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This is an exercise question from Spivak's calculus on manifolds chapter number 4 question 26.

Show that $\int_{C_{R,n}}d\theta=2\pi n$, and use stoke's theorem to conclude that $C_{R,n}\neq \partial c$ for any $2$ chain $c$ in $\mathbb{R}^2-0$ where $C_{R,n}$ is singular 1 cube $[0,1]\rightarrow \mathbb{R}^2-0$ defined as $t\mapsto (R\cos 2\pi n t,R\sin 2\pi n t)$

$d\theta$ is the $1$ form $$\frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy$$ I have tried the following.. I write $\tau$ for $C_{R,n}$.

We have $$\int_\tau d\theta=\int_{[0,1]}\tau^*(d\theta)$$

I am not very much sure how to compute $\tau^*(d\theta)$

$$\tau^*(d\theta)=\tau^*\left(\frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy\right)$$

$$=\left(\frac{-y}{x^2+y^2}\circ \tau\right) \tau^*(dx)+\left(\frac{-x}{x^2+y^2}\circ \tau\right) \tau^*(dy)$$

Now, $$\tau^*(dx)=\frac{\partial(R\cos 2\pi nt)}{\partial(t)}dt=-2\pi nR\sin 2\pi nt$$ $$\tau^*(dy)=\frac{\partial(R\sin 2\pi nt)}{\partial(t)}dt=2\pi nR\cos 2\pi nt$$

$$\left(\frac{-y}{x^2+y^2}\circ \tau\right):[0,1]\rightarrow \mathbb{R}; t\rightarrow -\frac{1}{R}\sin 2\pi n t$$

$$\left(\frac{x}{x^2+y^2}\circ \tau\right):[0,1]\rightarrow \mathbb{R}; t\rightarrow \frac{1}{R}\cos 2\pi n t$$

So, $$\tau^*(d\theta)=2\pi n \sin^2(2\pi nt)+2\pi n \cos^2(2\pi nt)=2\pi n$$

We have $$\int_\tau d\theta=\int_{[0,1]}\tau^*(d\theta)=\int_{[0,1]}2\pi n=2\pi n$$

Suppose we have $C_{R,n}\neq \partial c$ for some $2$ chain $c$ in $\mathbb{R}^2-0$ then

$$2\pi n=\int_{C_{R,n}}=\int_{\partial c}d\theta=\int_{c}d(d\theta)=\int_c 0=0$$

So, we have $2\pi n=0$ a contradiction.. Thus, there can be no $2$ chain $c$ such that $\partial c=C_{R,n}$

Please let me know if this is correct way of solving ...

Max Demirdilek
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kk lm
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1 Answers1

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To remove this from the list of unanswered questions, let me write a short answer.

Well done, the gist of your proposed solution is correct. The solution contains a few typos, however:

  • I am not familiar with your notation for partial derivatives. I would, for instance, write $$\frac{\partial}{\partial t}\Big( R\cos 2\pi nt \Big)dt$$ instead of

$$\frac{\partial(R\cos 2\pi nt)}{\partial(t)}dt.$$

  • When proving a proposition by contradiction, you better start off by assuming its negation. Explicitly, you should have written "Suppose we have $C_{R,n}=\partial c$ for some chain."

  • The integrand of one integral is missing. It should read $$2\pi n=\int_{C_{R,n}}d\theta.$$

Max Demirdilek
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  • OP was wondering why that is $2\pi n$ and not zero. In fact this inegral is one of the most ubiquitous homework exercises I am seeing. You might be interested in my answer here. Perhaps this seven year old question can be closed. – Kurt G. Mar 01 '23 at 10:02
  • OP was wondering whether their proposed solution is correct. That is what my answer is about. I merely posted my answer to remove this question from the unanswered list, since, while scrolling through unanswered questions, I had bumped into this question multiple times. I would be happy if this question was closed. – Max Demirdilek Mar 01 '23 at 10:13
  • I appreciate your efforts. – Kurt G. Mar 01 '23 at 10:21