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This is an exercise from Spivak's Calculus on Manifolds, problem 4-27.

Define the singular 1-cube $c_{R,n}:[0,1]\rightarrow \mathbb{R}^2 - \{0\}$ to be $c_{R,n}=(R\cos(2\pi nt), R\sin(2\pi nt))$. Geometrically, this is a circle that winds $n$ times around the origin. A prior problem was to show that if $c$ is a singular 1-cube in $\mathbb{R}^2 - \{0\}$ with $c(0)=c(1)$ (a closed curve), then there exists some $n$ with $c-c_{1,n}=\partial c^2$, for some 2-chain $c^2$. I have solved this problem.

Spivak defines this $n$ to be the winding number of the curve.

My question is: how do I use Stokes theorem to show this $n$ is unique? I think if you assume you have $n_1$ and $n_2$ that both work, then integrating the angle form over a difference of chains involving $n_1$ and $n_2$ works, but I can't quite work out the details.

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Consider the closed 1-form $d \theta= \frac{1}{x^2+y^2}(xdy-ydx)$ (which generates the first de Rham cohomology group of the punctured plane.)

Lemma: $\int_{c_{1,n}} \frac{1}{x^2+y^2}(xdy-ydx)= 2 \pi n.$

Proof: $\int_{c_{1,n}} \frac{1}{x^2+y^2}(xdy-ydx)= \int_{0}^{2 \pi n}(\cos^2 (2 \pi n \theta)+ \sin ^2(2 \pi n \theta)) d \theta =\int_{0}^{2 \pi n}d \theta =2\pi n.$

Applying the Stokes theorem for $\omega=d \theta$ on the 2-chain $c^2,$ we get

$$\int_{c^2}d \omega=\int_{\partial c^2}\omega.$$

$d\omega=0,$ then $\int_{\partial c^2}\omega=0$ implying that $\int_c \omega=\int_{c_{1,n}}\omega=\int_{c_{1,n}} d \theta=2 \pi n.$

If there exist $n_1,n_2$ s.t. $c-c_{1,n_1}=\partial c_1^2$ and $c-c_{1,n_2}=\partial c_2^2$ for some 2-chains $c^2_1,c_2^2,$ then by the same process, we will have $\int_c \omega=\int_{c_{1,n_1}} d \theta=\int_{c_{1,n_2}} d \theta.$ Hence, $2 \pi n_1=2 \pi n_2 \rightarrow n_1=n_2.$

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