TITLE: Is the metric on the circle, induced from the plane, not a flat one?
QUESTION [6 upvotes]: My question concerns the highlighted part posted below, from Wikipedia article. (Link to the revision at the time of this post.)
I'd say I can't detect the curvature of the unit circle if I go along the curved path within ${\mathbb R}^2$. The induced metric only measures the one-dimensional length and forgets about the plane. 
So why would the induced metric not be flat?

REPLY [9 votes]: It depends on which kind of "metric" you're looking at.
If we view the unit circle and the plane as Riemannian manifolds, then the Riemannian metric on the circle induced by its embedding in the plane is indeed flat. Here we're speaking about Riemannian metrics, which work only locally.
However, if you take those two Riemannian manifolds and derive a global distance function on each of them as the geodesic distance between two points, you make them into metric spaces. In the case of the plane, this produces the usual Euclidean distance.
However, as metric spaces, the metric on the unit circle is not the same as the metric it gets as a subset of $\mathbb R^2$.
Two opposite points on the unit circle has distance $\pi$ in the metric of geodesic distance within the circle, whereas their distance in the metric inherited from the metric space $\mathbb R^2$ is $2\neq \pi$.