Integration of forms on non-simply connected manifolds

Question

What I know is that closed forms are not exact on non-simply connected manifolds, so for instance, if $E$ is a closed form, then $dE = 0$ but $\int_\gamma E \neq 0$, where $\gamma$ is a non-contractible curve on the manifold. What I need to know is how to compute this integral on such a manifold. Thanks in advance.

Answer 1

A couple of corrections need to be made to your statement of the problem.

First, since you're integrating over curves, you have to assume that $E$ is a $1$-form, not just any differential form. (For $k>1$, simple connectedness is not sufficient to imply that every closed $k$-form is exact.)

Second, your statement that "if $E$ is a closed form, then $dE = 0$ but $\int_\gamma E \neq 0$, where $\gamma$ is a non-contractible curve on the manifold" is not true. Even if $\gamma$ is a non-contractible curve, there will be closed forms whose integrals over $\gamma$ are zero -- any exact form, for example.

To answer your main question: the most straightforward way to evaluate $\int_\gamma E$ is, as @Pedro said, to parametrize the curve and compute the integral directly. However, when $E$ is closed, there's another technique that can often be used to simplify the integral. If $\sigma$ is a smooth closed curve that's homotopic to $\gamma$ (provided all the curves in the homotopy are closed curves), then $\int_\gamma E = \int_\sigma E$. Often it's possible to find a much simpler curve that's homotopic to the original one, which makes the integral easier to compute. For example, on the punctured plane, if $\gamma$ is any curve with winding number $1$ around the origin, then $\int_\gamma E$ is equal to $\int_\sigma E$, where $\sigma$ is the unit circle traversed counterclockwise.