# Does there exist an isometry between a regular polygon and a circle?

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 $$I: \mathcal{M} \to S_1$$.

• 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. – Zach Teitler Apr 18 at 4:56
• What is a "fractional perimeter"? – Ben McKay Apr 18 at 7:00
• Any length metric on the circle inducing the usual topology is isometric to a circle of a unique radius. – Ben McKay Apr 18 at 7:02