4
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ 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. $\endgroup$ – Amitai Yuval Dec 25 '18 at 8:40
  • 1
    $\begingroup$ 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. $\endgroup$ – Amitai Yuval Dec 25 '18 at 8:47
  • $\begingroup$ 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 $M$ $\endgroup$ – user8268 Dec 25 '18 at 9:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.