1/3, -1/4)$,

\item The slope of $\overrightarrow{PQ}$
is $\frac{2(t-8)}{-7(2t-9)} \in (-1/3, -1/4)$.
\end{itemize}
Hence we can assume $P_1=Q$, and get
\[
h^0(F)/r \leq
\left(
	\Omega(\overrightarrow{OQ})+\Omega(\overrightarrow{QP})
	\right)/r
=\frac{16}{7}+\frac{16(t-4)}{49}
=\frac{16(t+3)}{49}.
\]

(2) Assume $t \in [1/8, 1/6)$.
Then the slope bounds on $\nu_{BN}(\iota_*F)$
are same as the case (1),
but the slope of $\overrightarrow{OQ}$
is in the interval $(-1/2, -1/3]$.
Hence we may take $P_1$ to be
$Q$ or $A$, where
$A$ is the point on the line $PQ$ with
slope $-1/3$.
The coordinates are given as
\[
Q=((2t-1)r, 2tr), \quad
A=\left(
	\frac{48(2t^2-10t+1)}{-8t+15}r,
	\frac{16(2t^2-10t+1)}{8t-15}r
	\right).
\]
Hence we have
\begin{align*}
&\Omega(\overrightarrow{OQ})+\Omega(\overrightarrow{QP})
=\frac{y_q}{5}+\frac{x_q}{25}
	+\frac{y_p-y_q}{7}+\frac{x_p-x_q}{49}
=\frac{12t+24}{25}r, \\
&\Omega(\overrightarrow{OA})+\Omega(\overrightarrow{AP})
=\frac{y_a}{12}
	+\frac{y_p-y_a}{7}+\frac{x_p-x_a}{49}
=\frac{1}{588}y_a+\frac{16(t+3)}{49}r.
\end{align*}
As a function on $t \in [1/8, 1/6)$, we have
\[
y_a(t) \leq \frac{458t-47}{41}r,
\]
and hence
\[
\Omega(\overrightarrow{OA})+\Omega(\overrightarrow{AP})
\leq \frac{85t}{246}r+\frac{481}{492}r.
\]
We conclude that
\[
h^0(F)/r \leq \max\left\{
	\frac{12t+24}{25},
	 \frac{85t}{246}+\frac{481}{492}
	\right\}.
\]

(3) Assume that $t \in [1/6. 1/4)$.
Then the slopes of
$\overrightarrow{OP}, \overrightarrow{OQ}, \overrightarrow{QP}$
are same as in the case (1),
but the slope of $\overrightarrow{OQ} \in (-1, -1/2]$ instead.
There are four possibilities of the point $P_1$,
say $Q, A, B$.
Here, $A, B$ are the points
on the line $PQ$ with slope
$-1/3, -1/2$, respectively.
The coordinates of $Q, A$
are same as in the case (2), and
\[
B=\left(
	\frac{32(2t^2-10t+1)}{-10t+31}r,
	\frac{16(2t^2-10t+1)}{10t-31}r
	\right).
\]
We therefore get
\begin{align*}
&\Omega(\overrightarrow{OQ})+\Omega(\overrightarrow{QP})
=\frac{y_q}{3}+\frac{x_q}{9}
	+\frac{y_p-y_q}{7}+\frac{x_p-x_q}{49}
=\frac{8(t+1)}{9}r, \\
&\Omega(\overrightarrow{OA})+\Omega(\overrightarrow{AP})
=\frac{y_a}{12}
	+\frac{y_p-y_a}{7}+\frac{x_p-x_a}{49}
=\frac{1}{588}y_a+\frac{16(t+3)}{49}r, \\
&\Omega(\overrightarrow{OB})+\Omega(\overrightarrow{BP})
=\frac{y_b}{8}
	+\frac{y_p-y_b}{7}+\frac{x_p-x_b}{49}
=\frac{9}{392}y_b+\frac{16(t+3)}{49}r.
\end{align*}
We can see that
$\Omega(\overrightarrow{OA})+\Omega(\overrightarrow{AP}) \leq
\Omega(\overrightarrow{OB})+\Omega(\overrightarrow{BP})$.
FUrthermore, as a function on $t \in [1/6, 1/4)$, we have
$y_b(t) \leq
\frac{100}{19}tr-\frac{31}{57}r$,
and hence we have
\[
\Omega(\overrightarrow{OB})+\Omega(\overrightarrow{BP})
\leq \frac{17}{38}tr+\frac{147}{152}r.
\]

(4) Assume that $t \in [1/4, 1/2]$.
In this case, the slope of $\overrightarrow{OQ}$
is bigger than or equal to $-1$.
Hence we may take $P_1=Q, A, B$, or $C$,
where $A, B$ are defined as in (3), and
$C$ is the point on the line $PQ$
with slope $-1$.
We have
\[
C=\left(
	-\frac{16(2t^2-10t+1)}{12t-47}r,
	\frac{16(2t^2-10t+1)}{12t-47}r
	\right).
\]
We have
\begin{align*}
&\Omega(\overrightarrow{OQ})+\Omega(\overrightarrow{QP})
=y_q+x_q
	+\frac{y_p-y_q}{7}+\frac{x_p-x_q}{49}
=4tr, \\
&\Omega(\overrightarrow{OC})+\Omega(\overrightarrow{CP})
=\frac{y_c}{4}
	+\frac{y_p-y_c}{7}+\frac{x_p-x_c}{49}
=\frac{25}{196}y_c+\frac{16(t+3)}{49}r,
\end{align*}
and $\Omega(\overrightarrow{OA})+\Omega(\overrightarrow{AP})$,
$\Omega(\overrightarrow{OB})+\Omega(\overrightarrow{CB})$
are as in (3).
Hence we can show that
\[
\Omega(\overrightarrow{OA})+\Omega(\overrightarrow{AP}),
\Omega(\overrightarrow{OB})+\Omega(\overrightarrow{BP}) \leq
\Omega(\overrightarrow{OC})+\Omega(\overrightarrow{CP}).
\]

On the other hand, as a function on $t \in [1/4, 1/2]$, we have
$y_c(t) \leq \frac{142}{41}tr-\frac{15}{41}r$, and so
\[
\Omega(\overrightarrow{OC})+\Omega(\overrightarrow{CP})
\leq \frac{63}{82}tr+\frac{153}{164}r.
\]

(5) Assume that $t \in [3/2, 15/8]$.
In this case, we have
\begin{itemize}
\item The slope of $\overrightarrow{OP}$
is $\frac{1}{t-4} \in (-1/2, -1/3)$,

\item The slope of $\overrightarrow{OQ}$
is $\frac{2t}{2t-1}>0$,

\item The slope of $\overrightarrow{PQ}$
is $\frac{2(t-8)}{-7(2t-9)} \in [-1/3, -1/4)$.
\end{itemize}
Hence we may choose $P_1$ as
$Q, B, C, D, E, F$, where
$B, C$ are defined as in (4),
$D$ is the point on the line $OQ$
such that the slope of $DP$ is $-1/3$,
and $E$ (resp. $F$) are the intersection points
of the lines $OD$ and $OB$ (resp. $C$).
Hence the computations of
$\Omega(\overrightarrow{OB})+\Omega(\overrightarrow{BP}),
\Omega(\overrightarrow{OC})+\Omega(\overrightarrow{CP})$,
and $\Omega(\overrightarrow{OQ})+\Omega(\overrightarrow{QP})$
are exactly same as in (4), and we can see that
\[
\Omega(\overrightarrow{OB})+\Omega(\overrightarrow{BP}),
\Omega(\overrightarrow{OC})+\Omega(\overrightarrow{CP}) \leq
\Omega(\overrightarrow{OQ})+\Omega(\overrightarrow{QP}).
\]

For $D, E$ and $F$, the coordinates are given as
\begin{align*}
&D=\left(
	\frac{16(2t-1)(t-1)}{8t-1}r,
	\frac{32t(t-1)}{8t-1}r
	\right), \quad
E=(-(32t-32)r, (32t-32)r), \\
&F=(-(8t-8)r, (8t-8)r).
\end{align*}
We get
\begin{align*}
\Omega(\overrightarrow{OD})+\Omega(\overrightarrow{DP})
&=y_d+x_d+\frac{y_p-y_d}{12}, \\
\Omega(\overrightarrow{OE})+\Omega(\overrightarrow{EP})
&=\frac{y_e}{8}+\frac{y_p-y_e}{12}=\frac{4}{3}tr \\
&=\frac{y_f}{4}+\frac{y_p-y_f}{12}
=\Omega(\overrightarrow{OF})+\Omega(\overrightarrow{FP}).
\end{align*}
We can also see that
$\Omega(\overrightarrow{OD})+\Omega(\overrightarrow{DP})
\leq \Omega(\overrightarrow{OQ})+\Omega(\overrightarrow{QP})=4tr$,
and hence we conclude that
\[
h^0(F)/r \leq 4t.
\]

(6) Assume $t \in \left(15/8, \frac{\sqrt{23}-1}{2} \right]$.
The only difference with (5) is that
the slope of $\overrightarrow{QP}$
is equal to $-1/3$ in the present case.
Hence we may choose $P_1$  to be
$D, E$, or $F$ appeared in (5).
It is easy to see that
\begin{align*}
&\Omega(\overrightarrow{OE})+\Omega(\overrightarrow{EP})=
\Omega(\overrightarrow{OF})+\Omega(\overrightarrow{FP})
\leq \Omega(\overrightarrow{OD})+\Omega(\overrightarrow{DP}), \\
&\Omega(\overrightarrow{OD})+\Omega(\overrightarrow{DP})
=\frac{4(46t^2-50t+11)}{3(8t-1)}r
\leq \frac{133t-114}{18}r.
\end{align*}

(7) Assume that
$t \in \left[\frac{\sqrt{23}-1}{2}, \frac{31}{16} \right]$.
Then we have
\begin{itemize}
\item The slope of $\overrightarrow{OP}$
is $\frac{1}{t-4} \in (-1/2, -1/3)$,

\item The slope of $\overrightarrow{OQ}$
is $\frac{12t-11t}{13t-16} >0$,

\item The slope of $\overrightarrow{PQ}$
is $-1/3$.
\end{itemize}
Hence we may choose $P_1$ to be
$Q, E$, or $F$, where
the points $E, F$ are defined as in (5).
We have
\[
Q=\left(
	\frac{16(13t-16)}{49}r,
	\frac{16(12t-11)}{49}r
	\right)
\]
and hence
\[
\Omega(\overrightarrow{OQ})+\Omega(\overrightarrow{QP})
=y_q+x_q
	+\frac{y_p-y_q}{12}
=\frac{236}{49}tr-\frac{148}{21}r.
\]
On the other hand, from the computations in (5),
we see that
\[
\Omega(\overrightarrow{OE})+\Omega(\overrightarrow{EP})
=\Omega(\overrightarrow{OF})+\Omega(\overrightarrow{FP})
=\frac{4}{3}tr
\leq \Omega(\overrightarrow{OQ})+\Omega(\overrightarrow{QP}).
\]
We conclude that
\[
h^0(F)/r \leq
\frac{236}{49}tr-\frac{148}{21}.
\]

(8) Assume $t \in [31/16, 2]$.
Then we have
\begin{itemize}
\item The slope of $\overrightarrow{OP}$
is $\frac{1}{t-4} \in (-1/2, -1/3)$,

\item The slope of $\overrightarrow{OQ}$
is $\frac{1}{4t-7} >0$,

\item The slope of $\overrightarrow{PQ}$
is $-1/3$.
\end{itemize}
Hence we may choose $P_1$ to be
$Q, E$, or $F$, where
the points $E, F$ are defined as in (5).
We have
\[
Q=\left(
	4(4t-7)r,
	4r
	\right)
\]
and hence
\[
\Omega(\overrightarrow{OQ})+\Omega(\overrightarrow{QP})
=y_q+x_q
	+\frac{y_p-y_q}{12}
=(16t-23)r.
\]
As in (7), we see that
\[
\Omega(\overrightarrow{OE})+\Omega(\overrightarrow{EP})
=\Omega(\overrightarrow{OF})+\Omega(\overrightarrow{FP})
=\frac{4}{3}tr
\leq \Omega(\overrightarrow{OQ})+\Omega(\overrightarrow{QP}),
\]
and we can conclude that
\[
h^0(F)/r \leq
16t-23.
\]

\end{proof}

\subsection{Strong (classical) BG inequality}
Using Proposition \ref{prop:clifford2},
we get the following (classical) BG type inequality on
a double cover CY3 $X$.
\begin{thm} \label{thm:BGonT2}
Let $X$ be a double cover CY3.
Let $F \in D^b(X)$ be a $\nu_{\alpha, 0}$-semistable object
for some $\alpha >0$, with $\mu_H(E) \in (-1, 1)$.
Then we have the inequality
\[
\frac{H\ch_2(F)}{H^3\ch_0(F)}
\leq \Xi \left(\left|\frac{H^2\ch_1(F)}{H^3\ch_0(F)} \right|\right).
\]
Here the function $\Xi$ is defined
as in Theorem \ref{thm:strongBG}, i.e.,
\[
\Xi(t)=\left\{ \begin{array}{ll}
t^2-t & (t \in [0, 1/4]) \\
3t/4-3/8 & (t \in [1/4, 1/2]) \\
t/4-1/8 & (t \in [1/2, 3/4]) \\
t^2-1/2 & (t \in [3/4, 1]).
\end{array} \right.
\]
\end{thm}

\begin{proof}
As in the proof of Theorem \ref{thm:strongBG},
the problem is reduced to proving the same statement
for tilt-semistable objects on $T$.
Assume that there exists a tilt semistable object $F$ on $T$
violating the inequality in the statement.
As before, we may assume that
$\mu(F) \in (0, 1/2]$ and $F|_C$ is slope semistable.
Then we have (cf.  equation (\ref{eq:rr}))
\begin{equation} \label{eq:rrT2}
\begin{aligned}
\ch_{2}(F)-H_{T}\ch_{1}(F)+10\ch_{0}(F)
&=\chi(F) \\
&\leq h^0(F|_C)+h^0(F^{\vee}(2H_{T})|_{C}),
\end{aligned}
\end{equation}
and
\begin{align*}
&\ch(F|_C)=(\ch_0(F), 4H\ch_1(F)) ,\\
&\ch(F^\vee(2H)|_C)=(\ch_0(F), 4(2H^2\ch_0(F)-H\ch_1(F))).
\end{align*}
By applying Proposition \ref{prop:clifford2}
to the right hand side of the inequality (\ref{eq:rrT2}),
we have

\[
\frac{\ch_{2}(F)}{H^2\ch_{0}(F)}
\leq
\begin{cases}
-\frac{143}{49}\mu_H(F)-\frac{23}{98}
& (\mu_H(F) \in (0, 1/16]) \\
-\frac{6}{49}\mu_H(F)-\frac{1081}{588}
& (\mu_{H}(F) \in [1/16, \frac{5-\sqrt{23}}{2}]) \\
-\frac{2701}{3528}\mu_H(F)-\frac{127}{882}
& (\mu_{H}(F) \in [ \frac{5-\sqrt{23}}{2}, 1/8)) \\
\frac{85}{984}\mu_H(F)-\frac{503}{1968}
& (\mu_{H}(F) \in [1/8, 217/1654]) \\
\frac{3}{25}\mu_H(F)-\frac{13}{50}
& (\mu_{H}(F) \in [217/1654, 1/6)) \\
\frac{17}{152}\mu_H(F)-\frac{157}{608}
& (\mu_{H}(F) \in [1/6, 107/604] \\
\frac{2}{9}\mu_H(F)-\frac{5}{18}
& (\mu_{H}(F) \in [107/604, 1/4)) \\
\frac{63}{328}\mu_H(F)-\frac{175}{656}
& (\mu_H(F) \in [1/4, 153/530]) \\
\mu_H(F)-\frac{1}{2}
& (\mu_H(F) \in [153/530, 1/2]),
\end{cases}
\]
which is a cotradiction.
\end{proof}

\begin{comment}
We are now able to prove the following theorem
for a double cover CY3.

\begin{thm} \label{thm:BGconj2}
Let $E$ be a BN stable object on $X$
with $\nu_{BN}(E) \in [0, 1/2]$.
Then we have
\begin{align*}
Q^{\Gamma}(E)&:=2(H\ch_{2}(E))(2H\ch_2(E)-3\Gamma H\ch_0(E)) \\
&\quad -6(H^2\ch_{1}(E))(\ch_{3}(E)-\Gamma\ch_1(E)) \geq 0.
\end{align*}
Here, we put $\Gamma:=1/3H^2$.
\end{thm}
\begin{proof}
We argue as in Theorem \ref{thm:bgconj}.
Let us consider the universal extension
\[
E \to \widetilde{E} \to \Hom(\mathcal{O}_X, E) \otimes \mathcal{O}_X[1].
\]
First assume that $\nu_{BN}(E)>0$.
Then $\widetilde{E}$ is BN semistable with
$\nu_{BN}(\widetilde{E})=\nu_{BN}(E)$.
By Theorem \ref{thm:BGonT2},
we have $\mu_H(\widetilde{E}) \notin (-1/4, 0)$.
If  $\mu_H(\widetilde{E}) \in [-1/2, -1/4]$,
Theorem \ref{thm:BGonT2} implies
\begin{equation} \label{eq:ub2}
\hom(\mathcal{O}_X, E) \leq
\frac{4}{3}H\ch_2(E)+H^2\ch_1(E)+\frac{1}{2}H^3\ch_0(E).
\end{equation}
If $\mu_H(\widetilde{E}) <-1/2$
or $\mu_H(\widetilde{E}) > 0$,
then the same inequality (\ref{eq:ub2}) clearly holds.

On the other hand, by the BN stability, we have
\begin{equation} \label{eq:lb2}
\hom(\mathcal{O}_X, E)
\geq \chi(\mathcal{O}_X, E)
=\ch_3(E)+\frac{11}{6}H^2\ch_1(E).
\end{equation}
By the inequalities (\ref{eq:ub2}) and (\ref{eq:lb2}),
we have
\begin{equation} \label{eq:qgamma}
\begin{aligned}
Q^\Gamma(E)
&\geq 4(H\ch_2(E))^2+6H^2\ch_1(E) \left(
-\frac{4}{3}H\ch_2(E)+\frac{5}{6}H^2\ch_1(E)-\ch_0(E)
\right) \\
&\quad +2\left(H^2\ch_1(E) \right)^2
-2H^3\ch_0(E)H\ch_2(E) \\
&=4b^2-8ab+7a^2-3ar-2br.
\end{aligned}
\end{equation}
Here, we put
$(r, a, b):=(H^3\ch_0(E), H^2\ch_1(E), H\ch_2(E))$.

By Theorem \ref{thm:BGonT2},
we know that $\mu_H(E) \notin [0, 1/2]$
as we assume $\nu_{BN}(E) >0$.
First assume $\mu_H(E) \geq 1$
or $\mu_H(E) <0$.
Then we have $a \geq r$.
We also have $a \geq 2b$
since $\nu_{BN}(E) \leq 1/2$.
Applying to the right hand side of the inequality (\ref{eq:qgamma}),
we have
\begin{align*}
Q^\Gamma(E)
&\geq 4b^2-10ab+4a^2 \\
&=2(2a-b)(a-2b) \geq0.
\end{align*}

Next assume $\mu_H(E) \in [1/2, 3/4]$.
Then by Theorem \ref{thm:BGonT2}, we have
\[
-r \geq -2a+8b.
\]
Hence
\begin{align*}
Q^\Gamma(E)
&\geq 4b^2-8ab+7a^2+3a(-2a+8b)+2b(-2a+8b) \\
&=20b^2+12ab+a^2 \geq 0.
\end{align*}

When $\mu_H(E) \in [3/4, 1)$, then
we have
$-r \geq -\frac{7}{5}a+\frac{4}{5}b$
by Theorem \ref{thm:BGonT2}.
Applying it to the inequality (\ref{eq:qgamma}), we get
\[
Q^\Gamma(E) \geq
\frac{14}{5}(a-2b)(a-b) \geq 0.
\]

Finally, consider the case
$\nu_{BN}(E)=0$.
As in the proof of Theorem \ref{thm:bgconj},
we can show that
$\mu_H(\widetilde{E}) \notin (-1/2, 0)$,
and hence $\hom(\mathcal{O}_X, E) \leq H^2\ch_1(E)+\ch_0(E)$.
Similarly, the inequality
$\hom(\mathcal{O}_X, E[2]) \leq H^2\ch_1(E)-\ch_0(E)$ holds.
Hence by the Riemann-Roch, we conclude that
\[
\ch_3(E)+\frac{11}{6}H^2\ch_1(E) \leq 2H^2\ch_1(E),
\]
so we have
\[
\ch_3(E) \leq \frac{1}{6}H^2\ch_1(E)
\leq \Gamma\ch_1(E)
\]
as desired.
\end{proof}
\end{comment}

\section{BG type inequality conjecture} \label{section:BGconj}
In this section, we will prove that
the strong BG inequality in Theorem \ref{thm:mainintro}
implies Theorem \ref{thm:main2intro}.
We work in the following general set up.
Let $X$ be a smooth projective Calabi-Yau threefold,
$H$ a nef and big divisor on $X$.
Let us put $d:=H^3$, $e:=H.\td_{X, 2}$.
We define the positive real number
$\delta_X=\delta_X(H)$ as follows:
\[
\delta_X:=\max\left\{
	\frac{4}{d},
	\frac{e}{d},
	\frac{26}{3d}-\frac{e}{d}-\frac{1}{3},
	\frac{57-7e}{13d},
	\frac{16-3e}{d}
	\right\}.
\]

It is easy to compute the number $\delta_X$ in the following cases:
\begin{itemize}
\item When $X$ is a triple cover CY3,
we have $\delta_X=25/18$.

\item When $X$ is a double cover CY3,
we have $\delta_X=13/6$.
\end{itemize}

We put the following assumption:
\begin{assump} \label{assump*}
Every object $E \in D^b(X)$, which is $\nu_{0, \alpha}$-semistable
for some $\alpha>0$, satisfies the strong BG inequality
(\ref{eq:strongBGintro}).
\end{assump}

\begin{prop} \label{prop:ch2-ch3}
Let $X$ be a smooth projective Calabi-Yau threefold,
H a nef and big divisor on $X$.
For a real number $\delta \geq \delta_X$,
define a $1$-cycle $\Gamma$ as
$\Gamma:=\delta H^2-\td_{X, 2}$.
Assume that Assumption \ref{assump*} holds,
that is, assume that
for every positive real number $\alpha >0$,
every $\nu_{0, \alpha}$-semistable object
satisfies the strong BG inequality (\ref{eq:strongBGintro}).
Then every BN stable object $E \in D^b(X)$
with $\nu_{BN}(E) \in [0, 1/2]$ satisfies the inequality
\begin{align*}
Q^\Gamma(E):=Q^\Gamma_{0, 0}(E)&=
	2(H\ch_{2}(E))(2H\ch_2(E)-3\Gamma H\ch_0(E)) \\
	&\quad -6(H^2\ch_{1}(E))(\ch_{3}(E)-\Gamma\ch_1(E)) \geq 0.
\end{align*}
\end{prop}
\begin{proof}
First assume that $\nu_{BN}(E) \in (0, 1/2]$.
Let us consider the universal extension
\[
E \to \widetilde{E} \to \Hom(\mathcal{O}_{X}, E) \otimes \mathcal{O}_X[1].
\]
By \cite[Lemma 2.12]{li19b},
$\widetilde{E}$ is BN semistable with
$\nu_{BN}(\widetilde{E})=\nu_{BN}(E)$.
By Assumption \ref{assump*},
we can see that
$\mu_H(\widetilde{E}) \notin (-1/4, 0]$.
Indeed, if otherwise, we have
\[
\frac{H\ch_2(\widetilde{E})}{H^3\ch_0(\widetilde{E})}
\leq \mu_H(\widetilde{E})^2
	+\mu_H(\widetilde{E})
< \frac{1}{2}\mu_H(\widetilde{E}).
\]
Dividing both sides by
$\mu_H(\widetilde{E}) (<0)$,
we get $\nu_{BN}(\widetilde{E}) > 1/2$,
a contradiction.
When $\mu_H(\widetilde{E}) \in [-1/2, -1/4]$,
using Assumption \ref{assump*}
to the object $\widetilde{E}$,
we have
\begin{equation} \label{eq:tildeBG}
\frac{H\ch_2(\widetilde{E})}{H^3\ch_0(\widetilde{E})}
\leq -\frac{3}{4}\mu_H(\widetilde{E})
	-\frac{3}{8}.
\end{equation}
Note that we have
$\ch_0(\widetilde{E})=\ch_0(E)-\hom(\mathcal{O}_X, E)$ and
$\ch_i(\widetilde{E})=\ch_i(E)$ for $i=1, 2$.
Note also that we have
$H^2\ch_1(E) \geq 0$ since $E \in \Coh^0(X)$.
Together with the assumption $\mu_H(\widetilde{E}) <0$,
we have $\ch_0(\widetilde{E}) <0$.
From these observations,
the inequality (\ref{eq:tildeBG})
is equivalent to the inequality
\begin{equation} \label{eq:lowerb}
\hom(\mathcal{O}_X, E) \leq
\frac{8}{3d}H\ch_2(E)+\frac{2}{d}H^2\ch_1(E)+\ch_0(E).
\end{equation}

On the other hand, the inequality
$\mu_H(\widetilde{E}) < -1/2$
is equivalent to
\[
\hom(\mathcal{O}_X, E) \leq
\frac{2}{d}H^2\ch_1(E)+\ch_0(E),
\]
which is stronger than (\ref{eq:lowerb})
since we have $H\ch_2(E) >0$
by our assumption $\nu_{BN}(E) >0$.
If $\mu_{H}(\widetilde{E})>0$,
the same inequality (\ref{eq:lowerb})
obviously holds
since $\ch_{0}(E)-\hom(\mathcal{O}_{X}, E)=\ch_{0}(\widetilde{E})>0$
and $H^2\ch_{1}(E) >0$.
Hence the inequality (\ref{eq:lowerb}) always holds.

On the other hand, by using the BN stability of
$E$ and $\mathcal{O}_{X}[1]$, we have
\begin{equation} \label{eq:upperb}
\hom(\mathcal{O}_{X}, E) \geq \chi(E)
=\ch_{3}(E)+\td_{X, 2}\ch_1(E).
\end{equation}

Combining the inequalities (\ref{eq:lowerb})
and (\ref{eq:upperb}), we get
\[
\ch_{3}(E)+\td_{X, 2}\ch_1(E) \leq
\frac{1}{d}H^3\ch_{0}(E)+\frac{2}{d}H^2\ch_{1}(E)+\frac{8}{3d}H\ch_2(E),
\]
and hence
\begin{equation} \label{eq:q}
\begin{aligned}
Q^{\Gamma}(E) &\geq
4(H\ch_{2}(E))^2
	-6\cdot\frac{\delta d-e}{d}H\ch_2(E)H^3\ch_0(E)
	+6\delta\left( H^2\ch_1(E) \right)^2 \\
&\quad -6H^2\ch_1(E)\left(
	\frac{1}{d}H^3\ch_0(E)+\frac{2}{d}H^2\ch_1(E)+\frac{8}{3d}H\ch_2(E)
	\right) \\
&=4b^2-\frac{16}{d}ab+6\left(\delta-\frac{2}{d}\right)a^2
	-6\left(\delta-\frac{e}{d}\right)rb-\frac{6}{d}ra.
\end{aligned}
\end{equation}
Here we put
$(r, a, b):=(H^3\ch_0(E), H^2\ch_1(E), \ch_2(E))$,
to simplify the notation.
Note that we have $a \geq 0$,
since $E \in \Coh^0(X)$.
Moreover, we also have $b \geq 0$
from the assumption $\nu_{BN}(E) \geq 0$.
Note also that by definition of $\delta_X$,
we have $\delta-2/d, \delta-e/d \geq 0$.

By Assumption \ref{assump*},
we know that $\mu_H(E) \notin [0, 1/2]$.
When $\mu_H(E) \notin [1/2, 1]$,
we have $r<a$.
Together with the inequality (\ref{eq:q}), we have
\begin{align*}
Q^\Gamma(E) & \geq
4b^2-\frac{16}{d}ab+6\left(\delta-\frac{2}{d}\right)a^2
	-6\left(\delta-\frac{e}{d}\right)ab-\frac{6}{d}a^2 \\
&=A_1a^2-B_1ab+4b^2,
\end{align*}
where we put
$A_1:=6(\delta-3/d),
B_1:=6(\delta-e/d)+16/d > 0$.
We can further compute as
\begin{equation} \label{eq:mucase1}
A_1a^2-B_1ab+4b^2
=(a-2b)\left(A_1a+(2A_1-B_1)b \right)
+(4A_1-2B_1+4)b^2.
\end{equation}
By the assumption $0< \nu_{BN}(E) \leq 1/2$
we have $0 < 2b \leq a$.
Moreover, we also have
$\delta \geq \delta_X \geq
\frac{26}{3d}-\frac{e}{d}-\frac{1}{3}$
by definition.
From these we can conclude that
the right hand side of the equality (\ref{eq:mucase1})
is non-negative,
and hence we have $Q^\Gamma(E) \geq 0$ as required.

When $\mu_H(E) \in [1/2, 3/4]$,
by Assumption \ref{assump*},
we have
\[
-r \geq -2a+8b.
\]
Combining with the inequality (\ref{eq:q}), we have
\begin{align*}
Q^\Gamma(E) & \geq
4b^2-\frac{16}{d}ab+6\left(\delta-\frac{2}{d} \right)a^2
+6\left(\delta-\frac{e}{d} \right)(-2a+8b)b
+\frac{6}{d}(-2a+8b)a \\
&=\left(4+48\left(\delta-\frac{e}{d} \right) \right)b^2
+\left(\frac{32}{d}-12\left(\delta-\frac{e}{d} \right) \right)ab
+6\left(\delta-\frac{e}{d} \right)a^2 \\
&=:C_2b^2-B_2ab+A_2a^2 \\
&=(a-2b)(A_2a+(2A_2-B_2)b)+(4A_2-2B_2+C_2)b^2,
\end{align*}
where the real numbers $A_2, B_2, C_2$
are defined so that the second equality holds.
Since $\delta \geq 4/d, 3/d$,
we can see that
$4A_2-2B_2+C_2 \geq0$.
Moreover, using the inequalities
$0 < 2b \leq a$ and
$\delta \geq -e/d+16/3d$,
we also obtain
$A_2a+(2A_2-B_2)b \geq 0$.
Hence we have $Q^\Gamma(E) \geq 0$.

Next consider the case when
$\mu(E) \in [3/4, 1]$.
By Assumption \ref{assump*},
we have
$b/r \leq 7a/4r-5/4$, equivalently,
\[
-r \geq -\frac{7}{5}a+\frac{4}{5}b.
\]
Together with the inequality (\ref{eq:q}), we have
\begin{align*}
5Q^\Gamma(E) &\geq
20b^2-\frac{80}{d}ab+30\left(\delta-\frac{2}{d}\right)a^2 \\
	&\quad +6\left(\delta-\frac{e}{d}\right)b\left(-7a+4b \right)
	+\frac{6}{d}a\left(-7a+4b \right) \\
&=\left(20+24\left(\delta-\frac{e}{d} \right) \right)b^2
-\left(42\left(\delta-\frac{e}{d} \right)-\frac{66}{d} \right)ab
+\left(30\delta-\frac{102}{d} \right)a^2 \\
&=:C_3b^2-B_3ab+A_3a^2 \\
&=(a-2b)(A_3a+(2A_3-B_3)b)+(4A_3-2B_3+C_3)b^2.
\end{align*}
Using the inequalities $a \geq 2b$
and
$\delta \geq -e/d+46/10d-1/3, (57-7e)/13d$,
we can show that $Q^\Gamma(E) \geq 0$.

The remaining case is when $\nu_{BN}(E)=0$.
The issue is that we do not know
whether $\widetilde{E}$ is BN semistable or not.
If it is $\nu_{\alpha, 0}$-semistable for some $\alpha>0$,
as in the case of $\nu_{BN}(E)>0$,
we have the inequality
\begin{equation} \label{eq:upbd2}
\hom(\mathcal{O}_X, E) \leq
\frac{2}{d}H^2\ch_1(E)+\ch_0(E).
\end{equation}

Assume that $\widetilde{E}$ is $\nu_{\alpha, 0}$-unstable
for all $\alpha>0$.
Then by the proof of \cite[Proposition 3.3]{li19b},
for each $0 < \delta \ll 1$, there exists
$\alpha_i>0$ and a filtration of $E$
such that each factor
$E_i$ is $\nu_{\alpha_i, 0}$-semistable
with $\nu_{BN}(E_i)<\delta$.
By Assumption \ref{assump*},
we must have
\[
\mu_H(E_i) \notin \left[
-\frac{3}{8\delta+6}, 0
\right].
\]
Taking a limit $\delta \to +0$,
we get
$\mu(\widetilde{E}) \notin [-1/2, 0]$, hence
the inequality (\ref{eq:upbd2}) holds.
Furthermore, by using the derived dual
(cf. proof of \cite[Proposition 3.3]{li19b}),
we also have
\[
\hom(\mathcal{O}_X, E[2]) \leq \frac{2}{3}H^2\ch_1(E)-\ch_0(E).
\]
Hence we get
\begin{align*}
\ch_3(E)+\td_{X, 2}\ch_1(E)
=\chi(E)
&\leq \hom(\mathcal{O}_X, E)+\hom(\mathcal{O}_X, E[2]) \\
&\leq \frac{4}{d}H^2\ch_1(E),
\end{align*}
from which we deduce
$Q^\Gamma(E) \geq 0$,
as we assume $\delta \geq 4/d$.
\end{proof}

\begin{cor} \label{cor:bgconj}
Let $X$ be a triple (resp. double) CY3.
We put $\gamma:=2/9$ (resp. $1/3$)
and $\Gamma:=\gamma H^2$.
Let $E$ be a BN stable object on $X$
with $\nu_{BN}(E) \in [0, 1/2]$.
Then we have
\begin{align*}
Q^\Gamma(E):=Q^\Gamma_{0, 0}(E)&=
	2(H\ch_{2}(E))(2H\ch_2(E)-3\Gamma H\ch_0(E)) \\
	&\quad -6(H^2\ch_{1}(E))(\ch_{3}(E)-\Gamma\ch_1(E)) \geq 0.
\end{align*}
\end{cor}
\begin{proof}
By Theorems \ref{thm:strongBG}, \ref{thm:BGonT2},
a triple/double CY3 satisfies
Assumption \ref{assump*}.
Furthermore, we can take the $1$-cycle $\Gamma$ to be
$\Gamma:=\delta_XH^2-\td_{X, 2}=\gamma H^2$.
\end{proof}

\section{Construction of Bridgeland stability conditions} \label{section:conststab}
The goal of this section is
to prove Theorem \ref{thm:main3intro}
in the introduction.
First let us recall the definition of
Bridgeland stability condition.

\begin{defin}[\cite{bri07}] \label{defin:stab}
Let $\mathcal{D}$ be a triangulated category.
Fix a lattice $\Lambda$ of finite rank
and a group homomorphism
$\cl \colon K(\mathcal{D}) \to \Lambda$.

A {\it stability condition} on $\mathcal{D}$
(with respect to $(\Lambda, \cl)$) is a pair
$(Z, \mathcal{A})$ consisting of
a group homomorphism
$Z \colon \Lambda \to \mathbb{C}$
and the heart of a bounded t-structure
$\mathcal{A} \subset \mathcal{D}$
satisfying the following axioms.
\begin{enumerate}
\item We have
$Z \circ \cl \left(\mathcal{A} \setminus \{0\} \right)
\subset \mathbb{H} \cup \mathbb{R}_{<0}$,
where $\mathbb{H}$ is the upper half plane.

\item Every non-zero object
in the heart $\mathcal{A}$
has a Harder-Narasimhan filtration
with respect to $\mu_Z$-stability.
Here we define a $Z$-slope function $\mu_Z$
as
\[
\mu_Z:=-\frac{\Re Z}{\Im Z}
\colon \mathcal{A}
\to \mathbb{R} \cup \{+\infty\},
\]
and define $\mu_Z$-stability
on the abelian category $\mathcal{A}$
in a usual way.

\item There exists a quadratic form $q$ on $\Lambda$
satisfying the following conditions.
\begin{itemize}
\item $q$ is negative definite on the kernel of $Z$,

\item For every $\mu_Z$-semistable object $E \in \mathcal{A}$,
we have $q(\cl(E)) \geq 0$.
\end{itemize}
\end{enumerate}

The group homomorphism $Z$ is called a {\it central charge},
and the axiom (3) is called the {\it support property}.
\end{defin}

Let $\Stab_\Lambda(\mathcal{D})$ be a set
of stability conditions on $\mathcal{D}$
with respect to $(\Lambda, \cl)$.
Then the set $\Stab_\Lambda(\mathcal{D})$ has
a structure of a complex manifold \cite{bri07}.
Moreover, there is an action of the group
$\widetilde{\GL^+}(2, \mathbb{R})$
on $\Stab_\Lambda(\mathcal{D})$,
where $\widetilde{\GL^+}(2, \mathbb{R})$
is the universal covering of
the group
\[
\GL^+(2, \mathbb{R}):=\{g \in \GL(2, \mathbb{R}) : \det(g)>0
\}.
\]

Let us consider the case when
$\mathcal{D}=D^b(X)$,
where $X$ is a double/triple cover CY3.
In this case, we fix a lattice $\Lambda$ to be
the image of the morphism
\[
\cl:=\left(H^3\ch_0, H^2\ch_1, H\ch_2, \ch_3 \right)
\colon K(X) \to H^{2*}(X, \mathbb{Q}).
\]
We simply denote as
$\Stab(X):=\Stab_{\Lambda}(D^b(X))$.
Following \cite{bms16, bmt14a},
we explain an explicit construction
of stability conditions on $D^b(X)$.
Let us recall several notions
from \cite{bms16}.
Fix real numbers
$\alpha, \beta \in \mathbb{R}$
with $\alpha>0$.

The heart corresponding to a stability condition
is constructed as a tilt of $\Coh^\beta(X)$.
Let us define a slope function
$\nu'_{\beta, \alpha}$
on $\Coh^\beta(X)$ as
\[
\nu'_{\beta, \alpha}:=
\frac{H\ch_2^\beta-\frac{1}{2}\alpha^2H^3\ch_0^\beta}{H^2\ch_1^\beta}
\colon \Coh^\beta(X) \to \mathbb{R} \cup \{+\infty\}.
\]
Compared with the function
$\nu_{\beta, \alpha}$
defined in Section \ref{section:preliminaries},
we have
\begin{equation} \label{eq:compare}
\nu'_{\beta, \alpha}=\nu_{\beta, \frac{1}{2}(\beta^2+\alpha^2)}-\beta.
\end{equation}

We define full subcategories
$\mathcal{T}'_{\beta, \alpha}, \mathcal{F}'_{\beta, \alpha}$
of $\Coh^\beta(X)$ as
\begin{align*}
&\mathcal{T}'_{\beta, \alpha}:=\left\langle
	T \in \Coh^\beta(X) :
	T \mbox{ is } \nu'_{\beta, \alpha} \mbox{-semistable with }
	\nu'_{\beta, \alpha}(E)>0
	\right\rangle, \\
&\mathcal{F}'_{\beta, \alpha}:=\left\langle
	F \in \Coh^\beta(X) :
	F \mbox{ is } \nu'_{\beta, \alpha} \mbox{-semistable with }
	\nu'_{\beta, \alpha}(E) \leq 0
	\right\rangle.
\end{align*}

Here, $\nu'_{\beta, \alpha}$-stability
is same as
$\nu_{\beta, \frac{1}{2}(\beta^2+\alpha^2)}$-stability,
and $\langle-\rangle$ denotes
the extension closure in the abelian category
$\Coh^\beta(X)$.
We now define the {\it double-tilted heart} as
\[
\mathcal{A}^{\beta, \alpha}:=
\left\langle
	\mathcal{F}'_{\beta, \alpha}[1],
	\mathcal{T}'_{\beta, \alpha}
	\right\rangle
\subset D^b(X).
\]

We also define a central charge function
$Z_{\beta, \alpha}^{a, b} \colon \Lambda \to \mathbb{C}$ as
\[
Z^{a, b}_{\alpha, \beta}:=
-\ch_3^{\beta}+bH\ch_2^\beta+aH^2\ch_1^\beta
+i\left(
	H\ch_2^\beta-\frac{1}{2}\alpha^2H^3\ch_0^\beta
	\right)
\]
for real numbers $a, b \in \mathbb{R}$.

Finally, for a real number $\gamma >0$,
define $U_\gamma$ to be a set of vectors
$(\alpha, \beta, a, b) \in \mathbb{R}^4$ satisfying
\begin{equation} \label{eq:Ugamma}
\alpha> 0, \quad
\alpha^2+\left(\beta-\lfloor\beta \rfloor-\frac{1}{2} \right)^2 > \frac{1}{4},
\quad a > \frac{1}{6}\alpha^2+\frac{1}{2}|b|\alpha+\gamma.
\end{equation}

\begin{thm}[cf. {\cite[Proposition 8.10]{bms16}}] \label{thm:familystab}
Let $X$ be a triple (resp. double) cover CY3,
and put $\gamma:=2/9$ (resp. 1/3).
Then there exists an injective continuous map
\[
U_\gamma \hookrightarrow \Stab(X), \quad
(\alpha, \beta, a, b) \mapsto
\left(Z_{\beta, \alpha}^{a, b}, \mathcal{A}_{\beta, \alpha} \right).
\]

Furthermore, the orbit
$\widetilde{\GL^+}(2, \mathbb{R}) \cdot U_\gamma$
forms an open subset in the space $\Stab(X)$
of stability conditions.
\end{thm}

We divide the proof of the above theorem
into several steps.
The arguments below are
essentially same as that in \cite[Section 8]{bms16}.

\begin{prop}[cf. {\cite[Theorem 8.6]{bms16}}] \label{prop:axiom12}
For every element $(\alpha, \beta, a, b) \in U_\gamma$
with $\alpha, \beta \in \mathbb{Q}$,
the pair
$\left(Z_{\beta, \alpha}^{a, b}, \mathcal{A}_{\beta, \alpha} \right)$
satisfies axioms (1) and (2)
in Definition \ref{defin:stab}.
\end{prop}

\begin{proof}
First we check the axiom (1)
in Definition \ref{defin:stab}
for the pair
$\left(Z_{\beta, \alpha}^{a, b}, \mathcal{A}_{\beta, \alpha} \right)$.
As in the proof of \cite[Theorem 8.6]{bms16},
it is enough to show the inequality
$Z_{\beta, \alpha}^{a, b}(F[1]) <0$
for every $\nu'_{\beta, \alpha}$-semistable object
$F$ with $\nu'_{\beta, \alpha}(F)=0$.
By the inequality
$\alpha^2+\left(\beta-\lfloor\beta \rfloor-\frac{1}{2} \right)^2
> \frac{1}{4}$
in (\ref{eq:Ugamma}),
we can apply Theorem \ref{thm:main2intro} to
the object $F$.
Noting the equation (\ref{eq:compare}),
we get
\begin{equation} \label{eq:ch3ineq}
\ch_3^\beta(F) \leq
\left(\gamma+\frac{1}{6}\alpha^2 \right)H^2\ch_1^\beta(F).
\end{equation}
Furthermore, together with the assumption
$H\ch_2^\beta(F)=\frac{1}{2}\alpha^2H^3\ch_0^\beta$,
the classical BG inequality (cf. \cite[Theorem 3.5]{bms16})
$\overline{\Delta}_H(F) \geq 0$ gives the inequality
\begin{equation} \label{eq:ch2ineq}
\left(H\ch_2^\beta(F) \right)^2
\leq \frac{1}{4}\alpha^2\left(H^2\ch_1^\beta(F) \right)^2.
\end{equation}

By the inequalities
(\ref{eq:ch3ineq}) and (\ref{eq:ch2ineq}),
we obtain
\begin{align*}
Z_{\beta, \alpha}^{a, b}(F[1])
&=\Re Z_{\beta, \alpha}^{a, b}(F[1]) \\
&\leq \left(\gamma+\frac{1}{6}\alpha^2 \right)H^2\ch_1^\beta(F)
	+\frac{1}{2}|b|\alpha H^2\ch_1^\beta(F)
	-aH^2\ch_1^\beta(F)
<0.
\end{align*}

Now the axiom (2) is also satisfied
since we assume $\alpha, \beta \in \mathbb{Q}$
(see the proof of \cite[Theorem 8.6]{bms16} for the detail).
\end{proof}

Next we discuss about the support property.
Let us put
\begin{align*}
\overline{\nabla}^{\alpha, \beta, \gamma}_H(E)
&:=3\gamma\alpha^2\left(H^3\ch_0^\beta(E) \right)^2
+2\left(H\ch^\beta_2(E) \right)\left(
	2H\ch^\beta_2(E)-3\gamma H^3 \ch^\beta_0(E)
	\right) \\
&\quad -6\left(H^2\ch^\beta_1(E) \right) \left(
	\ch^\beta_3(E)-\gamma H^2\ch_1^\beta(E)
	\right).
\end{align*}
for an object $E \in D^b(X)$.

\begin{prop}[cf. {\cite[Lemmas 8.5, 8.8]{bms16}}] \label{prop:support}
Fix an element $(\alpha, \beta, a, b) \in U_\gamma$.
Then there exists an interval
$I^{a, b}_{\alpha, \gamma} \subset \mathbb{R}$ such that
for every $K \in I^{a, b}_{\alpha, \gamma}$,
the quadratic form
$Q^{\alpha, \beta, \gamma}_K:=
K\overline{\Delta}_H
+\overline{\nabla}^{\alpha, \beta, \gamma}_H$
is negative definite on the kernel of
the central charge function
$Z_{\beta, \alpha}^{a, b}$.

Furthermore, if we assume
$(\alpha, \beta) \in \mathbb{Q}$,
then every $Z_{\beta, \alpha}^{a, b}$-semistable object $E$
satisfies the inequality
$Q^{\alpha, \beta, \gamma}_K(E) \geq 0$.
\end{prop}
\begin{proof}
Let us prove the first assertion.
The vectors
$\left(1, 0, \frac{1}{2}\alpha^2, \frac{1}{2}b\alpha \right)$
and $(0, 1, 0, a)$ forms the basis
the kernel of
$Z_{\beta, \alpha}^{a, b}$.
With respect to this basis,
the quadratic form
$K\overline{\Delta}_H
+\overline{\nabla}^{\alpha, \beta, \gamma}_H$
is represented by the matrix
\begin{equation} \label{eq:matrix}
\left(
\begin{array}{cc}
-\alpha^2K+\alpha^4 & -\frac{3}{2}b\alpha^2 \\
-\frac{3}{2}b\alpha^2 & K-6(a-\gamma)
\end{array}
\right).
\end{equation}

When $b=0$, the matrix (\ref{eq:matrix})
is negative definite if and only if
$K \in \left(\alpha^2, 6(a-\gamma) \right)
=:I^{a, b=0}_{\alpha, \gamma}$.
This interval is non-empty by (\ref{eq:Ugamma}).

When $b \neq 0$, we also need to require
the determinant of the matrix (\ref{eq:matrix})
to be positive, i.e.,
\begin{equation} \label{eq:detmatrix}
\alpha^2\left(
	-K^2+\left(6(a-\gamma)+\alpha^2 \right)K
	-6(a-\gamma)\alpha^2
-\frac{9}{4}b^2\alpha^2
\right)>0.
\end{equation}

The solution space $K \in I_{\alpha, \gamma}^{a, b}$
of the inequality (\ref{eq:detmatrix})
forms a non-empty open interval
since we have, by (\ref{eq:Ugamma}),
\[
a-\gamma > \frac{1}{6}\alpha^2+\frac{1}{2}|b|\alpha.
\]

We can prove the second assertion
as in \cite[Lemma 8.8]{bms16},
using the BG type inequality
obtained in Theorem \ref{thm:main2intro}.
\end{proof}

Finally we are able to prove Theorem \ref{thm:familystab}.

\begin{proof}[Proof of Theorem \ref{thm:familystab}]
By Propositions
\ref{prop:axiom12} and \ref{prop:support},
the pair
$\left(Z_{\beta, \alpha}^{a, b},
\mathcal{A}^{\beta, \alpha} \right)$
is a stability condition on $D^b(X)$
for every element
$(\alpha, \beta, a, b) \in U_\gamma$
with $\alpha, \beta \in \mathbb{Q}$.
We can deform them to the real parameters
$(\alpha, \beta)$ by the support property
in Proposition \ref{prop:support}.
See \cite[Proposition 8.10]{bms16}
for the precise proof.
\end{proof}