Curl-free field that is not conservative

Counterexample

One could guess that each $C^1$ vector field with zero curl on an open set is conservative. This is false. Let $U = \{(x, y, z) \in \mathbb{R}^3 : (x, y) \ne (0, 0)\}$, which is $\mathbb{R}^3$ minus the $z$-axis, and let $$F(x, y, z) = \left( \frac{-y}{x^2 + y^2}, \ \frac{x}{x^2 + y^2}, \ 0 \right).$$ Then $\nabla \times F = 0$ on $U$, but $F$ is not conservative on $U$.

Remarks

The third component of $\nabla \times F$ is $\partial_x \big( x / (x^2 + y^2) \big) - \partial_y \big( -y / (x^2 + y^2) \big) = \frac{y^2 - x^2}{(x^2 + y^2)^2} - \frac{y^2 - x^2}{(x^2 + y^2)^2} = 0$, and the other two components are zero. On the unit circle $\gamma(t) = (\cos t, \sin t, 0)$, $t \in [0, 2\pi]$, we have $F(\gamma(t)) \cdot \gamma'(t) = 1$. So the circulation is $2\pi \ne 0$, and path independence fails. The domain $U$ is not simply connected, so this does not contradict the theorem for convex or simply connected sets. Locally, $F = \nabla \theta$, where $\theta$ is the polar angle of $(x, y)$. But no continuous choice of $\theta$ exists on all of $U$.

Explanations

L1 human

The field circles around the $z$-axis, and its strength falls off with distance in just the right way to have no local swirl. But a loop around the axis picks up work $2\pi$, so no potential can exist. The hole in the domain causes this.