Are line integrals path-independent "up to homotopy" in a curl-free vector field?

Question

In a non-simply connected space like the punctured plane, there are curl-free fields that are not conservative. When I try to imagine paths that share endpoints but have different line integrals, they seem to not be homotopic. Intuitively, this leads me to believe that the line integrals of two paths (with common endpoints) that are homotopic will be the same.

Is this true? If so, does this result have a name, or is it an obvious case of a more general result? Or is wrong? My searches for "curl-free" and "homotopy" turn up very little.

Answer 1

Setting aside technical aspects for a minute, this is just the classical Stokes' theorem, one of its versions being:

Given a homotopy $H: [0,1]\times [0,1] \to \mathbb{R}^3$ and a vector field $F$, then $$\int_{\partial H} F = \int_H \nabla \times F, $$ where the integration $\int_{\partial H}$ refers to the sum of the work of $F$ on the curves $H|_{1 \times s}, H|_{(1-t) \times 1 }, H|_{0 \times (1-s)}, H|_{t \times 0}$.

Since the endpoints are fixed, the only relevant curves for the integration are $H|_{(1-t) \times 1 }$ and $H|_{t \times 0}$, which are $c_2(1-t)$ and $c_1(t)$ (those being the beginning and end of the homotopy, respectively). Then, we have that $$\int_{c_1} F -\int_{c_2}F = 0,$$ and the result follows.

One can prove this version of Stokes' theorem in a straightforward way by setting $G(s)=\int_{c_s}F$, differentiating $G$ and proceeding to use the fundamental theorem of calculus, differentiation under the integral, integration by parts and some linear algebra.

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If the machinery of differential forms is known, you can proceed to prove this result as follows: recast the vector field as a differential form $\omega$, and then apply Stokes' theorem to the pull-back of $d\omega$ under the homotopy $H$. We then have $$0 = \int_{Q} H^*d\omega =\int_QdH^*\omega=\int_{\partial Q} H^*\omega = \int_{c_1}H^*\omega - \int_{c_2} H^*\omega,$$ where $Q=[0,1]\times [0,1]$. (Note: $Q$ is actually a manifold with corners, so make sure that your definitions and your version of Stokes' theorem can handle that, at least in the simple case of the square.)

Answer 2

In general we don't talk about the curl of a vector field but about the differential $d \omega$ of a differential $1$-form $\omega$: when this vanishes the differential form is said to be closed, and it's true that the integral of a closed differential $1$-form along a path only depends on the path up to homotopy by Stokes' theorem.