Poincaré lemma for one-forms on convex sets

Statement

Let $U \subseteq \mathbb{R}^n$ be open and convex, and let $\omega = \sum_i f_i \, dx_i$ be a closed one-form on $U$ with $C^1$ coefficients. Then $\omega$ is exact. Explicitly, for a fixed $p \in U$, the function $$g(x) = \int_0^1 \sum_{i=1}^n f_i\big( p + t (x - p) \big) \, (x_i - p_i) \, dt$$ satisfies $dg = \omega$ on $U$.

Remarks

Proof: differentiate under the integral sign, and use $\partial_j f_i = \partial_i f_j$. The integrand of $\partial_j g$ becomes $\frac{d}{dt} \big( t \, f_j(p + t (x - p)) \big)$, which integrates to $f_j(x)$. In $\mathbb{R}^3$, this is the theorem that a curl-free field on a convex set is conservative. The lemma also holds for closed $k$-forms with $k \ge 1$ on star-shaped open sets. For one-forms, it holds on simply connected open sets. Some condition on $U$ is needed: the angle form on $\mathbb{R}^2 \setminus \{0\}$ is closed but not exact. See also the vortex field.

Explanations

L1 human

On a convex set, a one-form that passes the local test for being a derivative really is a derivative. The proof builds the function by integrating the form along straight segments from one base point.