# First proof of Poincaré Lemma

I know that a way of proving Poincare lemma is to use the homotopy invariance and contractibility of the Euclidean space. Is there is a way of doing it directly (without using the contractibility of $\mathbb{R}^n$)? What was the first proof of this statement ? I wish to know all the different ways of proving this lemma. Please provide references. Thanks !

• I know a proof which uses a linear operator $t$ on smooth forms s.t. $dt+td=id$: so, if $\omega$ is closed, then $dt\omega=\omega$ and $\theta:=t\omega$ gives the solution (i.e. proves that $\omega$ is exact). Would you like it in an answer? – Avitus Nov 10 '13 at 21:03
• @Avitus May be you mean $dt+td=i$. I think it would definitely be helpful if you could outline this or give a reference. – user90041 Nov 11 '13 at 2:06
• @Avitus The original reason why I asked this question was that if I expand the given form as linear combination of wedge product of the coordinate 1-forms, I can see that Poincare lemma is a statement about existence of a solution to a set of partial differential equations. If there is another proof of this lemma using the theory of PDEs, I would like to learn that as well. – user90041 Nov 11 '13 at 2:13
• According to Dieudonné's book on the history of algebraic and differential topology, it appears in Volterra, Opere mathematiche vol. I pp. 407-422 for the first time, but "in a different language" (whatever that's ought to mean) and the next appearance is E. Cartan's book Leçons sur le Invariants Intégraux 1922. I don't know these books, but I expect that at least the latter is readable and can easily be turned into a rigorous proof in modern language. – Ben Nov 11 '13 at 22:52

We want to show that on $\mathbb R^n$, all closed forms of degree $p\geq 1$ are exact. To do so we construct a linear operator

$$\alpha:\Omega^p(\mathbb R^n) \rightarrow \Omega^{p-1}(\mathbb R^n)$$

s.t. $$d\alpha+\alpha d=1.$$

Let $\omega$ be a closed $p$-form. Then, for any $x\in\mathbb R^n$ we define

$$(\alpha \omega)(x):=\int_0^1 t^{p-1}i_x\omega(tx)dt,$$

where $i_x$ is the interior product operator. Then (by Cartan's magic formula)

\begin{aligned} ((d\alpha + \alpha d)\omega)(x) & = \int_0^1 t^p\mathcal L_x\omega(tx)dt \\ & =(\text{use chain rule and pull-back definition of Lie derivative}) \\ & = \int_0^1\frac{d}{dt}(t^p\omega(tx))dt=\omega(x) \end{aligned}

and we are done. The diff. form $\theta:=t(\omega)$ is the exact form we need.

edit: For the chain rule step one wants to consider the pull-back $M_t^* \omega$ where $M_t : \mathbb R^n \to \mathbb R^n$ is scalar multiplication by $t$. i.e. $t^p \omega(tx) = M_t^* \omega (x)$.

• I am having trouble following your last chain of equalities. I do not quite see how you "use chain rule and pull-back definition of Lie derivative," nor how you get $t^p$ instead of $t^{p-1}$ in the first equality. – Lucky Feb 4 '18 at 17:41

We describe a linear functional $\alpha : \Omega^p \mathbb R^n \to \Omega^{p-1} \mathbb R^n$. The space alternating $p$-linear functions on $\mathbb R^n$ has dimension $n \choose p$, and you can write the basis as $dx_{i_1} \wedge dx_{i_2} \wedge \cdots \wedge dx_{i_p}$ where $1 \leq i_1 < i_2 < \cdots < i_p \leq n$. If $I = (i_1, i_2, \cdots, i_p)$ is such a multi-index let $dx_I = dx_{i_1} \wedge dx_{i_2} \wedge \cdots \wedge dx_{i_p}$. Let $I_1$ be the collection of multi-indices with $i_1=1$ and let $I_2$ be the collection of multi-indices with $i_1 > 1$.

Given a $p$-form $f dx_I$ with $f: \mathbb R^n \to \mathbb R$ we define $\alpha$ linearly, by $\alpha (f dx_I) = 0$ if $I \in I_2$. $\alpha (fdx_I) = \left(\int_0^{x_1} f dx_1\right) dx_{i_2} \wedge \cdots \wedge dx_{i_p}$ if $I \in I_1$. You can think of $\alpha$ is a type of `total contraction' of the form in the coordinate $x_1$-direction.

It's fairly direct to check that $$d(\alpha(\omega)) + \alpha(d \omega) = \omega - \pi^*(i^* \omega)$$ for every $p$-form $\omega$. Here $i : \mathbb R^{n-1} \to \mathbb R^n$ is the inclusion $i(x_2,\cdots,x_n) = (0,x_2,\cdots,x_n)$ and $\pi : \mathbb R^n \to \mathbb R^{n-1}$ is projection $\pi(x_1,x_2,\cdots,x_n) = (x_2,\cdots,x_n)$.

So this is a less technically-sophisticated argument than Avitus's but it might be a little simpler to follow conceptually. In the end you reduce showing a closed $p$-form on $\mathbb R^n$ is exact to solving the problem for one smaller $n$. Dimension $p=n$ is the base case, where the above formula starts the induction.

• This looks very similar to Bott&Tu's proof! – Bombyx mori May 22 '16 at 5:13

The above calculation in more detail. Define $$h_k:\Omega^p(\mathbb{R}^n)\rightarrow \Omega^{p-1}(\mathbb{R}^n)$$ by, $$\begin{eqnarray} h_k(\omega)(x)=\int_0^1t^{p-1}i_X\omega(tx)dt \end{eqnarray}$$ where $$X=\sum_ix^i\partial_i$$, this generates the one parameter group of diffeomorphisms $$\Phi_s:x^i\mapsto e^sx^i=:y^i$$. To see this note that $$X^i(t)=\frac{dx^i(t)}{dt}=x^i(t)$$, hence we solve this ODE to find the stated solution. Now, $$\begin{eqnarray} ((h_{k+1}\circ d_k+d_{k-1}\circ h_k)\omega)(x)&=&\int_0^1t^{p-1}\Big(d(i_X\omega)+i_X(d\omega)\Big)(tx)dt\\ &=&\int_0^1t^{p-1}(\mathcal{L}_X\omega)(tx)dt \end{eqnarray}$$ Then from the definition of the Lie derivative in terms of the pullback, $$\begin{eqnarray} (\mathcal{L}_X\omega)_x=\Big(\frac{d}{ds}\Big|_{s=0}(\Phi_s)^*\omega\Big)_x \end{eqnarray}$$ Now we can compute the pullback explicitly, $$\begin{eqnarray} [((\Phi_s)^*\omega)_x]_{k_1...k_p}=\frac{\partial y^{i_1}}{\partial x^{k_1}}...\frac{\partial y^{i_p}}{\partial x^{k_p}}[\omega_{\Phi_s(x)}]_{i_1...i_p}=e^{ps}[\omega_{e^sx}]_{k_1...k_p} \end{eqnarray}$$ Due to linearity we can restrict to the simple case where $$\omega(x)=f(x)dx^I$$. Therefore, $$\begin{eqnarray} (\mathcal{L}_X\omega)_x=\Big(\frac{d}{ds}\Big|_{s=0}(\Phi_s)^*\omega\Big)_x=p\omega_x+x^i\frac{\partial f}{\partial x^i}(x)dx^I \end{eqnarray}$$ Therefore, $$\begin{eqnarray} t^{p-1}(\mathcal{L}_X\omega)_{tx}&=&t^{p-1}\Big(p\omega_{tx}+tx^i\frac{\partial f}{\partial x^i}(tx)dx^I\Big)\\ &=&\frac{d}{dt}\Big(t^pf(tx)dx^I\Big)=\frac{d}{dt}(t^p\omega(tx)) \end{eqnarray}$$ Hence, $$\begin{eqnarray} ((h_{k+1}\circ d_k+d_{k-1}\circ h_k)\omega)(x)=t^p\omega(tx)|_{t=0}^1=\omega(x) \end{eqnarray}$$ Therefore since $$\omega$$ is closed, $$d\omega=0$$ so $$\mathcal{L}_X\omega=d(i_X\omega)$$ and hence $$d_{k-1}h_k\omega=\omega$$, i.e. $$\omega$$ is exact.