If metric circles on a surface are genuine circles, must the surface be sphere?

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?

• 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). Sep 20, 2022 at 20:08

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.
• "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 Sep 20, 2022 at 5:24
• In general your $S^1$ action is not defined on the whole surface. Sep 20, 2022 at 8:14
• 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. Sep 22, 2022 at 15:05