Motivation for the generalization of the de Rham isomorphism to twisted vector bundles

185 Views Asked by At

I am reading a paper that makes the following comment:

Let $(V,\nabla^V,h^v)$ be a vector bundle with a flat connection $\nabla^V$ and metric $h^V$. Then we can form the twisted de Rham complex $\Omega(M,V)$. We consider the sheaf $\mathcal{V}$ of parallel sections of $(V,\nabla^V)$. The de Rham isomorphism relates the sheaf cohomology of $\mathcal{V}$ to the cohomology of the twisted de Rham complex: $H^*(M,\mathcal{V}) = H^*(\Omega(M,V))$.

The notation previously established in the paper is $\Omega(M,V) := \bigoplus_k \Omega^k(M,V)$, with $\Omega^k(M,V) := \Gamma(\Lambda^k T^*M\otimes V)$.

I see that this is a generalization of the regular de Rham isomorphism (taking $V$ to be the trivial real line bundle). However, I am unsure why the relevant sheaf here is the sheaf of parallel sections. Is there an intuitive reason why parallel sections of a flat vector bundle are the right objects to be thinking about?

1

There are 1 best solutions below

0
On BEST ANSWER

After discussing with someone, I came to a satisfactory answer. The key fact is that the sheaf of parallel sections is a locally constant sheaf. This can be seen by trivializing $V$ in a neighborhood $U \subset M$. Then the connection determines a horizontal distribution of $V$ over $U$ (i.e. a subspace $H_x \subset V_x$ for each $x \in U$) and a parallel section is a vector valued function on $U$ whose values lie in $H_x$ for each $x$. The stalks at each $x\in U$ are then clearly all the same.

At a higher level, $\mathcal V$ and $(V,\nabla^V)$ each determine each other (see local system). Therefore it is natural to ask if their respective cohomology theories coincide.