MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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)


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.

share|cite|improve this question
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.