Suppose M is an arbitrary smooth manifold and D is its bundle of 1-densities.
On the category of finite-dimensional vector bundles over M and linear differential operators between them
there is a contravariant endofunctor that sends a vector bundle E to E*⊗D
and a differential operator f: E→F to the adjoint differential operator f*: F*⊗D→E*⊗D.
Applying this endofunctor to the standard de Rham (cochain) complex 0→Ω^0(M)→Ω^1(M)→⋯→Ω^n(M)→0
with morphisms being de Rham differentials we obtain another (chain) complex 0←Λ^0(M)⊗D←Λ^1(M)⊗D←⋯←Λ^n(M)⊗D←0
with morphisms being codifferentials.
Here Λ^k(M) denotes the bundle of k-polyvectors (kth exterior power of the tangent bundle).
What is the exact relationship between the homology of this complex and the usual singular (co)homology of M?
Using Hodge duality we can rewrite this complex as
0←Ω^n(M)⊗W←Ω^{n-1}(M)⊗W←⋯←Ω^0(M)⊗W←0,
where W is the orientation bundle.
It looks like the answer should be some standard fact from the 1950s,
therefore any references will be appreciated.
Best Answer
When you dualize the bundle of differential forms and multiply it with the line bundle of top forms, you get differential forms again, and not polyvector fields.