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:

  1. Is $\mathcal F$ always isomorphic to the associated graded of $\mathcal E$ with respect to some filtration?
  1. If not, is it true under the assumption that $\mathcal E$ is stable?
  • $\begingroup$ Just so that I understand. If you had a family of vector bundles, say of rank 2, with all isomorphic for $t\neq 0$ and at zero it is indecomposable, you want the special member isomorphic to the general? If false, will it give you an example for question 1? $\endgroup$
    – Mohan
    Jun 14, 2021 at 2:33
  • $\begingroup$ Yes, but frankly I don't think it should be true for arbitrary $\mathcal E$. But I do believe that it should be true if you impose some stability requirements on $\mathcal E$ (I certainly think that it should be true for very stable bundles, maybe also for just stable ones). $\endgroup$ Jun 14, 2021 at 13:42
  • $\begingroup$ It is possible to construct a family with general member trivial (direct sum of the structure sheaf) and special member indecomposable. $\endgroup$
    – Mohan
    Jun 14, 2021 at 14:13
  • $\begingroup$ If you restrict to only semistable bundles, then points of the coarse moduli space are S-equivalence classes which is kind of a filtration like you ask? $\endgroup$ Jun 14, 2021 at 19:20
  • $\begingroup$ Actually, I think if you assume the statement for stable $\mathcal{E}$, then the one for general $\mathcal{E}$ may follow by the following induction argument: consider the HN filtration on $\mathcal{E}$, and take its limit to the special fiber. $\endgroup$ Jun 15, 2021 at 13:17

2 Answers 2


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.

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    $\begingroup$ Sorry, I am missing the last point. Why does it let you reduce the statement to smaller rank? $\endgroup$ Jun 7, 2021 at 20:08
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    $\begingroup$ When you go modulo a non vanishing section you get a vector bundle of smaller rank. Since the section reduces to one on the special fiber, you get a family of smaller rank bundles and you can repeat. $\endgroup$
    – Mohan
    Jun 7, 2021 at 21:26
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    $\begingroup$ I am sorry, but I still don't understand. We are supposed to see associated graded with respect to a filtration. What you are proving is that there is a full filtration on the generic bundle which specializes to a filtration on the special bundle. Why is that the same statement? $\endgroup$ Jun 8, 2021 at 2:08
  • $\begingroup$ I am sorry a I do not understand what you mean. $\endgroup$
    – Mohan
    Jun 8, 2021 at 2:10
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    $\begingroup$ I do not see how to construct a filtration on the generic bundle such that the special one is the associated graded with respect to this filtration. You seem to be just constructing a filtration which is compatible with the degeneration. $\endgroup$ Jun 8, 2021 at 4:16

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.

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    $\begingroup$ I am sorry, I am a little confused: what do you mean by "the one corresponding to the generic point of that line"? Why is that a bundle defined over the original ground field? (I assume that by line you a line not passing through the origin, right?) $\endgroup$ Jun 12, 2021 at 11:33
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    $\begingroup$ Oops, I am sorry and you are right, $E$ is not defined over the base field. So this is not a counterexample! $\endgroup$
    – Angelo
    Jun 12, 2021 at 11:50

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