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Let $\gamma:I\to\mathbb{R}^2$ be a closed plane curve, for simplicity, a unit circle. Therefore, we have $$\gamma(\varphi) = (\cos \varphi, \sin \varphi).$$ What is the proper domain of $\varphi$? Wikipedia says it's $\varphi \in [0,2\pi]$ with $\gamma(0) = \gamma(2\pi)$. What is the advantage of this domain rather than a domain $\varphi\in[0,2\pi\rangle$ with the additional condition that $$\lim_{\varphi \to 2\pi}\gamma(\varphi)=\gamma(0)?$$ It seems to me that the second definition is much more natural since no point on the curve is repeating, and the usual angle variable in the polar cordinates uses this domain.

Fizikus
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  • The wikipedia page you linked to refers to curves in general, not necessarily closed ones, and in that case it does make sense to define it on a closed interval. See the respective notes about Jordan and closed curves. – dxiv Jan 07 '18 at 01:27
  • @dxiv If you scroll down the wiki page, you will find the definition of a closed curve exactly as I wrote it down. If you know of a reference which defines closed curve differently, I would very much appreciate it. – Fizikus Jan 07 '18 at 13:37
  • You seem bothered that $\gamma$ on $[0,2\pi]$ isn't injective at the endpoints. But this doesn't matter in practical situations, such as when you're computing line integrals around $\gamma$. Integrals can't distinguish between curves that differ on sets of measure zero. – symplectomorphic Jan 07 '18 at 22:28
  • @symplectomorphic So, the Stokes theorem remains equally valid regardless of the domain of the closed curve? – Fizikus Jan 07 '18 at 23:10
  • "regardless of the domain": of course not. $\gamma$ on $[0,2\pi)$ will give a different integral than $[0,\pi)$ (consider, say, arclength integrals). But $\gamma$ on $[0,2\pi]$ will give the same integral as $\gamma$ on $[0,2\pi)$. – symplectomorphic Jan 07 '18 at 23:32

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There is no advantage for the study of continuous curves: you can extend the continuous $\gamma \colon [0,2\pi) \to \mathbb{R}^2$ to a unique continuous map on $\Gamma \colon [0,2\pi] \to \mathbb{R}^2$. The problem for many applications (any in particular, for your intended application to Stokes's theorem) is that smoothness of $\gamma$ need not imply smoothness of $\Gamma$. For example, if $\gamma(t)=(x(t),y(t))$ has $y(t) \sim (t-2\pi) \sin(1/(t-2 \pi))$ as $t \to 2 \pi$, then $dy$ is not integrable over $\gamma$.

Ben McKay
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