# Understanding the homotopy operator for de Rham Cohomology

This is in John Lee's Smooth Manifold 2nd Edition, pg 444

For any smooth manifold $$M$$, there exists a linear map $$h:\Omega^p(M \times I ) \rightarrow \Omega^{p-1}(M)$$ such that $$h(dw)+d(hw) = i_1^*w - i^*_0w$$

The proof begins goes as follows. This is the same one here.

Let $$s$$ be the standard coordinate on $$\Bbb R$$ and let $$S$$ be the vector field on $$M \times \Bbb R$$ given by $$S_{(q,s)} = (0, \partial/\partial s|_s)$$. Let $$w$$ be a smooth $$p$$-form on $$M \times I$$. Define $$hw \in \Omega^{p-1}(M)$$ by $$hw = \int_0^1 i^*_t (S \lrcorner w) \, dt.$$ Specifically, given $$q \in M$$, this means, $$(hw)_q = \int_0^1 i^*_t\Big( (S \lrcorner w )_{(q,t)} \Big) \, dt$$

The notation $$S \lrcorner w$$ is interior multiplication. Otherwise denoted by $$\iota_Sw$$.

My question is, what does this integral even mean?

At each point $$q \in M$$, the integrand is a function of $$t$$ with values in the vector space $$\wedge^{p-1} (T^*q M)$$. How do we integrate over this?

My thoughts on the way to see this:

We work locally, in nhood $$U_q$$. For $$r \in U_q$$ our integrand is given by $$\sum a_I(r,t) dx^I$$ summing over indices $$I$$ increasing $$p-1$$ length subset of $$0$$ to $$n$$. Our new $$p-1$$ form is given by $$\sum \int a_I(r,t) \, dt \, d x^I$$

There is a problem however, that we have to show this is independent of the choice of coordinate.

• So the general fact is the following. Let $V$ be a (topological) vector space (one may need to specify some requirements, but finite-dimensional is certainly more than enough). Then for a path $\gamma:[0,1]\to V$ one can define the integral $\int_0^1\gamma(t)dt$, which is an element of $V$. This definition of integral has all the desired properties one could think of. – Amitai Yuval Dec 25 '18 at 8:40
• I guess you can imitate the definition of Riemann integrals of real-valued functions. Anyway, when you have convinced yourself that this integral is intrinsically well-defined, a rather comfortable way to compute it is in coordinates, as you write in your post. The above reasoning should help you rest assured that switching to a different basis does not change the integral. – Amitai Yuval Dec 25 '18 at 8:47
• You can also define $h$ via $\int_M h(\alpha)\wedge\beta=\int_{M\times I} \alpha\wedge p^*\beta$ where $p:M\times I\to M$ is the projection and $\beta$ is any compactly supported form on $M$ – user8268 Dec 25 '18 at 9:08