Direct proof^[4]

We shall prove the lemma for an open subset ${\displaystyle U\subset \mathbb {R} ^{n}}$ that is star-shaped or a cone over ${\displaystyle [0,1]}$; i.e., if ${\displaystyle x}$ is in ${\displaystyle U}$, then ${\displaystyle tx}$ is in ${\displaystyle U}$ for ${\displaystyle 0\leq t\leq 1}$. This case in particular covers the open ball case, since an open ball can be assumed to centered at the origin without loss of generality.

The trick is to consider differential forms on ${\displaystyle U\times [0,1]\subset \mathbb {R} ^{n+1}}$ (we use ${\displaystyle t}$ for the coordinate on ${\displaystyle [0,1]}$). First define the operator ${\displaystyle \pi _{*}}$ (called the fiber integration) for k-forms on ${\displaystyle U\times [0,1]}$ by

${\displaystyle \pi _{*}\left(\sum _{i_{1}<\cdots <i_{k-1}}\alpha _{i}dt\wedge dx^{i}+\sum _{j_{1}<\cdots <j_{k}}\beta _{j}dx^{j}\right)=\left(\int _{0}^{1}\alpha _{i}(\cdot ,t)\,dt\right)\,dx^{i}}$

where ${\displaystyle dx^{i}=dx_{i_{1}}\wedge \cdots \wedge dx_{i_{k}}}$, ${\displaystyle \alpha _{i}=\alpha _{i_{1},\dots ,i_{k}}}$ and similarly for ${\displaystyle dx^{j}}$ and ${\displaystyle \beta _{j}}$. Now, for ${\displaystyle \alpha =f\,dt\wedge dx^{i}}$, since ${\displaystyle d\alpha =-\sum _{l}{\frac {\partial f}{\partial x_{l}}}dt\wedge dx_{l}\wedge dx^{i}}$, using the differentiation under the integral sign, we have:

${\displaystyle \pi _{*}(d\alpha )=-d(\pi _{*}\alpha )=\alpha _{1}-\alpha _{0}-d(\pi _{*}\alpha )}$

where ${\displaystyle \alpha _{0},\alpha _{1}}$ denote the restrictions of ${\displaystyle \alpha }$ to the hyperplanes ${\displaystyle t=0,t=1}$ and they are zero since ${\displaystyle dt}$ is zero there. If ${\displaystyle \alpha =f\,dx^{j}}$, then a similar computation gives

${\displaystyle \pi _{*}(d\alpha )=\alpha _{1}-\alpha _{0}-d(\pi _{*}\alpha )}$.

Thus, the above formula holds for any ${\displaystyle k}$-form ${\displaystyle \alpha }$ on ${\displaystyle U\times [0,1]}$. Finally, let ${\displaystyle h(x,t)=tx}$ and then set ${\displaystyle J=\pi _{*}\circ h^{*}}$. Then, with the notation ${\displaystyle h_{t}=h(\cdot ,t)}$, we get: for any ${\displaystyle k}$-form ${\displaystyle \omega }$ on ${\displaystyle U}$,

${\displaystyle h_{1}^{*}\omega -h_{0}^{*}\omega =Jd\omega +dJ\omega ,}$

the formula known as the homotopy formula. The operator ${\displaystyle J}$ is called the homotopy operator (also called a chain homotopy). Now, if ${\displaystyle \omega }$ is closed, ${\displaystyle Jd\omega =0}$. On the other hand, ${\displaystyle h_{1}^{*}\omega =\omega }$ and ${\displaystyle h_{0}^{*}\omega =0}$. Hence,

${\displaystyle \omega =dJ\omega ,}$

which proves the Poincaré lemma.

The same proof in fact shows the Poincaré lemma for any contractible open subset U of a manifold. Indeed, given such a U, we have the homotopy ${\displaystyle h_{t}}$ with ${\displaystyle h_{1}=}$ the identity and ${\displaystyle h_{0}(U)=}$ a point. Approximating such ${\displaystyle h_{t}}$, we can assume ${\displaystyle h_{t}}$ is in fact smooth. The fiber integration ${\displaystyle \pi _{*}}$ is also defined for ${\displaystyle \pi :U\times [0,1]\to U}$. Hence, the same argument goes through.

Proof using Lie derivatives

Cartan's magic formula for Lie derivatives can be used to give a short proof of the Poincaré lemma. The formula states that the Lie derivative along a vector field ${\displaystyle \xi }$ is given as: ^[5]

${\displaystyle L_{\xi }=d\,i(\xi )+i(\xi )d}$

where ${\displaystyle i(\xi )}$ denotes the interior product; i.e., ${\displaystyle i(\xi )\omega =\omega (\xi ,\cdot )}$.

Let ${\displaystyle f_{t}:U\to U}$ be a smooth family of smooth maps for some open subset U of ${\displaystyle \mathbb {R} ^{n}}$ such that ${\displaystyle f_{t}}$ is defined for t in some closed interval I and ${\displaystyle f_{t}}$ is a diffeomorphism for t in the interior of I. Let ${\displaystyle \xi _{t}(x)}$ denote the tangent vectors to the curve ${\displaystyle f_{t}(x)}$; i.e., ${\displaystyle {\frac {d}{dt}}f_{t}(x)=\xi _{t}(f_{t}(x))}$. For a fixed t in the interior of I, let ${\displaystyle g_{s}=f_{t+s}\circ f_{t}^{-1}}$. Then ${\displaystyle g_{0}=\operatorname {id} ,\,{\frac {d}{ds}}g_{s}|_{s=0}=\xi _{t}}$. Thus, by the definition of a Lie derivative,

${\displaystyle (L_{\xi _{t}}\omega )(f_{t}(x))={\frac {d}{ds}}g_{s}^{*}\omega (f_{t}(x))|_{s=0}={\frac {d}{ds}}f_{t+s}^{*}\omega (x)|_{s=0}={\frac {d}{dt}}f_{t}^{*}\omega (x)}$.

That is,

${\displaystyle {\frac {d}{dt}}f_{t}^{*}\omega =f_{t}^{*}L_{\xi _{t}}\omega .}$

Assume ${\displaystyle I=[0,1]}$. Then, integrating both sides of the above and then using Cartan's formula and the differentiation under the integral sign, we get: for ${\displaystyle 0<t_{0}<t_{1}<1}$,

${\displaystyle f_{t_{1}}^{*}\omega -f_{t_{0}}^{*}\omega =d\int _{t_{0}}^{t_{1}}f_{t}^{*}i(\xi _{t})\omega \,dt+\int _{t_{0}}^{t_{1}}f_{t}^{*}i(\xi _{t})d\omega \,dt}$

where the integration means the integration of each coefficient in a differential form. Letting ${\displaystyle t_{0},t_{1}\to 0,1}$, we then have:

${\displaystyle f_{1}^{*}\omega -f_{0}^{*}\omega =dJ\omega +Jd\omega }$

with the notation ${\displaystyle J\omega =\int _{0}^{1}f_{t}^{*}i(\xi _{t})\omega \,dt.}$

Now, assume ${\displaystyle U}$ is an open ball with center ${\displaystyle x_{0}}$; then we can take ${\displaystyle f_{t}(x)=t(x-x_{0})+x_{0}}$. Then the above formula becomes:

${\displaystyle \omega =dJ\omega +Jd\omega }$,

which proves the Poincaré lemma when ${\displaystyle \omega }$ is closed.

A standard proof of the Poincaré lemma uses the homotopy invariance formula and can be found here, in the above section (cf. Integration along fibers#Example), Singer & Thorpe (1976, pp. 128–132) harvtxt error: no target: CITEREFSingerThorpe1976 (help), Lee (2012), Tu (2011) and Bott & Tu (1982).^[6][7][8] The local form of the homotopy operator is described in Edelen (2005) and the connection of the lemma with the Maurer-Cartan form is explained in Sharpe (1997).^[9][10]

Proof in the two-dimensional case

In two dimensions the Poincaré lemma can be proved directly for closed 1-forms and 2-forms as follows.^[11]

If ω = p dx + q dy is a closed 1-form on (a, b) × (c, d), then p_y = q_x. If ω = df then p = f_x and q = f_y. Set

${\displaystyle g(x,y)=\int _{a}^{x}p(t,y)\,dt,}$

so that g_x = p. Then h = f − g must satisfy h_x = 0 and h_y = q − g_y. The right hand side here is independent of x since its partial derivative with respect to x is 0. So

${\displaystyle h(x,y)=\int _{c}^{y}q(a,s)\,ds-g(a,y)=\int _{c}^{y}q(a,s)\,ds,}$

and hence

${\displaystyle f(x,y)=\int _{a}^{x}p(t,y)\,dt+\int _{c}^{y}q(a,s)\,ds.}$

Similarly, if Ω = r dx ∧ dy then Ω = d(a dx + b dy) with b_x − a_y = r. Thus a solution is given by a = 0 and

${\displaystyle b(x,y)=\int _{a}^{x}r(t,y)\,dt.}$