A $p$-form $\alpha$ is called closed if its exterior derivative vanishes, and exact if it is the derivative of a ($p-1$)-form, that is: \begin{align} \hbox{$\alpha$ closed } &\Longleftrightarrow d\alpha=0 \label{closed}\\ \hbox{$\alpha$ exact } &\Longleftrightarrow \exists\beta:\alpha=d\beta \label{exact} \end{align} Equivalently, $\alpha$ is closed if it lies in the kernel $\hbox{Ker}(d)$ of $d$, and is exact if it lies in the image $\hbox{Im}(d)$ of $d$.

Exact differentials such as $x\,dx+y\,dy=d(x^2+y^2)$ are also exact as 1-forms, and correspond (via $F=\FF\cdot d\rr$) to conservative vector fields, namely vector fields that are the gradient of some function, that is, $\FF=\grad f$. The Poincaré Lemma states that an exact differential form is also closed, which follows immediately from $d^2=0$, and which includes the vector calculus identities $\grad\times\grad f=\zero$ and $\grad\cdot\grad\times\FF=0$ as special cases.

The converse of the Poincaré Lemma 1) is not true in general, but requires additional assumptions about the topology of the underlying space. One version states that if $R$ is a simply connected region of $\RR^n$, then all closed $p$-forms on $R$ are also exact. (A simply connected region is one whose points can be connected by curves, and in which any closed curve can be shrunk to a point. Loosely speaking, a simply connected region “has no holes”.)

For example, any closed form in Euclidean $\RR^3$ is also exact, that is \begin{equation} d\alpha = 0 \Longrightarrow \exists\beta: \alpha=d\beta \label{poincare} \end{equation} for any $p$-form $\alpha$. If $\alpha$ is a 1-form, (\ref{poincare}) is nothing more than the curl test for conservative vector fields: If $\grad\times\FF=\zero$, then $\FF=\grad f$ for some potential function $f$. Similarly, if $\alpha$ is a 2-form, then (\ref{poincare}) corresponds to the statement that if $\grad\cdot\FF=0$, then $\FF=\grad\times\AA$ for some vector potential $\AA$.

Not surprisingly, the $(p-1)$-form $\beta$ in (\ref{exact}) is called a potential for the $p$-form $\alpha$. Potentials are not unique, since we can always add an exact $(p-1)$-form to $\beta$, that is \begin{equation} \alpha = d\beta \Longrightarrow \alpha = d(\beta+d\gamma) \end{equation} a special case of which is the addition of an integration constant to an indefinite integral, corresponding to \begin{equation} d(f+\hbox{constant}) = df \end{equation}

So the failure of closed forms to be exact is an indication of nontrivial topology!

For example, consider the unit circle, $\SSS^1$. The 1-form $d\phi$ is clearly closed, since there are no non-zero 2-forms on $\SSS^1$. We have \begin{equation} \int_{\SSS^1}d\phi = 2\pi \ne 0 \end{equation} which, as discussed in §5.7, tells us that “$d\phi$” can not be exact! But surely $d\phi$ is $d(\phi)$? This assertion is in fact false, and for a simple reason: There is no smooth function $\phi$ on the entire circle whose differential is $d\phi$. Rather, it takes two such functions, call them $\phi_1$ running from $0$ to $2\pi$ and $\phi_2$ running from $-\pi$ to $\pi$ with the endpoints excluded. Somewhat remarkably, $d\phi$ is well-defined everywhere, since $d\phi_1=d\phi_2$ wherever both are defined. The problem is that “$\phi$” is not well-defined. So $d\phi$ is not actually $d$ of any (single) function …

These ideas can be generalized into a characterization of topological structure known as de Rham cohomology. Any 1-form on $\SSS^1$ takes the form \begin{equation} \alpha = f\,d\phi \end{equation} so that \begin{equation} \int_{\SSS^1} \alpha = \int_0^{2\pi} f\,d\phi =: c_\alpha \end{equation} where the last equality defines the constant $c_\alpha$. But then \begin{equation} \gamma = \alpha-\frac{c_\alpha}{2\pi} \end{equation} does integrate to zero, and therefore can be (and in fact is) exact, that is, $\gamma = dg$, so that \begin{equation} \alpha = dg + \frac{c_\alpha}{2\pi} \end{equation} for some function $g$. We can therefore divide up the 1-forms on $\SSS^1$ into equivalence classes based on the value of $c_\alpha$ by considering two 1-forms to be equivalent if they differ by an exact differential such as $dg$. This amounts to considering the quotient space \begin{equation} \hbox{Ker}(d)\big/\hbox{Im}(d) = \{\hbox{closed forms}\}\big/\{\hbox{exact forms}\} \end{equation} which is called the (first) de Rham cohomology class. The fact that this space is 1-dimensional in this case, consisting of all real numbers $c_\alpha$, turns out to imply that there is one “hole”, a statement about the topology of the circle.