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A simple closed curve $\mathcal{C}$ in the plane is such that, going along the curve from a point $P$ thereon and getting back to it, the total angle has measure $2\pi$. So one can write $2\pi=\int_{\mathcal{C}} d\theta=\int_{\mathcal{C}}\dfrac{d\theta}{ds}ds=\int_{\mathcal{C}}C(s)ds$ where $C(s)$ is the curvature and $s$ the abscissa on the curve. Of course $C$ is continuous and $T$-periodic for $T=\int_{\mathcal{C}}ds$.

So let's consider the space $S(T)$ of all $T$-periodic continuous functions on $\mathbb{R}$ $f$ such that $\int_{0}^{T}f(t)dt=2\pi$.

Is there a bijection between the set of all simple closed planar curves of perimeter $T$ and $S(T)$?

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    $\begingroup$ Well, both sets have the size of a continuum, so, yes, there is a bijection. It is not clear what do you want from this bijection? $\endgroup$ Feb 20, 2020 at 15:57
  • $\begingroup$ I want to map the curve whose curvature is $s\mapsto f(s)$ to $f$. That way, to the circle of radius $R$ is associated the constant function $x\mapsto\frac{1}{R}$. $\endgroup$ Feb 20, 2020 at 16:30
  • $\begingroup$ Are you asking whether or not every function in $S(T)$ comes from the curvature of a simple closed curve? $\endgroup$ Feb 20, 2020 at 16:38
  • $\begingroup$ Yes, and the other way around too. $\endgroup$ Feb 20, 2020 at 16:41
  • $\begingroup$ The answer is no, i.e., the map which takes a curve to its curvature is not onto. You have the additional condition that the integral over the tangent vector is 0, where the tangent vector is $exp(i F)$, $F$ being the integral of $f$. The later condition is not always satisied. $\endgroup$
    – Sebastian
    Feb 20, 2020 at 17:07

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