# The Poincare Lemma

The Poincaré lemma is a generalization of the familiar fact that a curl-free vector field in ${\mathbb{R}^{3}}$ may be expressed as thee gradient of a function. In these notes we formulate and prove the Poincare lemma.

Closed and exact forms

A ${C^{1}}$ ${j}$-form ${\omega}$ on ${\mathbb{R}^{n}}$ is called exact if ${\omega=d\nu}$ for some ${C^{1}}$ ${j-1}$ form on ${\mathbb{R}^{n}}$ and is closed if ${d\omega=0}$. Since ${d(d\nu)=0}$ for any ${j-1}$ form, as we proved in a previous lecture. The Poincaré lemma asserts that the converse is true:

Thm 1 (Poincaré Lemma) Let ${\omega}$ be a closed ${C^{1}}$ ${j}$-form on ${\mathbb{R}^{n}}$ with ${j\ge1}$. Then ${\omega}$ is exact, i.e., there is a ${C^{1}}$ ${j-1}$-form ${\nu}$ such that ${\omega=d\nu}$ .

Stokes Theorem on parallelpipeds

In proving the Poincaré Lemma we will use Stokes theorem for integration over parallelpipeds in ${\mathbb{R}^{n}}$. Since this is technically different than the integration over oriented surfaces that we considered before, let us start by formulating the precise result here.

The unit ${j}$-cube is the set ${Q_{j}=[0,1]^{j}}$. A parameterized ${j}$-parallelpiped is the restriction to ${Q_{j}}$ of a ${C^{1}}$ map ${T:U\rightarrow\mathbb{R}^{n}}$ where ${U}$ is an open neighborhood of ${Q_{j}}$ in ${\mathbb{R}^{j}}$. We will use the notation ${T}$ to denote the parallelpiped as well as the ${C^{1}}$ map.

The boundary of the cube ${Q_{j}}$ is covered by ${2^{j}}$ parameterized ${j-1}$-parallelpipeds

$\displaystyle J_{\alpha}^{+}\left(s_{1},\ldots,s_{j-1}\right)=\left(s_{1},\ldots,s_{\alpha-1},1,s_{\alpha},\ldots,s_{j-1}\right),$

and

$\displaystyle J_{\alpha}^{-}\left(s_{1},\ldots,s_{j-1}\right)=\left(s_{1},\ldots,s_{\alpha-1},0,s_{\alpha},\ldots,s_{j-1}\right)$

for ${\alpha=1,\ldots,j}$. However, these maps are not oriented consistently. The correct definition of the oriented boundary of a parallelpiped is suggested by the following theorem

Thm 2 (Stoke’s Theorem for parallelpipeds) Let ${T}$ be a parameterized ${j}$-parallelpiped and let ${\omega}$ be a ${C^{1}}$ ${j-1}$ form defined on a neighborhood of ${T(Q_{j})}$. Then

$\displaystyle \int_{T}d\omega=\sum_{\alpha=1}^{j}\left(-1\right)^{\alpha-1}\left[\int_{T_{\alpha}^{+}}\omega-\int_{T_{\alpha}^{-}}\omega\right] \ \ \ \ \ (1)$

where ${T_{\alpha}^{+}=T\circ J_{\alpha}^{+}}$ and ${T_{\alpha}^{-}=T\circ J_{\alpha}^{-}}$.

To formulate the boundary of a parallelpiped ${T}$ we introduce the notion of a chain of parallelpipeds. A chain of ${j}$– parallelpipeds is a formal sum

$\displaystyle \mathcal{C}=\sum_{\alpha=1}^{N}m_{\alpha}S_{\alpha}$

where ${m_{\alpha}\in\mathbb{Z}}$ and ${S_{\alpha}}$ are ${j}$-parallelpipeds. The summation and multiplication by integers are not the usual summation of maps, instead the notation is a formal sum intended to indicate that

$\displaystyle \int_{\mathcal{C}}\omega=\sum_{\alpha=1}^{N}m_{\alpha}\int_{S_{\alpha}}\omega$

for any ${j}$-form ${\omega}$. Based on \vpageref{eq:stokesparallelpiped} we define the oriented boundary of ${T}$ to be the following chain of ${j-1}$-parallelpipeds

$\displaystyle \partial T=\sum_{\alpha=1}^{j}(-1)^{\alpha-1}\left(T_{\alpha}^{+}-T_{\alpha}^{-}\right).$

Proof: (Of Theorem 2) First suppose ${\omega=f(\mathbf{t})d\mathbf{t}_{\{\beta\}^{c}}}$ is a ${j-1}$ form defined on a neighborhood of ${Q_{j}}$ in ${\mathbb{R}^{j}}$, where

$\displaystyle d\mathbf{t}_{\left\{ \alpha\right\} ^{c}}=dt_{1}\wedge\cdots\wedge dt_{\beta-1}\wedge dt_{\beta+1}\wedge\cdots\wedge dt_{j}.$

Then

$\displaystyle d\omega=\left(-1\right)^{\beta-1}\frac{\partial f}{\partial t_{\beta}}dt_{1}\wedge\cdots\wedge dt_{j},$

and

$\displaystyle \begin{array}{rcl} \int_{Q_{j}}d\omega & = & \left(-1\right)^{\beta-1}\int_{0}^{1}\cdots\int_{0}^{1}\frac{\partial f}{\partial t_{\beta}}(t_{1},\ldots,t_{j})dt_{1}\cdots dt_{j}\\ & = & \left(-1\right)^{\beta-1}\int_{0}^{1}\cdots\int_{0}^{1}\left[f(t_{1},\ldots,t_{\beta-1},1,t_{\beta+1},\ldots,t_{j})\right.\\ & & \qquad\qquad\qquad\qquad\left.-f(t_{1},\ldots,t_{\beta-1},0,t_{\beta+1},\ldots,t_{j})\right]dt_{1}\cdots dt_{\beta-1}dt_{\beta+1}\cdots dt_{j}\\ & = & \left(-1\right)^{\beta-1}\left[\int_{J_{\beta}^{+}}\omega-\int_{J_{\beta}^{-}}\omega\right] \end{array}$

by the fundamental theorem of calculus. Since ${\int_{J_{\alpha}^{\pm}}\omega=0}$ for ${\alpha\neq\beta}$ we see that (1) holds for ${\omega}$.

The identity now follows for a general ${j-1}$ form on ${\mathbb{R}^{j}}$ by linearity and for form defined in a neighborhood of a ${j}$-parallelpiped in ${\mathbb{R}^{n}}$ by pulling the integrals back to ${\mathbb{R}^{j}}$. $\Box$

In the proof of the Poincaré lemma we will use affine parallelpipeds. Given ${\mathbf{a},\xi_{1},\ldots,\xi_{j}\in\mathbb{R}^{n}}$, we define the affine parallelpiped ${P}$ at ${\mathbf{a}}$ with sides ${\xi_{1},\ldots,\xi_{j}}$ to be the parallelpiped given by the map

$\displaystyle (t_{1},\ldots,t_{j})\mapsto\mathbf{a}+\sum_{\alpha=1}^{j}t_{\alpha}\mathbf{\xi}_{\alpha},$

and denote the parallelpiped by

$\displaystyle P=\left[\mathbf{a};\xi_{1},\ldots,\xi_{j}\right].$

The oriented boundary of an affine parallelpiped ${P}$ is a chain of affine parallelpipeds:

$\displaystyle \partial P=\sum_{\alpha=1}^{j}\left(-1\right)^{\alpha-1}\left(P_{\alpha}^{+}-P_{\alpha}^{-}\right),$

where

$\displaystyle P_{\alpha}^{+}=\left[\mathbf{a+\xi_{\alpha}};\xi_{1},\ldots,\xi_{\alpha-1},\xi_{\alpha+1},\ldots,\xi_{j}\right]$

and

$\displaystyle P_{\alpha}^{-}=\left[\mathbf{a};\xi_{1},\ldots,\xi_{\alpha-1},\xi_{\alpha+1},\ldots,\xi_{j}\right].$

Differentiation of forms

So far we have treated exterior differentiation of forms in a formal way. In the proof of the Poincaré Lemma we will need the following Lemma which shows that exterior differentiation can be expressed in terms of difference quotients:

Lemma 3 Let ${\omega}$ be a ${C^{1}}$ ${j}$-form defined on an open subset ${\mathbb{R}^{n}}$. Then

$\displaystyle \begin{array}{rcl} d\omega\left(\mathbf{x}\right)\left[\xi_{1},\ldots,\xi_{j+1}\right] & = & \lim_{\epsilon\rightarrow0}\frac{1}{\epsilon}\sum_{\alpha=1}^{j+1}\left(-1\right)^{\alpha-1}\left(\omega\left(\mathbf{x}+\epsilon\xi_{\alpha}\right)\left[\xi_{1},\ldots,\not\xi_{\alpha},\ldots,\xi_{j+1}\right]\right.\\ & & \qquad\qquad\qquad\qquad\qquad-\left.\omega\left(\mathbf{x}\right)\left[\xi_{1},\ldots,\not\xi_{\alpha},\ldots,\xi_{j+1}\right]\right). \end{array}$

Proof: By linearity it suffices to consider ${\omega=fdx_{S}}$ where ${dx_{S}=dx_{\beta_{1}}\wedge\cdots\wedge dx_{\beta_{j}}}$ is an elementary ${j}$-form and ${f\in C^{1}}$. Then

$\displaystyle \begin{array}{rcl} \mbox{R.H.S. } & = & \lim_{\epsilon\rightarrow0}\sum_{\alpha=1}^{j+1}\left(-1\right)^{\alpha-1}\frac{f\left(\mathbf{x}+\epsilon\xi_{\alpha}\right)-f\left(\mathbf{x}\right)}{\epsilon}dx_{S}\left[\xi_{1},\ldots,\not\xi_{\alpha},\ldots,\xi_{j+1}\right]\\ & = & \sum_{\alpha=1}^{j+1}\left(-1\right)^{\alpha-1}df\left(\mathbf{x}\right)\left[\xi_{\alpha}\right]dx_{S}\left[\xi_{1},\ldots,\not\xi_{\alpha},\ldots,\xi_{j+1}\right]\\ & = & df\left(\mathbf{x}\right)\wedge dx_{S}\left[\xi_{1},\ldots,\xi_{j+1}\right]. \end{array}$

$\Box$

Proof of the Poincaré Lemma

To fix ideas, first consider the case ${j=1}$. Suppose we are given a closed ${1}$-form ${\omega}$. We must construct a function ${\alpha}$ such that ${\omega=d\alpha}$. This is done by integrating ${\omega}$:

$\displaystyle \nu\left(\mathbf{x}\right)=\int_{[\mathbf{0;}\mathbf{x}]}\omega$

where ${[\mathbf{0};\mathbf{x}]}$ is the line segment from ${0}$ to ${\mathbf{x}}$ (which is the affine ${1}$-parallelpiped at ${0}$ with side ${\mathbf{x}}$). Explicitly ${\nu\left(\mathbf{x}\right)=\int_{0}^{1}\omega(t\mathbf{x})[\mathbf{x}]}$. Now consider the difference ${\alpha\left(\mathbf{x}+\mathbf{h}\right)-\alpha\left(\mathbf{x}\right)}$. The key observation is that, by Stoke’s Theorem,

$\displaystyle \nu\left(\mathbf{x}+\mathbf{h}\right)-\nu\left(\mathbf{x}\right)=\int_{[\mathbf{x};\mathbf{h}]}\omega. \ \ \ \ \ (2)$

Before we verify this identity, let us see how it implies that ${\omega=d\alpha}$. Replacing ${\mathbf{h}}$ by ${\epsilon\mathbf{h}}$ we see that

$\displaystyle \frac{\nu\left(\mathbf{x}+\epsilon\mathbf{h}\right)-\nu\left(\mathbf{x}\right)}{\epsilon}=\frac{1}{\epsilon}\int_{0}^{1}\omega\left(\mathbf{x}+\epsilon t\mathbf{h}\right)\left[\epsilon\mathbf{h}\right]dt=\int_{0}^{1}\omega\left(\mathbf{x}+\epsilon t\mathbf{h}\right)\left[\mathbf{h}\right]dt,$

since ${\omega\left(\mathbf{y}\right)\left[\mathbf{\epsilon}\mathbf{h}\right]=\epsilon\omega\left(\mathbf{y}\right)\left[\mathbf{h}\right]}$. Since ${\omega}$ is continuous it now follows, on taking ${\epsilon}$ to ${0}$, that

$\displaystyle \lim_{\epsilon\rightarrow0}\frac{\nu\left(\mathbf{x}+\epsilon\mathbf{h}\right)-\nu\left(\mathbf{x}\right)}{\epsilon}=\omega\left(\mathbf{x}\right)\left[\mathbf{h}\right].$

Thus the directional derivatives of ${\alpha}$ exist and agree with ${\omega}$. It follows that ${\omega=d\alpha}$.

To prove (2), consider the triangle in ${\mathbb{R}^{n}}$ bounded by the line segments ${[\mathbf{0};\mathbf{x}]}$, ${[\mathbf{0};\mathbf{x}+\mathbf{h}]}$, ${\left[\mathbf{x};\mathbf{h}\right]}$. (Draw a picture!) This triangle can be expressed as the image of the following parameterized ${2}$-parallelpiped,

$\displaystyle T(t,s)=t\left(\mathbf{x}+s\mathbf{h}\right).$

The boundary of ${T}$ is the chain

$\displaystyle \partial T=\left[\mathbf{x};\mathbf{h}\right]-\left[\mathbf{0};\mathbf{0}\right]-\left[\mathbf{0};\mathbf{x}+\mathbf{h}\right]+\left[\mathbf{0};\mathbf{x}\right].$

Note that the boundary component ${\left[\mathbf{0};\mathbf{0}\right]}$ is singular; it’s image is the single point ${\mathbf{0}}$ and ${\int_{\left[\mathbf{0};\mathbf{0}\right]}\eta=0}$ for any ${1}$-form ${\eta}$. Since ${d\omega=0}$ we have, by Stoke’s Theorem 2,

$\displaystyle 0=\int_{\left[\mathbf{x};\mathbf{h}\right]}\omega-\int_{[\mathbf{0};\mathbf{x}+\mathbf{h}]}\omega+\int_{\left[\mathbf{0};\mathbf{x}\right]}\omega$

which is (2).

The proof for ${j\ge2}$ is a direct generalization of the above. The triangle ${T}$ above is replaced by a cone over a parallelpiped, defined as follows. Let ${P=\left[\mathbf{a},\xi_{1},\ldots,\xi_{k}\right]}$ denote and affine ${k}$-parallelpiped in ${\mathbb{R}^{n}}$. The cone over ${P}$ is the parameterized ${k+1}$-parallelpiped ${\mathcal{C}P}$ given by the map

$\displaystyle (t,\mathbf{s})\in[0,1]\times Q_{k}\mapsto t\left(\mathbf{a}+\sum_{\alpha=1}^{k}s_{\alpha}\xi_{\alpha}\right).$

The boundary of ${\mathcal{C}P}$ is then the following chain

$\displaystyle \partial\mathcal{C}P=P-\left[\mathbf{0};\mathbf{0},\ldots,\mathbf{0}\right]+\sum_{\alpha=1}^{k}(-1)^{\alpha}\left[\mathcal{C}P_{\alpha}^{+}-\mathcal{C}P_{\alpha}^{-}\right]$

where ${\partial P=\sum_{\alpha}\left(-1\right)^{\alpha-1}\left[P_{\alpha}^{+}-P_{\alpha}^{-}\right]}$. Since ${\left[\mathbf{0};\mathbf{0},\ldots,\mathbf{0}\right]}$ is degenerate as above, we put this succinctly as

$\displaystyle \partial\mathcal{C}P=P-\mathcal{\mathcal{C}}\left(\partial P\right)$

where the cone over a chain of parallelpipeds is defined term by term.

Now let ${\omega}$ be a closed ${j}$-form in ${\mathbb{R}^{n}}$, and let ${\alpha}$ be the following ${j-1}$ form

$\displaystyle \nu\left(\mathbf{x}\right)\left[\xi_{1},\ldots,\xi_{j-1}\right]=\int_{0}^{1}\omega\left(t\mathbf{x}\right)\left[\mathbf{x},t\xi_{1},\ldots,t\xi_{j-1}\right]dt. \ \ \ \ \ (3)$

Then for any affine ${j-1}$ parallelpiped ${P=\left[\mathbf{a};\xi_{1},\ldots,\xi_{j-1}\right]}$ we have

$\displaystyle \begin{array}{rcl} \int_{P}\nu & = & \int_{Q_{j-1}}\nu\left(\mathbf{a}+\sum_{\alpha=1}^{j-1}s_{\alpha}\xi_{\alpha}\right)\left[\xi_{1},\ldots,\xi_{j-1}\right]d\mathbf{s}\\ & = & \int_{0}^{1}\int_{Q_{j-1}}\omega\left(t\left(\mathbf{a}+\sum_{\alpha=1}^{j-1}s_{\alpha}\xi_{\alpha}\right)\right)\left[\mathbf{a}+\sum_{\alpha=1}^{j-1}s_{\alpha}\xi_{\alpha},t\xi_{1},\ldots,t\xi_{j-1}\right]d\mathbf{s}dt\\ & = & \int_{\mathcal{C}P}\omega. \end{array}$

Thus, since ${d\omega=0}$, we have

$\displaystyle \int_{\widehat{P}}\omega=\int_{\mathcal{C}\left(\partial\widehat{P}\right)}\omega=\int_{\partial\widehat{P}}\nu$

for any affine ${j}$-parallelpiped ${\widehat{P}}$ by Stoke’s theorem and the identity ${\partial\mathcal{C}\widehat{P}=\widehat{P}-\mathcal{C}\left(\partial\widehat{P}\right)}$ derived above.

Now given ${\mathbf{a}}$, ${\xi_{1},\ldots,\xi_{j}\in\mathbb{R}^{n}}$ and ${\epsilon>0}$ let ${\widehat{P}_{\epsilon}=\left[\mathbf{a};\epsilon\xi_{1},\ldots,\epsilon\xi_{j}\right]}$. Then

$\displaystyle \frac{1}{\epsilon^{j}}\int_{\widehat{P}_{\epsilon}}\omega=\sum_{\alpha=1}^{j}\left(-1\right)^{\alpha}\frac{1}{\epsilon^{j}}\left[\int_{\widehat{P}_{\epsilon;\alpha}^{+}}\nu-\int_{\widehat{P}_{\epsilon;\alpha}^{-}}\nu\right].$

It is now a simple matter to check that

$\displaystyle \frac{1}{\epsilon^{j}}\int_{\widehat{P}_{\epsilon}}\omega=\int_{Q_{j}}\omega\left(\mathbf{a}+\epsilon\sum_{\alpha}s_{\alpha}\xi_{\alpha}\right)\left[\xi_{1},\ldots,\xi_{j}\right]d\mathbf{s}\rightarrow\omega\left(\mathbf{a}\right)\left[\xi_{1},\ldots,\xi_{j}\right]$

as ${\epsilon\rightarrow0}$, while

$\displaystyle \begin{array}{rcl} \lim_{\epsilon\rightarrow0}\frac{1}{\epsilon^{j}}\left[\int_{\widehat{P}_{\epsilon;\alpha}^{+}}\nu-\int_{\widehat{P}_{\epsilon;\alpha}^{-}}\nu\right] & = & \lim_{\epsilon\rightarrow0}\int_{Q_{j-1}}\frac{\nu\left(\widehat{P}_{\epsilon,\alpha}^{+}\left(\mathbf{s}\right)\right)\left[\xi_{1},\ldots,\not\xi_{\alpha},\ldots,\xi_{j}\right]-\nu\left(\widehat{P}_{\epsilon,\alpha}^{-}\left(\mathbf{s}\right)\right)\left[\xi_{1},\ldots,\not\xi_{\alpha},\ldots,\xi_{j}\right]}{\epsilon}d\mathbf{s}\\ & = & \lim_{\epsilon\rightarrow0}\frac{\nu\left(\mathbf{a}+\epsilon\xi_{\alpha}\right)\left[\xi_{1},\ldots,\not\xi_{\alpha},\ldots,\xi_{j}\right]-\nu\left(\mathbf{a}\right)\left[\xi_{1},\ldots,\not\xi_{\alpha},\ldots,\xi_{j}\right]}{\epsilon}. \end{array}$

It follows from Lemma (3) that ${\omega=d\nu}$.