Cohomology with Coefficients in the sheaf of distributions It just occurred to me that one could form the sheaf of distributions $F$ on any manifold where for an open set $U$ we have $F(U)$ is the algebra of distributions on $U.$ What does cohomology with coefficients in $F$ represent? Is there a good interpretation using differential forms or differential operators?
 A: So, I don't even know if you're still interested in this year-old question or what, but in case anyone else finds it via some web search and wants an answer, I'll post some thoughts I had on the question:
First of all, it's apparent that a distribution on the manifold $M$ restricts to a distribution on the submanifold $U \subset M$; just pull the distribution back along the inclusion. With a little checking one verifies that the sheaf axioms are satisfied and we obtain a sheaf $\mathcal{S}$ of (sets of) distributons on $M$.
The problem is that there's no convenient/natural way to give this the structure of a sheaf of abelian groups. We could use the Whitney sum of bundles and pass from the monoid structure thus determined to the Grothendieck group, but this employs the external direct sum of vector spaces pointwise and doesn't reflect the key fact that distributions are subbundles of $TM$.
If we instead take the inner sum of vector subspaces and attempt to define a sum of bundles by applying the inner sum pointwise, we run into a different problem. We get a commutative monoid out of the collection of vector subspaces of a given space with inner sum as the operation, surely. But, when trying to pass to the Grothendieck group, we find that the group obtained is trivial! Letting $V$ be any vector space and $M_1, M_2, N_1, N_2$ be any vector subspaces, we can set $K = M_1 + M_2 + N_1 + N_2$
and then we have
$$M_1 + N_2 + K = M_2 + N_1 + K.$$
So the Grothendieck group consists of exactly one equivalence class, containing all ordered pairs of vector subspaces of $V$.
Of course, there might still be some way to impose algebraic structure on $\mathcal{S}$, but I have to imagine it would be quite contrived.
