Differential 1-forms are closed if and only if they are exact

 Preliminary exam prep

The title refers to 1-forms in Euclidean $n$-space $\R^n$, for $n\geqslant 2$. This theorem is instructive to do in the case $n=2$, but we present it in general. We will use several facts, most importantly that the integral of a function $f:X\to Y$ over a curve $\gamma:[a,b]\to X$ is given by
$$\int_\gamma f\ dx_1\wedge \cdots \wedge dx_k = \int_a^b (f\circ \gamma)\ d(x_1\circ \gamma)\wedge \cdots \wedge d(x_n\circ \gamma),$$
where $x_1,\dots,x_n$ is some local frame on $X$. We will also use the fundamental theorem of calculus and one of its consequences, namely
$$\int_a^b \frac{\dy f}{\dy t}(t)\ dt = f(b)-f(a).$$

Theorem: A 1-form on $\R^n$ is closed if and only if it is exact, for $n\geqslant 2$.

Proof: Let $\omega = a_1dx_1+\cdots a_ndx_n\in \Omega^1_{\R^n}$ be a 1-form on $\R^n$. If there exists $\eta\in \Omega^0_{\R^n}$ such that $d\eta = \omega$, then $d\omega = d^2\eta = 0$, so the reverse direction is clear. For the forward direction, since $\omega$ is closed, we have
$$0 = d\omega = \sum_{i=1}^n \frac{\dy a_1}{\dy x_i}dx_i \wedge dx_1 + \cdots + \sum_{i=1}^n \frac{\dy a_n}{\dy x_n} dx_i\wedge dx_n \ \ \ \implies\ \ \ \frac{\dy a_i}{\dy x_j} = \frac{\dy a_j}{\dy x_i}\ \forall\ i\neq j.$$
Now fix some $(\textbf{x}_1,\dots,\textbf{x}_n)\in \R^n$, and define $f\in \Omega^0_{\R^n}$ by
$$f(\textbf x_1,\dots,\textbf x_n) = \int_{\gamma(\textbf x_1,\dots,\textbf x_n)}\omega,$$
for $\gamma$ the composition of the paths
$$\begin{array}{r c l} \gamma_1\ :\ [0,\textbf x_1] & \to & \R^n, \\ t & \mapsto & (t,0,\dots,0), \end{array} \hspace{5pt} \begin{array}{r c l} \gamma_2\ :\ [0,\textbf x_2] & \to & \R^n, \\ t & \mapsto & (\textbf x_1,t,0,\dots,0), \end{array} \hspace{5pt}\cdots\hspace{5pt} \begin{array}{r c l} \gamma_n\ :\ [0,\textbf x_n] & \to & \R^n, \\ t & \mapsto & (\textbf x_1,\dots,\textbf x_{n-1},t). \end{array}$$
By applying the definition of a pullback and the change of variables formula (use $s=\gamma_i(t)$ for every $i$),
$$\begin{align*} \int_{\gamma(\textbf x_1,\dots,\textbf x_n)}\omega  & = \sum_{i=1}^n \int_{\gamma_i} a_1 dx_1 + \cdots + \sum_{i=1}^n \int_{\gamma_i}a_n dx_n \\ & = \sum_{i=1}^n \int_{\gamma_i} a_1(x_1,\dots,x_n)\ dx_1 + \cdots + \sum_{i=1}^n \int_{\gamma_i}a_n(x_1,\dots,x_n)\ dx_n \\ & = \sum_{i=1}^n \int_0^{\textbf x_i} a_1(\gamma_i(t))\ d(x_1\circ \gamma_i)(t) + \cdots + \sum_{i=1}^n \int_0^{\textbf x_i}a_n(\gamma_i(t))\ d(x_n\circ \gamma_i)(t) \\ & = \int_0^{\textbf x_1} a_1(\gamma_1(t))\gamma'_1(t)\ dt + \cdots + \int_0^{\textbf x_n}a_n(\gamma_n(t))\gamma_n'(t)\ dt \\ & = \int_{(0,\dots,0)}^{(\textbf x_1,0,\dots,0)} a_1(s)\ ds + \cdots + \int_{(\textbf x_1,\dots,\textbf x_{n-1},0)}^{(\textbf x_1,\dots,\textbf x_n)}a_n(s)\ ds \\ & = \int_0^{\textbf x_1} a_1(s,0,\dots,0)\ ds + \cdots + \int_0^{\textbf x_n}a_n(\textbf x_1,\dots,\textbf x_{n-1}, s)\ ds. \end{align*}$$
To take the derivative of this, we consider the partial derivatives first. In the last variable, we have
$$\frac{\dy f}{\dy \textbf x_n} = \frac\dy{\dy \textbf x_n}\int_0^{\textbf x_n}a_n(\textbf x_1,\dots,\textbf x_{n-1}, s)\ ds = a_n(\textbf x_1,\dots,\textbf x_n) = a_n.$$
In the second-last variable, applying one of the identities from $\omega$ being closed, we have
$$\begin{align*} \frac{\dy f}{\dy \textbf x_{n-1}}  & = \frac\dy{\dy \textbf x_{n-1}}\int_0^{\textbf x_{n-1}}a_{n-1}(\textbf x_1,\dots,\textbf x_{n-2}, s,0)\ ds  +  \frac\dy{\dy \textbf  x_{n-1}}\int_0^{\textbf x_n}a_n(\textbf x_1,\dots,\textbf x_{n-1}, s)\ ds \\ & = a_{n-1}(\textbf x_1,\dots,\textbf x_{n-1},0) + \int_0^{\textbf x_n}\frac{\dy a_n}{\dy \textbf x_{n-1}}(\textbf x_1,\dots,\textbf x_{n-1}, s)\ ds \\ & = a_{n-1}(\textbf x_1,\dots,\textbf x_{n-1},0) + \int_0^{\textbf x_n}\frac{\dy a_{n-1}}{\dy s}(\textbf x_1,\dots,\textbf x_{n-1}, s)\ ds \\ & = a_{n-1}(\textbf x_1,\dots,\textbf x_{n-1},0) + a_{n-1}(\textbf x_1,\dots,\textbf x_n) - a_{n-1}(\textbf x_1,\dots,\textbf x_{n-1}, 0) \\ & = a_{n-1}(\textbf x_1,\dots, \textbf x_n) \\ & = a_{n-1}. \end{align*}$$
This pattern continues. For the other variables we have telescoping sums, and we compute the partial derivative in the first variable as an example:
$$\begin{align*} \frac{\dy f}{\dy \textbf x_1}  & = \frac{\dy}{\dy \textbf x_1}\int_0^{\textbf x_1}a_1(s,0,\dots,0)\ ds + \sum_{i=2}^n\frac\dy{\dy \textbf x_1}\int_0^{\textbf x_i}a_i(\textbf x_1,\dots,\textbf x_{i-1}, s,0,\dots,0)\ ds \\ & = a_1(\textbf x_1,0,\dots,0) + \sum_{i=2}^n \int_0^{\textbf x_i} \frac {\dy a_i}{\dy \textbf x_1} (\textbf x_1,\dots,\textbf x_{i-1}, s,0,\dots,0)\ ds \\ & = a_1(\textbf x_1,0,\dots,0) + \sum_{i=2}^n \int_0^{\textbf x_i} \frac {\dy a_1}{\dy s} (\textbf x_1,\dots,\textbf x_{i-1}, s,0,\dots,0)\ ds \\ & = a_1(\textbf x_1,0,\dots,0) + \sum_{i=2}^n \left(a_1(\textbf x_1,\dots,\textbf x_i, 0,\dots,0) - a_1(\textbf x_1,\dots,\textbf x_{i-1},0,\dots,0)\right) \\ & = a_1(\textbf x_1,\dots,\textbf x_n) \\ & = a_1. \end{align*}$$
Hence we get that
$$df = \frac{\dy f}{\dy x_1} dx_1 + \cdots + \frac{\dy f}{\dy x_n}dx_n = a_1dx_1 + \cdots + a_ndx_n = \omega,$$
so $\omega$ is exact. $\square$

References: Lee (Introduction to smooth manifolds, Chapter 11)