# Is there a stable reduction for a family of vector bundles?

I. General Question

Consider a one-parameter family of vector bundles $$E_t$$ on a smooth projective variety $$X$$ with fixed Chern character $$v$$. Suppose $$E_t$$ is Gieseker stable when $$t\neq 0$$ and $$E_0$$ is not Gieseker semi-stable. Is there a way to find the limit $$\lim_{t\to 0}E_t$$ in the Gieseker semi-stable moduli space $$\overline{\mathcal{M}}_v(X)$$?

Typically, the limit is a semi-stable coherent sheaf and not necessarily locally free. I'm wondering if there is a way that allows us "modify" the family and produce the semi-stable sheaf at special fiber after a suitable base change?

A counterpart that I have in mind is the Stable Reduction theorem [Harris-Morrison, Prop. 3.17] for curves, which provides us an algorithm to find the limiting stable curve for a flat family of curves with bad singularities at the special fiber. I'm not sure if "stable reduction" is too much to hope for in the realm of the vector bundles.

II. An Example

Here is an example I'm working on. On a smooth cubic threefold $$X$$, there is a rank-two vector bundle $$E$$ associated to an elliptic quintic curve $$C$$ on $$X$$ via Serre construction of the ideal sheaf $$I_C$$. One can show $$c_1(E)=0$$ and $$c_2(E)=2$$. When $$C$$ is projectively normal, $$E$$ is Gieseker stable, while when $$C$$ is not projectively normal (equivalently, $$C$$ is contained in a hyperplane), $$E$$ is not Gieseker semi-stable (in fact, it is slope semi-stable, though). See [Markushevich and Tikhomirov, Prop. 2.6]. Therefore, a smooth family $$C_t$$ of elliptic quintics on $$X$$ with $$C_0$$ contained in a hyperplane will produce a family of vector bundles $$E_t$$ in question.

So how to find the limit $$\lim_{t\to 0}E_t$$ in $$\overline{\mathcal{M}}_{2,0,2}(X)$$?

[Markushevich and Tikhomirov, Prop. 2.5] indicates there is a unique pair of lines $$L_1$$ and $$L_2$$ on $$X$$ associated to the non projectively normal curve $$C_0$$. On the other hand, [Druel, Theorem 3.5] and [Beauville, Prop. 6.2(a)] indicate the strict semi-stable sheaves have the form $$I_{M_1}\oplus I_{M_2}$$ for a pair of lines $$M_1,M_2$$ on $$X$$.

These results seem to suggest to us what the limit sheaf is. However, I still cannot find a geometric family of sheaves in $$\overline{\mathcal{M}}_{2,0,2}(X)$$ with $$I_{L_1}\oplus I_{L_2}$$ appears at time 0.

I appreciate it if anyone could help. Happy new year!

Consider the Harder-Narasimhan filtration of $$E_0$$. Assume for simplicity it has length 2: $$0 \to F_0 \to E_0 \to E_0/F_0 \to 0,$$ where $$F_0$$ and $$E_0/F_0$$ are semistable and the slope of $$F_0$$ is bigger than the slope of $$E_0/F_0$$. If $$B$$ is the parameter space of your family and $$i \colon X \to X \times B$$ is the embedding of the central fiber, consider the sheaf $$E'$$ on $$X \times B$$ defined by the exact sequence $$0 \to E' \to E \to i_*(E_0/F_0) \to 0.$$ Note that the new family $$E'$$ has $$E'_t = E_t$$ for $$t \ne 0$$. On the other hand, if you restrict this sequence to the central fiber and take into account the isomorphism $$L_1i^*(i_*G) \cong G$$ (because the normal bundle of the central fiber is trivial of rank 1), you get an exact sequence $$0 \to E_0/F_0 \to i^*E' \to E_0 \to E_0/F_0 \to 0,$$ which implies that $$E'$$ is flat over $$B$$ and $$E'_0$$ fits into an exact sequence $$0 \to E_0/F_0 \to E'_0 \to F_0 \to 0.$$ Now the sheaf $$E'_0$$ is "more semistable" than $$E_0$$, so iterating this procedure you can eventually modify your family to the one with semistable central fiber.
• Thanks, Sasha. Maybe this is trivial, but I'm confused on why there is a surjective morphism $E\to i_*(E_0/F_0)$ (even for the simplest case $E$ is constant structure sheaf $\mathcal{O}_{X\times B}/\pi^{*}\mathcal{O}_B$ over $B$, if it surjects onto the torsion sheaf $i_*\mathcal{O}_X$, then it seems to send sections to something discontinuous.) Jan 1, 2022 at 19:20
• This is the composition of the canonical morphism $E \to i_*i^*E$ with the pushforward of the morphism $i^*E = E_0 \to E_0/F_0$. Jan 1, 2022 at 19:32
• I am not sure how to answer this. Maybe the reason is that the authomorphism group of a stable bundle is $\mathbb{G}_{\mathrm{m}}$, hence connected, while in the curve case it is not. Jan 2, 2022 at 7:57