6.3 Poincaré's lemma

An exact form is always closed but the converse is not necessarily true. However, the following theorem provides the situation in which the converse is also true.

Theorem 6.3 · Poincaré's lemma

If a coordinate neighbourhood $U$ of a manifold $M$ is contractible to a point $p_0\in M$ , any closed $r$ -form on $U$ is also exact.

Proof

We assume $U$ is smoothly contractible to $p_0$ , that is, there exists a smooth map $F:U\times I\to U$ such that

$$F(x,0)=x,F(x,1)=p_0\text{f o r}x\in U.$$

Let us consider an $r$ -form $\eta \in {\Omega }^r(U\times I)$

$$\begin{array}{rl}& \eta =a_{i_1\dots i_r}(x,t)d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_r}\\ & +b_{j_1\dots j_{r-1}}(x,t)dt\wedge d{x}^{j_1}\wedge \cdots \wedge d{x}^{j_{r-1}}\end{array}\text{\hspace{1em}(6.28)}$$

where $x$ is the coordinate of $U$ and $t$ of $I$ . Define a map $P:{\Omega }^r(U\times I)\to {\Omega }^{r-1}(U)$ by

$$P\eta \equiv \left({\int }_0^1\mathrm{d}s{b}_{j_1\dots j_{r-1}}(x,s)\right)\mathrm{d}{x}^{j_1}\wedge \cdots \wedge \mathrm{d}{x}^{j_{r-1}}.\text{\hspace{1em}(6.29)}$$

Next, define a map $f_t:U\to U\times I$ by $f_t(x)=(x,t)$ . The pullback of the first term of (6.28) by $f_t^{\ast }$ is an element of ${\Omega }^r(U)$ ,

$$f_t^{\ast }\eta =a_{i_1\dots i_r}(x,t)\mathrm{d}{x}^{i_1}\wedge \cdots \wedge \mathrm{d}{x}^{i_r}\in {\Omega }^r(U).\text{\hspace{1em}(6.30)}$$

We now prove the following identity,

$$\mathrm{d}(P\eta )+P(\mathrm{d}\eta )=f_1^{\ast }\eta -f_0^{\ast }\eta .\text{\hspace{1em}(6.31)}$$

Each term of the LHS is calculated to be

$$\begin{array}{rl}\mathrm{d}P\eta & =\mathrm{d}\left({\int }_0^1\mathrm{d}s{b}_{j_1\dots j_{r-1}}\right)\mathrm{d}{x}^{j_1}\wedge \cdots \wedge \mathrm{d}{x}^{j_{r-1}}\\ & ={\int }_0^{1}ds\left(\frac{\partial b_{j_1\dots j_{r-1}}}{\partial x^{j_r}}\right)d{x}^{j_r}\wedge d{x}^{j_1}\wedge \cdots \wedge d{x}^{j_{r-1}}\end{array}$$

$$\begin{array}{rl}& P\mathrm{d}\eta =P[\left(\frac{\partial a_{i_1\dots i_r}}{\partial x^{i_{r+1}}}\right)\mathrm{d}{x}^{i_{r+1}}\wedge \mathrm{d}{x}^{i_1}\wedge \cdots \wedge \mathrm{d}{x}^{i_r}\\ & +\left(\frac{\partial a_{i_1\dots i_r}}{\partial t}\right)dt\wedge d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_r}\\ & +\left(\frac{\partial b_{j_1\dots j_{r-1}}}{\partial x^{j_r}}\right)d{x}^{j_r}\wedge dt\wedge d{x}^{j_1}\wedge \cdots \wedge d{x}^{j_{r-1}}]\\ & =\left[{\int }_0^{1}ds\left(\frac{\partial a_{i_1\dots i_r}}{\partial s}\right)\right]d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_r}\\ & -\left[{\int }_0^{1}ds\left(\frac{\partial b_{j_1\dots j_{r-1}}}{\partial x^{j_r}}\right)\right]d{x}^{j_r}\wedge d{x}^{j_1}\wedge \cdots \wedge d{x}^{j_{r-1}}.\end{array}$$

Collecting these results, we have

$$\begin{array}{rl}\mathrm{d}(P\eta )+P(\mathrm{d}\eta )& =\left[{\int }_0^1\mathrm{d}s\left(\frac{\partial a_{i_1\dots i_r}}{\partial s}\right)\right]\mathrm{d}{x}^{i_1}\wedge \cdots \wedge \mathrm{d}{x}^{i_r}\\ & =\left[a_{i_1\dots i_r}(x,1)-a_{i_1\dots i_r}(x,0)\right]d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_r}\\ & =f_1^{\ast }\eta -f_0^{\ast }\eta .\end{array}$$

Poincaré’s lemma readily follows from (6.31). Let $\omega$ be a closed $r$ -form on a contractible chart $U$ . We will show that $\omega$ is written as an exact form,

$$\omega =\mathrm{d}(-P{F}^{\ast }\omega ),\text{\hspace{1em}(6.32)}$$

$F$ being the smooth contraction map. In fact, if $\eta$ in (6.31) is replaced by $F^{\ast }\omega \in {\Omega }^r(U\times I)$ we have

$$\begin{array}{rl}\mathrm{d}P{F}^{\ast }\omega +P\mathrm{d}{F}^{\ast }\omega & =f_1^{\ast }\circ F^{\ast }\omega -f_0^{\ast }\circ F^{\ast }\omega \\ & ={\left(F\circ f_1\right)}^{\ast }\omega -{\left(F\circ f_0\right)}^{\ast }\omega \end{array}\text{\hspace{1em}(6.33)}$$

where use has been made of the relation $(f\circ g{)}^{\ast }=g^{\ast }\circ f^{\ast }$ . Clearly $F\circ f_1:U\to U$ is a constant map $x\mapsto p_0$ , hence $(F\circ f_1{)}^{\ast }=0$ . However, $F\circ f_0={\mathrm{id}}_U$ , hence $(F\circ f_0{)}^{\ast }:{\Omega }^r(U)\to {\Omega }^r(U)$ is the identity map. Thus, the RHS of (6.33) is simply $-\omega$ . The second term of the LHS vanishes since $\omega$ is closed; $\mathrm{d}{F}^{\ast }\omega =F^{\ast }\mathrm{d}\omega =0$ , where use has been made of (5.75). Finally, (6.33) becomes $\omega =-\mathrm{d}P{F}^{\ast }\omega$ , which proves the theorem.

Any closed form is exact at least locally. The de Rham cohomology group is regarded as an obstruction to the global exactness of closed forms.

Example 6.4

Since ${\mathbb{R}}^n$ is contractible, we have

$$H^r\left({\mathbb{R}}^n\right)=01\le r\le n.\text{\hspace{1em}(6.34)}$$

Note, however, that $H^0({\mathbb{R}}^n)=\mathbb{R}$ .

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