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I'm sorry if this has been asked before. I did my best to find it.

I have a nagging question about $S^1$ equipped with the "chord metric," hereafter referred to as the Riemannian circle, vs $S^1$ with the standard (extrinsic) metric inherited from its canonical embedding in $\mathbb{R}^2$.

If I am understanding the Whitney embedding theorem correctly, the Riemannian circle should isometrically embed into $\mathbb{R}^2$.

My question: What is this embedding?

I reasoned that since the diameter of the standard circle is $2$, a natural candidate for the embedding would be mapping the standard circle to a smaller circle, namely one of radius $\frac{2}{\pi}$, so that arc lengths in the smaller circle correspond to chord lengths in the larger circle. This would certainly preserve distances between antipodal points. The problem is that this seems to be false for other pairs of points: in the embedded circle arc length grows linearly with an angle $0\leq\theta\leq\pi$ (namely $l_1(\theta)=\frac{2\theta}{\pi}$), while the chord subtending $\theta$ in the standard circle has length $l_2(\theta)=2\sin(\frac{\theta}{2})$, which does not grow linearly.

On the other hand, this embedding seems to be the only "natural" choice: I am forced to conclude that if an embedding exists, it will distort the shape of the Riemannian circle, that is, the Riemannian circle does not actually "look like" a circle. What other homogeneous shape could it look like?

I suspect my intuition is incorrect. Does the Whitney Theorem not apply? I would be grateful if someone could point out where I'm going wrong.

Note: I did find this question on Quora, but I don't think it answers my question.

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    Whitney theorem tells only about an embedding of differential manifolds; it ignores the metric. – 2017-02-02
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    Presumably OP means one of the Nash Embedding Theorems. https://en.wikipedia.org/wiki/Nash_embedding_theorem – 2017-02-02
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    Anyway, Nash's Theorems apply to Riemannian metrics, not metrics in the sense of metric spaces (henceforth, 'metrics'). While any Riemannian metric induces a metric, not every metric on a smooth manifold arises from a Riemannian metric. Indeed, the chord metric cannot arise from a Riemannian metric, because the distance between any two points is strictly shorter than any curve that connects them. This makes using the term 'Riemannian circle' for the circle equipped with the chord metric badly misleading. – 2017-02-02
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    One thing that is worth taking away from the previous comments is that what a circle "looks like" is more profitably answered from an *intrinsic* viewpoint, namely the viewpoint of the Riemannian metric. By taking this viewpoint, no information is lost, because you can still study "extrinsic" questions such as metric properties of embeddings of the circle into the Euclidean plane. – 2017-02-03

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