# Limit of decomposable bundles

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.

• Please read the question. I start with a family of vector bundles parameterized by $B$.
– abx
Commented Dec 6, 2017 at 8:55
• It seems to me that there might be a counterexample with $X$ an elliptic curve; more precisly, a direct sum of two line bundles of the same degree can degenerate to an indecomposable bundle. To construct an explicit example, let $B=X$ be an elliptic curve. On $B \times X$ consider the reduced subscheme Z which is the union of the diagonal and $B \times \{0\}$. It seems likely that the (relative) Fourier transform of $\mathcal{O}_Z$ (viewed as a coherent sheaf) is such a bundle.
– naf
Commented Dec 6, 2017 at 10:16
• I don't see why semi-continuity contradicts what I suggested: For $b \neq 0$ the bundle $E_b$ is a direct sum of a trivial line bundle and a non-trivial (varying!) line bundle (of degree 0). The bundle $E_0$ is a non-trivial extension of trivial bundles.
– naf
Commented Dec 6, 2017 at 11:06
• I was thinking what about the inverse of your question? : If $E_0$ be irreducible(resp. decomposable), then all $E_t$ are irreducible?(resp. decomposable), something like proposition 3 of cambridge.org/core/services/aop-cambridge-core/content/view/… ,
– user21574
Commented Dec 6, 2017 at 20:47
• @ulrich: Yes, you are right of course. Sorry I was confused.
– abx
Commented Dec 8, 2017 at 14:10

What ulrich writes is correct. I will delete this answer if ulrich posts his answer. For me, the fastest way to think about this is via smoothness and irreducibility of the stack of locally free sheaves of specified rank and degree on a curve $X$.

For a fixed smooth, projective curve $X$ over a field $k$, there is an algebraic stack $\text{Bun}_{r,d}(X)$ parameterizing locally free $\mathcal{O}_X$-modules of rank $r$ and total degree $d$. This stack is smooth, since the obstruction group $\text{Ext}^2_{\mathcal{O}_X}(E,E)$ is zero. The stack is not quasi-compact (the slopes in the Harder-Narasimhan filtration of $E$ can be arbitrary integers subject to the constraint coming from $r$ and $d$). However, every quasi-compact open substack $V$ is integral in the sense that there is a representable, fppf (even smooth) $1$-morphism $p:V_0\to V$ from a (geometrically) integral $k$-scheme $V_0$. This follows, for instance, by writing every bounded family of locally free sheaves $E$ as quotients of some fixed locally free sheaf $\mathcal{O}(-N\cdot A)^{\oplus (r+1)}$ where $A$ is a fixed divisor class on $X$ of positive degree, and $N\gg 0$ is a sufficiently positive integer. In that case, $V_0$ is an open subset of an affine space bundle over the Jacobian of $X$.

For $X$ an elliptic curve, there is an open substack $U\subset \text{Bun}_{2,0}(X)$ parameterizing locally free sheaves of the form $E=L\oplus M$ where both $L$ and $M$ are invertible sheaves of degree $0$ such that $L\not\cong M$, e.g., $L\oplus \mathcal{O}_X$ where $L\not \cong \mathcal{O}_X$. The key to openness of this substack is the fact that every infinitesimal deformation of $L\oplus M$ is again of the form $L'\oplus M'$, since $$\text{Ext}^1_{\mathcal{O}_X}(L,L)\oplus \text{Ext}^1_{\mathcal{O}_X}(M,M)\to \text{Ext}^1_{\mathcal{O}_X}(L\oplus M, L\oplus M)$$ is surjective. This uses the fact that $\text{Ext}^1_{\mathcal{O}_X}(L,M)$ and $\text{Ext}^1_{\mathcal{O}_X}(M,L)$ are zero on a genus $1$ curve for invertible $\mathcal{O}_X$-modules $L$, $M$ of degree $0$ such that $L\not\cong M$.

Because of integrality of $V_0$, the open subset $U$ is "dense" in $\text{Bun}_{2,0}(X)$: the inverse image of $U$ in $V_0$ is a nonempty open in the integral scheme $V_0$. Since the nontrivial extension of $\mathcal{O}_X$ by $\mathcal{O}_X$ gives a geometric point of the stack, for any geometric point of $V_0$ mapping to that point, there is a smooth affine curve in $V_0$ that contains that point and whose intersection with the inverse image of $U$ is dense.

Edit addressing higher genus. For a smooth, projective $k$-curve $Y$ of genus $g\geq 1$ such that there exists a flat $k$-morphism $f:Y\to X$, the pullback by $f$ of the genus $1$ example gives an example on $Y$.

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.

• Very nice! Is there a reference for the construction you explain on $\Bbb{P}^2$?
– abx
Commented Dec 11, 2017 at 14:08
• @abx Thanks. This is culled from the general construction I used in my paper on deformations in char $p>0$, in Inventiones. There is also another construction in that paper for the plane. Commented Dec 11, 2017 at 14:36