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I have a question: How is the first Rham cohomology group computed? I revised the answer in 2, but I don't understand the proof.

Let M be the puntured plane and $\alpha \in \Omega^1(M)$ be a closed one differentiable form. I want to know if there exists a smooth function $u \in C^\infty (M)$ such that $\alpha = du$ or not. Let $P_0$ be a fixed point in M and $P=(x,y)\in M$ and C be a closed curve containing two points not containing the origin and $C_P$ the path along C connecting $P_0$ and $P$.

$$u(x,y) = \int_{C_P} \alpha$$

I read that this definition is path independent if

$$ \int_{C'} \alpha=0$$

where C' is a closed path containing the origin. I do not understand this particular condition. Why does the definition of u does not depend on the path if the last integral is zero?

Samuel QG
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1 Answers1

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This is a standard condition that is easy to understand intuitively. Consider $P_0 = (1,0)$ and take two paths from $P_0$ to $P_1 = (-1,0)$, one ($C_1$) going counterclockwise around the unit circle, the other ($C_2$) going clockwise around the unit circle. Then note that $$\int_{C_1}\alpha - \int_{C_2}\alpha = \int_C\alpha,$$ where $C$ is the unit circle, traversed once counterclockwise. So to have any chance at well-definedness of $u$, we certainly need to require that $\int_C\alpha = 0$.

The question is why this suffices. The point is that any closed curve $\Gamma$ whatsoever is homologous to some integer $n$ multiple of $C$. The requisite notion is that of the winding number of the closed curve $\Gamma$ around the origin. Basically what you want to draw is this: Project the points of $\Gamma$ radially onto $C$; as you go around $\Gamma$, following the projection gives you a path $\bar\Gamma$ going around $C$ (perhaps back and forth, perhaps multiple times), but it must start and end at the same point. The number of times this path traverses $C$ (counting counterclockwise as positive and clockwise as negative) will give you $n$.

For each revolution of $\Gamma$ around $C$, you can fill in a region between $C$ and the relevant portion of $\Gamma$ — which will itself be a closed path — and apply Green's/Stokes's Theorem to that region. Since $d\alpha = 0$, this will show you (adding up the pieces) that $\int_\Gamma\alpha = \int_{\bar\Gamma}\alpha = n\int_C\alpha$. So, with the hypothesis that $\int_C\alpha = 0$, we conclude that $\int_\alpha$ is in fact generally path-independent. (As before, take any two paths from $x_0$ to $x$ and one, followed by the reverse of the other, gives you a closed curve $\Gamma$.)

Ted Shifrin
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