Degeneration of vector bundles on an algebraic curve Let $X$ be a smooth projective irreducible curve over an algebraically closed field $k$. Let $\mathcal E$ be a vector bundle (say, of rank $n$), and let $\mathcal F$ be another vector bundle of rank $n$ which is a degeneration of $\mathcal E$ (i.e. there exists a d.v.r. $R$ with residue field $k$ and fraction field $K$ and a family of vector bundles on $X$ over $Spec(R)$ whose fiber the special point of $Spec(R)$ is $\mathcal F$ and whose fiber over the generic point of $Spec(R)$ is the pull-back of $\mathcal E$ to $X\times Spec(K)$).
One way to produce such a degeneration is to introduce a filtration on $\mathcal E$ whose associated graded is isomorphic to $\mathcal F$. My question is whether this is always the case, at least after some assumptions.
More precisely, the question is this:

*

*Is $\mathcal F$ always isomorphic to the associated graded of $\mathcal E$ with respect to some filtration?



*If not, is it true under the assumption that $\mathcal E$ is stable?

 A: This is correct. Proof is by induction on the rank.
For rank one, this is obvious since you get a morphism from $\operatorname{Spec}R\to\operatorname{Pic} X$ which by your assumption is generically constant and thus constant.
For rank greater than one, you may twist by a large line bundle from $X$ and assume that the special member is globally generated and the sections on the total space surjects onto the sections of the special member. Then easy to see that a general section is nowhere vanishing and take the quotient to reduce to a smaller rank.
A: I don't think this is true, even for stable bundles. Assume that the genus of $X$ is at least 2. Take a point $p$ of $X$, and a nontrivial extension of ${\cal O}(p)$ by ${\cal O}$, this is indecomposable and stable. Using the fact that the homomorphism ${\rm Ext}^1({\cal O(p), O}) = {\rm H}^1(X, {\cal O}(-p)) \to {\rm H}^1(X, {\cal O}) = {\rm Ext}^1({\cal O, O})$ induced by the embedding ${\cal O}(-p) \subseteq {\cal O}$ is an isomorphism, it is easily seen that ${\rm H}^0(X, E) = k$; so there is a unique embedding $\mathcal {O} \subseteq E$. This implies that if $E'$ is another such extension, then $E$ and $E'$ are isomorphic if and only if they differ by multiplication by a nonzero scalar in ${\rm Ext}^1({\cal O(p), O})$. There is a line in ${\rm H}^1(X, \cal O(-p))$ parametrizing non-trivial extensions of $\cal O(p)$ by $\cal O$, no two of which are isomorphic. These are all specializations of the one corresponding to the generic point of the line. Since they are all indecomposable, this gives a counterexample.
