Integration on the punctured plane

Question

Let $M = \mathbb{R}^2\backslash\{0\}$, $$ \alpha = \frac{x\cdot dy-y\cdot dx}{x^2+y^2}$$ and $\mathcal{C} = \{(f(e^{i\theta})\cos\theta,f(e^{i\theta})\sin\theta);\theta\in\mathbb{S}^1\}$, where $f:\mathbb{S}^1\to (0,1)$ is smooth. Let $j:\mathcal {C}\to M$ be the inclusion. Calculate $$\int_\mathcal{C}j^*\alpha.$$ I'm not sure if I'm doing this right, but I got: $$j^*\alpha = \frac{(x\circ j)d(y\circ j)-(y\circ j)d(x\circ j)}{(x\circ j)^2+(y\circ j)^2} =... = d\theta$$ So is the answer $\int_0^{ 2\pi}d\theta = 2\pi?$ Or is it$\int_0^{1}d\theta = 1$ since $f$ maps the circle on to $(0,1)$?

Answer 1

You did not obtain answers so far, because your description of the problem is somehow overloaded. The underlying domain of integration is an "abstract copy of $S^1$". This copy can be realized as ${\mathbb R}/(2\pi{\mathbb Z})$ or as geometric unit circle $\partial D$ in ${\mathbb R}^2$. At any rate the chosen parameter variable is $\theta$, and for a full performance through $S^1$ this $\theta$ should run over the interval $[0,2\pi]$. On the other hand it is true that this $\theta$ is not a 100% coordinate function on $S^1$, but we have control over the details $\ldots$

Understanding all this, your $f$, defined on the points of $\partial D$, should be considered as a function $$\rho(\theta):=f(e^{i\theta})\in\ ]0,1[\qquad(0\leq\theta\leq2\pi)\ .$$ At any rate, your $j^*\alpha=d\theta$ is correct; and $$\int_{\cal C} j^*\alpha=\int_0^{2\pi}d\theta=2\pi\ .$$ There is no question of an integral $\int_0^1d\theta$ here.

Maybe it is of some help to note that $\alpha=d\arg$, since $$\nabla\arg(x,y)\quad\left(=\nabla\arctan{y\over x}\right)\quad=\left({-y\over x^2+y^2}, \>{x\over x^2+y^2}\right)$$ is valid for all $(x,y)\ne(0,0)$.