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A moment's thought (or some careful examination of maps) reveals that any circle drawn on a globe is in fact a circle in real life. The same claim holds for circles drawn on flat surfaces (obviously).

Are these the only examples?

Formally, let $M$ be any Riemannian surface, (isometrically) embedded in $\mathbb{R}^n$. Moreover, suppose that any metric circle $C$ (i.e., a locus of points with constant geodesic distance from some fixed point) is in fact a circle in $\mathbb{R}^n$. Then must $M$ have constant nonnegative Gaussian curvature?

Narasimham
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Jacob Manaker
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    This is not quite the setting of your question, but I’ll comment anyhow that geodesic circles in the hyperbolic plane are Euclidean circles (but with different centers). – Ted Shifrin Sep 20 '22 at 20:08

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Yes!

First, note that such a surface must have nonnegative curvature. For: fix a point $p$. Working in geodesic polar coordinates about $p$, the group $\mathrm{SO}(\mathbb{R}^2)\cong S^1$ acts on any connected surface by rotating each metric circle. Since our metric circles are genuine circles, the resulting action is an isometry on some geodesic ball around $p$. But rotation by $90^\circ$ interchanges the principal curvatures; thus both curvatures are equal. Since the Gaussian curvature is the product of the two curvatures, it is a perfect square and so nonnegative.

Second, since genuine circles have constant geodesic curvature, this Q&A thread proves that $M$ must have constant curvature.

Jacob Manaker
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    "Since our metric circles are genuine circles, M is invariant under this action". I guess you wanted to say: "since our metric circles are genuine circles, $SO(2)$ acts by isometries". Because $M$ is invariant by definition of action. Maybe I am not understanding the point – A. J. Pan-Collantes Sep 20 '22 at 05:24
  • Why must the curvature be nonnegative? – Didier Sep 20 '22 at 07:54
  • In general your $S^1$ action is not defined on the whole surface. – Arctic Char Sep 20 '22 at 08:14
  • @AntonioJPan That's exactly what I meant; thanks for the proper phrasing. – Jacob Manaker Sep 21 '22 at 18:40
  • @Didier: If the two principal curvatures are $\kappa_1$ and $\kappa_2$, the Gaussian curvature is $\kappa_1\kappa_2$. But $\kappa_1=\kappa_2$ here and curvatures are real numbers. – Jacob Manaker Sep 21 '22 at 18:41
  • @ArcticChar: Thanks; I just need it on a neighborhood of $p$, and I've rewritten the sentence to reflect that. – Jacob Manaker Sep 21 '22 at 18:42
  • @JacobManaker I see you have edited. This was not explicit at all! – Didier Sep 21 '22 at 18:46
  • @Didier: https://hsm.stackexchange.com/questions/7247/in-a-popular-anecdote-who-took-20-minutes-to-decide-that-a-thing-was-obvious – Jacob Manaker Sep 21 '22 at 19:17
  • I'm not sure I understand the $SO(2)$ action. If we take two metric circles (of small radius, so they are both near $p$), how do we know that planes containing them (in $\mathbb{R}^3$) are parallel? If they aren't the $SO(2)$ actions on each metric circle are not compatible, and can't be patched together to get an $SO(2)$ action on the whole neighborhood of $p$. – Jason DeVito - on hiatus Sep 22 '22 at 02:53
  • @JasonDeVito: The action is through the coordinates patch – A. J. Pan-Collantes Sep 22 '22 at 07:06
  • @JasonDeVito: Huh? I'm not sure what patching is necessary. Any circle has a transitive, free action of $\mathrm{SO}(2)$. – Jacob Manaker Sep 22 '22 at 14:18
  • I agree each circle has a free transitive action which is isometric on each circle. It's not clear to me that each of these actions comes from one action on the small neighborhood. I am not saying you're wrong, I just don't see it. – Jason DeVito - on hiatus Sep 22 '22 at 15:00
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    In more detail: working in $\mathbb{R}^3$, take a point $p$ in the submanifold $M$. It seems you're trying to claim that the $S^1$ action on $\mathbb{R}^3$ which fixes $p$ and rotates the tangent plane at $p$ is an isometry of a small neighborhood $U\subseteq M$ of $p$. But it's not clear to me that the plane containing each metric circle about $p$ in $U$ is parallel to $T_pM \subseteq T_p \mathbb{R}^3$. If any one of those planes isn't parallel, the $S^1$ action need not map $M$ to itself. – Jason DeVito - on hiatus Sep 22 '22 at 15:05
  • @JasonDeVito I think what OP is considering is something like, in polar coordinates in a small neighbourhood: $\phi\cdot \exp_p(r \theta) = \exp_p(r(\theta+\phi))$. It is not clear to me that this defines an isometric action of the neighbourhood, I agree with you – Didier Sep 22 '22 at 18:23
  • @Didier: I see. Do you happen to know of an example of a surface $S\subseteq \mathbb{R}^3$ (with metric inherited from $\mathbb{R}^3$)together with an isometry $f:S\rightarrow S$ which does not extend to an isometry of $\mathbb{R}^3$? Bonus points if the isometry is in the identity component of $Isom(S)$. – Jason DeVito - on hiatus Sep 22 '22 at 18:44
  • @JasonDeVito If you allow drastically non smooth isometric embedding, an embedding of the flat torus would work. As for the smooth case, I have no idea right now – Didier Sep 22 '22 at 19:03