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

$$\begin{align*}\hom(\mathcal{O}(1), F)=0, \quad \operatorname{\mathrm{ext}}^2(\mathcal{O}(1), F)=\hom(F, \mathcal{O}(-1))=0 \end{align*}$$

and hence $0 \geq -\operatorname {\mathrm {ext}}^1\left (\mathcal {O}(1), F \right ) =\chi \left (\mathcal {O}(1), F \right )$ . By computing the right-hand side using the Riemann–Roch theorem, we get the inequality

$$\begin{align*}\frac{\operatorname{\mathrm{ch}}_{2}(F)}{H^2\operatorname{\mathrm{ch}}_{0}(F)} \leq 0. \end{align*}$$

On the other hand, again by the stability of F and the Serre duality, we also have

$$\begin{align*}\hom(F, \mathcal{O})=0, \quad \operatorname{\mathrm{ext}}^2(F, \mathcal{O})=\hom(\mathcal{O}, F(-2))=0, \end{align*}$$

which imply the inequality $0 \geq -\operatorname {\mathrm {ext}}^1(F, \mathcal {O})=\chi (F, \mathcal {O})$ . Hence, we obtain

$$\begin{align*}\frac{\operatorname{\mathrm{ch}}_2(F)}{H^2\operatorname{\mathrm{ch}}_0(F)} \leq \mu_H(F)-\frac{1}{2}. \end{align*}$$

Taking the minimum, the inequality (3) holds.

Proof.

1. (1) The first assertion follows from Lemma 3.3 and Proposition 2.5.

2. (2) For the second assertion, we can apply the arguments in [Reference Arcara and BertramAB13, Reference BridgelandBri08] by replacing the classical BG inequality with the stronger one, equation (3).

Proof. By the Grothendieck–Riemann–Roch theorem, we have

(5) $$ \begin{align} \operatorname{\mathrm{ch}}(\iota_{*}F)=(0, 6rH, r(\mu-36)). \end{align} $$

Suppose that there exists a positive integer $\alpha $ and a destabilizing sequence

$$\begin{align*}0 \to F_{2} \to \iota_{*}F \to F_{1} \to 0 \end{align*}$$

in $\operatorname {\mathrm {Coh}}^0(S)$ 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 $\operatorname {\mathrm {Coh}}^0(S)$ :

By taking the $\operatorname {\mathrm {Coh}}(S)$ -cohomology of the bottom row in the above diagram, we get the exact sequence

$$\begin{align*}0 \to Q/\mathcal{H}^{-1}(F_1) \to \iota_*F/T \to F_1 \to 0 \end{align*}$$

in $\operatorname {\mathrm {Coh}}(S)$ . 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

$$\begin{align*}6H^2\operatorname{\mathrm{ch}}_0(Q)=H\operatorname{\mathrm{ch}}_1(Q|_C) \geq H\operatorname{\mathrm{ch}}_1(Q/\mathcal{H}^{-1}(F_1)). \end{align*}$$

Note also that we have $\operatorname {\mathrm {ch}}_0(Q)=\operatorname {\mathrm {ch}}_0(\mathcal {H}^{-1}(F_1))$ . Now we have

(6) $$ \begin{align} \begin{aligned} \mu_H(Q)-\mu_H(\mathcal{H}^{-1}(F_1)) =\frac{H\operatorname{\mathrm{ch}}_1(Q/\mathcal{H}^{-1}(F_1))}{H^2\operatorname{\mathrm{ch}}_0(Q)} \leq 6. \end{aligned} \end{align} $$

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., [Reference LiLi19a, Lemma 2.9], [Reference MaciociaMaci14]), we know that for $0 < \epsilon \ll 1$ , we have

$$\begin{align*}Q \in \operatorname{\mathrm{Coh}}^{\beta_{2}-\epsilon}(S), \quad \mathcal{H}^{-1}(F_{1})[1] \in \operatorname{\mathrm{Coh}}^{\beta_{1}+\epsilon}(S), \end{align*}$$

which in particular imply

$$\begin{align*}\mu_{H}(Q)> \beta_{2}-\epsilon, \quad \mu_{H}(\mathcal{H}^{-1}(F_{1})) \leq \beta_{1}+\epsilon. \end{align*}$$

Combining with the inequality (6), we have the desired inequality

$$\begin{align*}\beta_{2}-\beta_{1} \leq 6.\\[-34pt] \end{align*}$$

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 equation (5) 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

$$\begin{align*}L_n:=\left\{ (n, y) : \frac{n^2-1}{2} < y < \frac{n^2}{2} \right\}, \quad n \in \mathbb{Z}. \end{align*}$$

When both of the end points $(\beta _i, \alpha _i)$ are on the curve $\Upsilon $ , we say that the wall W is of Type A; otherwise, we say it is of Type B.

First, assume that W is of Type A. By Lemma 3.6, 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:

• ○ When $t \in [0, 1/2]$ , the following is the first possible wall

  $$\begin{align*}y=(t-3)(x-t)+t-1/2. \end{align*}$$
  Note that, if $t \in [0, 2-\sqrt {14}/2]$ , it is negative at $x=0$ , hence $\iota _*F$ is BN stable.

• ○ When $t \in [3/2, 2]$ , one of the following is the first possible wall

  $$\begin{align*}y=(t-3)(x-t)+t-1/2, \quad y=(t-3)(x+4)+8. \end{align*}$$
  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 4 below.
  

  Figure 4 The first possible wall L when $t=3/2$ (left) and $t=23/12$ (right).

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 $\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 $E_{1}$ :

• ○ Since $\iota _{*}F$ is $\nu _{\alpha , 0}$ -stable for $\alpha $ sufficiently large, we have $\operatorname {\mathrm {ch}}_{0}(E_{1})>0$ .

• ○ $E_{1}$ satisfies the BG type inequality (4).

• ○ The point $p_{H}(E_{1})$ is below the line L.

Among all points satisfying the above three conditions, its slope becomes maximum at the point $(\alpha _{\max }, \beta _{\max })$ , hence we get the bound

$$\begin{align*}\nu_{BN}(E_{1}) \leq \frac{\alpha_{\max}}{\beta_{\max}}. \end{align*}$$

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

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

$$ \begin{align*} \hom(\mathcal{O}_{S}, F)=\chi(F) &=\int_{S}\operatorname{\mathrm{ch}}(F). (1, H, 1) \\ &=\operatorname{\mathrm{ch}}_{0}(F) + H\operatorname{\mathrm{ch}}_{1}(F) +\operatorname{\mathrm{ch}}_{2}(F). \end{align*} $$

Next, consider the case $\nu _{BN}(E) \in (-n-1, -n)$ , $n \in \mathbb {Z}_{>0}$ . Let

$$\begin{align*}(x_{F}, y_{F}):=p_{H}(F) =\left( \frac{H\operatorname{\mathrm{ch}}_{1}(F)}{H^2\operatorname{\mathrm{ch}}_{0}(F)}, \frac{\operatorname{\mathrm{ch}}_{2}(F)}{H^2\operatorname{\mathrm{ch}}_{0}(F)} \right). \end{align*}$$

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 $F, \mathcal {O}_S \in \operatorname {\mathrm {Coh}}^\beta (S)$ are $\nu _{\beta , \alpha }$ -semistable with

$$\begin{align*}\nu_{\beta, \alpha}(F) =\nu_{\beta, \alpha}(\mathcal{O}_{S}) =\alpha/\beta=\nu_{BN}(F). \end{align*}$$

Let us consider the exact triangle

$$\begin{align*}\operatorname{\mathrm{Hom}}(\mathcal{O}_{S}, F) \otimes \mathcal{O}_{S} \xrightarrow{ev} F \to \widetilde{F}:=\operatorname{\mathrm{Cone}}(ev). \end{align*}$$

Since the only Jordan–Hölder factor of $\operatorname {\mathrm {Hom}}(\mathcal {O}_S, F) \otimes \mathcal {O}_S$ with respect to $\nu _{\beta , \alpha }$ -stability is $\mathcal {O}_S$ , the evaluation map $ev$ must be injective in the category $\operatorname {\mathrm {Coh}}^{\beta }(S)$ . Hence, it follows that $\widetilde {F} \in \operatorname {\mathrm {Coh}}^{\beta }(S)$ , 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 $\mathcal {O}_{S}(-2n)[1] \in \operatorname {\mathrm {Coh}}^{\beta }(S)$ . 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

$$ \begin{align*} 0 &\leq \hom(\mathcal{O}_{S}(-2n), \widetilde{F}) \\ &=\chi(\mathcal{O}_{S}(-2n), \widetilde{F}) \\ &=\operatorname{\mathrm{ch}}_{2}(F)+(2n+1)H\operatorname{\mathrm{ch}}_{1}(F) +(2n+1)^2(\operatorname{\mathrm{ch}}_{0}(F)-\hom(\mathcal{O}_{S}, F)), \end{align*} $$

and so

$$\begin{align*}\hom(\mathcal{O}_{S}, F) \leq \operatorname{\mathrm{ch}}_{0}(F)+\frac{H\operatorname{\mathrm{ch}}_{1}(F)}{2n+1} +\frac{\operatorname{\mathrm{ch}}_{2}(F)}{(2n+1)^2} \end{align*}$$

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

$$\begin{align*}\hom(\mathcal{O}_S, F) \leq \operatorname{\mathrm{ch}}_0(F)+\frac{H\operatorname{\mathrm{ch}}_1(F)}{2n}+\frac{\operatorname{\mathrm{ch}}_2(F)}{(2n)^2} =\operatorname{\mathrm{ch}}_0(F)+\frac{1}{4n}H\operatorname{\mathrm{ch}}_{1}(F).\\[-40pt] \end{align*}$$

Proof. The proof is elementary and almost identical with that of [Reference LiLi19a, Lemma 4.11]. The key observations are the following:

• ○ The function $\Omega $ is linear with respect to both variables x and y as long as the slope $x/y$ is fixed.

• ○ The function $\Omega $ is upper semicontinuous.

We refer to [Reference LiLi19a, Lemma 4.11] for the details.

Proof of Proposition 3.1

Let F be a slope stable vector bundle on C of rank r and slope $\mu $ . Let

$$\begin{align*}0=E_{0} \subset E_{1} \subset \cdots \subset E_{m}=\iota_{*}F \end{align*}$$

be the HN filtration with respect to the BN stability, and define $P_{i}:=(\operatorname {\mathrm {ch}}_{2}(E_{i}), H\operatorname {\mathrm {ch}}_{1}(E_{i}))$ . We have an inequality

(7) $$ \begin{align} h^0(F) \leq \sum_{i=1}^m \Omega(\overrightarrow{P_{i-1}P_{i}}) \end{align} $$

by Lemma 3.8 (cf. [Reference LiLi19a, Equation (21)]). We will bound the RHS in the above inequality. Let us put

$$\begin{align*}P=(x_p, y_p):=(\operatorname{\mathrm{ch}}_2(\iota_{*}F), H\operatorname{\mathrm{ch}}_{1}(\iota_{*}F)) =(r(\mu-36), 12r), \end{align*}$$

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 3.7. We know that the HN polygon of $\iota _{*}F$ with respect to the BN stability is inside the triangle $OPQ$ . Hence, by Lemma 3.9, we may assume $m=2$ and the point $P_1$ satisfies one of the following conditions:

• ○ $P_1=Q$ ,

• ○ $P_1$ is on the line segment $OQ$ (resp. $PQ$ ) such that the slope of $P_1P$ (resp. $OP_1$ ) is $-1/n$ for some $n \in \mathbb {Z}_{>0}$ .

• ○ The lines $OP_1$ and $P_1P$ have slopes $-1/m, -1/n$ for some integers $m, n \in \mathbb {Z}_{>0}$ .

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 3.8, we have

$$\begin{align*}h^0(F)/r \leq \frac{12}{5}+\frac{12(t-3)}{25} =\frac{12t+24}{25}. \end{align*}$$

(1) Assume $t \in (2-\sqrt {14}/2, 1/6)$ . By Lemma 3.7, we have

• ○ The slope of $\overrightarrow {OP}$ is $\frac {1}{t-3} \in (-1/2, -1/3)$ ,

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

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

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 the same as the case (1). 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

$$\begin{align*}Q=((2t-1)r, 2tr), \quad A=\left( \frac{24(2t^2-8t+1)}{-6t+11}r, \frac{12(2t^2-8t+1)}{-6t+11}r \right). \end{align*}$$

First, consider the case of $P_1=Q \neq A$ . 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

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

As a function on $t \in [1/6, 1/4]$ , we have an inequality

$$\begin{align*}y_a(t) \leq \left(\frac{176}{19}t -\frac{23}{19}\right)r \end{align*}$$

since the equality holds for $t=1/4, 1/6$ , and $y^{\prime \prime }_a(t)> 0$ for $t < 11/6$ . Hence, we obtain

$$ \begin{align*} \Omega(\overrightarrow{OA})+\Omega(\overrightarrow{AP}) &\leq \frac{1}{200}\left( \frac{176}{19}t-\frac{23}{19} \right)r +\frac{12t+24}{25}r \\ &=\left(\frac{10}{19}t+\frac{145}{152}\right)r. \end{align*} $$

We conclude that

$$\begin{align*}h^0(F)/r \leq \max\left\{ \frac{8t+8}{9}, \frac{10}{19}t+\frac{145}{152} \right\}. \end{align*}$$

(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 the same as in equation (2), and we have

$$\begin{align*}B=(x_b, y_b)=\left( -\frac{12(2t^2-8t+1)}{8t-23}r, \frac{12(2t^2-8t+1)}{8t-23}r \right). \end{align*}$$

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

$$\begin{align*}y_b(t) \leq \left(\frac{82}{19}t-\frac{11}{19} \right)r \end{align*}$$

and hence

$$\begin{align*}\Omega(\overrightarrow{OB})+\Omega(\overrightarrow{BP}) \leq \frac{33}{38}tr+\frac{69}{76}r. \end{align*}$$

We can directly compute that

$$\begin{align*}\Omega(\overrightarrow{OA})+\Omega(\overrightarrow{AP}) =\frac{1}{200}y_a+\frac{12t+24}{25} \leq \frac{33}{38}tr+\frac{69}{76}r. \end{align*}$$

We conclude that

$$\begin{align*}h^0(F)/r \leq \max\left\{ 4t, \frac{33}{38}t+\frac{69}{76} \right\}. \end{align*}$$

(4) Assume $t \in [3/2, 11/6]$ . In this case, we have

• ○ The slope of $\overrightarrow {OP}$ is $\frac {1}{t-3} \in (-1, -1/2)$ ,

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

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

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

$$ \begin{align*} &Q=\left((2t-1)r, 2tr \right), \quad A=\left( -\frac{12(2t^2-8t+1)}{8t-23}r, \frac{12(2t^2-8t+1)}{8t-23}r \right), \\ &B=\left( \frac{12(2t-1)(t-1)}{6t-1}r, \frac{24t(t-1)}{6t-1}r \right), \quad C=\left( -12(t-1)r, 12(t-1)r \right). \end{align*} $$

Hence, we get

$$ \begin{align*} &\Omega(\overrightarrow{OQ})+\Omega(\overrightarrow{QP}) =\frac{4}{5}y_q+\frac{24}{25}x_q+\frac{12t+24}{25}r =4tr, \\ &\Omega(\overrightarrow{OA})+\Omega(\overrightarrow{AP}) =\frac{9}{100}y_a+\frac{12t+24}{25}r, \\ &\Omega(\overrightarrow{OB})+\Omega(\overrightarrow{BP}) =y_b+x_b+\frac{y_p-y_b}{8} =\frac{7}{8}y_b+x_b+\frac{3}{2}r, \\ &\Omega(\overrightarrow{OC})+\Omega(\overrightarrow{CP}) =\frac{y_c}{4}+\frac{y_p-y_c}{8} =\frac{3}{2}tr. \end{align*} $$

First, we can show that

$$\begin{align*}\frac{9}{100}y_a < \frac{4}{5}y_q+\frac{24}{25}x_q \end{align*}$$

for $t \in [3/2, 11/6]$ . On the other hand, it is easy to see

$$\begin{align*}\Omega(\overrightarrow{OC})+\Omega(\overrightarrow{CP}) \leq \Omega(\overrightarrow{OB})+\Omega(\overrightarrow{BP}) =\frac{90t^2-96t+21}{2(6t-1)}r \leq \left(\frac{231}{32}t-\frac{375}{64} \right)r. \end{align*}$$

Hence, we can conclude that

$$\begin{align*}h^0(F)/r \leq \max \left\{ 4t, \frac{231}{32}t-\frac{375}{64} \right\}. \end{align*}$$

(5) Assume $t \in (11/6, \sqrt {14}/2]$ . Then we have

• ○ The slope of $\overrightarrow {OP}$ is $\frac {1}{t-3} \in (-1, -1/2)$ ,

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

• ○ The slope of $\overrightarrow {QP}$ is $-1/2$ .

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

$$\begin{align*}\Omega(\overrightarrow{OC})+\Omega(\overrightarrow{CP}) \leq \Omega(\overrightarrow{OB})+\Omega(\overrightarrow{BP}) =\frac{90t^2-96t+21}{2(6t-1)}r \leq \left(\frac{233}{32}t-\frac{191}{32} \right)r. \end{align*}$$

(6) Assume $t \in [\sqrt {14}/2, 23/12]$ . In this case, we have

• ○ The slope of $\overrightarrow {OP}$ is $\frac {1}{t-3} \in (-1, -1/2)$ ,

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

• ○ The slope of $\overrightarrow {QP}$ is $-1/2$ .

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

$$\begin{align*}h^0(F)/r \leq \frac{192t-168}{25}. \end{align*}$$

(7) Assume $t \in [23/12, 2)$ . Then we have

• ○ The slope of $\overrightarrow {OP}$ is $\frac {1}{t-3} \in (-1, -1/2)$ ,

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

• ○ The slope of $\overrightarrow {QP}$ is $-1/2$ .

Hence, $P_{1}=Q$ or A, where A is the point on the line segment $PQ$ with slope $-1$ , i.e.,

$$\begin{align*}Q=((12t-20)r, 4r), \quad A=((-12t+12)r, (12t-12)r).\\[-15pt] \end{align*}$$

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

$$\begin{align*}h^0(F)/r \leq 12t-15.\\[-36pt] \end{align*}$$

Proof. Assume for contradiction that there is a tilt semistable object F violating the inequality (8). We may assume $\mu _H(F) \geq 0$ by replacing F with $F^\vee $ if necessary. First, observe that the following conditions hold:

• ○ 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 $ .

• ○ 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

  $$\begin{align*}b> a^2-a, \quad b > a^2-1/2.\\[-15pt] \end{align*}$$

Under these conditions, we can apply the arguments in [Reference LiLi19a, Proposition 5.2, Corollary 5.4]. 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

(9) $$ \begin{align} \frac{\operatorname{\mathrm{ch}}_2(F)}{H^2\operatorname{\mathrm{ch}}_0(F)}> \Xi \left(\frac{H\operatorname{\mathrm{ch}}_1(F)}{H^2\operatorname{\mathrm{ch}}_0(F)} \right). \end{align} $$

Furthermore, by the first paragraph in the proof of [Reference LiLi19a, Proposition 5.2], we may assume that

• ○ $\mu _{H_{T}}(F) \in (0, 1/2]$ ,

• ○ F is $\mu _{H_{T}}$ -stable coherent sheaf,

• ○ $F|_{C}, F^{\vee }(2H_{T})|_{C}$ are slope stable.

Using the Riemann–Roch and the vanishings

$$\begin{align*}\hom(\mathcal{O}_{T}, F(-2H_{T}))=0=\hom(\mathcal{O}_{T}, F^{\vee}) \end{align*}$$

(both follows from slope stability of F and the assumption on its slope), we have

(10) $$ \begin{align} \begin{aligned} \operatorname{\mathrm{ch}}_{2}(F)-H_{T}\operatorname{\mathrm{ch}}_{1}(F)+11\operatorname{\mathrm{ch}}_{0}(F) &=\chi(F) \\ &\leq h^0(F|_C)+h^0(F^{\vee}(2H_{T})|_{C}). \end{aligned} \end{align} $$

Note that we have

$$ \begin{align*} &\operatorname{\mathrm{ch}}(F |_C)=(\operatorname{\mathrm{ch}}_0(F), 2H\operatorname{\mathrm{ch}}_1(F)), \\ &\operatorname{\mathrm{ch}}(F^{\vee}(2H_{T})|_C)=(\operatorname{\mathrm{ch}}_0(F), 4H^2\operatorname{\mathrm{ch}}_0(F)-2H\operatorname{\mathrm{ch}}_1(F)). \end{align*} $$

Applying Proposition 3.1 to the RHS of equation (10), we get

(11) $$ \begin{align} \frac{\operatorname{\mathrm{ch}}_{2}(F)}{H^2\operatorname{\mathrm{ch}}_{0}(F)} \leq \begin{cases} -\frac{23}{25}\mu_H(F)-\frac{13}{75} & (\mu_H(F) \in (0, 1/12]) \\ -\frac{1}{5}\mu_H(F)-\frac{7}{30} & (\mu_{H}(F) \in [1/12, 2-\sqrt{14}/2]) \\ -\frac{641}{4800}\mu_H(F)-\frac{1157}{4800} & (\mu_{H}(F) \in [2-\sqrt{14}/2, 1/6)) \\ -\frac{421}{3648}\mu_H(F)-\frac{595}{2432} & (\mu_{H}(F) \in [1/6, 89/496]) \\ -\frac{95}{1728}\mu_H(F)-\frac{883}{3456} & (\mu_{H}(F) \in [89/496, 37/206]) \\ \frac{13}{27}\mu_H(F)-\frac{19}{54} & (\mu_{H}(F) \in [37/206, 1/4)) \\ \frac{109}{228}\mu_H(F)-\frac{53}{152} & (\mu_{H}(F) \in [1/4, 69/238]) \\ \mu_H(F)-\frac{1}{2} & (\mu_{H}(F) \in [69/238, 1/2]). \end{cases} \end{align} $$

In all cases, the inequalities (11) contradict the inequality (9).

Assumption 6.2. Every object $E \in D^b(X)$ , which is $\nu _{0, \alpha }$ -semistable for some $\alpha>0$ , satisfies the strong BG inequality (2).

Proof. First, assume that $\nu _{BN}(E) \in (0, 1/2]$ . Let us consider the universal extension

$$\begin{align*}E \to \widetilde{E} \to \operatorname{\mathrm{Hom}}(\mathcal{O}_{X}, E) \otimes \mathcal{O}_X[1]. \end{align*}$$

By [Reference LiLi19a, Lemma 2.12], $\widetilde {E}$ is BN semistable with $\nu _{BN}(\widetilde {E})=\nu _{BN}(E)$ . By Assumption 6.2, we can see that $\mu _H(\widetilde {E}) \notin (-1/4, 0]$ . Indeed, if otherwise, we have

$$\begin{align*}\frac{H\operatorname{\mathrm{ch}}_2(\widetilde{E})}{H^3\operatorname{\mathrm{ch}}_0(\widetilde{E})} \leq \mu_H(\widetilde{E})^2 +\mu_H(\widetilde{E}) < \frac{1}{2}\mu_H(\widetilde{E}). \end{align*}$$

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 6.2 to the object $\widetilde {E}$ , we have

(13) $$ \begin{align} \frac{H\operatorname{\mathrm{ch}}_2(\widetilde{E})}{H^3\operatorname{\mathrm{ch}}_0(\widetilde{E})} \leq -\frac{3}{4}\mu_H(\widetilde{E}) -\frac{3}{8}. \end{align} $$

Note that we have $\operatorname {\mathrm {ch}}_0(\widetilde {E})=\operatorname {\mathrm {ch}}_0(E)-\hom (\mathcal {O}_X, E)$ and $\operatorname {\mathrm {ch}}_i(\widetilde {E})=\operatorname {\mathrm {ch}}_i(E)$ for $i=1, 2$ . Note also that we have $H^2\operatorname {\mathrm {ch}}_1(E) \geq 0$ since $E \in \operatorname {\mathrm {Coh}}^0(X)$ . Together with the assumption $\mu _H(\widetilde {E}) <0$ , we have $\operatorname {\mathrm {ch}}_0(\widetilde {E}) <0$ . From these observations, the inequality (13) is equivalent to the inequality

(14) $$ \begin{align} \hom(\mathcal{O}_X, E) \leq \frac{8}{3d}H\operatorname{\mathrm{ch}}_2(E)+\frac{2}{d}H^2\operatorname{\mathrm{ch}}_1(E)+\operatorname{\mathrm{ch}}_0(E). \end{align} $$

On the other hand, the inequality $\mu _H(\widetilde {E}) < -1/2$ is equivalent to

$$\begin{align*}\hom(\mathcal{O}_X, E) \leq \frac{2}{d}H^2\operatorname{\mathrm{ch}}_1(E)+\operatorname{\mathrm{ch}}_0(E), \end{align*}$$

which is stronger than equation (14) since we have $H\operatorname {\mathrm {ch}}_2(E)>0$ by our assumption $\nu _{BN}(E)>0$ . If $\mu _{H}(\widetilde {E})>0$ , the same inequality (14) obviously holds since $\operatorname {\mathrm {ch}}_{0}(E)-\hom (\mathcal {O}_{X}, E)=\operatorname {\mathrm {ch}}_{0}(\widetilde {E})>0$ and $H^2\operatorname {\mathrm {ch}}_{1}(E)>0$ . Hence, the inequality (14) always holds.

On the other hand, by using the BN stability of E and $\mathcal {O}_{X}[1]$ , we have

(15) $$ \begin{align} \hom(\mathcal{O}_{X}, E) \geq \chi(E) =\operatorname{\mathrm{ch}}_{3}(E)+\operatorname{\mathrm{td}}_{X, 2}\operatorname{\mathrm{ch}}_1(E). \end{align} $$

Combining the inequalities (14) and (15), we get

$$\begin{align*}\operatorname{\mathrm{ch}}_{3}(E)+\operatorname{\mathrm{td}}_{X, 2}\operatorname{\mathrm{ch}}_1(E) \leq \frac{1}{d}H^3\operatorname{\mathrm{ch}}_{0}(E)+\frac{2}{d}H^2\operatorname{\mathrm{ch}}_{1}(E)+\frac{8}{3d}H\operatorname{\mathrm{ch}}_2(E), \end{align*}$$

and hence

(16) $$ \begin{align} \begin{aligned} Q^{\Gamma}(E) &\geq 4(H\operatorname{\mathrm{ch}}_{2}(E))^2 -6\cdot\frac{\delta d-e}{d}H\operatorname{\mathrm{ch}}_2(E)H^3\operatorname{\mathrm{ch}}_0(E) +6\delta\left( H^2\operatorname{\mathrm{ch}}_1(E) \right)^2 \\ &\quad -6H^2\operatorname{\mathrm{ch}}_1(E)\left( \frac{1}{d}H^3\operatorname{\mathrm{ch}}_0(E)+\frac{2}{d}H^2\operatorname{\mathrm{ch}}_1(E)+\frac{8}{3d}H\operatorname{\mathrm{ch}}_2(E) \right) \\ &=4b^2-\frac{16}{d}ab+6\left(\delta-\frac{2}{d}\right)a^2 -6\left(\delta-\frac{e}{d}\right)rb-\frac{6}{d}ra. \end{aligned} \end{align} $$

Here, we put $(r, a, b):=(H^3\operatorname {\mathrm {ch}}_0(E), H^2\operatorname {\mathrm {ch}}_1(E), \operatorname {\mathrm {ch}}_2(E))$ , to simplify the notation. Note that we have $a \geq 0$ since $E \in \operatorname {\mathrm {Coh}}^0(X)$ . 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 6.2, 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 (16), 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

(17) $$ \begin{align} A_1a^2-B_1ab+4b^2 =(a-2b)\left(A_1a+(2A_1-B_1)b \right) +(4A_1-2B_1+4)b^2. \end{align} $$

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 (17) is nonnegative, and hence we have $Q^\Gamma (E) \geq 0$ as required.

When $\mu _H(E) \in [1/2, 3/4]$ , by Assumption 6.2, we have

$$\begin{align*}-r \geq -2a+8b. \end{align*}$$

Combining with the inequality (16), 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 $4A_2-2B_2+C_2 \geq 0$ . 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 6.2, we have $b/r \leq 7a/4r-5/4$ , equivalently,

$$\begin{align*}-r \geq -\frac{7}{5}a+\frac{4}{5}b. \end{align*}$$

Together with the inequality (16), 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

(18) $$ \begin{align} \hom(\mathcal{O}_X, E) \leq \frac{2}{d}H^2\operatorname{\mathrm{ch}}_1(E)+\operatorname{\mathrm{ch}}_0(E). \end{align} $$

Assume that $\widetilde {E}$ is $\nu _{\alpha , 0}$ -unstable for all $\alpha>0$ . Then by the proof of [Reference LiLi19a, Proposition 3.3], 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 6.2, we must have

$$\begin{align*}\mu_H(E_i) \notin \left[ -\frac{3}{8\delta+6}, 0 \right]. \end{align*}$$

Taking a limit $\delta \to +0$ , we get $\mu (\widetilde {E}) \notin [-1/2, 0]$ , hence the inequality (18) holds. Furthermore, by using the derived dual (cf. proof of [Reference LiLi19a, Proposition 3.3]), we also have

$$\begin{align*}\hom(\mathcal{O}_X, E[2]) \leq \frac{2}{3}H^2\operatorname{\mathrm{ch}}_1(E)-\operatorname{\mathrm{ch}}_0(E). \end{align*}$$

Hence, we get

$$ \begin{align*} \operatorname{\mathrm{ch}}_3(E)+\operatorname{\mathrm{td}}_{X, 2}\operatorname{\mathrm{ch}}_1(E) =\chi(E) &\leq \hom(\mathcal{O}_X, E)+\hom(\mathcal{O}_X, E[2]) \\ &\leq \frac{4}{d}H^2\operatorname{\mathrm{ch}}_1(E), \end{align*} $$

from which we deduce $Q^\Gamma (E) \geq 0$ , as we assume $\delta \geq 4/d$ .

Proof. By Theorems 4.1 and 5.4, a triple/double CY3 satisfies Assumption 6.2. Furthermore, we can take the $1$ -cycle $\Gamma $ to be $\Gamma :=\delta _XH^2-\operatorname {\mathrm {td}}_{X, 2}=\gamma H^2$ .

Proof. First, we check the axiom (1) in Definition 7.1 for the pair $\left (Z_{\beta , \alpha }^{a, b}, \mathcal {A}_{\beta , \alpha } \right )$ . As in the proof of [Reference Bayer, Macrì and StellariBMS16, Theorem 8.6], it is enough to show the inequality $Z_{\beta , \alpha }^{a, b}(F[1]) <0$ for every $\nu ^{\prime }_{\beta , \alpha }$ -semistable object F with $\nu ^{\prime }_{\beta , \alpha }(F)=0$ . By the inequality $\alpha ^2+\left (\beta -\lfloor \beta \rfloor -\frac {1}{2} \right )^2> \frac {1}{4}$ in equation (20), we can apply Theorem 1.2 to the object F. Noting the equation (19), we get

(21) $$ \begin{align} \operatorname{\mathrm{ch}}_3^\beta(F) \leq \left(\gamma+\frac{1}{6}\alpha^2 \right)H^2\operatorname{\mathrm{ch}}_1^\beta(F). \end{align} $$

Furthermore, together with the assumption $H\operatorname {\mathrm {ch}}_2^\beta (F)=\frac {1}{2}\alpha ^2H^3\operatorname {\mathrm {ch}}_0^\beta $ , the classical BG inequality (cf. [Reference Bayer, Macrì and StellariBMS16, Theorem 3.5]) $\overline {\Delta }_H(F) \geq 0$ gives the inequality

(22) $$ \begin{align} \left(H\operatorname{\mathrm{ch}}_2^\beta(F) \right)^2 \leq \frac{1}{4}\alpha^2\left(H^2\operatorname{\mathrm{ch}}_1^\beta(F) \right)^2. \end{align} $$

By the inequalities (21) and (22), we obtain

$$ \begin{align*} Z_{\beta, \alpha}^{a, b}(F[1]) &=\Re Z_{\beta, \alpha}^{a, b}(F[1]) \\ &\leq \left(\gamma+\frac{1}{6}\alpha^2 \right)H^2\operatorname{\mathrm{ch}}_1^\beta(F) +\frac{1}{2}|b|\alpha H^2\operatorname{\mathrm{ch}}_1^\beta(F) -aH^2\operatorname{\mathrm{ch}}_1^\beta(F) <0. \end{align*} $$

Now the axiom (2) is also satisfied since we assume $\alpha , \beta \in \mathbb {Q}$ (see the proof of [Reference Bayer, Macrì and StellariBMS16, Theorem 8.6] for the detail).

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

(23) $$ \begin{align} \left( \begin{array}{cc} -\alpha^2K+\alpha^4 & -\frac{3}{2}b\alpha^2 \\ -\frac{3}{2}b\alpha^2 & K-6(a-\gamma) \end{array} \right). \end{align} $$

When $b=0$ , the matrix (23) 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 nonempty by equation (20).

When $b \neq 0$ , we also need to require the determinant of the matrix (23) to be positive, i.e.,

(24) $$ \begin{align} \alpha^2\left( -K^2+\left(6(a-\gamma)+\alpha^2 \right)K -6(a-\gamma)\alpha^2 -\frac{9}{4}b^2\alpha^2 \right)>0. \end{align} $$

The solution space $K \in I_{\alpha , \gamma }^{a, b}$ of the inequality (24) forms a nonempty open interval since we have, by equation (20),

$$\begin{align*}a-\gamma> \frac{1}{6}\alpha^2+\frac{1}{2}|b|\alpha. \end{align*}$$

We can prove the second assertion as in [Reference Bayer, Macrì and StellariBMS16, Lemma 8.8], using the BG type inequality obtained in Theorem 1.2.

Proof of Theorem 7.2

By Propositions 7.3 and 7.4, 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 7.4. See [Reference Bayer, Macrì and StellariBMS16, Proposition 8.10] for the precise proof.