Proof. Firstly, the pair $(\mathscr {A}_m, Z_t)$ satisfies the Harder–Narasimhan property because $\mathscr {A}_m$ is Artinian, meaning that every descending chain in $\mathscr {A}_m$ terminates. We prove next that $Z_t$ maps every object in $\mathscr {A}_m$ to the upper half-plane. This follows from the observation that the polynomial $\chi _t(\mathscr {O}_{\mathbb {P}^n})$ has simple roots $t= -1,\ldots ,-n$ . We show the details below.

It is sufficient to check that $Z_t$ maps the generators $\left \{ \mathscr {O}_{\mathbb {P}^n}(-m-n)[n], \ldots , \mathscr {O}_{\mathbb {P}^n}(-m)\right \}$ of $\mathscr {A}_m$ to the upper plane. More precisely, we will show that $\chi _t(\mathscr {O}_{\mathbb {P}^n}(-m-i)[i]) \geq 0$ , and if $\chi _t(\mathscr {O}_{\mathbb {P}^n}(-m-i)[i])=0$ , then $\chi ^{\prime }_t(\mathscr {O}_{\mathbb {P}^n}(-m-i)[i])<0$ .

The polynomial $\chi _t(\mathscr {O}_{\mathbb {P}^n}(-m))=(t+n-m)(t+n-m-1)\cdots (t+1-m)/n!$ has simple roots $t=m-1, m-2, \ldots , m-n$ . The sign of $\chi _t$ is given as follows:

1. 1. $\chi _t(\mathscr {O}_{\mathbb {P}^n}(-m))>0$ for

   $ \begin {array}{lc} t\in (m-n,m-n+1) ,\ldots ,(m-3,m-2), (m-1,\infty ),& \text {if}\quad n= \text {even,} \\ t\in (m-n+1,m-n+2) ,\ldots ,(m-3,m-2), (m-1,\infty ), & \text {if}\quad n= \text {odd.} \end {array} $

2. 2. $\chi _t(\mathscr {O}_{\mathbb {P}^n}(-m))<0$ for

   $ \begin {array}{lc} t\in (m-n+1,m-n+2) ,\ldots ,(m-4,m-3), (m-2,m-1),& \text {if}\quad n= \text {even,} \\ t\in (m-n,m-n+1) ,\ldots ,(m-4,m-3), (m-2,m-1), & \text {if}\quad n= \text {odd.} \end {array} $

• • For $\mathscr {O}_{\mathbb {P}^n}(-m)$ , we have $\chi _t(\mathscr {O}_{\mathbb {P}^n}(-m))>0$ when $t\in (m-1,m]$ .

• • For $\mathscr {O}_{\mathbb {P}^n}(-m-1)[1]$ , $\chi _t(\mathscr {O}_{\mathbb {P}^n}(-m-1)[1]=-\chi _t(\mathscr {O}_{\mathbb {P}^n}(-m-1))=-\chi _{t-1}(\mathscr {O}_{\mathbb {P}^n}(-m))>0$ when $t\in (m-1,m)$ . $\chi _t(\mathscr {O}_{\mathbb {P}^n}(-m-1)[1]=0$ when $t=m$ , and in this case, we have $\chi ^{\prime }_m(\mathscr {O}_{\mathbb {P}^n}(-m-1)[1])=-\chi ^{\prime }_m(\mathscr {O}_{\mathbb {P}^n}(-m-1))<0$ .

• • For $\mathscr {O}_{\mathbb {P}^n}(-m-i)[i]$ , $i=2,3,\ldots ,n$ , using the sign of $\chi _t$ above it is straightforward to check that

  $$\begin{align*}\begin{cases} \chi_t(\mathscr{O}_{\mathbb{P}^n}(-m-i)[i])>0, &\ \text{when} \ t\in (m-1,m),\\ \chi_t(\mathscr{O}_{\mathbb{P}^n}(-m-i)[i])=0, &\ \text{when}\ t=m.\\ \end{cases} \end{align*}$$

  The sign of $\chi ^{\prime }_m(\mathscr {O}_{\mathbb {P}^n}(-m-i))$ is alternating for $i=1,2,\ldots ,n$ . So the sign of $\chi ^{\prime }_m(\mathscr {O}_{\mathbb {P}^n}(-m-i))$ and $\chi ^{\prime }_m(\mathscr {O}_{\mathbb {P}^n}(-m))$ are different by $(-1)^i$ . On the other hand, the shift by $[i]$ will change the sign of $\chi ^{\prime }_t(\mathscr {O}_{\mathbb {P}^n}(-m-i))$ by $(-1)^i$ . So the sign of $\chi ^{\prime }_m(\mathscr {O}_{\mathbb {P}^n}(-m-i)[i])$ and $\chi ^{\prime }_m(\mathscr {O}_{\mathbb {P}^n}(-m))$ are different by $(-1)^i \cdot (-1)^i=1$ . This implies that the signs of $\chi ^{\prime }_m(\mathscr {O}_{\mathbb {P}^n}(-m-i))[i]$ are the same for all $i=0,1,\ldots ,n$ , and they are all negative.

Proof. By definition, we need to find numbers $C_t>0$ , such that $\displaystyle \frac {|Z_t(E)|}{||v(E)||}>C_t$ for any $t\in (0,1]$ . The category is $\mathscr {A}_1$ which is generated by these objects:

$$ \begin{align*}u_1:=\displaystyle \mathscr{O}_{\mathbb{P}^3}(-4)[3], u_2:=\mathscr{O}_{\mathbb{P}^3}(-3)[2], u_3:=\mathscr{O}_{\mathbb{P}^3}(-2)[1], u_4:=\mathscr{O}_{\mathbb{P}^3}(-1).\end{align*} $$

Figure 1

Support property.

Observe that for any $t\in (0,1]$ , the linear span $U:=\{a_1Z_t(u_1)+a_2Z_t(u_2)+a_3Z_t(u_3)+a_4Z_t(u_4)| a_i\geq 0\}$ is not the entire upper half-plane. For instance, Figure 1 shows the case when $t=0.9$ . We abuse the notation in Figure 1 that $u_1,\ldots , u_4$ denote the images $Z_t(u_1), \ldots , Z_t(u_4)$ in the upper half-plane when $t=0.9$ . The two black rays are passing through $Z(u_1)$ (the smallest phase among $Z_t(u_1), \ldots , Z_t(u_4)$ ) and $Z(u_4)$ (the biggest phase). So for $t=0.9$ and any object $u\in \mathscr {A}_t$ , $Z_t(u)$ must be above the two black rays. It is evident that the region above the black rays does not fill the entire upper half-plane. So we can find a line $\Gamma _t$ (the dotted line in Figure 1) such that all the objects in U have a non-zero projection to $\Gamma _t$ . For any object $\displaystyle u=\sum _{i=1}^4 a_iu_i$ ( $a_i\geq 0$ ), denote its projection to $\Gamma _t$ by $p_t(u)$ , and let $a_t:=min \{ |Z_t(p_t(u_i))| :i=1,2,3,4\}$ , $b_t:=max \{ ||(u_i)|| :i=1,2,3,4\}$ .

$$ \begin{align*}\frac{|Z_t(u)|}{||v(u)||}=\frac{|Z_t(\sum_{i=1}^4 a_iu_i)|}{||\sum_{i=1}^4 a_iu_i||}>\frac{|\sum_{i=1}^4a_ip_t(u_i)|}{\sum_{i=1}^4a_i||u_i||}\geq \frac{(\sum_{i=1}^4a_1)a_t}{(\sum_{i=1}^4a_1)b_t}=\frac{a_t}{b_t}=:C_t>0.\\[-47pt]\end{align*} $$

Proof. $\mathscr {A}'$ is the heart of a bounded t-structure since it is obtained from tilting $Coh(\mathbb {P}^3)$ (see [Reference Happel, Reiten and Smalø14]). The category $\mathscr {A}_t$ generated by the objects

$$ \begin{align*}\left\{ \mathscr{O}_{\mathbb{P}^3}(-n-3)[3], \mathscr{O}_{\mathbb{P}^3}(-n-2)[2], \mathscr{O}_{\mathbb{P}^3}(-n-1)[1], \mathscr{O}_{\mathbb{P}^3}(-n)\right\}\end{align*} $$

is also the heart of a bounded t-structure (see [Reference Macrì25, Lem. 3.14]). By [Reference Macrì and Schmidt27, Prop. 5.6], two hearts must coincide if one is contained in another. So it is sufficient to prove that the objects

$$ \begin{align*}\left\{ \mathscr{O}_{\mathbb{P}^3}(-n-3)[3], \mathscr{O}_{\mathbb{P}^3}(-n-2)[2], \mathscr{O}_{\mathbb{P}^3}(-n-1)[1], \mathscr{O}_{\mathbb{P}^3}(-n)\right\}\end{align*} $$

generating $\mathscr {A}_t$ are all contained in $\mathscr {A}^{\prime }_t$ .

We will prove that the objects $\left \{ \mathscr {O}_{\mathbb {P}^3}(-3)[3], \mathscr {O}_{\mathbb {P}^3}(-2)[2], \mathscr {O}_{\mathbb {P}^3}(-1)[1], \mathscr {O}_{\mathbb {P}^3}\right \}$ are in the category $\mathscr {A}^{\prime }_t$ where $t\in (-1,0]$ , and the proof for a general $t\in \mathbb {R}$ is analogous. We will keep track of the line bundles $\mathscr {O}_{\mathbb {P}^3}, \mathscr {O}_{\mathbb {P}^3}(-1), \mathscr {O}_{\mathbb {P}^3}(-2),\mathscr {O}_{\mathbb {P}^3}(-3)$ in $Coh(\mathbb {P}^{3})$ while we make the three time tiltings.

The line bundles $\mathscr {O}_{\mathbb {P}^3}(m)$ are Mumford stable for any $m\in \mathbb {Z}$ , so they are also twisted Mumford stable. It is straightforward to check that $\mu _t(\mathscr {O}_{\mathbb {P}^3}(k))=k+t+2$ for $k\in \mathbb {Z}$ , and $\mu _t(\mathscr {O}_{\mathbb {P}^3})>0$ $\mu _t(\mathscr {O}_{\mathbb {P}^3}(-1))>0$ $\mu _t(\mathscr {O}_{\mathbb {P}^3}(-2))<0$ $\mu _t(\mathscr {O}_{\mathbb {P}^3}(-3))<0$ for $t\in (-1,0]$ . This implies that the category $Coh^{-t-2}(\mathbb {P}^3)$ (for $t\in (-1,0]$ ) contains these objects:

$$ \begin{align*}\mathscr{O}_{\mathbb{P}^3}, \mathscr{O}_{\mathbb{P}^3}(-1), \mathscr{O}_{\mathbb{P}^3}(-2)[1], \mathscr{O}_{\mathbb{P}^3}(-3)[1].\end{align*} $$

Then, we use the slope function $\nu _t=\frac {ch^{-t-2}_2-1/6ch_0}{ch^{-t-2}_1}$ for the second tilt. It was proved in [Reference Macrì26] and [Reference Schmidt33] that line bundles and their shifts ( $\mathscr {O}_{\mathbb {P}^3}(m)$ , $\mathscr {O}_{\mathbb {P}^3}(m)[1]$ ) are tilt stable when they are in the category. A straightforward computation shows that $\displaystyle \nu _t(\mathscr {O}_{\mathbb {P}^3}(k))=\frac {k^2/2+(t+2)k+(t+2)^2/2-1/6}{k+t+2}$ for $k\in \mathbb {Z}$ , so $\mathscr {B}_t$ (a tilt of $Coh^{-t-2}(\mathbb {P}^3)$ ) contains the following objects for $t\in (-1,0]$ :

For the last tilt, we have that line bundles and their shifts are Bridgeland stable (see [Reference Macrì26]), that is, $\mathscr {O}_{\mathbb {P}^3}(m)$ , $\mathscr {O}_{\mathbb {P}^3}(m)[1]$ , $\mathscr {O}_{\mathbb {P}^3}(m)[2]$ are stable in the double tilt $\mathscr {B}_t$ .

Using the slope function $\displaystyle \lambda _t=\frac {\chi _t}{\chi ^{\prime }_t}=\frac {ch^{-t-2}_3-1/6ch^{-t-2}_1}{ch^{-t-2}_2-1/6ch_0}$ , the claim follows from a direct computation of those $\lambda _t(\mathscr {O}_{\mathbb {P}^3}(-i)[j])$ ’s in the last step. So as expected, the objects $\mathscr {O}_{\mathbb {P}^3}$ , $\mathscr {O}_{\mathbb {P}^3}(-1)[1]$ , $\mathscr {O}_{\mathbb {P}^3}(-2)[2]$ , $\mathscr {O}_{\mathbb {P}^3}(-3)[3]$ are in the category $\mathscr {A}^{\prime }_t$ .

Proof. Assume $t=a/b,$ where $a\in \mathbb {Z}$ , $b\in \mathbb {Z}^+$ , and $g.c.d.(a,b)=1$ .

Let $B\in \mathscr {B}_t$ be the quotient of the inclusion $A\hookrightarrow E$ , that is, defined by the short exact sequence $0\to A\to E\to B\to 0$ in $\mathscr {B}_t$ . Take the cohomologies in $Coh^{-t-2}(\mathbb {P}^3)$ , and we have a long exact sequence:

$$ \begin{align*}0\to\mathscr{H}^{-1}_{\beta}(A) \to \mathscr{H}^{-1}_{\beta}(E) \to \mathscr{H}^{-1}_{\beta}(B) \to \mathscr{H}^{0}_{\beta}(A) \to \mathscr{H}^{0}_{\beta}(E) \to \mathscr{H}^{0}_{\beta}(B)\to 0,\end{align*} $$

where $\mathscr {H}^{-1}_{\beta }$ and $\mathscr {H}^{0}_{\beta }$ denote the cohomologies in $Coh^{-t-2}(\mathbb {P}^3)$ . Since $E\in Coh(\mathbb {P}^3)$ is a sheaf, we have $\mathscr {H}^{-1}_{\beta }(E)=0$ . This implies that $E\cong \mathscr {H}^{0}_{\beta }(E)$ and $\mathscr {H}^{-1}_{\beta }(A)=0$ . So $A\cong \mathscr {H}^{0}_{\beta }(A)$ , and $A \in Coh^{-t-2}({\mathbb {P}^3})$ . The long exact sequence above is reduced to the following one:

$$ \begin{align*}0\to \mathscr{H}^{-1}_{\beta}(B) \to A \to E \to \mathscr{H}^{0}_{\beta}(B)\to 0.\end{align*} $$

This implies that $A\in Coh^{-t-2}(\mathbb {P}^3)\subset \mathscr {B}_t$ . So we have $\chi ^{\prime \prime }_t(A)\geq 0$ , and $\chi ^{\prime }_t(A)\geq 0$ . Moreover, we have $\chi ^{\prime }_t(B)\geq 0$ and $\chi ^{\prime }_t(A)+\chi ^{\prime }_t(B)=\chi ^{\prime }_t(E)=m$ . So $\chi ^{\prime }_t(A)\in [0,m]$ , and if $\chi ^{\prime }_t(A)=0,$ then $\chi ^{\prime \prime }_t(A)=0,$ otherwise, the tilt slope $\displaystyle \nu (A)=\frac {\chi ^{\prime }_t(A)}{\chi ^{\prime \prime }_t(A)}=0$ will make A shifted in $\mathscr {B}_t$ (i.e., $A[1]\in \mathscr {B}_t$ ). In this case, A is a sheaf supported on points and this violates the fact that E is Gieseker stable. So $\chi ^{\prime }_t(A)>0$ .

1. 1. Lower bound for $\displaystyle \frac {\chi ^{\prime \prime }_t(A)}{\chi ^{\prime }_t(A)}$ .

   $\displaystyle \frac {\chi ^{\prime \prime }_t(A)}{\chi ^{\prime }_t(A)}=\frac {ch_1+(t+2)ch_0}{\chi ^{\prime }_t(A)}\geq \frac {ch_1+(t+2)ch_0}{m}\geq \frac {1/b}{m}=\frac {1}{bm}$ or $\displaystyle \frac {\chi ^{\prime \prime }_t(A)}{\chi ^{\prime }_t(A)}=0$ . We choose $\displaystyle B_2=\frac {1}{bm}$ . The last inequality holds because $ch_0,ch_1\in \mathbb {Z}$ , and $ch_1+(t+2)ch_0>0$ .

2. 2. Upper bound for $\displaystyle \lambda _t(A)= \frac {\chi _t(A)}{\chi ^{\prime }_t(A)}$ .

   From the fact that $ch_0,ch_1\in \mathbb {Z}$ , $ch_2\in \frac {1}{2}\mathbb {Z}$ , and $\chi ^{\prime }_t(A)>0$ , we have $\displaystyle \chi ^{\prime }_t(A)=ch_2(A)+(t+2)ch_1(A)+\frac {(t+2)^2}{2}ch_0(A)-\frac {1}{6}ch_0(A)\geq \frac {1}{6b^2}>0$ .

   So we just need an upper bound for $\chi _t(A)$ . Consider the Harder–Narasimhan filtration of $A\in \mathscr {B}_t$ with respect to the central charge $Z_t=-\chi _t+i\cdot \chi ^{\prime }_t$ :

   $$ \begin{align*}0=A_0\subset A_1\subset \cdots\subset A_i\subset A_{i+1}\subset \cdots \subset A_{n-1}\subset A_n:=A.\end{align*} $$

   Let $i\in \mathbb {Z}$ be the number such that the semistable factor $\displaystyle \frac {A_i}{A_{i-1}}$ is the last one whose slope is positive, that is, $\displaystyle \lambda _t(\frac {A_i}{A_{i-1}})>0$ , and $\displaystyle \lambda _t( \frac {A_{i+1}}{A_{i}})\leq 0$ . If there is no such i, then $\chi ^{\prime }_t(E)\leq 0$ and we define $B_1$ to be $0$ .

   Similarly for the Harder–Narasimhan filtration for $E\in \mathscr {B}_t$ :

   $$ \begin{align*}0=E_0\subset E_1\subset \cdots\subset E_j\subset E_{j+1}\subset \cdots \subset E.\end{align*} $$

   Let j be the index such that the Harder–Narasimhan factor is the last one with a positive slope, that is, $\displaystyle \lambda _{t}(\frac {E_j}{E_{j-1}})>0$ and $\displaystyle \lambda _{t}(\frac {E_{j+1}}{E_{j}})\leq 0$ .

   In the short exact sequence $0\to A_i \to A \to \frac {A}{A_i}\to 0,$ we have $\chi _t(A_i)>0$ and $\displaystyle \chi _t(\frac {A}{A_i})\leq 0$ from how we choose i. More precisely, we have $\displaystyle \chi ^{\prime }_t(\frac {A_k}{A_{k-1}})\geq 0$ and $\displaystyle \chi _t(\frac {A_k}{A_{k-1}})>0$ for $k=1, 2,\ldots , i$ by the definition of i. Then $\displaystyle \chi _t(A_i)=\chi _t(A_0)+\chi _t(\frac {A_1}{A_0})+\cdots +\chi _t(\frac {A_i}{A_{i-1}})>0$ . Similarly, we have that $\displaystyle \chi _t(\frac {A}{A_i})=\chi _t(\frac {A_{i+1}}{A_i})+\chi _t(\frac {A_{i+2}}{A_{i+1}})+\cdots +\chi _t(\frac {A}{A_{n-1}})\leq 0$ because $\displaystyle \chi _t(\frac {A_k}{A_{k-1}})\leq 0$ for $k=i+1, \ldots , n$ . Therefore, we have $\chi _t(A)\leq \chi _t(A_i)$ , and it is sufficient to find an upper bound for $\chi _t(A_i)$ .

   Consider the inclusion $0\to A_i\xrightarrow {f} E$ in $\mathscr {B}_t$ , and the diagram

   

   $\phi $ is a zero map because the semistable factors of $A_i$ have $\lambda _t(A_i)>0$ and the semistable factors of $E/E_j$ have $\lambda _t\leq 0$ . So the morphism f lifts to a morphism $f_1$ from $A_i$ to $E_j$ , where they are both extended by semistable objects in $\mathscr {B}_t$ with positive slope $\lambda _t$ .

   Now, we make another tilt from $\mathscr {B}_t$ to $\mathscr {A}_t$ and the morphism $0\to A_i \to E_j$ stays the same in $\mathscr {A}_t$ because they are both generated by objects with positive slopes $\lambda _t$ . For a fixed E and t, the sub-object $E_j$ and its dimension vector are fixed in $\mathscr {A}_t$ . There are only finitely many choices of sub-objects of $E_j$ in $\mathscr {A}_t$ , and this implies that $\chi _t(A_i)$ is bounded from above for all $A\hookrightarrow E$ in $\mathscr {B}_t$ . So there exists $B_1\in \mathbb {R}$ such that $\displaystyle \frac {\chi _t(A)}{\chi ^{\prime }_t(A)}\geq B_1$ .

Proof. Let $\displaystyle \lambda _{t,u}:=\frac {\chi _t-\frac {u^2}{2}\chi ^{\prime \prime }_t}{\chi ^{\prime }_t}$ be the Bridgeland slope. The “if” part follows from the fact that $\lambda _{t,u}(E)=\frac {mt+\chi }{m}=t+{\chi }/{m}$ if E has class v, and the Bridgeland slope is equal to the twisted Mumford slope.

For the “only if” part, suppose the claim is not true, then for any fixed $u_0\gg 0$ , we can always find an $A_{u_0}$ , such that $\lambda _{t,u_0}(A_{u_0})\geq (>)\lambda _{t,u_0}(E)$ . More explicitly, we have

$$ \begin{align*}\lambda_{t,u_0}(A_{u_0})=\frac{\chi_t(A_{u_0})-\frac{u_0^2}{2}\chi^{\prime\prime}_t(A_{u_0})}{\chi^{\prime}_t(A_{u_0})}\geq (>)\frac{\chi_t(E)}{\chi^{\prime}_t(E)}=\frac{mt+\chi}{m}=t+\frac{\chi}{m}.\end{align*} $$

If $\chi ^{\prime \prime }_t(A_{u_0})\neq 0$ , then $\chi ^{\prime \prime }_t(A_{u_0})> 0$ because $A_{u_0}\in Coh^{-t-2}(\mathbb {P}^3)$ . The expression of $\lambda _{t,u}$ shows that $\displaystyle \lim _{u\to \infty }\lambda _{t,u}(A)=-\infty $ (for $\chi ^{\prime \prime }_t(A)\neq 0$ ). Using the boundedness results from Lemma 4.1, we actually have a universal bound and the following claim:

There exists a $u'\in \mathbb {R}$ such that for all $u>u'$ and $A\hookrightarrow E \in \mathscr {B}_t$ , we have $\lambda _{t,u}(A)<\lambda _{t,u}(E)$ . This violates the assumption that A destabilizes E. So it implies that if E is not Euler stable for all large u ( $u>u'$ ), then we must have $ch^{-t-2}_1(A)=\chi ^{\prime \prime }_t(A)=0$ for any destabilizing object $A\hookrightarrow E \in \mathscr {B}_t$ .

We then go back to the short exact sequence $0\to A\to E\to B\to 0$ in $\mathscr {B}_t$ with its long exact sequence in $Coh^{-t-2}(\mathbb {P}^3)$ :

where the morphism $A\to E$ factors through $Q \in Coh^{-t-2}(\mathbb {P}^3)$ . From the fact that $\chi ^{\prime \prime }_t(A)=ch^{-t-2}_1(A)=\chi ^{\prime \prime }_t(E)=ch^{-t-2}_1(E)=0$ , and all the objects in the diagram has $ch^{-t-2}_1=\chi ^{\prime \prime }_t\geq 0$ (because they are in $Coh^{-t-2}(\mathbb {P}^3)$ ), we have $\chi ^{\prime \prime }_t(\mathscr {H}^{-1}_{\beta }(B))=\chi ^{\prime \prime }_t(Q)=\chi ^{\prime \prime }_t(\mathscr {H}^{0}_{\beta }(B))=0$ .

This implies that $\displaystyle \nu (\mathscr {H}^{-1}_{\beta }(B))=\frac {\chi ^{\prime }_t(\mathscr {H}^{-1}_{\beta }(B))}{\chi ^{\prime \prime }_t(\mathscr {H}^{-1}_{\beta }(B))=0}=\infty $ which contradicts with the fact that it is shifted from $Coh^{-t-2}(\mathbb {P}^3)$ to $\mathscr {B}_t$ . So $\mathscr {H}^{-1}_{\beta }(B)=0$ , and the short exact sequence $0\to A\to E\to B\to 0$ is indeed in $Coh^{-t-2}(\mathbb {P}^3)\subset \mathscr {B}_t$ .

Similarly, consider its long exact sequence of cohomologies in $Coh(\mathbb {P}^3)$ as

$$ \begin{align*}0\to \mathscr{H}^{-1}(B)\to A \to E \to \mathscr{H}^{0}(B)\to 0.\end{align*} $$

From the fact that $ch_0(A)=ch_0(E)=0$ and $ch^{-t-2}_1(A)=ch^{-t-2}_1(E)=0$ , we have $ch_0(\mathscr {H}^{-1}(B))=ch_0(\mathscr {H}^{0}(B))$ , and $ch^{-t-2}_1(\mathscr {H}^{-1}(B))=ch^{-t-2}_1(\mathscr {H}^{0}(B))$ . In particular, $\displaystyle \frac {ch^{-t-2}_1(\mathscr {H}^{-1}(B))}{ch_0(\mathscr {H}^{-1}(B))}=\frac {ch^{-t-2}_1(\mathscr {H}^{0}(B))}{ch_0(\mathscr {H}^{0}(B))}$ , which is a contradiction unless one of $\mathscr {H}^{-1}(B)$ and $\mathscr {H}^{0}(B)$ is zero and the nonzero object has its twisted Mumford slope $\infty $ . So $\mathscr {H}^{-1}(B)=0$ and $\mathscr {H}^{0}(B)\neq 0$ with $\displaystyle \frac {ch^{-t-2}_1(\mathscr {H}^{0}(B))}{ch_0(\mathscr {H}^{0}(B))}=\infty $ .

This shows that B is actually a sheaf, and the short exact sequence $0 \to A\to E\to B\to 0$ is indeed in $Coh(\mathbb {P}^3)$ . Now, A is a sub-sheaf of E and they are both one-dimensional. We have that the slope $\lambda _{t,u}$ for the class v coincides with the Mumford slope, so the assumption $\lambda _{t,u}(A)\geq (>)\lambda _{t,u}(E)$ is equivalent to E being Mumford (Gieseker) unstable which is a contradiction.

Proof. In $Coh^{\beta }(\mathbb {P}^3),$ we have the short exact sequence:

$$ \begin{align*}0\to \mathscr{H}^{-1}(E)[1]\to E \to \mathscr{H}^{0}(E) \to 0.\end{align*} $$

Denote $(\mathscr {T}_t,\mathscr {F}_t)$ to be the torsion pair on $Coh(\mathbb {P}^3)$ defined by $\mu _t$ .

Since $ch_0(E)=ch_1(E)=0$ , we have that $ch_0(\mathscr {H}^{-1}(E))=ch_0(\mathscr {H}^{0}(E))$ and $ch_1(\mathscr {H}^{-1}(E))=ch_0(\mathscr {H}^{0}(E))$ . This implies that $\mu _t(\mathscr {H}^{0}(E))=\mu _t(\mathscr {H}^{-1}(E))$ , but it contradicts with $\mathscr {H}^{-1}(E)\in \mathscr {F}_t$ and $\mathscr {H}^{0}(E)\in \mathscr {T}_t$ . So one of $\mathscr {H}^{-1}(E)$ and $\mathscr {H}^{0}(E)$ is zero, and then the twisted Mumford slope $\mu _t$ of the non-zero object will be infinity. This shows that $\mathscr {H}^{-1}(E)=0$ , and $E\cong \mathscr {H}^{0}(E)$ which is a sheaf.

Proof. E fits into the short exact sequence in $\mathscr {B}_t$ :

$$ \begin{align*}0\to \mathscr{H}^{-1}_{\beta}(E)[1] \to E \to \mathscr{H}^{0}_{\beta}(E) \to 0.\end{align*} $$

It is sufficient to show that $\mathscr {H}^{-1}_{\beta }(E)=0$ , and then the claim would follow from Lemma 4.4.

Suppose $\mathscr {H}^{-1}_{\beta }(E)\neq 0$ , then $\displaystyle \nu (\mathscr {H}^{-1}_{\beta }(E))=\frac {\chi ^{\prime }_t(\mathscr {H}^{-1}_{\beta }(E))}{\chi ^{\prime \prime }_t(\mathscr {H}^{-1}_{\beta }(E))}\leq 0$ and $\chi ^{\prime \prime }_t(\mathscr {H}^{-1}_{\beta }(E))>0$ .

Consider the $\sigma _{t,u}$ slope of $\mathscr {H}^{-1}_{\beta }(B)[1]$ :

$$ \begin{align*}\lambda_{t,u}(\mathscr{H}^{-1}_{\beta}(E)[1])=\frac{\chi_t(\mathscr{H}^{-1}_{\beta}(E)[1])-\frac{u^2}{2}\chi^{\prime\prime}_t(\mathscr{H}^{-1}_{\beta}(E)[1])}{\chi^{\prime}_t(\mathscr{H}^{-1}_{\beta}(E)[1])},\end{align*} $$

$\displaystyle \lim _{u\to \infty }\lambda _{t,u}(\mathscr {H}^{-1}_{\beta }(E)[1])=+\infty $ so E is not stable for $u\gg 0$ unless $\chi ^{\prime \prime }_t(\mathscr {H}^{-1}_{\beta }(E)[1])=0$ . But this implies $\displaystyle \nu _t(\mathscr {H}^{-1}_{\beta }(E))=\frac {\chi ^{\prime }_t(\mathscr {H}^{-1}_{\beta }(E))}{\chi ^{\prime \prime }_t(\mathscr {H}^{-1}_{\beta }(E))=0}=+\infty $ and $\mathscr {H}^{-1}_{\beta }(E)\in \mathscr {B}_t$ which is a contradiction. So $\mathscr {H}^{-1}_{\beta }(E)=0$ and this proves the claim.

Proof. Consider the short exact sequence in $\mathscr {B}_t$ as

$$ \begin{align*}0\to \mathscr{H}_{\beta}^{-1}(E)[1] \to E \to \mathscr{H}_{\beta}^{0}(E)\to 0,\end{align*} $$

we must have $\mathscr {H}_{\beta }^{-1}(E)\neq 0$ and $\mathscr {H}_{\beta }^{0}(E)\neq 0,$ otherwise, Lemma 4.4 will tell us E is a sheaf.

$\mathscr {H}_{\beta }^{-1}(E)\in Coh^{-t-2}(\mathbb {P}^3)$ implies $\chi ^{\prime \prime }_t(\mathscr {H}_{\beta }^{-1}(E))>0$ . The claim follows from the observation that:

$$ \begin{align*}\lambda_{t,u}(\mathscr{H}_{\beta}^{-1}(E)[1])=\frac{\chi_t(\mathscr{H}_{\beta}^{-1}(E)[1])-\frac{u^2}{2}\chi^{\prime\prime}_t(\mathscr{H}_{\beta}^{-1}(E)[1])}{\chi^{\prime}_t(\mathscr{H}_{\beta}^{-1}(E)[1])}>\lambda_{t,u}(E)=\frac{mt+\chi}{m}\end{align*} $$

for $u\gg 0$ . We can also find the critical point for $u_E$ , and this is when the equal sign holds, that is,

$$ \begin{align*}u_E:=\frac{(t+\frac{\chi}{m})\chi^{\prime}_t(\mathscr{H}_{\beta}^{-1}(E)[1])-\chi_t(\mathscr{H}_{\beta}^{-1}(E)[1])}{-\chi^{\prime\prime}_t(\mathscr{H}_{\beta}^{-1}(E)[1])}.\\[-50pt]\end{align*} $$

Proof. It follows from Propositions 4.3 and 4.5, and Lemma 4.6. We take $B_E$ to be the maximum of $u_E$ and the bound in Proposition 4.3.

Proof. E fits into the short exact sequence in $\mathscr {B}_t$ as

$$ \begin{align*}0\to \mathscr{H}^{-1}_{\beta}(E)[1] \to E \to \mathscr{H}^{0}_{\beta}(E) \to 0,\end{align*} $$

where $\mathscr {H}^{-1}_{\beta }(E)$ and $\mathscr {H}^{0}_{\beta }(E)$ are its cohomologies in $Coh^{-t-2}(\mathbb {P}^3)$ .

Assume $ch_0(\mathscr {H}^{0}_{\beta }(E))=R$ , $ch_1(\mathscr {H}^{0}_{\beta }(E))=C$ and $ch_2(\mathscr {H}^{0}_{\beta }(E))=D$ then it follows that $ch_0(\mathscr {H}^{-1}_{\beta }(E))=R$ , $ch_1(\mathscr {H}^{-1}_{\beta }(E))=C$ and $ch_2(\mathscr {H}^{-1}_{\beta }(E))=D-m$ .

From the definition of $Coh^{-t-2}(\mathbb {P}^3)$ , we have the numerical results that

$$ \begin{align*} \left\{ \begin{array}{ll} \chi^{\prime}_t(\mathscr{H}^{0}_{\beta}(E))\geq 0 & \chi^{\prime\prime}_t(\mathscr{H}^{0}_{\beta}(E))\geq 0 \\ \chi^{\prime}_t(\mathscr{H}^{-1}_{\beta}(E))\leq 0 & \chi^{\prime\prime}_t(\mathscr{H}^{-1}_{\beta}(E))\geq 0.\\ \end{array} \right. \end{align*} $$

A direct computation shows that

$$ \begin{align*} \left\{ \begin{array}{l} \chi^{\prime\prime}_t(\mathscr{H}^{-1}_{\beta}(E))=\chi^{\prime\prime}_t(\mathscr{H}^{0}_{\beta}(E))=C+(t+2)R, \\ \chi^{\prime}_t(\mathscr{H}^{0}_{\beta}(E))=\frac{(t+2)^2}{2}R+(t+2)C+D-\frac{1}{6}R,\\ \chi^{\prime}_t(\mathscr{H}^{-1}_{\beta}(E))=\frac{(t+2)^2}{2}R+(t+2)C+D-\frac{1}{6}R-m.\\ \end{array} \right. \end{align*} $$

Observe that, for $R\neq 0$ , the vertical line $\chi ^{\prime \prime }_t(\mathscr {H}^{k}_{\beta }(E))=0$ ( $k=0$ or $-1$ ) in a $(t,s)$ -plane is the axis of symmetry of the parabola $f_k(t)=\chi ^{\prime }_t(\mathscr {H}^{k}_{\beta }(E))$ .

The claim will follow from the following results:

1. 1. If $R>0$ , then $f_1(t):=\chi ^{\prime }_t(\mathscr {H}^{-1}_{\beta }(E))$ , and $f_2(t):=\chi ^{\prime }_t(\mathscr {H}^{0}_{\beta }(E))$ are two parabolas and $f_2(t)$ is a shift upward by m from $f_1(t)$ as shown in Figure 3.

   Figure 3

   Function $f_i(t)=\chi_t'(\mathcal{H}^i_{\beta}(E))$ for $R>0$

   
   Figure 4

   Function $f_i(t)=\chi_t'(\mathcal{H}^i_{\beta}(E))$ for $R=0$

   

   The inequality $\chi ^{\prime \prime }_t(\mathscr {H}^{-1}_{\beta }(E))=\chi ^{\prime \prime }_t(\mathscr {H}^{0}_{\beta }(E))=C+(t+2)R\geq 0$ corresponds to the region to the right of the dotted line. So the region for t satisfying $f_1\leq 0$ and $f_2\geq 0$ is the interval $[A,B]$ . A direct computation shows that $\displaystyle |AB|<\sqrt {\frac {2m}{R}}\leq \sqrt {2m}.$

2. 2. If $R<0$ , then the proof is similar to case $(1)$ .

3. 3. If $R=0$ , then $C>0$ . We have in this case

   $$ \begin{align*} \left\{ \begin{array}{l} \chi^{\prime\prime}_t(\mathscr{H}^{-1}_{\beta}(E))=\chi^{\prime\prime}_t(\mathscr{H}^{0}_{\beta}(E))=C, \\ \chi^{\prime}_t(\mathscr{H}^{0}_{\beta}(E))=(t+2)C+D,\\ \chi^{\prime}_t(\mathscr{H}^{-1}_{\beta}(E))=(t+2)C+D-m.\\ \end{array} \right. \end{align*} $$

   The picture is shown as Figure 4. A simple computation shows $|AB|=\frac {m}{C}\leq m$ .

   Finally, when $m\geq 3$ , $\sqrt {2m}<m$ . So for all the possible cases, we have $|AB|\leq m$ , and the claim follows.