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I've been trying for days to understand the statement, content, and proof of the Poincaré Lemma.

In hindsight, I think the Poincaré Lemma first appeared (secretly) in my first course in multivariable calculus:

  1. If a vector field $P \mathbf i + Q \mathbf j$ on $\mathbf R^2$ is conservative, then we must have $\partial Q/\partial x = \partial P/\partial y$ by the equality of mixed partials.
  2. The interesting question is whether or not the converse holds.
  3. In my first course, we were told that "yes, the converse is true in $\mathbf R^2$."

High-tech translation: Convert $P \mathbf i + Q \mathbf j$ to $P\,dx + Q\,dy$ using the standard Riemannian metric. Then "conservative vector field" becomes "exact 1-form" and "$\partial Q/\partial x = \partial P/\partial y$" becomes "$P\,dx + Q\,dy$ is closed". Thus, the 3 statements above become

  1. $d^2=0$.
  2. Is $H^1(\mathbf R^2)=0$?
  3. Yes, $H^1(\mathbf R^2)=0$.

Some years later, I learned about Theorem 4-11 in Spivak's Calculus on Manifolds, which states that every closed form on a star-shaped open set in $\mathbf R^n$ is exact. I still don't understand the mysterious $I$ operator in his proof. He defines $I:\Omega^k(\mathbf R^n) \to \Omega^{k-1}(\mathbf R^n)$ and does a bunch of manipulation to show that $Id+dI=\text{identity}$. Explicitly, if $\omega=\sum_{i_1<\cdots<i_k}\omega_{i_1,\ldots,i_k}\,dx^{i_1}\cdots dx^{i_k}\in\Omega^k(\mathbf R^n)$, then $$ I\omega(x^1,\ldots,x^n)=\sum_{i_1<\cdots<i_k}\sum_{\alpha=1}^k(-1)^{\alpha-1}\left(\int_0^1t^{k-1}\omega_{i_1,\ldots,i_k}(tx^1,\ldots,tx^n)\,dt\right)x^{i_\alpha}\,dx^{i_1}\cdots\widehat{dx^{i_\alpha}}\cdots dx^{i_k}.$$ The star-shaped hypothesis is crucial for this proof because we need $\omega$ to be defined along line segments from the origin.

More recently, I learned from Bott and Tu (Chapter 1 §4) that the Poincaré Lemma is the statement that $$H^k(\mathbf R^n)=\begin{cases}\mathbf R,&k=0,\\0,&k\neq0.\end{cases}$$ Proof:

  1. Let $\pi:\mathbf R^{n+1}\to\mathbf R^n$ be the projection onto the first $n$ coordinates. Let $s:\mathbf R^n\to\mathbf R^{n+1}$ be the zero section.
  2. $\pi s=1\implies s^*\pi^*=1 \text{ on } \Omega^*(\mathbf R^n) \implies s^*\pi^*=1 \text{ on } H^*(\mathbf R^n)$.
  3. $1-\pi^*s^*=\pm(dK\pm Kd) \text{ on } \Omega^*(\mathbf R^{n+1})\implies \pi^*s^*=1\text{ on }H^*(\mathbf R^{n+1})$.

Again, the main content of the proof is to construct $K:\Omega^k(\mathbf R^{n+1})\to\Omega^{k-1}(\mathbf R^{n+1})$ and check that it's a homotopy operator. Again, I don't know what $K$ is actually doing.

A few days ago, I read here and here that there is yet another formulation. Given a homotopy $H:M\times[0,1]\to N$, define $i_t:M\to M\times[0,1]:x\mapsto(x,t)$ and $f_t=H\circ i_t$. Then we again construct a homotopy operator $I:\Omega^k(M\times [0,1])\to\Omega^{k-1}(M)$ and eventually show that $(H\circ i_0)^*=(H\circ i_1)^*:H^k(N)\to H^k(M)$.

In each manifestation of the Poincaré Lemma, there's some kind of homotopy operator being constructed. But they seem to go between different domains and codomains in each case. There's something slightly different about each formulation and I can't seem to put them all under one big result.

I was flipping through Hatcher earlier (for something else) and came across the prism operator, which he uses to prove the homotopy invariance of homology. The idea struck me as very similar to the proof of the Poincaré Lemma: The longest part of the proof was constructing the homotopy operator and checking that it is in fact a homotopy operator. Both proofs give the same intuitive "feel" that crossing with $[0,1]$ or $\mathbf R$ shouldn't change the homology/cohomology. Is one just the dual of the other? Or am I hallucinating?

Main questions:

  1. How do all the different formulations of the Poincaré Lemma fit under one big result?

  2. Is Hatcher's prism proof the dual of these Poincaré Lemma proofs?

Not-so-important questions that I could start a new thread for:

  1. What is the $I$ operator in Spivak doing? (I know what it does to 1-forms, but for higher forms it looks like voodoo and I don't know how one would come up with that formula.)

  2. What is the $K$ operator in Bott and Tu doing?

  3. How does one weaken the star-shaped hypothesis to just being contractible? (This might just follow from the answers to the first four questions.)

whippedcream
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  • It's because you are combining two statements in one: 1/ $\mathbb{R}^n$ (or more generally a star-shaped domain) is contractible, meaning its identity map is homotopic to a constant map; 2/ homotopic maps induce equal maps on cohomology. You can read the proofs of these two statements separately (especially the second one), it should make things a bit clearer. – Najib Idrissi Jun 23 '15 at 11:39
  • I thought about the same thing when I studied Poincare's lemma from do Carmo's Differential forms and applications. Yes, the proof of Poincare's lemma involves the Prism operator, and they are exactly the same algebraically. – Henry Jul 01 '15 at 14:08
  • The condition of "contractible" can be weakened of course. For instance, we only need certain homology groups to be trivial. See my question http://math.stackexchange.com/questions/1257348/poincar%C3%A9-lemma-on-a-space-with-trivial-homology-group – Henry Jul 01 '15 at 14:09

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