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.

A PDF version of these notes is available here.

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}.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s