# Differential Calculus and the De Rham Homotopy Operator

Suppose $M$ is a smooth manifold, $\Omega^k(M)$ the vector space (or $C^\infty(M)$-module in case this is better suited) of $k$-differentialforms on $M$ and

$$I: \Omega^k(M\times \mathbb{R}) \to \Omega^{k-1}(M)$$ the De Rham homotopy operator defined by $I(\omega):= \int_0^1i_{\partial_t}\omega(t)dt$. (Where $t$ is assumed to be the global coordinate of $\mathbb{R}$ and $\partial_t$ its appropriate tangent coordinate)

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I would like to know, how $I$ interacts with the operators of differential calculus:

First we have the well known equation: $$d\circ I + I \circ d = i_1^* − i_0^*$$

where $i_j^*$ is the pullback defined by the inclusion $i_j : M \to M \times \mathbb{R}; x \mapsto (x,j)$. This is the 'interaction' of $I$ with the exterior differential $d$.

1.) How does $I$ interacts with the interior product $i_X: \Omega^k(M \times \mathbb{R}) \to \Omega^{k-1}(M \times \mathbb{R})$ for a vector field $X$ on $M$ (or $M \times \mathbb{R}$) or (via inclusion) with $i_X: \Omega^k(M) \to \Omega^{k-1}(M)$ ?

2.) How does $I$ interact with the exterior product?

Edit: If someone knows a different definition of a homotopy operator, with a known behavior related to the interior product $i_X$, I would like to know it, too.

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I think (1.) comes down to the question how integration of differential forms interact with the interior product. Is there something like a 'partial integration' rule for the integral and the interior product? Like $i_X \int \omega =i_X \omega − \int i_X \omega$? (... Just a thought ...) –  Mark.Neuhaus Aug 19 '12 at 21:56
Bott and Tu do a bunch of calculations with the homotopy operator in the first chapter of their book, "Differential Forms in Algebraic Topology". Have you looked there? –  Paul Siegel Aug 20 '12 at 14:49
No. But I will. Thanks –  Mark.Neuhaus Aug 20 '12 at 15:43
Consider the case where $M$ is just $\mathbb{R}^n$ and just do the computation, and then you will see the answers to your questions. –  Dan Lee Aug 20 '12 at 19:36

If $X$ is a vector field on $M$ (time independent) then $i_X \circ I = - I \circ i_X$. If $X$ is also time dependent you can play with the cases $\omega$ exact etc.