8
$\begingroup$

Given a vector bundle $E \to M$ with connection $\nabla$, we get a twisted de Rham sequence using the exterior covariant derivative: $$0 \to \mathcal{E} \xrightarrow{d^\nabla} \Omega^1_M \otimes \mathcal{E} \xrightarrow{d^\nabla} \Omega^2_M \otimes \mathcal{E} \xrightarrow{d^\nabla} \cdots$$ Here, I am using $\mathcal{E}$ to denote the sheaf of smooth sections of $E$. We say that the connection $\nabla$ is flat if $(d^\nabla)^2 = 0$, and in this case we get an actual complex. In this situation, the sheaf $$\mathcal{L} := \mathrm{Ker}\left(\mathcal{E} \xrightarrow{d^\nabla} \Omega^1_M \otimes \mathcal{E} \right) $$ of parallel sections is a local system and we can use this de Rham complex (a soft resolution) to compute its sheaf cohomology and cohomology with compact support.

But what if the connection is not flat? We obviously no longer have a complex, but the differential operators are sheaf homomorphisms, so we still have a sequence of sheaves, and their kernels and images are not necessarily acyclic.

Can we use sheaf cohomology or D-module theory to say anything interesting in this case?

$\endgroup$
6
  • $\begingroup$ If the connection is not flat you do not have a complex. $\endgroup$ Feb 24, 2017 at 1:33
  • $\begingroup$ Yep that's obviously the case (it was a typo at the end of my post) but I am curious what sheaf cohomology can say about the the sheaf of parallel sections and the sequence of operators. $\endgroup$
    – ಠ_ಠ
    Feb 24, 2017 at 1:37
  • $\begingroup$ It seems like even if you had an answer, you would never be incorporating the wedge-product. The fact that we seem to need a sheaf of abelian groups is the part that's bothering me here. $\endgroup$
    – cheyne
    Aug 17, 2017 at 22:39
  • 1
    $\begingroup$ What is your main object of interest, this particular sequence of differential operators or the sheaf of flat sections? At least in the case when its stalks have constant dimension, the sheaf of flat sections will have a soft resolution by differential operators, just not the naive twisted de Rham complex. $\endgroup$ Aug 18, 2017 at 4:44
  • 1
    $\begingroup$ This is probably known to you but one can always induce a morphism of $D$-modules $\mathcal{D}_M \otimes_{\mathcal{O}_M} \mathcal{E} \to \mathcal{D}_M \otimes_{\mathcal{O}_M} \Omega^1_M \otimes \mathcal{E}$ then this complex can be studied with D-module theory. There's actually an equivalence between derived categories of complexes of vector bundles and differential operators with complexes of D-modules when one sets everything up correctly. $\endgroup$ Sep 7, 2017 at 22:57

0

Your Answer

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