Generalized Poincaré Lemma

Question

I'm reading the proof of an improved version of Poincaré's Lemma on Ana Cannas da Silva's Lectures on Symplectic Geometry, page 40. I am terribly confused. Here's the setup: $U_0$ is a tubular neighborhood of the zero section of the normal bundle $NX$ of a submanifold $X\subseteq M$. She defines $\rho_t\colon U_0 \to U_0$ by $\rho_t(x,v) = (x,tv)$, for $0 \leq t \leq 1$, and then we have a homotopy operator $$Q\omega = \int_0^1 \rho_t^*(\iota_{v_t}\omega)\,{\rm d}t,$$where $v_t$ is, at the point $\rho_t(x,v)$, the vector tangent to the curve $\rho_s(x,v)$ at $s=t$.

This doesn't look correct to me at $t=0$. If $t \neq 0$, then $\rho_t$ is a diffeomorphism with $\rho_t^{-1} = \rho_{1/t}$, since we have $v_t(x,v) = (x, v/t)$, and this does not have any extension for $t=0$. Indeed, $\rho_0(x,v) = (x,0)$ is just a submersion. But she says that if $x \in X$, then $v_t(x,0) = 0$ because $\rho_t(x,0) = (x,0)$ is a constant curve, which I agree. But $v_0$ doesn't seem to be defined away from $X$, since we would have in general $$v_t(x,v) = \frac{{\rm d}}{{\rm d}s}\bigg|_{s=t} \rho_s\rho_{t}^{-1}(x,v),$$and so $\rho_t$ needs to have an inverse. And even if you consider the above integral as an improper integral, why should it converge?

So:

How can the proof go on if $v_t$ is not continuous? How to make sense of this integral, since what happens near $0$ is exactly what we're interested in?

Answer 1

First, a small point: none of the $\rho_t$ with $t < 1$ need to be diffeomorphisms. The issue is that all we know about $U_0$ is that it's convex. So, if $(x,v)\in U_0$ and $t\leq 1$, then $(x,tv)\in U_0$, but there is no reason that $(x,v/t)\in U_0$. In particular, there is no reason $\rho_t$ should be surjective for $t<1$.

Now, a more important point. I believe that we simply have $v_t(x,v) = v$. Believing this for a moment, everything would make perfect sense at the zero section (where $v=0$).

So, why is $v_t(x,v) = v$? Well, it is by definition \begin{align*} x_t(x,v) &= \frac{d}{ds}|_{s=t} \rho_s(x,v)\\ &= \frac{d}{ds}|_{s=t} (x,sv)\\ &= (x,v),\end{align*} where is last line follows simply by noting that the curve lies entirely within a fiber of a vector bundle - that is, we can compute as if we're in a vector space.

Answer 2

If we let $\sigma_t= \rho_t^*(\iota_{v_t}\omega)$, then $$\sigma_t((x,v);u)=\omega(\rho_t(x,v); (0,v), d\rho_t(x,v)u),$$ thus $\sigma_t$ is smooth at zero as well. And this is what matters, not $v_t$. This issue is alluded to in McDuff's book.

One thing to be noticed is that the main objective of this "dance" is to have the form to be zero on the submanifold. If this was not required, then the result that Ana Cannas proves on that page is trivial from homotopy invariance alone, or essentially just proving it again.