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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.

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  • $\begingroup$ Sergei Merkulov and his grad students uses operadic techniques and A-infinity ideas to study polyvector fields. Maybe his work can get you closer to what you seek. $\endgroup$
    – Mikael Vejdemo-Johansson
    Commented Feb 28, 2010 at 7:40

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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.

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    $\begingroup$ I never claimed that I get polyvector fields. My statement was about polyvector fields twisted by D. Certainly, applying Hodge duality we can map this bundle isomorphically to the bundle of differential forms of complementary degree, but they still have to be twisted by the orientation bundle, so that the complex now looks like this: 0→Ω^n(M)⊗W→Ω^{n-1}(M)⊗W→⋯→Ω^0(M)⊗W→0, where W is the orientation bundle. Perhaps this is nothing else but the Poincaré duality expressed in this language. Either way of writing down this complex is fine with me. $\endgroup$
    – Dmitri Pavlov
    Commented Feb 28, 2010 at 8:53
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    $\begingroup$ No, the differential will be raising the degrees of the forms twisted with orientations, not lowering them. What you get in this way is simply the de Rham complex with coefficients in the orientations (notice that the bundle of orientations has a natural flat connection). So your `de Rham homology' are simply the cohomology of the orientation local system (which are of course isomorphic to the Borel-Moore homology with constant coefficients, by Poincare duality). $\endgroup$
    – Leonid Positselski
    Commented Feb 28, 2010 at 9:12
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    $\begingroup$ Oops, I accidentally reversed the directions of arrows. With this correction everything becomes trivial. $\endgroup$
    – Dmitri Pavlov
    Commented Feb 28, 2010 at 9:45

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