I'm having some trouble with 1-form $$\omega = a(x,y)dx+b(x,y)dy$$ with $a$ and $b$ defined on $A\subset\mathbb{R}^2$. I know that if the 1-form is closed in a simply connected space, i.e. $$\frac{\partial a}{\partial y} = \frac{\partial b}{\partial x}$$ the 1-form is also exact, so $\exists f:A\mapsto\mathbb{R}$ such that $$\frac{\partial f}{\partial x}(x,y) = a(x,y)\quad\text{and}\quad\frac{\partial f}{\partial y}(x,y) = b(x,y)$$ If $A$ is not simply connected, I should study how $A$ is defined. Now I have some question for you:
- If $A = \mathbb{R}^2 - \{(x_0,y_0)\}$, is it sufficient to affirm that $\omega$, defined on $A$, closed and $C^1(A)$, is exact if and only if the line integral of $\omega$ on a closed curve that surround $(x_0,y_0)$ is $0$? I know by theory that every line integral on a closed curve must be $0$, but I found some exercises that prove it only for one curve. And if I have more than one point not in $\mathbb{R}^2$, can I generalize what I said before taking a curve that surround all the point? Or I have to verify individually?
- Suppose that $A=\{(x,y) \in \mathbb{R}^2 | y\neq f(x)\}$ such that $y=f(x)$ splits $\mathbb{R}^2$ in $n$ parts, $A_1,\dots,A_n$, such that $A_i \cap A_j=\emptyset,\,\forall i\neq j$ and $\bigcup_{i=1}^n A_i = A$ (so $A$ is not connected, and so not simply connected), but every $A_i$ is simply connected. Can I affirm that if $\omega$ is closed in $A_i$ and $C^1(A_i)$ for all $i$, it exists the primitive that is defined locally on $A_i$? How do I find that primitive?
You should be able to show that if the closed form $\omega$ satisfies $\int_C \omega = 0$ for $C$ a circle (of any radius) centered at $(x_0,y_0)$, then $\omega$ is in fact exact on $\Bbb R^2-\{(x_0,y_0)\}$. (Your proof might not be 100% rigorous, but should be convincing, if you use Green's Theorem and some pictures.)
If you remove $n$ points, you'll need the corresponding fact for (little) circles $C_i$ around each of the $n$ points. Having one curve that goes around all $n$ points will obviously not suffice, because knowing that large integral vanishes won't tell you that the individual ones do, and so you'll have lots of non-exactness.