de Rham cohomology vs. iterated tangent bundles? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T18:13:05Z http://mathoverflow.net/feeds/question/65829 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/65829/de-rham-cohomology-vs-iterated-tangent-bundles de Rham cohomology vs. iterated tangent bundles? jakob 2011-05-24T08:50:34Z 2011-05-25T10:48:06Z <p>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.</p> <ol> <li><p>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 $T^n M \to \mathbf R$? </p></li> <li><p>Can de Rham cohomology be expressed in terms of (the simplicial abelian group of) functions $T^\bullet M \to \mathbf R$?</p></li> </ol> <p>Thank you. </p> http://mathoverflow.net/questions/65829/de-rham-cohomology-vs-iterated-tangent-bundles/65847#65847 Answer by quim for de Rham cohomology vs. iterated tangent bundles? quim 2011-05-24T13:13:56Z 2011-05-24T15:32:28Z <p><strong>EDIT</strong>: I think an answer to your <strong>first</strong> question is explained in the papers:</p> <ul> <li>P.-A. Meyer, <em>Qu'est ce qu'une différentielle d'ordre $n$</em>, Exposition. Math. 7 (1989), 249–264.</li> <li>Laksov, Dan; Thorup, Anders, <em>These are the differentials of order $n$</em>. Trans. Amer. Math. Soc. 351 (1999), no. 4, 1293–1353. Freely available <a href="http://www.ams.org/journals/tran/1999-351-04/S0002-9947-99-02120-0/S0002-9947-99-02120-0.pdf" rel="nofollow">online</a>.</li> </ul> <p>Quote from the second:</p> <blockquote> <p>The higher order differentials were part of the <em>folklore</em> of mathematics up to the end of the previous century, and formulas like <code>$d^2f=f'_xd^2x+f'_yd^2y+{f''_{x^2}}dx^2+2{f''_{xy}}dxdy+{f''_{y^2}}dy^2$</code> 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.</p> <p>(...)</p> <p>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...</p> </blockquote> <p>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 <code>$d(\omega\cdot\pi)=d\omega\cdot\pi + \omega\cdot d\pi$</code>.</p>