1

I have a couple of question regarding the following proof of Poincare's lemma in cohomology. The proof is as follows:

We want to prove tht $\int_{\mathbb{R}^{n}} :H_{c}^{n}\left ( \mathbb{R}^{n} \right )\rightarrow \mathbb{R}$ is an isomorphism. So we want to show that for $\int _{\mathbb{R}^{n}}\omega=0 $ for some $\omega \in \Omega _{c}^{n}\left ( \mathbb{R}^{n} \right )$that we must have $\omega \in B _{c}^{n}\left ( \mathbb{R}^{n} \right )$. We will think of $\mathbb{R}^{n}=S^{n}-\left \{ p \right \}$. We denote by $i:\mathbb{R}^{n}\rightarrow S^{n}$ the canonical embedding. Then for some $\omega \in \Omega _{c}^{n}\left ( \mathbb{R}^{n} \right )$ we have that $i_{*}\omega \in \Omega ^{n}\left (S^{n} \right )$. The condition

$ \int _{S^{n}} i^{*}\omega=\int _{\mathbb{R}^{n}}\omega=0$

implies that $i^{*} \omega=d \eta$ for some $\eta \in \Omega ^{n-1}\left ( S^{n} \right )$. We take an open contractible neighborhood U of p in $S^{n}$ so that $\omega$ vanishes on U. Then $\eta$ is closed and exact, since U is contractible. So one can find $\mu \in \Omega^{n-2}\left ( U \right )$ so that $\eta=d \mu$ on U.One can then pick a bump function $\rho$ on $S^{n}$ which vanishes on $S^{n}-U$ and which is identically 1 near p. Then $\eta - d \rho \mu$ is an (n-1) form on $S^{n}$ that vanishes near p, and thus defines a compactly supported (n-1) form on $\mathbb{R}^{n}$, such that

$d\left ( \eta - d\left ( \rho \mu \right ) \right )=d \eta=\omega$

My first question is: I get that $d\left ( \eta - d\left ( \rho \mu \right ) \right )=i^{*}\omega$ is compactly supported on $S^{n}$. But how can we also have that $d\left ( \eta - d\left ( \rho \mu \right ) \right )=\omega$ ? What I mean is that $S^{n}$ and $\mathbb{R}^{n}$ are two different spaces and there is a mapping, i, between them. I would expect to see a mapping of the form $d\left ( \eta - d\left ( \rho \mu \right ) \right )$ from $S^{n}$ coordinates into $\mathbb{R}^{n}$ coordinates. Is the lack of such a mapping just a convenient abuse of the notation?

My second question is: why do we need to construct $d\left ( \eta - d\left ( \rho \mu \right ) \right )=d \eta=i^{*}\omega$ at all, when we already know near the start of the proof that $i^{*}\omega=d \eta$ ?

Ashwin Iyengar
  • 1,855
Rory Cornish
  • 263
  • You may want to remove the notational ambiguity. It says that $\eta - d(\rho\mu)$ defines a compactly supported $(n-1)$-form on $\mathbb R^n$. Make this statement precise and show that its differential is actually $\omega$ (and not $i_\omega$ or $i^\omega$; which one do you actually mean?). Notice that $\eta$ does not necessarily define a compactly supported $(n-1)$-form on $\mathbb R^n$. – Claudius Apr 03 '17 at 12:56
  • Hi. Thanks for your advice. You have basically picked up on the point I was trying to raise, and the source of my confusion. The above is a "proof" that I found in some lecture notes. My confusion is that near the start of the proof it says : $i^{*} \omega=d \eta$, but at the end of the proof it says $d\left ( \eta - d\left ( \rho \mu \right ) \right )=d \eta= \omega$, and I don't see that the latter is possible given the former. I think it may be simply an abuse of notation. I think the abuse is because I think $\mathbb{R}^{n}$ is diffeomorphic to $S^{n}- {p}$, is that right? – Rory Cornish Apr 04 '17 at 12:44
  • My confusion, and I suspect your confusion is that on the one hand the proof says $d \eta$ belongs to $S^{n}- {p}$, but on the other hand it says that $d \eta$ belongs to $\mathbb{R}^{n}$. – Rory Cornish Apr 04 '17 at 12:48
  • I think I have got it now. As Ashwin pointed out $\eta$ need not be compact in $S^{n}$. But by construction, $d\left ( \eta -d\left ( \rho \mu \right ) \right )$ is zero near p, and so must be compactly supported in $S^{n}$. I think that is why the construction is needed. And I am almost certain the proof IS using abuse of notation as it is thinking of $S^{n}$ as essentially the same thing as $\mathbb{R}^{n}$ and $\omega$; as essentially the same thing as $i^{*}\omega$ due the fact that these two spaces are related by diffeomorphism. – Rory Cornish Apr 04 '17 at 14:07
  • I made a little mistake above. I meant to say that $\eta -d\left ( \rho \mu \right ) $ is compactly supported. Also that $d\left ( \eta -d\left ( \rho \mu \right ) \right )=d\left ( \eta \right )+d \circ d \left ( \rho \mu \right )=d\left ( \eta \right )+0=d\left ( \eta \right )=i^{}\omega$. And so $i^{}\omega\epsilon B_{c}^{n}\left ( S^{n} \right )$. And thus $\omega\epsilon B_{c}^{n}\left ( \mathbb{R}^{n} \right )$ – Rory Cornish Apr 04 '17 at 14:29

0 Answers0