de Rham cohomology vs. iterated tangent bundles?

2011-05-24T08:50:34Z

I have two related questions. Here $M$ is a real smooth manifold, $TM$ is its tangent bundle, $T^n M := T ... TM$ is the $n$-th iterated tangent bundle.

Fiberwise linear smooth functions $TM \to \mathbf R$ are the same as smooth one-forms on $M$. Is there a handy generalization of this to $n$-forms and some functions $T^n M \to \mathbf R$?
Can de Rham cohomology be expressed in terms of (the simplicial abelian group of) functions $T^\bullet M \to \mathbf R$?

Thank you.

Answer by quim for de Rham cohomology vs. iterated tangent bundles?

2011-05-24T13:13:56Z

EDIT: I think an answer to your first question is explained in the papers:

P.-A. Meyer, Qu'est ce qu'une différentielle d'ordre $n$, Exposition. Math. 7 (1989), 249–264.
Laksov, Dan; Thorup, Anders, These are the differentials of order $n$. Trans. Amer. Math. Soc. 351 (1999), no. 4, 1293–1353. Freely available online.

Quote from the second:

The higher order differentials were part of the folklore of mathematics up to the end of the previous century, and formulas like $d^2f=f'_xd^2x+f'_yd^2y+{f''_{x^2}}dx^2+2{f''_{xy}}dxdy+{f''_{y^2}}dy^2$ can be found in most classical calculus books. (...) the higher order differentials vanished (...) because the extensive user of exterior differentials led mathematicians to believe that $d^2$ should always be zero.

(...)

Let $C_n:=\mathcal{C}^\infty(T^nX)$. The differential of $\mathcal{C}^\infty$- functions on $T^nX$ can be viewed as a k-linear map $d:C_n\rightarrow C_{n+1}$, and we obtain a sequence of linear maps...

Then $\Omega^n$ is defined as a suitable submodule of $C_n$, and there is a product $\Omega^p \otimes \Omega^n \rightarrow \Omega^{p+n}$ such that $d(\omega\cdot\pi)=d\omega\cdot\pi + \omega\cdot d\pi$ .

de Rham cohomology vs. iterated tangent bundles? - MathOverflow

de Rham cohomology vs. iterated tangent bundles?

Answer by quim for de Rham cohomology vs. iterated tangent bundles?