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Let

$$ \omega=\frac{-y\; dx}{x^{2}+y^{2}}+\frac{x\; dy}{x^{2}+y^{2}} \in \Omega^{1}(\mathbb{R}^{2}\backslash\{0\}) $$

I understand that the form $\frac{-y}{x^{2}+y^{2}}dx+\frac{x}{x^{2}+y^{2}}dy$ is closed but not exact on $\mathbb{R}^{2}\backslash\{0\}$, and I am somewhat familiar with its relationship with the winding number.

I encountered the assertion that (I'm pretty sure I'm recalling it correctly), if one restricts to a path, $\gamma:[0,1]\rightarrow \mathbb{R}^{2}\backslash\{0\}$ on which $\gamma(t)=(x(t),y(t))$ satisfies $y(t)\ge 0$ for all $t$, then any path, $\alpha$ satisfying these requirements whose endpoints coincide with those of $\gamma$ also satisfies $\int_{\gamma}\omega=\int_{\alpha}\omega$.

Someone told me about a homotopy-based proof, but I'm not very good at proving when functions are homotopic, so I'm a little reluctant to tread down that path.

If one makes a coordinate change to polar coordinates, $\omega$ becomes $d\theta$ (right?).

What is the significance of this fact when $\gamma$ is forced to obey the conditions above (i.e. $y$ component is non-negative)?

I sincerely apologize for the vagueness of my question, this is my first time posting on here, a bit nervous asking for help. I just find differential forms and the like are hard to parse, partly because the notes provided tend to suppress a lot of notation.

If anyone else could explain the situation to me I'd be always grateful.Right now I'm talking to myself only as if I were a five year old.

user227187
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    You don't need to be anxious when posting a question. ;) – Pedro Mar 29 '15 at 04:52
  • It's a good question, but about a subject I know less than nothing about. You just might want to consider breaking it up into a few paragraphs, it's a little hard to parse. – pjs36 Mar 29 '15 at 05:08

2 Answers2

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Short answer: You probably want to look into de Rham cohomology. The idea of branch cuts in complex analysis is also related, and avoids having to deal with forms on manifolds in general. Still, what's going on is really topological in nature, and I haven't seen a complex analysis textbook that really treats the matter well while still staying within a standard introductory complex analysis course.

Long answer: The underlying idea is if $X$ is contractible, then any closed $\eta\in \Omega^p(X)$ is also exact. I don't know how much topology you're familiar with, but one of the common situations encountered is that of $X$ star-shaped: there exists a point $x_0\in X$ such the line segment from any $x\in X$ to $x_0$ lies entirely in $X$. Basically, to construct the form $\theta$ with $d\theta = \eta$, we just integrate along each of those line segments; the fact that $d\eta$ means that (because of Stokes' theorem) this operation is well-defined and gives a smooth function on $X$.

So, let's consider $\gamma(t) = (x(t), y(t))$ above with $y\geq 0$. Then $\gamma$ lies entirely within the region $$D = \{(x, y):\, y \geq 0, (x, y)\not = (0, 0)\},$$ which is star-shaped. (Take $x_0$ to be a point of the form $(0, y)$. Also, I'm being sloppy here in addressing issues of compactness and smoothness, but I don't have space here to deal with it properly. We also ) That means that $\eta = d\theta$ for some $\theta$ on $D$. For any two paths $\alpha, \alpha'$ both running from $p$ to $p'$, we can formed a closed loop $\beta$ by taking $\alpha$ from $p$ to $p'$, then $\alpha'$ from $p'$ to $p$. Thus by Stokes' theorem, $$\int_\alpha \eta - \int_\alpha' \eta = \int_\beta \eta = \int_\beta d\theta = \int_{\delta \beta} \theta = 0.$$ On the other hand, there isn't a function $\zeta$ on the entirety of $\mathbb{R}^2\setminus (0, 0)$ that satisfies $\eta = d\zeta$. The obstruction is, as you mentioned, that $\int_{S^1} \eta \not= 0$. If there were such a function $\zeta$, then we would have to $\int_{S^1} \eta = 0$ by the same argument above.

anomaly
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You have figured out that $w = d\theta$ when written in polar coordinate, and that is sufficient to show the path independent property.

Note that if the curve $\gamma$ satisfies $y(t)\ge 0$, then $\gamma$ does not pass throught the line $L = \{(0,t): t < 0\}$. In particular, $\theta$ is a well defined function on $\mathbb R^2 \setminus L$ and so

$$\int_{\gamma} \omega = \int_{\gamma} d\theta = \theta (\gamma(1)) - \theta(\gamma(0)). $$

By the fundamental theorem of calculus. In particular, it is independent of curve (depends only on the endpoints).