4
$\begingroup$

Let $X_{t}$ be a family of compact complex manifolds over the disk $\mathbb{D}\subset \mathbb{C}.$ Formally, $X_{t}$ is the fiber over $t\in\mathbb{D}$ of a proper, holomorphic submersion of a complex manifold $\mathcal{X}$ onto $\mathbb{D}.$ Suppose $E_{t}$ is a family of holomorphic vector bundles over $X_{t}$ equipped with a family of holomorphic flat connections $\nabla^{t}.$

If $\mathbb{V}_{t}$ denotes the local system prescribed by $\nabla^{t},$ we have the inclusion of sheaves $0\rightarrow \mathbb{V}_{t}\rightarrow \mathcal{E}_{t}$ where $\mathcal{E}_{t}$ is the sheaf of germs of holomorphic sections of $E_{t}.$

Suppose that for all $t\neq 0$ in the disk, the induced map $H^{1}(X_{t}, \mathbb{V}_{t})\rightarrow H^{1}(X_{t}, \mathcal{E}_{t})$ is an isomorphism.

Question: Does it follow that the induced map over zero $H^{1}(X_{0}, \mathbb{V}_{0})\rightarrow H^{1}(X_{0}, \mathcal{E}_{0})$ is an isomorphism?

If it makes life any better, I am happy to assume that the dimension of $H^{1}(X_{t}, \mathbb{V}_{t})$ is constant over all of $\mathbb{D}.$

I suspect the answer is no, but haven't unearthed a counterexample as of yet. Thank you for any insight you can provide.

$\endgroup$
2
  • $\begingroup$ One can come up with counterexamples where you actually have an isomonodromic deformation, but that seems to be more than what you are asking. We can take X_t to be an elliptic curve (independent of t), and E_t to be a line bundle. Then we have the following cases: 1. E_t is not trivial: both cohomology groups are zero; 2. E_t is trivial, but connection is not: first cohomology group is zero, second is one-dimensional 3. E_t and connection are both trivial: first cohomology group is two-dimensional, second is one-dimensional. So your map is an isomorphism in case 1, but not in cases 2 and 3... $\endgroup$
    – t3suji
    May 18, 2017 at 18:31
  • $\begingroup$ so can just take a family of line bundles on your elliptic curve that is trivial for t=0 and non-trivial for t\ne 0. $\endgroup$
    – t3suji
    May 18, 2017 at 18:35

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.