TITLE: Limit of decomposable bundles
QUESTION [10 upvotes]: Let $(E_b)_{b\in B}$ be a family of vector bundles on a smooth projective variety $X$, parameterized by a smooth curve $B$. Let $\mathrm{o}\in B$. Assume that $E_b$ is decomposable (= direct sum of two lower rank bundles) for $b\neq \mathrm{o}$. Can we conclude that  $E_{\mathrm{o}}$ is decomposable? If not, what would be a counter-example?
Background : For $X=\mathbb{P}^n$, $n\geq 2$, I believe this is an open problem (see Mohan's answer to this MO question). I am primarily interested in the case of rank 2 vector bundles on a curve.

REPLY [3 votes]: Though Jason Starr has given an excellent answer above, let me just describe another path which might be of interest.
Constructing deformations as required by abx on the projective space $P$ is equivalent to finding a vector bundle $G$ on $P$ with a nilpotent endomorphism $t$ and a map $\phi:E\to G$ such that $\phi(E)+t(G)=G$. Then you can write down a family on $P\times A$ where  $A=\mathbb{A}^1=\mathrm{Spec} k[t]$ (same $t$ intentionally) by treating $G$ as a sheaf over $X\times A$ supported on $t^n=0$ for some $n$. Then, you have a surjection on $P\times A$, $E\to G$, using $\phi$ and the kernel is a vector bundle whose generic fibers over $A$ are $E$. In general, if $G$ is not a direct sum of line bundles and $E$ is, you get a family whose general member is decomposable and special member is not.
On $P=\mathbb{P}^2$, this is easy to do and for higher dimensional case (at least in characteristic zero) this is an open question. Here is one of the simplest such. Take the usual resolution of the ideal sheaf $I$ of a point and take the push-out $\mathcal{O}(-2)\to \mathcal{O}$, by a quadric not passing through the point to get $0\to \mathcal{O}\to G\to I\to 0$, a rank two vector bundle with a nilpotent endomorphism, got by sending $G\to I\to \mathcal{O}\to G$ and $\phi:\mathcal{O}(-1)^2\to G$.
This bundle gives you a family of rank two vector bundles on $P$ with general member $\mathcal{O}(-1)^2$ and special member indecomposable and thus unstable.
Finally, if $M$ is a rank 2 unstable non-split bundle on $P$, one may assume after twisting, that we have an exact sequence $0\to \mathcal{O}(k)\to M\to J\to 0$, $J$ defining a zero dimensional non-empty scheme. If we restrict this to a curve not passing through $V(J)$, $M$ splits if and only if it splits as $\mathcal{O}(k)\oplus\mathcal{O}$. If we take the curve $C$ to be of sufficiently large degree, one can make sure the map $H^0(M)\to H^0(M|C)$ to be an isomorphism and then $M|C$ can not be split.