Is the following true ? If so, is there a quick proof of it ? (Perhaps using the uniqueness of the graded object associated to a Jordan-Holder filtration or maybe otherwise)


Suppose $E$ is an $\omega$-semistable bundle with slope $\mu$ over a compact Kahler manifold $(X,\omega)$. There are only finitely many (upto isomorphism) $\omega$-semistable subbundles of $E$ having slope equal to $\mu$.



Number of semistable subbundles of a semistable bundle
That is not true. The issue has to do with nontrivial extensions between semistable bundles of the same slope. If you have a compact Kähler manifold where every semistable sheaf of slope $0$ has vanishing $H^1$, then I suspect that it is true that there are only finitely many isomorphism classes occuring for semistable subbundles of the same slope. However, for manifolds where semistable sheaves of slope $0$ can have large $H^1,$ there are counterexamples. The proposition below gives one class of counterexamples.


Let $k$ be a field, e.g., the field of complex numbers. Let $X$ be a smooth, projective, geometrically connected $k$-curve of genus $g.$


**Lemma.** There exists a short exact sequence of $\mathcal{O}\_X$-modules,$$e:\ \ 0\to \mathcal{O}\_X \xrightarrow{q} E \xrightarrow{p} H^1(X,\mathcal{O}\_X)\otimes\_k \mathcal{O}\_X \to 0,$$ such that the connecting map of the long exact sequence, $$ \partial\_e:H^1(X,\mathcal{O}\_X)\otimes\_k H^0(X,\mathcal{O}\_X)\to H^1(X,\mathcal{O}\_X),$$ is the identity map. The locally free $\mathcal{O}\_X$-module $E$ is semistable of slope $0$ and rank $g+1.$


**Proof.** Choose a $k$-basis, $$e\_1,\dots,e\_g\in \text{Ext}^1\_k(\mathcal{O}\_X,\mathcal{O}\_X) = H^1(X,\mathcal{O}\_X).$$ Each element $e\_i$ gives a Yoneda extension class of a short exact sequence, $$e\_i: \ \ 0 \to \mathcal{O}\_X \xrightarrow{q\_i} E\_i \xrightarrow{p\_i} \mathcal{O}\_X \to 0.$$ Each element $E\_i$ is semistable (not polystable) of slope $0$ and of rank $2.$ Moreover, the connecting map $\delta\_{e\_i}$ has image equal to the span of $e\_i$ in $H^1(X,\mathcal{O}\_X).$ 


The direct sum $F= E\_1\oplus \dots \oplus E\_g$ is semistable of slope $0$ and rank $2g.$ There is a short exact sequence, $$\oplus e\_i:\ \ 0 \to \bigoplus\_{i=1}^g \mathcal{O}\_X \xrightarrow{\oplus q\_i} \bigoplus\_{i=1}^g E\_i \xrightarrow{\oplus p\_i} \bigoplus\_{i=1}^g \mathcal{O}\_X \to 0. $$ The connecting map of this short exact sequence is the direct sum of the connecting maps $\delta\_{e\_i}.$ For the first term in the short exact sequence, there is also a "summing" surjective $\mathcal{O}\_X$-module homomorphism, $$\Sigma:\bigoplus\_{i=1}^g \mathcal{O}\_X \to \mathcal{O}\_X.$$ Denote the kernel by $K=\text{Ker}(\Sigma).$ Define $E$ to be the quotient of $F$ by $K.$ Thus, there is a short exact sequence, $$e:\ \ 0 \to \mathcal{O}\_X \xrightarrow{q} E \xrightarrow{\oplus p\_i} \bigoplus\_{i=1}^g \mathcal{O}\_X \to 0. $$ Again, since the first and third terms are semistable of slope $0$, also the middle term $E$ is semistable of slope $,$ and of rank $g+1.$ Finally, the connecting map $\delta\_e$ is the composition of the direct sum of the summands $\partial\_{e\_i}$ and the summing map. Thus, the connecting map is an isomorphism. **QED**


Now assume that $g\geq 1.$ For every nonzero element $a\in H^1(X,\mathcal{O}\_X),$ consider the subbundle $E\_a$ of $E$ that is the pullback by the following map, $$a:\mathcal{O}\_X\to H^1(X,\mathcal{O}\_X)\otimes\_k\mathcal{O}\_X.$$ In other words, there is a commutative diagram of short exact sequences, $$\begin{array}{cccccccccc} e\_a: & 0 & \to & \mathcal{O}\_X & \xrightarrow{q} & E\_a & \to & \mathcal{O}\_X & \to & 0 \\ & & & \text{Id}~\downarrow & & \downarrow{f\_a} & & \downarrow{a} \\ e: & 0 & \to & \mathcal{O}\_X & \xrightarrow{q} & E & \xrightarrow{\oplus p\_i} & H^1(X,\mathcal{O}\_X)\otimes\_k\mathcal{O}\_X & \to & 0 \end{array}.$$ By the Snake Lemma, $f\_a$ is injective with locally free cokernel, i.e., $f\_a(E\_a)$ is a subbundle of $E$ isomorphic to $E\_a$, a semistable bundle of slope $0,$ and of rank $2.$ Note that the image of $f\_a$ equals the image of $f\_{\lambda\cdot a}$ for every $\lambda\in k^\times.$ Thus, the subsheaf of $E$ depends only on the $k$-point $[a]=[\text{Span}(a)]$ in the projective $k$-space $\mathbb{P}H^1(X,\mathcal{O}\_X) \cong \mathbb{P}^{g-1}\_k$ parameterizing rank $1$ subspaces of $H^1(X,\mathcal{O}\_X).$


**Proposition.** Every semistable subbundle of $E$ of slope $0$ and rank $2$ equals $E\_a$ for a unique $k$-point $[a]\in \mathbb{P}H^1(X,\mathcal{O}\_X).$ For $k$-points $[a]$ and $[b],$ the $\mathcal{O}\_X$-module $E\_a$ is isomorphic to $E\_b$ if and only if $[a]$ equals $[b].$


**Proof.** For every semistable, rank $2$ subsheaf $G$ of slope $0$, consider the short exact sequence, $$0\to (G\cap q(\mathcal{O}\_X)) \to G \to p(G) \to 0.$$ As a subsheaf of the locally free sheaf $H^1(X,\mathcal{O}\_X)\otimes\_k \mathcal{O}\_X$, the sheaf $p(G)$ is torsion-free, hence locally free (since $X$ is a smooth curve). As a quotient of a semistable sheaf of slope $0$, also $p(G)$ has slope $\geq 0$ and rank equal to either $1$ or $2$, depending on whether the rank of the subsheaf $G\cap q(\mathcal{O}\_X)$ of $q(\mathcal{O}\_X)$ equals $1$ or $0.$ Since $H^1(X,\mathcal{O}\_X)\otimes\_k \mathcal{O}\_X$ is semistable of slope $0$, also $p(G)$ has slope $\leq 0.$ Thus, $p(G)$ has slope $0.$ It follows that $p(G)$ is a direct summand of $H^1(X,\mathcal{O}\_X)\otimes\_k \mathcal{O}\_X.$ Since this sheaf is polystable, $p(G)$ is of the form $V\otimes\_k \mathcal{O}\_X$ for a $k$-subspace $V$ of $H^1(X,\mathcal{O}\_X)$ of rank $1$ or $2$. Finally, if the rank of $V$ is $2$, then $G\cap q(\mathcal{O}\_X)$ is zero, so that $G\to p(G)$ is an isomorphism. But then $V=H^0(X,G)$ is a $k$-subspace of $H^0(X,E)$ of rank $2$, whereas $k=H^0(X,q(\mathcal{O}\_X))\to H^0(X,E)$ is an isomorphism of $k$-vector spaces by the construction of $E$. This is a contradiction. Therefore $p(G)$ has rank $1$. This forces the subsheaf $G\cap q(\mathcal{O}\_X)$ of $q(\mathcal{O}\_X)$ to be of rank $1$ and of slope $0$, thus equal to the entire sheaf $q(\mathcal{O}\_X).$ Thus $V$ equals $\text{Span}(a)$ for some nonzero $a$, and $G$ equals $E\_a.$


Associated to the short exact sequence $e\_a$, consider the long exact sequence of cohomology, $$0 \to H^0(X,\mathcal{O}\_X) \to H^0(X,E\_a) \to H^0(X,\mathcal{O}\_X) \xrightarrow{\partial\_a} H^1(X,\mathcal{O}\_X).$$ Since $a$ is nonzero, the first map above is an isomorphism, and the last map above is an injection. Thus, the subsheaf $q(\mathcal{O}\_X)$ is the intrinsic subsheaf $H^0(X,E\_a)\otimes\_k \mathcal{O}\_X,$ the cokernel by this intrinsic sheaf is isomorphic to $\mathcal{O}\_X,$ and the connecting map $\delta$ is identified with the intrinsic connecting map, $$\partial\_{E\_a}:H^0(X,E\_a/H^0(X,E\_a)\otimes\_k\mathcal{O}\_X) \to H^1(X,H^0(X,E\_a)\otimes\_k \mathcal{O}\_X), \text{ i.e.,}$$
$$\partial\_{E\_a}:H^0(X,E\_a/H^0(X,E\_a)\otimes\_k\mathcal{O}\_X) \to H^0(X,E\_a)\otimes\_k H^1(X,\mathcal{O}\_X).$$ Of course the image of $\partial\_a$ in $H^1(X,\mathcal{O}\_X)\setminus \{0\}$ gives a well-defined and intrinsic element in the projective space, $$\mathbb{P} H^1(X,\mathcal{O}\_X) \cong \mathbb{P}^{g-1}\_k.$$ Thus, the sheaf $E\_a$ is isomorphic to $E\_b$ if and only if the elements $[a],[b]\in \mathbb{P} H^1(X,\mathcal{O}\_X)$ are equal. **QED**


Finally, if $k$ is infinite and if $g\geq 2,$ then there are infinitely many $k$-points of $\mathbb{P}^{g-1}\_k.$ Thus, there are infinitely many isomorphism classes of rank-$2$, slope-$0$ semistable subsheaves $f\_a(E\_a)$ of the semistable, rank-$(g+1)$, slope-$0$ sheaf $E$.


