A closed curve $\gamma$ in a manifold $X$ is defined to be a smooth map $\gamma: S^1 \to X$, and I am tasked with finding an explicit formula for the line integral $\oint_{\gamma}\omega = \int_{S^1} \gamma^*\omega$ (here $X = \mathbf{R}^k$).
What I have tried so far:
By definition we have that $\omega = \sum_{i=1}^kf_idx^i$ for $f_i$ smooth, and the pullback by definition is given by $\gamma^*(\omega) = \sum_{i=1}^k(f_i \circ \gamma)d\gamma_i$.
Where I am having conceptual difficulty is computing the $d\gamma^i$'s here. I know that $S^1$ is a one dimensional manifold and so one of my potential solutions has been to think of each point of $S^1$ as being parameterized by $\theta$ and so abstractly $\gamma(\theta) = (\gamma^1(\theta), …, \gamma^k(\theta))$ which would make the expression above $\sum_{i=1}^k(f_i \circ \gamma)\frac{\partial \gamma^i}{\partial \theta}d\theta$.
On the other hand, we know that each point of $S^1$ is specified by two components $(\cos\theta,\sin\theta)$ and so I could think of $\gamma(\cos\theta,\sin\theta) = (\gamma^1(\cos\theta,\sin\theta), …, \gamma^k(\cos\theta,\sin\theta))$.
I guess my question boils to to how I should be thinking about $S^1$ to make sense of differentials of the $\gamma^i$'s. Any help would be appreciated.
Best Answer
Based on context, I assume $\omega$ is a $1$-form on $X = \Bbb{R}^k$. In this case, $\int_{\Bbb{R}^k} \omega$ makes no sense. You probably meant $\int_{\gamma}\omega$; because a $1$-form needs to be integrated over a "1-dimensional object".
For the purposes of calculation, I think it's best to parametrize $S^1$ by the trigonometric map you suggested: $\alpha: [0,2\pi] \to S^1$, $\alpha(t) = (\cos t, \sin t)$. Of course, this map is not a diffeomorphism, nor a homeomorphism, but if you restrict to $(0,2\pi)$, it's a diffeomorphism onto its image, which is $S^1$ without a single point, so for the purposes of integration, this makes no difference.
Also, for integration over $S^1$, which is a manifold, one has to prescribe an orientation, but spheres have a natural orientation, and the map $\alpha$ defined above is indeed positively oriented with respect to that orientation. So, we have: \begin{align} \int_{\gamma}\omega &= \int_{S^1} \gamma^* \omega \\ &= \int_{\alpha[0,2\pi]} \gamma^* \omega \\ &= \int_{[0,2\pi]} \alpha^* \gamma^* \omega \tag{$\alpha$ is positively oriented}\\ &= \int_{[0,2\pi]} (\gamma \circ \alpha)^* \omega \end{align} Now, if $\omega = \sum_{i=1}^k f_i \, dx^i$, and if we denote $t$ to be the coordinate on $[0,2\pi]$, then \begin{align} (\gamma \circ \alpha)^* \omega &= \sum_{i=1}^k f_i \circ (\gamma \circ \alpha) \cdot d(\gamma^i \circ \alpha) \\ &= \sum_{i=1}^k (f_i \circ \gamma \circ \alpha) \cdot (\gamma^i \circ \alpha)' \, dt \end{align} Hence, \begin{align} \int_{\gamma} \omega &= \sum_{i=1}^k \int_{[0,2\pi]} (f_i \circ \gamma \circ \alpha) \cdot (\gamma^i \circ \alpha)' \, dt \end{align} Or in more common notation, \begin{align} \int_{\gamma} \omega &= \sum_{i=1}^k \int_{0}^{2\pi} f_i(\gamma(\cos t, \sin t)) \cdot \left(\dfrac{d}{dt}\gamma^i(\cos t, \sin t) \right) \, dt \end{align}
By the way, while the definition of a closed curve in $X$ as being a map $\gamma:S^1 \to X$ is of course correct, I think for actual calculations, it is much clearer to think of it instead as a curve $\tilde{\gamma}:[a,b] \to X$ such that $\tilde{\gamma}(a) = \tilde{\gamma}(b)$. In other words, you think of the curve as having already been pulled-back to an interval. In the above discussion, we would then have $\tilde{\gamma} = \gamma \circ \alpha$, and the formulas would look much nicer: \begin{align} \int_{\tilde{\gamma}} \omega &= \sum_{i=1}^k\int_a^b (f_i \circ \tilde{\gamma})(t) \cdot (\tilde{\gamma}^i)'(t) \, dt \end{align} Or of course, if you're willing to abuse notation further and denote the curve $\tilde{\gamma}(\cdot)$ by $x(\cdot)$, we get the very classical and memorable expression that if $\omega = \sum_{i=1}^k f_i \, dx^i$ then \begin{align} \int_{\tilde{\gamma}} \sum_{i=1}^k f_i \, dx^i &= \sum_{i=1}^k \int_a^b f_i(x(t)) \cdot \dfrac{dx^i}{dt}\, dt. \end{align}