3
$\begingroup$

After edit:

How do we show that for every (holomorphic) vector bundle over a curve, is it possible to deform it to another one which is decomposable (into line bundles)?

Before edit:

I am not sure how much obvious or wrong is the following question:

For every (holomorphic) vector bundle over a complex projective variety, is it possible to deform it to another one which is decomposable (into line bundles)? I feel the answer is yes (since obstruction lies in the ext group and you can deform any element of ext group to zero) but I don't know how easy is a precise proof!

Is this true at least over curves?

Improve this question
$\endgroup$
14
  • $\begingroup$ Your idea of deforming extension classes will work if the vector bundle has a filtration with line bundle quotients. Otherwise, I don't think it's possible in general, the most basic obstructions being the Chern classes, which stay constant in families of vector bundles. $\endgroup$
    – Piotr Achinger
    Commented May 29, 2013 at 15:14
  • $\begingroup$ @Piotr: I basically need this over curves. That might simplify the situation a lot. $\endgroup$
    – Mohammad Farajzadeh-Tehrani
    Commented May 29, 2013 at 15:17
  • 1
    $\begingroup$ Yes, it is true over curves. $\endgroup$
    – Angelo
    Commented May 29, 2013 at 15:19
  • 2
    $\begingroup$ In general this is false. However, over curves this should be true, at least for rank $2$ vector bundles. In fact, assume that $E$ is a rank $2$ vector bundle on $X$ and take a sufficiently very ample divisor $L$ such that $E \times L$ is generated by global sections. $\endgroup$
    – Francesco Polizzi
    Commented May 29, 2013 at 15:27
  • 2
    $\begingroup$ Then one has a short exact sequence $0 \to \mathcal{O}_E \to E \otimes L \to M \to $, where $M$ is another line bundle (here one uses the fact that we are on a curve). Now deforming the extension class one shows that $E \otimes L$ can be deformed to $\mathcal{O}_E \oplus M$, hence $E$ can be deformed to $L^{-1} \oplus (M \otimes L^{-1})$. $\endgroup$
    – Francesco Polizzi
    Commented May 29, 2013 at 15:28

2 Answers 2

Reset to default
6
$\begingroup$

To expand my comment above:

This is not possible in general and Chern classes are a possible obstruction. For an easy example, take the cotangent bundle $\Omega$ of $\mathbb{P}^2$. The Euler sequence $$ 0 \to \Omega \to \mathcal{O}(-1)^3 \to \mathcal{O} \to 0 $$ shows that $ch(\Omega) = (1-H)^3 / 1 = 1 - 3H + 3H^2$, where $H$ is the hyperplane class. If it was possible to deform $\Omega$ to $\Omega'$ sitting inside an extension $$ 0 \to \mathcal{O}(a) \to \Omega' \to \mathcal{O}(b) \to 0 $$ then we would have $1 - 3H + 3H^2 = ch(\Omega) = ch(\Omega') = (1 + aH)(1 + bH)$, which is impossible.

On the other hand, if the given vector bundle $E$ has a filtration with line bundle quotients then I think it is possible. Assume $E$ sits in an extension $$ 0\to E_1 \to E \to E_0 \to 0. $$ To deform $E$ to $E_0 \oplus E_1$, as you said we need to "deform the $Ext$ class". To make sense of this, we first need to understand how $Ext$ classes and extensions correspond. Given $\xi\in Ext^1(E_0, E_1)$, the corresponding extension is obtained as a certain pushout. Replacing $\xi$ by $t\cdot \xi$ where $t$ is the coordinate of $\mathbb{A}^1$, we easily do it in a family (or even construct a "universal" family over $X \times Ext^1(E_0, E_1)$).

(to be continued, sorry, have to go)

Improve this answer
$\endgroup$
3
$\begingroup$

Francesco, in comments, shows that any vector bundle on a curve degenerates to a direct sum of a line bundles. (By the way, I observe the convention that "degeneration" means moving towards the special fiber and "deformation" means moving away from it; you are doing degeneration.)

In the comments, the OP asks whether the moduli space of vector bundles on a curve with fixed rank and Chern character is connected. The answer is yes. Actually, there are a lot of subtleties in talking about this moduli space (stability issues), so I'll just directly answer the question about what can be connected to what in families over connected bases.

By Francesco's argument, we can degenerate from any vector bundle to a direct sum of line bundles. We need to show that, if $r=s$ and $\sum_{i=1}^r \deg L_i = \sum_{j=1}^s \deg M_j$, then we can build a path from $\bigoplus_{i=1}^r L_i$ to $\bigoplus_{j=1}^s M_j$.

Step 1 On a curve, for any two ample line bundles $L_1$ and $L_2$, there is a degeneration from $L_1 \oplus L_2$ to $\mathcal{O} \oplus (L_1 \otimes L_2)$.

Proof Let $f_1$ and $f_2$ be sections of $L_1$ and $L_2$ with disjoint zero locus. Then $$0 \to \mathcal{O} \stackrel{\begin{pmatrix} f \\ g \end{pmatrix}}{\longrightarrow} L_1 \oplus L_2 \stackrel{\begin{pmatrix} \otimes g & -f \otimes \end{pmatrix}}{\longrightarrow} L_1 \otimes L_2 \to 0$$ is exact. As in Francesco's argument, this shows we can degenerate from $L_1 \oplus L_2$ to $\mathcal{O} \oplus (L_1 \otimes L_2)$.

Step 2 On a curve, for any $r$ line bundles $L_1$, $L_2$ ..., $L_r$, there is a path connecting $\bigoplus L_i$ to $\mathcal{O}^{\oplus(r-1)} \oplus \bigotimes L_i$

Proof Choose $D$ large enough that $\mathcal{O}(D)$ and $(\bigotimes L_i)(D)$ and all the $L_i(D)$ are ample. Recursively using Step 1 gives a family from $\bigoplus L_i(D)$ to $\mathcal{O}^{\oplus(r-1)} \oplus \bigotimes \left( L_i(D) \right)$. Step 1 also gives a path to this point from $\mathcal{O}(D)^{\oplus (r-1)} \oplus \left( \bigoplus L_i \right)(D)$. So there is a path connecting $\mathcal{O}^{\oplus(r-1)} \oplus \bigotimes \left( L_i(D) \right)$ and $\mathcal{O}(D)^{\oplus (r-1)} \oplus \left( \bigoplus L_i \right)(D)$. Tensor that path with $\mathcal{O}(-D)$ to get a path joining $\bigoplus L_i$ to $\mathcal{O}^{\oplus(r-1)} \oplus \bigotimes L_i$.

So we can join $\bigoplus_{i=1}^r L_i$ to $\mathcal{O}^{\oplus (r-1)} \oplus \bigotimes L_i$ and we can do similarly with the $M$'s. If $\sum \deg L_i = \sum \deg M_i =d$ then, since $\mathrm{Pic}^d(X)$ is connected, we can find a path from $\bigotimes L_i$ to $\bigotimes M_i$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Here's another argument. For any bounded family of vector bundles on $X$ of rank $r$ with fixed determinant $L\vee$, for every ample invertible sheaf $A^\vee$, for all integers $n$ sufficiently positive, every vector bundle $E$ in the family fits into a short exact sequence $$0\to L\otimes A^{\otimes n(r+1)} \to (A^{\otimes n})^{\oplus (r+1)} \to E \to 0.$$ Thus, there is a surjective morphism to the moduli space from an open subset of $\text{Hom}_{\mathcal{O}_X}(L\otimes A^{\otimes n(r+1)},(A^n)^{\oplus(r+1)})$ -- the open set parameterizing injective maps with semistable cokernel. $\endgroup$
    – Jason Starr
    Commented May 29, 2013 at 22:08
  • $\begingroup$ Correction: Replace "semistable" in last sentence of previous comment by "isomorphic to locally free sheaf in the bounded family. For the corresponding set to be open, please assume the original bounded family is versal (which I am sure is a hypothesis anyway). $\endgroup$
    – Jason Starr
    Commented May 30, 2013 at 17:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .