In order to define the question in a meaningful fashion, I am referring to a smooth manifold $\mathcal{M}$ within an $\epsilon$-neighborhood of a regular polygon $\mathcal{P}$ satisfying $$\max\{\|x-p\|: x \in \mathcal{M} \cap B_{\delta}, p \in \mathcal{P} \cap B_{\delta}\} < \varepsilon$$ for a circular $\delta$-neighborhood $B_{\delta}$. I have already proved that $\mathcal{M}$ is diffeomorphic to the unit circle $S_1$; however, I now wish to examine the possibility of an isometry $I: \mathcal{M} \to S_1$ where the metric on $\mathcal{M}$ is simply taken as a fractional perimeter instead of any ambient distance.

I believe that the existence of $I$ is impossible as even if I were to produce a conformal map, I do not consider any map to preserve the area between $\mathcal{M}$ and $S_1$. As local areas were not preserved w.r.t the metric, the notion of magnitude for a vector $v \in \mathcal{T}_p(\mathcal{M})$ would not be preserved. Finally, the first fundamental form would not be preserved, so no isometry could exist.

Do you all have suggestions as to how I may prove or disprove the existence of such an isometry $I: \mathcal{M} \to S_1$.

  • $\begingroup$ What is the area of a one dimensional manifold? Is $\mathcal{M}$ supposed to be a curve, or the region enclosed by the curve? What is $S_2$? (a sphere?) I’m not completely sure what is the barrier to any smooth one-dimensional manifold being isometric to a circle. The concerns about “local area” and conformal maps are very unclear to me. $\endgroup$ Apr 18, 2021 at 4:56
  • 1
    $\begingroup$ What is a "fractional perimeter"? $\endgroup$
    – Ben McKay
    Apr 18, 2021 at 7:00
  • 2
    $\begingroup$ Any length metric on the circle inducing the usual topology is isometric to a circle of a unique radius. $\endgroup$
    – Ben McKay
    Apr 18, 2021 at 7:02


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