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Dmitri Pavlov
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Suppose $M$M is an arbitrary smooth manifold and $D$D is its bundle of $1$1-densities. On the category of finite-dimensional vector bundles over $M$M and linear differential operators between them. There there is a contravariant endofunctor that sends a vector bundle $E$ to $E*⊗D$ and a differential operator $f: E→F$E to the adjoint differential operator $f*: F*⊗D→E*⊗D$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 $$0→Ω^0(M)→Ω^1(M)→⋯→Ω^n(M)→0$$0→Ω^0(M)→Ω^1(M)→⋯→Ω^n(M)→0 with morphisms being de Rham differentials we obtain another (chain) complex $$0←Λ^0(M)⊗D←Λ^1(M)⊗D←⋯←Λ^n(M)⊗D←0$$0←Λ^0(M)⊗D←Λ^1(M)⊗D←⋯←Λ^n(M)⊗D←0 with morphisms being codifferentials. Here $Λ^k(M)$Λ^k(M) denotes the bundle of $k$k-polyvectors ($k$thkth exterior power of the tangent bundle).

What is the exact relationship between the homology of this complex and the usual singular (co)homology of $M$M?

Using Hodge duality we can rewrite this complex as $$0←Ω^n(M)⊗W←Ω^{n-1}(M)⊗W←⋯←Ω^0(M)⊗W←0$$0←Ω^n(M)⊗W←Ω^{n-1}(M)⊗W←⋯←Ω^0(M)⊗W←0, where $W$W is the orientation bundle.

It looks like the answer should be some standard fact from the 1950s, therefore any references will be appreciated.

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 ($k$th 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.

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.

Suppose M$M$ is an arbitrary smooth manifold and D$D$ is its bundle of 1$1$-densities. On the category of finite-dimensional vector bundles over M$M$ and linear differential operators between them there. There is a contravariant endofunctor that sends a vector bundle E$E$ to E⊗D and a differential operator f: E→F to the adjoint differential operator f: F⊗D→E⊗D$E*⊗D$ 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$$0→Ω^0(M)→Ω^1(M)→⋯→Ω^n(M)→0$$ 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$$0←Λ^0(M)⊗D←Λ^1(M)⊗D←⋯←Λ^n(M)⊗D←0$$ with morphisms being codifferentials. Here Λ^k(M)$Λ^k(M)$ denotes the bundle of k$k$-polyvectors (kth$k$th 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$M$?

Using Hodge duality we can rewrite this complex as 0←Ω^n(M)⊗W←Ω^{n-1}(M)⊗W←⋯←Ω^0(M)⊗W←0,$$0←Ω^n(M)⊗W←Ω^{n-1}(M)⊗W←⋯←Ω^0(M)⊗W←0$$ where W$W$ is the orientation bundle.

It looks like the answer should be some standard fact from the 1950s, therefore any references will be appreciated.

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.

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 ($k$th 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.

deleted 83 characters in body
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Dmitri Pavlov
  • 37.8k
  • 4
  • 97
  • 183

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^n0←Ω^n(M)⊗W→Ω^⊗W←Ω^{n-1}(M)⊗W→⋯→Ω^0⊗W←⋯←Ω^0(M)⊗W→0⊗W←0, where W is the orientation bundle. Perhaps this is nothing else but the Poincaré duality expressed in this language?

It looks like the answer should be some standard fact from the 1950s, therefore any references will be appreciated.

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. Perhaps this is nothing else but the Poincaré duality expressed in this language?

It looks like the answer should be some standard fact from the 1950s, therefore any references will be appreciated.

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.

added 214 characters in body
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Dmitri Pavlov
  • 37.8k
  • 4
  • 97
  • 183
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Source Link
Dmitri Pavlov
  • 37.8k
  • 4
  • 97
  • 183
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